ANSWER-NEW-Vector Mechanics for Engineers Statics (Ferdinand P. Beer)-ANSWER.pdf

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ww.elsolucionario. SOLUTION MANUAL


CHAPTER 2





www.elsolucionario.net PROBLEM 2.1 Two forces are applied at point B of beam AB. Determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule.

(a)

Parallelogram law:

(b)

Triangle rule:

We measure:

R = 3.30 kN, α = 66.6°

R = 3.30 kN

66.6° 

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SOLUTION

www.elsolucionario.net PROBLEM 2.2 The cable stays AB and AD help support pole AC. Knowing that the tension is 120 lb in AB and 40 lb in AD, determine graphically the magnitude and direction of the resultant of the forces exerted by the stays at A using (a) the parallelogram law, (b) the triangle rule.

We measure: (a)

Parallelogram law:

(b)

Triangle rule:

We measure:


SOLUTION

α = 51.3° β = 59.0°

R = 139.1 lb,

γ = 67.0°

R = 139.1 lb

67.0° 




www.elsolucionario.net PROBLEM 2.3 Two structural members B and C are bolted to bracket A. Knowing that both members are in tension and that P = 10 kN and Q = 15 kN, determine graphically the magnitude and direction of the resultant force exerted on the bracket using (a) the parallelogram law, (b) the triangle rule.

(a)

Parallelogram law:

(b)

Triangle rule:

We measure:


SOLUTION

α = 21.2°

R = 20.1 kN,

R = 20.1 kN

21.2° 


www.elsolucionario.net PROBLEM 2.4 Two structural members B and C are bolted to bracket A. Knowing that both members are in tension and that P = 6 kips and Q = 4 kips, determine graphically the magnitude and direction of the resultant force exerted on the bracket using (a) the parallelogram law, (b) the triangle rule.

(a)

Parallelogram law:

(b)

Triangle rule:

We measure:


SOLUTION

R = 8.03 kips, α = 3.8°

R = 8.03 kips

3.8° 


www.elsolucionario.net PROBLEM 2.5 A stake is being pulled out of the ground by means of two ropes as shown. Knowing that α = 30°, determine by trigonometry (a) the magnitude of the force P so that the resultant force exerted on the stake is vertical, (b) the corresponding magnitude of the resultant.

Using the triangle rule and the law of sines: (a) (b)

120 N P = sin 30° sin 25°

P = 101.4 N 

30° + β + 25° = 180°

β = 180° − 25° − 30° = 125° 120 N R = sin 30° sin125°

R = 196.6 N 



SOLUTION

www.elsolucionario.net PROBLEM 2.6 A trolley that moves along a horizontal beam is acted upon by two forces as shown. (a) Knowing that α = 25°, determine by trigonometry the magnitude of the force P so that the resultant force exerted on the trolley is vertical. (b) What is the corresponding magnitude of the resultant?

Using the triangle rule and the law of sines: (a) (b)

1600 N P = sin 25° sin 75°

P = 3660 N 

25° + β + 75° = 180°

β = 180° − 25° − 75° = 80° 1600 N R = sin 25° sin 80°

R = 3730 N 



SOLUTION

www.elsolucionario.net PROBLEM 2.7 A trolley that moves along a horizontal beam is acted upon by two forces as shown. Determine by trigonometry the magnitude and direction of the force P so that the resultant is a vertical force of 2500 N.

Using the law of cosines:

Using the law of sines:

P 2 = (1600 N)2 + (2500 N)2 − 2(1600 N)(2500 N) cos 75° P = 2596 N

sin α sin 75° = 1600 N 2596 N α = 36.5°

P is directed 90° − 36.5° or 53.5° below the horizontal.

P = 2600 N

53.5° 



SOLUTION

www.elsolucionario.net PROBLEM 2.8 A telephone cable is clamped at A to the pole AB. Knowing that the tension in the left-hand portion of the cable is T1 = 800 lb, determine by trigonometry (a) the required tension T2 in the right-hand portion if the resultant R of the forces exerted by the cable at A is to be vertical, (b) the corresponding magnitude of R.

Using the triangle rule and the law of sines: (a)

75° + 40° + α = 180°

α = 180° − 75° − 40° = 65°

(b)

T2 800 lb = sin 65° sin 75°

T2 = 853 lb 

800 lb R = sin 65° sin 40°

R = 567 lb 



SOLUTION

www.elsolucionario.net PROBLEM 2.9 A telephone cable is clamped at A to the pole AB. Knowing that the tension in the right-hand portion of the cable is T2 = 1000 lb, determine by trigonometry (a) the required tension T1 in the left-hand portion if the resultant R of the forces exerted by the cable at A is to be vertical, (b) the corresponding magnitude of R.


75° + 40° + β = 180°

β = 180° − 75° − 40° = 65°

(b)

T1 1000 lb = sin 75° sin 65°

T1 = 938 lb 

1000 lb R = sin 75° sin 40°

R = 665 lb 



SOLUTION

www.elsolucionario.net PROBLEM 2.10 Two forces are applied as shown to a hook support. Knowing that the magnitude of P is 35 N, determine by trigonometry (a) the required angle α if the resultant R of the two forces applied to the support is to be horizontal, (b) the corresponding magnitude of R.

SOLUTION Using the triangle rule and law of sines: sin α sin 25° = 50 N 35 N sin α = 0.60374

α = 37.138° (b)

α = 37.1° 

α + β + 25° = 180° β = 180° − 25° − 37.138° = 117.862° R 35 N = sin117.862° sin 25°

R = 73.2 N 



(a)

www.elsolucionario.net PROBLEM 2.11 A steel tank is to be positioned in an excavation. Knowing that α = 20°, determine by trigonometry (a) the required magnitude of the force P if the resultant R of the two forces applied at A is to be vertical, (b) the corresponding magnitude of R.


β + 50° + 60° = 180° β = 180° − 50° − 60° = 70°

(b)

425 lb P = sin 70° sin 60°

P = 392 lb 

425 lb R = sin 70° sin 50°

R = 346 lb 



SOLUTION

www.elsolucionario.net PROBLEM 2.12 A steel tank is to be positioned in an excavation. Knowing that the magnitude of P is 500 lb, determine by trigonometry (a) the required angle α if the resultant R of the two forces applied at A is to be vertical, (b) the corresponding magnitude of R.


(α + 30°) + 60° + β = 180°

β = 180° − (α + 30°) − 60° β = 90° − α sin (90° − α ) sin 60° 425 lb

=

500 lb

90° − α = 47.402°

(b)

R 500 lb = sin (42.598° + 30°) sin 60°

α = 42.6°  R = 551 lb 



SOLUTION

www.elsolucionario.net PROBLEM 2.13 A steel tank is to be positioned in an excavation. Determine by trigonometry (a) the magnitude and direction of the smallest force P for which the resultant R of the two forces applied at A is vertical, (b) the corresponding magnitude of R.

SOLUTION

(a)

P = (425 lb) cos 30°

(b)

R = (425 lb)sin 30°

P = 368 lb



R = 213 lb 



The smallest force P will be perpendicular to R.

www.elsolucionario.net PROBLEM 2.14 For the hook support of Prob. 2.10, determine by trigonometry (a) the magnitude and direction of the smallest force P for which the resultant R of the two forces applied to the support is horizontal, (b) the corresponding magnitude of R.

SOLUTION

(a)

P = (50 N)sin 25°

(b)

R = (50 N) cos 25°

P = 21.1 N



R = 45.3 N 



The smallest force P will be perpendicular to R.

www.elsolucionario.net PROBLEM 2.15 Solve Problem 2.2 by trigonometry. PROBLEM 2.2 The cable stays AB and AD help support pole AC. Knowing that the tension is 120 lb in AB and 40 lb in AD, determine graphically the magnitude and direction of the resultant of the forces exerted by the stays at A using (a) the parallelogram law, (b) the triangle rule.


SOLUTION

8 10 α = 38.66°

tan α =

6 10 β = 30.96°

tan β =

Using the triangle rule:

Using the law of cosines:

α + β + ψ = 180° 38.66° + 30.96° + ψ = 180° ψ = 110.38° R 2 = (120 lb)2 + (40 lb) 2 − 2(120 lb)(40 lb) cos110.38° R = 139.08 lb

Using the law of sines:

sin γ sin110.38° = 40 lb 139.08 lb

γ = 15.64° φ = (90° − α ) + γ φ = (90° − 38.66°) + 15.64° φ = 66.98°

R = 139.1 lb

67.0° 


www.elsolucionario.net PROBLEM 2.16 Solve Problem 2.4 by trigonometry. PROBLEM 2.4 Two structural members B and C are bolted to bracket A. Knowing that both members are in tension and that P = 6 kips and Q = 4 kips, determine graphically the magnitude and direction of the resultant force exerted on the bracket using (a) the parallelogram law, (b) the triangle rule.

SOLUTION Using the force triangle and the laws of cosines and sines:

γ = 180° − (50° + 25°)


We have:

= 105°

Then

R 2 = (4 kips) 2 + (6 kips)2 − 2(4 kips)(6 kips) cos105°

= 64.423 kips 2 R = 8.0264 kips

And

4 kips 8.0264 kips = sin(25° + α ) sin105° sin(25° + α ) = 0.48137 25° + α = 28.775° α = 3.775° R = 8.03 kips

3.8° 


www.elsolucionario.net PROBLEM 2.17 For the stake of Prob. 2.5, knowing that the tension in one rope is 120 N, determine by trigonometry the magnitude and direction of the force P so that the resultant is a vertical force of 160 N. PROBLEM 2.5 A stake is being pulled out of the ground by means of two ropes as shown. Knowing that α = 30°, determine by trigonometry (a) the magnitude of the force P so that the resultant force exerted on the stake is vertical, (b) the corresponding magnitude of the resultant.


SOLUTION

Using the laws of cosines and sines: P 2 = (120 N) 2 + (160 N)2 − 2(120 N)(160 N) cos 25° P = 72.096 N

And

sin α sin 25° = 120 N 72.096 N sin α = 0.70343 α = 44.703° P = 72.1 N

44.7° 


www.elsolucionario.net PROBLEM 2.18 For the hook support of Prob. 2.10, knowing that P = 75 N and α = 50°, determine by trigonometry the magnitude and direction of the resultant of the two forces applied to the support. PROBLEM 2.10 Two forces are applied as shown to a hook support. Knowing that the magnitude of P is 35 N, determine by trigonometry (a) the required angle α if the resultant R of the two forces applied to the support is to be horizontal, (b) the corresponding magnitude of R.

SOLUTION Using the force triangle and the laws of cosines and sines: We have

β = 180° − (50° + 25°)

Then

R 2 = (75 N) 2 + (50 N) 2 − 2(75 N)(50 N) cos 105° R = 10, 066.1 N 2 R = 100.330 N 2

and

Hence:

sin γ sin105° = 75 N 100.330 N sin γ = 0.72206 γ = 46.225°

γ − 25° = 46.225° − 25° = 21.225°

R = 100.3 N

21.2° 



= 105°

www.elsolucionario.net PROBLEM 2.19 Two forces P and Q are applied to the lid of a storage bin as shown. Knowing that P = 48 N and Q = 60 N, determine by trigonometry the magnitude and direction of the resultant of the two forces.


γ = 180° − (20° + 10°) = 150°

R 2 = (48 N)2 + (60 N)2 − 2(48 N)(60 N) cos150° R = 104.366 N

and

Hence:

48 N 104.366 N = sin α sin150° sin α = 0.22996 α = 13.2947°

φ = 180° − α − 80° = 180° − 13.2947° − 80° = 86.705° R = 104.4 N

86.7° 



Then

www.elsolucionario.net PROBLEM 2.20 Two forces P and Q are applied to the lid of a storage bin as shown. Knowing that P = 60 N and Q = 48 N, determine by trigonometry the magnitude and direction of the resultant of the two forces.


γ = 180° − (20° + 10°) = 150° R 2 = (60 N)2 + (48 N) 2 − 2(60 N)(48 N) cos 150° R = 104.366 N

and

Hence:

60 N 104.366 N = sin α sin150° sin α = 0.28745 α = 16.7054°

φ = 180° − α − 180° = 180° − 16.7054° − 80° = 83.295° R = 104.4 N

83.3° 



Then

www.elsolucionario.net PROBLEM 2.21 Determine the x and y components of each of the forces shown.

80-N Force:

120-N Force:

150-N Force:

Fx = + (80 N) cos 40°

Fx = 61.3 N 

Fy = + (80 N) sin 40°

Fy = 51.4 N 

Fx = + (120 N) cos 70°

Fx = 41.0 N 

Fy = +(120 N) sin 70°

Fy = 112.8 N 

Fx = −(150 N) cos 35°

Fx = −122. 9 N 

Fy = +(150 N) sin 35°

Fy = 86.0 N 



SOLUTION


40-lb Force:

50-lb Force:

60-lb Force:

Fx = + (40 lb) cos 60°

Fx = 20.0 lb 

Fy = −(40 lb)sin 60°

Fy = −34.6 lb 

Fx = −(50 lb)sin 50°

Fx = −38.3 lb 

Fy = −(50 lb) cos 50°

Fy = −32.1 lb 

Fx = + (60 lb) cos 25°

Fx = 54.4 lb 

Fy = +(60 lb)sin 25°

Fy = 25.4 lb 



SOLUTION


SOLUTION

OA = (600) 2 + (800) 2 = 1000 mm OB = (560)2 + (900) 2 = 1060 mm OC = (480) 2 + (900)2 = 1020 mm

800-N Force:

424-N Force:

408-N Force:

Fx = + (800 N)

800 1000

Fx = +640 N 

Fy = +(800 N)

600 1000

Fy = +480 N 

Fx = −(424 N)

560 1060

Fx = −224 N 

Fy = −(424 N)

900 1060

Fy = −360 N 

Fx = + (408 N)

480 1020

Fx = +192.0 N 

Fy = −(408 N)

900 1020

Fy = −360 N 



Compute the following distances:


SOLUTION

OA = (24 in.)2 + (45 in.)2 = 51.0 in. OB = (28 in.) 2 + (45 in.) 2 = 53.0 in. OC = (40 in.) 2 + (30 in.) 2 = 50.0 in.

102-lb Force:

106-lb Force:

200-lb Force:

Fx = −102 lb

24 in. 51.0 in.

Fx = −48.0 lb 

Fy = +102 lb

45 in. 51.0 in.

Fy = +90.0 lb 

Fx = +106 lb

28 in. 53.0 in.

Fx = +56.0 lb 

Fy = +106 lb

45 in. 53.0 in.

Fy = +90.0 lb 

Fx = −200 lb

40 in. 50.0 in.

Fx = −160.0 lb 

Fy = −200 lb

30 in. 50.0 in.

Fy = −120.0 lb 



Compute the following distances:

www.elsolucionario.net PROBLEM 2.25 The hydraulic cylinder BD exerts on member ABC a force P directed along line BD. Knowing that P must have a 750-N component perpendicular to member ABC, determine (a) the magnitude of the force P, (b) its component parallel to ABC.

(a)

750 N = P sin 20° P = 2192.9 N

(b)

P = 2190 N 

PABC = P cos 20° = (2192.9 N) cos 20°

PABC = 2060 N 



SOLUTION

www.elsolucionario.net PROBLEM 2.26 Cable AC exerts on beam AB a force P directed along line AC. Knowing that P must have a 350-lb vertical component, determine (a) the magnitude of the force P, (b) its horizontal component.

(a)

P=

Py cos 55°

350 lb cos 55° = 610.21 lb =

(b)

P = 610 lb 

Px = P sin 55°

= (610.21 lb)sin 55° = 499.85 lb

Px = 500 lb 



SOLUTION

www.elsolucionario.net PROBLEM 2.27 Member BC exerts on member AC a force P directed along line BC. Knowing that P must have a 325-N horizontal component, determine (a) the magnitude of the force P, (b) its vertical component.

SOLUTION BC = (650 mm)2 + (720 mm) 2  650  Px = P    970 

(a)

or

 970  P = Px    650   970  = 325 N    650  = 485 N P = 485 N 

(b)

 720  Py = P    970   720  = 485 N    970  = 360 N Py = 970 N 



= 970 mm

www.elsolucionario.net PROBLEM 2.28 Member BD exerts on member ABC a force P directed along line BD. Knowing that P must have a 240-lb vertical component, determine (a) the magnitude of the force P, (b) its horizontal component.

(a)

P=

(b)

Px =

Py sin 40°

=

Py tan 40°

240 lb sin 40°

=

240 lb tan 40°

or P = 373 lb  or Px = 286 lb 



SOLUTION

www.elsolucionario.net PROBLEM 2.29 The guy wire BD exerts on the telephone pole AC a force P directed along BD. Knowing that P must have a 720-N component perpendicular to the pole AC, determine (a) the magnitude of the force P, (b) its component along line AC.

(a)

37 Px 12 37 = (720 N) 12 = 2220 N

P=

P = 2.22 kN 

(b)

35 Px 12 35 = (720 N) 12 = 2100 N

Py =

Py = 2.10 kN 



SOLUTION

www.elsolucionario.net PROBLEM 2.30 The hydraulic cylinder BC exerts on member AB a force P directed along line BC. Knowing that P must have a 600-N component perpendicular to member AB, determine (a) the magnitude of the force P, (b) its component along line AB.

SOLUTION 180° = 45° + α + 90° + 30° α = 180° − 45° − 90° − 30° = 15° Px P P P= x cos α 600 N = cos15° = 621.17 N

cos α =

P = 621 N 

(b)

tan α =

Py Px

Py = Px tan α = (600 N) tan15° = 160.770 N Py = 160.8 N 



(a)

www.elsolucionario.net PROBLEM 2.31 Determine the resultant of the three forces of Problem 2.23. PROBLEM 2.23 Determine the x and y components of each of the forces shown.

SOLUTION

Force

x Comp. (N)


Components of the forces were determined in Problem 2.23: y Comp. (N)

800 lb

+640

+480

424 lb

–224

–360

408 lb

+192

–360

Rx = +608

Ry = −240

R = Rx i + Ry j = (608 lb)i + (−240 lb) j Ry tan α = Rx 240 608 α = 21.541° 240 N R= sin(21.541°) = 653.65 N =

R = 654 N

21.5° 



SOLUTION Components of the forces were determined in Problem 2.21: x Comp. (N)

y Comp. (N)

80 N

+61.3

+51.4

120 N

+41.0

+112.8

150 N

–122.9

+86.0

Rx = −20.6

Ry = + 250.2


Force

R = Rx i + Ry j = ( −20.6 N)i + (250.2 N) j tan α =

Ry Rx

250.2 N 20.6 N tan α = 12.1456 tan α =

α = 85.293° R=

250.2 N sin 85.293°

R = 251 N

85.3° 



SOLUTION x Comp. (lb)

y Comp. (lb)

40 lb

+20.00

–34.64

50 lb

–38.30

–32.14

60 lb

+54.38

+25.36

Rx = +36.08

Ry = −41.42

R = Rx i + Ry j = ( +36.08 lb)i + (−41.42 lb) j tan α =

Ry Rx

41.42 lb 36.08 lb tan α = 1.14800 tan α =

α = 48.942° R=

41.42 lb sin 48.942°

R = 54.9 lb

48.9° 



Force


SOLUTION

Force

x Comp. (lb)


Components of the forces were determined in Problem 2.24: y Comp. (lb)

102 lb

−48.0

+90.0

106 lb

+56.0

+90.0

200 lb

−160.0

−120.0

Rx = −152.0

Ry = 60.0

R = Rx i + Ry j = ( −152 lb)i + (60.0 lb) j tan α =

Ry Rx

60.0 lb 152.0 lb tan α = 0.39474 tan α =

α = 21.541° R=

60.0 lb sin 21.541°

R = 163.4 lb

21.5° 


www.elsolucionario.net PROBLEM 2.35 Knowing that α = 35°, determine the resultant of the three forces shown.

SOLUTION Fx = +(100 N) cos 35° = +81.915 N

100-N Force:

Fy = −(100 N)sin 35° = −57.358 N Fx = +(150 N) cos 65° = +63.393 N

150-N Force:


Fy = −(150 N) sin 65° = −135.946 N Fx = −(200 N) cos 35° = −163.830 N

200-N Force:

Fy = −(200 N)sin 35° = −114.715 N

Force

x Comp. (N)

y Comp. (N)

100 N

+81.915

−57.358

150 N

+63.393

−135.946

200 N

−163.830

−114.715

Rx = −18.522

Ry = −308.02

R = Rx i + Ry j = (−18.522 N)i + (−308.02 N) j tan α =

Ry Rx

308.02 18.522 α = 86.559° =

R=

308.02 N sin 86.559

R = 309 N

86.6° 


www.elsolucionario.net PROBLEM 2.36 Knowing that the tension in rope AC is 365 N, determine the resultant of the three forces exerted at point C of post BC.

SOLUTION

Cable force AC:

500-N Force:

200-N Force:

and


Determine force components: 960 = −240 N 1460 1100 = −275 N Fy = −(365 N) 1460 Fx = −(365 N)

24 = 480 N 25 7 Fy = (500 N) = 140 N 25 Fx = (500 N)

4 = 160 N 5 3 Fy = −(200 N) = −120 N 5 Fx = (200 N)

Rx = ΣFx = −240 N + 480 N + 160 N = 400 N Ry = ΣFy = −275 N + 140 N − 120 N = −255 N R = Rx2 + Ry2 = (400 N) 2 + (−255 N) 2 = 474.37 N

Further:

255 400 α = 32.5°

tan α =

R = 474 N

32.5° 



SOLUTION 60-lb Force:

Fx = (60 lb) cos 20° = 56.382 lb Fy = (60 lb)sin 20° = 20.521 lb

80-lb Force:

Fx = (80 lb) cos 60° = 40.000 lb

120-lb Force:


Fy = (80 lb)sin 60° = 69.282 lb Fx = (120 lb) cos 30° = 103.923 lb Fy = −(120 lb)sin 30° = −60.000 lb

and

Rx = ΣFx = 200.305 lb Ry = ΣFy = 29.803 lb R = (200.305 lb)2 + (29.803 lb) 2 = 202.510 lb

Further:

tan α =

29.803 200.305

α = tan −1

29.803 200.305

R = 203 lb

= 8.46°

8.46° 



SOLUTION 60-lb Force:

Fx = (60 lb) cos 20° = 56.382 lb Fy = (60 lb) sin 20° = 20.521 lb Fx = (80 lb) cos 95° = −6.9725 lb Fy = (80 lb)sin 95° = 79.696 lb

120-lb Force:

Fx = (120 lb) cos 5 ° = 119.543 lb Fy = (120 lb)sin 5° = 10.459 lb

Then

Rx = ΣFx = 168.953 lb Ry = ΣFy = 110.676 lb

and

R = (168.953 lb) 2 + (110.676 lb)2 = 201.976 lb

110.676 168.953 tan α = 0.65507 α = 33.228° tan α =

R = 202 lb

33.2° 



80-lb Force:

www.elsolucionario.net PROBLEM 2.39 For the collar of Problem 2.35, determine (a) the required value of α if the resultant of the three forces shown is to be vertical, (b) the corresponding magnitude of the resultant.

SOLUTION Rx = ΣFx

(1)

Ry = ΣFy

= −(100 N) sin α − (150 N)sin (α + 30°) − (200 N)sin α Ry = −(300 N) sin α − (150 N)sin (α + 30°)

(a)

(2)

For R to be vertical, we must have Rx = 0. We make Rx = 0 in Eq. (1): −100 cos α + 150cos (α + 30°) = 0 −100cos α + 150 (cos α cos 30° − sin α sin 30°) = 0 29.904cos α = 75sin α 29.904 75 = 0.39872

tan α =

α = 21.738° (b)

α = 21.7° 

Substituting for α in Eq. (2): Ry = −300sin 21.738° − 150sin 51.738° = −228.89 N R = | Ry | = 228.89 N

R = 229 N 



= (100 N) cos α + (150 N) cos (α + 30°) − (200 N) cos α Rx = −(100 N) cos α + (150 N) cos (α + 30°)

www.elsolucionario.net PROBLEM 2.40 For the post of Prob. 2.36, determine (a) the required tension in rope AC if the resultant of the three forces exerted at point C is to be horizontal, (b) the corresponding magnitude of the resultant.

Rx = ΣFx = − Rx = −

48 TAC + 640 N 73

R y = ΣFy = − Ry = −

(a)

960 24 4 TAC + (500 N) + (200 N) 1460 25 5

1100 7 3 TAC + (500 N) − (200 N) 1460 25 5

55 TAC + 20 N 73

(2)

For R to be horizontal, we must have R y = 0. Set R y = 0 in Eq. (2):

−

55 TAC + 20 N = 0 73 TAC = 26.545 N

(b)

(1)

TAC = 26.5 N 

Substituting for TAC into Eq. (1) gives 48 (26.545 N) + 640 N 73 Rx = 622.55 N Rx = −

R = Rx = 623 N

R = 623 N 



SOLUTION

www.elsolucionario.net PROBLEM 2.41 A hoist trolley is subjected to the three forces shown. Knowing that α = 40°, determine (a) the required magnitude of the force P if the resultant of the three forces is to be vertical, (b) the corresponding magnitude of the resultant.

SOLUTION ΣFx = P + (200 lb)sin 40° − (400 lb) cos 40°

Rx = P − 177.860 lb Ry =

ΣFy = (200 lb) cos 40° + (400 lb) sin 40°

Ry = 410.32 lb

(a)

(2)

For R to be vertical, we must have Rx = 0. Set

Rx = 0 in Eq. (1)

0 = P − 177.860 lb P = 177.860 lb

(b)

(1)

P = 177.9 lb 

Since R is to be vertical: R = Ry = 410 lb

R = 410 lb 



Rx =

www.elsolucionario.net PROBLEM 2.42 A hoist trolley is subjected to the three forces shown. Knowing that P = 250 lb, determine (a) the required value of α if the resultant of the three forces is to be vertical, (b) the corresponding magnitude of the resultant.

SOLUTION Rx =

ΣFx = 250 lb + (200 lb)sin α − (400 lb) cos α

Rx = 250 lb + (200 lb)sin α − (400 lb) cos α Ry =

ΣFy = (200 lb) cos α + (400 lb)sin α

For R to be vertical, we must have Rx = 0. Rx = 0 in Eq. (1)

Set

0 = 250 lb + (200 lb)sin α − (400 lb) cos α

(400 lb) cos α = (200 lb) sin α + 250 lb 2 cos α = sin α + 1.25 4cos 2 α = sin 2 α + 2.5sin α + 1.5625 4(1 − sin 2 α ) = sin 2 α + 2.5sin α + 1.5625 0 = 5sin 2 α + 2.5sin α − 2.4375

Using the quadratic formula to solve for the roots gives sin α = 0.49162

α = 29.447°

or (b)

α = 29.4° 

Since R is to be vertical: R = Ry = (200 lb) cos 29.447° + (400 lb) sin 29.447°

R = 371 lb 



(a)

(1)

www.elsolucionario.net PROBLEM 2.43 Two cables are tied together at C and are loaded as shown. Determine the tension (a) in cable AC, (b) in cable BC.

SOLUTION


Free-Body Diagram 1100 960 α = 48.888° 400 tan β = 960 β = 22.620° tan α =

Force Triangle Law of sines: TAC TBC 15.696 kN = = sin 22.620° sin 48.888° sin108.492°

(a)

TAC =

15.696 kN (sin 22.620°) sin108.492°

TAC = 6.37 kN 

(b)

TBC =

15.696 kN (sin 48.888°) sin108.492°

TBC = 12.47 kN 



3 2.25 α = 53.130° 1.4 tan β = 2.25 β = 31.891° tan α =


SOLUTION Free-Body Diagram

Law of sines:

Force-Triangle

TAC TBC 660 N = = sin 31.891° sin 53.130° sin 94.979°

(a)

TAC =

660 N (sin 31.891°) sin 94.979°

TAC = 350 N 

(b)

TBC =

660 N (sin 53.130°) sin 94.979°

TBC = 530 N 


www.elsolucionario.net PROBLEM 2.45 Knowing that α = 20°, determine the tension (a) in cable AC, (b) in rope BC.

SOLUTION

Law of sines:

Force Triangle

TAC T 1200 lb = BC = sin 110° sin 5° sin 65°

(a)

TAC =

1200 lb sin 110° sin 65°

TAC = 1244 lb 

(b)

TBC =

1200 lb sin 5° sin 65°

TBC = 115.4 lb 



Free-Body Diagram

www.elsolucionario.net PROBLEM 2.46 Knowing that α = 55° and that boom AC exerts on pin C a force directed along line AC, determine (a) the magnitude of that force, (b) the tension in cable BC.

SOLUTION

Law of sines:

Force Triangle

FAC T 300 lb = BC = sin 35° sin 50° sin 95°

(a)

FAC =

300 lb sin 35° sin 95°

(b)

TBC =

300 lb sin 50° sin 95°

FAC = 172.7 lb  TBC = 231 lb 



Free-Body Diagram

www.elsolucionario.net PROBLEM 2.47 Two cables are tied together at C and loaded as shown. Determine the tension (a) in cable AC, (b) in cable BC.

SOLUTION Free-Body Diagram 1.4 4.8 α = 16.2602°

tan α =

1.6 3 β = 28.073°

Force Triangle Law of sines: TAC TBC 1.98 kN = = sin 61.927° sin 73.740° sin 44.333°

(a)

TAC =

1.98 kN sin 61.927° sin 44.333°

TAC = 2.50 kN 

(b)

TBC =

1.98 kN sin 73.740° sin 44.333°

TBC = 2.72 kN 



tan β =

www.elsolucionario.net PROBLEM 2.48 Two cables are tied together at C and are loaded as shown. Knowing that P = 500 N and α = 60°, determine the tension in (a) in cable AC, (b) in cable BC.

SOLUTION

Law of sines:

Force Triangle

TAC T 500 N = BC = sin 35° sin 75° sin 70°

(a)

TAC =

500 N sin 35° sin 70°

TAC = 305 N 

(b)

TBC =

500 N sin 75° sin 70°

TBC = 514 N 



Free-Body Diagram

www.elsolucionario.net PROBLEM 2.49 Two forces of magnitude TA = 8 kips and TB = 15 kips are applied as shown to a welded connection. Knowing that the connection is in equilibrium, determine the magnitudes of the forces TC and TD.

SOLUTION

ΣFx = 0

15 kips − 8 kips − TD cos 40° = 0

TD = 9.1379 kips ΣFy = 0

TD sin 40° − TC = 0 (9.1379 kips) sin 40° − TC = 0

TC = 5.8737 kips

TC = 5.87 kips  TD = 9.14 kips 



Free-Body Diagram

www.elsolucionario.net PROBLEM 2.50 Two forces of magnitude TA = 6 kips and TC = 9 kips are applied as shown to a welded connection. Knowing that the connection is in equilibrium, determine the magnitudes of the forces TB and TD.

SOLUTION

Σ Fx = 0

TB − 6 kips − TD cos 40° = 0

Σ Fy = 0

TD sin 40° − 9 kips = 0

(1)

9 kips sin 40° TD = 14.0015 kips TD =

Substituting for TD into Eq. (1) gives: TB − 6 kips − (14.0015 kips) cos 40° = 0 TB = 16.7258 kips TB = 16.73 kips  TD = 14.00 kips 



Free-Body Diagram

www.elsolucionario.net PROBLEM 2.51 Two cables are tied together at C and loaded as shown. Knowing that P = 360 N, determine the tension (a) in cable AC, (b) in cable BC.

SOLUTION

(a)

ΣFx = 0: −

(b)

ΣFy = 0:

12 4 TAC + (360 N) = 0 13 5

TAC = 312 N 

5 3 (312 N) + TBC + (360 N) − 480 N = 0 13 5 TBC = 480 N − 120 N − 216 N

TBC = 144 N 



Free Body: C

www.elsolucionario.net PROBLEM 2.52 Two cables are tied together at C and loaded as shown. Determine the range of values of P for which both cables remain taut.

SOLUTION Free Body: C 12 4 TAC + P = 0 13 5 TAC =

ΣFy = 0:

Substitute for TAC from (1):

13 P 15

(1)

5 3 TAC + TBC + P − 480 N = 0 13 5

3  5  13   13  15  P + TBC + 5 P − 480 N = 0    TBC = 480 N −

14 P 15

(2)

From (1), TAC ⬎ 0 requires P ⬎ 0. From (2), TBC ⬎ 0 requires

14 P ⬍ 480 N, P ⬍ 514.29 N 15 0 ⬍ P ⬍ 514 N 

Allowable range:



ΣFx = 0: −

www.elsolucionario.net PROBLEM 2.53 A sailor is being rescued using a boatswain’s chair that is suspended from a pulley that can roll freely on the support cable ACB and is pulled at a constant speed by cable CD. Knowing that α = 30° and β = 10° and that the combined weight of the boatswain’s chair and the sailor is 900 N, determine the tension (a) in the support cable ACB, (b) in the traction cable CD.

Free-Body Diagram

ΣFx = 0: TACB cos 10° − TACB cos 30° − TCD cos 30° = 0 TCD = 0.137158TACB

(1)

ΣFy = 0: TACB sin 10° + TACB sin 30° + TCD sin 30° − 900 = 0 0.67365TACB + 0.5TCD = 900

(a)

Substitute (1) into (2):

0.67365 TACB + 0.5(0.137158 TACB ) = 900 TACB = 1212.56 N

(b)

From (1):

(2)

TACB = 1213 N 

TCD = 0.137158(1212.56 N)

TCD = 166.3 N 



SOLUTION

www.elsolucionario.net PROBLEM 2.54 A sailor is being rescued using a boatswain’s chair that is suspended from a pulley that can roll freely on the support cable ACB and is pulled at a constant speed by cable CD. Knowing that α = 25° and β = 15° and that the tension in cable CD is 80 N, determine (a) the combined weight of the boatswain’s chair and the sailor, (b) in tension in the support cable ACB.

SOLUTION


Free-Body Diagram

ΣFx = 0: TACB cos 15° − TACB cos 25° − (80 N) cos 25° = 0 TACB = 1216.15 N

ΣFy = 0: (1216.15 N) sin 15° + (1216.15 N) sin 25°

+ (80 N) sin 25° − W = 0 W = 862.54 N

(a)

W = 863 N 

(b) TACB = 1216 N 


www.elsolucionario.net PROBLEM 2.55 Two forces P and Q are applied as shown to an aircraft connection. Knowing that the connection is in equilibrium and that P = 500 lb and Q = 650 lb, determine the magnitudes of the forces exerted on the rods A and B.

SOLUTION

Resolving the forces into x- and y-directions: R = P + Q + FA + FB = 0

Substituting components:

R = −(500 lb) j + [(650 lb) cos 50°]i

− [(650 lb) sin 50°] j + FB i − ( FA cos 50°)i + ( FA sin 50°) j = 0

In the y-direction (one unknown force): −500 lb − (650 lb)sin 50° + FA sin 50° = 0

Thus,

FA =

500 lb + (650 lb) sin 50° sin 50°

= 1302.70 lb

In the x-direction: Thus,

FA = 1303 lb 

(650 lb) cos 50° + FB − FA cos 50° = 0 FB = FA cos 50° − (650 lb) cos50° = (1302.70 lb) cos 50° − (650 lb) cos 50° = 419.55 lb

FB = 420 lb 



Free-Body Diagram

www.elsolucionario.net PROBLEM 2.56 Two forces P and Q are applied as shown to an aircraft connection. Knowing that the connection is in equilibrium and that the magnitudes of the forces exerted on rods A and B are FA = 750 lb and FB = 400 lb, determine the magnitudes of P and Q.



Resolving the forces into x- and y-directions: R = P + Q + FA + FB = 0

Substituting components:

R = − Pj + Q cos 50°i − Q sin 50° j − [(750 lb) cos 50°]i + [(750 lb)sin 50°] j + (400 lb)i

In the x-direction (one unknown force): Q cos 50° − [(750 lb) cos 50°] + 400 lb = 0

(750 lb) cos 50° − 400 lb cos 50° = 127.710 lb

Q=

In the y-direction:

− P − Q sin 50° + (750 lb) sin 50° = 0 P = −Q sin 50° + (750 lb) sin 50° = −(127.710 lb)sin 50° + (750 lb) sin 50° = 476.70 lb

P = 477 lb; Q = 127.7 lb 


www.elsolucionario.net PROBLEM 2.57 Two cables tied together at C are loaded as shown. Knowing that the maximum allowable tension in each cable is 800 N, determine (a) the magnitude of the largest force P that can be applied at C, (b) the corresponding value of α.

SOLUTION Force Triangle

Force triangle is isosceles with 2 β = 180° − 85° β = 47.5° P = 2(800 N)cos 47.5° = 1081 N

(a)

P = 1081 N 

Since P ⬎ 0, the solution is correct. (b)

α = 180° − 50° − 47.5° = 82.5°

α = 82.5° 



Free-Body Diagram: C

www.elsolucionario.net PROBLEM 2.58 Two cables tied together at C are loaded as shown. Knowing that the maximum allowable tension is 1200 N in cable AC and 600 N in cable BC, determine (a) the magnitude of the largest force P that can be applied at C, (b) the corresponding value of α.

SOLUTION

(a)

Law of cosines:

Force Triangle

P 2 = (1200 N) 2 + (600 N) 2 − 2(1200 N)(600 N) cos 85° P = 1294.02 N

Since P ⬎ 1200 N, the solution is correct. P = 1294 N 

(b)

Law of sines: sin β sin 85° = 1200 N 1294.02 N β = 67.5° α = 180° − 50° − 67.5°

α = 62.5° 



Free-Body Diagram

www.elsolucionario.net PROBLEM 2.59 For the situation described in Figure P2.45, determine (a) the value of α for which the tension in rope BC is as small as possible, (b) the corresponding value of the tension. PROBLEM 2.45 Knowing that α = 20°, determine the tension (a) in cable AC, (b) in rope BC.


To be smallest, TBC must be perpendicular to the direction of TAC . (a) (b)

Thus,

α = 5°

α = 5.00°

TBC = (1200 lb) sin 5°



TBC = 104.6 lb 



Free-Body Diagram

www.elsolucionario.net PROBLEM 2.60 For the structure and loading of Problem 2.46, determine (a) the value of α for which the tension in cable BC is as small as possible, (b) the corresponding value of the tension.

TBC must be perpendicular to FAC to be as small as possible.

Free-Body Diagram: C

Force Triangle is a right triangle

To be a minimum, TBC must be perpendicular to FAC . (a)

We observe:

α = 90° − 30°

α = 60.0° 

TBC = (300 lb)sin 50°

(b) or

TBC = 229.81 lb

TBC = 230 lb 



SOLUTION

www.elsolucionario.net PROBLEM 2.61 For the cables of Problem 2.48, it is known that the maximum allowable tension is 600 N in cable AC and 750 N in cable BC. Determine (a) the maximum force P that can be applied at C, (b) the corresponding value of α.

SOLUTION

(a)

Law of cosines

Force Triangle

P 2 = (600) 2 + (750)2 − 2(600)(750) cos (25° + 45°) P = 784.02 N

(b)

Law of sines

P = 784 N 

sin β sin (25° + 45°) = 600 N 784.02 N

β = 46.0°

∴ α = 46.0° + 25°

α = 71.0° 



Free-Body Diagram

www.elsolucionario.net PROBLEM 2.62 A movable bin and its contents have a combined weight of 2.8 kN. Determine the shortest chain sling ACB that can be used to lift the loaded bin if the tension in the chain is not to exceed 5 kN.


h 0.6 m

(1)


tan α =

Isosceles Force Triangle

Law of sines: sin α =

1 2

(2.8 kN) TAC

TAC = 5 kN sin α =

1 2

(2.8 kN)

5 kN α = 16.2602°

From Eq. (1): tan16.2602° =

h 0.6 m

∴ h = 0.175000 m

Half length of chain = AC = (0.6 m) 2 + (0.175 m)2 = 0.625 m

Total length:

= 2 × 0.625 m

1.250 m 


www.elsolucionario.net PROBLEM 2.63 Collar A is connected as shown to a 50-lb load and can slide on a frictionless horizontal rod. Determine the magnitude of the force P required to maintain the equilibrium of the collar when (a) x = 4.5 in., (b) x = 15 in.

SOLUTION Free Body: Collar A

Force Triangle P 50 lb = 4.5 20.5

(b)

Free Body: Collar A

P = 10.98 lb 

Force Triangle P 50 lb = 15 25

P = 30.0 lb 





(a)

www.elsolucionario.net PROBLEM 2.64 Collar A is connected as shown to a 50-lb load and can slide on a frictionless horizontal rod. Determine the distance x for which the collar is in equilibrium when P = 48 lb.


N 2 = (50) 2 − (48) 2 = 196 N = 14.00 lb

Similar Triangles x 48 lb = 20 in. 14 lb x = 68.6 in. 



Free Body: Collar A

www.elsolucionario.net PROBLEM 2.65 Three forces are applied to a bracket as shown. The directions of the two 150-N forces may vary, but the angle between these forces is always 50°. Determine the range of values of α for which the magnitude of the resultant of the forces acting at A is less than 600 N.

SOLUTION


Combine the two 150-N forces into a resultant force Q:

Q = 2(150 N) cos 25° = 271.89 N

Equivalent loading at A:

Using the law of cosines: (600 N) 2 = (500 N) 2 + (271.89 N)2 + 2(500 N)(271.89 N) cos(55° + α ) cos(55° + α ) = 0.132685

Two values for α :

55° + α = 82.375 α = 27.4°

or

55° + α = −82.375° 55° + α = 360° − 82.375° α = 222.6° 27.4° < α < 222.6 

For R < 600 lb:


www.elsolucionario.net PROBLEM 2.66 A 200-kg crate is to be supported by the rope-and-pulley arrangement shown. Determine the magnitude and direction of the force P that must be exerted on the free end of the rope to maintain equilibrium. (Hint: The tension in the rope is the same on each side of a simple pulley. This can be proved by the methods of Ch. 4.)

Free-Body Diagram: Pulley A  5  ΣFx = 0: − 2 P   + P cos α = 0  281  cos α = 0.59655 α = ±53.377°

For α = +53.377°:  16  ΣFy = 0: 2 P   + P sin 53.377° − 1962 N = 0  281  P = 724 N

53.4° 

For α = −53.377°:  16  ΣFy = 0: 2 P   + P sin(−53.377°) − 1962 N = 0  281  P = 1773

53.4° 



SOLUTION

www.elsolucionario.net PROBLEM 2.67 A 600-lb crate is supported by several rope-andpulley arrangements as shown. Determine for each arrangement the tension in the rope. (See the hint for Problem 2.66.)

SOLUTION Free-Body Diagram of Pulley ΣFy = 0: 2T − (600 lb) = 0 T=

1 (600 lb) 2 T = 300 lb 

(b)

ΣFy = 0: 2T − (600 lb) = 0 T=

1 (600 lb) 2 T = 300 lb 

(c)

ΣFy = 0: 3T − (600 lb) = 0 1 T = (600 lb) 3 T = 200 lb 

(d)

ΣFy = 0: 3T − (600 lb) = 0 1 T = (600 lb) 3 T = 200 lb 

(e)

ΣFy = 0: 4T − (600 lb) = 0 T=

1 (600 lb) 4 T = 150.0 lb 





(a)

www.elsolucionario.net PROBLEM 2.68 Solve Parts b and d of Problem 2.67, assuming that the free end of the rope is attached to the crate. PROBLEM 2.67 A 600-lb crate is supported by several rope-and-pulley arrangements as shown. Determine for each arrangement the tension in the rope. (See the hint for Problem 2.66.)

SOLUTION

(b)

ΣFy = 0: 3T − (600 lb) = 0 1 T = (600 lb) 3 T = 200 lb 

(d)

ΣFy = 0: 4T − (600 lb) = 0 T=

1 (600 lb) 4 T = 150.0 lb 





Free-Body Diagram of Pulley and Crate

www.elsolucionario.net PROBLEM 2.69 A load Q is applied to the pulley C, which can roll on the cable ACB. The pulley is held in the position shown by a second cable CAD, which passes over the pulley A and supports a load P. Knowing that P = 750 N, determine (a) the tension in cable ACB, (b) the magnitude of load Q.

SOLUTION Free-Body Diagram: Pulley C ΣFx = 0: TACB (cos 25° − cos 55°) − (750 N) cos 55° = 0

(a)

TACB = 1292.88 N

Hence:

ΣFy = 0: TACB (sin 25° + sin 55°) + (750 N) sin 55° − Q = 0

(b) 

(1292.88 N)(sin 25° + sin 55°) + (750 N) sin 55° − Q = 0 Q = 2219.8 N

or

Q = 2220 N 



TACB = 1293 N 

www.elsolucionario.net PROBLEM 2.70 An 1800-N load Q is applied to the pulley C, which can roll on the cable ACB. The pulley is held in the position shown by a second cable CAD, which passes over the pulley A and supports a load P. Determine (a) the tension in cable ACB, (b) the magnitude of load P.

SOLUTION Free-Body Diagram: Pulley C ΣFx = 0: TACB (cos 25° − cos 55°) − P cos 55° = 0 P = 0.58010TACB

or

(1)

1.24177TACB + 0.81915 P = 1800 N

or (a)

(2)

Substitute Equation (1) into Equation (2): 1.24177TACB + 0.81915(0.58010TACB ) = 1800 N TACB = 1048.37 N

Hence:

TACB = 1048 N 

(b)

P = 0.58010(1048.37 N) = 608.16 N

Using (1),

P = 608 N 



ΣFy = 0: TACB (sin 25° + sin 55°) + P sin 55° − 1800 N = 0

www.elsolucionario.net PROBLEM 2.71 Determine (a) the x, y, and z components of the 900-N force, (b) the angles θx, θy, and θz that the force forms with the coordinate axes.

SOLUTION Fh = F cos 65°

(a)

Fx = Fh sin 20° = (380.36 N)sin 20° Fx = −130.091 N,

Fx = −130.1 N 

Fy = F sin 65° = (900 N)sin 65° Fy = +815.68 N,

Fy = +816 N 

Fz = Fh cos 20° = (380.36 N) cos 20° Fz = +357.42 N

(b)

cos θ x = cos θ y =

cos θ z =

Fx −130.091 N = 900 N F Fy F

=

+815.68 N 900 N

Fz +357.42 N = 900 N F

Fz = +357 N 

θ x = 98.3°  θ y = 25.0° 

θ z = 66.6° 



= (900 N) cos 65° Fh = 380.36 N

www.elsolucionario.net PROBLEM 2.72 Determine (a) the x, y, and z components of the 750-N force, (b) the angles θx, θy, and θz that the force forms with the coordinate axes.

SOLUTION Fh = F sin 35°

(a)

Fx = Fh cos 25° = (430.18 N) cos 25° Fx = +389.88 N,

Fx = +390 N 

Fy = F cos 35°



= (750 N) cos 35° Fy = +614.36 N,

Fy = +614 N 

Fz = Fh sin 25° = (430.18 N)sin 25° Fz = +181.802 N

(b)

cos θ x = cos θ y =

cos θ z =

Fx +389.88 N = 750 N F Fy F

=

+614.36 N 750 N

Fz +181.802 N = 750 N F

Fz = +181.8 N 

θ x = 58.7°  θ y = 35.0° 

θ z = 76.0° 



= (750 N)sin 35° Fh = 430.18 N

www.elsolucionario.net PROBLEM 2.73 A gun is aimed at a point A located 35° east of north. Knowing that the barrel of the gun forms an angle of 40° with the horizontal and that the maximum recoil force is 400 N, determine (a) the x, y, and z components of that force, (b) the values of the angles θx, θy, and θz defining the direction of the recoil force. (Assume that the x, y, and z axes are directed, respectively, east, up, and south.)

SOLUTION Recoil force

F = 400 N

∴ FH = (400 N) cos 40°

(a)

Fx = − FH sin 35° = −(306.42 N)sin 35° = −175.755 N

Fx = −175.8 N 

Fy = − F sin 40° = −(400 N)sin 40° = −257.12 N

Fy = −257 N 

Fz = + FH cos 35° = +(306.42 N) cos35° = +251.00 N

(b)

cos θ x =

Fx −175.755 N = 400 N F −257.12 N 400 N

cos θ y =

Fy

cos θ z =

Fz 251.00 N = 400 N F

F

=

Fz = +251 N 

θ x = 116.1°  θ y = 130.0° 

θ z = 51.1° 



= 306.42 N

www.elsolucionario.net PROBLEM 2.74 Solve Problem 2.73, assuming that point A is located 15° north of west and that the barrel of the gun forms an angle of 25° with the horizontal. PROBLEM 2.73 A gun is aimed at a point A located 35° east of north. Knowing that the barrel of the gun forms an angle of 40° with the horizontal and that the maximum recoil force is 400 N, determine (a) the x, y, and z components of that force, (b) the values of the angles θx, θy, and θz defining the direction of the recoil force. (Assume that the x, y, and z axes are directed, respectively, east, up, and south.)

SOLUTION Recoil force

F = 400 N

∴ FH = (400 N) cos 25°

(a)

Fx = + FH cos15° = + (362.52 N) cos15° = +350.17 N

Fx = +350 N 

Fy = − F sin 25° = −(400 N)sin 25° = −169.047 N

Fy = −169.0 N 

Fz = + FH sin15° = +(362.52 N)sin15° = +93.827 N

(b)

cos θ x = cos θ y = cos θ z =

Fx +350.17 N = 400 N F Fy F

=

−169.047 N 400 N

Fz +93.827 N = 400 N F

Fz = +93.8 N 

θ x = 28.9°  θ y = 115.0°  θ z = 76.4° 



= 362.52 N

www.elsolucionario.net PROBLEM 2.75

SOLUTION 56 ft 65 ft = 0.86154

cos θ y =

From triangle AOB:

θ y = 30.51° Fx = − F sin θ y cos 20°

(a)

= −(3900 lb)sin 30.51° cos 20° Fx = −1861 lb  Fy = + F cos θ y = (3900 lb)(0.86154)



Fz = + (3900 lb)sin 30.51° sin 20°

(b)

cos θ x =

From above:

Fx 1861 lb =− = − 0.4771 3900 lb F

θ y = 30.51° cos θ z =

Fz 677 lb =+ = + 0.1736 3900 lb F

Fy = +3360 lb  Fz = +677 lb 

θ x = 118.5°  θ y = 30.5°  θ z = 80.0° 



Cable AB is 65 ft long, and the tension in that cable is 3900 lb. Determine (a) the x, y, and z components of the force exerted by the cable on the anchor B, (b) the angles θ x , θ y , and θ z defining the direction of that force.



Cable AC is 70 ft long, and the tension in that cable is 5250 lb. Determine (a) the x, y, and z components of the force exerted by the cable on the anchor C, (b) the angles θx, θy, and θz defining the direction of that force.

SOLUTION AC = 70 ft OA = 56 ft F = 5250 lb

In triangle AOB:

cos θ y =



56 ft 70 ft

θ y = 36.870° FH = F sin θ y = (5250 lb) sin 36.870° = 3150.0 lb

(a)

(b)

Fx = − FH sin 50° = −(3150.0 lb)sin 50° = −2413.04 lb

Fx = −2413 lb 

Fy = + F cos θ y = + (5250 lb) cos 36.870° = +4200.0 lb

Fy = +4200 lb 

Fz = − FH cos 50° = −3150cos50° = −2024.8 lb

Fz = −2025 lb 

cos θ x =

From above:

Fx −2413.04 lb = 5250 lb F

θ y = 36.870° θz =

Fz −2024.8 lb = 5250 lb F

θ x = 117.4°  θ y = 36.9°  θ z = 112.7° 


www.elsolucionario.net PROBLEM 2.77 The end of the coaxial cable AE is attached to the pole AB, which is strengthened by the guy wires AC and AD. Knowing that the tension in wire AC is 120 lb, determine (a) the components of the force exerted by this wire on the pole, (b) the angles θx, θy, and θz that the force forms with the coordinate axes.

(a)

Fx = (120 lb) cos 60° cos 20° Fx = 56.382 lb

Fx = +56.4 lb 

Fy = −(120 lb)sin 60° Fy = −103.923 lb

Fy = −103.9 lb 

Fz = −(120 lb) cos 60° sin 20° Fz = −20.521 lb

(b)

Fz = −20.5 lb 

cos θ x =

Fx 56.382 lb = F 120 lb

cos θ y =

Fy

cos θ z =

F

=

−103.923 lb 120 lb

Fz −20.52 lb = F 120 lb

θ x = 62.0°  θ y = 150.0°  θ z = 99.8° 



SOLUTION

www.elsolucionario.net PROBLEM 2.78 The end of the coaxial cable AE is attached to the pole AB, which is strengthened by the guy wires AC and AD. Knowing that the tension in wire AD is 85 lb, determine (a) the components of the force exerted by this wire on the pole, (b) the angles θx, θy, and θz that the force forms with the coordinate axes.

(a)

Fx = (85 lb)sin 36° sin 48°

= 37.129 lb

Fx = 37.1 lb 

Fy = −(85 lb) cos 36°

= −68.766 lb

Fy = −68.8 lb 

Fz = (85 lb)sin 36° cos 48° Fz = 33.4 lb 

= 33.431 lb

(b)

cos θ x =

Fx 37.129 lb = F 85 lb

cos θ y =

Fy

cos θ z =

Fz 33.431 lb = F 85 lb

F

=

−68.766 lb 85 lb

θ x = 64.1°  θ y = 144.0°  θ z = 66.8° 



SOLUTION

www.elsolucionario.net PROBLEM 2.79 Determine the magnitude and direction of the force F = (690 lb)i + (300 lb)j – (580 lb)k.

SOLUTION F = (690 N)i + (300 N) j − (580 N)k

F = Fx2 + Fy2 + Fz2 = (690 N) 2 + (300 N)2 + (−580 N) 2

cos θ x =

Fx 690 N = F 950 N

θ x = 43.4° 

cos θ y =

Fy

θ y = 71.6° 

cos θ z =

F

=

300 N 950 N

Fz −580 N = F 950 N

θ z = 127.6° 



F = 950 N 

= 950 N

www.elsolucionario.net PROBLEM 2.80 Determine the magnitude and direction of the force F = (650 N)i − (320 N)j + (760 N)k.

SOLUTION F = (650 N)i − (320 N) j + (760 N)k F = Fx2 + Fy2 + Fz2

F = 1050 N 

cos θ x =

Fx 650 N = F 1050 N

θ x = 51.8° 

cos θ y =

Fy

−320 N 1050 N

θ y = 107.7° 

cos θ z =

Fz 760 N = F 1050 N

θ z = 43.6° 

F

=



= (650 N) 2 + (−320 N)2 + (760 N)2

www.elsolucionario.net PROBLEM 2.81 A force acts at the origin of a coordinate system in a direction defined by the angles θx = 75° and θz = 130°. Knowing that the y component of the force is +300 lb, determine (a) the angle θy, (b) the other components and the magnitude of the force.

SOLUTION cos 2 θ x + cos 2 θ y + cos 2 θ z = 1 cos 2 (75°) + cos 2 θ y + cos 2 (130°) = 1 cos θ y = ±0.72100

(b)

Since Fy ⬎ 0, we choose cos θ y ⫽⫹0.72100

∴ θ y = 43.9° 

Fy = F cos θ y 300 lb = F (0.72100) F = 416.09 lb

F = 416 lb 

Fx = F cos θ x = 416.09 lb cos 75° Fz = F cos θ z = 416.09 lb cos130°

Fx = +107.7 lb  Fz = −267 lb 



(a)

www.elsolucionario.net PROBLEM 2.82 A force acts at the origin of a coordinate system in a direction defined by the angles θy = 55° and θz = 45°. Knowing that the x component of the force is −500 N, determine (a) the angle θx, (b) the other components and the magnitude of the force.

SOLUTION cos 2 θ x + cos 2 θ y + cos 2 θ z = 1 cos 2 θ x + cos 2 55° + cos 2 45° = 1 cos θ x = ±0.41353

(a) (b)

Since Fy ⬍ 0, we choose cos θ x ⫽⫺0.41353

∴ θ x = 114.4° 

Fx = F cos θ x F = 1209.10 N

F = 1209.1 N 

Fy = F cos θ y = 1209.10 N cos55°

Fy = +694 N 

Fz = F cos θ z = 1209.10 N cos 45°

Fz = +855 N 



−500 N = F (−0.41353)

www.elsolucionario.net PROBLEM 2.83 A force F of magnitude 230 N acts at the origin of a coordinate system. Knowing that θx = 32.5°, Fy = −60 N, and Fz > 0, determine (a) the components Fx and Fz, (b) the angles θy and θz.

SOLUTION (a)

We have Fx = F cos θ x = (230 N) cos 32.5°

Then:

Fx = −194.0 N 

Fx = 193.980 N

F 2 = Fx2 + Fy2 + Fz2

Hence: (b)

(230 N) 2 = (193.980 N) 2 + (−60 N) 2 + Fz2

Fz = + (230 N)2 − (193.980 N)2 − (−60 N)2

Fz = 108.0 N 

Fz = 108.036 N −60 N = = − 0.26087 F 230 N F 108.036 N cos θ z = z = = 0.46972 230 N F cos θ y =

Fy

θ y = 105.1°  θ z = 62.0° 



So:

www.elsolucionario.net PROBLEM 2.84 A force F of magnitude 210 N acts at the origin of a coordinate system. Knowing that Fx = 80 N, θz = 151.2°, and Fy < 0, determine (a) the components Fy and Fz, (b) the angles θx and θy.

SOLUTION Fz = F cos θ z = (210 N) cos151.2° = −184.024 N

Then: So: Hence:

F 2 = Fx2 + Fy2 + Fz2 (210 N) 2 = (80 N) 2 + ( Fy ) 2 + (184.024 N)2 Fy = − (210 N) 2 − (80 N) 2 − (184.024 N) 2

= −61.929 N

(b)

Fz = −184.0 N 

Fy = −62.0 lb 

cos θ x =

Fx 80 N = = 0.38095 F 210 N

θ x = 67.6° 

cos θ y =

Fy

θ y = 107.2° 

F

=

61.929 N = −0.29490 210 N



(a)

www.elsolucionario.net PROBLEM 2.85 In order to move a wrecked truck, two cables are attached at A and pulled by winches B and C as shown. Knowing that the tension in cable AB is 2 kips, determine the components of the force exerted at A by the cable.

Cable AB:

λAB

 AB (−46.765 ft)i + (45 ft) j + (36 ft)k = = 74.216 ft AB

TAB = TAB λAB =

−46.765i + 45 j + 36k 74.216

(TAB ) x = −1.260 kips  (TAB ) y = +1.213 kips  (TAB ) z = +0.970 kips 



SOLUTION

www.elsolucionario.net PROBLEM 2.86 In order to move a wrecked truck, two cables are attached at A and pulled by winches B and C as shown. Knowing that the tension in cable AC is 1.5 kips, determine the components of the force exerted at A by the cable.

Cable AB:

λAC

 AC (−46.765 ft)i + (55.8 ft) j + (−45 ft)k = = 85.590 ft AC

TAC = TAC λAC = (1.5 kips)

−46.765i + 55.8 j − 45k 85.590

(TAC ) x = −0.820 kips  (TAC ) y = +0.978 kips  (TAC ) z = −0.789 kips 



SOLUTION

www.elsolucionario.net PROBLEM 2.87 Knowing that the tension in cable AB is 1425 N, determine the components of the force exerted on the plate at B.

SOLUTION

BA = (900 mm) 2 + (600 mm) 2 + (360 mm) 2 = 1140 mm TBA = TBA λ BA  BA = TBA BA 1425 N [ −(900 mm)i + (600 mm) j + (360 mm)k ] TBA = 1140 mm = −(1125 N)i + (750 N) j + (450 N)k

(TBA ) x = −1125 N, (TBA ) y = 750 N, (TBA ) z = 450 N 



 BA = −(900 mm)i + (600 mm) j + (360 mm)k

www.elsolucionario.net PROBLEM 2.88 Knowing that the tension in cable AC is 2130 N, determine the components of the force exerted on the plate at C.

SOLUTION

CA = (900 mm)2 + (600 mm)2 + (920 mm) 2 = 1420 mm TCA = TCA λ CA  CA = TCA CA 2130 N TCA = [−(900 mm)i + (600 mm) j − (920 mm)k ] 1420 mm = −(1350 N)i + (900 N) j − (1380 N)k (TCA ) x = −1350 N, (TCA ) y = 900 N, (TCA ) z = −1380 N 



 CA = −(900 mm)i + (600 mm) j − (920 mm)k

www.elsolucionario.net PROBLEM 2.89 A frame ABC is supported in part by cable DBE that passes through a frictionless ring at B. Knowing that the tension in the cable is 385 N, determine the components of the force exerted by the cable on the support at D.

 DB = (480 mm)i − (510 mm) j + (320 mm)k DB = (480 mm)2 + (510 mm 2 ) + (320 mm) 2 = 770 mm F = F λ DB  DB =F DB 385 N = [(480 mm)i − (510 mm)j + (320 mm)k ] 770 mm = (240 N)i − (255 N) j + (160 N)k Fx = +240 N, Fy = −255 N, Fz = +160.0 N 



SOLUTION

www.elsolucionario.net PROBLEM 2.90 For the frame and cable of Problem 2.89, determine the components of the force exerted by the cable on the support at E. PROBLEM 2.89 A frame ABC is supported in part by cable DBE that passes through a frictionless ring at B. Knowing that the tension in the cable is 385 N, determine the components of the force exerted by the cable on the support at D.

 EB = (270 mm)i − (400 mm) j + (600 mm)k EB = (270 mm)2 + (400 mm) 2 + (600 mm)2 = 770 mm F = F λ EB  EB =F EB 385 N = [(270 mm)i − (400 mm)j + (600 mm)k ] 770 mm F = (135 N)i − (200 N) j + (300 N)k Fx = +135.0 N, Fy = −200 N, Fz = +300 N 



SOLUTION

www.elsolucionario.net PROBLEM 2.91 Find the magnitude and direction of the resultant of the two forces shown knowing that P = 600 N and Q = 450 N.

P = (600 N)[sin 40° sin 25°i + cos 40° j + sin 40° cos 25°k ] = (162.992 N)i + (459.63 N) j + (349.54 N)k Q = (450 N)[cos 55° cos 30°i + sin 55° j − cos 55° sin 30°k ] = (223.53 N)i + (368.62 N) j − (129.055 N)k R =P+Q = (386.52 N)i + (828.25 N) j + (220.49 N)k R = (386.52 N)2 + (828.25 N)2 + (220.49 N) 2

R = 940 N 

= 940.22 N

cos θ x =

Rx 386.52 N = R 940.22 N

θ x = 65.7° 

cos θ y =

Ry

θ y = 28.2° 

cos θ z =

Rz 220.49 N = R 940.22 N

R

=

828.25 N 940.22 N

θ z = 76.4° 



SOLUTION


P = (450 N)[sin 40° sin 25°i + cos 40° j + sin 40° cos 25°k ] = (122.244 N)i + (344.72 N) j + (262.154 N)k Q = (600 N)[cos 55° cos 30°i + sin 55° j − cos 55° sin 30°k ] = (298.04 N)i + (491.49 N)j − (172.073 N)k R =P+Q = (420.28 N)i + (836.21 N) j + (90.081 N)k R = (420.28 N) 2 + (836.21 N) 2 + (90.081 N) 2 = 940.21 N

R = 940 N 

cos θ x =

Rx 420.28 = R 940.21

θ x = 63.4° 

cos θ y =

Ry

θ y = 27.2° 

cos θ z =

Rz 90.081 = R 940.21

R

=

836.21 940.21

θ z = 84.5° 



SOLUTION

www.elsolucionario.net PROBLEM 2.93 Knowing that the tension is 425 lb in cable AB and 510 lb in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables.

 AB = (40 in.)i − (45 in.) j + (60 in.)k AB = (40 in.)2 + (45 in.)2 + (60 in.) 2 = 85 in.  AC = (100 in.)i − (45 in.) j + (60 in.)k AC = (100 in.)2 + (45 in.)2 + (60 in.)2 = 125 in.   (40 in.)i − (45 in.) j + (60 in.)k  AB TAB = TAB λAB = TAB = (425 lb)   AB 85 in.   TAB = (200 lb)i − (225 lb) j + (300 lb)k   (100 in.)i − (45 in.) j + (60 in.)k  AC TAC = TAC λAC = TAC = (510 lb)   125 in. AC   TAC = (408 lb)i − (183.6 lb) j + (244.8 lb)k R = TAB + TAC = (608)i − (408.6 lb) j + (544.8 lb)k

Then: and

R = 912.92 lb

R = 913 lb 

cos θ x =

608 lb = 0.66599 912.92 lb

cos θ y =

408.6 lb = −0.44757 912.92 lb

cos θ z =

544.8 lb = 0.59677 912.92 lb

θ x = 48.2°  θ y = 116.6°  θ z = 53.4° 



SOLUTION

www.elsolucionario.net PROBLEM 2.94 Knowing that the tension is 510 lb in cable AB and 425 lb in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables.

 AB = (40 in.)i − (45 in.) j + (60 in.)k AB = (40 in.)2 + (45 in.)2 + (60 in.) 2 = 85 in.  AC = (100 in.)i − (45 in.) j + (60 in.)k AC = (100 in.)2 + (45 in.)2 + (60 in.)2 = 125 in.   (40 in.)i − (45 in.) j + (60 in.)k  AB TAB = TAB λAB = TAB = (510 lb)   AB 85 in.   TAB = (240 lb)i − (270 lb) j + (360 lb)k   (100 in.)i − (45 in.) j + (60 in.)k  AC TAC = TAC λAC = TAC = (425 lb)   125 in. AC   TAC = (340 lb)i − (153 lb) j + (204 lb)k R = TAB + TAC = (580 lb)i − (423 lb) j + (564 lb)k

Then: and

R = 912.92 lb

R = 913 lb 

cos θ x =

580 lb = 0.63532 912.92 lb

θ x = 50.6° 

cos θ y =

−423 lb = −0.46335 912.92 lb

θ y = 117.6° 

cos θ z =

564 lb = 0.61780 912.92 lb

θ z = 51.8° 



SOLUTION

www.elsolucionario.net PROBLEM 2.95 For the frame of Problem 2.89, determine the magnitude and direction of the resultant of the forces exerted by the cable at B knowing that the tension in the cable is 385 N. PROBLEM 2.89 A frame ABC is supported in part by cable DBE that passes through a frictionless ring at B. Knowing that the tension in the cable is 385 N, determine the components of the force exerted by the cable on the support at D.

SOLUTION


 BD = −(480 mm)i + (510 mm) j − (320 mm)k BD = (480 mm) 2 + (510 mm) 2 + (320 mm) 2 = 770 mm  BD FBD = TBD λ BD = TBD BD (385 N) [−(480 mm)i + (510 mm) j − (320 mm)k ] = (770 mm) = −(240 N)i + (255 N) j − (160 N)k  BE = −(270 mm)i + (400 mm) j − (600 mm)k BE = (270 mm) 2 + (400 mm) 2 + (600 mm) 2 = 770 mm  BE FBE = TBE λ BE = TBE BE (385 N) [−(270 mm)i + (400 mm) j − (600 mm)k ] = (770 mm) = −(135 N)i + (200 N) j − (300 N)k

R = FBD + FBE = −(375 N)i + (455 N) j − (460 N)k

R = (375 N)2 + (455 N)2 + (460 N)2 = 747.83 N

R = 748 N 

cos θ x =

−375 N 747.83 N

θ x = 120.1° 

cos θ y =

455 N 747.83 N

θ y = 52.5° 

cos θ z =

−460 N 747.83 N

θ z = 128.0° 


www.elsolucionario.net PROBLEM 2.96 For the cables of Problem 2.87, knowing that the tension is 1425 N in cable AB and 2130 N in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables.

SOLUTION (use results of Problem 2.87)

(TAB ) x = +1125 N (TAB ) y = −750 N (TAB ) z = − 450 N TAC = −TCA

(use results of Problem 2.88)

(TAC ) x = +1350 N (TAC ) y = −900 N (TAC ) z = +1380 N

Resultant:

Rx = ΣFx = +1125 + 1350 = +2475 N Ry = ΣFy = −750 − 900 = −1650 N Rz = ΣFz = −450 + 1380 = + 930 N R = Rx2 + Ry2 + Rz2 = (+2475)2 + (−1650) 2 + (+930) 2

= 3116.6 N cos θ x = cos θ y = cos θ z =

R = 3120 N 

Rx +2475 = R 3116.6

θ x = 37.4° 

−1650 3116.6

θ y = 122.0° 

Ry R

=

Rz + 930 = R 3116.6

θ z = 72.6° 



TAB = −TBA

www.elsolucionario.net PROBLEM 2.97 The boom OA carries a load P and is supported by two cables as shown. Knowing that the tension in cable AB is 183 lb and that the resultant of the load P and of the forces exerted at A by the two cables must be directed along OA, determine the tension in cable AC.

Cable AB:

TAB = 183 lb

TAB TAB Cable AC:

TAC TAC

Load P:


SOLUTION

 AB (−48 in.)i + (29 in.) j + (24 in.)k = TAB λ AB = TAB = (183 lb) AB 61 in. = −(144 lb)i + (87 lb) j + (72 lb)k  AC (−48 in.)i + (25 in.) j + (−36 in.)k = TAC λ AC = TAC = TAC AC 65 in. 48 25 36 = − TAC i + TAC j − TAC k 65 65 65

P = Pj

For resultant to be directed along OA, i.e., x-axis

Rz = 0: ΣFz = (72 lb) −

36 ′ =0 TAC 65

TAC = 130.0 lb 


www.elsolucionario.net PROBLEM 2.98 For the boom and loading of Problem. 2.97, determine the magnitude of the load P. PROBLEM 2.97 The boom OA carries a load P and is supported by two cables as shown. Knowing that the tension in cable AB is 183 lb and that the resultant of the load P and of the forces exerted at A by the two cables must be directed along OA, determine the tension in cable AC.

See Problem 2.97. Since resultant must be directed along OA, i.e., the x-axis, we write

Ry = 0: ΣFy = (87 lb) +

25 TAC − P = 0 65

TAC = 130.0 lb from Problem 2.97.

Then

(87 lb) +

25 (130.0 lb) − P = 0 65

P = 137.0 lb 



SOLUTION

www.elsolucionario.net PROBLEM 2.99 A container is supported by three cables that are attached to a ceiling as shown. Determine the weight W of the container, knowing that the tension in cable AB is 6 kN.

SOLUTION

The forces applied at A are:

TAB , TAC , TAD , and W

where W = W j. To express the other forces in terms of the unit vectors i, j, k, we write  AB = − (450 mm)i + (600 mm) j AB = 750 mm  AC = + (600 mm) j − (320 mm)k AC = 680 mm  AD = + (500 mm)i + (600 mm) j + (360 mm)k AD = 860 mm  AB (−450 mm)i + (600 mm) j and TAB = λ ABTAB = TAB = TAB AB 750 mm

 45 60  j  TAB = − i +  75 75  TAC = λ AC TAC = TAC

TAD = λ ADTAD = TAD

 AC (600 mm)i − (320 mm) j = TAC AC 680 mm 32   60 =  j − k  TAC 68 68    AD (500 mm)i + (600 mm) j + (360 mm)k = TAD AD 860 mm 60 36   50 j + k  TAD = i+ 86 86   86



Free-Body Diagram at A:

www.elsolucionario.net PROBLEM 2.99 (Continued)

Equilibrium condition:

ΣF = 0: ∴ TAB + TAC + TAD + W = 0

Substituting the expressions obtained for TAB , TAC , and TAD ; factoring i, j, and k; and equating each of the coefficients to zero gives the following equations: From i: From j: From k:

45 50 TAB + TAD = 0 75 86

(1)

60 60 60 TAB + TAC + TAD − W = 0 75 68 86

(2)

32 36 TAC + TAD = 0 68 86

(3)

−

−

TAC = 6.1920 kN TAC = 5.5080 kN

W = 13.98 kN 



Setting TAB = 6 kN in (1) and (2), and solving the resulting set of equations gives

www.elsolucionario.net PROBLEM 2.100 A container is supported by three cables that are attached to a ceiling as shown. Determine the weight W of the container, knowing that the tension in cable AD is 4.3 kN.

SOLUTION

45 50 TAB + TAD = 0 75 86

(1)

60 60 60 TAB + TAC + TAD − W = 0 75 68 86

(2)

32 36 TAC + TAD = 0 68 86

(3)

−

−

Setting TAD = 4.3 kN into the above equations gives TAB = 4.1667 kN TAC = 3.8250 kN

W = 9.71 kN 



See Problem 2.99 for the figure and analysis leading to the following set of linear algebraic equations:


SOLUTION

The forces applied at A are:

TAB , TAC , TAD , and P

where P = Pj. To express the other forces in terms of the unit vectors i, j, k, we write  AB = − (4.20 m)i − (5.60 m) j AB = 7.00 m  AC = (2.40 m)i − (5.60 m) j + (4.20 m)k AC = 7.40 m  AD = − (5.60 m)j − (3.30 m)k AD = 6.50 m  AB and TAB = TAB λ AB = TAB = ( − 0.6i − 0.8 j)TAB AB  AC TAC = TAC λ AC = TAC = (0.32432i − 0.75676 j + 0.56757k )TAC AC  AD TAD = TAD λ AD = TAD = (− 0.86154 j − 0.50769k )TAD AD



Three cables are used to tether a balloon as shown. Determine the vertical force P exerted by the balloon at A knowing that the tension in cable AD is 481 N.


Equilibrium condition:

ΣF = 0: TAB + TAC + TAD + Pj = 0

Substituting the expressions obtained for TAB , TAC , and TAD and factoring i, j, and k: (− 0.6TAB + 0.32432TAC )i + (−0.8TAB − 0.75676TAC − 0.86154TAD + P) j + (0.56757TAC − 0.50769TAD )k = 0

Equating to zero the coefficients of i, j, k: − 0.6TAB + 0.32432TAC = 0

(1)

− 0.8TAB − 0.75676TAC − 0.86154TAD + P = 0

(2)

0.56757TAC − 0.50769TAD = 0

(3)

TAC = 430.26 N TAD = 232.57 N

P = 926 N 



Setting TAD = 481 N in (2) and (3), and solving the resulting set of equations gives


SOLUTION See Problem 2.101 for the figure and analysis leading to the linear algebraic Equations (1), (2), and (3). − 0.6TAB + 0.32432TAC = 0

(1)

− 0.8TAB − 0.75676TAC − 0.86154TAD + P = 0

(2)

0.56757TAC − 0.50769TAD = 0

(3)

From Eq. (1):

TAB = 0.54053TAC

From Eq. (3):

TAD = 1.11795TAC

Substituting for TAB and TAD in terms of TAC into Eq. (2) gives − 0.8(0.54053TAC ) − 0.75676TAC − 0.86154(1.11795TAC ) + P = 0 2.1523TAC = P ; P = 800 N 800 N 2.1523 = 371.69 N

TAC =

Substituting into expressions for TAB and TAD gives TAB = 0.54053(371.69 N) TAD = 1.11795(371.69 N) TAB = 201 N, TAC = 372 N, TAD = 416 N 



Three cables are used to tether a balloon as shown. Knowing that the balloon exerts an 800-N vertical force at A, determine the tension in each cable.


SOLUTION The forces applied at A are: TAB , TAC , TAD and W

where P = Pj. To express the other forces in terms of the unit vectors i, j, k, we write  AB = − (36 in.)i + (60 in.) j − (27 in.)k AB = 75 in.  AC = (60 in.) j + (32 in.)k AC = 68 in.  AD = (40 in.)i + (60 in.) j − (27 in.)k AD = 77 in.  AB and TAB = TAB λAB = TAB AB = (− 0.48i + 0.8 j − 0.36k )TAB  AC TAC = TAC λAC = TAC AC = (0.88235 j + 0.47059k )TAC  AD TAD = TAD λAD = TAD AD = (0.51948i + 0.77922 j − 0.35065k )TAD Equilibrium Condition with

W = − Wj ΣF = 0: TAB + TAC + TAD − Wj = 0



A crate is supported by three cables as shown. Determine the weight of the crate knowing that the tension in cable AB is 750 lb.


Substituting the expressions obtained for TAB , TAC , and TAD and factoring i, j, and k: (−0.48TAB + 0.51948TAD )i + (0.8TAB + 0.88235TAC + 0.77922TAD − W ) j + (−0.36TAB + 0.47059TAC − 0.35065TAD )k = 0

Equating to zero the coefficients of i, j, k: −0.48TAB + 0.51948TAD = 0 0.8TAB + 0.88235TAC + 0.77922TAD − W = 0 −0.36TAB + 0.47059TAC − 0.35065TAD = 0

Substituting TAB = 750 lb in Equations (1), (2), and (3) and solving the resulting set of equations, using conventional algorithms for solving linear algebraic equations, gives: TAC = 1090.1 lb W = 2100 lb 


TAD = 693 lb


www.elsolucionario.net PROBLEM 2.104 A crate is supported by three cables as shown. Determine the weight of the crate knowing that the tension in cable AD is 616 lb.

See Problem 2.103 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: − 0.48TAB + 0.51948TAD = 0

(1)

0.8TAB + 0.88235TAC + 0.77922TAD − W = 0

(2)

− 0.36TAB + 0.47059TAC − 0.35065TAD = 0

(3)

Substituting TAD = 616 lb in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms, gives: TAB = 667.67 lb TAC = 969.00 lb

W = 1868 lb 



SOLUTION

www.elsolucionario.net PROBLEM 2.105 A crate is supported by three cables as shown. Determine the weight of the crate knowing that the tension in cable AC is 544 lb.

See Problem 2.103 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: − 0.48TAB + 0.51948TAD = 0

(1)

0.8TAB + 0.88235TAC + 0.77922TAD − W = 0

(2)

− 0.36TAB + 0.47059TAC − 0.35065TAD = 0

(3)

Substituting TAC = 544 lb in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms, gives: TAB = 374.27 lb TAD = 345.82 lb

W = 1049 lb 



SOLUTION

www.elsolucionario.net PROBLEM 2.106 A 1600-lb crate is supported by three cables as shown. Determine the tension in each cable.

See Problem 2.103 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: −0.48TAB + 0.51948TAD = 0

(1)

0.8TAB + 0.88235TAC + 0.77922TAD − W = 0

(2)

−0.36TAB + 0.47059TAC − 0.35065TAD = 0

(3)

Substituting W = 1600 lb in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms, gives TAB = 571 lb  TAC = 830 lb  TAD = 528 lb 



SOLUTION



Three cables are connected at A, where the forces P and Q are applied as shown. Knowing that Q = 0, find the value of P for which the tension in cable AD is 305 N.

SOLUTION



ΣFA = 0: TAB + TAC + TAD + P = 0 where P = Pi  AB = −(960 mm)i − (240 mm)j + (380 mm)k AB = 1060 mm  AC = −(960 mm)i − (240 mm) j − (320 mm)k AC = 1040 mm   AD = −(960 mm)i + (720 mm) j − (220 mm)k AD = 1220 mm  AB 19   48 12 = TAB  − i − j + k  TAB = TAB λAB = TAB AB 53 53 53    AC 3 4   12 TAC = TAC λAC = TAC = TAC  − i − j − k  AC  13 13 13  305 N [( −960 mm)i + (720 mm) j − (220 mm)k ] TAD = TAD λAD = 1220 mm = −(240 N)i + (180 N) j − (55 N)k

Substituting into ΣFA = 0, factoring i, j, k , and setting each coefficient equal to φ gives: i: P =

48 12 TAB + TAC + 240 N 53 13

(1)

j:

12 3 TAB + TAC = 180 N 53 13

(2)

k:

19 4 TAB − TAC = 55 N 53 13

(3)

Solving the system of linear equations using conventional algorithms gives: TAB = 446.71 N TAC = 341.71 N

P = 960 N 



SOLUTION We assume that TAD = 0 and write ΣFA = 0: TAB + TAC + Qj + (1200 N)i = 0  AB = −(960 mm)i − (240 mm)j + (380 mm)k AB = 1060 mm  AC = −(960 mm)i − (240 mm) j − (320 mm)k AC = 1040 mm  AB  48 12 19  TAB = TAB λAB = TAB =  − i − j + k  TAB AB  53 53 53   AC  12 3 4  TAC = TAC λAC = TAC =  − i − j − k  TAC AC  13 13 13  Substituting into ΣFA = 0, factoring i, j, k , and setting each coefficient equal to φ gives: i: −

48 12 TAB − TAC + 1200 N = 0 53 13

(1)

j: −

12 3 TAB − TAC + Q = 0 53 13

(2)

k:

19 4 TAB − TAC = 0 53 13

(3)

Solving the resulting system of linear equations using conventional algorithms gives: TAB = 605.71 N TAC = 705.71 N Q = 300.00 N

0 ⱕ Q ⬍ 300 N 

Note: This solution assumes that Q is directed upward as shown (Q ⱖ 0), if negative values of Q are considered, cable AD remains taut, but AC becomes slack for Q = −460 N. 



Three cables are connected at A, where the forces P and Q are applied as shown. Knowing that P = 1200 N, determine the values of Q for which cable AD is taut.

www.elsolucionario.net PROBLEM 2.109 A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AC is 60 N, determine the weight of the plate.

SOLUTION

Free Body A: ΣF = 0: TAB + TAC + TAD + Pj = 0

We have:  AB = −(320 mm)i − (480 mm)j + (360 mm)k AB = 680 mm  AC = (450 mm)i − (480 mm) j + (360 mm)k AC = 750 mm  AD = (250 mm)i − (480 mm) j − ( 360 mm ) k AD = 650 mm

Thus: TAB TAC TAD

 AB  8 12 9  = TAB λ AB = TAB =  − i − j + k  TAB AB  17 17 17   AC = TAC λAC = TAC = ( 0.6i − 0.64 j + 0.48k ) TAC AC  AD  5 9.6 7.2  = TAD λAD = TAD = i− j− k TAD AD  13 13 13 

Dimensions in mm

Substituting into the Eq. ΣF = 0 and factoring i, j, k : 5  8   − TAB + 0.6TAC + TAD  i 13  17  9.6  12  TAD + P  j +  − TAB − 0.64TAC − 13  17  7.2  9  TAD  k = 0 +  TAB + 0.48TAC − 13  17 



We note that the weight of the plate is equal in magnitude to the force P exerted by the support on Point A.

www.elsolucionario.net PROBLEM 2.109 (Continued) Setting the coefficient of i, j, k equal to zero: i:

−

8 5 TAB + 0.6TAC + TAD = 0 17 13

(1)

j:

−

12 9.6 TAB − 0.64TAC − TAD + P = 0 7 13

(2)

9 7.2 TAB + 0.48TAC − TAD = 0 17 13

(3)

8 5 TAB + 36 N + TAD = 0 17 13

(1′)

9 7.2 TAB + 28.8 N − TAD = 0 17 13

(3′)

k:

−

Multiply (1′) by 9, (3′) by 8, and add: 554.4 N −

12.6 TAD = 0 TAD = 572.0 N 13

Substitute into (1′) and solve for TAB : TAB =

17  5  36 + × 572   8  13 

TAB = 544.0 N

Substitute for the tensions in Eq. (2) and solve for P: 12 9.6 (544 N) + 0.64(60 N) + (572 N) 17 13 = 844.8 N

P=

Weight of plate = P = 845 N 



Making TAC = 60 N in (1) and (3):

www.elsolucionario.net PROBLEM 2.110 A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AD is 520 N, determine the weight of the plate.

SOLUTION

8 5 TAB + 0.6TAC + TAD = 0 17 13

(1)

12 9.6 TAB + 0.64 TAC − TAD + P = 0 17 13

(2)

9 7.2 TAB + 0.48TAC − TAD = 0 17 13

(3)

− −

Making TAD = 520 N in Eqs. (1) and (3): 8 TAB + 0.6TAC + 200 N = 0 17

(1′)

9 TAB + 0.48TAC − 288 N = 0 17

(3′)

−

Multiply (1′) by 9, (3′) by 8, and add: 9.24TAC − 504 N = 0 TAC = 54.5455 N

Substitute into (1′) and solve for TAB : TAB =

17 (0.6 × 54.5455 + 200) TAB = 494.545 N 8

Substitute for the tensions in Eq. (2) and solve for P: 12 9.6 (494.545 N) + 0.64(54.5455 N) + (520 N) 17 13 Weight of plate = P = 768 N  = 768.00 N

P=



See Problem 2.109 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below:

www.elsolucionario.net PROBLEM 2.111 A transmission tower is held by three guy wires attached to a pin at A and anchored by bolts at B, C, and D. If the tension in wire AB is 630 lb, determine the vertical force P exerted by the tower on the pin at A.

SOLUTION

We write

ΣF = 0: TAB + TAC + TAD + Pj = 0  AB = −45i − 90 j + 30k AB = 105 ft  AC = 30i − 90 j + 65k AC = 115 ft  AD = 20i − 90 j − 60k AD = 110 ft  AB TAB = TAB λ AB = TAB AB 6 2   3 =  − i − j + k  TAB 7 7   7  AC TAC = TAC λAC = TAC AC 18 13   6 = i− j + k  TAC 23 23 23    AD TAD = TAD λAD = TAD AD 9 6  2 =  i − j − k  TAD  11 11 11 

Substituting into the Eq. ΣF = 0 and factoring i, j, k : 6 2  3   − TAB + TAC + TAD  i 23 11  7  6 18 9   +  − TAB − TAC − TAD + P  j 23 11  7  13 6 2  +  TAB + TAC − TAD  k = 0 23 11 7  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 117


Free Body A:


Setting the coefficients of i, j, k , equal to zero: i:

3 6 2 − TAB + TAC + TAD = 0 7 23 11

(1)

j:

6 18 9 − TAB − TAC − TAD + P = 0 7 23 11

(2)

k:

2 13 6 TAB + TAC − TAD = 0 7 23 11

(3)

6 2 TAC + TAD = 0 23 11

(1′)

18 9 TAC − TAD + P = 0 23 11

(2′)

13 6 TAC − TAD = 0 23 11

(3′)

−270 lb + −540 lb −

180 lb +

Solving,

TAC = 467.42 lb TAD = 814.35 lb P = 1572.10 lb

P = 1572 lb 



Set TAB = 630 lb in Eqs. (1) – (3):

www.elsolucionario.net PROBLEM 2.112 A transmission tower is held by three guy wires attached to a pin at A and anchored by bolts at B, C, and D. If the tension in wire AC is 920 lb, determine the vertical force P exerted by the tower on the pin at A.

SOLUTION

3 6 2 − TAB + TAC + TAD = 0 7 23 11

(1)

6 18 9 − TAB − TAC − TAD + P = 0 7 23 11

(2)

2 13 6 TAB + TAC − TAD = 0 7 23 11

(3)

Substituting for TAC = 920 lb in Equations (1), (2), and (3) above and solving the resulting set of equations using conventional algorithms gives:

Solving,

3 2 − TAB + 240 lb + TAD = 0 7 11

(1′)

6 9 − TAB − 720 lb − TAD + P = 0 7 11

(2′)

2 6 TAB + 520 lb − TAD = 0 7 11

(3′)

TAB = 1240.00 lb TAD = 1602.86 lb P = 3094.3 lb

P = 3090 lb 




www.elsolucionario.net PROBLEM 2.113 In trying to move across a slippery icy surface, a 180-lb man uses two ropes AB and AC. Knowing that the force exerted on the man by the icy surface is perpendicular to that surface, determine the tension in each rope.

SOLUTION

30   16 N= N i+ j 34   34 and W = W j = −(180 lb) j

TAC = TAC λ AC = TAC

 AC ( −30 ft)i + (20 ft) j − (12 ft)k = TAC AC 38 ft 6   15 10 = TAC  − i + j − k   19 19 19 

TAB = TAB λ AB = TAB

 AB (−30 ft)i + (24 ft) j + (32 ft)k = TAB AB 50 ft 12 16   15 = TAB  − i + j+ k 25 25   25

Equilibrium condition: ΣF = 0 TAB + TAC + N + W = 0



Free-Body Diagram at A


Substituting the expressions obtained for TAB , TAC , N, and W; factoring i, j, and k; and equating each of the coefficients to zero gives the following equations: 15 15 16 TAB − TAC + N =0 25 19 34

(1)

From j:

12 10 30 TAB + TAC + N − (180 lb) = 0 25 19 34

(2)

From k:

16 6 TAB − TAC = 0 25 19

(3)

From i:

−

Solving the resulting set of equations gives:


TAB = 31.7 lb; TAC = 64.3 lb 


www.elsolucionario.net PROBLEM 2.114 Solve Problem 2.113, assuming that a friend is helping the man at A by pulling on him with a force P = −(60 lb)k. PROBLEM 2.113 In trying to move across a slippery icy surface, a 180-lb man uses two ropes AB and AC. Knowing that the force exerted on the man by the icy surface is perpendicular to that surface, determine the tension in each rope.

SOLUTION

15 15 16 TAB − TAC + N =0 25 19 34

(1)

12 10 30 TAB + TAC + N − (180 lb) = 0 25 19 34

(2)

16 6 TAB − TAC − (60 lb) = 0 25 19

(3)

−

Solving the resulting set of equations simultaneously gives: TAB = 99.0 lb  TAC = 10.55 lb 



Refer to Problem 2.113 for the figure and analysis leading to the following set of equations, Equation (3) being modified to include the additional force P = ( −60 lb)k.

www.elsolucionario.net PROBLEM 2.115 For the rectangular plate of Problems 2.109 and 2.110, determine the tension in each of the three cables knowing that the weight of the plate is 792 N.

See Problem 2.109 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below. Setting P = 792 N gives: 8 5 TAB + 0.6TAC + TAD = 0 17 13

(1)

12 9.6 TAB − 0.64TAC − TAD + 792 N = 0 17 13

(2)

9 7.2 TAB + 0.48TAC − TAD = 0 17 13

(3)

− −

Solving Equations (1), (2), and (3) by conventional algorithms gives TAB = 510.00 N

TAB = 510 N 

TAC = 56.250 N

TAC = 56.2 N 

TAD = 536.25 N

TAD = 536 N 



SOLUTION



For the cable system of Problems 2.107 and 2.108, determine the tension in each cable knowing that P = 2880 N and Q = 0.

SOLUTION ΣFA = 0: TAB + TAC + TAD + P + Q = 0

Where

P = Pi and Q = Qj  AB = −(960 mm)i − (240 mm) j + (380 mm)k AB = 1060 mm  AC = −(960 mm)i − (240 mm) j − (320 mm)k AC = 1040 mm  AD = −(960 mm)i + (720 mm) j − (220 mm)k AD = 1220 mm  AB 19   48 12 TAB = TAB λAB = TAB = TAB  − i − j + k  AB 53 53   53  AC 3 4   12 TAC = TAC λAC = TAC = TAC  − i − j − k  AC  13 13 13   AD 36 11   48 = TAD  − i + TAD = TAD λAD = TAD j− k AD 61 61   61

Substituting into ΣFA = 0, setting P = (2880 N)i and Q = 0, and setting the coefficients of i, j, k equal to 0, we obtain the following three equilibrium equations: i: −

48 12 48 TAB − TAC − TAD + 2880 N = 0 53 13 61

(1)

j: −

12 3 36 TAB − TAC + TAD = 0 53 13 61

(2)

k:

19 4 11 TAB − TAC − TAD = 0 53 13 61

(3)



Solving the system of linear equations using conventional algorithms gives: TAB = 1340.14 N TAC = 1025.12 N TAD = 915.03 N

TAB = 1340 N  TAC = 1025 N 


TAD = 915 N 



SOLUTION See Problem 2.116 for the analysis leading to the linear algebraic Equations (1), (2), and (3) below: −

48 12 48 TAB − TAC − TAD + P = 0 53 13 61

(1)

−

12 3 36 TAB − TAC + TAD + Q = 0 53 13 61

(2)

19 4 11 TAB − TAC − TAD = 0 53 13 61

(3)

Setting P = 2880 N and Q = 576 N gives: −

48 12 48 TAB − TAC − TAD + 2880 N = 0 53 13 61

(1′)

12 3 36 TAB − TAC + TAD + 576 N = 0 53 13 61

(2′)

19 4 11 TAB − TAC − TAD = 0 53 13 61

(3′)

−

Solving the resulting set of equations using conventional algorithms gives: TAB = 1431.00 N TAC = 1560.00 N TAD = 183.010 N

TAB = 1431 N  TAC = 1560 N  TAD = 183.0 N 



For the cable system of Problems 2.107 and 2.108, determine the tension in each cable knowing that P = 2880 N and Q = 576 N.


SOLUTION See Problem 2.116 for the analysis leading to the linear algebraic Equations (1), (2), and (3) below: 

−

48 12 48 TAB − TAC − TAD + P = 0 53 13 61

(1)

−

12 3 36 TAB − TAC + TAD + Q = 0 53 13 61

(2)

19 4 11 TAB − TAC − TAD = 0 53 13 61

(3)

Setting P = 2880 N and Q = −576 N gives: −

48 12 48 TAB − TAC − TAD + 2880 N = 0 53 13 61

(1′)

12 3 36 TAB − TAC + TAD − 576 N = 0 53 13 61

(2′)

19 4 11 TAB − TAC − TAD = 0 53 13 61

(3′)

−

Solving the resulting set of equations using conventional algorithms gives: TAB = 1249.29 N TAC = 490.31 N TAD = 1646.97 N

TAB = 1249 N  TAC = 490 N  TAD = 1647 N 



For the cable system of Problems 2.107 and 2.108, determine the tension in each cable knowing that P = 2880 N and Q = −576 N. (Q is directed downward).

www.elsolucionario.net PROBLEM 2.119 For the transmission tower of Problems 2.111 and 2.112, determine the tension in each guy wire knowing that the tower exerts on the pin at A an upward vertical force of 2100 lb.

SOLUTION

3 6 2 − TAB + TAC + TAD = 0 7 23 11

(1)

6 18 9 − TAB − TAC − TAD + P = 0 7 23 11

(2)

2 13 6 TAB + TAC − TAD = 0 7 23 11

(3)

Substituting for P = 2100 lb in Equations (1), (2), and (3) above and solving the resulting set of equations using conventional algorithms gives: 3 6 2 − TAB + TAC + TAD = 0 7 23 11

(1′)

6 18 9 − TAB − TAC − TAD + 2100 lb = 0 7 23 11

(2′)

2 13 6 TAB + TAC − TAD = 0 7 23 11

(3′)

TAB = 841.55 lb TAC = 624.38 lb TAD = 1087.81 lb

TAB = 842 lb  TAC = 624 lb  TAD = 1088 lb 




www.elsolucionario.net PROBLEM 2.120 A horizontal circular plate weighing 60 lb is suspended as shown from three wires that are attached to a support at D and form 30° angles with the vertical. Determine the tension in each wire.

ΣFx = 0: −TAD (sin 30°)(sin 50°) + TBD (sin 30°)(cos 40°) + TCD (sin 30°)(cos 60°) = 0

Dividing through by sin 30° and evaluating: −0.76604TAD + 0.76604TBD + 0.5TCD = 0

(1)

ΣFy = 0: −TAD (cos 30°) − TBD (cos 30°) − TCD (cos 30°) + 60 lb = 0 TAD + TBD + TCD = 69.282 lb

or

(2)

ΣFz = 0: TAD sin 30° cos 50° + TBD sin 30° sin 40° − TCD sin 30° sin 60° = 0



or

0.64279TAD + 0.64279TBD − 0.86603TCD = 0

(3)

Solving Equations (1), (2), and (3) simultaneously: TAD = 29.5 lb  TBD = 10.25 lb 







TCD = 29.5 lb 



SOLUTION

www.elsolucionario.net PROBLEM 2.121 Cable BAC passes through a frictionless ring A and is attached to fixed supports at B and C, while cables AD and AE are both tied to the ring and are attached, respectively, to supports at D and E. Knowing that a 200-lb vertical load P is applied to ring A, determine the tension in each of the three cables.

SOLUTION

Since TBAC = tension in cable BAC, it follows that TAB = TAC = TBAC

TAB = TBAC λ AB = TBAC

(−17.5 in.)i + (60 in.) j 60   −17.5 = TBAC  i+ j 62.5 in. 62.5   62.5

TAC = TBAC λ AC = TBAC

(60 in.)i + (25 in.)k 25   60 = TBAC  j + k  65 in. 65   65

TAD = TAD λ AD = TAD

(80 in.)i + (60 in.) j 3  4 = TAD  i + j  100 in. 5  5

TAE = TAE λ AE = TAE

(60 in.) j − (45 in.)k 3  4 = TAE  j − k  75 in. 5  5



Free Body Diagram at A:


Substituting into ΣFA = 0, setting P = ( −200 lb) j, and setting the coefficients of i, j, k equal to φ , we obtain the following three equilibrium equations: 17.5 4 TBAC + TAD = 0 62.5 5

From

i: −

From

3 4  60 60  j:  +  TBAC + TAD + TAE − 200 lb = 0 62.5 65 5 5  

From

k:

(1)

25 3 TBAC − TAE = 0 65 5

(2) (3)

Solving the system of linear equations using convential acgorithms gives:


TBAC = 76.7 lb; TAD = 26.9 lb; TAE = 49.2 lb 


www.elsolucionario.net PROBLEM 2.122 Knowing that the tension in cable AE of Prob. 2.121 is 75 lb, determine (a) the magnitude of the load P, (b) the tension in cables BAC and AD. PROBLEM 2.121 Cable BAC passes through a frictionless ring A and is attached to fixed supports at B and C, while cables AD and AE are both tied to the ring and are attached, respectively, to supports at D and E. Knowing that a 200-lb vertical load P is applied to ring A, determine the tension in each of the three cables.

Refer to the solution to Problem 2.121 for the figure and analysis leading to the following set of equilibrium equations, Equation (2) being modified to include Pj as an unknown quantity: −

17.5 4 TBAC + TAD = 0 62.5 5

(1)

60  3 4  60  62.5 + 65  TBAC + 5 TAD + 5 TAE − P = 0   25 3 TBAC − TAE = 0 65 5

(2)

(3)

Substituting for TAE = 75 lb and solving simultaneously gives: P = 305 lb; TBAC = 117.0 lb; TAD = 40.9 lb 





SOLUTION

www.elsolucionario.net PROBLEM 2.123 A container of weight W is suspended from ring A. Cable BAC passes through the ring and is attached to fixed supports at B and C. Two forces P = Pi and Q = Qk are applied to the ring to maintain the container in the position shown. Knowing that W = 376 N, determine P and Q. (Hint: The tension is the same in both portions of cable BAC.)

TAB = T λ AB  AB =T AB (−130 mm)i + (400 mm) j + (160 mm)k =T 450 mm 40 16   13 j+ k =T − i + 45 45   45


SOLUTION Free-Body A:

TAC = T λ AC  AC =T AC ( −150 mm)i + (400 mm) j + (−240 mm)k =T 490 mm 40 24   15 = T − i + j− k 49 49   49 ΣF = 0: TAB + TAC + Q + P + W = 0

Setting coefficients of i, j, k equal to zero: i: −

13 15 T − T +P=0 45 49

0.59501T = P

(1)

j: +

40 40 T + T −W = 0 45 49

1.70521T = W

(2)

k: +

16 24 T − T +Q =0 45 49

0.134240 T = Q

(3)



Data:

W = 376 N 1.70521T = 376 N T = 220.50 N 0.59501(220.50 N) = P

P = 131.2 N  Q = 29.6 N 


0.134240(220.50 N) = Q


www.elsolucionario.net PROBLEM 2.124 For the system of Problem 2.123, determine W and Q knowing that P = 164 N. PROBLEM 2.123 A container of weight W is suspended from ring A. Cable BAC passes through the ring and is attached to fixed supports at B and C. Two forces P = Pi and Q = Qk are applied to the ring to maintain the container in the position shown. Knowing that W = 376 N, determine P and Q. (Hint: The tension is the same in both portions of cable BAC.)

Refer to Problem 2.123 for the figure and analysis resulting in Equations (1), (2), and (3) for P, W, and Q in terms of T below. Setting P = 164 N we have: Eq. (1):

0.59501T = 164 N

Eq. (2):

1.70521(275.63 N) = W

Eq. (3):

0.134240(275.63 N) = Q

T = 275.63 N W = 470 N  Q = 37.0 N 



SOLUTION

www.elsolucionario.net PROBLEM 2.125 Collars A and B are connected by a 525-mm-long wire and can slide freely on frictionless rods. If a force P = (341 N)j is applied to collar A, determine (a) the tension in the wire when y = 155 mm, (b) the magnitude of the force Q required to maintain the equilibrium of the system.

For both Problems 2.125 and 2.126:

Free-Body Diagrams of Collars: ( AB) 2 = x 2 + y 2 + z 2

Here

(0.525 m) 2 = (0.20 m) 2 + y 2 + z 2 y 2 + z 2 = 0.23563 m 2

or Thus, when y given, z is determined, Now

λAB

 AB = AB

1 (0.20i − yj + zk )m 0.525 m = 0.38095i − 1.90476 yj + 1.90476 zk =

Where y and z are in units of meters, m. From the F.B. Diagram of collar A:

ΣF = 0: N x i + N z k + Pj + TAB λ AB = 0

Setting the j coefficient to zero gives

P − (1.90476 y )TAB = 0 P = 341 N

With

TAB =

341 N 1.90476 y

Now, from the free body diagram of collar B:

ΣF = 0: N x i + N y j + Qk − TAB λAB = 0

Setting the k coefficient to zero gives

Q − TAB (1.90476 z ) = 0

And using the above result for TAB , we have

Q = TAB z =

341 N (341 N)( z ) (1.90476 z ) = (1.90476) y y



SOLUTION


Then from the specifications of the problem, y = 155 mm = 0.155 m z 2 = 0.23563 m 2 − (0.155 m) 2 z = 0.46 m

and 341 N 0.155(1.90476) = 1155.00 N

TAB =

(a)

TAB = 1155 N 

or and 341 N(0.46 m)(0.866) (0.155 m) = (1012.00 N)

Q = 1012 N 

or



Q=

(b)

www.elsolucionario.net PROBLEM 2.126 Solve Problem 2.125 assuming that y = 275 mm. PROBLEM 2.125 Collars A and B are connected by a 525-mm-long wire and can slide freely on frictionless rods. If a force P = (341 N)j is applied to collar A, determine (a) the tension in the wire when y = 155 mm, (b) the magnitude of the force Q required to maintain the equilibrium of the system.

SOLUTION


From the analysis of Problem 2.125, particularly the results: y 2 + z 2 = 0.23563 m 2 341 N TAB = 1.90476 y 341 N Q= z y

With y = 275 mm = 0.275 m, we obtain: z 2 = 0.23563 m 2 − (0.275 m) 2 z = 0.40 m

and TAB =

(a)

341 N = 651.00 (1.90476)(0.275 m) TAB = 651 N 

or and Q=

(b)

341 N(0.40 m) (0.275 m) Q = 496 N 

or


www.elsolucionario.net PROBLEM 2.127 Two structural members A and B are bolted to a bracket as shown. Knowing that both members are in compression and that the force is 15 kN in member A and 10 kN in member B, determine by trigonometry the magnitude and direction of the resultant of the forces applied to the bracket by members A and B.

SOLUTION Using the force triangle and the laws of cosines and sines, we have

γ = 180° − (40° + 20°) = 120°

Then

R 2 = (15 kN) 2 + (10 kN)2 = 475 kN 2 R = 21.794 kN

and

Hence:

10 kN 21.794 kN = sin α sin120°  10 kN  sin α =   sin120°  21.794 kN  = 0.39737 α = 23.414

φ = α + 50° = 73.414

R = 21.8 kN

73.4° 



− 2(15 kN)(10 kN) cos120°

www.elsolucionario.net PROBLEM 2.128 Member BD exerts on member ABC a force P directed along line BD. Knowing that P must have a 300-lb horizontal component, determine (a) the magnitude of the force P, (b) its vertical component.

P sin 35° = 300 lb

(a)

P=

(b)

Vertical component

300 lb sin 35°

P = 523 lb 

Pv = P cos 35°

= (523 lb) cos 35°

Pv = 428 lb 



SOLUTION

www.elsolucionario.net PROBLEM 2.129 Determine (a) the required tension in cable AC, knowing that the resultant of the three forces exerted at Point C of boom BC must be directed along BC, (b) the corresponding magnitude of the resultant.

Using the x and y axes shown: Rx = ΣFx = TAC sin10° + (50 lb) cos 35° + (75 lb) cos 60°

= TAC sin10° + 78.458 lb

(1)

Ry = ΣFy = (50 lb)sin 35° + (75 lb)sin 60° − TAC cos10° Ry = 93.631 lb − TAC cos10°

(a)

(2)

Set Ry = 0 in Eq. (2): 93.631 lb − TAC cos10° = 0 TAC = 95.075 lb

(b)

TAC = 95.1 lb 

Substituting for TAC in Eq. (1): Rx = (95.075 lb)sin10° + 78.458 lb = 94.968 lb R = Rx

R = 95.0 lb 



SOLUTION



W = mg = (200 kg)(9.81 m/s 2 ) = 1962 N

Law of sines:

TAC TBC 1962 N = = sin 15° sin 105° sin 60°

(a)

TAC =

(1962 N) sin 15° sin 60°

TAC = 586 N 

(b)

TBC =

(1962 N) sin 105° sin 60°

TBC = 2190 N 



Free-Body Diagram

www.elsolucionario.net PROBLEM 2.131 A welded connection is in equilibrium under the action of the four forces shown. Knowing that FA = 8 kN and FB = 16 kN, determine the magnitudes of the other two forces.

SOLUTION

ΣFx = 0:

With


Free-Body Diagram of Connection

3 3 FB − FC − FA = 0 5 5

FA = 8 kN FB = 16 kN FC =

4 4 (16 kN) − (8 kN) 5 5

Σ Fy = 0: − FD +

With FA and FB as above:

FC = 6.40 kN 

3 3 FB − FA = 0  5 5

3 3 FD = (16 kN) − (8 kN)  5 5

FD = 4.80 kN 


www.elsolucionario.net PROBLEM 2.132 Two cables tied together at C are loaded as shown. Determine the range of values of Q for which the tension will not exceed 60 lb in either cable.

ΣFx = 0: −TBC − Q cos 60° + 75 lb = 0

Free-Body Diagram

TBC = 75 lb − Q cos 60°

(1)

ΣFy = 0: TAC − Q sin 60° = 0 TAC = Q sin 60°

(2)

TAC ⱕ 60 lb:

Requirement:

Q sin 60° ⱕ 60 lb

From Eq. (2):

Q ⱕ 69.3 lb TBC ⱕ 60 lb:

Requirement: From Eq. (1):

75 lb − Q sin 60° ⱕ 60 lb Q ⱖ 30.0 lb

30.0 lb ⱕ Q ⱕ 69.3 lb 



SOLUTION

www.elsolucionario.net PROBLEM 2.133 A horizontal circular plate is suspended as shown from three wires that are attached to a support at D and form 30° angles with the vertical. Knowing that the x component of the force exerted by wire AD on the plate is 110.3 N, determine (a) the tension in wire AD, (b) the angles θ x, θ y, and θ z that the force exerted at A forms with the coordinate axes.

(a)

Fx = F sin 30° sin 50° = 110.3 N (Given) F=

(b)

cos θ x =

110.3 N = 287.97 N sin 30° sin 50°

F = 288 N 

Fx 110.3 N = = 0.38303 F 287.97 N

θ x = 67.5° 

Fy = F cos 30° = 249.39 cos θ y =

Fy F

=

249.39 N = 0.86603 287.97 N

θ y = 30.0° 

Fz = − F sin 30° cos 50° = −(287.97 N)sin 30°cos 50° = −92.552 N cos θ z =

Fz −92.552 N = = −0.32139 F 287.97 N

θ z = 108.7° 



SOLUTION

www.elsolucionario.net PROBLEM 2.134 A force acts at the origin of a coordinate system in a direction defined by the angles θy = 55° and θz = 45°. Knowing that the x component of the force is −500 lb, determine (a) the angle θx, (b) the other components and the magnitude of the force.

SOLUTION (a)

We have (cos θ x ) 2 + (cos θ y )2 + (cos θ z ) 2 = 1  (cos θ y ) 2 = 1 − (cos θ y ) 2 − (cos θ z ) 2

Since Fx ⬍ 0, we must have cos θ x ⬍ 0. Thus, taking the negative square root, from above, we have

(b)

θ x = 114.4° 

Then Fx 500 lb = = 1209.10 lb cos θ x 0.41353

F = 1209 lb 

Fy = F cos θ y = (1209.10 lb) cos 55°

Fy = 694 lb 

Fz = F cos θ z = (1209.10 lb) cos 45°

Fz = 855 lb 

F=

and



cos θ x = − 1 − (cos 55) 2 − (cos 45) 2 = 0.41353


SOLUTION

= (400 N)[0.60402i + 0.76604 j − 0.21985] = (241.61 N)i + (306.42 N) j − (87.939 N)k R = P+Q = (174.367 N)i + (456.42 N) j + (163.011 N)k R = (174.367 N)2 + (456.42 N)2 + (163.011 N) 2 = 515.07 N

R = 515 N 

cos θ x =

Rx 174.367 N = = 0.33853 515.07 N R

θ x = 70.2° 

cos θ y =

Ry

θ y = 27.6° 

cos θ z =

Rz 163.011 N = = 0.31648 R 515.07 N

R

=

456.42 N = 0.88613 515.07 N

θ z = 71.5° 



P = (300 N)[− cos 30° sin15°i + sin 30° j + cos 30° cos15°k ] = − (67.243 N)i + (150 N) j + (250.95 N)k Q = (400 N)[cos 50° cos 20°i + sin 50° j − cos 50° sin 20°k ]


SOLUTION See Problem 2.101 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: − 0.6TAB + 0.32432TAC = 0

(1)

− 0.8TAB − 0.75676TAC − 0.86154TAD + P = 0

(2)

0.56757TAC − 0.50769TAD = 0

(3)

Substituting TAC = 444 N in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms gives TAB = 240 N TAD = 496.36 N

P = 956 N 



Three cables are used to tether a balloon as shown. Determine the vertical force P exerted by the balloon at A knowing that the tension in cable AC is 444 N.

www.elsolucionario.net PROBLEM 2.137 Collars A and B are connected by a 25-in.-long wire and can slide freely on frictionless rods. If a 60-lb force Q is applied to collar B as shown, determine (a) the tension in the wire when x = 9 in., (b) the corresponding magnitude of the force P required to maintain the equilibrium of the system.

SOLUTION

A:

B:

λAB

 AB − xi − (20 in.) j + zk = = AB 25 in.

ΣF = 0: Pi + N y j + N z k + TAB λ AB = 0

Collar A:

Substitute for λAB and set coefficient of i equal to zero: P−

Collar B:

TAB x =0 25 in.

(1)

ΣF = 0: (60 lb)k + N x′ i + N y′ j − TAB λ AB = 0

Substitute for λAB and set coefficient of k equal to zero: 60 lb − x = 9 in.

(a)

From Eq. (2): (b)

From Eq. (1):

TAB z =0 25 in.

(2)

(9 in.)2 + (20 in.) 2 + z 2 = (25 in.) 2 z = 12 in. 60 lb − TAB (12 in.) 25 in. P=

(125.0 lb)(9 in.)  25 in.

TAB = 125.0 lb  P = 45.0 lb 



Free-Body Diagrams of Collars:

www.elsolucionario.net PROBLEM 2.138 Collars A and B are connected by a 25-in.-long wire and can slide freely on frictionless rods. Determine the distances x and z for which the equilibrium of the system is maintained when P = 120 lb and Q = 60 lb.

SOLUTION

P=

TAB x =0 25 in.

(1)

60 lb −

TAB z =0 25 in.

(2)

For P = 120 lb, Eq. (1) yields

TAB x = (25 in.)(20 lb)

(1′)

From Eq. (2):

TAB z = (25 in.)(60 lb)

(2′)

x =2 z

Dividing Eq. (1′) by (2′), Now write

x 2 + z 2 + (20 in.) 2 = (25 in.) 2

(3) (4)

Solving (3) and (4) simultaneously, 4 z 2 + z 2 + 400 = 625 z 2 = 45 z = 6.7082 in.

From Eq. (3):

x = 2 z = 2(6.7082 in.) = 13.4164 in. x = 13.42 in., z = 6.71 in. 



See Problem 2.137 for the diagrams and analysis leading to Equations (1) and (2) below:

www.elsolucionario.net PROBLEM 2F1 Two cables are tied together at C and loaded as shown. Draw the freebody diagram needed to determine the tension in AC and BC.

SOLUTION


Free-Body Diagram of Point C:




www.elsolucionario.net PROBLEM 2.F2 A chairlift has been stopped in the position shown. Knowing that each chair weighs 250 N and that the skier in chair E weighs 765 N, draw the free-body diagrams needed to determine the weight of the skier in chair F.

SOLUTION

WE = 250 N + 765 N = 1015 N 8.25 = 30.510° 14 10 = tan −1 = 22.620° 24

θ AB = tan −1 θ BC

Use this free body to determine TAB and TBC.


θCD = tan −1

1.1 = 10.3889° 6

Use this free body to determine TCD and WF. Then weight of skier WS is found by

WS = WF − 250 N 





Free-Body Diagram of Point B:

www.elsolucionario.net PROBLEM 2.F3 Two cables are tied together at A and loaded as shown. Draw the freebody diagram needed to determine the tension in each cable.

Free-Body Diagram of Point A:





SOLUTION

www.elsolucionario.net PROBLEM 2.F4 The 60-lb collar A can slide on a frictionless vertical rod and is connected as shown to a 65-lb counterweight C. Draw the free-body diagram needed to determine the value of h for which the system is in equilibrium.

Free-Body Diagram of Point A:





SOLUTION

www.elsolucionario.net PROBLEM 2.F5 A 36-lb triangular plate is supported by three cables as shown. Draw the free-body diagram needed to determine the tension in each wire.

SOLUTION


Free-Body Diagram of Point D:




www.elsolucionario.net PROBLEM 2.F6 A 70-kg cylinder is supported by two cables AC and BC, which are attached to the top of vertical posts. A horizontal force P, perpendicular to the plane containing the posts, holds the cylinder in the position shown. Draw the free-body diagram needed to determine the magnitude of P and the force in each cable.

SOLUTION

W = (70 kg)(9.81 m/s 2 ) = 686.7 N  W = −(686.7 N) j




www.elsolucionario.net PROBLEM 2.F7 Three cables are connected at point D, which is located 18 in. below the T-shaped pipe support ABC. The cables support a 180-lb cylinder as shown. Draw the free-body diagram needed to determine the tension in each cable.

SOLUTION

 



Free-Body Diagram of Point D:

www.elsolucionario.net PROBLEM 2.F8

SOLUTION Free-Body Diagram of Ring A:

W = (100 kg)(9.81 m/s 2 ) = 981 N  W = −(681 N) j





A 100-kg container is suspended from ring A, to which cables AC and AE are attached. A force P is applied to end F of a third cable that passes over a pulley at B and through ring A and then is attached to a support at D. Draw the free-body diagram needed to determine the magnitude of P. (Hint: The tension is the same in all portions of cable FBAD.)

CHAPTER 3





www.elsolucionario.net PROBLEM 3.1 A 20-lb force is applied to the control rod AB as shown. Knowing that the length of the rod is 9 in. and that α = 25°, determine the moment of the force about Point B by resolving the force into horizontal and vertical components.

SOLUTION Free-Body Diagram of Rod AB: x = (9 in.) cos 65°


= 3.8036 in. y = (9 in.)sin 65° = 8.1568 in.

F = Fx i + Fy j

rA/B

= (20 lb cos 25°)i + ( −20 lb sin 25°) j = (18.1262 lb)i − (8.4524 lb) j  = BA = ( −3.8036 in.)i + (8.1568 in.)j

M B = rA /B × F = (−3.8036i + 8.1568 j) × (18.1262i − 8.4524 j) = 32.150k − 147.852k = −115.702 lb-in.

M B = 115.7 lb-in.




www.elsolucionario.net PROBLEM 3.2 A 20-lb force is applied to the control rod AB as shown. Knowing that the length of the rod is 9 in. and that α = 25°, determine the moment of the force about Point B by resolving the force into components along AB and in a direction perpendicular to AB.

SOLUTION Free-Body Diagram of Rod AB:

θ = 90° − (65° − 25°)


= 50°

Q = (20 lb) cos 50° = 12.8558 lb M B = Q (9 in.) = (12.8558 lb)(9 in.) = 115.702 lb-in.

M B = 115.7 lb-in.




www.elsolucionario.net PROBLEM 3.3 A 20-lb force is applied to the control rod AB as shown. Knowing that the length of the rod is 9 in. and that the moment of the force about B is 120 lb · in. clockwise, determine the value of α.

SOLUTION Free-Body Diagram of Rod AB:


α = θ − 25°

Q = (20 lb) cos θ

and Therefore,

M B = (Q )(9 in.)

120 lb-in. = (20 lb)(cos θ )(9 in.) 120 lb-in. cos θ = 180 lb-in.

or

θ = 48.190°

Therefore,

α = 48.190° − 25°

α = 23.2° 


www.elsolucionario.net PROBLEM 3.4 A crate of mass 80 kg is held in the position shown. Determine (a) the moment produced by the weight W of the crate about E, (b) the smallest force applied at B that creates a moment of equal magnitude and opposite sense about E.

(a)

By definition, We have

W = mg = 80 kg(9.81 m/s 2 ) = 784.8 N

ΣM E : M E = (784.8 N)(0.25 m) M E = 196.2 N ⋅ m

(b)



For the force at B to be the smallest, resulting in a moment (ME) about E, the line of action of force FB must be perpendicular to the line connecting E to B. The sense of FB must be such that the force produces a counterclockwise moment about E. Note: We have

d = (0.85 m) 2 + (0.5 m) 2 = 0.98615 m

ΣM E : 196.2 N ⋅ m = FB (0.98615 m) FB = 198.954 N

and

 0.85 m   = 59.534°  0.5 m 

θ = tan −1 

FB = 199.0 N

or

59.5° 



SOLUTION

www.elsolucionario.net PROBLEM 3.5 A crate of mass 80 kg is held in the position shown. Determine (a) the moment produced by the weight W of the crate about E, (b) the smallest force applied at A that creates a moment of equal magnitude and opposite sense about E, (c) the magnitude, sense, and point of application on the bottom of the crate of the smallest vertical force that creates a moment of equal magnitude and opposite sense about E.

SOLUTION

W = mg = (80 kg)(9.81 m/s 2 ) = 784.8 N

(a)

We have

M E = rH /EW = (0.25 m)(784.8 N) = 196.2 N ⋅ m

(b)

(c)

or M E = 196.2 N ⋅ m



For FA to be minimum, it must be perpendicular to the line joining Points A and E. Then with FA directed as shown, we have (− M E ) = rA/E ( FA )min .

Where

rA /E = (0.35 m)2 + (0.5 m)2 = 0.61033 m

then

196.2 N ⋅ m = (0.61033 m)( FA )min

or

( FA ) min = 321 N

Also

tan φ =

0.35 m 0.5 m

or

φ = 35.0°

(FA ) min = 321 N

35.0° 

For Fvertical to be minimum, the perpendicular distance from its line of action to Point E must be maximum. Thus, apply (Fvertical)min at Point D, and then (− M E ) = rD / E ( Fvertical ) min

or (Fvertical )min = 231 N

196.2 N ⋅ m = (0.85 m)( Fvertical ) min

at Point D 



First note. . .

www.elsolucionario.net PROBLEM 3.6 A 300-N force P is applied at Point A of the bell crank shown. (a) Compute the moment of the force P about O by resolving it into horizontal and vertical components. (b) Using the result of part (a), determine the perpendicular distance from O to the line of action of P.

SOLUTION

x = (0.2 m) cos 40°

(a)


= 0.153209 m y = (0.2 m)sin 40° = 0.128558 m ∴ rA /O = (0.153209 m)i + (0.128558 m) j

Fx = (300 N)sin 30°

= 150 N Fy = (300 N) cos 30° = 259.81 N F = (150 N)i + (259.81 N) j M O = rA/ O × F = (0.153209i + 0.128558 j) m × (150i + 259.81j) N = (39.805k − 19.2837k ) N ⋅ m = (20.521 N ⋅ m)k

(b)

M O = 20.5 N ⋅ m



M O = Fd 20.521 N ⋅ m = (300 N)(d )

d = 0.068403 m

d = 68.4 mm 


www.elsolucionario.net PROBLEM 3.7 A 400-N force P is applied at Point A of the bell crank shown. (a) Compute the moment of the force P about O by resolving it into components along line OA and in a direction perpendicular to that line. (b) Determine the magnitude and direction of the smallest force Q applied at B that has the same moment as P about O.

SOLUTION

(a)

Portion OA of crank:

θ = 90° − 30° − 40° θ = 20°

= 136.81 N M O = rO /A S = (0.2 m)(136.81 N) = 27.362 N ⋅ m

(b)

M O = 27.4 N ⋅ m



Smallest force Q must be perpendicular to OB.

Portion OB of crank:

M O = rO /B Q M O = (0.120 m)Q 27.362 N ⋅ m = (0.120 m)Q

Q = 228 N

42.0° 



S = P sin θ = (400 N) sin 20°

www.elsolucionario.net PROBLEM 3.8 It is known that a vertical force of 200 lb is required to remove the nail at C from the board. As the nail first starts moving, determine (a) the moment about B of the force exerted on the nail, (b) the magnitude of the force P that creates the same moment about B if α = 10°, (c) the smallest force P that creates the same moment about B.

(a)

M B = rC/B FN

We have

= (4 in.)(200 lb) = 800 lb ⋅ in.

or MB = 800 lb ⋅ in.  (b)

By definition,

M B = rA/B P sin θ

θ = 10° + (180° − 70°) = 120°

Then

800 lb ⋅ in. = (18 in.) × P sin120°

or P = 51.3 lb  (c)

For P to be minimum, it must be perpendicular to the line joining Points A and B. Thus, P must be directed as shown. Thus

M B = dPmin d = rA/B

or or

800 lb ⋅ in. = (18 in.)Pmin Pmin = 44.4 lb

Pmin = 44.4 lb

20° 



SOLUTION

www.elsolucionario.net PROBLEM 3.9 It is known that the connecting rod AB exerts on the crank BC a 500-lb force directed down and to the left along the centerline of AB. Determine the moment of the force about C.

SOLUTION M C = y1 ( FAB ) x + x1 ( FAB ) y  7   24  = (2.24 in.)  × 500 lb  + (1.68 in.)  × 500 lb   25   25  = 1120 lb ⋅ in.

(a) MC = 1.120 kip ⋅ in.



Using (b): M C = y2 ( FAB ) x  7  = (8 in.)  × 500 lb   25  = 1120 lb ⋅ in.

(b) M C = 1.120 kip ⋅ in.





Using (a):

www.elsolucionario.net PROBLEM 3.10 It is known that the connecting rod AB exerts on the crank BC a 500-lb force directed down and to the left along the centerline of AB. Determine the moment of the force about C.

SOLUTION Using (a):  7   24  = −(2.24 in.)  × 500 lb  + (1.68 in.)  × 500 lb   25   25  = +492.8 lb ⋅ in.

(a) MC = 493 lb ⋅ in.



Using (b): M C = y2 ( FAB ) x  7  = (3.52 in.)  × 500 lb   25  = +492.8 lb ⋅ in.

(b) M C = 493 lb ⋅ in.





M C = − y1 ( FAB ) x + x1 ( FAB ) y



A winch puller AB is used to straighten a fence post. Knowing that the tension in cable BC is 1040 N and length d is 1.90 m, determine the moment about D of the force exerted by the cable at C by resolving that force into horizontal and vertical components applied (a) at Point C, (b) at Point E.

SOLUTION (a)

Slope of line: Then

and

Then

EC =

0.875 m 5 = 1.90 m + 0.2 m 12

12 (TAB ) 13 12 = (1040 N) 13 = 960 N 5 TABy = (1040 N) 13 = 400 N TABx =

(a)

M D = TABx (0.875 m) − TABy (0.2 m) = (960 N)(0.875 m) − (400 N)(0.2 m) = 760 N ⋅ m

(b)

We have

or M D = 760 N ⋅ m

M D = TABx ( y ) + TABx ( x) = (960 N)(0) + (400 N)(1.90 m) = 760 N ⋅ m



(b) or M D = 760 N ⋅ m






It is known that a force with a moment of 960 N · m about D is required to straighten the fence post CD. If d = 2.80 m, determine the tension that must be developed in the cable of winch puller AB to create the required moment about Point D.

SOLUTION

Slope of line:

EC =

0.875 m 7 = 2.80 m + 0.2 m 24

Then

TABx =

24 TAB 25

and

TABy =

7 TAB 25

We have

M D = TABx ( y ) + TABy ( x) 24 7 TAB (0) + TAB (2.80 m) 25 25 = 1224 N

960 N ⋅ m =

TAB

or TAB = 1224 N 


www.elsolucionario.net PROBLEM 3.13 It is known that a force with a moment of 960 N · m about D is required to straighten the fence post CD. If the capacity of winch puller AB is 2400 N, determine the minimum value of distance d to create the specified moment about Point D.

The minimum value of d can be found based on the equation relating the moment of the force TAB about D: M D = (TAB max ) y (d )

where

M D = 960 N ⋅ m (TAB max ) y = TAB max sin θ = (2400 N) sin θ

Now

sin θ =

0.875 m (d + 0.20)2 + (0.875)2 m

 0.875 960 N ⋅ m = 2400 N   ( d + 0.20) 2 + (0.875) 2 

or

(d + 0.20) 2 + (0.875) 2 = 2.1875d

or

(d + 0.20) 2 + (0.875) 2 = 4.7852d 2

or

  (d )  

3.7852d 2 − 0.40d − 0.8056 = 0

Using the quadratic equation, the minimum values of d are 0.51719 m and − 0.41151 m. Since only the positive value applies here, d = 0.51719 m or d = 517 mm  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 173


SOLUTION

www.elsolucionario.net PROBLEM 3.14 A mechanic uses a piece of pipe AB as a lever when tightening an alternator belt. When he pushes down at A, a force of 485 N is exerted on the alternator at B. Determine the moment of that force about bolt C if its line of action passes through O.

SOLUTION M C = rB/C × FB

Noting the direction of the moment of each force component about C is clockwise, M C = xFBy + yFBx

where

and

x = 120 mm − 65 mm = 55 mm y = 72 mm + 90 mm = 162 mm FBx = FBy =

65 (65) + (72) 2 2

72 (65) + (72) 2 2

(485 N) = 325 N (485 N) = 360 N

M C = (55 mm)(360 N) + (162)(325 N) = 72450 N ⋅ m = 72.450 N ⋅ m

or M C = 72.5 N ⋅ m





We have

www.elsolucionario.net PROBLEM 3.15 Form the vector products B × C and B′ × C, where B = B′, and use the results obtained to prove the identity sin α cos β =

1 1 sin (α + β ) + sin (α − β ). 2 2

SOLUTION Note:

B = B(cos β i + sin β j) B′ = B(cos β i − sin β j)

By definition,

Now

| B × C | = BC sin (α − β )

(1)

| B′ × C | = BC sin (α + β )

(2)

B × C = B(cos β i + sin β j) × C (cos α i + sin α j)

= BC (cos β sin α − sin β cos α )k

and

(3)

B′ × C = B(cos β i − sin β j) × C (cos α i + sin α j)

= BC (cos β sin α + sin β cos α ) k

(4)

Equating the magnitudes of B × C from Equations (1) and (3) yields: BC sin(α − β ) = BC (cos β sin α − sin β cos α )

(5)

Similarly, equating the magnitudes of B′ × C from Equations (2) and (4) yields: BC sin(α + β ) = BC (cos β sin α + sin β cos α )

(6)

Adding Equations (5) and (6) gives: sin(α − β ) + sin(α + β ) = 2cos β sin α

or sin α cos β =

1 1 sin(α + β ) + sin(α − β )  2 2



C = C (cos α i + sin α j)

www.elsolucionario.net PROBLEM 3.16 The vectors P and Q are two adjacent sides of a parallelogram. Determine the area of the parallelogram when (a) P = −7i + 3j − 3k and Q = 2i + 2j + 5k, (b) P = 6i − 5j − 2k and Q = −2i + 5j − k.

SOLUTION (a)

We have

A = |P × Q|

where

P = −7i + 3j − 3k Q = 2i + 2 j + 5k

= [(15 + 6)i + ( −6 + 35) j + ( −14 − 6)k ] = (21)i + (29) j(−20)k

A = (20) 2 + (29) 2 + (−20) 2

(b)

We have

A = |P × Q|

where

P = 6i − 5 j − 2k

or A = 41.0 

Q = −2i + 5 j − 1k

Then

i j k P × Q = 6 −5 −2 −2 5 −1 = [(5 + 10)i + (4 + 6) j + (30 − 10)k ] = (15)i + (10) j + (20)k

A = (15) 2 + (10) 2 + (20) 2

or A = 26.9 



Then

i j k P × Q = −7 3 −3 2 2 5

www.elsolucionario.net PROBLEM 3.17 A plane contains the vectors A and B. Determine the unit vector normal to the plane when A and B are equal to, respectively, (a) i + 2j − 5k and 4i − 7j − 5k, (b) 3i − 3j + 2k and −2i + 6j − 4k.

SOLUTION (a)

A×B |A × B|

We have

λ=

where

A = 1i + 2 j − 5k B = 4i − 7 j − 5k

= (−10 − 35)i + (20 + 5) j + (−7 − 8)k = 15(3i − 1j − 1k )

and

|A × B | = 15 (−3)2 + (−1) 2 + (−1)2 = 15 11 λ=

(b)

15(−3i − 1j − 1k ) 15 11

or λ =

1 11

(−3i − j − k ) 

A×B |A × B|

We have

λ=

where

A = 3i − 3 j + 2k B = −2i + 6 j − 4k

Then

i j k A × B = 3 −3 2 −2 6 −4 = (12 − 12)i + (−4 + 12) j + (18 − 6)k = (8 j + 12k )

and

|A × B| = 4 (2) 2 + (3) 2 = 4 13 λ=

4(2 j + 3k ) 4 13

or λ =

1 13

(2 j + 3k ) 



Then

i j k A × B = 1 +2 −5 4 −7 −5

www.elsolucionario.net PROBLEM 3.18 A line passes through the Points (20 m, 16 m) and (−1 m, −4 m). Determine the perpendicular distance d from the line to the origin O of the system of coordinates.

SOLUTION d AB = [20 m − ( −1 m)]2 + [16 m − ( −4 m)]2 = 29 m

Assume that a force F, or magnitude F(N), acts at Point A and is directed from A to B.

where

F = F λ AB λ AB = =

By definition, where Then

rB − rA d AB 1 (21i + 20 j) 29


Then

M O = | rA × F | = dF rA = −(1 m)i − (4 m) j M O = [ −(−1 m)i − (4 m) j] ×

F [(21 m)i + (20 m) j] 29 m

F [−(20)k + (84)k ] 29  64  =  F k N ⋅ m  29  =

Finally,

 64   29 F  = d ( F )   64 d= m 29

d = 2.21 m 


www.elsolucionario.net PROBLEM 3.19 Determine the moment about the origin O of the force F = 4i − 3j + 5k that acts at a Point A. Assume that the position vector of A is (a) r = 2i + 3j − 4k, (b) r = −8i + 6j − 10k, (c) r = 8i − 6j + 5k.

SOLUTION MO = r × F

= (15 − 12)i + (−16 − 10) j + (−6 − 12)k

(b)

i j k M O = −8 6 −10 4 −3 5 = (30 − 30)i + ( −40 + 40) j + (24 − 24)k

(c)

M O = 3i − 26 j − 18k 

MO = 0 

i j k M O = 8 −6 5 4 −3 5 = (−30 + 15)i + (20 − 40) j + (−24 + 24)k

M O = −15i − 20 j 

Note: The answer to Part (b) could have been anticipated since the elements of the last two rows of the determinant are proportional.



(a)

i j k M O = 2 3 −4 4 −3 5

www.elsolucionario.net PROBLEM 3.20 Determine the moment about the origin O of the force F = 2i + 3j − 4k that acts at a Point A. Assume that the position vector of A is (a) r = 3i − 6j + 5k, (b) r = i − 4j − 2k, (c) r = 4i + 6j − 8k.

SOLUTION MO = r × F

= (24 − 15)i + (10 + 12) j + (9 + 12)k

(b)

i j k M O = 1 −4 −2 2 3 −4 = (16 + 6)i + (−4 + 4) j + (3 + 8)k

(c)

M O = 9i + 22 j + 21k 

M O = 22i + 11k 

i j k M O = 4 6 −8 2 3 −4 = (−24 + 24)i + ( −16 + 16) j + (12 − 12)k

MO = 0 

Note: The answer to Part (c) could have been anticipated since the elements of the last two rows of the determinant are proportional.



(a)

i j k M O = 3 −6 5 2 3 −4

www.elsolucionario.net PROBLEM 3.21 The wire AE is stretched between the corners A and E of a bent plate. Knowing that the tension in the wire is 435 N, determine the moment about O of the force exerted by the wire (a) on corner A, (b) on corner E.

(a)

AE = (0.21 m) 2 + (−0.16 m) 2 + (0.12 m) 2 = 0.29 m  AE FA = FA λ AE = F AE 0.21i − 0.16 j + 0.12k = (435 N) 0.29 = (315 N)i − (240 N) j + (180 N)k rA/O = −(0.09 m)i + (0.16 m) j i j k M O = −0.09 0.16 0 315 −240 180 = 28.8i + 16.20 j + (21.6 − 50.4)k

(b)

M O = (28.8 N ⋅ m)i + (16.20 N ⋅ m) j − (28.8 N ⋅ m)k 

FE = −FA = −(315 N)i + (240 N) j − (180 N)k rE / O = (0.12 m)i + (0.12 m)k i j k M O = 0.12 0 0.12 −315 240 −180 = −28.8i + (−37.8 + 21.6) j + 28.8k

M O = −(28.8 N ⋅ m)i − (16.20 N ⋅ m) j + (28.8 N ⋅ m)k 



SOLUTION  AE = (0.21 m)i − (0.16 m) j + (0.12 m)k

www.elsolucionario.net PROBLEM 3.22 A small boat hangs from two davits, one of which is shown in the figure. The tension in line ABAD is 82 lb. Determine the moment about C of the resultant force RA exerted on the davit at A.

We have

R A = 2FAB + FAD

where and

FAB = −(82 lb) j  AD 6i − 7.75 j − 3k FAD = FAD = (82 lb) AD 10.25 FAD = (48 lb)i − (62 lb) j − (24 lb)k

Thus

R A = 2FAB + FAD = (48 lb)i − (226 lb) j − (24 lb)k

Also

rA/C = (7.75 ft) j + (3 ft)k

Using Eq. (3.21):

i j k M C = 0 7.75 3 48 − 226 −24 = (492 lb ⋅ ft)i + (144.0 lb ⋅ ft) j − (372 lb ⋅ ft)k M C = (492 lb ⋅ ft)i + (144.0 lb ⋅ ft) j − (372 lb ⋅ ft)k 



SOLUTION

www.elsolucionario.net PROBLEM 3.23 A 6-ft-long fishing rod AB is securely anchored in the sand of a beach. After a fish takes the bait, the resulting force in the line is 6 lb. Determine the moment about A of the force exerted by the line at B.

SOLUTION

Then

Txz = (6 lb) cos 8° = 5.9416 lb


We have

Tx = Txz sin 30° = 2.9708 lb Ty = TBC sin 8° = − 0.83504 lb Tz = Txz cos 30° = −5.1456 lb

Now

M A = rB/A × TBC

where

rB/A = (6sin 45°) j − (6cos 45°)k =

Then

or

6 ft 2

(j − k)

i j k MA = 0 1 −1 2 2.9708 −0.83504 −5.1456 6 6 6 (−5.1456 − 0.83504)i − (2.9708) j − (2.9708)k = 2 2 2 6

M A = −(25.4 lb ⋅ ft)i − (12.60 lb ⋅ ft) j − (12.60 lb ⋅ ft)k




www.elsolucionario.net PROBLEM 3.24 A precast concrete wall section is temporarily held by two cables as shown. Knowing that the tension in cable BD is 900 N, determine the moment about Point O of the force exerted by the cable at B.

SOLUTION  BD BD

where F = 900 N

 BD = −(1 m)i − (2 m) j + (2 m)k BD = (−1 m) 2 + ( −2 m) 2 + (2 m) 2 =3m

−i − 2 j + 2k 3 = −(300 N)i − (600 N) j + (600 N)k = (2.5 m)i + (2 m) j

F = (900 N)

rB /O M O = rB /O × F

i j k = 2.5 2 0 −300 −600 600 = 1200i − 1500 j + ( −1500 + 600)k

MO = (1200 N ⋅ m)i − (1500 N ⋅ m) j − (900 N ⋅ m)k 



F=F

www.elsolucionario.net PROBLEM 3.25 A 200-N force is applied as shown to the bracket ABC. Determine the moment of the force about A.

We have

M A = rC/A × FC

where

rC/A = (0.06 m)i + (0.075 m) j FC = −(200 N) cos 30° j + (200 N)sin 30°k

Then

i j k M A = 200 0.06 0.075 0 0 − cos 30° sin 30° = 200[(0.075sin 30°)i − (0.06sin 30°) j − (0.06 cos 30°)k ]

or M A = (7.50 N ⋅ m)i − (6.00 N ⋅ m) j − (10.39 N ⋅ m)k 



SOLUTION

www.elsolucionario.net PROBLEM 3.26 The 6-m boom AB has a fixed end A. A steel cable is stretched from the free end B of the boom to a Point C located on the vertical wall. If the tension in the cable is 2.5 kN, determine the moment about A of the force exerted by the cable at B.

SOLUTION First note

d BC = (−6)2 + (2.4) 2 + (−4) 2 2.5 kN (−6i + 2.4 j − 4k ) 7.6

Then

TBC =

We have

M A = rB/A × TBC

where

rB/A = (6 m)i

Then

M A = (6 m)i ×

2.5 kN (−6i + 2.4 j − 4k ) 7.6

or M A = (7.89 kN ⋅ m) j + (4.74 kN ⋅ m)k 



= 7.6 m

www.elsolucionario.net PROBLEM 3.27 In Prob. 3.21, determine the perpendicular distance from point O to wire AE. PROBLEM 3.21 The wire AE is stretched between the corners A and E of a bent plate. Knowing that the tension in the wire is 435 N, determine the moment about O of the force exerted by the wire (a) on corner A, (b) on corner E.

SOLUTION

M O = (28.8 N ⋅ m)i + (16.20 N ⋅ m) j − (28.8 N ⋅ m)k M O = (28.8) 2 + (16.20) 2 + (28.8) 2 = 43.8329 N ⋅ m

But

M O = FA d

or

MO FA 43.8329 N ⋅ m d= 435 N = 0.100765 m d=

d = 100.8 mm 



From the solution to Prob. 3.21

www.elsolucionario.net PROBLEM 3.28 In Prob. 3.21, determine the perpendicular distance from point B to wire AE. PROBLEM 3.21 The wire AE is stretched between the corners A and E of a bent plate. Knowing that the tension in the wire is 435 N, determine the moment about O of the force exerted by the wire (a) on corner A, (b) on corner E.

SOLUTION

FA = (315 N)i − (240 N) j + (180 N)k rA/B = −(0.210 m)i M B = rA /B × FA = −0.21i × (315i − 240 j + 180k ) = 50.4k + 37.8 j M B = (50.4) 2 + (37.8) 2 = 63.0 N ⋅ m M B = FA d

or

MB FA 63.0 N ⋅ m d= 435 N = 0.144829 m

d=

d = 144.8 mm 



From the solution to Prob. 3.21

www.elsolucionario.net PROBLEM 3.29 In Problem 3.22, determine the perpendicular distance from point C to portion AD of the line ABAD. PROBLEM 3.22 A small boat hangs from two davits, one of which is shown in the figure. The tension in line ABAD is 82 lb. Determine the moment about C of the resultant force RA exerted on the davit at A.

SOLUTION

From Problem 3.22:


First compute the moment about C of the force FDA exerted by the line on D: FDA = − FAD = −(48 lb) i + (62 lb) j + (24 lb)k

M C = rD/C × FDA = + (6 ft)i × [−(48 lb)i + (62 lb) j + (24 lb)k ] = −(144 lb ⋅ ft) j + (372 lb ⋅ ft)k

M C = (144) 2 + (372)2 = 398.90 lb ⋅ ft

Then

M C = FDA d

Since

FDA = 82 lb d= =

MC FDA 398.90 lb ⋅ ft 82 lb

d = 4.86 ft 


www.elsolucionario.net PROBLEM 3.30 In Prob. 3.23, determine the perpendicular distance from point A to a line drawn through points B and C. PROBLEM 3.23 A 6-ft-long fishing rod AB is securely anchored in the sand of a beach. After a fish takes the bait, the resulting force in the line is 6 lb. Determine the moment about A of the force exerted by the line at B.

From the solution to Prob. 3.23: M A = −(25.4 lb ⋅ ft)i − (12.60 lb ⋅ ft) j − (12.60 lb ⋅ ft)k

M A = (−25.4) 2 + (−12.60) 2 + (−12.60)2 = 31.027 lb ⋅ ft M A = TBC d

MA TBC 31.027 lb ⋅ ft = 6 lb = 5.1712 ft

d=

or

d = 5.17 ft 



SOLUTION

www.elsolucionario.net PROBLEM 3.31 In Prob. 3.23, determine the perpendicular distance from point D to a line drawn through points B and C. PROBLEM 3.23 A 6-ft-long fishing rod AB is securely anchored in the sand of a beach. After a fish takes the bait, the resulting force in the line is 6 lb. Determine the moment about A of the force exerted by the line at B.


SOLUTION

AB = 6 ft TBC = 6 lb

We have

| M D | = TBC d

where d = perpendicular distance from D to line BC. M D = rB /D × TBC

rB /D = (6sin 45° ft) j = (4.2426 ft)

TBC : (TBC ) x = (6 lb) cos8° sin 30° = 2.9708 lb

(TBC ) y = −(6 lb) sin 8° = −0.83504 lb (TBC ) z = −(6 lb) cos8° cos 30° = −5.1456 lb TBC = (2.9708 lb)i − (0.83504 lb) j − (5.1456 lb)k i j k MD = 0 4.2426 0 2.9708 −0.83504 −5.1456 = −(21.831 lb ⋅ ft)i − (12.6039 lb ⋅ ft) | M D | = ( −21.831) 2 + (−12.6039) 2 = 25.208 lb ⋅ ft 25.208 lb ⋅ ft = (6 lb)d

d = 4.20 ft 


www.elsolucionario.net PROBLEM 3.32 In Prob. 3.24, determine the perpendicular distance from point O to cable BD. PROBLEM 3.24 A precast concrete wall section is temporarily held by two cables as shown. Knowing that the tension in cable BD is 900 N, determine the moment about Point O of the force exerted by the cable at B.

SOLUTION From the solution to Prob. 3.24 we have

M O = (1200) 2 + (−1500) 2 + ( −900)2 = 2121.3 N ⋅ m M O = Fd

MO F 2121.3 N ⋅ m = 900 N

d=

d = 2.36 m 



M O = (1200 N ⋅ m)i − (1500 N ⋅ m) j − (900 N ⋅ m)k

www.elsolucionario.net PROBLEM 3.33 In Prob. 3.24, determine the perpendicular distance from point C to cable BD. PROBLEM 3.24 A precast concrete wall section is temporarily held by two cables as shown. Knowing that the tension in cable BD is 900 N, determine the moment about Point O of the force exerted by the cable at B.

SOLUTION

F = −(300 N)i − (600 N) j + (600 N)k rB /C = (2 m) j M C = rB /C × F = (2 m) j × ( −300 Ni − 600 Nj + 600 Nk ) = (600 N ⋅ m)k + (1200 N ⋅ m)i M C = (600)2 + (1200) 2 = 1341.64 N ⋅ m M C = Fd

MC F 1341.64 N ⋅ m = 900 N

d=

d = 1.491 m 



From the solution to Prob. 3.24 we have

www.elsolucionario.net PROBLEM 3.34 Determine the value of a that minimizes the perpendicular distance from Point C to a section of pipeline that passes through Points A and B.

SOLUTION Assuming a force F acts along AB, |M C | = |rA / C × F| = F ( d )


d = perpendicular distance from C to line AB

where

F = λ AB F =

(24 ft) i + (24 ft) j − (28 ft) k (24) 2 + (24) 2 + (28) 2 ft

F

F (6) i + (6) j − (7) k 11 = (3 ft)i − (10 ft) j − (a − 10 ft)k =

rA/C

i j k F M C = 3 −10 10a 11 6 6 −7 = [(10 + 6a)i + (81 − 6a) j + 78 k ] |M C | = |rA/C × F 2 |

Since

or

F 11 |rA/C × F 2 | = ( dF ) 2

1 (10 + 6a) 2 + (81 − 6a) 2 + (78)2 = d 2 121

Setting

d da

(d 2 ) = 0 to find a to minimize d: 1 [2(6)(10 + 6a) + 2(−6)(81 − 6a)] = 0 121

Solving

a = 5.92 ft

or a = 5.92 ft 


www.elsolucionario.net PROBLEM 3.35 Given the vectors P = 3i − j + 2k, Q = 4i + 5j − 3k, and S = −2i + 3j − k, compute the scalar products P · Q, P · S, and Q · S.

SOLUTION P ⋅ Q = (3i − j + 2k ) ⋅ (4i + 5 j − 3k ) = (3)(4) + (−1)(5) + (2)(−3) = 12 − 5 − 6 P ⋅ Q = +1 

P ⋅ S = −11  Q ⋅ S = (4i + 5 j − 3k ) ⋅ (−2i + 3j − k ) = (4)(−2) + (5)(3) + ( −3)(−1) = −8 + 15 + 3 Q ⋅ S = +10 



P ⋅ S = (3i − j + 2k ) ⋅ (−2i + 3j − k ) = (3)(−2) + (−1)(3) + (2)( −1) = −6 − 3 − 2

www.elsolucionario.net PROBLEM 3.36 Form the scalar product B · C and use the result obtained to prove the identity cos (α − β) = cos α cos β + sin α sin β .

SOLUTION B = B cos α i + B sin α j

(1)

C = C cos β i + C sin β j

(2)

By definition: B ⋅ C = BC cos(α − β )

From (1) and (2): B ⋅ C = ( B cos α i + B sin α j) ⋅ (C cos β i + C sin β j) = BC (cos α cos β + sin α sin β )

(4)

Equating the right-hand members of (3) and (4), cos(α − β ) = cos α cos β + sin α sin β 



(3)

www.elsolucionario.net PROBLEM 3.37 Consider the volleyball net shown. Determine the angle formed by guy wires AB and AC.

First note:

AB = (−6.5)2 + (−8)2 + (2) 2 = 10.5 ft AC = (0) 2 + (−8) 2 + (6)2 = 10 ft

and

By definition, or

or

 AB = −(6.5 ft)i − (8 ft) j + (2 ft)k  AC = −(8 ft) j + (6 ft)k   AB ⋅ AC = ( AB )( AC ) cos θ (−6.5i − 8 j + 2k ) ⋅ (−8 j + 6k ) = (10.5)(10) cos θ (−6.5)(0) + ( −8)( −8) + (2)(6) = 105cos θ cos θ = 0.72381

or θ = 43.6° 



SOLUTION

www.elsolucionario.net PROBLEM 3.38 Consider the volleyball net shown. Determine the angle formed by guy wires AC and AD.

First note:

AC = (0)2 + (−8) 2 + (6) 2 = 10 ft AD = (4) 2 + ( −8) 2 + (1) 2

and

By definition, or

= 9 ft  AC = −(8 ft)j + (6 ft)k  AD = (4 ft)i − (8 ft) j + (1 ft)k   AC ⋅ AD = ( AC )( AD ) cos θ (−8 j + 6k ) ⋅ (4i − 8 j + k ) = (10)(9) cos θ (0)(4) + ( −8)( −8) + (6)(1) = 90 cos θ

or

cos θ = 0.77778

or θ = 38.9° 



SOLUTION

www.elsolucionario.net PROBLEM 3.39 Three cables are used to support a container as shown. Determine the angle formed by cables AB and AD.

SOLUTION First note:

AB = (450 mm)2 + (600 mm)2 AD = (−500 mm) 2 + (600 mm) 2 + (360 mm) 2

and

By definition,

= 860 mm  AB = (450 mm)i + (600 mm) j  AD = (−500 mm)i + (600 mm) j + (360 mm)k   AB ⋅ AD = ( AB)( AD ) cos θ

(450i + 600 j) ⋅ (−500i − 600 j + 360k ) = (750)(860) cos θ (450)(−500) + (600)(600) + (0)(360) = (750)(860) cos θ cos θ = 0.20930

or

θ = 77.9° 



= 750 mm

www.elsolucionario.net PROBLEM 3.40 Three cables are used to support a container as shown. Determine the angle formed by cables AC and AD.


AC = (600 mm) 2 + (−320 mm) 2 AD = (−500 mm) 2 + (600 mm) 2 + (360 mm) 2

and

By definition,

= 860 mm  AC = (600 mm)j + (−320 mm)k  AD = (−500 mm)i + (600 mm) j + (360 mm)k   AC ⋅ AD = ( AC )( AD ) cos θ (600 j − 320k ) ⋅ (−500i + 600 j + 360k ) = (680)(860) cos θ 0(−500) + (600)(600) + (−320)(360) = (680)(860) cos θ cos θ = 0.41860

θ = 65.3° 



= 680 mm

www.elsolucionario.net PROBLEM 3.41 The 20-in. tube AB can slide along a horizontal rod. The ends A and B of the tube are connected by elastic cords to the fixed point C. For the position corresponding to x = 11 in., determine the angle formed by the two cords (a) using Eq. (3.32), (b) applying the law of cosines to triangle ABC.

SOLUTION (a)

Using Eq. (3.32):

CA = (11) 2 + ( −12) 2 + (24) 2 = 29 in.  CB = 31i − 12 j + 24k CB = (31) 2 + ( −12) 2 + (24) 2 = 41 in.   CA ⋅ CB cos θ = (CA)(CB) (11i − 12 j + 24k ) ⋅ (31i − 12 j + 24k ) = (29)(41) (11)(31) + (−12)(−12) + (24)(24) = (29)(41) = 0.89235

(b)

θ = 26.8° 

Law of cosines: ( AB) 2 = (CA) 2 + (CB) 2 − 2(CA)(CB ) cos θ (20) 2 = (29)2 + (41) 2 − 2(29)(41) cos θ cos θ = 0.89235

θ = 26.8° 



 CA = 11i − 12 j + 24k

www.elsolucionario.net PROBLEM 3.42 Solve Prob. 3.41 for the position corresponding to x = 4 in. PROBLEM 3.41 The 20-in. tube AB can slide along a horizontal rod. The ends A and B of the tube are connected by elastic cords to the fixed point C. For the position corresponding to x = 11 in., determine the angle formed by the two cords (a) using Eq. (3.32), (b) applying the law of cosines to triangle ABC.

SOLUTION (a)

Using Eq. (3.32):

CA = (4) 2 + (−12) 2 + (24) 2 = 27.129 in.  CB = 24i − 12 j + 24k CB = (24) 2 + (−12)2 + (24)2 = 36 in.   CA ⋅ CB cos θ = (CA)(CB) (4i − 12 j + 24k ) ⋅ (24i − 12 j + 24k ) = (27.129)(36) = 0.83551

(b)

θ = 33.3° 

Law of cosines: ( AB) 2 = (CA) 2 + (CB )2 − 2(CA)(CB ) cos θ (20) 2 = (27.129)2 + (36) 2 − 2(27.129)(36) cos θ cos θ = 0.83551

θ = 33.3° 



 CA = 4i − 12 j + 24k

www.elsolucionario.net PROBLEM 3.43 Ropes AB and BC are two of the ropes used to support a tent. The two ropes are attached to a stake at B. If the tension in rope AB is 540 N, determine (a) the angle between rope AB and the stake, (b) the projection on the stake of the force exerted by rope AB at Point B.

SOLUTION BA = (−3) 2 + (3)2 + (−1.5) 2 = 4.5 m


First note:

BD = (−0.08) 2 + (0.38)2 + (0.16) 2 = 0.42 m

λ BD

(a)

TBA (−3i + 3j − 1.5k ) 4.5 T = BA (−2i + 2 j − k ) 3  1 BD ( −0.08i + 0.38 j + 0.16k ) = = BD 0.42 1 = (−4i + 19 j + 8k ) 21

TBA =

Then

We have or

or

TBA ⋅ λ BD = TBA cos θ TBA 1 ( −2i + 2 j − k ) ⋅ (−4i + 19 j + 8k ) = TBA cos θ 3 21 1 [(−2)( −4) + (2)(19) + (−1)(8)] 63 = 0.60317

cos θ =

or θ = 52.9°  (b)

We have

(TBA ) BD = TBA ⋅ λ BD = TBA cos θ = (540 N)(0.60317)

or (TBA ) BD = 326 N 


www.elsolucionario.net PROBLEM 3.44 Ropes AB and BC are two of the ropes used to support a tent. The two ropes are attached to a stake at B. If the tension in rope BC is 490 N, determine (a) the angle between rope BC and the stake, (b) the projection on the stake of the force exerted by rope BC at Point B.

SOLUTION BC = (1) 2 + (3) 2 + (−1.5) 2 = 3.5 m


First note:

BD = (−0.08) 2 + (0.38)2 + (0.16) 2 = 0.42 m TBC (i + 3j − 1.5k ) 3.5 T = BC (2i + 6 j − 3k ) 7  1 BD ( −0.08i + 0.38 j + 0.16k ) = = BD 0.42 1 = (−4i + 19 j + 8k ) 21

TBC =

λBD

(a)

TBC ⋅ λBD = TBC cos θ TBC 1 (2i + 6 j − 3k ) ⋅ (−4i + 19 j + 8k ) = TBC cos θ 7 21 1 [(2)( −4) + (6)(19) + (−3)(8)] 147 = 0.55782

cos θ =

θ = 56.1°  (b)

(TBC ) BD = TBC ⋅ λBD = TBC cos θ = (490 N)(0.55782) (TBC ) BD = 273 N 


www.elsolucionario.net PROBLEM 3.45 Given the vectors P = 4i − 2 j + 3k , Q = 2i + 4 j − 5k , and S = S x i − j + 2k , determine the value of S x for which the three vectors are coplanar.

SOLUTION If P, Q, and S are coplanar, then P must be perpendicular to (Q × S). P ⋅ (Q × S) = 0

(or, the volume of a parallelepiped defined by P, Q, and S is zero). Then

−2 3 4 −5 = 0 −1 2

32 + 10S x − 6 − 20 + 8 − 12S x = 0

Sx = 7 


or

4 2 Sx


www.elsolucionario.net PROBLEM 3.46 Determine the volume of the parallelepiped of Fig. 3.25 when (a) P = 4i − 3j + 2k, Q = −2i − 5j + k, and S = 7i + j − k, (b) P = 5i − j + 6k, Q = 2i + 3j + k, and S = −3i − 2j + 4k.

SOLUTION Volume of a parallelepiped is found using the mixed triple product. Vol. = P ⋅ (Q × S) 4 −3 2 = −2 −5 1 in.3 7 1 −1 = (20 − 21 − 4 + 70 + 6 − 4) = 67

or Volume = 67.0  (b)

Vol. = P ⋅ (Q × S) 5 −1 6 = 2 3 1 in.3 −3 −2 4 = (60 + 3 − 24 + 54 + 8 + 10) = 111

or Volume = 111.0 



(a)

www.elsolucionario.net PROBLEM 3.47 Knowing that the tension in cable AB is 570 N, determine the moment about each of the coordinate axes of the force exerted on the plate at B.

SOLUTION

BA = (−900) 2 + (600)2 + (360) 2 = 1140 mm  BA FB = FB BA −900i + 600 j + 360k = (570 N) 1140 = −(450 N)i + (300 N) j + (180 N)k rB = (0.9 m)i M O = rB × FB = 0.9i × (−450i + 300 j + 180k ) = 270k − 162 j MO = M x i + M y j + M z k = −(162 N ⋅ m) j + (270 N ⋅ m)k M x = 0, M y = −162.0 N ⋅ m, M z = +270 N ⋅ m 

Therefore,



BA = (−900 mm)i + (600 mm) j + (360 mm)k

www.elsolucionario.net PROBLEM 3.48 Knowing that the tension in cable AC is 1065 N, determine the moment about each of the coordinate axes of the force exerted on the plate at C.

SOLUTION

CA = ( −900)2 + (600) 2 + (−920) 2 = 1420 mm  CA FC = FC CA −900i + 600 j − 920k = (1065 N) 1420 = −(675 N)i + (450 N) j − (690 N)k rC = (0.9 m)i + (1.28 m)k

Using Eq. (3.19): i j k 0.9 0 1.28 M O = rC × FC = −675 450 −690 M O = −(576 N ⋅ m)i − (243 N ⋅ m) j + (405 N ⋅ m)k

But Therefore,

MO = M x i + M y j + M z k M x = −576 N ⋅ m, M y = −243 N ⋅ m, M z = +405 N ⋅ m 



 CA = (−900 mm)i + (600 mm) j + (−920 mm)k

www.elsolucionario.net PROBLEM 3.49 A small boat hangs from two davits, one of which is shown in the figure. It is known that the moment about the z-axis of the resultant force RA exerted on the davit at A must not exceed 279 lb⋅ft in absolute value. Determine the largest allowable tension in line ABAD when x = 6 ft.

SOLUTION R A = 2TAB + TAD


First note:

Also note that only TAD will contribute to the moment about the z-axis. Now

AD = (6) 2 + (−7.75) 2 + (−3) 2 = 10.25 ft  AD =T AD T (6i − 7.75 j − 3k ) = 10.25

Then

TAD

Now

M z = k ⋅ (rA/C × TAD )

where

rA/C = (7.75 ft) j + (3 ft)k

Then for Tmax ,

0 0 1 Tmax 279 = 0 7.75 3 10.25 6 −7.75 −3 =

Tmax | − (1)(7.75)(6)| 10.25

or Tmax = 61.5 lb 


www.elsolucionario.net PROBLEM 3.50 For the davit of Problem 3.49, determine the largest allowable distance x when the tension in line ABAD is 60 lb.

SOLUTION

=

60 lb x 2 + ( −7.75) 2 + (−3)2


From the solution of Problem 3.49, TAD is now  AD TAD = T AD ( xi − 7.75 j − 3k )

Then M z = k ⋅ (rA / C × TAD ) becomes 279 = 279 =

60 x 2 + (−7.75) 2 + ( −3) 2 60 x 2 + 69.0625

0 0 1 0 7.75 3 x −7.75 −3

| − (1)(7.75)( x) |

279 x 2 + 69.0625 = 465 x 0.6 x 2 + 69.0625 = x

Squaring both sides:

0.36 x 2 + 24.8625 = x 2 x 2 = 38.848

x = 6.23 ft 


www.elsolucionario.net PROBLEM 3.51 A farmer uses cables and winch pullers B and E to plumb one side of a small barn. If it is known that the sum of the moments about the x-axis of the forces exerted by the cables on the barn at Points A and D is equal to 4728 lb ⋅ ft, determine the magnitude of TDE when TAB = 255 lb.

The moment about the x-axis due to the two cable forces can be found using the z components of each force acting at their intersection with the xy plane (A and D). The x components of the forces are parallel to the x-axis, and the y components of the forces intersect the x-axis. Therefore, neither the x or y components produce a moment about the x-axis. We have

ΣM x : (TAB ) z ( y A ) + (TDE ) z ( yD ) = M x

where

(TAB ) z = k ⋅ TAB = k ⋅ (TAB λ AB )   − i − 12 j + 12k   = k ⋅  255 lb   17    = 180 lb

(TDE ) z = k ⋅ TDE = k ⋅ (TDE λDE )   1.5i − 14 j + 12k   = k ⋅ TDE   18.5    = 0.64865TDE y A = 12 ft yD = 14 ft M x = 4728 lb ⋅ ft (180 lb)(12 ft) + (0.64865TDE )(14 ft) = 4728 lb ⋅ ft

and

TDE = 282.79 lb

or TDE = 283 lb 



SOLUTION

www.elsolucionario.net PROBLEM 3.52 Solve Problem 3.51 when the tension in cable AB is 306 lb. PROBLEM 3.51 A farmer uses cables and winch pullers B and E to plumb one side of a small barn. If it is known that the sum of the moments about the x-axis of the forces exerted by the cables on the barn at Points A and D is equal to 4728 lb ⋅ ft, determine the magnitude of TDE when TAB = 255 lb.

The moment about the x-axis due to the two cable forces can be found using the z components of each force acting at the intersection with the xy plane (A and D). The x components of the forces are parallel to the x-axis, and the y components of the forces intersect the x-axis. Therefore, neither the x or y components produce a moment about the x-axis. We have

ΣM x : (TAB ) z ( y A ) + (TDE ) z ( yD ) = M x

Where

(TAB ) z = k ⋅ TAB = k ⋅ (TAB λAB )   − i − 12 j + 12k   = k ⋅ 306 lb   17    = 216 lb (TDE ) z = k ⋅ TDE = k ⋅ (TDE λDE )   1.5i − 14 j + 12k   = k ⋅ TDE   18.5    = 0.64865TDE y A = 12 ft yD = 14 ft M x = 4728 lb ⋅ ft (216 lb)(12 ft) + (0.64865TDE )(14 ft) = 4728 lb ⋅ ft

and

or TDE = 235 lb 

TDE = 235.21 lb



SOLUTION

www.elsolucionario.net PROBLEM 3.53 A single force P acts at C in a direction perpendicular to the handle BC of the crank shown. Knowing that Mx = +20 N · m and My = −8.75 N · m, and Mz = −30 N · m, determine the magnitude of P and the values of φ and θ.

SOLUTION rC = (0.25 m)i + (0.2 m) sin θ j + (0.2 m) cos θ k i j M O = rC × P = 0.25 0.2sin θ 0 − P sin φ

k 0.2 cos θ P cos φ

Expanding the determinant, we find M x = (0.2) P(sin θ cos φ + cos θ sin φ )

Dividing Eq. (3) by Eq. (2) gives:

M x = (0.2) P sin(θ + φ )

(1)

M y = −(0.25) P cos φ

(2)

M z = −(0.25) P sin φ

(3)

tan φ =

Mz My

(4)

−30 N ⋅ m −8.75 N ⋅ m φ = 73.740

tan φ =

φ = 73.7° 

Squaring Eqs. (2) and (3) and adding gives: M y2 + M z2 = (0.25) 2 P 2

or

P = 4 M y2 + M z2

(5)

P = 4 (8.75) 2 + (30) 2 = 125.0 N

P = 125.0 N 

Substituting data into Eq. (1): (+20 N ⋅ m) = 0.2 m(125.0 N) sin(θ + φ ) (θ + φ ) = 53.130° and (θ + φ ) = 126.87° θ = −20.6° and θ = 53.1°

Q = 53.1° 



P = − P sin φ j + P cos φ k

www.elsolucionario.net PROBLEM 3.54 A single force P acts at C in a direction perpendicular to the handle BC of the crank shown. Determine the moment Mx of P about the x-axis when θ = 65°, knowing that My = −15 N · m and Mz = −36 N · m.

SOLUTION See the solution to Prob. 3.53 for the derivation of the following equations:

P = 4 M y2 + M z2

(1) (4) (5)

Substituting for known data gives: tan φ =

−36 N ⋅ m −15 N ⋅ m

φ = 67.380° P = 4 ( −15) 2 + (−36) 2 P = 156.0 N M x = 0.2 m(156.0 N) sin(65° + 67.380°) = 23.047 N ⋅ m M x = 23.0 N ⋅ m 



M x = (0.2) P sin(θ + φ ) M tan φ = z My

www.elsolucionario.net PROBLEM 3.55 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable AE at A is 55 N, determine the moment of that force about the line joining Points D and B.

SOLUTION TAE = TAE


First note:

 AE AE

AE = (0.9)2 + (−0.6) 2 + (0.2) 2 = 1.1 m

Then

Also,

Then

Now where

Then

55 N (0.9i − 0.6 j + 0.2k ) 1.1 = 5[(9 N)i − (6 N) j + (2 N)k ]

TAE =

DB = (1.2) 2 + ( −0.35) 2 + (0)2

λ DB

= 1.25 m  DB = DB 1 (1.2i − 0.35 j) = 1.25 1 (24i − 7 j) = 25

M DB = λ DB ⋅ (rA/D × TAE ) rA/D = −(0.1 m) j + (0.2 m)k

M DB

24 −7 0 1 (5) 0 −0.1 0.2 = 25 9 2 −6 1 = (−4.8 − 12.6 + 28.8) 5

or M DB = 2.28 N ⋅ m 


www.elsolucionario.net PROBLEM 3.56 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable CF at C is 33 N, determine the moment of that force about the line joining Points D and B.

SOLUTION TCF = TCF


First note:

 CF CF

CF = (0.6)2 + (−0.9) 2 + (−0.2) 2 = 1.1 m

33 N (0.6i − 0.9 j + 0.2k ) 1.1 = 3[(6 N)i − (9 N) j − (2 N)k ]

Then

TCF =

Also,

DB = (1.2) 2 + ( −0.35) 2 + (0)2

= 1.25 m  DB = DB 1 (1.2i − 0.35 j) = 1.25 1 (24i − 7 j) = 25

Then

λ DB

Now

M DB = λ DB ⋅ (rC/D × TCF )

where

rC/D = (0.2 m) j − (0.4 m)k

Then

M DB

24 −7 0 1 (3) 0 0.2 −0.4 = 25 6 −9 −2 =

3 (−9.6 + 16.8 − 86.4) 25

or M DB = −9.50 N ⋅ m 


www.elsolucionario.net PROBLEM 3.57 The 23-in. vertical rod CD is welded to the midpoint C of the 50-in. rod AB. Determine the moment about AB of the 235-lb force P.

SOLUTION


 AB = (32 in.)i − (30 in.) j − (24 in.)k AB = (32)2 + ( −30) 2 + ( −24) 2 = 50 in.  AB = 0.64i − 0.60 − 0.48k λ AB = AB

We shall apply the force P at Point G: rG /B = (5 in.)i + (30 in.)k  DG = (21 in.)i − (38 in.) j + (18 in.)k

DG = (21) 2 + (−38)2 + (18) 2 = 47 in.  DG 21i − 38 j + 18k P=P = (235 lb) 47 DG P = (105 lb)i − (190 lb) j + (90 lb)k

The moment of P about AB is given by Eq. (3.46): M AB

0.64 −0.60 −0.48 0 30 in. = λ AB ⋅ (rG /B × P) = 5 in. 105 lb −190 lb 90 lb

M AB = 0.64[0 − (30 in.)(−190 lb)] − 0.60[(30 in.)(105 lb) − (5 in.)(90 lb)] − 0.48[(5 in.)( −190 lb) − 0] = +2484 lb ⋅ in. M AB = +207 lb ⋅ ft  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 217

www.elsolucionario.net PROBLEM 3.58 The 23-in. vertical rod CD is welded to the midpoint C of the 50-in. rod AB. Determine the moment about AB of the 174-lb force Q.

SOLUTION


 AB = (32 in.)i − (30 in.) j − (24 in.)k AB = (32)2 + ( −30) 2 + ( −24) 2 = 50 in.  AB = 0.64i − 0.60 j − 0.48k λ AB = AB

We shall apply the force Q at Point H: rH /B = −(32 in.)i + (17 in.) j  DH = −(16 in.)i − (21 in.) j − (12 in.)k

DH = (16) 2 + (−21) 2 + (−12)2 = 29 in.  DH −16i − 21j − 12k Q= = (174 lb) DH 29 Q = −(96 lb)i − (126 lb) j − (72 lb)k

The moment of Q about AB is given by Eq. (3.46): M AB

0.64 −0.60 −0.48 = λ AB ⋅ (rH /B × Q) = −32 in. 17 in. 0 −96 lb −126 lb −72 lb

M AB = 0.64[(17 in.)(−72 lb) − 0] − 0.60[(0 − (−32 in.)( −72 lb)] − 0.48[(−32 in.)(−126 lb) − (17 in.)(−96 lb)] = −2119.7 lb ⋅ in. M AB = 176.6 lb ⋅ ft  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 218

www.elsolucionario.net PROBLEM 3.59 The frame ACD is hinged at A and D and is supported by a cable that passes through a ring at B and is attached to hooks at G and H. Knowing that the tension in the cable is 450 N, determine the moment about the diagonal AD of the force exerted on the frame by portion BH of the cable.


SOLUTION M AD = λ AD ⋅ (rB/A × TBH )

Where

and

1 λ AD = (4i − 3k ) 5 rB/A = (0.5 m)i d BH = (0.375)2 + (0.75) 2 + (−0.75) 2

= 1.125 m

Then

Finally,

450 N (0.375i + 0.75 j − 0.75k ) 1.125 = (150 N)i + (300 N) j − (300 N)k

TBH =

MAD

4 0 −3 1 0 = 0.5 0 5 150 300 −300 1 = [(−3)(0.5)(300)] 5

or M AD = − 90.0 N ⋅ m 


www.elsolucionario.net PROBLEM 3.60 In Problem 3.59, determine the moment about the diagonal AD of the force exerted on the frame by portion BG of the cable. PROBLEM 3.59 The frame ACD is hinged at A and D and is supported by a cable that passes through a ring at B and is attached to hooks at G and H. Knowing that the tension in the cable is 450 N, determine the moment about the diagonal AD of the force exerted on the frame by portion BH of the cable.


SOLUTION M AD = λ AD ⋅ (rB/A × TBG )

Where

and

1 λ AD = (4i − 3k ) 5 rB/A = (0.5 m) j BG = (−0.5) 2 + (0.925)2 + (−0.4)2

= 1.125 m

Then

Finally,

450 N (−0.5i + 0.925 j − 0.4k ) 1.125 = −(200 N)i + (370 N) j − (160 N)k

TBG =

MAD

4 0 −3 1 = 0.5 0 0 5 −200 370 −160 1 = [(−3)(0.5)(370)] 5

M AD = −111.0 N ⋅ m 


www.elsolucionario.net PROBLEM 3.61 A regular tetrahedron has six edges of length a. A force P is directed as shown along edge BC. Determine the moment of P about edge OA.

SOLUTION

From triangle OBC:

M OA = λOA ⋅ (rC /O × P)

(OA) x =

a 2

(OA) z = (OA) x tan 30° =

Since

a 1  a  = 2 3  2 3

(OA) 2 = (OA) 2x + (OA) 2y + (OAz )2 2

or


We have

 a  a a =   + (OA) 2y +   2 2 3

2

2

(OA) y = a 2 −

2 a2 a2 − =a 4 12 3

a 2 a i+a j+ k 2 3 2 3

Then

rA/O =

and

λOA = i +

1 2

2 1 j+ k 3 2 3

P = λ BC P =

(a sin 30°)i − (a cos 30°)k P ( P) = (i − 3k ) a 2

rC /O = ai

M OA

1 2 = 1

2 3 0

2 3 P (a)   0 2

1

0

− 3

=

1

aP  2  aP  −  (1)(− 3) = 2  3 2

M OA =

aP 2




www.elsolucionario.net PROBLEM 3.62 A regular tetrahedron has six edges of length a. (a) Show that two opposite edges, such as OA and BC, are perpendicular to each other. (b) Use this property and the result obtained in Problem 3.61 to determine the perpendicular distance between edges OA and BC.

SOLUTION For edge OA to be perpendicular to edge BC,   OA ⋅ BC = 0 From triangle OBC:

(OA) x =

a 2


(a)

a 1  a  = 2 3  2 3   a   a  OA =   i + (OA) y j +  k 2 2 3  BC = ( a sin 30°) i − (a cos 30°) k

(OA) z = (OA) x tan 30° =

and

=

Then

or

so that (b)

We have M OA

a a 3 a i− k = (i − 3 k ) 2 2 2

a  a   a  i + (OA) y j +   k  ⋅ (i − 3k ) = 0 2 2 3  2 a2 a2 + (OA) y (0) − =0 4 4   OA ⋅ BC = 0

  OA is perpendicular to BC.    = Pd , with P acting along BC and d the perpendicular distance from OA to BC.

From the results of Problem 3.57, M OA = Pa 2

Pa 2

= Pd

or d =

a 2




www.elsolucionario.net PROBLEM 3.63 Two forces F1 and F2 in space have the same magnitude F. Prove that the moment of F1 about the line of action of F2 is equal to the moment of F2 about the line of action of F1 .

SOLUTION

First note that

F1 = F1λ1 and F2 = F2 λ 2

Now, by definition,

M1 = λ 1 ⋅ (rB /A × F2 ) = λ 1 ⋅ (rB /A × λ 2 ) F2 M 2 = λ 2 ⋅ (rA/B × F1 ) = λ 2 ⋅ (rA/B × λ 1 ) F1

Since

F1 = F2 = F

and rA /B = −rB /A

M1 = λ 1 ⋅ (rB /A × λ 2 ) F M 2 = λ 2 ⋅ (−rB /A × λ 1 ) F

Using Equation (3.39): so that

λ 1 ⋅ (rB /A × λ 2 ) = λ 2 ⋅ (−rB /A × λ 1 ) M 2 = λ 1⋅ (rB /A × λ 2 ) F 

M12 = M 21 



Let M1 = moment of F2 about the line of action of F1 and M 2 = moment of F1 about the line of action of F2 .

www.elsolucionario.net PROBLEM 3.64 In Problem 3.55, determine the perpendicular distance between cable AE and the line joining Points D and B. PROBLEM 3.55 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable AE at A is 55 N, determine the moment of that force about the line joining Points D and B.

From the solution to Problem 3.55:

ΤAE = 55 N TAE = 5[(9 N)i − (6 N) j + (2 N)k ] | M DB | = 2.28 N ⋅ m λ DB =

1 (24i − 7 j) 25

Based on the discussion of Section 3.11, it follows that only the perpendicular component of TAE will contribute to the moment of TAE about line DB. Now

(TAE )parallel = TAE ⋅ λ DB = 5(9i − 6 j + 2k ) ⋅

1 (24i − 7 j) 25

1 = [(9)(24) + (−6)(−7)] 5 = 51.6 N

Also, so that

TAE = (TAE ) parallel + (TAE ) perpendicular (TAE )perpendicular = (55) 2 + (51.6)2 = 19.0379 N

Since λ DB and (TAE )perpendicular are perpendicular, it follows that M DB = d (TAE ) perpendicular

or

2.28 N ⋅ m = d (19.0379 N) d = 0.119761

d = 0.1198 m 



SOLUTION

www.elsolucionario.net PROBLEM 3.65 In Problem 3.56, determine the perpendicular distance between cable CF and the line joining Points D and B. PROBLEM 3.56 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable CF at C is 33 N, determine the moment of that force about the line joining Points D and B.

ΤCF = 33 N

From the solution to Problem 3.56:

TCF = 3[(6 N)i − (9 N) j − (2 N)k ] | M DB | = 9.50 N ⋅ m λ DB =

1 (24i − 7 j) 25

Based on the discussion of Section 3.11, it follows that only the perpendicular component of TCF will contribute to the moment of TCF about line DB. Now

(TCF )parallel = TCF ⋅ λ DB = 3(6i − 9 j − 2k ) ⋅

1 (24i − 7 j) 25

3 [(6)(24) + (−9)( −7)] 25 = 24.84 N =

Also, so that

TCF = (TCF ) parallel + (TCF ) perpendicular (TCF )perpendicular = (33) 2 − (24.84) 2 = 21.725 N

Since λ DB and (TCF )perpendicular are perpendicular, it follows that | M DB | = d (TCF ) perpendicular

or

9.50 N ⋅ m = d × 21.725 N

or d = 0.437 m 



SOLUTION

www.elsolucionario.net PROBLEM 3.66 In Prob. 3.57, determine the perpendicular distance between rod AB and the line of action of P. PROBLEM 3.57 The 23-in. vertical rod CD is welded to the midpoint C of the 50-in. rod AB. Determine the moment about AB of the 235-lb force P.

SOLUTION


 AB = (32 in.)i − (30 in.) j − (24 in.)k AB = (32)2 + (−30) 2 + ( −24) 2 = 50 in.  AB = 0.64i − 0.60 j − 0.48k λ AB = AB

λP =

P 105i − 190 j + 90k = P 235

Angle θ between AB and P: cos θ = λ AB ⋅ λ P = (0.64i − 0.60 j − 0.48k ) ⋅

105i − 190 j + 90k 235

= 0.58723 ∴ θ = 54.039°

The moment of P about AB may be obtained by multiplying the projection of P on a plane perpendicular to AB by the perpendicular distance d from AB to P: M AB = ( P sin θ )d

From the solution to Prob. 3.57: M AB = 207 lb ⋅ ft = 2484 lb ⋅ in. We have

2484 lb ⋅ in. = (235 lb)(sin 54.039) d d = 13.06 in. 


www.elsolucionario.net PROBLEM 3.67 In Prob. 3.58, determine the perpendicular distance between rod AB and the line of action of Q. PROBLEM 3.58 The 23-in. vertical rod CD is welded to the midpoint C of the 50-in. rod AB. Determine the moment about AB of the 174-lb force Q.

SOLUTION


 AB = (32 in.)i − (30 in.) j − (24 in.)k AB = (32)2 + (−30) 2 + ( −24) 2 = 50 in.  AB λ AB = = 0.64i − 0.60 j − 0.48k AB

λQ =

Q −96i − 126 j − 72k = Q 174

Angle θ between AB and Q: cos θ = λ AB ⋅ λQ = (0.64i − 0.60 j − 0.48k ) ⋅

(−96i − 126 j − 72k ) 174

= 0.28000 ∴ θ = 73.740°

The moment of Q about AB may be obtained by multiplying the projection of Q on a plane perpendicular to AB by the perpendicular distance d from AB to Q: M AB = (Q sin θ )d

From the solution to Prob. 3.58: M AB = 176.6 lb ⋅ ft = 2119.2 lb ⋅ in. 2119.2 lb ⋅ in. = (174 lb)(sin 73.740°) d d = 12.69 in. 


www.elsolucionario.net PROBLEM 3.68 In Problem 3.59, determine the perpendicular distance between portion BH of the cable and the diagonal AD.

SOLUTION From the solution to Problem 3.59:

TBH = 450 N

TBH = (150 N)i + (300 N) j − (300 N)k | M AD | = 90.0 N ⋅ m 1 λ AD = (4i − 3k ) 5

Based on the discussion of Section 3.11,  it follows that only the perpendicular component of TBH will contribute to the moment of TBH about line AD. Now

(TBH )parallel = TBH ⋅ λ AD 1 = (150i + 300 j − 300k ) ⋅ (4i − 3k ) 5 1 = [(150)(4) + (−300)(−3)] 5 = 300 N

Also, so that

TBH = (TBH ) parallel + (TBH )perpendicular (TBH )perpendicular = (450) 2 − (300) 2 = 335.41 N

Since λ AD and (TBH )perpendicular are perpendicular, it follows that M AD = d (TBH ) perpendicular

or

90.0 N ⋅ m = d (335.41 N) d = 0.26833 m

d = 0.268 m 



PROBLEM 3.59 The frame ACD is hinged at A and D and is supported by a cable that passes through a ring at B and is attached to hooks at G and H. Knowing that the tension in the cable is 450 N, determine the moment about the diagonal AD of the force exerted on the frame by portion BH of the cable.

www.elsolucionario.net PROBLEM 3.69 In Problem 3.60, determine the perpendicular distance between portion BG of the cable and the diagonal AD.

SOLUTION From the solution to Problem 3.60:

ΤBG = 450 N TBG = −(200 N)i + (370 N) j − (160 N)k | M AD | = 111 N ⋅ m 1 λ AD = (4i − 3k ) 5

Based on the discussion of Section 3.11,  it follows that only the perpendicular component of TBG will contribute to the moment of TBG about line AD. Now

(TBG ) parallel = TBG ⋅ λ AD 1 = ( −200i + 370 j − 160k ) ⋅ (4i − 3k ) 5 1 = [(−200)(4) + (−160)(−3)] 5 = −64 N

Also, so that

TBG = (TBG ) parallel + (TBG )perpendicular (TBG ) perpendicular = (450) 2 − ( −64) 2 = 445.43 N

Since λ AD and (TBG ) perpendicular are perpendicular, it follows that M AD = d (TBG ) perpendicular

or

111 N ⋅ m = d (445.43 N)

d = 0.24920 m

d = 0.249 m 



PROBLEM 3.60 In Problem 3.59, determine the moment about the diagonal AD of the force exerted on the frame by portion BG of the cable.

www.elsolucionario.net PROBLEM 3.70 A plate in the shape of a parallelogram is acted upon by two couples. Determine (a) the moment of the couple formed by the two 21-lb forces, (b) the perpendicular distance between the 12-lb forces if the resultant of the two couples is zero, (c) the value of α if the resultant couple is 72 lb ⋅ in. clockwise and d is 42 in.

SOLUTION (a)

M1 = d1 F1

We have

d1 = 16 in.

where

F1 = 21 lb M1 = (16 in.)(21 lb)

(b)

(c)

336 lb ⋅ in. − d 2 (12 lb) = 0

d 2 = 28.0 in. 

M total = M1 + M 2

We have or



M1 + M 2 = 0

We have or

or M1 = 336 lb ⋅ in.

−72 lb ⋅ in. = 336 lb ⋅ in. − (42 in.)(sin α )(12 lb) sin α = 0.80952

α = 54.049°

and

or α = 54.0° 



= 336 lb ⋅ in.

www.elsolucionario.net PROBLEM 3.71 Four 1-in.-diameter pegs are attached to a board as shown. Two strings are passed around the pegs and pulled with the forces indicated. (a) Determine the resultant couple acting on the board. (b) If only one string is used, around which pegs should it pass and in what directions should it be pulled to create the same couple with the minimum tension in the string? (c) What is the value of that minimum tension?

SOLUTION

M = 470 lb ⋅ in.

(b)



With only one string, pegs A and D, or B and C should be used. We have 6 8

tan θ =

θ = 36.9°

90° − θ = 53.1°

Direction of forces:

(c)

With pegs A and D:

θ = 53.1° 

With pegs B and C:

θ = 53.1° 

The distance between the centers of the two pegs is 82 + 62 = 10 in.

Therefore, the perpendicular distance d between the forces is 1  d = 10 in. + 2  in.  2  = 11 in. M = Fd

We must have

470 lb ⋅ in. = F (11 in.)

F = 42.7 lb 



M = (35 lb)(7 in.) + (25 lb)(9 in.) = 245 lb ⋅ in. + 225 lb ⋅ in.

(a)

www.elsolucionario.net PROBLEM 3.72 Four pegs of the same diameter are attached to a board as shown. Two strings are passed around the pegs and pulled with the forces indicated. Determine the diameter of the pegs knowing that the resultant couple applied to the board is 485 lb·in. counterclockwise.

SOLUTION M = d AD FAD + d BC FBC d = 1.250 in. 


485 lb ⋅ in. = [(6 + d ) in.](35 lb) + [(8 + d ) in.](25 lb)


www.elsolucionario.net PROBLEM 3.73 A piece of plywood in which several holes are being drilled successively has been secured to a workbench by means of two nails. Knowing that the drill exerts a 12-N·m couple on the piece of plywood, determine the magnitude of the resulting forces applied to the nails if they are located (a) at A and B, (b) at B and C, (c) at A and C.

(a)

M = Fd 12 N ⋅ m = F (0.45 m) F = 26.7 N 

(b)

M = Fd 12 N ⋅ m = F (0.24 m) F = 50.0 N 

(c)

M = Fd

d = (0.45 m)2 + (0.24 m)2 = 0.510 m

12 N ⋅ m = F (0.510 m) F = 23.5 N 



SOLUTION

www.elsolucionario.net PROBLEM 3.74 Two parallel 40-N forces are applied to a lever as shown. Determine the moment of the couple formed by the two forces (a) by resolving each force into horizontal and vertical components and adding the moments of the two resulting couples, (b) by using the perpendicular distance between the two forces, (c) by summing the moments of the two forces about Point A.

SOLUTION We have where

ΣM B : − d1C x + d 2 C y = M

d1 = (0.270 m) sin 55° = 0.22117 m d 2 = (0.270 m) cos 55° = 0.154866 m C x = (40 N) cos 20° = 37.588 N C y = (40 N) sin 20° = 13.6808 N M = −(0.22117 m)(37.588 N)k + (0.154866 m)(13.6808 N)k = −(6.1946 N ⋅ m)k

(b)

We have

We have



or M = 6.19 N ⋅ m



or M = 6.19 N ⋅ m



M = Fd (−k ) = 40 N[(0.270 m)sin(55° − 20°)](−k )

= −(6.1946 N ⋅ m)k

(c)

or M = 6.19 N ⋅ m

ΣM A : Σ(rA × F ) = rB/A × FB + rC/A × FC = M

i j k sin 55° 0 M = (0.390 m)(40 N) cos 55° − cos 20° − sin 20° 0 i j k + (0.660 m)(40 N) cos 55° sin 55° 0 cos 20° sin 20° 0 = (8.9478 N ⋅ m − 15.1424 N ⋅ m)k = −(6.1946 N ⋅ m)k



(a)

www.elsolucionario.net PROBLEM 3.75 The two shafts of a speed-reducer unit are subjected to couples of magnitude M1 = 15 lb·ft and M2 = 3 lb·ft, respectively. Replace the two couples with a single equivalent couple, specifying its magnitude and the direction of its axis.

SOLUTION M1 = (15 lb ⋅ ft)k


M 2 = (3 lb ⋅ ft)i

M = M12 + M 22 = (15) 2 + (3) 2 = 15.30 lb ⋅ ft 15 tan θ x = =5 3

θ x = 78.7° θ y = 90°

θ z = 90° − 78.7° = 11.30° M = 15.30 lb ⋅ ft; θ x = 78.7°, θ y = 90.0°, θ z = 11.30° 


www.elsolucionario.net PROBLEM 3.76 Replace the two couples shown with a single equivalent couple, specifying its magnitude and the direction of its axis.

Replace the couple in the ABCD plane with two couples P and Q shown: P = (50 N)

CD  160 mm  = (50 N)   = 40 N CG  200 mm 

Q = (50 N)

CF  120 mm  = (50 N)   = 30 N CG  200 mm 

Couple vector M1 perpendicular to plane ABCD: M1 = (40 N)(0.24 m) − (30 N)(0.16 m) = 4.80 N ⋅ m

Couple vector M2 in the xy plane: M 2 = −(12.5 N)(0.192 m) = −2.40 N ⋅ m

144 mm θ = 36.870° 192 mm M1 = (4.80 cos 36.870°) j + (4.80 sin 36.870°)k = 3.84 j + 2.88k

tan θ =

M 2 = −2.40 j M = M1 + M 2 = 1.44 j + 2.88k M = 3.22 N ⋅ m; θ x = 90.0°, θ y = 53.1°, θ z = 36.9°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 236


SOLUTION

www.elsolucionario.net PROBLEM 3.77 Solve Prob. 3.76, assuming that two 10-N vertical forces have been added, one acting upward at C and the other downward at B. PROBLEM 3.76 Replace the two couples shown with a single equivalent couple, specifying its magnitude and the direction of its axis.

Replace the couple in the ABCD plane with two couples P and Q shown. P = (50 N)

CD  160 mm  = (50 N)   = 40 N CG  200 mm 

Q = (50 N)

CF  120 mm  = (50 N)   = 30 N CG  200 mm 

Couple vector M1 perpendicular to plane ABCD. M1 = (40 N)(0.24 m) − (30 N)(0.16 m) = 4.80 N ⋅ m

144 mm θ = 36.870° 192 mm M1 = (4.80cos 36.870°) j + (4.80sin 36.870°)k = 3.84 j + 2.88k

tan θ =

M 2 = −(12.5 N)(0.192 m) = −2.40 N ⋅ m = −2.40 j M 3 = rB /C × M 3 ; rB /C = (0.16 m)i + (0.144 m) j − (0.192 m)k = (0.16 m)i + (0.144 m) j − (0.192 m)k × (−10 N) j = −1.92i − 1.6k



SOLUTION


M = M1 + M 2 + M 3 = (3.84 j + 2.88k ) − 2.40 j + (−1.92i − 1.6k ) = −(1.92 N ⋅ m)i + (1.44 N ⋅ m) j + (1.28 N ⋅ m)k M = ( −1.92) 2 + (1.44) 2 + (1.28) 2 = 2.72 N ⋅ m

M = 2.72 N ⋅ m 

cos θ x = −1.92/2.72 cos θ y = 1.44/2.72

θ x = 134.9° θ y = 58.0° θ z = 61.9° 


cos θ z = 1.28/2.72


www.elsolucionario.net PROBLEM 3.78 If P = 0, replace the two remaining couples with a single equivalent couple, specifying its magnitude and the direction of its axis.

SOLUTION M = M1 + M 2 ; F1 = 16 lb, F2 = 40 lb M 2 = rE/B × F2 ; rE/B = (15 in.)i − (5 in.) j d DE = (0) 2 + (5) 2 + (10) 2 = 5 5 in. F2 =

40 lb 5 5

(5 j − 10k )

= 8 5[(1 lb) j − (2 lb)k ]

i j k M 2 = 8 5 15 −5 0 0 1 −2 = 8 5[(10 lb ⋅ in.)i + (30 lb ⋅ in.) j + (15 lb ⋅ in.)k ]

M = −(480 lb ⋅ in.)k + 8 5[(10 lb ⋅ in.)i + (30 lb ⋅ in.) j + (15 lb ⋅ in.)k ] = (178.885 lb ⋅ in.)i + (536.66 lb ⋅ in.) j − (211.67 lb ⋅ in.)k M = (178.885) 2 + (536.66) 2 + (−211.67) 2 = 603.99 lb ⋅ in

M = 604 lb ⋅ in. 

M = 0.29617i + 0.88852 j − 0.35045k M cos θ x = 0.29617 λ axis =

cos θ y = 0.88852

θ x = 72.8° θ y = 27.3° θ z = 110.5° 

cos θ z = −0.35045



M1 = rC × F1 = (30 in.)i × [−(16 lb) j] = −(480 lb ⋅ in.)k

www.elsolucionario.net PROBLEM 3.79 If P = 20 lb, replace the three couples with a single equivalent couple, specifying its magnitude and the direction of its axis.

SOLUTION

16-lb force:

M1 = −(480 lb ⋅ in.)k

40-lb force:

M 2 = 8 5[(10 lb ⋅ in.)i + (30 lb ⋅ in.) j + (15 lb ⋅ in.)k ]

P = 20 lb

M 3 = rC × P = (30 in.)i × (20 lb)k = (600 lb ⋅ in.) j M = M1 + M 2 + M 3 = −(480)k + 8 5 (10i + 30 j + 15k ) + 600 j = (178.885 lb ⋅ in.)i + (1136.66 lb ⋅ in.) j − (211.67 lb ⋅ in.)k M = (178.885) 2 + (113.66) 2 + (211.67)2 M = 1170 lb ⋅ in. 

= 1169.96 lb ⋅ in. M = 0.152898i + 0.97154 j − 0.180921k M cos θ x = 0.152898 λ axis =

cos θ y = 0.97154

θ x = 81.2° θ y = 13.70° θ z = 100.4° 

cos θ z = −0.180921



From the solution to Problem. 3.78:

www.elsolucionario.net PROBLEM 3.80 In a manufacturing operation, three holes are drilled simultaneously in a workpiece. If the holes are perpendicular to the surfaces of the workpiece, replace the couples applied to the drills with a single equivalent couple, specifying its magnitude and the direction of its axis.

SOLUTION M = M1 + M 2 + M 3

M = (0) 2 + (−4.4956) 2 + (0.22655) 2 = 4.5013 N ⋅ m

M = 4.50 N ⋅ m 

M = −(0.99873j + 0.050330k ) M cos θ x = 0

λaxis =

cos θ y = −0.99873

θ x = 90.0°, θ y = 177.1°, θ z = 87.1° 

cos θ z = 0.050330



= (1.5 N ⋅ m)(− cos 20° j + sin 20°k ) − (1.5 N ⋅ m) j + (1.75 N ⋅ m)(− cos 25° j + sin 25°k ) = −(4.4956 N ⋅ m) j + (0.22655 N ⋅ m)k

www.elsolucionario.net PROBLEM 3.81 A 260-lb force is applied at A to the rolled-steel section shown. Replace that force with an equivalent force-couple system at the center C of the section.

SOLUTION AB = (2.5 in.) 2 + (6.0 in.)2 = 6.50 in. 2.5 in. 5 = 6.5 in. 13 6.0 in. 12 cos α = = 6.5 in. 13

α = 22.6°

F = − F sin α i − F cos α j 5 12 = −(260 lb) i − (260 lb) j 13 13 = −(100.0 lb)i − (240 lb) j M C = rA /C × F

= (2.5i + 4.0 j) × (−100.0i − 240 j) = 400k − 600k = −(200 lb ⋅ in.)k F = 260 lb

67.4°; M C = 200 lb ⋅ in.





sin α =

www.elsolucionario.net PROBLEM 3.82 A 30-lb vertical force P is applied at A to the bracket shown, which is held by screws at B and C. (a) Replace P with an equivalent force-couple system at B. (b) Find the two horizontal forces at B and C that are equivalent to the couple obtained in part a.

SOLUTION (a)

M B = (30 lb)(5 in.) = 150.0 lb ⋅ in.

(b)

B=C =



150 lb ⋅ in. = 50.0 lb 3.0 in. B = 50.0 lb

; C = 50.0 lb





F = 30.0 lb , M B = 150.0 lb ⋅ in.

www.elsolucionario.net PROBLEM 3.83 The force P has a magnitude of 250 N and is applied at the end C of a 500-mm rod AC attached to a bracket at A and B. Assuming α = 30° and β = 60°, replace P with (a) an equivalent force-couple system at B, (b) an equivalent system formed by two parallel forces applied at A and B.

SOLUTION (a)

ΣF : F = P or F = 250 N

Equivalence requires

60°

ΣM B : M = −(0.3 m)(250 N) = −75 N ⋅ m

The equivalent force-couple system at B is FB = 250 N



We require


(b)

M B = 75.0 N ⋅ m

60°

Equivalence then requires ΣFx : 0 = FA cos φ + FB cos φ FA = − FB

or cos φ = 0

ΣFy : − 250 = − FA sin φ − FB sin φ

Now if

FA = − FB  −250 = 0, reject. cos φ = 0

or and Also,

φ = 90° FA + FB = 250 ΣM B : − (0.3 m)(250 N) = (0.2m) FA

or

FA = −375 N

and

FB = 625 N FA = 375 N

60°

FB = 625 N

60.0° 


www.elsolucionario.net PROBLEM 3.84 Solve Problem 3.83, assuming α = β = 25°. PROBLEM 3.83 The force P has a magnitude of 250 N and is applied at the end C of a 500-mm rod AC attached to a bracket at A and B. Assuming α = 30° and β = 60°, replace P with (a) an equivalent forcecouple system at B, (b) an equivalent system formed by two parallel forces applied at A and B.

(a)

Equivalence requires ΣF : FB = P or FB = 250 N

25.0°

ΣM B : M B = −(0.3 m)[(250 N)sin 50°] = −57.453 N ⋅ m

The equivalent force-couple system at B is FB = 250 N

(b)

25.0°

M B = 57.5 N ⋅ m



We require

Equivalence requires M B = d AE Q (0.3 m)[(250 N) sin 50°] = [(0.2 m) sin 50°]Q Q = 375 N

Adding the forces at B:

FA = 375 N

25.0°

FB = 625 N

25.0° 



SOLUTION

www.elsolucionario.net PROBLEM 3.85 The 80-N horizontal force P acts on a bell crank as shown. (a) Replace P with an equivalent force-couple system at B. (b) Find the two vertical forces at C and D that are equivalent to the couple found in part a.

SOLUTION (a)

ΣF : FB = F = 80 N

Based on

or

FB = 80.0 N



M B = 4.00 N ⋅ m



ΣM : M B = Fd B

or (b)

If the two vertical forces are to be equivalent to MB, they must be a couple. Further, the sense of the moment of this couple must be counterclockwise. Then with FC and FD acting as shown, ΣM : M D = FC d 4.0000 N ⋅ m = FC (0.04 m) FC = 100.000 N

or FC = 100.0 N 

ΣFy : 0 = FD − FC FD = 100.000 N

or FD = 100.0 N 



= 80 N (0.05 m) = 4.0000 N ⋅ m

www.elsolucionario.net PROBLEM 3.86 A dirigible is tethered by a cable attached to its cabin at B. If the tension in the cable is 1040 N, replace the force exerted by the cable at B with an equivalent system formed by two parallel forces applied at A and C.

Require the equivalent forces acting at A and C be parallel and at an angle of α with the vertical. Then for equivalence, ΣFx : (1040 N)sin 30° = FA sin α + FB sin α

(1)

ΣFy : −(1040 N) cos 30° = − FA cos α − FB cos α

(2)

Dividing Equation (1) by Equation (2), ( FA + FB ) sin α (1040 N) sin 30° = −(1040 N) cos 30° −( FA + FB ) cos α

Simplifying yields α = 30°. Based on ΣM C : [(1040 N) cos 30°](4 m) = ( FA cos 30°)(10.7 m) FA = 388.79 N FA = 389 N

or

60.0° 

Based on ΣM A : − [(1040 N) cos 30°](6.7 m) = ( FC cos 30°)(10.7 m) FC = 651.21 N FC = 651 N

or

60.0° 



SOLUTION

www.elsolucionario.net PROBLEM 3.87 Three control rods attached to a lever ABC exert on it the forces shown. (a) Replace the three forces with an equivalent force-couple system at B. (b) Determine the single force that is equivalent to the force-couple system obtained in part a, and specify its point of application on the lever.

SOLUTION

First note that the two 90-N forces form a couple. Then F = 216 N

where and

θ

θ = 180° − (60° + 55°) = 65° M = ΣM B = (0.450 m)(216 N) cos 55° − (1.050 m)(90 N) cos 20° = −33.049 N ⋅ m

The equivalent force-couple system at B is F = 216 N

(b)

65.0°; M = 33.0 N ⋅ m



The single equivalent force F′ is equal to F. Further, since the sense of M is clockwise, F′ must be applied between A and B. For equivalence, ΣM B : M = aF ′ cos 55°

where a is the distance from B to the point of application of F′. Then −33.049 N ⋅ m = − a(216 N) cos 55° a = 0.26676 m

or

F ′ = 216 N

65.0° applied to the lever 267 mm to the left of B 



(a)

www.elsolucionario.net PROBLEM 3.88 A hexagonal plate is acted upon by the force P and the couple shown. Determine the magnitude and the direction of the smallest force P for which this system can be replaced with a single force at E.

SOLUTION From the statement of the problem, it follows that ΣM E = 0 for the given force-couple system. Further, for Pmin, we must require that P be perpendicular to rB/E . Then

+ (0.2 m)sin 30° × 300 N − (0.4 m) Pmin = 0

or

Pmin = 300 N Pmin = 300 N

30.0° 



ΣM E : (0.2 sin 30° + 0.2)m × 300 N

www.elsolucionario.net PROBLEM 3.89 A force and couple act as shown on a square plate of side a = 25 in. Knowing that P = 60 lb, Q = 40 lb, and α = 50°, replace the given force and couple by a single force applied at a point located (a) on line AB, (b) on line AC. In each case determine the distance from A to the point of application of the force.

SOLUTION Replace the given force-couple system with an equivalent forcecouple system at A. Px = (60 lb)(cos 50°) = 38.567 lb

M A = Py a − Qa = (45.963 lb)(25 in.) − (40 lb)(25 in.) = 149.075 lb ⋅ in.

(a)

Equating moments about A gives: 149.075 lb ⋅ in. = (45.963 lb) x x = 3.24 in. P = 60.0 lb

(b)

50.0°; 3.24 in. from A 

149.075 lb ⋅ in. = (38.567 lb) y y = 3.87 in. P = 60.0 lb

50.0°; 3.87 in. below A 



Py = (60 lb)(sin 50°) = 45.963 lb

www.elsolucionario.net PROBLEM 3.90 The force and couple shown are to be replaced by an equivalent single force. Knowing that P = 2Q, determine the required value of α if the line of action of the single equivalent force is to pass through (a) Point A, (b) Point C.

SOLUTION

(a)

We must have M A = 0

sin α =

Q Q 1 = = P 2Q 2

α = 30.0°  (b)

MC = 0

We must have

( P sin α )a − ( P cos α ) a − Q (a) = 0 sin α − cos α =

Q Q 1 = = P 2Q 2

sin α = cos α +

1 2

(1)

1 4 1 1 − cos 2 α = cos 2 α + cos α + 4 sin 2 α = cos 2 α + cos α +

2 cos 2 α + cos α − 0.75 = 0

(2)

Solving the quadratic in cos α : cos α =

−1 ± 7 4

α = 65.7° or 155.7°

Only the first value of α satisfies Eq. (1), therefore α = 65.7° 



( P sin α )a − Q (a ) = 0

www.elsolucionario.net PROBLEM 3.91 The shearing forces exerted on the cross section of a steel channel can be represented by a 900-N vertical force and two 250-N horizontal forces as shown. Replace this force and couple with a single force F applied at Point C, and determine the distance x from C to line BD. (Point C is defined as the shear center of the section.)

SOLUTION

M H = (0.18)(250 N) = 45 N ⋅ m

Then or

M H = x(900 N) 45 N ⋅ m = x(900 N) x = 0.05 m F = 900 N

x = 50.0 mm 



Replace the 250-N forces with a couple and move the 900-N force to Point C such that its moment about H is equal to the moment of the couple

www.elsolucionario.net PROBLEM 3.92 A force and a couple are applied as shown to the end of a cantilever beam. (a) Replace this system with a single force F applied at Point C, and determine the distance d from C to a line drawn through Points D and E. (b) Solve part a if the directions of the two 360-N forces are reversed.

(a)

We have

ΣF : F = (360 N) j − (360 N) j − (600 N)k

or F = −(600 N)k  ΣM D : (360 N)(0.15 m) = (600 N)(d )

and

d = 0.09 m

or d = 90.0 mm below ED  (b)

We have from part a:

F = −(600 N)k 

ΣM D : − (360 N)(0.15 m) = −(600 N)(d )

and

d = 0.09 m

or d = 90.0 mm above ED 



SOLUTION


SOLUTION We have

d AB = (−64)2 + (−128) 2 + (16) 2 = 144 ft

Then

TAB =

Now

288 lb ( −64i − 128 j + 16k ) 144 = (32 lb)( −4i − 8 j + k )

M = M O = rA / O × TAB

= 128 j × 32(−4i − 8 j + k ) = (4096 lb ⋅ ft)i + (16,384 lb ⋅ ft)k

The equivalent force-couple system at O is F = −(128.0 lb)i − (256 lb) j + (32.0 lb)k  M = (4.10 kip ⋅ ft)i + (16.38 kip ⋅ ft)k 



An antenna is guyed by three cables as shown. Knowing that the tension in cable AB is 288 lb, replace the force exerted at A by cable AB with an equivalent force-couple system at the center O of the base of the antenna.


SOLUTION We have

d AD = ( −64) 2 + (−128)2 + (−128) 2 = 192 ft

Then

Now

270 lb (−64i − 128 j + 128k ) 192 = (90 lb)(−i − 2 j − 2k )

TAD =

M = M O = rA/O × TAD

= 128 j × 90(−i − 2 j − 2k ) = −(23, 040 lb ⋅ ft)i + (11,520 lb ⋅ ft)k

The equivalent force-couple system at O is F = −(90.0 lb)i − (180.0 lb) j − (180.0 lb)k  M = −(23.0 kip ⋅ ft)i + (11.52 kip ⋅ ft)k 



An antenna is guyed by three cables as shown. Knowing that the tension in cable AD is 270 lb, replace the force exerted at A by cable AD with an equivalent force-couple system at the center O of the base of the antenna.

www.elsolucionario.net PROBLEM 3.95 A 110-N force acting in a vertical plane parallel to the yz-plane is applied to the 220-mm-long horizontal handle AB of a socket wrench. Replace the force with an equivalent forcecouple system at the origin O of the coordinate system.

We have

ΣF : PB = F

where

PB = 110 N[− (sin15°) j + (cos15°)k ]

= −(28.470 N) j + (106.252 N)k or F = −(28.5 N) j + (106.3 N)k 

We have where

ΣM O : rB/O × PB = M O rB /O = [(0.22 cos 35°)i + (0.15) j − (0.22sin 35°)k ] m

= (0.180213 m)i + (0.15 m) j − (0.126187 m)k

i j k 0.180213 0.15 0.126187 N ⋅ m = M O 0 −28.5 106.3 M O = [(12.3487)i − (19.1566) j − (5.1361)k ] N ⋅ m or M O = (12.35 N ⋅ m)i − (19.16 N ⋅ m)j − (5.13 N ⋅ m)k 



SOLUTION

www.elsolucionario.net PROBLEM 3.96 An eccentric, compressive 1220-N force P is applied to the end of a cantilever beam. Replace P with an equivalent force-couple system at G.

SOLUTION We have ΣF : − (1220 N)i = F

Also, we have ΣM G : rA/G × P = M i j k 1220 0 −0.1 −0.06 N ⋅ m = M −1 0 0 M = (1220 N ⋅ m)[(−0.06)(−1) j − ( −0.1)( −1)k ]

or M = (73.2 N ⋅ m) j − (122 N ⋅ m)k 



F = − (1220 N)i 


SOLUTION We have ΣF : PAB = FC

where PAB = λAB PAB

=

(33 mm)i + (990 mm) j − (594 mm)k (175 N) 1155.00 mm

or FC = (5.00 N)i + (150.0 N) j − (90.0 N)k  We have

ΣM C : rB/C × PAB = M C i j k M C = 5 0.683 −0.860 0 N ⋅ m 1 30 −18

= (5){(− 0.860)(−18)i − (0.683)(−18) j + [(0.683)(30) − (0.860)(1)]k}

or M C = (77.4 N ⋅ m)i + (61.5 N ⋅ m) j + (106.8 N ⋅ m)k 



To keep a door closed, a wooden stick is wedged between the floor and the doorknob. The stick exerts at B a 175-N force directed along line AB. Replace that force with an equivalent force-couple system at C.

www.elsolucionario.net PROBLEM 3.98 A 46-lb force F and a 2120-lb⋅in. couple M are applied to corner A of the block shown. Replace the given force-couple system with an equivalent force-couple system at corner H.

We have Then

Also

d AJ = (18) 2 + (−14) 2 + (−3) 2 = 23 in. 46 lb (18i − 14 j − 3k ) 23 = (36 lb)i − (28 lb) j − (6 lb)k

F=

d AC = (−45) 2 + (0) 2 + (−28) 2 = 53 in. 2120 lb ⋅ in. ( −45i − 28k ) 53 = −(1800 lb ⋅ in.)i − (1120 lb ⋅ in.)k

Then

M=

Now

M ′ = M + rA/H × F

where

Then

rA/H = (45 in.)i + (14 in.) j i j k M ′ = (−1800i − 1120k ) + 45 14 0 36 −28 −6

= (−1800i − 1120k ) + {[(14)(−6)]i + [−(45)( −6)]j + [(45)(−28) − (14)(36)]k} = (−1800 − 84)i + (270) j + (−1120 − 1764)k = −(1884 lb ⋅ in.)i + (270 lb ⋅ in.)j − (2884 lb ⋅ in.)k = −(157 lb ⋅ ft)i + (22.5 lb ⋅ ft) j − (240 lb ⋅ ft)k F′ = (36.0 lb)i − (28.0 lb) j − (6.00 lb)k 

The equivalent force-couple system at H is

M ′ = −(157.0 lb ⋅ ft)i + (22.5 lb ⋅ ft) j − (240 lb ⋅ ft)k 



SOLUTION


SOLUTION (a)

ΣM Bz : M 2 z = 0

We have

k ⋅ (rH /B × F1 ) + M 1z = 0

where

(1)

rH /B = (0.31 m)i − (0.0233) j F1 = λ EH F1 (0.06 m)i + (0.06 m) j − (0.07 m)k (77 N) 0.11 m = (42 N)i + (42 N) j − (49 N)k =

M1z = k ⋅ M1 M1 = λEJ M 1 =

− di + (0.03 m) j − (0.07 m)k d 2 + 0.0058 m

(31 N ⋅ m)

Then from Equation (1), 0 0 1 ( −0.07 m)(31 N ⋅ m) =0 0.31 −0.0233 0 + 2 + 0.0058 d −49 42 42

Solving for d, Equation (1) reduces to (13.0200 + 0.9786) −

from which

2.17 N ⋅ m d 2 + 0.0058

d = 0.1350 m

=0

or d = 135.0 mm 



A 77-N force F1 and a 31-N ⋅ m couple M1 are applied to corner E of the bent plate shown. If F1 and M1 are to be replaced with an equivalent force-couple system (F2, M2) at corner B and if (M2)z = 0, determine (a) the distance d, (b) F2 and M2.


(b)

F2 = F1 = (42i + 42 j − 49k ) N

or F2 = (42.0 N)i + (42.0 N) j − (49.0 N)k 

M 2 = rH /B × F1 + M1 i j k (0.1350)i + 0.03j − 0.07k (31 N ⋅ m) = 0.31 −0.0233 0 + 0.155000 42 42 −49 = (1.14170i + 15.1900 j + 13.9986k ) N ⋅ m + (−27.000i + 6.0000 j − 14.0000k ) N ⋅ m M 2 = − (25.858 N ⋅ m)i + (21.190 N ⋅ m) j


or M 2 = − (25.9 N ⋅ m)i + (21.2 N ⋅ m) j 


www.elsolucionario.net PROBLEM 3.100 A 2.6-kip force is applied at Point D of the cast iron post shown. Replace that force with an equivalent force-couple system at the center A of the base section.

 DE = −(12 in.) j − (5 in.)k; DE = 13.00 in.  DE F = (2.6 kips) DE F = (2.6 kips)

−12 j − 5k 13 F = −(2.40 kips) j − (1.000 kip)k 

M A = rD /A × F

where

rD /A = (6 in.)i + (12 in.) j i j k 12 in. 0 M A = 6 in. 0 −2.4 kips −1.0 kips M A = −(12.00 kip ⋅ in.)i + (6.00 kip ⋅ in.) j − (14.40 kip ⋅ in.)k 



SOLUTION


SOLUTION

(a)

(a) We have

ΣFY : − 300 N − 200 N = Ra

or R a = 500 N  and

ΣM A : − 400 N ⋅ m − (200 N)(3 m) = M a

or M a = 1000 N ⋅ m (b) We have



ΣFY : 200 N + 300 N = Rb

or R b = 500 N  and

ΣM A : − 400 N ⋅ m + (300 N)(3 m) = M b

or M b = 500 N ⋅ m (c) We have



ΣFY : − 200 N − 300 N = Rc

or R c = 500 N  and

ΣM A : 400 N ⋅ m − (300 N)(3 m) = M c

or M c = 500 N ⋅ m





A 3-m-long beam is subjected to a variety of loadings. (a) Replace each loading with an equivalent forcecouple system at end A of the beam. (b) Which of the loadings are equivalent?


(d) We have

ΣFY : − 500 N = Rd

or R d = 500 N  and

ΣM A : 400 N ⋅ m − (500 N)(3 m) = M d

or M d = 1100 N ⋅ m (e) We have



ΣFY : 300 N − 800 N = Re

or R e = 500 N  ΣM A : 400 N ⋅ m + 1000 N ⋅ m − (800 N)(3 m) = M e

or M e = 1000 N ⋅ m (f ) We have



ΣFY : − 300 N − 200 N = R f

or R f = 500 N  and

ΣM A : 400 N ⋅ m − (200 N)(3 m) = M f

or M f = 200 N ⋅ m (g) We have



ΣFY : − 800 N + 300 N = Rg

or R g = 500 N  and

ΣM A : 1000 N ⋅ m + 400 N ⋅ m + (300 N)(3 m) = M g

or M g = 2300 N ⋅ m (h) We have



ΣFY : − 250 N − 250 N = Rh

or R h = 500 N  and

ΣM A : 1000 N ⋅ m + 400 N ⋅ m − (250 N)(3 m) = M h

or M h = 650 N ⋅ m (b)



Therefore, loadings (a) and (e) are equivalent.



and

www.elsolucionario.net PROBLEM 3.102 A 3-m-long beam is loaded as shown. Determine the loading of Prob. 3.101 that is equivalent to this loading.

SOLUTION

ΣFY : − 200 N − 300 N = R R = 500 N

or and

ΣM A : 500 N ⋅ m + 200 N ⋅ m − (300 N)(3 m) = M M = 200 N ⋅ m

or

Problem 3.101 equivalent force-couples at A:

Case

 R

 M

(a)

500 N

1000 N⋅m

(b)

500 N

500 N⋅m

(c)

500 N

500 N⋅m

(d)

500 N

1100 N⋅m

(e)

500 N

1000 N⋅m

(f )

500 N

200 N⋅m

(g)

500 N

2300 N⋅m

(h)

500 N

650 N⋅m Equivalent to case (f ) of Problem 3.101 



We have

www.elsolucionario.net PROBLEM 3.103 Determine the single equivalent force and the distance from Point A to its line of action for the beam and loading of (a) Prob. 3.101a, (b) Prob. 3.101b, (c) Prob. 3.102.

SOLUTION For equivalent single force at distance d from A: (a)

We have

ΣFY : − 300 N − 200 N = R

or R = 500 N  ΣM C : − 400 N ⋅ m + (300 N)(d )

and

− (200 N)(3 − d ) = 0

(b)

We have

ΣFY : 200 N + 300 N = R

or R = 500 N  ΣM C : − 400 N ⋅ m − (200 N)(d )

and

+ (300 N)(3 − d ) = 0

or d = 1.000 m 

(c)

We have

ΣFY : − 200 N − 300 N = R

or R = 500 N  ΣM C : 500 N ⋅ m + 200 N ⋅ m

and

+ (200 N)( d ) − (300 N)(3 − d ) = 0

or d = 0.400 m 



or d = 2.00 m 


SOLUTION First note that the force-couple system at F cannot be equivalent because of the direction of the force [The force of the other four systems is (10 lb)i]. Next, move each of the systems to the origin O; the forces remain unchanged. A: M A = ΣM O = (5 lb ⋅ ft) j + (15 lb ⋅ ft)k + (2 ft)k × (10 lb)i

= (25 lb ⋅ ft) j + (15 lb ⋅ ft)k D : M D = ΣM O = −(5 lb ⋅ ft) j + (25 lb ⋅ ft)k

+ [(4.5 ft)i + (1 ft) j + (2 ft)k ] × 10 lb)i = (15 lb ⋅ ft)i + (15 lb ⋅ ft)k G : M G = ΣM O = (15 lb ⋅ ft)i + (15 lb ⋅ ft) j I : M I = ΣM I = (15 lb ⋅ ft) j − (5 lb ⋅ ft)k

+ [(4.5 ft)i + (1 ft) j] × (10 lb) j = (15 lb ⋅ ft) j − (15 lb ⋅ ft)k

The equivalent force-couple system is the system at corner D.





Five separate force-couple systems act at the corners of a piece of sheet metal, which has been bent into the shape shown. Determine which of these systems is equivalent to a force F = (10 lb)i and a couple of moment M = (15 lb ⋅ ft)j + (15 lb ⋅ ft)k located at the origin.

www.elsolucionario.net PROBLEM 3.105 Three horizontal forces are applied as shown to a vertical cast iron arm. Determine the resultant of the forces and the distance from the ground to its line of action when (a) P = 200 N, (b) P = 2400 N, (c) P = 1000 N.

SOLUTION

RD = +200 N − 600 N − 400 N = −800 N M D = −(200 N)(0.450 m) + (600 N)(0.300 m) + (400 N)(0.1500 m)

= +150.0 N ⋅ m y=

M D 150 N ⋅ m = = 0.1875 m R 800 N R = 800 N

; y = 187.5 mm 

(b)

RD = +2400 N − 600 N − 400 N = +1400 N M D = −(2400 N)(0.450 m) + (600 N)(0.300 m) + (400 N)(0.1500 m)

= −840 N ⋅ m y=

M D 840 N ⋅ m = = 0.600 m R 1400 N R = 1400 N

; y = 600 mm 



(a)


(c) RD = +1000 − 600 − 400 = 0 M D = −(1000 N)(0.450 m) + (600 N)(0.300 m) + (400 N)(0.1500 m)

= −210 N ⋅ m

∴ y = ∞ System reduces to a couple.




M D = 210 N ⋅ m


www.elsolucionario.net PROBLEM 3.106 Three stage lights are mounted on a pipe as shown. The lights at A and B each weigh 4.1 lb, while the one at C weighs 3.5 lb. (a) If d = 25 in., determine the distance from D to the line of action of the resultant of the weights of the three lights. (b) Determine the value of d so that the resultant of the weights passes through the midpoint of the pipe.


SOLUTION

For equivalence, ΣFy : − 4.1 − 4.1 − 3.5 = − R or R = 11.7 lb ΣFD : − (10 in.)(4.1 lb) − (44 in.)(4.1 lb) −[(4.4 + d ) in.](3.5 lb) = −( L in.)(11.7 lb) 375.4 + 3.5d = 11.7 L (d , L in in.)

or

d = 25 in.

(a) We have

375.4 + 3.5(25) = 11.7 L or

L = 39.6 in.

The resultant passes through a point 39.6 in. to the right of D.



L = 42 in.

(b) We have

375.4 + 3.5d = 11.7(42)

or d = 33.1 in. 


www.elsolucionario.net PROBLEM 3.107 The weights of two children sitting at ends A and B of a seesaw are 84 lb and 64 lb, respectively. Where should a third child sit so that the resultant of the weights of the three children will pass through C if she weighs (a) 60 lb, (b) 52 lb.

(a)

For the resultant weight to act at C, Then

ΣM C = 0 WC = 60 lb

(84 lb)(6 ft) − 60 lb(d ) − 64 lb(6 ft) = 0 d = 2.00 ft to the right of C 

(b)

For the resultant weight to act at C, Then

ΣM C = 0 WC = 52 lb

(84 lb)(6 ft) − 52 lb(d ) − 64 lb(6 ft) = 0 d = 2.31 ft to the right of C 



SOLUTION

www.elsolucionario.net PROBLEM 3.108 A couple of magnitude M = 54 lb ⋅ in. and the three forces shown are applied to an angle bracket. (a) Find the resultant of this system of forces. (b) Locate the points where the line of action of the resultant intersects line AB and line BC.

SOLUTION (a)

We have

ΣF : R = (−10 j) + (30 cos 60°)i + 30 sin 60° j + (−45i ) = −(30 lb)i + (15.9808 lb) j

or R = 34.0 lb First reduce the given forces and couple to an equivalent force-couple system (R , M B ) at B. We have

ΣM B : M B = (54 lb ⋅ in) + (12 in.)(10 lb) − (8 in.)(45 lb) = −186 lb ⋅ in.

Then with R at D, or and with R at E, or

ΣM B : −186 lb ⋅ in = a(15.9808 lb)

a = 11.64 in. ΣM B : −186 lb ⋅ in = C (30 lb)

C = 6.2 in.

The line of action of R intersects line AB 11.64 in. to the left of B and intersects line BC 6.20 in. below B.





(b)

28.0° 

www.elsolucionario.net PROBLEM 3.109 A couple M and the three forces shown are applied to an angle bracket. Find the moment of the couple if the line of action of the resultant of the force system is to pass through (a) Point A, (b) Point B, (c) Point C.

SOLUTION In each case, we must have M1R = 0 M AB = ΣM A = M + (12 in.)[(30 lb) sin 60°] − (8 in.)(45 lb) = 0 M = +48.231 lb ⋅ in.

(b)



M = 240 lb ⋅ in.



M BR = ΣM B = M + (12 in.)(10 lb) − (8 in.)(45 lb) = 0 M = +240 lb ⋅ in.

(c)

M = 48.2 lb ⋅ in.

M CR = ΣM C = M + (12 in.)(10 lb) − (8 in.)[(30 lb) cos 60°] = 0 M =0

M=0 



(a)

www.elsolucionario.net PROBLEM 3.110 A 32-lb motor is mounted on the floor. Find the resultant of the weight and the forces exerted on the belt, and determine where the line of action of the resultant intersects the floor.

SOLUTION We have

R = (181.244 lb)i + (38.0 lb) j

or R = 185.2 lb We have

11.84° 

ΣM O : ΣM O = xRy

− [(140 lb) cos 30°][(4 + 2 cos 30°)in.] − [(140 lb) sin 30°][(2 in.)sin 30°] − (60 lb)(2 in.) = x(38.0 lb)

x=

and

1 (− 694.97 − 70.0 − 120) in. 38.0

x = −23.289 in.

Or resultant intersects the base (x-axis) 23.3 in. to the left of the vertical centerline (y-axis) of the motor.





ΣF : (60 lb)i − (32 lb) j + (140 lb)(cos 30°i + sin 30° j) = R

www.elsolucionario.net PROBLEM 3.111 A machine component is subjected to the forces and couples shown. The component is to be held in place by a single rivet that can resist a force but not a couple. For P = 0, determine the location of the rivet hole if it is to be located (a) on line FG, (b) on line GH.

SOLUTION


We have

First replace the applied forces and couples with an equivalent force-couple system at G. ΣFx : 200cos 15° − 120 cos 70° + P = Rx

Thus,

Rx = (152.142 + P) N

or

ΣFy : − 200sin 15° − 120sin 70° − 80 = Ry

Ry = −244.53 N

or

ΣM G : − (0.47 m)(200 N) cos15° + (0.05 m)(200 N)sin15° + (0.47 m)(120 N) cos 70° − (0.19 m)(120 N)sin 70° − (0.13 m)( P N) − (0.59 m)(80 N) + 42 N ⋅ m + 40 N ⋅ m = M G

M G = −(55.544 + 0.13P) N ⋅ m

or

(1)

Setting P = 0 in Eq. (1): Now with R at I,

ΣM G : − 55.544 N ⋅ m = − a(244.53 N)

a = 0.227 m

or and with R at J,

ΣM G : − 55.544 N ⋅ m = −b(152.142 N)

b = 0.365 m

or (a)

The rivet hole is 0.365 m above G.



(b)

The rivet hole is 0.227 m to the right of G.




www.elsolucionario.net PROBLEM 3.112 Solve Problem 3.111, assuming that P = 60 N. PROBLEM 3.111 A machine component is subjected to the forces and couples shown. The component is to be held in place by a single rivet that can resist a force but not a couple. For P = 0, determine the location of the rivet hole if it is to be located (a) on line FG, (b) on line GH.

SOLUTION See the solution to Problem 3.111 leading to the development of Equation (1):


M G = −(55.544 + 0.13P) N ⋅ m Rx = (152.142 + P) N

and

P = 60 N

For we have

Rx = (152.142 + 60) = 212.14 N M G = −[55.544 + 0.13(60)] = −63.344 N ⋅ m

Then with R at I,

ΣM G : −63.344 N ⋅ m = −a(244.53 N) a = 0.259 m

or and with R at J,

ΣM G : −63.344 N ⋅ m = −b(212.14 N) b = 0.299 m

or (a)

The rivet hole is 0.299 m above G.



(b)

The rivet hole is 0.259 m to the right of G.




www.elsolucionario.net PROBLEM 3.113 A truss supports the loading shown. Determine the equivalent force acting on the truss and the point of intersection of its line of action with a line drawn through Points A and G.

SOLUTION We have

R = ΣF R = (240 lb)(cos 70°i − sin 70° j) − (160 lb) j

R = −(147.728 lb)i − (758.36 lb) j R = Rx2 + Ry2 = (147.728) 2 + (758.36) 2 = 772.62 lb  Ry    Rx   −758.36  = tan −1    −147.728  = 78.977°

θ = tan −1 

or

We have

ΣM A = dRy

where

ΣM A = −[240 lb cos 70°](6 ft) − [240 lbsin 70°](4 ft)

R = 773 lb

79.0° 

− (160 lb)(12 ft) + [300 lb cos 40°](6 ft) − [300 lb sin 40°](20 ft) − (180 lb)(8 ft) = −7232.5 lb ⋅ ft −7232.5 lb ⋅ ft −758.36 lb = 9.5370 ft

d=

or

d = 9.54 ft to the right of A 



+ (300 lb)(− cos 40°i − sin 40° j) − (180 lb) j

www.elsolucionario.net PROBLEM 3.114 Four ropes are attached to a crate and exert the forces shown. If the forces are to be replaced with a single equivalent force applied at a point on line AB, determine (a) the equivalent force and the distance from A to the point of application of the force when α = 30°, (b) the value of α so that the single equivalent force is applied at Point B.

SOLUTION

(a)

For equivalence,

ΣFx : −100 cos 30° + 400 cos 65° + 90 cos 65° = Rx

or

Rx = 120.480 lb

ΣFy : 100 sin α + 160 + 400 sin 65° + 90 sin 65° = Ry

or

Ry = (604.09 + 100sin α ) lb

With α = 30°,

Ry = 654.09 lb

Then

R = (120.480) 2 + (654.09) 2

= 665 lb

(1)

654.09 120.480 or θ = 79.6°

tan θ =

ΣM A : (46 in.)(160 lb) + (66 in.)(400 lb) sin 65°

Also

or

+ (26 in.)(400 lb) cos 65° + (66 in.)(90 lb)sin 65° + (36 in.)(90 lb) cos 65° = d (654.09 lb) ΣM A = 42, 435 lb ⋅ in. and d = 64.9 in.

79.6°  

and R is applied 64.9 in. to the right of A. (b)

R = 665 lb

We have d = 66 in. Then or Using Eq. (1):

ΣM A : 42, 435 lb ⋅ in = (66 in.) Ry Ry = 642.95 lb

642.95 = 604.09 + 100sin α

or α = 22.9° 



We have

www.elsolucionario.net PROBLEM 3.115 Solve Prob. 3.114, assuming that the 90-lb force is removed. PROBLEM 3.114 Four ropes are attached to a crate and exert the forces shown. If the forces are to be replaced with a single equivalent force applied at a point on line AB, determine (a) the equivalent force and the distance from A to the point of application of the force when α = 30°, (b) the value of α so that the single equivalent force is applied at Point B.

(a)

For equivalence,

ΣFx : − (100 lb) cos 30° + (400 lb)sin 25° = Rx

or

Rx = 82.445 lb


SOLUTION

ΣFy : 160 lb + (100 lb) sin 30° + (400 lb) cos 25° = Ry

or

Ry = 572.52 lb R = (82.445)2 + (572.52) = 578.43 lb

tan θ =

572.52 82.445

or θ = 81.806°

ΣM A : (46 in.)(160 lb) + (66 in.)(400 lb) cos 25° + (26 in.)(400 lb)sin 25° = d (527.52 lb) d = 62.3 in. R = 578 lb

(b)

81.8° and is applied 62.3 in. to the right of A.



We have d = 66.0 in. For R applied at B, ΣM A : Ry (66 in.) = (160 lb)(46 in.) + (66 in.)(400 lb) cos 25° + (26 in.)(400 lb)sin 25° Ry = 540.64 lb

ΣFY : 160 lb + (100 lb)sin α + (400 lb) cos 25° = 540.64 lb

α = 10.44° 


www.elsolucionario.net PROBLEM 3.116 Four forces act on a 700 × 375-mm plate as shown. (a) Find the resultant of these forces. (b) Locate the two points where the line of action of the resultant intersects the edge of the plate.

SOLUTION R = ΣF = (−400 N + 160 N − 760 N)i + (600 N + 300 N + 300 N) j = −(1000 N)i + (1200 N) j R = (1000 N) 2 + (1200 N)2

= 1562.09 N  1200 N  tan θ =  −   1000 N  = −1.20000 θ = −50.194°

(b)

R = 1562 N

50.2° 

M CR = Σr × F = (0.5 m)i × (300 N + 300 N) j = (300 N ⋅ m)k (300 N ⋅ m)k = xi × (1200 N) j x = 0.25000 m x = 250 mm (300 N ⋅ m) = yj × ( −1000 N)i

y = 0.30000 m y = 300 mm

Intersection 250 mm to right of C and 300 mm above C 



(a)

www.elsolucionario.net PROBLEM 3.117 Solve Problem 3.116, assuming that the 760-N force is directed to the right. PROBLEM 3.116 Four forces act on a 700 × 375-mm plate as shown. (a) Find the resultant of these forces. (b) Locate the two points where the line of action of the resultant intersects the edge of the plate.

SOLUTION = ( −400N + 160 N + 760 N)i + (600 N + 300 N + 300 N) j = (520 N)i + (1200 N) j

R = (520 N) 2 + (1200 N) 2 = 1307.82 N

 1200 N  tan θ =   = 2.3077  520 N  θ = 66.5714°

R = 1308 N

66.6° 

M CR = Σr × F

(b)

= (0.5 m)i × (300 N + 300 N) j = (300 N ⋅ m)k (300 N ⋅ m)k = xi × (1200 N) j x = 0.25000 m

or

x = 0.250 mm (300 N ⋅ m)k = [ x′i + (0.375 m) j] × [(520 N)i + (1200 N) j] = (1200 x′ − 195)k

x′ = 0.41250 m

or

x′ = 412.5 mm

Intersection 412 mm to the right of A and 250 mm to the right of C 



R = ΣF

(a)

www.elsolucionario.net PROBLEM 3.118 As follower AB rolls along the surface of member C, it exerts a constant force F perpendicular to the surface. (a) Replace F with an equivalent force-couple system at Point D obtained by drawing the perpendicular from the point of contact to the x-axis. (b) For a = 1 m and b = 2 m, determine the value of x for which the moment of the equivalent force-couple system at D is maximum.

SOLUTION The slope of any tangent to the surface of member C is dy d   x2 = b 1 − 2 dx dx   a

  −2b   = 2 x   a


(a)

Since the force F is perpendicular to the surface,  dy  tan α = −    dx 

−1

=

a2  1  2b  x 

For equivalence, ΣF : F = R ΣM D : ( F cos α )( y A ) = M D

where cos α =

2bx (a ) + (2bx)2 2 2

 x2  y A = b 1 − 2   a   x3  2 Fb 2  x − 2  a   MD = 4 2 2 a + 4b x

Therefore, the equivalent force-couple system at D is R=F

 a2  tan −1     2bx 

 x3  2 Fb2  x − 2  a   M=  a 4 + 4b 2 x 2 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 282


(b)

To maximize M, the value of x must satisfy

dM =0 dx

a = 1 m, b = 2 m

where for M=

8F ( x − x3 ) 1 + 16 x 2

dM = 8F dx

1  1 + 16 x 2 (1 − 3x 2 ) − ( x − x3 )  (32 x)(1 + 16 x 2 ) −1/ 2  2  =0 (1 + 16 x 2 ) (1 + 16 x 2 )(1 − 3x 2 ) − 16 x( x − x3 ) = 0

x2 =

−3 ± 9 − 4(32)(−1) = 0.136011 m 2 2(32)

Using the positive value of x2:

x = 0.36880 m

and − 0.22976 m 2

or x = 369 mm 



32 x 4 + 3x 2 − 1 = 0

or

www.elsolucionario.net PROBLEM 3.119 As plastic bushings are inserted into a 60-mm-diameter cylindrical sheet metal enclosure, the insertion tools exert the forces shown on the enclosure. Each of the forces is parallel to one of the coordinate axes. Replace these forces with an equivalent force-couple system at C.

For equivalence,

ΣF :

R = FA + FB + FC + FD = −(17 N) j − (12 N) j − (16 N)k − (21 N)i = −(21 N)i − (29 N) j − (16 N)k

ΣM C : M = rA /C × FA + rB /C × FB + rD /C × FD

M = [(0.11 m) j − (0.03 m)k ] × [−(17 N)] j + [(0.02 m)i + (0.11 m) j − (0.03 m)k ] × [ −(12 N)]j + [(0.03 m)i + (0.03 m) j − (0.03 m)k ] × [ −(21 N)]i = −(0.51 N ⋅ m)i + [−(0.24 N ⋅ m)k − (0.36 N ⋅ m)i] + [(0.63 N ⋅ m)k + (0.63 N ⋅ m) j]

∴

The equivalent force-couple system at C is R = −(21.0 N)i − (29.0 N) j − (16.00 N)k  M = −(0.870 N ⋅ m)i + (0.630 N ⋅ m) j + (0.390 N ⋅ m)k 



SOLUTION


SOLUTION Equivalent force-couple at each pulley: Pulley B:

R B = (145 N)(− cos 20° j + sin 20°k ) − 215 Nj = − (351.26 N) j + (49.593 N)k M B = − (215 N − 145 N)(0.075 m)i = − (5.25 N ⋅ m)i

Pulley C:

R C = (155 N + 240 N)(− sin10° j − cos10°k ) = − (68.591 N) j − (389.00 N)k M C = (240 N − 155 N)(0.075 m)i = (6.3750 N ⋅ m)i

Then

R = R B + R C = − (419.85 N) j − (339.41)k

or R = (420 N) j − (339 N)k 

M A = M B + M C + rB/ A × R B + rC/ A × R C i j k 0 0 N⋅m = − (5.25 N ⋅ m)i + (6.3750 N ⋅ m)i + 0.225 0 −351.26 49.593 i j k + 0.45 0 0 N⋅m 0 −68.591 −389.00 = (1.12500 N ⋅ m)i + (163.892 N ⋅ m) j − (109.899 N ⋅ m)k

or M A = (1.125 N ⋅ m)i + (163.9 N ⋅ m) j − (109.9 N ⋅ m)k 



Two 150-mm-diameter pulleys are mounted on line shaft AD. The belts at B and C lie in vertical planes parallel to the yz-plane. Replace the belt forces shown with an equivalent forcecouple system at A.

www.elsolucionario.net PROBLEM 3.121 Four forces are applied to the machine component ABDE as shown. Replace these forces with an equivalent force-couple system at A.

R = −(50 N) j − (300 N)i − (120 N)i − (250 N)k R = −(420 N)i − (50 N)j − (250 N)k rB = (0.2 m)i rD = (0.2 m)i + (0.16 m)k rE = (0.2 m)i − (0.1 m) j + (0.16 m)k M RA = rB × [−(300 N)i − (50 N) j] + rD × (−250 N)k + r × ( − 120 N)i i j k i j k = 0.2 m 0 0 + 0.2 m 0 0.16 m −300 N −50 N 0 0 0 −250 N i j k + 0.2 m −0.1 m 0.16 m −120 N 0 0 = −(10 N ⋅ m)k + (50 N ⋅ m) j − (19.2 N ⋅ m) j − (12 N ⋅ m)k

Force-couple system at A is R = −(420 N)i − (50 N) j − (250 N)k M RA = (30.8 N ⋅ m) j − (220 N ⋅ m)k 



SOLUTION

www.elsolucionario.net PROBLEM 3.122 While using a pencil sharpener, a student applies the forces and couple shown. (a) Determine the forces exerted at B and C knowing that these forces and the couple are equivalent to a forcecouple system at A consisting of the force R = (2.6 lb)i + Ry j − (0.7 lb)k and the couple M RA = M x i + (1.0 lb · ft)j − (0.72 lb · ft)k. (b) Find the corresponding values of Ry and M x .

(a)

From the statement of the problem, equivalence requires ΣF : B + C = R or

ΣFx : Bx + C x = 2.6 lb

(1)

ΣFy : − C y = R y

(2)

ΣFz : − C z = −0.7 lb or C z = 0.7 lb

and

ΣM A : (rB/A × B + M B ) + rC/A × C = M AR

or

 1.75  ΣM x : (1 lb ⋅ ft) +  ft  (C y ) = M x  12 

(3)

 3.75   1.75   3.5  ΣM y :  ft  ( Bx ) +  ft  (C x ) +  ft  (0.7 lb) = 1 lb ⋅ ft  12   12   12 

or Using Eq. (1):

3.75Bx + 1.75C x = 9.55 3.75Bx + 1.75(2.6 Bx ) = 9.55

or

Bx = 2.5 lb

and

C x = 0.1 lb  3.5  ΣM z : −  ft  (C y ) = −0.72 lb ⋅ ft  12  C y = 2.4686 lb

or

B = (2.50 lb)i C = (0.1000 lb)i − (2.47 lb) j − (0.700 lb)k 

(b)

Eq. (2)  Using Eq. (3):

Ry = −2.47 lb   1.75  1+   (2.4686) = M x  12 

or M x = 1.360 lb ⋅ ft 



SOLUTION

www.elsolucionario.net PROBLEM 3.123 A blade held in a brace is used to tighten a screw at A. (a) Determine the forces exerted at B and C, knowing that these forces are equivalent to a force-couple system at A consisting of R = −(30 N)i + Ry j + Rz k and M RA = − (12 N · m)i. (b) Find the corresponding values of Ry and Rz . (c) What is the orientation of the slot in the head of the screw for which the blade is least likely to slip when the brace is in the position shown?

SOLUTION Equivalence requires or Equating the i coefficients: Also, or

Equating coefficients:

ΣF : R = B + C −(30 N)i + Ry j + Rz k = − Bk + (−C x i + C y j + C z k ) i : − 30 N = −C x

or C x = 30 N

ΣM A : M RA = rB/A × B + rC/A × C −(12 N ⋅ m)i = [(0.2 m)i + (0.15 m)j] × (− B)k +(0.4 m)i × [−(30 N)i + C y j + C z k ]

i : − 12 N ⋅ m = −(0.15 m) B k : 0 = (0.4 m)C y

or or

B = 80 N Cy = 0

j: 0 = (0.2 m)(80 N) − (0.4 m)C z or

C z = 40 N

B = −(80.0 N)k C = −(30.0 N)i + (40.0 N)k 

(b)

Now we have for the equivalence of forces −(30 N)i + Ry j + Rz k = −(80 N)k + [(−30 N)i + (40 N)k ]

Equating coefficients:

j: R y = 0

Ry = 0 

k : Rz = −80 + 40

(c)

or

Rz = −40.0 N 

First note that R = −(30 N)i − (40 N)k. Thus, the screw is best able to resist the lateral force Rz when the slot in the head of the screw is vertical.





(a)

www.elsolucionario.net PROBLEM 3.124 In order to unscrew the tapped faucet A, a plumber uses two pipe wrenches as shown. By exerting a 40-lb force on each wrench, at a distance of 10 in. from the axis of the pipe and in a direction perpendicular to the pipe and to the wrench, he prevents the pipe from rotating, and thus avoids loosening or further tightening the joint between the pipe and the tapped elbow C. Determine (a) the angle θ that the wrench at A should form with the vertical if elbow C is not to rotate about the vertical, (b) the force-couple system at C equivalent to the two 40-lb forces when this condition is satisfied.

We first reduce the given forces to force-couple systems at A and B, noting that | M A | = | M B | = (40 lb)(10 in.) = 400 lb ⋅ in.

We now determine the equivalent force-couple system at C. R = (40 lb)(1 − cos θ )i − (40 lb) sin θ j

(1)

M CR = M A + M B + (15 in.)k × [−(40 lb) cos θ i − (40 lb)sin θ j] + (7.5 in.)k × (40 lb)i = + 400 − 400 − 600cos θ j + 600sin θ i + 300 j = (600 lb ⋅ in.)sin θ i + (300 lb ⋅ in.)(1 − 2 cos θ ) j

(a)

(2)

For no rotation about vertical, y component of M CR must be zero. 1 − 2cos θ = 0 cos θ = 1/2

θ = 60.0°  (b)

For θ = 60.0° in Eqs. (1) and (2), R = (20.0 lb)i − (34.641 lb) j; M CR = (519.62 lb ⋅ in.)i R = (20.0 lb)i − (34.6 lb) j; M CR = (520 lb ⋅ in.)i 



SOLUTION


PROBLEM 3.124 In order to unscrew the tapped faucet A, a plumber uses two pipe wrenches as shown. By exerting a 40-lb force on each wrench, at a distance of 10 in. from the axis of the pipe and in a direction perpendicular to the pipe and to the wrench, he prevents the pipe from rotating, and thus avoids loosening or further tightening the joint between the pipe and the tapped elbow C. Determine (a) the angle θ that the wrench at A should form with the vertical if elbow C is not to rotate about the vertical, (b) the force-couple system at C equivalent to the two 40-lb forces when this condition is satisfied.

SOLUTION The equivalent force-couple system at C for θ = 60° was obtained in the solution to Prob. 3.124: R = (20.0 lb)i − (34.641 lb) j M CR = (519.62 lb ⋅ in.)i

The equivalent force-couple system at D is made of R and M RD where M RD = M CR + rC /D × R = (519.62 lb ⋅ in.)i + (25.0 in.) j × [(20.0 lb)i − (34.641 lb) j] = (519.62 lb ⋅ in.)i − (500 lb ⋅ in.)k

Equivalent force-couple at D: R = (20.0 lb)i − (34.6 lb) j; M CR = (520 lb ⋅ in.)i − (500 lb ⋅ in.)k 

(a) (b)

Since M RD has no component along the y-axis, the plumber’s action will neither loosen nor tighten the joint between pipe CD and elbow.



, the plumber’s action will tend to tighten Since the x component of M RD is the joint between elbow and pipe DE.





Assuming θ = 60° in Prob. 3.124, replace the two 40-lb forces with an equivalent force-couple system at D and determine whether the plumber’s action tends to tighten or loosen the joint between (a) pipe CD and elbow D, (b) elbow D and pipe DE. Assume all threads to be right-handed.

www.elsolucionario.net PROBLEM 3.126 As an adjustable brace BC is used to bring a wall into plumb, the force-couple system shown is exerted on the wall. Replace this force-couple system with an equivalent force-couple system at A if R = 21.2 lb and M = 13.25 lb · ft.

We have where

or We have where

ΣF : R = R A = RλBC

λBC =

(42 in.)i − (96 in.) j − (16 in.)k 106 in.

RA =

21.2 lb (42i − 96 j − 16k ) 106

R A = (8.40 lb)i − (19.20 lb) j − (3.20 lb)k



ΣM A : rC/A × R + M = M A

rC/A = (42 in.)i + (48 in.)k =

1 (42i + 48k )ft 12

= (3.5 ft)i + (4.0 ft)k R = (8.40 lb)i − (19.50 lb) j − (3.20 lb)k M = −λBC M −42i + 96 j + 16k (13.25 lb ⋅ ft) 106 = −(5.25 lb ⋅ ft)i + (12 lb ⋅ ft) j + (2 lb ⋅ ft)k =

Then

i j k 3.5 0 4.0 lb ⋅ ft + (−5.25i + 12 j + 2k ) lb ⋅ ft = M A 8.40 −19.20 −3.20 M A = (71.55 lb ⋅ ft)i + (56.80 lb ⋅ ft)j − (65.20 lb ⋅ ft)k

or M A = (71.6 lb ⋅ ft)i + (56.8 lb ⋅ ft)j − (65.2 lb ⋅ ft)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 291


SOLUTION

www.elsolucionario.net PROBLEM 3.127 Three children are standing on a 5 × 5-m raft. If the weights of the children at Points A, B, and C are 375 N, 260 N, and 400 N, respectively, determine the magnitude and the point of application of the resultant of the three weights.

We have

ΣF : FA + FB + FC = R −(375 N) j − (260 N) j − (400 N) j = R −(1035 N) j = R or

We have

R = 1035 N 

ΣM x : FA ( z A ) + FB ( z B ) + FC ( zC ) = R ( z D ) (375 N)(3 m) + (260 N)(0.5 m) + (400 N)(4.75 m) = (1035 N)(z D ) z D = 3.0483 m

We have

or

z D = 3.05 m 

ΣM z : FA ( x A ) + FB ( xB ) + FC ( xC ) = R ( xD ) 375 N(1 m) + (260 N)(1.5 m) + (400 N)(4.75 m) = (1035 N)( xD ) xD = 2.5749 m

or

xD = 2.57 m 



SOLUTION

www.elsolucionario.net PROBLEM 3.128 Three children are standing on a 5 × 5-m raft. The weights of the children at Points A, B, and C are 375 N, 260 N, and 400 N, respectively. If a fourth child of weight 425 N climbs onto the raft, determine where she should stand if the other children remain in the positions shown and the line of action of the resultant of the four weights is to pass through the center of the raft.

We have


SOLUTION

ΣF : FA + FB + FC = R −(375 N) j − (260 N) j − (400 N) j − (425 N) j = R R = −(1460 N) j

We have

ΣM x : FA ( z A ) + FB ( z B ) + FC ( zC ) + FD ( z D ) = R ( z H ) (375 N)(3 m) + (260 N)(0.5 m) + (400 N)(4.75 m) + (425 N)(z D ) = (1460 N)(2.5 m) z D = 1.16471 m

We have

or

z D = 1.165 m 

ΣM z : FA ( x A ) + FB ( xB ) + FC ( xC ) + FD ( xD ) = R ( xH ) (375 N)(1 m) + (260 N)(1.5 m) + (400 N)(4.75 m) + (425 N)(xD ) = (1460 N)(2.5 m) xD = 2.3235 m

or

xD = 2.32 m 




Four signs are mounted on a frame spanning a highway, and the magnitudes of the horizontal wind forces acting on the signs are as shown. Determine the magnitude and the point of application of the resultant of the four wind forces when a = 1 ft and b = 12 ft.

SOLUTION We have

Assume that the resultant R is applied at Point P whose coordinates are (x, y, 0). Equivalence then requires ΣFz : − 105 − 90 − 160 − 50 = − R

or R = 405 lb  ΣM x : (5 ft)(105 lb) − (1 ft)(90 lb) + (3 ft)(160 lb) + (5.5 ft)(50 lb) = − y (405 lb)

or

y = −2.94 ft

ΣM y : (5.5 ft)(105 lb) + (12 ft)(90 lb) + (14.5 ft)(160 lb) + (22.5 ft)(50 lb) = − x(405 lb) x = 12.60 ft

or

R acts 12.60 ft to the right of member AB and 2.94 ft below member BC.





SOLUTION Since R acts at G, equivalence then requires that ΣM G of the applied system of forces also be zero. Then at G : ΣM x : − (a + 3) ft × (90 lb) + (2 ft)(105 lb)

+ (2.5 ft)(50 lb) = 0

or a = 0.722 ft  

ΣM y : − (9 ft)(105 ft) − (14.5 − b) ft × (90 lb) + (8 ft)(50 lb) = 0

 or b = 20.6 ft 



Four signs are mounted on a frame spanning a highway, and the magnitudes of the horizontal wind forces acting on the signs are as shown. Determine a and b so that the point of application of the resultant of the four forces is at G.

www.elsolucionario.net PROBLEM 3.131* A group of students loads a 2 × 3.3-m flatbed trailer with two 0.66 × 0.66 × 0.66-m boxes and one 0.66 × 0.66 × 1.2-m box. Each of the boxes at the rear of the trailer is positioned so that it is aligned with both the back and a side of the trailer. Determine the smallest load the students should place in a second 0.66 × 0.66 × 1.2-m box and where on the trailer they should secure it, without any part of the box overhanging the sides of the trailer, if each box is uniformly loaded and the line of action of the resultant of the weights of the four boxes is to pass through the point of intersection of the centerlines of the trailer and the axle. (Hint: Keep in mind that the box may be placed either on its side or on its end.)

For the smallest weight on the trailer so that the resultant force of the four weights acts over the axle at the intersection with the center line of the trailer, the added 0.66 × 0.66 × 1.2-m box should be placed adjacent to one of the edges of the trailer with the 0.66 × 0.66-m side on the bottom. The edges to be considered are based on the location of the resultant for the three given weights. We have

ΣF : − (224 N) j − (392 N) j − (176 N) j = R R = −(792 N) j

We have

ΣM z : − (224 N)(0.33 m) − (392 N)(1.67 m) − (176 N)(1.67 m) = ( −792 N)( x) xR = 1.29101 m

We have

ΣM x : (224 N)(0.33 m) + (392 N)(0.6 m) + (176 N)(2.0 m) = (792 N)( z ) z R = 0.83475 m

From the statement of the problem, it is known that the resultant of R from the original loading and the lightest load W passes through G, the point of intersection of the two center lines. Thus, ΣM G = 0. Further, since the lightest load W is to be as small as possible, the fourth box should be placed as far from G as possible without the box overhanging the trailer. These two requirements imply (0.33 m ≤ x ≤ 1 m) (1.5 m ≤ z ≤ 2.97 m)



SOLUTION

www.elsolucionario.net PROBLEM 3.131* (Continued) xL = 0.33 m

With at

G : ΣM z : (1 − 0.33) m × WL − (1.29101 − 1) m × (792 N) = 0 WL = 344.00 N

or

Now we must check if this is physically possible, at

G : ΣM x : ( z L − 1.5) m × 344 N) − (1.5 − 0.83475) m × (792 N) = 0 z L = 3.032 m

or which is not acceptable.

z L = 2.97 m:

With

G : ΣM x : (2.97 − 1.5) m × WL − (1.5 − 0.83475) m × (792 N) = 0 WL = 358.42 N

or Now check if this is physically possible, at

G : ΣM z : (1 − xL ) m × (358.42 N) − (1.29101 − 1) m × (792 N) = 0

or

xL = 0.357 m ok! WL = 358 N 

The minimum weight of the fourth box is

And it is placed on end A (0.66 × 0.66-m side down) along side AB with the center of the box 0.357 m from side AD.





at

www.elsolucionario.net PROBLEM 3.132*

PROBLEM 3.131* A group of students loads a 2 × 3.3-m flatbed trailer with two 0.66 × 0.66 × 0.66-m boxes and one 0.66 × 0.66 × 1.2-m box. Each of the boxes at the rear of the trailer is positioned so that it is aligned with both the back and a side of the trailer. Determine the smallest load the students should place in a second 0.66 × 0.66 × 1.2-m box and where on the trailer they should secure it, without any part of the box overhanging the sides of the trailer, if each box is uniformly loaded and the line of action of the resultant of the weights of the four boxes is to pass through the point of intersection of the centerlines of the trailer and the axle. (Hint: Keep in mind that the box may be placed either on its side or on its end.)

SOLUTION First replace the three known loads with a single equivalent force R applied at coordinate ( xR , 0, z R ). Equivalence requires ΣFy : − 224 − 392 − 176 = − R

or

R = 792 N

ΣM x : (0.33 m)(224 N) + (0.6 m)(392 N) + (2 m)(176 N) = z R (792 N)

or

z R = 0.83475 m

ΣM z : − (0.33 m)(224 N) − (1.67 m)(392 N) − (1.67 m)(176 N) = xR (792 N)

or

xR = 1.29101 m

From the statement of the problem, it is known that the resultant of R and the heaviest loads WH passes through G, the point of intersection of the two center lines. Thus, ΣM G = 0

Further, since WH is to be as large as possible, the fourth box should be placed as close to G as possible while keeping one of the sides of the box coincident with a side of the trailer. Thus, the two limiting cases are xH = 0.6 m or

z H = 2.7 m



Solve Problem 3.131 if the students want to place as much weight as possible in the fourth box and at least one side of the box must coincide with a side of the trailer.

www.elsolucionario.net PROBLEM 3.132* (Continued) Now consider these two possibilities. With xH = 0.6 m at

G : ΣM z : (1 − 0.6) m × WH − (1.29101 − 1) m × (792 N) = 0 WH = 576.20 N

or Checking if this is physically possible at or

G : ΣM x : ( z H − 1.5) m × (576.20 N) − (1.5 − 0.83475) m × (792 N) = 0 z H = 2.414 m

which is acceptable.

at or

G : ΣM x : (2.7 − 1.5) m × WH − (1.5 − 0.83475) m × (792 N) = 0 WH = 439 N

Since this is less than the first case, the maximum weight of the fourth box is WH = 576 N 

and it is placed with a 0.66 × 1.2-m side down, a 0.66-m edge along side AD, and the center 2.41 m from side DC.





With z H = 2.7 m

www.elsolucionario.net PROBLEM 3.133* A piece of sheet metal is bent into the shape shown and is acted upon by three forces. If the forces have the same magnitude P, replace them with an equivalent wrench and determine (a) the magnitude and the direction of the resultant force R, (b) the pitch of the wrench, (c) the axis of the wrench.

SOLUTION

(

)

We have ΣF : − Pj + Pj + Pk = R R = Pk

or

 5   ΣM O : − (aP ) j +  −( aP )i +  aP  k  = M OR 2    5   M OR = aP  −i − j + k  2  

or (a)

Then for the wrench, R=P 

and

λ axis =

R =k R

cos θ x = 0 cos θ y = 0 cos θ z = 1

or (b)

θ x = 90° θ y = 90° θ z = 0°



Now M1 = λ axis ⋅ M OR 5   = k ⋅ aP  −i − j + k  2   5 = aP 2

Then

P=

M1 25 aP = R P

or P =

5 a  2



First reduce the given forces to an equivalent force-couple system R, M OR at the origin.

www.elsolucionario.net PROBLEM 3.133* (Continued) (c)

The components of the wrench are (R , M1 ), where M1 = M1λ axis , and the axis of the wrench is assumed to intersect the xy-plane at Point Q, whose coordinates are (x, y, 0). Thus, we require M z = rQ × R R

where

M z = M O × M1

Then 5  5  aP  −i − j + k  − aPk = ( xi + yj) + Pk 2  2 

Equating coefficients: or

j: − aP = − xP or

y = −a x=a

The axis of the wrench is parallel to the z-axis and intersects the xy-plane at

x = a, y = −a. 



i : − aP = yP

www.elsolucionario.net PROBLEM 3.134* Three forces of the same magnitude P act on a cube of side a as shown. Replace the three forces by an equivalent wrench and determine (a) the magnitude and direction of the resultant force R, (b) the pitch of the wrench, (c) the axis of the wrench.

SOLUTION Force-couple system at O:

M OR = aj × Pi + ak × Pj + ai × Pk

= − Pak − Pai − Paj M OR = − Pa (i + j + k )

Since R and M OR have the same direction, they form a wrench with M1 = M OR . Thus, the axis of the wrench is the diagonal OA. We note that cos θ x = cos θ y = cos θ z =

a a 3

=

1 3

R = P 3 θ x = θ y = θ z = 54.7° M1 = M OR = − Pa 3 Pitch = p =

M 1 − Pa 3 = = −a R P 3

(a)

R = P 3 θ x = θ y = θ z = 54.7° 

(b)

– a

(c)

Axis of the wrench is diagonal OA.



R = Pi + Pj + Pk = P(i + j + k )

www.elsolucionario.net PROBLEM 3.135* The forces and couples shown are applied to two screws as a piece of sheet metal is fastened to a block of wood. Reduce the forces and the couples to an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the xz-plane.


SOLUTION

First, reduce the given force system to a force-couple at the origin. We have

ΣF : − (10 lb) j − (11 lb) j = R R = − (21 lb) j

We have

ΣM O : Σ(rO × F ) + ΣM C = M OR M OR

i j k i j k = 0 0 20 lb ⋅ in. + 0 0 −15 lb ⋅ in. − (12 lb ⋅ in) j 0 −10 0 0 −11 0

= (35 lb ⋅ in.)i − (12 lb ⋅ in.) j R = − (21 lb) j

(a) (b)

We have

or R = − (21.0 lb) j 

R R = (− j) ⋅ [(35 lb ⋅ in.)i − (12 lb ⋅ in.) j]

M1 = λ R ⋅ M OR

λR =

= 12 lb ⋅ in. and M1 = −(12 lb ⋅ in.) j

and pitch

p=

M 1 12 lb ⋅ in. = = 0.57143 in. R 21 lb

or

p = 0.571 in. 


www.elsolucionario.net PROBLEM 3.135* (Continued)

(c)

We have

M OR = M1 + M 2 M 2 = M OR − M1 = (35 lb ⋅ in.)i

We require

M 2 = rQ/O × R

(35 lb ⋅ in.)i = ( xi + zk ) × [ −(21 lb) j] 35i = −(21x)k + (21z )i

From i:

From k:

35 = 21z z = 1.66667 in. 0 = − 21x z=0




The axis of the wrench is parallel to the y-axis and intersects the xz-plane at x = 0, z = 1.667 in.


www.elsolucionario.net PROBLEM 3.136* The forces and couples shown are applied to two screws as a piece of sheet metal is fastened to a block of wood. Reduce the forces and the couples to an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the xz-plane.

SOLUTION

We have

ΣF : − (20 N)i − (15 N) j = R

We have

ΣM O : Σ(rO × F ) + ΣM C = M OR

R = 25 N

M OR = −20 N(0.1 m)j − (4 N ⋅ m)i − (1 N ⋅ m)j

= −(4 N ⋅ m)i − (3 N ⋅ m) j R = −(20.0 N)i − (15.00 N)j 

(a) (b)

We have

R R = (−0.8i − 0.6 j) ⋅ [−(4 N ⋅ m)i − (3 N ⋅ m)j]

M1 = λR ⋅ M OR

λ=

= 5 N⋅m

Pitch:

p=

M1 5 N ⋅ m = = 0.200 m R 25 N

or p = 0.200 m  (c)

From above, note that M1 = M OR

Therefore, the axis of the wrench goes through the origin. The line of action of the wrench lies in the xy-plane with a slope of y=

3 x  4



First, reduce the given force system to a force-couple system.

www.elsolucionario.net PROBLEM 3.137* Two bolts at A and B are tightened by applying the forces and couples shown. Replace the two wrenches with a single equivalent wrench and determine (a) the resultant R, (b) the pitch of the single equivalent wrench, (c) the point where the axis of the wrench intersects the xz-plane.


SOLUTION

First, reduce the given force system to a force-couple at the origin. We have

ΣF : − (84 N) j − (80 N)k = R

and

ΣM O : Σ(rO × F ) + ΣM C = M OR

R = 116 N

i j k i j k 0.6 0 0.1 + 0.4 0.3 0 + (−30 j − 32k ) N ⋅ m = M OR 0 84 0 0 0 80 M OR = − (15.6 N ⋅ m)i + (2 N ⋅ m) j − (82.4 N ⋅ m)k R = − (84.0 N) j − (80.0 N)k 

(a) (b)

We have

M1 = λ R ⋅ M OR

λR =

R R

−84 j − 80k ⋅ [− (15.6 N ⋅ m)i + (2 N ⋅ m) j − (82.4 N ⋅ m)k ] 116 = 55.379 N ⋅ m =−

and Then pitch

M1 = M1λR = − (40.102 N ⋅ m) j − (38.192 N ⋅ m)k

p=

M 1 55.379 N ⋅ m = = 0.47741 m R 116 N

or p = 0.477 m 



(c)

We have

M OR = M1 + M 2 M 2 = M OR − M1 = [(−15.6i + 2 j − 82.4k ) − (40.102 j − 38.192k )] N ⋅ m = − (15.6 N ⋅ m)i + (42.102 N ⋅ m) j − (44.208 N ⋅ m)k

We require

M 2 = rQ/O × R (−15.6i + 42.102 j − 44.208k ) = ( xi + zk ) × (84 j − 80k ) = (84 z )i + (80 x) j − (84 x)k

or From k:

or

−15.6 = 84 z z = − 0.185714 m

z = − 0.1857 m −44.208 = −84 x x = 0.52629 m

x = 0.526 m

The axis of the wrench intersects the xz-plane at x = 0.526 m

y = 0 z = − 0.1857 m 



From i:


SOLUTION

First, reduce the given force system to a force-couple at the origin at B. (a)

We have

15   8 ΣF : − (26.4 lb)k − (17 lb)  i + j  = R  17 17  R = − (8.00 lb)i − (15.00 lb) j − (26.4 lb)k 

and We have

R = 31.4 lb ΣM B : rA/B × FA + M A + M B = M BR

M RB

i j k 15   8 0 − 220k − 238  i + j  = 264i − 220k − 14(8i + 15 j) = 0 −10  17 17  0 0 − 26.4

M RB = (152 lb ⋅ in.)i − (210 lb ⋅ in.)j − (220 lb ⋅ in.)k

(b)

We have

R R −8.00i − 15.00 j − 26.4k = ⋅ [(152 lb ⋅ in.)i − (210 lb ⋅ in.) j − (220 lb ⋅ in.)k ] 31.4 = 246.56 lb ⋅ in.

M1 = λ R ⋅ M OR

λR =



Two bolts at A and B are tightened by applying the forces and couples shown. Replace the two wrenches with a single equivalent wrench and determine (a) the resultant R, (b) the pitch of the single equivalent wrench, (c) the point where the axis of the wrench intersects the xz-plane.


and Then pitch (c)

We have

M1 = M1λR = − (62.818 lb ⋅ in.)i − (117.783 lb ⋅ in.) j − (207.30 lb ⋅ in.)k p=

M 1 246.56 lb ⋅ in. = = 7.8522 in. 31.4 lb R

or p = 7.85 in. 

M RB = M1 + M 2 M 2 = M RB − M1 = (152i − 210 j − 220k ) − ( − 62.818i − 117.783j − 207.30k ) = (214.82 lb ⋅ in.)i − (92.217 lb ⋅ in.) j − (12.7000 lb ⋅ in.)k

We require

M 2 = rQ/B × R

= (15 z )i − (8 z ) j + (26.4 x) j − (15 x)k

From i:

214.82 = 15 z

z = 14.3213 in.

From k:

−12.7000 = −15 x

x = 0.84667 in.

The axis of the wrench intersects the xz-plane at

x = 0.847 in. y = 0 z = 14.32 in. 



i j k 214.82i − 92.217 j − 12.7000k = x 0 z −8 −15 −26.4

www.elsolucionario.net PROBLEM 3.139* A flagpole is guyed by three cables. If the tensions in the cables have the same magnitude P, replace the forces exerted on the pole with an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the xz-plane.

(a)

First reduce the given force system to a force-couple at the origin. We have

ΣF : Pλ BA + P λDC + P λDE = R  4 3  3 4   −9 4 12   R = P  j − k  +  i − j  +  i − j + k   5 5 5 5 25 5 25       

R=

R=

We have

3P (2i − 20 j − k )  25

3P 27 5 P (2) 2 + (20) 2 + (1)2 = 25 25 ΣM : Σ(rO × P) = M OR

3P  4P  4P 12 P   −4 P  3P  −9 P (24a) j ×  j− k  + (20a) j ×  i− j  + (20a) j ×  i− j+ k  = M OR 5 5 5 5 25 5 25       M OR =

24 Pa ( −i − k ) 5



SOLUTION


We have

M1 = λ R ⋅ M OR

where

λR =

Then

M1 =

25 1 R 3P (2i − 20 j − k ) (2i − 20 j − k ) = = R 25 27 5 P 9 5

p=

and pitch

1 9 5

(2i − 20 j − k ) ⋅

M 1 −8Pa  25  −8a =  = R 15 5  27 5 P  81

M1 = M1λ R =

(c)

Then

M 2 = M OR − M1 =

24 Pa −8 Pa (−i − k ) = 5 15 5

or p = − 0.0988a 

−8Pa  1  8 Pa ( −2i + 20 j + k )   (2i − 20 j − k ) = 675 15 5  9 5 

24 Pa 8Pa 8 Pa ( −i − k ) − (−2i + 20 j + k ) = (−430i − 20 j − 406k ) 5 675 675 M 2 = rQ/O × R

We require

 8Pa   3P   675  (−403i − 20 j − 406k ) = ( xi + zk ) ×  25  (2i − 20 j − k )      3P  =  [20 zi + ( x + 2 z ) j − 20 xk ]  25 

From i:

8(− 403)

Pa  3P  = 20 z   675  25 

From k:

8(−406)

Pa  3P  = −20 x   x = 2.0049a 675  25 

z = −1.99012a

The axis of the wrench intersects the xz-plane at x = 2.00a, z = −1.990a 



(b)

www.elsolucionario.net PROBLEM 3.140* Two ropes attached at A and B are used to move the trunk of a fallen tree. Replace the forces exerted by the ropes with an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the yz-plane.

SOLUTION (a)

(

)

First replace the given forces with an equivalent force-couple system R, M OR at the origin. We have


d AC = (6) 2 + (2) 2 + (9) 2 = 11 m d BD = (14) 2 + (2)2 + (5) 2 = 15 m

Then 1650 N = (6i + 2 j + 9k ) 11 = (900 N)i + (300 N) j + (1350 N)k

TAC =

and 1500 N = (14i + 2 j + 5k ) 15 = (1400 N)i + (200 N) j + (500 N)k

TBD =

Equivalence then requires ΣF : R = TAC + TBD = (900i + 300 j + 1350k ) +(1400i + 200 j + 500k ) = (2300 N)i + (500 N) j + (1850 N)k ΣM O : M OR = rA × TAC + rB × TBD = (12 m)k × [(900 N)i + (300 N)j + (1350 N)k ] + (9 m)i × [(1400 N)i + (200 N)j + (500 N)k ] = −(3600)i + (10,800 − 4500) j + (1800)k = −(3600 N ⋅ m)i + (6300 N ⋅ m)j + (1800 N ⋅ m)k

The components of the wrench are (R , M1 ), where R = (2300 N)i + (500 N) j + (1850 N)k




www.elsolucionario.net PROBLEM 3.140* (Continued)  (b)

We have R = 100 (23)2 + (5) 2 + (18.5) 2 = 2993.7 N

Let λ axis =

Then

1 R (23i + 5 j + 18.5k ) = R 29.937

M1 = λ axis ⋅ M OR 1 (23i + 5 j + 18.5k ) ⋅ (−3600i + 6300 j + 1800k ) 29.937 1 [(23)( −36) + (5)(63) + (18.5)(18)] = 0.29937 = −601.26 N ⋅ m

Finally,

P=

M1 −601.26 N ⋅ m = R 2993.7 N

or P = − 0.201 m  (c)

We have

M1 = M 1 λ axis = (−601.26 N ⋅ m) ×

1 (23i + 5 j + 18.5k ) 29.937

or

M1 = −(461.93 N ⋅ m)i − (100.421 N ⋅ m) j − (371.56 N ⋅ m)k

Now

M 2 = M OR − M1 = (−3600i + 6300 j + 1800k ) − ( −461.93i − 100.421j − 371.56k ) = − (3138.1 N ⋅ m)i + (6400.4 N ⋅ m)j + (2171.6 N ⋅ m)k

For equivalence:



=


Thus, we require

M 2 = rP × R

r = ( yj + zk )

Substituting: −3138.1i + 6400.4 j + 2171.6k =

i j k 0 y z 2300 500 1850

Equating coefficients: j : 6400.4 = 2300 z

or

k : 2171.6 = −2300 y or

y = − 0.944 m

y = −0.944 m

z = 2.78 m




The axis of the wrench intersects the yz-plane at

z = 2.78 m


www.elsolucionario.net PROBLEM 3.141* Determine whether the force-and-couple system shown can be reduced to a single equivalent force R. If it can, determine R and the point where the line of action of R intersects the yz-plane. If it cannot be so reduced, replace the given system with an equivalent wrench and determine its resultant, its pitch, and the point where its axis intersects the yz-plane.

SOLUTION

d DA = (−12) 2 + (9) 2 + (8) 2 = 17 in. d ED = (−6) 2 + (0)2 + (−8)2 = 10 in.

Then 34 lb = (−12i + 9 j + 8k ) 17 = −(24 lb)i + (18 lb) j + (16 lb)k

FDA =

and 30 lb = (−6i − 8k ) 10 = −(18 lb)i − (24 lb)k

FED =

Then ΣF : R = FDA + FED = (−24i + 18 j + 16k + ( −18i − 24k ) = −(42 lb)i + (18 lb)j − (8 lb)k

For the applied couple d AK = ( −6) 2 + (−6) 2 + (18)2 = 6 11 in.

Then M=

160 lb ⋅ in.

( −6i − 6 j + 18k ) 6 11 160 = [−(1 lb ⋅ in.)i − (1 lb ⋅ in.)j + (3 lb ⋅ in.)k ] 11

To be able to reduce the original forces and couple to a single equivalent force, R and M must be perpendicular. Thus ?

R ⋅ M =0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 315


First determine the resultant of the forces at D. We have


Substituting (−42i + 18 j − 8k ) ⋅ 160

or

11

160 11

?

( −i − j + 3k ) = 0 ?

[(−42)(−1) + (18)(−1) + (−8)(3)] = 0 

0 =0

or

R and M are perpendicular so that the given system can be reduced to the single equivalent force. R = −(42.0 lb)i + (18.00 lb) j − (8.00 lb)k



M = rP/D × R

Thus, we require where


Then for equivalence,

rP/D = −(12 in.)i + [( y − 3)in.] j + ( z in.)k

Substituting: i j k ( −i − j + 3k ) = −12 ( y − 3) z 11 18 −42 −8 = [( y − 3)( −8) − ( z )(18)]i

160

+ [( z )(−42) − (−12)(−8)]j + [( −12)(18) − ( y − 3)(−42)]k

Equating coefficients: j: − k:

160

= − 42 z − 96 or 11 480 = −216 + 42( y − 3) or 11

The line of action of R intersects the yz-plane at

x=0

z = −1.137 in. y = 11.59 in.

y = 11.59 in. z = −1.137 in.




www.elsolucionario.net PROBLEM 3.142* Determine whether the force-and-couple system shown can be reduced to a single equivalent force R. If it can, determine R and the point where the line of action of R intersects the yz-plane. If it cannot be so reduced, replace the given system with an equivalent wrench and determine its resultant, its pitch, and the point where its axis intersects the yz-plane.

SOLUTION First, reduce the given force system to a force-couple at the origin. ΣF : FA + FG = R  (40 mm)i + (60 mm) j − (120 mm)k  R = (50 N)k + 70 N   140 mm   = (20 N)i + (30 N) j − (10 N)k

and We have

R = 37.417 N ΣM O : Σ(rO × F) + ΣM C = M OR

M OR = [(0.12 m) j × (50 N)k ] + {(0.16 m)i × [(20 N)i + (30 N) j − (60 N)k ]}  (160 mm)i − (120 mm) j  + (10 N ⋅ m)   200 mm    (40 mm)i − (120 mm) j + (60 mm)k  + (14 N ⋅ m)   140 mm   M 0R = (18 N ⋅ m)i − (8.4 N ⋅ m) j + (10.8 N ⋅ m)k

To be able to reduce the original forces and couples to a single equivalent force, R and M must be perpendicular. Thus, R ⋅ M = 0. Substituting ?

(20i + 30 j − 10k ) ⋅ (18i − 8.4 j + 10.8k ) = 0 ?

or

(20)(18) + (30)(−8.4) + (−10)(10.8) = 0

or

0=0



R and M are perpendicular so that the given system can be reduced to the single equivalent force. R = (20.0 N)i + (30.0 N) j − (10.00 N)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 317


We have

www.elsolucionario.net PROBLEM 3.142* (Continued)  Then for equivalence,

Thus, we require

M OR = rp × R rp = yj + zk

i j k 18i − 8.4 j + 10.8k = 0 y z 20 30 −10

Equating coefficients: j: − 8.4 = 20 z k:

or

z = −0.42 m

10.8 = −20 y or

y = −0.54 m

The line of action of R intersects the yz-plane at

x=0

y = −0.540 m z = −0.420 m





Substituting:

www.elsolucionario.net PROBLEM 3.143* Replace the wrench shown with an equivalent system consisting of two forces perpendicular to the y-axis and applied respectively at A and B.

SOLUTION Express the forces at A and B as B = Bx i + Bz k

Then, for equivalence to the given force system, ΣFx : Ax + Bx = 0

(1)

ΣFz : Az + Bz = R

(2)

ΣM x : Az ( a) + Bz ( a + b) = 0

(3)

ΣM z : − Ax (a) − Bx (a + b) = M

(4)

Bx = − Ax

From Equation (1), Substitute into Equation (4):

− Ax ( a) + Ax ( a + b) = M M M and Bx = − Ax = b b

From Equation (2),

Bz = R − Az

and Equation (3),

Az a + ( R − Az )(a + b) = 0

 a Az = R 1 +   b

and

a  Bz = R − R 1 +   b a Bz = − R b

M A=  b

Then

a    i + R 1 + b  k    

M B = −  b

 a  i −  b Rk    



A = Ax i + Az k

www.elsolucionario.net PROBLEM 3.144* Show that, in general, a wrench can be replaced with two forces chosen in such a way that one force passes through a given point while the other force lies in a given plane.

First, choose a coordinate system so that the xy-plane coincides with the given plane. Also, position the coordinate system so that the line of action of the wrench passes through the origin as shown in Figure a. Since the orientation of the plane and the components (R, M) of the wrench are known, it follows that the scalar components of R and M are known relative to the shown coordinate system. A force system to be shown as equivalent is illustrated in Figure b. Let A be the force passing through the given Point P and B be the force that lies in the given plane. Let b be the x-axis intercept of B. The known components of the wrench can be expressed as R = Rx i + Ry j + Rz k and M = M x i + M y j + M z k

while the unknown forces A and B can be expressed as A = Ax i + Ay j + Az k and B = Bx i + Bz k

Since the position vector of Point P is given, it follows that the scalar components (x, y, z) of the position vector rP are also known. Then, for equivalence of the two systems, ΣFx : Rx = Ax + Bx

(1)

ΣFy : Ry = Ay

(2)

ΣFz : Rz = Az + Bz

(3)

ΣM x : M x = yAz − zAy

(4)

ΣM y : M y = zAx − xAz − bBz

(5)

ΣM z : M z = xAy − yAx

(6)

Based on the above six independent equations for the six unknowns ( Ax , Ay , Az , Bx , Bz , b), there exists a unique solution for A and B. Ay = Ry 

From Equation (2),



SOLUTION


1 Ax =   ( xRy − M z )   y

Equation (6):

1 Bx = Rx −   ( xRy − M z )   y

Equation (1):

Equation (4):

1 Az =   ( M x + zRy )   y

Equation (3):

1 Bz = Rz −   ( M x + zRy )   y b=

( xM x + yM y + zM z ) ( M x − yRz + zRy )




Equation (5):


www.elsolucionario.net PROBLEM 3.145* Show that a wrench can be replaced with two perpendicular forces, one of which is applied at a given point.

SOLUTION

See Figures a and b. We have

R = Rj and M = Mj

and are known.

The unknown forces A and B can be expressed as A = Ax i + Ay j + Az k and B = Bx i + By j + Bz k

The distance a is known. It is assumed that force B intersects the xz-plane at (x, 0, z). Then for equivalence, ΣFx :

0 = Ax + Bx

(1)

ΣFy :

R = Ay + By

(2)

ΣFz :

0 = Az + Bz

(3)

0 = − zBy

(4)

ΣM x :

ΣM y : M = − aAz − xBz + zBx

(5)

ΣM z :

(6)

0 = aAy + xBy

Since A and B are made perpendicular, A ⋅ B = 0 or

There are eight unknowns:

Ax Bx + Ay B y + Az Bz = 0

(7)

Ax , Ay , Az , Bx , By , Bz , x, z

But only seven independent equations. Therefore, there exists an infinite number of solutions. 0 = − zBy

Next, consider Equation (4): If By = 0, Equation (7) becomes

Ax Bx + Az Bz = 0 Ax2 + Az2 = 0

Using Equations (1) and (3), this equation becomes



First, observe that it is always possible to construct a line perpendicular to a given line so that the constructed line also passes through a given point. Thus, it is possible to align one of the coordinate axes of a rectangular coordinate system with the axis of the wrench while one of the other axes passes through the given point.

www.elsolucionario.net PROBLEM 3.145* (Continued) Since the components of A must be real, a nontrivial solution is not possible. Thus, it is required that By ≠ 0, so that from Equation (4), z = 0. To obtain one possible solution, arbitrarily let Ax = 0. (Note: Setting Ay , Az , or Bz equal to zero results in unacceptable solutions.) 0 = Bx

(1)′

R = Ay + By

(2)

0 = Az + Bz

(3)

M = − aAz − xBz

(5)′

0 = aAy + xBy

(6)

Ay By + Az Bz = 0

(7)′

Then Equation (2) can be written

Ay = R − By

Equation (3) can be written

Bz = − Az

Equation (6) can be written

x=−

aAy By

Substituting into Equation (5)′,  R − By M = − aAz −  − a  By  M Az = − By aR

or

  ( − Az )  

(8)

Substituting into Equation (7)′,  M  M  ( R − By ) By +  − By  By  = 0  aR  aR 

or

By =

a 2 R3 a2 R2 + M 2

Then from Equations (2), (8), and (3), a2 R2 RM 2 = 2 2 2 2 a R +M a R + M2  M  a 2 R3 aR 2 M Az = − = −   aR  a 2 R 2 + M 2  a2 R2 + M 2 Ay = R −

Bz =

2

aR 2 M a2 R2 + M 2



The defining equations then become


In summary, A=

RM ( Mj − aRk ) a R2 + M 2



B=

aR 2 (aRj + Mk ) a2 R2 + M 2



2

Which shows that it is possible to replace a wrench with two perpendicular forces, one of which is applied at a given point. Lastly, if R > 0 and M > 0, it follows from the equations found for A and B that Ay > 0 and B y > 0.





From Equation (6), x < 0 (assuming a > 0). Then, as a consequence of letting Ax = 0, force A lies in a plane parallel to the yz-plane and to the right of the origin, while force B lies in a plane parallel to the yz-plane but to the left to the origin, as shown in the figure below.

www.elsolucionario.net PROBLEM 3.146* Show that a wrench can be replaced with two forces, one of which has a prescribed line of action.

First, choose a rectangular coordinate system where one axis coincides with the axis of the wrench and another axis intersects the prescribed line of action (AA′). Note that it has been assumed that the line of action of force B intersects the xz-plane at Point P(x, 0, z). Denoting the known direction of line AA′ by

λ A = λx i + λ y j + λz k it follows that force A can be expressed as A = Aλ A = A(λx i + λ y j + λz k )

Force B can be expressed as B = Bx i + B y j + Bz k

Next, observe that since the axis of the wrench and the prescribed line of action AA′ are known, it follows that the distance a can be determined. In the following solution, it is assumed that a is known. Then for equivalence, ΣFx : 0 = Aλx + Bx

(1)

ΣFy : R = Aλ y + By

(2)

ΣFz : 0 = Aλz + Bz

(3)

ΣM x : 0 = − zBy

(4)

ΣM y : M = − aAλz + zBx − xBz

(5)

ΣM x : 0 = − aAλ y + xB y

(6)

Since there are six unknowns (A, Bx, By, Bz, x, z) and six independent equations, it will be possible to obtain a solution.



SOLUTION

www.elsolucionario.net PROBLEM 3.146* (Continued) Case 1: Let z = 0 to satisfy Equation (4). Now Equation (2): Equation (3): Equation (6):

Aλ y = R − B y Bz = − Aλz x=−

aAλ y By

 a = −  By 

  ( R − By )  

Substitution into Equation (5):   a M = − aAλz −  −    By 1 M  A=−  B λz  aR  y

   ( R − By )(− Aλz )    

R=− By =

Then

1 M  B λ + By λz  aR  y y

λz aR 2 λz aR − λ y M

MR R = λz aR − λ y M λ − aR λ y z M λx MR Bx = − Aλx = λz aR − λ y M A=−

Bz = − Aλz =

λz MR λz aR − λ y M A=

In summary,

B=

and

 R  x = a 1 −   By     λz aR − λ y M = a 1 − R   λ aR 2  z 

P λA  aR λy − λz M

R (λ M i + λz aRj + λz M k )  λz aR − λ y M x

    

or x =

λy M λz R



Note that for this case, the lines of action of both A and B intersect the x-axis. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 326


Substitution into Equation (2):


Case 2: Let By = 0 to satisfy Equation (4). Now Equation (2):

A=

R

λy

Equation (1):

λ Bx = − R  x  λy 

   

Equation (3):

λ Bz = − R  z  λy 

   

Equation (6):

aAλ y = 0

which requires a = 0

 λ M = z − R  x   λ y

 λ   − x − R  z    λ y 

 M   or λz x − λx z =    R 

  λy 

This last expression is the equation for the line of action of force B. In summary,  R A=  λy 

  λA  



 R B=  λy 

  ( −λ x i − λx k )  



Assuming that λx , λ y , λz ⬎ 0, the equivalent force system is as shown below.

Note that the component of A in the xz-plane is parallel to B.



Substitution into Equation (5):

www.elsolucionario.net PROBLEM 3.147 A 300-N force is applied at A as shown. Determine (a) the moment of the 300-N force about D, (b) the smallest force applied at B that creates the same moment about D.

SOLUTION (a)

Fx = (300 N) cos 25°

= 126.785 N F = (271.89 N)i + (126.785 N) j  r = DA = −(0.1 m)i − (0.2 m) j MD = r × F M D = [−(0.1 m)i − (0.2 m) j] × [(271.89 N)i + (126.785 N) j]

= −(12.6785 N ⋅ m)k + (54.378 N ⋅ m)k = (41.700 N ⋅ m)k M D = 41.7 N ⋅ m

(b)



The smallest force Q at B must be perpendicular to  DB at 45°  M D = Q ( DB) 41.700 N ⋅ m = Q (0.28284 m)

Q = 147.4 N

45.0° 



= 271.89 N Fy = (300 N) sin 25°

www.elsolucionario.net PROBLEM 3.148 The tailgate of a car is supported by the hydraulic lift BC. If the lift exerts a 125-lb force directed along its centerline on the ball and socket at B, determine the moment of the force about A.

SOLUTION First note

dCB = (12.0 in.) 2 + (2.33 in.) 2

Then

and


= 12.2241 in. 12.0 in. 12.2241 in. 2.33 in. sin θ = 12.2241 in.

cos θ =

FCB = FCB cos θ i − FCB sin θ j

=

125 lb [(12.0 in.) i − (2.33 in.) j] 12.2241 in.

Now

M A = rB/A × FCB

where

rB/A = (15.3 in.) i − (12.0 in. + 2.33 in.) j

= (15.3 in.) i − (14.33 in.) j

Then

M A = [(15.3 in.)i − (14.33 in.) j] ×

125 lb (12.0i − 2.33j) 12.2241 in.

= (1393.87 lb ⋅ in.)k = (116.156 lb ⋅ ft)k

or M A = 116.2 lb ⋅ ft




www.elsolucionario.net PROBLEM 3.149 The ramp ABCD is supported by cables at corners C and D. The tension in each of the cables is 810 N. Determine the moment about A of the force exerted by (a) the cable at D, (b) the cable at C.

(a)

We have

M A = rE/A × TDE

where

rE/A = (2.3 m) j TDE = λ DE TDE =

(0.6 m)i + (3.3 m) j − (3 m)k (0.6) 2 + (3.3)2 + (3) 2 m

(810 N)

= (108 N)i + (594 N) j − (540 N)k i j k MA = 0 2.3 0 N⋅m 108 594 −540 = −(1242 N ⋅ m)i − (248.4 N ⋅ m)k

or M A = −(1242 N ⋅ m)i − (248 N ⋅ m)k  (b)

We have

M A = rG/A × TCG

where

rG/A = (2.7 m)i + (2.3 m) j TCG = λ CGTCG =

−(.6 m)i + (3.3 m) j − (3 m)k (.6) 2 + (3.3) 2 + (3) 2 m

(810 N)

= −(108 N)i + (594 N) j − (540 N)k i j k M A = 2.7 2.3 0 N⋅m −108 594 −540 = −(1242 N ⋅ m)i + (1458 N ⋅ m) j + (1852 N ⋅ m)k

or M A = −(1242 N ⋅ m)i + (1458 N ⋅ m) j + (1852 N ⋅ m)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 330


SOLUTION


SOLUTION First note

 AB = AB(sin 37° j − cos 37°k ) CD = CD(− cos 40° cos 55° j + sin 40° j − cos 40° sin 55°k )

Now

  AB ⋅ CD = ( AB)(CD ) cos θ

or

AB(sin 37° j − cos 37°k ) ⋅ CD (− cos 40° cos 55°i + sin 40° j − cos 40° sin 55°k ) = (AB)(CD) cos θ

or

cos θ = (sin 37°)(sin 40°) + (− cos 37°)(− cos 40° sin 55°) = 0.88799

or θ = 27.4° 



Section AB of a pipeline lies in the yz-plane and forms an angle of 37° with the z-axis. Branch lines CD and EF join AB as shown. Determine the angle formed by pipes AB and CD.



To lift a heavy crate, a man uses a block and tackle attached to the bottom of an I-beam at hook B. Knowing that the moments about the y and the z axes of the force exerted at B by portion AB of the rope are, respectively, 120 N ⋅ m and −460 N ⋅ m, determine the distance a.

SOLUTION First note

 BA = (2.2 m)i − (3.2 m) j − ( a m)k

Now

M D = rA/D × TBA

where

rA/D = (2.2 m)i + (1.6 m) j TBA =

Then

i j k TBA MD = 2.2 1.6 0 d BA 2.2 −3.2 − a =

Thus

TBA (2.2i − 3.2 j − ak ) (N) d BA

TBA {−1.6a i + 2.2a j + [(2.2)(−3.2) − (1.6)(2.2)]k} d BA

M y = 2.2

TBA a d BA

M z = −10.56

Then forming the ratio

TBA d BA

(N ⋅ m) (N ⋅ m)

My Mz T

2.2 dBA (N ⋅ m) 120 N ⋅ m BA = −460 N ⋅ m −10.56 TdBA (N ⋅ m)

or a = 1.252 m 

BA


www.elsolucionario.net PROBLEM 3.152 To loosen a frozen valve, a force F of magnitude 70 lb is applied to the handle of the valve. Knowing that θ = 25°, Mx = −61 lb ⋅ ft, and M z = − 43 lb ⋅ ft, determine φ and d.

We have

ΣM O : rA/O × F = M O

where

rA/O = −(4 in.)i + (11 in.) j − (d )k F = F (cos θ cos φ i − sin θ j + cos θ sin φ k )

For

F = 70 lb, θ = 25° F = (70 lb)[(0.90631cos φ )i − 0.42262 j + (0.90631sin φ )k ] i M O = (70 lb) −4 −0.90631cos φ

j k 11 −d in. −0.42262 0.90631sin φ

= (70 lb)[(9.9694sin φ − 0.42262d ) i + (−0.90631d cos φ + 3.6252sin φ ) j + (1.69048 − 9.9694cos φ )k ] in.

and

M x = (70 lb)(9.9694sin φ − 0.42262d ) in. = −(61 lb ⋅ ft)(12 in./ft)

(1)

M y = (70 lb)(−0.90631d cos φ + 3.6252sin φ ) in.

(2)

M z = (70 lb)(1.69048 − 9.9694 cos φ ) in. = − 43 lb ⋅ ft(12 in./ft)

(3)

 634.33   = 24.636°  697.86 

From Equation (3):

φ = cos −1 

or

From Equation (1):

 1022.90  d =  = 34.577 in.  29.583 

or d = 34.6 in. 

φ = 24.6° 



SOLUTION

www.elsolucionario.net PROBLEM 3.153 The tension in the cable attached to the end C of an adjustable boom ABC is 560 lb. Replace the force exerted by the cable at C with an equivalent forcecouple system (a) at A, (b) at B.

SOLUTION Based on

ΣF : FA = T = 560 lb FA = 560 lb

or

20.0° 

ΣM A : M A = (T sin 50°)(d A ) = (560 lb)sin 50°(18 ft) = 7721.7 lb ⋅ ft M A = 7720 lb ⋅ ft

or (b)

Based on

ΣF : FB = T = 560 lb FB = 560 lb

or



20.0° 

ΣM B : M B = (T sin 50°)(d B ) = (560 lb) sin 50°(10 ft) = 4289.8 lb ⋅ ft M B = 4290 lb ⋅ ft

or





(a)

www.elsolucionario.net PROBLEM 3.154 While tapping a hole, a machinist applies the horizontal forces shown to the handle of the tap wrench. Show that these forces are equivalent to a single force, and specify, if possible, the point of application of the single force on the handle.

Since the forces at A and B are parallel, the force at B can be replaced with the sum of two forces with one of the forces equal in magnitude to the force at A except with an opposite sense, resulting in a force-couple. We have FB = 2.9 lb − 2.65 lb = 0.25 lb, where the 2.65-lb force is part of the couple. Combining the two parallel forces, M couple = (2.65 lb)[(3.2 in. + 2.8 in.) cos 25°] = 14.4103 lb ⋅ in.

and

M couple = 14.4103 lb ⋅ in.

A single equivalent force will be located in the negative z direction. Based on

ΣM B : −14.4103 lb ⋅ in. = [(0.25 lb) cos 25°](a )

a = 63.600 in.

F′ = (0.25 lb)(cos 25°i + sin 25°k ) F′ = (0.227 lb)i + (0.1057 lb)k and is applied on an extension of handle BD at a distance of 63.6 in. to the right of B.





SOLUTION

www.elsolucionario.net PROBLEM 3.155 Replace the 150-N force with an equivalent force-couple system at A.

SOLUTION ΣF : F = (150 N)(− cos 35° j − sin 35°k ) = −(122.873 N) j − (86.036 N)k ΣMA : M = rD/A × F

where

Then

rD/A = (0.18 m)i − (0.12 m) j + (0.1 m)k i j k −0.12 0.1 N ⋅ m M = 0.18 0 −122.873 −86.036 = [( −0.12)(−86.036) − (0.1)(−122.873)]i + [−(0.18)(−86.036)]j + [(0.18)(−122.873)]k = (22.6 N ⋅ m)i + (15.49 N ⋅ m) j − (22.1 N ⋅ m)k

The equivalent force-couple system at A is F = −(122.9 N) j − (86.0 N)k  M = (22.6 N ⋅ m)i + (15.49 N ⋅ m) j − (22.1 N ⋅ m)k 



Equivalence requires

www.elsolucionario.net PROBLEM 3.156 A beam supports three loads of given magnitude and a fourth load whose magnitude is a function of position. If b = 1.5 m and the loads are to be replaced with a single equivalent force, determine (a) the value of a so that the distance from support A to the line of action of the equivalent force is maximum, (b) the magnitude of the equivalent force and its point of application on the beam.

For equivalence, ΣFy : −1300 + 400

a − 400 − 600 = − R b

a  R =  2300 − 400  N b  

or

ΣM A :

a a 400  − a (400) − (a + b)(600) = − LR 2  b

L=

or

Then with

(1)

1000a + 600b − 200 2300 − 400

b = 1.5 m

L=

a2 b

a b

10a + 9 −

4 2 a 3

8 23 − a 3

(2)

where a, L are in m. (a)

Find value of a to maximize L. 8  8   4  8   10 − a  23 − a  − 10a + 9 − a 2  −  dL  3  3   3  3  = 2 da 8    23 − 3 a   



SOLUTION


230 −

or

16a 2 − 276a + 1143 = 0

276 ± (−276) 2 − 4(16)(1143) 2(16)

a=

or

a = 10.3435 m and a = 6.9065 m AB = 9 m, a must be less than 9 m

a = 6.91 m 

6.9065 1.5

or R = 458 N 

Using Eq. (1),

R = 2300 − 400

and using Eq. (2),

4 10(6.9065) + 9 − (6.9065)2 3 L= = 3.16 m 8 23 − (6.9065) 3

R is applied 3.16 m to the right of A. 



Then

Since (b)

184 80 64 2 80 32 a− a+ a + a + 24 − a 2 = 0 3 3 9 3 9

or

www.elsolucionario.net PROBLEM 3.157 A mechanic uses a crowfoot wrench to loosen a bolt at C. The mechanic holds the socket wrench handle at Points A and B and applies forces at these points. Knowing that these forces are equivalent to a force-couple system at C consisting of the force C = (8 lb)i + (4 lb)k and the couple M C = (360 lb · in.)i, determine the forces applied at A and at B when Az = 2 lb.

We have

ΣF :

A+B=C

or

Fx : Ax + Bx = 8 lb Bx = −( Ax + 8 lb)

(1)

ΣFy : Ay + By = 0 Ay = − By

or

(2)

ΣFz : 2 lb + Bz = 4 lb Bz = 2 lb

or

(3)

ΣM C : rB/C × B + rA/C × A = M C

We have

i 8 Bx

j 0 By

k i 2 + 8 2 Ax

j 0 Ay

k 8 2

lb ⋅ in. = (360 lb ⋅ in.)i

(2 By − 8 Ay )i + (2 Bx − 16 + 8 Ax − 16) j

or

+ (8By + 8 Ay )k = (360 lb ⋅ in.)i

From

i-coefficient: j-coefficient: k-coefficient:

2 By − 8 Ay = 360 lb ⋅ in. −2 Bx + 8 Ax = 32 lb ⋅ in. 8 By + 8 Ay = 0

(4) (5) (6)



SOLUTION


From Equations (2) and (4):

2 By − 8(− By ) = 360 By = 36 lb

From Equations (1) and (5):

Ay = 36 lb

2(− Ax − 8) + 8 Ax = 32 Ax = 1.6 lb

From Equation (1):

Bx = −(1.6 + 8) = −9.6 lb A = (1.600 lb)i − (36.0 lb) j + (2.00 lb)k 


B = −(9.60 lb)i + (36.0 lb) j + (2.00 lb)k 


www.elsolucionario.net PROBLEM 3.158 A concrete foundation mat in the shape of a regular hexagon of side 12 ft supports four column loads as shown. Determine the magnitudes of the additional loads that must be applied at B and F if the resultant of all six loads is to pass through the center of the mat.

SOLUTION From the statement of the problem, it can be concluded that the six applied loads are equivalent to the resultant R at O. It then follows that


ΣM O = 0 or ΣM x = 0 ΣM z = 0

For the applied loads:

Then

ΣM x = 0: (6 3 ft) FB + (6 3 ft)(10 kips) − (6 3 ft)(20 kips) − (6 3 ft) FF = 0 FB − FF = 10

or

(1)

ΣM z = 0: (12 ft)(15 kips) + (6 ft) FB − (6 ft)(10 kips) − (12 ft)(30 kips) − (6 ft)(20 kips) + (6 ft) FF = 0 FB + FF = 60

or

(2)

Then Eqs. (1) + (2) 

FB = 35.0 kips 

and

FF = 25.0 kips 




CHAPTER 4





www.elsolucionario.net PROBLEM 4.1 Two crates, each of mass 350 kg, are placed as shown in the bed of a 1400-kg pickup truck. Determine the reactions at each of the two (a) rear wheels A, (b) front wheels B.

SOLUTION

W = (350 kg)(9.81 m/s 2 ) = 3.4335 kN Wt = (1400 kg)(9.81 m/s 2 ) = 13.7340 kN

(a)

Rear wheels:

ΣM B = 0: W (1.7 m + 2.05 m) + W (2.05 m) + Wt (1.2 m) − 2 A(3 m) = 0

(3.4335 kN)(3.75 m) + (3.4335 kN)(2.05 m) + (13.7340 kN)(1.2 m) − 2 A(3 m) = 0

A = +6.0659 kN

(b)

Front wheels:

A = 6.07 kN 

ΣFy = 0: − W − W − Wt + 2 A + 2 B = 0 −3.4335 kN − 3.4335 kN − 13.7340 kN + 2(6.0659 kN) + 2B = 0 B = +4.2346 kN

B = 4.23 kN 



Free-Body Diagram:

www.elsolucionario.net PROBLEM 4.2 Solve Problem 4.1, assuming that crate D is removed and that the position of crate C is unchanged. PROBLEM 4.1 Two crates, each of mass 350 kg, are placed as shown in the bed of a 1400-kg pickup truck. Determine the reactions at each of the two (a) rear wheels A, (b) front wheels B.

SOLUTION

W = (350 kg)(9.81 m/s 2 ) = 3.4335 kN Wt = (1400 kg)(9.81 m/s 2 ) = 13.7340 kN

(a)

Rear wheels:

ΣM B = 0: W (1.7 m + 2.05 m) + Wt (1.2 m) − 2 A(3 m) = 0 (3.4335 kN)(3.75 m) + (13.7340 kN)(1.2 m) − 2 A(3 m) = 0 A = + 4.8927 kN

(b)

Front wheels:

A = 4.89 kN 

ΣM y = 0: − W − Wt + 2 A + 2 B = 0 −3.4335 kN − 13.7340 kN + 2(4.8927 kN) + 2B = 0 B = +3.6911 kN

B = 3.69 kN 



Free-Body Diagram:

www.elsolucionario.net PROBLEM 4.3 A T-shaped bracket supports the four loads shown. Determine the reactions at A and B (a) if a = 10 in., (b) if a = 7 in.

SOLUTION Free-Body Diagram:

ΣFx = 0: Bx = 0 ΣM B = 0: (40 lb)(6 in.) − (30 lb)a − (10 lb)(a + 8 in.) + (12 in.) A = 0

(40a − 160) 12

A=

(1)

− (10 lb)(a + 20 in.) + (12 in.) B y = 0 By =

Bx = 0, B =

Since (a)

(b)

(1400 + 40a) 12 (1400 + 40a ) 12

(2)

For a = 10 in.,

Eq. (1):

A=

(40 × 10 − 160) = +20.0 lb 12

Eq. (2):

B=

(1400 + 40 × 10) = +150.0 lb 12

B = 150.0 lb 

Eq. (1):

A=

(40 × 7 − 160) = +10.00 lb 12

A = 10.00 lb 

Eq. (2):

B=

(1400 + 40 × 7) = +140.0 lb 12

B = 140.0 lb 

A = 20.0 lb 

For a = 7 in.,



ΣM A = 0: − (40 lb)(6 in.) − (50 lb)(12 in.) − (30 lb)(a + 12 in.)

www.elsolucionario.net PROBLEM 4.4 For the bracket and loading of Problem 4.3, determine the smallest distance a if the bracket is not to move. PROBLEM 4.3 A T-shaped bracket supports the four loads shown. Determine the reactions at A and B (a) if a = 10 in., (b) if a = 7 in.

SOLUTION

For no motion, reaction at A must be downward or zero; smallest distance a for no motion corresponds to A = 0. ΣM B = 0: (40 lb)(6 in.) − (30 lb)a − (10 lb)(a + 8 in.) + (12 in.) A = 0 A=

(40a − 160) 12

A = 0: (40a − 160) = 0

a = 4.00 in. 



Free-Body Diagram:

www.elsolucionario.net PROBLEM 4.5 A hand truck is used to move two kegs, each of mass 40 kg. Neglecting the mass of the hand truck, determine (a) the vertical force P that should be applied to the handle to maintain equilibrium when α = 35°, (b) the corresponding reaction at each of the two wheels.

SOLUTION

W = mg = (40 kg)(9.81 m/s 2 ) = 392.40 N a1 = (300 mm)sinα − (80 mm)cosα a2 = (430 mm)cosα − (300 mm)sinα b = (930 mm)cosα

From free-body diagram of hand truck, Dimensions in mm

ΣM B = 0: P(b) − W ( a2 ) + W (a1 ) = 0

(1)

ΣFy = 0: P − 2W + 2 B = 0

(2)

α = 35°

For

a1 = 300sin 35° − 80 cos 35° = 106.541 mm a2 = 430 cos 35° − 300sin 35° = 180.162 mm b = 930cos 35° = 761.81 mm

(a)

From Equation (1): P(761.81 mm) − 392.40 N(180.162 mm) + 392.40 N(106.54 mm) = 0 P = 37.921 N

(b)

or P = 37.9 N 

From Equation (2): 37.921 N − 2(392.40 N) + 2 B = 0

or B = 373 N 



Free-Body Diagram:

www.elsolucionario.net PROBLEM 4.6 Solve Problem 4.5 when α = 40°. PROBLEM 4.5 A hand truck is used to move two kegs, each of mass 40 kg. Neglecting the mass of the hand truck, determine (a) the vertical force P that should be applied to the handle to maintain equilibrium when α = 35°, (b) the corresponding reaction at each of the two wheels.

SOLUTION

W = mg = (40 kg)(9.81 m/s 2 ) W = 392.40 N a1 = (300 mm)sinα − (80 mm)cosα a2 = (430 mm)cosα − (300 mm)sinα b = (930 mm)cosα

From F.B.D.: ΣM B = 0: P(b) − W ( a2 ) + W (a1 ) = 0 P = W ( a2 − a1 )/b

(1)

ΣFy = 0: − W − W + P + 2 B = 0 B =W −

For

1 P 2

(2)

α = 40°: a1 = 300sin 40° − 80 cos 40° = 131.553 mm a2 = 430 cos 40° − 300sin 40° = 136.563 mm b = 930cos 40° = 712.42 mm

(a)

From Equation (1):

P=

392.40 N (0.136563 m − 0.131553 m) 0.71242 m

P = 2.7595 N

(b)

From Equation (2):

B = 392.40 N −

P = 2.76 N 

1 (2.7595 N) 2

B = 391 N 



Free-Body Diagram:

www.elsolucionario.net PROBLEM 4.7 A 3200-lb forklift truck is used to lift a 1700-lb crate. Determine the reaction at each of the two (a) front wheels A, (b) rear wheels B.

SOLUTION


Free-Body Diagram:

(a)

Front wheels:

ΣM B = 0: (1700 lb)(52 in.) + (3200 lb)(12 in.) − 2 A(36 in.) = 0 A = +1761.11 lb

(b)

Rear wheels:

A = 1761 lb 

ΣFy = 0: − 1700 lb − 3200 lb + 2(1761.11 lb) + 2 B = 0 B = +688.89 lb

B = 689 lb 


www.elsolucionario.net PROBLEM 4.8 For the beam and loading shown, determine (a) the reaction at A, (b) the tension in cable BC.

SOLUTION

(a)

Reaction at A:

ΣFx = 0: Ax = 0 ΣMB = 0: (15 lb)(28 in.) + (20 lb)(22 in.) + (35 lb)(14 in.) + (20 lb)(6 in.) − Ay (6 in.) = 0 Ay = +245 lb

(b)

Tension in BC:

A = 245 lb 

ΣM A = 0: (15 lb)(22 in.) + (20 lb)(16 in.) + (35 lb)(8 in.) − (15 lb)(6 in.) − FBC (6 in.) = 0 FBC = +140.0 lb

Check:

FBC = 140.0 lb 

ΣFy = 0: − 15 lb − 20 lb = 35 lb − 20 lb + A − FBC = 0 −105 lb + 245 lb − 140.0 = 0 0 = 0 (Checks)



Free-Body Diagram:

www.elsolucionario.net PROBLEM 4.9 For the beam and loading shown, determine the range of the distance a for which the reaction at B does not exceed 100 lb downward or 200 lb upward.

SOLUTION Assume B is positive when directed .

ΣM D = 0: (300 lb)(8 in. − a ) − (300 lb)(a − 2 in.) − (50 lb)(4 in.) + 16 B = 0 −600a + 2800 + 16B = 0 (2800 + 16B) 600

(1)

[2800 + 16( −100)] 1200 = = 2 in. 600 600

a ≥ 2.00 in. 

a=

For B = 100 lb = −100 lb, Eq. (1) yields: a≥

For B = 200 = +200 lb, Eq. (1) yields: a≤

Required range:

[2800 + 16(200)] 6000 = = 10 in. 600 600 2.00 in. ≤ a ≤ 10.00 in.

a ≤ 10.00 in. 





Sketch showing distance from D to forces.

www.elsolucionario.net PROBLEM 4.10 The maximum allowable value of each of the reactions is 180 N. Neglecting the weight of the beam, determine the range of the distance d for which the beam is safe.

SOLUTION

B = By ΣM A = 0: (50 N) d − (100 N)(0.45 m − d ) − (150 N)(0.9 m − d ) + B(0.9 m − d ) = 0 50d − 45 + 100d − 135 + 150d + 0.9 B − Bd = 0 d=

180 N ⋅ m − (0.9 m) B 300 A − B

(1)

ΣM B = 0: (50 N)(0.9 m) − A(0.9 m − d ) + (100 N)(0.45 m) = 0 45 − 0.9 A + Ad + 45 = 0 (0.9 m) A − 90 N ⋅ m A

(2)

d≥

180 − (0.9)180 18 = = 0.15 m 300 − 180 120

d ≥ 150.0 mm 

d≤

(0.9)180 − 90 72 = = 0.40 m 180 180

d=

Since B ≤ 180 N, Eq. (1) yields

Since A ≤ 180 N, Eq. (2) yields

Range:

150.0 mm ≤ d ≤ 400 mm

d ≤ 400 mm 





ΣFx = 0: Bx = 0

www.elsolucionario.net PROBLEM 4.11 Three loads are applied as shown to a light beam supported by cables attached at B and D. Neglecting the weight of the beam, determine the range of values of Q for which neither cable becomes slack when P = 0.

ΣM B = 0: (3.00 kN)(0.500 m) + TD (2.25 m) − Q (3.00 m) = 0 Q = 0.500 kN + (0.750) TD

(1)

ΣM D = 0: (3.00 kN)(2.75 m) − TB (2.25 m) − Q(0.750 m) = 0 Q = 11.00 kN − (3.00) TB

(2)

For cable B not to be slack, TB ≥ 0, and from Eq. (2), Q ≤ 11.00 kN

For cable D not to be slack, TD ≥ 0, and from Eq. (1), Q ≥ 0.500 kN

For neither cable to be slack, 0.500 kN ≤ Q ≤ 11.00 kN 



SOLUTION

www.elsolucionario.net PROBLEM 4.12 Three loads are applied as shown to a light beam supported by cables attached at B and D. Knowing that the maximum allowable tension in each cable is 4 kN and neglecting the weight of the beam, determine the range of values of Q for which the loading is safe when P = 0.

ΣM B = 0: (3.00 kN)(0.500 m) + TD (2.25 m) − Q (3.00 m) = 0 Q = 0.500 kN + (0.750) TD

(1)

ΣM D = 0: (3.00 kN)(2.75 m) − TB (2.25 m) − Q(0.750 m) = 0 Q = 11.00 kN − (3.00) TB

(2)

For TB ≤ 4.00 kN, Eq. (2) yields Q ≥ 11.00 kN − 3.00(4.00 kN)

Q ≥ −1.000 kN

For TD ≤ 4.00 kN, Eq. (1) yields Q ≤ 0.500 kN + 0.750(4.00 kN)

Q ≤ 3.50 kN

For loading to be safe, cables must also not be slack. Combining with the conditions obtained in Problem 4.11, 0.500 kN ≤ Q ≤ 3.50 kN 



SOLUTION

www.elsolucionario.net PROBLEM 4.13 For the beam of Problem 4.12, determine the range of values of Q for which the loading is safe when P = 1 kN. PROBLEM 4.12 Three loads are applied as shown to a light beam supported by cables attached at B and D. Knowing that the maximum allowable tension in each cable is 4 kN and neglecting the weight of the beam, determine the range of values of Q for which the loading is safe when P = 0.

ΣM B = 0: (3.00 kN)(0.500 m) − (1.000 kN)(0.750 m) + TD (2.25 m) − Q(3.00 m) = 0 Q = 0.250 kN + 0.75 TD

(1)

ΣM D = 0: (3.00 kN)(2.75 m) + (1.000 kN)(1.50 m) − TB (2.25 m) − Q (0.750 m) = 0 Q = 13.00 kN − 3.00 TB

(2)

For the loading to be safe, cables must not be slack and tension must not exceed 4.00 kN. Making 0 ≤ TB ≤ 4.00 kN in Eq. (2), we have 13.00 kN − 3.00(4.00 kN) ≤ Q ≤ 13.00 kN − 3.00(0) 1.000 kN ≤ Q ≤ 13.00 kN

(3)

Making 0 ≤ TD ≤ 4.00 kN in Eq. (1), we have 0.250 kN + 0.750(0) ≤ Q ≤ 0.250 kN + 0.750(4.00 kN) 0.250 kN ≤ Q ≤ 3.25 kN

(4) 1.000 kN ≤ Q ≤ 3.25 kN 

Combining Eqs. (3) and (4),



SOLUTION

www.elsolucionario.net PROBLEM 4.14 For the beam of Sample Problem 4.2, determine the range of values of P for which the beam will be safe, knowing that the maximum allowable value of each of the reactions is 30 kips and that the reaction at A must be directed upward.

SOLUTION ΣFx = 0: Bx = 0

B = By ΣM A = 0: − P(3 ft) + B(9 ft) − (6 kips)(11 ft) − (6 kips)(13 ft) = 0 P = 3B − 48 kips

(1)

P = 1.5 A + 6 kips

(2)

Since B ≤ 30 kips, Eq. (1) yields P ≤ (3)(30 kips) − 48 kips

P ≤ 42.0 kips 

Since 0 ≤ A ≤ 30 kips, Eq. (2) yields 0 + 6 kips ≤ P ≤ (1.5)(30 kips)1.6 kips 6.00 kips ≤ P ≤ 51.0 kips



Range of values of P for which beam will be safe: 6.00 kips ≤ P ≤ 42.0 kips





ΣM B = 0: − A(9 ft) + P (6 ft) − (6 kips)(2 ft) − (6 kips)(4 ft) = 0

www.elsolucionario.net PROBLEM 4.15 The bracket BCD is hinged at C and attached to a control cable at B. For the loading shown, determine (a) the tension in the cable, (b) the reaction at C.

SOLUTION

Ty Tx

=

0.18 m 0.24 m

3 Ty = Tx 4

(a)

(1)

ΣM C = 0: Tx (0.18 m) − (240 N)(0.4 m) − (240 N)(0.8 m) = 0 Tx = +1600 N

From Eq. (1):

Ty =

3 (1600 N) = 1200 N 4

T = Tx2 + Ty2 = 16002 + 12002 = 2000 N

(b)

T = 2.00 kN 

ΣFx = 0: Cx − Tx = 0 Cx − 1600 N = 0 C x = +1600 N

C x = 1600 N

ΣFy = 0: C y − Ty − 240 N − 240 N = 0 C y − 1200 N − 480 N = 0 C y = +1680 N

C y = 1680 N

α = 46.4° C = 2320 N

C = 2.32 kN

46.4° 



At B:

www.elsolucionario.net PROBLEM 4.16 Solve Problem 4.15, assuming that a = 0.32 m. PROBLEM 4.15 The bracket BCD is hinged at C and attached to a control cable at B. For the loading shown, determine (a) the tension in the cable, (b) the reaction at C.

At B:

Ty Tx

=

0.32 m 0.24 m

4 Ty = Tx 3 ΣM C = 0: Tx (0.32 m) − (240 N)(0.4 m) − (240 N)(0.8 m) = 0 Tx = 900 N

From Eq. (1):

Ty =

4 (900 N) = 1200 N 3

T = Tx2 + Ty2 = 9002 + 12002 = 1500 N

T = 1.500 kN 

ΣFx = 0: C x − Tx = 0 C x − 900 N = 0 C x = +900 N

C x = 900 N

ΣFy = 0: C y − Ty − 240 N − 240 N = 0 C y − 1200 N − 480 N = 0 C y = +1680 N

C y = 1680 N

α = 61.8° C = 1906 N

C = 1.906 kN

61.8° 



SOLUTION

www.elsolucionario.net PROBLEM 4.17 The lever BCD is hinged at C and attached to a control rod at B. If P = 100 lb, determine (a) the tension in rod AB, (b) the reaction at C.

SOLUTION

(a)


Free-Body Diagram:

ΣM C = 0: T (5 in.) − (100 lb)(7.5 in.) = 0 T = 150.0 lb 

(b)

3 ΣFx = 0: C x + 100 lb + (150.0 lb) = 0 5 C x = −190 lb

C x = 190 lb

4 ΣFy = 0: C y + (150.0 lb) = 0 5

C y = −120 lb

C y = 120 lb

α = 32.3°

C = 225 lb

C = 225 lb

32.3° 


www.elsolucionario.net PROBLEM 4.18 The lever BCD is hinged at C and attached to a control rod at B. Determine the maximum force P that can be safely applied at D if the maximum allowable value of the reaction at C is 250 lb.

SOLUTION


Free-Body Diagram:

ΣM C = 0: T (5 in.) − P (7.5 in.) = 0 T = 1.5P 3 ΣFx = 0: P + C x + (1.5P) = 0 5 C x = −1.9 P ΣFy = 0: C y +

C x = 1.9 P

4 (1.5P) = 0 5

C y = −1.2 P

C y = 1.2 P

C = C x2 + C y2 = (1.9 P) 2 + (1.2 P) 2 C = 2.2472 P

For C = 250 lb, 250 lb = 2.2472P P = 111.2 lb

P = 111.2 lb




www.elsolucionario.net PROBLEM 4.19 Two links AB and DE are connected by a bell crank as shown. Knowing that the tension in link AB is 720 N, determine (a) the tension in link DE, (b) the reaction at C.

SOLUTION Free-Body Diagram: ΣM C = 0: FAB (100 mm) − FDE (120 mm) = 0

(a)

For

(1)

FAB = 720 N FDE =

(b)

5 FAB 6

5 (720 N) 6

FDE = 600 N 

3 ΣFx = 0: − (720 N) + C x = 0 5 C x = +432 N 4 ΣFy = 0: − (720 N) + C y − 600 N = 0 5 C y = +1176 N C = 1252.84 N α = 69.829° C = 1253 N

69.8° 



FDE =

www.elsolucionario.net PROBLEM 4.20 Two links AB and DE are connected by a bell crank as shown. Determine the maximum force that may be safely exerted by link AB on the bell crank if the maximum allowable value for the reaction at C is 1600 N.

SOLUTION See solution to Problem 4.15 for F.B.D. and derivation of Eq. (1). 5 FAB 6

(1)

3 ΣFx = 0: − FAB + C x = 0 5 ΣFy = 0: −

Cx =

3 FAB 5

4 FAB + C y − FDE = 0 5

4 5 − FAB + C y − FAB = 0 5 6 49 Cy = FAB 30

C = C x2 + C y2 1 (49) 2 + (18) 2 FAB 30 C = 1.74005FAB =

For C = 1600 N, 1600 N = 1.74005FAB

FAB = 920 N 



FDE =

www.elsolucionario.net PROBLEM 4.21 Determine the reactions at A and C when (a) α = 0, (b) α = 30°.

SOLUTION (a)

α =0 From F.B.D. of member ABC: ΣM C = 0: (300 N)(0.2 m) + (300 N)(0.4 m) − A(0.8 m) = 0

or

A = 225 N 

ΣFy = 0: C y + 225 N = 0 C y = −225 N or C y = 225 N ΣFx = 0: 300 N + 300 N + C x = 0 C x = −600 N or C x = 600 N

Then

C = C x2 + C y2 = (600) 2 + (225) 2 = 640.80 N

and

θ = tan −1 

 Cy  −1  −225   = tan   = 20.556°  −600   Cx 

or (b)

C = 641 N

20.6° 

α = 30° From F.B.D. of member ABC: ΣM C = 0: (300 N)(0.2 m) + (300 N)(0.4 m) − ( A cos 30°)(0.8 m) + ( A sin 30°)(20 in.) = 0 A = 365.24 N

or

A = 365 N

60.0° 

ΣFx = 0: 300 N + 300 N + (365.24 N) sin 30° + C x = 0 C x = −782.62



A = 225 N


ΣFy = 0: C y + (365.24 N) cos 30° = 0 C y = −316.31 N or C y = 316 N

Then

C = C x2 + C y2 = (782.62) 2 + (316.31) 2 = 884.12 N

and

θ = tan −1 

 Cy  −1  −316.31   = tan   = 22.007°  −782.62   Cx  C = 884 N

22.0° 


or


www.elsolucionario.net PROBLEM 4.22 Determine the reactions at A and B when (a) α = 0, (b) α = 90°, (c) α = 30°.

SOLUTION (a)

α =0 ΣM A = 0: B(20 in.) − 75 lb(10 in.) = 0 ΣFx = 0: Ax = 0 + ΣFy = 0: Ay − 75 lb + 37.5 lb = 0

Ay = 37.5 lb A = B = 37.5 lb 

(b)

α = 90° ΣM A = 0: B(12 in.) − 75 lb(10 in.) = 0

B = 62.5 lb ΣFx = 0: Ax − B = 0

Ax = 62.5 lb ΣFy = 0: Ay − 75 lb = 0

Ay = 75 lb A = Ax2 + Ay2 = (62.5 lb) 2 + (75 lb) 2 = 97.6 lb

75 62.5 θ = 50.2°

tan θ =

A = 97.6 lb

50.2°; B = 62.51 lb



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B = 37.5 lb

www.elsolucionario.net PROBLEM 4.22 (Continued) (c)

α = 30° ΣM A = 0: ( B cos 30°)(20 in.) + ( B sin 30°)(12 in.) − (75 lb)(10 in.) = 0 B = 32.161 lb

ΣFx = 0: Ax − (32.161) sin 30° = 0 Ax = 16.0805 lb

ΣFy = 0: Ay + (32.161) cos 30° − 75 = 0 Ay = 47.148 lb A = Ax2 + Ay2

= (16.0805) 2 + (47.148) 2 47.148 16.0805 θ = 71.2°

tan θ =

A = 49.8 lb

71.2°; B = 32.2 lb

60.0° 



= 49.8 lb

www.elsolucionario.net PROBLEM 4.23 Determine the reactions at A and B when (a) h = 0, (b) h = 200 mm.

SOLUTION

ΣM A = 0: ( B cos 60°)(0.5 m) − ( B sin 60°)h − (150 N)(0.25 m) = 0 37.5 B= 0.25 − 0.866h

(a)

(1)

When h = 0, B=

From Eq. (1):

37.5 = 150 N 0.25

B = 150.0 N

30.0° 

ΣFy = 0: Ax − B sin 60° = 0 Ax = (150)sin 60° = 129.9 N

A x = 129.9 N

ΣFy = 0: Ay − 150 + B cos 60° = 0 Ay = 150 − (150) cos 60° = 75 N

A y = 75 N

α = 30° A = 150.0 N

(b)

A = 150.0 N

30.0° 

When h = 200 mm = 0.2 m, From Eq. (1):

B=

37.5 = 488.3 N 0.25 − 0.866(0.2)

B = 488 N

30.0° 

ΣFx = 0: Ax − B sin 60° = 0 Ax = (488.3) sin 60° = 422.88 N

A x = 422.88 N

ΣFy = 0: Ay − 150 + B cos 60° = 0 Ay = 150 − (488.3) cos 60° = −94.15 N

A y = 94.15 N

α = 12.55° A = 433.2 N

A = 433 N

12.55° 



Free-Body Diagram:

www.elsolucionario.net PROBLEM 4.24 A lever AB is hinged at C and attached to a control cable at A. If the lever is subjected to a 75-lb vertical force at B, determine (a) the tension in the cable, (b) the reaction at C.

SOLUTION Free-Body Diagram: x AC = (10 in.) cos 20° = 9.3969 in. y AC = (10 in.)sin 20° = 3.4202 in.  yDA = 12 in. − 3.4202 in. = 8.5798 in.  yDA  −1  8.5798   = tan   = 42.397°  9.3969   x AC 

α = tan −1 

β = 90° − 20° − 42.397° = 27.603° Equilibrium for lever: ΣM C = 0: TAD cos 27.603°(10 in.) − (75 lb)[(15 in.)cos 20°] = 0

(a)

TAD = 119.293 lb

TAD = 119.3 lb 

ΣFx = 0: C x + (119.293 lb) cos 42.397° = 0

(b)

C x = −88.097 lb

ΣFy = 0: C y − 75 lb − (119.293 lb) sin 42.397° = 0 C y = 155.435

Thus,

C = C x2 + C y2 = (−88.097) 2 + (155.435) 2 = 178.665 lb

and

θ = tan −1

Cy Cx

= tan −1

155.435 = 60.456° 88.097

C = 178.7 lb

60.5° 



Geometry:

www.elsolucionario.net PROBLEM 4.25 For each of the plates and loadings shown, determine the reactions at A and B.


ΣM A = 0: B(20 in.) − (50 lb)(4 in.) − (40 lb)(10 in.) = 0 B = +30 lb

B = 30.0 lb 

ΣFx = 0: Ax + 40 lb = 0 Ax = −40 lb

A x = 40.0 lb

ΣFy = 0: Ay + B − 50 lb = 0 Ay + 30 lb − 50 lb = 0 Ay = +20 lb

A y = 20.0 lb

α = 26.56° A = 44.72 lb

A = 44.7 lb

26.6° 

  



(a)


(b)

Free-Body Diagram:

ΣM A = 0: ( B cos 30°)(20 in.) − (40 lb)(10 in.) − (50 lb)(4 in.) = 0 B = 34.64 lb

B = 34.6 lb

60.0° 

ΣFx = 0: Ax − B sin 30° + 40 lb

Ax = −22.68 lb

A x = 22.68 lb

ΣFy = 0: Ay + B cos 30° − 50 lb = 0 Ay + (34.64 lb) cos 30° − 50 lb = 0 Ay = +20 lb

A y = 20.0 lb

α = 41.4° A = 30.2 lb

A = 30.24 lb

41.4° 



Ax − (34.64 lb) sin 30° + 40 lb = 0

www.elsolucionario.net PROBLEM 4.26 For each of the plates and loadings shown, determine the reactions at A and B.

(a)

Free-Body Diagram:

ΣM B = 0: A(20 in.) + (50 lb)(16 in.) − (40 lb)(10 in.) = 0 A = +20 lb

A = 20.0 lb 

ΣFx = 0: 40 lb + Bx = 0 Bx = −40 lb

B x = 40 lb

ΣFy = 0: A + By − 50 lb = 0 20 lb + By − 50 lb = 0 By = +30 lb

α = 36.87°

B = 50 lb

B y = 30 lb B = 50.0 lb

36.9° 



SOLUTION


(b)

ΣM A = 0: − ( A cos 30°)(20 in.) − (40 lb)(10 in.) + (50 lb)(16 in.) = 0 A = 23.09 lb

A = 23.1 lb

60.0° 

ΣFx = 0: A sin 30° + 40 lb + Bx = 0

Bx = −51.55 lb

B x = 51.55 lb

ΣFy = 0: A cos 30° + By − 50 lb = 0 (23.09 lb) cos 30° + By − 50 lb = 0

By = +30 lb

B y = 30 lb

α = 30.2° B = 59.64 lb

B = 59.6 lb

30.2° 



(23.09 lb) sin 30° + 40 lb + 8 x = 0

www.elsolucionario.net PROBLEM 4.27 A rod AB hinged at A and attached at B to cable BD supports the loads shown. Knowing that d = 200 mm, determine (a) the tension in cable BD, (b) the reaction at A.

SOLUTION

(a)

Move T along BD until it acts at Point D. ΣM A = 0: (T sin 45°)(0.2 m) + (90 N)(0.1 m) + (90 N)(0.2 m) = 0 T = 190.919 N

(b)

T = 190.9 N 

ΣFx = 0: Ax − (190.919 N) cos 45° = 0 Ax = +135.0 N

A x = 135.0 N

ΣFy = 0: Ay − 90 N − 90 N + (190.919 N) sin 45° = 0 Ay = +45.0 N

A y = 45.0 N A = 142.3 N

18.43°



Free-Body Diagram:

www.elsolucionario.net PROBLEM 4.28 A rod AB, hinged at A and attached at B to cable BD, supports the loads shown. Knowing that d = 150 mm, determine (a) the tension in cable BD, (b) the reaction at A.

SOLUTION

tan α =

(a)

10 ; α = 33.690° 15

Move T along BD until it acts at Point D.

ΣM A = 0: (T sin 33.690°)(0.15 m) − (90 N)(0.1 m) − (90 N)(0.2 m) = 0

T = 324.50 N

(b)

T = 324 N 

ΣFx = 0: Ax − (324.50 N) cos 33.690° = 0 Ax = +270 N

A x = 270 N

ΣFy = 0: Ay − 90 N − 90 N + (324.50 N) sin 33.690° = 0 Ay = 0

A = 270 N





Free-Body Diagram:

www.elsolucionario.net PROBLEM 4.29 A force P of magnitude 90 lb is applied to member ACDE, which is supported by a frictionless pin at D and by the cable ABE. Since the cable passes over a small pulley at B, the tension may be assumed to be the same in portions AB and BE of the cable. For the case when a = 3 in., determine (a) the tension in the cable, (b) the reaction at D.

SOLUTION

(a)

ΣM D = 0: (90 lb)(9 in.) −

5 12 T (9 in.) − T (7 in.) + T (3 in.) = 0 13 13

T = 117 lb

(b)

ΣFx = 0: Dx − 117 lb −

T = 117.0 lb 

5 (117 lb) + 90 = 0 13

Dx = +72 lb

ΣFy = 0: D y +

12 (117 lb) = 0 13 Dy = −108 lb

D = 129.8 lb

56.3° 





Free-Body Diagram:

www.elsolucionario.net PROBLEM 4.30 Solve Problem 4.29 for a = 6 in. PROBLEM 4.29 A force P of magnitude 90 lb is applied to member ACDE, which is supported by a frictionless pin at D and by the cable ABE. Since the cable passes over a small pulley at B, the tension may be assumed to be the same in portions AB and BE of the cable. For the case when a = 3 in., determine (a) the tension in the cable, (b) the reaction at D.

SOLUTION

(a)

ΣM D = 0: (90 lb)(6 in.) −

5 12 T (6 in.) − T (7 in.) + T (6 in.) = 0 13 13

T = 195 lb

(b)

T = 195.0 lb 

ΣFx = 0: Dx − 195 lb −

5 (195 lb) + 90 = 0 13

Dx = +180 lb

ΣFy = 0: D y +

12 (195 lb) = 0 13 Dy = −180 lb

D = 255 lb

45.0° 



Free-Body Diagram:

www.elsolucionario.net PROBLEM 4.31 Neglecting friction, determine the tension in cable ABD and the reaction at support C.

SOLUTION

ΣM C = 0: T (0.25 m) − T (0.1 m) − (120 N)(0.1 m) = 0 ΣFx = 0: C x − 80 N = 0

C x = +80 N

ΣFy = 0: C y − 120 N + 80 N = 0

C y = +40 N

T = 80.0 N  C x = 80.0 N C y = 40.0 N C = 89.4 N

26.6° 



Free-Body Diagram:

www.elsolucionario.net PROBLEM 4.32 Neglecting friction and the radius of the pulley, determine (a) the tension in cable ADB, (b) the reaction at C.

SOLUTION Free-Body Diagram: Dimensions in mm

Distance:

AD = (0.36) 2 + (0.150) 2 = 0.39 m

Distance:

BD = (0.2)2 + (0.15)2 = 0.25 m

Equilibrium for beam: ΣM C = 0:

(a)

 0.15   0.15  (120 N)(0.28 m) −  T  (0.36 m) −  T  (0.2 m) = 0  0.39   0.25 

T = 130.000 N

or

T = 130.0 N 

 0.36   0.2  ΣFx = 0: C x +   (130.000 N) +  0.25  (130.000 N) = 0 0.39    

(b)

C x = − 224.00 N  0.15   0.15  (130.00 N) +  ΣFy = 0: C y +    (130.00 N) − 120 N = 0  0.39   0.25 

C y = − 8.0000 N

Thus,

C = C x2 + C y2 = (−224) 2 + (− 8) 2 = 224.14 N

and

θ = tan −1

Cy Cx

= tan −1

8 = 2.0454° 224

C = 224 N

2.05° 



Geometry:

www.elsolucionario.net PROBLEM 4.33 Rod ABC is bent in the shape of an arc of circle of radius R. Knowing the θ = 30°, determine the reaction (a) at B, (b) at C.


ΣM D = 0: C x ( R) − P( R) = 0

ΣFx = 0: C x − B sin θ = 0 P − B sin θ = 0 B = P/sin θ B=

P sin θ

θ

ΣFy = 0: C y + B cos θ − P = 0 C y + ( P/sin θ ) cos θ − P = 0 1   C y = P 1 −  tan θ  

For θ = 30°, (a) (b)

B = P/sin 30° = 2 P Cx = + P

B = 2P

60.0° 

Cx = P

C y = P(1 − 1/tan 30°) = − 0.732/P

C y = 0.7321P C = 1.239P

36.2° 



Cx = + P

www.elsolucionario.net PROBLEM 4.34 Rod ABC is bent in the shape of an arc of circle of radius R. Knowing that θ = 60°, determine the reaction (a) at B, (b) at C.

SOLUTION See the solution to Problem 4.33 for the free-body diagram and analysis leading to the following expressions: 1   C y = P 1 − tan θ   P B= sin θ

For θ = 60°, (a) (b)

B = P/sin 60° = 1.1547 P Cx = + P

B = 1.155P

30.0° 

Cx = P

C y = P(1 − 1/tan 60°) = + 0.4226 P

C y = 0.4226 P C = 1.086P

22.9° 



Cx = + P

www.elsolucionario.net PROBLEM 4.35 A movable bracket is held at rest by a cable attached at C and by frictionless rollers at A and B. For the loading shown, determine (a) the tension in the cable, (b) the reactions at A and B.

SOLUTION

(a)

ΣFy = 0: T − 600 N = 0 T = 600 N 

(b)

ΣFx = 0: B − A = 0

∴ B=A

Note that the forces shown form two couples. ΣM = 0: (600 N)(600 mm) − A(90 mm) = 0 A = 4000 N ∴ B = 4000 N A = 4.00 kN

; B = 4.00 kN





Free-Body Diagram:

www.elsolucionario.net PROBLEM 4.36 A light bar AB supports a 15-kg block at its midpoint C. Rollers at A and B rest against frictionless surfaces, and a horizontal cable AD is attached at A. Determine (a) the tension in cable AD, (b) the reactions at A and B.

SOLUTION

W = (15 kg)(9.81 m/s 2 ) = 147.150 N

(a)

ΣFx = 0: TAD − 105.107 N = 0 TAD = 105.1 N 

(b)

ΣFy = 0: A − W = 0 A − 147.150 N = 0 A = 147.2 N 

ΣM A = 0: B(350 mm) − (147.150 N) (250 mm) = 0 B = 105.107 N B = 105.1 N





Free-Body Diagram:

www.elsolucionario.net PROBLEM 4.37 A light bar AD is suspended from a cable BE and supports a 50-lb block at C. The ends A and D of the bar are in contact with frictionless vertical walls. Determine the tension in cable BE and the reactions at A and D.

SOLUTION

ΣFx = 0:

A= D

ΣFy = 0:

TBE = 50.0 lb 

We note that the forces shown form two couples. ΣM = 0: A(8 in.) − (50 lb)(3 in.) = 0 A = 18.75 lb A = 18.75 lb

D = 18.75 lb





Free-Body Diagram:

www.elsolucionario.net PROBLEM 4.38 A light rod AD is supported by frictionless pegs at B and C and rests against a frictionless wall at A. A vertical 120-lb force is applied at D. Determine the reactions at A, B, and C.

SOLUTION

ΣFx = 0: A cos 30° − (120 lb) cos 60° = 0 A = 69.28 lb

A = 69.3 lb



ΣM B = 0: C (8 in.) − (120 lb)(16 in.) cos 30° + (69.28 lb)(8 in.)sin 30° = 0 C = 173.2 lb

C = 173.2 lb

60.0° 

B = 34.6 lb

60.0° 

ΣM C = 0: B(8 in.) − (120 lb)(8 in.) cos 30° + (69.28 lb)(16 in.) sin 30° = 0 B = 34.6 lb

Check:

ΣFy = 0: 173.2 − 34.6 − (69.28)sin 30° − (120)sin 60° = 0 0 = 0 (check) 



Free-Body Diagram:

www.elsolucionario.net PROBLEM 4.39 Bar AD is attached at A and C to collars that can move freely on the rods shown. If the cord BE is vertical (α = 0), determine the tension in the cord and the reactions at A and C.

SOLUTION

ΣFy = 0: − T cos 30° + (80 N) cos 30° = 0 T = 80 N

T = 80.0 N 

ΣM C = 0: ( A sin 30°)(0.4 m) − (80 N)(0.2 m) − (80 N)(0.2 m) = 0 A = + 160 N

A = 160.0 N

30.0° 

C = 160.0 N

30.0° 

ΣM A = 0: (80 N)(0.2 m) − (80 N)(0.6 m) + (C sin 30°)(0.4 m) = 0 C = + 160 N



Free-Body Diagram:

www.elsolucionario.net PROBLEM 4.40 Solve Problem 4.39 if the cord BE is parallel to the rods (α = 30°). PROBLEM 4.39 Bar AD is attached at A and C to collars that can move freely on the rods shown. If the cord BE is vertical (α = 0), determine the tension in the cord and the reactions at A and C.

SOLUTION

ΣFy = 0: − T + (80 N) cos 30° = 0 T = 69.282 N

T = 69.3 N 

ΣM C = 0: − (69.282 N) cos 30°(0.2 m) − (80 N)(0.2 m) + ( A sin 30°)(0.4 m) = 0 A = + 140.000 N

A = 140.0 N

30.0° 

C = 180.0 N

30.0° 

ΣM A = 0: + (69.282 N) cos 30°(0.2 m) − (80 N)(0.6 m) + (C sin 30°)(0.4 m) = 0 C = + 180.000 N



Free-Body Diagram:

www.elsolucionario.net PROBLEM 4.41 The T-shaped bracket shown is supported by a small wheel at E and pegs at C and D. Neglecting the effect of friction, determine the reactions at C, D, and E when θ = 30°.

SOLUTION Free-Body Diagram: ΣFy = 0: E cos 30° − 20 − 40 = 0 60 lb = 69.282 lb cos 30°

E = 69.3 lb

60.0° 

ΣM D = 0: (20 lb)(4 in.) − (40 lb)(4 in.) − C (3 in.) + E sin 30°(3 in.) = 0 −80 − 3C + 69.282(0.5)(3) = 0 C = 7.9743 lb

C = 7.97 lb



D = 42.6 lb



ΣFx = 0: E sin 30° + C − D = 0 (69.282 lb)(0.5) + 7.9743 lb − D = 0 D = 42.615 lb



E=

www.elsolucionario.net PROBLEM 4.42 The T-shaped bracket shown is supported by a small wheel at E and pegs at C and D. Neglecting the effect of friction, determine (a) the smallest value of θ for which the equilibrium of the bracket is maintained, (b) the corresponding reactions at C, D, and E.

SOLUTION Free-Body Diagram: ΣFy = 0: E cos θ − 20 − 40 = 0 60 cos θ

(1)

ΣM D = 0: (20 lb)(4 in.) − (40 lb)(4 in.) − C (3 in.)  60  + sin θ  3 in. = 0  cos θ  1 C = (180 tan θ − 80) 3

(a)

For C = 0,

180 tan θ = 80 tan θ =

From Eq. (1):

E=

4 θ = 23.962° 9

θ = 24.0° 

60 = 65.659 cos 23.962°

ΣFx = 0: −D + C + E sin θ = 0 D = (65.659) sin 23.962 = 26.666 lb

(b)

C = 0 D = 26.7 lb

E = 65.7 lb

66.0° 



E=

www.elsolucionario.net PROBLEM 4.43 Beam AD carries the two 40-lb loads shown. The beam is held by a fixed support at D and by the cable BE that is attached to the counterweight W. Determine the reaction at D when (a) W = 100 lb, (b) W = 90 lb.

SOLUTION W = 100 lb

(a) From F.B.D. of beam AD:

ΣFx = 0: Dx = 0

Dy = −20.0 lb

or

D = 20.0 lb 

ΣM D = 0: M D − (100 lb)(5 ft) + (40 lb)(8 ft) + (40 lb)(4 ft) = 0 M D = 20.0 lb ⋅ ft

or M D = 20.0 lb ⋅ ft



W = 90 lb

(b) From F.B.D. of beam AD:

ΣFx = 0: Dx = 0 ΣFy = 0: D y + 90 lb − 40 lb − 40 lb = 0 Dy = −10.00 lb

or

D = 10.00 lb 

ΣM D = 0: M D − (90 lb)(5 ft) + (40 lb)(8 ft) + (40 lb)(4 ft) = 0 M D = −30.0 lb ⋅ ft

or M D = 30.0 lb ⋅ ft





ΣFy = 0: D y − 40 lb − 40 lb + 100 lb = 0

www.elsolucionario.net PROBLEM 4.44 For the beam and loading shown, determine the range of values of W for which the magnitude of the couple at D does not exceed 40 lb ⋅ ft.

SOLUTION For Wmin , From F.B.D. of beam AD:

M D = − 40 lb ⋅ ft ΣM D = 0: (40 lb)(8 ft) − Wmin (5 ft) + (40 lb)(4 ft) − 40 lb ⋅ ft = 0

For Wmax , From F.B.D. of beam AD:

M D = 40 lb ⋅ ft ΣM D = 0: (40 lb)(8 ft) − Wmax (5 ft) + (40 lb)(4 ft) + 40 lb ⋅ ft = 0 Wmax = 104.0 lb

or 88.0 lb ≤ W ≤ 104.0 lb 



Wmin = 88.0 lb

www.elsolucionario.net PROBLEM 4.45 An 8-kg mass can be supported in the three different ways shown. Knowing that the pulleys have a 100-mm radius, determine the reaction at A in each case.

SOLUTION  

(a)

ΣFx = 0: Ax = 0 ΣFy = 0: Ay − W = 0



A y = 78.480 N

ΣM A = 0: M A − W (1.6 m) = 0



M A = + (78.480 N)(1.6 m)



M A = 125.568 N ⋅ m

A = 78.5 N

 

(b)

M A = 125.6 N ⋅ m

ΣFx = 0: Ax − W = 0

A x = 78.480

ΣFy = 0: Ay − W = 0

A y = 78.480

A = (78.480 N) 2 = 110.987 N



45°

ΣM A = 0: M A − W (1.6 m) = 0



M A = + (78.480 N)(1.6 m)

 

A = 111.0 N





(c)

M A = 125.568 N ⋅ m M A = 125.6 N ⋅ m

45°



ΣFx = 0: Ax = 0 ΣFy = 0: Ay − 2W = 0

 

Ay = 2W = 2(78.480 N) = 156.960 N ΣM A = 0: M A − 2W (1.6 m) = 0 M A = + 2(78.480 N)(1.6 m)

A = 157.0 N

M A = 251.14 N ⋅ m M A = 251 N ⋅ m





W = mg = (8 kg)(9.81 m/s2 ) = 78.480 N

www.elsolucionario.net PROBLEM 4.46 A tension of 20 N is maintained in a tape as it passes through the support system shown. Knowing that the radius of each pulley is 10 mm, determine the reaction at C.

SOLUTION

ΣFx = 0: C x + (20 N) = 0

C x = −20 N

ΣFy = 0: C y − (20 N) = 0

C y = +20 N

C = 28.3 N

45.0° 

ΣM C = 0: M C + (20 N)(0.160 m) + (20 N) (0.055 m) = 0 M C = −4.30 N ⋅ m

M C = 4.30 N ⋅ m





Free-Body Diagram:

www.elsolucionario.net PROBLEM 4.47 Solve Problem 4.46, assuming that 15-mm-radius pulleys are used. PROBLEM 4.46 A tension of 20 N is maintained in a tape as it passes through the support system shown. Knowing that the radius of each pulley is 10 mm, determine the reaction at C.

SOLUTION

ΣFx = 0: C x + (20 N) = 0

C x = −20 N

ΣFy = 0: C y − (20 N) = 0

C y = +20 N

C = 28.3 N

45.0° 

ΣM C = 0: M C + (20 N) (0.165 m) + (20 N) (0.060 m) = 0 M C = −4.50 N ⋅ m

M C = 4.50 N ⋅ m





Free-Body Diagram:

www.elsolucionario.net PROBLEM 4.48 The rig shown consists of a 1200-lb horizontal member ABC and a vertical member DBE welded together at B. The rig is being used to raise a 3600-lb crate at a distance x = 12 ft from the vertical member DBE. If the tension in the cable is 4 kips, determine the reaction at E, assuming that the cable is (a) anchored at F as shown in the figure, (b) attached to the vertical member at a point located 1 ft above E.

SOLUTION

M E = 0: M E + (3600 lb) x + (1200 lb) (6.5 ft) − T (3.75 ft) = 0 M E = 3.75T − 3600 x − 7800

(a)

(1)

For x = 12 ft and T = 4000 lbs, M E = 3.75(4000) − 3600(12) − 7800 = 36, 000 lb ⋅ ft ΣFx = 0 ∴ Ex = 0 ΣFy = 0:

E y − 3600 lb − 1200 lb − 4000 = 0

E y = 8800 lb

E = 8.80 kips ; M E = 36.0 kip ⋅ ft



   PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 396


Free-Body Diagram:


ΣM E = 0: M E + (3600 lb)(12 ft) + (1200 lb)(6.5 ft) = 0

 



M E = −51, 000 lb ⋅ ft ΣFx = 0 ∴ Ex = 0 ΣFy = 0:

E y − 3600 lb − 1200 lb = 0

E y = 4800 lb

E = 4.80 kips ; M E = 51.0 kip ⋅ ft





(b)

www.elsolucionario.net PROBLEM 4.49 For the rig and crate of Prob. 4.48, and assuming that cable is anchored at F as shown, determine (a) the required tension in cable ADCF if the maximum value of the couple at E as x varies from 1.5 to 17.5 ft is to be as small as possible, (b) the corresponding maximum value of the couple.

SOLUTION

M E = 0: M E + (3600 lb) x + (1200 lb)(6.5 ft) − T (3.75 ft) = 0 M E = 3.75T − 3600 x − 7800

(1)

For x = 1.5 ft, Eq. (1) becomes ( M E )1 = 3.75T − 3600(1.5) − 7800

(2)

For x = 17.5 ft, Eq. (1) becomes ( M E ) 2 = 3.75T − 3600(17.5) − 7800

(a)

For smallest max value of |M E |, we set ( M E )1 = − ( M E )2 3.75T − 13, 200 = −3.75T + 70,800

(b)

T = 11.20 kips 

From Equation (2), then M E = 3.75(11.20) − 13.20

|M E | = 28.8 kip ⋅ ft 



Free-Body Diagram:

www.elsolucionario.net PROBLEM 4.50 A 6-m telephone pole weighing 1600 N is used to support the ends of two wires. The wires form the angles shown with the horizontal axis and the tensions in the wires are, respectively, T1 = 600 N and T2 = 375 N. Determine the reaction at the fixed end A.

SOLUTION


Free-Body Diagram:

ΣFx = 0: Ax + (375 N) cos 20° − (600 N) cos10° = 0 Ax = +238.50 N ΣFy = 0: Ay − 1600 N − (600 N)sin10° − (375 N) sin 20° = 0 Ay = +1832.45 N A = 238.502 + 1832.452 1832.45 θ = tan −1 238.50

A = 1848 N

82.6° 

ΣM A = 0: M A + (600 N) cos10°(6 m) − (375 N) cos 20°(6 m) = 0 M A = −1431.00 N ⋅ m

M A = 1431 N ⋅ m




www.elsolucionario.net PROBLEM 4.51 A vertical load P is applied at end B of rod BC. (a) Neglecting the weight of the rod, express the angle θ corresponding to the equilibrium position in terms of P, l, and the counterweight W. (b) Determine the value of θ corresponding to equilibrium if P = 2W.

SOLUTION

(a)


Free-Body Diagram:

Triangle ABC is isosceles. We have θ  θ  CD = ( BC ) cos   = l cos   2 2

θ  ΣM C = 0: P(l cos θ ) − W  l cos  = 0 2  Setting cos θ = 2 cos 2

θ 2

− 1:

θ θ   Pl  2 cos 2 − 1 − Wl cos = 0 2  2  cos 2

θ

θ 1 W  −  cos − = 0 2  2P  2 2 cos

θ 2

=

 1 W W2  ±  8 +  4 P P2  

1 W

θ = 2cos −1  

4 P  

±

 W2 + 8    2  P 



(b)

For P = 2W ,

cos

θ

cos

θ

2

2

θ 2

=

 1 11 1 + 8  = 1 ± 33  ±  8 42 4 

(

= 0.84307 and cos = 32.534°

θ = 65.1°

θ 2

θ 2

)

= −0.59307 = 126.375°

θ = 252.75° (discard)


θ = 65.1° 


www.elsolucionario.net PROBLEM 4.52 A vertical load P is applied at end B of rod BC. (a) Neglecting the weight of the rod, express the angle θ corresponding to the equilibrium position in terms of P, l, and the counterweight W. (b) Determine the value of θ corresponding to equilibrium if P = 2W.

SOLUTION Triangle ABC is isosceles. We have CD = ( BC ) cos

θ 2

= l cos

θ 2

θ  ΣM C = 0: W  l cos  − P(l sin θ ) = 0 2  Setting sin θ = 2sin

θ

θ

θ

θ

W − 2 P sin

(b)

For P = 2W ,

sin

θ 2

θ 2

or

θ

cos : Wl cos − 2 Pl sin cos = 0 2 2 2 2 2

θ 2

=

θ 2

=0

θ = 2sin −1 

W     2P 

W W = = 0.25 2 P 4W

θ = 29.0° 

= 14.5° = 165.5° θ = 331° (discard)



(a)

www.elsolucionario.net PROBLEM 4.53 A slender rod AB, of weight W, is attached to blocks A and B, which move freely in the guides shown. The blocks are connected by an elastic cord that passes over a pulley at C. (a) Express the tension in the cord in terms of W and θ. (b) Determine the value of θ for which the tension in the cord is equal to 3W.

SOLUTION (a)

From F.B.D. of rod AB:

T=

W cos θ 2(cosθ − sin θ )

Dividing both numerator and denominator by cos θ, T=

(b)

For T = 3W ,

or

3W =

W 2

1    1 − tan θ   

or T =

( W2 )

(1 − tan θ )



( W2 )

(1 − tan θ ) 1 1 − tan θ = 6

5  

θ = tan −1   = 39.806° 6

or

θ = 39.8° 



 1   ΣM C = 0: T (l sin θ ) + W   cos θ  − T (l cos θ ) = 0  2  

www.elsolucionario.net PROBLEM 4.54 Rod AB is acted upon by a couple M and two forces, each of magnitude P. (a) Derive an equation in θ, P, M, and l that must be satisfied when the rod is in equilibrium. (b) Determine the value of θ corresponding to equilibrium when M = 150 N · m, P = 200 N, and l = 600 mm.


(a)

From free-body diagram of rod AB:

or sinθ + cosθ =

(b)

M  Pl

For M = 150 lb ⋅ in., P = 20 lb, and l = 6 in., sin θ + cos θ =

150 lb ⋅ in. 5 = = 1.25 (20 lb)(6 in.) 4 sin 2 θ + cos 2 θ = 1

Using identity

sin θ + (1 − sin 2 θ )1/2 = 1.25 (1 − sin 2 θ )1/2 = 1.25 − sin θ 1 − sin 2 θ = 1.5625 − 2.5sin θ + sin 2 θ 2sin 2 θ − 2.5sin θ + 0.5625 = 0

Using quadratic formula sin θ = =

or

−( −2.5) ± (625) − 4(2)(0.5625) 2(2) 2.5 ± 1.75 4

sin θ = 0.95572 and sin θ = 0.29428 θ = 72.886° and θ = 17.1144° or θ = 17.11° and θ = 72.9° 



ΣM C = 0: P(l cos θ ) + P(l sin θ ) − M = 0

www.elsolucionario.net PROBLEM 4.55 Solve Sample Problem 4.5, assuming that the spring is unstretched when θ = 90°.


T = tension in spring = ks

where

s = deformation of spring = rβ F = kr β

or

ΣM 0 = 0: W (l cos β ) − F (r ) = 0 Wl cos β − kr 2 β = 0 cos β =

For


From F.B.D. of assembly:

kr 2 β Wl

k = 250 lb/in. r = 3 in. l = 8 in. W = 400 lb cos β =

or

(250 lb/in.)(3 in.)2 β (400 lb)(8 in.)

cos β = 0.703125β

Solving numerically,

β = 0.89245 rad

or

β = 51.134°

Then

θ = 90° + 51.134° = 141.134°

or θ = 141.1° 


www.elsolucionario.net PROBLEM 4.56 A slender rod AB, of weight W, is attached to blocks A and B that move freely in the guides shown. The constant of the spring is k, and the spring is unstretched when θ = 0. (a) Neglecting the weight of the blocks, derive an equation in W, k, l, and θ that must be satisfied when the rod is in equilibrium. (b) Determine the value of θ when W = 75 lb, l = 30 in., and k = 3 lb/in.

SOLUTION

Spring force:

Fs = ks = k (l − l cos θ ) = kl (1 − cos θ )

l  ΣM D = 0: Fs (l sin θ ) − W  cos θ  = 0 2 

(a)

kl (1 − cos θ )l sin θ − kl (1 − cos θ ) tan θ −

(b)

For given values of

W l cos θ = 0 2

W =0 2

or (1 − cos θ ) tan θ =

W  2kl

W = 75 lb l = 30 in. k = 3 lb/in. (1 − cos θ ) tan θ = tan θ − sin θ 75 lb = 2(3 lb/in.)(30 in.) = 0.41667


θ = 49.710°

or

θ = 49.7° 



Free-Body Diagram:

www.elsolucionario.net PROBLEM 4.57 A vertical load P is applied at end B of rod BC. The constant of the spring is k, and the spring is unstretched when θ = 60°. (a) Neglecting the weight of the rod, express the angle θ corresponding to the equilibrium position terms of P, k, and l. (b) Determine the value of θ 1 corresponding to equilibrium if P = 4 kl.

SOLUTION

(a)

Triangle ABC is isosceles. We have θ  θ  AB = 2( AD ) = 2l sin   ; CD = l cos   2 2

Elongation of spring:

x = ( AB)θ − ( AB )θ = 60°

θ  = 2l sin   − 2l sin 30° 2  θ 1 T = k x = 2kl  sin −  2 2 

θ  ΣM C = 0: T  l cos  − P(l sin θ ) = 0 2 



Free-Body Diagram:


θ θ θ  θ 1  2kl  sin −  l cos − Pl  2sin cos  = 0 2 2 2 2 2   cos

θ 2

=0

2(kl − P ) sin

or

θ = 180° (trivial)

sin

θ 2

θ 2

− kl = 0 =

1 2

kl

kl − P

1 2

 

θ = 2sin −1  kl /( kl − P)   (b)

For P =

1 kl , 4

sin

θ 2

θ

1 2 3 4

kl kl

=

2 3

θ = 83.6° 

= 41.8°


2

=


www.elsolucionario.net PROBLEM 4.58 A collar B of weight W can move freely along the vertical rod shown. The constant of the spring is k, and the spring is unstretched when θ = 0. (a) Derive an equation in θ, W, k, and l that must be satisfied when the collar is in equilibrium. (b) Knowing that W = 300 N, l = 500 mm, and k = 800 N/m, determine the value of θ corresponding to equilibrium.

First note:

T = ks

where

k = spring constant s = elongation of spring l = −l cos θ l (1 − cos θ ) = cos θ kl T= (1 − cos θ ) cos θ

(a)

From F.B.D. of collar B: or

(b)

For

ΣFy = 0: T sin θ − W = 0 kl (1 − cos θ )sin θ − W = 0 cos θ

or tan θ − sin θ =

W  kl

W = 3 lb l = 6 in. k = 8 lb/ft 6 in. l= = 0.5 ft 12 in./ft tan θ − sin θ =


3 lb = 0.75 (8 lb/ft)(0.5 ft)

θ = 57.957°

or

θ = 58.0° 



SOLUTION

www.elsolucionario.net PROBLEM 4.59 Eight identical 500 × 750-mm rectangular plates, each of mass m = 40 kg, are held in a vertical plane as shown. All connections consist of frictionless pins, rollers, or short links. In each case, determine whether (a) the plate is completely, partially, or improperly constrained, (b) the reactions are statically determinate or indeterminate, (c) the equilibrium of the plate is maintained in the position shown. Also, wherever possible, compute the reactions.

1.

Three non-concurrent, non-parallel reactions: (a)

Plate: completely constrained

(b)

Reactions: determinate

(c)

Equilibrium maintained A = C = 196.2 N

2.



(b)


(c)

Equilibrium maintained B = 0, C = D = 196.2 N

3.

Four non-concurrent, non-parallel reactions: (a)


(b)

Reactions: indeterminate

(c)

Equilibrium maintained A x = 294 N

,

D x = 294 N

( A y + D y = 392 N )

4.

Three concurrent reactions (through D): (a)

Plate: improperly constrained

(b)


(c)

No equilibrium

(ΣM D ≠ 0)



SOLUTION


5.

Two reactions: (a)

Plate: partial constraint

(b)


(c)

Equilibrium maintained C = D = 196.2 N



(b)


(c)

Equilibrium maintained B = 294 N

7.

8.

, D = 491 N

53.1°

Two reactions: (a)

Plate: improperly constrained

(b)

Reactions determined by dynamics

(c)

No equilibrium

(ΣFy ≠ 0)



(b)


(c)

Equilibrium maintained B = D y = 196.2 N

(C + D x = 0)



6.


SOLUTION 1.


Bracket: complete constraint

(b)


(c)

Equilibrium maintained A = 120.2 lb

2.

3.

56.3°, B = 66.7 lb

Four concurrent, reactions (through A): (a)

Bracket: improper constraint

(b)


(c)

No equilibrium

(ΣM A ≠ 0)

Two reactions: (a)

Bracket: partial constraint

(b)


(c)

Equilibrium maintained A = 50 lb , C = 50 lb

4.



(b)


(c)

Equilibrium maintained A = 50 lb , B = 83.3 lb

36.9°, C = 66.7 lb



The bracket ABC can be supported in the eight different ways shown. All connections consist of smooth pins, rollers, or short links. For each case, answer the questions listed in Problem 4.59, and, wherever possible, compute the reactions, assuming that the magnitude of the force P is 100 lb.


5.

6.



(b)


(c)

Equilibrium maintained

(ΣM C = 0) A y = 50 lb



(b)


(c)

Equilibrium maintained A x = 66.7 lb

B x = 66.7 lb

( A y + B y = 100 lb )



(b)


(c)

Equilibrium maintained A = C = 50 lb

8.

Three concurrent, reactions (through A) (a)

Bracket: improper constraint

(b)


(c)

No equilibrium

(ΣM A ≠ 0)



7.

www.elsolucionario.net PROBLEM 4.61 Determine the reactions at A and B when a = 150 mm.

SOLUTION

Force triangle

80 mm 80 mm = a 150 mm β = 28.072°

tan β =

A=

320 N sin 28.072°

B=

320 N tan 28.072°

A = 680 N

28.1° 

B = 600 N







Free-Body Diagram:

www.elsolucionario.net PROBLEM 4.62 Determine the value of a for which the magnitude of the reaction at B is equal to 800 N.

SOLUTION

Force triangle

tan β =

80 mm a

a=

80 mm tan β

(1)

From force triangle: tan β =

From Eq. (1):

a=

320 N = 0.4 800 N

80 mm 0.4

a = 200 mm 



Free-Body Diagram:

www.elsolucionario.net PROBLEM 4.63 Using the method of Sec. 4.7, solve Problem 4.22b. PROBLEM 4.22 Determine the reactions at A and B when (a) α = 0, (b) α = 90°, (c) α = 30°.



(Three-force body)

The line of action at A must pass through C, where B and the 75-lb load intersect. In triangle ACE:

Force triangle

tan θ =

10 in. 12 in.

θ = 39.806°

B = (75 lb) tan 39.806° = 62.5 lb 75 lb A= = 97.6° cos 39.806° A = 97.6 lb

50.2°; B = 62.5 lb






www.elsolucionario.net PROBLEM 4.64 A 500-lb cylindrical tank, 8 ft in diameter, is to be raised over a 2-ft obstruction. A cable is wrapped around the tank and pulled horizontally as shown. Knowing that the corner of the obstruction at A is rough, find the required tension in the cable and the reaction at A.

SOLUTION


Free-Body Diagram:

Force triangle cos α =

GD 2 ft = = 0.5 AG 4 ft

α = 60°

1 ( β = 60°) 2 T = (500 lb) tan 30° T = 289 lb

θ = α = 30°

A=

500 lb cos 30°

A = 577 lb

60.0° 


www.elsolucionario.net PROBLEM 4.65 For the frame and loading shown, determine the reactions at A and C.

Since member AB is acted upon by two forces, A and B, they must be colinear, have the same magnitude, and be opposite in direction for AB to be in equilibrium. The force B acting at B of member BCD will be equal in magnitude but opposite in direction to force B acting on member AB. Member BCD is a three-force body with member forces intersecting at E. The F.B.D.’s of members AB and BCD illustrate the above conditions. The force triangle for member BCD is also shown. The angle β is found from the member dimensions:  6 in.   = 30.964°  10 in. 

β = tan −1 

Applying the law of sines to the force triangle for member BCD, 30 lb B C = = sin(45° − β ) sin β sin135°

or

30 lb B C = = sin14.036° sin 30.964° sin135°

A= B=

(30 lb)sin 30.964° = 63.641 lb sin14.036°

or and

C=

A = 63.6 lb

45.0° 

C = 87.5 lb

59.0° 

(30 lb) sin135° = 87.466 lb sin14.036°

or



SOLUTION

www.elsolucionario.net PROBLEM 4.66 For the frame and loading shown, determine the reactions at C and D.

SOLUTION Since BD is a two-force member, the reaction at D must pass through Points B and D. Free-Body Diagram:


(Three-force body)

Reaction at C must pass through E, where the reaction at D and the 150-lb load intersect. Triangle CEF:

tan β =

4.5 ft 3 ft

β = 56.310°

Triangle ABE:

tan γ =

1 2

γ = 26.565°


www.elsolucionario.net PROBLEM 4.66 (Continued) Force Triangle

150 lb C D = = sin 29.745° sin116.565° sin 33.690°

C = 270.42 lb, D = 167.704 lb C = 270 lb

56.3°; D = 167.7 lb

26.6° 



Law of sines:

www.elsolucionario.net PROBLEM 4.67 Determine the reactions at B and D when b = 60 mm.

SOLUTION Since CD is a two-force member, the line of action of reaction at D must pass through Points C and D. Free-Body Diagram:


(Three-force body)

Reaction at B must pass through E, where the reaction at D and the 80-N force intersect. 220 mm 250 mm β = 41.348°

tan β =

Force triangle

Law of sines: 80 N B D = = sin 3.652° sin 45° sin131.348° B = 888.0 N D = 942.8 N 

B = 888 N

41.3°

D = 943 N

45.0° 


www.elsolucionario.net PROBLEM 4.68 Determine the reactions at B and D when b = 120 mm.

SOLUTION Since CD is a two-force member, line of action of reaction at D must pass through C and D

.


Free-Body Diagram: (Three-force body)

Reaction at B must pass through E, where the reaction at D and the 80-N force intersect. 280 mm 250 mm β = 48.24°

tan β =

Force triangle

Law of sines:

80 N B D = = sin 3.24° sin135° sin 41.76° B = 1000.9 N D = 942.8 N 

B = 1001 N

48.2° D = 943 N

45.0° 


www.elsolucionario.net PROBLEM 4.69 A T-shaped bracket supports a 300-N load as shown. Determine the reactions at A and C when α = 45°.



(Three-force body)

The line of action of C must pass through E, where A and the 300-N force intersect. Triangle ABE is isosceles:

EA = AB = 400 mm

In triangle CEF: tan θ =

CF CF 150 mm = = EF EA + AF 700 mm

θ = 12.0948°



Law of sines:

A C 300 N = = sin 32.905° sin135° sin12.0948°

C = 1012 N

77.9° 


A = 778 N ;


www.elsolucionario.net PROBLEM 4.70 A T-shaped bracket supports a 300-N load as shown. Determine the reactions at A and C when α = 60°.

SOLUTION


Free-Body Diagram:

EA = (400 mm) tan 30° = 230.94 mm

In triangle CEF:

tan θ =

CF CF = EF EA + AF

150 230.94 + 300 θ = 15.7759°

tan θ =



Law of sines:

A C 300 N = = sin 44.224° sin120° sin15.7759° A = 770 N C = 956 N

A = 770 N ;

C = 956 N

74.2° 



Force Triangle

www.elsolucionario.net PROBLEM 4.71 A 40-lb roller, of diameter 8 in., which is to be used on a tile floor, is resting directly on the subflooring as shown. Knowing that the thickness of each tile is 0.3 in., determine the force P required to move the roller onto the tiles if the roller is (a) pushed to the left, (b) pulled to the right.

SOLUTION Geometry: For each case as roller comes into contact with tile, 3.7 in. 4 in. α = 22.332°

 (a)



Roller pushed to left (three-force body): Forces must pass through O.

Law of sines:

Force Triangle

40 lb P = ; P = 24.87 lb sin 37.668° sin 22.332° P = 24.9 lb

(b)

30.0° 

Roller pulled to right (three-force body): Forces must pass through O.

Law of sines:

40 lb P = ; P = 15.3361 lb sin 97.668° sin 22.332° P = 15.34 lb

30.0° 

   Force Triangle 





α = cos −1

www.elsolucionario.net PROBLEM 4.72 One end of rod AB rests in the corner A and the other end is attached to cord BD. If the rod supports a 40-lb load at its midpoint C, find the reaction at A and the tension in the cord.

SOLUTION



The line of action of reaction at A must pass through E, where T and the 40-lb load intersect. Force triangle EF 23 = AF 12 α = 62.447° 5 EH tan β = = DH 12 β = 22.620°

tan α =

A T 40 lb = = sin 67.380° sin 27.553° sin 85.067°

A = 37.1 lb

62.4° 

T = 18.57 lb 


www.elsolucionario.net PROBLEM 4.73 A 50-kg crate is attached to the trolley-beam system shown. Knowing that a = 1.5 m, determine (a) the tension in cable CD, (b) the reaction at B.

SOLUTION Three-force body: W and TCD intersect at E. 0.7497 m 1.5 m β = 26.556°


tan β =

Three forces intersect at E. W = (50 kg) 9.81 m/s 2 = 490.50 N

Law of sines:

Force triangle

TCD 490.50 N B = = sin 61.556° sin 63.444° sin 55° TCD = 498.99 N B = 456.96 N TCD = 499 N 

(a)

B = 457 N

(b)

26.6° 


www.elsolucionario.net PROBLEM 4.74 Solve Problem 4.73, assuming that a = 3 m. PROBLEM 4.73 A 50-kg crate is attached to the trolley-beam system shown. Knowing that a = 1.5 m, determine (a) the tension in cable CD, (b) the reaction at B.

SOLUTION W and TCD intersect at E. Free-Body Diagram:


Three-Force Body

AE 0.301 m = AB 3m β = 5.7295°

tan β =

Three forces intersect at E.

Force Triangle W = (50 kg) 9.81 m/s 2 = 490.50 N

Law of sines:

TCD 490.50 N B = = sin 29.271° sin 95.730° sin 55° TCD = 998.18 N B = 821.76 N TCD = 998 N 

(a)

B = 822 N

(b)

5.73° 


www.elsolucionario.net PROBLEM 4.75 Determine the reactions at A and B when β = 50°.

SOLUTION Free-Body Diagram: (Three-force body) Reaction A must pass through Point D where the 100-N force and B intersect.


In right Δ BCD:

α = 90° − 75° = 15° BD = 250 tan 75° = 933.01 mm

In right Δ ABD:

Dimensions in mm

AB 150 mm = BD 933.01 mm γ = 9.1333°

tan γ =

Force Triangle Law of sines: 100 N A B = = sin 9.1333° sin15° sin155.867° A = 163.1 N; B = 257.6 N A = 163.1 N

74.1° B = 258 N

65.0° 


www.elsolucionario.net PROBLEM 4.76 Determine the reactions at A and B when β = 80°.



(Three-force body)

Reaction A must pass through D where the 100-N force and B intersect. In right triangle BCD:

α = 90° − 75° = 15° BD = BC tan 75° = 250 tan75° BD = 933.01 mm

In right triangle ABD:

tan γ =

AB 150 mm = BD 933.01 mm

γ = 9.1333°



Law of sines:



A = 163.1 N

55.9° 



B = 258 N

65.0° 



100 N A B = = sin 9.1333° sin15° sin155.867°

www.elsolucionario.net PROBLEM 4.77 Knowing that θ = 30°, determine the reaction (a) at B, (b) at C.

SOLUTION Free-Body Diagram: (Three-force body) Reaction at C must pass through D where force P and reaction at B intersect. ( 3 − 1) R R = 3 −1 β = 36.2°

tan β =


In Δ CDE:

Force Triangle

Law of sines:

P B C = = sin 23.8° sin126.2° sin 30° B = 2.00 P C = 1.239 P

(a)

B = 2P

(b)

C = 1.239P

60.0°  36.2° 


www.elsolucionario.net PROBLEM 4.78 Knowing that θ = 60°, determine the reaction (a) at B, (b) at C.

SOLUTION Reaction at C must pass through D where force P and reaction at B intersect. tan β =

R−

=1−


R 1


In ΔCDE:

R 3

3 β = 22.9°

Force Triangle

Law of sines:

P B C = = sin 52.9° sin 67.1° sin 60° B = 1.155P C = 1.086 P

(a)

B = 1.155P

30.0° 

(b)

C = 1.086P

22.9° 


www.elsolucionario.net PROBLEM 4.79 Using the method of Section 4.7, solve Problem 4.23. PROBLEM 4.23 Determine the reactions at A and B when (a) h = 0, (b) h = 200 mm.

SOLUTION  

Free-Body Diagram:

(a)

h=0 Reaction A must pass through C where the 150-N weight and B interect.

(b)

A = 150.0 N

30.0° 

B = 150.0 N

30.0° 

h = 200 mm



55.662 250 β = 12.5521°

tan β =

 Law of sines:

A B 150 N = = sin17.4480° sin 60° sin102.552° A = 433.24 N B = 488.31 N

  Force Triangle

A = 433 N B = 488 N

12.55°  30.0° 





Force triangle is equilateral.

www.elsolucionario.net PROBLEM 4.80 Using the method of Section 4.7, solve Problem 4.24. PROBLEM 4.24 A lever AB is hinged at C and attached to a control cable at A. If the lever is subjected to a 75-lb vertical force at B, determine (a) the tension in the cable, (b) the reaction at C.

SOLUTION 

Reaction at C must pass through E, where the 75-lb force and T intersect.

 


9.3969 in. 8.5798 in. α = 47.602°

tan α =

 

14.0954 in. 24.870 in. β = 29.543°



tan β =

 

Force Triangle

      

Law of sines:

  

75 lb T C = = sin18.0590° sin 29.543° sin132.398°

Free-Body Diagram:

 Dimensions in in.

(a)



(b)

T = 119.3 lb  C = 178.7 lb

60.5° 


www.elsolucionario.net PROBLEM 4.81 Member ABC is supported by a pin and bracket at B and by an inextensible cord attached at A and C and passing over a frictionless pulley at D. The tension may be assumed to be the same in portions AD and CD of the cord. For the loading shown and neglecting the size of the pulley, determine the tension in the cord and the reaction at B.


Reaction at B must pass through D. 7 in. 12 in. α = 30.256° 7 in. tan β = 24 in. β = 16.26°


tan α =

Force Triangle

Law of sines:

T T − 72 lb B = = sin 59.744° sin13.996° sin106.26 T (sin13.996°) = (T − 72 lb)(sin 59.744°) T (0.24185) = (T − 72)(0.86378) T = 100.00 lb sin 106.26° sin 59.744° = 111.14 lb

T = 100.0 lb 

B = (100 lb)

B = 111.1 lb

30.3° 


www.elsolucionario.net PROBLEM 4.82 Member ABC is supported by a pin and bracket at B and by an inextensible cord attached at A and C and passing over a frictionless pulley at D. The tension may be assumed to be the same in portions AD and CD of the cord. For the loading shown and neglecting the size of the pulley, determine the tension in the cord and the reaction at B.


Reaction at B must pass through D. tan α =

120 ; α = 36.9° 160

T T − 75 N B = = 4 3 5 3T = 4T − 300; T = 300 N 5 5 B = T = (300 N) = 375 N 4 4

B = 375 N

36.9°



Force Triangle

www.elsolucionario.net PROBLEM 4.83 A thin ring of mass 2 kg and radius r = 140 mm is held against a frictionless wall by a 125-mm string AB. Determine (a) the distance d, (b) the tension in the string, (c) the reaction at C.


The force T exerted at B must pass through the center G of the ring, since C and W intersect at that point. Thus, points A, B, and G are in a straight line. (a) From triangle ACG:

d = ( AG ) 2 − (CG )2 = (265 mm)2 − (140 mm) 2 = 225.00 mm d = 225 mm 

Force Triangle W = (2 kg)(9.81 m/s 2 ) = 19.6200 N

Law of sines:

T C 19.6200 N = = 265 mm 140 mm 225.00 mm T = 23.1 N 

(b)

C = 12.21 N

(c)





(Three-force body)

www.elsolucionario.net PROBLEM 4.84 A uniform rod AB of length 2R rests inside a hemispherical bowl of radius R as shown. Neglecting friction, determine the angle θ corresponding to equilibrium.

SOLUTION Based on the F.B.D., the uniform rod AB is a three-force body. Point E is the point of intersection of the three forces. Since force A passes through O, the center of the circle, and since force C is perpendicular to the rod, triangle ACE is a right triangle inscribed in the circle. Thus, E is a point on the circle. Note that the angle α of triangle DOA is the central angle corresponding to the inscribed angle θ of triangle DCA.

α = 2θ


The horizontal projections of AE , ( x AE ), and AG , ( x AG ), are equal. x AE = x AG = x A

or

( AE ) cos 2θ = ( AG ) cos θ

and

(2 R) cos 2θ = R cos θ

Now then or

cos 2θ = 2 cos 2 θ − 1 4 cos 2 θ − 2 = cos θ 4 cos 2 θ − cos θ − 2 = 0

Applying the quadratic equation, cos θ = 0.84307 and cos θ = − 0.59307

θ = 32.534° and θ = 126.375° (Discard)

or θ = 32.5° 


www.elsolucionario.net PROBLEM 4.85 A slender rod BC of length L and weight W is held by two cables as shown. Knowing that cable AB is horizontal and that the rod forms an angle of 40° with the horizontal, determine (a) the angle θ that cable CD forms with the horizontal, (b) the tension in each cable.


(a)


(Three-force body)

The line of action of TCD must pass through E, where TAB and W intersect. CF EF L sin 40° = 1 L cos 40° 2

tan θ =

= 2 tan 40° = 59.210°

θ = 59.2°  (b)

Force Triangle

TAB = W tan 30.790° = 0.59588W

TAB = 0.596W 



W cos 30.790°  = 1.16408W

TCD =

TCD = 1.164W 




www.elsolucionario.net PROBLEM 4.86 A slender rod of length L and weight W is attached to a collar at A and is fitted with a small wheel at B. Knowing that the wheel rolls freely along a cylindrical surface of radius R, and neglecting friction, derive an equation in θ, L, and R that must be satisfied when the rod is in equilibrium.

SOLUTION Free-Body Diagram (Three-force body) Reaction B must pass through D where B and W intersect. Note that ΔABC and ΔBGD are similar.

In Δ ABC:

(CE ) 2 + ( BE )2 = ( BC )2 (2 L cos θ ) 2 + ( L sin θ )2 = R 2 2

R 2 2   = 4cos θ + sin θ L 2

R 2 2   = 4cos θ + 1 − cos θ L 2

R 2   = 3cos θ + 1 L 2  1 cos 2 θ =  R  − 1  3  L  



AC = AE = L cos θ

www.elsolucionario.net PROBLEM 4.87 Knowing that for the rod of Problem 4.86, L = 15 in., R = 20 in., and W = 10 lb, determine (a) the angle θ corresponding to equilibrium, (b) the reactions at A and B.

SOLUTION

2  1 cos 2 θ =  R  − 1 3  L  

For L = 15 in., R = 20 in., and W = 10 lb, 2  1  20 in.  cos θ =   − 1 ; θ = 59.39° 3  15 in.     2

(a)

In Δ ABC:

θ = 59.4° 

BE L sin θ 1 = = tan θ CE 2 L cos θ 2 1 tan α = tan 59.39° = 0.8452 2 α = 40.2° tan α =

Force Triangle

A = W tan α = (10 lb) tan 40.2° = 8.45 lb W (10 lb) B= = = 13.09 lb cos α cos 40.2° A = 8.45 lb

(b)

B = 13.09 lb



49.8° 



See the solution to Problem 4.86 for the free-body diagram and analysis leading to the following equation:

www.elsolucionario.net PROBLEM 4.88 Rod AB is bent into the shape of an arc of circle and is lodged between two pegs D and E. It supports a load P at end B. Neglecting friction and the weight of the rod, determine the distance c corresponding to equilibrium when a = 20 mm and R = 100 mm.


slope of ED is

45°;

slope of HC is

45°.

Also

DE = 2 a

and

a 1 DH = HE =   DE = 2 2  

For triangles DHC and EHC,

sin β =

a 2

R

=

a 2R

Now

c = R sin(45° − β )

For

a = 20 mm and sin β =


Since yED = xED = a,

R = 100 mm

20 mm

2(100 mm) = 0.141421 β = 8.1301°

and

c = (100 mm) sin(45° − 8.1301°) = 60.00 mm

or c = 60.0 mm 


www.elsolucionario.net PROBLEM 4.89 A slender rod of length L is attached to collars that can slide freely along the guides shown. Knowing that the rod is in equilibrium, derive an expression for the angle θ in terms of the angle β.

SOLUTION

Free-Body Diagram: tan β =

xGB y AB

where y AB = L cos θ

and xGB =

tan β =

1 L sin θ 2 1 2

L sin θ

L cos θ 1 = tan θ 2

or tan θ = 2 tan β 



As shown in the free-body diagram of the slender rod AB, the three forces intersect at C. From the force geometry:

www.elsolucionario.net PROBLEM 4.90 An 8-kg slender rod of length L is attached to collars that can slide freely along the guides shown. Knowing that the rod is in equilibrium and that β = 30°, determine (a) the angle θ that the rod forms with the vertical, (b) the reactions at A and B.

SOLUTION (a)

As shown in the free-body diagram of the slender rod AB, the three forces intersect at C. From the geometry of the forces:

tan β =


Free-Body Diagram: xCB yBC

where xCB =

1 L sin θ 2

and yBC = L cos θ tan β =

1 tan θ 2

or

tan θ = 2 tan β

For

β = 30° tan θ = 2 tan 30° = 1.15470 θ = 49.107°

or

W = mg = (8 kg)(9.81 m/s2 ) = 78.480 N

(b) From force triangle:

A = W tan β = (78.480 N) tan 30° = 45.310 N

and

θ = 49.1° 

B=

or

W 78.480 N = = 90.621 N cos β cos 30°

A = 45.3 N

or B = 90.6 N



60.0° 


www.elsolucionario.net PROBLEM 4.91 A 200-mm lever and a 240-mm-diameter pulley are welded to the axle BE that is supported by bearings at C and D. If a 720-N vertical load is applied at A when the lever is horizontal, determine (a) the tension in the cord, (b) the reactions at C and D. Assume that the bearing at D does not exert any axial thrust.

SOLUTION

We have six unknowns and six equations of equilibrium. —OK ΣM C = 0: (−120k ) × ( Dx i + Dy j) + (120 j − 160k ) × T i + (80k − 200i ) × (−720 j) = 0

−120 Dx j + 120 D y i − 120T k − 160Tj + 57.6 × 103 i + 144 × 103 k = 0

Equating to zero the coefficients of the unit vectors: k:

−120T + 144 × 103 = 0

i:

120 Dy + 57.6 × 103 = 0

j: − 120 Dx − 160(1200 N) = 0

(b)

ΣFx = 0:

C x + Dx + T = 0

ΣFy = 0:

C y + Dy − 720 = 0

ΣFz = 0:

Cz = 0

(a) T = 1200 N  Dy = −480 N Dx = −1600 N C x = 1600 − 1200 = 400 N C y = 480 + 720 = 1200 N

C = (400 N)i + (1200 N) j; D = −(1600 N)i − (480 N) j 



Dimensions in mm

www.elsolucionario.net PROBLEM 4.92 Solve Problem 4.91, assuming that the axle has been rotated clockwise in its bearings by 30° and that the 720-N load remains vertical. PROBLEM 4.91 A 200-mm lever and a 240-mmdiameter pulley are welded to the axle BE that is supported by bearings at C and D. If a 720-N vertical load is applied at A when the lever is horizontal, determine (a) the tension in the cord, (b) the reactions at C and D. Assume that the bearing at D does not exert any axial thrust.

SOLUTION

We have six unknowns and six equations of equilibrium. ΣM C = 0: (−120k ) × ( Dx i + Dy j) + (120 j − 160k ) × T i + (80k − 173.21i ) × (−720 j) = 0 −120 Dx j + 120 D y i −120T k −160T j + 57.6 × 103 i + 124.71 × 103 k = 0

Equating to zero the coefficients of the unit vectors, k : − 120T + 124.71 × 103 = 0 i:

T = 1039 N 

120 Dy + 57.6 × 103 = 0 Dy = −480 N

j: − 120 Dx − 160(1039.2)

(b)

T = 1039.2 N

ΣFx = 0:

C x + Dx + T = 0

ΣFy = 0:

C y + Dy − 720 = 0

ΣFz = 0:

Cz = 0

Dx = −1385.6 N C x = 1385.6 − 1039.2 = 346.4 C y = 480 + 720 = 1200 N

C = (346 N)i + (1200 N) j D = −(1386 N)i − (480 N) j 



Dimensions in mm

www.elsolucionario.net PROBLEM 4.93 A 4 × 8-ft sheet of plywood weighing 40 lb has been temporarily propped against column CD. It rests at A and B on small wooden blocks and against protruding nails. Neglecting friction at all surfaces of contact, determine the reactions at A, B, and C.


ΣM A = 0: rB /A × ( B y j + Bz k ) + rC /A × C k + rG /A × ( −40 lb) j = 0

i j 5 0 0 By

k i j k i j k 0 + 4 4sin 60° −4cos 60° + 2 2sin 60° −2 cos 60° = 0 Bz 0 C −40 0 0 0

(4C sin 60° − 80 cos 60°) i + (−5Bz − 4C ) j + (5B y − 80)k = 0

Equating the coefficients of the unit vectors to zero, i:

4C sin 60° − 80 cos 60° = 0

j:

−5Bz − 4C = 0

k:

5B y − 80 = 0

ΣFy = 0:

Ay + B y − 40 = 0

ΣFz = 0:

Az + Bz + C = 0

C = 11.5470 lb Bz = 9.2376 lb B y = 16.0000 lb Ay = 40 − 16.0000 = 24.000 lb Az = 9.2376 − 11.5470 = −2.3094 lb

A = (24.0 lb)j − (2.31 lb)k ; B = (16.00 lb) j − (9.24 lb)k ; C = (11.55 lb)k 



We have five unknowns and six equations of equilibrium. Plywood sheet is free to move in x direction, but equilibrium is maintained (ΣFx = 0).

www.elsolucionario.net PROBLEM 4.94 Two tape spools are attached to an axle supported by bearings at A and D. The radius of spool B is 1.5 in. and the radius of spool C is 2 in. Knowing that TB = 20 lb and that the system rotates at a constant rate, determine the reactions at A and D. Assume that the bearing at A does not exert any axial thrust and neglect the weights of the spools and axle.

SOLUTION

We have six unknowns and six equations of equilibrium. ΣM A = 0: (4.5i + 1.5k ) × (−20 j) + (10.5i + 2 j) × (−TC k ) + (15i) × ( Dx i + Dy j + Dz k ) = 0 −90k + 30i + 10.5TC j − 2TC i + 15D y k − 15Dz j = 0

Equate coefficients of unit vectors to zero:  i: 30 − 2TC = 0  j: 10.5TC − 15Dz = 0 10.5(15) − 15Dz = 0  k: −90 + 15 Dy = 0 ΣFx = 0:

Dx = 0

ΣFy = 0:

Ay + D y − 20 lb = 0

Ay = 20 − 6 = 14 lb

ΣFz = 0:

Az + Dz − 15 lb = 0

Az = 15 − 10.5 = 4.5 lb

TC = 15 lb Dz = 10.5 lb Dy = 6 lb

A = (14.00 lb) j + (4.50 lb)k ; D = (6.00 lb) j + (10.50 lb)k 



Free-Body Diagram:


SOLUTION We replace TB and TB′ by their resultant (−180 N)j and TC and TC′ by their resultant (−300 N)k.

Dimensions in mm

We have five unknowns and six equations of equilibrium. Axle AD is free to rotate about the x-axis, but equilibrium is maintained (ΣMx = 0). ΣM A = 0: (150i ) × (−180 j) + (250i) × ( −300k ) + (450i ) × ( Dy j + Dz k ) = 0 −27 × 103 k + 75 × 103 j + 450 Dy k − 450 Dz j = 0

Equating coefficients of j and k to zero, j:

75 × 103 − 450 Dz = 0

Dz = 166.7 N

k:

− 27 × 103 + 450 Dy = 0

Dy = 60.0 N

ΣFx = 0:

Ax = 0

ΣFy = 0:

Ay + Dy − 180 N = 0

ΣFz = 0: Az + Dz − 300 N = 0

Ay = 180 − 60 = 120.0 N Az = 300 − 166.7 = 133.3 N

A = (120.0 N) j + (133.3 N)k ; D = (60.0 N) j + (166.7 N)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 452


Two transmission belts pass over a double-sheaved pulley that is attached to an axle supported by bearings at A and D. The radius of the inner sheave is 125 mm and the radius of the outer sheave is 250 mm. Knowing that when the system is at rest, the tension is 90 N in both portions of belt B and 150 N in both portions of belt C, determine the reactions at A and D. Assume that the bearing at D does not exert any axial thrust.

www.elsolucionario.net PROBLEM 4.96 Solve Problem 4.95, assuming that the pulley rotates at a constant rate and that TB = 104 N, T′B = 84 N, TC = 175 N.

SOLUTION

Dimensions in mm

We have six unknowns and six equations of equilibrium. —OK ΣM A = 0: (150i + 250k ) × (−104 j) + (150i − 250k ) × ( −84 j) + (250i + 125 j) × (−175k ) + (250i − 125 j) × (−TC ) + 450i × ( D y j + Dz k ) = 0 −150(104 + 84)k + 250(104 − 84)i + 250(175 + TC′ ) j − 125(175 − TC′ ) + 450 D y k − 450 Dz j = 0

Equating the coefficients of the unit vectors to zero, i : 250(104 − 84) − 125(175 − TC′ ) = 0

175 = TC′ = 40

j: 250(175 + 135) − 450 Dz = 0

Dz = 172.2 N

k : − 150(104 + 84) + 450 D y = 0

Dy = 62.7 N

TC′ = 135;



PROBLEM 4.95 Two transmission belts pass over a double-sheaved pulley that is attached to an axle supported by bearings at A and D. The radius of the inner sheave is 125 mm and the radius of the outer sheave is 250 mm. Knowing that when the system is at rest, the tension is 90 N in both portions of belt B and 150 N in both portions of belt C, determine the reactions at A and D. Assume that the bearing at D does not exert any axial thrust.


ΣFx = 0:

Ax = 0

ΣFy = 0:

Ay − 104 − 84 + 62.7 = 0

Ay = 125.3 N

ΣFz = 0:

Az − 175 − 135 + 172.2 = 0

Az = 137.8 N


A = (125.3 N) j + (137.8 N)k; D = (62.7 N) j + (172.2 N)k 


www.elsolucionario.net PROBLEM 4.97 Two steel pipes AB and BC, each having a mass per unit length of 8 kg/m, are welded together at B and supported by three wires. Knowing that a = 0.4 m, determine the tension in each wire.

SOLUTION


W1 = 0.6m′g W2 = 1.2m′g

ΣM D = 0: rA/D × TA j + rE/D × (−W1 j) + rF/D × (−W2 j) + rC/D × TC j = 0 (−0.4i + 0.6k ) × TA j + (−0.4i + 0.3k ) × (−W1 j) + 0.2i × (−W2 j) + 0.8i × TC j = 0 −0.4TAk − 0.6TA i + 0.4W1k + 0.3W1i − 0.2W2 k + 0.8TC k = 0

Equate coefficients of unit vectors to zero: 1 1 i : − 0.6TA + 0.3W1 = 0; TA = W1 = 0.6m′g = 0.3m′g 2 2 k : − 0.4TA + 0.4W1 − 0.2W2 + 0.8TC = 0 −0.4(0.3m′g ) + 0.4(0.6m′g ) − 0.2(1.2m′g ) + 0.8TC = 0 TC =

(0.12 − 0.24 − 0.24)m′g = 0.15m′g 0.8

ΣFy = 0: TA + TC + TD − W1 − W2 = 0 0.3m′g + 0.15m′g + TD − 0.6m′g − 1.2m′g = 0 TD = 1.35m′g m′g = (8 kg/m)(9.81m/s 2 ) = 78.48 N/m TA = 0.3m′g = 0.3 × 78.45

TA = 23.5 N 

TB = 0.15m′g = 0.15 × 78.45

TB = 11.77 N 

TC = 1.35m′g = 1.35 × 78.45

TC = 105.9 N 


www.elsolucionario.net PROBLEM 4.98 For the pipe assembly of Problem 4.97, determine (a) the largest permissible value of a if the assembly is not to tip, (b) the corresponding tension in each wire.

SOLUTION


W1 = 0.6m′g W2 = 1.2m′g

ΣM D = 0: rA/D × TA j + rE/D × (−W1 j) + rF/D × (−W2 j) + rC/D × TC j = 0 (− ai + 0.6k ) × TA j + (− ai + 0.3k ) × (−W1 j) + (0.6 − a)i × (−W2 j) + (1.2 − a)i × TC j = 0 −TA ak − 0.6TA i + W1ak + 0.3W1i − W2 (0.6 − a)k + TC (1.2 − a )k = 0

Equate coefficients of unit vectors to zero: 1 1 i : − 0.6TA + 0.3W1 = 0; TA = W1 = 0.6m′g = 0.3m′g 2 2 k : − TA a + W1a − W2 (0.6 − a ) + TC (1.2 − a) = 0 −0.3m′ga + 0.6m′ga − 1.2m′g (0.6 − a ) + TC (1.2 − a) = 0 TC =

(a)

0.3a − 0.6a + 1.2(0.6 − a) 1.2 − a

For maximum a and no tipping, TC = 0.

−0.3a + 1.2(0.6 − a) = 0 −0.3a + 0.72 − 1.2a = 0 1.5a = 0.72

a = 0.480 m 



(b)

Reactions:

m′g = (8 kg/m) 9.81 m/s 2 = 78.48 N/m TA = 0.3m′g = 0.3 × 78.48 = 23.544 N

TA = 23.5 N 

ΣFy = 0: TA + TC + TD − W1 − W2 = 0 TA + 0 + TD − 0.6m′g − 1.2m′g = 0 TD = 117.7 N 


TD = 1.8m′g − TA = 1.8 × 78.48 − 23.544 = 117.72


www.elsolucionario.net PROBLEM 4.99 The 45-lb square plate shown is supported by three vertical wires. Determine the tension in each wire.

SOLUTION


Free-Body Diagram:

ΣM B = 0: rC /B × TC j + rA/B × TA j + rG /B × (−45 lb) j = 0  [−(20 in.)i + (15 in.)k ] × TC j + (20 in.)k × TA j

+ [−(10 in.)i + (10 in.)k ] × [−(45 lb)j] = 0 −20TC k − 15TC i − 20TAi + 450k + 450i = 0

Equating to zero the coefficients of the unit vectors, k:

−20TC + 450 = 0

TC = 22.5 lb 

i:

−15(22.5) − 20TA + 450 = 0

TA = 5.625 lb 

ΣFy = 0:

TA + TB + TC − 45 lb = 0

5.625 lb + TB + 22.5 lb − 45 lb = 0

TB = 16.875 lb  TA = 5.63 lb; TB = 16.88 lb; TC = 22.5 lb 


www.elsolucionario.net PROBLEM 4.100 The table shown weighs 30 lb and has a diameter of 4 ft. It is supported by three legs equally spaced around the edge. A vertical load P of magnitude 100 lb is applied to the top of the table at D. Determine the maximum value of a if the table is not to tip over. Show, on a sketch, the area of the table over which P can act without tipping the table.

SOLUTION r = 2 ft b = r sin 30° = 1 ft

We shall sum moments about AB. (b + r )C + (a − b) P − bW = 0

1 C = [30 − (a − 1)100] 3

If table is not to tip, C ≥ 0. [30 − ( a − 1)100] ≥ 0 30 ≥ (a − 1)100

a − 1 ≤ 0.3 a ≤ 1.3 ft a = 1.300 ft

Only ⊥ distance from P to AB matters. Same condition must be satisfied for each leg. P must be located in shaded area for no tipping.



(1 + 2)C + (a − 1)100 − (1)30 = 0

www.elsolucionario.net PROBLEM 4.101 An opening in a floor is covered by a 1 × 1.2-m sheet of plywood of mass 18 kg. The sheet is hinged at A and B and is maintained in a position slightly above the floor by a small block C. Determine the vertical component of the reaction (a) at A, (b) at B, (c) at C.

rB/A = 0.6i rC/A = 0.8i + 1.05k rG/A = 0.3i + 0.6k

W = mg = (18 kg)9.81 W = 176.58 N ΣM A = 0: rB/A × Bj + rC/A × Cj + rG/A × (−Wj) = 0 (0.6i ) × Bj + (0.8i + 1.05k ) × Cj + (0.3i + 0.6k ) × ( −Wj) = 0 0.6 Bk + 0.8Ck − 1.05Ci − 0.3Wk + 0.6Wi = 0

Equate coefficients of unit vectors to zero:  0.6  i : 1.05C + 0.6W = 0 C =  176.58 N = 100.90 N  1.05  k : 0.6 B + 0.8C − 0.3W = 0 0.6 B + 0.8(100.90 N) − 0.3(176.58 N) = 0 B = −46.24 N

ΣFy = 0: A + B + C − W = 0 A − 46.24 N + 100.90 N + 176.58 N = 0

A = 121.92 N

(a ) A = 121.9 N (b) B = −46.2 N (c) C = 100.9 N 



SOLUTION

www.elsolucionario.net PROBLEM 4.102 Solve Problem 4.101, assuming that the small block C is moved and placed under edge DE at a point 0.15 m from corner E. PROBLEM 4.101 An opening in a floor is covered by a 1 × 1.2-m sheet of plywood of mass 18 kg. The sheet is hinged at A and B and is maintained in a position slightly above the floor by a small block C. Determine the vertical component of the reaction (a) at A, (b) at B, (c) at C.

rB/A = 0.6i rC/A = 0.65i + 1.2k rG/A = 0.3i + 0.6k

W = mg = (18 kg) 9.81 m/s 2 W = 176.58 N ΣM A = 0: rB/A × Bj + rC/A × Cj + rG/A × (−Wj) = 0 0.6i × Bj + (0.65i + 1.2k ) × Cj + (0.3i + 0.6k ) × (−Wj) = 0 0.6 Bk + 0.65Ck − 1.2Ci − 0.3Wk + 0.6Wi = 0

Equate coefficients of unit vectors to zero:  0.6  C = 176.58 N = 88.29 N  1.2 

i : −1.2C + 0.6W = 0 k : 0.6 B + 0.65C − 0.3W = 0

0.6 B + 0.65(88.29 N) − 0.3(176.58 N) = 0 B = −7.36 N

ΣFy = 0: A + B + C − W = 0 A − 7.36 N + 88.29 N − 176.58 N = 0

A = 95.648 N

(a ) A = 95.6 N (b) B = − 7.36 N (c) C = 88.3 N 



SOLUTION

www.elsolucionario.net PROBLEM 4.103 The rectangular plate shown weighs 80 lb and is supported by three vertical wires. Determine the tension in each wire.

SOLUTION


Free-Body Diagram:

ΣM B = 0: rA/B × TA j + rC/B × TC j + rG/B × (−80 lb) j = 0 (60 in.)k × TA j + [(60 in.)i + (15 in.)k ] × TC j + [(30 in.)i + (30 in.)k ] × ( −80 lb) j = 0

−60TAi + 60TC k − 15TC i − 2400k + 2400i = 0

Equating to zero the coefficients of the unit vectors, i:

60TA − 15(40) + 2400 = 0

TA = 30.0 lb 

k:

60TC − 2400 = 0

TC = 40.0 lb 

ΣFy = 0:

TA + TB + TC − 80 lb = 0 30 lb + TB + 40 lb − 80 lb = 0

TB = 10.00 lb 


www.elsolucionario.net PROBLEM 4.104 The rectangular plate shown weighs 80 lb and is supported by three vertical wires. Determine the weight and location of the lightest block that should be placed on the plate if the tensions in the three wires are to be equal.

SOLUTION


Free-Body Diagram:

Let −Wb j be the weight of the block and x and z the block’s coordinates. Since tensions in wires are equal, let TA = TB = TC = T ΣM 0 = 0: (rA × Tj) + (rB × Tj) + (rC × Tj) + rG × (−Wj) + ( xi + zk ) × ( −Wb j) = 0

or

(75 k ) × Tj + (15 k ) × Tj + (60i + 30k ) × Tj + (30i + 45k ) × (−Wj) + ( xi + zk ) × (−Wb j) = 0

or

−75T i − 15T i + 60T k − 30T i − 30W k + 45W i − Wb × k + Wb z i = 0

Equate coefficients of unit vectors to zero: i:

−120T + 45W + Wb z = 0

(1)

k:

60T − 30W − Wb x = 0

(2)

ΣFy = 0:

3T − W − Wb = 0

(3)

Eq. (1) + 40 Eq. (3):

5W + ( z − 40)Wb = 0

(4)

Eq. (2) – 20 Eq. (3):

−10W − ( x − 20)Wb = 0

(5)

Also,



Solving Eqs. (4) and (5) for Wb /W and recalling that 0 ≤ x ≤ 60 in., 0 ≤ z ≤ 90 in., Eq. (4):

Wb 5 5 = ≥ = 0.125 W 40 − z 40 − 0

Eq. (5):

Wb 10 10 = ≥ = 0.5 W 20 − x 20 − 0

Thus, (Wb ) min = 0.5W = 0.5(80) = 40 lb

(Wb ) min = 40.0 lb 

Making Wb = 0.5W in Eqs. (4) and (5): z = 30.0 in. 

−10W − ( x − 20)(0.5W ) = 0

x = 0 in. 


5W + ( z − 40)(0.5W ) = 0


www.elsolucionario.net PROBLEM 4.105 A 2.4-m boom is held by a ball-and-socket joint at C and by two cables AD and AE. Determine the tension in each cable and the reaction at C.

SOLUTION Five unknowns and six equations of equilibrium, but equilibrium is maintained


Free-Body Diagram: (ΣMAC = 0).

rB = 1.2k

TAD TAE

rA = 2.4k  AD = −0.8i + 0.6 j − 2.4k AD = 2.6 m  AE = 0.8i + 1.2 j − 2.4k AE = 2.8 m  AD TAD = = (−0.8i + 0.6 j − 2.4k ) AD 2.6  AE TAE = = (0.8i + 1.2 j − 2.4k ) AE 2.8

ΣM C = 0: rA × TAD + rA × TAE + rB × (−3 kN) j = 0 i j k i j k TAD T 0 0 2.4 0 2.4 AE + 1.2k × (−3.6 kN) j = 0 + 0 2.6 2.8 0.8 1.2 −2.4 −0.8 0.6 −2.4

Equate coefficients of unit vectors to zero: i : − 0.55385TAD − 1.02857TAE + 4.32 = 0

(1)

j : − 0.73846TAD + 0.68671TAE = 0 TAD = 0.92857TAE

From Eq. (1):

(2)

−0.55385(0.92857)TAE − 1.02857TAE + 4.32 = 0 1.54286TAE = 4.32 TAE = 2.800 kN

TAE = 2.80 kN 



From Eq. (2):

TAD = 0.92857(2.80) = 2.600 kN

TAD = 2.60 kN 

0.8 0.8 (2.6 kN) + (2.8 kN) = 0 2.6 2.8 0.6 1.2 (2.6 kN) + (2.8 kN) − (3.6 kN) = 0 ΣFy = 0: C y + 2.6 2.8 2.4 2.4 (2.6 kN) − (2.8 kN) = 0 ΣFz = 0: C z − 2.6 2.8 ΣFx = 0: C x −

Cx = 0 C y = 1.800 kN C z = 4.80 kN


C = (1.800 kN) j + (4.80 kN)k 


www.elsolucionario.net PROBLEM 4.106 Solve Problem 4.105, assuming that the 3.6-kN load is applied at Point A. PROBLEM 4.105 A 2.4-m boom is held by a ball-and-socket joint at C and by two cables AD and AE. Determine the tension in each cable and the reaction at C.

SOLUTION Five unknowns and six equations of equilibrium, but equilibrium is maintained


Free-Body Diagram: (ΣMAC = 0).

 AD = −0.8i + 0.6 j − 2.4k AD = 2.6 m  AE = 0.8i + 1.2 j − 2.4k AE = 2.8 m  AD TAD (−0.8i + 0.6 j − 2.4k ) TAD = = AD 2.6  AE TAE (0.8i + 1.2 j − 2.4k ) TAE = = AE 2.8

ΣM C = 0: rA × TAD + rA × TAE + rA × (−3.6 kN) j

Factor rA : or Coefficient of i:

rA × (TAD + TAE − (3.6 kN) j) TAD + TAE − (3 kN) j = 0

−

(Forces concurrent at A)

TAD T (0.8) + AE (0.8) = 0 2.6 2.8 TAD =

Coefficient of j:

2.6 TAE 2.8

(1)

TAD T (0.6) + AE (1.2) − 3.6 kN = 0 2.6 2.8 2.6  0.6  1.2 TAE  TAE − 3.6 kN = 0 + 2.8  2.6  2.8  0.6 + 1.2  TAE   = 3.6 kN  2.8  TAE = 5.600 kN

TAE = 5.60 kN 



From Eq. (1):

TAD =

2.6 (5.6) = 5.200 kN 2.8

TAD = 5.20 kN 

0.8 0.8 (5.2 kN) + (5.6 kN) = 0 2.6 2.8 0.6 1.2 ΣFy = 0: C y + (5.2 kN) + (5.6 kN) − 3.6 kN = 0 2.6 2.8 2.4 2.4 ΣFz = 0: Cz − (5.2 kN) − (5.6 kN) = 0 2.6 2.8 ΣFx = 0: C x −

Cx = 0 Cy = 0 Cz = 9.60 kN C = (9.60 kN)k 


Note: Since the forces and reaction are concurrent at A, we could have used the methods of Chapter 2.


www.elsolucionario.net PROBLEM 4.107 A 10-ft boom is acted upon by the 840-lb force shown. Determine the tension in each cable and the reaction at the ball-and-socket joint at A.

SOLUTION We have five unknowns and six equations of equilibrium, but equilibrium is maintained (ΣM x = 0).


Free-Body Diagram:

 BD = (−6 ft)i + (7 ft) j + (6 ft)k BD = 11 ft  BE = (−6 ft)i + (7 ft) j − (6 ft)k BE = 11 ft  BD TBD TBD = TBD = (−6i + 7 j + 6k ) BD 11  BE TBE TBE = TBE = (−6i + 7 j − 6k ) BE 11

ΣM A = 0: rB × TBD + rB × TBE + rC × ( −840 j) = 0 6i ×

TBD T (−6i + 7 j + 6k ) + 6i × BE (−6i + 7 j − 6k ) + 10i × (−840 j) = 0 11 11 42 36 42 36 TBD k − TBD j + TBE k + TBE j − 8400k = 0 11 11 11 11

Equate coefficients of unit vectors to zero: i: − k:

36 36 TBD + TBE = 0 TBE = TBD 11 11

42 42 TBD + TBE − 8400 = 0 11 11  42  2  TBD  = 8400  11 

TBD = 1100 lb  TBE = 1100 lb 



ΣFx = 0: Ax −

6 6 (1100 lb) − (1100 lb) = 0 11 11 Ax = 1200 lb

ΣFy = 0: Ay +

7 7 (1100 lb) + (1100 lb) − 840 lb = 0 11 11 Ay = −560 lb

ΣFz = 0: Az +

6 6 (1100 lb) − (1100 lb) = 0 11 11 A = (1200 lb)i − (560 lb) j 


Az = 0


www.elsolucionario.net PROBLEM 4.108 A 12-m pole supports a horizontal cable CD and is held by a ball and socket at A and two cables BE and BF. Knowing that the tension in cable CD is 14 kN and assuming that CD is parallel to the x-axis (φ = 0), determine the tension in cables BE and BF and the reaction at A.

SOLUTION

There are five unknowns and six equations of equilibrium. The pole is free to rotate about the y-axis, but equilibrium is maintained under the given loading (ΣM y = 0).   Resolve BE and BF into components:  BE = (7.5 m)i − (8 m) j + (6 m)k  BF = (7.5 m)i − (8 m) j − (6 m)k

BE = 12.5 m BF = 12.5 m

Express TBE and TBF in terms of components: TBE = TBE TBF = TBF

 BE = TBE (0.60i − 0.64 j + 0.48k ) BE  BF = TBF (0.60i − 0.64 j − 0.48k ) BF

(1) (2)



Free-Body Diagram:


ΣM A = 0: rB/A × TBE + rB/A × TBF + rC/A × (−14 kN)i = 0 8 j × TBE (0.60i − 0.64 j + 0.48k ) + 8 j × TBF (0.60i − 0.64 j − 0.48k ) + 12 j × (−14i ) = 0 −4.8 TBE k + 3.84 TBE i − 4.8TBF k − 3.84TBF i + 168k = 0

Equating the coefficients of the unit vectors to zero, i:

3.84TBE − 3.84TBF = 0

TBE = TBF

k:

−4.8TBE − 4.8TBF + 168 = 0

ΣFx = 0:

Ax + 2(0.60)(17.50 kN) − 14 kN = 0

Ax = 7.00 kN

ΣFy = 0:

Ay − z (0.64) (17.50 kN) = 0

Ay = 22.4 kN

ΣFz = 0:

Az + 0 = 0

TBE = TBF = 17.50 kN 

Az = 0

Because of the symmetry, we could have noted at the outset that TBF = TBE and eliminated one unknown.



A = −(7.00 kN)i + (22.4 kN) j 

www.elsolucionario.net PROBLEM 4.109 Solve Problem 4.108, assuming that cable CD forms an angle φ = 25° with the vertical xy plane. PROBLEM 4.108 A 12-m pole supports a horizontal cable CD and is held by a ball and socket at A and two cables BE and BF. Knowing that the tension in cable CD is 14 kN and assuming that CD is parallel to the x-axis (φ = 0), determine the tension in cables BE and BF and the reaction at A.

SOLUTION

 BE = (7.5 m)i − (8 m) j + (6 m)k BE = 12.5 m  BF = (7.5 m)i − (8 m)j − (6 m)k BF = 12.5 m  BE TBE = TBE = TBE (0.60i − 0.64 j + 0.48k ) BE  BF TBF = TBF = TBF (0.60i − 0.64 j − 0.48k ) BF ΣM A = 0: rB/A × TBE + rB/A × TBF + rC/A × TCD = 0 8 j × TBE (0.60i − 0.64 j + 0.48k ) + 8 j × TBF (0.60i − 0.64 j − 0.48k ) + 12 j × (19 kN)(− cos 25° i + sin 25°k ) = 0 −4.8TBE k + 3.84TBE i − 4.8TBF k − 3.84TBF i + 152.6 k − 71.00 i = 0

Equating the coefficients of the unit vectors to zero,

Solving simultaneously,

i: 3.84TBE − 3.84TBF + 71.00 = 0;

TBF − TBE = 18.4896

k: − 4.8TBE − 4.8 TBF + 152.26 = 0;

TBF + TBE = 31.721

TBE = 6.6157 kN;

TBF = 25.105 kN TBE = 6.62 kN; TBF = 25.1 kN 

ΣFx = 0: Ax + (0.60) (TBF + TBE ) − 14 cos 25° = 0 Ax = 12.6883 − 0.60(31.7207) Ax = −6.34 kN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 473


Free-Body Diagram:


ΣFy = 0: Ay − (0.64) (TBF + TBE ) = 0 Ay = 0.64(31.721) Ay = 20.3 kN ΣFz = 0: Az − 0.48(TBF − TBE ) + 14sin 25° = 0 Az = 0.48(18.4893) − 5.9167 Az = 2.96 kN


A = −(6.34 kN)i + (20.3 kN) j + (2.96 kN) k 


www.elsolucionario.net PROBLEM 4.110 A 48-in. boom is held by a ball-and-socket joint at C and by two cables BF and DAE; cable DAE passes around a frictionless pulley at A. For the loading shown, determine the tension in each cable and the reaction at C.

SOLUTION

Five unknowns and six equations of equilibrium, but equilibrium is maintained (ΣMAC = 0). T = Tension in both parts of cable DAE. rB = 30k rA = 48k  AD = −20i − 48k AD = 52 in.  AE = 20 j − 48k AE = 52 in.  BF = 16i − 30k BF = 34 in.  AD T T TAD = T = (−20i − 48k ) = ( −5i − 12k ) AD 52 13  AE T T TAE = T = (20 j − 48k ) = (5 j − 12k ) AE 52 13  T BF TBF TBF = TBF = (16i − 30k ) = BF (8i − 15k ) BF 34 17

ΣM C = 0: rA × TAD + rA × TAE + rB × TBF + rB × (−320 lb) j = 0 i j k i j k i j k T T T + 0 0 48 + 0 0 30 BF + (30k ) × (−320 j) = 0 0 0 48 13 13 17 −5 0 −12 0 5 −12 8 0 −15

Coefficient of i:

−

240 T + 9600 = 0 13

T = 520 lb



Free-Body Diagram:


Coefficient of j:

−

240 240 T+ TBD = 0 13 17 TBD =

17 17 T = (520) TBD = 680 lb 13 13

ΣF = 0: TAD + TAE + TBF − 320 j + C = 0 −

20 8 (520) + (680) + Cx = 0 52 17

−200 + 320 + Cx = 0

Coefficient of j:

20 (520) − 320 + C y = 0 52 200 − 320 + C y = 0

Coefficient of k:

Cx = −120 lb

−

C y = 120 lb

48 48 30 (520) − (520) − (680) + Cz = 0 52 52 34 −480 − 480 − 600 + Cz = 0 Cz = 1560 lb

Answers: TDAE = T

TDAE = 520 lb  TBD = 680 lb  C = −(120.0 lb)i + (120.0 lb) j + (1560 lb)k 



Coefficient of i:

www.elsolucionario.net PROBLEM 4.111 Solve Problem 4.110, assuming that the 320-lb load is applied at A. PROBLEM 4.110 A 48-in. boom is held by a ball-andsocket joint at C and by two cables BF and DAE; cable DAE passes around a frictionless pulley at A. For the loading shown, determine the tension in each cable and the reaction at C.

SOLUTION

Five unknowns and six equations of equilibrium, but equilibrium is maintained (ΣMAC = 0). T = tension in both parts of cable DAE. rB = 30k rA = 48k  AD = −20i − 48k AD = 52 in.  AE = 20 j − 48k AE = 52 in.  BF = 16i − 30k BF = 34 in.  AD T T TAD = T = (−20i − 48k ) = ( −5i − 12k ) AD 52 13  AE T T TAE = T = (20 j − 48k ) = (5 j − 12k ) AE 52 13  T BF TBF TBF = TBF = (16i − 30k ) = BF (8i − 15k ) BF 34 17

ΣM C = 0: rA × TAD + rA × TAE + rB × TBF + rA × ( −320 lb) j = 0 i j k i j k i j k T T T + 0 0 48 + 0 0 30 BF + 48k × (−320 j) = 0 0 0 48 13 13 17 −5 0 −12 0 5 −12 8 0 −15

Coefficient of i:

−

240 T + 15,360 = 0 13

T = 832 lb



Free-Body Diagram:


Coefficient of j:

−

240 240 T+ TBD = 0 13 17 TBD =

17 17 T = (832) 13 13

TBD = 1088 lb

ΣF = 0: TAD + TAE + TBF − 320 j + C = 0 −

Coefficient of i:

20 8 (832) + (1088) + C x = 0 52 17

−320 + 512 + Cx = 0 20 (832) − 320 + C y = 0 52 320 − 320 + C y = 0 −

Cy = 0

48 48 30 (832) − (852) − (1088) + Cz = 0 52 52 34 −768 − 768 − 960 + Cz = 0

Answers:

TDAE = T

Cz = 2496 lb TDAE = 832 lb  TBD = 1088 lb  C = −(192.0 lb)i + (2496 lb)k 



Coefficient of j:

Coefficient of k:

Cx = −192 lb



WC = 600 lb WG = 200 lb

We have five unknowns (TDE , TDF , Ax , Ay , Az ) and five equilibrium equations. The boom is free to spin about the AB axis, but equilibrium is maintained, since ΣM AB = 0.  BH = (30 ft)i − (22.5 ft) j BH = 37.5 ft We have  8.8 DE = (13.8 ft)i − (22.5 ft) j + (6.6 ft)k 12 = (13.8 ft)i − (16.5 ft) j + (6.6 ft)k DE = 22.5 ft  DF = (13.8 ft)i − (16.5 ft) j − (6.6 ft)k DF = 22.5 ft



A 600-lb crate hangs from a cable that passes over a pulley B and is attached to a support at H. The 200-lb boom AB is supported by a ball-and-socket joint at A and by two cables DE and DF. The center of gravity of the boom is located at G. Determine (a) the tension in cables DE and DF, (b) the reaction at A.


TBH = TBH

Thus:

TDE = TDE TDF = TDF

(a)

 BH 30i − 22.5 j = (600 lb) = (480 lb)i − (360 lb) j BH 37.5  DE TDE = (13.8i − 16.5 j + 6.6k ) DE 22.5  DF TDE = (13.8i − 16.5 j − 6.6k ) DF 22.5

ΣM A = 0: (rJ × WC ) + (rK × WG ) + (rH × TBH ) + (rE × TDE ) + (rF × TDF ) = 0 − (12i ) × (−600 j) − (6i ) × (−200 j) + (18i ) × (480i − 360 j)

7200k + 1200k − 6480k + 4.84(TDE − TDF )i

or

+

58.08 82.5 (TDE − TDF ) j − (TDE + TDF )k = 0 22.5 22.5

Equating to zero the coefficients of the unit vectors, i or j:

TDE − TDF = 0

k : 7200 + 1200 − 6480 −

TDE = TDF *

82.5 (2TDE ) = 0 22.5

TDE = 261.82 lb TDE = TDF = 262 lb 

(b)

 13.8  ΣFx = 0: Ax + 480 + 2   (261.82) = 0  22.5   16.5  ΣFy = 0: Ay − 600 − 200 − 360 − 2   (261.82) = 0  22.5  ΣFz = 0: Az = 0

Ax = −801.17 lb Ay = 1544.00 lb A = −(801 lb)i + (1544 lb) j 

*Remark: The fact is that TDE = TDF could have been noted at the outset from the symmetry of structure with respect to xy plane.



i j k i j k TDE TDF 5 0 6.6 + 5 0 + −6.6 = 0 22.5 22.5 13.8 −16.5 6.6 13.8 −16.5 −6.6


SOLUTION rB/A (960 − 180)i = 780i

Dimensions in mm

 960  450 rG/A =  − 90  i + k 2  2  = 390i + 225k rC/A = 600i + 450k

T = Tension in cable DCE

 CD = −690i + 675 j − 450k  CE = 270i + 675 j − 450k

CD = 1065 mm CE = 855 mm

T (−690i + 675 j − 450k ) 1065 T TCE = (270i + 675 j − 450k ) 855 W = −mgi = −(100 kg)(9.81 m/s 2 ) j = −(981 N) j

TCD =

ΣM A = 0: rC/A × TCD + rC/A × TCE + rG/A × (−Wj) + rB/A × B = 0 i j k i j k T T 600 0 450 450 + 600 0 1065 855 270 675 −450 −690 675 −450 i + 390 0

j 0 −981

k i 225 + 780 0

0

j 0

k 0 =0

By

Bz



A 100-kg uniform rectangular plate is supported in the position shown by hinges A and B and by cable DCE that passes over a frictionless hook at C. Assuming that the tension is the same in both parts of the cable, determine (a) the tension in the cable, (b) the reactions at A and B. Assume that the hinge at B does not exert any axial thrust.


Coefficient of i:

−(450)(675)

T T − (450)(675) + 220.73 × 103 = 0 1065 855 T = 344.64 N

Coefficient of j:

(−690 × 450 + 600 × 450)

T = 345 N 

344.64 344.64 + (270 × 450 + 600 × 450) − 780 Bz = 0 1065 855 Bz = 185.516 N

Coefficient of k: (600)(675)

344.64 344.64 + (600)(675) − 382.59 × 103 + 780 By = 0 By = 113.178 N 1065 855 B = (113.2 N) j + (185.5 N)k 

690 270 (344.64) + (344.64) = 0 1065 855

Coefficient of i:

Ax −

Ax = 114.5 N

Coefficient of j:

Ay + 113.178 +

675 675 (344.64) + (344.64) − 981 = 0 1065 855

Ay = 377 N

Coefficient of k:

Az + 185.516 −

450 450 (344.64) − (344.64) = 0 1065 855

Az = 141.5 N

A = (114.5 N)i + (377 N) j + (144.5 N)k 



ΣF = 0: A + B + TCD + TCE + W = 0

www.elsolucionario.net PROBLEM 4.114 Solve Problem 4.113, assuming that cable DCE is replaced by a cable attached to Point E and hook C. PROBLEM 4.113 A 100-kg uniform rectangular plate is supported in the position shown by hinges A and B and by cable DCE that passes over a frictionless hook at C. Assuming that the tension is the same in both parts of the cable, determine (a) the tension in the cable, (b) the reactions at A and B. Assume that the hinge at B does not exert any axial thrust.

SOLUTION


See solution to Problem 4.113 for free-body diagram and analysis leading to the following: CD = 1065 mm CE = 855 mm

T (−690i + 675 j − 450k ) 1065 T (270i + 675 j − 450k ) TCE = 855 W = −mgi = −(100 kg)(9.81 m/s 2 ) j = −(981 N)j

TCD =

Now,

ΣM A = 0: rC/A × TCE + rG/A × (−W j) + rB/A × B = 0 i j k i j k i j T 600 0 450 0 225 + 780 0 + 390 855 270 675 −450 0 −981 0 0 By

Coefficient of i:

−(450)(675)

k 0 =0 Bz

T + 220.73 × 103 = 0 855 T = 621.31 N

Coefficient of j: Coefficient of k:

(270 × 450 + 600 × 450) (600)(675)

T = 621 N 

621.31 − 780 Bz = 0 Bz = 364.74 N 855

621.31 − 382.59 × 103 + 780 By = 0 By = 113.186 N 855 B = (113.2 N)j + (365 N)k 



ΣF = 0: A + B + TCE + W = 0 270 (621.31) = 0 855

Coefficient of i:

Ax +

Coefficient of j:

Ay + 113.186 +

Coefficient of k:

Az + 364.74 −

675 (621.31) − 981 = 0 855

450 (621.31) = 0 855

Ay = 377.3 N Az = −37.7 N  A = −(196.2 N)i + (377 N)j − (37.7 N)k 






Ax = −196.2 N


www.elsolucionario.net PROBLEM 4.115 The rectangular plate shown weighs 75 lb and is held in the position shown by hinges at A and B and by cable EF. Assuming that the hinge at B does not exert any axial thrust, determine (a) the tension in the cable, (b) the reactions at A and B.

SOLUTION rB/A = (38 − 8)i = 30i rE/A = (30 − 4)i + 20k


rG/A

= 26i + 20k 38 = i + 10k 2 = 19i + 10k

 EF = 8i + 25 j − 20k EF = 33 in.  AE T T =T = (8i + 25 j − 20k ) AE 33 ΣM A = 0: rE/A × T + rG/A × (−75 j) + rB/A × B = 0 i j k i j k i j T + 19 0 10 + 30 0 26 0 20 33 8 25 −20 0 −75 0 0 By −(25)(20)

Coefficient of i: Coefficient of j: Coefficient of k:

(160 + 520) (26)(25)

T + 750 = 0: 33

k 0 =0 Bz T = 49.5 lb 

49.5 − 30 Bz = 0: Bz = 34 lb 33

49.5 − 1425 + 30 By = 0: By = 15 lb 33

B = (15.00 lb)j + (34.0 lb)k 

    PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 485




ΣF = 0: A + B + T − (75 lb)j = 0  Ax +


Ay + 15 +

8 (49.5) = 0 33

25 (49.5) − 75 = 0 33

Az + 34 −

20 (49.5) = 0 33

Ax = −12.00 lb Ay = 22.5 lb Az = −4.00 lb  A = −(12.00 lb)i + (22.5 lb)j − (4.00 lb)k 





www.elsolucionario.net PROBLEM 4.116 Solve Problem 4.115, assuming that cable EF is replaced by a cable attached at points E and H. PROBLEM 4.115 The rectangular plate shown weighs 75 lb and is held in the position shown by hinges at A and B and by cable EF. Assuming that the hinge at B does not exert any axial thrust, determine (a) the tension in the cable, (b) the reactions at A and B.

SOLUTION rB/A = (38 − 8)i = 30i rE/A = (30 − 4)i + 20k


= 26i + 20k

38 i + 10k 2 = 19i + 10k

rG/A =

 EH = −30i + 12 j − 20k EH = 38 in.  EH T T=T = (−30i + 12 j − 20k ) EH 38 ΣM A = 0: rE/A × T + rG/A × (−75 j) + rB/A × B = 0

i j k i j k i j T + 19 0 10 + 30 0 26 0 20 38 −30 12 −20 0 −75 0 0 By −(12)(20)


(−600 + 520) (26)(12)

T + 750 = 0 38

T = 118.75

k 0 =0 Bz T = 118.8lb 

118.75 − 30 Bz = 0 Bz = −8.33lb 38

118.75 − 1425 + 30 By = 0 B y = 15.00 lb 38

B = (15.00 lb)j − (8.33 lb)k 



ΣF = 0: Ax −


Ay + 15 +

A + B + T − (75 lb)j = 0 30 (118.75) = 0 38

12 (118.75) − 75 = 0 38

Az − 8.33 −

20 (118.75) = 0 38

Ax = 93.75 lb Ay = 22.5 lb Az = 70.83 lb

A = (93.8 lb)i + (22.5 lb)j + (70.8 lb)k 





www.elsolucionario.net PROBLEM 4.117 A 20-kg cover for a roof opening is hinged at corners A and B. The roof forms an angle of 30° with the horizontal, and the cover is maintained in a horizontal position by the brace CE. Determine (a) the magnitude of the force exerted by the brace, (b) the reactions at the hinges. Assume that the hinge at A does not exert any axial thrust.

SOLUTION F = F (cos 75°)i + F (sin 75°) j F = F (0.25882i + 0.96593j) W = mg = 20 kg(9.81 m/s 2 ) = 196.2 N rA /B = 0.6k rC /B = 0.9i + 0.6k rG /B = 0.45i + 0.3k F = F (0.25882i + 0.96593j) ΣM B = 0: rG/B × (−196.2 j) + rC/B × F + rA/B × A = 0

(a)

i j k i j k i 0.45 0 0.3 + 0.9 0 0.6 F + 0 0 −196.2 0 0.25882 +0.96593 0 Ax

j 0 Ay

k 0.6 = 0 0

Coefficient of i :

+58.86 − 0.57956 F − 0.6 Ay = 0

(1)

Coefficient of j:

+0.155292 F + 0.6 Ax = 0

(2)

Coefficient of k:

−88.29 + 0.86934 F = 0:

From Eq. (2):

+58.86 − 0.57956(101.56) − 0.6 Ay = 0

From Eq. (3):

+0.155292(101.56) + 0.6 Ax = 0

F = 101.56 N Ay = 0 Ax = −26.286 N F = (101.6 N) 

(b) Coefficient of i: Coefficient of j:

ΣF : A + B + F − Wj = 0 26.286 + Bx + 0.25882(101.56) = 0 Bx = 0 B y + 0.96593(101.56) − 196.2 = 0 B y = 98.1 N Bz = 0

Coefficient of k:

A = −(26.3 N)i; B = (98.1 N) j 



Force exerted by CE:

www.elsolucionario.net PROBLEM 4.118 The bent rod ABEF is supported by bearings at C and D and by wire AH. Knowing that portion AB of the rod is 250 mm long, determine (a) the tension in wire AH, (b) the reactions at C and D. Assume that the bearing at D does not exert any axial thrust.

SOLUTION ΔABH is equilateral.

Free-Body Diagram:


Dimensions in mm

rH/C = −50i + 250 j rD/C = 300i rF/C = 350i + 250k T = T (sin 30°) j − T (cos 30°)k = T (0.5 j − 0.866k ) ΣM C = 0: rH/C × T + rD × D + rF/C × (−400 j) = 0 i j k i j −50 250 0 T + 300 0 0 0.5 −0.866 0 Dy

Coefficient i:

k i j k 0 + 350 0 250 = 0 Dz 0 −400 0

−216.5T + 100 × 103 = 0 T = 461.9 N

Coefficient of j:

T = 462 N 

−43.3T − 300 Dz = 0 −43.3(461.9) − 300 Dz = 0

Dz = −66.67 N



Coefficient of k:

−25T + 300 D y − 140 × 103 = 0 −25(461.9) + 300 Dy − 140 × 103 = 0

D y = 505.1 N D = (505 N) j − (66.7 N)k 

ΣF = 0: C + D + T − 400 j = 0

Coefficient i:

Cx = 0

Cx = 0

Coefficient j:

C y + (461.9)0.5 + 505.1 − 400 = 0 C y = −336 N

Coefficient k:

Cz − (461.9)0.866 − 66.67 = 0

C = −(336 N) j + (467 N)k 


Cz = 467 N


www.elsolucionario.net PROBLEM 4.119 Solve Problem 4.115, assuming that the hinge at B is removed and that the hinge at A can exert couples about axes parallel to the y and z axes. PROBLEM 4.115 The rectangular plate shown weighs 75 lb and is held in the position shown by hinges at A and B and by cable EF. Assuming that the hinge at B does not exert any axial thrust, determine (a) the tension in the cable, (b) the reactions at A and B.

SOLUTION rG/A = (0.5 × 38)i + 10k = 19i + 10k  AE = 8i + 25 j − 20k AE = 33 in.  AE T T =T = (8i + 25 j − 20k ) AE 33

ΣM A = 0: rE/A × T + rG/A × (−75 j) + ( M A ) y j + ( M A ) z k = 0 i j k i j k T + 19 0 10 + ( M A ) y j + ( M A ) z k = 0 26 0 20 33 8 25 −20 0 −75 0 −(20)(25)


(160 + 520) (26)(25)

T + 750 = 0 33

T = 49.5 lb 

49.5 + ( M A ) y = 0 ( M A ) y = −1020 lb ⋅ in. 33

49.5 − 1425 + ( M A ) z = 0 33

( M A ) z = 450 lb ⋅ in.

ΣF = 0: A + T − 75 j = 0 Ax +

Coefficient of i:

8 (49.5) = 0 33

M A = −(1020 lb ⋅ in.)j + (450 lb ⋅ in.)k 

Ax = 12.00 lb

Coefficient of j:

Ay +

25 (49.5) − 75 = 0 33

Ay = 37.5 lb

Coefficient of k:

Az −

20 (49.5) = 0 33

Az = 30.0 lb A = −(12.00 lb)i + (37.5 lb)j + (30.0 lb)k 



rE/A = (30 − 4)i + 20k = 26i + 20k

www.elsolucionario.net PROBLEM 4.120 Solve Problem 4.118, assuming that the bearing at D is removed and that the bearing at C can exert couples about axes parallel to the y and z axes. PROBLEM 4.118 The bent rod ABEF is supported by bearings at C and D and by wire AH. Knowing that portion AB of the rod is 250 mm long, determine (a) the tension in wire AH, (b) the reactions at C and D. Assume that the bearing at D does not exert any axial thrust.


ΔABH is equilateral.


Dimensions in mm

rH/C = −50i + 250 j rF/C = 350i + 250k T = T (sin 30°) j − T (cos 30°)k = T (0.5 j − 0.866k ) ΣM C = 0: rF/C × (−400 j) + rH/C × T + ( M C ) y j + ( M C ) z k = 0 i j k i j k 350 0 250 + −50 250 0 T + (M C ) y j + (M C ) z k = 0 0 −400 0 0 0.5 −0.866

Coefficient of i: Coefficient of j:

+100 × 103 − 216.5T = 0 T = 461.9 N

T = 462 N 

−43.3(461.9) + ( M C ) y = 0 ( M C ) y = 20 × 103 N ⋅ mm (M C ) y = 20.0 N ⋅ m



Coefficient of k:

−140 × 103 − 25(461.9) + ( M C ) z = 0 ( M C ) z = 151.54 × 103 N ⋅ mm (M C ) z = 151.5 N ⋅ m ΣF = 0: C + T − 400 j = 0

M C = (20.0 N ⋅ m)j + (151.5 N ⋅ m)k 


Cx = 0 C y + 0.5(461.9) − 400 = 0 C y = 169.1 N C z − 0.866(461.9) = 0 C z = 400 N

C = (169.1 N)j + (400 N)k 





www.elsolucionario.net PROBLEM 4.121 The assembly shown is welded to collar A that fits on the vertical pin shown. The pin can exert couples about the x and z axes but does not prevent motion about or along the y-axis. For the loading shown, determine the tension in each cable and the reaction at A.

SOLUTION

First note:

−(0.08 m)i + (0.06 m) j

TCF = λ CF TCF =

(0.08) 2 + (0.06)2 m

TCF

= TCF (−0.8i + 0.6 j) TDE = λ DE TDE =

(0.12 m)j − (0.09 m)k (0.12) 2 + (0.09)2 m

TDE

= TDE (0.8 j − 0.6k )

(a)

From F.B.D. of assembly: ΣFy = 0: 0.6TCF + 0.8TDE − 480 N = 0 0.6TCF + 0.8TDE = 480 N

or

(1)

ΣM y = 0: − (0.8TCF )(0.135 m) + (0.6TDE )(0.08 m) = 0 TDE = 2.25TCF

or

(2)

Substituting Equation (2) into Equation (1), 0.6TCF + 0.8[(2.25)TCF ] = 480 N TCF = 200.00 N TCF = 200 N 

or 

and from Equation (2):

TDE = 2.25(200.00 N) = 450.00 TDE = 450 N 

or



Free-Body Diagram:


(b)

From F.B.D. of assembly: ΣFz = 0: Az − (0.6)(450.00 N) = 0

Az = 270.00 N

ΣFx = 0: Ax − (0.8)(200.00 N) = 0

Ax = 160.000 N

or A = (160.0 N)i + (270 N)k  ΣM x = 0:

MAx + (480 N)(0.135 m) − [(200.00 N)(0.6)](0.135 m) − [(450 N)(0.8)](0.09 m) = 0

M Ax = −16.2000 N ⋅ m ΣM z = 0:

MAz − (480 N)(0.08 m) + [(200.00 N)(0.6)](0.08 m) + [(450 N)(0.8)](0.08 m) = 0

or M A = −(16.20 N ⋅ m)i 



M Az = 0

www.elsolucionario.net PROBLEM 4.122 The assembly shown is used to control the tension T in a tape that passes around a frictionless spool at E. Collar C is welded to rods ABC and CDE. It can rotate about shaft FG but its motion along the shaft is prevented by a washer S. For the loading shown, determine (a) the tension T in the tape, (b) the reaction at C.

Free-Body Diagram: rA/C = 4.2 j + 2k rE/C = 1.6i − 2.4 j ΣM C = 0: rA/C × (−6 j) + rE/C × T (i + k ) + ( M C ) y j + ( M C ) z k = 0 (4.2 j + 2k ) × (−6 j) + (1.6i − 2.4 j) × T (i + k ) + ( M C ) y j + ( M C ) z k = 0


12 − 2.4T = 0

T = 5.00 lb 

−1.6(5 lb) + (M C ) y = 0 ( M C ) y = 8 lb ⋅ in. 2.4(5 lb) + ( M C ) z = 0 ( M C ) z = −12 lb ⋅ in. M C = (8.00 lb ⋅ in.)j − (12.00 lb ⋅ in.)k 

ΣF = 0: Cx i + C y j + C z k − (6 lb)j + (5 lb)i + (5 lb)k = 0

Equate coefficients of unit vectors to zero. Cx = −5 lb C y = 6 lb Cz = −5 lb  C = −(5.00 lb)i + (6.00 lb)j − (5.00 lb)k 





SOLUTION

www.elsolucionario.net PROBLEM 4.123 The rigid L-shaped member ABF is supported by a ball-and-socket joint at A and by three cables. For the loading shown, determine the tension in each cable and the reaction at A.



rB/A = 12i rF/A = 12 j − 8k rD/A = 12i − 16k rE/A = 12i − 24k rF/A = 12i − 32k  BG = −12i + 9k BG = 15 in. λ BG = −0.8i + 0.6k

 DH = −12i + 16 j; DH = 20 in.; λDH = −0.6i + 0.8 j  FJ = −12i + 16 j; FJ = 20 in.; λFJ = −0.6i + 0.8 j

ΣM A = 0: rB/A × TBG λBG + rDH × TDH λDH + rF/A × TFJ λFJ +rF/A × (−24 j) + rE/A × ( −24 j) = 0 i j k i j k i j k 12 0 0 TBG + 12 0 −16 TDH + 12 0 −32 TFJ −0.8 0 0.6 −0.6 0.8 0 −0.6 0.8 0 i j k i j k + 12 0 −8 + 12 0 −24 = 0 0 −24 0 0 −24 0

Coefficient of i:

+12.8TDH + 25.6TFJ − 192 − 576 = 0

(1)

Coefficient of k:

+9.6TDH + 9.6TFJ − 288 − 288 = 0

(2)

9.6TFJ = 0

TFJ = 0 

3 4

Eq. (1) − Eq. (2):



12.8TDH − 268 = 0

TDH = 60 lb 

−7.2TBG + (16 × 0.6)(60.0 lb) = 0

TBG = 80.0 lb 

From Eq. (1): Coefficient of j: ΣF = 0:

A + TBG λ BG + TDH λ DH + TFJ − 24 j − 24 j = 0

Coefficient of i:

Ax + (80)( −0.8) + (60.0)(−0.6) = 0

Coefficient of j:

Ay + (60.0)(0.8) − 24 − 24 = 0

Coefficient of k:

Az + (80.0)(+0.6) = 0

Ax = 100.0 lb Ay = 0 Az = −48.0 lb A = (100.0 lb)i − (48.0 lb) j 


Note: The value Ay = 0 can be confirmed by considering ΣM BF = 0.


www.elsolucionario.net PROBLEM 4.124 Solve Problem 4.123, assuming that the load at C has been removed. PROBLEM 4.123 The rigid L-shaped member ABF is supported by a ball-and-socket joint at A and by three cables. For the loading shown, determine the tension in each cable and the reaction at A.



rB/A = 12i rB/A = 12i − 16k rE/A = 12i − 24k rF /A = 12i − 32k

 BG = −12i + 9k ; BG = 15 in.; λ BG = −0.8i + 0.6k  DH = −12i + 16 j; DH = 20 in.; λ DH = −0.6i + 0.8 j  FJ = −12i + 16 j; FJ = 20 in.; λ FJ = −0.6i + 0.8 j ΣM A = 0: rB/A × TBG λ BG + rD/A × TDH λ DH + rF /A × TFJ λ FJ + rE/A × (−24 j) = 0 i j k i j k i j k i j k 12 0 0 TBG + 12 0 −16 TDH + 12 0 −32 TFJ + 12 0 −24 = 0 −0.8 0 0.6 −0.6 0.8 0 −0.6 0.8 0 0 −24 0 i:

+ 12.8TDH + 25.6TFJ − 576 = 0

(1)

k:

+9.6TDH + 9.6TFJ − 288 = 0

(2)



Multiply Eq. (1) by

3 4

and subtract Eq. (2):

From Eq. (1):

9.6TFJ − 144 = 0

TFJ = 15.00 lb 

12.8TDH + 25.6(15.00) − 576 = 0

TDH = 15.00 lb 

j:

−7.2TBG + (16)(0.6)(15) + (32)(0.6)(15) = 0 − 7.2TBG + 432 = 0

TBG = 60.0 lb 

ΣF = 0: A + TBG λ BG + TDAλ DH + TFJ λ FJ − 24 j = 0

j:

Ay + (15)(0.8) + (15)(0.8) − 24 = 0

k:

Az + (60)(0.6) = 0

Ax = 66.0 lb Ay = 0 Az = −36.0 lb A = (66.0 lb)i − (36.0 lb)k 



i : Ax + (60)(−0.8) + (15)( −0.6) + (15)(−0.6) = 0

www.elsolucionario.net PROBLEM 4.125 The rigid L-shaped member ABC is supported by a ball-and-socket joint at A and by three cables. If a 1.8-kN load is applied at F, determine the tension in each cable.


We have

Thus,

a = 210 mm  CD = (240 mm)j − (320 mm)k CD = 400 mm  BD = −(420 mm)i + (240 mm)j − (320 mm)k BD = 580 mm  BE = (420 mm)i − (320 mm)k BE = 528.02 mm

TCD TBD TBE

 CD = TCD = TCD (0.6 j − 0.8k ) CD  BD = TBD = TBD (−0.72414i + 0.41379 j − 0.55172k ) BD  BE = TBE = TBE (0.79542i − 0.60604k ) BE

ΣM A = 0: (rC × TCD ) + (rB × TBD ) + (rB × TBE ) + (rW × W) = 0



In this problem:

Dimensions in mm


rC = −(420 mm)i + (320 mm)k

Noting that

rB = (320 mm)k rW = − ai + (320 mm)k

and using determinants, we write i j k i j k −420 0 320 TCD + 0 0 320 TBD 0 0.6 −0.8 −0.72414 0.41379 −0.55172 i j k i j k + 0 0 320 TBE + −a 0 320 = 0 0.79542 0 −0.60604 0 −1.8 0

i:

−192TCD − 132.413TBD + 576 = 0

(1)

j:

−336TCD − 231.72TBD + 254.53TBE = 0

(2)

k:

−252TCD + 1.8a = 0

(3)

Recalling that a = 210 mm, Eq. (3) yields TCD =

From Eq. (1):

1.8(210) = 1.500 kN 252

−192(1.5) − 132.413TBD + 576 = 0 TBD = 2.1751 kN

From Eq. (2):

TCD = 1.500 kN 

TBD = 2.18 kN 

−336(1.5) − 231.72(2.1751) + 254.53TBE = 0 TBE = 3.9603 kN 



TBE = 3.96 kN 



Equating to zero the coefficients of the unit vectors,

www.elsolucionario.net PROBLEM 4.126 Solve Problem 4.125, assuming that the 1.8-kN load is applied at C. PROBLEM 4.125 The rigid L-shaped member ABC is supported by a ball-and-socket joint at A and by three cables. If a 1.8-kN load is applied at F, determine the tension in each cable.

SOLUTION

−192TCD − 132.413TBD + 576 = 0

(1)

−336TCD − 231.72TBD + 254.53TBE = 0

(2)

−252TCD + 1.8a = 0

(3)

In this problem, the 1.8-kN load is applied at C and we have a = 420 mm. Carrying into Eq. (3) and solving for TCD , TCD = 3.00

From Eq. (1): From Eq. (2):

−(192)(3) − 132.413TBD + 576 = 0 −336(3) − 0 + 254.53TBE = 0 

TCD = 3.00 kN  TBD = 0  TBE = 3.96 kN 



See solution of Problem 4.125 for free-body diagram and derivation of Eqs. (1), (2), and (3):

www.elsolucionario.net PROBLEM 4.127 The assembly shown consists of an 80-mm rod AF that is welded to a cross consisting of four 200-mm arms. The assembly is supported by a ball-and-socket joint at F and by three short links, each of which forms an angle of 45° with the vertical. For the loading shown, determine (a) the tension in each link, (b) the reaction at F.


SOLUTION

rE/F = −200 i + 80 j TB = TB (i − j) / 2

rB/F = 80 j − 200k

TC = TC (− j + k ) / 2 rC/F = 200i + 80 j TD = TD (− i + j) / 2

rD/E = 80 j + 200k

ΣM F = 0: rB/F × TB + rC/F × TC + rD/F × TD + rE/F × (− Pj) = 0 i j k i j k i j k i j k TC TB TD 0 80 −200 + 200 80 0 + 0 80 200 + −200 80 0 = 0 2 2 2 1 −1 0 0 −1 1 0 −1 −1 0 −P 0

Equate coefficients of unit vectors to zero and multiply each equation by 2. i:

−200 TB + 80 TC + 200 TD = 0

(1)

j:

−200 TB − 200 TC − 200 TD = 0

(2)

k:

−80 TB − 200 TC + 80 TD + 200 2 P = 0

(3)

−80 TB − 80 TC − 80 TD = 0

(4)

80 (2): 200

Eqs. (3) + (4):

−160TB − 280TC + 200 2 P = 0

Eqs. (1) + (2):

−400TB − 120TC = 0 TB = −

(5)

120 TC − 0.3TC 400

(6)



Eqs. (6)

−160( −0.3TC ) − 280TC + 200 2 P = 0

(5):

−232TC + 200 2 P = 0 TC = 1.2191P

TC = 1.219 P 

From Eq. (6):

TB = −0.3(1.2191P) = − 0.36574 = P

From Eq. (2):

− 200(− 0.3657 P) − 200(1.2191P) − 200Tθ D = 0 TD = − 0.8534 P

TD = − 0.853P 

F + TB + TC + TD − Pj = 0

i : Fx +

(− 0.36574 P) 2

−

Fx = − 0.3448P j: Fy −

k : Fz +

=0

2

Fx = − 0.345P

(− 0.36574 P) 2

Fy = P

( − 0.8534 P)

−

(1.2191P) 2

−

(− 0.8534 P) 2

− 200 = 0

Fy = P

(1.2191P) 2

=0

Fz = − 0.8620 P Fz = − 0.862 P

F = − 0.345P i + Pj − 0.862Pk 



ΣF = 0:

TB = −0.366 P 

www.elsolucionario.net PROBLEM 4.128 The uniform 10-kg rod AB is supported by a ball-and-socket joint at A and by the cord CG that is attached to the midpoint G of the rod. Knowing that the rod leans against a frictionless vertical wall at B, determine (a) the tension in the cord, (b) the reactions at A and B.

Free-Body Diagram: Five unknowns and six equations of equilibrium, but equilibrium is maintained (ΣMAB = 0). W = mg = (10 kg) 9.81m/s 2 W = 98.1 N  GC = − 300i + 200 j − 225k GC = 425 mm  GC T T =T = (− 300i + 200 j − 225k ) GC 425 rB/ A = − 600i + 400 j + 150 mm rG/ A = − 300i + 200 j + 75 mm ΣMA = 0: rB/ A × B + rG/ A × T + rG/ A × (− W j) = 0 i j k i j k i j k T − 600 400 150 + − 300 200 75 + − 300 200 75 425 B 0 0 − 300 200 − 225 0 − 98.1 0

Coefficient of i : (−105.88 − 35.29)T + 7357.5 = 0 T = 52.12 N

T = 52.1 N 

 



SOLUTION


Coefficient of j : 150 B − (300 × 75 + 300 × 225)

52.12 =0 425 B = 73.58 N

B = (73.6 N)i 

ΣF = 0: A + B + T − W j = 0

Coefficient of i : Coefficient of j:

Ax = −36.8 N 

Ay + 52.15

200 − 98.1 = 0 425

Ay = 73.6 N 

Az − 52.15

225 =0 425

Az = 27.6 N 


Coefficient of k :

300 =0 425

Ax + 73.58 − 52.15


www.elsolucionario.net PROBLEM 4.129 Three rods are welded together to form a “corner” that is supported by three eyebolts. Neglecting friction, determine the reactions at A, B, and C when P = 240 lb, a = 12 in., b = 8 in., and c = 10 in.

SOLUTION

ΣM O = 0: rA/O × A + rB/O × B + rC/O × C = 0 i 12 0

j 0 Ay

k i 0 + 0 Az Bx

j k i 8 0 + 0 0 Bz Cx

j k 0 10 = 0 Cy 0

(−12 Az j + 12 Ay k ) + (8 Bz i − 8 Bx k ) + (−10 C y i + 10 Cx j) = 0

From i-coefficient:

Bz = 1.25C y

or j-coefficient:

k-coefficient:

or

(3)

ΣF = 0: A + B + C − P = 0  ( Bx + Cx )i + ( Ay + C y − 240 lb) j + ( Az + Bz )k = 0 

From i-coefficient:

Bx + Cx = 0 C x = − Bx

or j-coefficient: or

(2)

12 Ay − 8 Bx = 0 Bx = 1.5 Ay

or

(1)

−12 Az + 10 C x = 0 Cx = 1.2 Az

or



8 Bz − 10 C y = 0

(4)

Ay + C y − 240 lb = 0 Ay + C y = 240 lb

(5)



From F.B.D. of weldment:


k-coefficient:

Az + Bz = 0 Az = − Bz

or

(6)

Substituting Cx from Equation (4) into Equation (2), − Bz = 1.2 Az

(7)

Using Equations (1), (6), and (7), Cy =

Bz − Az 1  Bx  Bx = = = 1.25 1.25 1.25  1.2  1.5

(8)

From Equations (3) and (8): 1.5 Ay 1.5

or C y = Ay

and substituting into Equation (5), 2 Ay = 240 lb Ay = C y = 120 lb

(9)

Using Equation (1) and Equation (9), Bz = 1.25(120 lb) = 150.0 lb

Using Equation (3) and Equation (9), Bx = 1.5(120 lb) = 180.0 lb

From Equation (4):

Cx = −180.0 lb

From Equation (6):

Az = −150.0 lb

Therefore,

A = (120.0 lb) j − (150.0 lb)k 



B = (180.0 lb)i + (150.0 lb)k 



C = −(180.0 lb)i + (120.0 lb) j 



Cy =

www.elsolucionario.net PROBLEM 4.130 Solve Problem 4.129, assuming that the force P is removed and is replaced by a couple M = +(600 lb ⋅ in.)j acting at B. PROBLEM 4.129 Three rods are welded together to form a “corner” that is supported by three eyebolts. Neglecting friction, determine the reactions at A, B, and C when P = 240 lb, a = 12 in., b = 8 in., and c = 10 in.

SOLUTION

ΣM O = 0: rA/O × A + rB/O × B + rC/O × C + M = 0 i 12 0

j 0 Ay

k i 0 + 0 Az Bx

j k i 8 0 + 0 0 Bz Cx

j k 0 10 + (600 lb ⋅ in.) j = 0 Cy 0

(−12 Az j + 12 Ay k ) + (8 Bz j − 8 Bx k ) + ( −10C y i + 10C x j) + (600 lb ⋅ in.) j = 0

From i-coefficient:

8 Bz − 10 C y = 0 C y = 0.8Bz

or j-coefficient:

(1)

−12 Az + 10 Cx + 600 = 0 Cx = 1.2 Az − 60

or k-coefficient:

(2)

12 Ay − 8 Bx = 0 Bx = 1.5 Ay

or

(3)



ΣF = 0: A + B + C = 0 



( Bx + C x )i + ( Ay + C y ) j + ( Az + Bz )k = 0 

From i-coefficient:

C x = − Bx

(4)

j-coefficient:

C y = − Ay

(5)

k-coefficient:

Az = − Bz

(6)



From F.B.D. of weldment:


Substituting Cx from Equation (4) into Equation (2), B  Az = 50 −  x   1.2 

(7)

2 C y = 0.8 Bz = − 0.8 Az =   Bx − 40 3

(8)

Using Equations (1), (6), and (7),

From Equations (3) and (8): C y = Ay − 40

Substituting into Equation (5),

2 Ay = 40

From Equation (5):

C y = −20.0 lb

Equation (1):

Bz = −25.0 lb

Equation (3):

Bx = 30.0 lb

Equation (4):

Cx = −30.0 lb

Equation (6):

Az = 25.0 lb A = (20.0 lb) j + (25.0 lb)k 

Therefore, 

 B = (30.0 lb)i − (25.0 lb)k 



C = − (30.0 lb)i − (20.0 lb) j 



Ay = 20.0 lb


SOLUTION From F.B.D. of pipe assembly ABCD: ΣFx = 0: Bx = 0 ΣM D ( x -axis) = 0: (48 N)(2.5 m) − Bz (2 m) = 0 Bz = 60.0 N

and B = (60.0 N)k  ΣM D ( z -axis) = 0: C y (3 m) − 90 N ⋅ m = 0 C y = 30.0 N ΣM D ( y -axis) = 0: − C z (3 m) − (60.0 N)(4 m) + (48 N)(4 m) = 0 Cz = −16.00 N

and C = (30.0 N) j − (16.00 N)k  ΣFy = 0: Dy + 30.0 = 0 Dy = −30.0 N ΣFz = 0: Dz − 16.00 N + 60.0 N − 48 N = 0 Dz = 4.00 N

and D = − (30.0 N) j + (4.00 N)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 513


In order to clean the clogged drainpipe AE, a plumber has disconnected both ends of the pipe and inserted a power snake through the opening at A. The cutting head of the snake is connected by a heavy cable to an electric motor that rotates at a constant speed as the plumber forces the cable into the pipe. The forces exerted by the plumber and the motor on the end of the cable can be represented by the wrench F = −(48 N)k , M = −(90 N ⋅ m)k. Determine the additional reactions at B, C, and D caused by the cleaning operation. Assume that the reaction at each support consists of two force components perpendicular to the pipe.

www.elsolucionario.net PROBLEM 4.132 Solve Problem 4.131, assuming that the plumber exerts a force F = −(48 N)k and that the motor is turned off (M = 0). PROBLEM 4.131 In order to clean the clogged drainpipe AE, a plumber has disconnected both ends of the pipe and inserted a power snake through the opening at A. The cutting head of the snake is connected by a heavy cable to an electric motor that rotates at a constant speed as the plumber forces the cable into the pipe. The forces exerted by the plumber and the motor on the end of the cable can be represented by the wrench F = −(48 N)k , M = −(90 N ⋅ m)k. Determine the additional reactions at B, C, and D caused by the cleaning operation. Assume that the reaction at each support consists of two force components perpendicular to the pipe.

From F.B.D. of pipe assembly ABCD: ΣFx = 0: Bx = 0 ΣM D ( x -axis) = 0: (48 N)(2.5 m) − Bz (2 m) = 0 Bz = 60.0 N

and B = (60.0 N)k 

ΣM D ( z -axis) = 0: C y (3 m) − Bx (2 m) = 0 Cy = 0 ΣM D ( y -axis) = 0: C z (3 m) − (60.0 N)(4 m) + (48 N)(4 m) = 0 Cz = −16.00 N

and C = − (16.00 N)k 

ΣFy = 0: Dy + C y = 0 Dy = 0 ΣFz = 0: Dz + Bz + C z − F = 0 Dz + 60.0 N − 16.00 N − 48 N = 0 Dz = 4.00 N

and D = (4.00 N)k 



SOLUTION

www.elsolucionario.net PROBLEM 4.133 The 50-kg plate ABCD is supported by hinges along edge AB and by wire CE. Knowing that the plate is uniform, determine the tension in the wire.


SOLUTION Free-Body Diagram: W = mg = (50 kg)(9.81 m/s 2 ) W = 490.50 N  CE = − 240i + 600 j − 400k CE = 760 mm  CE T = (− 240i + 600 j − 400k ) T =T CE 760  AB 480i − 200 j 1 λ AB = = = (12i − 5 j) AB 520 13 ΣMAB = 0: λ AB ⋅ (rE/ A × T ) + λ AB ⋅ (rG/ A × − W j) = 0 rE/ A = 240i + 400 j; rG/ A = 240i − 100 j + 200k

12 0 12 0 −5 −5 1 T 240 400 0 + 240 −100 200 =0 13 × 20 13 − 240 600 − 400 0 −W 0 (−12 × 400 × 400 − 5 × 240 × 400)

T + 12 × 200W = 0 760 T = 0.76W = 0.76(490.50 N)

T = 373 N 


www.elsolucionario.net PROBLEM 4.134 Solve Problem 4.133, assuming that wire CE is replaced by a wire connecting E and D. PROBLEM 4.133 The 50-kg plate ABCD is supported by hinges along edge AB and by wire CE. Knowing that the plate is uniform, determine the tension in the wire.


SOLUTION Free-Body Diagram: Dimensions in mm

W = mg = (50 kg)(9.81 m/s 2 ) W = 490.50 N  DE = − 240i + 400 j − 400k DE = 614.5 mm  DE T = (240i + 400 j − 400k ) T =T DE 614.5  AB 480i − 200 j 1 λ AB = = = (12i − 5 j) AB 520 13 rE/ A = 240i + 400 j; rG/ A = 240i − 100 j + 200k

12 −5 0 12 5 0 1 T 240 400 0 + 240 −100 200 =0 13 × 614.5 13 240 400 − 400 0 −W 0 (−12 × 400 × 400 − 5 × 240 × 400)

T + 12 × 200 × W = 0 614.5 T = 0.6145W = 0.6145(490.50 N) T = 301 N 


www.elsolucionario.net PROBLEM 4.135 Two rectangular plates are welded together to form the assembly shown. The assembly is supported by ball-and-socket joints at B and D and by a ball on a horizontal surface at C. For the loading shown, determine the reaction at C.

SOLUTION −(6 in.)i − (9 in.) j + (12 in.)k


λ BD =

First note:

(6) 2 + (9) 2 + (12)2 in.

1 (− 6i − 9 j + 12k ) 16.1555 = − (6 in.)i =

rA/B

P = (80 lb)k rC/D = (8 in.)i C = (C ) j

From the F.B.D. of the plates: ΣM BD = 0: λ BD ⋅ (rA/B × P ) + λ BD ⋅ ( rC/D × C ) = 0

−6 −9 12 −6 −9 12  6(80)   C (8)  −1 0 0  + 1 0 0  =0  16.1555  16.1555  0 0 1 0 1 0 ( −9)(6)(80) + (12)(8)C = 0 C = 45.0 lb

or C = (45.0 lb) j 




Two 2 × 4-ft plywood panels, each of weight 12 lb, are nailed together as shown. The panels are supported by ball-and-socket joints at A and F and by the wire BH. Determine (a) the location of H in the xy plane if the tension in the wire is to be minimum, (b) the corresponding minimum tension.

SOLUTION Free-Body Diagram:  AF = 4i − 2 j − 4k

AF = 6 ft

1 λ AF = (2i − j − 2k ) 3 rG1/A = 2i − j rG2 /A = 4i − j − 2k rB/A = 4i

ΣM AF = 0: λ AF ⋅ (rG1 /A × ( −12 j) + λ AF ⋅ (rG2 /A × (−12 j)) + λ AF ⋅ (rB /A × T ) = 0 2 −1 −2 2 −1 −2 1 1 + 4 −1 −2 + λ AF ⋅ (rB/A × T) = 0 2 −1 0 3 3 0 −12 0 0 −12 0 1 1 (2 × 2 × 12) + (−2 × 2 × 12 + 2 × 4 × 12) + λ AF ⋅ (rB/A × T) = 0 3 3 λ AF ⋅ (rB/A × T) = −32 or T ⋅ (λ A/F × rB/A ) = −32

(1)



Projection of T on (λ AF × rB/A ) is constant. Thus, Tmin is parallel to 1 1 λ AF × rB/A = (2i − j − 2k ) × 4i = (−8 j + 4k ) 3 3

Corresponding unit vector is

1 5

(−2 j + k ). Tmin = T ( −2 j + k )

(2)

5

T 1  (−2 j + k ) ⋅  (2i − j − 2k ) × 4i  = −32 5 3  T 1 (−2 j + k ) ⋅ ( −8 j + 4k ) = −32 3 5 T (16 + 4) = −32 3 5

T =−

3 5(32) = 4.8 5 20


From Eq. (1):

1

T = 10.7331 lb

From Eq. (2):

Tmin = T ( −2 j + k )

1 5

= 4.8 5( −2 j + k ) Tmin

1

5 = −(9.6 lb)j + (4.8 lb k )

Since Tmin has no i component, wire BH is parallel to the yz plane, and x = 4 ft.

(a) (b)

x = 4.00 ft;

y = 8.00 ft 

Tmin = 10.73 lb 


www.elsolucionario.net PROBLEM 4.137 Solve Problem 4.136, subject to the restriction that H must lie on the y-axis.


PROBLEM 4.136 Two 2 × 4-ft plywood panels, each of weight 12 lb, are nailed together as shown. The panels are supported by ball-andsocket joints at A and F and by the wire BH. Determine (a) the location of H in the xy plane if the tension in the wire is to be minimum, (b) the corresponding minimum tension.

SOLUTION Free-Body Diagram:  AF = 4i − 2 j − 4k 1 λ AF = (2i − j − 2k ) 3 rG1/A = 2i − j rG2 /A = 4i − j − 2k rB/A = 4i ΣMAF = 0: λ AF ⋅ (rG/A × (−12 j) + λ AF ⋅ (rG2 /A × (−12 j)) + λ AF ⋅ (rB/A × T ) = 0 2 −1 2 2 −1 −2 1 1 2 −1 0 + 4 −1 −2 + λ AF ⋅ (rB/A × T) = 0 3 3 0 −12 0 0 −12 0 1 1 (2 × 2 × 12) + (−2 × 2 × 12 + 2 × 4 × 12) + λ AF ⋅ (rB/A × T) = 0 3 3 λ AF ⋅ (rB/A × T) = −32  BH = −4i + yj − 4k BH = (32 + y 2 )1/2  BH −4i + yj − 4k =T T=T BH (32 + y 2 )1/ 2

(1)



From Eq. (1): λ AF

2 −1 −2 T ⋅ (rB/A × T ) = 4 0 0 = −32 3(32 + y 2 )1/2 −4 y −4

(−16 − 8 y )T = −3 × 32(32 + y 2 )1/2

T = 96

(32 + y 2 )1/2 8 y + 16

(2)

(8y +16) 12 (32 + y 2 ) −1/ 2 (2 y ) + (32 + y 2 )1/2 (8) dT = 0: 96 dy (8 y + 16) 2 (8 y + 16) y = (32 + y 2 )8 8 y 2 + 16 y = 32 × 8 + 8 y 2

From Eq. (2):

T = 96

(32 + 162 )1/ 2 = 11.3137 lb 8 × 16 + 16

x = 0 ft; y = 16.00 ft  Tmin = 11.31 lb 



Numerator = 0:

www.elsolucionario.net PROBLEM 4.138 The frame ACD is supported by ball-and-socket joints at A and D and by a cable that passes through a ring at B and is attached to hooks at G and H. Knowing that the frame supports at Point C a load of magnitude P = 268 N, determine the tension in the cable.



λ AD λ AD TBG

TBH

 AD (1 m)i − (0.75 m)k = = AD 1.25 m = 0.8i − 0.6k  BG = TBG BG −0.5i + 0.925 j − 0.4k = TBG 1.125  BH = TBH BH 0.375i + 0.75 j − 0.75k = TBH 1.125



rB /A = (0.5 m)i; rC/A = (1 m)i; P = −(268 N)j

To eliminate the reactions at A and D, we shall write ΣM AD = 0:

λ AD ⋅ (rB /A × TBG ) + λ AD ⋅ (rB /A × TBH ) + λ AD ⋅ (rC /A × P) = 0

(1)

Substituting for terms in Eq. (1) and using determinants, −0.6 −0.6 −0.6 0.8 0 0.8 0 0.8 0 TBG TBH + 0.5 + 1 0.5 0 0 0 0 0 0 =0 1.125 1.125 −0.5 0.925 −0.4 0.375 0.75 −0.75 0 −268 0

Multiplying all terms by (–1.125), 0.27750TBG + 0.22500TBH = 180.900 TBG = TBH = T (0.27750 + 0.22500)T = 180.900

T = 360 N 



For this problem,

(2)

www.elsolucionario.net PROBLEM 4.139 Solve Prob. 4.138, assuming that cable GBH is replaced by a cable GB attached at G and B.


λ AD λ AD TBG

TBH

 AD (1 m)i − (0.75 m)k = = AD 1.25 m = 0.8i − 0.6k  BG = TBG BG −0.5i + 0.925 j − 0.4k = TBG 1.125  BH = TBH BH 0.375i + 0.75 j − 0.75k = TBH 1.125



PROBLEM 4.138 The frame ACD is supported by ball-andsocket joints at A and D and by a cable that passes through a ring at B and is attached to hooks at G and H. Knowing that the frame supports at Point C a load of magnitude P = 268 N, determine the tension in the cable.


rB /A = (0.5 m)i; rC/A = (1 m)i; P = −(268 N)j

To eliminate the reactions at A and D, we shall write ΣM AD = 0: λ AD ⋅ (rB /A × TBG ) + λ AD ⋅ (rB /A × TBH ) + λ AD ⋅ (rC /A × P) = 0

(1)

Substituting for terms in Eq. (1) and using determinants, −0.6 −0.6 −0.6 0.8 0 0.8 0 0.8 0 TBG TBH + 0.5 + 1 0.5 0 0 0 0 0 0 =0 1.125 1.125 −0.5 0.925 −0.4 0.375 0.75 −0.75 0 −268 0

Multiplying all terms by (–1.125), 0.27750TBG + 0.22500TBH = 180.900

(2)

Thus, Eq. (2) reduces to 0.27750TBG = 180.900

TBG = 652 N 



For this problem, TBH = 0.

www.elsolucionario.net PROBLEM 4.140 The bent rod ABDE is supported by ball-and-socket joints at A and E and by the cable DF. If a 60-lb load is applied at C as shown, determine the tension in the cable.



 DF = −16i + 11j − 8k DF = 21 in.  DE T T=T = (−16i + 11j − 8k ) DF 21 rD/E = 16i rC/E = 16i − 14k  EA 7i − 24k = λ EA = EA 25 ΣM EA = 0: λ EA ⋅ (rB/E × T) + λ EA ⋅ (rC/E ⋅ (− 60 j)) = 0 7 0 −24 7 0 −24 1 T 16 0 0 + 16 0 −14 =0 21 × 25 25 0 −60 0 −16 11 −8 −

24 × 16 × 11 −7 × 14 × 60 + 24 × 16 × 60 T+ =0 21 × 25 25 201.14 T + 17,160 = 0 T = 85.314 lb

T = 85.3 lb 


www.elsolucionario.net PROBLEM 4.141 Solve Problem 4.140, assuming that cable DF is replaced by a cable connecting B and F.

SOLUTION Free-Body Diagram: rB/ A = 9i


rC/ A = 9i + 10k  BF = −16i + 11j + 16k BF = 25.16 in.  BF T = (−16i + 11j + 16k ) T =T BF 25.16  AE 7i − 24k λ AE = = AE 25

ΣMAE = 0: λ AF ⋅ (rB/ A × T) + λ AE ⋅ (rC/ A ⋅ (− 60 j)) = 0 7 0 −24 7 0 −24 1 T 9 0 0 +9 0 10 =0 25 × 25.16 25 −16 11 16 0 −60 0 −

24 × 9 × 11 24 × 9 × 60 + 7 × 10 × 60 T+ =0 25 × 25.16 25 94.436 T − 17,160 = 0

T = 181.7 lb 


www.elsolucionario.net PROBLEM 4.142 A gardener uses a 60-N wheelbarrow to transport a 250-N bag of fertilizer. What force must she exert on each handle?

SOLUTION



ΣM A = 0: (2 F )(1 m) − (60 N)(0.15 m) − (250 N)(0.3 m) = 0 F = 42.0 N 





Free-Body Diagram:

www.elsolucionario.net PROBLEM 4.143 The required tension in cable AB is 200 lb. Determine (a) the vertical force P that must be applied to the pedal, (b) the corresponding reaction at C.

SOLUTION

BC = 7 in.

(a)

ΣM C = 0: P(15 in.) − (200 lb)(6.062 in.) = 0 P = 80.83 lb

(b)

P = 80.8 lb 

ΣFy = 0: C x − 200 lb = 0

C x = 200 lb

ΣFy = 0: C y − P = 0 C y − 80.83 lb = 0

C y = 80.83 lb

α = 22.0° C = 215.7 lb C = 216 lb

22.0° 



Free-Body Diagram:

www.elsolucionario.net PROBLEM 4.144 A lever AB is hinged at C and attached to a control cable at A. If the lever is subjected to a 500-N horizontal force at B, determine (a) the tension in the cable, (b) the reaction at C.

Triangle ACD is isosceles with  C = 90° + 30° = 120°  A =  D =


SOLUTION

1 (180° − 120°) = 30°. 2

Thus, DA forms angle of 60° with the horizontal axis. (a)

We resolve FAD into components along AB and perpendicular to AB.

ΣM C = 0: ( FAD sin 30°)(250 mm) − (500 N)(100 mm) = 0

(b)

FAD = 400 N 

ΣFx = 0: − (400 N) cos 60° + C x − 500 N = 0

C x = +300 N

ΣFy = 0: − (400 N) sin 60° + C y = 0

C y = +346.4 N C = 458 N

49.1° 


www.elsolucionario.net PROBLEM 4.145 A force P of magnitude 280 lb is applied to member ABCD, which is supported by a frictionless pin at A and by the cable CED. Since the cable passes over a small pulley at E, the tension may be assumed to be the same in portions CE and ED of the cable. For the case when a = 3 in., determine (a) the tension in the cable, (b) the reaction at A.

Free-Body Diagram: (a)

ΣM A = 0: − (280 lb)(8 in.) 7 T (12 in.) 25 24 − T (8 in.) = 0 25

T (12 in.) −

(12 − 11.04)T = 840

(b)

ΣFx = 0:

T = 875 lb 

7 (875 lb) + 875 lb + Ax = 0 25 Ax = −1120

ΣFy = 0: Ay − 280 lb − Ay = +1120

A x = 1120 lb

24 (875 lb) = 0 25 A y = 1120 lb A = 1584 lb

45.0° 



SOLUTION

www.elsolucionario.net PROBLEM 4.146 Two slots have been cut in plate DEF, and the plate has been placed so that the slots fit two fixed, frictionless pins A and B. Knowing that P = 15 lb, determine (a) the force each pin exerts on the plate, (b) the reaction at F.

SOLUTION

(a) (b)

ΣFx = 0: 15 lb − B sin 30° = 0

B = 30.0 lb

60.0° 

ΣM A = 0: − (30 lb)(4 in.) + B sin 30°(3 in.) + B cos 30°(11 in.) − F (13 in.) = 0 −120 lb ⋅ in. + (30 lb) sin 30°(3 in.) + (30 lb) cos 30°(11 in.) − F (13 in.) = 0 F = + 16.2145 lb

(a)

F = 16.21 lb 

ΣFy = 0: A − 30 lb + B cos 30° − F = 0 A − 30 lb + (30 lb) cos 30° − 16.2145 lb = 0 A = + 20.23 lb

A = 20.2 lb 



Free-Body Diagram:

www.elsolucionario.net PROBLEM 4.147 Knowing that the tension in wire BD is 1300 N, determine the reaction at the fixed support C of the frame shown.

T = 1300 N 5 Tx = T 13 = 500 N 12 Ty = T 13 = 1200 N ΣM x = 0: C x − 450 N + 500 N = 0

C x = −50 N

ΣFy = 0: C y − 750 N − 1200 N = 0 C y = +1950 N

C x = 50 N C y = 1950 N

C = 1951 N

88.5° 

ΣM C = 0: M C + (750 N)(0.5 m) + (4.50 N)(0.4 m) − (1200 N)(0.4 m) = 0 M C = −75.0 N ⋅ m

M C = 75.0 N ⋅ m





SOLUTION

www.elsolucionario.net PROBLEM 4.148 The spanner shown is used to rotate a shaft. A pin fits in a hole at A, while a flat, frictionless surface rests against the shaft at B. If a 60-lb force P is exerted on the spanner at D, find the reactions at A and B.



(Three-force body)

The line of action of A must pass through D, where B and P intersect. 3sin 50° 3cos 50° + 15 = 0.135756 α = 7.7310°

tan α =

60 lb sin 7.7310° = 446.02 lb 60 lb B= tan 7.7310° = 441.97 lb A=

Force triangle

A = 446 lb

7.73° 

B = 442 lb






www.elsolucionario.net PROBLEM 4.149 Rod AB is supported by a pin and bracket at A and rests against a frictionless peg at C. Determine the reactions at A and C when a 170-N vertical force is applied at B.

SOLUTION

The reaction at A must pass through D where C and the 170-N force intersect. 160 mm 300 mm α = 28.07°

tan α =

We note that triangle ABD is isosceles (since AC = BC) and, therefore,  CAD = α = 28.07°

Also, since CD ⊥ CB, reaction C forms angle α = 28.07° with the horizontal axis.

Force triangle We note that A forms angle 2α with the vertical axis. Thus, A and C form angle 180° − (90° − α ) − 2α = 90° − α

Force triangle is isosceles, and we have A = 170 N C = 2(170 N)sin α = 160.0 N A = 170.0 N

33.9°;

C = 160.0 N

28.1° 




www.elsolucionario.net PROBLEM 4.150 The 24-lb square plate shown is supported by three vertical wires. Determine (a) the tension in each wire when a = 10 in., (b) the value of a for which the tension in each wire is 8 lb.

SOLUTION rB/A = ai + 30k


rC/A = 30i + ak rG/A = 15i + 15k

By symmetry, B = C. ΣM A = 0: rB/A × Bj + rC × Cj + rG/A × (−W j) = 0 (ai + 30k ) × Bj + (30i + ak ) × Bj + (15i + 15k ) × (−W j) = 0 Bak − 30 Bi + 30 Bk − Bai − 15Wk + 15W i = 0

Equate coefficient of unit vector i to zero: i : − 30 B − Ba + 15W = 0

B=

15W 30 + a

C=B=

15W 30 + a

(1)

ΣFy = 0: A + B + C − W = 0  15W  A+ 2  − W = 0;  30 + a 

(a)

For

a = 10 in.

From Eq. (1):

C=B=

From Eq. (2):

A=

A=

aW 30 + a

(2)

15(24 lb) = 9.00 lb 30 + 10

10(24 lb) = 6.00 lb 30 + 10

A = 6.00 lb; B = C = 9.00 lb 



(b)

For tension in each wire = 8 lb, From Eq. (1):

8 lb =

15(24 lb) 30 + a a = 15.00 in. 


30 in. + a = 45


www.elsolucionario.net PROBLEM 4.151 Frame ABCD is supported by a ball-and-socket joint at A and by three cables. For a = 150 mm, determine the tension in each cable and the reaction at A.


TDG = λ DG TDG =

−(0.48 m)i + (0.14 m)j (0.48) 2 + (0.14) 2 m

TDG

−0.48i + 0.14 j TDG 0.50 T = DG (24i + 7 j) 25 TBE = λ BE TBE =

−(0.48 m)i + (0.2 m)k (0.48)2 + (0.2)2 m


=

TBE

−0.48i + 0.2k TBE 0.52 T = BE (−12 j + 5k ) 13 =

From F.B.D. of frame ABCD:  7  ΣM x = 0:  TDG  (0.3 m) − (350 N)(0.15 m) = 0  25  TDG = 625 N 

or  24   5  ΣM y = 0:  × 625 N  (0.3 m) −  TBE  (0.48 m) = 0 13  25   

TBE = 975 N 

or  7  ΣM z = 0: TCF (0.14 m) +  × 625 N  (0.48 m) − (350 N)(0.48 m) = 0  25 

TCF = 600 N 

or



ΣFx = 0: Ax + TCF + (TBE ) x + (TDG ) x = 0  12   24  Ax − 600 N −  × 975 N  −  × 625 N  = 0  13   25  Ax = 2100 N

ΣFy = 0: Ay + (TDG ) y − 350 N = 0  7  Ay +  × 625 N  − 350 N = 0 25   Ay = 175.0 N

 5  Az +  × 975 N  = 0 13   Az = −375 N A = (2100 N)i + (175.0 N) j − (375 N)k 

Therefore,



ΣFz = 0: Az + (TBE ) z = 0

www.elsolucionario.net PROBLEM 4.152 The pipe ACDE is supported by ball-and-socket joints at A and E and by the wire DF. Determine the tension in the wire when a 640-N load is applied at B as shown.

Free-Body Diagram: Dimensions in mm

 AE = 480i + 160 j − 240k AE = 560 mm  AE 480i + 160 j − 240k λ AE = = AE 560 6i + 2 j − 3k λ AE = 7 rB/A = 200i rD/A = 480i + 160 j  DF = −480i + 330 j − 240k ; DF = 630 mm  DF −480i + 330 j − 240k −16i + 11j − 8k = TDF = TDF TDF = TDF DF 630 21 ΣM AE = λ AE ⋅ (rD/A × TDF ) + λ AE ⋅ (rB/A × (−600 j)) = 0 6 2 −3 6 2 −3 TDF 1 480 160 0 0 0 =0 + 200 21 × 7 7 0 −640 0 −16 11 −8 3 × 200 × 640 −6 × 160 × 8 + 2 × 480 × 8 − 3 × 480 × 11 − 3 × 160 × 16 TDF + =0 21 × 7 7 −1120TDF + 384 × 103 = 0 TDF = 342.86 N

TDF = 343 N 



SOLUTION


SOLUTION (a)

ΣM A = 0: − Pa + (C sin 45°)2a + (cos 45°)a = 0 3

C 2

=P

C=

2 P 3

 2  1 ΣFx = 0: Ax −  P  3  2  

Ax =

P 3

 2  1 P ΣFy = 0: Ay − P +   3  2  

Ay =

2P 3

C = 0.471P

A = 0.745P

(b)

45° 

63.4° 

ΣM C = 0: +Pa − ( A cos 30°)2a + ( A sin 30°)a = 0

A(1.732 − 0.5) = P

A = 0.812 P A = 0.812 P

ΣFx = 0: (0.812 P )sin 30° + C x = 0

60.0° 

Cx = −0.406 P

ΣFy = 0: (0.812 P) cos 30° − P + C y = 0 C y = −0.297 P C = 0.503P

36.2° 



A force P is applied to a bent rod ABC, which may be supported in four different ways as shown. In each case, if possible, determine the reactions at the supports.


(c)

ΣM C = 0: + Pa − ( A cos 30°)2a + ( A sin 30°)a = 0

A(1.732 + 0.5) = P

A = 0.448P A = 0.448P

ΣFx = 0: − (0.448P) sin 30° + Cx = 0

60.0° 

Cx = 0.224 P

ΣFy = 0: (0.448 P) cos 30° − P + C y = 0 C y = 0.612 P C = 0.652 P

69.9° 

 (d)

Force T exerted by wire and reactions A and C all intersect at Point D.

Equilibrium is not maintained. Rod is improperly constrained. 



ΣM D = 0: Pa = 0

www.elsolucionario.net PROBLEM 4.F1 For the frame and loading shown, draw the free-body diagram needed to determine the reactions at A and E when α = 30°.

SOLUTION







Free-Body Diagram of Frame:

www.elsolucionario.net PROBLEM 4.F2 Neglecting friction, draw the free-body diagram needed to determine the tension in cable ABD and the reaction at C when θ = 60°.

SOLUTION


Free-Body Diagram of Member ACD:




www.elsolucionario.net PROBLEM 4.F3 Bar AC supports two 400-N loads as shown. Rollers at A and C rest against frictionless surfaces and a cable BD is attached at B. Draw the free-body diagram needed to determine the tension in cable BD and the reactions at A and C.

SOLUTION

 Note: By similar triangles yB 0.15 m = 0.25 m 0.5 m

yB = 0.075 m 



Free-Body Diagram of Bar AC:

www.elsolucionario.net PROBLEM 4.F4 Draw the free-body diagram needed to determine the tension in each cable and the reaction at D.

SOLUTION





Free-Body Diagram of Member ABCD:

www.elsolucionario.net PROBLEM 4.F5 A 4 × 8-ft sheet of plywood weighing 34 lb has been temporarily placed among three pipe supports. The lower edge of the sheet rests on small collars at A and B and its upper edge leans against pipe C. Neglecting friction on all surfaces, draw the free-body diagram needed to determine the reactions at A, B, and C.

SOLUTION





Free-Body Diagram of Plywood sheet:

www.elsolucionario.net PROBLEM 4.F6 Two transmission belts pass over sheaves welded to an axle supported by bearings at B and D. The sheave at A has a radius of 2.5 in. and the sheave at C has a radius of 2 in. Knowing that the system rotates at a constant rate, draw the free-body diagram needed to determine the tension T and the reactions at B and D. Assume that the bearing at D does not exert any axial thrust and neglect the weights of the sheaves and axle.

SOLUTION


Free-Body Diagram of axle-sheave system:




www.elsolucionario.net PROBLEM 4.F7 The 6-m pole ABC is acted upon by a 455-N force as shown. The pole is held by a ball-and-socket joint at A and by two cables BD and BE. Draw the free-body diagram needed to determine the tension in each cable and the reaction at A.

SOLUTION


Free-Body Diagram of Pole:






CHAPTER 5





www.elsolucionario.net PROBLEM 5.1 Locate the centroid of the plane area shown.

A, in 2

x , in

y , in

x A,in 3

y A,in 3

1

8

0.5

4

4

32

2

3

2.5

2.5

7.5

7.5

Σ

11

11.5

39.5

X ΣA= xA X (11 in 2 ) = 11.5 in 3

X = 1.045 in. 

Y ΣA=ΣyA Y (11) = 39.5

Y = 3.59 in. 





SOLUTION


SOLUTION For the area as a whole, it can be concluded by observation that 2 (72 mm) 3

or Y = 48.0 mm 


Y =

Dimensions in mm

A, mm 2

x , mm

x A, mm3

1

1 × 30 × 72 = 1080 2

20

21,600

2

1 × 48 × 72 = 1728 2

46

79,488

Σ

Then X A = Σ x A

2808

101,088 X (2808) = 101, 088

or X = 36.0 mm 



Then

A, mm 2

x , mm

y , mm

xA, mm3

yA, mm3

1

126 × 54 = 6804

9

27

61,236

183,708

2

1 × 126 × 30 = 1890 2

30

64

56,700

120,960

3

1 × 72 × 48 = 1728 2

48

−16

82,944

−27,648

Σ

10,422

200,880

277,020

X ΣA = Σ xA X (10, 422 m 2 ) = 200,880 mm 2

and

or X = 19.27 mm 

Y Σ A = Σ yA Y (10, 422 m 2 ) = 270, 020 mm3

or

Y = 26.6 mm 



SOLUTION


Then

A, in 2

x , in

y , in

x A, in 3

y A, in 3

1

1 (12)(6) = 36 2

4

4

144

144

2

(6)(3) = 18

9

7.5

162

135

Σ

54

306

279

XA = Σ xA X (54) = 306

X = 5.67 in. 

YA = Σ yA Y (54) = 279

Y = 5.17 in. 



SOLUTION


By symmetry, X = Y A, in 2

Component I

Quarter circle

II

Square

Σ

π 4

x , in.

x A, in 3

(10) 2 = 78.54

4.2441

333.33

−(5)2 = −25

2.5

−62.5

53.54

270.83

X Σ A = Σ x A: X (53.54 in 2 ) = 270.83 in 3 X = 5.0585 in.

X = Y = 5.06 in. 



SOLUTION


Then

A, in 2

x , in.

y , in.

x A, in 3

y A, in 3

1

14 × 20 = 280

7

10

1960

2800

2

−π (4) 2 = −16π

6

12

–301.59

–603.19

Σ

229.73

1658.41

2196.8

X=

Σ xA 1658.41 = ΣA 229.73

X = 7.22 in. 

Y =

Σ y A 2196.8 = Σ A 229.73

Y = 9.56 in. 



SOLUTION


SOLUTION

Component

A, in 2

y , in.

y A, in 3

I

Rectangle

(3)(6) = 18

1.5

27.0

II

Semicircle

2.151

−13.51

Σ

−

π 2

(2) 2 = −6.28

13.49

11.72 Y ΣA=ΣyA Y (11.72 in.2 ) = 13.49 in 3 Y = 1.151 in.

X =0 Y = 1.151 in. 



By symmetry, X = 0


1

Σ

Then

x , mm

y , mm

x A, mm3

y A, mm3

(60)(120) = 7200

–30

60

−216 × 103

432 × 103

(60) 2 = 2827.4

25.465

95.435

72.000 × 103

269.83 × 103

(60) 2 = −2827.4

–25.465

25.465

72.000 × 103

−72.000 × 103

−72.000 × 103

629.83 × 103

π

2

3

A, mm 2

4 −

π 4

7200 XA = Σ x A

X (7200) = −72.000 × 103

YA = Σ y A

Y (7200) = 629.83 × 103

X = −10.00 mm  Y = 87.5 mm 



SOLUTION


A, mm 2

x , mm

y , mm

x A, mm3

y A, mm3

1

1 (120)(75) = 4500 2

80

25

360 × 103

112.5 × 103

2

(75)(75) = 5625

157.5

37.5

885.94 × 103

210.94 × 103

163.169

43.169

−720.86 × 103

−190.716 × 103

525.08 × 103

132.724 × 103

3 Σ

Then

−

π 4

(75) 2 = −4417.9

5707.1 XA = Σx A

X (5707.1) = 525.08 × 103

X = 92.0 mm 

YA = Σ y A

Y (5707.1) = 132.724 × 103

Y = 23.3 mm 



SOLUTION


SOLUTION

A, in 2 −

1

2

π (12) 2

2

2

Σ

Then

π (8) 2

y , in.

yA, in 3

= −100.531

3.3953

–341.33

= 226.19

5.0930

1151.99

125.659

Y =

810.66

Σ y A 810.66 in 3 = Σ A 125.66 in 2

or Y = 6.45 in. 



X = 0 

First note that symmetry implies


SOLUTION


X =0 


 A, m 2 4 × 4.5 × 3 = 18 3

1

2

y, m

−

π

Σ

2

(1.8) 2 = −5.0894

12.9106

yA, m3

1.2

21.6

0.76394

−3.8880 17.7120

 Then

Y =

Σ y A 17.7120 m3 = Σ A 12.9106 m 2

or Y = 1.372 m 



 Area mm2

x , mm

y , mm

xA, mm3

yA, mm3

1

1 (200)(480) = 32 × 103 3

360

60

11.52 × 106

1.92 × 106

2

1 − (50)(240) = 4 × 103 3

180

15

−0.72 × 106

−0.06 × 106

Σ

28 × 103

10.80 × 106

1.86 × 106

 X Σ A = Σ x A:

X (28 × 103 mm 2 ) = 10.80 × 106 mm3 X = 385.7 mm

Y Σ A = Σ y A:

X = 386 mm 

Y (28 × 103 mm 2 ) = 1.86 × 106 mm3 Y = 66.43 mm

Y = 66.4 mm 



SOLUTION


Then

A, mm 2

x , mm

y , mm

xA, mm3

yA, mm3

1

(15)(80) = 1200

40

7.5

48 × 103

9 × 103

2

1 (50)(80) = 1333.33 3

60

30

80 × 103

40 × 103

Σ

2533.3

128 × 103

49 × 103

X A = Σ xA X (2533.3) = 128 × 103

X = 50.5 mm 

YA = Σ yA Y (2533.3) = 49 × 103

Y = 19.34 mm 



SOLUTION


Dimensions in in.

Then

A, in 2

x , in.

y , in.

xA, in 3

yA, in 3

1

2 (4)(8) = 21.333 3

4.8

1.5

102.398

32.000

2

1 − (4)(8) = −16.0000 2

5.3333

1.33333

85.333

−21.333

Σ

5.3333

17.0650

10.6670

X ΣA = Σ xA X (5.3333 in 2 ) = 17.0650 in 3

and

or X = 3.20 in. 

Y Σ A = Σ yA Y (5.3333 in 2 ) = 10.6670 in 3

or

Y = 2.00 in. 



SOLUTION


SOLUTION

A, mm 2

1

Then

π 2

× 47 × 26 = 1919.51

2

1 × 94 × 70 = 3290 2

Σ

5209.5

x , mm

y , mm

x A, mm3

y A, mm3

0

11.0347

0

21,181

−15.6667

−23.333

−51,543

−76,766

−51,543

−55,584

X=

Σ x A −51,543 = ΣA 5209.5

X = −9.89 mm 

Y =

Σ y A −55,584 = ΣA 5209.5

Y = −10.67 mm 



Dimensions in mm

www.elsolucionario.net PROBLEM 5.16 Determine the x coordinate of the centroid of the trapezoid shown in terms of h1, h2, and a.

A

x

xA

1

1 h1a 2

1 a 3

1 h1a 2 6

2

1 h2 a 2

2 a 3

2 h2 a2 6

Σ

1 a(h1 + h 2 ) 2

X=


SOLUTION

1 2 a ( h1 + 2h 2 ) 6

1 2 Σ x A 6 a ( h1 + 2h 2 ) = 1 a (h1 + h 2 ) ΣA 2

1 h1 + 2h 2 X= a  3 h1 + h 2


www.elsolucionario.net PROBLEM 5.17 For the plane area of Problem 5.5, determine the ratio a/r so that the centroid of the area is located at point B.

SOLUTION

Area

x

xA

I

Quarter circle

1 2 πr 4

4r 3π

1 3 r 3

II

Square

−a 2

1 a 2

1 − a3 2

π

Σ

4

r 2 − a2

X Σ A = Σ x A:

1 π  1 X  r 2 − a 2  = r 3 − a3 2 4  3

Set X = a :

1 π  1 a  r 2 − a 2  = r 3 − a3 2 4  3


By symmetry, X = Y . For centroid to be at B, X = a.

1 3 1 3 r − a 3 2

1 3 π 2 1 a − r a + r3 = 0 2 4 3

Divide by

1 3 r : 2

3

a π a 2 r − 2 r + 3 =0  

a = 0.508  r


www.elsolucionario.net PROBLEM 5.18 Determine the y coordinate of the centroid of the shaded area in terms of r1, r2, and α.

SOLUTION

y2 =

From Figure 5.8A:

Similarly,

=

2 cos α r2 π 3 ( 2 −α )

y1 =

2 cos α r1 3 ( π2 − α )

Σ yA =

Then

π 2 sin ( 2 − α ) r2 π 3 ( 2 −α )

=


First, determine the location of the centroid. π  A2 =  − α  r22 2 

π  A1 =  − α  r12 2 

2 cos α  π   2 cos α  π  2 − α  r22  − r1 π r2   − α  r1  3 ( π2 − α )  2 3 2 α     ( 2 − ) 

(

)

2 3 3 r2 − r1 cos α 3

π  π  ΣA =  − α  r22 −  − α  r12 2 2    

and

π  =  − α  r22 − r12 2  

(

)

Y ΣA = Σ yA

Now

 π  2  Y  − α  r22 − r12  = r23 − r13 cos α   2  3

(

)

(

)

Y =

2  r23 − r13  3  r22 − r12

  2 cos α        π − 2α 


www.elsolucionario.net PROBLEM 5.19 Show that as r1 approaches r2, the location of the centroid approaches that for an arc of circle of radius (r1 + r2 )/2.

First, determine the location of the centroid. y2 =

From Figure 5.8A:

Similarly,

=

2 cos α r2 3 ( π2 − α )

y1 =

2 cos α r1 3 ( π2 − α )

Σ yA =

Then

π 2 sin ( 2 − α ) r2 π 3 ( 2 −α )

=

π  A2 =  − α  r22 2 

π  A1 =  − α  r12 2  

2 cos α  π   2 cos α  π  2 r2 − α  r22  − r1 π   − α  r1  3 ( π2 − α )  2 3 2     ( 2 − α ) 

(

)

2 3 3 r2 − r1 cos α 3

π  π  ΣA =  − α  r22 −  − α  r12 2 2    

and

π  =  − α  r22 − r12 2  

(

)

Y ΣA = Σ yA

Now

 π  2  Y  − α  r22 − r12  = r23 − r13   2  3 2  r3 − r3 Y =  22 12 3  r2 − r1

(

)

(

) cos α   2cos α       π − 2α 



SOLUTION


Using Figure 5.8B, Y of an arc of radius Y = =

Now

r23 − r13 r22

−

r12

= =

1 (r1 + r2 ) is 2 sin( π − α ) 1 (r1 + r2 ) π 2 2 ( 2 −α) 1 cos α (r1 + r2 ) π 2 ( 2 −α)

(

(r2 − r1 ) r22 + r1r2 + r12 (r2 − r1 )(r2 + r1 )

(1)

)

r22 + r1r2 + r12 r2 + r1

r2 = r + Δ

Let

r=

Then

and

In the limit as Δ

r23 − r13 r22 − r12


r1 = r − Δ 1 (r1 + r2 ) 2

=

(r + Δ) 2 + (r + Δ)(r − Δ)( r − Δ) 2 (r + Δ ) + (r − Δ )

=

3r 2 + Δ 2 2r

0 (i.e., r1 = r2 ), then r23 − r13 r22

So that

−

r12

=

3 r 2

=

3 1 × ( r1 + r2 ) 2 2

Y =

2 3 cos α × (r1 + r2 ) π 3 4 −α 2

or Y = ( r1 + r2 )

cos α  π − 2α

which agrees with Equation (1).


www.elsolucionario.net PROBLEM 5.20 The horizontal x-axis is drawn through the centroid C of the area shown, and it divides the area into two component areas A1 and A2. Determine the first moment of each component area with respect to the x-axis, and explain the results obtained.

Length of BD: BD = 0.48 in. + (1.44 in. − 0.48 in.)

0.84in. = 0.48 + 0.56 = 1.04 in. 0.84 in. × 0.60 in.

Area above x-axis (consider two triangular areas): 1  1  Q1 = Σ y A = (0.28 in.)  (0.84 in.)(1.04 in.)  + (0.56 in.)  (0.84 in.)(0.48 in.)  2 2     = 0.122304 in 3 + 0.112896 in 3 Q1 = 0.2352 in 3 

Area below x-axis: 1  1  Q2 = Σ yA = −(0.40 in.)  (0.60 in.)(1.44 in.)  − (0.20 in.)  (0.60 in.)  2  2  = −0.1728 in 3 − 0.0624 in 3 Q2 = −0.2352 in 3  |Q| = |Q2 |, since C is centroid and thus, Q = Σ y A = 0



SOLUTION

www.elsolucionario.net PROBLEM 5.21 The horizontal x-axis is drawn through the centroid C of the area shown, and it divides the area into two component areas A1 and A2. Determine the first moment of each component area with respect to the x-axis, and explain the results obtained.

SOLUTION


Dimensions in mm

Area above x-axis (Area A1): Q1 = Σ y A = (25)(20 × 80) + (7.5)(15 × 20) = 40 × 103 + 2.25 × 103 Q1 = 42.3 × 103 mm3 

Area below x-axis (Area A2): Q2 = Σ y A = (−32.5)(65 × 20)

Q2 = −42.3 × 103 mm3 

|Q1| = |Q2 |, since C is centroid and thus, Q = Σ y A = 0 


www.elsolucionario.net PROBLEM 5.22 A composite beam is constructed by bolting four plates to four 60 × 60 × 12-mm angles as shown. The bolts are equally spaced along the beam, and the beam supports a vertical load. As proved in mechanics of materials, the shearing forces exerted on the bolts at A and B are proportional to the first moments with respect to the centroidal x-axis of the red-shaded areas shown, respectively, in parts a and b of the figure. Knowing that the force exerted on the bolt at A is 280 N, determine the force exerted on the bolt at B.


SOLUTION

From the problem statement, F is proportional to Qx . (Qx ) B FA (Qx ) A

Therefore,

FA FB = , or (Qx ) A (Qx ) B

For the first moments,

12   (Qx ) A =  225 +  (300 × 12) 2 

FB =

= 831, 600 mm3 12   (Qx ) B = (Qx ) A + 2  225 −  (48 × 12) + 2(225 − 30)(12 × 60) 2  = 1,364,688 mm3

Then

FB =

1,364, 688 (280 N) 831, 600

or FB = 459 N 


www.elsolucionario.net PROBLEM 5.23 The first moment of the shaded area with respect to the x-axis is denoted by Qx. (a) Express Qx in terms of b, c, and the distance y from the base of the shaded area to the x-axis. (b) For what value of y is Qx maximum, and what is that maximum value?

SOLUTION Shaded area: Qx = yA = Qx =

(a) (b)

For Qmax , For y = 0,

1 (c + y )[b(c − y )] 2 1 b (c 2 − y 2 ) 2

dQ = 0 or dy (Qx ) =



1 b(−2 y ) = 0 2

1 2 bc 2

y=0  (Qx ) =

1 2 bc  2



A = b (c − y )

www.elsolucionario.net PROBLEM 5.24 A thin, homogeneous wire is bent to form the perimeter of the figure indicated. Locate the center of gravity of the wire figure thus formed.

SOLUTION

L, mm

Then

x , mm

y , mm

yL, mm 2

yL, mm 2

1

302 + 722 = 78

15

36

1170.0

2808.0

2

482 + 722 = 86.533

54

36

4672.8

3115.2

39

72

3042.0

5616.0

8884.8

11,539.2

3

78

Σ

242.53 X ΣL = Σx L X (242.53) = 8884.8

or X = 36.6 mm  and

Y ΣL = Σ yL Y (242.53) = 11,539.2

or Y = 47.6 mm 



First note that because the wire is homogeneous, its center of gravity will coincide with the centroid of the corresponding line.


L, mm 1

Then

722 + 482 = 86.533

x , mm

y , mm

xL, mm 2

yL, mm 2

36

−24

3115.2

−2076.8

2

132

72

18

9504.0

2376.0

3

1262 + 302 = 129.522

9

69

1165.70

8937.0

4

54

−54

27

−2916.0

5

54

−27

0

−1458.0

Σ

456.06

9410.9

1458.0 0 10,694.2

X ΣL = Σx L X (456.06) = 9410.9

or X = 20.6 mm 

Y ΣL = Σ y L Y (456.06) = 10, 694.2

or Y = 23.4 mm 



SOLUTION


Then

y , in.

xL, in 2

yL, in 2

3

80.498

40.249

7.5

36

22.5

9

9

54

54.0

3

6

7.5

18

22.5

5

6

3

6

18

36.0

6

6

0

3

0

18.0

Σ

37.416

206.50

193.249

L, in.

x , in.

1

122 + 62 = 13.4164

6

2

3

12

3

6

4

X ΣL = Σx L Y ΣL =Σ y L

X (37.416) = 206.50 Y (37.416) = 193.249

X = 5.52 in.  Y = 5.16 in. 



SOLUTION


SOLUTION

x , in.

L, in. 1

Then

1 π (10) = 15.7080 2

2(10)

π

= 6.3662

yL, in 2

100

2

5

0

0

3

5

2.5

12.5

4

5

5

25

5

5

7.5

37.5

Σ

35.708 X ΣL = Σx L


By symmetry, X = Y .

175 X (35.708) = 175

X = Y = 4.90 in. 


www.elsolucionario.net PROBLEM 5.28 The homogeneous wire ABCD is bent as shown and is attached to a hinge at C. Determine the length L for which portion BCD of the wire is horizontal.

SOLUTION First note that for equilibrium, the center of gravity of the wire must lie on a vertical line through C. Further, because the wire is homogeneous, the center of gravity of the wire will coincide with the centroid of the corresponding line. Thus,

Then

L + (−40 mm)(80 mm) + (−40 mm)(100 mm) = 0 2 L2 = 14, 400 mm 2

L = 120.0 mm 



X = 0 so that Σ X L = 0

www.elsolucionario.net PROBLEM 5.29 The homogeneous wire ABCD is bent as shown and is attached to a hinge at C. Determine the length L for which portion AB of the wire is horizontal.

WI = 80w

WII = 100w

WIII = Lw

Σ M C = 0: (80 w)(32) + (100 w)(14) − ( Lw)(0.4 L) = 0

L2 = 9900

L = 99.5 mm 



SOLUTION

www.elsolucionario.net PROBLEM 5.30 The homogeneous wire ABC is bent into a semicircular arc and a straight section as shown and is attached to a hinge at A. Determine the value of θ for which the wire is in equilibrium for the indicated position.

First note that for equilibrium, the center of gravity of the wire must lie on a vertical line through A. Further, because the wire is homogeneous, its center of gravity will coincide with the centroid of the corresponding line. Thus, X =0

so that

Σx L = 0

Then

 1   2r   − 2 r cos θ  (r ) +  π − r cos θ  (π r ) = 0    

or

cos θ =

4 1 + 2π = 0.54921

or θ = 56.7° 



SOLUTION

www.elsolucionario.net PROBLEM 5.31 A uniform circular rod of weight 8 lb and radius 10 in. is attached to a pin at C and to the cable AB. Determine (a) the tension in the cable, (b) the reaction at C.

SOLUTION For quarter circle,

2r

π

 2r  ΣM C = 0: W   − Tr = 0 π  2 2 T = W   = (8 lb)   π  π 

(b)

T = 5.09 lb 

ΣFx = 0: T − C x = 0 5.09 lb − C x = 0

C x = 5.09 lb

ΣFy = 0: C y − W = 0

C y = 8 lb

C y − 8 lb = 0

C = 9.48 lb

57.5° 



(a)

r =

www.elsolucionario.net PROBLEM 5.32 Determine the distance h for which the centroid of the shaded area is as far above line BB′ as possible when (a) k = 0.10, (b) k = 0.80.

A

y

yA

1

1 ba 2

1 a 3

1 2 a b 6

2

1 − (kb)h 2

1 h 3

1 − kbh 2 6

Σ

b (a − kh) 2


SOLUTION

b 2 ( a − kh 2 ) 6

Y Σ A = Σ yA

Then

b  b Y  ( a − kh)  = (a 2 − kh 2 ) 2   6 Y =

or

(1)

dY 1 −2kh(a − kh) − (a 2 − kh 2 )(− k ) = =0 dh 3 ( a − kh) 2

and or

a 2 − kh 2 3(a − kh)

2h(a − kh) − a 2 + kh 2 = 0

(2)

Simplifying Eq. (2) yields kh 2 − 2ah + a 2 = 0



Then

2a ± (−2a) 2 − 4( k )(a 2 ) 2k a = 1 ± 1 − k  k

h=

Note that only the negative root is acceptable since h < a. Then (a)

k = 0.10 h=

(b)

a  1 − 1 − 0.10   0.10 

or h = 0.513a 

k = 0.80 a  1 − 1 − 0.80   0.80 

or h = 0.691a 


h=


www.elsolucionario.net PROBLEM 5.33 Knowing that the distance h has been selected to maximize the distance y from line BB′ to the centroid of the shaded area, show that y = 2h/3.

SOLUTION See solution to Problem 5.32 for analysis leading to the following equations: a 2 − kh 2 3(a − kh)

(1)

2h(a − kh) − a 2 + kh 2 = 0

(2)

Rearranging Eq. (2) (which defines the value of h which maximizes Y ) yields a 2 − kh 2 = 2h(a − kh)

Then substituting into Eq. (1) (which defines Y ), Y =

1 × 2h(a − kh) 3( a − kh)

or Y =

2 h  3



Y =

www.elsolucionario.net PROBLEM 5.34 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and h.

SOLUTION h x a

y=

We have

dA = (h − y )dx

and

x  = h 1 −  dx  a


xEL = x 1 (h + y ) 2 h x = 1 +  2 a

yEL =



A = dA =

Then

and

x y

a

=



EL dA =



EL dA

=

0

a 0

h2 2



a 0

a

 x x2  1  h 1 −  dx = h  x −  = ah a 2 a    0 2 a

 x 2 x3    x  1 x  h 1 −  dx  = h  −  = a 2 h   a   2 3a  0 6 h x   x  1 +   h 1 −  dx   a   a   2



a 0

x2 − 1  2  a

a

 h2  x3  1 = − = ah2 dx x   2 2 3a  0 3  

1  1 xA = xEL dA: x  ah  = a 2 h 2  6

x=

2 a  3

1  1 y A = yEL dA: y  ah  = ah2 2  3

y=

2 h  3







SOLUTION y = h(1 − kx3 )

For x = a, y = 0. 0 = h(1 − k a3 )

∴ k=

1 a3

xEL = x,

yEL =



a

A = dA =

x



EL dA

=

yEL dA =

=



a

0





0

xydx =

1 y dA = ydx 2 ydx =



a

0



h2 2

a

0

a

  x3  x4  3 h 1 − 3  dx = h  x − 3  = ah 4a  0 4  a   a

  x2 x4  x5  3 h  x − 3  dx = h  − 3  = a 2 h a    2 5a  0 10

a 0


 x3  y = h 1 − 3   a 

1  1  y  ydx = 2 2  



a 0

 x3  h2 h 2 1 − 3  dx = 2  a 



a 0

2 x3 x 6 1 − 3 + 6 a a 

  dx 

a

 x4 x7  9 2 ah x − 3 + 6  = 28 2a 7a  0 

3  3 xA = xEL dA: x  ah  = a 2 h  4  10

x=

2 a  5

3  9 2 yA = yEL dA: y  ah  = ah  4  28

y=

3 h  7







SOLUTION y1 : h = ka 2

At (a, h),

k=

or

h a2

m=

or

xEL = x

Now

yEL =

1 ( y1 + y2 ) 2

h  h dA = ( y2 − y1 )dx =  x − 2 x 2  dx a a   h = 2 (ax − x 2 ) dx a

and



A = dA =

Then

and

h a

 



a 0

a

h h a 1  1 (ax − x 2 )dx = 2  x 2 − x3  = ah 2 3 0 6 a a 2 a

h a 1  1 h xEL dA = x  2 (ax − x 2 ) dx = 2  x3 − x 4  = a 2 h 0 a 4  0 12 a 3 1 1 2 yEL dA = y2 − y12 dx ( y1 + y2 )[( y2 − y1 ) dx] = 2 2



a

 (



=

1 2



)

a 0

h 2 2 h2 4   2 x − 4 x dx a a  a

1 h2  a 2 3 1 5  =  x − x  2 a4  3 5 0 1 = ah 2 15



y2 : h = ma


1  1 xA = xEL dA: x  ah  = a 2 h  6  12

x=

1 a  2

1  1 yA = yEL dA: y  ah  = ah 2  6  15

y=

2 h  5







www.elsolucionario.net PROBLEM 5.37 Determine by direct integration the centroid of the area shown.

SOLUTION y1 = k x 2

a = k a 2 , thus, k =

But

y2 = a x ,

y1 =

1 a2

x2 a


x , k

y2 =

xEL = x  x2  dA = ( y2 − y1 ) dx =  ax −  dx a  



A = dA =



a

x2   ax −  dx 0 a   a

2 1 x3  = a x3/2 −  = a 2 3a  0 3 3

x

EL dA

=



a 0

 x2  x  a x −  dx = a  



a 0

 a x 

1  3 3 xA = xEL dA: x  a 2  = a  3  20



a

3/2

2 x3  x4  3 3 −  dx =  a x5/2 −  = a a  4a  0 20 5

x=

9a 20 y=x=

By symmetry,

9a  20



SOLUTION b 2 a − x2 a

dA = (b − y ) dx

and

xEL

)

(

b a − a 2 − x 2 dx a =x =

1 ( y + b) 2 b = a + a 2 − x2 2a

yEL =

(



A = dA =

Then



To integrate, let

x = a sin θ :

Then

A=



π /2 0

a 0

) )

(

b a − a 2 − x 2 dx a

a 2 − x 2 = a cos θ , dx = a cos θ dθ

b (a − a cos θ )(a cos θ dθ ) a π /2

=

b 2 2θ   2 θ  a sin θ − a  + sin  a 4   0 2

 π = ab 1 −  4 

and

x

EL dA

a

)

(

b  x  a − a 2 − x 2 dx  a  

=



=

b  a 2 1 2 2 3/2    x + ( a − x )   a  2 3  0

=

1 3 ab 6

0

π /2



y=

For the element (EL) shown,


y

EL dA

=



a 0

b2 = 2 2a =

) (

(

)

b b  a + a 2 − x 2  a − a 2 − x 2 dx  2a a   a



a 0

b 2  x3  ( x ) dx = 2   2a  3  0 2

1 2 ab 6



  π  1 x  ab 1 −   = a 2 b 4  6  

or x =

2a  3(4 − π )



  π  1 y  ab 1 −   = ab 2 4  6  

or y =

2b  3(4 − π )

xA = xEL dA:


yA = yEL dA:



SOLUTION x =0 

First note that symmetry implies For the element (EL) shown, yEL =

2r

(Figure 5.8B)



A = dA =

Then

and

So



yEL dA =



r2 r1



r2 r1

 r2 π rdr = π   2

r2

 π 2 2 r2 − r1  =  r1 2 r2

(

2 1  (π rdr ) = 2  r 3  = r23 − r13 π  3  r1 3

2r

(

)

π  2 yA = yEL dA: y  r22 − r12  = r23 − r13 2   3



(

)

(

)

)

or y =

4 r23 − r13  3π r22 − r12



π dA = π rdr

www.elsolucionario.net PROBLEM 5.40 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and b.

SOLUTION b 2 x a2 b y2 = k2 x 4 but b = k2 a 4 y2 = 4 x 4 a 4 b  x  dA = ( y2 − y1 )dx = 2  x 2 − 2  dx a  a  xEL = x but b = k1a 2

1 ( y1 + y2 ) 2 b  x4 = 2  x 2 + 2 2a  a

y1 =

yEL =



A = dA =

b a2



  

a 0

x4 2 − x  a2 

  dx 

a



b  x3 x5  = 2  − 2 a  3 5a  0 2 = ba 15 a b  x4  xEL dA = x 2  x 2 − 2  dx 0 a  a  b a x5  = 2  x3 − 2  dx a 0 a 





a

 x4 x6  −  2  4 6a  0

=

b a2

=

1 2 a b 12



y1 = k1 x 2




yEL dA = =



a 0

b 2a 2

b2 2a 4



 2 x4  b  x + 2  2 a a 

 2 x4   x − 2  dx a  

a 0

x8  4  x − 4  dx a   a

b 2  x5 x9  2 2 = 4  − 4 = ab 2a  5 9a  0 45  2  1 xA = xEL dA: x  ba  = a 2b  15  12

5 x= a  8

 2  2 2 yA = yEL dA: y  ba  = ab   15  45

1 y = b  3










SOLUTION y =b 

First note that symmetry implies y=b

y1 : b = ka 2 y1 =

Then

k=

or

b a2


x = a,

At

b 2 x a2

y2 : b = 2b − ca 2 c=

or

 x2 y2 = b  2 − 2 a 

Then

  

  x2 dA = ( y2 − y1 )dx2 = b  2 − 2 a    x2  = 2b 1 − 2  dx  a 

Now

 b 2  − 2 x  dx   a

xEL = x

and

 

A = dA

Then

and

b a2



a

0

 x2 2b 1 − 2  a

a

  x3  4  dx = 2b  x − 2  = ab 3a  0 3   a

   x2 x2   x4  1 xEL dA = x  2b 1 − 2  dx  = 2b  − 2  = a 2b 0   a    2 4a  0 2 4  1 xA = xEL dA: x  ab  = a 2b 3  2



a



3 x= a  8



xEL = x

yEL =



L

A = dA =



xEL dA =



L

0



0

1 y dA = y dx 2 L

  x x2  x 2 2 x3  5 h 1 + − 2 2  dx = h  x + − = hL 2 L 2L 3 L 0 6 L   

 x x2  xh 1 + − 2 2  dx = h L L  



L

0

x2 x3  − 2 2  dx  x + L L  

L

 x 2 1 x3 2 x 4  1 = h + − = hL2 2   2 3 L 4 L 0 3 5  1 xA = xEL dA: x  hL  = hL2 6  3



A=



5 hL 6

yEL =



=

h2 2

h2 = 2





L 0

2 L  5

 x x2  y = h 1 + − 2 2  L L  

1 y 2

1 2 h2 yEL dA = y dx = 2 2

x=

2

x x2  1 + − 2 2  dx L L  

L 0

x2 x4 x x2 x3  1 + 2 + 4 4 + 2 − 4 2 − 4 3  dx L L L L L   L

 4 x3 4 x5 x 2 4 x3 x 4  x + + 4 + − 2 − 3  = h2 L  2 10 L 3 5 3 L L L L  0

5  4 yA = yEL dA: y  hL  = h 2 L  6  10



y=

12 h  25



SOLUTION


SOLUTION For y1 at x = a,

y = 2b, 2b = ka 2 , or k =

Then

y1 =

By observation,

b x  y2 = − ( x + 2b) = b  2 −  a a 

2b a2

Now

xEL = x

and for 0 ≤ x ≤ a,

yEL =

1 b y1 = 2 x 2 2 a

For a ≤ x ≤ 2a,

yEL =

1 b x x  y2 =  2 −  and dA = y2 dx = b  2 −  dx a a 2 2  



A = dA =

Then



a 0

and dA = y1 dx =

2b 2 x dx + a2



2a a

2b 2 x dx a2

x  b  2 −  dx a  2a

2  a 2b  x3  x  7 = 2   + b  −  2 −   = ab a   6 a  3 0  2  0 a

and



xEL dA =



a 0

 2b  x  2 x 2 dx  + a 



2a a

a

  x  x b  2 −  dx  a    2a

 2b  x 4  x3  = 2   + b  x2 −  3a  0 a  4 0 

1 2 1  2   (2a ) − (a )3   a b + b  (2a) 2 − (a) 2  +  2 3 a   7 2 = a b 6 =



2b 2 x a2




yEL dA =



a 0

b 2  2b 2  x x dx  + a 2  a 2 



2a 0

b x   x   2 −  b  2 −  dx  2 a   a  2a

3 2b 2  x5  b2  a  x  = 4   + −  2 −   a   a  5  0 2  3  a 17 2 ab = 30 a

Hence,

7  7 xA = xEL dA: x  ab  = a 2b 6  6



 7  17 2 yA = yEL dA: y  ab  = ab  6  30

y=

17 b  35




x =a 



SOLUTION For y2 at x = a,

y = b, a = kb 2 , or k =

Then

y2 =

Now

xEL = x a , 2

yEL =

a

x1/2

y2 b x1/ 2 = 2 2 a x1/2

dA = y2 dx = b

For

a ≤ x ≤ a, 2

yEL =

a

dx

1 b  x 1 x1/ 2  ( y1 + y2 ) =  − +  2 2a 2 a 

 x1/2 x 1  dA = ( y2 − y1 )dx = b  − +  dx  a a 2

Then



A = dA =



a /2

b 0

x1/ 2 a

dx +



 x1/2 x 1  b  − +  dx a/ 2  a a 2 a

a

a/ 2  2 x3/2 x 2 1  b  2 3/2  = x + b − + x   a  3 0  3 a 2a 2  a/2 3/ 2 3/ 2 2 b  a  a  3/ 2 =   + ( a ) −    3 a  2   2   2  1  2  a   1   a    + b −  (a ) −    +  (a) −     2a   2   2   2     13 ab = 24



and for 0 ≤ x ≤

b

a b2


and



xEL dA =



a/2 0

 x1/ 2  x  b dx  + a  



  x1/ 2 x 1   x b  − +   dx a/2   a a 2   a

a

a/ 2  2 x5/2 x3 x 4  b  2 5/2  = + − +  x b   a  5 0  5 a 3a 4  a/2 5/2 5/2 2 b  a  a  5/2 =   + ( a ) −    5 a  2   2   3 2  1  a  1  a    + b  −  ( a )3 −    +  ( a ) 2 −      2   4   2     3a  71 2 a b = 240



a/2 0

+



b x1/ 2 2 a

 x1/ 2  dx  b a  




yEL dA =

b  x 1 x1/2    x1/ 2 x 1   − +  dx   − +  b  a/ 2 2  a 2 a    a a 2    a

a

a/ 2 3 b2  1 2  b 2  x 2 1  x 1       x = + − − 2a  2  0 2  2a 3a  a 2      a/2

Hence,

=

3 2 2 2 b  a  a  b a 1 −   + ( a ) 2 −    −   4a  2   2   6a  2 2 

=

11 2 ab 48

 13  71 2 xA = xEL dA : x  ab  = a b  24  240



 13  11 2 yA = yEL dA: y  ab  = ab  24  48



x=

17 a = 0.546a  130

y=

11 b = 0.423b  26


www.elsolucionario.net PROBLEM 5.45 A homogeneous wire is bent into the shape shown. Determine by direct integration the x coordinate of its centroid.

SOLUTION First note that because the wire is homogeneous, its center of gravity coincides with the centroid of the corresponding line. xEL = r cos θ

Then

L = dL =

and





xEL dL =

7π /4

π

/4

and dL = rdθ 7π / 4

π

/4

3 r dθ = r[θ ]π7π/ 4/ 4 = π r 2

r cos θ (rdθ )

= r 2 [sin θ ] π7π/4/ 4  1 1  = r2  − −  2 2  = −r 2 2

Thus

3  xL = x dL : x  π r  = − r 2 2 2 



x =−

2 2 r  3π



Now

www.elsolucionario.net PROBLEM 5.46 A homogeneous wire is bent into the shape shown. Determine by direct integration the x coordinate of its centroid.

SOLUTION First note that because the wire is homogeneous, its center of gravity coincides with the centroid of the corresponding line. xEL = a cos3 θ

Now

and dL = dx 2 + dy 2

x = a cos3 θ : dx = −3a cos 2 θ sin θ dθ

where


y = a sin 3 θ : dy = 3a sin 2 θ cos θ dθ dL = [(−3a cos 2 θ sin θ dθ ) 2 + (3a sin 2 θ cos θ dθ )2 ]1/ 2

Then

= 3a cos θ sin θ (cos 2 θ + sin 2 θ )1/ 2 dθ = 3a cos θ sin θ dθ



L = dL =

and

x

EL dL

=

3 a 2

=



π /2 0



π /2 0

π /2

1  3a cos θ sin θ dθ = 3a  sin 2 θ  2 0

a cos3θ (3a cos θ sin θ dθ ) π /2

 1  = 3a 2  − cos5 θ   5 0

Hence,

=

3 2 a 5

3  3 xL = xEL dL : x  a  = a 2 2  5



x=

2 a  5

Alternative Solution:  x x = a cos3 θ  cos 2 θ =   a  y y = a sin θ  sin θ =   a 3

x a  

2/3

2/3

2/3

2

 y +  a

2/3

= 1 or

y = (a 2/3 − x 2/3 )3/2



dy = (a 2/3 − x 2/3 )1/ 2 (− x −1/3 ) dx

Now

xEL = x

and

 dy  dL = 1 +    dx 

2

{

dx = 1 + ( a 2/3 − x 2/3 )1/2 (− x −1/3 ) 



L = dL =

Then

and

Hence



xEL dL =



a 0



a 0

2

}

1/2

dx a

a1/3 3 3  dx = a1/3  x 2/3  = a 1/3 x 2 0 2

a  a1/3  3 3  x  1/3 dx  = a1/3  x5/3  = a 2 5 0 5 x 

3  3 xL = xEL dL : x  a  = a 2 2  5



x=

2 a  5



Then

www.elsolucionario.net PROBLEM 5.47* A homogeneous wire is bent into the shape shown. Determine by direct integration the x coordinate of its centroid. Express your answer in terms of a.

SOLUTION First note that because the wire is homogeneous, its center of gravity will coincide with the centroid of the corresponding line. We have at x = a,

y = a, a = ka3/2 , or k =

Then

y=

a

a

x3/ 2


1

1

dy 3 1/2 = x dx 2 a

and Now

xEL = x

and

 dy  dL = 1 +   dx  dx 

2

1/2

2   3   x1/2   = 1 +    2 a   1 = 4a + 9 x dx 2 a



L = dL =

Then



a

1

0

2 a

dx

4a + 9 x dx a

1 2 1  × (4a + 9 x)3/ 2   3 9 2 a 0 a = [(13)3/2 − 8] 27 = 1.43971a =

and

x

EL dL

=



a 0

 1 x 2 a

 4a + 9 x dx  



Use integration by parts with u=x du = dx

Then



dv = 4a + 9 x dx 2 v= (4a + 9 x)3/ 2 27

a 1   2 3/2  xEL dL =   x × (4a + 9 x)  − 2 a   27 0



a

0

 2 (4a + 9 x)3/ 2 dx  27  a

=

(13)3/2 2 1 2  (4a + 9 x)5/2  a − 27 27 a  45 0

=

2 a2   3/ 2 5/2 (13) − [(13) − 32] 27  45 



xL = xEL dL : x (1.43971a ) = 0.78566a 2

or x = 0.546a 



= 0.78566a 2


SOLUTION

y = a sin

πx L

1 y, dA = y dx 2 L /2 L /2 πx sin A= y dx = a dx 0 0 L yEL =





L /2

L π x  A = a   − cos  L  0 π  

Setting u =

πx L

, we have x =

Integrating by parts,

L /2



xEL dA =



L

u , dx =

L

π


xEL = x,

0

π

xydx =



L /2

aL

=

π

xa sin

0

πx L

du,



π /2 

L  L  L  π u  a sin u  π du  = a  π       



xEL dA =



L  xEL dA = a   [−u cos u ]π0 /2 + π  



yEL dA =

0

2

=

dx



L /2

0

a2 L 2π 2

1 2 1 y dx = a 2 2 2



π /2 1 0

2



L /2

0



π /2 0

sin 2

2



π /2 0

u sin x du

2  aL cos u du = 2  π

πx L

dx =

a2 L 2π



π /2 0

sin 2 u du π /2

(1 − cos 2u )du =

2  aL  aL xA = xEL dA: x  =  2 π  π



 aL  1 2 yA = yEL dA: y  = a L π  8



a2 L  1  u − sin 2u   4π  2 0

1 = a2 L 8 x= y=

π 8

L

π



a 


www.elsolucionario.net PROBLEM 5.49* Determine by direct integration the centroid of the area shown.

SOLUTION 2 2 r cos θ = aeθ cos θ 3 3 2 2 θ = r sin θ = ae sin θ 3 3

xEL = yEL

dA =

and

1 1 (r )(rdθ ) = a 2 e2θ d θ 2 2



A = dA =

Then



π 0

π

1 2 2θ 1 1  a e dθ = a 2  e 2θ  2 2 2 0

1 2 2π a (e − 1) 4 = 133.623a 2 =

and

x

EL dA =



π 0

2 θ 1  ae cos θ  a 2 e2θ dθ  3 2 

1 = a3 3



π 0

e3θ cos θ dθ

To proceed, use integration by parts, with u = e3θ

du = 3e3θ dθ

and

dv = cos θ dθ

Then

e

3θ



then

du = 3e3θ dθ

dv = sin θ dθ , then

Then

e

3θ

v = sin θ

cos θ dθ = e3θ sin θ − sin θ (3e3θ dθ ) u = e3θ

Now let

and

v = − cos θ

cos θ dθ = e3θ sin θ − 3  −e3θ cos θ − ( − cos θ )(3e3θ dθ )   





We have


so that



e3θ cos θ dθ =



π

 1  e3θ (sin θ + 3cos θ )  xEL dA = a 3  3  10 0 =



Also,

e3θ (sin θ + 3cos θ ) 10

yEL dA =

a3 (−3e3π − 3) = −1239.26a3 30



π 0

2 θ 1  ae sin θ  a 2 e2θ dθ  3 2  

1 = a3 3



π 0

e3θ sin θ dθ

u = e3θ

and



dv = sin θ dθ

Then

e

so that



3θ

and

v = − cos θ



sin θ dθ = −e3θ cos θ − (− cos θ )(3e3θ dθ )

e3θ sin θ dθ =



e3θ (− cos θ + 3sin θ ) 10 π

 1  e3θ (− cos θ + 3sin θ )  yEL dA = a3  3  10 0 =

Hence,

du = 3e3θ dθ

a 3 3π (e + 1) = 413.09a 3 30



or x = −9.27a 



or

xA = xEL dA: x (133.623a 2 ) = −1239.26a 3 yA = yEL dA : y (133.623a 2 ) = 413.09a3

y = 3.09a 



Use integration by parts, as above, with

www.elsolucionario.net PROBLEM 5.50 Determine the centroid of the area shown when a = 2 in.

SOLUTION xEL = x

We have

yEL =



A = dA =

Then

x 

EL dA

=

yEL dA =



a



a

1

1




 1 dA = ydx = 1 −  dx x 

and

and

1 1 1 y = 1 −  2 2 x

a

1

1  dx 2 a 1 − x  2 = [ x − ln x]1 = ( a − ln a − 1) in   a

  a2  1    x 2 1 − a +  in 3 x 1 −  dx  =  − x  =  x   2 2  1  2 1  1   1   1 1 −  1 −  dx  = 2  x   x  2



a

1

2 1  1 − x + 2  dx x  

a

=

1 1 1 1 x − 2ln x −  =  a − 2ln a −  in 3 2  x 1 2  a



xA = xEL dA: x =



yA = yEL dA: y =

a2 2

−a+

1 2

in. a − ln a − 1 a − 2ln a − 1a 2(a − ln a − 1)

in.

Find x and y when a = 2 in. We have

x=

and

y=

1 2

(2) 2 − 2 +

1 2

2 − ln 2 − 1 2 − 2ln 2 − 12 2(2 − ln 2 − 1)

or

x = 1.629 in. 

or y = 0.1853 in. 


www.elsolucionario.net PROBLEM 5.51 Determine the value of a for which the ratio x / y is 9.

SOLUTION xEL = x yEL =

1 1 1 y = 1 −  2 2 x

 1 dA = y dx = 1 −  dx x 

and



A = dA =

Then



a

1

1  dx = [ x − ln x]1a 1 −  x 2 

= (a − ln a − 1) in 2

and

x

EL dA

=



a

1

a

  1    x 2 x 1 −  dx  =  − x  x   2  1

 a2 1 =  − a +  in 3 2  2



yEL dA =



a

1

1  1   1   1 1 −  1 −  dx  = 2  x   x  2



a

1

2 1  1 − x + 2  dx x  

a

=

1 1 x − 2ln x −  2  x 1

=

1 1 a − 2ln a −  in 3 2  a



xA = xEL dA: x =



yA = yEL dA: y =

a2 2

−a+

1 2

in. a − ln a − 1 a − 2ln a − 1a 2(a − ln a − 1)

in.



We have


Find a so that

x = 9. y x xA = = y yA

We have

Then or

1 2

1 2

a2 − a +

1 2

( a − 2 ln a − a1 )

x y

EL dA EL dA

=9

a 3 − 11a 2 + a + 18a ln a + 9 = 0

Using trial and error or numerical methods, and ignoring the trivial solution a = 1 in., we find


a = 1.901 in. and a = 3.74 in. 


www.elsolucionario.net PROBLEM 5.52 Determine the volume and the surface area of the solid obtained by rotating the area of Problem 5.1 about (a) the x-axis, (b) the y-axis.

SOLUTION From the solution of Problem 5.1, we have A = 11 in 2 Σ yA = 39.5 in 3

Applying the theorems of Pappus-Guldinus, we have (a)

Rotation about the x-axis: Volume = 2π yarea A = 2π Σ yA = 2π (39.5 in 3 )

or Volume = 248 in 3 

Area = 2π yline L = 2π Σ( yline ) L = 2π ( y2 L2 + y3 L3 + y4 L4 + y5 L5 + y6 L6 + y7 L7 + y8 L8 ) = 2π [(1)(2) + (2)(3) + (2.5)(1) + (3)(3) + (5.5)(5) + (8)(1) + (4)(8)]

or Area = 547 in 2  (b)

Rotation about the y-axis: Volume = 2π xarea A = 2π Σ xA = 2π (11.5 in 3 )

or Volume = 72.3 in 3 

Area = 2π xline L = 2π Σ( xline ) L = 2π ( x1 L1 + x2 L2 + x3 L3 + x4 L4 + x5 L5 + x6 L6 + x7 L7 ) = 2π [(0.5)(1) + (1)(2) + (2.5)(3) + (4)(1) + (2.5)(3) + (1)(5) + (0.5)(1)]

or Area = 169.6 in 2 



Σ xA = 11.5 in 3

www.elsolucionario.net PROBLEM 5.53 Determine the volume and the surface area of the solid obtained by rotating the area of Problem 5.2 about (a) the line y = 72 mm, (b) the x-axis.

SOLUTION From the solution of Problem 5.2, we have

y = 48 mm


Rotation about the line y = 72 mm: Volume = 2π (72 − y ) A = 2π (72 − 48)(2808)

Volume = 423 × 103 mm3 

Area = 2π yline L = 2π Σ( yline ) L = 2π ( y1 L1 + y3 L3 )

where y1 and y3 are measured with respect to line y = 72 mm. Area = 2π (36) 

(

)

482 + 722 + (36)

(

)

302 + 722   Area = 37.2 × 103 mm 2 

(b)

Rotation about the x-axis: Volume = 2π yarea A = 2π (48)(2808)

Volume = 847 × 103 mm3 

Area = 2π yline L = 2π Σ( yline ) L = 2π ( y1 L1 + y2 L2 + y3 L3 ) = 2π (36) 

(

)

482 + 722 + (72)(78) + (36)

(

)

302 + 722   Area = 72.5 × 103 mm 2 



A = 2808 mm 2 x = 36 mm

www.elsolucionario.net PROBLEM 5.54 Determine the volume and the surface area of the solid obtained by rotating the area of Problem 5.8 about (a) the line x = −60 mm, (b) the line y = 120 mm.

SOLUTION From the solution of Problem 5.8, we have A = 7200 mm 2 Σ x A = −72 × 103 mm3


Rotation about line x = −60 mm: Volume = 2π ( x + 60) A = 2π (Σ xA + 60 A) = 2π [−72 × 103 + 60(7200)]

Volume = 2.26 × 106 mm3 

Area = 2π xline L = 2π Σ( xline ) L = 2π ( x1 L1 + x2 L2 + x3 L3 )

  2(60)  π (60)   2(60)  π (60)  = 2π  60 − +  60 + + (60)(120)      π  2   π  2   

where x1 , x2 , x3 are measured with respect to line x = −60 mm. (b)

Area = 116.3 × 103 mm 2 

Rotation about line y = 120 mm: Volume = 2π (120 − y ) A = 2π (120 A −Σ yA) = 2π [120(7200) − 629.83 × 103 ]

Volume = 1.471 × 106 mm3 

Area = 2π yline L = 2π Σ( yline ) L = 2π ( y1 L1 + y2 L2 + y4 L4 )

where y1 , y2 , y4 are measured with respect to line y = 120 mm.   2(60)  π (60)   2(60)  π (60)  Area = 2π 120 − + + (60)(120)      π  2   π  2    Area = 116.3 × 103 mm 2 



Σ y A = 629.83 × 103 mm3

www.elsolucionario.net PROBLEM 5.55 Determine the volume of the solid generated by rotating the parabolic area shown about (a) the x-axis, (b) the axis AA′.

SOLUTION First, from Figure 5.8a, we have

4 ah 3 2 y= h 5

A=

Applying the second theorem of Pappus-Guldinus, we have Rotation about the x-axis: Volume = 2π yA

 2  4  = 2π  h  ah   5  3 

(b)

or Volume =

16 π ah 2  15

or Volume =

16 2 πa h  3

Rotation about the line AA′: Volume = 2π (2a) A

4  = 2π (2a)  ah  3 



(a)

www.elsolucionario.net PROBLEM 5.56 Determine the volume and the surface area of the chain link shown, which is made from a 6-mm-diameter bar, if R = 10 mm and L = 30 mm.

SOLUTION The area A and circumference C of the cross section of the bar are A=

π 4

d 2 and C = π d .

V = 2(Vside ) + 2(Vend ) = 2( AL) + 2(π RA) = 2( L + π R) A

or

π  V = 2[30 mm + π (10 mm)]  (6 mm) 2  4  = 3470 mm3

For the area A,

or V = 3470 mm3 

A = 2( Aside ) + 2( Aend ) = 2(CL) + 2(π RC ) = 2( L + π R)C

or

A = 2[30 mm + π (10 mm)][π (6 mm)] = 2320 mm 2

or A = 2320 mm 2 



Also, the semicircular ends of the link can be obtained by rotating the cross section through a horizontal semicircular arc of radius R. Now, applying the theorems of Pappus-Guldinus, we have for the volume V,

www.elsolucionario.net PROBLEM 5.57 Verify that the expressions for the volumes of the first four shapes in Figure 5.21 on Page 264 are correct.

SOLUTION Following the second theorem of Pappus-Guldinus, in each case, a specific generating area A will be rotated about the x-axis to produce the given shape. Values of y are from Figure 5.8a. Hemisphere: the generating area is a quarter circle. We have

(2)

2 V = π a3  3

 4a  π  V = 2π y A = 2π   ha   3π  4 

2 or V = π a 2 h  3

Paraboloid of revolution: the generating area is a quarter parabola. We have

(4)

or

Semiellipsoid of revolution: the generating area is a quarter ellipse. We have

(3)

 4a  π 2  V = 2π y A = 2π   a   3π  4 

 3  2  V = 2π y A = 2π  a  ah   8  3 

1 or V = π a 2 h  2

Cone: the generating area is a triangle. We have

 a  1  V = 2π y A = 2π   ha   3  2 

1 or V = π a 2 h  3



(1)

www.elsolucionario.net PROBLEM 5.58 Determine the volume and weight of the solid brass knob shown, knowing that the specific weight of brass is 0.306 lb/in3.

SOLUTION

Area, in2 1

π 4

(0.75) 2 = 0.4418

y , in.

y A, in 3

0.8183

0.3615

2

(0.5)(0.75) = 0.375

0.25

0.0938

3

(1.25)(0.75) = 0.9375

0.625

0.5859

4

−π (0.75) 2 = −0.4418 4

0.9317

Σ


Volume of knob is obtained by rotating area at left about the x-axis. Consider area as made of components shown below.

−0.4116 0.6296

V = 2π Σ y A = 2π (0.6296 in 3 ) = 3.9559 in 3

V = 3.96 in 3 

W = γ V = (0.306 lb/in 3 )(3.9559 in 3 )

W = 1.211 lb 


www.elsolucionario.net PROBLEM 5.59 Determine the total surface area of the solid brass knob shown.

Area is obtained by rotating lines shown about the x-axis.

1 2 3 4

π 2

π 2

L, in.

y , in.

yL, in 2

0.5

0.25

0.1250

(0.75) = 1.1781

0.9775

1.1516

(0.75) = 1.1781

0.7725

0.9101

0.25

0.1250

0.5

Σ

2.3117

A = 2π Σ y L = 2π (2.3117 in 2 )

A = 14.52 in 2 



SOLUTION

www.elsolucionario.net PROBLEM 5.60 The aluminum shade for the small high-intensity lamp shown has a uniform thickness of 1 mm. Knowing that the density of aluminum is 2800 kg/m3, determine the mass of the shade.

The mass of the lamp shade is given by

m = ρV = ρ At

where A is the surface area and t is the thickness of the shade. The area can be generated by rotating the line shown about the x-axis. Applying the first theorem of Pappus Guldinus, we have A = 2π yL = 2π Σ yL = 2π ( y1 L1 + y2 L2 + y3 L3 + y4 L4 )

or

13 mm  13 + 16  2 2 A = 2π  (13 mm) +   mm × (32 mm) + (3 mm)  2   2  16 + 28  2 2 +  mm × (8 mm) + (12 mm)  2    28 + 33  mm × (28 mm) 2 + (5 mm) 2  +   2   = 2π (84.5 + 466.03 + 317.29 + 867.51) = 10,903.4 mm 2

Then

m = ρ At = (2800 kg/m3 )(10.9034 × 10−3 m 2 )(0.001 m) m = 0.0305 kg 

or



SOLUTION

www.elsolucionario.net PROBLEM 5.61 The escutcheon (a decorative plate placed on a pipe where the pipe exits from a wall) shown is cast from brass. Knowing that the density of brass is 8470 kg/m3, determine the mass of the escutcheon.

SOLUTION The mass of the escutcheon is given by m = (density)V , where V is the volume. V can be generated by rotating the area A about the x-axis. From the figure:

L1 = 752 − 12.52 = 73.9510 m L2 =

37.5 = 76.8864 mm tan 26°

12.5 = 9.5941° 75 26° − 9.5941° = 8.2030° = 0.143168 rad α= 2

φ = sin −1

Area A can be obtained by combining the following four areas:

Applying the second theorem of Pappus-Guldinus and using Figure 5.8a, we have V = 2π yA = 2π Σ yA

Seg.

A, mm 2

y , mm

yA, mm3

1

1 (76.886)(37.5) = 1441.61 2

1 (37.5) = 12.5 3

18,020.1

2

−α (75) 2 = −805.32

2(75)sin α sin (α + φ ) = 15.2303 3α

−12,265.3

3

1 − (73.951)(12.5) = −462.19 2

1 (12.5) = 4.1667 3

−1925.81

4

−(2.9354)(12.5) = −36.693

1 (12.5) = 6.25 2

−229.33

Σ

3599.7



a = L2 − L1 = 2.9324 mm


Then

V = 2π Σ yA = 2π (3599.7 mm3 ) = 22, 618 mm3 m = (density)V

or m = 0.1916 kg 


= (8470 kg/m3 )(22.618 × 10−6 m3 ) = 0.191574 kg


www.elsolucionario.net PROBLEM 5.62 A 34 - in.-diameter hole is drilled in a piece of 1-in.-thick steel; the hole is then countersunk as shown. Determine the volume of steel removed during the countersinking process.

SOLUTION

V = 2π x A

3 1  1   1 1 1  = 2π  +   in. ×  × in. × in. 8 3 4 2 4 4      

V = 0.0900 in 3 



The required volume can be generated by rotating the area shown about the y-axis. Applying the second theorem of Pappus-Guldinus, we have

www.elsolucionario.net PROBLEM 5.63 Knowing that two equal caps have been removed from a 10-in.-diameter wooden sphere, determine the total surface area of the remaining portion.

SOLUTION The surface area can be generated by rotating the line shown about the y-axis. Applying the first theorem of Pappus-Guldinus, we have A = 2 π XL = 2 π Σ x L = 2π (2 x1 L1 + x2 L2 ) tan α =

4 3

or

α = 53.130°

Then

x2 =

5 in. × sin 53.130° π 53.130° × 180 °

= 4.3136 in.

and

π   L2 = 2  53.130° ×  (5 in.) 180°   = 9.2729 in.  3   A = 2π  2  in.  (3 in.) + (4.3136 in.)(9.2729 in.)   2   A = 308 in 2 

or



Now

www.elsolucionario.net PROBLEM 5.64 Determine the capacity, in liters, of the punch bowl shown if R = 250 mm.

SOLUTION The volume can be generated by rotating the triangle and circular sector shown about the y-axis. Applying the second theorem of Pappus-Guldinus and using Figure 5.8a, we have V = 2π xA = 2π Σ xA = 2π ( x1 A1 + x2 A2 )

 R3 R3  = 2π  +   16 3 2 3  3 3 π R3 8 3 3 = π (0.25 m)3 8 = 0.031883 m3 =

Since

103 l = 1 m3 V = 0.031883 m3 ×

103 l 1 m3

V = 31.9 l 



 1 1   1 1 3   2 R sin 30°   π 2   = 2π  × R   × R × R +  R    2   3 × π6   6    3 2   2 2

www.elsolucionario.net PROBLEM 5.65* The shade for a wall-mounted light is formed from a thin sheet of translucent plastic. Determine the surface area of the outside of the shade, knowing that it has the parabolic cross section shown.

SOLUTION First note that the required surface area A can be generated by rotating the parabolic cross section through π radians about the y-axis. Applying the first theorem of Pappus-Guldinus, we have A = π xL x = 100 mm, 250 = k (100)

and

2

y = 250 mm


Now at

or k = 0.025 mm −1

xEL = x 2

 dy  dL = 1 +   dx  dx 

where

dy = 2kx dx

Then

dL = 1 + 4k 2 x 2 dx

We have

xL = xEL dL =





100 0

x

( 1 + 4k x dx ) 2 2

100

1 1  (1 + 4k 2 x 2 )3/2  xL =  2 3  4k 0 1 1 = [1 + 4(0.025)2 (100) 2 ]3/2 − (1)3/2 12 (0.025) 2

{

}

= 17,543.3 mm 2

Finally,

A = π (17,543.3 mm 2 )

or A = 55.1 × 103 mm 2 


www.elsolucionario.net PROBLEM 5.66 For the beam and loading shown, determine (a) the magnitude and location of the resultant of the distributed load, (b) the reactions at the beam supports.

(a)

1 RI = (1100 N/m) (6 m) = 2200 N 3 RII = (900 N/m) (6 m) = 5400 N R = RI + RII = 2200 + 5400 = 7600 N XR = Σ xR :

X (7600) = (2200)(1.5) + (5400)(3) X = 2.5658 m

(b)

R = 7.60 kN ,

X = 2.57 m 

Σ M A = 0: B (6 m) − (7600 N)(2.5658 m) = 0 B = 3250.0 N

B = 3.25 kN  Σ Fy = 0: A + 3250.0 N − 7600 N = 0 A = 4350.0 N

A = 4.35 kN 



SOLUTION

www.elsolucionario.net PROBLEM 5.67 For the beam and loading shown, determine (a) the magnitude and location of the resultant of the distributed load, (b) the reactions at the beam supports.

SOLUTION

1 (150lb/ft)(9 ft) = 675 lb 2 1 RII = (120 lb/ft)(9ft) = 540lb 2 R = RI + RII = 675 + 540 = 1215 lb XR = Σ x R: X (1215) = (3)(675) + (6)(540)

X = 4.3333 ft R = 1215 lb

(a) (b)

Reactions:

X = 4.33 ft 

Σ M A = 0: B (9 ft) − (1215 lb) (4.3333 ft) = 0 B = 585.00 lb

B = 585 lb 

Σ Fy = 0: A + 585.00 lb − 1215 lb = 0 A = 630.00 lb

A = 630 lb 



RI =

www.elsolucionario.net PROBLEM 5.68 Determine the reactions at the beam supports for the given loading.

SOLUTION


First replace the given loading by the loadings shown below. Both loading are equivalent since they are both defined by a linear relation between load and distance and have the same values at the end points.

1 (900 N/m) (1.5 m) = 675 N 2 1 R2 = (400 N/m) (1.5 m) = 300 N 2 R1 =

Σ M A = 0: − (675 N)(1.4 m) + (300 N)(0.9 m) + B(2.5 m) = C B = 270 N

B = 270 N 

Σ Fy = 0: A − 675 N + 300 N + 270 N = 0 A = 105.0 N

A = 105.0 N 



SOLUTION R I = (200 lb/ft)(15 ft) R I = 3000 lb 1 (200 lb/ft)(6 ft) 2 R II = 600 lb R II =

B = 740 lb

B = 740 lb 

ΣFy = 0: A + 740 lb − 3000 lb − 600 lb = 0 A = 2860 lb

A = 2860 lb 



ΣM A = 0: − (3000 lb)(1.5 ft) − (600 lb)(9 ft + 2 ft) + B(15 ft) = 0


SOLUTION R I = (200 lb/ft)(4 ft) = 800 lb R II =

1 (150 lb/ft)(3 ft) = 225 lb 2

A = 575 lb 

ΣM A = 0: M A − (800 lb)(2 ft) + (225 lb)(5 ft) = 0 M A = 475 lb ⋅ ft





ΣFy = 0: A − 800 lb + 225 lb = 0


SOLUTION 1 (4 kN/m)(6 m) 2 = 12 kN RII = (2 kN/m)(10 m) RI =

= 20 kN

A = 32.0 kN 

ΣM A = 0: M A − (12 kN)(2 m) − (20 kN)(5 m) = 0 M A = 124.0 kN ⋅ m





ΣFy = 0: A − 12 kN − 20 kN = 0


SOLUTION

We have

RI = (6 m)(300 N/m) = 1800 N RII =

Then

2 (6 m)(1200 N/m) = 4800 N 3

ΣFx = 0: Ax = 0 ΣFy = 0: Ay + 1800 N − 4800 N = 0

or

Ay = 3000 N

A = 3.00 kN 

 15  ΣM A = 0: M A + (3 m)(1800 N) −  m  (4800 N) = 0  4 

or

M A = 12.6 kN ⋅ m

M A = 12.60 kN ⋅ m





First replace the given loading with the loading shown below. The two loadings are equivalent because both are defined by a parabolic relation between load and distance and the values at the end points are the same.


We have

Then

1 RI = (12 ft)(200 lb/ft) = 800 lb 3 1 RII = (6 ft)(100 lb/ft) = 200 lb 3 ΣFx = 0: Ax = 0 ΣFy = 0: Ay − 800 lb − 200 lb = 0

or

Ay = 1000 lb

A = 1000 lb 

ΣM A = 0: M A − (3 ft)(800 lb) − (16.5 ft)(200 lb) = 0

or

M A = 5700 lb ⋅ ft

M A = 5700 lb ⋅ ft





SOLUTION

www.elsolucionario.net PROBLEM 5.74 Determine the reactions at the beam supports for the given loading when wO = 400 lb/ft.

1 1 wO (12 ft) = (400 lb/ft)(12 ft) = 2400 lb 2 2 1 RII = (300 lb/ft)(12 ft) = 1800 lb 2 RI =

ΣM B = 0: (2400 lb)(1 ft) − (1800 lb)(3 ft) + C (7 ft) = 0 C = 428.57 lb

C = 429 lb 

ΣFy = 0: B + 428.57 lb − 2400 lb − 1800 lb = 0 B = 3771 lb

B = 3770 lb 



SOLUTION

www.elsolucionario.net PROBLEM 5.75 Determine (a) the distributed load wO at the end A of the beam ABC for which the reaction at C is zero, (b) the corresponding reaction at B.

For wO ,

1 wO (12 ft) = 6 wO 2 1 RII = (300 lb/ft)(12 ft) = 1800 lb 2 RI =

(a)

For C = 0,

ΣM B = 0: (6 wO )(1 ft) − (1800 lb)(3 ft) = 0

(b)

Corresponding value of R I :

wO = 900 lb/ft 

RI = 6(900) = 5400 lb ΣFy = 0: B − 5400 lb − 1800 lb = 0

B = 7200 lb 



SOLUTION

www.elsolucionario.net PROBLEM 5.76 Determine (a) the distance a so that the vertical reactions at supports A and B are equal, (b) the corresponding reactions at the supports.

SOLUTION (a)

Then

1 (a m)(1800 N/m) = 900a N 2 1 RII = [(4 − a ) m](600 N/m) = 300(4 − a) N 2 RI =

ΣFy = 0: Ay − 900a − 300(4 − a ) + B y = 0

or

Ay + B y = 1200 + 600a

Now

Ay = B y  Ay = B y = 600 + 300a (N)

Also,

(1)

 a  ΣM B = 0: − (4 m) Ay +  4 −  m  [(900a) N] 3   1  +  (4 − a ) m  [300(4 − a) N] = 0 3 

or

Ay = 400 + 700a − 50a 2

Equating Eqs. (1) and (2),

600 + 300a = 400 + 700a − 50a 2

or

a 2 − 8a + 4 = 0

(2)

8 ± (−8) 2 − 4(1)(4) 2

Then

a=

or

a = 0.53590 m

Now

a≤4m

a = 7.4641 m a = 0.536 m 



We have


(b)

We have From Eq. (1):

ΣFx = 0: Ax = 0 Ay = By = 600 + 300(0.53590) A = B = 761 N 


= 761 N


www.elsolucionario.net PROBLEM 5.77 Determine (a) the distance a so that the reaction at support B is minimum, (b) the corresponding reactions at the supports.

SOLUTION

We have

Then or Then (b)

From Eq. (1): and

1 (a m)(1800 N/m) = 900a N 2 1 RII = [(4 − a )m](600 N/m) = 300(4 − a) N 2 RI =

a  8+ a  ΣM A = 0: −  m  (900a N) −  m  [300(4 − a)N] + (4 m) By = 0 3    3 

By = 50a 2 − 100a + 800 dBy da

= 100a − 100 = 0

By = 50(1)2 − 100(1) + 800 = 750 N

(1) or a = 1.000 m  B = 750 N



ΣFx = 0: Ax = 0 ΣFy = 0: Ay − 900(1) N − 300(4 − 1) N + 750 N = 0

or

Ay = 1050 N

A = 1050 N 



(a)

www.elsolucionario.net PROBLEM 5.78 A beam is subjected to a linearly distributed downward load and rests on two wide supports BC and DE, which exert uniformly distributed upward loads as shown. Determine the values of wBC and wDE corresponding to equilibrium when wA = 600 N/m.

We have

1 (6 m)(600 N/m) = 1800 N 2 1 RII = (6 m)(1200 N/m) = 3600 N 2 RBC = (0.8 m) (wBC N/m) = (0.8wBC ) N RI =

RDE = (1.0 m) (wDE N/m) = ( wDE ) N

Then

ΣM G = 0: − (1 m)(1800 N) − (3 m)(3600 N) + (4 m)( wDE N) = 0 wDE = 3150 N/m 

or and or

ΣFy = 0: (0.8wBC ) N − 1800 N − 3600 N + 3150 N = 0 wBC = 2812.5 N/m

wBC = 2810 N/m 



SOLUTION

www.elsolucionario.net PROBLEM 5.79 A beam is subjected to a linearly distributed downward load and rests on two wide supports BC and DE, which exert uniformly distributed upward loads as shown. Determine (a) the value of wA so that wBC = wDE, (b) the corresponding values of wBC and wDE.

SOLUTION

1 (6 m)( wA N/m) ⋅ (3 wA ) N 2 1 RII = (6 m)(1200 N/m) = 3600 N 2 RBC = (0.8 m) (wBC N/m) = (0.8wBC ) N RI =

We have

RDE = (1 m) (wDE N/m) = ( wDE ) N ΣFy = 0: (0.8wBC ) N − (3wA ) N − 3600 N + ( wDE ) N = 0

Then or

0.8wBC + wDE = 3600 + 3wA 5 wBC = wDE  wBC = wDE = 2000 + wA 3

Now

(1)

ΣM G = 0: −(1 m)(3wA N) − (3 m)(3600 N) + (4 m)( wDE N) = 0

Also,

wDE = 2700 +

or

3 wA 4

(2)

Equating Eqs. (1) and (2), 5 3 2000 + wA = 2700 + wA 3 4

or (b) Eq. (1) 

wA =

8400 N/m 11

wA = 764 N/m 

5  8400  wBC = wDE = 2000 +  3  11 

or wBC = wDE = 3270 N/m 



(a)

www.elsolucionario.net PROBLEM 5.80 The cross section of a concrete dam is as shown. For a 1-m-wide dam section, determine (a) the resultant of the reaction forces exerted by the ground on the base AB of the dam, (b) the point of application of the resultant of part a, (c) the resultant of the pressure forces exerted by the water on the face BC of the dam.

SOLUTION Consider free body made of dam and section BDE of water. (Thickness = 1 m)


(a)

p = (3 m)(10 kg/m3 )(9.81 m/s 2 ) W1 = (1.5 m)(4 m)(1 m)(2.4 × 103 kg/m3 )(9.81 m/s 2 ) = 144.26 kN 1 W2 = (2 m)(3 m)(1 m)(2.4 × 103 kg/m3 )(9.81 m/s 2 ) = 47.09 kN 3 2 W3 = (2 m)(3 m)(1 m)(103 kg/m3 )(9.81 m/s 2 ) = 39.24 kN 3 1 1 P = Ap = (3 m)(1 m)(3 m)(103 kg/m3 )(9.81 m/s 2 ) = 44.145 kN 2 2 ΣFx = 0: H − 44.145 kN = 0 H = 44.145 kN

H = 44.1 kN



ΣFy = 0: V − 141.26 − 47.09 − 39.24 = 0 V = 227.6 kN

V = 228 kN





1 (1.5 m) = 0.75 m 2 1 x2 = 1.5 m + (2 m) = 2 m 4 5 x3 = 1.5 m + (2 m) = 2.75 m 8 x1 =

ΣM A = 0: xV − Σ xW + P (1 m) = 0

x(227.6 kN) − (141.26 kN)(0.75 m) − (47.09 kN)(2 m) − (39.24 kN)(2.75 m) + (44.145 kN)(1 m) = 0 x(227.6 kN) − 105.9 − 94.2 − 107.9 + 44.145 = 0 x(227.6) − 263.9 = 0

(b)

Resultant of face BC: Consider free body of section BDE of water.

− R = 59.1 kN R = 59.1 kN

41.6° 41.6° 



x = 1.159 m (to right of A) 

www.elsolucionario.net PROBLEM 5.81 The cross section of a concrete dam is as shown. For a 1-m-wide dam section, determine (a) the resultant of the reaction forces exerted by the ground on the base AB of the dam, (b) the point of application of the resultant of part a, (c) the resultant of the pressure forces exerted by the water on the face BC of the dam.

SOLUTION Consider free body made of dam and triangular section of water shown. (Thickness = 1 m.)


(a)

p = (7.2 m)(103 kg/m3 )(9.81m/s 2 ) 2 (4.8 m)(7.2 m)(1 m)(2.4 × 103 kg/m3 )(9.81 m/s 2 ) 3 = 542.5 kN 1 W2 = (2.4 m)(7.2 m)(1 m)(2.4 × 103 kg/m3 )(9.81 m/s 2 ) 2 = 203.4 kN 1 W3 = (2.4 m)(7.2 m)(1 m)(103 kg/m3 )(9.81 m/s 2 ) 2 = 84.8 kN 1 1 P = Ap = (7.2 m)(1 m)(7.2 m)(103 kg/m3 )(9.81 m/s 2 ) 2 2 = 254.3 kN W1 =

ΣFx = 0: H − 254.3 kN = 0

H = 254 kN



ΣFy = 0: V − 542.5 − 203.4 − 84.8 = 0 V = 830.7 kN

V = 831 kN 



5 x1 = (4.8 m) = 3 m 8 1 x2 = 4.8 + (2.4) = 5.6 m 3 2 x3 = 4.8 + (2.4) = 6.4 m 3

(b)

ΣM A = 0: xV − Σ xW + P(2.4 m) = 0 x(830.7 kN) − (3 m)(542.5 kN) − (5.6 m)(203.4 kN) − (6.4 m)(84.8 kN) + (2.4 m)(254.3 kN) = 0 x(830.7) − 1627.5 − 1139.0 − 542.7 + 610.3 = 0 x(830.7) − 2698.9 = 0

(c)

Resultant on face BC: Direct computation: P = ρ gh = (103 kg/m3 )(9.81 m/s 2 )(7.2 m) P = 70.63 kN/m 2 BC = (2.4) 2 + (7.2) 2 = 7.589 m θ = 18.43° 1 PA 2 1 = (70.63 kN/m 2 )(7.589 m)(1 m) 2

R=

R = 268 kN

18.43° 

Alternate computation: Use free body of water section BCD.

− R = 268 kN R = 268 kN

18.43° 18.43° 



x = 3.25 m (to right of A) 

www.elsolucionario.net PROBLEM 5.82 An automatic valve consists of a 9 × 9-in. square plate that is pivoted about a horizontal axis through A located at a distance h = 3.6 in. above the lower edge. Determine the depth of water d for which the valve will open.

SOLUTION Since valve is 9 in. wide, w = 9 p = 9γ h, where all dimensions are in inches. w1 = 9γ (d − 9), w2 = 9γ d 1 1 (9 in.) w1 = (9)(9γ )(d − 9) 2 2 1 1 PII = (9 in.) w2 = (9)(9γ d ) 2 2

Valve opens when B = 0. ΣM A = 0: PI (6 in. − 3.6 in.) − PII (3.6 in. − 3 in.) = 0

1  1   2 (9)(9γ )( d − 9)  (2.4) −  2 (9)(9γ d )  (0.6) = 0     (d − 9)(2.4) − d (0.6) = 0 1.8d − 21.6 = 0

d = 12.00 in. 



PI =

www.elsolucionario.net PROBLEM 5.83 An automatic valve consists of a 9 × 9-in. square plate that is pivoted about a horizontal axis through A. If the valve is to open when the depth of water is d = 18 in., determine the distance h from the bottom of the valve to the pivot A.

SOLUTION Since valve is 9 in. wide, w = 9 p = 9γ h, where all dimensions are in inches. w1 = 9γ ( d − 9) w2 = 9γ d


For d = 18 in., w1 = 9γ (18 − 9) = 81γ w2 = 9γ (18) = 162γ 1 1 (9)(9γ )(18 − 9) = (729γ ) 2 2 1 PII = (9)(9γ )(18) = 729γ 2 PI =

Valve opens when B = 0. ΣM A = 0: P1 (6 − h) − PII ( h − 3) = 0 1 729γ (6 − h) − 729( h − 3) = 0 2 1 3− h −h +3= 0 2 6 − 1.5h = 0

h = 4.00 in. 


www.elsolucionario.net PROBLEM 5.84 The 3 × 4-m side AB of a tank is hinged at its bottom A and is held in place by a thin rod BC. The maximum tensile force the rod can withstand without breaking is 200 kN, and the design specifications require the force in the rod not to exceed 20 percent of this value. If the tank is slowly filled with water, determine the maximum allowable depth of water d in the tank.

SOLUTION Consider the free-body diagram of the side.

Now where

P=

1 1 Ap = A( ρ gd ) 2 2

ΣM A = 0: hT −

d P=0 3

h=3m

Then for d max , (3 m)(0.2 × 200 × 103 N) −

d max 3

1  3 3 2  2 (4 m × d max ) × (10 kg/m × 9.81 m/s × d max )  = 0   3 120 N ⋅ m − 6.54d max N/m 2 = 0

or

d max = 2.64 m 

or



We have

www.elsolucionario.net PROBLEM 5.85 The 3 × 4-m side of an open tank is hinged at its bottom A and is held in place by a thin rod BC. The tank is to be filled with glycerine, whose density is 1263 kg/m3. Determine the force T in the rod and the reactions at the hinge after the tank is filled to a depth of 2.9 m.

SOLUTION Consider the free-body diagram of the side.

Then

1 1 Ap = A( ρ gd ) 2 2 1 = [(2.9 m)(4 m)] [(1263 kg/m3 )(9.81 m/s 2 )(2.9 m)] 2 = 208.40 kN

P=

ΣFy = 0: Ay = 0

 2.9  m  (208.4 kN) = 0 ΣM A = 0: (3 m)T −   3  T = 67.151 kN

or ΣFx = 0:

T = 67.2 kN



Ax + 208.40 kN − 67.151 kN = 0 Ax = −141.249 kN

or

A = 141.2 kN





We have

www.elsolucionario.net PROBLEM 5.86 The friction force between a 6 × 6-ft square sluice gate AB and its guides is equal to 10 percent of the resultant of the pressure forces exerted by the water on the face of the gate. Determine the initial force needed to lift the gate if it weighs 1000 lb.

SOLUTION Consider the free-body diagram of the gate. 1 1 ApI = [(6 × 6) ft 2 ][(62.4 lb/ft 3 )(9 ft)] 2 2 = 10,108.8 lb

PI =

1 1 ApII = [(6 × 6) ft 2 ][(62.4 lb/ft 3 )(15 ft)] 2 2 = 16,848 lb

PII =

Then

F = 0.1P = 0.1( PI + PII ) = 0.1(10,108.8 + 16,848) lb = 2695.7 lb

Finally

ΣFy = 0: T − 2695.7 lb − 1000 lb = 0

or T = 3.70 kips 



Now

www.elsolucionario.net PROBLEM 5.87 A tank is divided into two sections by a 1 × 1-m square gate that is hinged at A. A couple of magnitude 490 N · m is required for the gate to rotate. If one side of the tank is filled with water at the rate of 0.1 m3/min and the other side is filled simultaneously with methyl alcohol (density ρma = 789 kg/m3) at the rate of 0.2 m3/min, determine at what time and in which direction the gate will rotate.

SOLUTION Consider the free-body diagram of the gate.

Then

dW = d MA =

Now

P=

0.1 m3 / min × t (min) = 0.25t (m) (0.4 m)(1 m) 0.2 m3 / min × t (min) = t (m) (0.2 m)(1 m) 1 1 Ap = A( ρ gh) so that 2 2

1 PW = [(0.25t ) m × (1 m)][(103 kg/m3 )(9.81 m/s 2 )(0.25t ) m] 2 = 306.56t 2 N 1 PMA = [(t ) m × (1 m)][(789 kg/m3 )(9.81 m/s 2 )(t ) m] 2 = 3870t 2 N

Now assume that the gate will rotate clockwise and when d MA ≤ 0.6 m. When rotation of the gate is impending, we require 1 1    ΣM A : M R =  0.6 m − d MA  PMA −  0.6 m − dW 3 3   

  PW 

Substituting 1  1    490 N ⋅ m =  0.6 − t  m × (3870t 2 ) N −  0.6 − × 0.25t  m × (306.56t 2 ) N 3 3    



First note V = A base d and V = rt.


Simplifying

1264.45t 3 − 2138.1t 2 + 490 = 0

Solving (positive roots only) t = 0.59451 min and t = 1.52411 min

Now check assumption using the smaller root. We have d MA = (t ) m = 0.59451 m < 0.6 m

∴ t = 0.59451min = 35.7 s 


and the gate rotates clockwise. 


www.elsolucionario.net PROBLEM 5.88 A prismatically shaped gate placed at the end of a freshwater channel is supported by a pin and bracket at A and rests on a frictionless support at B. The pin is located at a distance h = 0.10 m below the center of gravity C of the gate. Determine the depth of water d for which the gate will open.

SOLUTION

a=

d − (0.25 − h) 3

and

b=

2 8 d  (0.4) −   3 15  3 

Now

a 8 = b 15

so that

d − (0.25 − h) 3 2 (0.4) − 158 d3 3

( )

=

8 15

Simplifying yields 289 70.6 d + 15h = 45 12

(1)

Alternative solution: Consider a free body consisting of a 1-m thick section of the gate and the triangular section BDE of water above the gate. Now

1 1 Ap′ = (d × 1 m)( ρ gd ) 2 2 1 = ρ gd 2 (N) 2 1 8  W ′ = ρ gV = ρ g  × d × d × 1 m  2 15   4 = ρ gd 2 (N) 15 P′ =



First note that when the gate is about to open (clockwise rotation is impending), By 0 and the line of action of the resultant P of the pressure forces passes through the pin at A. In addition, if it is assumed that the gate is homogeneous, then its center of gravity C coincides with the centroid of the triangular area. Then


Then with By = 0 (as explained above), we have 2 1  8   4  d  1  ΣM A = 0:  (0.4) −  d    ρ gd 2  −  − (0.25 − h)   ρ gd 2  = 0 3 3 15 15 3 2        

Simplifying yields

289 70.6 d + 15h = 45 12

as above. Find d:

289 70.6 d + 15(0.10) = 45 12

or d = 0.683 m 


Substituting into Eq. (1),

h = 0.10 m


www.elsolucionario.net PROBLEM 5.89 A prismatically shaped gate placed at the end of a freshwater channel is supported by a pin and bracket at A and rests on a frictionless support at B. Determine the distance h if the gate is to open when d = 0.75 m.

SOLUTION

a=

d − (0.25 − h) 3

and

b=

2 8 d (0.4) −   3 15  3 

Now

a 8 = b 15

so that

d − (0.25 − h) 3 2 (0.4) − 158 d3 3

( )

=

8 15

Simplifying yields 289 70.6 d + 15h = 45 12

(1)

Alternative solution: Consider a free body consisting of a 1-m thick section of the gate and the triangular section BDE of water above the gate. Now

1 1 Ap′ = (d × 1 m)( ρ gd ) 2 2 1 2 = ρ gd (N) 2 1 8  W ′ = ρ gV = ρ g  × d × d × 1 m   2 15  4 = ρ gd 2 (N) 15 P′ =



First note that when the gate is about to open (clockwise rotation is impending), By 0 and the line of action of the resultant P of the pressure forces passes through the pin at A. In addition, if it is assumed that the gate is homogeneous, then its center of gravity C coincides with the centroid of the triangular area. Then


Then with By = 0 (as explained above), we have 2 1  8   4  d  1  ΣM A = 0:  (0.4) −  d    ρ gd 2  −  − (0.25 − h)   ρ gd 2  = 0 3  15    15  3  2  3

Simplifying yields

289 70.6 d + 15h = 45 12

as above. Find h:

289 70.6 (0.75) + 15h = 45 12

or h = 0.0711 m 



d = 0.75 m


www.elsolucionario.net PROBLEM 5.90 The square gate AB is held in the position shown by hinges along its top edge A and by a shear pin at B. For a depth of water d = 3.5 ft, determine the force exerted on the gate by the shear pin.

SOLUTION First consider the force of the water on the gate. We have 1 Ap 2 1 = A(γ h) 2

Then

1 (1.8 ft) 2 (62.4 lb/ft 3 )(1.7 ft) 2 = 171.850 lb

PI =

1 (1.8 ft)2 (62.4 lb/ft 3 ) × (1.7 + 1.8cos 30°) ft 2 = 329.43 lb

PII =

Now or

1  2  ΣM A = 0:  LAB  PI +  LAB  PII − LAB FB = 0 3  3  1 2 (171.850 lb) + (329.43 lb) − FB = 0 3 3 FB = 276.90 lb

or

FB = 277 lb

30.0° 



P=

www.elsolucionario.net PROBLEM 5.91 A long trough is supported by a continuous hinge along its lower edge and by a series of horizontal cables attached to its upper edge. Determine the tension in each of the cables, at a time when the trough is completely full of water.

SOLUTION


Consider free body consisting of 20-in. length of the trough and water. l = 20-in. length of free body π  W = γ v = γ  r 2l  4   PA = γ r P=

1 1 1 PA rl = (γ r )rl = γ r 2l 2 2 2

1  ΣM A = 0: Tr − Wr − P  r  = 0 3   π  4r Tr −  γ r 2 l   4  3π

  1 2  1   −  2 γ r l  3 r  = 0    

1 1 1 T = γ r 2l + γ r 2l = γ r 2l 3 6 2

Data:

γ = 62.4 lb/ft 3 r =

Then

T=

24 20 ft = 2 ft l = ft 12 12

1  20  (62.4 lb/ft 3 )(2 ft)2  ft  2  12 

= 208.00 lb

T = 208 lb 


www.elsolucionario.net PROBLEM 5.92 A 0.5 × 0.8-m gate AB is located at the bottom of a tank filled with water. The gate is hinged along its top edge A and rests on a frictionless stop at B. Determine the reactions at A and B when cable BCD is slack.

SOLUTION First consider the force of the water on the gate. P=

1 1 Ap = A( ρ gh) 2 2


We have

1 PI = [(0.5 m)(0.8 m)] × [(103 kg/m3 )(9.81 m/s 2 )(0.45 m)] 2 = 882.9 N

so that

1 PII = [(0.5 m)(0.8 m)] × [(103 kg/m3 )(9.81 m/s 2 )(0.93 m)] 2 = 1824.66 N

Reactions at A and B when T = 0: We have

ΣM A = 0:

1 2 (0.8 m)(882.9 N) + (0.8 m)(1824.66 N) − (0.8 m)B = 0 3 3 B = 1510.74 N

or or

B = 1511 N

53.1° 

A = 1197 N

53.1° 

ΣF = 0: A + 1510.74 N − 882.9 N − 1824.66 N = 0

or


www.elsolucionario.net PROBLEM 5.93 A 0.5 × 0.8-m gate AB is located at the bottom of a tank filled with water. The gate is hinged along its top edge A and rests on a frictionless stop at B. Determine the minimum tension required in cable BCD to open the gate.

SOLUTION First consider the force of the water on the gate.

so that

P=

1 1 Ap = A( ρ gh) 2 2

1 PI = [(0.5 m)(0.8 m)] × [(103 kg/m3 )(9.81 m/s 2 )(0.45 m)] 2 = 882.9 N


We have

1 PII = [(0.5 m)(0.8 m)] × [(103 kg/m3 )(9.81 m/s 2 )(0.93 m)] 2 = 1824.66 N

T to open gate: First note that when the gate begins to open, the reaction at B Then

or

ΣM A = 0:

0.

1 2 (0.8 m)(882.9 N) + (0.8 m)(1824.66 N) 3 3  8  − (0.45 + 0.27)m ×  T  = 0  17 

235.44 + 973.152 − 0.33882 T = 0

or T = 3570 N 


www.elsolucionario.net PROBLEM 5.94 A 4 × 2-ft gate is hinged at A and is held in position by rod CD. End D rests against a spring whose constant is 828 lb/ft. The spring is undeformed when the gate is vertical. Assuming that the force exerted by rod CD on the gate remains horizontal, determine the minimum depth of water d for which the bottom B of the gate will move to the end of the cylindrical portion of the floor.

SOLUTION First determine the forces exerted on the gate by the spring and the water when B is at the end of the cylindrical portion of the floor. 2 θ = 30° 4

Then

xSP = (3 ft) tan 30°

and

FSP = kxSP


sin θ =

We have

= 828 lb/ft × 3 ft × tan30° = 1434.14 lb

Assume

d ≥ 4 ft

We have

P=

1 1 Ap = A(γ h) 2 2

1 PI = [(4 ft)(2 ft)] × [(62.4 lb/ft 3 )( d − 4) ft] 2 = 249.6(d − 4) lb

Then

1 PII = [(4 ft)(2 ft)] × [(62.4 lb/ft 3 )(d − 4 + 4cos 30°)] 2 = 249.6(d − 0.53590°) lb

For d min so that the gate opens, W = 0 Using the above free-body diagrams of the gate, we have 4  8  ΣM A = 0:  ft  [249.6(d − 4) lb] +  ft  [249.6( d − 0.53590) lb] 3  3  −(3 ft)(1434.14 lb) = 0

or

(332.8d − 1331.2) + (665.6d − 356.70) − 4302.4 = 0 d = 6.00 ft

or d ≥ 4 ft  assumption correct

d = 6.00 ft 


www.elsolucionario.net PROBLEM 5.95 Solve Problem 5.94 if the gate weighs 1000 lb. PROBLEM 5.94 A 4 × 2-ft gate is hinged at A and is held in position by rod CD. End D rests against a spring whose constant is 828 lb/ft. The spring is undeformed when the gate is vertical. Assuming that the force exerted by rod CD on the gate remains horizontal, determine the minimum depth of water d for which the bottom B of the gate will move to the end of the cylindrical portion of the floor.

SOLUTION First determine the forces exerted on the gate by the spring and the water when B is at the end of the cylindrical portion of the floor. sin θ =

2 θ = 30° 4

Then

xSP = (3 ft) tan 30°

and

FSP = kxSP = 828 lb/ft × 3 ft × tan30°


We have

= 1434.14 lb

Assume

d ≥ 4 ft

We have

P=

1 1 Ap = A(γ h) 2 2

1 PI = [(4 ft)(2 ft)] × [(62.4 lb/ft 3 )( d − 4) ft] 2 = 249.6(d − 4) lb 1 PII = [(4 ft)(2 ft)] × [(62.4 lb/ft 3 )(d − 4 + 4cos 30°)] 2 = 249.6(d − 0.53590°) lb

Then

For d min so that the gate opens, W = 1000 lb Using the above free-body diagrams of the gate, we have 4  8  ΣM A = 0:  ft  [249.6( d − 4) lb] +  ft  [249.6( d − 0.53590) lb] 3  3  − (3 ft)(1434.14 lb) − (1 ft)(1000 lb) = 0

or

(332.8d − 1331.2) + (665.6d − 356.70) − 4302.4 − 1000 = 0 d = 7.00 ft

or d ≥ 4 ft  assumption correct

d = 7.00 ft 


www.elsolucionario.net PROBLEM 5.96 A hemisphere and a cone are attached as shown. Determine the location of the centroid of the composite body when (a) h = 1.5a, (b) h = 2a.

V

y

yV

Cone I

1 2 πa h 3

h 4

1 π a 2 h2 12

Hemisphere II

2 3 πa 3

−

3a 8

1 − π a4 4

1 V = π a 2 ( h + 2a ) 3 1 Σ yV = π a 2 (h 2 − 3a 2 ) 12

(a)

For h = 1.5a,

1 7 V = π a 2 (1.5a + 2a ) = π a 2 3 6 Σ yV =

1 1 π a 2 [(1.5a) 2 − 3a 2 ] = − π a 4 12 16

1 3 7  Y V = Σ yV : Y  π a3  = − π a 4 Y = − a 16 56 6 

Centroid is 0.0536a below base of cone.  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 666


SOLUTION


(b)

For h = 2a,

1 4 V = π a 2 (2a + 2a) = π a 3 3 3 Σ yV =

1 1 π a 2 [(2a) 2 − 3a 2 ] = π a 4 12 12

1 4  1 Y V = Σ yV : Y  π a 3  = π a 4 Y = a 16 3  12


Centroid is 0.0625a above base of cone. 


www.elsolucionario.net PROBLEM 5.97 Consider the composite body shown. Determine (a) the value of x when h = L/2, (b) the ratio h/L for which x = L.

V

x

xV

Rectangular prism

Lab

1 L 2

1 2 L ab 2

Pyramid

1 b a h 3  2 

1 h 4

1 1   abh  L + h  6 4  

1   ΣV = ab  L + h  6   1  1   Σ xV = ab 3L2 + h  L + h   6  4  

Then

X ΣV = Σ xV

Now so that

  1  1  1  X  ab  L + h   = ab  3L2 + hL + h 2  6 6 4     

h 1 h2   1 h 1  X 1 + L 3 = + +    L 4 L2   6 L  6 

or

(a)

L+

X = ? when h =

Substituting

(1)

1 L. 2

h 1 = into Eq. (1), L 2 2  1  1  1   1  1  1   X 1 +    = L 3 +   +     6  2   6   2  4  2  

or

X=

57 L 104

X = 0.548L 



SOLUTION


(b)

h = ? when X = L. L


or

1+

1 h 1 1 h 1 h2 = + + 6 L 2 6 L 24 L2 h2 = 12 L2

h =2 3  L


or

h 1 h2   1 h 1  L 1 + = L 3 + +    L 4 L2   6 L  6 


www.elsolucionario.net PROBLEM 5.98 Determine the y coordinate of the centroid of the body shown.

SOLUTION

V

y

Cone

1 2 πa h 3

1 − h 4

2

Cylinder

1 a −π   b = − π a 2 b 4 2

1 − b 2

Σ

We have Then

π 12

a 2 (4 h − 3b)


First note that the values of Y will be the same for the given body and the body shown below. Then

yV −

1 π a2 h2 12

1 2 2 πa b 8 −

π 24

a 2 (2 h 2 − 3b 2 )

Y ΣV = Σ yV

π π  Y  a 2 (4h − 3b)  = − a 2 (2h 2 − 3b 2 ) 24  12 

or Y = −

2h 2 − 3b 2  2(4h − 3b)


www.elsolucionario.net PROBLEM 5.99 Determine the z coordinate of the centroid of the body shown. (Hint: Use the result of Sample Problem 5.13.)

SOLUTION

V

1 2 πa h 6

Half cone

Half cylinder

Σ

−

π a

2

24

π

1 − a3h 6

4 a 2a =−   3π  2  3π

1 3 ab 12

−

π

b = − a 2b 2  2  8

π

zV

z

−

a

−

a 2 (4h − 3b)

1 3 a (2h − b) 12

From Sample Problem 5.13: We have Then

Z ΣV = Σ zV 1 π  Z  a 2 (4h − 3b)  = − a3 (2h − b) 12  24 

or Z = −

a  4h − 2b   π  4h − 3b 



First note that the body can be formed by removing a half cylinder from a half cone, as shown.

www.elsolucionario.net PROBLEM 5.100 For the machine element shown, locate the y coordinate of the center of gravity.

SOLUTION For half-cylindrical hole,

4(1.25) 3π = 1.470 in.

yIII = 2 −

For half-cylindrical plate,

r = 2 in. 4(2) zIV = 7 + = 7.85 in. 3π

V, in 3

y , in.

z , in.

y V , in 4

z V , in 4

3.5

–7.875

73.50

I

Rectangular plate

(7)(4)(0.75) = 21.0

–0.375

II

Rectangular plate

(4)(2)(1) = 8.0

1.0

2

8.000

16.00

III

–(Half cylinder)

(1.25) 2 (1) = 2.454

1.470

2

–3.607

–4.908

IV

Half cylinder

(2) 2 (0.75) = 4.712

–0.375

–1.767

36.99

V

–(Cylinder)

−π (1.25) 2 (0.75) = −3.682

1.381

–25.77

Σ

27.58

–3.868

95.81

−

π 2

π 2

–0.375

–7.85 7

Y ΣV = Σ yV Y (27.58 in 3 ) = −3.868 in 4

Y = −0.1403 in. 



r = 1.25 in.

www.elsolucionario.net PROBLEM 5.101 For the machine element shown, locate the y coordinate of the center of gravity.

SOLUTION

V , mm3

x , mm

y , mm

x V, mm 4

y V, mm 4

I

(100)(18)(90) = 162, 000

50

9

8,100,000

1,458,000

II

(16)(60)(50) = 48, 000

92

48

4,416,000

2,304,000

III

π (12)2 (10) = 4523.9

105

54

475,010

244,290

IV

−π (13) 2 (18) = −9556.7

28

9

–267,590

–86,010

Σ

204,967.2

12,723,420

3,920,280

We have

Y ΣV = Σ yV Y (204,967.2 mm3 ) = 3,920, 280 mm 4

or Y = 19.13 mm 



First assume that the machine element is homogeneous so that its center of gravity will coincide with the centroid of the corresponding volume.

www.elsolucionario.net PROBLEM 5.102 For the machine element shown, locate the x coordinate of the center of gravity.

SOLUTION

V , mm3

x , mm

y , mm

xV , mm 4

yV, mm 4

I

(100)(18)(90) = 162, 000

50

9

8,100,000

1,458,000

II

(16)(60)(50) = 48, 000

92

48

4,416,000

2,304,000

III

π (12)2 (10) = 4523.9

105

54

475,010

244,290

IV

−π (13) 2 (18) = −9556.7

28

9

–267,590

–86,010

Σ

204,967.2

12,723,420

3,920,280

We have

X ΣV = Σ xV X (204,967.2 mm3 ) = 12, 723, 420 mm 4

X = 62.1 mm 



First assume that the machine element is homogeneous so that its center of gravity will coincide with the centroid of the corresponding volume.

www.elsolucionario.net PROBLEM 5.103 For the machine element shown, locate the z coordinate of the center of gravity.

SOLUTION For half-cylindrical hole,

4(1.25) 3π = 1.470 in.

yIII = 2 −

For half-cylindrical plate,

r = 2 in. 4(2) zIV = 7 + = 7.85 in. 3π

V, in 3

Now

y , in.

z , in.

yV , in 4

z V , in 4

3.5

–7.875

73.50

I

Rectangular plate

(7)(4)(0.75) = 21.0

–0.375

II

Rectangular plate

(4)(2)(1) = 8.0

1.0

2

8.000

16.00

III

–(Half cylinder)

(1.25) 2 (1) = 2.454

1.470

2

–3.607

–4.908

IV

Half cylinder

(2) 2 (0.75) = 4.712

–0.375

–1.767

36.99

V

–(Cylinder)

−π (1.25) 2 (0.75) = −3.682

1.381

–25.77

Σ

27.58

–3.868

95.81

−

π 2

π 2

–0.375

–7.85 7

Z ΣV = zV Z (27.58 in 3 ) = 95.81 in 4

Z = 3.47 in. 



r = 1.25 in.

www.elsolucionario.net PROBLEM 5.104 For the machine element shown, locate the x coordinate of the center of gravity.

SOLUTION

V, mm3

x , mm z , mm

x V , mm 4

z V , mm 4

19

45

649.8 × 103

1539 × 103

46.5

20

292.17 × 103

125.664 × 103

−π (12)2 (10) = −4.5239 × 103

38

20

Rectangular prism

(30)(10)(24) = 7.2 × 103

5

78

36 × 103

561.6 × 103

Triangular prism

1 (30)(9)(24) = 3.24 × 103 2

13

78

42.12 × 103

252.72 × 103

Σ

46.399 × 103

848.18 × 103

2388.5 × 103

I

Rectangular plate

II

Half cylinder

III

–(Cylinder)

IV V

(10)(90)(38) = 34.2 × 103

π 2

(20)2 (10) = 6.2832 × 103

−171.908 × 103 −90.478 × 103

X ΣV = Σ xV X=

Σ xV 848.18 × 103 mm4 = ΣV 46.399 × 103 mm3

X = 18.28 mm 



Dimensions in mm

www.elsolucionario.net PROBLEM 5.105 For the machine element shown, locate the z coordinate of the center of gravity.

SOLUTION

V, mm3

x V , mm 4

z V , mm 4

19

45

649.8 × 103

1539 × 103

46.5

20

292.17 × 103

125.664 × 103

−π (12)2 (10) = −4.5239 × 103

38

20

(30)(10)(24) = 7.2 × 103

5

78

36 × 103

561.6 × 103

Triangular prism

1 (30)(9)(24) = 3.24 × 103 2

13

78

42.12 × 103

252.72 × 103

Σ

46.399 × 103

848.18 × 103

2388.5 × 103

I

Rectangular plate

II

Half cylinder

III

–(Cylinder)

IV Rectangular prism V

x , mm z , mm

(10)(90)(38) = 34.2 × 103

π 2

(20)2 (10) = 6.2832 × 103

−171.908 × 103 −90.478 × 103

Z ΣV = Σ zV Z=

Σ zV 2388.5 × 103 mm4 = ΣV 46.399 × 103 mm3

Z = 51.5 mm 



Dimensions in mm

www.elsolucionario.net PROBLEM 5.106 Locate the center of gravity of the sheet-metal form shown.

SOLUTION

4(80) = 33.953 mm 3π 2(60) zII = − = −38.197 mm zI =

π

A, mm 2

I

π 2

(80) 2 = 10, 053

II

π (60)(160) = 30,159

Σ

40,212

x , mm

z , mm

xA, mm3

zA, mm3

0

33.953

0

341.33 × 103

60

−38.197

1809.54 × 103 −1151.98 × 103 1809.54 × 103

−810.65 × 103

X Σ A = Σ xA: X (40, 212) = 1809.54 × 103

X = 45.0 mm 

Z Σ A = Σ zA: Z (40, 212) = −810.65 × 103

Z = −20.2 mm 



Y = 80.0 mm 

By symmetry,


SOLUTION

1 yI = 0.18 + (0.12) = 0.22 m 3 1 zI = (0.2 m) 3 2 × 0.18 0.36 = m xII = yII =

π

xIV

I II

IV Σ

A, m 2

x, m

y, m

z, m

x A, m3

yA, m3

1 (0.2)(0.12) = 0.012 2

0

0.22

0.2 3

0

0.00264

0.0008

0.36

0.36

π

π

0.1

0.00648

0.00648

0.005655

0.26

0

0.1

0.00832

0

0.0032

0.31878

0

0.1

–0.001258

0

–0.000393

0.013542

0.00912

0.009262

π

(0.18)(0.2) = 0.018π

2

(0.16)(0.2) = 0.032

III −

π

4 × 0.05 = 0.34 − 3π = 0.31878 m

π 2

(0.05) 2 = −0.00125π

0.096622

z A, m3



First assume that the sheet metal is homogeneous so that the center of gravity of the form will coincide with the centroid of the corresponding area.


X ΣV = Σ xV : X (0.096622 m 2 ) = 0.013542 m3

or X = 0.1402 m 

Y ΣV = Σ yV : Y (0.096622 m 2 ) = 0.00912 m3

or Y = 0.0944 m 

Z ΣV = Σ zV : Z (0.096622 m 2 ) = 0.009262 m3

or Z = 0.0959 m 


We have


www.elsolucionario.net PROBLEM 5.108 A window awning is fabricated from sheet metal of uniform thickness. Locate the center of gravity of the awning.

First, assume that the sheet metal is homogeneous so that the center of gravity of the awning coincides with the centroid of the corresponding area. (4)(25) = 14.6103 in. 3π (4)(25) 100 in. zII = zVI = = 3π 3π (2)(25) yIV = 4 + = 19.9155 in. yII = yVI = 4 +

π

zIV =

(2)(25)

π

AII = AVI = AIV = A, in

2

I

(4)(25) = 100

II

490.87

III

(4)(34) = 136

IV

1335.18

V

(4)(25) = 100

VI Σ

490.87

y , in.

2

π

2 19.9155

in.

(25) 2 = 490.87 in 2

(25)(34) = 1335.18 in 2 zA, in 3

12.5

200

1250

100 3π

14.6103

π

yA, in 3

π 12.5

2

4

50

z , in.

100 3π 25 50

14.6103

2

π

=

7171.8 272

3400

26,591

21,250

200

1250

7171.8 41,607

2652.9

5208.3

5208.3 37,567



SOLUTION


X = 17.00 in. 

Now, symmetry implies Y Σ A = Σ yA: Y (2652.9 in 2 ) = 41,607 in 3

or Y = 15.68 in. 

Z Σ A = Σ zA: Z (2652.9 in 2 ) = 37,567 in 3

or Z = 14.16 in. 


and


www.elsolucionario.net PROBLEM 5.109 A thin sheet of plastic of uniform thickness is bent to form a desk organizer. Locate the center of gravity of the organizer.

First assume that the plastic is homogeneous so that the center of gravity of the organizer will coincide with the centroid of the corresponding area. Now note that symmetry implies Z = 30.0 mm 

x2 = 6 −

2× 6

x4 = 36 + x8 = 58 −

π

= 2.1803 mm

2×6

π

2× 6

π

x10 = 133 +

= 39.820 mm = 54.180 mm

2×6

π

= 136.820 mm

y2 = y4 = y8 = y10 = 6 − y6 = 75 +

2×5

π

2×6

π

= 2.1803 mm

= 78.183 mm

A2 = A4 = A8 = A10 =

π

× 6 × 60 = 565.49 mm 2 2 A6 = π × 5 × 60 = 942.48 mm 2



SOLUTION


A, mm 2

We have

x , mm

1

(74)(60) = 4440

2

565.49

3

(30)(60) = 1800

4

565.49

5

(69)(60) = 4140

42

6

942.48

7

(69)(60) = 4140

8

565.49

9

y , mm

0

xA, mm3

yA, mm3

0

190,920

1233

1233

43

2.1803

2.1803

21

0

37,800

0

39.820

2.1803

22,518

1233

40.5

173,880

167,670

47

78.183

44,297

73,686

52

40.5

215,280

167,670

54.180

2.1803

30,638

1233

(75)(60) = 4500

95.5

0

429,750

0

10

565.49

136.820

2.1803

77,370

1233

Σ

22,224.44

1,032,766

604,878

X Σ A = Σ xA: X (22, 224.44 mm 2 ) = 1,032, 766 mm3

or X = 46.5 mm 

Y Σ A = Σ yA: Y (22, 224.44 mm 2 ) = 604,878 mm3

or Y = 27.2 mm 



PROBLEM 5.109 (Continued)

www.elsolucionario.net PROBLEM 5.110 A wastebasket, designed to fit in the corner of a room, is 16 in. high and has a base in the shape of a quarter circle of radius 10 in. Locate the center of gravity of the wastebasket, knowing that it is made of sheet metal of uniform thickness.

By symmetry,

X =Z

For III (Cylindrical surface),

x= A=

For IV (Quarter-circle bottom),

x= A=

2r

π π 2

=

2(10)

rh =

π π 2

= 6.3662 in.

(10)(16) = 251.33 in 2

4r 4(10) = = 4.2441 in. 3π 3π

π

A, in 2

4

r2 =

π

4

(10) 2 = 78.540 in 2

x , in.

x , in.

xA, in 3

yA, in 3

I

(10)(16) = 160

5

8

800

1280

II

(10)(16) = 160

0

8

0

1280

III

251.33

6.3662

8

4.2441

0

IV Σ X Σ A = Σ xA:

78.540

649.87

1600.0 333.33 2733.3

2010.6 0 4570.6

X (649.87 in 2 ) = 2733.3 in 3 X = 4.2059 in.

Y Σ A = Σ yA:


SOLUTION

X = Z = 4.21 in. 

Y (649.87 in 2 ) = 4570.6 in 3 Y = 7.0331 in.

Y = 7.03 in. 


www.elsolucionario.net PROBLEM 5.111 A mounting bracket for electronic components is formed from sheet metal of uniform thickness. Locate the center of gravity of the bracket.

SOLUTION

4(0.625) 3π = 1.98474 in.

zV = 2.25 − AV = −

π

(0.625) 2 2 = −0.61359 in 2

A, in 2

x , in.

I

(2.5)(6) = 15

1.25

II

(1.25)(6) = 7.5

2.5

III

(0.75)(6) = 4.5

2.875

IV

5 −   (3) = − 3.75 4

1.0

0

3.75

V

− 0.61359

1.0

0

1.98474 0.61359

Σ

22.6364

We have

y , in.

z , in.

0

xA, in 3

yA, in 3

zA, in 3

0

45

3

18.75

–0.625

3

18.75

–4.6875

22.5

–1.25

3

12.9375

–5.625

13.5

3.75

46.0739

0

–14.0625

0

–1.21782

10.3125

65.7197

X ΣA = Σ xA X (22.6364 in 2 ) = 46.0739 in 3

or

X = 2.04 in. 

Y ΣA = Σ yA Y (22.6364 in 2 ) = −10.3125 in 3

or Y = − 0.456 in. 

Z ΣA = Σ zA Z (22.6364 in 2 ) = 65.7197 in 3

or

Z = 2.90 in. 



First, assume that the sheet metal is homogeneous so that the center of gravity of the bracket coincides with the centroid of the corresponding area. Then (see diagram)

www.elsolucionario.net PROBLEM 5.112 An 8-in.-diameter cylindrical duct and a 4 × 8-in. rectangular duct are to be joined as indicated. Knowing that the ducts were fabricated from the same sheet metal, which is of uniform thickness, locate the center of gravity of the assembly.

SOLUTION Assume that the body is homogeneous so that its center of gravity coincides with the centroid of the area.

1 2

A, in 2

x , in.

y , in.

xA, in 3

yA, in 3

π (8)(12) = 96π

0

6

0

576π

10

−128

−160π

4(4) 16 =− 3π 3π

12

− 42.667

96π

6

12

576

1152

6

8

576

768

4(4) 16 = 3π 3π

8

− 42.667

− 64π

6

10

288

480

6

10

288

480

1514.6

4287.4

−

π 2

π

3

Then

(4) 2 = 8π

4

2 (8)(12) = 96

5

(8)(12) = 96

6

−

2(4)

(8)(4) = −16π

π

(4) 2 = − 8π

7

2 (4)(12) = 48

8

(4)(12) = 48

Σ

539.33

π −

=

8

π

X =

Σ xA 1514.67 = in. 539.33 ΣA

or X = 2.81 in. 

Y =

Σ yA 4287.4 = in. 539.33 ΣA

or Y = 7.95 in. 



By symmetry, z = 0.

www.elsolucionario.net PROBLEM 5.113 An elbow for the duct of a ventilating system is made of sheet metal of uniform thickness. Locate the center of gravity of the elbow.

SOLUTION First, assume that the sheet metal is homogeneous so that the center of gravity of the duct coincides with the centroid of the corresponding area. Also, note that the shape of the duct implies Y = 38.0 mm  xI = zI = 400 − xII = 400 − zII = 300 −

2

π 2

π

2

π

(400) = 145.352 mm

(200) = 272.68 mm (200) = 172.676 mm

xIV = zIV = 400 −

4 (400) = 230.23 mm 3π

4 (200) = 315.12 mm 3π 4 zV = 300 − (200) = 215.12 mm 3π

xV = 400 −

Also note that the corresponding top and bottom areas will contribute equally when determining x and z . Thus,

A, mm 2

x , mm

z , mm

xA, mm3

zA, mm3

(400)(76) = 47,752

145.352

145.352

6,940,850

6,940,850

(200)(76) = 23,876

272.68

172.676

6,510,510

4,122,810

III

100(76) = 7600

200

350

1,520,000

2,660,000

IV

π  2   (400)2 = 251,327 4

230.23

230.23

57,863,020

57,863,020

V

π  −2   (200)2 = −62,832 4

315.12

215.12

–19,799,620

–13,516,420

VI

−2(100)(200) = −40,000

300

350

–12,000,000

–14,000,000

Σ

227,723

41,034,760

44,070,260

I II

π 2

π 2



Note that


X Σ A = Σ xA: X (227, 723 mm 2 ) = 41,034, 760 mm3

or X = 180.2 mm 

Z Σ A = Σ zA: Z (227, 723 mm 2 ) = 44,070, 260 mm3

or Z = 193.5 mm 


We have


www.elsolucionario.net PROBLEM 5.114 Locate the center of gravity of the figure shown, knowing that it is made of thin brass rods of uniform diameter.

SOLUTION

L, in.

y , in.

z , in.

yL, in 2

zL, in 2

AB

302 + 162 = 34

15

0

510

0

AD

302 + 162 = 34

15

8

510

272

AE

302 + 162 = 34

15

0

510

0

BDE

π (16) = 50.265

0

0

512

Σ

152.265

1530

784

2(16)

π

= 10.186

Y ΣL = Σ y L : Y (152.265 in.) = 1530 in 2 Y = 10.048 in.



Y = 10.05 in. 

Z ΣL = Σ z L : Z (152.265 in.) = 784 in 2  Z = 5.149 in. 



Z = 5.15 in. 



X =0 

By symmetry,

www.elsolucionario.net PROBLEM 5.115 Locate the center of gravity of the figure shown, knowing that it is made of thin brass rods of uniform diameter.

SOLUTION

AB 2 = (1 m) 2 + (0.6 m) 2 + (1.5 m) 2 AB = 1.9 m

L, m

x, m

y, m

z, m

xL, m 2

yL, m 2

ΣL , m

AB

1.9

0.5

0.75

0.3

0.95

1.425

0.57

BD

0.6

1.0

0

0.3

0.60

0

0.18

DO

1.0

0.5

0

0

0.50

0

0

OA

1.5

0

0.75

0

0

1.125

0

Σ

5.0

2.05

2.550

0.75

X Σ L = Σ x L : X (5.0 m) = 2.05 m 2

X = 0.410 m 

Y Σ L = Σ y L : Y (5.0 m) = 2.55 m 2

Y = 0.510 m 

Z Σ L = Σ z L : Z (5.0 m) = 0.75 m 2

Z = 0.1500 m 



Uniform rod:

www.elsolucionario.net PROBLEM 5.116 A thin steel wire of uniform cross section is bent into the shape shown. Locate its center of gravity.

SOLUTION

x2 = z2 =

1 2

We have

π 2

2 × 2.4

π

=

4.8

π

m

L, m

x, m

y, m

z, m

xL, m 2

yL, m 2

zL, m 2

2.6

1.2

0.5

0

3.12

1.3

0

π

5.76

0

5.76

× 2.4 = 1.2π

4.8

π

0

4.8

3

2.4

0

0

1.2

0

0

2.88

4

1.0

0

0.5

0

0

0.5

0

Σ

9.7699

8.88

1.8

8.64

X Σ L = Σ x L : X (9.7699 m) = 8.88 m 2

or X = 0.909 m 

Y Σ L = Σ y L : Y (9.7699 m) = 1.8 m 2

or Y = 0.1842 m 

Z Σ L = Σ z L : Z (9.7699 m) = 8.64 m 2

or Z = 0.884 m 



First assume that the wire is homogeneous so that its center of gravity will coincide with the centroid of the corresponding line.

www.elsolucionario.net PROBLEM 5.117 The frame of a greenhouse is constructed from uniform aluminum channels. Locate the center of gravity of the portion of the frame shown.

SOLUTION

2×3

x8 = x9 =

π

y8 = y9 = 5 + L, ft

6

π

2×3

π

y , ft

ft = 6.9099 ft

z , ft

xL, ft 2

yL, ft 2

zL, ft 2

1

2

3

0

1

6

0

2

2

3

1.5

0

2

4.5

0

6

3

5

3

2.5

0

15

12.5

0

4

5

3

2.5

2

15

12.5

10

5

8

0

4

2

0

32

16

6

2

3

5

1

6

10

2

7

3

1.5

5

2

4.5

15

6

6.9099

0

9

32.562

0

π

6.9099

2

9

32.562

9.4248

0

8

1

0

16

2

8 9 10 Σ

We have

x , ft

=

π 2

π 2

× 3 = 4.7124 × 3 = 4.7124

2

6

π 6

39.4248

69

163.124

53.4248

X Σ L = Σ x L : X (39.4248 ft) = 69 ft 2

or X = 1.750 ft 

Y Σ L = Σ y L : Y (39.4248 ft) = 163.124 ft 2

or

Z Σ L = Σ z L : Z (39.4248 ft) = 53.4248 ft 2

or Z = 1.355 ft 

Y = 4.14 ft 



First assume that the channels are homogeneous so that the center of gravity of the frame will coincide with the centroid of the corresponding line.

www.elsolucionario.net PROBLEM 5.118 Three brass plates are brazed to a steel pipe to form the flagpole base shown. Knowing that the pipe has a wall thickness of 8 mm and that each plate is 6 mm thick, determine the location of the center of gravity of the base. (Densities: brass = 8470 kg/m3; steel = 7860 kg/m3.)

SOLUTION Since brass plates are equally spaced, we note that the center of gravity lies on the y-axis.

V=

π

[(0.064 m) 2 − (0.048 m) 2 ](0.192 m) 4 = 270.22 × 10−6 m3

Steel pipe:

m = ρ V = (7860 kg/m3 )(270.22 × 10−6 m3 ) = 2.1239 kg

Each brass plate:

1 (0.096 m)(0.192 m)(0.006 m) = 55.296 × 10−6 m3 2 m = ρ V = (8470 kg/m3 )(55.296 × 10−6 m3 ) = 0.46836 kg

V=

Flagpole base: Σm = 2.1239 kg + 3(0.46836 kg) = 3.5290 kg Σ y m = (0.096 m)(2.1239 kg) + 3[(0.064 m)(0.46836 kg)] = 0.29382 kg ⋅ m Y Σm = Σ y m :

Y (3.5290 kg) = 0.29382 kg ⋅ m Y = 0.083259 m

Y = 83.3 mm above the base 



x =z =0 

Thus,

www.elsolucionario.net PROBLEM 5.119 A brass collar, of length 2.5 in., is mounted on an aluminum rod of length 4 in. Locate the center of gravity of the composite body. (Specific weights: brass = 0.306 lb/in3, aluminum = 0.101 lb/in3)

Aluminum rod:


SOLUTION

W = γV π  = (0.101 lb/in 3 )  (1.6 in.)2 (4 in.)  4  = 0.81229 lb

Brass collar:

W = γV

π

= (0.306 lb/in.3 ) [(3 in.) 2 − (1.6 in.) 2 ](2.5 in.) 4 = 3.8693 lb yW (lb ⋅ in.)

Component

W(lb)

Rod

0.81229

2

1.62458

Collar

3.8693

1.25

4.8366

Σ

4.6816

y (in.)

6.4612

Y ΣW = Σ y W : Y (4.6816 lb) = 6.4612 lb ⋅ in. Y = 1.38013 in.

Y = 1.380 in. 


www.elsolucionario.net PROBLEM 5.120 A bronze bushing is mounted inside a steel sleeve. Knowing that the specific weight of bronze is 0.318 lb/in3 and of steel is 0.284 lb/in3, determine the location of the center of gravity of the assembly.

SOLUTION

Now

W = ( ρ g )V  π   yI = 0.20 in. WI = (0.284 lb/in 3 )   [(1.82 − 0.752 ) in 2 ](0.4 in.)  = 0.23889 lb  4    π   yII = 0.90 in. WII = (0.284 lb/in 3 )   [(1.1252 − 0.752 ) in 2 ](1 in.)  = 0.156834 lb 4     π   yIII = 0.70 in. WIII = (0.318 lb/in 3 )   [(0.752 − 0.52 ) in 2 ](1.4 in.)  = 0.109269 lb  4  

We have

Y ΣW = Σ yW (0.20 in.)(0.23889 lb) + (0.90 in.)(0.156834 lb) + (0.70 in.)(0.109269 lb) Y = 0.23889 lb + 0.156834 lb + 0.109269 lb Y = 0.526 in. 

or

(above base)



X =Z =0 

First, note that symmetry implies

www.elsolucionario.net PROBLEM 5.121 A scratch awl has a plastic handle and a steel blade and shank. Knowing that the density of plastic is 1030 kg/m3 and of steel is 7860 kg/m3, locate the center of gravity of the awl.

SOLUTION

5 xI = (12.5 mm) = 7.8125 mm 8  2π  3 WI = (1030 kg/m3 )   (0.0125 m) 3   = 4.2133 × 10−3 kg xII = 52.5 mm π  WII = (1030 kg/m3 )   (0.025 m) 2 (0.08 m) 4 −3 = 40.448 × 10 kg xIII = 92.5 mm − 25 mm = 67.5 mm π  WIII = −(1030 kg/m3 )   (0.0035 m) 2 (0.05 m) 4 −3 = −0.49549 × 10 kg xIV = 182.5 mm − 70 mm = 112.5 mm π  WIV = (7860 kg/m3 )   (0.0035 m) 2 (0.14 m) 2 = 10.5871 × 10−3 kg 4 1 xV = 182.5 mm + (10 mm) = 185 mm 4 π  WV = (7860 kg/m3 )   (0.00175 m) 2 (0.01 m) = 0.25207 × 10−3 kg 3



Y = Z = 0 



W, kg

xW, kg ⋅ mm

I

4.123 × 10−3

II

40.948 × 10−3

52.5

2123.5 × 10−3

III

−0.49549 × 10−3

67.5

−33.447 × 10−3

IV

10.5871 × 10−3

112.5

1191.05 × 10−3

V

0.25207 × 10−3

185

46.633 × 10−3

Σ

55.005 × 10−3

7.8125

32.916 × 10−3

3360.7 × 10−3

X ΣW = Σ xW : X (55.005 × 10−3 kg) = 3360.7 × 10−3 kg ⋅ mm X = 61.1 mm 

or

(from the end of the handle)



We have

x, mm

www.elsolucionario.net PROBLEM 5.122 Determine by direct integration the values of x for the two volumes obtained by passing a vertical cutting plane through the given shape of Figure 5.21. The cutting plane is parallel to the base of the given shape and divides the shape into two volumes of equal height. A hemisphere

SOLUTION Choose as the element of volume a disk of radius r and thickness dx. Then dV = π r 2 dx, xEL = x

The equation of the generating curve is x 2 + y 2 = a 2 so that r 2 = a 2 − x 2 and then

V1 =

Component 1:

=

and



1

xEL dV =



a/2 0

a/2

 x3  π (a − x )dx = π  a 2 x −  3 0  2

2

11 3 πa 24



a/2 0

x π (a 2 − x 2 ) dx  a/2

 x2 x4  = π a2 −  2 4 0  7 = π a4 64

Now

x1V1 =

x 1

EL dV :

 11  7 x1  π a 3  = π a 4  24  64

or

x1 =

21 a  88

Component 2: a

 x3  π (a − x )dx = π  a 2 x −  V2 = a /2 3  a/2 



a

2

2

  a 3  a3   2 a  ( 2 )    2 = π   a (a ) −  − a   −  3   2 3      5 = π a3 24



dV = π ( a 2 − x 2 ) dx


and



2

xEL dV =



a

 x 2 x4  −  x π ( a 2 − x 2 )dx  = π  a 2 a/2 2 4  a/2  a

2 4    a a 2 ( (a ) 4   2 ( 2 )   2 (a) 2 )  = π a − −  − a 2 4   2 4       9 = π a4 64

x2V2 =



2

 5  9 xEL dV : x2  π a3  = π a 4  24  64

or

x2 =

27 a  40


Now


www.elsolucionario.net PROBLEM 5.123 Determine by direct integration the values of x for the two volumes obtained by passing a vertical cutting plane through the given shape of Figure 5.21. The cutting plane is parallel to the base of the given shape and divides the shape into two volumes of equal height. A semiellipsoid of revolution

SOLUTION Choose as the element of volume a disk of radius r and thickness dx. Then dV = π r 2 dx, xEL = x x2 y2 + = 1 so that h2 a 2

The equation of the generating curve is

a2 2 (h − x 2 )dx 2 h

dV = π

and then

V1 =

Component 1:

=

and

a2 2 (h − x 2 ) 2 h



1

xEL dV =



h/2 0

h/2

a2 a2  x3  π 2 (h 2 − x 2 )dx = π 2  h 2 x −  3 0 h h 

11 2 πa h 24



h/2

0

 a2  x π 2 (h 2 − x 2 ) dx   h 

a2 =π 2 h =

Now

Component 2:

x1V1 = V2 =


r2 =

h/2

 2 x2 x4  −  h 4 0  2

7 π a 2 h2 64

x 1



EL dV :

h h/2

 11  7 x1  π a 2 h  = π a 2 h 2 24   64

or

x1 =

21 h  88

h

π

a2 2 a2  x3  (h − x 2 )dx = π 2  h 2 x −  2 3  h/2 h h 

  h 3  a 2  2 ( h)3   2  h  ( 2 )   = π 2 h ( h ) −  − h  − 3   2 3  h     5 = π a2h 24 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 701


and



2

xEL dV =



 a2  x π 2 (h 2 − x 2 ) dx  h/2  h  h

a2 =π 2 h

h

 2 x2 x4  −  h 4  h/2  2

  h 2 h 4  ( a 2   2 ( h) 2 ( h) 4   2 ( 2 ) 2 )  = π 2 h − −   − h 2 4  2 4  h     9 = π a 2 h2 64 x2V2 =



 5  9 xEL dV : x2  π a 2 h  = π a 2 h 2 2  24  64

or

x2 =

27 h  40


Now


www.elsolucionario.net PROBLEM 5.124 Determine by direct integration the values of x for the two volumes obtained by passing a vertical cutting plane through the given shape of Figure 5.21. The cutting plane is parallel to the base of the given shape and divides the shape into two volumes of equal height. A paraboloid of revolution

SOLUTION Choose as the element of volume a disk of radius r and thickness dx. Then dV = π r 2 dx, xEL = x

h 2 a2 y so that r 2 = (h − x). 2 h a a2 (h − x)dx h

dV = π

and then

V1 =

Component 1:



h/2

0

π

a2 =π h


The equation of the generating curve is x = h −

a2 ( h − x) dx h h/2

 x2  hx −   2 0 

3 = π a2 h 8

and



1

xEL dV =



h/2

0

 a2  x π (h − x)dx   h  h/2

a 2  x 2 x3  1 2 2 =π h −  = π a h h  2 3 0 12

Now

x1V1 =

x 1

EL dV :

3  1 x1  π a 2 h  = π a 2 h 2 8  12

or

x1 =

2 h  9

Component 2: h

a2 a2  x2  V2 = hx π (h − x)dx = π −   h/2 h h  2  h/2



h

  h 2  ( h) 2    h  ( 2 )   a2  =π   h( h ) −   − h  − h  2   2 2     1 2 = πa h 8



and



2

xEL dV =



a2 =π h = x2V2 =

  h 2 ( h )3   2 ( h )3   ( 2 )   ( h)  − − 2  h − h   2 3 2 3      

1 π a 2 h2 12



2

1  1 xEL dV : x2  π a 2 h  = π a 2 h 2 8   12

or

x2 =

2 h  3


Now

h

 a2  a 2  x 2 x3  x π ( h − x)dx  = π h −  h/2 h  2 3  h/2  h  h


www.elsolucionario.net PROBLEM 5.125 Locate the centroid of the volume obtained by rotating the shaded area about the x-axis.

SOLUTION y =0 


z =0 

Choose as the element of volume a disk of radius r and thickness dx. Then dV = π r 2 dx, xEL = x


r = kx1/3

Now

dV = π k 2 x 2/3 dx

so that at x = h, y = a,

a = kh1/3

or

k=

a h1/3 a 2 2/3 x dx h 2/3

dV = π

Then

V=

and



h 0

π

a 2 2/3 x dx h 2/3

a2 h 2/3

=π

h

 3 5/3  5 x   0

3 = π a2 h 5

Also

x

EL dV

=



h 0

h  a 2 2/3  a 2  3 8/3  x  π 2/3 x dx  = π 2/3  x  h 8 0  h 

3 = π a 2 h2 8

Now



xV = xdV :

3  3 x  π a 2 h  = π a 2 h2 5  8

5 or x = h  8



SOLUTION y =0 


z =0 

Choose as the element of volume a disk of radius r and thickness dx. dV = π r 2 dx, xEL = x

Then

2

 1 dV = π 1 −  dx x   2 1  = π 1 − + 2  dx x x  

1 so that x

V=

Then

3



1

 

π 1 −


Now r = 1 −

3

2 1  1  +  dx = π  x − 2 ln x −  x x2  x 1 

 1  1  = π  3 − 2ln 3 −  − 1 − 2 ln1 −   3  1   = (0.46944π ) m3

and

x

EL dV

=



3

1

3

 x2    2 1   x π 1 − + 2  dx  = π  − 2 x + ln x  x x     2 1

  32  13   = π   − 2(3) + ln 3 −  − 2(1) + ln1    2  2   = (1.09861π ) m

Now



xV = xEL dV : x (0.46944π m3 ) = 1.09861π m 4

or x = 2.34 m 


www.elsolucionario.net PROBLEM 5.127 Locate the centroid of the volume obtained by rotating the shaded area about the line x = h.

SOLUTION x =h 


z =0 

Choose as the element of volume a disk of radius r and thickness dx. Then

Now x 2 =

Then

and Let Then

h2 2 h 2 (a − y 2 ) so that r = h − a − y2 . 2 a a dV = π V=



(

h2 a − a2 − y 2 a2 a 0

π

) dy 2

(

h2 a − a2 − y2 a2

) dy 2

y = a sin θ  dy = a cos θ dθ h2 a2 h2 =π 2 a



= π ah 2



V =π



π /2 0

π /2 0

π /2 0

(

a − a 2 − a 2 sin 2 θ

) a cosθ dθ 2

 a 2 − 2a (a cos θ ) + (a 2 − a 2 sin 2 θ )  a cos θ dθ   (2cos θ − 2 cos 2 θ − sin 2 θ cos θ ) dθ π /2

  θ sin 2θ  1 3  = π ah  2sin θ − 2  + − sin θ  4  3 2  0 2

  π  1 = π ah 2  2 − 2  2  −    2  3  = 0.095870π ah 2



dV = π r 2 dy, yEL = y


and



yEL dV =

0

(

 h2 y π 2 a − a 2 − y 2  a

( 2a y − 2ay

) dy  2

)

=π

h2 a2



=π

h2 a2

1 4  2 2 2 2 2 3/2  a y + 3 a (a − y ) − 4 y   0

=π

h2 a2

 2 2 1 4   2 2 3/2     a (a ) − a  −  a( a )   4  3  

=

a 0

2

a 2 − y 2 − y 3 dy a

1 π a 2 h2 12



y (0.095870π ah2 ) =

yV = yEL dV :

1 π a2 h2 12

or y = 0.869a 


Now



a


www.elsolucionario.net PROBLEM 5.128* Locate the centroid of the volume generated by revolving the portion of the sine curve shown about the x-axis.

SOLUTION y =0 


z =0 

Choose as the element of volume a disk of radius r and thickness dx.

r = b sin

Now

πx 2a

dV = π b 2 sin 2

so that

V=

Then



2a a


dV = π r 2 dx, xEL = x

Then

πx 2a

dx

π b 2 sin 2

πx 2a

dx 2a

 x sin π x  = π b  − π2 a  2 a   2 a 2

= π b 2 ( 22a ) − ( a2 )  1 = π ab 2 2

and

x

EL dV

=



2a a

πx   x  π b 2 sin 2 dx 2a  

Use integration by parts with u=x

dV = sin 2

du = dx

V=

x − 2

πx 2a sin πax 2π a



Then



2a    2a  x sin π x   x sin πax    xEL dV = π b   x  − 2π   − − 2π a  dx      a 2    2 a a  a    2a  2 π x    2a   a   1 2 a 2  = π b   2a   − a    −  x + 2 cos   a a  2π  2   4    2  



2

 3   1 a2 1 a 2   = π b 2  a 2  −  (2a) 2 + 2 − (a) 2 + 2   4 2π 2π    2   4 3 1  = π a 2b 2  − 2  4 π 

Now

1  xV = xEL dV : x  π ab 2  = 0.64868π a 2b 2 2  



or x = 1.297a 



= 0.64868π a 2b 2

www.elsolucionario.net PROBLEM 5.129* Locate the centroid of the volume generated by revolving the portion of the sine curve shown about the y-axis. (Hint: Use a thin cylindrical shell of radius r and thickness dr as the element of volume.)

SOLUTION x =0 


z =0

Choose as the element of volume a cylindrical shell of radius r and thickness dr.

Now so that Then

dV = (2π r )( y )(dr ), y = b sin



2a a

1 y 2

πr 2a

dV = 2π br sin V=

yEL =


Then

πr 2a

dr

2π br sin

πr 2a

dr

Use integration by parts with u = rd

dv = sin

dr 2a 2a πr v = − cos 2a π

du = dr

Then

πr

2a    2a π r  V = 2π b  ( r )  − cos   − π 2a   a   



2a π r    π cos 2a  dr    

2a  a

2a  2a  4a 2 π r    = 2π b  − [ (2a)(−1) ] +  2 sin   2a  a   π π 

 4a 2 4 a 2 − 2 V = 2π b  π  π 1  = 8a 2 b  1 −  π  

  

= 5.4535a 2b



Also

y

EL dV

π r  πr  1 2π br sin b sin dr    2 a  2a  2a  2a πr r sin 2 dr = π b2 a 2a =



2a 



Use integration by parts with u=r

dv = sin 2

du = dr



r − 2

dr 2a sin πar 2π a

2a    2a  r sin π r   r sin πar    yEL dV = π b  ( r )  − 2π   − − 2π a  dr      a 2    2 a a  a    2a  2 π r   a2  2a   a   r 2  = π b  (2a)   − ( a)    −  + 2 cos   a a   2   2    4 2π   



2

 3  (2a) 2 (a ) 2 a2 a 2   = π b2  a 2 −  + 2− + 2  4 2π 2π    2  4 3 1  = π a 2b 2  − 2  4 π  = 2.0379a 2 b 2

Now



yV = yEL dV : y (5.4535a 2b) = 2.0379a 2b 2

or y = 0.374b 



Then

v=

πr

www.elsolucionario.net PROBLEM 5.130* Show that for a regular pyramid of height h and n sides (n = 3, 4, ) the centroid of the volume of the pyramid is located at a distance h/4 above the base.

SOLUTION Choose as the element of a horizontal slice of thickness dy. For any number N of sides, the area of the base of the pyramid is given by Abase = kb 2

where k = k ( N ); see note below. Using similar triangles, we have s h− y = b h b (h − y ) h

dV = Aslice dy = ks 2 dy = k

Then

V=

and

h



k 0


s=

or

b2 (h − y ) 2 dy h2

b2 b2 2 ( h y ) dy k − = h2 h2

h

 1 3  − 3 (h − y )   0

1 = kb 2 h 3 yEL = y

Also, so that



yEL dV =



h

0

=k

Now

Note:

 b2  b2 y  k 2 ( h − y ) 2 dy  = k 2 h  h 

b2 h2



h

0

( h 2 y − 2hy 2 + y 3 ) dy

h

1 2 2 1 2 2 2 3 1 4   2 h y − 3 hy + 4 y  = 12 kb h  0

1  1 yV = yEL dV : y  kb 2 h  = kb 2 h 2 3  12



1 Abase = N  × b × 2 N = b2 4 tan πN

b 2 tan πN

or y =

1 h Q.E.D.  4

   

= k ( N )b 2


www.elsolucionario.net PROBLEM 5.131 Determine by direct integration the location of the centroid of one-half of a thin, uniform hemispherical shell of radius R.

SOLUTION x = 0


The element of area dA of the shell shown is obtained by cutting the shell with two planes parallel to the xy plane. Now dA = (π r )( Rdθ ) 2r

yEL = −


π

r = R sin θ

where

dA = π R 2 sin θ dθ 2R yEL = − sin θ

so that

π

A=

Then



π /2 0

π R 2 sin θ dθ = π R 2 [− cos θ ]π0 /2

= π R2

and

y

EL dA

=



π /2  0

2R  sin θ  (π R 2 sin θ dθ ) − π   π /2

 θ sin 2θ  = −2 R3  − 4  0 2

=−

Now Symmetry implies

π 2

R3



yA = yEL dA: y (π R 2 ) = −

π 2

R3

1 or y = − R  2 1 z = − R  2

z=y


www.elsolucionario.net PROBLEM 5.132 The sides and the base of a punch bowl are of uniform thickness t. If t < < R and R = 250 mm, determine the location of the center of gravity of (a) the bowl, (b) the punch.

SOLUTION (a)

Bowl: x =0 


z =0 

for the coordinate axes shown below. Now assume that the bowl may be treated as a shell; the center of gravity of the bowl will coincide with the centroid of the shell. For the walls of the bowl, an element of area is obtained by rotating the arc ds about the y-axis. Then


dAwall = (2π R sin θ )( R dθ ) ( yEL ) wall = − R cos θ

and

Awall =

Then

π /2

π

/6

2π R 2 sin θ dθ π /2

= 2π R 2 [ − cos θ ]π /6 = π 3R2

 π = π

ywall Awall = ( yEL ) wall dA

and

/2

/6

(− R cos θ )(2π R 2 sin θ dθ )

= π R3 [cos 2 θ ]ππ /2 /6 3 = − π R3 4

π

By observation,

Abase =

Now

y ΣA = Σ yA

or or

4

R2 ,

ybase = −

3 R 2

 π 3 π 3    y  π 3R 2 + R 2  = − π R3 + R 2  − R   4 4 4    2  y = − 0.48763R R = 250 mm

y = −121.9 mm 



(b)

Punch: x =0 


z =0 

and that because the punch is homogeneous, its center of gravity will coincide with the centroid of the corresponding volume. Choose as the element of volume a disk of radius x and thickness dy. Then dV = π x 2 dy,

Now

yEL = y

x2 + y 2 = R2 dV = π ( R 2 − y 2 )dy

so that

V=

Then



0 − 3/2 R

π ( R 2 − y 2 ) dy


0

1   = π  R 2 y − y3  3 − 

3/2 R

3   3  1 3   3 2  R − − R  = π 3R3 = −π R  −      2  3  2   8  

and

y

EL dV

=



0 − 3/2 R

(

)

( y ) π R 2 − y 2 dy    0

1  1 = π  R2 y 2 − y4  2 4 − 

3/2 R

4    1   1 3 3 15 2 R − − R   = − π R4 = −π  R  −     2  2  4 2   64   2

Now

15 3  yV = yEL dV : y  π 3 R3  = − π R 4 64 8 



y =−

or

5 8 3

R R = 250 mm

y = − 90.2 mm 


www.elsolucionario.net PROBLEM 5.133 Locate the centroid of the section shown, which was cut from a thin circular pipe by two oblique planes.

SOLUTION x =0 


dV = ( y2 − y1 )ta dθ ,

Now

y1 = =

and

h 3

2a

z+

h 6

yEL =

1 ( y1 + y2 ) z EL = z 2 y2 = −

h ( z + a) 6a

=

2h 3

2a

z+

2 h 3

h ( − z + 2a ) 3a

z = a cos θ h h (−a cos θ + 2a) − (a cos θ + a ) 3a 6a h = (1 − cos θ ) 2

Then

( y2 − y1 ) =

and

( y1 + y2 ) =

h h (a cos θ + a) + (− a cos θ + 2a) 6a 3a h = (5 − cos θ ) 6 aht h dV = (1 − cos θ )dθ yEL = (5 − cos θ ), z EL = a cos θ 2 12



Assume that the pipe has a uniform wall thickness t and choose as the element of volume a vertical strip of width adθ and height ( y2 − y1 ). Then


and

V =2



π

=2



π

aht (1 − cos θ )dθ = aht[θ − sin θ ]π0 0 2 = π aht

Then

y

EL dV

h  aht  (5 − cos θ )  (1 − cos θ ) dθ  12 2  

0

=

ah 2 t 12

=

ah 2 t  θ sin 2θ  5θ − 6sin θ + + 12  2 4  0

=

11 π ah2 t 24



π 0

(5 − 6 cos θ + cos 2 θ ) dθ

z

EL dV

=2



π 0

 aht  a cos θ  (1 − cos θ ) dθ   2  π

θ sin 2θ   = a ht sin θ − − 2 4  0  1 = − π a 2 ht 2 2



11 π ah 2t 24

Now

yV = yEL dV : y (π aht ) =

and

1 zV = zEL dV : z (π aht ) = − π a 2 ht 2



or y =

11 h  24

1 or z = − a  2



π

www.elsolucionario.net PROBLEM 5.134* Locate the centroid of the section shown, which was cut from an elliptical cylinder by an oblique plane.

SOLUTION

Choose as the element of volume a vertical slice of width zx, thickness dz, and height y. Then dV = 2 xy dz , x=

and

y=−

Then

V=

Then



h/2 h h z+ = (b − z ) b 2 2b

 a 2  h  b − z 2   (b − z )  dz 2  −b  b   2b  b

z = b sin θ V=

1 , z EL = z 24

a 2 b − z2 b

Now

Let

yEL =

ah b2

π /2

π

= abh

/2

dz = b cos θ dθ (b cos θ )[b(1 − sin θ )] b cos θ dθ

π /2

π

− /2

(cos 2 θ − sin θ cos 2 θ ) dθ π /2

 θ sin 2θ 1  = abh  + + cos3 θ  4 3 2  −π /2 1 V = π abh 2



x=0 



y

EL dV

Now so that Then

Also,



=

1 ah 2 4 b3



b

−b

z = b sin θ

Let Then

1 h   a 2  h   b − z 2   (b − z )  dz  × (b − z )   2  −b  2 2b   b   2b   b

=



dz = b cos θ dθ

1 ah 2 4 b3 1 = abh 2 4

sin 2 θ cos 2 θ =

y

EL dV

z

EL dV

π /2

π

yEL dV =

sin 2 θ =

(b − z ) 2 b 2 − z 2 dz

− /2

[b(1 − sin θ )]2 (b cos θ ) × (b cos θ dθ )

π /2

π

− /2

(cos 2 θ − 2sin θ cos 2 θ + sin 2 θ cos 2 θ ) dθ

1 1 (1 − cos 2θ ) cos 2 θ = (1 + cos 2θ ) 2 2 1 (1 − cos 2 2θ ) 4 π /2



=

1 abh 2 4

=

 θ sin 2θ 1 abh 2  + 4 4  2

=

5 π abh2 32

π

− /2  

cos 2 θ − 2sin θ cos 2 θ +

1  (1 − cos 2 2θ )  dθ 4 

1 1  θ sin 4θ  1 3  + 3 cos θ + 4 θ − 4  2 + 8  

π /2

    −π /2

 a 2 h   z 2 a − z 2  (b − z )  dz  −b  b  2b   ah b z (b − z ) b 2 − z 2 dz = 2 b −b =



b



z = b sin θ

Let Then

z

EL dV

=

ah b2

π /2

π

− /2

= ab 2 h

Using

sin 2 θ cos 2 θ =

z

EL dV

dz = b cos θ dθ (b sin θ )[b(1 − sin θ )](b cos θ ) × (b cos θ dθ )

π /2

π

− /2

(sin θ cos 2 θ − sin 2 θ cos 2 θ ) dθ

1 (1 − cos 2 2θ ) from above, 4

= ab 2 h

π /2

π



− /2  

1  sin θ cos 2 θ − (1 − cos 2 2θ )  dθ 4 

 1 1 1  θ sin 4θ = ab 2 h  − cos3 θ − θ +  + 4 4 2 8  3

π /2

1  = − π ab 2 h  8   −π /2



and


Now

1  5 yV = yEL dV : y  π abh  = π abh 2 2   32

or y =

and

1 1  z V = z EL dV : z  π abh  = − π ab 2 h 8 2 

1 or z = − b  4






5 h  16


www.elsolucionario.net PROBLEM 5.135 After grading a lot, a builder places four stakes to designate the corners of the slab for a house. To provide a firm, level base for the slab, the builder places a minimum of 3 in. of gravel beneath the slab. Determine the volume of gravel needed and the x coordinate of the centroid of the volume of the gravel. (Hint: The bottom surface of the gravel is an oblique plane, which can be represented by the equation y = a + bx + cz.)

SOLUTION

For x = 0 and z = 0,

y = − 3 in., and therefore, −

For x = 30 ft and z = 0,


For x = 0 and z = 50 ft,

3 1 ft = a, or a = − ft 12 4

5 1 1 ft = − ft + b(30 ft), or b = − 12 4 180


6 1 1 ft = − ft + c(50 ft), or c = − 12 4 200

Therefore,

y=−

Now

x=

1 1 1 ft − x− z 4 180 200

x

EL dV

V

A volume element can be chosen as dV = | y| dx dz 1 1 1  1+ x+ z dx dz 4  45 50 

or

dV =

and

xEL = x



The centroid can be found by integration. The equation for the bottom of the gravel is y = a + bx + cz , where the constants a, b, and c can be determined as follows:


Then

x

EL dV

=

50

30

0

0

 

x 1 1  1+ x+ z dx dz 4  45 50  30

50  x 2

z 2 1 3 x + x  dz  + 100  0  2 135

1 = 4



=

1 4



=

1 9  650 z + z 2  4  2 0

0 50 0

(650 + 9 z ) dz 50

= 10937.5 ft 4



V dV =

50

30

0

0

 

=

1 4



=

1 4



1 1 1  1+ x+ z dx dz  4 45 50  30

50  0

1 2 z   x + 90 x + 50 x  dz  0

50  0

3   40 + z  dz 5   50

1 3  =  40 z + z 2  4 10  0 = 687.50 ft 3

Then

x =

x

EL dV

V

=

10937.5ft 4 = 15.9091 ft 687.5 ft 3 V = 688 ft 3 

Therefore,

x = 15.91 ft 



The volume is

www.elsolucionario.net PROBLEM 5.136 Determine by direct integration the location of the centroid of the volume between the xz plane and the portion shown of the surface y = 16h(ax − x2)(bz − z2)/a2b2.


x=

a  2

z=

b  2

Choose as the element of volume a filament of base dx × dz and height y. Then dV = y dx dz ,

or Then

dV =

yEL =

1 y 2

16h (ax − x 2 )(bz − z 2 ) dx dz 2 2 a b b

a

0

0

V=



V=

16h a 2b2

16h (ax − x 2 )(bz − z 2 ) dx dz a 2b2



a

1  a (bz − z 2 )  x 2 − x3  dz 0 3 0 z b

b

16h  a 1 1   b = 2 2  ( a 2 ) − ( a )3   z 2 − z 3  3 3 0 a b 2  2 8ah  b 2 1 3  (b) − (b)  3 3b 2  2  4 = abh 9 =



SOLUTION


y

EL dV

b

a

0

0

=



=

128h 2 a 4b 4

1  16h   16h  ( ax − x 2 )(bz − z 2 )   2 2 (ax − x 2 )(bz − z 2 ) dx dz   2 2 2 a b a b 

128h 2 = 2 4 a b

Now

b

a

0

0

 

(a 2 x 2 − 2ax3 + x 4 )(b 2 z 2 − 2bz 3 + z 4 )dx dz a

 a2 a 1  (b z − 2bz + z )  x3 − x 4 + x5  dz 0 2 5 0 3 b

2 2

3

4

b

=

128h 2 a 4b4

 a2 a 4 1 5   b2 3 b 4 1 5  3 − ( a ) (a) + (a)   z − z + z   2 5 z 5 0 3  3

=

64ah 2 15b 4

 b3 3 b 4 1 5  32 abh 2  (b) − (b) + (b)  = 3 2 5 225  

4  32 yV = yEL dV : y  abh  = abh 2 9  225



8 or y = 25 h 



and


A, in 2

1

Then

π 2

(38)2 = 2268.2

2

−20 × 16 = 320

Σ

1948.23

x , in.

y , in.

x A, in 3

y A, in 3

0

16.1277

0

36,581

3200

−2560

3200

34,021

−10

8

X=

Σ xA 3200 = Σ A 1948.23

X = 1.643 in. 

Y =

Σ y A 34, 021 = Σ A 1948.23

Y = 17.46 in. 



SOLUTION


Then

A, mm 2

x , mm

y , mm

x A, mm3

y A, mm3

1

2 (75)(120) = 6000 3

28.125

48

168,750

288,000

2

1 − (75)(60) = −2250 2

25

20

–56,250

–45,000

Σ

3750

112,500

243,000

X ΣA = ΣxA X (3750 mm 2 ) = 112,500 mm3

and

or X = 30.0 mm 

Y ΣA = Σ yA Y (3750 mm 2 ) = 243, 000 mm3

or Y = 64.8 mm 



SOLUTION

www.elsolucionario.net PROBLEM 5.139 The frame for a sign is fabricated from thin, flat steel bar stock of mass per unit length 4.73 kg/m. The frame is supported by a pin at C and by a cable AB. Determine (a) the tension in the cable, (b) the reaction at C.

SOLUTION

L, m

x, m

xL, m 2

1

1.35

0.675

0.91125

2

0.6

0.3

0.18

3

0.75

0

0

4

0.75

0.2

0.15

1.07746

1.26936

5 Σ

Then

π 2

(0.75) = 1.17810

4.62810

2.5106

X ΣL = Σx L X (4.62810) = 2.5106

or

X = 0.54247 m

The free-body diagram of the frame is then where

W = (m′Σ L) g = 4.73 kg/m × 4.62810 m × 9.81 m/s 2 = 214.75 N



First note that because the frame is fabricated from uniform bar stock, its center of gravity will coincide with the centroid of the corresponding line.


Equilibrium then requires 3  ΣM C = 0: (1.55 m)  TBA  − (0.54247 m)(214.75 N) = 0 5 

(a)

TBA = 125.264 N

or

or TBA = 125.3 N 

3 ΣFx = 0: C x − (125.264 N) = 0 5 C x = 75.158 N

or ΣFy = 0: C y +

4 (125.264 N) − (214.75 N) = 0 5 C y = 114.539 N

or

C = 137.0 N

Then

56.7° 



(b)


SOLUTION For the element (EL) shown, or k = x=

Then

xEL yEL

a 1/3 y dy h1/3 1 1 a 1/3 = x= y 2 2 h1/3 =y



A = dA =

Then

Hence

a 1/3 y h1/3

dA = x dy =

Now

and

h a3


at x = a, y = h, h = ka 3



xEL dA =



yEL dA =

 

h 0

h 0



h 0

( )

a 1/3 3 a y dy = y 4/3 1/3 1/3 4h h

h

= 0

3 ah 4 h

1 a 1/3  a 1/3  1 a  3 5/3  3 y  1/3 y dy  = y  = a2h 1/3 2/3  2h h  2 h 5  0 10 h

3 a 3  a   y  1/3 y1/3 dy  = 1/3  y 7/3  = ah 2 7 7 h  h  0



3  3 x  ah  = a 2 h  4  10

x=

2 a  5



3  3 y  ah  = ah2 4  7

y=

4 h  7

xA = xEL dA: yA = yEL dA:



SOLUTION xEL = x

We have

a x x2  1 y = 1 − + 2  L L  2 2  x x2  dA = y dx = a 1 − + 2  dx L L  

yEL =



0

2L

  x x2  x2 x3  a 1 − + 2  dx = a  x − + 2 2 L 3L  0 L L   




A = dA =

Then

2L

8 = aL 3

and

x



Hence,

EL dA

2L

2L

   x 2 x3 x x2   x4  x  a 1 − + 2  dx  = a  − + 2 L L    2 3L 4 L  0  

=



=

10 2 aL 3

yEL dA =



0

2L 0

a x x2    x x2   1 − + 2   a 1 − + 2  dx  L L    L L   2 EL 

x x2 x3 x 4  1 − 2 + 3 2 − 2 3 + 4  dx L L L L  

=

a2 2



=

a2 2

 x 2 x3 x4 x5  + 2 − 3 + 4 x − L L 2L 5L 0 

=

11 2 a L 5

0

2L

 8  10 xA = xEL dA: x  aL  = aL2 3  3



 1  11 yA = yEL dA: y  a  = a 2 8  5



x= y=

5 L  4

33 a  40


www.elsolucionario.net PROBLEM 5.142 Three different drive belt profiles are to be studied. If at any given time each belt makes contact with one-half of the circumference of its pulley, determine the contact area between the belt and the pulley for each design.

SOLUTION

SOLUTION

AC = π yL = π Σ yL

where the individual lengths are the lengths of the belt cross section that are in contact with the pulley. AC = π [2( y1 L1 ) + y2 L2 ]

(a)

   0.125    0.125 in.  = π  2  3 − + [(3 − 0.125) in.](0.625 in.)  in.    2    cos 20°     AC = 8.10 in 2 

or AC = π [2( y1 L1 )]

(b)

 0.375    0.375 in.  = 2π  3 − 0.08 − in. 2    cos 20°   AC = 6.85 in 2 

or AC = π [2( y1 L1 )]

(c)

 2(0.25)   = π  3 − in. [π (0.25 in.)] π    AC = 7.01 in 2 

or



Applying the first theorem of Pappus-Guldinus, the contact area AC of a belt is given by


SOLUTION

Then

1 (3ft)(480 lb/ft) = 720 lb 2 1 R II = (6 ft)(600 lb/ft) = 1800 lb 2 R III = (2 ft)(600 lb/ft) = 1200 lb R I=

ΣFx = 0: Bx = 0 ΣM B = 0: (2 ft)(720 lb) − (4 ft)(1800 lb) + (6 ft)C y − (7 ft)(1200 lb) = 0

or

C y = 2360 lb

C = 2360 lb 

ΣFy = 0: −720 lb + B y − 1800 lb + 2360 lb − 1200 lb = 0

or

B y = 1360 lb

B = 1360 lb 



We have

www.elsolucionario.net PROBLEM 5.144 The beam AB supports two concentrated loads and rests on soil that exerts a linearly distributed upward load as shown. Determine the values of ω A and ω B corresponding to equilibrium.

1 RI = ω A (1.8 m) = 0.9ω A 2 1 RII = ωB (1.8 m) = 0.9ωB 2 ΣM D = 0: (24 kN)(1.2 − a) − (30 kN)(0.3 m) − (0.9ω A )(0.6 m) = 0

(1)

24(1.2 − 0.6) − (30)(0.3) − 0.54ωa = 0

For a = 0.6 m,

14.4 − 9 − 0.54ω A = 0 ΣFy = 0: − 24 kN − 30 kN + 0.9(10 kN/m) + 0.9ωB = 0

ω A = 10.00 kN/m  ωB = 50.0 kN/m 



SOLUTION

www.elsolucionario.net PROBLEM 5.145 The base of a dam for a lake is designed to resist up to 120 percent of the horizontal force of the water. After construction, it is found that silt (that is equivalent to a liquid of density ρ s = 1.76 × 103 kg/m3 ) is settling on the lake bottom at the rate of 12 mm/year. Considering a 1-m-wide section of dam, determine the number of years until the dam becomes unsafe.

SOLUTION First determine force on dam without the silt, 1 1 Apw = A( ρ gh) 2 2 1 = [(6.6 m)(1 m)][(103 kg/m3 )(9.81 m/s 2 )(6.6 m)] 2 = 213.66 kN = 1.2 Pw = (1.5)(213.66 kN) = 256.39 kN

Pallow


Pw =

Next determine the force P′ on the dam face after a depth d of silt has settled. We have

1 Pw′ = [(6.6 − d ) m × (1 m)][(103 kg/m3 )(9.81 m/s 2 )(6.6 − d ) m] 2 = 4.905(6.6 − d ) 2 kN ( Ps ) I = [d (1 m)][(103 kg/m3 )(9.81 m/s 2 )(6.6 − d ) m] = 9.81(6.6d − d 2 ) kN 1 ( Ps )II = [ d (1 m)][(1.76 × 103 kg/m3 )(9.81 m/s 2 )(d ) m] 2 = 8.6328d 2 kN P′ = Pw′ + ( Ps ) I + ( Ps ) II = [4.905(43.560 − 13.2000d + d 2 ) + 9.81(6.6d − d 2 ) + 8.6328d 2 ] kN = [3.7278d 2 + 213.66] kN

Now it’s required that P′ = Pallow to determine the maximum value of d. (3.7278d 2 + 213.66) kN = 256.39 kN

or Finally,

d = 3.3856 m 3.3856 m = 12 × 10−3

m ×N year

or N = 282 years 


www.elsolucionario.net PROBLEM 5.146 Determine the location of the centroid of the composite body shown when (a) h = 2b, (b) h = 2.5b.

V

x

xV

Cylinder I

π a 2b

1 b 2

1 2 2 πa b 2

Cone II

1 2 πa h 3

b+

1 h 4

1 2  1  π a hb + h 3 4  

1   V = π a2  b + h  3   1 1 1  Σ xV = π a 2  b 2 + hb + h 2  2 3 12  

(a)

For h = 2b,

1   5 V = π a 2 b + (2b)  = π a 2 b 3   3 1 1 1  Σ xV = π a 2  b 2 + (2b)b + (2b) 2  3 12 2  1 2 1 3 = π a 2b 2  + +  = π a 2 b2  2 3 3 2

5  3 XV = Σ xV : X  π a 2b  = π a 2 b 2 3  2

X=

9 b 10

Centroid is

1 b 10

to left of base of cone. 



SOLUTION


(b)

For h = 2.5b,

1   V = π a 2 b + (2.5b)  = 1.8333π a 2 b 3   1 1 1  Σ xV = π a 2  b 2 + (2.5b)b + (2.5b) 2  3 12 2  = π a 2b 2 [0.5 + 0.8333 + 0.52083] = 1.85416π a 2b 2

XV = Σ xV : X (1.8333π a 2 b) = 1.85416π a 2b 2

X = 1.01136b

Centroid is 0.01136b to right of base of cone. 


Note: Centroid is at base of cone for h = 6b = 2.449b. 



SOLUTION

1 yI = − (1.2) = −0.4 m 3 1 zI = (3.6) = 1.2 m 3 4(1.8) 2.4 xIII = − =− m π 3π

A, m 2

x, m

y, m

z, m

xA, m3

I

1 (3.6)(1.2) = 2.16 2

1.5

−0.4

1.2

II

(3.6)(1.7) = 6.12

0.75

0.4 0.8

III Σ

We have

π 2

(1.8)2 = 5.0894

−

2.4

π

13.3694

yA, m3

zA, m3

3.24

−0.864

2.592

1.8

4.59

2.448

11.016

1.8

−3.888

4.0715

3.942

5.6555

9.1609 22.769

X ΣV = Σ xV : X (13.3694 m 2 ) = 3.942 m3

or X = 0.295 m 

Y ΣV = Σ yV : Y (13.3694 m 2 ) = 5.6555 m3

or Y = 0.423 m 

Z ΣV = Σ zV : Z (13.3694 m 2 ) = 22.769 m3

or Z = 1.703 m 



First, assume that the sheet metal is homogeneous so that the center of gravity of the form will coincide with the centroid of the corresponding area.


SOLUTION y =0 

and

z =0 

We have

y = k ( X − h) 2

At x = 0, y = a,

a = k ( − h) 2

or

k=



a h2

Choose as the element of volume a disk of radius r and thickness dx. Then dV = π r 2 dx, X EL = x r=

Now

a ( x − h) 2 2 h

so that

dV = π

Then

V=



a2 ( x − h) 4 dx h4 h 0

π

a2 π a2 4 x h dx ( − ) = [( x − h)5 ]0h 5 h4 h4

1 = π a2 h 5

and



 a2  x π 4 ( x − h) 4 dx  0  h  2 a h 5 ( x − 4hx 4 + 6h 2 x3 − 4h3 x 2 + h4 x)dx =π 4 h 0

xEL dV =



h



h

=π =

Now

a2  1 6 4 5 3 2 4 4 3 3 1 4 2  x − hx + h x − h x + h x  5 2 3 2 h4  6 0

1 π a 2 h2 30

π  π 2 2 xV = xEL dV : x  a 2 h  = a h 5  30



or

x=

1 h  6




CHAPTER 6





www.elsolucionario.net PROBLEM 6.1 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression.

SOLUTION

ΣFy = 0: By = 0

By = 0

ΣM C = 0: − Bx (3.2 m) − (48 kN)(7.2 m) = 0 Bx = −108 kN B x = 108 kN ΣFx = 0: C − 108 kN + 48 kN = 0 C = 60 kN

C = 60 kN

Free body: Joint B: FAB FBC 108 kN = = 5 4 3 FAB = 180.0 kN T  FBC = 144.0 kN T 

Free body: Joint C:

FAC FBC 60 kN = = 13 12 5 FAC = 156.0 kN C  FBC = 144 kN (checks) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 743


Free body: Entire truss:


SOLUTION Reactions: ΣM A = 0: C = 1260 lb ΣFx = 0: A x = 0

Joint B: FAB FBC 300 lb = = 12 13 5 FAB = 720 lb T  FBC = 780 lb C 

Joint A: ΣFy = 0: − 960 lb −

4 FAC = 0 5 FAC = 1200 lb

FAC = 1200 lb C 

3 ΣFx = 0: 720 lb − (1200 lb) = 0 (checks) 5



ΣFy = 0: A y = 960 lb


SOLUTION AB = 32 + 1.252 = 3.25 m BC = 32 + 42 = 5 m


Reactions: ΣM A = 0: (84 kN)(3 m) − C (5.25 m) = 0 C = 48 kN

ΣFx = 0: Ax − C = 0 A x = 48 kN

ΣFy = 0: Ay = 84 kN = 0 A y = 84 kN

Joint A: ΣFx = 0: 48 kN −

12 FAB = 0 13 FAB = +52 kN

ΣFy = 0: 84 kN −

FAB = 52.0 kN T 

5 (52 kN) − FAC = 0 13 FAC = +64.0 kN

FAC = 64.0 kN T 

Joint C:

FBC 48 kN = 5 3

FBC = 80.0 kN C 



SOLUTION Free body: Truss:

Free body: Joint B:

FAB 5

=

FBC 300 lb = 2 1 FAB = 671 lb T

FBC = 600 lb C 

Free body: Joint C: ΣFy = 0:

3 FAC + 600 lb = 0 5 FAC = −1000 lb

ΣFx = 0:

4 ( −1000 lb) + 600 lb + FCD = 0 5

FAC = 1000 lb C  FCD = 200 lb T 

From symmetry: FAD = FAC = 1000 lb C , FAE = FAB = 671 lb T , FDE = FBC = 600 lb C 



C = D = 600 lb

From the symmetry of the truss and loading, we find


SOLUTION


Reactions:

ΣM D = 0: Fy (24) − (4 + 2.4)(12) − (1)(24) = 0 Fy = 4.2 kips

ΣFx = 0: Fx = 0 ΣFy = 0: D − (1 + 4 + 1 + 2.4) + 4.2 = 0 D = 4.2 kips

Joint A: ΣFx = 0: FAB = 0

FAB = 0 

ΣFy = 0 : −1 − FAD = 0 FAD = −1 kip

Joint D:

ΣFy = 0: − 1 + 4.2 +

ΣFx = 0:

8 FBD = 0 17 FBD = −6.8 kips

15 (−6.8) + FDE = 0 17 FDE = +6 kips

FAD = 1.000 kip C 

FBD = 6.80 kips C 

FDE = 6.00 kips T 



Joint E:

ΣFy = 0 : FBE − 2.4 = 0 FBE = +2.4 kips

FBE = 2.40 kips T 


Truss and loading symmetrical about cL.



AD = 52 + 122 = 13 ft BCD = 122 + 162 = 20 ft

Reactions:

ΣFx = 0: Dx = 0 ΣM E = 0: D y (21 ft) − (693 lb)(5 ft) = 0 ΣFy = 0: 165 lb − 693 lb + E = 0

Joint D:

D y = 165 lb E = 528 lb

ΣFx = 0:

5 4 FAD + FDC = 0 13 5

(1)

ΣFy = 0:

12 3 FAD + FDC + 165 lb = 0 13 5

(2)

Solving Eqs. (1) and (2) simultaneously, FAD = −260 lb

FAD = 260 lb C 

FDC = +125 lb

FDC = 125 lb T 



SOLUTION


Joint E: ΣFx = 0:

5 4 FBE + FCE = 0 13 5

(3)

ΣFy = 0:

12 3 FBE + FCE + 528 lb = 0 13 5

(4)

Solving Eqs. (3) and (4) simultaneously, FBE = −832 lb

FBE = 832 lb C 

FCE = +400 lb

FCE = 400 lb T 

Force polygon is a parallelogram (see Fig. 6.11, p. 209). FAC = 400 lb T  FBC = 125.0 lb T 

Joint A:

ΣFx = 0:

5 4 (260 lb) + (400 lb) + FAB = 0 13 5 FAB = −420 lb

ΣFy = 0:

FAB = 420 lb C 

12 3 (260 lb) − (400 lb) = 0 13 5 0 = 0 (Checks)



Joint C:


SOLUTION

ΣFx = 0: Cx + 2(5 kN) = 0

Cx = −10 kN

 C x = 10 kN

ΣM C = 0: D(2 m) − (5 kN)(8 m) − (5 kN)(4 m) = 0  D = +30 kN D = 30 kN  ΣFy = 0: C y + 30 kN = 0 C y = −30 kN C y = 30 kN

Free body: Joint A: FAB FAD 5 kN = = 4 1 17

FAB = 20.0 kN T 





FAD = 20.6 kN C 

Free body: Joint B: ΣFx = 0: 5 kN +

1 5

FBD = 0 FBD = −5 5 kN

ΣFy = 0: 20 kN − FBC −

2 5

FBD = 11.18 kN C 

(−5 5 kN) = 0 FBC = +30 kN

FBC = 30.0 kN T 

 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 751


Free body: Entire truss:


Free body: Joint C:

ΣFx = 0: FCD − 10 kN = 0 FCD = +10 kN

FCD = 10.00 kN T 


ΣFy = 0: 30 kN − 30 kN = 0 (checks)



SOLUTION Reactions: ΣM C = 0: A x = 16 kN ΣFy = 0: A y = 9 kN C = 16 kN

Joint E: FBE FDE 3 kN = = 5 4 3 FBE = 5.00 kN T  FDE = 4.00 kN C 

Joint B: ΣFx = 0:

4 (5 kN) − FAB = 0 5 FAB = +4 kN

FAB = 4.00 kN T 

3 ΣFy = 0: − 6 kN − (5 kN) − FBD = 0 5 FBD = −9 kN

FBD = 9.00 kN C 

Joint D: ΣFy = 0: − 9 kN +

3 FAD = 0 5 FAD = +15 kN

FAD = 15.00 kN T 

4 ΣFx = 0: − 4 kN − (15 kN) − FCD = 0 5 FCD = −16 kN

FCD = 16.00 kN C 



ΣFx = 0:

www.elsolucionario.net PROBLEM 6.9 Determine the force in each member of the Gambrel roof truss shown. State whether each member is in tension or compression.

SOLUTION Free body: Truss: ΣFx = 0: H x = 0

A = Hy =

1 total load 2

A = H y = 1200 lb

Free body: Joint A: FAB FAC 900 lb = = 5 4 3

FAB = 1500 lb C  FAC = 1200 lb T 

Free body: Joint C: BC is a zero-force member. FBC = 0

FCE = 1200 lb T 

Free body: Joint B: ΣFx = 0:

24 4 4 FBD + FBE + (1500 lb) = 0 25 5 5 24 FBD + 20 FBE = −30, 000 lb

or ΣFy = 0:

(1)

7 3 3 FBD − FBE + (1500) − 600 = 0 25 5 5 7 FBD − 15FBE = −7,500 lb

or

(2)



Because of the symmetry of the truss and loading,


Multiply Eq. (1) by 3, Eq. (2) by 4, and add: 100 FBD = −120, 000 lb

FBD = 1200 lb C 

Multiply Eq. (1) by 7, Eq. (2) by –24, and add: 500 FBE = −30,000 lb

FBE = 60.0 lb C 

Free body: Joint D: 24 24 FDF = 0 (1200 lb) + 25 25 FDF = −1200 lb

FDF = 1200 lb C 

7 7 (1200 lb) − (−1200 lb) − 600 lb − FDE = 0 25 25

ΣFy = 0:

FDE = 72.0 lb

FDE = 72.0 lb T 

Because of the symmetry of the truss and loading, we deduce that FEF = FBE

FEF = 60.0 lb C 

FEG = FCE

FEG = 1200 lb T 

FFG = FBC

FFG = 0 

FFH = FAB

FFH = 1500 lb C 

FGH = FAC

FGH = 1200 lb T 

Note: Compare results with those of Problem 6.11.



ΣFx = 0:

www.elsolucionario.net PROBLEM 6.10 Determine the force in each member of the Howe roof truss shown. State whether each member is in tension or compression.

SOLUTION


Free body: Truss:

ΣFx = 0: H x = 0

Because of the symmetry of the truss and loading, A = Hy =

1 total load 2

A = H y = 1200 lb

Free body: Joint A: FAB FAC 900 lb = = 5 4 3

FAB = 1500 lb C  FAC = 1200 lb T 

Free body: Joint C: BC is a zero-force member. FBC = 0

FCE = 1200 lb T 



Free body: Joint B:

or

FBD + FBE = −1500 lb ΣFy = 0:

or

4 4 4 FBD + FBC + (1500 lb) = 0 5 5 5

(1)

3 3 3 FBD − FBE + (1500 lb) − 600 lb = 0 5 5 5

FBD − FBE = −500 lb

(2)

Add Eqs. (1) and (2):

2 FBD = −2000 lb

FBD = 1000 lb C 

Subtract Eq. (2) from Eq. (1):

2 FBE = −1000 lb

FBE = 500 lb C 

Free Body: Joint D: 4 4 (1000 lb) + FDF = 0 5 5

ΣFx = 0:

FDF = −1000 lb

FDF = 1000 lb C 

3 3 (1000 lb) − ( −1000 lb) − 600 lb − FDE = 0 5 5

ΣFy = 0:

FDE = +600 lb

FDE = 600 lb T 

Because of the symmetry of the truss and loading, we deduce that FEF = FBE

FEF = 500 lb C 

FEG = FCE

FEG = 1200 lb T 

FFG = FBC

FFG = 0 

FFH = FAB

FFH = 1500 lb C 

FGH = FAC

FGH = 1200 lb T 



ΣFx = 0:

www.elsolucionario.net PROBLEM 6.11 Determine the force in each member of the Pratt roof truss shown. State whether each member is in tension or compression.

SOLUTION


Free body: Truss:

ΣFx = 0: Ax = 0

Due to symmetry of truss and load, Ay = H =

1 total load = 21 kN 2

Free body: Joint A:

FAB FAC 15.3 kN = = 37 35 12 FAB = 47.175 kN FAC = 44.625 kN

FAB = 47.2 kN C  FAC = 44.6 kN T 

Free body: Joint B:

From force polygon:

FBD = 47.175 kN, FBC = 10.5 kN

FBC = 10.50 kN C  FBD = 47.2 kN C 



Free body: Joint C:

ΣFy = 0:

3 FCD − 10.5 = 0 5

ΣFx = 0: FCE +

FCD = 17.50 kN T 

4 (17.50) − 44.625 = 0 5 FCE = 30.625 kN

Free body: Joint E:

DE is a zero-force member.

FCE = 30.6 kN T  FDE = 0 


Truss and loading symmetrical about cL .


www.elsolucionario.net PROBLEM 6.12 Determine the force in each member of the Fink roof truss shown. State whether each member is in tension or compression.

SOLUTION Free body: Truss: ΣFx = 0: A x = 0

Because of the symmetry of the truss and loading, 1 total load 2

A y = G = 6.00 kN

Free body: Joint A: F FAB 4.50 kN = AC = 2.462 2.25 1 FAB = 11.08 kN C  FAC = 10.125 kN

FAC = 10.13 kN T 


3 2.25 2.25 FBC + FBD + (11.08 kN) = 0 5 2.462 2.462

(1)

F 4 11.08 kN FBC + BD + − 3 kN = 0 5 2.462 2.462

(2)

ΣFy = 0: −

Multiply Eq. (2) by –2.25 and add to Eq. (1): 12 FBC + 6.75 kN = 0 5

FBC = −2.8125

FBC = 2.81 kN C 

Multiply Eq. (1) by 4, Eq. (2) by 3, and add: 12 12 FBD + (11.08 kN) − 9 kN = 0 2.462 2.462 FBD = −9.2335 kN

FBD = 9.23 kN C 



Ay = G =


Free body: Joint C:

ΣFy = 0:

4 4 FCD − (2.8125 kN) = 0 5 5 FCD = 2.8125 kN,

FCD = 2.81 kN T 

3 3 ΣFx = 0: FCE − 10.125 kN + (2.8125 kN) + (2.8125 kN) = 0 5 5 FCE = +6.7500 kN

FCE = 6.75 kN T  FCD = 2.81 kN T 

FDF = FBD

FDF = 9.23 kN C 

FEF = FBC

FEF = 2.81 kN C 

FEG = FAC

FEG = 10.13 kN T 

FFG = FAB

FFG = 11.08 kN C 


Because of the symmetry of the truss and loading, we deduce that FDE = FCD


www.elsolucionario.net PROBLEM 6.13 Using the method of joints, determine the force in each member of the double-pitch roof truss shown. State whether each member is in tension or compression.

SOLUTION

ΣM A = 0: H (18 m) − (2 kN)(4 m) − (2 kN)(8 m) − (1.75 kN)(12 m) − (1.5 kN)(15 m) − (0.75 kN)(18 m) = 0

H = 4.50 kN ΣFx = 0: Ax = 0 ΣFy = 0: Ay + H − 9 = 0 Ay = 9 − 4.50

A y = 4.50 kN

Free body: Joint A:

FAB

FAC 3.50 kN = 2 1 5 FAB = 7.8262 kN C =

FAB = 7.83 kN C  FAC = 7.00 kN T 



Free body: Truss:



2 5

FBD +

2 5

(7.8262 kN) +

1 2

FBC = 0

FBD + 0.79057 FBC = −7.8262 kN

or ΣFy = 0:

1 5

FBD +

1 5

(7.8262 kN) −

1 2

(1)

FBC − 2 kN = 0

FBD − 1.58114 FBC = −3.3541

or

(2)

Multiply Eq. (1) by 2 and add Eq. (2): 3FBD = −19.0065 FBD = −6.3355 kN

FBD = 6.34 kN C 

2.37111FBC = −4.4721 FBC = 1.886 kN C 

FBC = −1.8861 kN

Free body: Joint C: 2

ΣFy = 0:

5

FCD −

1 2

(1.8861 kN) = 0

FCD = +1.4911 kN ΣFx = 0: FCE − 7.00 kN + FCE

1

FCD = 1.491 kN T 

(1.8861 kN) +

2 = +5.000 kN

1 5

(1.4911 kN) = 0 FCE = 5.00 kN T 

Free body: Joint D: ΣFx = 0:

2 5

FDF +

1 2

FDE +

2 5

(6.3355 kN) −

1 5

(1.4911 kN) = 0

FDF + 0.79057 FDE = −5.5900 kN

or ΣFy = 0:

or

1 5

FDF −

1 2

FDE +

1 5

(6.3355 kN) −

FDF − 0.79057 FDE = −1.1188 kN


2 5

(1.4911 kN) − 2 kN = 0

(2)

2 FDF = −6.7088 kN FDF = −3.3544 kN


(1)

FDF = 3.35 kN C 

1.58114 FDE = −4.4712 kN FDE = −2.8278 kN

FDE = 2.83 kN C 





Free body: Joint F: ΣFx = 0:

1

FFG +

2

2

(3.3544 kN) = 0

5

FFG = −4.243 kN ΣFy = 0: −FEF − 1.75 kN +

1 5

FFG = 4.24 kN C  (3.3544 kN) −

1 2

FEF = 2.750 kN

(−4.243 kN) = 0 FEF = 2.75 kN T 

ΣFx = 0:

1

FGH −

2

ΣFy = 0: −

1 2

1 2

(4.243 kN) = 0

FGH −

1 2

FEG −

1 2

(1)

(4.243 kN) − 1.5 kN = 0

FGH + FEG = −6.364 kN

or

(2)

2 FGH = −10.607 FGH = −5.303


2

FEG +

FGH − FEG = −4.243 kN

or


1

FGH = 5.30 kN C 

2 FEG = −2.121 kN FEG = −1.0605 kN

FEG = 1.061 kN C 

Free body: Joint H:

FEH 3.75 kN = 1 1

We can also write

FGH 2

=

FEH = 3.75 kN T 

3.75 kN 1

FGH = 5.30 kN C (Checks)



Free body: Joint G:

www.elsolucionario.net PROBLEM 6.14 The truss shown is one of several supporting an advertising panel. Determine the force in each member of the truss for a wind load equivalent to the two forces shown. State whether each member is in tension or compression.

SOLUTION Free body: Entire truss: A = +2250 N

A = 2250 N

ΣFy = 0: 2250 N + Fy = 0 Fy = −2250 N Fy = 2250 N ΣFx = 0: −800 N − 800 N + Fx = 0 Fx = +1600 N Fx = 1600 N

Joint D: 800 N FDE FBD = = 8 15 17 FBD = 1700 N C  FDE = 1500 N T 

Joint A: 2250 N FAB FAC = = 15 17 8 FAB = 2250 N C  FAC = 1200 N T 

Joint F: ΣFx = 0: 1600 N − FCF = 0 FCF = +1600 N

FCF = 1600 N T 

ΣFy = 0: FEF − 2250 N = 0 FEF = +2250 N

FEF = 2250 N T 



ΣM F = 0: (800 N)(7.5 m) + (800 N)(3.75 m) − A(2 m) = 0


ΣFx = 0:

8 FCE − 1200 N + 1600 N = 0 17 FCE = −850 N

ΣFy = 0: FBC +

15 FCE = 0 17

FBC = −

15 15 FCE = − ( −850 N) 17 17 FBC = +750 N

Joint E:

FCE = 850 N C 

ΣFx = 0: − FBE − 800 N +

FBC = 750 N T 

8 (850 N) = 0 17 FBE = −400 N

ΣFy = 0: 1500 N − 2250 N +

FBE = 400 N C 

15 (850 N) = 0 17 0 = 0 (checks)



Joint C:

www.elsolucionario.net PROBLEM 6.15 Determine the force in each of the members located to the left of line FGH for the studio roof truss shown. State whether each member is in tension or compression.


ΣFx = 0: A x = 0

Ay = L =

1 total load 2

A y = L = 1200 lb

Zero-Force Members: Examining joints C and H, we conclude that BC, EH, and GH are zero-force members. Thus, FBC = FEH = 0



Also,

FCE = FAC

(1)

Free body: Joint A:

FAB

FAC 1000 lb = 2 1 5 FAB = 2236 lb C =

FAB = 2240 lb C  FAC = 2000 lb T  FCE = 2000 lb T 

From Eq. (1): Free body: Joint B: ΣFx = 0:

2 5

FBD +

2 5

FBE +

2 5

(2236 lb) = 0

FBD + FBE = −2236 lb

or ΣFy = 0:

1 5

FBD −

1 5

FBE +

1

FBD − FBE = −1342 lb

or

(2)

5

(2236 lb) − 400 lb = 0

(3)



Because of symmetry of loading,


2 FBD = −3578 lb


FBD = 1789 lb C 

Subtract Eq. (3) from Eq. (1): 2 FBE = − 894 lb

FBE = 447 lb C 

Free body: Joint E: ΣFx = 0:

2 5

FEG +

2 5

(447 lb) − 2000 lb = 0 FEG = 1789 lb T 

ΣFy = 0: FDE +

1 5

(1789 lb) −

1 5

FDE = − 600 lb

(447 lb) = 0 FDE = 600 lb C 

ΣFx = 0:

2 5

FDF +

2 5

FDG +

2 5

(1789 lb) = 0

FDF + FDG = −1789 lb

or ΣFy = 0:

1

FDF −

1

FDG +

(4) 1

5 5 5 + 600 lb − 400 lb = 0 FDF − FDG = −2236 lb

or Add Eqs. (4) and (5):

(1789 lb)

(5)

2 FDF = − 4025 lb FDF = 2010 lb C 

Subtract Eq. (5) from Eq. (4): 2 FDG = 447 lb

FDG = 224 lb T 



Free body: Joint D:

www.elsolucionario.net PROBLEM 6.16 Determine the force in member FG and in each of the members located to the right of FG for the studio roof truss shown. State whether each member is in tension or compression.

SOLUTION Reaction at L: Because of the symmetry of the loading, 1 total load, 2

L = 1200 lb

(See F.B. diagram to the left for more details.) Free body: Joint L: 9 = 26.57° 18 3 β = tan −1 = 9.46° 18 FJL FKL 1000 lb = = sin 63.43° sin 99.46° sin17.11°

α = tan −1

FKL = 3352.7 lb C

FJL = 3040 lb T  FKL = 3350 lb C 

Free body: Joint K: ΣFx = 0: −

or

2 5

FIK −

2 5

2

FJK −

5

(3352.7 lb) = 0

FIK + FJK = − 3352.7 lb ΣFy = 0:

or

1 5

FIK −

1 5

FJK +

(1) 1 5

(3352.7) − 400 = 0

FIK − FJK = − 2458.3 lb

(2)

2 FIK = − 5811.0


FIK = − 2905.5 lb

FIK = 2910 lb C 

Subtract Eq. (2) from Eq. (1): 2 FJK = − 894.4 FJK = − 447.2 lb

FJK = 447 lb C 



L=


Free body: Joint J: 2

ΣFx = 0: −

13 3

ΣFy = 0:

FIJ +

13

6

FIJ −

37 1 37

6

FGJ +

FGJ −

37 1 37

2

(3040 lb) −

(3040 lb) −

5

1 5

(447.2) = 0

(447.2) = 0

(3) (4)

Multiply Eq. (4) by 6 and add to Eq. (3): 16 13

FIJ −

8

(447.2) = 0

5

FIJ = 360.54 lb

FIJ = 361 lb T 

Multiply Eq. (3) by 3, Eq. (4) by 2, and add: 16 37

( FGJ − 3040) −

8

(447.2) = 0

5

FGJ = 2431.7 lb

FGJ = 2430 lb T 

Free body: Joint I: 2

ΣFx = 0: −

5

2

FFI −

5

2

FGI −

5

2

(2905.5) +

13

(360.54) = 0

FFI + FGI = − 2681.9 lb

or ΣFy = 0:

1 5

FFI −

1 5

FGI +

1 5

(5) (2905.5) −

3 13

(360.54) − 400 = 0

FFI − FGI = −1340.3 lb

or

(6)

2 FFI = −4022.2



FFI = −2011.1 lb

FFI = 2010 lb C 

2 FGI = −1341.6 lb

FGI = 671 lb C 

Free body: Joint F: From

ΣFx = 0: FDF = FFI = 2011.1 lb C

 1  2011.1 lb  = 0 ΣFy = 0: FFG − 400 lb + 2   5  FFG = + 1400 lb

FFG = 1400 lb T 



−

www.elsolucionario.net PROBLEM 6.17 Determine the force in each of the members located to the left of FG for the scissors roof truss shown. State whether each member is in tension or compression.

SOLUTION

ΣFx = 0: A x = 0 ΣM L = 0: (1 kN)(12 m) + (2 kN)(10 m) + (2 kN)(8 m) + (1 kN)(6 m) − Ay (12 m) = 0 A y = 4.50 kN

FBC = 0 

We note that BC is a zero-force member: Also,

FCE = FAC

Free body: Joint A:

ΣFx = 0:

1

ΣFy = 0:

1

(1)

2

2

FAB +

2

FAB +

1

5

5

FAC = 0

(2)

FAC + 3.50 kN = 0

(3)

Multiply Eq. (3) by –2 and add Eq. (2): −

1 2

FAB − 7 kN = 0

FAB = 9.90 kN C 



Free Body: Truss:


Subtract Eq. (3) from Eq. (2): 1 5

FAC − 3.50 kN = 0 FAC = 7.826 kN FCE = FAC = 7.826 kN

From Eq. (1):

FAC = 7.83 kN T  FCE = 7.83 kN T 

Free body: Joint B: 1 2

FBD +

1 2

(9.90 kN) − 2 kN = 0 FBD = −7.071 kN

1

ΣFx = 0: FBE +

2

(9.90 − 7.071) kN = 0 FBE = −2.000 kN

Free body: Joint E:

ΣFx = 0:

2 5

FBD = 7.07 kN C 

FBE = 2.00 kN C 

( FEG − 7.826 kN) + 2.00 kN = 0 FEG = 5.590 kN

ΣFy = 0: FDE −

1 5

FEG = 5.59 kN T 

(7.826 − 5.590) kN = 0 FDE = 1.000 kN

FDE = 1.000 kN T 


2 5

( FDF + FDG ) +

1 2

(7.071 kN)

FDF + FDG = −5.590 kN

or ΣFy = 0:

or

1 5

( FDF − FDG ) +

1 2

(4)

(7.071 kN) = 2 kN − 1 kN = 0

FDE − FDG = −4.472

(5)


2 FDF = −10.062 kN

FDF = 5.03 kN C 


2 FDG = −1.1180 kN

FDG = 0.559 kN C 



ΣFy = 0:

www.elsolucionario.net PROBLEM 6.18 Determine the force in member FG and in each of the members located to the right of FG for the scissors roof truss shown. State whether each member is in tension or compression.

SOLUTION


Free body: Truss:

ΣM A = 0: L(12 m) − (2 kN)(2 m) − (2 kN)(4 m) − (1 kN)(6 m) = 0 L = 1.500 kN

Angles: tan α = 1 tan β =

1 2

α = 45° β = 26.57°

Zero-force members: Examining successively joints K, J, and I, we note that the following members to the right of FG are zeroforce members: JK, IJ, and HI. FHI = FIJ = FJK = 0 

Thus, We also note that

and

FGI = FIK = FKL

(1)

FHJ = FJL

(2)



Free body: Joint L:

FJL F 1.500 kN = KL = sin116.57° sin 45° sin18.43° FJL = 4.2436 kN

FJL = 4.24 kN C  FKL = 3.35 kN T 

FGI = FIK = FKL

FGI = FIK = 3.35 kN T 

From Eq. (2):

FHJ = FJL = 4.2436 kN

FHJ = 4.24 kN C 

FGH FFH 4.2436 = = sin108.43° sin18.43° sin 53.14°

FFH = 5.03 kN C 

Free body: Joint H:

FGH = 1.677 kN T 

Free body: Joint F: ΣFx = 0: − FDF cos 26.57° − (5.03 kN) cos 26.57° = 0

FDF = −5.03 kN

ΣFy = 0: − FFG − 1 kN + (5.03 kN) sin 26.57° − ( −5.03 kN)sin 26.57° = 0

FFG = 3.500 kN

FFG = 3.50 kN T 



From Eq. (1):

www.elsolucionario.net PROBLEM 6.19 Determine the force in each member of the Warren bridge truss shown. State whether each member is in tension or compression.

SOLUTION ΣFx = 0: Ax = 0

Free body: Truss: Due to symmetry of truss and loading,

Free body: Joint A:

1 total load = 6 kips 2

FAB FAC 6 kips = = 5 3 4

FAB = 7.50 kips C  FAC = 4.50 kips T 



Free body: Joint B:

FBC FBD 7.5 kips = = 5 6 5

FBC = 7.50 kips T  FBD = 9.00 kips C 


4 4 (7.5) + FCD − 6 = 0 5 5

FCD = 0  3 ΣFx = 0: FCE − 4.5 − (7.5) = 0 5

FCE = +9 kips

FCE = 9.00 kips T 

Truss and loading is symmetrical about cL.



Ay = G =

www.elsolucionario.net PROBLEM 6.20 Solve Problem 6.19 assuming that the load applied at E has been removed. PROBLEM 6.19 Determine the force in each member of the Warren bridge truss shown. State whether each member is in tension or compression.


ΣFx = 0: Ax = 0 ΣM G = 0: 6(36) − Ay (54) = 0 A y = 4 kips

Free body: Joint A:

FAB FAC 4 kips = = 5 3 4

FAB = 5.00 kips C  FAC = 3.00 kips T 

Free body Joint B:

FBC FBD 5 kips = = 5 6 5

FBC = 5.00 kips T  FBD = 6.00 kips C 

Free body Joint C: ΣM y = 0:

4 4 (5) + FCD − 6 = 0 5 5

3 3 ΣFx = 0: FCE + (2.5) − (5) − 3 = 0 5 5

FCD = 2.50 kips T  FCE = 4.50 kips T 



ΣFy = 0: 4 − 6 + G = 0 G = 2 kips


Free body: Joint D: 4 4 ΣFy = 0: − (2.5) − FDE = 0 5 5

FDE = −2.5 kips

FDE = 2.50 kips C 

3 3 ΣFx = 0: FDF + 6 − (2.5) − (2.5) = 0 5 5

Free body: Joint F:

FDF = −3 kips

FDF = 3.00 kips C 

FEF FFG 3 kips = = 5 5 6

FEF = 2.50 kips T  FFG = 2.50 kips C 

Free body: Joint G:

FEG 2 kips = 3 4

Also,

FFG 2 kips = 5 4

FEG = 1.500 kips T 

FFG = 2.50 kips C (Checks)





www.elsolucionario.net PROBLEM 6.21 Determine the force in each member of the Pratt bridge truss shown. State whether each member is in tension or compression.

SOLUTION Free body: Truss: ΣFz = 0: A x = 0 ΣM A = 0 : H (36 ft) − (4 kips)(9 ft)

H = 6 kips ΣFy = 0: Ay + 6 kips − 12 kips = 0

A y = 6 kips

Free body: Joint A: FAB FAC 6 kips = = 5 3 4 FAB = 7.50 kips C  FAC = 4.50 kips T 

Free body: Joint C: ΣFx = 0:

FCE = 4.50 kips T 

ΣFy = 0:

FBC = 4.00 kips T 

Free body: Joint B: ΣFy = 0: −

4 4 FBE + (7.50 kips) − 4.00 kips = 0  5 5 FBE = 2.50 kips T 

ΣFx = 0:

8 3 (7.50 kips) + (2.50 kips) + FBD = 0  5 5 FBD = −6.00 kips

FBD = 6.00 kips C 



− (4 kips)(18 ft) − (4 kips)(27 ft) = 0


Free body: Joint D: We note that DE is a zero-force member:

FDE = 0  FDF = 6.00 kips C 

Also, From symmetry:

FEF = 2.50 kips T 

FEG = FCE


FFG = FBC

FFG = 4.00 kips T 

FFH = FAB

FFH = 7.50 kips C 

FGH = FAC

FGH = 4.50 kips T 


FFE = FBE


www.elsolucionario.net PROBLEM 6.22 Solve Problem 6.21 assuming that the load applied at G has been removed. PROBLEM 6.21 Determine the force in each member of the Pratt bridge truss shown. State whether each member is in tension or compression.

SOLUTION Free body: Truss: ΣFx = 0: A x = 0 ΣM A = 0: H (36 ft) − (4 kips)(9 ft) − (4 kips)(18 ft) = 0 ΣFy = 0: A y + 5.00 kips

We note that DE and FG are zero-force members. FDE = 0,

Therefore,

FFG = 0 

Also,

FBD = FDF

(1)

and

FEG = FGH

(2)

Free body: Joint A: FAB FAC 5 kips = = 5 3 4 FAB = 6.25 kips C  FAC = 3.75 kips T  


FCE = 3.75 kips T 

ΣFy = 0:

FBC = 4.00 kips T 


4 4 (6.25 kips) − 4.00 kips − FBE = 0 5 5 FBE = 1.250 kips T 

3 3 ΣFx = 0: FBD + (6.25 kips) + (1.250 kips) = 0  5 5 FBD = −4.50 kips

FBD = 4.50 kips C 



H = 3.00 kips


Free body: Joint F: We recall that FFG = 0, and from Eq. (1) that FDF = FBD

FDF = 4.50 kips C 

FEF FFH 4.50 kips = = 5 5 6 FEF = 3.75 kips T  FFH = 3.75 kips C 

FGH 3.00 kips = 3 4 FGH = 2.25 kips T 

Also, FFH 3.00 kips = 5 4 FFH = 3.75 kips C (checks)

From Eq. (2): FEG = FGH




Free body: Joint H:


SOLUTION Free body: Joint A: F FAB 1.2 kN = AC = 2.29 2.29 1.2

FAB = 2.29 kN T  FAC = 2.29 kN C 

Free body: Joint F: FDF F 1.2 kN = EF = 2.29 2.29 2.1

FDF = 2.29 kN T  FEF = 2.29 kN C 

Free body: Joint D: FBD FDE 2.29 kN = = 2.21 0.6 2.29

FBD = 2.21 kN T  FDE = 0.600 kN C 


4 2.21 (2.29 kN) = 0 FBE + 2.21 kN − 5 2.29 FBE = 0 

3 0.6 (2.29 kN) = 0 ΣFy = 0: − FBC − (0) − 5 2.29 FBC = − 0.600 kN

FBC = 0.600 kN C 



The portion of truss shown represents the upper part of a power transmission line tower. For the given loading, determine the force in each of the members located above HJ. State whether each member is in tension or compression.


Free body: Joint C: ΣFx = 0: FCE +

2.21 (2.29 kN) = 0 2.29

FCE = − 2.21 kN ΣFy = 0: − FCH − 0.600 kN − FCH = −1.200 kN

FCE = 2.21 kN C  0.6 (2.29 kN) = 0 2.29 FCH = 1.200 kN C 

Free body: Joint E: ΣFx = 0: 2.21 kN −

2.21 4 (2.29 kN) − FEH = 0 2.29 5

ΣFy = 0: − FEJ − 0.600 kN − FEJ = −1.200 kN

0.6 (2.29 kN) − 0 = 0 2.29 FEJ = 1.200 kN C 



FEH = 0 

www.elsolucionario.net PROBLEM 6.24 For the tower and loading of Problem 6.23 and knowing that FCH = FEJ = 1.2 kN C and FEH = 0, determine the force in member HJ and in each of the members located between HJ and NO. State whether each member is in tension or compression. PROBLEM 6.23 The portion of truss shown represents the upper part of a power transmission line tower. For the given loading, determine the force in each of the members located above HJ. State whether each member is in tension or compression.

Free body: Joint G: FGH F 1.2 kN = GI = 3.03 3.03 1.2

FGH = 3.03 kN T  FGI = 3.03 kN C 

Free body: Joint L: FJL F 1.2 kN = KL = 3.03 3.03 1.2

FJL = 3.03 kN T  FKL = 3.03 kN C 

Free body: Joint J: ΣFx = 0: − FHJ +

2.97 (3.03 kN) = 0 3.03 FHJ = 2.97 kN T 

0.6 (3.03 kN) = 0 3.03 = −1.800 kN FJK = 1.800 kN C 

Fy = 0: − FJK − 1.2 kN − FJK

Free body: Joint H: ΣFx = 0:

4 2.97 (3.03 kN) = 0 FHK + 2.97 kN − 5 3.03 FHK = 0  0.6 3 (3.03) kN − (0) = 0 3.03 5 = −1.800 kN FHI = 1.800 kN C 

ΣFy = 0: − FHI − 1.2 kN − FHI



SOLUTION


Free body: Joint I: ΣFx = 0: FIK +

2.97 (3.03 kN) = 0 3.03

FIK = −2.97 kN ΣFy = 0: − FIN − 1.800 kN − FIN = −2.40 kN

FIK = 2.97 kN C  0.6 (3.03 kN) = 0 3.03 FIN = 2.40 kN C 

Free body: Joint K: ΣFx = 0: −

4 2.97 (3.03 kN) = 0 FKN + 2.97 kN − 5 3.03

ΣFy = 0: − FKO −

0.6 3 (3.03 kN) − 1.800 kN − (0) = 0 3.03 5

FKO = −2.40 kN

FKO = 2.40 kN C 



FKN = 0 

www.elsolucionario.net PROBLEM 6.25 Solve Problem 6.23 assuming that the cables hanging from the right side of the tower have fallen to the ground. PROBLEM 6.23 The portion of truss shown represents the upper part of a power transmission line tower. For the given loading, determine the force in each of the members located above HJ. State whether each member is in tension or compression.

Zero-Force Members: Considering joint F, we note that DF and EF are zero-force members: FDF = FEF = 0 

Considering next joint D, we note that BD and DE are zero-force members: FBD = FDE = 0 

Free body: Joint A: F FAB 1.2 kN = AC = 2.29 2.29 1.2

FAB = 2.29 kN T  FAC = 2.29 kN C 


4 2.21 (2.29 kN) = 0 FBE − 5 2.29 FBE = 2.7625 kN

ΣFy = 0: − FBC −

FBE = 2.76 kN T 

0.6 3 (2.29 kN) − (2.7625 kN) = 0 2.29 5

FBC = −2.2575 kN

FBC = 2.26 kN C 



SOLUTION


Free body: Joint C: ΣFx = 0: FCE +

2.21 (2.29 kN) = 0 2.29 FCE = 2.21 kN C 

ΣFy = 0: − FCH − 2.2575 kN − FCH = −2.8575 kN

0.6 (2.29 kN) = 0 2.29 FCH = 2.86 kN C 

Free body: Joint E: ΣFx = 0: −

4 4 FEH − (2.7625 kN) + 2.21 kN = 0 5 5

3 3 ΣFy = 0: − FEJ + (2.7625 kN) − (0) = 0 5 5 FEJ = +1.6575 kN

FEJ = 1.658 kN T 



FEH = 0 

www.elsolucionario.net PROBLEM 6.26 Determine the force in each of the members connecting joints A through F of the vaulted roof truss shown. State whether each member is in tension or compression.

SOLUTION Free body: Truss: ΣFx = 0: A x = 0

− Ay (6a) = 0 A y = 5.40 kN

Free body: Joint A: FAB

FAC 4.20 kN = 2 1 5 FAB = 9.3915 kN =

FAB = 9.39 kN C  FAC = 8.40 kN T 


2

ΣFy = 0:

1

5

5

FBD +

1

FBD −

1

2

2

FBC +

2

FBC +

1

5

5

(9.3915) = 0

(1)

(9.3915) − 2.4 = 0

(2)

Add Eqs. (1) and (2): 3 5

FBD +

3 5

(9.3915 kN) − 2.4 kN = 0

FBD = − 7.6026 kN

FBD = 7.60 kN C 

Multiply Eq. (2) by –2 and add Eq. (1): 3 2

FB + 4.8 kN = 0

FBC = −2.2627 kN

FBC = 2.26 kN C 



ΣM K = 0: (1.2 kN)6a + (2.4 kN)5a + (2.4 kN)4a + (1.2 kN)3a



1

ΣFy = 0:

2

5

5

4

FCD +

17 1

FCD +

17

FCE +

1

FCE −

1

2

2

(2.2627) − 8.40 = 0

(3)

(2.2627) = 0

(4)

Multiply Eq. (4) by −4 and add Eq. (1): −

7 5

FCD +

5

(2.2627) − 8.40 = 0

2

FCD = − 0.1278 kN

FCD = 0.128 kN C 

7 17

FCE +

3 2

(2.2627) − 2(8.40) = 0

FCE = 7.068 kN

FCE = 7.07 kN T 


2

1

+ ΣFy = 0:

FDF +

5

5

1 5 +

(0.1278) = 0

FDF −

2 5

1 2 FDE + (7.6026) 1.524 5

(5)

1.15 1 FDE + (7.6026) 1.524 5

(0.1278) − 2.4 = 0

(6)

Multiply Eq. (5) by 1.15 and add Eq. (6): 3.30 5

FDF +

3.30 5

(7.6026) +

3.15 5

(0.1278) − 2.4 = 0

FDF = − 6.098 kN

FDF = 6.10 kN C 

Multiply Eq. (6) by –2 and add Eq. (5): 3.30 3 FDE − (0.1278) + 4.8 = 0 1.524 5 FDE = − 2.138 kN

FDE = 2.14 kN C 



Multiply Eq. (1) by 2 and subtract Eq. (2):


Free body: Joint E: ΣFx = 0:

0.6 4 1 FEF + ( FEH − FCE ) + (2.138) = 0 2.04 1.524 17

(7)

ΣFy = 0:

1.95 1 1.15 FEF + ( FEH − FCE ) − (2.138) = 0 2.04 1.524 17

(8)

Multiply Eq. (8) by 4 and subtract Eq. (7): FEF = 2.23 kN T 


7.2 FEF − 7.856 kN = 0 2.04


www.elsolucionario.net PROBLEM 6.27 Determine the force in each member of the truss shown. State whether each member is in tension or compression.


G = 42 kips ΣFx = 0: Fx + 15 kips = 0

Fx = 15 kips ΣFy = 0: Fy − 40 kips + 42 kips = 0

Fy = 2 kips

Free body: Joint F: ΣFx = 0:

1 5

FDF − 15 kips = 0 FDF = 33.54 kips

ΣFy = 0: FBF − 2 kips +

2 5

FDF = 33.5 kips T 

(33.54 kips) = 0

FBF = − 28.00 kips

FBF = 28.0 kips C 


5

ΣFy = 0:

2

29

29

FAB +

5

FAB −

6

61

61

FBD + 15 kips = 0

(1)

FBD + 28 kips = 0

(2)



ΣM F = 0: G (20 ft) − (15 kips)(16 ft) − (40 kips)(15 ft) = 0


Multiply Eq. (1) by 6, Eq. (2) by 5, and add: 40 29

FAB + 230 kips = 0 FAB = −30.96 kips

FAB = 31.0 kips C 

Multiply Eq. (1) by 2, Eq. (2) by –5, and add: 40 61

FBD − 110 kips = 0 FBD = 21.48 kips

FBD = 21.5 kips T 

Free body: Joint D: 5

2

FAD −

5

(33.54) +

6 61

(21.48) = 0 FAD = 15.09 kips T 

1

ΣFx = 0: FDE +

5

(15.09 − 33.54) −

5

(21.48) = 0

61


Free body: Joint A: 5

ΣFx = 0:

29

ΣFy = 0: −

1

FAC +

2 29

5

FAC −

FAE +

2 5

5

FAE +

29

(30.36) −

2 29

1 5

(15.09) = 0

2

(30.96) −

5

(15.09) = 0

(3) (4)

Multiply Eq. (3) by 2 and add Eq. (4): 8 29

FAC +

12 29

(30.96) −

4 5

FAC = −28.27 kips,

(15.09) = 0 FAC = 28.3 kips C 

Multiply Eq. (3) by 2, Eq. (4) by 5, and add: −

8 5

FAE +

20 29

(30.96) −

12 5

(15.09) = 0 FAE = 9.50 kips T 



2

ΣFy = 0:


Free body: Joint C: From force triangle: FCE 61

=

FCG 28.27 kips = 8 29

FCE = 41.0 kips T  FCG = 42.0 kips C 

Free body: Joint G: ΣFx = 0: (Checks) 


ΣFy = 0: 42 kips − 42 kips = 0

FEG = 0 


www.elsolucionario.net PROBLEM 6.28 Determine the force in each member of the truss shown. State whether each member is in tension or compression.

SOLUTION ΣFx = 0:

Ex = 0

ΣM F = 0:

E y = 45 kips

ΣFy = 0:

F = 60 kips

Joint D: FCD FDH 15 kips = = 12 13 5 FCD = 36.0 kips T  FDH = 39.0 kips C 

Joint H: ΣF = 0:

FCH = 0 

ΣF = 0:

FGH = 39.0 kips C 

ΣF = 0:

FCG = 0 

ΣF = 0:

FBC = 36.0 kips T 

ΣF = 0:

FBG = 0 

ΣF = 0:

FFG = 39.0 kips C 

ΣF = 0:

FBF = 0 

Joint C:

Joint G:

Joint B:

ΣF = 0:

FAB = 36.0 kips T 



Reactions:


Joint A: AE = 122 + 152 = 19.21 ft tan 38.7° =

36 kips FAF

FAF = 45.0 kips C 

sin 38.7° =

36 kips FAE

FAE = 57.6 kips T 

Joint E: ΣFx = 0: + (57.6 kips) sin 38.7° + FEF = 0 FEF = 36.0 kips C 

ΣFy = 0: (57.6 kips) cos 38.7° − 45 kips = 0 (Checks)



FEF = −36.0 kips


SOLUTION Truss of Problem 6.31a: Starting with triangle HDI and adding two members at a time, we obtain successively joints A, E, J, and B, but cannot go further. Thus, this truss is not a simple truss. 

Truss of Problem 6.32a: Starting with triangle ABC and adding two members at a time, we obtain joints D, E, G, F, and H, but cannot go further. Thus, this truss is not a simple truss. 

Truss of Problem 6.33a: Starting with triangle ABD and adding two members at a time, we obtain successively joints H, G, F, E, I, C, and J, thus completing the truss. Therefore, this is a simple truss. 



Determine whether the trusses of Problems 6.31a, 6.32a, and 6.33a are simple trusses.

www.elsolucionario.net PROBLEM 6.30 Determine whether the trusses of Problems 6.31b, 6.32b, and 6.33b are simple trusses.

Truss of Problem 6.31b: Starting with triangle CGM and adding two members at a time, we obtain successively joints B, L, F, A, K, J, then H, D, N, I, E, O, and P, thus completing the truss. Therefore, this truss is a simple truss. 

Truss of Problem 6.32b: Starting with triangle ABC and adding two members at a time, we obtain successively joints E, D, F, G, and H, but cannot go further. Thus, this truss is not a simple truss. 

Truss of Problem 6.33b: Starting with triangle GFH and adding two members at a time, we obtain successively joints D, E, C, A, and B, thus completing the truss. Therefore, this is a simple truss. 



SOLUTION



For the given loading, determine the zero-force members in each of the two trusses shown.

SOLUTION Truss (a):

FB: Joint B: FBJ = 0 FB: Joint D: FDI = 0 FB: Joint E: FEI = 0 FB: Joint I : FAI = 0 (a)

FB: Joint F : FFK = 0 FB: Joint G: FGK = 0 FB: Joint K : FCK = 0

AI , BJ , CK , DI , EI , FK , GK 

The zero-force members, therefore, are Truss (b):

FB: Joint K : FFK = 0 FB: Joint O : FIO = 0

(b)

The zero-force members, therefore, are

FK and IO 

All other members are either in tension or compression.



SOLUTION Truss (a):

FB: Joint B: FBC = 0 FB: Joint C: FCD = 0 FB: Joint J : FIJ = 0 FB: Joint I : FIL = 0 FB: Joint N : FMN = 0

(a)

FB: Joint M : FLM = 0 BC , CD, IJ , IL, LM , MN 

The zero-force members, therefore, are Truss (b):

FB: Joint C: FBC = 0 FB: Joint B: FBE = 0 FB: Joint G: FFG = 0 FB: Joint F : FEF = 0 FB: Joint E: FDE = 0 FB: Joint I : FIJ = 0

(b)

FB: Joint M : FMN = 0 FB: Joint N : FKN = 0 BC , BE , DE , EF , FG , IJ , KN , MN 






SOLUTION Truss (a):

Note: Reaction at F is vertical ( Fx = 0). Joint G:

ΣF = 0,

FDG = 0 

Joint D:

ΣF = 0,

FDB = 0 

Joint F :

ΣF = 0,

FFG = 0 

Joint G:

ΣF = 0,

FGH = 0 

Joint J :

ΣF = 0,

FIJ = 0 

Joint I :

ΣF = 0,

FHI = 0 





Truss (b):

ΣF = 0,

FAC = 0 

Joint C:

ΣF = 0,

FCE = 0 

Joint E:

ΣF = 0,

FEF = 0 

Joint F :

ΣF = 0,

FFG = 0 

Joint G:

ΣF = 0,

FGH = 0 


Joint A:


www.elsolucionario.net PROBLEM 6.34 Determine the zero-force members in the truss of (a) Problem 6.26, (b) Problem 6.28.

SOLUTION Truss of Problem 6.26: FB : Joint I : FIJ = 0 FB : Joint J : FGJ = 0 FB : Joint G: FGH = 0 GH , GJ , IJ 


(b)

Truss of Problem 6.28: FB : Joint B: FBF = 0 FB : Joint B: FBG = 0 FB : Joint C: FCG = 0 FB : Joint C: FCH = 0 BF , BG, CG, CH 




(a)

www.elsolucionario.net PROBLEM 6.35* The truss shown consists of six members and is supported by a short link at A, two short links at B, and a ball and socket at D. Determine the force in each of the members for the given loading.

SOLUTION Free body: Truss: Dx = Bx

and

Dy = B y

ΣM z = 0: − A(10 ft) − (400 lb)(24 ft) = 0 A = −960 lb ΣFx = 0: Bx + Dx + A = 0 2 Bx − 960 lb = 0, Bx = 480 lb ΣFy = 0: B y + Dy − 400 lb = 0

2 By = 400 lb By = +200 lb B = (480 lb)i + (200 lb) j 

Thus, Free body: C: FCA = FAC FCB = FBC FCD = FCD

 CA FAC = ( −24i + 10 j) CA 26  CB FBC = (−24i + 7k ) CB 25  CD FCD = (−24i − 7k ) CD 25

ΣF = 0: FCA + FCB + FCD − (400 lb) j = 0



From symmetry:


Substituting for FCA , FCB , FCD , and equating to zero the coefficients of i, j, k : 24 24 FAC − ( FBC + FCD ) = 0 26 25

i:

−

j:

10 FAC − 400 lb = 0 26

k:

7 ( FBC − FCD ) = 0 FCD = FBC 25

(1) FAC = 1040 lb T 

Substitute for FAC and FCD in Eq. (1): −

24 24 (10.40 lb) − (2 FBC ) = 0 FBC = −500 lb 26 25

FBC = FCD = 500 lb C 

FBC FBA FBD

 CB = (500 lb) = −(480 lb)i + (140 lb)k CB  BA FAB = FAB = (10 j − 7k ) BA 12.21 = − FBD k

ΣF = 0: FBA + FBD + FBC + (480 lb)i + (200 lb) j = 0

Substituting for FBA , FBD , FBC and equating to zero the coefficients of j and k: j:

10 FAB + 200 lb = 0 FAB = −244.2 lb 12.21

k: −

7 FAB − FBD + 140 lb = 0 12.21 FBD = −

From symmetry:

7 (−244.2 lb) + 140 lb = +280 lb 12.21

FAD = FAB

FAB = 244 lb C 

FBD = 280 lb T  FAD = 244 lb C 



Free body: B:

www.elsolucionario.net PROBLEM 6.36* The truss shown consists of six members and is supported by a ball and socket at B, a short link at C, and two short links at D. Determine the force in each of the members for P = (−2184 N)j and Q = 0.

Free body: Truss: From symmetry: Dx = Bx and Dy = By ΣFx = 0: 2 Bx = 0 Bx = Dx = 0 ΣFz = 0: Bz = 0 ΣM c z = 0: − 2 By (2.8 m) + (2184 N)(2 m) = 0 By = 780 N B = (780 N) j 

Thus, Free body: A:  AB FAB FAB = FAB = ( −0.8i − 4.8 j + 2.1k ) AB 5.30  AC FAC = FAC = FAC (2i − 4.8 j) AC 5.20  AD FAD = FAD = FAD (+0.8i − 4.8 j − 2.1k ) AD 5.30 ΣF = 0: FAB + FAC + FAD − (2184 N) j = 0



SOLUTION


Substituting for FAB , FAC , FAD , and equating to zero the coefficients of i, j, k : i:

−

0.8 2 ( FAB + FAD ) + FAC = 0 5.30 5.20

(1)

j:

−

4.8 4.8 ( FAB + FAD ) − FAC − 2184 N = 0 5.30 5.20

(2)

2.1 ( FAB − FAD ) = 0 5.30

k:

FAD = FAB

Multiply Eq. (1) by –6 and add Eq. (2):  16.8  −  FAC − 2184 N = 0, FAC = −676 N  5.20 

FAC = 676 N C 

 0.8   2  −  2 FAB +   (−676 N) = 0, FAB = −861.25 N  5.30   5.20 

Free body: B:

FAB = FAD = 861 N C 

 AB = −(130 N)i − (780 N) j + (341.25 N)k FAB = (861.25 N) AB  2.8i − 2.1k  FBC = FBC   = FBC (0.8i − 0.6k ) 3.5   FBD = − FBD k ΣF = 0: FAB + FBC + FBD + (780 N) j = 0

Substituting for FAB , FBC , FBD and equating to zero the coefficients of i and k , i: k:

−130 N + 0.8FBC = 0 FBC = +162.5 N 341.25 N − 0.6 FBC − FBD = 0 FBD = 341.25 − 0.6(162.5) = +243.75 N

From symmetry:

FBC = 162.5 N T 

FCD = FBC

FBD = 244 N T  FCD = 162.5 N T 



Substitute for FAC and FAD in Eq. (1):

www.elsolucionario.net PROBLEM 6.37* The truss shown consists of six members and is supported by a ball and socket at B, a short link at C, and two short links at D. Determine the force in each of the members for P = 0 and Q = (2968 N)i.

Free body: Truss: From symmetry: Dx = Bx and Dy = By ΣFx = 0: 2 Bx + 2968 N = 0 Bx = Dx = −1484 N ΣM cz ′ = 0: − 2 By (2.8 m) − (2968 N)(4.8 m) = 0 By = −2544 N B = −(1484 N)i − (2544 N) j 

Thus, Free body: A: FAB = FAB

 AB AB

FAB (−0.8i − 4.8 j + 2.1k ) 5.30  AC FAC = FAC = (2i − 4.8 j) AC 5.20  AD = FAD AD FAD = (−0.8i − 4.8 j − 2.1k ) 5.30 =

FAC FAD

ΣF = 0: FAB + FAC + FAD + (2968 N)i = 0



SOLUTION


Substituting for FAB , FAC , FAD , and equating to zero the coefficients of i, j, k , 0.8 2 FAC + 2968 N = 0 ( FAB + FAD ) + 5.30 5.20

(1)

4.8 4.8 FAC = 0 ( FAB + FAD ) − 5.30 5.20

(2)

i:

−

j:

−

k:

2.1 ( FAB − FAD ) = 0 5.30

FAD = FAB

Multiply Eq. (1) by –6 and add Eq. (2):  16.8  −  FAC − 6(2968 N) = 0, FAC = −5512 N  5.20 

FAC = 5510 N C 

 4.8   4.8  −  2 FAB −   (−5512 N) = 0, FAB = +2809 N  5.30   5.20  FAB = FAD = 2810 N T 

Free body: B: FAB FBC FBD

 BA = (2809 N) = (424 N)i + (2544 N) j − (1113 N)k BA  2.8 i − 2.1k  = FBC   = FBC (0.8i − 0.6k ) 3.5   = − FBD k

ΣF = 0: FAB + FBC + FBD − (1484 N)i − (2544 N) j = 0

Substituting for FAB , FBC , FBD and equating to zero the coefficients of i and k , i:

+24 N + 0.8FBC − 1484 N = 0, FBC = +1325 N

k:

−1113 N − 0.6 FBC − FBD = 0

From symmetry:

FBC = 1325 N T 

FBD = −1113 N − 0.6(1325 N) = −1908 N,

FBD = 1908 N C 

FCD = FBC

FCD = 1325 N T 



Substitute for FAC and FAD in Eq. (2):

www.elsolucionario.net PROBLEM 6.38* The truss shown consists of nine members and is supported by a ball and socket at A, two short links at B, and a short link at C. Determine the force in each of the members for the given loading.

SOLUTION Free body: Truss: Az = Bz = 0 ΣFx = 0: Ax = 0 ΣM BC = 0: Ay (6 ft) + (1600 lb)(7.5 ft) = 0 Ay = −2000 lb

A = −(2000 lb)j 

By = C

From symmetry:

ΣFy = 0: 2 By − 2000 lb − 1600 lb = 0 By = 1800 lb

B = (1800 lb) j 

ΣF = 0: FAB + FAC + FAD − (2000 lb) j = 0

Free body: A:

FAB

i+k

+ FAC

2

i−k 2

+ FAD (0.6 i + 0.8 j) − (2000 lb) j = 0

Factoring i, j, k and equating their coefficient to zero, 1 2

FAB +

1

FAC + 0.6 FAD = 0

(1)

0.8FAD − 2000 lb = 0

FAD = 2500 lb T 

2

1 2

FAB −

1 2

FAC = 0

FAC = FAB



From symmetry:


Substitute for FAD and FAC into Eq. (1): 2 2

Free body: B:

FAB + 0.6(2500 lb) = 0, FAB = −1060.7 lb,

FBA FBC

FAB = FAC = 1061 lb C 

 BA i+k = FAB = + (1060.7 lb) = (750 lb)(i + k ) BA 2 = − FBC k

FBD = FBD (0.8 j − 0.6k )  BE FBE (7.5i + 8 j − 6k ) FBE = FBE = BE 12.5 ΣF = 0: FBA + FBC + FBD + FBE + (1800 lb) j = 0

i:

 7.5  750 lb +   FBE = 0, FBE = −1250 lb  12.5 

FBE = 1250 lb C 

j:

 8  0.8 FBD +   (−1250 lb) + 1800 lb = 0  12.5 

FBD = 1250 lb C 

k:

750 lb − FBC − 0.6( −1250 lb) −

6 ( −1250 lb) = 0 12.5 FBC = 2100 lb T  FBD = FCD = 1250 lb C 

From symmetry: Free body: D: ΣF = 0: FDA + FDB + FDC + FDE i = 0

We now substitute for FDA , FDB , FDC and equate to zero the coefficient of i. Only FDA contains i and its coefficient is −0.6 FAD = −0.6(2500 lb) = −1500 lb

i:

−1500 lb + FDE = 0

FDE = 1500 lb T 



Substituting for FBA , FBC , FBD , and FBE and equating to zero the coefficients of i, j, k ,

www.elsolucionario.net PROBLEM 6.39* The truss shown consists of nine members and is supported by a ball and socket at B, a short link at C, and two short links at D. (a) Check that this truss is a simple truss, that it is completely constrained, and that the reactions at its supports are statically determinate. (b) Determine the force in each member for P = (−1200 N)j and Q = 0.

SOLUTION Free body: Truss: + (0.6i − 0.75k ) × (−1200 j) = 0 −1.8 Ck + 1.8 D y k − 1.8 Dz j + 3D y i − 720k − 900i = 0

Equate to zero the coefficients of i, j, k : i:

3Dy − 900 = 0, D y = 300 N

j:

Dz = 0,

D = (300 N) j 

k:

1.8C + 1.8(300) − 720 = 0

C = (100 N) j 

ΣF = 0: B + 300 j + 100 j − 1200 j = 0

B = (800 N) j 

Free body: B: ΣF = 0: FBA + FBC + FBE + (800 N) j = 0, with  BA FAB (0.6i + 3j − 0.75k ) FBA = FAB = BA 3.15 FBC = FBC i

FBE = − FBE k

Substitute and equate to zero the coefficient of j, i, k : j:

 3    FAB + 800 N = 0, FAB = −840 N,  3.315 

i:

 0.6    (−840 N) + FBC = 0  3.15 

FBC = 160.0 N T 

k:

 0.75  −  (−840 N) − FBE = 0  3.15 

FBE = 200 N T 

FAB = 840 N C 



ΣM B = 0: 1.8i × Cj + (1.8i − 3k ) × ( D y j + Dk )


Free body: C: ΣF = 0: FCA + FCB + FCD + FCE + (100 N) j = 0, with  F CA FCA = FAC = AC (−1.2i + 3j − 0.75 k ) CA 3.317

FCB = −(160 N) i FCD = − FCD k FCE = FCE

 F CE = CE (−1.8i − 3k ) CE 3, 499

Substitute and equate to zero the coefficient of j, i, k :  3    FAC + 100 N = 0, FAC = −110.57 N  3.317  −

i: k:

−

FAC = 110.6 N C 

1.2 1.8 FCE = 0, FCE = −233.3 (−110.57) − 160 − 3.317 3.499

FCE = 233 N C 

0.75 3 (−110.57) − FCD − (−233.3) = 0 3.317 3.499

FCD = 225 N T 

Free body: D: ΣF = 0: FDA + FDC + FDE + (300 N) j = 0, with  F DA FDA = FAD = AD (−1.2i + 3j + 2.25k ) DA 3.937 FDC = FCD k = (225 N)k

FDE = − FDE i

Substitute and equate to zero the coefficient of j, i, k : j:

 3    FAD + 300 N = 0,  3.937 

i:

 1.2  −  ( −393.7 N) − FDE = 0  3.937 

k:

 2.25    (−393.7 N) + 225 N = 0 (Checks)  3.937 

FAD = −393.7 N,

FAD = 394 N C  FDE = 120.0 N T 

Free body: E: Member AE is the only member at E which does not lie in the xz plane. Therefore, it is a zero-force member. FAE = 0 



j:

www.elsolucionario.net PROBLEM 6.40* Solve Problem 6.39 for P = 0 and Q = (−900 N)k. PROBLEM 6.39* The truss shown consists of nine members and is supported by a ball and socket at B, a short link at C, and two short links at D. (a) Check that this truss is a simple truss, that it is completely constrained, and that the reactions at its supports are statically determinate. (b) Determine the force in each member for P = (−1200 N)j and Q = 0.

SOLUTION

ΣM B = 0: 1.8i × Cj + (1.8i − 3k ) × ( D y j + Dz k ) + (0.6i + 3j − 0.75k ) × (−900N)k = 0 1.8 Ck + 1.8D y k − 1.8Dz j + 3D y i + 540 j − 2700i = 0

Equate to zero the coefficient of i, j, k : 3Dy − 2700 = 0

Dy = 900 N

−1.8Dz + 540 = 0

Dz = 300 N

1.8C + 1.8 Dy = 0

Thus,

C = − D y = −900 N

C = −(900 N) j D = (900 N) j + (300 N)k ΣF = 0: B − 900 j + 900 j + 300k − 900k = 0

 B = (600 N)k 

Free body: B:

Since B is aligned with member BE, FAB = FBC = 0,

FBE = 600 N T 

Free body: C: 

ΣF = 0: FCA + FCD + FCE − (900 N) j = 0, with



Free body: Truss:




FCA FCD

 F CA = FAC = AC ( −1.2i + 3j − 0.75k )  CA 3.317  F CE = − FCD k FCE = FCE = CE (−1.8i − 3k ) CE 3.499

Substitute and equate to zero the coefficient of j, i, k:  3    FAC − 900 N = 0,  3.317 

FAC = 995.1 N

FAC = 995 N T 

FCE = −699.8 N

FCE = 700 N C 

i:

−

1.2 1.8 FCE = 0, (995.1) − 3.317 3.499

k:

−

0.75 3 (995.1) − FCD − ( −699.8) = 0 3.317 3.499

FCD = 375 N T 

Free body: D: ΣF = 0: FDA + FDE + (375 N)k +(900 N)j + (300 N) k = 0  F DA FDA = FAD = AD (−1.2i + 3j + 2.25k ) DA 3.937

with

FDE = − FDE i

and

Substitute and equate to zero the coefficient j, i, k: j:

 3    FAD + 900 N = 0, FAD = −1181.1 N  3.937 

i:

 1.2  −  (−1181.1 N) − FDE = 0  3.937 

k:

FAD = 1181 N C  FDE = 360 N T 

 2.25    (−1181.1 N + 375 N + 300 N = 0)  3.937 

(Checks)

Free body: E: Member AE is the only member at E which does not lie in the xz plane. Therefore, it is a zero-force member. FAE = 0 



j:

www.elsolucionario.net PROBLEM 6.41* The truss shown consists of 18 members and is supported by a ball and socket at A, two short links at B, and one short link at G. (a) Check that this truss is a simple truss, that it is completely constrained, and that the reactions at its supports are statically determinate. (b) For the given loading, determine the force in each of the six members joined at E.

SOLUTION Check simple truss.


(a)

(1) Start with tetrahedron BEFG. (2) Add members BD, ED, GD joining at D. (3) Add members BA, DA, EA joining at A. (4) Add members DH, EH, GH joining at H. (5) Add members BC, DC, GC joining at C. Truss has been completed: It is a simple truss. Free body: Truss: Check constraints and reactions. Six unknown reactions—ok; however, supports at A and B constrain truss to rotate about AB and support at G prevents such a rotation. Thus, Truss is completely constrained and reactions are statically determinate. Determination of reactions: ΣM A = 0: 11i × ( By j + Bz k ) + (11i − 9.6k ) × G j + (10.08 j − 9.6k ) × (275i + 240k ) = 0 11By k − 11By j + 11Gk + 9.6Gi − (10.08)(275)k + (10.08)(240)i − (9.6)(275) j = 0

Equate to zero the coefficient of i, j, k: i : 9.6G + (10.08)(240) = 0 G = −252 lb

G = (−252 lb) j 

j: − 11Bz − (9.6)(275) = 0 Bz = −240 lb k : 11By + 11(−252) − (10.08)(275) = 0, By = 504 lb

B = (504 lb) j − (240 lb)k 



ΣF = 0: A + (504 lb) j − (240 lb)k − (252 lb) j + (275 lb)i + (240 lb)k = 0 A = −(275 lb)i − (252 lb) j 

Zero-force members. The determination of these members will facilitate our solution. FB: C: Writing

ΣFx = 0, ΣFy = 0, ΣFz = 0

yields FBC = FCD = FCG = 0 

FB: F: Writing

ΣFx = 0, ΣFy = 0, ΣFz = 0

yields FBF = FEF = FFG = 0 

FB: H: Writing

Az = 0, writing ΣFz = 0 ΣFy = 0

yields FAD = 0  yields FDH = 0 

FB: D: Since FAD = FCD = FDH = 0, we need consider only members DB, DE, and DG. Since FDE is the only force not contained in plane BDG, it must be zero. Simple reasonings show that the other two forces are also zero. FBD = FDE = FDG = 0 

The results obtained for the reactions at the supports and for the zero-force members are shown on the figure below. Zero-force members are indicated by a zero (“0”).

(b)

Force in each of the members joined at E. FDE = FEF = 0 

We already found that Free body: A:

ΣFy = 0

yields FAE = 252 lb T 

Free body: H:

ΣFz = 0

yields FEH = 240 lb C 



FB: A: Since


ΣF = 0: FEB + FEG + (240 lb)k − (252 lb) j = 0

Free body: E:

F FBE (11i − 10.08 j) + EG (11i − 9.6k ) + 240k − 252 j = 0 14.92 14.6

Equate to zero the coefficient of y and k:  10.08  j: −   FBE − 252 = 0  14.92 

FBE = 373 lb C 

 9.6  −  FEG + 240 = 0  14.6 

FEG = 365 lb T 


k:


www.elsolucionario.net PROBLEM 6.42* The truss shown consists of 18 members and is supported by a ball and socket at A, two short links at B, and one short link at G. (a) Check that this truss is a simple truss, that it is completely constrained, and that the reactions at its supports are statically determinate. (b) For the given loading, determine the force in each of the six members joined at G.

SOLUTION

(b)

Force in each of the members joined at G. We already know that FCG = FDG = FFG = 0 

Free body: H: Free body: G:

ΣFx = 0

yields FGH = 275 lb C 

ΣF = 0: FGB + FGE + (275 lb)i − (252 lb) j = 0

FBG F (−10.08 j + 9.6k ) + EG (−11i + 9.6k ) + 275i − 252 j = 0 13.92 14.6

Equate to zero the coefficient of i, j, k:  11  i: −   FEG + 275 = 0  14.6 

FEG = 365 lb T 

 10.08  j: −   FBG − 252 = 0  13.92 

FBG = 348 lb C 

 9.6   9.6  k:   (−348) +   (365) = 0  13.92   14.6 

(Checks)



See solution to Problem 6.41 for part (a) and for reactions and zero-force members.

www.elsolucionario.net PROBLEM 6.43 Determine the force in members CD and DF of the truss shown.

SOLUTION

ΣM J = 0: (12 kN)(4.8 m) + (12 kN)(2.4 m) − B(9.6 m) = 0 B = 9.00 kN ΣFy = 0: 9.00 kN − 12.00 kN − 12.00 kN + J = 0 J = 15.00 kN

Member CD: ΣFy = 0: 9.00 kN + FCD = 0

FCD = 9.00 kN C 

Member DF: ΣM C = 0: FDF (1.8 m) − (9.00 kN)(2.4 m) = 0

FDF = 12.00 kN T 



Reactions:

www.elsolucionario.net PROBLEM 6.44 Determine the force in members FG and FH of the truss shown.

SOLUTION

ΣM J = 0: (12 kN)(4.8 m) + (12 kN)(2.4 m) − B (9.6 m) = 0 B = 9.00 kN ΣFy = 0: 9.00 kN − 12.00 kN − 12.00 kN + J = 0 J = 15.00 kN

Member FG:

3 ΣFy = 0: − FFG − 12.00 kN + 15.00 kN = 0 5 FFG = 5.00 kN T 

Member FH: ΣM G = 0: (15.00 kN)(2.4 m) − FFH (1.8 m) = 0 FFH = 20.0 kN T 



Reactions:

www.elsolucionario.net PROBLEM 6.45 A Warren bridge truss is loaded as shown. Determine the force in members CE, DE, and DF.

SOLUTION Free body: Truss: ΣFx = 0: k x = 0 ΣM A = 0: k y (62.5 ft) − (6000 lb)(12.5 ft)

k = k y = 3600 lb  ΣFy = 0: A + 3600 lb − 6000 lb − 6000 lb = 0

A = 8400 lb 

We pass a section through members CE, DE, and DF and use the free body shown. ΣM D = 0: FCE (15 ft) − (8400 lb)(18.75 ft) + (6000 lb)(6.25 ft) = 0 FCE = +8000 lb ΣFy = 0: 8400 lb − 6000 lb −

FCE = 8000 lb T  15 FDE = 0 16.25

FDE = +2600 lb

FDE = 2600 lb T 

ΣM E = 0: 6000 lb (12.5 ft) − (8400 lb)(25 ft) − FDF (15 ft) = 0 FDF = −9000 lb

FDF = 9000 lb C 



− (6000 lb)(25 ft) = 0

www.elsolucionario.net PROBLEM 6.46 A Warren bridge truss is loaded as shown. Determine the force in members EG, FG, and FH.

SOLUTION See solution of Problem 6.45 for free-body diagram of truss and determination of reactions: A = 8400 lb

k = 3600 lb 

We pass a section through members EG, FG, and FH, and use the free body shown.

FEG = +7500 lb ΣFy = 0:

FEG = 7500 lb T 

15 FFG + 3600 lb = 0 16.25 FFG = −3900 lb

FFG = 3900 lb C 

ΣM G = 0: FFH (15 ft) + (3600 lb)(25 ft) = 0 FFH = −6000 lb

FFH = 6000 lb C 



ΣM F = 0: (3600 lb)(31.25 ft) − FEG (15 ft) = 0

www.elsolucionario.net PROBLEM 6.47 Determine the force in members DF, EF, and EG of the truss shown.

tan β =

3 4

Reactions: A=N=0

Member DF:

3 ΣM E = 0: + (16 kN)(6 m) − FDF (4 m) = 0 5 FDF = + 40 kN

Member EF:

+ ΣF = 0: (16 kN) sin β − FEF cos β = 0 FEF = 16 tan β = 16(0.75) = 12 kN

Member EG:

FDF = 40.0 kN T 

ΣM F = 0: (16 kN)(9 m) +

FEF = 12.00 kN T 

4 FEG (3 m) = 0 5 FEG = −60 kN

FEG = 60.0 kN C 



SOLUTION

www.elsolucionario.net PROBLEM 6.48 Determine the force in members GI, GJ, and HI of the truss shown.

SOLUTION

Reactions:

Member GI:

+ ΣF = 0: (16 kN)sin β + FGI sin β = 0 FGI = −16 kN

Member GJ:

ΣM I = 0: − (16 kN)(9 m) −

4 FGJ (3 m) = 0 5 FGJ = −60 kN

Member HI:

ΣM G = 0: − (16 kN)(9 m) +

FGI = 16.00 kN C 

FGJ = 60.0 kN C 

3 FHI (4 m) = 0 5 FHI = +60 kN

FHI = 60.0 kN T 



A=N=0

www.elsolucionario.net PROBLEM 6.49 Determine the force in members AD, CD, and CE of the truss shown.

SOLUTION

ΣM k = 0: 36(2.4) − B(13.5) + 20(9) + 20(4.5) = 0 ΣFx = 0: −36 + K x = 0

B = 26.4 kN

K x = 36 kN

ΣFy = 0: 26.4 − 20 − 20 + K y = 0 K y = 13.6 kN

ΣM C = 0: 36(1.2) − 26.4(2.25) − FAD (1.2) = 0 FAD = −13.5 kN

 8  ΣM A = 0:  FCD  (4.5) = 0 17  

FAD = 13.5 kN C  FCD = 0 

 15  ΣM D = 0:  FCE  (2.4) − 26.4(4.5) = 0  17  FCE = +56.1 kN

FCE = 56.1 kN T 



Reactions:

www.elsolucionario.net PROBLEM 6.50 Determine the force in members DG, FG, and FH of the truss shown.

SOLUTION See the solution to Problem 6.49 for free-body diagram and analysis to determine the reactions at the supports B and K. K x = 36.0 kN

;

K y = 13.60 kN

ΣM F = 0: 36(1.2) − 26.4(6.75) + 20(2.25) − FDG (1.2) = 0 FDG = −75 kN

FDG = 75.0 kN C 

 8  ΣM D = 0: − 26.4(4.5) +  FFG  (4.5) = 0 17   FFG = +56.1 kN



 15 ΣM G = 0: 20(4.5) − 26.4(9) +  FFH  17

  (2.4) = 0   FFH = +69.7 kN 



FFG = 56.1 kN T 

FFH = 69.7 kN T 



B = 26.4 kN ;

www.elsolucionario.net PROBLEM 6.51 A stadium roof truss is loaded as shown. Determine the force in members AB, AG, and FG.

We pass a section through members AB, AG, and FG, and use the free body shown.

ΣM G = 0:

 40   41 FAB  (6.3 ft) − (1.8 kips)(14 ft)   − (0.9 kips)(28 ft) = 0 FAB = 8.20 kips T 

FAB = +8.20 kips

3  ΣM D = 0: −  FAG  (28 ft) + (1.8 kips)(28 ft) 5  + (1.8 kips)(14 ft) = 0 FAG = 4.50 kips T 

FAG = +4.50 kips ΣM A = 0: − FFG (9 ft) − (1.8 kips)(12 ft) − (1.8 kips)(26 ft) − (0.9 kips)(40 ft) = 0

FFG = 11.60 kips C 

FFG = −11.60 kips



SOLUTION

www.elsolucionario.net PROBLEM 6.52 A stadium roof truss is loaded as shown. Determine the force in members AE, EF, and FJ.

We pass a section through members AE, EF, and FJ, and use the free body shown.

  8 ΣM F = 0:  FAE  (9 ft) − (1.8 kips)(12 ft) − (1.8 kips)(26 ft) − (0.9 kips)(40 ft) = 0  82 + 92    FAE = +17.46 kips

FAE = 17.46 kips T 

ΣM A = 0: − FEF (9 ft) − (1.8 kips)(12 ft) − (1.8 kips)(26 ft) − (0.9 kips)(40 ft) = 0 FEF = −11.60 kips

FEF = 11.60 kips C 

ΣM E = 0: − FFJ (8 ft) − (0.9 kips)(8 ft) − (1.8 kips)(20 ft) − (1.8 kips)(34 ft) − (0.9 kips)(48 ft) = 0 FFJ = −18.45 kips

FFJ = 18.45 kips C 



SOLUTION

www.elsolucionario.net PROBLEM 6.53 Determine the force in members CD and DF of the truss shown.

tan α =

5 α = 22.62° 12

sin α =

5 12 cos α = 13 13

Member CD: ΣM I = 0: FCD (9 m) + (10 kN)(9 m) + (10 kN)(6 m) + (10 kN)(3 m) = 0 FCD = 20.0 kN C 

FCD = −20 kN

Member DF: ΣM C = 0: ( FDF cos α )(3.75 m) + (10 kN)(3 m) + (10 kN)(6 m) + (10 kN)(9 m) = 0 FDF cos α = −48 kN

 12  FDF   = −48 kN  13 

FDF = −52.0 kN

FDF = 52.0 kN C 



SOLUTION

www.elsolucionario.net PROBLEM 6.54 Determine the force in members CE and EF of the truss shown.

Member CE: ΣM F = 0: FCE (2.5 m) − (10 kN)(3 m) − (10 kN)(6 m) = 0 FCE = 36.0 kN T 

FCE = +36 kN

Member EF: ΣM I = 0: FEF (6 m) + (10 kN)(6 m) + (10 kN)(3 m) = 0 FEF = 15.00 kN C 

FEF = −15 kN



SOLUTION

www.elsolucionario.net PROBLEM 6.55 The truss shown was designed to support the roof of a food market. For the given loading, determine the force in members FG, EG, and EH.

SOLUTION Reactions at supports. Because of the symmetry of the loading, Ay = O =

1 total load 2

A = O = 4.48 kN 

We pass a section through members FG, EG, and EH, and use the free body shown. Slope FG = Slope FI = Slope EG =

1.75 m 6m

5.50 m 2.4 m

ΣM E = 0: (0.6 kN)(7.44 m) + (1.24 kN)(3.84 m)

− (4.48 kN)(7.44 m)  6  − FFG  (4.80 m) = 0 6.25   FFG = −5.231 kN

FFG = 5.23 kN C 

ΣM G = 0: FEH (5.50 m) + (0.6 kN)(9.84 m)

+ (1.24 kN)(6.24 m) + (1.04 kN)(2.4 m) −(4.48 kN)(9.84 m) = 0

ΣFy = 0:

FEH = 5.08 kN T 

5.50 1.75 (−5.231 kN) + 4.48 kN − 0.6 kN − 1.24 kN − 1.04 kN = 0 FEG + 6.001 6.25 FEG = −0.1476 kN

FEG = 0.1476 kN C 



Ax = 0,

www.elsolucionario.net PROBLEM 6.56 The truss shown was designed to support the roof of a food market. For the given loading, determine the force in members KM, LM, and LN.

SOLUTION Because of symmetry of loading,

O=

1 load 2

O = 4.48 kN 

 3.84  ΣM M = 0:  FLN  (3.68 m)  4  + (4.48 kN − 0.6 kN)(3.6 m) = 0 FLN = −3.954 kN

FLN = 3.95 kN C 

ΣM L = 0: − FKM (4.80 m) − (1.24 kN)(3.84 m) + (4.48 kN − 0.6 kN)(7.44 m) = 0 FKM = +5.022 kN

ΣFy = 0:

FKM = 5.02 kN T 

4.80 1.12 (−3.954 kN) − 1.24 kN − 0.6 kN + 4.48 kN = 0 FLM + 6.147 4 FLM = −1.963 kN

FLM = 1.963 kN C 



We pass a section through KM, LM, LN, and use free body shown.

www.elsolucionario.net PROBLEM 6.57 A Polynesian, or duopitch, roof truss is loaded as shown. Determine the force in members DF, EF, and EG.

SOLUTION

ΣFx = 0: N x = 0 ΣM N = 0: (200 lb)(8a ) + (400 lb)(7a + 6a + 5a )+(350 lb)(4a ) + (300 lb)(3a + 2a + a ) − A(8a ) = 0

A = 1500 lb 

ΣFy = 0: 1500 lb − 200 lb − 3(400 lb) − 350 lb − 3(300 lb) − 150 lb + N y = 0 N y = 1300 lb N = 1300 lb  We pass a section through DF, EF, and EG, and use the free body shown. (We apply FDF at F.) ΣM E = 0: (200 lb)(18 ft) + (400 lb)(12 ft) + (400 lb)(6 ft) − (1500 lb)(18 ft)   18 FDF  (4.5 ft) = 0 −  182 + 4.52    FDF = −3711 lb

FDF = 3710 lb C 

ΣM A = 0: FEF (18 ft) − (400 lb)(6 ft) − (400 lb)(12 ft) = 0 FEF = 400 lb T 

FEF = +400 lb

ΣM F = 0: FEG (4.5 ft) − (1500 lb)(18 ft)+(200 lb)(18 ft) + (400 lb)(12 ft) + (400 lb)(6 ft) = 0 FEG = +3600 lb

FEG = 3600 lb T 



Free body: Truss:

www.elsolucionario.net PROBLEM 6.58 A Polynesian, or duopitch, roof truss is loaded as shown. Determine the force in members HI, GI, and GJ.

SOLUTION See solution of Problem 6.57 for reactions:

A = 1500 lb ,

N = 1300 lb 

(We apply FHI at H.)

  6 ΣM G = 0:  FHI  (8.5 ft) + (1300 lb)(24 ft) − (300 lb)(6 ft)  2  2  6 +4  − (300 lb)(12 ft) − (300 lb)(18 ft) − (150 lb)(24 ft) = 0 FHI = −2375.4 lb

FHI = 2375 lb C 

ΣM I = 0: (1300 lb)(18 ft) − (300 lb)(6 ft) − (300 lb)(12 ft) − (150 lb)(18 ft) − FGJ (4.5 ft) = 0 FGJ = +3400 lb

ΣFx = 0: −

FGJ = 3400 lb T 

4 6 FGI − (−2375.4 lb) − 3400 lb = 0 2 5 6 + 42 FGI = −1779.4 lb

FGI = 1779 lb C 



We pass a section through HI, GI, and GJ, and use the free body shown.

www.elsolucionario.net PROBLEM 6.59 Determine the force in members DE and DF of the truss shown when P = 20 kips.

SOLUTION Reactions: C = K = 2.5P

7.5 18 β = 22.62°

Member DE: ΣM A = 0: (2.5 P )(6 ft) − FDE (12 ft) = 0

FDE = +1.25 P

For P = 20 kips,

FDE = +1.25(20) = +25 kips


Member DF: ΣM E = 0: P (12 ft) − (2.5 P )(6 ft) − FDF cos β (5 ft) = 0 12 P − 15 P − FDF cos 22.62°(5 ft) = 0

FDF = −0.65 P

For P = 20 kips,

FDF = −0.65(20) = −13 kips FDF = 13.00 kips C 



tan β =

www.elsolucionario.net PROBLEM 6.60 Determine the force in members EG and EF of the truss shown when P = 20 kips.

SOLUTION

Reactions:

Member EG: ΣM F = 0: P (18 ft) − 2.5 P (12 ft) − P (6 ft) + FEG (7.5 ft) = 0

FEG = +0.8 P;

For P = 20 kips,


FEG = 0.8(20) = +16 kips

Member EF: ΣM A = 0: 2.5 P (6 ft) − P (12 ft) + FEF sin 51.34°(12 ft) = 0

FEF = −0.320 P;

For P = 20 kips,

FEF = −0.320(20) = −6.4 kips FEF = 6.40 kips C 



C = K = 2.5P 7.5 tan α = 6 α = 51.34°

www.elsolucionario.net PROBLEM 6.61 Determine the force in members EH and GI of the truss shown. (Hint: Use section aa.)

SOLUTION

ΣFx = 0: Ax = 0

ΣM P = 0: 12(45) + 12(30) + 12(15) − Ay (90) = 0 A y = 12 kips ΣFy = 0: 12 − 12 − 12 − 12 + P = 0

P = 24 kips

ΣM G = 0: − (12 kips)(30 ft) − FEH (16 ft) = 0

FEH = −22.5 kips ΣFx = 0: FGI − 22.5 kips = 0

FEH = 22.5 kips C  FGI = 22.5 kips T 



Reactions:

www.elsolucionario.net PROBLEM 6.62 Determine the force in members HJ and IL of the truss shown. (Hint: Use section bb.)

SOLUTION See the solution to Problem 6.61 for free body diagram and analysis to determine the reactions at supports A and P.

A x = 0; A y = 12.00 kips ;

P = 24.0 kips

FHJ = −33.75 kips

FHJ = 33.8 kips C 

ΣFx = 0: 33.75 kips − FIL = 0

FIL = +33.75 kips

FIL = 33.8 kips T 



ΣM L = 0: FHJ (16 ft) − (12 kips)(15 ft) + (24 kips)(30 ft) = 0

www.elsolucionario.net PROBLEM 6.63 Determine the force in members DG and FI of the truss shown. (Hint: Use section aa.)

ΣM F = 0: FDG (4 m) − (5 kN)(3 m) = 0

FDG = +3.75 kN

FDG = 3.75 kN T 

ΣFy = 0: − 3.75 kN − FFI = 0 FFI = −3.75 kN

FFI = 3.75 kN C 



SOLUTION

www.elsolucionario.net PROBLEM 6.64 Determine the force in members GJ and IK of the truss shown. (Hint: Use section bb.)

ΣM I = 0: FGJ (4 m) − (5 kN)(6 m) − (5 kN)(3 m) = 0

FGJ = +11.25 kN

FGJ = 11.25 kN T 

ΣFy = 0: − 11.25 kN − FIK = 0 FIK = −11.25 kN

FIK = 11.25 kN C 



SOLUTION

www.elsolucionario.net PROBLEM 6.65 The diagonal members in the center panels of the truss shown are very slender and can act only in tension; such members are known as counters. Determine the forces in the counters that are acting under the given loading.

SOLUTION

ΣFx = 0: Fx = 0

ΣM H = 0: 4.8(3a) + 4.8(2a) + 4.8a − 2.4a − Fy (2a) = 0 Fy = +13.20 kips

F = 13.20 kips 

ΣFy = 0: H + 13.20 kips − 3(4.8 kips) − 2(2.4 kips) = 0 H = +6.00 kips

H = 6.00 kips 

Free body: ABF: We assume that counter BG is acting. ΣFy = 0: −

9.6 FBG + 13.20 − 2(4.8) = 0 14.6 FBG = +5.475

FBG = 5.48 kips T 

Since BG is in tension, our assumption was correct. Free body: DEH: We assume that counter DG is acting. ΣFy = 0: −

9.6 FDG + 6.00 − 2(2.4) = 0 14.6 FDG = 1.825 kips T 

FDG = +1.825

Since DG is in tension, O.K.



Free body: Truss:

www.elsolucionario.net PROBLEM 6.66 The diagonal members in the center panels of the truss shown are very slender and can act only in tension; such members are known as counters. Determine the forces in the counters that are acting under the given loading.

SOLUTION

ΣFx = 0: Fx = 0

ΣM G = 0: − Fy a + 4.8(2a) + 4.8a − 2.4a − 2.4(2a) = 0 Fy = 7.20

F = 7.20 kips 

Free body: ABF: We assume that counter CF is acting. ΣFy = 0:

9.6 FCF + 7.20 − 2(4.8) = 0 14.6 FCF = +3.65

FCF = 3.65 kips T 

Since CF is in tension, O.K. Free body: DEH: We assume that counter CH is acting. ΣFy = 0:

9.6 FCH − 2(2.4 kips) = 0 14.6 FCH = +7.30

FCH = 7.30 kips T 

Since CH is in tension, O.K.



Free body: Truss:

www.elsolucionario.net PROBLEM 6.67 The diagonal members in the center panels of the power transmission line tower shown are very slender and can act only in tension; such members are known as counters. For the given loading, determine (a) which of the two counters listed below is acting, (b) the force in that counter.

SOLUTION Free body: Portion ABDFEC of tower. We assume that counter CJ is acting and show the forces exerted by that counter and by members CH and EJ.

ΣFx = 0:

4 FCJ − 2(1.2 kN)sin 20° = 0 FCJ = +1.026 kN 5

Since CJ is found to be in tension, our assumption was correct. Thus, the answers are (a) CJ



(b) 1.026 kN T 



Counters CJ and HE

www.elsolucionario.net PROBLEM 6.68 The diagonal members in the center panels of the power transmission line tower shown are very slender and can act only in tension; such members are known as counters. For the given loading, determine (a) which of the two counters listed below is acting, (b) the force in that counter.

SOLUTION Free body: Portion of tower shown.

We assume that counter IO is acting and show the forces exerted by that counter and by members IN and KO. ΣFx = 0 :

4 FIO − 4(1.2 kN)sin 20° = 0 FIO = +2.05 kN 5

Since IO is found to be in tension, our assumption was correct. Thus, the answers are (a) IO



(b) 2.05 kN T 



Counters IO and KN

www.elsolucionario.net PROBLEM 6.69 Classify each of the structures shown as completely, partially, or improperly constrained; if completely constrained, further classify as determinate or indeterminate. (All members can act both in tension and in compression.)

SOLUTION Structure (a) m = 16 Number of joints: n = 10 Reaction components: r=4 m + r = 20, 2 n = 20

(a)

m + r = 2n 

Thus,

To determine whether the structure is actually completely constrained and determinate, we must try to find the reactions at the supports. We divide the structure into two simple trusses and draw the free-body diagram of each truss.

This is a properly supported simple truss – O.K.

This is an improperly supported simple truss. (Reaction at C passes through B. Thus, Eq. ΣM B = 0 cannot be satisfied.) Structure is improperly constrained. 

Structure (b) m = 16 n = 10 r=4 m + r = 20, 2n = 20

(b)

m + r = 2n 

Thus,



Number of members:


We must again try to find the reactions at the supports dividing the structure as shown.

Both portions are simply supported simple trusses. Structure is completely constrained and determinate. 

Structure (c) m = 17 n = 10 r=4 m + r = 21, 2n = 20

m + r > 2n 

Thus,

This is a simple truss with an extra support which causes reactions (and forces in members) to be indeterminate. Structure is completely constrained and indeterminate. 



(c)


SOLUTION Structure (a):


Nonsimple truss with r = 4, m = 16, n = 10, so m + r = 20 = 2n, but we must examine further.

FBD Sections:

FBD

I:

ΣM A = 0 

T1

II:

ΣFx = 0 

T2

I:

ΣFx = 0



Ax

I:

ΣFy = 0



Ay

II:

ΣM E = 0 

Cy

II:

ΣFy = 0 

Ey

Since each section is a simple truss with reactions determined, structure is completely constrained and determinate. 

Structure (b): Nonsimple truss with r = 3, m = 16, n = 10, so

m + r = 19 < 2n = 20

Structure is partially constrained. 



Structure (c):




Simple truss with r = 3, m = 17, n = 10, m + r = 20 = 2n, but the horizontal reaction forces Ax and E x are collinear and no equilibrium equation will resolve them, so the structure is improperly constrained and indeterminate.



SOLUTION


Structure (a):

Nonsimple truss with r = 4, m = 12, n = 8 so r + m = 16 = 2n. Check for determinacy:

One can solve joint F for forces in EF, FG and then solve joint E for E y and force in DE. This leaves a simple truss ABCDGH with r = 3, m = 9, n = 6 so r + m = 12 = 2n

Structure is completely constrained and determinate. 

Structure (b):

Simple truss (start with ABC and add joints alphabetically to complete truss) with r = 4, m = 13, n = 8 so

r + m = 17 > 2n = 16

Constrained but indeterminate 



Structure (c):

Nonsimple truss with r = 3, m = 13, n = 8 so r + m = 16 = 2n. To further examine, follow procedure in part (a) above to get truss at left. Since F1 ≠ 0 (from solution of joint F), ΣM A = aF1

≠ 0 and there is no equilibrium.


Structure is improperly constrained. 



Structure (a) Number of members: m = 12 Number of joints: n=8 Reaction components:

Thus,

r =3 m + r = 15,

2 n = 16

m + r < 2n

 Structure is partially constrained. 

Structure (b) m = 13, n = 8 r =3 m + r = 16, 2n = 16

Thus,

m + r = 2n



To verify that the structure is actually completely constrained and determinate, we observe that it is a simple truss (follow lettering to check this) and that it is simply supported by a pin-and-bracket and a roller. Thus, structure is completely constrained and determinate. 



SOLUTION


Structure (c) m = 13, n = 8 r=4 m + r = 17, 2n = 16

Thus,

m + r > 2n



Structure is completely constrained and indeterminate. 


This result can be verified by observing that the structure is a simple truss (follow lettering to check this), therefore it is rigid, and that its supports involve four unknowns.



Structure (a):

Rigid truss with r = 3, m = 14, n = 8, so

r + m = 17 > 2n = 16 so completely constrained but indeterminate 

Structure (b):

Simple truss (start with ABC and add joints alphabetically), with r = 3, m = 13, n = 8, so r + m = 16 = 2n

so completely constrained and determinate 

Structure (c):

Simple truss with r = 3, m = 13, n = 8, so r + m = 16 = 2n, but horizontal reactions ( Ax and Dx ) are collinear, so cannot be resolved by any equilibrium equation. Structure is improperly constrained. 



SOLUTION


SOLUTION Structure (a): No. of joints

n = 8 m + r = 16 = 2n

No. of reaction components

r = 4 unknows = equations


m = 12

No. of members

FBD of EH:

ΣM H = 0

FDE ; ΣFx = 0

FGH ; ΣFy = 0

Hy

Then ABCDGF is a simple truss and all forces can be determined. This example is completely constrained and determinate. 

Structure (b): No. of members

m = 12

No. of joints

n = 8 m + r = 15 < 2n = 16


r = 3 unknows < equations

partially constrained  Note: Quadrilateral DEHG can collapse with joint D moving downward; in (a), the roller at F prevents this action. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 854


Structure (c): No. of members

m = 13

No. of joints

n = 8 m + r = 17 > 2n = 16


r = 4 unknows > equations


completely constrained but indeterminate 


www.elsolucionario.net PROBLEM 6.75 Determine the force in member BD and the components of the reaction at C.

SOLUTION We note that BD is a two-force member. The force it exerts on ABC, therefore, is directed along line BD.

BD = (24) 2 + (10) 2 = 26 in.

 10  ΣM C = 0: (160 lb)(30 in.) −  FBD  (16 in.) = 0  26  FBD = +780 lb

ΣM x = 0: C x +

FBD = 780 lb T 

24 (780 lb) = 0 26 C x = −720 lb

ΣFy = 0: C y − 160 lb +

C x = 720 lb



10 (780 lb) = 0 26 C y = −140.0 lb

C y = 140.0 lb 



Free body: ABC:

www.elsolucionario.net PROBLEM 6.76 Determine the force in member BD and the components of the reaction at C.


SOLUTION We note that BD is a two-force member. The force it exerts on ABC, therefore, is directed along line BD. Free body: ABC:

Attaching FBD at D and resolving it into components, we write ΣM C = 0:

 450  FBD  (240 mm) = 0 (400 N)(135 mm) +   510  FBD = −255 N

ΣFx = 0: C x +

FBD = 255 N C 

240 (−255 N) = 0 510 C x = +120.0 N

ΣFy = 0: C y − 400 N +

C x = 120.0 N



450 (−255 N) = 0 510

C y = +625 N

C y = 625 N 


www.elsolucionario.net PROBLEM 6.77 Determine the components of all forces acting on member ABCD of the assembly shown.

SOLUTION Free body: Entire assembly: ΣM B = 0: D (120 mm) − (480 N)(80 mm) = 0

D = 320 N  ΣFx = 0: Bx + 480 N = 0



ΣFy = 0: By + 320 N = 0 B y = 320 N 

Free body: Member ABCD: ΣM A = 0: (320 N)(200 mm) − C (160 mm) − (320 N)(80 mm) − (480 N)(40 mm) = 0

C = 120.0 N  ΣFx = 0: Ax − 480 N = 0 A x = 480 N



ΣFy = 0: Ay − 320 N − 120 N + 320 N = 0 A y = 120.0 N 



B x = 480 N

www.elsolucionario.net PROBLEM 6.78 Determine the components of all forces acting on member ABD of the frame shown.

SOLUTION

ΣM D = 0: (300 lb) − (12 ft) + (450 lb)(4 ft) − E (6 ft) = 0

E = +900 lb

E = 900 lb



D x = 900 lb



ΣFx = 0: Dx − 900 lb = 0

ΣFy = 0: Dy − 300 lb − 450 lb = 0 D y = 750 lb 

Free body: Member ABD: We note that BC is a two-force member and that B is directed along BC. ΣM A = 0: (750 lb)(16 ft) − (900 lb)(6 ft) − B (8 ft) = 0

B = +825 lb

B = 825 lb 

ΣFx = 0: Ax + 900 lb = 0 Ax = −900 lb

A x = 900 lb



ΣFy = 0: Ay + 750 lb − 825 lb = 0 Ay = +75 lb

A y = 75.0 lb 



Free body: Entire frame:

www.elsolucionario.net PROBLEM 6.79 For the frame and loading shown, determine the components of all forces acting on member ABC.

SOLUTION Free body: Entire frame: ΣM E = 0: − Ax (4) − (20 kips)(5) = 0 Ax = −25 kips,

A x = 25.0 kips



Ay = 20 kips

A y = 20.0 kips 

Free body: Member ABC: Note: BE is a two-force member, thus B is directed along line BE and By =

2 Bx . 5

ΣM C = 0: (25 kips)(4 ft) − (20 kips)(10 ft) + Bx (2 ft) + By (5 ft) = 0 −100 kip ⋅ ft + Bx (2 ft) +

2 Bx (5 ft) = 0 5

Bx = 25 kips

By =

B x = 25.0 kips

2 2 ( Bx ) = (25) = 10 kips 5 5



B y = 10.00 kips 

ΣFx = 0: C x − 25 kips − 25 kips = 0 C x = 50 kips

C x = 50.0 kips



ΣFy = 0: C y + 20 kips − 10 kips = 0 C y = −10 kips

C y = 10.00 kips 



ΣFy = 0: Ay − 20 kips = 0

www.elsolucionario.net PROBLEM 6.80 Solve Problem 6.79 assuming that the 20-kip load is replaced by a clockwise couple of magnitude 100 kip ⋅ ft applied to member EDC at Point D.

PROBLEM 6.79 For the frame and loading shown, determine the components of all forces acting on member ABC.

SOLUTION Free body: Entire frame:

ΣFy = 0: Ay = 0 ΣM E = 0: − Ax (4 ft) − 100 kip ⋅ ft = 0 A x = 25.0 kips

A = 25.0 kips



B x = 25.0 kips



Free body: Member ABC: Note: BE is a two-force member, thus B is directed along line BE and By =

2 Bx . 5

ΣM C = 0: (25 kips)(4 ft) + Bx (2 ft) + By (5 ft) = 0 100 kip ⋅ ft + Bx (2 ft) +

2 Bx (5 ft) = 0 5

Bx = −25 kips

By =

2 2 Bx = ( −25) = −10 kips; 5 5

ΣFx = 0: − 25 kips + 25 kips + C x = 0

B y = 10.00 kips  Cx = 0

ΣFy = 0: +10 kips + C y = 0 C y = −10 kips

C y = 10 kips C = 10.00 kips 



Ax = −25 kips

www.elsolucionario.net PROBLEM 6.81 Determine the components of all forces acting on member ABCD when θ = 0.

SOLUTION Free body: Entire assembly:

A = +375 N

A = 375 N



B x = 375 N



ΣFx = 0: Bx + 375 N = 0

Bx = −375 N ΣFy = 0: By − 150 N = 0 By = +150 N

B y = 150 N 

Free body: Member ABCD: We note that D is directed along DE, since DE is a two-force member. ΣM C = 0: D (300) − (150 N)(100) + (375 N)(200) = 0

D = −200 N

D = 200 N 

ΣFx = 0: C x + 375 − 375 = 0 Cx = 0

ΣFy = 0: C y + 150 − 200 = 0 C y = +50.0 N

C = 50.0 N 



ΣM B = 0: A(200) − (150 N)(500) = 0

www.elsolucionario.net PROBLEM 6.82 Determine the components of all forces acting on member ABCD when θ = 90°.

SOLUTION Free body: Entire assembly:

A = +150.0 N

A = 150.0 N



ΣFx = 0: Bx + 150 − 150 = 0

Bx = 0 ΣFy = 0: By = 0

B = 0 

Free body: Member ABCD: We note that D is directed along DE, since DE is a two-force member. ΣM C = 0: D (300) + (150 N)(200) = 0

D = −100.0 N

D = 100.0 N 

ΣFx = 0: C x + 150 N = 0 C x = −150 N

Cx = 150.0 N



ΣFy = 0: C y − 100 N = 0 C y = +100.0 N

C y = 100.0 N 



ΣM B = 0: A(200) − (150 N)(200) = 0

www.elsolucionario.net PROBLEM 6.83 Determine the components of the reactions at A and E if a 750-N force directed vertically downward is applied (a) at B, (b) at D.

SOLUTION Free body: Entire frame: The following analysis is valid for both parts (a) and (b) since position of load on its line of action is immaterial. ΣM E = 0: − (750 N)(80 mm) − Ax (200 mm) = 0 A x = 300 N

ΣFx = 0: E x − 300 N = 0 E x = 300 N E x = 300 N

ΣFy = 0: Ay + E y − 750 N = 0

(a)

(1)

Load applied at B. Free body: Member CE: CE is a two-force member. Thus, the reaction at E must be directed along CE. Ey 300 N

From Eq. (1):

=

75 mm 250 mm

E y = 90 N

Ay + 90 N − 750 N = 0

Ay = 660 N

Thus, reactions are

(b)

A x = 300 N

,

A y = 660 N 

E x = 300 N

,

E y = 90.0 N 

Load applied at D. Free body: Member AC: AC is a two-force member. Thus, the reaction at A must be directed along AC. Ay 300 N

=

125 mm 250 mm

Ay = 150 N



Ax = − 300 N


Ay + E y − 750 N = 0

From Eq. (1):

150 N + E y − 750 N = 0

E y = 600 N E y = 600 N A x = 300 N

,

E x = 300 N

,

A y = 150.0 N  E y = 600 N 


Thus, reactions are


www.elsolucionario.net PROBLEM 6.84 Determine the components of the reactions at A and E if a 750-N force directed vertically downward is applied (a) at B, (b) at D.

SOLUTION Free-body: Entire frame: The following analysis is valid for both parts (a) and (b) since position of load on its line of action is immaterial. ΣM E = 0: − (750 N)(240 mm) − Ax (400 mm) = 0 A x = 450 N

ΣFx = 0: E x − 450 N = 0 E x = 450 N E x = 450 N

ΣFy = 0: Ay + E y − 750 N = 0 (a)

(1)

Load applied at B. Free body: Member CE: CE is a two-force member. Thus, the reaction at E must be directed along CE. Ey 450 N

From Eq. (1):

=

240 mm ; E y = 225 N 480 mm

Ay + 225 − 750 = 0; A y = 525 N

Thus, reactions are A x = 450 N

, ,

E y = 225 N 

A x = 450 N

,

A y = 150.0 N 

E x = 450 N

,

E x = 450N

(b)

A y = 525 N 

Load applied at D. Free body: Member AC: AC is a two-force member. Thus, the reaction at A must be directed along AC. Ay 450 N

From Eq. (1):

=

160 mm 480 mm

A y = 150.0 N

Ay + E y − 750 N = 0 150 N + E y − 750 N = 0

E y = 600 N E y = 600 N Thus, reactions are

E y = 600 N 



Ax = −450 N

www.elsolucionario.net PROBLEM 6.85 Determine the components of the reactions at A and E if the frame is loaded by a clockwise couple of magnitude 36 N · m applied (a) at B, (b) at D.

SOLUTION Free body: Entire frame: The following analysis is valid for both parts (a) and (b) since the point of application of the couple is immaterial. Ax = − 180 N

A x = 180 N

ΣFx = 0: − 180 N + E x = 0 E x = 180 N E x = 180 N

ΣFy = 0: Ay + E y = 0 (a)

(1)

Couple applied at B. Free body: Member CE: AC is a two-force member. Thus, the reaction at E must be directed along EC. Ey 180 N

From Eq. (1):

=

0.075 m 0.25 m

E y = 54 N

Ay + 54 N = 0 Ay = − 54 N A y = 54.0 N

Thus, reactions are

(b)

A x = 180.0 N

,

A y = 54.0 N 

E x = 180.0 N

,

E y = 54.0 N 

Couple applied at D. Free body: Member AC: AC is a two-force member. Thus, the reaction at A must be directed along EC. Ay 180 N

=

0.125 m 0.25 m

Ay = 90 N



ΣM E = 0: −36 N ⋅ m − Ax (0.2 m) = 0


From Eq. (1):

Ay + E y = 0 90 N + E y = 0

E y = − 90 N E y = 90 N Thus, reactions are A x = 180.0 N

,

E y = 90.0 N 


E x = −180.0 N

A y = 90.0 N 

,


www.elsolucionario.net PROBLEM 6.86 Determine the components of the reactions at A and E if the frame is loaded by a clockwise couple of magnitude 36 N · m applied (a) at B, (b) at D.

SOLUTION Free body: Entire frame: The following analysis is valid for both parts (a) and (b) since the point of application of the couple is immaterial. Ax = − 90 N

A x = 90.0 N

ΣFx = 0: −90 +E x = 0 E x = 90 N

E x = 90.0 N

ΣFy = 0: Ay + E y = 0 (a)

(1)

Couple applied at B. Free body: Member CE: AC is a two-force member. Thus, the reaction at E must be directed along EC. Ey 90 N

From Eq. (1):

=

0.24 m ; E y = 45.0 N 0.48 m

Ay + 45 N = 0 Ay = − 45 N A y = 45.0 N

Thus, reactions are

(b)

A x = 90.0 N

,

A y = 45.0 N 

E x = 90.0 N

,

E y = 45.0 N 

Couple applied at D. Free body: Member AC: AC is a two-force member. Thus, the reaction at A must be directed along AC. Ay 90 N

=

0.16 m ; A y = 30 N 0.48 m



ΣM E = 0: −36 N ⋅ m − Ax (0.4 m) = 0


From Eq. (1):

Ay + E y = 0 30 N + E y = 0

E y = − 30 N E y = 30 N A x = 90.0 N

Thus, reactions are

,

E y = 30.0 N 


E x = − 90.0 N

A y = 30.0 N 

,


www.elsolucionario.net PROBLEM 6.87 Determine all the forces exerted on member AI if the frame is loaded by a clockwise couple of magnitude 1200 lb · in. applied (a) at Point D, (b) at Point E.


ΣM H = 0: I (48 in.) − 1200 lb ⋅ in. = 0

I = +25.0 lb

(a) and (b)

I = 25.0 lb 

We note that AB, BC, and FG are two-force members.

Free body: Member AI: tan α = (a)

20 5 = 48 12

α = 22.6°

Couple applied at D. ΣFy = 0: −

ΣM G = 0:

5 A + 25 lb = 0 13 A = +65.0 lb

A = 65.0 lb

12 (65 lb)(40 in.) − C (20 in.) = 0 13 C = +120 lb

ΣFx = 0: −

12 (65 lb) + 120 lb + G = 0 13 G = −60.0 lb

22.6° 

C = 120 lb

G = 60 lb







Location of couple is immaterial.


(b)

Couple applied at E. ΣFy = 0: −

ΣM G = 0:

5 A + 25 lb = 0 13 A = +65.0 lb

A = 65.0 lb

22.6° 

12 (65 lb) + (40 in.) − C (20 in.) − 1200 lb ⋅ in. = 0 13 C = +60.0 lb C = 60.0 lb 12 (65 lb) + 60 lb + G = 0 13

G=0 


ΣFx = 0: −




www.elsolucionario.net PROBLEM 6.88 Determine all the forces exerted on member AI if the frame is loaded by a 40-lb force directed horizontally to the right and applied (a) at Point D, (b) at Point E.

SOLUTION Free body: Entire frame: ΣM H = 0: I (48 in.) − (40 lb)(30 in.) = 0

I = +25.0 lb

(a) and (b)

I = 25.0 lb 

We note that AB, BC, and FG are two-force members.

Free body: Member AI: tan α = (a)

20 5 = 48 12

α = 22.6°

Force applied at D. ΣFy = 0: −

ΣM G = 0:

5 A + 25 lb = 0 13 A = +65.0 lb

12 (65 lb)(40 in.) − C (20 in.) = 0 13 C = +120.0 lb

ΣFx = 0: −

12 (65 lb) + 120 lb + G = 0 13 G = −60.0 lb

A = 65.0 lb

22.6° 

C = 120.0 lb



G = 60.0 lb





Location of 40-lb force on its line of action DE is immaterial.


(b)

Force applied at E. ΣFy = 0: −

ΣM G = 0:

5 A + 25 lb = 0 13 A = +65.0 lb

A = 65.0 lb

22.6° 

12 (65 lb)(40 in.) − C (20 in.) − (40 lb)(10 in.) = 0 13 C = +100.0 lb C = 100.0 lb

ΣFx = 0: −



12 (65 lb) + 100 lb + 40 lb + G = 0 13

G = 80.0 lb




G = −80.0 lb


www.elsolucionario.net PROBLEM 6.89 Determine the components of the reactions at A and B, (a) if the 500-N load is applied as shown, (b) if the 500-N load is moved along its line of action and is applied at Point F.

SOLUTION

Analysis is valid for either parts (a) or (b), since position of 100-lb load on its line of action is immaterial.

ΣM A = 0: By (10) − (100 lb)(6) = 0 By = + 60 lb ΣFy = 0: Ay + 60 − 100 = 0

Ay = + 40 lb

ΣFx = 0: Ax + Bx = 0

(a)

(1)

Load applied at E. Free body: Member AC:

Since AC is a two-force member, the reaction at A must be directed along CA. We have Ax 40 lb = 10 in. 5 in.

From Eq. (1):

A x = 80.0 lb

,

A y = 40.0 lb 

B x = 80.0 lb

,

B y = 60.0 lb 

− 80 + Bx = 0 Bx = + 80 lb

Thus,





(b)

Load applied at F. Free body: Member BCD: Since BCD is a two-force member (with forces applied at B and C only), the reaction at B must be directed along CB. We have, therefore, Bx = 0

The reaction at B is From Eq. (1):

Ax + 0 = 0

Ax = 0

A y = 40.0 lb 

Ax = 0


The reaction at A is

B y = 60.0 lb 

Bx = 0


www.elsolucionario.net PROBLEM 6.90 (a) Show that when a frame supports a pulley at A, an equivalent loading of the frame and of each of its component parts can be obtained by removing the pulley and applying at A two forces equal and parallel to the forces that the cable exerted on the pulley. (b) Show that if one end of the cable is attached to the frame at a Point B, a force of magnitude equal to the tension in the cable should also be applied at B.


SOLUTION First note that, when a cable or cord passes over a frictionless, motionless pulley, the tension is unchanged. ΣM C = 0: rT1 − rT2 = 0 T1 = T2

(a)

Replace each force with an equivalent force-couple.

(b)

Cut the cable and replace the forces on pulley with equivalent pair of forces at A as above.




www.elsolucionario.net PROBLEM 6.91 A 3-ft-diameter pipe is supported every 16 ft by a small frame like that shown. Knowing that the combined weight of the pipe and its contents is 500 lb/ft and assuming frictionless surfaces, determine the components (a) of the reaction at E, (b) of the force exerted at C on member CDE.

SOLUTION Free body: 16-ft length of pipe:


W = (500 lb/ft)(16 ft) = 8 kips

Force Triangle

B D 8 kips = = 3 5 4 B = 6 kips D = 10 kips

Determination of CB = CD.

   We note that horizontal projection of BO + OD = horizontal projection of CD

Thus,

3 4 r + r = (CD ) 5 5 8 CB = CD = r = 2(1.5 ft) = 3 ft 4

Free body: Member ABC:

ΣM A = 0: Cx h − (6 kips)(h − 3) Cx =

h−3 (6 kips) h

Cx =

9−3 (6 kips) = 4 kips 9

(1)

For h = 9 ft,



Free body: Member CDE: From above, we have

Cx = 4.00 kips



ΣM E = 0: (10 kips)(7 ft) − (4 kips)(6 ft) − C y (8 ft) = 0 C y = +5.75 kips, 3 ΣFx = 0: − 4 kips + (10 kips) + Ex = 0 5 Ex = −2 kips,

C y = 5.75 kips 

E x = 2.00 kips



4 ΣFy = 0: 5.75 kips − (10 kips) + E y = 0, 5


E y = 2.25 kips 


www.elsolucionario.net PROBLEM 6.92 Solve Problem 6.91 for a frame where h = 6 ft.

PROBLEM 6.91 A 3-ft-diameter pipe is supported every 16 ft by a small frame like that shown. Knowing that the combined weight of the pipe and its contents is 500 lb/ft and assuming frictionless surfaces, determine the components (a) of the reaction at E, (b) of the force exerted at C on member CDE.

SOLUTION

For h = 6 ft, C x =

h−3 6−3 (6 kips) = = 3 kips h 6

Free body: Member CDE: From above, we have

Cx = 3.00 kips



ΣM E = 0: (10 kips)(7 ft) − (3 kips)(6 ft) − C y (8 ft) = 0

C y = +6.50 kips, 3 ΣFx = 0: −3 kips + (10 kips) + Ex = 0 5 Ex = −3.00 kips, 4 ΣFy = 0: 6.5 kips − (10 kips) + E y = 0 5 E y = 1.500 kips

C y = 6.50 kips 

E x = 3.00 kips



E y = 1.500 kips 



See solution of Problem 6.91 for derivation of Eq. (1).

www.elsolucionario.net PROBLEM 6.93 Knowing that the pulley has a radius of 0.5 m, determine the components of the reactions at A and E.

SOLUTION FBD Frame:

ΣM A = 0: (7 m) E y − (4.5 m)(700 N) = 0 ΣFy = 0: Ay − 700 N + 450 N = 0

A y = 250 N 

Ax = E x

Dimensions in m

FBD Member ABC: ΣM C = 0: (1 m)(700 N) − (1 m)(250 N) − (3 m) Ax = 0 A x = 150.0 N so E x = 150.0 N

 





ΣFx = 0: Ax − E x = 0

E y = 450 N 

www.elsolucionario.net PROBLEM 6.94 Knowing that the pulley has a radius of 50 mm, determine the components of the reactions at B and E.

SOLUTION Free body: Entire assembly: ΣM E = 0 : − (300 N)(350 mm) − Bx (150 mm) = 0 Bx = −700 N

B x = 700 N

E x = 700 N

E x = 700 N

ΣFy = 0: By + E y − 300 N = 0

(1)

Free body: Member ACE:

ΣM C = 0: (700 N)(150 mm) − (300 N)(50 mm) − E y (180 mm) = 0 E y = 500 N From Eq. (1):

E y = 500 N

By + 500 N − 300 N = 0 By = −200 N

B y = 200 N

Thus, reactions are B x = 700 N

,

B y = 200 N 

E x = 700 N

,

E y = 500 N 



ΣFx = 0: − 700 N + E x = 0


SOLUTION (a)

Free body: Trailer: (We shall denote by A, B, C the reaction at one wheel.) ΣM A = 0: − (2400 lb)(2 ft) + D (11 ft) = 0

D = 436.36 lb

ΣFy = 0: 2 A − 2400 lb + 436.36 lb = 0 A = 981.82 lb

A = 982 lb 

Free body: Truck. ΣM B = 0: (436.36 lb)(3 ft) − (2900 lb)(5 ft) + 2C (9 ft) = 0

C = 732.83 lb

C = 733 lb 

ΣFy = 0: 2B − 436.36 lb − 2900 lb + 2(732.83 lb) = 0 B = 935.35 lb

(b)

B = 935 lb 

Additional load on truck wheels. Use free body diagram of truck without 2900 lb. ΣM B = 0: (436.36 lb)(3 ft) + 2C (9 ft) = 0

C = −72.73 lb

ΔC = −72.7 lb 

ΣFy = 0: 2B − 436.36 lb − 2(72.73 lb) = 0 B = 290.9 lb

ΔB = +291 lb 



A trailer weighing 2400 lb is attached to a 2900-lb pickup truck by a ball-and-socket truck hitch at D. Determine (a) the reactions at each of the six wheels when the truck and trailer are at rest, (b) the additional load on each of the truck wheels due to the trailer.


PROBLEM 6.95 A trailer weighing 2400 lb is attached to a 2900-lb pickup truck by a ball-and-socket truck hitch at D. Determine (a) the reactions at each of the six wheels when the truck and trailer are at rest, (b) the additional load on each of the truck wheels due to the trailer.

SOLUTION (a)

We small first find the additional reaction Δ at each wheel due to the trailer. Free body diagram: (Same Δ at each truck wheel) ΣM A = 0: − (2400 lb)(2 ft) + 2 Δ (14 ft) + 2 Δ (23 ft) = 0

Δ = 64.86 lb ΣFy = 0: 2 A − 2400 lb + 4(64.86 lb) = 0;

A = 1070 lb;

A = 1070 lb

Free body: Truck: (Trailer loading only) ΣM D = 0: 2 Δ (12 ft) + 2 Δ (3 ft) − 2T (1.7 ft) = 0

T = 8.824Δ = 8.824(64.86 lb) T = 572.3 lb

T = 572 lb 

Free body: Truck: (Truck weight only) ΣM B = 0: − (2900 lb)(5 ft) + 2C ′(9 ft) = 0

C ′ = 805.6 lb

C′ = 805.6 lb



In order to obtain a better weight distribution over the four wheels of the pickup truck of Problem 6.95, a compensating hitch of the type shown is used to attach the trailer to the truck. The hitch consists of two bar springs (only one is shown in the figure) that fit into bearings inside a support rigidly attached to the truck. The springs are also connected by chains to the trailer frame, and specially designed hooks make it possible to place both chains in tension. (a) Determine the tension T required in each of the two chains if the additional load due to the trailer is to be evenly distributed over the four wheels of the truck. (b) What are the resulting reactions at each of the six wheels of the trailer-truck combination?

www.elsolucionario.net PROBLEM 6.96 (Continued) ΣFy = 0: 2B′ − 2900 lb + 2(805.6 lb) = 0 B ′ = 644.4 lb

B′ = 644.4 lb

Actual reactions: B = B′ + Δ = 644.4 lb + 64.86 = 709.2 lb

B = 709 lb 

C = C ′ + Δ = 805.6 lb + 64.86 = 870.46 lb

C = 870 lb  A = 1070 lb 


From part a:




The cab and motor units of the front-end loader shown are connected by a vertical pin located 2 m behind the cab wheels. The distance from C to D is 1 m. The center of gravity of the 300-kN motor unit is located at Gm, while the centers of gravity of the 100-kN cab and 75-kN load are located, respectively, at Gc and Gl. Knowing that the machine is at rest with its brakes released, determine (a) the reactions at each of the four wheels, (b) the forces exerted on the motor unit at C and D.

SOLUTION (a)

Free body: Entire machine:

A = Reaction at each front wheel B = Reaction at each rear wheel ΣM A = 0: 75(3.2 m) − 100(1.2 m) + 2 B (4.8 m) − 300(5.6 m) = 0

2 B = 325 kN

B = 162.5 kN 

ΣFy = 0: 2 A + 325 − 75 − 100 − 300 = 0 2 A = 150 kN

A = 75.0 kN 


www.elsolucionario.net PROBLEM 6.97 (Continued) (b)

Free body: Motor unit: ΣM D = 0: C (1 m) + 2 B (2.8 m) − 300(3.6 m) = 0

C = 1080 − 5.6 B

B = 162.5 kN, C = 1080 − 5.6(162.5) = 170 kN

ΣFx = 0: Dx − 170 = 0







ΣFy = 0: 2(162.5) − Dy − 300 = 0

C = 170.0 kN



D x = 170.0 kN



D y = 25.0 kN 



Recalling

(1)

www.elsolucionario.net PROBLEM 6.98 Solve Problem 6.97 assuming that the 75-kN load has been removed.

SOLUTION (a)

Free body: Entire machine: A = Reaction at each front wheel B = Reaction at each rear wheel ΣM A = 0: 2 B (4.8 m) − 100(1.2 m) − 300(5.6 m) = 0

2 B = 375 kN

B = 187.5 kN 

ΣFy = 0: 2 A + 375 − 100 − 300 = 0 2 A = 25 kN

(b)

A = 12.50 kN 

Free body: Motor unit: See solution of Problem 6.97 for free body diagram and derivation of Eq. (1). With B = 187.5 kN, we have C = 1080 − 5.6(187.5) = 30 kN

ΣFx = 0: Dx − 30 = 0

ΣFy = 0: 2(187.5) − Dy − 300 = 0

C = 30.0 kN



D x = 30.0 kN



D y = 75.0 kN 



PROBLEM 6.97 The cab and motor units of the front-end loader shown are connected by a vertical pin located 2 m behind the cab wheels. The distance from C to D is 1 m. The center of gravity of the 300-kN motor unit is located at Gm, while the centers of gravity of the 100-kN cab and 75-kN load are located, respectively, at Gc and Gl. Knowing that the machine is at rest with its brakes released, determine (a) the reactions at each of the four wheels, (b) the forces exerted on the motor unit at C and D.

www.elsolucionario.net PROBLEM 6.99 For the frame and loading shown, determine the components of all forces acting on member ABE.

SOLUTION FBD Frame:

ΣM E = 0: (1.8 m) Fy − (2.1 m)(12 kN) = 0

ΣFy = 0: − E y + 14.00 kN − 12 kN = 0 E y = 2 kN E y = 2.00 kN  FBD member BCD:

ΣM B = 0: (1.2 m)C y − (12 kN)(1.8 m) = 0 C y = 18.00 kN But C is ⊥ ACF, so Cx = 2C y ; Cx = 36.0 kN ΣFx = 0: − Bx + C x = 0

Bx = C x = 36.0 kN

Bx = 36.0 kN

on BCD

ΣFy = 0: − By + 18.00 kN − 12 kN = 0 By = 6.00 kN on BCD B x = 36.0 kN

On ABE:



B y = 6.00 kN  FBD member ABE: ΣM A = 0: (1.2 m)(36.0 kN) − (0.6 m)(6.00 kN) + (0.9 m)(2.00 kN) − (1.8 m)( E x ) = 0

ΣFx = 0: − 23.0 kN + 36.0 kN − Ax = 0

ΣFy = 0: − 2.00 kN + 6.00 kN − Ay = 0

E x = 23.0 kN



A x = 13.00 kN



A y = 4.00 kN 



Fy = 14.00 kN

www.elsolucionario.net PROBLEM 6.100 For the frame and loading shown, determine the components of all forces acting on member ABE.

PROBLEM 6.99 For the frame and loading shown, determine the components of all forces acting on member ABE.

SOLUTION

ΣM F = 0: (1.2 m)(2400 N) − (4.8 m) E y = 0

E y = 600 N 

FBD member BC: Cy =

4.8 8 Cx = Cx 5.4 9

ΣM C = 0: (2.4 m) By − (1.2 m)(2400 N) = 0 By = 1200 N B y = 1200 N 

On ABE:

ΣFy = 0: −1200 N + C y − 2400 N = 0 C y = 3600 N 9 Cx = C y 8

so ΣFx = 0: − Bx + C x = 0

Bx = 4050 N

C x = 4050 N

on BC B x = 4050 N

On ABE:





FBD Frame:


FBD member AB0E: ΣM A = 0: a (4050 N) − 2 aE x = 0 E x = 2025 N ΣFx = 0 : − Ax + (4050 − 2025) N = 0



A x = 2025 N



A y = 1800 N 


ΣFy = 0: 600 N + 1200 N − Ay = 0

E x = 2025 N


www.elsolucionario.net PROBLEM 6.101 For the frame and loading shown, determine the components of all forces acting on member ABD.

SOLUTION

Σ M A = 0: E (12 in.) − (360 lb)(15 in.) − (240 lb)(33 in.) = 0

E = +1110 lb

E = +1110 lb

ΣFx = 0: Ax + 1110 lb = 0 Ax = −1110 lb

A x = 1110 lb



ΣFy = 0: Ay − 360 lb − 240 lb = 0 Ay = +600 lb

A y = 600 lb 

Free body: Member CDE:

Σ M C = 0: (1110 lb)(24 in.) − Dx (12 in.) = 0 Dx = +2220 lb





Free body: Member ABD: D x = 2220 lb

From above:



ΣM B = 0: Dy (18 in.) − (600 lb)(6 in.) = 0 Dy = +200 lb

D y = 200 lb 

ΣFx = 0: Bx + 2220 lb − 1110 lb = 0 Bx = −1110 lb

B x = 1110 lb



ΣFy = 0: By + 200 lb + 600 lb = 0 B y = 800 lb 


By = −800 lb


www.elsolucionario.net PROBLEM 6.102 Solve Problem 6.101 assuming that the 360-lb load has been removed. PROBLEM 6.101 For the frame and loading shown, determine the components of all forces acting on member ABD.

SOLUTION

ΣM A = 0: E (12 in.) − (240 lb)(33 in.) = 0 E = +660 lb

E = 660 lb

ΣFx = 0: Ax + 660 lb = 0 Ax = −660 lb

A x = 660 lb



ΣFy = 0: Ay − 240 lb = 0 Ay = +240 lb

A y = 240 lb 


ΣM C = 0: (660 lb)(24 in.) − Dx (12 in.) = 0 Dx = +1320 lb



Free body diagram of entire frame.


Free body: Member ABD: D x = 1320 lb

From above:



ΣM B = 0: Dy (18 in.) − (240 lb)(6 in.) = 0 Dy = +80 lb

D y = 80.0 lb 

ΣFx = 0: Bx + 1320 lb − 660 lb = 0 Bx = −660 lb

B x = 660 lb



ΣFy = 0: By + 80 lb + 240 lb = 0 B y = 320 lb 


By = −320 lb


www.elsolucionario.net PROBLEM 6.103 For the frame and loading shown, determine the components of the forces acting on member CDE at C and D.


Ay = 25 lb

A y = 25 lb 

ΣM F = 0: Ax (6.928 + 2 × 3.464) − (25 lb)(12 in.) = 0 Ax = 21.651 lb

A y = 21.65 lb

ΣFx = 0: F − 21.651 lb = 0 F = 21.651 lb

F = 21.65 lb

Free body: Member CDE: ΣM C = 0: Dy (4 in.) − (25 lb)(10 in.) = 0 Dy = +62.5 lb

D y = 62.5 lb 

ΣFy = 0: −C y + 62.5 lb − 25 lb = 0 C y = +37.5 lb

C y = 37.5 lb 

Free body: Member ABD: ΣM B = 0: Dx (3.464 in.) + (21.65 lb)(6.928 in.) −(25 lb)(4 in.) − (62.5 lb)(2 in.) = 0 Dx = +21.65 lb

Return to free body: Member CDE: From above: Dx = +21.65 lb

D x = 21.7 lb



C x = 21.7 lb



ΣFx = 0: C x − 21.65 lb C x = +21.65 lb



ΣM y = 0: Ay − 25 lb = 0

www.elsolucionario.net PROBLEM 6.104 For the frame and loading shown, determine the components of the forces acting on member CFE at C and F.

SOLUTION

ΣM D = 0: (40 lb)(13 in.) + Ax (10 in.) = 0 Ax = −52 lb,

A x = 52 lb



Free body: Member ABF: ΣM B = 0: − (52 lb)(6 in.) + Fx (4 in.) = 0 Fx = +78 lb

Free body: Member CFE: Fx = 78.0 lb

From above:



ΣM C = 0: (40 lb)(9 in.) − (78 lb)(4 in.) − Fy (4 in.) = 0 Fy = +12 lb

Fy = 12.00 lb 

ΣFx = 0: C x − 78 lb = 0 C x = +78 lb ΣFy = 0: − 40 lb + 12 lb + C y = 0; C y = +28 lb

C x = 78.0 lb



C y = 28.0 lb 




www.elsolucionario.net PROBLEM 6.105 For the frame and loading shown, determine the components of all forces acting on member ABD.

SOLUTION Free body: Entire frame: Dimensions in mm

ΣM A = 0: F (500) − (3 kN)(600) − (2 kN)(1000) = 0 F = 7.60 kN

ΣFx = 0: Ax + 7.60 N = 0, Ax = −7.60 N

A x = 7.60 kN



ΣFy = 0: Ay − 3 kN − 2 kN = 0 Ay = +5 kN

A y = 5.00 kN 

Free body: Member CDE: ΣM C = 0: D y (400) − (3 kN)(200) − (2 kN)(600) = 0 Dy = +4.50 kN

Free body: Member ABD: D y = 4.50 kN 

From above:

ΣM B = 0: Dx (200) − (4.50 kN)(400) − (5 kN)(400) = 0 Dx = +19 kN

D x = 19.00 kN



B x = 11.40 kN



ΣFx = 0: Bx + 19 kN − 7.60 kN = 0 Bx = −11.40 kN ΣFy = 0: By + 5 kN − 4.50 kN = 0 By = −0.50 kN

B y = 0.500 kN 



F = +7.60 kN

www.elsolucionario.net PROBLEM 6.106 Solve Problem 6.105 assuming that the 3-kN load has been removed. PROBLEM 6.105 For the frame and loading shown, determine the components of all forces acting on member ABD.

SOLUTION Free body: Entire frame: ΣM A = 0: F (500) − (2 kN)(1000) = 0 F = +4 kN

F = 4.00 kN

ΣFx = 0: Ax + 4 kN = 0, Ax = − 4 kN

A x = 4.00 kN



ΣFy = 0: Ay − 2 kN = 0, Ay = +2 kN

A y = 2.00 kN 

Free body: Member CDE: ΣM C = 0: D y (400) − (2 kN)(600) = 0 Dy = +3.00 kN

Free body: Member ABD: D y = 3.00 kN 

From above:

ΣM B = 0: Dx (200) − (3 kN)(400) − (2 kN)(400) = 0 Dx = +10.00 kN

D x = 10.00 kN



ΣFx = 0: Bx + 10 kN − 4 kN = 0 Bx = −6 kN

B x = 6.00 kN



ΣFy = 0: By + 2 kN − 3 kN = 0 By = +1 kN

B y = 1.000 kN 



Dimensions in mm

www.elsolucionario.net PROBLEM 6.107 Determine the reaction at F and the force in members AE and BD.

SOLUTION



ΣM C = 0: Fy (9 in.) − (450 lb)(24 in.) = 0 Fy = 1200 lb

Fy = 1200 lb

Free body: Member DEF: ΣM J = 0: (1200 lb)(4.5 in.) − Fx (18 in.) = 0 Fx = 300 lb 3  ΣM D = 0: − Fx (24 in.) −  FAE  (12 in.) = 0 5  10 10 FAE = − Fx = − (300 lb) 3 3 FAE = −1000 lb ΣFy = 0: 1200 lb + FBD =

Fx = 300 lb



FAE = 1000 lb C 

4 4 (−1000 lb) − FBD = 0 5 5

5 (1200 lb) − 1000 lb = +500 lb 4

FBD = 500 lb T 


www.elsolucionario.net PROBLEM 6.108 For the frame and loading shown, determine the reactions at A, B, D, and E. Assume that the surface at each support is frictionless.

SOLUTION Free body: Entire frame: A − B + (1000 lb)sin 30° = 0 A − B + 500 = 0

(1)

ΣFy = 0: D + E − (1000 lb) cos 30° = 0 D + E − 866.03 = 0

(2)

Free body: Member ACE: ΣM C = 0: − A(6 in.) + E (8 in.) = 0 E=

3 A 4

(3)

Free body: Member BCD: ΣM C = 0: − D(8 in.) + B (6 in.) = 0 D=

3 B 4

(4)

Substitute E and D from Eqs. (3) and (4) into Eq. (2): −

3 3 A + B − 866.06 = 0 4 4 A + B − 1154.71 = 0

(5)

From Eq. (1): A − B + 500 = 0

(6)

Eqs. (5) + (6): 2 A − 654.71 = 0

A = 327.4 lb

A = 327 lb



Eqs. (5) − (6): 2 B − 1654.71 = 0

B = 827.4 lb

B = 827 lb



From Eq. (4): D =

3 (827.4) 4

D = 620.5 lb

D = 621 lb 

From Eq. (3): E =

3 (327.4) 4

E = 245.5 lb

E = 246 lb 



ΣFx = 0:

www.elsolucionario.net PROBLEM 6.109 The axis of the three-hinge arch ABC is a parabola with the vertex at B. Knowing that P = 112 kN and Q = 140 kN, determine (a) the components of the reaction at A, (b) the components of the force exerted at B on segment AB.

Free body: Segment AB: ΣM A = 0: Bx (3.2 m) − B y (8 m) − P(5 m) = 0

(1)

Bx (2.4 m) − By (6 m) − P (3.75 m) = 0

(2)

0.75 (Eq. 1):

Free body: Segment BC: ΣM C = 0: Bx (1.8 m) + B y (6 m) − Q(3 m) = 0


(3)

4.2 Bx − 3.75P − 3Q = 0 Bx = (3.75 P + 3Q)/4.2

From Eq. (1):

(3.75P + 3Q)

(4)

3.2 − 8B y − 5 P = 0 4.2 B y = (− 9P + 9.6Q )/33.6

(5)

given that P = 112 kN and Q = 140 kN. (a)

Reaction at A. Considering again AB as a free body, ΣFx = 0: Ax − Bx = 0;

Ax = Bx = 200 kN

A x = 200 kN



ΣFy = 0: Ay − P − B y = 0 Ay − 112 kN − 10 kN = 0 Ay = + 122 kN

A y = 122.0 kN 



SOLUTION


(b)

Force exerted at B on AB. From Eq. (4):

Bx = (3.75 × 112 + 3 × 140)/4.2 = 200 kN

B x = 200 kN

From Eq. (5):



B y = (− 9 × 112 + 9.6 × 140)/33.6 = + 10 kN


B y = 10.00 kN 


www.elsolucionario.net PROBLEM 6.110 The axis of the three-hinge arch ABC is a parabola with the vertex at B. Knowing that P = 140 kN and Q = 112 kN, determine (a) the components of the reaction at A, (b) the components of the force exerted at B on segment AB.

Free body: Segment AB: ΣM A = 0: Bx (3.2 m) − B y (8 m) − P(5 m) = 0

(1)

Bx (2.4 m) − By (6 m) − P (3.75 m) = 0

(2)

0.75 (Eq. 1):

Free body: Segment BC: ΣM C = 0: Bx (1.8 m) + B y (6 m) − Q(3 m) = 0


(3)

4.2 Bx − 3.75P − 3Q = 0 Bx = (3.75 P + 3Q)/4.2 (3.75P + 3Q)

From Eq. (1):

(4)

3.2 − 8B y − 5 P = 0 4.2 B y = (− 9P + 9.6Q )/33.6

(5)

given that P = 140 kN and Q = 112 kN. (a)

Reaction at A. ΣFx = 0: Ax − Bx = 0;

Ax = Bx = 205 kN

A x = 205 kN



ΣFy = 0: Ay − P − B y = 0 Ay − 140 kN − (− 5.5 kN) = 0 Ay = 134.5 kN

A y = 134.5 kN 



SOLUTION


(b)

Force exerted at B on AB. From Eq. (4):

Bx = (3.75 × 140 + 3 × 112)/4.2 = 205 kN

B x = 205 kN

From Eq. (5):



B y = (− 9 × 140 + 9.6 × 112)/33.6 = − 5.5 kN


B y = 5.50 kN 


www.elsolucionario.net PROBLEM 6.111 Members ABC and CDE are pin-connected at C and supported by four links. For the loading shown, determine the force in each link.

SOLUTION

I FBD I: FBD II: FBDs combined:

II

aC y − a

1

M D = 0: aC y − a

1

ΣM B = 0:

ΣM G = 0:

aP − a

FAF = 0 FAF = 2C y

2

2 1 2

FEH = 0 FEH = 2C y 1

FAF − a

2

FEH = 0 P = Cy =

P 2

1 2

2C y +

1

2C y

2

so FAF =

2 P C  2

FEH =

2 P T  2

 FBD I:

ΣFy = 0:

1 2

FAF +

1 2

FBG − P + C y = 0

P 1 P + FBG − P + = 0 2 2 2 FBG = 0 

FBD II:

ΣFy = 0: − C y +

1 2

FDG −

1 2

FEH = 0 −

P 1 P + FDG − = 0 2 2 2

FDG = 2 P C 



Member FBDs:


SOLUTION

I

II

FBD I:

ΣM I = 0: 2aC y + aC x − aP = 0 2C y + C x = P

FBD II:

ΣM J = 0: 2aC y − aC x = 0 2C y − C x = 0

Solving, FBD I:

Cx =

P P ; C y = as shown. 2 4

ΣFx = 0: −

1 2

FBG + C x = 0 FBG = C x 2

ΣFy = 0: FAF − P +

FBD II:

ΣFx = 0: − Cx + ΣFy = 0: − C y +

1 2

1  2  P P + = 0  2  2  4 FDG = 0 FDG = Cx 2

1  2  P P P P  + FEH = 0 FEH = − = −   4 2 4 2 2 

FBG =

P 2

P 4

C 

P

C 

FAF = FDG =

C 

FEH =

2

P T  4



Member FBDs:


SOLUTION

I

II

From FBD I:

ΣM J = 0:

a 3a a Cx + C y − P = 0 Cx + 3C y = P 2 2 2

FBD II:

ΣM K = 0:

a 3a C x − C y = 0 Cx − 3C y = 0 2 2

Solving,

Cx =

FBD I:

P P ; C y = as drawn. 2 6

ΣM B = 0: aC y − a ΣFx = 0: −

1 2 FAG = 0 FAG = 2C y = P 6 2

FAG =

1 1 2 2 FAG + FBF − Cx = 0 FBF = FAG + Cx 2 = P+ P 6 2 2 2 FBF =

FBD II:

ΣM D = 0: a

2 P C  6

1 2 FEH + aC y = 0 FEH = − 2C y = − P 6 2

ΣFx = 0: C x −

2 2 P C  3

FEH =

2 P T  6

1 1 2 2 FDI + FEH = 0 FDI = FEH + C x 2 = − P+ P 6 2 2 2 FDI =

2 P C  3



Member FBDs:

www.elsolucionario.net PROBLEM 6.114 Members ABC and CDE are pin-connected at C and supported by the four links AF, BG, DG, and EH. For the loading shown, determine the force in each link.

SOLUTION

Free body: CDE: ΣM J = 0: C x (4a) + C y (2a) = 0

C y = −2Cx

(1)

Free body: ABC: ΣM K = 0: Cx (2a) + C y (4a) − P(3a) = 0

Substituting for Cy from Eq. (1):

Cx (2a ) − 2Cx (4a) − P(3a ) = 0 1 Cx = − P 2

ΣFx = 0:

1 2

 1  C y = −2  − P  = + P  2 

FBG + C x = 0 P  1  FBG = − 2Cx = − 2  − P  = + , 2  2 

FBG =

P 2

T 



We consider members ABC and CDE:


ΣFy = 0: − FAF −

1 2

FBG − P + C y = 0 FAF = −

1 2

P 2

−P+P=−

P 2

FAF =

P 2

C 

Free body: CDE: 1 2

FDG − C y = 0

FDG = 2C y = + 2 P ΣFx = 0: − FEH − Cx −

1 2

FDG = 2 P T 

FDG = 0 P  1  1 FEH = −  − P  − 2P = − 2 2  2 

FEH =

P 2

C 



ΣFy = 0:

www.elsolucionario.net PROBLEM 6.115 Solve Problem 6.112 assuming that the force P is replaced by a clockwise couple of moment M0 applied to member CDE at D. PROBLEM 6.112 Members ABC and CDE are pin-connected at C and supported by four links. For the loading shown, determine the force in each link.

SOLUTION Free body: Member ABC: ΣM J = 0: C y (2a) + Cx ( a) = 0 Cx = − 2C y ΣM K = 0: C y (2a ) − Cx (a ) − M 0 = 0 C y (2a) − (− 2C y )(a) − M 0 = 0 Cx = − 2C y : D

ΣFx = 0:

2

D

+ Cx = 0;

2

−

M0

D=

M0 䉰 4a M Cx = − 0 䉰 2a Cy =

M0 =0 2a FDG =

2a

M0 2a

T 

ΣM D = 0: E (a) − C y ( a) + M 0 = 0 M E (a) −  0  4a

  (a) + M 0 = 0 

E=−

3 M0 4 a

3 M0 4 a

C 

FBG =

M0

T 

FAF =

M0 4a

FEH =

Return to free body of ABC: ΣFx = 0:

B 2

+ Cx = 0; B=

B 2

−

M0 =0 2a

M0 2a

ΣM B = 0: A( a) + C y (a); A(a) +

2a

M0 (a ) = 0 4a A=−

M0 4a

C 




www.elsolucionario.net PROBLEM 6.116 Solve Problem 6.114 assuming that the force P is replaced by a clockwise couple of moment M0 applied at the same point. PROBLEM 6.114 Members ABC and CDE are pin-connected at C and supported by the four links AF, BG, DG, and EH. For the loading shown, determine the force in each link.

SOLUTION ΣM J = 0: C x (4a) + C y (2a) = 0

C y = −2Cx

(1)

Free body: ABC: ΣM K = 0: C x (2a) + C y (4a) − M 0 = 0

Substituting for Cy from Eq. (1): Cx (2a ) − 2Cx (4a) − M 0 = 0 M0 6a  M C y = −2  − 0  6a Cx = −

ΣFx = 0:

1 2

M0   = + 3a 

FBG + C x = 0 2M 0 6a

FBG = − 2C x = + ΣFy = 0: − FAF −

1 2

FBG =

2M 0 6a

T 

M0 6a

T 

FBG + C y = 0 FAF = −

1 2

2M 0 M 0 M + =+ 0 6a 3a 6a

FAF =



Free body: CDE: (same as for Problem 6.114)


Free body: CDE:

(Use F.B. diagram of Problem 6.114.) ΣFy = 0:

1 2

FDG − C y = 0 FDG = 2C y = +

ΣFx = 0: − FEH − C x −

1 2

2M 0 3a

FDG =

2M 0 3a

T 

M0 6a

C 

FDG = 0

  1   2M 0  −   2   3a

 M  = − 0 , 6a 

FEH =


 M FEH = −  − 0  6a


www.elsolucionario.net PROBLEM 6.117 Four beams, each of length 3a, are held together by single nails at A, B, C, and D. Each beam is attached to a support located at a distance a from an end of the beam as shown. Assuming that only vertical forces are exerted at the connections, determine the vertical reactions at E, F, G, and H.

We shall draw the free body of each member. Force P will be applied to member AFB. Starting with member AED, we shall express all forces in terms of reaction E. Member AFB: ΣM D = 0: A(3a) + E (a ) = 0 E 3 ΣM A = 0: − D (3a ) − E (2a ) = 0 A=−

D=−

2E 3

Member DHC:  2E  ΣM C = 0:  −  (3a) − H (a) = 0  3  H = − 2E

(1)

 2E  ΣM H = 0:  −  (2a) + C (a ) = 0  3  C=+

4E 3

Member CGB:  4E  ΣM B = 0: +   (3a ) − G (a ) = 0  3  G = + 4E

(2)

 4E  ΣM G = 0: +   (2a) + B(a ) = 0  3  B=−

8E 3



SOLUTION


Member AFB: ΣFy = 0: F − A − B − P = 0

 E   8E  F − −  − − −P =0  3  3  F = P − 3E

(3)

ΣM A = 0: F ( a) − B(3a) = 0

 8E  ( P − 3E )(a) −  −  (3a ) = 0  3  P 5

E=

P  5

Substitute E = − P5 into Eqs. (1), (2), and (3).  P H = −2 E = − 2  −   5

H =+

2P 5

H=

2P  5

 P G = + 4E = 4  −   5

G=−

4P 5

G=

4P  5

 P F = P − 3E = P − 3  −   5

F =+

8P 5

F=

8P  5





P − 3E + 8 E = 0; E = −

www.elsolucionario.net PROBLEM 6.118 Four beams, each of length 2a, are nailed together at their midpoints to form the support system shown. Assuming that only vertical forces are exerted at the connections, determine the vertical reactions at A, D, E, and H.

Note that, if we assume P is applied to EG, each individual member FBD looks like so

2 Fleft = 2 Fright = Fmiddle

Labeling each interaction force with the letter corresponding to the joint of its application, we see that B = 2 A = 2F C = 2B = 2D G = 2C = 2 H P + F = 2G ( = 4C = 8 B = 16 F ) = 2 E

From P + F = 16 F ,

F=

P 15

so A =



P  15

D=

2P  15

H=

4P  15

E=

8P  15



SOLUTION

www.elsolucionario.net PROBLEM 6.119 Each of the frames shown consists of two L-shaped members connected by two rigid links. For each frame, determine the reactions at the supports and indicate whether the frame is rigid.

SOLUTION Member FBDs:

I

II

FBD I:

ΣM A = 0: aF1 − 2aP = 0 F1 = 2 P;

FBD II:

ΣM B = 0: − aF2 = 0 F2 = 0

ΣFy = 0:

Ay − P = 0 A y = P

ΣFx = 0: Bx + F1 = 0, Bx = − F1 = −2 P

B x = 2P

ΣFy = 0: B y = 0

FBD I:

ΣFx = 0: − Ax − F1 + F2 = 0

so B = 2P Ax = F2 − F1 = 0 − 2 P



A x = 2P

so A = 2.24P

26.6° 

Frame is rigid.  (b)

FBD left:

FBD whole:

I FBD I:

ΣM E = 0:

FBD II:

ΣM B = 0:

II a a 5a P + Ax − Ay = 0 Ax − 5 Ay = − P 2 2 2 3aP + aAx − 5aAy = 0 Ax − 5 Ay = −3P

This is impossible unless P = 0.

not rigid 



(a)


(c)

Member FBDs:

I FBD I:

II

ΣFy = 0: A − P = 0

A = P 

ΣM D = 0: aF1 − 2aA = 0 F1 = 2 P ΣFx = 0: F2 − F1 = 0 F2 = 2 P  ΣM B = 0: 2aC − aF1 = 0 C =

F1 =P 2

C = P 

ΣFx = 0: F1 − F2 + Bx = 0 Bx = P − P = 0  ΣFx = 0: By + C = 0 B y = − C = − P

B=P 

Frame is rigid. 



FBD II:


SOLUTION Member FBDs:

I FBD II: FBD I:

ΣFy = 0: B y = 0

II ΣM B = 0: aF2 = 0

F2 = 0

ΣM A = 0: aF2 − 2aP = 0, but F2 = 0

so P = 0 (b)

not rigid for P ≠ 0 

Member FBDs:

Note: Seven unknowns ( Ax , Ay , Bx , B y , F1 , F2 , C ), but only six independent equations System is statically indeterminate.  

System is, however, rigid. 



(a)


FBD whole:

FBD right:

I FBD I:

ΣM A = 0: 5aBy − 2aP = 0 ΣFy = 0: Ay − P +

FBD II: FBD I:

II

ΣM C = 0:

2 P=0 5

a 5a Bx − B y = 0 Bx = 5 B y 2 2

ΣFx = 0: Ax + Bx = 0

Ax = − Bx

By =

2 P  5

Ay =

3 P 5

B x = 2P A x = 2P A = 2.09P

16.70° 

B = 2.04P

11.31° 

System is rigid. 



(c)


SOLUTION

(a)

FBD whole:

ΣM A = 0: − 2aP − ΣFx = 0: Ax − ΣFy = 0:

a 4 1 B + 5a B = 0 B = 2.06 P 4 17 17

4 17

Ay − P +

B=0 1 17

B = 2.06P

14.04° 

A = 2.06P

14.04° 

A x = 2P

B = 0 Ay =

P 2

rigid  (b)

FBD whole:

Since B passes through A,

ΣM A = 2aP = 0 only if P = 0.

no equilibrium if P ≠ 0

not rigid 



Note: In all three cases, the right member has only three forces acting, two of which are parallel. Thus, the third force, at B, must be parallel to the link forces.


FBD whole:

ΣM A = 0: 5a

1 17

ΣFx = 0: Ax +

B+

4 17

ΣFy = 0: Ay − P +

3a 4 17 B − 2aP = 0 B = P 4 17 4

B=0 1 17

B = 1.031P

14.04° 

A = 1.250P

36.9° 

Ax = − P

B=0

Ay = P −

P 3P = 4 4

System is rigid. 



(c)


SOLUTION We note that BD is a two-force member. Free body: Member ABC: We have the components: Px = (400 N) sin 30° = 200 N Py = (400 N) cos 30° = 346.41 N (FBD ) x =

25 FBD 65

(FBD ) y =

60 FBD 65

ΣM C = 0: ( FBD ) x (45) + ( FBD ) y (30) − Px (45 + 300sin 30°) − Py (30 + 300cos 30°) = 0

 25   60   FBD  (45) +  FBD  (30) = (200)(195) + (346.41)(289.81)  65   65  45 FBD = 139.39 × 103 FBD = 3097.6 N

(a)

Vertical component of force exerted on shearing blade at D. ( FBD ) y =

60 60 (3097.6 N) = 2859.3 N FBD = 65 65

(FBD ) y = 2860 N 



The shear shown is used to cut and trim electronic circuit board laminates. For the position shown, determine (a) the vertical component of the force exerted on the shearing blade at D, (b) the reaction at C.


(b)

Returning to FB diagram of member ABC, ΣFx = 0: ( FBD ) x − Px − C x = 0 Cx = ( FBD ) x − Px =

25 FBD − Px 65

25 (3097.6) − 200 65 Cx = +991.39 =

C x = 991.39 N

ΣFy = 0: ( FBD ) y − Py − C y = 0

C y = +2512.9 N C = 2295 N

60 60 (3097.6) − 346.41 FBD − Py = 65 65 C y = 2512.9 C = 2700 N

68.5° 



C y = ( FBD ) y − Py =

www.elsolucionario.net PROBLEM 6.123 The press shown is used to emboss a small seal at E. Knowing that P = 250 N, determine (a) the vertical component of the force exerted on the seal, (b) the reaction at A.

SOLUTION FBD Stamp D:


ΣFy = 0: E − FBD cos 20° = 0, E = FBD cos 20°

FBD ABC:

ΣM A = 0: (0.2 m)(sin 30°)( FBD cos 20°) + (0.2 m)(cos 30°)( FBD sin 20°) − [(0.2 m)sin 30° + (0.4 m) cos15°](250 N) = 0 FBD = 793.64 N C

and, from above, E = (793.64 N) cos 20°

(a)

E = 746 N 

ΣFx = 0: Ax − (793.64 N)sin 20° = 0 A x = 271.44 N

ΣFy = 0: Ay + (793.64 N) cos 20° − 250 N = 0 A y = 495.78 N

so

(b)

A = 565 N

61.3° 


www.elsolucionario.net PROBLEM 6.124 The press shown is used to emboss a small seal at E. Knowing that the vertical component of the force exerted on the seal must be 900 N, determine (a) the required vertical force P, (b) the corresponding reaction at A.

SOLUTION FBD Stamp D:

(a) FBD ABC:

ΣM A = 0: [(0.2 m)(sin 30°)](957.76 N) cos 20° + [(0.2 m)(cos 30°)](957.76 N) sin 20° − [(0.2 m)sin 30° + (0.4 m) cos15°]P = 0 P = 301.70 N,

(b)

P = 302 N 

ΣFx = 0: Ax − (957.76 N)sin 20° = 0 A x = 327.57 N

ΣFy = 0: −Ay + (957.76 N) cos 20° − 301.70 N = 0 A y = 598.30 N

so

A = 682 N

61.3° 



ΣFy = 0: 900 N − FBD cos 20° = 0, FBD = 957.76 N C

www.elsolucionario.net PROBLEM 6.125 Water pressure in the supply system exerts a downward force of 135 N on the vertical plug at A. Determine the tension in the fusible link DE and the force exerted on member BCE at B.

SOLUTION Free body: Entire linkage:

C = + 135 N

Free body: Member BCE: ΣFx = 0: Bx = 0 ΣM B = 0: (135 N)(6 mm) − TDE (10 mm) = 0 TDE = 81.0 N  ΣFy = 0: 135 + 81 − By = 0 By = + 216 N

B = 216 N 



+ ΣFy = 0: C − 135 = 0

www.elsolucionario.net PROBLEM 6.126 An 84-lb force is applied to the toggle vise at C. Knowing that θ = 90°, determine (a) the vertical force exerted on the block at D, (b) the force exerted on member ABC at B.

SOLUTION


We note that BD is a two-force member. Free body: Member ABC: BD = (7) 2 + (24)2 = 25 in.

We have

( FBD ) x =

7 24 FBD , ( FBD ) y = FBD 25 25

ΣM A = 0: ( FBD ) x (24) + ( FBD ) y (7) − 84(16) = 0

 7   24   FBD  (24) +  FBD  (7) = 84(16) 25 25     336 FBD = 1344 25 FBD = 100 lb tan α =

24 α = 73.7° 7 FBD = 100.0 lb

(b)

Force exerted at B.

(a)

Vertical force exerted on block. ( FBD ) y =

73.7° 

24 24 FBD = (100 lb) = 96 lb 25 25 (FBD ) y = 96.0 lb 


www.elsolucionario.net PROBLEM 6.127 Solve Problem 6.126 when θ = 0. PROBLEM 6.126 An 84-lb force is applied to the toggle vise at C. Knowing that θ = 90°, determine (a) the vertical force exerted on the block at D, (b) the force exerted on member ABC at B.

SOLUTION


We note that BD is a two-force member. Free body: Member ABC: BD = (7) 2 + (24)2 = 25 in.

We have

( FBD ) x =

7 24 FBD , ( FBD ) y = FBD 25 25

ΣM A = 0: ( FBD ) x (24) + ( FBD ) y (7) − 84(40) = 0  7   24   FBD  (24) +  FBD  (7) = 84(40) 25 25     336 FBD = 3360 25 FBD = 250 lb tan α =

24 α = 73.7° 7 FBD = 250.0 lb

(b)

Force exerted at B.

(a)

Vertical force exerted on block. ( FBD ) y =

73.7° 

24 24 FBD = (250 lb) = 240 lb 25 25 (FBD ) y = 240 lb 


www.elsolucionario.net PROBLEM 6.128 For the system and loading shown, determine (a) the force P required for equilibrium, (b) the corresponding force in member BD, (c) the corresponding reaction at C.

SOLUTION FBD I:

I: ΣM C = 0: R( FBD sin 30°) − [ R(1 − cos 30°)](100 N) − R(50 N) = 0 FBD = 126.795 N ΣFx = 0: −C x + (126.795 N) cos 30° = 0

(b) FBD = 126.8 N T  C x = 109.808 N

ΣFy = 0: C y + (126.795 N) sin 30° − 100 N − 50 N = 0 C y = 86.603 N

II:

so (c) C = 139.8 N

38.3° 

FBD II: ΣM A = 0: aP − a[(126.795 N) cos 30°] = 0

(a ) P = 109.8 N





Member FBDs:

www.elsolucionario.net PROBLEM 6.129 The Whitworth mechanism shown is used to produce a quick-return motion of Point D. The block at B is pinned to the crank AB and is free to slide in a slot cut in member CD. Determine the couple M that must be applied to the crank AB to hold the mechanism in equilibrium when (a) α = 0, (b) α = 30°.

SOLUTION Free body: Member CD: ΣM C = 0: B(0.5 m) − (1200 N)(0.7 m) = 0 B = 1680 N

Free body: Crank AB: ΣM A = 0: M − (1680 N)(0.1 m) = 0 M = 168.0 N ⋅ m

(b)

M = 168.0 N ⋅ m



Geometry: AB = 100 mm, α = 30° tan θ =

50 θ = 5.87° 486.6

BC = (50)2 + (486.6) 2 = 489.16 mm

Free body: Member CD:

ΣM C = 0: B (0.48916) − (1200 N)(0.7) cos 5.87° = 0 B = 1708.2 N



(a)


Free body: Crank AB: ΣM A = 0: M − ( B sin 65.87°)(0.1 m) = 0 M = 155.9 N ⋅ m




M = (1708.2 N)(0.1 m)sin 65.87° M = 155.90 N ⋅ m


www.elsolucionario.net PROBLEM 6.130 Solve Problem 6.129 when (a) α = 60°, (b) α = 90°. PROBLEM 6.129 The Whitworth mechanism shown is used to produce a quick-return motion of Point D. The block at B is pinned to the crank AB and is free to slide in a slot cut in member CD. Determine the couple M that must be applied to the crank AB to hold the mechanism in equilibrium when (a) α = 0, (b) α = 30°.

SOLUTION Geometry:


(a)

AB = 100 mm, α = 60° BE = 100 cos 60° = 86.60

CE = 400 + 100sin 60° = 450 86.60 tan θ = θ = 10.89° 450 BC = (86.60)2 + (450) 2 = 458.26 mm


ΣM C = 0: B (0.45826 m) − (1200 N)(0.7 m) cos10.89° = 0 B = 1800.0 N

Free body: Crank AB: ΣM A = 0: M − ( B sin 40.89°)(0.1 m) = 0 M = (1800.0 N)(0.1 m)sin 40.89° M = 117.83 N ⋅ m

M = 117.8 N ⋅ m





(b)

Free body: Member CD: 0.1 m 0.4 m θ = 14.04°

tan θ =

BC = (0.1) 2 + (0.4) 2 = 0.41231 m ΣM C = 0: B(0.41231) − (1200 N)(0.7 cos14.04°) = 0 B = 1976.4 N

Free body: Crank AB:

M = (1976.4 N)(0.1 m) sin14.04° M = 47.948 N ⋅ m

M = 47.9 N ⋅ m





ΣM A = 0: M − ( B sin14.04°)(0.1 m) = 0

www.elsolucionario.net PROBLEM 6.131 A couple M of magnitude 1.5 kN ⋅ m is applied to the crank of the engine system shown. For each of the two positions shown, determine the force P required to hold the system in equilibrium.

SOLUTION FBDs:

50 mm 175 mm 2 = 7

Dimensions in mm

Note:

tan θ =

FBD whole:

ΣM A = 0: (0.250 m)C y − 1.5 kN ⋅ m = 0 C y = 6.00 kN

FBD piston:


(a)

ΣFy = 0: C y − FBC sin θ = 0 FBC =

Cy

sin θ

=

6.00 kN sinθ

ΣFx = 0: FBC cos θ − P = 0 P = FBC cos θ =

6.00 kN = 7 kips tan θ P = 21.0 kN





(b)

FBDs:

Dimensions in mm

2 as above 7

Note:

tan θ =

FBD whole:

ΣM A = 0: (0.100 m)C y − 1.5 kN ⋅ m = 0 C y = 15 kN

ΣFx = 0:

Cy

sin θ

FBC cos θ − P = 0 P = FBC cos θ =

Cy

tan θ

=

15 kN 2/7

P = 52.5 kN





ΣFy = 0: C y − FBC sin θ = 0 FBC =

www.elsolucionario.net PROBLEM 6.132 A force P of magnitude 16 kN is applied to the piston of the engine system shown. For each of the two positions shown, determine the couple M required to hold the system in equilibrium.

SOLUTION FBDs:

Note:

FBD piston:

50 mm 175 mm 2 = 7

Dimensions in mm

tan θ =

ΣFx = 0: FBC cos θ − P = 0 FBC =

P cos θ

ΣFy = 0: C y − FBC sin θ = 0 C y = FBC sin θ = P tan θ =

FBD whole:


(a)

2 P 7

ΣM A = 0: (0.250 m)C y − M = 0

2 M = (0.250 m)   (16 kN) 7 = 1.14286 kN ⋅ m

M = 1143 N ⋅ m





(b)

FBDs:

Dimensions in mm

tan θ =

Note: FBD piston, as above:

C y = P tan θ =

2 P 7

2 ΣM A = 0: (0.100 m)C y − M = 0 M = (0.100 m) (16 kN) 7 M = 0.45714 kN ⋅ m

M = 457 N ⋅ m





FBD whole:

2 as above 7

www.elsolucionario.net PROBLEM 6.133 The pin at B is attached to member ABC and can slide freely along the slot cut in the fixed plate. Neglecting the effect of friction, determine the couple M required to hold the system in equilibrium when θ = 30°.

SOLUTION Free body: Member ABC:


ΣM C = 0: (25 lb)(13.856 in.) − B(3 in.) = 0 B = +115.47 lb ΣFy = 0: − 25 lb + C y = 0 C y = +25 lb ΣFx = 0: 115.47 lb − Cx = 0 Cx = +115.47 lb


β = sin −1

5.196 ; β = 40.505° 8

CD cos β = (8 in.) cos 40.505° = 6.083 in. ΣM D = 0: M − (25 lb)(5.196 in.) − (115.47 lb)(6.083 in.) = 0 M = +832.3 lb ⋅ in. M = 832 lb ⋅ in. 


www.elsolucionario.net PROBLEM 6.134 The pin at B is attached to member ABC and can slide freely along the slot cut in the fixed plate. Neglecting the effect of friction, determine the couple M required to hold the system in equilibrium when θ = 60°.

SOLUTION Free body: Member ABC:


ΣM C = 0: (25 lb)(8 in.) − B(5.196 in.) = 0 B = +38.49 lb ΣFx = 0: 38.49 lb − C x = 0 Cx = +38.49 lb ΣFy = 0: − 25 lb + C y = 0 C y = +25 lb

Free body: Member CD: 3 8

β = sin −1 ; β = 22.024° CD cos β = (8 in.) cos 22.024° = 7.416 in. ΣM D = 0: M − (25 lb)(3 in.) − (38.49 lb)(7.416 in.) = 0 M = +360.4 lb ⋅ in.

M = 360 lb ⋅ in.




www.elsolucionario.net PROBLEM 6.135 Rod CD is attached to the collar D and passes through a collar welded to end B of lever AB. Neglecting the effect of friction, determine the couple M required to hold the system in equilibrium when θ = 30°.

SOLUTION FBD DC: D y sin 30° − (150 N) cos 30° = 0


ΣFx′ = 0:

Dy = (150 N) ctn 30° = 259.81 N

FBD machine: ΣM A = 0: (0.100 m)(150 N) + d (259.81 N) − M = 0 d = b − 0.040 m b= so

0.030718 m tan 30

b = 0.053210 m d = 0.0132100 m M = 18.4321 N ⋅ m M = 18.43 N ⋅ m.




www.elsolucionario.net PROBLEM 6.136 Rod CD is attached to the collar D and passes through a collar welded to end B of lever AB. Neglecting the effect of friction, determine the couple M required to hold the system in equilibrium when θ = 30°.

SOLUTION B ⊥ CD

Note: FBD DC:

ΣFx′ = 0: D y sin 30° − (300 N) cos 30° = 0 Dy =

300 N = 519.62 N tan 30°

ΣM A = 0:

0.200 m 519.62 N − M = 0 sin 30°

M = 207.85 N ⋅ m

M = 208 N ⋅ m





FBD machine:

www.elsolucionario.net PROBLEM 6.137 Two rods are connected by a frictionless collar B. Knowing that the magnitude of the couple MA is 500 lb · in., determine (a) the couple MC required for equilibrium, (b) the corresponding components of the reaction at C.

(a)


SOLUTION

Free body: Rod AB & collar: ΣM A = 0: ( B cos α )(6 in.) + ( B sin α )(8 in.) − M A = 0 B = (6cos 21.8° + 8sin 21.8°) − 500 = 0 B = 58.535 lb

Free body: Rod BC: +ΣM C = 0: M C − B = 0 M C = B = (58.535 lb)(21.541 in.) = 1260.9 lb ⋅ in. M C = 1261 lb ⋅ in.

(b)



ΣFx = 0: Cx + B cos α = 0 Cx = − B cos α = −(58.535 lb) cos 21.8° = −54.3 lb C x = 54.3 lb



ΣFy = 0: C y − B sin α = 0 C y = B sin α = (58.535 lb) sin 21.8° = +21.7 lb C y = 21.7 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 943

www.elsolucionario.net PROBLEM 6.138 Two rods are connected by a frictionless collar B. Knowing that the magnitude of the couple MA is 500 lb · in., determine (a) the couple MC required for equilibrium, (b) the corresponding components of the reaction at C.

(a)


SOLUTION

Free body: Rod AB: ΣM A = 0: B(10 in.) − 500 lb ⋅ in. = 0 B = 50.0

Free body: Rod BC & collar: ΣM C = 0: M C − (0.6 B)(20 in.) − (0.8 B)(8 in.) = 0 M C − (30 lb)(20 in.) − (40 lb)(8 in.) = 0 M C = 920 lb ⋅ in.

(b)

M C = 920 lb ⋅ in.



C x = 30.0 lb



ΣFx = 0: Cx + 0.6 B = 0 Cx = −0.6 B = −0.6(50.0 lb) = −30.0 lb

ΣFy = 0: C y − 0.8B = 0 C y = 0.8 B = 0.8(50.0 lb) = +40.0 lb C y = 40.0 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 944

www.elsolucionario.net PROBLEM 6.139 Two hydraulic cylinders control the position of the robotic arm ABC. Knowing that in the position shown the cylinders are parallel, determine the force exerted by each cylinder when P = 160 N and Q = 80 N.

SOLUTION

ΣM B = 0:

4 FAE (150 mm) − (160 N)(600 mm) = 0 5

FAE = +800 N

FAE = 800 N T 

3 ΣFx = 0: − (800 N) + Bx − 80 N = 0 5

Bx = +560 N 4 ΣFy = 0: − (800 N) + By − 160 N = 0 5

By = +800 N

Free body: Member BDF: ΣM F = 0: (560 N)(400 mm) − (800 N)(300 mm) − FDG = −100 N

4 FDG (200 mm) = 0 5

FDG = 100.0 N C 





SOLUTION Free body: Member ABC: ΣM B = 0:

4 (600 N)(150 mm) − P(600 mm) = 0 5

P = +120 N ΣM C = 0:

P = 120.0 N 

4 (600)(750 mm) − By (600 mm) = 0 5

By = +600 N

Free body: Member BDF: ΣM F = 0: Bx (400 mm) − (600 N)(300 mm) 4 − (50 N)(200 mm) = 0 5

Bx = +470 N

Return to free body: Member ABC: 3 ΣFx = 0: − (600 N) + 470 N − Q = 0 5

Q = +110 N

Q = 110.0 N





Two hydraulic cylinders control the position of the robotic arm ABC. In the position shown, the cylinders are parallel and both are in tension. Knowing the FAE = 600 N and FDG = 50 N, determine the forces P and Q applied at C to arm ABC.

www.elsolucionario.net PROBLEM 6.141 The tongs shown are used to apply a total upward force of 45 kN on a pipe cap. Determine the forces exerted at D and F on tong ADF.

SOLUTION

By symmetry,

FBD ADF:

A = B = 22.5 kN

ΣM F = 0: (75 mm) D − (100 mm)(22.5 kN) = 0 D = 30.0 kN



ΣFx = 0: Fx − D = 0 Fx = D = 30 kN ΣFy = 0: 22.5 kN − Fy = 0 Fy = 22.5 kN F = 37.5 kN

so

36.9° 



FBD whole:

www.elsolucionario.net PROBLEM 6.142 If the toggle shown is added to the tongs of Problem 6.141 and a single vertical force is applied at G, determine the forces exerted at D and F on tong ADF.

SOLUTION Free body: Toggle: By symmetry,

Ay =

1 (45 kN) = 22.5 kN 2

Ax 22.5 kN = 22 mm 55 mm Ax = 56.25 kN

Free body: Tong ADF: ΣFy = 0: 22.5 kN − Fy = 0 Fy = +22.5 kN ΣM F = 0 : D (75 mm) − (22.5 kN)(100 mm) − (56.25 kN)(160 mm) = 0 D = +150 kN

D = 150.0 kN



ΣFx = 0: 56.25 kN − 150 kN + Fx = 0 Fx = 93.75 kN F = 96.4 kN

13.50° 



AG is a two-force member.

www.elsolucionario.net PROBLEM 6.143 A small barrel weighing 60 lb is lifted by a pair of tongs as shown. Knowing that a = 5 in., determine the forces exerted at B and D on tong ABD.

SOLUTION We note that BC is a two-force member.

Bx By = 15 5

Bx = 3By

ΣM D = 0: By (3 in.) + 3By (5 in.) − (60 lb)(9 in.) = 0 By = 30 lb Bx = 3By : Bx = 90 lb ΣFx = 0: −90 lb + Dx = 0

D x = 90 lb

ΣFy = 0: 60 lb − 30 lb − D y = 0

D y = 30 lb B = 94.9 lb

18.43° 

D = 94.9 lb

18.43° 



Free body: Tong ABD:



A 39-ft length of railroad rail of weight 44 lb/ft is lifted by the tongs shown. Determine the forces exerted at D and F on tong BDF.

SOLUTION Free body: Rail: W = (39 ft)(44 lb/ft) = 1716 lb

Free body: Upper link:

By symmetry,

1 E y = Fy = W = 858 lb 2

By symmetry,

1 ( FAB ) y = ( FAC ) y = W = 858 lb 2

Since AB is a two-force member, ( FAB ) x ( FAB ) y = 9.6 6

Free Body: Tong BDF:

( FAB ) x =

9.6 (858) = 1372.8 lb 6

ΣM D = 0: (Attach FAB at A.) Fx (8) − ( FAB ) x (18) − Fy (0.8) = 0 Fx (8) − (1372.8 lb)(18) − (858 lb)(0.8) = 0 Fx = +3174.6 lb

F = 3290 lb

15.12° 

ΣFx = 0: − Dx + ( FAB ) x + Fx = 0 Dx = ( FAB ) x + Fx = 1372.8 + 3174.6 = 4547.4 lb ΣFy = 0:

Dy + ( FAB ) y − Fy = 0 Dy = 0

D = 4550 lb




www.elsolucionario.net PROBLEM 6.145 Determine the magnitude of the gripping forces produced when two 300-N forces are applied as shown.

SOLUTION We note that AC is a two-force member. FBD handle CD: 2.8 8.84

A

 1  + (30 mm)  A = 0  8.84 

A = 2863.6 8.84 N

Dimensions in mm

FBD handle AB: ΣM B = 0: (132 mm)(300 N) − (120 mm)

1 8.84

(2863.6 8.84 N)

+ (36 mm) F = 0 F = 8.45 kN  Dimensions in mm



ΣM D = 0: − (126 mm)(300 N) − (6 mm)

www.elsolucionario.net PROBLEM 6.146 The compound-lever pruning shears shown can be adjusted by placing pin A at various ratchet positions on blade ACE. Knowing that 300-lb vertical forces are required to complete the pruning of a small branch, determine the magnitude P of the forces that must be applied to the handles when the shears are adjusted as shown.

SOLUTION

( FAB ) x ( FAB ) y = 0.65 in. 0.55 in. 11 ( FAB ) y = ( FAB ) x 13

(1)

Free body: Blade ACE:

ΣM C = 0: (300 lb)(1.6 in.) − ( FAB ) x (0.5 in.) − ( FAB ) y (1.4 in.) = 0

Use Eq. (1):

( FAB ) x (0.5 in.) +

11 ( FAB ) x (1.4 in.) = 480 lb ⋅ in. 13

1.6846( FAB ) x = 480 ( FAB ) y =

11 (284.9 lb) 13

( FAB ) x = 284.9 lb ( FAB ) y = 241.1 lb

Free body: Lower handle:

ΣM D = 0: (241.1 lb)(0.75 in.) − (284.9 lb)(0.25 in.) − P(3.5 in.) = 0 P = 31.3 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 952


We note that AB is a two-force member.

www.elsolucionario.net PROBLEM 6.147 The pliers shown are used to grip a 0.3-in.-diameter rod. Knowing that two 60-lb forces are applied to the handles, determine (a) the magnitude of the forces exerted on the rod, (b) the force exerted by the pin at A on portion AB of the pliers.

SOLUTION Free body: Portion AB: (a)

ΣM A = 0: Q (1.2 in.) − (60 lb)(9.5 in.) = 0

(b)

ΣFx = 0: Q(sin 30°) + Ax = 0 (475 lb)(sin 30°) + Ax = 0 Ax = −237.5 lb

A x = 237.5 lb

ΣFy = 0: − Q (cos 30°) + Ay − 60 lb = 0 −(475 lb)(cos 30°) + Ay − 60 lb = 0 Ay = +471.4 lb

A y = 471.4 lb A = 528 lb

63.3° 



Q = 475 lb 

www.elsolucionario.net PROBLEM 6.148 In using the bolt cutter shown, a worker applies two 300-N forces to the handles. Determine the magnitude of the forces exerted by the cutter on the bolt.

SOLUTION

I FBD I:

FBD handle BC:

II

Dimensions in mm

ΣFx = 0: Bx = 0

FBD II:

ΣM C = 0: (12 mm)By − (448 mm)300 N = 0 By = 11, 200.0 N

Then

FBD I:

ΣM A = 0: (96 mm) By − (24 mm) F = 0

F = 4 By F = 44,800 N = 44.8 kN 



FBD cutter AB:

www.elsolucionario.net PROBLEM 6.149 The specialized plumbing wrench shown is used in confined areas (e.g., under a basin or sink). It consists essentially of a jaw BC pinned at B to a long rod. Knowing that the forces exerted on the nut are equivalent to a clockwise (when viewed from above) couple of magnitude 135 lb ⋅ in., determine (a) the magnitude of the force exerted by pin B on jaw BC, (b) the couple M0 that is applied to the wrench.

Free body: Jaw BC: This is a two-force member. Cy 1.5 in.

Free body: Nut:

=

Cx C y = 2.4 C x in.

5 8

ΣFx = 0: Bx = C x

(1)

ΣFy = 0: By = C y = 2.4 C x

(2)

ΣFx = 0: C x = Dx ΣM = 135 lb ⋅ in.

C x (1.125 in.) = 135 lb ⋅ in. C x = 120 lb

(a)

Eq. (1):

Bx = C x = 120 lb

Eq. (2):

By = C y = 2.4(120 lb) = 288 lb

(

B = Bx2 + By2

(b)

)

1/ 2

= (1202 + 2882 )1/2

B = 312 lb 

Free body: Rod: ΣM D = 0: − M 0 + By (0.625 in.) − Bx (0.375 in.) = 0 − M 0 + (288)(0.625) − (120)(0.375) = 0 M 0 = 135.0 lb ⋅ in.





SOLUTION

www.elsolucionario.net PROBLEM 6.150 Determine the force P that must be applied to the toggle CDE to maintain bracket ABC in the position shown.

SOLUTION

Free body: Joint D: ( FCD ) x ( FCD ) y = 30 150

Similarly,

( FCD ) y = 5( FCD ) x Dimensions in mm

( FDE ) y = 5( FDE ) x ΣFy = 0: ( FDE ) y = ( FCD ) y

It follows that

( FDE ) x = ( FCD ) x ΣFx = 0: P − ( FDE ) x − ( FCD ) x = 0 ( FDE ) x = ( FCD ) x =

Also,

1 P 2

1  ( FDE ) y = ( FCD ) y = 5  P  = 2.5P 2 


1  ΣM A = 0: (2.5 P )(300) +  P  (450) − (910 N)(150) = 0 2  (750 + 225) P = (910 N)(150) P = 140.0 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 956


We note that CD and DE are two-force members.

www.elsolucionario.net PROBLEM 6.151 Determine the force P that must be applied to the toggle CDE to maintain bracket ABC in the position shown.

SOLUTION


We note that CD and DE are two-force members. Free body: Joint D: ( FCD ) x ( FCD ) y = 30 150

Similarly,

( FCD ) y = 5( FCD ) x

( FDE ) y = 5( FDE ) x

Dimensions in mm

ΣFy = 0: ( FDE ) y = ( FCD ) y It follows that

( FDE ) x = ( FCD ) x Σ Fx = 0: ( FDE ) x + ( FCD ) x − P = 0 ( FDE ) x = ( FCD ) x =

Also,

1 P 2

1  ( FDE ) y = ( FCD ) y = 5  P  = 2.5P 2 


1  ΣM A = 0: (2.5P)(300) −  P  (450) − (910 N)(150) = 0 2  (750 − 225) P = (910 N)(150) P = 260 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 957

www.elsolucionario.net PROBLEM 6.152 A 45-lb shelf is held horizontally by a self-locking brace that consists of two parts EDC and CDB hinged at C and bearing against each other at D. Determine the force P required to release the brace.

SOLUTION Free body: Shelf:

By = 28.125 lb

Free body: Portion ECB:

ΣM E = 0: − Bx (7.5 in.) − P(7.5 in.) − (28.125 lb)(10 in.) = 0 Bx = −37.5 − P

Free body: Portion CDB:

ΣM C = 0: − (28.125 lb)(4.6 in.) − Bx (2.2 in.) = 0 −(28.125 lb)(4.6 in.) − ( −37.5 − P)(2.2 in.) = 0 P = 21.3 lb P = 21.3 lb





ΣM A = 0: By (10 in.) − (45 lb)(6.25 in.) = 0

www.elsolucionario.net PROBLEM 6.153 The telescoping arm ABC is used to provide an elevated platform for construction workers. The workers and the platform together have a mass of 200 kg and have a combined center of gravity located directly above C. For the position when θ = 20°, determine (a) the force exerted at B by the single hydraulic cylinder BD, (b) the force exerted on the supporting carriage at A.

SOLUTION b = (2.4 m) cos 20° = 2.2553 m c = (2.4 m) sin 20° = 0.8208 m d = b − 0.5 = 1.7553 m e = c + 0.9 = 1.7208 m tan β =

e 1.7208 = ; β = 44.43° d 1.7553

Free body: Arm ABC: We note that BD is a two-force member.

W = (200 kg)(9.81 m/s 2 ) = 1.962 kN (a)

ΣM A = 0: (1.962 kN)(4.6985 m) − FBD sin 44.43°(2.2553 m) + FBD cos 44.43(0.8208 m) = 0 9.2185 − FBD (0.9927) = 0: FBD = 9.2867 kN FBD = 9.29 kN

(b)

44.4° 

ΣFx = 0: Ax − FBD cos β = 0 Ax = (9.2867 kN) cos 44.43° = 6.632 kN

A x = 6.632 kN

ΣFy = 0: Ay − 1.962 kN + FBD sin β = 0 Ay = 1.962 kN − (9.2867 kN)sin 44.43° = −4.539 kN A y = 4.539 kN A = 8.04 kN

34.4° 



a = (5 m) cos 20° = 4.6985 m

Geometry:

www.elsolucionario.net PROBLEM 6.154 The telescoping arm ABC can be lowered until end C is close to the ground, so that workers can easily board the platform. For the position when θ = −20°, determine (a) the force exerted at B by the single hydraulic cylinder BD, (b) the force exerted on the supporting carriage at A.

SOLUTION b = (2.4 m) cos 20° = 2.2552 m c = (2.4 m) sin 20° = 0.8208 m d = b − 0.5 = 1.7553 m e = 0.9 − c = 0.0792 m tan β =

e 0.0792 = ; β = 2.584° d 1.7552

Free body: Arm ABC: We note that BD is a two-force member. W = (200 kg)(9.81 m/s 2 ) W = 1962 N = 1.962 kN (a)

ΣM A = 0: (1.962 kN)(4.6985 m) − FBD sin 2.584°(2.2553 m) − FBD cos 2.584°(0.8208 m) = 0 9.2185 − FBD (0.9216) = 0 FBD = 10.003 kN FBD = 10.00 kN

(b)

2.58° 

ΣFx = 0: Ax − FBD cos β = 0 Ax = (10.003 kN)cos 2.583° = 9.993 kN

A x = 9.993 kN

ΣFy = 0: Ay − 1.962 kN + FBD sin β = 0 Ay = 1.962 kN − (10.003 kN)sin 2.583° = −1.5112 kN A y = 1.5112 kN A = 10.11 kN

8.60° 



a = (5 m) cos 20° = 4.6985 m

Geometry:

www.elsolucionario.net PROBLEM 6.155 The bucket of the front-end loader shown carries a 3200-lb load. The motion of the bucket is controlled by two identical mechanisms, only one of which is shown. Knowing that the mechanism shown supports one-half of the 3200-lb load, determine the force exerted (a) by cylinder CD, (b) by cylinder FH.

SOLUTION Free body: Bucket: (one mechanism)

FAB = 1500 lb Note: There are two identical support mechanisms. Free body: One arm BCE: 8 20 β = 21.8°

tan β =

ΣM E = 0: (1500 lb)(23 in.) + FCD cos 21.8°(15 in.) − FCD sin 21.8°(5 in.) = 0 FCD = −2858 lb

FCD = 2.86 kips C 

Free body: Arm DFG:

ΣM G = 0: (1600 lb)(75 in.) + FFH sin 45°(24 in.) − FFH cos 45°(6 in.) = 0 FFH = −9.428 kips

FFH = 9.43 kips C 



ΣM D = 0: (1600 lb)(15 in.) − FAB (16 in.) = 0

www.elsolucionario.net PROBLEM 6.156 The motion of the bucket of the front-end loader shown is controlled by two arms and a linkage that are pin-connected at D. The arms are located symmetrically with respect to the central, vertical, and longitudinal plane of the loader; one arm AFJ and its control cylinder EF are shown. The single linkage GHDB and its control cylinder BC are located in the plane of symmetry. For the position and loading shown, determine the force exerted (a) by cylinder BC, (b) by cylinder EF.

SOLUTION Free body: Bucket

FGH = 4091 lb

Free body: Arm BDH ΣM D = 0: − (4091 lb)(24 in.) − FBC (20 in.) = 0 FBC = −4909 lb FBC = 4.91 kips C 

Free body: Entire mechanism (Two arms and cylinders AFJE)

Note: Two arms thus 2 FEF 18 in. 65 in. β = 15.48°

tan β =

ΣM A = 0: (4500 lb)(123 in.) + FBC (12 in.) + 2 FEF cos β (24 in.) = 0 (4500 lb)(123 in.) − (4909 lb)(12 in.) + 2 FEF cos 15.48°(24 in.) = 0 FEF = −10.690 lb




ΣM J = 0: (4500 lb)(20 in.) − FGH (22 in.) = 0

www.elsolucionario.net PROBLEM 6.157 The motion of the backhoe bucket shown is controlled by the hydraulic cylinders AD, CG, and EF. As a result of an attempt to dislodge a portion of a slab, a 2-kip force P is exerted on the bucket teeth at J. Knowing that θ = 45°, determine the force exerted by each cylinder.

SOLUTION

ΣM H = 0:

(Dimensions in inches) 4 3 FCG (10) + FCG (10) + P cos θ (16) + P sin θ (8) = 0 5 5 FCG = −

P (16 cos θ + 8 sin θ ) 14

(1)

Free body: Arm ABH and bucket: (Dimensions in inches)

ΣM B = 0:

4 3 FAD (12) + FAD (10) + P cos θ (86) − P sin θ (42) = 0 5 5 FAD = −

P (86 cos θ − 42 sin θ ) 15.6

(2)

Free body: Bucket and arms IEB + ABH: Geometry of cylinder EF:

16 in. 40 in. β = 21.801°

tan β =



Free body: Bucket:


ΣM I = 0: FEF cos β (18 in.) + P cos θ (28 in.) − P sin θ (120 in.) = 0 P(120 sin θ − 28 cos θ ) cos 21.8°(18) P = (120 sin θ − 28 cos θ ) 16.7126

FEF =

θ = 45°

From Eq. (1):

FCG = −

2 (16 cos 45° + 8 sin 45°) = −2.42 kips 14

FCG = 2.42 kips C 

From Eq. (2):

FAD = −

2 (86 cos 45° − 42 sin 45°) = −3.99 kips 15.6

FAD = 3.99 kips C 

From Eq. (3):

FEF =

2 (120 sin 45° − 28 cos 45°) = +7.79 kips 16.7126

FEF = 7.79 kips T 



For P = 2 kips,

(3)

www.elsolucionario.net PROBLEM 6.158 Solve Problem 6.157 assuming that the 2-kip force P acts horizontally to the right (θ = 0). PROBLEM 6.157 The motion of the backhoe bucket shown is controlled by the hydraulic cylinders AD, CG, and EF. As a result of an attempt to dislodge a portion of a slab, a 2-kip force P is exerted on the bucket teeth at J. Knowing that θ = 45°, determine the force exerted by each cylinder.

SOLUTION

ΣM H = 0:

(Dimensions in inches) 4 3 FCG (10) + FCG (10) + P cos θ (16) + P sin θ (8) = 0 5 5

FCG = −

P (16 cos θ + 8 sin θ ) 14

(1)

Free body: Arm ABH and bucket: (Dimensions in inches)

ΣM B = 0:

4 3 FAD (12) + FAD (10) + P cos θ (86) − P sin θ (42) = 0 5 5

FAD = −

P (86 cos θ − 42 sin θ ) 15.6

(2)

Free body: Bucket and arms IEB + ABH: Geometry of cylinder EF:

16 in. 40 in. β = 21.801°

tan β =



Free body: Bucket:


ΣM I = 0: FEF cos β (18 in.) + P cos θ (28 in.) − P sin θ (120 in.) = 0 P(120 sin θ − 28 cos θ ) cos 21.8°(18) P = (120 sin θ − 28 cos θ ) 16.7126

FEF =

θ =0

From Eq. (1):

FCG = −

2 (16 cos 0 + 8 sin 0) = −2.29 kips 14

From Eq. (2):

FAD = −

2 (86 cos 0 − 42 sin 0) = −11.03 kips 15.6

FAD = 11.03 kips C 

From Eq. (3):

FEF =

2 (120 sin 0 − 28 cos 0) = −3.35 kips 16.7126


FCG = 2.29 kips C 



For P = 2 kips,

(3)

www.elsolucionario.net PROBLEM 6.159 In the planetary gear system shown, the radius of the central gear A is a = 18 mm, the radius of each planetary gear is b, and the radius of the outer gear E is (a + 2b). A clockwise couple of magnitude MA = 10 N ⋅ m is applied to the central gear A and a counterclockwise couple of magnitude MS = 50 N ⋅ m is applied to the spider BCD. If the system is to be in equilibrium, determine (a) the required radius b of the planetary gears, (b) the magnitude ME of the couple that must be applied to the outer gear E.

SOLUTION By symmetry,

F1 = F2 = F3 = F

ΣM A = 0: 3(rA F ) − 10 N ⋅ m = 0,

ΣM C = 0: rB ( F − F4 ) = 0,

FBD Gear C:

F=

10 N⋅m 3rA


FBD Central Gear:

F4 = F

ΣFx′ = 0: Cx′ = 0 ΣFy′ = 0: C y′ − 2 F = 0,

C y′ = 2 F

Gears B and D are analogous, each having a central force of 2F.

ΣM A = 0: 50 N ⋅ m − 3(rA + rB )2F = 0

FBD Spider:

50 N ⋅ m − 3(rA + rB )

20 N⋅m = 0 rA

rA + rB r = 2.5 = 1 + B , rA rA

rB = 1.5rA

Since rA = 18 mm, FBD Outer Gear:

rB = 27.0 mm 

(a)

ΣM A = 0: 3(rA + 2rB ) F − M E = 0 3(18 mm + 54 mm)

10 N ⋅ m − ME = 0 54 mm

M E = 40.0 N ⋅ m

(b)




www.elsolucionario.net PROBLEM 6.160 The gears D and G are rigidly attached to shafts that are held by frictionless bearings. If rD = 90 mm and rG = 30 mm, determine (a) the couple M0 that must be applied for equilibrium, (b) the reactions at A and B.

(a)

Projections on yz plane. Free body: Gear G:

ΣM G = 0: 30 N ⋅ m − J (0.03 m) = 0; J = 1000 N Free body: Gear D:

ΣM D = 0: M 0 − (1000 N)(0.09 m) = 0 M 0 = 90 N ⋅ m (b)

M 0 = (90.0 N ⋅ m)i 

Gear G and axle FH:

ΣM F = 0: H (0.3 m) − (1000 N)(0.18 m) = 0 H = 600 N

ΣFy = 0: F + 600 − 1000 = 0 F = 400 N

Gear D and axle CE:

ΣM C = 0: (1000 N)(0.18 m) − E (0.3 m) = 0 E = 600 N

ΣFy = 0: 1000 − C − 600 = 0 C = 400 N



SOLUTION


Free body: Bracket AE:

ΣFy = 0: A − 400 + 400 = 0

A = 0 

ΣM A = 0: M A + (400 N)(0.32 m) − (400 N)(0.2 m) = 0 M A = −48 N ⋅ m

M A = −(48.0 N ⋅ m)i 

Free body: Bracket BH:

ΣFy = 0: B − 600 + 600 = 0

B=0 

ΣM B = 0: M B + (600 N)(0.32 m) − (600 N)(0.2 m) = 0 M B = −(72.0 N ⋅ m)i 


M B = −72 N ⋅ m



SOLUTION We recall from Figure 4.10 that a universal joint exerts on members it connects a force of unknown direction and a couple about an axis perpendicular to the crosspiece. Free body: Shaft DF:

ΣM x = 0: M C cos30° − 500 lb ⋅ in. = 0 M C = 577.35 lb ⋅ in. Free body: Shaft BC: We use here x′, y ′, z with x′ along BC .

ΣM C = 0: − M R i′ − (577.35 lb ⋅ in.)i′ + (−5 in.)i′ × ( By j ′ + Bz k ) = 0



Two shafts AC and CF, which lie in the vertical xy plane, are connected by a universal joint at C. The bearings at B and D do not exert any axial force. A couple of magnitude 500 lb ⋅ in. (clockwise when viewed from the positive x-axis) is applied to shaft CF at F. At a time when the arm of the crosspiece attached to shaft CF is horizontal, determine (a) the magnitude of the couple that must be applied to shaft AC at A to maintain equilibrium, (b) the reactions at B, D, and E. (Hint: The sum of the couples exerted on the crosspiece must be zero.)


Equate coefficients of unit vectors to zero: i:

M A − 577.35 lb ⋅ in. = 0

j:

Bz = 0

k:

By = 0

M A = 577.35 lb ⋅ in. M A = 577 lb ⋅ in.  B=0

ΣF = 0:

B + C = 0,

B=0

since B = 0,

C=0

Return to free body of shaft DF.

ΣM D = 0

(Note that C = 0 and M C = 577.35 lb ⋅ in. )

(577.35 lb ⋅ in.)(cos 30°i + sin 30° j) − (500 lb ⋅ in.)i

+(6 in.)i × ( Ex i + E y j + Ez k ) = 0


(500 lb ⋅ in.)i + (288.68 lb ⋅ in.) j − (500 lb ⋅ in.)i

+ (6 in.) E y k − (6 in.) Ez j = 0 Equate coefficients of unit vectors to zero: j:

288.68 lb ⋅ in. − (6 in.)Ez = 0 Ez = 48.1 lb

k:

Ey = 0 ΣF = 0: C + D + E = 0

0 + Dy j + Dz k + Ex i + (48.1 lb)k = 0 i:

Ex = 0

j:

Dy = 0

k:

Dz + 48.1 lb = 0

Dz = −48.1 lb B=0 

Reactions are:

D = −(48.1 lb)k  E = (48.1 lb)k 


www.elsolucionario.net PROBLEM 6.162* Solve Problem 6.161 assuming that the arm of the crosspiece attached to shaft CF is vertical.

SOLUTION Free body: Shaft DF.

Σ M x = 0: M C − 500 lb ⋅ in. = 0 M C = 500 lb ⋅ in. Free body: Shaft BC:

We resolve −(520 lb ⋅ in.)i into components along x′ and y ′ axes:

−MC = −(500 lb ⋅ in.)(cos30°i′ + sin 30° j′) ΣMC = 0: M Ai′ − (500 lb ⋅ in.)(cos30°i′ + sin 30° j′) + (5 in.)i′ × ( By′ j′ + Bz k ) = 0 M A i′ − (433 lb ⋅ in.)i ′ − (250 lb ⋅ in.) j + (5 in.) By′k − (5 in.) Bz j′ = 0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 972


PROBLEM 6.161 Two shafts AC and CF, which lie in the vertical xy plane, are connected by a universal joint at C. The bearings at B and D do not exert any axial force. A couple of magnitude 500 lb · in. (clockwise when viewed from the positive x-axis) is applied to shaft CF at F. At a time when the arm of the crosspiece attached to shaft CF is horizontal, determine (a) the magnitude of the couple that must be applied to shaft AC at A to maintain equilibrium, (b) the reactions at B, D, and E. (Hint: The sum of the couples exerted on the crosspiece must be zero.)


Equate to zero coefficients of unit vectors:

i′: M A − 433 lb ⋅ in. = 0

M A = 433 lb ⋅ in. 

j′: − 250 lb ⋅ in. − (5 in.) Bz = 0

Bz = −50 lb

k : B y′ = 0 B = −(50 lb)k

Reactions at B. ΣF = 0: B − C = 0 −(50 lb)k − C = 0

C = −(50 lb)k

Return to free body of shaft DF.

ΣM D = 0: (6 in.)i × ( Ex i + E y j + Ez k ) − (4 in.)i × (−50 lb)k (6 in.) E y k − (6 in.) Ez j − (200 lb ⋅ in.) j = 0 k : Ey = 0 j: −(6 in.) Ez − 200 lb ⋅ in. = 0

Ez = −33.3 lb

ΣF = 0: C + D + E = 0

−(50 lb)k + Dy j + Dz k + Ex i − (33.3 lb)k = 0 i : Ex = 0 k : −50 lb − 33.3 lb + Dz = 0

Dz = 83.3 lb B = −(50 lb)k 

Reactions are

D = (83.3 lb)k  E = −(33.3 lb)k 



− (500 lb ⋅ in.)i + (500 lb ⋅ in.)i = 0



The large mechanical tongs shown are used to grab and lift a thick 7500-kg steel slab HJ. Knowing that slipping does not occur between the tong grips and the slab at H and J, determine the components of all forces acting on member EFH. (Hint: Consider the symmetry of the tongs to establish relationships between the components of the force acting at E on EFH and the components of the force acting at D on DGJ.)

SOLUTION Free body: Pin A:

T = W = mg = (7500 kg)(9.81 m/s 2 ) = 73.575 kN ΣFx = 0: ( FAB ) x = ( FAC ) x 1 ΣFy = 0: ( FAB ) y = ( FAC ) y = W 2

Also,

( FAC ) x = 2( FAC ) y = W

Free body: Member CDF: 1 Σ M D = 0: W (0.3) + W (2.3) − Fx (1.8) − Fy (0.5 m) = 0 2

or

1.8Fx + 0.5Fy = 1.45W

(1)



ΣFx = 0: Dx − Fx − W = 0 Ex − Fx = W

or

(2)

1 ΣFy = 0: Fy − Dy + W = 0 2 1 E y − Fy = W 2

or

(3)

Free body: Member EFH: 1 ΣM E = 0: Fx (1.8) + Fy (1.5) − H x (2.3) + W (1.8 m) = 0 2

(4)

ΣFx = 0: Ex + Fx − H x = 0 Ex + Fx = H x

or

2 Fx = H x − W

Subtract Eq. (2) from Eq. (5): Subtract Eq. (4) from 3 × (1):

3.6Fx = 5.25W − 2.3H x

Add Eq. (7) to 2.3 × Eq. (6):

8.2Fx = 2.95W Fx = 0.35976W

(5) (6) (7)

(8)

Substitute from Eq. (8) into Eq. (1):

(1.8)(0.35976W ) + 0.5Fy = 1.45W 0.5Fy = 1.45W − 0.64756W = 0.80244W Fy = 1.6049W

(9)


Ex − 0.35976W = W ; Ex = 1.35976W


1 E y − 1.6049W = W 2

From Eq. (5):

H x = Ex + Fx = 1.35976W + 0.35976W = 1.71952W

Recall that

1 Hy = W 2

E y = 2.1049W



1.8Fx + 1.5Fy = 2.3H x − 0.9W

or


Since all expressions obtained are positive, all forces are directed as shown on the free-body diagrams. Substitute

W = 73.575 kN

E y = 154.9 kN 

E x = 100.0 kN Fx = 26.5 kN

Fy = 118.1 kN 



H y = 36.8 kN 


H x = 126.5 kN



SOLUTION Free Body: Truss:

F = 2025 N

ΣFx = 0: Ex + 900 N + 900 N = 0 Ex = −1800 N E x = 1800 N ΣFy = 0: E y + 2025 N = 0 E y = −2025 N E y = 2025 N FAB = FBD = 0 

We note that AB and BD are zero-force members. Free body: Joint A: FAC FAD 900 N = = 2.25 3.75 3

FAC = 675 N T  FAD = 1125 N C 

Free body: Joint D: FCD FDE 1125 N = = 3 2.23 3.75

FCD = 900 N T  FDF = 675 N C 

Free body: Joint E:

ΣFx = 0: FEF − 1800 N = 0

FEF = 1800 N T 

ΣFy = 0: FCE − 2025 N = 0

FCE = 2025 N T 



ΣM E = 0: F (3 m) − (900 N)(2.25 m) − (900 N)(4.5 m) = 0


Free body: Joint F: ΣFy = 0:

2.25 FCF + 2025 N − 675 N = 0 3.75

FCF = −2250 N 3 (−2250 N) − 1800 N = 0 (Checks) 3.75


ΣFx = −

FCF = 2250 N C 


www.elsolucionario.net PROBLEM 6.165 Using the method of joints, determine the force in each member of the roof truss shown. State whether each member is in tension or compression.


From symmetry of loading: Ay = E =

1 2

total load

A y = E = 3.6 kN We note that DF is a zero-force member and that EF is aligned with the load. Thus,

FDF = 0  FEF = 1.2 kN C  Free body: Joint A: FAB FAC 2.4 kN = = 13 12 5

FAB = 6.24 kN C  FAC = 2.76 kN T 

Free body: Joint B: ΣFx = 0: Σ Fy = 0:

3 12 12 FBC + FBD + (6.24 kN) = 0 3.905 13 13 −

2.5 5 5 FBC + FBD + (6.24 kN) − 2.4 kN = 0 3.905 13 13

(1) (2)



ΣFx = 0: A x = 0


Multiply Eq. (1) by 2.5, Eq. (2) by 3, and add: 45 45 FBD + (6.24 kN) − 7.2 kN = 0, FBD = −4.16 kN, 13 13

FBD = 4.16 kN C 

Multiply Eq. (1) by 5, Eq. (2) by –12, and add: 45 FBC + 28.8 kN = 0, FBC = −2.50 kN, 3.905

FBC = 2.50 kN C 


5 2.5 (2.50 kN) = 0 FCD − 5.831 3.905

FCD = 1.867 kN T  3 3 (2.50 kN) + (1.867 kN) = 0 3.905 5.831

FCE = 2.88 kN T Free body: Joint E: ΣFy = 0:

5 FDE + 3.6 kN − 1.2 kN = 0 7.81

FDE = −3.75 kN ΣFx = 0: − FCE −

FDE = 3.75 kN C 

6 (−3.75 kN) = 0 7.81

FCE = +2.88 kN FCE = 2.88 kN T

(Checks)



ΣFx = 0: FCE − 5.76 kN +

www.elsolucionario.net PROBLEM 6.166 A Howe scissors roof truss is loaded as shown. Determine the force in members DF, DG, and EG.

SOLUTION Reactions at supports. Because of symmetry of loading, Ay = L =

1 1 (total load) = (9.60 kips) = 4.80 kips 2 2 A = L = 4.80 kips 

We pass a section through members DF, DG, and EG, and use the free body shown. We slide FDF to apply it at F: ΣM G = 0: (0.8 kip)(24 ft) + (1.6 kips)(16 ft) + (1.6 kips)(8 ft) 8 FDF

− (4.8 kips)(24 ft) −

82 + 3.52

(6 ft) = 0

FDF = −10.48 kips, FDF = 10.48 kips C  ΣM A = 0: − (1.6 kips)(8 ft) − (1.6 kips)(16 ft) −

2.5FDG 8 + 2.5 2

2

(16 ft) −

8FDG 82 + 2.52

(7 ft) = 0

FDG = −3.35 kips, FDG = 3.35 kips C  ΣM D = 0: (0.8 kips)(16 ft) + (1.6 kips)(8 ft) − (4.8 kips)(16 ft) −

8 FEG 82 + 1.52

(4 ft) = 0

FEG = +13.02 kips, FEG = 13.02 kips T 





Ax = 0,

www.elsolucionario.net PROBLEM 6.167 A Howe scissors roof truss is loaded as shown. Determine the force in members GI, HI, and HJ.

SOLUTION Reactions at supports. Because of symmetry of loading, 1 (Total load) 2 1 = (9.60 kips) 2

Ax = 0,

= 4.80 kips

A = L = 4.80 kips 

We pass a section through members GI, HI, and HJ, and use the free body shown.

ΣM H = 0: −

16 FGI 162 + 32

(4 ft) + (4.8 kips)(16 ft) − (0.8 kip)(16 ft) − (1.6 kips)(8 ft) = 0

FGI = +13.02 kips

FGI = 13.02 kips T 

ΣM L = 0: (1.6 kips)(8 ft) − FHI (16 ft) = 0 FHI = +0.800 kips

FHI = 0.800 kips T  We slide FHG to apply it at H. ΣM I = 0:

8FHJ 82 + 3.52

(4 ft) + (4.8 kips)(16 ft) − (1.6 kips)(8 ft) − (0.8 kip)(16 ft) = 0

FHJ = −13.97 kips

FHJ = 13.97 kips C 



Ay = L =

www.elsolucionario.net PROBLEM 6.168 Rod CD is fitted with a collar at D that can be moved along rod AB, which is bent in the shape of an arc of circle. For the position when θ = 30°, determine (a) the force in rod CD, (b) the reaction at B.

SOLUTION

(a)

ΣM C = 0: (15 in.)(20 lb − By ) = 0 ΣFy = 0: − 20 lb + FCD sin 30° − 20 lb = 0

(b)

B y = 20 lb FCD = 80.0 lb T 

ΣFx = 0: (80 lb) cos30° − Bx = 0 B x = 69.282 lb so B = 72.1 lb

16.10° 



FBD:

www.elsolucionario.net PROBLEM 6.169 For the frame and loading shown, determine the components of all forces acting on member ABC.


ΣFx = 0: Ax + 18 kN = 0 A x = 18.00 kN



ΣM E = 0: −(18 kN)(4 m) − Ay (3.6 m) = 0 Ay = −20 kN

A y = 20.0 kN 

ΣFy = 0: − 20 kN + F = 0 F = +20 kN

F = 20 kN

Free body: Member ABC Note: BE is a two-force member, thus B is directed along line BE.

ΣM C = 0: B(4 m) − (18 kN)(6 m) + (20 kN)(3.6 m) = 0 B = 9 kN

B = 9.00 kN



Cx = 9.00 kN



ΣFx = 0: Cx − 18 kN + 9 kN = 0 Cx = 9 kN ΣFy = 0: C y − 20 kN = 0 C y = 20 kN

C y = 20.0 kN 



Ax = −18 kN

www.elsolucionario.net PROBLEM 6.170 Knowing that each pulley has a radius of 250 mm, determine the components of the reactions at D and E.

SOLUTION

ΣM E = 0: (4.8 kN)(4.25 m) − Dx (1.5 m) = 0 Dx = +13.60 kN

D x = 13.60 kN



E x = 13.60 kN



ΣFx = 0: Ex + 13.60 kN = 0 Ex = −13.60 kN Σ Fy = 0: Dy + E y − 4.8 kN = 0

(1)

Free body: Member ACE:

ΣM A = 0 : (4.8 kN)(2.25 m) + E y (4 m) = 0 E y = −2.70 kN From Eq. (1):

E y = 2.70 kN 

Dy − 2.70 kN − 4.80 kN = 0 Dy = +7.50 kN

D y = 7.50 kN 



Free body: Entire assembly:

www.elsolucionario.net PROBLEM 6.171 For the frame and loading shown, determine the components of the forces acting on member DABC at B and D.


H = 25 kN H = 25 kN

Free body: Member BEH:

ΣM F = 0: Bx (0.5 m) − (25 kN)(0.2 m) = 0 Bx = +10 kN

Free body: Member DABC:

B x = 10.00 kN

From above:



ΣM D = 0: − By (0.8 m) + (10 kN + 12 kN)(0.5 m) = 0 By = +13.75 kN

B y = 13.75 kN 

ΣFx = 0: − Dx + 10 kN + 12 kN = 0 Dx = + 22 kN

D x = 22.0 kN



ΣFy = 0: − Dy + 13.75 kN = 0 Dy = +13.75 kN

D y = 13.75 kN 



ΣM G = 0: H (0.6 m) − (12 kN)(1 m) − (6 kN)(0.5 m) = 0

www.elsolucionario.net PROBLEM 6.172 For the frame and loading shown, determine (a) the reaction at C, (b) the force in member AD.

Free body: Member ABC: ΣM C = 0: + (100 lb)(45 in.) +

4 4 FAD (45 in.) + FBE (30 in.) = 0 5 5

3FAD + 2 FBF = − 375 lb

(1)

Free Body: Member DEF: ΣM F = 0:

4 4 FAD (30 in.) + FBF (15 in.) = 0 5 5

FBE = −2FAD (b)

(2)


3FAD + 2(− 2FAD ) = − 375 lb FAD = + 375 lb From Eq. (2):

FBE = − 2 FAD = − 2(375 lb) FBE = −750 lb

(a)

FAD = 375 lb T 

FBE = 750 lb C.

Return to free body of member ABC. 4 4 ΣFx = 0: C x + 100 lb + FAD + FBE = 0 5 5 C x + 100 +

4 4 (375) + (−750) = 0 5 5

Cx = + 200 lb Cx = 200 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 987


SOLUTION


3 3 ΣFy = 0: C y − FAD − FBF = 0 5 5 3 3 C y − (375) − ( − 750) = 0 5 5

C y = − 225 lb C y = 225 lb

α = 48.37° C = 301 lb

48.4° 


C = 301.0 lb


www.elsolucionario.net PROBLEM 6.173 The control rod CE passes through a horizontal hole in the body of the toggle system shown. Knowing that link BD is 250 mm long, determine the force Q required to hold the system in equilibrium when β = 20°.

SOLUTION We note that BD is a two-force member. Free body: Member ABC:

Since BD = 250,

θ = sin −1


Dimensions in mm

33.404 ; θ = 7.679° 250

Σ M C = 0: ( FBD sin θ )187.94 − ( FBD cos θ )68.404 + (100 N)328.89 = 0 FBD [187.94sin 7.679° − 68.404 cos 7.679°] = 32,889 FBD = 770.6 N

ΣFx = 0: (770.6 N) cos 7.679° = Cx = 0 Cx = + 763.7 N Member CQ:

ΣFx = 0: Q = Cx = 763.7 N Q = 764 N




www.elsolucionario.net PROBLEM 6.174 Determine the magnitude of the gripping forces exerted along line aa on the nut when two 50-lb forces are applied to the handles as shown. Assume that pins A and D slide freely in slots cut in the jaws.

SOLUTION FBD jaw AB:

ΣFx = 0: Bx = 0

A=


ΣM B = 0: (0.5 in.)Q − (1.5 in.) A = 0 Q 3

ΣFy = 0: A + Q − By = 0 By = A + Q =

FBD handle ACE:

4Q 3

By symmetry and FBD jaw DE, Q 3 Ex = Bx = 0 D= A=

E y = By =

4Q 3

ΣM C = 0: (5.25 in.)(50 lb) + (0.75 in.)

Q 4Q − (0.75 in.) =0 3 3 Q = 350 lb 


www.elsolucionario.net PROBLEM 6.175 Knowing that the frame shown has a sag at B of a = 1 in., determine the force P required to maintain equilibrium in the position shown.

SOLUTION We note that AB and BC are two-force members. Free body: Toggle: P 2 Cy

Cx = 10 in. a Cx =

10 10 P 5 P Cy = ⋅ = a a 2 a


ΣM E = 0: Cx (6 in.) − C y (20 in.) − (50 lb)(10 in.) = 0 5P P (b) − (20) = 500 2 a 30   P  − 10  = 500  a 

For

(1)

a = 1.0 in.  30  P  − 10  = 500 1   20 P = 500

P = 25.0 lb 



Cy =

By symmetry,

www.elsolucionario.net PROBLEM 6.F1 For the frame and loading shown, draw the free-body diagram(s) needed to determine the forces acting on member ABC at B and C.

SOLUTION We note that BD is a two-force member.







Free body: ABC

www.elsolucionario.net PROBLEM 6.F2 For the frame and loading shown, draw the free-body diagram(s) needed to determine all forces acting on member GBEH.

SOLUTION We note that AG, DEF, and DH are two-force members.

Note:

Sum moments about A to determine By.

Note:

Sum moments about E to determine FDH (which becomes zero).

Free body: DEF

Sum forces in y to determine Ey (which also becomes zero). Free body: GBEH

Note:

With By, Ey, and FDH previously determined, apply three equilibrium equations to determine three remaining forces on GBEH.



Free body: ABC

www.elsolucionario.net PROBLEM 6.F3 For the frame and loading shown, draw the free-body diagram(s) needed to determine the reactions at B and F.

SOLUTION We note that BC is two-force member.





Free body: ACF

www.elsolucionario.net PROBLEM 6.F4 Knowing that the surfaces at A and D are frictionless, draw the free-body diagram(s) needed to determine the forces exerted at B and C on member BCE.

SOLUTION

Note:

Sum moments about H to relate Cx and Cy.

Free body: BCE

Note:

Sum moments about B to relate Cx and Cy ; combine with relation from free body ACD to determine Cx and Cy ; finally, sum forces in x and y to determine Bx and By.





Free body: ACD

www.elsolucionario.net PROBLEM 6.F5 The position of member ABC is controlled by the hydraulic cylinder CD. Knowing that θ = 30°, draw the free-body diagram(s) needed to determine the force exerted by the hydraulic cylinder on pin C, and the reaction at B.

SOLUTION We note that CD is a two-force member.


Free body: ABC

Note: To find θ, consider geometry of triangle BCD:

Law of cosines:

(CD)2 = (0.5)2 + (1.5)2 − 2(0.5)(1.5) cos 60°

CD = 1.32288 m Law of sines:

sin q sin 60° = 0.5 m 1.32288 m

θ = 19.1066° 


www.elsolucionario.net PROBLEM 6.F6 Arm ABC is connected by pins to a collar at B and to crank CD at C. Neglecting the effect of friction, draw the free-body diagram(s) needed to determine the couple M to hold the system in equilibrium when θ = 30°.

SOLUTION

Note:

Sum forces in x to determine Cx ; sum moments about B to determine Cy.

Free body: CD

Note: Sum moments about D to determine M.



Free body: ABC

www.elsolucionario.net PROBLEM 6.F7 Since the brace shown must remain in position even when the magnitude of P is very small, a single safety spring is attached at D and E. The spring DE has a constant of 50 lb/in. and an unstretched length of 7 in. Knowing that l = 10 in. and that the magnitude of P is 800 lb, draw the free-body diagram(s) needed to determine the force Q required to release the brace.

Free body: ABC

Note:

Sum moments about C to relate Q to A.

Spring force. Unstretched length: Stretched length:

 0 = 7 in.  = 10 in. k = 50 lb/in.

F = kx = k ( −  0 ) = 50(10 − 7) = 150 lb

Free body: ADB

Note:

Sum moments about B to determine A; use relation from free body ABC to determine Q.





SOLUTION

www.elsolucionario.net PROBLEM 6.F8 A log weighing 800 lb is lifted by a pair of tongs as shown. Draw the free-body diagram(s) needed to determine the forces exerted at E and F on tong DEF.

We note that AC and BD are two-force members. Free body: AB

Note:

Free body: LOG

Sum moments about A to determine FBD.

Free body: DEF

Note: Sum moments about E to find Fx; sum forces in x and y to find Ex and Ey.

Note: Sum moments about G to determine Fy.  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 999


SOLUTION



CHAPTER 7





www.elsolucionario.net PROBLEM 7.1 Determine the internal forces (axial force, shearing force, and bending moment) at Point J of the structure indicated. Frame and loading of Problem 6.75.

SOLUTION From Problem 6.75:

C x = 720 lb

C y = 140 lb

ΣFx = 0: F − 720 lb = 0 F = +720 lb

F = 720 lb



ΣFy = 0: V − 140 lb = 0 V = 140 lb

V = 140.0 lb 

Σ M J = 0: M − (140 lb)(8 in.) = 0 M = +1120 lb ⋅ in.

M = 1120 lb ⋅ in.





FBD of JC:

www.elsolucionario.net PROBLEM 7.2 Determine the internal forces (axial force, shearing force, and bending moment) at Point J of the structure indicated. Frame and loading of Problem 6.78.


D x = 900 lb D y = 750 lb

ΣM J = 0: − M + (750 lb)(4 ft) − (900 lb)(1.5 ft) = 0 M = +1650 lb ⋅ ft

M = 1650 lb ⋅ ft



ΣF = 0: −V + (750 lb) cos 20.56° − (900 lb)sin 20.56° = 0 V = +386.2 lb

V = 386 lb

69.4° 

ΣF = 0: F − (750 lb)sin 20.56° + (900 lb) cos 20.56° = 0 F = +1106.1 lb

F = 1106 lb

20.6° 



FBD of JD:

www.elsolucionario.net PROBLEM 7.3 Determine the internal forces at Point J when α = 90°.

SOLUTION

ΣM A = 0: B y = 0  12  ΣFy = 0: A   − 780 N = 0  13  A = 845 N ΣFx = 0: (845 N)

A = 845 N

5 + Bx = 0 13 Bx = −325 N

B x = 325 N

FBD BJ: ΣF = 0: 125 N − F = 0

F = 125.0 N

67.4° 

ΣF = 0: V − 300 N = 0

V = 300 N

22.6° 

Σ M = 0: (325 N)(0.480 m) − M = 0

M = +156 N ⋅ m

M = 156.0 N ⋅ m





Reactions (α = 90°)

www.elsolucionario.net PROBLEM 7.4 Determine the internal forces at Point J when α = 0.

SOLUTION ΣM A = 0: (780 N)(0.720 m) + By (0.3 m) = 0

By = −1872 N

B y = 1872 N

 12  ΣFy = 0: A   − 1872 N = 0  13  A = 2028 N

 5 ΣFx = 0: (2028 N)   + 780 N + Bx = 0  13  Bx = −1560 N

B x = 1560 N

FBD BJ: ΣF = 0: 1728 N + 600 N − F = 0 F = +2328 N

F = 2330 N

67.4° 

V = 720 N

22.6° 

ΣF = 0: 720 N − 1440 N + V = 0 V = +720 N

BJ = 4802 + 2002 = 520 mm ΣM J = 0: (1440 N)(0.520 m) − (720 N)(0.520 m) − M = 0 M = +374.4 N ⋅ m

Alternate

M = 374 N ⋅ m



Computation of M using B x + B y : ΣM J = 0: (1560 N)(0.48 m) − (1872 N)(0.2 m) − M = 0 M = +374.4 N ⋅ m

M = 374 N ⋅ m





Reactions (α = 0)

www.elsolucionario.net PROBLEM 7.5 Knowing that the turnbuckle has been tightened until the tension in wire AD is 850 N, determine the internal forces at point indicated: Point J.

 160  ΣFx = 0: − V +   (850 N) = 0  340 

V = +400 N

V = 400 N



 300  ΣFy = 0: F −   (850 N) = 0  340 

F = +750 N

F = 750 N 

 300   160  ΣM J = 0: M −   (850 N)(120 mm) −  340  (850 N)(100 mm) = 0 340    

M = +130 N ⋅ m

M = 130 N ⋅ m





SOLUTION

www.elsolucionario.net PROBLEM 7.6 Knowing that the turnbuckle has been tightened until the tension in wire AD is 850 N, determine the internal forces at point indicated: Point K.

SOLUTION

AD = 1602 + 3002 = 340 mm

On portion KBA:  160  ΣFx = 0: − V +   (850 N) = 0  340 

V = +400 N

V = 400 N



 300  ΣFy = 0: F −   (850 N) = 0  340 

F = +750 N

F = 750 N 

 300   160  (850 N)(120 mm) −  ΣM J = 0: M −    (850 N)(200 mm) = 0  340   340 

M = +170 N ⋅ m

M = 170.0 N ⋅ m

Internal forces acting on KCD are equal and opposite F = 750 N , V = 400 N

, M = 170.0 N ⋅ m





Free body AK:

www.elsolucionario.net PROBLEM 7.7 Two members, each consisting of a straight and a quartercircular portion of rod, are connected as shown and support a 75-lb load at A. Determine the internal forces at Point J.

SOLUTION Free body: Entire frame ΣM C = 0: (75 lb)(12 in.) − F (9 in.) = 0

ΣFx = 0: Cx = 0 ΣFy = 0: C y − 75 lb − 100 lb = 0 C = 175 lb

C y = + 175 lb

Free body: Member BEDF ΣM B = 0: D (12 in.) − (100 lb)(15 in.) = 0 D = 125 lb

ΣFx = 0: Bx = 0 ΣFy = 0: By + 125 lb − 100 lb = 0 B = 25 lb

By = − 25 lb

Free body: BJ ΣFx = 0: F − (25 lb)sin 30° = 0 F = 12.50 lb

30.0° 

ΣFy = 0: V − (25 lb) cos 30° = 0 V = 21.7 lb

60.0° 

ΣM J = 0: − M + (25 lb)(3 in.) = 0 M = 75.0 lb ⋅ in.





F = 100 lb

www.elsolucionario.net PROBLEM 7.8 Two members, each consisting of a straight and a quartercircular portion of rod, are connected as shown and support a 75-lb load at A. Determine the internal forces at Point K.

SOLUTION Free body: Entire frame ΣM C = 0: (75 lb)(12 in.) − F (9 in.) = 0

ΣFx = 0: Cx = 0 ΣFy = 0: C y − 75 lb − 100 lb = 0 C = 175 lb

C y = + 175 lb

Free body: Member BEDF ΣM B = 0: D (12 in.) − (100 lb)(15 in.) = 0 D = 125 lb

ΣFx = 0: Bx = 0 ΣFy = 0: By + 125 lb − 100 lb = 0 By = − 25 lb

B = 25 lb

Free body: DK We found in Problem 7.11 that D = 125 lb on BEDF. D = 125 lb on DK.

Thus ΣFx = 0: F − (125 lb) cos 30° = 0

F = 108.3 lb

60.0° 



F = 100 lb


ΣFy = 0: V − (125 lb)sin 30° = 0 V = 62.5 lb

30.0° 

ΣM K = 0: M − (125 lb)d = 0 M = (125 lb)d = (125 lb)(0.8038 in.) = 100.5 lb ⋅ in.




M = 100.5 lb ⋅ in.


www.elsolucionario.net PROBLEM 7.9 A semicircular rod is loaded as shown. Determine the internal forces at Point J.

SOLUTION

ΣM B = 0: Ax (2r ) = 0 Ax = 0

ΣFx′ = 0: V − (120 N) cos 60° = 0 V = 60.0 N



F = 103.9 N



 FBD AJ: ΣFy′ = 0: F + (120 N)sin 60° = 0 F = −103.923 N

ΣM J = 0: M − [(0.180 m)sin 60°](120 N) = 0 M = 18.7061 M = 18.71 



FBD Rod:

www.elsolucionario.net PROBLEM 7.10 A semicircular rod is loaded as shown. Determine the internal forces at Point K.

SOLUTION

ΣFy = 0: By − 120 N = 0 B y = 120 N ΣM A = 0: 2rBx = 0 B x = 0 ΣFx′ = 0: V − (120 N) cos 30° = 0 V = 103.923 N V = 103.9 N



F = 60.0 N



M = 10.80 N ⋅ m



FBD BK: ΣFy′ = 0: F + (120 N)sin 30° = 0 F = − 60 N

ΣM K = 0: M − [(0.180 m)sin 30°](120 N) = 0



FBD Rod:

www.elsolucionario.net PROBLEM 7.11 A semicircular rod is loaded as shown. Determine the internal forces at Point J knowing that θ = 30°.

SOLUTION 4  3  ΣM A = 0: r  C  + r  C  − 2r (280 N) = 0 5  5  C = 400 N

ΣFx = 0: − Ax +

4 (400 N) = 0 5 A x = 320 N

3 ΣFy = 0: Ay + (400 N) − 280 N = 0 5 A y = 40.0 N

FBD AJ:

ΣFx′ = 0: F − (320 N) sin 30° − (40.0 N) cos 30° = 0 F = 194.641 N F = 194.6 N

60.0° 

ΣFy ′ = 0: V − (320 N) cos 30° + (40 N) sin 30° = 0 V = 257.13 N V = 257 N

30.0° 

ΣM 0 = 0: (0.160 m)(194.641 N) − (0.160 m)(40.0 N) − M = 0 M = 24.743 M = 24.7 N ⋅ m





FBD AB:

www.elsolucionario.net PROBLEM 7.12 A semicircular rod is loaded as shown. Determine the magnitude and location of the maximum bending moment in the rod.

SOLUTION

4  3  ΣM A = 0:  FCD  (0.16 m) +  FCD  (0.16 m) 5  5  − (280 N)(0.32 m) = 0

ΣFx = 0: Ax +

FCD = 400 N

A x = 320 N

4 (400 N) = 0 5

Ax = − 320 N 3 + ΣFy = 0: Ay + (400 N) − 280 N = 0 5 Ay = + 40.0 N

A y = 40.0 N

Free body: AJ (For θ < 90°) ΣM J = 0: (320 N)(0.16 m) sin θ − (40.0 N)(0.16 m)(1 − cos θ ) − M = 0 M = 51.2sin θ + 6.4 cos θ − 6.4

(1)

For maximum value between A and C: dM = 0: 51.2 cos θ − 6.4sin θ = 0 dθ tan θ =

51.2 =8 6.4

θ = 82.87° 

Carrying into (1): M = 51.2sin 82.87° + 6.4cos82.87° − 6.4 = + 45.20 N ⋅ m 



Free body: Rod ACB


Free body: BJ (For θ > 90°) ΣM J = 0: M − (280 N)(0.16 m)(1 − cos φ ) = 0 M = (44.8 N ⋅ m)(1 − cos φ )

Largest value occurs for φ = 90°, that is, at C, and is M C = 44.8 N ⋅ m

We conclude that for

θ = 82.9° 


M max = 45.2 N ⋅ m


www.elsolucionario.net PROBLEM 7.13 The axis of the curved member AB is a parabola with vertex at A. If a vertical load P of magnitude 450 lb is applied at A, determine the internal forces at J when h = 12 in., L = 40 in., and a = 24 in.

SOLUTION ΣFy = 0: − 450 lb + By = 0

Free body AB

B y = 450 lb


ΣM A = 0: Bx (12 in.) − (450 lb)(40 in.) = 0 B x = 1500 lb

ΣFx = 0: A = 1500 lb

y = k x2

Parabola:

12 in. = k (40 in.) 2

At B: Equation of parabola:

y = 0.0075 x 2 slope =

At J:

k = 0.0075

dy = 0.015 x dx

xJ = 24 in.

y J = 0.0075(24) 2 = 4.32 in.

slope = 0.015(24) = 0.36, tan θ = 0.36, θ = 19.8°

Free body AJ

ΣM J = 0: (450 lb)(24 in.) − (1500 lb)(4.32 in.) − M = 0 M = 4320 lb ⋅ in.

M = 4320 lb ⋅ in.




www.elsolucionario.net PROBLEM 7.13 (Continued) ΣF = 0: F −(450 lb) sin19.8° − (1500 lb) cos19.8° = 0 F = +1563.8 lb

F = 1564 lb

19.8° 

V = 84.7 lb

70.2° 

ΣF = 0: −V − (450 lb) cos19.8° + (1500 lb)sin19.8° = 0


V = +84.71 lb


www.elsolucionario.net PROBLEM 7.14 Knowing that the axis of the curved member AB is a parabola with vertex at A, determine the magnitude and location of the maximum bending moment.

SOLUTION Parabola

y = kx 2

At B:

h = kL2

Equation of parabola

y = hx 2/L2

Σ M B = 0: P( L) − A(h) = 0

Free body AJ


k = h /L2

At J: xJ = a

A = PL /h

y J = ha 2/L2

Σ M J = 0: Pa − ( PL /h)(ha 2 /L2 ) + M = 0

 a2  M = P  − a   L 

For maximum:

dM  2a  = P − 1 = 0 da  L  M max occurs at: a =

 ( L /2)2 L  PL M max = P  − =− L 2 4  

|M |max =

1 L  2

1 PL  4


www.elsolucionario.net PROBLEM 7.15 Knowing that the radius of each pulley is 200 mm and neglecting friction, determine the internal forces at Point J of the frame shown.

SOLUTION FBD Frame with pulley and cord: ΣM A = 0: (1.8 m) Bx − (2.6 m)(360 N)


− (0.2 m)(360 N) = 0 B x = 560 N

FBD BE: Note: Cord forces have been moved to pulley hub as per Problem 6.91. ΣM E = 0: (1.4 m)(360 N) + (1.8 m)(560 N) − (2.4 m) B y = 0 B y = 630 N

FBD BJ: 3 ΣFx′ = 0: F + 360 N − (630 N − 360 N) 5 4 − (560 N) = 0 5 F = 250 N

ΣFy′ = 0: V +

36.9° 

4 3 (630 N − 360 N) − (560 N) = 0 5 5 V = 120.0 N

53.1° 

ΣM J = 0: M + (0.6 m)(360 N) + (1.2 m)(560 N) − (1.6 m)(630 N) = 0 M = 120.0 N ⋅ m




www.elsolucionario.net PROBLEM 7.16 Knowing that the radius of each pulley is 200 mm and neglecting friction, determine the internal forces at Point K of the frame shown.

SOLUTION Free body: frame and pulleys − (360 N)(2.6 m) = 0 Bx = −560 N

B x = 560 N

A x = +920 N

ΣFx = 0: Ax − 560 N − 360 N = 0 Ax = +920 N ΣFy = 0: Ay + By − 360 N = 0 Ay + By = 360 N

(1)

Free body: member AE We recall that the forces applied to a pulley may be applied directly to the axle of the pulley. Σ M E = 0: − Ay (2.4 m) − (360 N)(1.8 m) = 0 Ay = −270 N

From (1):

A y = 270 N

By = 360 N + 270 N By = 630 N

B y = 630 N



Σ M A = 0: − Bx (1.8 m) − (360 N)(0.2 m)


Free body: AK ΣFx = 0: 920 N − 360 N − F = 0 F = +560 N

F = 560 N



ΣFy = 0: 360 N − 270 N − V = 0 V = +90.0 N

V = 90.0 N 

Σ M K = 0: (270 N)(1.6 m) − (360 N)(1 m) − M = 0 M = 72.0 N ⋅ m




M = +72.0 N ⋅ m


www.elsolucionario.net PROBLEM 7.17 A 5-in.-diameter pipe is supported every 9 ft by a small frame consisting of two members as shown. Knowing that the combined weight of the pipe and its contents is 10 lb/ft and neglecting the effect of friction, determine the magnitude and location of the maximum bending moment in member AC.

SOLUTION

4 ΣFx = 0: D − (90 lb) = 0 5

D = 72 lb

3 ΣFy = 0: E − (90 lb) = 0 5

E = 54 lb

Free body: Frame ΣM B = 0: − Ay (18.75 in.) + (72 lb)(2.5 in.) + (54 lb)(8.75 in.) = 0 A y = 34.8 lb

Ay = + 34.8 lb

4 3 ΣFy = 0: By + 34.8 lb − (72 lb) − (54 lb) = 0 5 5 B y = 55.2 lb

By = + 55.2 lb 3 4 ΣFx = 0: Ax + Bx − (72 lb) + (54 lb) = 0 5 5 Ax + Bx = 0

(1)

Free body: Member AC ΣM C = 0: (72 lb)(2.5 in.) − (34.8 lb)(12 in.) − Ax (9 in.) = 0 Ax = − 26.4 lb

A x = 26.4 lb

B x = 26.4 lb

Bx = − Ax = + 26.4 lb

From (1):



Free body: 10-ft section of pipe


Free body: Portion AJ For x ≤ 12.5 in. ( AJ ≤ AD ) : 3 4 ΣM J = 0: (26.4 lb) x − (34.8 lb) x + M = 0 5 5 M = 12 x M max = 150 lb ⋅ in. for x = 12.5 in. M max = 150.0 lb ⋅ in. at D For x > 12.5 in.( AJ > AD ):

M = 900 − 60 x M max = 150 lb ⋅ in. for x = 12.5 in. M max = 150.0 lb ⋅ in. at D 

Thus:



3 4 ΣM J = 0: (26.4 lb) x − (34.8 lb) x + (72 lb)( x − 12.5) + M = 0 5 5

www.elsolucionario.net PROBLEM 7.18 For the frame of Problem 7.17, determine the magnitude and location of the maximum bending moment in member BC. PROBLEM 7.17 A 5-in.-diameter pipe is supported every 9 ft by a small frame consisting of two members as shown. Knowing that the combined weight of the pipe and its contents is 10 lb/ft and neglecting the effect of friction, determine the magnitude and location of the maximum bending moment in member AC.

SOLUTION

4 ΣFx = 0: D − (90 lb) = 0 5

D = 72 lb

3 ΣFy = 0: E − (90 lb) = 0 5

E = 54 lb

Free body: Frame ΣM B = 0: − Ay (18.75 in.) + (72 lb)(2.5 in.) + (54 lb)(8.75 in.) = 0 A y = 34.8 lb

Ay = + 34.8 lb

4 3 ΣFy = 0: By + 34.8 lb − (72 lb) − (54 lb) = 0 5 5 B y = 55.2 lb

By = + 55.2 lb 3 4 ΣFx = 0: Ax + Bx − (72 lb) + (54 lb) = 0 5 5 Ax + Bx = 0

(1)

Free body: Member AC ΣM C = 0: (72 lb)(2.5 in.) − (34.8 lb)(12 in.) − Ax (9 in.) = 0 Ax = − 26.4 lb

From (1):

A x = 26.4 lb

B x = 26.4 lb

Bx = − Ax = + 26.4 lb



Free body: 10-ft section of pipe


Free body: Portion BK For x ≤ 8.75 in.( BK ≤ BE ): 3 4 ΣM K = 0: (55.2 lb) x − (26.4 lb) x − M = 0 5 5 M = 12 x M max = 105.0 lb ⋅ in. for

x = 8.75 in.

M max = 105.0 lb ⋅ in. at E For x > 8.75 in.( BK > BE ):

M = 472.5 − 42 x M max = 105.0 lb ⋅ in. for

x = 8.75 in.

M max = 105.0 lb ⋅ in. at E 

Thus



3 4 ΣM K = 0: (55.2 lb) x − (26.4 lb) x − (54 lb)( x − 8.75 in.) − M = 0 5 5

www.elsolucionario.net PROBLEM 7.19 Knowing that the radius of each pulley is 150 mm, that α = 20°, and neglecting friction, determine the internal forces at (a) Point J, (b) Point K.

Tension in cable = 500 N. Replace cable tension by forces at pins A and B. Radius does not enter computations: (cf. Problem 6.90) (a)

Free body: AJ

ΣFx = 0: 500 N − F = 0 F = 500 N

F = 500 N



ΣFy = 0: V − 500 N = 0 V = 500 N

V = 500 N 

ΣM J = 0: (500 N)(0.6 m) = 0 M = 300 N ⋅ m

(b)

M = 300 N ⋅ m



V = 171.0 N



Free body: ABK

ΣFx = 0: 500 N − 500 N + (500 N)sin 20° − V = 0 V = 171.01 N



SOLUTION


ΣFy = 0: − 500 N − (500 N) cos 20° + F = 0 F = 969.8 N

F = 970 N 

ΣM K = 0: (500 N)(1.2 m) − (500 N)sin 20°(0.9 m) − M = 0 M = 446.1 N ⋅ m




M = 446 N ⋅ m


www.elsolucionario.net PROBLEM 7.20 Knowing that the radius of each pulley is 150 mm, that α = 30°, and neglecting friction, determine the internal forces at (a) Point J, (b) Point K.

Tension in cable = 500 N. Replace cable tension by forces at pins A and B. Radius does not enter computations: (cf. Problem 6.90) (a)

Free body: AJ: ΣFx = 0: 500 N − F = 0 F = 500 N

F = 500 N



ΣFy = 0: V − 500 N = 0 V = 500 N

V = 500 N 

ΣM J = 0: (500 N)(0.6 m) = 0 M = 300 N ⋅ m

(b)

M = 300 N ⋅ m



V = 250 N



FBD: Portion ABK:

ΣFx = 0: 500 N − 500 N + (500 N)sin 30° − V ΣFy = 0: − 500 N − (500 N) cos 30° + F = 0 ΣM K = 0: (500 N)(1.2 m) − (500 N)sin 30°(0.9 m) − M = 0

F = 933 N  M = 375 N ⋅ m





SOLUTION

www.elsolucionario.net PROBLEM 7.21 A force P is applied to a bent rod that is supported by a roller and a pin and bracket. For each of the three cases shown, determine the internal forces at Point J.

(a)

FBD Rod: ΣFx = 0:

Ax = 0

ΣM D = 0: aP − 2aAy = 0

Ay =

P 2

ΣFx = 0: V = 0 

FBD AJ: ΣFy = 0:

P −F =0 2

F=

P  2

ΣM J = 0: M = 0 

(b)

FBD Rod:

4  3  ΣM A = 0: 2a  D  + 2a  D  − aP = 0 5  5  ΣFx = 0: Ax −

4 5 P=0 5 14

ΣFy = 0: Ay − P +

3 5 P=0 5 14

D=

5P 14

Ax =

2P 7

Ay =

11P 14



SOLUTION


FBD AJ:



 

2 P −V = 0 7

ΣFy = 0:

11P −F =0 14

ΣM J = 0: a

2P −M =0 7

ΣM A = 0:

a  4D  − aP = 0 2  5 

V=

2P 7

F= M=



11P  14

2 aP 7



FBD Rod: D=

5P 2

Ax −

4 5P =0 5 2

Ax = 2 P

ΣFy = 0: Ay − P −

3 5P =0 5 2

Ay =

ΣFx = 0:

5P 2

FBD AJ: ΣFx = 0:

2P − V = 0

ΣFy = 0:

5P −F =0 2

ΣM J = 0: a(2 P) − M = 0

V = 2P F=



5P  2

M = 2aP







(c)

ΣFx = 0:

www.elsolucionario.net PROBLEM 7.22 A force P is applied to a bent rod that is supported by a roller and a pin and bracket. For each of the three cases shown, determine the internal forces at Point J.

(a)

FBD Rod:

ΣM D = 0: aP − 2aA = 0 A=

ΣFx = 0: V −

FBD AJ:

P =0 2

V=

ΣFy = 0:

P 2



F=0 

ΣM J = 0: M − a

(b)

P 2

P =0 2

M=

aP 2



FBD Rod: ΣM D = 0: aP −

a4  A =0 2  5  A=

5P 2



SOLUTION


FBD AJ:

(c)

ΣFx = 0:

3 5P −V = 0 5 2

ΣFy = 0:

4 5P −F =0 5 2

V=

3P 2



F = 2P  M=

3 aP 2



V=

3P 14



FBD Rod: 3  4  ΣM D = 0: aP − 2a  A  − 2a  A  = 0 5  5 

 3 5P  ΣFx = 0: V −  =0  5 14  ΣFy = 0:

4 5P −F =0 5 14

 3 5P  ΣM J = 0: M − a  =0  5 14 

5P 14

F= M=

2P 7

3 aP 14

 



A=

www.elsolucionario.net PROBLEM 7.23 A quarter-circular rod of weight W and uniform cross section is supported as shown. Determine the bending moment at Point J when θ = 30°.

SOLUTION FBD Rod: ΣFx = 0: A x = 0

FBD AJ:

π

W − rAy = 0

α = 15°, weight of segment = W r=

r

α

ΣFy ′ = 0:

sin α =

r π

Ay =

2W

π


2r

ΣM B = 0:

30° W = 90° 3

sin15° = 0.9886r

12

2W

π

cos 30° −

W cos 30° − F = 0 3 F=

W 3  2 1  −  2 π 3

2W  W  ΣM 0 = M + r  F −  + r cos15° 3 = 0 π  

M = 0.0557 Wr




www.elsolucionario.net PROBLEM 7.24 A quarter-circular rod of weight W and uniform cross section is supported as shown. Determine the bending moment at Point J when θ = 30°.

SOLUTION FBD Rod: 2r

π

W =0 B=

α = 15° =

FBD BJ:

r=

r π

2W

π


ΣM A = 0: rB −

π 12

sin15° = 0.98862r

12

Weight of segment = W

30° W = 90° 3

ΣFy′ = 0: F −

W 2W cos 30° − sin 30° = 0 3 π  3 1 F= + W  6 π   

ΣM 0 = 0: rF − ( r cos15°)

W −M =0 3  3 1  cos15°  M = rW  + − 0.98862 Wr  6 π   3   

M = 0.289Wr




www.elsolucionario.net PROBLEM 7.25 For the rod of Problem 7.23, determine the magnitude and location of the maximum bending moment. PROBLEM 7.23 A quarter-circular rod of weight W and uniform cross section is supported as shown. Determine the bending moment at Point J when θ = 30°.

SOLUTION ΣFx = 0: Ax = 0 ΣM B = 0:

2r

π

W − rAy = 0

α=

θ 2

Ay =

2W

r=

r

,


2α π

=

F=

FBD AJ:

2W  ΣM 0 = 0: M +  F − π 

M=

But, so

sin α cos α =

M=

2W

π

4α

π 2W

π

W cos 2α +

α

4α

2

ΣFx′ = 0: − F −

π

π

2W

π

sin α

W

cos 2α = 0

(1 − 2α ) cos 2α =

2W

π

(1 − θ ) cos θ

4α   r + (r cos α ) π W = 0 

(1 + θ cos θ − cos θ ) r −

4αW r

π

α

sin α cos α

1 1 sin 2α = sin θ 2 2 2r

π

W (1 − cos θ + θ cos θ − sin θ )

dM 2rW (sin θ − θ sin θ + cos θ − cos θ ) = 0 = dθ π

for

(1 − θ )sin θ = 0

dM = 0 for θ = 0, 1, nπ ( n = 1, 2,) dθ

Only 0 and 1 in valid range At

θ = 0 M = 0, at θ = 1 rad at

θ = 57.3°

M = M max = 0.1009Wr 



FBD Rod:

www.elsolucionario.net PROBLEM 7.26 For the rod of Problem 7.24, determine the magnitude and location of the maximum bending moment. PROBLEM 7.24 A quarter-circular rod of weight W and uniform cross section is supported as shown. Determine the bending moment at Point J when θ = 30°.

SOLUTION FBD Bar: 2r

π

W =0

B=

θ

so

α= r=

2

r

α

2W

π 0 ≤α ≤

π 4

sin α


2α π

2

= ΣFx′ = 0: F −

4α

π

π

W

W cos 2α −

F=

2W

=

2W

π π

4α

2W

π

sin 2α = 0

(sin 2α + 2α cos 2α ) (sin θ + θ cos θ )

FBD BJ: ΣM 0 = 0: rF − (r cos α )

M=

But,

sin α cos α =

4α

π

W −M =0

r  4α Wr (sin θ + θ cos θ ) −  sin α cos α  W π α  π 2

1 1 sin 2α = sin θ 2 2



ΣM A = 0: rB −


so

M=

or

M=

2Wr

π 2

π

(sin θ + θ cos θ − sin θ )

Wrθ cos θ

dM 2 = Wr (cos θ − θ sin θ ) = 0 at θ tan θ = 1 dθ π

Solving numerically

θ = 0.8603 rad M = 0.357 Wr

and



at θ = 49.3° 


(Since M = 0 at both limits, this is the maximum)


www.elsolucionario.net PROBLEM 7.27 A half section of pipe rests on a frictionless horizontal surface as shown. If the half section of pipe has a mass of 9 kg and a diameter of 300 mm, determine the bending moment at Point J when θ = 90°.

SOLUTION For half section

m = 9 kg W = mg = (9)(9.81) = 88.29 N

Portion JC: 1 Weight = W = 44.145 N 2

x=

2r

π

=

2(150)

π

x = 95.49 mm ΣM J = 0: (44.145 N)(0.15 m) − (44.145 N)(0.09549 m) − M = 0 M = +2.406 N ⋅ m

M = 2.41 N ⋅ m





From Fig. 5.8B:

www.elsolucionario.net PROBLEM 7.28 A half section of pipe rests on a frictionless horizontal surface as shown. If the half section of pipe has a mass of 9 kg and a diameter of 300 mm, determine the bending moment at point J when θ = 90°.

SOLUTION m = 9 kg

For half section

W = mg = (9 kg)(9.81 m/s 2 ) = 88.29 N

Free body JC Weight of portion JC,

1 = W = 44.145 N 2

From Fig. 5.8B: x=

2r

π

=

2(150)

π

= 95.49 mm

ΣM J = 0: M − (44.145 N)(0.09549 m) = 0 M = +4.2154 N ⋅ m

M = 4.22 N ⋅ m





r = 150 mm

www.elsolucionario.net PROBLEM 7.29 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.

SOLUTION

ΣFy = 0: − wx − V = 0

 x ΣM1 = 0: wx   + M = 0 2

1 M = − wx 2 2

|V |max = wL 

| M |max =

1 2 wL  2



V = − wx


SOLUTION

ΣFy = 0: −

1 x2 w0 −V = 0 2 L

1 x2  x ΣFy = 0:  w0  +M =0 L  3 2

By similar D’s


1 1 x 1 x2 wx =  w0  x = w0 2 2 L 2 L

1 x2 V = − w0 2 L 1 x3 M = − w0 6 L

|V |max =

1 w0 L  2

| M |max =

1 w0 L2  6



SOLUTION Free body: entire beam  2L  L ΣM D = 0: P  − P   − AL = 0   3  3

A = P /3

ΣFy = 0:

P − P + P + Dy = 0 3 Dy = − P /3

(a)

D = P /3

Shear and bending moment. Since the loading consists of concentrated loads, the shear diagram is made of horizontal straight-line segments and the B. M. diagram is made of oblique straight-line segments. Just to the right of A: ΣFy = 0: − V1 +

P =0 3

ΣM1 = 0: M 1 −

P (0) = 0 3

V1 = + P /3 M1 = 0

Just to the right of B: ΣFy = 0: − V2 + ΣM 2 = 0: M 2 −

P − P = 0, 3 PL + P(0) = 0 3  3 

V2 = −2 P /3 M 2 = + PL /9

Just to the right of C: ΣFy = 0:

P − P + P − V3 = 0 3

ΣM 3 = 0: M 3 −

P  2L  L + P − P(0) = 0   3 3  3

V3 = + P /3 M 3 = − PL /9



ΣFx = 0: Dx = 0


Just to the left of D: P =0 3

ΣM 4 = 0: − M 4 −

P (0) = 0 3

V4 = +

P 3

M4 = 0

|V |max = 2 P /3; | M |max = PL /9 

(b)



ΣFy = 0: V4 −


SOLUTION (a)

Shear and bending moment. Just to the right of A: V1 = + P;

M1 = 0

ΣFy = 0: P − P − V2 = 0; L ΣM 2 = 0: M 2 − P   = 0; 2

V2 = 0 M 2 = + PL /2

Just to the left of C: ΣFy = 0: P − P − V3 = 0, L ΣM 3 = 0: M 3 + P   − PL = 0, 2

V3 = 0 M 3 = + PL /2

|V |max = P 

(b)

| M |max = PL /2 



Just to the right of B:


SOLUTION (a)

FBD Beam:

ΣM C = 0: LAy − M 0 = 0

Ay =

M0 L

ΣFy = 0: − Ay + C = 0 M0 L


C=

Along AB: ΣFy = 0: −

M0 −V = 0 L V =−

ΣM J = 0: x

M0 L

M0 +M =0 L

Straight with

M =−

M0 x L

M =−

M0 at B 2

Along BC: ΣFy = 0: −

M0 −V = 0 L

ΣM K = 0: M + x

Straight with (b)

M=

V =−

M0 − M0 = 0 L

M0 at B 2

M0 L x  M = M 0 1 −  L 

M = 0 at C |V |max = M 0 /L 

From diagrams:

| M |max =

M0 at B  2


www.elsolucionario.net PROBLEM 7.34 For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.

SOLUTION Free body: Portion AJ ΣFy = 0: − P − V = 0 ΣM J = 0: M + Px − PL = 0

M = P( L − x) 

The V and M diagrams are obtained by plotting the functions V and M.

|V |max = P 

(b)

| M |max = PL 



(a)

V = −P 


SOLUTION (a)

Just to the right of A: ΣFy = 0 V1 = +15 kN M1 = 0

Just to the left of C: V2 = +15 kN M 2 = +15 kN ⋅ m


Just to the right of C: V3 = +15 kN M 3 = +5 kN ⋅ m

Just to the right of D: V4 = −15 kN M 4 = +12.5 kN ⋅ m

Just to the right of E: V5 = −35 kN M 5 = +5 kN ⋅ m

At B:

(b)

M B = −12.5 kN ⋅ m

|V |max = 35.0 kN

| M |max = 12.50 kN ⋅ m 



SOLUTION

ΣM A = 0: B(3.2 m) − (40 kN)(0.6 m) − (32 kN)(1.5 m) − (16 kN)(3 m) = 0 B = +37.5 kN

B = 37.5 kN 

ΣFx = 0: Ax = 0 ΣFy = 0: Ay + 37.5 kN − 40 kN − 32 kN − 16 kN = 0 Ay = +50.5 kN

(a)

A = 50.5 kN 

Shear and bending moment

Just to the right of A:

V1 = 50.5 kN

M1 = 0 

Just to the right of C: ΣFy = 0: 50.5 kN − 40 kN − V2 = 0 ΣM 2 = 0: M 2 − (50.5 kN)(0.6 m) = 0

V2 = +10.5 kN  M 2 = +30.3 kN ⋅ m 

Just to the right of D: ΣFy = 0: 50.5 − 40 − 32 − V3 = 0 ΣM 3 = 0: M 3 − (50.5)(1.5) + (40)(0.9) = 0

V3 = −21.5 kN  M 3 = +39.8 kN ⋅ m 



Free body: Entire beam


Just to the right of E: ΣFy = 0: V4 + 37.5 = 0

V4 = −37.5 kN 

ΣM 4 = 0: − M 4 + (37.5)(0.2) = 0 VB = M B = 0

At B:

M 4 = +7.50 kN ⋅ m 



| M |max = 39.8 kN ⋅ m 



|V |max = 50.5 kN 

(b)


SOLUTION

ΣM A = 0: E (6 ft) − (6 kips)(2 ft) − (12 kips)(4 ft) − (4.5 kips)(8 ft) = 0 E = +16 kips

E = 16 kips 

ΣFx = 0: Ax = 0 ΣFy = 0: Ay + 16 kips − 6 kips − 12 kips − 4.5 kips = 0 Ay = +6.50 kips

(a)

A = 6.50 kips 

Shear and bending moment Just to the right of A: V1 = +6.50 kips M1 = 0



Just to the right of C: ΣFy = 0: 6.50 kips − 6 kips − V2 = 0 ΣM 2 = 0: M 2 − (6.50 kips)(2 ft) = 0

V2 = +0.50 kips  M 2 = +13 kip ⋅ ft 

Just to the right of D: ΣFy = 0: 6.50 − 6 − 12 − V3 = 0 ΣM 3 = 0: M 3 − (6.50)(4) − (6)(2) = 0

V3 = +11.5 kips  M 3 = +14 kip ⋅ ft 





Just to the right of E: ΣFy = 0: V4 − 4.5 = 0 ΣM 4 = 0: − M 4 − (4.5)2 = 0 VB = M B = 0

At B:

V4 = +4.5 kips  M 4 = −9 kip ⋅ ft 



| M |max = 14.00 kip ⋅ ft 



|V |max = 11.50 kips 

(b)


SOLUTION

ΣM C = 0: (120 lb)(10 in.) − (300 lb)(25 in.) + E (45 in.) − (120 lb)(60 in.) = 0 E = +300 lb

E = 300 lb 

ΣFx = 0: Cx = 0 ΣFy = 0: C y + 300 lb − 120 lb − 300 lb − 120 lb = 0 C y = +240 lb

(a)

C = 240 lb 

Shear and bending moment Just to the right of A: ΣFy = 0: − 120 lb − V1 = 0

V1 = −120 lb, M 1 = 0 

Just to the right of C: ΣFy = 0: 240 lb − 120 lb − V2 = 0 ΣM C = 0: M 2 + (120 lb)(10 in.) = 0

V2 = +120 lb  M 2 = −1200 lb ⋅ in. 

Just to the right of D: ΣFy = 0: 240 − 120 − 300 − V3 = 0 ΣM 3 = 0: M 3 + (120)(35) − (240)(25) = 0,

V3 = −180 lb  M 3 = +1800 lb ⋅ in. 





Just to the right of E: +ΣFy = 0: V4 − 120 lb = 0 ΣM 4 = 0: − M 4 − (120 lb)(15 in.) = 0

V4 = +120 lb  M 4 = −1800 lb ⋅ in.  VB = M B = 0 

At B:

| M |max = 1800 lb ⋅ in. 



|V |max = 180.0 lb 

(b)


SOLUTION

ΣM A = 0: B(5 m) − (60 kN)(2 m) − (50 kN)(4 m) = 0 B = +64.0 kN

B = 64.0 kN 

ΣFx = 0: Ax = 0 ΣFy = 0: Ay + 64.0 kN − 6.0 kN − 50 kN = 0 Ay = +46.0 kN

(a)

A = 46.0 kN 

Shear and bending-moment diagrams. From A to C: ΣFy = 0: 46 − V = 0 ΣM y = 0: M − 46 x = 0

V = +46 kN  M = (46 x)kN ⋅ m 

From C to D: ΣFy = 0: 46 − 60 − V = 0

V = −14 kN 

ΣM j = 0: M − 46 x + 60( x − 2) = 0 M = (120 − 14 x)kN ⋅ m

For

x = 2 m: M C = +92.0 kN ⋅ m



For

x = 3 m: M D = +78.0 kN ⋅ m







From D to B: ΣFy = 0: V + 64 − 25μ = 0 V = (25μ − 64)kN μ ΣM j = 0: 64 μ − (25μ )   − M = 0 2

M = (64μ − 12.5μ 2 )kN ⋅ m

μ = 0: VB = −64 kN

MB = 0  |V |max = 64.0 kN 

(b)

| M |max = 92.0 kN ⋅ m 



For


SOLUTION

ΣM B = 0: (24 kN)(9 m) − C (6 m) + (24 kN)(4.5 m) = 0

C = 54 kN 

ΣFy = 0: 54 − 24 − 24 + B = 0 B = −6 kN

B = 6 kN

From A to C: ΣFy = 0: − 24 − V = 0

V = −24 kN

ΣM1 = 0: (24)( x) + M = 0

M = (−24 x)kN ⋅ m

From C to D: ΣFy = 0: − 24 − 8( x − 3) − V + 54 = 0 V = (−8 x + 54)kN  x −3 ΣM 2 = 0: (24)( x) + 8( x − 3)   − (54)( x − 3) + M = 0  2 

M = ( −4 x 2 + 54 x − 198)kN ⋅ m

From D to B: ΣFy = 0: V − 6 = 0 ΣM 3 = 0: − M − (6)(9 − x) = 0

V = +6 kN M = (6 x − 54)kN ⋅ m





|V |max = 30.0 kN 

(b)


| M |max = 72.0 kN ⋅ m 



SOLUTION (a)

By symmetry: Ay = B = 8 kips +

1 (4 kips)(5 ft) A y = B = 18 kips 2

Along AC:

ΣM J = 0: M − x(18 kips) M = (18 kips) x M = 36 kip ⋅ ft at C ( x = 2 ft)

Along CD: ΣFy = 0: 18 kips − 8 kips − (4 kips/ft) x1 − V = 0 V = 10 kips − (4 kips/ft) x1 V = 0 at x1 = 2.5 ft (at center) ΣM K = 0: M +

x1 (4 kips/ft) x1 + (8 kips) x1 − (2 ft + x1 )(18 kips) = 0 2

M = 36 kip ⋅ ft + (10 kips/ft) x1 − (2 kips/ft) x12 M = 48.5 kip ⋅ ft at x1 = 2.5 ft

Complete diagram by symmetry (b)

|V |max = 18.00 kips 

From diagrams:

|M |max = 48.5 kip ⋅ ft 



ΣFy = 0 : 18 kips − V = 0 V = 18 kips


SOLUTION

ΣM A = 0: B(10 ft) − (15 kips)(3 ft) − (12 kips)(6 ft) = 0 B = +11.70 kips

B = 11.70 kips 

ΣFx = 0: Ax = 0 ΣFy = 0: Ay − 15 − 12 + 11.70 = 0 Ay = +15.30 kips

(a)

A = 15.30 kips 

Shear and bending-moment diagrams From A to C: ΣFy = 0: 15.30 − 2.5 x − V = 0 V = (15.30 − 2.5 x) kips

 x ΣM J = 0: M + (2.5 x)   − 15.30 x = 0 2 M = 15.30 x − 1.25 x 2

For x = 0:

VA = +15.30 kips

For x = 6 ft:

VC = +0.300 kip

MA = 0  M C = +46.8 kip ⋅ ft 





From C to B: ΣFy = 0: V + 11.70 = 0

V = −11.70 kips 

ΣM J = 0: 11.70μ − M = 0 M = (11.70μ ) kip ⋅ ft M C = +46.8 kip ⋅ ft 

For μ = 4 ft: For μ = 0:

MB = 0 

(b)

|V |max = 15.30 kips 

     |M |max = 46.8 kip ⋅ ft 





www.elsolucionario.net PROBLEM 7.43 Assuming the upward reaction of the ground on beam AB to be uniformly distributed and knowing that P = wa, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.

SOLUTION Free body: Entire beam ΣFy = 0: wg (4a) − 2wa − wa = 0 wg =

(a)

3 w  4

Shear and bending-moment diagrams

ΣFy = 0:

3 wx − wx − V = 0 4 1 V = − wx 4

ΣM J = 0 : M + ( wx)

x 3 x wx − =0 2  4  2

1 M = − wx 2 8

For x = 0:

VA = M A = 0  1 VC = − wa 4

For x = a :

1 M C = − wa 2  8

From C to D: ΣFy = 0:

3 wx − wa − V = 0 4 3  V =  x − aw 4  

a 3  x  ΣM J = 0: M + wa  x −  − wx   = 0 2 4 2  3 a  M = wx 2 − wa  x −  8 2 

(1)



From A to C:


For x = a :

1 VC = − wa 4

1 M C = − wa 2  8

For x = 2a :

1 VD = + wa 2

MD = 0 

Because of the symmetry of the loading, we can deduce the values of V and M for the right-hand half of the beam from the values obtained for its left-hand half. |V |max =

To find | M |max , we differentiate Eq. (1) and set

dM dx

1 wa  2

= 0:

dM 3 4 = wx − wa = 0, x = a dx 4 3 2

3 4  wa 2 4 1 M = w  a  − wa 2  −  = − 8 3  6 3 2 |M |max =

1 2 wa  6

Bending-moment diagram consists of four distinct arcs of parabola.



(b)

www.elsolucionario.net PROBLEM 7.44 Solve Problem 7.43 knowing that P = 3wa. PROBLEM 7.43 Assuming the upward reaction of the ground on beam AB to be uniformly distributed and knowing that P = wa, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.

SOLUTION Free body: Entire beam ΣFy = 0: wg (4a ) − 2wa − 3wa = 0

(a)

5 w  4

Shear and bending-moment diagrams From A to C: ΣFy = 0:

5 wx − wx − V = 0 4 1 V = + wx 4

ΣM J = 0 : M + ( wx)

x 5 x − wx =0 2  4  2

1 M = + wx 2 8

For x = 0:

VA = M A = 0  1 VC = + wa 4

For x = a :

1 M C = + wa 2  8

From C to D: ΣFy = 0:

5 wx − wa − V = 0 4 5  V =  x − aw 4  

a 5  x  ΣM J = 0: M + wa  x −  − wx   = 0 2 4 2  M=

5 2 a  wx − wa  x −  8 2 

(1)



wg =


1 1 VC = + wa, M C = + wa 2  4 8

For x = a : For x = 2a :

3 VD = + wa, M D = + wa 2  2

Because of the symmetry of the loading, we can deduce the values of V and M for the right-hand half of the beam from the values obtained for its left-hand half. |V |max =

To find |M |max , we differentiate Eq. (1) and set

dM dx

3 wa  2

= 0:

dM 5 = wx − wa = 0 dx 4 4 x = a < a (outside portion CD) 5

The maximum value of |M | occurs at D: |M |max = wa 2 

Bending-moment diagram consists of four distinct arcs of parabola.



(b)

www.elsolucionario.net PROBLEM 7.45 Assuming the upward reaction of the ground on beam AB to be uniformly distributed and knowing that a = 0.3 m, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.

SOLUTION (a)

FBD Beam:

ΣFy = 0: w(1.5 m) − 2(3.0 kN) = 0 w = 4.0 kN/m

Along AC: ΣFy = 0: (4.0 kN/m) x − V = 0


V = (4.0 kN/m) x x (4.0 kN/m) x = 0 2 M = (2.0 kN/m) x 2

ΣM J = 0: M −

Along CD: ΣFy = 0: (4.0 kN/m) x − 3.0 kN − V = 0 V = (4.0 kN/m) x − 3.0 kN ΣM K = 0: M + ( x − 0.3 m)(3.0 kN) −

x (4.0 kN/m) x = 0 2

M = 0.9 kN ⋅ m − (3.0 kN) x + (2.0 kN/m) x 2

Note: V = 0 at x = 0.75 m, where M = −0.225 kN ⋅ m Complete diagrams using symmetry. |V |max = 1.800 kN 

(b)

|M |max = 0.225 kN ⋅ m 


www.elsolucionario.net PROBLEM 7.46 Solve Problem 7.45 knowing that a = 0.5 m. PROBLEM 7.45 Assuming the upward reaction of the ground on beam AB to be uniformly distributed and knowing that a = 0.3 m, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.

SOLUTION

ΣFy = 0: wg (1.5 m) − 3 kN − 3 kN = 0

(a)

wg = 4 kN/m 

Shear and bending moment From A to C: ΣFy = 0: 4 x − V = 0 V = (4 x) kN ΣM J = 0: M − (4 x)

x = 0, M = (2 x 2 ) kN ⋅ m 2

For x = 0: For x = 0.5 m: From C to D:

VA = M A = 0  VC = 2 kN,

M C = 0.500 kN ⋅ m 

ΣFy = 0: 4 x − 3 kN − V = 0 V = (4 x − 3) kN ΣM J = 0: M + (3 kN)( x − 0.5) − (4 x)

x =0 2

M = (2 x 2 − 3 x + 1.5) kN ⋅ m

For x = 0.5 m:

VC = −1.00 kN,

For x = 0.75 m:

VC = 0,

For x = 1.0 m:

VC = 1.00 kN,

M C = 0.500 kN ⋅ m  M C = 0.375 kN ⋅ m  M C = 0.500 kN ⋅ m 





From D to B: ΣFy = 0: V + 4μ = 0 V = −(4μ ) kN ΣM J = 0: (4 μ )

μ 2

− M = 0, M = 2μ 2

For μ = 0: For μ = 0.5 m:

VB = M B = 0  VD = −2 kN,

M D = 0.500 kN ⋅ m 

(b)

|V |max = 2.00 kN 



|M |max = 0.500 kN ⋅ m 

       





www.elsolucionario.net PROBLEM 7.47 Assuming the upward reaction of the ground on beam AB to be uniformly distributed, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.

SOLUTION Free body: Entire beam ΣFy = 0: wg (12 ft) − (3 kips/ft)(6 ft) = 0 wg = 1.5 kips/ft 

Shear and bending-moment diagrams from A to C: ΣFy = 0: 1.5 x − V = 0 ΣM J = 0: M − (1.5 x)

V = (1.5 x)kips x 2

M = (0.75 x 2 )kip ⋅ ft

For x = 0:

VA = M A = 0 

For x = 3 ft: VC = 4.5 kips,

M C = 6.75 kip ⋅ ft 

From C to D: ΣFy = 0: 1.5 x − 3( x − 3) − V = 0 V = (9 − 1.5 x)kips ΣM J = 0: M + 3( x − 3)

x−3 x − (1.5 x) = 0 2 2

M = [0.75 x 2 − 1.5( x − 3)2 ]kip ⋅ ft

For x = 3 ft: VC = 4.5 kips,

M C = 6.75 kip ⋅ ft 

For x = 6 ft: VcL = 0,

McL= 13.50 kip ⋅ ft 

For x = 9 ft: VD = −4.5 kips,

M D = 6.75 kip ⋅ ft  VB = M B = 0 

At B:

|V |max = 4.50 kips 

(b)

|M |max = 13.50 kip ⋅ ft 

Bending-moment diagram consists or three distinct arcs of parabola, all located above the x axis. Thus: M ≥ 0 everywhere 



(a)

www.elsolucionario.net PROBLEM 7.48 Assuming the upward reaction of the ground on beam AB to be uniformly distributed, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.

SOLUTION

ΣFy = 0: wg (12 ft) − (3 kips/ft)(6 ft) = 0

(a)

wg = 1.5 kips/ft 

Shear and bending-moment diagrams. From A to C: ΣFy = 0: 1.5 x − 3x − V = 0 V = (−1.5 x) kips x x − (1.5 x) = 0 2 2 2 M = ( −0.75 x ) kip ⋅ ft

ΣM J = 0: M + (3x)

For x = 0:

VA = M A = 0 

For x = 3 ft:

VC = −4.5 kips M C = −6.75 kip ⋅ ft 

From C to D: ΣFy = 0: 1.5 x = 9 − V = 0, V = (1.5 x − 9) kips ΣM J = 0: M + 9( x − 1.5) − (1.5 x)

x =0 2

M = 0.75 x 2 − 9 x + 13.5

For x = 3 ft:

VC = −4.5 kips,

For x = 6 ft:

VC = 0,

M C = −6.75 kip ⋅ ft  M C = −13.50 kip ⋅ ft 





For x = 9 ft:

VD = 4.5 kips, M D = −6.75 kip ⋅ ft  VB = M B = 0 

At B: (b)

|V |max = 4.50 kips 

   Bending-moment diagram consists of three distinct arcs of parabola.

M ≤ 0 everywhere 

Since entire diagram is below the x axis:



|M |max = 13.50 kip ⋅ ft 

www.elsolucionario.net PROBLEM 7.49 Draw the shear and bending-moment diagrams for the beam AB, and determine the maximum absolute values of the shear and bending moment.

SOLUTION Reactions: ΣM A = 0: By (0.4) − (120)(0.2) = 0 B y = 60 N

ΣFx = 0:

Bx = 0 A = 180 N


ΣFy = 0:

Equivalent loading on straight part of beam AB.

From A to C: ΣFy = 0: V = +180 N ΣM1 = 0: M = +180 x

From C to B: ΣFy = 0: 180 − 120 − V = 0 V = 60 N ΣM x = 0: − (180 N)( x) + 24 N ⋅ m + (120 N)( x − 0.2) + M = 0 M = +60 x

|V |max = 180.0 N  |M |max = 36.0 N ⋅ m  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1072


SOLUTION Free body: Entire beam ΣM A = 0: B(0.9 m) − (400 N)(0.3 m) − (400 N)(0.6 m) B = +800 N

B = 800 N 

ΣFx = 0: Ax = 0 ΣFy = 0: Ay + 800 N − 3(400 N) = 0 Ay = +400 N

A = 400 N 

We replace the loads by equivalent force-couple systems at C, D, and E. We consider successively the following F-B diagrams. V5 = −400 N

V1 = +400 N M1 = 0

M 5 = +180 N ⋅ m

V2 = +400 N

V6 = −400 N

M 2 = +60 N ⋅ m

M 6 = +60 N ⋅ m

V3 = 0

V7 = −800 N

M 3 = +120 N ⋅ m

M 7 = +120 N ⋅ m V8 = −800 N

V4 = 0 M 4 = +120 N ⋅ m

M8 = 0



− (400 N)(0.9 m) = 0


(b)

|V |max = 800 N 

|M |max = 180.0 N ⋅ m 


  



SOLUTION

∴

Dy 7

=

Dx 10

 7  ΣM H = 0: (50 lb)(26 in.) + Dx (4 in.) −  Dx  (20 in.) + (100 lb)(10 in.) − (50 lb)(6 in.) = 0  10 

2000 + (4 − 14) Dx = 0 Dy =

D x = 200 lb

7 7 Dx = (200) 10 10

ΣFx = 0: 200 lb − H x = 0

D y = 140 lb

H x = 200 lb

ΣFy = 0: 140 − 50 − 100 − 50 + H y = 0

H y = 60 lb

Equivalent loading on straight part of beam AB.



Slope of cable CD is


From A to E: ΣFy = 0: V = −50 lb ΣM1 = 0: M = −50 x

From E to F: ΣFy = 0: −50 + 140 − V = 0 V = +90 lb ΣM 2 = 0: (50 lb) x − (140 lb)( x − 6) − 800 lb ⋅ in. + M = 0 M = −40 + 90 x ΣFy = 0: −50 + 140 − 100 − V = 0 V = −10 lb ΣM 3 = 0: 50 x − 140( x − 6) + 100( x − 16) − 800 + M = 0 M = 1560 − 10 x

From G to B: ΣFy = 0: V − 50 = 0 V = 50 lb ΣM 4 = 0: − M − (50)(32 − x) = 0 M = −1600 + 50 x

|V |max = 90.0 lb; |M |max = 1400 lb ⋅ in. 



From F to G:


ΣFG = 0: T (9 in.) − (45 lb)(9 in.) − (120 lb)(21 in.) = 0

T = 325 lb

ΣFx = 0: − 325 lb + Gx = 0

G x = 325 lb

ΣFy = 0: G y − 45 lb − 120 lb = 0

G y = 165 lb

Equivalent loading on straight part of beam AB

From A to E: ΣFy = 0: V = +165 lb ΣM1 = 0: + 1625 lb ⋅ in. − (165 lb) x + M = 0 M = −1625 + 165 x

From E to F: ΣFy = 0: 165 − 45 − V = 0 V = +120 lb ΣM 2 = 0 : + 1625 lb ⋅ in. − (165 lb) x + (45 lb)( x − 9) + M = 0 M = −1220 + 120 x



SOLUTION


From F to B:

ΣFy = 0 V = +120 lb ΣM 3 = 0: − (120)(21 − x) − M = 0 M = −2520 + 120 x

|M |max = 1625lb ⋅ in. 



|V |max = 165.0 lb 

www.elsolucionario.net PROBLEM 7.53 Two small channel sections DF and EH have been welded to the uniform beam AB of weight W = 3 kN to form the rigid structural member shown. This member is being lifted by two cables attached at D and E. Knowing that θ = 30° and neglecting the weight of the channel sections, (a) draw the shear and bending-moment diagrams for beam AB, (b) determine the maximum absolute values of the shear and bending moment in the beam.

SOLUTION FBD Beam + channels: (a)

T1 = T2 = T

By symmetry:

T= T1x =

3 3 3

kN

2 3 3 T1 y = kN 2 M = (0.5 m)

FBD Beam:

3

2 3 = 0.433 kN ⋅ m

kN

With cable force replaced by equivalent force-couple system at F and G Shear Diagram:

V is piecewise linear  dV   dx = −0.6 kN/m  with 1.5 kN  

discontinuities at F and H. VF − = −(0.6 kN/m)(1.5 m) = 0.9 kN + V increases by 1.5 kN to +0.6 kN at F

VG = 0.6 kN − (0.6 kN/m)(1 m) = 0

Finish by invoking symmetry



ΣFy = 0: 2T sin 60° − 3 kN = 0


Moment diagram: M is piecewise parabolic  dM   dx decreasing with V   

with discontinuities of .433 kN at F and H. 1 M F − = − (0.9 kN)(1.5 m) 2 = −0.675 kN ⋅ m

M increases by 0.433 kN m to –0.242 kN ⋅ m at F+ M G = −0.242 kN ⋅ m +

1 (0.6 kN)(1 m) 2

= 0.058 kN ⋅ m |V |max = 900 N 

(b)

− + at F and G

|M |max = 675 N ⋅ m 

at F and G



Finish by invoking symmetry

www.elsolucionario.net PROBLEM 7.54 Solve Problem 7.53 when θ = 60°. PROBLEM 7.53 Two small channel sections DF and EH have been welded to the uniform beam AB of weight W = 3 kN to form the rigid structural member shown. This member is being lifted by two cables attached at D and E. Knowing that θ = 30° and neglecting the weight of the channel sections, (a) draw the shear and bending-moment diagrams for beam AB, (b) determine the maximum absolute values of the shear and bending moment in the beam.

SOLUTION Free body: Beam and channels E y = Dy

Ex = Dx = Dy tan θ

Thus:

(1)

ΣFy = 0: D y + E y − 3 kN = 0

D y = E y = 1.5 kN 

D x = (1.5 kN) tan θ

From (1):

E = (1.5 kN) tan θ



We replace the forces at D and E by equivalent force-couple systems at F and H, where M 0 = (1.5 kN tan θ )(0.5 m) = (750 N ⋅ m) tan θ

(2)

We note that the weight of the beam per unit length is w=

(a)

W 3 kN = = 0.6 kN/m = 600 N/m L 5m

Shear and bending moment diagrams From A to F: ΣFy = 0: − V − 600 x = 0 V = (−600 x) N ΣM J = 0: M + (600 x)

For x = 0:

x = 0, M = (−300 x 2 ) N ⋅ m 2 VA = M A = 0 

For x = 1.5 m:

VF = −900 N, M F = −675 N ⋅ m 



From symmetry:


From F to H: ΣFy = 0: 1500 − 600 x − V = 0 



V = (1500 − 600 x) N 

 ΣM J = 0: M + (600 x)



x − 1500( x − 1.5) − M 0 = 0 2

M = M 0 − 300 x 2 + 1500( x − 1.5) N ⋅ m 

For x = 1.5 m:

VF = +600 N, M F = ( M 0 − 675) N ⋅ m 

For x = 2.5 m:

VG = 0, M G = ( M 0 − 375) N ⋅ m 

From G to B, The V and M diagrams will be obtained by symmetry,

Making θ = 60° in Eq. (2):

|V |max = 900 N 

M 0 = 750 tan 60° = 1299 N ⋅ m M = 1299 − 675 = 624 N ⋅ m 

Thus, just to the right of F:

M G = 1299 − 375 = 924 N ⋅ m 

and

(b)

|V |max = 900 N 

  |M |max = 924 N ⋅ m 





(b)

www.elsolucionario.net PROBLEM 7.55 For the structural member of Problem 7.53, determine (a) the angle θ for which the maximum absolute value of the bending moment in beam AB is as small as possible, (b) the corresponding value of | M |max. (Hint: Draw the bending-moment diagram and then equate the absolute values of the largest positive and negative bending moments obtained.)

SOLUTION See solution of Problem 7.50 for reduction of loading or beam AB to the following:

M 0 = (750 N ⋅ m) tan θ 

where [Equation (2)] The largest negative bending moment occurs Just to the left of F:  1.5 m  ΣM1 = 0: M 1 + (900 N)  =0  2 

M1 = −675 N ⋅ m 

The largest positive bending moment occurs At G: ΣM 2 = 0: M 2 − M 0 + (1500 N)(1.25 m − 1 m) = 0 M 2 = M 0 − 375 N ⋅ m 

Equating M 2 and − M 1 : M 0 − 375 = +675 M 0 = 1050 N ⋅ m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1083


PROBLEM 7.53 Two small channel sections DF and EH have been welded to the uniform beam AB of weight W = 3 kN to form the rigid structural member shown. This member is being lifted by two cables attached at D and E. Knowing that θ = 30° and neglecting the weight of the channel sections, (a) draw the shear and bendingmoment diagrams for beam AB, (b) determine the maximum absolute values of the shear and bending moment in the beam.


(a)

From Equation (2):

tan θ =

1050 = 1.400 750

θ = 54.5° 

|M |max = 675 N ⋅ m 


(b)


www.elsolucionario.net PROBLEM 7.56 For the beam of Problem 7.43, determine (a) the ratio k = P/wa for which the maximum absolute value of the bending moment in the beam is as small as possible, (b) the corresponding value of |M |max. (See hint for Problem 7.55.) PROBLEM 7.43 Assuming the upward reaction of the ground on beam AB to be uniformly distributed and knowing that P = wa, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.

SOLUTION Free body: Entire beam

wg = wg

Setting

w

w (2 + k ) 4

=α

(1)

k = 4α − 2

We have

(2)

Minimum value of B.M. For M to have negative values, we must have wg < w. We verify that M will then be negative and keep decreasing in the portion AC of the beam. Therefore, M min will occur between C and D. From C to D: a   x ΣM J = 0: M + wa  x −  − α wx   = 0 2  2 M=

We differentiate and set

dM = 0: dx

1 w(α x 2 − 2ax + a 2 ) 2

αx − a = 0

xmin =

a

α

(3) (4)

Substituting in (3): 1 2 1 2  wa  − + 1 2 α α   α 1 − = − wa 2 2α

M min = M min

(5)



ΣFy = 0: wg (4a) − 2wa − kwa = 0


Maximum value of bending moment occurs at D  3a  ΣM D = 0: M D + wa   − (2α wa)a = 0  2  3  M max = M D = wa 2  2α −  2 

(6)

Equating − M min and M max : wa 2

1−α 3  = wa 2  2α −  2α 2 

α= (a)

Substitute in (2):

(b)

Substitute for α in (5):

2 + 20 8

α=

1+ 5 = 0.809 4

k = 4(0.809) − 2

|M |max = − M min = − wa 2

Substitute for α in (4):

k = 1.236 

1 − 0.809 2(0.809)

|M |max = 0.1180wa 2  xmin =

a 1.236a  0.809



4α 2 − 2α − 1 = 0

www.elsolucionario.net PROBLEM 7.57 For the beam of Problem 7.45, determine (a) the distance a for which the maximum absolute value of the bending moment in the beam is as small as possible, (b) the corresponding value of |M |max. (See hint for Problem 7.55.) PROBLEM 7.45 Assuming the upward reaction of the ground on beam AB to be uniformly distributed and knowing that a = 0.3 m, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.

SOLUTION Force per unit length exerted by ground: 6 kN = 4 kN/m 1.5 m

The largest positive bending moment occurs Just to the left of C: ΣM1 = 0: M 1 = (4a)

a 2

M 1 = 2a 2 

The largest negative bending moment occurs

At the center line: ΣM 2 = 0: M 2 + 3(0.75 − a ) − 3(0.375) = 0

Equating M1 and − M 2 :

M 2 = −(1.125 − 3a) 

2a 2 = 1.125 − 3a a 2 + 1.5a − 0.5625 = 0

(a)

Solving the quadratic equation:

(b)

Substituting:

a = 0.31066, |M |max = M 1 = 2(0.31066) 2

a = 0.311 m  |M |max = 193.0 N ⋅ m 



wg =

www.elsolucionario.net PROBLEM 7.58 For the beam and loading shown, determine (a) the distance a for which the maximum absolute value of the bending moment in the beam is as small as possible, (b) the corresponding value of |M |max. (See hint for Problem 7.55.)

SOLUTION Free body: Entire beam ΣFx = 0: Ax = 0 ΣM E = 0: − Ay (2.4) + (3)(1.8) + 3(1) − (2)a = 0 5 A = 3.5 kN − a  6

Free body: AC 5   ΣM C = 0: M C −  3.5 − a  (0.6 m) = 0, 6  

M C = +2.1 −

a  2

Free body: AD 5   ΣM D = 0: M D −  3.5 − a  (1.4 m) + (3 kN)(0.8 m) = 0 6   M D = +2.5 − ΣM E = 0: − M E − (2 kN)a = 0

Free body: EB

7 a  6

M E = −2a 

We shall assume that M C > M D and, thus, that M max = M C . We set M max = | M min |

or

M C = | M E | = 2.1 −

a = 2a 2

a = 0.840 m 

| M |max = M C = | M E | = 2a = 2(0.840)

|M |max = 1.680 N ⋅ m 

We must check our assumption. M D = 2.5 −

7 (0.840) = 1.520 N ⋅ m 6

Thus, M C > M D , O.K. The answers are

(a)

a = 0.840 m 

(b)

|M |max = 1.680 N ⋅ m 



5 Ay = 3.5 kN − a 6

www.elsolucionario.net PROBLEM 7.59 A uniform beam is to be picked up by crane cables attached at A and B. Determine the distance a from the ends of the beam to the points where the cables should be attached if the maximum absolute value of the bending moment in the beam is to be as small as possible. (Hint: Draw the bending-moment diagram in terms of a, L, and the weight w per unit length, and then equate the absolute values of the largest positive and negative bending moments obtained.)

SOLUTION

To the left of A:

 x ΣM1 = 0: M + wx   = 0 2 1 M = − wx 2 2 1 2 M A = − wa 2

Between A and B: 1  1  ΣM 2 = 0: M −  wL  ( x − a) + ( wx)  x  = 0 2  2  1 1 1 M = − wx 2 + wLx − wLa 2 2 2

At center C:

x=

L 2 2

1 L 1 L 1 M C = − w   + wL   − wLa 2 2 2 2 2



w = weight per unit length


We set

| M A| = | M C | :

1 1 1 1 1 1 − wa 2 = wL2 − wLa + wa 2 = wL2 − wLa 2 8 2 2 8 2 a 2 + La − 0.25L2 = 0 1 1 ( L ± L2 + L2 ) = ( 2 − 1) L 2 2 1 2 = w(0.207 L) = 0.0214 wL2 2

a=

a = 0.207 L 


M max


www.elsolucionario.net PROBLEM 7.60 Knowing that P = Q = 150 lb, determine (a) the distance a for which the maximum absolute value of the bending moment in beam AB is as small as possible, (b) the corresponding value of |M |max . (See hint for Problem 7.55.)

SOLUTION Free body: Entire beam ΣM A = 0: Da − (150)(30) − (150)(60) = 0 D=

13,500 a



ΣM C = 0: − M C − (150)(30) +

13,500 (a − 30) = 0 a

 45  M C = 9000 1 −  a  



Free body: DB ΣM D = 0: − M D − (150)(60 − a) = 0 M D = −150(60 − a)

(a)

We set  45  M C = − M D : 9000 1 −  = 150(60 − a) a  

M max = | M min | or 60 −

2700 = 60 − a a

a 2 = 2700 a = 51.96 in.

(b)



a = 52.0 in. 

|M |max = − M D = 150(60 − 51.96) | M | max = 1206 lb ⋅ in. 



Free body: CB

www.elsolucionario.net PROBLEM 7.61 Solve Problem 7.60 assuming that P = 300 lb and Q = 150 lb. PROBLEM 7.60 Knowing that P = Q = 150 lb, determine (a) the distance a for which the maximum absolute value of the bending moment in beam AB is as small as possible, (b) the corresponding value of |M |max . (See hint for Problem 7.55.)

SOLUTION Free body: Entire beam ΣM A = 0: Da − (300)(30) − (150)(60) = 0 18,000  a

Free body: CB ΣM C = 0: − M C − (150)(30) +

18,000 (a − 30) = 0 a

 40  M C = 13,500 1 −   a  

Free body: DB ΣM D = 0: − M D − (150)(60 − a) = 0 M D = −150(60 − a) 

(a)

We set  40  M max = | M min | or M C = − M D : 13,500 1 −  = 150(60 − a) a   90 −

3600 = 60 − a a

a 2 + 30a − 3600 = 0 a=

−30 + 15.300 = 46.847 2 a = 46.8 in. 

(b)

|M |max = − M D = 150(60 − 46.847) |M | max = 1973 lb ⋅ in. 



D=

www.elsolucionario.net PROBLEM 7.62* In order to reduce the bending moment in the cantilever beam AB, a cable and counterweight are permanently attached at end B. Determine the magnitude of the counterweight for which the maximum absolute value of the bending moment in the beam is as small as possible and the corresponding value of |M |max . Consider (a) the case when the distributed load is permanently applied to the beam, (b) the more general case when the distributed load may either be applied or removed.

SOLUTION M due to distributed load: ΣM J = 0: − M −

x wx = 0 2


1 M = − wx 2 2

M due to counter weight: ΣM J = 0: − M + xw = 0 M = wx

(a)

Both applied: M = Wx −

w 2 x 2

dM W = W − wx = 0 at x = dx w

And here M =

W2 > 0 so Mmax; Mmin must be at x = L 2w

So M min = WL − so

1 2 wL . For minimum | M |max set M max = − M min , 2 W2 1 = − WL + wL2 2w 2

or

W 2 + 2 wLW − w2 L2 = 0

W = − wL ± 2w2 L2 (need + )

W = ( 2 − 1) wL = 0.414 wL 



w may be removed M max =

Without w,

With w (see Part a)

W 2 ( 2 − 1) 2 2 = wL 2w 2

M max = 0.0858 wL2 

M = Wx M max = WL at A M = Wx −

w 2 x 2

W2 W at x = 2w w 1 2 = WL − wL at x = L 2

M max = M min

For minimum M max , set M max (no w) = − M min (with w) WL = − WL +

1 2 1 wL → W = wL → 2 4

With

M max =

1 2 wL  4

W=

1 wL  4



(b)

www.elsolucionario.net PROBLEM 7.63 Using the method of Section 7.6, solve Problem 7.29. PROBLEM 7.29 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.

SOLUTION Free body: Entire beam ΣFy = 0: wL − B = 0 B = wL

L ΣM B = 0: ( wL)   − M B = 0 2 MB =

wL2 2

Moment diagram (b)

|V |max = wL;

|M |max =

wL2  2



Shear diagram


SOLUTION Free body: Entire beam 1 ΣFy = 0: B − ( w0 )( L) = 0 2 1 w0 L 2

Shear diagram ΣM B = 0:

1 L ( w0 )( L)   − M B = 0 2 3 MB =

1 w0 L2 6

Moment diagram |V |max = w0 L /2 

(b)

| M |max = w0 L2 /6 



B=


SOLUTION Free body: Entire beam  2L  L ΣM D = 0: P  − P   − AL = 0   3  3

Shear diagram We note that VA = A = + P /3 |V |max = 2 P /3 

Bending diagram We note that M A = 0 |M |max = PL /9 



A = P /3


SOLUTION Free body: Entire beam ΣFy = 0: C = 0 1 PL 2

Shear diagram At A:

VA = + P |V |max = P 

Moment diagram At A:

MA = 0 |M |max =

1 PL  2



ΣM C = 0: M C =


SOLUTION Free body: Entire beam ΣFy = 0: A = C

A=C =

M0 L

VA = −

M0 L

Shear diagram At A:

|V | max =

M0  L

Bending-moment diagram MA = 0

At A:

At B, M increases by M0 on account of applied couple. |M | max = M 0 /2 



ΣM C = 0: Al − M 0 = 0


SOLUTION Free body: Entire beam ΣFy = 0: B − P = 0 B=P

MB = 0

Shear diagram VA = − P

At A:

|V | max = P 

Bending-moment diagram M A = M 0 = PL

At A:

|M | max = PL 



Σ M B = 0: M B − M 0 + PL = 0


SOLUTION Reactions

ΣFx = 0

Ax = 0

ΣM D = 0: + 3 kN ⋅ m + (6 kN)(3.5 m) + (3 kN)(2 m) − 1.2 kN ⋅ m − Ay (4.5 m) = 0

(b)

|V |max = 6.40 kN;

Ay = + 6.4 kN

|M |max = 4.00 kN ⋅ m 



Ay = + 6.4 kN


SOLUTION Reactions Σ M A = 0: 12 kN ⋅ m − 3(5 kN)(4 m) + E (8 m) = 0 E = 6 kN Σ Fy = 0


A = 9 kN

(b)

|V |max = 9.00 kN;

|M |max = 14.00 kN ⋅ m 



SOLUTION Free body: Beam ΣFx = 0: Ax = 0 ΣM B = 0: (60 kN)(3 m) + (50 kN)(1 m) − Ay (5 m) = 0

ΣFy = 0: B + 46.0 kN − 60 kN − 50 kN = 0 B = + 64.0 kN 䉰

Shear diagram At A:

VA = Ay = + 46.0 kN |V |max = 64.0 kN 

Bending-moment diagram At A:

MA = 0 |M |max = 92.0 kN ⋅ m 

Parabola from D to B. Its slope at D is same as that of straight-line segment CD since V has no discontinuity at D.



Ay = + 46.0 kN 䉰


SOLUTION

Free body: Entire beam Σ M B = 0: (24 kN)(9 m) − C (6 m) + (24 kN)(4.5 m) = 0


C = 54 kN

Shear diagram Σ Fy = 0: 54 − 24 − 24 − B = 0 B = 6 kN

(b)

|V |max = 30.0 kN;

|M |max = 72.0 kN ⋅ m 



SOLUTION Reactions at supports. Because of the symmetry: A= B=

1 (8 + 8 + 4 × 5) kips 2

Shear diagram At A:

VA = + 18 kips |V |max = 18.00 kips 


MA = 0 |M |max = 48.5 kip ⋅ ft 

Discontinuities in slope at C and D, due to the discontinuities of V.



A = B = 18 kips 䉰


SOLUTION Free body: Beam Σ Fx = 0: Ax = 0 Σ M B = 0: (12 kips)(4 ft) + (15 kips)(7 ft) − Ay (10 ft) = 0

Σ Fy = 0: B + 15.3 − 15 − 12 = 0 B = + 11.7 kips 䉰

Shear diagram At A:

VA = Ay = 15.3 kips |V |max = 15.30 kips 


MA = 0 |M |max = 46.8 kip ⋅ ft 



Ay = + 15.3 kips 䉰


SOLUTION

Reactions

A = +25 kips

A = 25 kips

ΣFy = 0: 25 − 16 − 45 − 8 + Dy = 0 Dy = +44,

Dy = 44 kips

Σ Fx = 0: Dx = 0

(b)

|V |max = 36.0 kips;

|M |max = 120.0 kip ⋅ ft 



Σ M D = 0: (16)(9) + (45)(4) − (8)(3) − A(12) = 0


SOLUTION Reactions Σ Fx = 0: Bx = 0 Σ M E = 0: (20 lb/in.)(9 in.)(40.5 in.) + (125 in.)(24 in.) + (125 in.)(12 in.) − B(36 in.) = 0 By = +327.5 lb

B y = 327.5 lb

E = +102.5 lb

(b)

|V |max = 180.0 lb;

E = 102.5 lb

|M |max = 1230 lb ⋅ in. 



Σ Fy = 0: −(20 lb/in.)(9 in.) − 125 lb − 125 lb + 327.5 lb + E = 0

www.elsolucionario.net PROBLEM 7.77 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the magnitude and location of the maximum absolute value of the bending moment.

SOLUTION

Reactions

C = +30 kN

C = 30 kN

Σ Fy = 0: A − 90 kN + 30 kN = 0 A = 60 kN

75.0 kN ⋅ m, 4.00 m from A 

(b)



Σ M A = 0: + 45 kN ⋅ m − (90 kN)(3 m) + C (7.5 m) = 0


SOLUTION

ΣM A = 0: (8.75)(1.75) − B(2.5) = 0

ΣFy = 0: A = 2.625 kN

Similar D’s x 2.5 − x 2.5 = = 2.625  3.625   6.25

}

Add numerators and denominators x = 1.05 m

1.378 kN ⋅ m, 1.050 m from A 

(b)



B = 6.125 kN

www.elsolucionario.net PROBLEM 7.79 For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the magnitude and location of the maximum absolute value of the bending moment.

SOLUTION Free body: Beam ΣM A = 0: B(4 m) − (100 kN)(2 m) − 20 kN ⋅ m = 0 B = + 55 kN 䉰 ΣFx = 0: Ax = 0

Ay = + 45 kN 䉰

Shear diagram VA = Ay = + 45 kN

At A:

To determine Point C where V = 0: VC − VA = − wx 0 − 45 kN = − (25 kN ⋅ m) x x = 1.8 m 䉰

We compute all areas bending-moment Bending-moment diagram At A:

MA = 0

At B:

M B = − 20 kN ⋅ m | M |max = 40.5 kN ⋅ m 





1.800 m from A 

Single arc of parabola



ΣFy = 0: Ay + 55 − 100 = 0

www.elsolucionario.net PROBLEM 7.80 Solve Problem 7.79 assuming that the 20-kN ⋅ m couple applied at B is counterclockwise. PROBLEM 7.79 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the magnitude and location of the maximum absolute value of the bending moment.

SOLUTION Free body: Beam ΣM A = 0: B(4 m) − (100 kN)(2 m) − 20 kN ⋅ m = 0 B = + 45 kN 䉰 ΣFy = 0: Ay + 45 − 100 = 0 Ay = + 55 kN 䉰

Shear diagram VA = Ay = + 55 kN

At A:

To determine Point C where V = 0 : VC − VA = − wx 0 − 55 kN = − (25 kN/m) x x = 2.20 m 䉰

We compute all areas bending-moment Bending-moment diagram At A:

MA = 0

At B:

M B = + 20 kN ⋅ m | M |max = 60.5 kN ⋅ m 





2.20 m from A 

Single arc of parabola



ΣFx = 0: Ax = 0

www.elsolucionario.net PROBLEM 7.81 For the beam shown, draw the shear and bending-moment diagrams, and determine the magnitude and location of the maximum absolute value of the bending moment, knowing that (a) M = 0, (b) M = 24 kip · ft.

SOLUTION Free body: Beam ΣFx = 0: Ax = 0 ΣM B = 0: (16 kips)(6 ft) − Ay (8 ft) − M = 0

(1) 

1 ΣFy = 0: B + 12 − M − 16 = 0 8 1 B = 4 kips + M 8

(a)

(2) 

M = 0:

Load diagram Making M = 0 in. (1) and (2). Ay = +12 kips B = +4 kips VA = Ay = +12 kips

Shear diagram

To determine Point D where V = 0: VD − VA = − wx 0 − 12 kips = −(4 kips/ft)x

x = 3 ft 

We compute all areas B. M. Diagram At A:

MA = 0 | M |max = 18.00 kip ⋅ ft,  3.00 ft from A 

Parabola from A to C



1 Ay = 12 kips − M 8


(b)

M = 24 kip ⋅ ft

Load diagram Making M = 24 kip ⋅ ft in (1) and (2) 1 A = 12 − (24) = +9 kips 8 1 B = 4 + (24) = +7 kips 8

Shear diagram

VA = Ay = +9 kips

To determine Point D where V = 0: VD = VA = − wx x = 2.25 ft 

B. M. Diagram At A:

M A = +24 kip ⋅ ft |M |max = 34.1 kip ⋅ ft,  2.25 ft from A 



Parabola from A to C.



0 − 9 kips = −(4 kips/ft) x

www.elsolucionario.net PROBLEM 7.82 For the beam shown, draw the shear and bending-moment diagrams, and determine the magnitude and location of the maximum absolute value of the bending moment, knowing that (a) P = 6 kips, (b) P = 3 kips.

SOLUTION Free body: Beam ΣFx = 0: Ax = 0 ΣM A = 0: C (6 ft) − (12 kips)(3 ft) − P(8 ft) = 0 4 P 3

(1) 

4   ΣFy = 0: Ay +  6 + P  − 12 − P = 0 3   1 Ay = 6 kips − P 3

(a)

(2) 

P = 6 kips.

Load diagram Substituting for P in Eqs. (2) and (1): 1 Ay = 6 − (6) = 4 kips 3 4 C = 6 + (6) = 14 kips 3 VA = Ay = +4 kips

Shear diagram

To determine Point D where V = 0: VD − VA = − wx 0 − 4 kips = (2 kips/ft)x

x = 2 ft 

We compute all areas Bending-moment diagram At A:

MA = 0 |M |max = 12.00 kip ⋅ ft, at C 

Parabola from A to C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1115


C = 6 kips +


(b)

P = 3 kips

Load diagram Substituting for P in Eqs. (2) and (1): 1 A = 6 − (3) = 5 kips 3 4 C = 6 + (3) = 10 kips 3

Shear diagram

VA = Ay = +5 kips

To determine D where V = 0: VD − VA = − wx x = 2.5 ft 


MA = 0 |M |max = 6.25 kip ⋅ ft  2.50 ft from A 



Parabola from A to C.



0 − (5 kips) = −(2 kips/ft) x

www.elsolucionario.net PROBLEM 7.83 (a) Draw the shear and bending-moment diagrams for beam AB, (b) determine the magnitude and location of the maximum absolute value of the bending moment.

SOLUTION Reactions at supports Because of symmetry of load 1 (300 × 8 + 300) 2 A = B = 1350 lb 䉰

Load diagram for AB The 300-lb force at D is replaced by an equivalent force-couple system at C.

Shear diagram VA = A = 1350 lb

At A:

To determine Point E where V = 0: VE − VA = − wx 0 − 1350 lb = − (300 lb/ft) x x = 4.50 ft 䉰


MA = 0

Note 600 − lb ⋅ ft drop at C due to couple |M |max = 3040 lb ⋅ ft 





4.50 ft from A 



A= B=

www.elsolucionario.net PROBLEM 7.84 Solve Problem 7.83 assuming that the 300-lb force applied at D is directed upward. PROBLEM 7.83 (a) Draw the shear and bending-moment diagrams for beam AB, (b) determine the magnitude and location of the maximum absolute value of the bending moment.

SOLUTION Reactions at supports Because of symmetry of load: 1 (300 × 8 − 300) 2 A = B = 1050 lb 䉰

Load diagram The 300-lb force at D is replaced by an equivalent force-couple system at C.

Shear diagram VA = A = 1050 lb

At A:

To determine Point E where V = 0: VE − VA = − wx 0 − 1050 lb = − (300 lb/ft) x x = 3.50 ft 䉰

We compute all areas Bending-moment diagram MA = 0

At A:

Note 600 − lb ⋅ ft increase at C due to couple |M |max = 1838 lb ⋅ ft 





3.50 ft from A 



A= B=

www.elsolucionario.net PROBLEM 7.85 For the beam and loading shown, (a) write the equations of the shear and bending-moment curves, (b) determine the magnitude and location of the maximum bending moment.

π x dv = − w = w0 cos 2 L dx  2L  π x V = − wdx = − w0   sin 2 L + C1  π 



(1)

dM  2L  π x = V = − w0   sin 2 L + C1 dx  π  2

πx  2L  M = Vdx = + w0   cos 2 L + C1 x + C2 π  



(2)

Boundary conditions At x = 0:

V = C1 = 0 C1 = 0

At x = 0:

 2L  M = + w0   cos(0) + C2 = 0  π 

2

 2L  C2 = − w0    π 

2

 2L  π x V = − w0   sin 2 L   π 

Eq. (1)

2

πx  2L   −1 + cos  M = w0    2 L   π   2

4  2L  |M max | = w0  | −1 + 0| = 2 w0 L2   π  π 

M max at x = L :



SOLUTION


SOLUTION

dV x = − w = − w0 dx L



V = − w0

(1)

dM 1 x2 = V = − w0 + C1 2 dx L

Eq. (7.3):

M=

(a)

x 1 x2 dx = − w0 + C1 2 L L



 1  x2 1 x3 + C1  dx = − w0 = + C1 x + C2  − w0 L 6 L  2 

(2)

Boundary conditions At

x = 0: M = 0 = 0 + 0 + C2

C2 = 0

1 x = L : M = 0 = − w0 L2 + C1 L 6

C1 =

1 w0 L 6

Substituting C1 and C2 into (1) and (2): 1 x2 1 V = − w0 + w0 L L 6 2 1 x3 1 M = − w0 + w0 Lx 6 L 6

(b)

V= M=

 1 x2  1 w0 L  − 2   2 3 L 

 x x3  1 w0 L2  − 3   6 L L 

Max moment occurs when V = 0: 1− 3

x2 =0 L2

M max =

Note: At x = 0,

x 1 = L 3

 1  1 3  1 − w0 L2    6  3  3  

A = VA =

1 1 w0 L   2 3

x = 0.577 L 

M max = 0.0642w0 L2  A=

1 w0 L  6



Eq. (7.1):


SOLUTION (a)

We note that at Load:

B( x = L): VB = 0, M B = 0

(1)

x 1 x   4x  w( x) = w0 1 −  − w0   = w0 1 −  L 3 L      3L 

VB − VC = −

L



V ( x) = w0

x



w( x) dx 0 − V ( x) = −



L x

w( x)dx

L

4x  1 − 3L  dx  

x

L

  2 x2  2L 2x2  w L x = w0  x − = − − +    0 3L  x 3 3L    V ( x) =

w0 (2 x 2 − 3Lx + L2 ) 3L

(2) 

Bending moment: We use to Eq. (7.4) between C ( x = x) and B( x = L) : M B − MC = =



L x

w0 3L

V ( x)dx 0 − M ( x)



L x

(2 x 2 − 3Lx + L2 )dx L

w 2 3  M ( x) = − 0  x3 − Lx 2 + L2 x  3L  3 2  w = − 0 [4 x3 − 9 Lx 2 + 6 L2 x]Lx 18L w = − 0 [(4 L3 − 9 L3 + 6 L3 ) − (4 x3 − 9 Lx 2 + 6 L2 x)] 18L M ( x) =

w0 (4 x3 − 9 Lx 2 + 6 L2 x − L3 ) 18L

(3) 



Shear: We use Eq. (7.2) between C ( x = x) and B( x = L):


(b)

Maximum bending moment dM =V = 0 dx

Eq. (2):

2 x 2 − 3Lx + L2 = 0 x=

Carrying into (3):

|M |max =

w0 L2 , At 72 w0 L2 18

At

x=

L 2



x =0


Note:

M max =

3− 9−8 L L= 4 2


www.elsolucionario.net PROBLEM 7.88 For the beam and loading shown, (a) write the equations of the shear and bending-moment curves, (b) determine the maximum bending moment.

SOLUTION (a)

Reactions at supports:

1 W A = B = W , where = Total load 2 L W=



L

wdx = w0

0



πx 1 − sin  dx L  

L 0

2  = w0 L 1 −   π 1 1 2  VA = A = W = w0 L 1 −  2 2  π

Thus

MA = 0

(1)

πx  w( x) = w0 1 − sin L  

Load: Shear: From Eq. (7.2):

V ( x ) − VA = −



x

w( x) dx

0

= − w0



πx 1 − sin L  dx  

x 0

Integrating and recalling first of Eqs. (1), x

1 2 L πx   V ( x) − w0 L 1 −  = − w0  x + cos  π 2 L 0  π  V ( x) =

πx 1 2 L L   w0 L 1 −  − w0  2 + cos  + w0 π π 2 L   π 

πx L L V ( x) = w0  − x − cos π 2 L  

(2) 



L

πx L  = w0  x + cos  x L 0 


Bending moment: From Eq. (7.4) and recalling that M A = 0. M ( x) − M A =



x

V ( x) dx

0 x

2 L 1 πx L = w0  x − x 2 −   sin  2 L  π   2 0

M ( x) =

(b)

1  2 L2 πx w0  Lx − x 2 − 2 sin  2  L  π

(3) 

Maximum bending moment dM = V = 0. dx L 2

as we may check from (2):

π L L L L V   = w0  − − cos  = 0 π 2 2 2 2    From (3):

2 L2 2 L2 π L 1 L M   = w0  − − 2 sin  4 π 2 2 2  2 1 8   = w0 L2 1 − 2  8  π 

= 0.0237 w0 L2 M max = 0.0237 w0 L2 , at

x=

L 2





This occurs at x =

www.elsolucionario.net PROBLEM 7.89 The beam AB is subjected to the uniformly distributed load shown and to two unknown forces P and Q. Knowing that it has been experimentally determined that the bending moment is +800 N ⋅ m at D and +1300 N ⋅ m at E, (a) determine P and Q, (b) draw the shear and bending-moment diagrams for the beam.

SOLUTION (a)

Free body: Portion AD ΣFx = 0: C x = 0 ΣM D = 0: −C y (0.3 m) + 0.800 kN ⋅ m + (6 kN)(0.45 m) = 0 C = 11.667 kN 

Free body: Portion EB ΣM E = 0: B(0.3 m) − 1.300 kN ⋅ m = 0 B = 4.333 kN 

Free body: Entire beam ΣM D = 0: (6 kN)(0.45 m) − (11.667 kN)(0.3 m) − Q (0.3 m) + (4.333 kN)(0.6 m) = 0 Q = 6.00 kN  ΣM y = 0: 11.667 kN + 4.333 kN −6 kN − P − 6 kN = 0 P = 4.00 kN 

Load diagram

(b)

Shear diagram At A: VA = 0 |V |max = 6 kN 



C y = +11.667 kN


Bending-moment diagram MA = 0

At A:

|M |max = 1300 N ⋅ m 

We check that M D = +800 N ⋅ m and M E = +1300 N ⋅ m

As given: M C = −900 N ⋅ m 


At C:


www.elsolucionario.net PROBLEM 7.90 Solve Problem 7.89 assuming that the bending moment was found to be +650 N ⋅ m at D and +1450 N ⋅ m at E. PROBLEM 7.89 The beam AB is subjected to the uniformly distributed load shown and to two unknown forces P and Q. Knowing that it has been experimentally determined that the bending moment is +800 N ⋅ m at D and +1300 N ⋅ m at E, (a) determine P and Q, (b) draw the shear and bendingmoment diagrams for the beam.

SOLUTION 

(a)

Free body: Portion AD ΣFx = 0: C x = 0

C y = +11.167 kN

C = 11.167 kN 

Free body: Portion EB ΣM E = 0: B(0.3 m) − 1.450 kN ⋅ m = 0

 

B = 4.833 kN 



Free body: Entire beam ΣM D = 0: (6 kN)(0.45 m) − (11.167 kN)(0.3 m) −Q (0.3 m) + (4.833 kN)(0.6 m) = 0 Q = 7.50 kN 

 

ΣM y = 0: 11.167 kN + 4.833 kN − 6 kN − P − 7.50 kN = 0 P = 2.50 kN 

Load diagram

 

(b)

Shear diagram VA = 0

At A:

|V |max = 6 kN 



ΣM D = 0: −C (0.3 m) + 0.650 kN ⋅ m + (6 kN)(0.45 m) = 0

www.elsolucionario.net 

PROBLEM 7.90 (Continued)

 Bending-moment diagram MA = 0

At A:

|M |max = 1450 N ⋅ m 

We check that M D = +650 N ⋅ m and M E = +1450 N ⋅ m

As given: M C = −900 N ⋅ m 

At C:





www.elsolucionario.net PROBLEM 7.91* The beam AB is subjected to the uniformly distributed load shown and to two unknown forces P and Q. Knowing that it has been experimentally determined that the bending moment is +6.10 kip ⋅ ft at D and +5.50 kip ⋅ ft at E, (a) determine P and Q, (b) draw the shear and bending-moment diagrams for the beam.

SOLUTION (a)

Free body: Portion DE ΣM E = 0: 5.50 kip ⋅ ft − 6.10 kip ⋅ ft + (1 kip)(2 ft) − VD (4 ft) = 0

VD = +0.350 kip

VE = −0.650 kip

Free body: Portion AD ΣM A = 0: 6.10 kip ⋅ ft − P(2ft) − (1 kip)(2 ft) − (0.350 kip)(4 ft) = 0

P = 1.350 kips  ΣFx = 0: Ax = 0 ΣFy = 0: Ay − 1 kip − 1.350 kip − 0.350 kip = 0

Ay = +2.70 kips

A = 2.70 kips 

Free body: Portion EB ΣM B = 0: (0.650 kip)(4 ft) + (1 kip)(2 ft) + Q(2 ft) − 5.50 kip ⋅ ft = 0

Q = 0.450 kip  ΣFy = 0: B − 0.450 − 1 − 0.650 = 0

B = 2.10 kips 



ΣFy = 0: 0.350 kip − 1kip − VE = 0

www.elsolucionario.net PROBLEM 7.91* (Continued) (b)

Load diagram

Shear diagram VA = A = +2.70 kips

At A:

To determine Point G where V = 0, we write VG − VC = − wμ

μ = 3.40 ft  |V |max = 2.70 kips at A 

We next compute all areas

Bending-moment diagram At A: M A = 0 Largest value occurs at G with AG = 2 + 3.40 = 5.40 ft |M |max = 6.345 kip ⋅ ft 

5.40 ft from A  Bending-moment diagram consists of 3 distinct arcs of parabolas.



0 − 0.85 kips = −(0.25 kip/ft)μ

www.elsolucionario.net PROBLEM 7.92* Solve Problem 7.91 assuming that the bending moment was found to be +5.96 kip ⋅ ft at D and +6.84 kip ⋅ ft at E. PROBLEM 7.91* The beam AB is subjected to the uniformly distributed load shown and to two unknown forces P and Q. Knowing that it has been experimentally determined that the bending moment is +6.10 kip ⋅ ft at D and +5.50 kip ⋅ ft at E, (a) determine P and Q, (b) draw the shear and bendingmoment diagrams for the beam.

SOLUTION (a)

Free body: Portion DE ΣM E = 0: 6.84 kip ⋅ ft − 5.96 kip ⋅ ft + (1 kip)(2 ft) − VD (4 ft) = 0

ΣFy = 0: 0.720 kip − 1kip − VE = 0 VE = −0.280 kip

Free body: Portion AD ΣM A = 0: 5.96 kip ⋅ ft − P (2 ft) − (1 kip)(2 ft) − (0.720 kip)(4 ft) = 0

P = 0.540 kip  ΣFx = 0: Ax = 0 ΣFy = 0: Ay − 1 kip − 0.540 kip − 0.720 kip = 0 Ay = +2.26 kips

A = 2.26 kips 

Free body: Portion EB ΣM B = 0: (0.280 kip)(4 ft) + (1 kip)(2 ft) + Q(2 ft) − 6.84 kip ⋅ ft = 0

Q = 1.860 kips  ΣFy = 0: B − 1.860 − 1 − 0.280 = 0

B = 3.14 kips 



VD = +0.720 kip


(b)

Load diagram

Shear diagram VA = A = +2.26 kips

At A:

VG − VC = − wμ 0 − (1.22 kips) = −(0.25 kip/ft)μ

μ = 4.88 ft  |V |max = 3.14 kips at B 

We next compute all areas Bending-moment diagram At A: M A = 0 Largest value occurs at G with AG = 2 + 4.88 = 6.88 ft

|M |max = 6.997 kip ⋅ ft 

6.88 ft from A  Bending-moment diagram consists of 3 distinct arcs of parabolas.



To determine Point G where V = 0, we write

www.elsolucionario.net PROBLEM 7.93 Three loads are suspended as shown from the cable ABCDE. Knowing that dC = 3 m, determine (a) the components of the reaction at E, (b) the maximum tension in the cable.

SOLUTION

ΣM A = 0: E y (16 m) − (2 kN)(4 m) − (3 kN)(8 m) − (4 kN)(12 m) = 0 E y = + 5 kN

E y = 5.00 kN 

Portion CDE:

ΣM C = 0: (5 kN)(8 m) − (4 kN)(4 m) − E x (3 m) = 0 E x = 8 kN

(b)

E x = 8.00 kN



Maximum tension occurs in DE: Tm = Ex2 + E y2 = 82 + 52

Tm = 9.43 kN 



(a)

www.elsolucionario.net PROBLEM 7.94 Knowing that the maximum tension in cable ABCDE is 13 kN, determine the distance dC.

SOLUTION

From force triangle Portion CDE:

Ex2 + 52 = 132 Ex = 12 kN

ΣM C = 0: (5 kN)(8 m) − (12 kN)hC − (4 kN)(4 m) = 0 hC = 2.00 m 



Maximum tension of 13 kN occurs in DE. See solution of Problem 7.93 for the determination of E y = 5.00 kN

www.elsolucionario.net PROBLEM 7.95 If dC = 8 ft, determine (a) the reaction at A, (b) the reaction at E.

SOLUTION Free body: Portion ABC ΣM C = 0 2 Ax − 16 Ay + 300(8) = 0 Ax = 8 Ay − 1200

Free body: Entire cable

ΣM E = 0: + 6 Ax + 32 Ay − (300 lb + 200 lb + 300 lb)16 ft = 0 3 Ax + 16 Ay − 6400 = 0

Substitute from Eq. (1): 3(8 Ay − 1200) + 16 Ay − 6400 = 0

Eq. (1)

A y = 250 lb

Ax = 8(250) − 1200

A x = 800 lb

ΣFx = 0: − Ax + Ex = 0 − 800 lb + Ex = 0

E x = 800 lb

ΣFy = 0: 250 + E y − 300 − 200 − 300 = 0

E y = 550 lb

(a)

A = 838 lb

(b)

E = 971 lb

17.35°  34.5° 



(1)

www.elsolucionario.net PROBLEM 7.96 If dC = 4.5 ft, determine (a) the reaction at A, (b) the reaction at E.

SOLUTION Free body: Portion ABC ΣM C = 0: − 1.5 Ax − 16 Ay + 300 × 8 = 0


(2400 − 16 Ay )

(1)

1.5

ΣM E = 0: + 6 Ax + 32 Ay − (300 lb + 200 lb + 300 lb)16 ft = 0 3 Ax + 16 Ay − 6400 = 0

Substitute from Eq. (1):

3(2400 − 16 Ay ) 1.5

+ 16 Ay − 6400 = 0 Ay = −100 lb A y = 100 lb

Thus Ay acts downward Eq. (1)

(2400 − 16( −100)) = 2667 lb 1.5

A x = 2667 lb

ΣFx = 0: − Ax + Ex = 0 − 2667 + Ex = 0

E x = 2667 lb

Ax =

ΣFy = 0: Ay + E y − 300 − 200 − 300 = 0 −100 lb + E y − 800 lb = 0

E y = 900 lb

A = 2670 lb

(a)

E = 2810 lb

(b)

2.10°  18.65° 



Ax =

www.elsolucionario.net PROBLEM 7.97 Knowing that dC = 3 m, determine (a) the distances dB and dD (b) the reaction at E.

SOLUTION Free body: Portion ABC ΣM C = 0: 3 Ax − 4 Ay + (5 kN)(2 m) = 0 4 10 Ay − 3 3

(1)


Ax =


ΣM E = 0: 4 Ax − 10 Ay + (5 kN)(8 m) + (5 kN)(6 m) + (10 kN)(3 m) = 0 4 Ax − 10 Ay + 100 = 0

Substitute from Eq. (1): 10  4 4  Ay −  − 10 Ay + 100 = 0 3 3 Ay = +18.571 kN

Eq. (1)

A y = 18.571 kN

4 10 (18.511) − = +21.429 kN 3 3

A x = 21.429 kN

ΣFx = 0: − Ax + Ex = 0 − 21.429 + Ex = 0

E x = 21.429 kN

ΣFy = 0: 18.571 kN + E y + 5 kN + 5 kN + 10 kN = 0

E y = 1.429 kN

Ax =

E = 21.5 kN

(b)

3.81° 



(a)

Portion AB ΣM B = 0: (18.571 kN)(2 m) − (21.429 kN)d B = 0 d B = 1.733 m 

Portion DE Geometry h = (3 m) tan 3.8° = 0.199 m d D = 4 m + 0.199 m d D = 4.20 m 





www.elsolucionario.net PROBLEM 7.98 Determine (a) distance dC for which portion DE of the cable is horizontal, (b) the corresponding reactions at A and E.

SOLUTION

(b)

ΣFy = 0: Ay − 5 kN − 5 kN − 10 kN = 0

A y = 20 kN

ΣM A = 0: E (4 m) − (5 kN)(2 m) − (5 kN)(4 m) − (10 kN)(7 m) = 0 E = +25 kN

E = 25.0 kN

ΣFx = 0: − Ax + 25 kN = 0

A x = 25 kN A = 32.0 kN

(a)



38.7° 

Free body: Portion ABC

ΣM C = 0: (25 kN)dC − (20 kN)(4 m) + (5 kN)(2 m) = 0 25dC − 70 = 0

dC = 2.80 m 




www.elsolucionario.net PROBLEM 7.99 An oil pipeline is supported at 6-ft intervals by vertical hangers attached to the cable shown. Due to the combined weight of the pipe and its contents the tension in each hanger is 400 lb. Knowing that dC = 12 ft, determine (a) the maximum tension in the cable, (b) the distance dD.

SOLUTION FBD Cable: Hanger forces at A and F act on the supports, so A y and Fy act on the cable. ΣM F = 0: (6 ft + 12 ft + 18 ft + 24 ft)(400 lb) − (30 ft)Ay − (5 ft) Ax = 0 Ax + 6 Ay = 4800 lb

ΣM C = 0: (7 ft)Ax − (12 ft)Ay + (6 ft)(400 lb) = 0

(2)

Solving (1) and (2) A x = 800 lb Ay =

From FBD Cable:

2000 lb 3

ΣFx = 0: − 800 lb + Fx = 0 Fx = 800 lb

FBD DEF: ΣFy = 0:

200 lb − 4 (400 lb) + Fy = 0 3

Fy =

Since Ax = Fx and Fy > Ay ,

 2800  lb  Tmax = TEF = (800 lb) 2 +   3 

2800 lb 3

2



FBD ABC:

(1)


(a)

Tmax = 1229.27 lb,

Tmax = 1229 lb   2800  ΣM D = 0: (12 ft)  lb  − d D (800 lb) − (6 ft)(400 lb) = 0  3  d D = 11.00 ft 


(b)


www.elsolucionario.net PROBLEM 7.100 Solve Problem 7.99 assuming that dC = 9 ft. PROBLEM 7.99 An oil pipeline is supported at 6-ft intervals by vertical hangers attached to the cable shown. Due to the combined weight of the pipe and its contents the tension in each hanger is 400 lb. Knowing that dC = 12 ft, determine (a) the maximum tension in the cable, (b) the distance dD.

SOLUTION FBD CDEF:

Fx − 2 Fy = −800 lb

(1)

EM A = 0: (30 ft)Fy − (5 ft)Fx

FBD Cable:

− (6 ft)(1 + 2 + 3 + 4)(400 lb) = 0 Fx − 6 Fy = −4800 lb

(2)

Solving (1) and (2), Fx = 1200 lb

, Fy = 1000 lb

ΣFx = 0: − Ax + 1200 lb = 0, A x = 1200 lb

Point F:

ΣFy = 0: Ay + 1000 lb − 4(400 lb) = 0,

A y = 600 lb



ΣM C = 0: (18 ft)Fy − (9 ft)Fy − (6 ft + 12 ft)(400 lb) = 0


Since Ax = Ay

and Fy > Ay , Tmax = TEF

Tmax = (1 kip) 2 + (1.2 kips)2 Tmax = 1.562 kips 

(a) FBD DEF:

ΣM D = 0: (12 ft)(1000 lb) − d D (1200 lb)

d D = 8.00 ft 

(b)



− (6 ft)(400 lb) = 0

www.elsolucionario.net PROBLEM 7.101 Knowing that mB = 70 kg and mC = 25 kg, determine the magnitude of the force P required to maintain equilibrium.

SOLUTION Free body: Portion CD ΣM C = 0: D y (4 m) − Dx (3 m) = 0 3 Dx 4


ΣM A = 0:

3 Dx (14 m) − WB (4 m) − WC (10 m) − P(5 m) = 0 4

ΣM B = 0:

3 Dx (10 m) − Dx (5 m) − WC (6 m) = 0 4

(1)

Free body: Portion BCD

Dx = 2.4WC

For

mB = 70 kg mC = 25 kg

g = 9.81 m/s 2 :

WB = 70 g

Eq. (2):

Dx = 2.4WC = 2.4(25g ) = 60 g

Eq. (1):

(2)

WC = 25 g

3 60 g (14) − 70 g (4) − 25 g (10) − 5P = 0 4 100 g − 5 P = 0: P = 20 g P = 20(9.81) = 196.2 N

P = 196.2 N 



Dy =

www.elsolucionario.net PROBLEM 7.102 Knowing that mB = 18 kg and mC = 10 kg, determine the magnitude of the force P required to maintain equilibrium.

SOLUTION Free body: Portion CD ΣM C = 0: D y (4 m) − Dx (3 m) = 0 3 Dx 4


ΣM A = 0:

3 Dx (14 m) − WB (4 m) − WC (10 m) − P(5 m) = 0 4

ΣM B = 0:

3 Dx (10 m) − Dx (5 m) − WC (6 m) = 0 4

(1)

Free body: Portion BCD

Dx = 2.4WC

For

mB = 18 kg mC = 10 kg

g = 9.81 m/s 2 :

WB = 18 g

Eq. (2):

Dx = 2.4WC = 2.4(10 g ) = 24 g

Eq. (1):

3 24 g (14) − (18 g )(4) − (10 g )(10) − 5P = 0 4

(2)

WC = 10 g

80 g − 5P : P = 16 g P = 16(9.81) = 156.96 N

P = 157.0 N 



Dy =

www.elsolucionario.net PROBLEM 7.103 Cable ABC supports two loads as shown. Knowing that b = 21 ft, determine (a) the required magnitude of the horizontal force P, (b) the corresponding distance a.

Free body: ABC ΣM A = 0: P(12 ft) − (140 lb) a − (180 lb)b = 0

(1)

ΣM B = 0: P(9 ft) − (180 lb)(b − a) = 0

(2)

Free body: BC

Data

b = 21 ft

Equate (3) and (4) through P : Eq. (3):

P=

20 a + 180 3

Eq. (1): 21P − 140a − 180(21) = 0

P=

Eq. (2): 9 P − 180(21 − a) = 0

P = −20a + 420

(4)

(b)

a = 9.00 ft 

(a)

P = 240 lb 

20 a + 180 = −20a + 420 3

20 (9.00) + 180 3

(3)



SOLUTION

www.elsolucionario.net PROBLEM 7.104 Cable ABC supports two loads as shown. Determine the distances a and b when a horizontal force P of magnitude 200 lb is applied at A.

Free body: ABC ΣM A = 0: P(12 ft) − (140 lb) a − (180 lb)b = 0

(1)

ΣM B = 0: P(9 ft) − (180 lb)(b − a) = 0

(2)

Free body: BC

Data

P = 200 lb

Eq. (1): 200(21) − 140a − 180b = 0

(3)

Eq. (2): 200(9) + 180a − 180b = 0

(4)

(3) − (4) : 2400 − 320a = 0

a = 7.50 ft 

Eq. (2): (200 lb)(9 ft) − (180 lb)(b − 7.50 ft) = 0

b = 17.50 ft 



SOLUTION

www.elsolucionario.net PROBLEM 7.105 If a = 3 m, determine the magnitudes of P and Q required to maintain the cable in the shape shown.

SOLUTION Free body: Portion DE

Ey =

5 Ex 4

Free body: Portion CDE ΣM C = 0:

5 Ex (8 m) − E x (7 m) − P(2 m) = 0 4 Ex =

2 P 3

(1)

Free body: Portion BCDE

ΣM B = 0:

5 Ex (12 m) − E x (5 m) − (120 kN)(4 m) = 0 4 10 Ex − 480 = 0; Ex = 48 kN

Eq. (1):

48 kN =

2 P 3

P = 72.0 kN 



ΣM D = 0: E y (4 m) − Ex (5 m) = 0



ΣM A = 0:

5 Ex (16 m) − Ex (2 m) + P(3 m) − Q (4 m) − (120 kN)(8 m) = 0 4 (48 kN)(20 m − 2 m) + (72 kN)(3 m) − Q (4 m) − 960 kN ⋅ m = 0 Q = 30.0 kN 


4Q = 120


www.elsolucionario.net PROBLEM 7.106 If a = 4 m, determine the magnitudes of P and Q required to maintain the cable in the shape shown.

SOLUTION Free body: Portion DE ΣM D = 0: E y (4 m) − Ex (6 m) = 0 3 Ex 2

Free body: Portion CDE ΣM C = 0:

3 Ex (8 m) − E x (8 m) − P(2 m) = 0 2 Ex =

1 P 2

(1)

Free body: Portion BCDE

ΣM B = 0:

3 Ex (12 m) − E x (6 m) + (120 kN)(4 m) = 0 2 12 Ex = 480

Eq (1):

Ex =

1 P; 2

Ex = 40 kN 40 kN =

1 P 2

P = 80.0 kN 



Ey =



ΣM A = 0:

3 Ex (16 m) − Ex (2 m) + P (4 m) − Q (4 m) − (120 kN)(8 m) = 0 2 (40 kN)(24 m − 2 m) + (80 kN)(4 m) − Q (4 m) − 960 kN ⋅ m = 0 Q = 60.0 kN 


4Q = 240


www.elsolucionario.net PROBLEM 7.107 A transmission cable having a mass per unit length of 0.8 kg/m is strung between two insulators at the same elevation that are 75 m apart. Knowing that the sag of the cable is 2 m, determine (a) the maximum tension in the cable, (b) the length of the cable.

SOLUTION w = (0.8 kg/m)(9.81 m/s 2 ) = 7.848 N/m W = (7.848 N/m)(37.5 m) W = 294.3 N 1  ΣM B = 0: T0 (2 m) − W  37.5 m  = 0 2  1 T0 (2 m) − (294.3 N) (37.5 m) = 0 2 T0 = 2759 N Tm2 = (294.3 N) 2 + (2759 N)2

(b)

 sB = xB 1 +  

Tm = 2770 N 

2  2  yB    +  3  xB   

 2  2 m 2  = 37.5 m 1+  +    3  37.5 m   = 37.57 m

Length = 2sB = 2(37.57 m)

Length = 75.14 m 



(a)

www.elsolucionario.net PROBLEM 7.108 The total mass of cable ACB is 20 kg. Assuming that the mass of the cable is distributed uniformly along the horizontal, determine (a) the sag h, (b) the slope of the cable at A.

SOLUTION

Σ M D = 0: Ax (4.5 m) − (196.2 N)(4 m) − (1471.5 N)(6 m) = 0 Ax = 2136.4 N

Free body: Entire cable Σ M D = 0: Ay (8 m) − (196.2 N)(4 m) = 0 Ay = 98.1 N

(a)

Free body: Portion AC Σ Fx = 0: T0 = Ax = 2136.4 N Σ M A = 0: T0 h − (98.1 N)(2 m) = 0 (2136.4 N)h − 196.2 N ⋅ m = 0 h = 0.09183 m

(b)

tan θ A =

h = 91.8 mm 

Ax 98.1 N = Ay 2136.4 N

tan θ A = 0.045918

θ A = 2.629°

θ A = 2.63° 



Free body: Entire frame

www.elsolucionario.net PROBLEM 7.109 The center span of the Verrazano-Narrows Bridge consists of two uniform roadways suspended from four cables. The uniform load supported by each cable is w = 10.8 kips/ft along the horizontal. Knowing that the span L is 4260 ft and that the sag h is 390 ft, determine (a) the maximum tension in each cable, (b) the length of each cable.

SOLUTION

yB =

At B:

390 ft =

wxB2 2T0 (10.8 kips/ft)(2130 ft) 2 2T0

T0 = 62,819 kips

Tm = T02 + w2 xB2 = (62,819 kips) 2 + (10.8 kips/ft) 2 (2130 ft) 2 = 62,8192 + (23, 004) 2 = 66,898 kips

(b)

Tm = 66,900 kips 

Length of cable  sB = xB 1 +  

2 4 2  yB  2  yB   −      3  xB  5  xB    2 4  2  390 ft  2  390 ft    = (2130 ft) 1 +   −    = 2176.65 ft 5  2130 ft    3  2130 ft   

Total length:

2 S B = 2(2176.65 ft) = 4353.3 ft

L = 4353 ft 



(a)

www.elsolucionario.net PROBLEM 7.110 The center span of the Verrazano-Narrows Bridge consists of two uniform roadways suspended from four cables. The design of the bridge allows for the effect of extreme temperature changes that cause the sag of the center span to vary from hw = 386 ft in winter to hs = 394 ft in summer. Knowing that the span is L = 4260 ft, determine the change in length of the cables due to extreme temperature changes.

SOLUTION Eq. 7.10:

 2  y 2 2  y 4  sB = xB 1 +  B  −  B  +   3  xB  5  x B    

Winter:

yB = h = 386 ft, xB =

 2  386 2 2  386 4  sB = (2130) 1 +  −  +  = 2175.715 ft   5  2130   3  2130  

Summer:

yB = h = 394 ft, xB =  sB = (2130) 1 + 

1 L = 2130 ft 2

2 4  2  394  2  394  − +  = 2177.59 ft     3  2130  5  2130  

Δ = 2( Δ sB ) = 2(2177.59 ft − 2175.715 ft) = 2(1.875 ft)

Change in length = 3.75 ft 



1 L = 2130 ft 2

www.elsolucionario.net PROBLEM 7.111 Each cable of the Golden Gate Bridge supports a load w = 11.1 kips/ft along the horizontal. Knowing that the span L is 4150 ft and that the sag h is 464 ft, determine (a) the maximum tension in each cable, (b) the length of each cable.

SOLUTION

At B:

(a)

yB =

wxB2 2T0

T0 =

wxB2 (11.1 kip/ft)(2075 ft)2 = 2 yB 2(464 ft)

T0 = 51.500 kips W = wxB = (11.1 kips/ft)(2075 ft) = 23.033 kips Tm = T02 + W 2 = (51.500 kips)2 + (23.033 kips) 2 Tm = 56, 400 kips 

(b)

 sB = xB 1 +  

2 4  2  yB  2  yB   − +      3  xB  5  y B   

yB 464 ft = = 0.22361 xB 2075 ft

2  2  sB = (2075 ft) 1 + (0.22361) 2 − (0.22361) 4 +  = 2142.1 ft 5  3 

Length = 2sB = 2(2142.1 ft)

Length = 4284 ft 



Eq. (7.8) Page 386:

www.elsolucionario.net PROBLEM 7.112 Two cables of the same gauge are attached to a transmission tower at B. Since the tower is slender, the horizontal component of the resultant of the forces exerted by the cables at B is to be zero. Knowing that the mass per unit length of the cables is 0.4 kg/m, determine (a) the required sag h, (b) the maximum tension in each cable.

SOLUTION W = wxB

Horiz. comp. = T0 =

(a) Cable AB:

w(45 m) 2 2h

xB = 30 m, T0 =

Equate T0 = T0

wxB2 2 yB

xB = 45 m T0 =

Cable BC:

yB =0 2

yB = 3 m

w(30 m) 2 2(3 m)

w (45 m) 2 w (30 m) 2 = 2h 2(3 m)

h = 6.75 m 

Tm2 = T02 + W 2

(b) Cable AB:

w = (0.4 kg/m)(9.81 m/s) = 3.924 N/m xB = 45 m, T0 =

yB = h = 6.75 m

wxB2 (3.924 N/m)(45 m)2 = = 588.6 N 2 yB 2(6.75 m)

W = wxB = (3.924 N/m)(45 m) = 176.58 N Tm2 = (588.6 N) 2 + (176.58 N)2 Tm = 615 N 

For AB:



ΣM B = 0: T0 yB − ( wxB )


Cable BC:

xB = 30 m, T0 =

yB = 3 m

wxB2 (3.924 N/m)(30 m)2 = = 588.6 N (Checks) 2 yB 2(3 m)

W = wxB = (3.924 N/m)(30 m) = 117.72 N Tm2 = (588.6 N) 2 + (117.72 N) 2 Tm = 600 N 


For BC:


www.elsolucionario.net PROBLEM 7.113 A 50.5-m length of wire having a mass per unit length of 0.75 kg/m is used to span a horizontal distance of 50 m. Determine (a) the approximate sag of the wire, (b) the maximum tension in the wire. [Hint: Use only the first two terms of Eq. (7.10).]

SOLUTION First two terms of Eq. 7.10 1 (50.5 m) = 25.25 m, 2 1 xB = (50 m) = 25 m 2 yB = h  sB = xB 1 +  

2 2  yB      3  xB   

 2  y 2  25.25 m = 25 m 1 −  B    3  xB     2

 yB  3   = 0.01  = 0.015 x 2  B yB = 0.12247 xB 2

h = 0.12247 25m h = 3.0619 m

(b)

h = 3.06 m 

Free body: Portion CB w = (0.75 kg/m) (9.81m) = 7.3575 N/m W = sB w = (25.25 m) (7.3575 N/m) W = 185.78 N ΣM 0 = 0: T0 (3.0619 m) − (185.78 N) (12.5 m) = 0 T0 = 758.4 N Bx = T0 = 758.4 N + ΣFy = 0: By − 185.78 N = 0 By = 185.78 N

Tm = Bx2 + By2 = (758.4 N) 2 + (185.78 N) 2

Tm = 781 N 



sB =

(a)

www.elsolucionario.net PROBLEM 7.114 A cable of length L + Δ is suspended between two points that are at the same elevation and a distance L apart. (a) Assuming that Δ is small compared to L and that the cable is parabolic, determine the approximate sag in terms of L and Δ. (b) If L = 100 ft and Δ = 4 ft, determine the approximate sag. [Hint: Use only the first two terms of Eq. (7.10).

SOLUTION Eq. 7.10

(First two terms)

(a)

 2  y 2  sB = xB 1 +  B    3  xB     xB = L/2 1 ( L + Δ) 2 yB = h  1 L ( L + Δ) = 1 + 2 2 

2 2 h     3  L2   

Δ 4 h2 3 = ; h 2 = LΔ; 2 3 L 8

(b)

L = 100 ft, h = 4 ft.

h= h=

3 (100)(4); 8

3 LΔ  8

h = 12.25 ft 



sB =

www.elsolucionario.net PROBLEM 7.115 The total mass of cable AC is 25 kg. Assuming that the mass of the cable is distributed uniformly along the horizontal, determine the sag h and the slope of the cable at A and C.

SOLUTION Cable:

m = 25 kg W = 25 (9.81)

Block:


= 245.25 N m = 450 kg W = 4414.5 N ΣM B = 0: (245.25) (2.5) + (4414.5)(3) − C x (2.5) = 0 C x = 5543 N ΣFx = 0: Ax = C x = 5543 N ΣM A = 0: C y (5) − (5543) (2.5) − (245.25) (2.5) = 0 C y = 2894 N

+ ΣFy = 0: C y − Ay − 245.25 N = 0 2894 N − Ay − 245.25 N = 0

Point A:

tan θ A =

Ay

Point C:

tan θC =

Cy

Free body: Half cable

Ax

Cx

A y = 2649 N

=

2649 = 0.4779; 5543

θ A = 25.5° 

=

2894 = 0.5221; 5543

θ A = 27.6° 

W = (12.5 kg) g = 122.6 ΣM 0 = 0: (122.6 N) (1.25 M) + (2649 N) (2.5 m) − (5543 N) yd = 0 yd = 1.2224 m; sag = h = 1.25 m − 1.2224 m h = 0.0276 m = 27.6 mm 


www.elsolucionario.net PROBLEM 7.116 Cable ACB supports a load uniformly distributed along the horizontal as shown. The lowest Point C is located 9 m to the right of A. Determine (a) the vertical distance a, (b) the length of the cable, (c) the components of the reaction at A.

SOLUTION Free body: Portion AC ΣFy = 0: Ay − 9 w = 0 A y = 9w

ΣM A = 0: T0 a − (9 w)(4.5 m) = 0

(1)


Free body: Portion CB ΣFy = 0: By − 6w = 0 B y = 6w


ΣM A = 0: 15w (7.5 m) − 6 w (15 m) − T0 (2.25 m) = 0 T0 = 10 w

(a) Eq. (1):

T0 a − (9w)(4.5 m) = 0 10wa = (9w)(4.5) = 0

a = 4.05 m 



(b)

Length = AC + CB Portion AC:

x A = 9 m, s AC s AC

Portion CB:

y A = a = 4.05 m;

y A 4.05 = = 0.45 9 xA

 2  y 2 2  y 4  = xB 1 +  A  −  B  +  5  xA   3  xA     2  2  = 9 m 1+ 0.452 − 0.454 +   = 10.067 m 5  3 

xB = 6 m,

yB = 4.05 − 2.25 = 1.8 m;

yB = 0.3 xB

Total length = 10.067 m + 6.341 m

(c)

Total length = 16.41 m 

Components of reaction at A. Ay = 9 w = 9(60 kg/m)(9.81 m/s 2 ) = 5297.4 N Ax = T0 = 10 w = 10(60 kg/m)(9.81 m/s 2 ) = 5886 N

A x = 5890 N



A y = 5300 N 



2 2   sCB = 6 m 1 + 0.32 − 0.34 +   = 6.341 m 3 5  

www.elsolucionario.net PROBLEM 7.117 Each cable of the side spans of the Golden Gate Bridge supports a load w = 10.2 kips/ft along the horizontal. Knowing that for the side spans the maximum vertical distance h from each cable to the chord AB is 30 ft and occurs at midspan, determine (a) the maximum tension in each cable, (b) the slope at B.

SOLUTION FBD AB: ΣM A = 0: (1100 ft)TBy − (496 ft)TBx − (550 ft)W = 0 

(1)

FBD CB: ΣM C = 0: (550 ft)TBy − (278 ft)TBx − (275 ft)

W =0 2

11TBy − 5.56TBx = 2.75W

Solving (1) and (2)

TBy = 28, 798 kips

Solving (1) and (2

TBx = 51, 425 kips  Tmax = TB = TB2x + TB2y

So that

tan θ B =

TBy TBx

(2)

 (a) (b)

Tmax = 58,900 kips 

θ B = 29.2° 



11TBy − 4.96TBx = 5.5W

www.elsolucionario.net PROBLEM 7.118 A steam pipe weighting 45 lb/ft that passes between two buildings 40 ft apart is supported by a system of cables as shown. Assuming that the weight of the cable system is equivalent to a uniformly distributed loading of 5 lb/ft, determine (a) the location of the lowest Point C of the cable, (b) the maximum tension in the cable.

SOLUTION Note:

xB − x A = 40 ft

or

x A = xB − 40 ft

Use Eq. 7.8 Point A:

yA =

wx A2 w ( xB − 40) 2 ; 9= 2T0 2T0

(1)

Point B:

yB =

wxB2 wx 2 ; 4= B 2T0 2T0

(2)

Dividing (1) by (2): (b)

9 ( xB − 40) 2 ; xB = 16 ft = 4 xB2

Point C is 16.00 ft to left of B 

Maximum slope and thus Tmax is at A x A = xB − 40 = 16 − 40 = −24 ft yA =

wx A2 (50 lb/ft)(−24 ft) 2 ; 9 ft = ; T0 = 1600 lb 2T0 2T0

WAC = (50 lb/ft)(24 ft) = 1200 lb Tmax = 2000 lb 



(a)

www.elsolucionario.net PROBLEM 7.119* A cable AB of span L and a simple beam A′B′ of the same span are subjected to identical vertical loadings as shown. Show that the magnitude of the bending moment at a point C′ in the beam is equal to the product T0h, where T0 is the magnitude of the horizontal component of the tension force in the cable and h is the vertical distance between Point C and the chord joining the points of support A and B.

SOLUTION ΣM B = 0: LACy + aT0 − ΣM B loads = 0

(1)

(Where ΣM B loads includes all applied loads) x  ΣM C = 0: xACy −  h − a  T0 − ΣM C left = 0 L  

(2)

FBD AC: (Where ΣM C left includes all loads left of C) x x (1) − (2): hT0 − ΣM B loads + ΣM C left = 0 L L

(3)

ΣM B = 0: LABy − ΣM B loads = 0

(4)

ΣM C = 0: xABy − ΣM C left − M C = 0

(5)

FBD Beam:

FBD AC:

Comparing (3) and (6)

x x (4) − (5): − ΣM B loads + ΣM C left + M C = 0 L L M C = hT0

(6) Q.E.D.



FBD Cable:

www.elsolucionario.net PROBLEM 7.120 Making use of the property established in Problem 7.119, solve the problem indicated by first solving the corresponding beam problem. PROBLEM 7.94 Knowing that the maximum tension in cable ABCDE is 13 kN, determine the distance dC.

SOLUTION Free body: beam AE ΣM E = 0: − A(16) + 2(12) + 3(8) + 4(4) = 0

E = 5 kN


ΣFy = 0: 4 − 2 − 3 − 4 + E = 0

A = 4 kN

At E: At C:

Tm2 = T02 + E 2 M C = T0 hC ;

132 = T02 + 52

T0 = 12 kN

24 kN ⋅ m = (12 kN)hC hC = dC = 2.00 m

2.00 m 


www.elsolucionario.net PROBLEM 7.121 Making use of the property established in Problem 7.119, solve the problem indicated by first solving the corresponding beam problem. PROBLEM 7.97 (a) Knowing that dC = 3 m, determine the distances dB and dD .


SOLUTION

ΣM B = 0: A(10 m) − (5 kN)(8 m) − (5 kN)(6 m) − (10 kN)(3 m) = 0 A = 10 kN

Geometry: dC = 1.6 m + hC 3 m = 1.6 m + hC hC = 1.4 m

Since M = T0h, h is proportional to M, thus h hB h = C = D ; M B MC M D

hB hD 1.4 m = = 20 kN ⋅ m 30 kN ⋅ m 30 kN ⋅ m

 20  hB = 1.4   = 0.9333 m  30 

 30  hD = 1.4   = 1.4 m  30 

d B = 0.8 m + 0.9333 m

d D = 2.8 m + 1.4 m

d B = 1.733 m 

d D = 4.20 m 


www.elsolucionario.net PROBLEM 7.122 Making use of the property established in Problem 7.119, solve the problem indicated by first solving the corresponding beam problem. PROBLEM 7.99 An oil pipeline is supported at 6-ft intervals by vertical hangers attached to the cable shown. Due to the combined weight of the pipe and its contents the tension in each hanger is 400 lb. Knowing that dC = 12 ft, determine (a) the maximum tension in the cable.

SOLUTION 1 (4 × 400) = 800 lb 2


A= B=

Geometry dC = hC + 3 ft 12 ft = hC + 3 ft hC = 9 ft d D = hD + 2 ft

At C:

M C = T0 hC 7200 lb ⋅ ft = T0 (9 ft)

At D:

T0 = 800 lb

M D = T0 hD 7200 lb ⋅ ft = (800 lb)hD

Eq. (1):

(1)

h0 = 9 ft

d D = 9 ft + 2 ft

d D = 11.00 ft 


www.elsolucionario.net PROBLEM 7.123 Making use of the property established in Problem 7.119, solve the problem indicated by first solving the corresponding beam problem. PROBLEM 7.100 An oil pipeline is supported at 6-ft intervals by vertical hangers attached to the cable shown. Due to the combined weight of the pipe and its contents the tension in each hanger is 400 lb. Knowing that dC = 9 ft, determine (b) the distance dD.

SOLUTION 1 (4 × 400) 2 A = B = 800 lb

At any point:


A= B=

M = T0 h

We note that since M C = M D , we have hC = hD Geometry dC = hC + 3 ft 9 ft = hC + 3 ft hC = 6 ft

and

hD = 6 ft

d D = hD + 2 ft = 6 ft + 2 ft d D = 8.00 ft 


www.elsolucionario.net PROBLEM 7.124* Show that the curve assumed by a cable that carries a distributed load w(x) is defined by the differential equation d2y/dx2 = w(x)/T0, where T0 is the tension at the lowest point.

SOLUTION FBD Elemental segment:

So

Ty ( x + Δ x ) T0

−

Ty ( x )

Ty

But

So

In lim : Δ x →0

T0

T0 dy dx

− x +Δx

Δx

dy dx

x

=

w( x) Δx T0

=

dy dx

=

w( x) T0

d 2 y w( x) = T0 dx 2

Q.E.D.



ΣFy = 0: Ty ( x + Δ x) − Ty ( x) − w( x)Δ x = 0

www.elsolucionario.net PROBLEM 7.125* Using the property indicated in Problem 7.124, determine the curve assumed by a cable of span L and sag h carrying a distributed load w = w0 cos (π x/L), where x is measured from mid-span. Also determine the maximum and minimum values of the tension in the cable. PROBLEM 7.124 Show that the curve assumed by a cable that carries a distributed load w(x) is defined by the differential equation d2y/dx2 = w(x)/T0, where T0 is the tension at the lowest point.

w( x) = w0 cos

πx L

From Problem 7.124

πx d 2 y w( x) w0 = = cos 2 T0 T0 L dx So

  dy = 0  using dx 0  

πx dy W0 L = sin dx T0π L

y=

But

w L2 L y  = h = 0 2 T0π 2

And

T0 = Tmin

w0 L2 π  1 − cos 2  so T0 = 2 π h   Tmin =

so

Tmax = TA = TB :

TBy =

w0 L2  πx 1 − cos [using y (0) = 0]  2  L  T0π 

TBy T0

=

w0 L

dy dx

= x = L/2

π 2h



w0 L T0π 2 + T02 = TB = TBy

π

w0 L2

w0 L

π

 L  1+   πh 

2





SOLUTION

www.elsolucionario.net PROBLEM 7.126* If the weight per unit length of the cable AB is w0/cos2 θ, prove that the curve formed by the cable is a circular arc. (Hint: Use the property indicated in Problem 7.124.) PROBLEM 7.124 Show that the curve assumed by a cable that carries a distributed load w(x) is defined by the differential equation d2y/dx2 = w(x)/T0, where T0 is the tension at the lowest point.

SOLUTION Elemental Segment: w ( x)dx =

w0 cos 2 θ

ds

dx = cos θ ds, so w( x) =

But

w0 cos3 θ

From Problem 7.119

w0 d 2 y w( x) = = 2 T0 dx T0 cos3 θ

In general

d 2 y d  dy  d dθ =   = (tan θ ) = sec2 θ 2 dx dx dx dx dx  

So

or Giving r =

w0 w0 dθ = = dx T0 cos3 θ sec 2 θ T0 cos θ T0 cos θ dθ = dx = rdθ cos θ w0 T0 = constant. So curve is circular arc w0

Q.E.D.

*For large sag, it is not appropriate to approximate ds by dx.



Load on segment*

www.elsolucionario.net PROBLEM 7.127 A 20-m chain of mass 12 kg is suspended between two points at the same elevation. Knowing that the sag is 8 m, determine (a) the distance between the supports, (b) the maximum tension in the chain.

SOLUTION mass/meter = (12 kg)/(20 m) = 0.6 kg/m w = (0.6 kg/m)(9.81 m/s 2 ) = 5.886 N/m

yB2 − sB2 = c 2 ; (8 + c)2 − 102 = c 2 64 + 16c + c 2 − 100 = c 2 16c = 36 c = 2.25 m

Eq. 7.18:

Tm = wyB = (5.886 N/m)(8 m + 2.25 m) Tm = 60.33 N

Eq. 7.15:

Tm = 60.3 N 

xB x ; 10 = (2.25 m) sinh B c c xB xB sinh = 4.444; = 2.197 c c sB = c sinh

xB = 2.197(2.25 m) = 4.944 m; L = 2 xB = 2(4.944 m) = 9.888 m L = 9.89 m 



Eq. 7.17:

www.elsolucionario.net PROBLEM 7.128 A 600-ft-long aerial tramway cable having a weight per unit length of 3.0 lb/ft is suspended between two points at the same elevation. Knowing that the sag is 150 ft, find (a) the horizontal distance between the supports, (b) the maximum tension in the cable.

SOLUTION

Given: Unit mass = 3.0 lb/ft h = 150 ft sB = 300 ft

Then,

yB = h + c = 150 ft + c yB2

− sB2 = c 2 ; (150 + c) 2 − (300) 2 = c 2

1502 + 300c + c 2 − 3002 = c 2 c = 225 ft sB = c sinh

xB x ; 300 = 225sinh B 225 c xB = 247.28 ft

span = L = 2 xB = 2(247.28 ft) Tm = wyB = (3 lb/ft)(150 ft + 225 ft)

L = 495 ft  Tm = 1125 lb 



Length = 600 ft

www.elsolucionario.net PROBLEM 7.129 A 40-m cable is strung as shown between two buildings. The maximum tension is found to be 350 N, and the lowest point of the cable is observed to be 6 m above the ground. Determine (a) the horizontal distance between the buildings, (b) the total mass of the cable.

SOLUTION

Tm = 350 N h = 14 m − 6 m = 8 m yB = h + c = 8 m + c

Eq. 7.17:

yB2 − sB2 = c 2 ; (8 + c)2 − 202 = c 2 64 + 16c + c 2 − 400 = c 2 c = 21.0 m xB x ; 20 = (21.0)sinh B 21.0 c

Eq. 7.15:

sB = c sinh

(a)

xB = 17.7933 m; L = 2 xB

(b)

Eq. 7.18:

L = 35.6 m 

Tm = wyB ; 350 N = w(8 + 21.0) w = 12.0690 N/m W = 2sB w = (40 m)(12.0690 N/m) = 482.76 N W 482.76 N m= = g 9.81 m/s 2

Total mass = 49.2 kg 



sB = 20 m

www.elsolucionario.net PROBLEM 7.130 A 200-ft steel surveying tape weighs 4 lb. If the tape is stretched between two points at the same elevation and pulled until the tension at each end is 16 lb, determine the horizontal distance between the ends of the tape. Neglect the elongation of the tape due to the tension.

SOLUTION

sB = 100 ft

Eq. 7.18: Eq. 7.17: Eq. 7.15:

Tm = wyB ; 16 lb = (0.02 lb/ft) yB ;

yB = 800 ft

yB2 − sB2 = c 2 ; (800)2 − (100) 2 = c 2 ; c = 793.73 ft sB = c sinh

xB x ; 100 = 793.73 sinh B c c

xB = 0.12566; xB = 99.737 ft c L = 2 xB = 2(99.737 ft)

L = 199.5 ft 



 4 lb  w=  = 0.02 lb/ft Tm = 16 m  200 ft 

www.elsolucionario.net PROBLEM 7.131 A 20-m length of wire having a mass per unit length of 0.2 kg/m is attached to a fixed support at A and to a collar at B. Neglecting the effect of friction, determine (a) the force P for which h = 8 m, (b) the corresponding span L.

SOLUTION


FBD Cable:

20 m   sT = 20 m  so sB = = 10 m  2   w = (0.2 kg/m)(9.81 m/s 2 ) = 1.96200 N/m hB = 8 m yB2 = (c + hB )2 = c 2 + sB2

So

c=

sB2 − hB2 2hB

c=

(10 m)2 − (8 m)2 2(8 m)

= 2.250 m

Now

xB s → xB = c sinh −1 B c c   10 m = (2.250 m)sinh −1   2.250 m 

sB = c sinh

xB = 4.9438 m P = T0 = wc = (1.96200 N/m)(2.250 m) L = 2 xB = 2(4.9438 m)

(a) (b)

P = 4.41 N



L = 9.89 m 


www.elsolucionario.net PROBLEM 7.132 A 20-m length of wire having a mass per unit length of 0.2 kg/m is attached to a fixed support at A and to a collar at B. Knowing that the magnitude of the horizontal force applied to the collar is P = 20 N, determine (a) the sag h, (b) the span L.

SOLUTION

sT = 20 m, w = (0.2 kg/m)(9.81 m/s 2 ) = 1.96200 N/m P = T0 = wc c = c=

P w

20 N = 10.1937 m 1.9620 N/m

yB2 = (hB + c) 2 = c 2 + sB2 h 2 + 2ch − sB2 = 0 sB =

20 m = 10 m 2

h 2 + 2(10.1937 m)h − 100 m 2 = 0 h = 4.0861 m sB = c sinh

(a)

h = 4.09 m 

xA s  10 m  → xB = c sinh −1 B = (10.1937 m)sinh −1   c c  10.1937 m  = 8.8468 m L = 2 xB = 2(8.8468 m)

(b)

L = 17.69 m 



FBD Cable:

www.elsolucionario.net PROBLEM 7.133 A 20-m length of wire having a mass per unit length of 0.2 kg/m is attached to a fixed support at A and to a collar at B. Neglecting the effect of friction, determine (a) the sag h for which L = 15 m, (b) the corresponding force P.

SOLUTION


FBD Cable:

20 m = 10 m 2 w = (0.2 kg/m)(9.81 m/s 2 ) = 1.96200 N/m L = 15 m

sT = 20 m → sB =

L xB = c sinh 2 c c 7.5 m 10 m = c sinh c

sB = c sinh

Solving numerically:

c = 5.5504 m x yB = c cosh  B  c yB = 11.4371 m

  7.5   = (5.5504) cosh  5.5504    

hB = yB − c = 11.4371 m − 5.5504 m

(a) P = wc = (1.96200 N/m)(5.5504 m)

(b)

hB = 5.89 m 

P = 10.89 N




www.elsolucionario.net PROBLEM 7.134 Determine the sag of a 30-ft chain that is attached to two points at the same elevation that are 20 ft apart.

SOLUTION


L = 20 ft

c = 6.1647 ft yB = c cosh

xB c

= (6.1647 ft)cosh

10 ft 6.1647 ft

= 16.2174 ft hB = yB − c = 16.2174 ft − 6.1647 ft

hB = 10.05 ft 



30 ft = 15 ft 2 L xB = = 10 ft 2 x sB = c sinh B c 10 ft 15 ft = c sinh c sB =

www.elsolucionario.net PROBLEM 7.135 A 10-ft rope is attached to two supports A and B as shown. Determine (a) the span of the rope for which the span is equal to the sag, (b) the corresponding angle θB.


SOLUTION

Note: Since L = h, xB =

Eq. 7.16:

L h = 2 2

xB c h /2 h + c = c cosh c h 1 h + 1 = cosh   c 2 c yB = cosh

Solve for h/c:

h = 4.933 c

Eq. 7.16:

y = c cosh

x c

dy x = sinh dx c

At B:

Substitute

tan θ B =

dy dx

= sinh B

xB c

h 1 h 1  xB = : tan θ B = sinh  = sinh  × 4.933   2 2 c 2  tan θ B = 5.848

θ B = 80.3° 



Eq. 7.17:

Recall that

xB 1 h = c sinh   c 2 c 1  5 ft = c sinh  × 4.933  2  5 ft = c(5.848) c = 0.855 sB = c sinh

h = 4.933 c h = 4.933(0.855) = 4.218


h = 4.22 ft 


www.elsolucionario.net PROBLEM 7.136 A 90-m wire is suspended between two points at the same elevation that are 60 m apart. Knowing that the maximum tension is 300 N, determine (a) the sag of the wire, (b) the total mass of the wire.

SOLUTION

sB = 45 m

Eq. 7.16:

xB c 30 45 = c sinh ; c = 18.494 m c sB = c sinh

yB = c cosh

xB c

yB = (18.494) cosh

30 18.494

yB = 48.652 m yB = h + c 48.652 = h + 18.494 h = 30.158 m

Eq. 7.18:

h = 30.2 m 

Tm = wyB 300 N = w(48.652 m) w = 6.166 N/m

Total weight of cable

W = w(Length) = (6.166 N/m)(90 m) = 554.96 N

Total mass of cable

m=

W 554.96 N = = 56.57 kg 9.81 m/s g

m = 56.6 kg 



Eq. 7.17:

www.elsolucionario.net PROBLEM 7.137 A cable weighing 2 lb/ft is suspended between two points at the same elevation that are 160 ft apart. Determine the smallest allowable sag of the cable if the maximum tension is not to exceed 400 lb.

SOLUTION

Eq. 7.16:

Tm = wyB ; 400 lb = (2 lb/ft) yB ;

yB = 200 ft

xB c 80 ft 200 ft = c cosh c yB = c cosh

Solve for c: c = 182.148 ft and c = 31.592 ft yB = h + c; h = yB − c

For

c = 182.148 ft;

h = 200 − 182.147 = 17.852 ft 

For

c = 31.592 ft;

h = 200 − 31.592 = 168.408 ft 

For

Tm ≤ 400 lb:

smallest h = 17.85 ft 



Eq. 7.18:

www.elsolucionario.net PROBLEM 7.138 A uniform cord 50 in. long passes over a pulley at B and is attached to a pin support at A. Knowing that L = 20 in. and neglecting the effect of friction, determine the smaller of the two values of h for which the cord is in equilibrium.

SOLUTION

b = 50 in. − 2sB

Length of overhang:


Weight of overhang equals max. tension Tm = TB = wb = w(50 in. − 2sB )

Eq. 7.15:

sB = c sinh

xB c

Eq. 7.16:

yB = c cosh

xB c

Eq. 7.18:

Tm = wyB w(50 in. − 2sB ) = wyB x  x  w  50 in. − 2c sinh B  = wc cosh B c  c  xB = 10: 50 − 2c sinh

Solve by trial and error: For c = 5.549 in.

c = 5.549 in.

and

10 10 = c cosh c c

c = 27.742 in.

10 in. = 17.277 in. 5.549 in. yB = h + c; 17.277 in. = h + 5.549 in. yB = (5.549 in.) cosh

h = 11.728 in.

For c = 27.742 in.

h = 11.73 in. 

10 in. = 29.564 in. 27.742 in. yB = h + c; 29.564 in. = h + 27.742 in. yB = (27.742 in.) cosh

h = 1.8219 in.

h = 1.822 in. 


www.elsolucionario.net PROBLEM 7.139 A motor M is used to slowly reel in the cable shown. Knowing that the mass per unit length of the cable is 0.4 kg/m, determine the maximum tension in the cable when h = 5 m.

SOLUTION

xB c L hB + c = c cosh 2c 5m   5 m = c  cosh − 1 c   yB = c cosh


c = 3.0938 m yB = hB + c = 5 m + 3.0938 m Tmax

= 8.0938 m = TB = wyB = (0.4 kg/m)(9.81 m/s 2 )(8.0938 m) Tmax = 31.8 N 



w = 0.4 kg/m L = 10 m hB = 5 m

www.elsolucionario.net PROBLEM 7.140 A motor M is used to slowly reel in the cable shown. Knowing that the mass per unit length of the cable is 0.4 kg/m, determine the maximum tension in the cable when h = 3 m.

w = 0.4 kg/m, L = 10 m, hB = 3 m xB L = c cosh c 2c 5m   − 1 3 m = c  c cosh c   yB = hB + c = c cosh


c = 4.5945 m yB = hB + c = 3 m + 4.5945 m Tmax

= 7.5945 m = TB = wyB = (0.4 kg/m)(9.81 m/s 2 )(7.5945 m) Tmax = 29.8 N 



SOLUTION

www.elsolucionario.net PROBLEM 7.141 The cable ACB has a mass per unit length of 0.45 kg/m. Knowing that the lowest point of the cable is located at a distance a = 0.6 m below the support A, determine (a) the location of the lowest Point C, (b) the maximum tension in the cable.

SOLUTION xB − x A = 12 m

Note:

Point A:

y A = c cosh

− xA 12 − xB ; c + 0.6 = c cosh c c

(1)

Point B:

yB = c cosh

xB x ; c + 2.4 = c cosh B c c

(2)

From (1):

12 xB  c + 0.6  − = cosh −1   c c  c 

(3)

From (2):

xB  c + 2.4  = cosh −1   c  c 

(4)

Add (3) + (4):

12  c + 0.6   c + 2.4  = cosh −1  + cosh −1    c  c   c  c = 13.6214 m

Solve by trial and error: Eq. (2):

13.6214 + 2.4 = 13.6214 cosh cosh

xB = 1.1762; c

xB c

xB = 0.58523 c

xB = 0.58523(13.6214 m) = 7.9717 m

Point C is 7.97 m to left of B  yB = c + 2.4 = 13.6214 + 2.4 = 16.0214 m

Eq. 7.18:

Tm = wyB = (0.45 kg/m)(9.81 m/s 2 )(16.0214 m) Tm = 70.726 N

Tm = 70.7 N 



− x A = 12 m − xB

or,

www.elsolucionario.net PROBLEM 7.142 The cable ACB has a mass per unit length of 0.45 kg/m. Knowing that the lowest point of the cable is located at a distance a = 2 m below the support A, determine (a) the location of the lowest Point C, (b) the maximum tension in the cable.

SOLUTION Note: xB − x A = 12 m

or

Point A:

y A = c cosh

− xA 12 − xB ; c + 2 = c cosh c c

(1)

Point B:

yB = c cosh

xB x ; c + 3.8 = c cosh B c c

(2)

12 xB c+2 − = cosh −1   c c  c 

From (1):

(3)

From (2):

xB  c + 3.8  = cosh −1   c  c 

Add (3) + (4):

12 c+2  c + 3.8  = cosh −1  + cosh −1    c  c   c  c = 6.8154 m

Solve by trial and error: Eq. (2):

(4)

6.8154 m + 3.8 m = (6.8154 m) cosh cosh

xB c

xB xB = 1.5576 = 1.0122 c c xB = 1.0122(6.8154 m) = 6.899 m

Point C is 6.90 m to left of B  yB = c + 3.8 = 6.8154 + 3.8 = 10.6154 m

Eq. (7.18):

Tm = wyB = (0.45 kg/m)(9.81 m/s 2 )(10.6154 m) Tm = 46.86 N

Tm = 46.9 N 



− x A = 12 m − xB

www.elsolucionario.net PROBLEM 7.143 A uniform cable weighing 3 lb/ft is held in the position shown by a horizontal force P applied at B. Knowing that P = 180 lb and θA = 60°, determine (a) the location of Point B, (b) the length of the cable.

SOLUTION T0 = P = cw

Eq. 7.18:

c=

c = 60 ft

P cos 60° cw = = 2cw 0.5

Tm = w(h + c)

Eq. 7.18:

2cw = w(h + c) 2c = h + c h = b = c

b = 60.0 ft 

xA c x h + c = c cosh A c y A = c cosh

Eq. 7.16:

(60 ft + 60 ft) = (60 ft) cosh cosh

xA =2 60 m

xA 60

xA = 1.3170 60 m

x A = 79.02 ft

(b)

Eq. 7.15:

a = 79.0 ft 

xB 79.02 ft = (60 ft)sinh 60 ft c s A = 103.92 ft s A = c sinh

length = s A

s A = 103.9 ft 



Tm =

At A:

(a)

P 180 lb = w 3 lb/ft

www.elsolucionario.net PROBLEM 7.144 A uniform cable weighing 3 lb/ft is held in the position shown by a horizontal force P applied at B. Knowing that P = 150 lb and θA = 60°, determine (a) the location of Point B, (b) the length of the cable.

SOLUTION Eq. 7.18:

T0 = P = cw c=

P cos 60° cw = = 2 cw 0.5 Tm = w(h + c)

Eq. 7.18:

2cw = w(h + c) 2c = h + c h = c = b

Eq. 7.16:

xA c xA h + c = c cosh c y A = c cosh

(50 ft + 50 ft) = (50 ft) cosh cosh

xA =2 c

xA c

xA = 1.3170 c

x A = 1.3170(50 ft) = 65.85 ft

(b)

Eq. 7.15:

b = 50.0 ft 

s A = c sinh

a = 65.8 ft 

xA 65.85 ft = (50 ft)sinh 50 ft c

s A = 86.6 ft

length = s A = 86.6 ft 



Tm =

At A:

(a)

P 150 lb = = 50 ft w 3 lb/ft

www.elsolucionario.net PROBLEM 7.145 To the left of Point B the long cable ABDE rests on the rough horizontal surface shown. Knowing that the mass per unit length of the cable is 2 kg/m, determine the force F when a = 3.6 m.

xD = a = 3.6 m h = 4 m xD c a h + c = c cosh c 3.6 m   − 1 4 m = c  cosh c   yD = c cosh

Solving numerically: Then

c = 2.0712 m yB = h + c = 4 m + 2.0712 m = 6.0712 m F = Tmax = wyB = (2 kg/m)(9.81 m/s 2 )(6.0712 m)

F = 119.1 N





SOLUTION

www.elsolucionario.net PROBLEM 7.146 To the left of Point B the long cable ABDE rests on the rough horizontal surface shown. Knowing that the mass per unit length of the cable is 2 kg/m, determine the force F when a = 6 m.

xD = a = 6 m h = 4 m xD c a h + c = c cosh c 6m   − 1 4 m = c  cosh c   y D = c cosh


c = 5.054 m yB = h + c = 4 m + 5.054 m = 9.054 m F = TD = wyD = (2 kg/m)(9.81 m/s 2 )(9.054 m)

F = 177.6 N





SOLUTION

www.elsolucionario.net PROBLEM 7.147* The 10-ft cable AB is attached to two collars as shown. The collar at A can slide freely along the rod; a stop attached to the rod prevents the collar at B from moving on the rod. Neglecting the effect of friction and the weight of the collars, determine the distance a.

SOLUTION

Point A:

x dy x = sinh y = c cosh ; c dx c xA dy tan θ = = sinh c dx A xA = sinh(tan (90° − θ )) c x A = c sinh(tan (90° − θ ))

Length of cable = 10 ft

(1)

10 ft = AC + CB xA x + c sinh B c c xB 10 xA sinh = − sinh c c c 10 = c sinh

x  10 xB = c sinh −1  − sinh A  c  c x x y A = c cosh A yB = c cosh B c c

In Δ ABD:

tan θ =

yB − y A xB + x A

(2) (3) (4)



Collar at A: Since μ = 0, cable ⬜ rod


Method of solution: For given value of θ, choose trial value of c and calculate: From Eq. (1): xA Using value of xA and c, calculate: From Eq. (2): xB From Eq. (3): yA and yB Substitute values obtained for xA, xB, yA, yB into Eq. (4) and calculate θ Choose new trial value of θ and repeat above procedure until calculated value of θ is equal to given value of θ. For θ = 30° c = 1.803 ft x A = 2.3745 ft xB = 3.6937 ft y A = 3.606 ft yB = 7.109 ft a = yB − y A = 7.109 ft − 3.606 ft = 3.503 ft

a = 3.50 ft 



Result of trial and error procedure:

www.elsolucionario.net PROBLEM 7.148* Solve Problem 7.147 assuming that the angle θ formed by the rod and the horizontal is 45°. PROBLEM 7.147 The 10-ft cable AB is attached to two collars as shown. The collar at A can slide freely along the rod; a stop attached to the rod prevents the collar at B from moving on the rod. Neglecting the effect of friction and the weight of the collars, determine the distance a.

SOLUTION

Point A:


Collar at A: Since μ = 0, cable ⬜ rod

x dy x = sinh y = c cosh ; c dx c x dy tan θ = = sinh A c dx A xA = sinh(tan (90° − θ )) c x A = c sinh(tan (90° − θ ))

Length of cable = 10 ft

(1)

10 ft = AC + CB x x 10 = c sinh A + c sinh B c c xB 10 xA sinh = − sinh c c c x  10 xB = c sinh −1  − sinh A  c  c y A = c cosh

xA c

yB = c cosh

(2) xB c

(3)



In Δ ABD:

tan θ =

yB − y A xB + x A

(4)

Method of solution: For given value of θ, choose trial value of c and calculate: From Eq. (1): xA Using value of xA and c, calculate: From Eq. (2): xB From Eq. (3): yA and yB Substitute values obtained for xA, xB, yA, yB into Eq. (4) and calculate θ Choose new trial value of θ and repeat above procedure until calculated value of θ is equal to given value of θ. Result of trial and error procedure: c = 1.8652 ft x A = 1.644 ft xB = 4.064 ft y A = 2.638 ft yB = 8.346 ft a = yB − y A = 8.346 ft − 2.638 ft = 5.708 ft

a = 5.71 ft 



For θ = 45°

www.elsolucionario.net PROBLEM 7.149 Denoting by θ the angle formed by a uniform cable and the horizontal, show that at any point (a) s = c tan θ, (b) y = c sec θ.

SOLUTION dy x = sinh dx c x s = c sinh = c tan θ c

tan θ =

(a)

(b)

Q.E.D.

Also

y 2 = s 2 + c 2 (cosh 2 x = sinh 2 x + 1)

so

y 2 = c 2 (tan 2 θ + 1)c 2 sec2 θ y = c secθ

Q.E.D.


and


www.elsolucionario.net PROBLEM 7.150* (a) Determine the maximum allowable horizontal span for a uniform cable of weight per unit length w if the tension in the cable is not to exceed a given value Tm. (b) Using the result of part a, determine the maximum span of a steel wire for which w = 0.25 lb/ft and Tm = 8000 lb.

SOLUTION

Tm = wyB

(a)

xB c  x 1   cosh B xB  c  c 

  = wxB   

We shall find ratio

( ) for when T xB c

m


= wc cosh

is minimum

2        d Tm x xB  1 1 = wxB  =0 sinh B −  cosh   xB c  xB  c   xB  d     c   c   c    xB sinh c = 1 x xB cosh B c c xB c = tanh c xB

Solve by trial and error for: sB = c sinh

Eq. 7.17:

xB = c sinh(1.200): c

xB = 1.200 c

(1)

sB = 1.509 c yB2 − sB2 = c 2   s 2  yB2 = c 2 1 +  B   = c 2 (1 + 1.5092 )   c   yB = 1.810c



Eq. 7.18:

Tm = wyB = 1.810 wc Tm c= 1.810 w

Eq. (1): Span: (b)

xB = 1.200c = 1.200

Tm T = 0.6630 m w 1.810 w

L = 2 xB = 2(0.6630)

Tm w

L = 1.326

Tm  w

For w = 0.25lb/ft and Tm = 8000 lb 8000 lb 0.25lb/ft = 42, 432ft

L = 8.04 miles 



L = 1.326

www.elsolucionario.net PROBLEM 7.151* A cable has a mass per unit length of 3 kg/m and is supported as shown. Knowing that the span L is 6 m, determine the two values of the sag h for which the maximum tension is 350 N.

SOLUTION

L =h+c 2c w = (3 kg/m)(9.81 m/s 2 ) = 29.43 N/m

Tmax = wymax Tmax w 350 N = = 11.893 m 29.43 N/m

ymax = ymax c cosh


3m = 11.893 m c c1 = 0.9241 m c2 = 11.499 m h = ymax − c h1 = 11.893 m − 0.9241 m

h1 = 10.97 m 

h2 = 11.893 m − 11.499 m

h2 = 0.394 m 



ymax = c cosh

www.elsolucionario.net PROBLEM 7.152* Determine the sag-to-span ratio for which the maximum tension in the cable is equal to the total weight of the entire cable AB.

SOLUTION

y B = 2 sB L 2c L tanh 2c L 2c hB c

c cosh

= 2c sinh =

L 2c

1 2

1 = 0.549306 2 y −c L = B = cosh − 1 = 0.154701 2c c = tanh −1

h

B hB 0.5(0.154701) = cL = = 0.14081 0.549306 L 2 ( 2c )

hB = 0.1408  L



Tmax = wyB = 2wsB

www.elsolucionario.net PROBLEM 7.153* A cable of weight per unit length w is suspended between two points at the same elevation that are a distance L apart. Determine (a) the sag-to-span ratio for which the maximum tension is as small as possible, (b) the corresponding values of θB and Tm.

L 2c L L L  = w  cosh − sinh  2 2 2 c c c 

Tmax = wyB = wc cosh

(a)

dTmax dc

For

min Tmax ,

dTmax =0 dc tanh

L 2c L = → = 1.1997 2c L 2c yB L = cosh = 1.8102 2c c h yB = − 1 = 0.8102 c c 0.8102 h  1 h  2c   = = 0.3375 =   L  2 c  L   2(1.1997)

T0 = wc Tmax = wc cosh

(b)

But

T0 = Tmax cos θ B

So

θ B = sec −1 

 yB  c

Tmax = wyB = w

L 2c

h = 0.338  L

Tmax L yB = cosh = T0 c 2c

Tmax = sec θ B T0

 −1  = sec (1.8102) = 56.46° 

yB  2c  L  L  = w(1.8102) 2(1.1997) c  L  2  

θ B = 56.5°  Tmax = 0.755wL 



SOLUTION

www.elsolucionario.net PROBLEM 7.154 Determine the internal forces at Point J of the structure shown.


SOLUTION FBD ABC: ΣM D = 0: (0.375 m)(400 N) − (0.24 m)C y = 0 C y = 625 N

ΣM B = 0: − (0.45 m)Cx + (0.135 m)(400 N) = 0 C x = 120 N

FBD CJ:

ΣFy = 0: 625 N − F = 0 ΣFx = 0: 120 N − V = 0

F = 625 N  V = 120.0 N



M = 27.0 N ⋅ m



ΣM J = 0: M − (0.225 m)(120 N) = 0


www.elsolucionario.net PROBLEM 7.155 Determine the internal forces at Point K of the structure shown.

FBD AK: ΣFx = 0: V = 0 V=0 

ΣFy = 0: F − 400 N = 0 F = 400 N 

ΣM K = 0: (0.135 m)(400 N) − M = 0 M = 54.0 N ⋅ m





SOLUTION

www.elsolucionario.net PROBLEM 7.156 An archer aiming at a target is pulling with a 45-lb force on the bowstring. Assuming that the shape of the bow can be approximated by a parabola, determine the internal forces at Point J.

SOLUTION

By symmetry


FBD Point A: T1 = T2

3  ΣFx = 0: 2  T1  − 45 lb = 0 T1 = T2 = 37.5 lb 5 

Curve CJB is parabolic: x = ay 2 FBD BJ: At B : x = 8 in.

y = 32 in. 8 in. 1 a= = 2 128 in. (32 in.) x=

y2 128

Slope of parabola = tan θ =

At J: So

dx 2 y y = = dy 128 64

 16   = 14.036°  64 

θ J = tan −1  α = tan −1

4 − 14.036° = 39.094° 3

ΣFx′ = 0: V − (37.5 lb) cos(39.094°) = 0

V = 29.1 lb

14.04° 



ΣFy ′ = 0: F + (37.5 lb) sin (39.094°) = 0 F = − 23.647

F = 23.6 lb

76.0° 

3  4  ΣMJ = 0: M + (16 in.)  (37.5 lb)  + [(8 − 2) in.]  (37.5 lb)  = 0 5 5    




M = 540 lb ⋅ in.


www.elsolucionario.net PROBLEM 7.157 Knowing that the radius of each pulley is 200 mm and neglecting friction, determine the internal forces at Point J of the frame shown.

SOLUTION Free body: Frame and pulleys

Bx = − 520 N

B x = 520 N

䉰

A x = 520 N

䉰

ΣFx = 0: Ax − 520 N = 0 Ax = + 520 N ΣFy = 0: Ay + By − 360 N = 0 Ay + By = 360 N

(1)

Free body: Member AE ΣME = 0: − Ay (2.4 m) − (360 N)(1.6 m) = 0 Ay = − 240 N

From (1):

A y = 240 N 䉰

By = 360 N + 240 N B y = 600 N 䉰

By = + 600 N

Free body: BJ We recall that the forces applied to a pulley may be applied directly to its axle. ΣFy = 0:

3 4 (600 N) + (520 N) 5 5 3 − 360 N − (360 N) − F = 0 5 F = + 200 N

F = 200 N

36.9° 



ΣMA = 0: − Bx (1.8 m) − (360 N)(2.6 m) = 0


ΣFx = 0:

4 3 4 (600 N) − (520 N) − (360 N) + V = 0 5 5 5 V = + 120.0 N

V = 120.0 N

53.1° 

ΣM J = 0: (520 N)(1.2 m) − (600 N)(1.6 m) + (360 N)(0.6 m) + M = 0 M = 120.0 N ⋅ m




M = + 120.0 N ⋅ m


www.elsolucionario.net PROBLEM 7.158 For the beam shown, determine (a) the magnitude P of the two upward forces for which the maximum absolute value of the bending moment in the beam is as small as possible, (b) the corresponding value of |M |max .

SOLUTION By symmetry:

Ay = B = 60 kips − P

Along AC: ΣM J = 0: M − x(60 kips − P) = 0 x = 2 ft


M = (60 kips − P) x M = 120 kips ⋅ ft − (2 ft) P at

Along CD: ΣM K = 0: M + ( x − 2 ft)(60 kips) − x(60 kips − P) = 0 M = 120 kip ⋅ ft − Px M = 120 kip ⋅ ft − (4 ft) P at

x = 4 ft

Along DE: ΣM L = 0: M − ( x − 4 ft) P + ( x − 2 ft)(60 kips) − x(60 kips − P) = 0 M = 120 kip ⋅ ft − (4 ft) P (const)

Complete diagram by symmetry For minimum |M |max , set M max = − M min 120 kip ⋅ ft − (2 ft) P = − [120 kip ⋅ ft − (4 ft) P]

M min = 120 kip ⋅ ft − (4 ft) P

(a)

P = 40.0 kips 

(b)

| M |max = 40.0 kip ⋅ ft 



SOLUTION ΣM A = 0: (8)(2) + (4)(3.2) − 4C = 0 C = 7.2 kN

ΣFy = 0: A = 4.8 kN

Shear diagram

Similar Triangles:


(a)

x 3.2 − x 3.2 = = ; x = 2.4 m 4.8 1.6 6.4 ↑ Add num. & den.

Bending-moment diagram

(b)

|V | max = 7.20 kN 



|M | max = 5.76 kN ⋅ m 



SOLUTION Free body: Beam ΣM B = 0: 6 kip ⋅ ft + 12 kip ⋅ ft + (2 kips)(18 ft) + (3 kips)(12 ft) + (4 kips)(6 ft) − Ay (24 ft) = 0 Ay = + 4.75 kips 䉰

Shear diagram At A:

VA = Ay = + 4.75 kips |V |max = 4.75 kips 


M A = − 6 kip ⋅ ft | M |max = 39.0 kip ⋅ ft 



ΣFx = 0: Ax = 0


SOLUTION Free body: Entire beam ΣM A = 0: (6 kips)(10 ft) − (9 kips)(7 ft) + 8(4 ft) = 0 B = + 0.75 kips B = 0.75 kips

ΣFy = 0: A + 0.75 kips − 9 kips + 6 kips = 0


A = + 2.25 kips A = 2.25 kips

M max = 12.00 kip ⋅ ft  6.00 ft from A


www.elsolucionario.net PROBLEM 7.162 The beam AB, which lies on the ground, supports the parabolic load shown. Assuming the upward reaction of the ground to be uniformly distributed, (a) write the equations of the shear and bending-moment curves, (b) determine the maximum bending moment.

SOLUTION ΣFy = 0: wg L −

(a)



L 0

4 w0 L2

( Lx − x 2 ) dx = 0

4 w0  1 2 1 3  2  LL − 3 L  = 3 w0 L L2  2  2w wg = 0 3 x dx so d ξ = L L

 x  x 2  2 net load w = 4 w0  −    − w0  L  L   3

Define

ξ=

or

 1  w = 4 w0  − + ξ − ξ 2  6    1  4w0 L  − + ξ − ξ 2  d ξ  6  1 1  2 1 = 0 + 4w0 L  ξ + ξ 2 − ξ 3  V = w0 L (ξ − 3ξ 2 + 2ξ 3 )  6 2 3 3   x ξ 2 M = M 0 + Vdx = 0 + w0 L2 (ξ − 3ξ 2 + 2ξ 3 )d ξ 0 0 3 2 1  1 1 = w0 L2  ξ 2 − ξ 3 + ξ 4  = w0 L2 (ξ 2 − 2ξ 3 + ξ 4 )  3 2 2  3  V = V (0) −



ξ

0



(b)

Max M occurs where

V =0



1 − 3ξ + 2ξ 2 = 0

1 1  M  ξ =  = w0 L2 2 3 

ξ=

1 2

2  1 2 1  w0 L − + =   48  4 8 16 

M max =

w0 L2 at center of beam  48



wg L =

www.elsolucionario.net PROBLEM 7.163 Two loads are suspended as shown from the cable ABCD. Knowing that d B = 1.8 m, determine (a) the distance dC , (b) the components of the reaction at D, (c) the maximum tension in the cable.

SOLUTION ΣFx = 0: − Ax + Dx = 0

FBD Cable:

Ax = Dx

ΣM A = 0: (10 m)Dy − (6 m)(10 kN) − (3 m)(6 kN) = 0 D y = 7.8 kN


ΣFy = 0: Ay − 6 kN − 10 kN + 7.8 kN = 0 A y = 8.2 kN

FDB AB:

ΣM B = 0: (1.8 m) Ax − (3 m)(8.2 kN) = 0 Ax =

From above

Dx = Ax =

41 kN 3

41 kN 3

FBD CD:  41  kN  = 0 ΣM C = 0: (4 m)(7.8 kN) − dC   3  dC = 2.283 m dC = 2.28 m 

(a)

D x = 13.67 kN

(b)



D y = 7.80 kN 

Since Ax = Bx and Ay > By , max T is TAB 2

 41  kN  + (8.2 kN)2 TAB = Ax2 + Ay2 =   3  Tmax = 15.94 kN 

(c)


www.elsolucionario.net PROBLEM 7.164 A wire having a mass per unit length of 0.65 kg/m is suspended from two supports at the same elevation that are 120 m apart. If the sag is 30 m, determine (a) the total length of the wire, (b) the maximum tension in the wire.

SOLUTION

Solve by trial and error: Eq. 7.15:

xB c 60 30m + c = c cosh c yB = c cosh

c = 64.459 m sB = c sin h

xB c

sB = (64.456 m) sinh

60 m 64.459 m

sB = 69.0478 m

Length = 2 sB = 2(69.0478 m) = 138.0956 m Eq. 7.18:

L = 138.1 m 

Tm = wyB = w(h + c) = (0.65 kg/m)(9.81 m/s 2 )(30 m + 64.459 m) Tm = 602.32 N

Tm = 602 N 



Eq. 7.16:

www.elsolucionario.net PROBLEM 7.165 A counterweight D is attached to a cable that passes over a small pulley at A and is attached to a support at B. Knowing that L = 45 ft and h = 15 ft, determine (a) the length of the cable from A to B, (b) the weight per unit length of the cable. Neglect the weight of the cable from A to D.

SOLUTION Given:

L = 45 ft h = 15 ft TA = 80 lb xB = 22.5 ft

By symmetry:

TB = TA = Tm = 80 lb

We have

yB = c cosh

and

yB = h + c = 15 + c

or Solve by trial for c: (a)

c cosh

22.5 = 15 + c c

cosh

22.5 15 = +1 c c


Then

xB 22.5 = c cosh c c

c = 18.9525 ft sB = c sinh

xB c

= (18.9525 ft) sinh

22.5 18.9525

= 28.170 ft

Length = 2sB = 2(28.170 ft) = 56.3 ft  (b)

Tm = wyB = w(h + c) 80 lb = w(15 ft + 18.9525 ft)

w = 2.36 lb/ft 


CHAPTER 8





www.elsolucionario.net PROBLEM 8.1 Determine whether the block shown is in equilibrium and find the magnitude and direction of the friction force when θ = 25° and P = 750 N.

SOLUTION Assume equilibrium: ΣFx = 0: F + (1200 N) sin 25° − (750 N) cos 25° = 0 F = +172.6 N

F = 172.6 N

N = 1404.5 N

Maximum friction force:

Fm = μ s N = 0.35(1404.5 N) = 491.6 N

Since F < Fm ,

block is in equilibrium  F = 172.6 N

Friction force:

25.0° 



ΣFy = 0: N − (1200 N) cos 25° − (750 N)sin 25° = 0


SOLUTION Assume equilibrium: ΣFx = 0: F + (1200 N) sin 30° − (150 N) cos 30° = 0 F = 470.1 N

ΣFy = 0: N − (1200 N) cos 30° − (150 N)sin 30° = 0 N = 1114.2 N

(a)

Maximum friction force:

Fm = μ s N = 0.35(1114.2 N) = 390.0 N

Since F is (b)

and F > Fm ,

Actual friction force:

block moves down  F = Fk = μ k N = 0.25(1114.2 N) = 279 N

F = 279 N

30.0° 



F = − 470.1 N

www.elsolucionario.net PROBLEM 8.3 Determine whether the block shown is in equilibrium and find the magnitude and direction of the friction force when P = 100 lb.

SOLUTION Assume equilibrium: ΣFx = 0: F + (45 lb)sin 30° − (100 lb) cos 40° = 0

ΣFy = 0: N − (45 lb) cos 30° − (100 lb)sin 40° = 0 N = 103.2 lb

(a)

Maximum friction force: Fm = μ s N = 0.40(103.2 lb) = 41.30 lb

We note that F > Fm . Thus, (b)

block moves up 

Actual friction force: F = Fk = μk N = 0.30(103.2 lb) = 30.97 lb, 

F = 31.0 lb

30.0° 



F = +54.0 lb

www.elsolucionario.net PROBLEM 8.4 Determine whether the block shown is in equilibrium and find the magnitude and direction of the friction force when P = 60 lb.

SOLUTION Assume equilibrium:

F = +23.46 lb ΣFy = 0: N − (45 lb) cos 30° − (60 lb) sin 40° = 0 N = 77.54 lb

(a)

Maximum friction force: Fm = μs N = 0.40(77.54 lb) = 31.02 lb

We check that F < Fm . Thus, (b)

Thus

block is in equilibrium 

F = +23.46 lb

F = 23.5 lb

30.0° 

Note: We have Fk = μk N = 0.30(77.54) = 23.26 lb. Thus F > Fk . If block originally in motion, it will keep moving with Fk = 23.26 lb.



ΣFx = 0: F + (45 lb) sin 30° − (60 lb) cos 40° = 0

www.elsolucionario.net PROBLEM 8.5 Determine the smallest value of P required to (a) start the block up the incline, (b) keep it moving up, (c) prevent it from moving down.

SOLUTION (a)

To start block up the incline:

μs = 0.40

From force triangle: P 45 lb = sin 51.80° sin 28.20°

(b)

P = 74.8 lb 

To keep block moving up:

μk = 0.30 φk = tan −1 0.30 = 16.70°

From force triangle: P 45 lb = ° sin 46.70 sin 33.30°

P = 59.7 lb 



φs = tan −1 0.40 = 21.80°


(c)

To prevent block from moving down:

From force triangle: P = 6.76 lb 


P 45 lb = sin 8.20° sin 71.80°


www.elsolucionario.net PROBLEM 8.6 Knowing that the coefficient of friction between the 25-kg block and the incline is μ s = 0.25, determine (a) the smallest value of P required to start the block moving up the incline, (b) the corresponding value of β.

SOLUTION FBD block (Impending motion up) = (25 kg)(9.81 m/s 2 ) = 245.25 N

φs = tan −1 μs = tan −1 (0.25) = 14.04°

(a)

(Note: For minimum P, P ^ R so β = φs .) Then P = W sin (30° + φs ) = (245.25 N)sin 44.04° Pmin = 170.5 N 

(b)

We have β = φs

β = 14.04° 



W = mg

www.elsolucionario.net PROBLEM 8.7 The 80-lb block is attached to link AB and rests on a moving belt. Knowing that μs = 0.25 and μk = 0.20, determine the magnitude of the horizontal force P that should be applied to the belt to maintain its motion (a) to the right, (b) to the left.

SOLUTION We note that link AB is a two-force member, since there is motion between belt and block μk = 0.20 and φk = tan −1 0.20 = 11.31° (a)

Belt moves to right

Force triangle: R 80 lb = sin 120° sin 48.69° R = 92.23 lb

Free body: Belt ΣFx = 0: P − (92.23 lb) sin 11.31° P = 18.089 lb

(b)

P = 18.09 lb



P = 14.34 lb



Belt moves to left

Free body: Block

Force triangle: R 80 lb = sin 60° sin 108.69° R = 73.139 lb

Free body: Belt ΣFx = 0: (73.139 lb)sin 11.31° − P = 0 P = 14.344 lb



Free body: Block

www.elsolucionario.net PROBLEM 8.8 The coefficients of friction between the block and the rail are μs = 0.30 and μk = 0.25. Knowing that θ = 65°, determine the smallest value of P required (a) to start the block moving up the rail, (b) to keep it from moving down.

SOLUTION (a)

To start block up the rail:

μ s = 0.30

Force triangle: P 500 N = sin 51.70° sin (180° − 25° − 51.70°)

(b)

P = 403 N 

To prevent block from moving down:

Force triangle: P 500 N = sin 18.30° sin (180° − 25° − 18.30°)

P = 229 N 



φs = tan −1 0.30 = 16.70°

www.elsolucionario.net PROBLEM 8.9 Considering only values of θ less than 90°, determine the smallest value of θ required to start the block moving to the right when (a) W = 75 lb, (b) W = 100 lb.

SOLUTION

φs = tan −1 μs = 14.036° W 30 lb = sin φs sin(θ − φs ) sin(θ − φs ) =

or

sin(θ − 14.036°) =

W sin 14.036° 30 lb W 123.695 lb

(a)

W = 75 lb:

θ = 14.036° + sin −1

75 lb 123.695 lb

θ = 51.4° 

(b)

W = 100 lb: θ = 14.036° + sin −1

100 lb 123.695 lb

θ = 68.0° 



FBD block (Motion impending):

www.elsolucionario.net PROBLEM 8.10 Determine the range of values of P for which equilibrium of the block shown is maintained.

SOLUTION FBD block:


(Impending motion down):

φs = tan −1 μs = tan −1 0.25 P = (500 N) tan (30° − tan −1 0.25) = 143.03 N

(Impending motion up):

P = (500 N) tan (30° + tan −1 0.25) = 483.46 N 143.0 N ≤ P ≤ 483 N 

Equilibrium is maintained for


www.elsolucionario.net PROBLEM 8.11 The 20-lb block A and the 30-lb block B are supported by an incline that is held in the position shown. Knowing that the coefficient of static friction is 0.15 between the two blocks and zero between block B and the incline, determine the value of θ for which motion is impending.

SOLUTION Since motion impends, F = μ s N between A + B

Impending motion: ΣFy = 0: N1 = 20cos θ ΣFx = 0: T − 20sin θ − μ s N1 = 0 T = 20sin θ + 0.15(20 cos θ ) T = 20sin θ + 3cos θ

(1)

Free body: Block B

Impending motion: ΣFx = 0: T − 30sin θ + μ s N1 = 0 T = 30sin θ − μ s N1 = 30sin θ − 0.15(20 cos θ ) T = 30sin θ − 3cos θ

Eq. (1)-Eq. (2):

(2)

20sin θ + 3cos θ − 30sin θ + 3cos θ = 0 6 cos θ = 10sin θ : tan θ =

6 ; θ = 30.96° 10

θ = 31.0° 



Free body: Block A

www.elsolucionario.net PROBLEM 8.12 The 20-lb block A and the 30-lb block B are supported by an incline that is held in the position shown. Knowing that the coefficient of static friction is 0.15 between all surfaces of contact, determine the value of θ for which motion is impending.

SOLUTION Since motion impends, F = μ s N at all surfaces.

Impending motion: ΣFy = 0: N1 = 20cos θ ΣFx = 0: T − 20sin θ − μ s N1 = 0 T = 20sin θ + 0.15(20 cos θ ) T = 20sin θ + 3cos θ

(1)

Free body: Block B

Impending motion: ΣFy = 0: N 2 − 30 cos θ − N1 = 0

N 2 = 30 cos θ + 20cos θ = 50 cos θ F2 = μ s N 2 = 0.15(50 cos θ ) = 6cos θ ΣFx = 0: T − 30sin θ + μ s N1 + μs N 2 = 0 T = 30sin θ − 0.15(20 cos θ ) − 0.15(50cos θ ) T = 30sin θ − 3cos θ − 7.5cos θ

Eq. (1) subtracted by Eq. (2):

(2)

20sin θ + 3cos θ − 30sin θ + 3cos θ + 7.5cos θ = 0 13.5cos θ = 10sin θ , tan θ =

13.5° 10

θ = 53.5° 



Free body: Block A

www.elsolucionario.net PROBLEM 8.13 The coefficients of friction are μs = 0.40 and μk = 0.30 between all surfaces of contact. Determine the smallest force P required to start the 30-kg block moving if cable AB (a) is attached as shown, (b) is removed.

SOLUTION (a)

Free body: 20-kg block W1 = (20 kg)(9.81 m/s 2 ) = 196.2 N F1 = μs N1 = 0.4(196.2 N) = 78.48 N

ΣF = 0: T − F1 = 0 T = F1 = 78.48 N

Free body: 30-kg block N 2 = 196.2 N + 294.3 N = 490.5 N F2 = μ s N 2 = 0.4(490.5 N) = 196.2 N

ΣF = 0: P − F1 − F2 − T = 0 P = 78.48 N + 196.2 N + 78.48 N = 353.2 N

(b)

P = 353 N



P = 196.2 N



Free body: Both blocks Blocks move together W = (50 kg)(9.81 m/s 2 ) = 490.5 N

ΣF = 0: P − F = 0 P = μs N = 0.4(490.5 N) = 196.2 N



W2 = (30 kg)(9.81 m/s 2 ) = 294.3 N

www.elsolucionario.net PROBLEM 8.14 The coefficients of friction are μs = 0.40 and μk = 0.30 between all surfaces of contact. Determine the smallest force P required to start the 30-kg block moving if cable AB (a) is attached as shown, (b) is removed.

SOLUTION (a)

Free body: 20-kg block W1 = (20 kg)(9.81 m/s 2 ) = 196.2 N F1 = μs N1 = 0.4(196.2 N) = 78.48 N

ΣF = 0: T − F1 = 0 T = F1 = 78.48 N

Free body: 30-kg block N 2 = 196.2 N + 294.3 N = 490.5 N F2 = μ s N 2 = 0.4(490.5 N) = 196.2 N

ΣF = 0: P − F1 − F2 = 0 P = 78.48 N + 196.2 N = 274.7 N

(b)

P = 275 N



P = 196.2 N



Free body: Both blocks Blocks move together W = (50 kg)(9.81 m/s 2 ) = 490.5 N

ΣF = 0: P − F = 0 P = μ s N = 0.4(490.5 N) = 196.2 N



W2 = (30 kg)(9.81 m/s 2 ) = 294.3 N

www.elsolucionario.net PROBLEM 8.15 A 40-kg packing crate must be moved to the left along the floor without tipping. Knowing that the coefficient of static friction between the crate and the floor is 0.35, determine (a) the largest allowable value of α, (b) the corresponding magnitude of the force P.

SOLUTION (a)

Free-body diagram If the crate is about to tip about C, contact between crate and ground is only at C and the reaction R is applied at C. As the crate is about to slide, R must form with the vertical an angle

φs = tan −1 μ s = tan −1 Since the crate is a 3-force body. P must pass through E where R and W intersect. EF =

CF 0.4 m = = 1.1429 m tan θ s 0.35

EH = EF − HF = 1.1429 − 0.5 = 0.6429 m EH 0.6429 m tan α = = 0.4 m HB

α = 58.11° (b)

α = 58.1° 

Force Triangle P W = sin19.29° sin128.82°

P = 0.424 W

P = 0.424(40 kg)(9.81 m/s 2 ),

P = 166.4 N 

Note: After the crate starts moving, μ s should be replaced by the lower value μk . This will yield a larger value of α . 



0.35 = 19.29°

www.elsolucionario.net PROBLEM 8.16 A 40-kg packing crate is pulled by a rope as shown. The coefficient of static friction between the crate and the floor is 0.35. If α = 40°, determine (a) the magnitude of the force P required to move the crate, (b) whether the crate will slide or tip.

SOLUTION Force P for which sliding is impending (We assume that crate does not tip) W = (40 kg)(9.81 m/s 2 ) = 392.4 N

N = W − P sin 40°

(1)

ΣFx = 0: 0.35 N − P cos 40° = 0 0.35(W − P sin 40°) − P cos 40° = 0

Substitute for N from Eq. (1):

P=

0.35W 0.35sin 40° + cos 40°

P = 0.3532W 

Force P for which crate rotates about C (We assume that crate does not slide) ΣM C = 0: ( P sin 40°)(0.8 m) + ( P cos 40°)(0.5 m) −W (0.4 m) = 0 P=

0.4 W = 0.4458W 0.8sin 40° + 0.5cos 40°

 Crate will first slide 

P = 0.3532(392.4 N)

P = 138.6 N 



ΣFy = 0: N − W + P sin 40° = 0

www.elsolucionario.net PROBLEM 8.17 A 120-lb cabinet is mounted on casters that can be locked to prevent their rotation. The coefficient of static friction between the floor and each caster is 0.30. If h = 32 in., determine the magnitude of the force P required to move the cabinet to the right (a) if all casters are locked, (b) if the casters at B are locked and the casters at A are free to rotate, (c) if the casters at A are locked and the casters at B are free to rotate.

SOLUTION FBD cabinet: Note: for tipping, N A = FA = 0


ΣM B = 0: (12 in.)W − (32 in.)Ptip = 0 Ptip = 2.66667

(a)

All casters locked. Impending slip: FA = μ s N A FB = μ s N B

ΣFy = 0 : N A + N B − W = 0 N A + NB = W

So

W = 120 lb

FA + FB = μ sW

μ s = 0.3

ΣFx = 0: P − FA − FB = 0 P = FA + FB = μ sW P = 0.3(120 lb) ( P = 0.3W < Ptip

(b)

Casters at A free, so

FA = 0

Impending slip:

FB = μ s N B

or P = 36.0 lb



OK)

ΣFx = 0: P − FB = 0 P = FB = μ s N B

NB =

P

μs



ΣM A = 0: (32 in.)P + (12 in.)W − (24 in.)N B = 0 8 P + 3W − 6

P = 0 P = 0.25W 0.3

( P = 0.25W < Ptip

OK)

P = 0.25(120 lb)

(c)

Casters at B free, so

FB = 0

Impending slip:

FA = μ s N A

or P = 30.0 lb



ΣFx = 0: P − FA = 0 P = FA = μ s N A P

μs

=

P 0.3

ΣM B = 0: (12 in.)W − (32 in.)P − (24 in.)N A = 0 3W − 8P − 6

( P < Ptip

P =0 0.3 P = 0.107143W = 12.8572

OK) P = 12.86 lb







NA =

www.elsolucionario.net PROBLEM 8.18 A 120-lb cabinet is mounted on casters that can be locked to prevent their rotation. The coefficient of static friction between the floor and each caster is 0.30. Assuming that the casters at both A and B are locked, determine (a) the force P required to move the cabinet to the right, (b) the largest allowable value of h if the cabinet is not to tip over.

SOLUTION FBD cabinet: N A + NB = W

Impending slip: FA = μ s N A FB = μ s N B

So FA + FB = μ sW ΣFx = 0: P − FA − FB = 0

W = 120 lb μ s = 0.3

P = FA + FB = μ sW P = 0.3(120 lb) = 141.26 N

P = 36.0 lb

(b)

For tipping,



N A = FA = 0 ΣM B = 0: hP − (12 in.)W = 0 hmax = (12 in.)

W 1 12 in. = (12 in.) = P 0.3 μs hmax = 40.0 in. 



ΣFy = 0: N A + N B − W = 0

(a)

www.elsolucionario.net PROBLEM 8.19 Wire is being drawn at a constant rate from a spool by applying a vertical force P to the wire as shown. The spool and the wire wrapped on the spool have a combined weight of 20 lb. Knowing that the coefficients of friction at both A and B are μs = 0.40 and μk = 0.30, determine the required magnitude of the force P.

SOLUTION Since spool is rotating FA = μk N A

FB = μk N B

3P − 6 μ k ( N A + N B ) = 0

(1)

ΣFx = 0: FA − N B = 0 N B = μk N A

(2)

ΣFy = 0: P + N A + FB − 20 lb = 0 P + N A + μ k N B − 20 = 0

Substitute for N B from (2):

P + N A + μ k N A − 20 = 0 NA =

Substitute from (2) into (1):

20 − P 1 + μk 2

(3)

3P − 6 μ k ( N A + μ k N A ) = 0 NA =

(3)=(4):

20 − P P = 2 1 + μk 2( μ k + μ k2 )

Substitute μk = 0.30:

20 − P 1 + (0.3) 2

1 P 2 μ k (1 + μk )

(4)

P 2(0.3) (1.03)

20 − P = 1.3974 P; 2.3974 P = 20;

P = 8.34 lb 



ΣM G = 0: P(3 in.) − FA (6 in.) − FB (6 in.) = 0

www.elsolucionario.net PROBLEM 8.20 Solve Problem 8.19 assuming that the coefficients of friction at B are zero. PROBLEM 8.19 Wire is being drawn at a constant rate from a spool by applying a vertical force P to the wire as shown. The spool and the wire wrapped on the spool have a combined weight of 20 lb. Knowing that the coefficients of friction at both A and B are μs = 0.40 and μk = 0.30, determine the required magnitude of the force P.

SOLUTION Since spool is rotating FA = μ k N A

P = 2 FA = 2 μk N A

(1)

ΣFy = 20: P − 20 lb + N A = 0 N A = 20 − P

Substitute for N A from (2) into (1)

(2)

P = 2 μk (20 − P)

Substitute μk = 0.30: P = 2(0.3)(20 − P) 1.667 P = 20 − P 2.667 P = 20 P = 7.50 lb

P = 7.50 lb 



ΣM G = 0: P(3 in.) − FA (6 in.) = 0

www.elsolucionario.net PROBLEM 8.21 The hydraulic cylinder shown exerts a force of 3 kN directed to the right on Point B and to the left on Point E. Determine the magnitude of the couple M required to rotate the drum clockwise at a constant speed.

SOLUTION Free body: Drum ΣM C = 0: M − (0.25 m)( FL + FR ) = 0 M = (0.25 m)( FL + FR )

Since drum is rotating FL = μk N L = 0.3N L FR = μk N R = 0.3 N R

Free body: Left arm ABL

ΣM A = 0: (3 kN)(0.15 m) + FL (0.15 m) − N L (0.45 m) = 0 0.45 kN ⋅ m + (0.3N L )(0.15 m) − N L (0.45 m) = 0 0.405 N L = 0.45 N L = 1.111 kN FL = 0.3N L = 0.3(1.111 kN) = 0.3333 kN

Free body: Right arm DER

(2)

ΣM D = 0: (3 kN)(0.15 m) − FR (0.15 m) − N R (0.45 m) = 0 0.45 kN ⋅ m − (0.3N R )(0.15 m) − N R (0.45 m) = 0 0.495 N R = 0.45 N R = 0.9091 kN FR = μ k N R = 0.3(0.9091 kN) = 0.2727 kN

Substitute for FL and FR into (1):

(3)

M = (0.25 m)(0.333 kN + 0.2727 kN) M = 0.1515 kN ⋅ m M = 151.5 N ⋅ m 



(1)

www.elsolucionario.net PROBLEM 8.22 A couple M of magnitude 100 N ⋅ m is applied to the drum as shown. Determine the smallest force that must be exerted by the hydraulic cylinder on joints B and E if the drum is not to rotate.

SOLUTION Free body: Drum ΣM C = 0: 100 N ⋅ m − (0.25 m)( FL + FR ) = 0

(1)

Since motion impends FL = μ s N L = 0.4 N L FR = μ s N R = 0.4 N R

Free body: Left arm ABL ΣM A = 0: T (0.15 m) + FL (0.15 m) − N L (0.45 m) = 0 0.15T + (0.4 N L )(0.15 m) − N L (0.45 m) = 0 0.39 N L = 0.15T ; N L = 0.38462T FL = 0.4 N L = 0.4(0.38462T ) FL = 0.15385T

(2)

Free body: Right arm DER ΣM D = 0: T (0.15 m) − FR (0.15 m) − N R (0.45 m) = 0 0.15T − (0.4 N R )(0.15 m) − N R (0.45 m) = 0 0.51N R = 0.15T ; N R = 0.29412T FR = 0.4 N R = 0.4(0.29412T ) FR = 0.11765T

(3)

Substitute for FL and FR into Eq. (1): 0.15385T + 0.11765T = 400 T = 1473.3 N T = 1.473 kN  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1244


FL + FR = 400 N

www.elsolucionario.net PROBLEM 8.23 A slender rod of length L is lodged between peg C and the vertical wall and supports a load P at end A. Knowing that the coefficient of static friction between the peg and the rod is 0.15 and neglecting friction at the roller, determine the range of values of the ratio L/a for which equilibrium is maintained.

SOLUTION FBD rod:

 a ΣM B = 0: PL sin θ − NC   sin θ

 =0 

NC =

PL 2 sin θ a

(1)

ΣFy = 0: NC sin θ − μ s NC cos θ − P = 0 NC (sin θ − μ cos θ ) = P

Substitute for NC from Eq. (1), and solve for a/L. a = sin 2 θ (sin θ − μ s cos θ ) L

For θ = 30° and μ s = 0.15:

(2)

a = sin 2 30°(sin 30° − 0.15cos 30°) L a L = 0.092524 = 10.808 L a

For motion of B impending downward, reverse sense of friction force FC . To do this we make μ s = −0.15 in. Eq. (2). Eq. (2):

a = sin 2 30°(sin 30° − (−0.15) cos 30°) L a L = 0.15748 = 6.350 L a 6.35 ≤

Range of values of L/a for equilibrium:

L ≤ 10.81  a



Free-body diagram: For motion of B impending upward:

www.elsolucionario.net PROBLEM 8.24 Solve Problem 8.23 assuming that the coefficient of static friction between the peg and the rod is 0.60. PROBLEM 8.23 A slender rod of length L is lodged between peg C and the vertical wall and supports a load P at end A. Knowing that the coefficient of static friction between the peg and the rod is 0.15 and neglecting friction at the roller, determine the range of values of the ratio L/a for which equilibrium is maintained.

SOLUTION Free-body diagram: For motion of B impending upward  =0 

NC =

PL 2 sin θ a

(1)

ΣFy = 0: NC sin θ − μ s NC cos θ − P = 0 NC (sin θ − μ s cos θ ) = P

Substitute for NC from (1), and solve for

a L a = sin 2 θ (sin θ − μ s cos θ ) L

For θ = 30° and μ s = 0.60:

a = sin 2 30°(sin 30° − 0.60 cos 30°) L

(2)

a = −0.0049 < 0 L

Thus, slipping of B upward does not occur for motion of B impending downward, reverse sense of friction force FC. To do this we make μC = −0.60 in Eq. (2). a = sin 2 30°(sin 30° − (−0.60) cos 30°) L a L = 0.2459 = 3.923 L a L /a ≥ 3.92 

Range of L/a for equilibrium:



 a ΣM B = 0: PL sin θ − N C   sin θ

www.elsolucionario.net PROBLEM 8.25 A 6.5-m ladder AB leans against a wall as shown. Assuming that the coefficient of static friction μ s is zero at B, determine the smallest value of μ s at A for which equilibrium is maintained.

SOLUTION


Free body: Ladder Three-force body. Line of action of A must pass through D, where W and B intersect.

At A:

μ s = tan φs =

1.25 m = 0.2083 6m

μ s = 0.208 


www.elsolucionario.net PROBLEM 8.26 A 6.5-m ladder AB leans against a wall as shown. Assuming that the coefficient of static friction μ s is the same at A and B, determine the smallest value of μ s for which equilibrium is maintained.

SOLUTION

Motion impending: FA = μ s N A FB = μ s N B ΣM A = 0: W (1.25 m) − N B (6 m) − μ s N B (2.5 m) = 0 NB =

1.25W 6 + 2.5μ s

(1)

ΣFy = 0: N A + μ s N B − W = 0 N A = W − μs N B NA = W −

1.25μ sW 6 + 2.5μ s

(2)

ΣFx = 0: μ s N A − N B = 0

Substitute for NA and NB from Eqs. (1) and (2):

μsW −

1.25μ s2W 1.25W = 6 + 2.5μs 6 + 2.5μ s

6μ s + 2.5μ s2 − 1.25μ s2 = 1.25 1.25μ s2 + 6μ s − 1.25 = 0

μ s = 0.2 and

μs = −5 (Discard) μs = 0.200 



Free body: Ladder

www.elsolucionario.net PROBLEM 8.27 The press shown is used to emboss a small seal at E. Knowing that the coefficient of static friction between the vertical guide and the embossing die D is 0.30, determine the force exerted by the die on the seal.

SOLUTION

Dimensions in mm ΣM A = 0: FBD cos 20°(100) + FBD sin 20°(173.205) − (250 N)(100 + 386.37) = 0 FBD = 793.64 N

Free body: Die D

φs = tan −1 μs = tan −1 0.3 = 16.6992°

Force triangle: D 793.64 N = sin 53.301° sin 106.6992° D = 664.35 N

D = 664 N 

On seal:



Free body: Member ABC

www.elsolucionario.net PROBLEM 8.28 The machine base shown has a mass of 75 kg and is fitted with skids at A and B. The coefficient of static friction between the skids and the floor is 0.30. If a force P of magnitude 500 N is applied at corner C, determine the range of values of θ for which the base will not move.

SOLUTION Free-body: Machine base m = (75 kg)(9.81 m/s 2 ) = 735.75 N

Assume sliding impends FB = μs N B

ΣFy = 0: N A + N B − W − P sin θ = 0 ( N A + N B ) = W + P sin θ

(1)

ΣFx = 0: FA + FB − P cos θ = 0

μs ( N A + N B ) = P cos θ = 0 Eq. (2) : Eq. (1)

μs =

(2)

P cos θ W + P sin θ

μsW + μs P sin θ = P cos θ 0.30(735.75 N) + 0.30(500 N) sinθ = 500 cosθ 500 cos θ − 150sin θ = 220.73 Solve for θ : θ = 48.28°

Assume tipping about B impends:

∴ NA = 0

ΣM B = 0: P sin θ (0.4 m) − P cos θ (0.5 m) − W (0.2 m) = 0 500 sin θ (0.4) − 500 cos θ (0.5) − 735.75(0.2 m) = 0 200sin θ − 250cos θ = 147.15

Solve for θ :

θ = 78.70° 48.3° ≤ θ ≤ 78.7° 

Range for no motion:



FA = μs N A

www.elsolucionario.net PROBLEM 8.29 The 50-lb plate ABCD is attached at A and D to collars that can slide on the vertical rod. Knowing that the coefficient of static friction is 0.40 between both collars and the rod, determine whether the plate is in equilibrium in the position shown when the magnitude of the vertical force applied at E is (a) P = 0, (b) P = 20 lb.

SOLUTION (a)

P=0

ΣM D = 0: N A (2 ft) − (50 lb)(3 ft) = 0

ΣFx = 0: N D = N A = 75 lb ΣFy = 0: FA + FD − 50 lb = 0 FA + FD = 50 lb

But:

( FA ) m = μs N A = 0.40(75 lb) = 30 lb (FD ) m = μs N D = 0.40(75 lb) = 30 lb

(b)

Thus:

( FA ) m + ( FD )m = 60 lb

and

( FA ) m + ( FD ) m > FA + FD

P = 20 lb

Plate is in equilibrium 

ΣM D = 0: N A (2 ft) − (50 lb)(3 ft) + (20 lb)(5 ft) = 0 N A = 25 lb

ΣFx = 0: N D = N A = 25 lb ΣFy = 0: FA + FD − 50 lb + 20 lb = 0 FA + FD = 30 lb

But:

( FA ) m = μs N A = 0.4(25 lb) = 10 lb (FD ) m = μ s N D = 0.4(25 lb) = 10 lb

Thus: and

( FA ) m + ( FD )m = 20 lb FA + FD > ( FA ) m + ( FD ) m

Plate moves downward 



N A = 75 lb

www.elsolucionario.net PROBLEM 8.30 In Problem 8.29, determine the range of values of the magnitude P of the vertical force applied at E for which the plate will move downward. PROBLEM 8.29 The 50-lb plate ABCD is attached at A and D to collars that can slide on the vertical rod. Knowing that the coefficient of static friction is 0.40 between both collars and the rod, determine whether the plate is in equilibrium in the position shown when the magnitude of the vertical force applied at E is (a) P = 0, (b) P = 20 lb.

SOLUTION We shall consider the following two cases: (1)

0 < P < 30 lb


ΣM D = 0: N A (2 ft) − (50 lb)(3 ft) + P (5 ft) = 0 N A = 75 lb − 2.5P

(Note: N A ≥ 0 and directed

for P ≤ 30 lb as assumed here) ΣFx = 0: N A = N D ΣFy = 0: FA + FD + P − 50 = 0 FA + FD = 50 − P

But:

( FA ) m = ( F0 ) m = μ s N A = 0.40(75 − 2.5P ) = 30 − P

Plate moves if: or (2)

FA + FD > ( FA ) m + ( FD ) m

50 − P > (30 − P) + (30 − P)

P > 10 lb 

30 lb < P < 50 lb ΣM D = 0: − N A (2 ft) − (50 lb)(3 ft) + P (5 ft) = 0 N A = 2.5P − 75

(Note: NA > and directed

for P > 30 lb as assumed) ΣFx = 0: N A = N D ΣFy = 0: FA + FD + P − 50 = 0 FA + FD = 50 − P



But:

( FA ) m = ( FD ) m = μ s N A = 0.40(2.5 P − 75) = P − 30 lb

Plate moves if:

FA + FD > ( FA ) m + ( FD ) m

50 − P > ( P − 30) + ( P − 30)

P<

110 = 36.7 lb  3

10.00 lb < P < 36.7 lb 

Thus, plate moves downward for:


(Note: For P > 50 lb, plate is in equilibrium)


www.elsolucionario.net PROBLEM 8.31 A rod DE and a small cylinder are placed between two guides as shown. The rod is not to slip downward, however large the force P may be; i.e., the arrangement is said to be self-locking. Neglecting the weight of the cylinder, determine the minimum allowable coefficients of static friction at A, B, and C.

SOLUTION

Since cylinder is a two-force body, R B and R C have the same line of action. Thus φB = φC : From triangle OBC:

φB + φC = θ Thus:

φB = φC =

θ 2

For no sliding, we must have tan φB ≤ ( μ B ) s , tan φC ≤ ( μC ) s Therefore:

( μ B ) s ≥ tan

θ 2

( μc ) s ≥ tan

,

θ 2



We also note that RB and RC are indeterminate. Free body: Rod Since RB is indeterminate, it may be as large as necessary to satisfy equation Σ Fy = 0, no matter how large P is or how small φ A is. Therefore ( μ A ) s may have any value 



Free body: Cylinder

www.elsolucionario.net PROBLEM 8.32 A 500-N concrete block is to be lifted by the pair of tongs shown. Determine the smallest allowable value of the coefficient of static friction between the block and the tongs at F and G.


SOLUTION Free body: Members CA, AB, BD C y = Dy =

By symmetry:

1 (500) = 250 N 2

Since CA is a two-force member, Cy Cx 250 N = = 90 mm 75 mm 75 mm C x = 300 N

ΣFx = 0: Dx = C x Dx = 300 N

Free body: Tong DEF ΣM E = 0: (300 N)(105 mm) + (250 N)(135 mm) + (250 N)(157.5 mm) − Fx (360 mm) = 0 Fx = +290.625 N

Minimum value of μ s :

μs =

Fy Fx

=

250 N 290.625 N

μs = 0.860 


www.elsolucionario.net PROBLEM 8.33 The 100-mm-radius cam shown is used to control the motion of the plate CD. Knowing that the coefficient of static friction between the cam and the plate is 0.45 and neglecting friction at the roller supports, determine (a) the force P required to maintain the motion of the plate, knowing that the plate is 20 mm thick, (b) the largest thickness of the plate for which the mechanism is self-locking (i.e., for which the plate cannot be moved however large the force P may be).

SOLUTION

Impending motion:

F = μs N

ΣM A = 0: QR − NR sin θ + ( μs N ) R cos θ = 0 N=

Free body: Plate

ΣFx = 0 P = μ s N

Q sin θ − μ s cos θ

(1) (2)

Geometry in ΔABD with R = 100 mm and d = 20 mm R−d R 80 mm = 100 mm = 0.8

cos θ =

sin θ = 1 − cos 2 θ = 0.6



Free body: Cam


(a)

Eq. (1) using

Q = 60 N and μ s = 0.45 60 N 0.6 − (0.45)(0.8) 60 = = 250 N 0.24

N=

Eq. (2)

P = μ s N = (0.45)(250 N) P = 112.5 N 

(b)

For P = ∞, N = ∞. Denominator is zero in Eq. (1). sin θ − μ s cos θ = 0 tan θ = μ s = 0.45 R−d R 100 − d cos 24.23 = 100 cos θ =

d = 8.81 mm 



θ = 24.23°

www.elsolucionario.net PROBLEM 8.34 A safety device used by workers climbing ladders fixed to high structures consists of a rail attached to the ladder and a sleeve that can slide on the flange of the rail. A chain connects the worker’s belt to the end of an eccentric cam that can be rotated about an axle attached to the sleeve at C. Determine the smallest allowable common value of the coefficient of static friction between the flange of the rail, the pins at A and B, and the eccentric cam if the sleeve is not to slide down when the chain is pulled vertically downward.

SOLUTION Free body: Cam

ND =

6P 0.8 − 3μ s

(1)

Free body: Sleeve and cam ΣFx = 0: N D − N A − N B = 0 N A + NB = ND

(2)

ΣFy = 0: FA + FB + FD − P = 0

μs ( N A + N B + N D ) = P

or

μs (2 N D ) = P


ND =

P 2μ s

(3) (4)

Equate expressions for N D from Eq. (1) and Eq. (4): P 6P = ; 0.8 − 3μ s = 12 μs 2μs 0.8 − 3μ s

μs =

0.8 15

μs = 0.0533 

(Note: To verify that contact at pins A and B takes places as assumed, we shall check that N A > 0 and N B = 0.)



ΣM C = 0: N D (0.8 in.) − μ s N D (3in.) − P (6 in.) = 0


From Eq. (4):

ND =

P P = = 9.375P 2μs 2(0.0533)

From free body of cam and sleeve: ΣM B = 0: N A (8 in.) − N D (4 in.) − P(9 in.) = 0 8 N A = (9.375 P)(4) + 9 P N A = 5.8125P > 0 OK

From Eq. (2):

N A + NB = ND 5.8125P + N B = 9.375 P


N B = 3.5625P > 0 OK


www.elsolucionario.net PROBLEM 8.35 To be of practical use, the safety sleeve described in Problem 8.34 must be free to slide along the rail when pulled upward. Determine the largest allowable value of the coefficient of static friction between the flange of the rail and the pins at A and B if the sleeve is to be free to slide when pulled as shown in the figure, assuming (a) θ = 60°, (b) θ = 50°, (c) θ = 40°.

SOLUTION Note the cam is a two-force member.

We assume contact between rail and pins as shown. ΣM C = 0: FA (3 in.) + FB (3 in.) − N A (4 in.) − N B (4 in.) = 0

But

FA = μ s N A FB = μ s N B

We find

3μ s ( N A + N B ) − 4( N A + N B ) = 0

μs =

4 = 1.33333 3

We now verify that our assumption was correct. ΣFx = 0: N A − N B + P cos θ = 0

N B − N A = P cos θ

(1)

ΣFy = 0: − FA − FB + P sin θ = 0

μs N A + μ s N B = P sin θ N A + NB =


P sin θ

μs

(2)

 sin θ  2 N B = P  cos θ +  > 0 OK μs  



Free body: Sleeve


 sin θ  2N A = P  − cos θ   μs 

Subtract Eq. (1) from Eq. (2): NA > 0 only if

sin θ

μs

− cos θ > 0 tan θ > μ s = 1.33333

θ = 53.130° (a)

For case (a): Condition is satisfied, contact takes place as shown. Answer is correct.

μs = 1.333  But for (b) and (c): θ < 53.130° and our assumption is wrong, N A is directed to left.

N A + N B = P cos θ

(3)

ΣFy = 0: − FA − FA + P sin θ = 0

μs ( N A + N B ) = P sin θ

(4)

Divide Eq. (4) by Eq. (3):

μs = tan θ (b)

(c)

(5)

We make θ = 50° in Eq. (5):

μs = tan 50°

μs = 1.192 

μs = tan 40°

μs = 0.839 

We make θ = 40° in Eq. (5):



ΣFx = 0: − NA − N B + P cos θ = 0

www.elsolucionario.net PROBLEM 8.36 Two 10-lb blocks A and B are connected by a slender rod of negligible weight. The coefficient of static friction is 0.30 between all surfaces of contact, and the rod forms an angle θ = 30°. with the vertical. (a) Show that the system is in equilibrium when P = 0. (b) Determine the largest value of P for which equilibrium is maintained.

SOLUTION FBD block B: Since P = 2.69 lb to initiate motion,

equilibrium exists with P = 0 

(b)

For Pmax , motion impends at both surfaces: ΣFy = 0: N B − 10 lb − FAB cos 30° = 0

Block B:

N B = 10 lb +

Impending motion:

3 FAB 2

(1)

FB = μs N B = 0.3N B ΣFx = 0: FB − FAB sin 30° = 0 FAB = 2 FB = 0.6 N B

Solving Eqs. (1) and (2):

N B = 10 lb +

(2)

3 (0.6 N B ) = 20.8166 lb 2

FBD block A: FAB = 0.6 N B = 12.4900 lb

Then Block A:

ΣFx = 0: FAB sin 30° − N A = 0 NA =

Impending motion:

1 1 FAB = (12.4900 lb) = 6.2450 lb 2 2

FA = μs N A = 0.3(6.2450 lb) = 1.8735 lb ΣFy = 0: FA + FAB cos 30° − P − 10 lb = 0 3 FAB − 10 lb 2 3 = 1.8735 lb + (12.4900 lb) − 10 lb 2 = 2.69 lb

P = FA +

P = 2.69 lb 



(a)

www.elsolucionario.net PROBLEM 8.37 Bar AB is attached to collars that can slide on the inclined rods shown. A force P is applied at Point D located at a distance a from end A. Knowing that the coefficient of static friction μs between each collar and the rod upon which it slides is 0.30 and neglecting the weights of the bar and of the collars, determine the smallest value of the ratio a/L for which equilibrium is maintained.

SOLUTION FBD bar and collars: Impending motion:

φs = tan −1 μs

Neglect weights: 3-force FBD and  ACB = 90° so

AC =

a cos (45° + φs )

= l sin (45° − φs ) a = sin (45° − 16.6992°) cos (45° + 16.6992°) l a = 0.225  l



= tan −1 0.3 = 16.6992°

www.elsolucionario.net PROBLEM 8.38 Two identical uniform boards, each of weight 40 lb, are temporarily leaned against each other as shown. Knowing that the coefficient of static friction between all surfaces is 0.40, determine (a) the largest magnitude of the force P for which equilibrium will be maintained, (b) the surface at which motion will impend.

SOLUTION

Assume impending motion at C, so FC = μ s N C = 0.4 N C

FBD II:

ΣM B = 0: (6 ft)N C − (8 ft)FC − (3 ft)(40 lb) = 0 [6 ft − 0.4(8 ft)]N C = (3 ft)(40 lb)

or

N C = 42.857 lb

and

FC = 0.4 N C = 17.143 lb ΣFx = 0: N B − FC = 0

N B = FC = 17.143 lb ΣM y = 0: − FB − 40 lb + N C = 0

FB = N C − 40 lb = 2.857 lb

Check for motion at B: FBD I:

FB 2.857 lb = = 0.167 < μ s , OK, no motion. N B 17.143 lb ΣM A = 0: (8 ft)N B + (6 ft)FB − (3 ft)(P + 40 lb) = 0

(8 ft)(17.143 lb) + (6 ft)(2.857 lb) − 40 lb 3 ft = 11.429 lb

P=



Board FBDs:


Check for slip at A (unlikely because of P): ΣFx = 0: FA − N B = 0 or

FA = N B = 17.143 lb

ΣFy = 0: N A − P − 40 lb + FB = 0

or

NA = 11.429 lb + 40 lb − 2.857 lb = 48.572 lb

Then

FA 17.143 lb = = 0.353 < μ s N A 48.572 lb

OK, no slip  assumption is correct.

(a)

Pmax = 11.43 lb 

(b)

Motion impends at C 



Therefore

www.elsolucionario.net PROBLEM 8.39 Knowing that the coefficient of static friction between the collar and the rod is 0.35, determine the range of values of P for which equilibrium is maintained when θ = 50° and M = 20 N ⋅ m.

SOLUTION Free body member AB: ΣM A = 0: 20 N ⋅ m − FBC cos 50°(0.1 m) = 0 FBC = 311.145 N

Motion of C impending upward: ΣFx = 0: (311.145 N) cos 50° − N = 0 N = 200 N ΣFy = 0: (311.145 N) sin 50° − P − (0.35)(200 N) = 0 P = 168.351 N 

Motion of C impending downward: ΣFx = 0: (311.145 N) cos 50° − N = 0 N = 200 N ΣFy = 0: (311.145 N) sin 50° − P + (0.35)(200 N) = 0 P = 308.35 N  168.4 N ≤ P ≤ 308 N 

Range of P:



BC is a two-force member.

www.elsolucionario.net PROBLEM 8.40 Knowing that the coefficient of static friction between the collar and the rod is 0.40, determine the range of values of M for which equilibrium is maintained when θ = 60° and P = 200 N.

SOLUTION Free body member AB: ΣM A = 0: M − FBC cos 60°(0.1 m) = 0 M = 0.05FBC

(1)

Motion of C impending upward: ΣFx = 0: FBC cos 60° − N = 0 N = 0.5FBC ΣFy = 0: FBC sin 60° − 200 N − (0.40)(0.5FBC ) = 0 FBC = 300.29 N

Eq. (1):

M = 0.05(300.29) M = 15.014 N ⋅ m 

Motion of C impending downward: ΣFx = 0: FBC cos 60° − N = 0 N = 0.5FBC ΣFy = 0: FBC sin 60° − 200 N + (0.40)(0.5 FBC ) = 0 FBC = 187.613 N

Eq. (1):

M = 0.05(187.613) M = 9.381 N ⋅ m  9.38 N ⋅ m ≤ M ≤ 15.01 N ⋅ m 

Range of M:



BC is a two-force member.

www.elsolucionario.net PROBLEM 8.41 A 10-ft beam, weighing 1200 lb, is to be moved to the left onto the platform. A horizontal force P is applied to the dolly, which is mounted on frictionless wheels. The coefficients of friction between all surfaces are μ s = 0.30 and μk = 0.25, and initially x = 2 ft. Knowing that the top surface of the dolly is slightly higher than the platform, determine the force P required to start moving the beam. (Hint: The beam is supported at A and D.)

SOLUTION


FBD beam:

ΣM A = 0: N D (8ft) − (1200 lb)(5ft) = 0 N D = 750 lb

ΣFy = 0: N A − 1200 + 750 = 0 N A = 450 lb

( FA ) m = μ s N A = 0.3(450) = 135.0 lb ( FD ) m = μ s N D = 0.3(750) = 225 lb

Since ( FA ) m < ( FD ) m , sliding first impends at A with FA = ( FA ) m = 135 lb ΣFx = 0: FA − FD = 0 FD = FA = 135.0 lb

FBD dolly: From FBD of dolly: ΣFx = 0: FD − P = 0 P = FD = 135.0 lb

P = 135.0 lb 


www.elsolucionario.net PROBLEM 8.42 (a) Show that the beam of Problem 8.41 cannot be moved if the top surface of the dolly is slightly lower than the platform. (b) Show that the beam can be moved if two 175-lb workers stand on the beam at B and determine how far to the left the beam can be moved. PROBLEM 8.41 A 10-ft beam, weighing 1200 lb, is to be moved to the left onto the platform. A horizontal force P is applied to the dolly, which is mounted on frictionless wheels. The coefficients of friction between all surfaces are μ s = 0.30 and μk = 0.25, and initially x = 2 ft. Knowing that the top surface of the dolly is slightly higher than the platform, determine the force P required to start moving the beam. (Hint: The beam is supported at A and D.)

(a)

Beam alone ΣM C = 0: N B (8 ft) − (1200 lb)(3 ft) = 0 N B = 450 lb

ΣFy = 0: NC + 450 − 1200 = 0 N C = 750 lb

( FC ) m = μ s NC = 0.3(750) = 225 lb (FB ) m = μ s N B = 0.3(450) = 135 lb

Since ( FB ) m < ( FC )m , sliding first impends at B, and

(b)

Beam cannot be moved 

Beam with workers standing at B

ΣM C = 0: N B (10 − x) − (1200)(5 − x) − 350(10 − x) = 0 NB =

9500 − 1550 x 10 − x

ΣM B = 0: (1200)(5) − NC (10 − x) = 0 NC =

6000 10 − x



SOLUTION


Check that beam starts moving for x = 2 ft: For x = 2 ft:

9500 − 1550(2) = 800 lb 10 − 2 6000 NC = = 750 lb 10 − 2 ( FC ) m = μs NC = 0.3(750) = 225 lb NB =

( FB ) m = μs N B = 0.3(800) = 240 lb

Since ( FC ) m < ( FB )m , sliding first impends at C,

Beam moves 

How far does beam move? Beam will stop moving when

But

FC = μk NC = 0.25

and

( FB ) m = μs N B = 0.30

Setting FC = ( FB ) m :

6000 1500 = 10 − x 10 − x 9500 − 1550 x 2850 − 465 x = 10 − x 10 − x

1500 = 2850 − 465 x

x = 2.90ft 

(Note: We have assumed that, once started, motion is continuous and uniform (no acceleration).)



FC = ( FB ) m

www.elsolucionario.net PROBLEM 8.43 Two 8-kg blocks A and B resting on shelves are connected by a rod of negligible mass. Knowing that the magnitude of a horizontal force P applied at C is slowly increased from zero, determine the value of P for which motion occurs, and what that motion is, when the coefficient of static friction between all surfaces is (a) μs = 0.40, (b) μs = 0.50.

SOLUTION (a)

μs = 0.40:

Assume blocks slide to right.

FB = μ s N B ΣFy = 0: N A + N B − 2W = 0 N A + N B = 2W ΣFx = 0: P − FA − FB = 0 P = FA + FB = μ s ( N A + N B ) = μ s (2W )

(1)

P = 0.40(2)(78.48 N) = 62.78 N ΣM B = 0: P (0.1 m) − ( N A − W )(0.09326 m) + FA (0.2 m) = 0 (62.78)(0.1) − ( N A − 78.48)(0.09326) + (0.4)( N B )(0.2) = 0 0.17326 N A = 1.041 N A = 6.01 N > 0 OK

System slides: P = 62.8 N  (b)

μs = 0.50: See part a. Eq. (1):

P = 0.5(2)(78.48 N) = 78.48 N ΣM B = 0: P(0.1 m) + ( N A − W )(0.09326 m) + FA (0.2 m) = 0 (78.48)(0.1) + ( N A − 78.48)(0.09326) + (0.5) N A (0.2) = 0 0.19326 N A = −0.529 N A = −2.73 N < 0 uplift, rotation about B



W = mg = (8 kg)(9.81 m/s 2 ) = 78.48 N FA = μ s N A


For N A = 0:

ΣM B = 0: P(0.1 m) − W (0.09326 m) = 0 P = (78.48 N)(0.09326 m)/(0.1) = 73.19


System rotates about B: P = 73.2 N 


www.elsolucionario.net PROBLEM 8.44 A slender steel rod of length 225 mm is placed inside a pipe as shown. Knowing that the coefficient of static friction between the rod and the pipe is 0.20, determine the largest value of θ for which the rod will not fall into the pipe.

SOLUTION Motion of rod impends down at A and to left at B. FA = μ s N A FB = μs N B

N A − N B sin θ + μ s N B cos θ = 0 N A = N B (sin θ − μ s cos θ )

(1)

ΣFy = 0: FA + N B cos θ + FB sin θ − W = 0

μs N A + N B cos θ + μs N B sin θ − W = 0

(2)

Substitute for NA from Eq. (1) into Eq. (2):

μs N B (sin θ − μ s cos θ ) + N B cos θ + μs N B sin θ − W = 0 NB =

W (1 −

 75 ΣM A = 0: N B   cos θ

μs2 ) cos θ

+ 2μ s sin θ

=

W (1 − 0.2 ) cos θ + 2(0.2) sin θ 2

(3)

  − W (112.5cos θ ) = 0 

Substitute for N B from Eq. (3), cancel W, and simplify to find 9.6 cos3 θ + 4sin θ cos 2 θ − 6.6667 = 0 cos3 θ (2.4 + tan θ ) = 1.6667

θ = 35.8° 

Solve by trial & error:



ΣFx = 0: N A − N B sin θ + FB cos θ = 0

www.elsolucionario.net PROBLEM 8.45 In Problem 8.44, determine the smallest value of θ for which the rod will not fall out the pipe. PROBLEM 8.44 A slender steel rod of length 225 mm is placed inside a pipe as shown. Knowing that the coefficient of static friction between the rod and the pipe is 0.20, determine the largest value of θ for which the rod will not fall into the pipe.

SOLUTION Motion of rod impends up at A and right at B. FA = μ s N A

FB = μ s N B

N A − N B sin θ − μ s N B cos θ = 0 N A = N B (sin θ + μ s cos θ )

(1)

ΣFy = 0: −FA + N B cos θ − FB sin θ − W = 0 − μ s N A + N B cos θ − μs N B sin θ − W = 0

(2)

Substitute for N A from Eq. (1) into Eq. (2): − μ s N B (sin θ + μ s cos θ ) + N B cos θ − μ s N B sin θ − W = 0 NB =

W W = (1 − μ s2 ) cos θ − 2μ s sin θ (1 − 0.22 ) cos θ − 2(0.2)sin θ

(3)

 75  ΣM A = 0: N B   − W (112.5cos θ ) = 0  cos θ 

Substitute for NB from Eq. (3), cancel W, and simplify to find 9.6 cos3 θ − 4sin θ cos 2 θ − 6.6667 = 0 cos3 θ (2.4 − tan θ ) = 1.6667

θ = 20.5° 

Solve by trial + error:



ΣFx = 0: N A − N B sin θ − FB cos θ = 0

www.elsolucionario.net PROBLEM 8.46 Two slender rods of negligible weight are pin-connected at C and attached to blocks A and B, each of weight W. Knowing that θ = 80° and that the coefficient of static friction between the blocks and the horizontal surface is 0.30, determine the largest value of P for which equilibrium is maintained.

SOLUTION

FAC = P sin 20° = 0.34202 P FBC = P cos 20° = 0.93969 P ΣFy = 0: N A − W − FAC sin 30° = 0 N A = W + 0.34202 P sin 30° = W + 0.171010 P

or FBD block A:

ΣFx = 0: FA − FAC cos 30° = 0

or

FA = 0.34202 P cos 30° = 0.29620 P

For impending motion at A:

FA = μ s N A

Then

NA =

FA

μs

: W + 0.171010 P =

0.29620 P 0.3

P = 1.22500W

or

ΣFy = 0: N B − W − FBC cos 30° = 0 N B = W + 0.93969 P cos 30° = W + 0.81380 P ΣFx = 0: FBC sin 30° − FB = 0 FB = 0.93969 P sin 30° = 0.46985P



FBD pin C:


FBD block B: For impending motion at B:

FB = μ s N B

Then

NB =

FB

μs

: W + 0.81380 P =

0.46985P 0.3

P = 1.32914W

or

Pmax = 1.225W 


Thus, maximum P for equilibrium


www.elsolucionario.net PROBLEM 8.47 Two slender rods of negligible weight are pin-connected at C and attached to blocks A and B, each of weight W. Knowing that P = 1.260W and that the coefficient of static friction between the blocks and the horizontal surface is 0.30, determine the range of values of θ, between 0 and 180°, for which equilibrium is maintained.

SOLUTION AC and BC are two-force members Free body: Joint C

From Force triangle: FAB = P sin(θ − 60°) = 1.26W sin(θ − 60°)

(1)

FBC = P cos(θ − 60°) = 1.26W cos(θ − 60°)

(2)

We shall, in turn, seek θ corresponding to impending motion of each block For motion of A impending to left from solution of Prob. 8.46:

FAC = 0.419W

FAC = 0.419W = 1.26W sin(θ − 60°)

EQ (1):

sin(θ − 60°) = 0.33254 θ − 60° = 19.423°

θ = 79.42°



For motion of B impending to right. from solution of Prob. 8.46: FBC = 1.249W Eq. (2):

FBC = 1.249W = 1.26W cos(θ − 60°) cos(θ − 60°) = 0.99127

θ − 60° = ±7.58° θ − 60° = +7.58°

θ = 67.6° 

θ − 60° = −7.58°

θ = 52.4° 



Force triangle:


For motion of A impending to right

γ = 180° − 60° − 16.7° = 103.3° − FAB W = sin16.7° sin103.3° FAB = −0.29528W

Law of sines:

Note: Direction of + FAB is kept same as in free body of Joint C. FAB = −0.29528W = 1.26W sin(θ − 60°)

Eq. (1):

sin(θ − 60°) = −0.23435 (θ − 60°) = −13.553°

θ = 46.4° 

A moves to right

No motion 46.4°

No motion for:

B moves to right 52.4°

No motion 67.6°

A moves to left 79.4°

θ

46.4° ≤ θ ≤ 52.4° and 67.6° ≤ θ ≤ 79.4°





Summary:

www.elsolucionario.net PROBLEM 8.48 The machine part ABC is supported by a frictionless hinge at B and a 10° wedge at C. Knowing that the coefficient of static friction at both surfaces of the wedge is 0.20, determine (a) the force P required to move the wedge, (b) the components of the corresponding reaction at B.

SOLUTION

φs = tan −1 0.20 = 11.31°

ΣM B = 0 (1800 N)(0.35 m) − R cos 21.31°(0.6 m) = 0 RC = 1127.1 N

Force triangle:

Free body: Wedge

(a)

Law of sines:

P 1127.1 N = sin(11.31° + 21.31°) sin 78.69° P = 619.6 N

(b)

P = 620 N



B x = 1390 N



Return to part ABC: ΣFx = 0: Bx + 1800 N − RC sin 21.31° = 0 Bx + 1800 N − (1127.1 N) sin 21.31° Bx = −1390.4 N ΣFy = 0: By + RC cos 21.31° = 0 By + (1127.1 N) cos 21.31° = 0 By = −1050 N

B y = 1050 N 



Free body: Part ABC

www.elsolucionario.net PROBLEM 8.49 Solve Problem 8.48 assuming that the force P is directed to the right. PROBLEM 8.48 The machine part ABC is supported by a frictionless hinge at B and a 10° wedge at C. Knowing that the coefficient of static friction at both surfaces of the wedge is 0.20, determine (a) the force P required to move the wedge, (b) the components of the corresponding reaction at B.

SOLUTION

φs = tan −1 0.20 = 11.31°


Free body: Part ABC

Σ M B = 0: (1800 N)(0.35 m) − RC cos1.31°(0.6 m) = 0 RC = 1050.3 N

Free body: Wedge

Force triangle:

γ = 90° − 11.31° = 78.69° (a)

Law of sines:

P 1050.3 N = sin (11.31° + 1.31°) sin 78.69° P = 234 N

(b)

Return to Part ABC:

P = 234 N



Σ Fx = 0 : Bx + 1800 N + RC sin1.31° = 0 Bx + 1800 N + (1050.3 N) sin1.31° = 0 Bx = −1824 N

B x = 1824 N



Σ Fy = 0 : By + RC cos1.31° = 0 By + (1050.3 N) cos1.31° = 0 By = −1050 N

B y = 1050 N 


www.elsolucionario.net PROBLEM 8.50 The elevation of the end of the steel beam supported by a concrete floor is adjusted by means of the steel wedges E and F. The base plate CD has been welded to the lower flange of the beam, and the end reaction of the beam is known to be 100 kN. The coefficient of static friction is 0.30 between two steel surfaces and 0.60 between steel and concrete. If the horizontal motion of the beam is prevented by the force Q, determine (a) the force P required to raise the beam, (b) the corresponding force Q.

SOLUTION Free body: Beam and plate CD (100 kN) cos16.7° R1 = 104.4 kN

Free body: Wedge E

P 104.4 kN = sin 43.4° sin 63.3°

(a)

P = 80.3 kN



φs = tan −1 0.3 = 16.7°

(b)

Q = (100 kN) tan16.7°

Q = 30 kN



Free body: Wedge F (To check that it does not move.) Since wedge F is a two-force body, R2 and R3 are colinear Thus But

θ = 26.7° φconcrete = tan −1 0.6 = 31.0° > θ

OK



R1 =

www.elsolucionario.net PROBLEM 8.51 The elevation of the end of the steel beam supported by a concrete floor is adjusted by means of the steel wedges E and F. The base plate CD has been welded to the lower flange of the beam, and the end reaction of the beam is known to be 100 kN. The coefficient of static friction is 0.30 between two steel surfaces and 0.60 between steel and concrete. If the horizontal motion of the beam is prevented by the force Q, determine (a) the force P required to raise the beam, (b) the corresponding force Q.

Free body: Wedge F

φs = tan −1 0.30 = 16.7°

(a)

P = (100 kN) tan 26.7° + (100 kN) tan φs P = 50.29 kN + 30 kN P = 80.29 kN R1 =

P = 80.3 kN



Q = 50.3 kN



(100 kN) = 111.94 kN cos 26.7°

Free body: Beam, plate, and wedge E

(b)

Q = W tan 26.7° = (100 kN) tan 26.7° Q = 50.29 kN



SOLUTION

www.elsolucionario.net PROBLEM 8.52 Two 10° wedges of negligible weight are used to move and position the 400-lb block. Knowing that the coefficient of static friction is 0.25 at all surfaces of contact, determine the smallest force P that should be applied as shown to one of the wedges.

SOLUTION Free body: Block and top wedge

φs = tan −1 0.25 = 14.04°

Law of sines:

Free body: Lower wedge

Law of sine:


Force triangle

R2 400 lb = sin104.04° sin 51.92° R2 = 493 lb

Force triangle

P 493 lb = sin(14.04° + 24.04°) sin 75.96° P = 313.4 N

P = 313 lb




www.elsolucionario.net PROBLEM 8.53 Two 10° wedges of negligible weight are used to move and position the 400-lb block. Knowing that the coefficient of static friction is 0.25 at all surfaces of contact, determine the smallest force P that should be applied as shown to one of the wedges.

SOLUTION Free body: Block

φs = tan 0.25 = 14.04°

Law of sines:

Free body: Wedge

Law of sines:


Force triangle

R2 400 lb = sin104.04° sin 61.92° R2 = 439.8 lb

Force triangle

P 439.8 lb = sin(24.04° + 14.04°) sin 65.96° P = 297.0 lb

P = 297 lb




www.elsolucionario.net PROBLEM 8.54 Block A supports a pipe column and rests as shown on wedge B. Knowing that the coefficient of static friction at all surfaces of contact is 0.25 and that θ = 45°, determine the smallest force P required to raise block A.

SOLUTION

φs = tan −1 μs = tan −1 0.25 = 14.036°


FBD block A:

R2 3 kN = sin 104.036° sin 16.928° R2 = 10.0000 kN

FBD wedge B:

P 10.0000 kN = sin 73.072° sin 75.964° P = 9.8611 kN

P = 9.86 kN




www.elsolucionario.net PROBLEM 8.55 Block A supports a pipe column and rests as shown on wedge B. Knowing that the coefficient of static friction at all surfaces of contact is 0.25 and that θ = 45°, determine the smallest force P for which equilibrium is maintained.

SOLUTION

φs = tan −1 μs = tan −1 0.25 = 14.036°


FBD block A:

R2 3 kN = sin(75.964°) sin(73.072°) R2 = 3.0420 kN

FBD wedge B:

P 3.0420 kN = sin 16.928° sin 104.036° P = 0.91300 kN

P = 913 N




www.elsolucionario.net PROBLEM 8.56 Block A supports a pipe column and rests as shown on wedge B. The coefficient of static friction at all surfaces of contact is 0.25. If P = 0, determine (a) the angle θ for which sliding is impending, (b) the corresponding force exerted on the block by the vertical wall.

SOLUTION Free body: Wedge B (a)

φs = tan −1 0.25 = 14.04°

Since wedge is a two-force body, R 2 and R 3 must be equal and opposite. Therefore, they form equal angles with vertical

and

θ − φs = φs θ = 2φs = 2(14.04°) θ = 28.1°  Free body: Block A

R1 = (3 kN)sin 14.04° = 0.7278 kN

(b)

R1 = 728 N

Force exerted by wall:

14.04° 



β = φs

www.elsolucionario.net PROBLEM 8.57 A wedge A of negligible weight is to be driven between two 100-lb plates B and C. The coefficient of static friction between all surfaces of contact is 0.35. Determine the magnitude of the force P required to start moving the wedge (a) if the plates are equally free to move, (b) if plate C is securely bolted to the surface.

SOLUTION (a)

With plates equally free to move Free body: Plate B

φs = tan −1 μs = tan −1 0.35 = 19.2900°


Force triangle:

α = 180° − 124.29° − 19.29° = 36.42° Law of sines:

R1 100 lb = sin 19.29° sin 36.42° R1 = 55.643 lb

Free body: Wedge A Force triangle:

By symmetry,

R3 = R1 = 55.643 lb

β = 19.29° + 15° = 34.29°

(b)

Then

P = 2 R1 sin β

or

P = 2(55.643) sin 34.29°

P = 62.7 lb 

With plate C bolted The free body diagrams of plate B and wedge A (the only members to move) are same as above. Answer is thus the same. P = 62.7 lb 


www.elsolucionario.net PROBLEM 8.58 A 10° wedge is used to split a section of a log. The coefficient of static friction between the wedge and the log is 0.35. Knowing that a force P of magnitude 600 lb was required to insert the wedge, determine the magnitude of the forces exerted on the wood by the wedge after insertion.

SOLUTION

φs = tan −1 μs = tan −1 0.35 = 19.29°

By symmetry:

R1 = R2 ΣFy = 0: 2 R1 sin(5° + φs ) − 600 lb = 0

or

R1 = R2 =

300 lb = 729.30 lb sin (5°+19.29°)

When P is removed, the vertical components of R1 and R2 vanish, leaving the horizontal components R1x = R2 x = R1 cos(5° + φs ) = (729.30 lb) cos (5° + 19.29°)

R1x = R2 x = 665 lb 

(Note that φs > 5°, so wedge is self-locking.)



FBD wedge (impending motion ):

www.elsolucionario.net PROBLEM 8.59 A 10° wedge is to be forced under end B of the 5-kg rod AB. Knowing that the coefficient of static friction is 0.40 between the wedge and the rod and 0.20 between the wedge and the floor, determine the smallest force P required to raise end B of the rod.

SOLUTION FBD AB: W = mg W = (5 kg)(9.81 m/s 2 ) W = 49.050 N


φs1 = tan −1 ( μ s )1 = tan −1 0.40 = 21.801° ΣM A = 0: rR1 cos(10° + 21.801°) − rR1 sin(10° + 21.801°) −

2r

π

(49.050 N) = 0

R1 = 96.678 N

FBD wedge:

φs 2 = tan −1 ( μ s ) 2 = tan −1 0.20 = 11.3099 P 96.678 N = sin(43.111°) sin 78.690°

P = 67.4 N




www.elsolucionario.net PROBLEM 8.60 The spring of the door latch has a constant of 1.8 lb/in. and in the position shown exerts a 0.6-lb force on the bolt. The coefficient of static friction between the bolt and the strike plate is 0.40; all other surfaces are well lubricated and may be assumed frictionless. Determine the magnitude of the force P required to start closing the door.

SOLUTION Free body: Bolt

μs = 0.40 φs = tan −1 0.40

Force triangle:

From force triangle, P = (0.6 lb) tan 66.801° P = 1.39997 lb

P = 1.400 lb 



= 21.801°

www.elsolucionario.net ––

PROBLEM 8.61 In Problem 8.60, determine the angle that the face of the bolt should form with the line BC if the force P required to close the door is to be the same for both the position shown and the position when B is almost at the strike plate. PROBLEM 8.60 The spring of the door latch has a constant of 1.8 lb/in. and in the position shown exerts a 0.6-lb force on the bolt. The coefficient of static friction between the bolt and the strike plate is 0.40; all other surfaces are well lubricated and may be assumed frictionless. Determine the magnitude of the force P required to start closing the door.

SOLUTION For position shown in figure: P = 1.400 lb


From Prob. 8.60:

For position when B reaches strike plate: Free body: Bolt

μs = 0.40 φs = tan −1 0.40 = 21.801° F = 0.6 lb + k x 3  = 0.6 lb + (1.8 lb/in.)  in.  8  = 1.275 lb

Force triangle: From force triangle, 1.400 lb = 1.09804 1.275 lb α + φs = 47.675°

tan(α − φs ) =

α = 47.675° − 21.801° = 25.874° θ = 180° − 90° − α = 90° − α = 90° − 25.874°

θ = 64.1° 


www.elsolucionario.net PROBLEM 8.62 A 5° wedge is to be forced under a 1400-lb machine base at A. Knowing that the coefficient of static friction at all surfaces is 0.20, (a) determine the force P required to move the wedge, (b) indicate whether the machine base will move.

SOLUTION Free body: Machine base ΣM B = 0: (1400 lb)(20 in) − Ay (70 in) = 0 Ay = 400 lb ΣFy = 0: Ay + By − 1400 lb = 0

Free body: Wedge (Assume machine will not move)

μs = 0.20, φs = tan 0.20 = 11.31° We know that

Ay = 400 lb

Force triangle:

P = (400 lb)(tan11.31° + tan16.31°) = 197.0 lb

(a) (b)

P = 197.0 lb



Total maximum friction force at A and B: Fm = μcW = 0.20(1400 lb) = 280 lb

Since

P < Fm :

Machine will not move 



By = 1400 − 400 = 1000 lb

www.elsolucionario.net PROBLEM 8.63 Solve Problem 8.62 assuming that the wedge is to be forced under the machine base at B instead of A. PROBLEM 8.62 A 5° wedge is to be forced under a 1400-lb machine base at A. Knowing that the coefficient of static friction at all surfaces is 0.20, (a) determine the force P required to move the wedge, (b) indicate whether the machine base will move.

SOLUTION See solution to Prob. 8.62 for F.B.D. of machine base and determination of B y = 1000 lb. (Assume machine will not move)

μs = 0.20

φs = tan −1 0.20 = 11.31°


Free body: Wedge

Force triangle:

P = (1000 lb)(tan 11.31° + tan16.31°) = 493 lb

Total maximum friction force at A and B: Fm = μsW = 0.20(1400 lb) = 280 lb < 493 lb

(b)

Since P > Fm ,

(a)

We then have

machine will move with wedge  P = Fm ,

P = 280 lb




www.elsolucionario.net PROBLEM 8.64 A 15° wedge is forced under a 50-kg pipe as shown. The coefficient of static friction at all surfaces is 0.20. (a) Show that slipping will occur between the pipe and the vertical wall. (b) Determine the force P required to move the wedge.

SOLUTION Free body: Pipe ΣM B = 0: Wr sin θ + FA r (1 + sin θ ) − N A r cos θ = 0

Assume slipping at A: NA cos θ − μs NA (1 + sin θ ) = W sin θ NA =

W sin θ cos θ − μs (1 + sin θ )

W sin 15° cos 15° − (0.20)(1 + sin 15°) = 0.36241W

NA =

ΣFx = 0: − FB − W sin θ − FA sin θ + N A cos θ = 0 FB = N A cos θ − μs N A sin θ − W sin θ FB = 0.36241W cos 15° − 0.20(0.36241W )sin 15° − W sin 15° FB = 0.072482W ΣFy = 0: N B − W cos θ − FA cos θ − NA sin θ = 0 N B = N A sin θ + μs N A cos θ + W cos θ N B = (0.36241W )sin 15° + 0.20(0.36241W ) cos 15° + W cos 15° N B = 1.12974W

Maximum available: (a) (b)

FB = μs N B = 0.22595W

We note that FB < Fmax

No slip at B  ΣFy = 0: N 2 − N B cos θ + FB sin θ = 0

Free body: Wedge

N 2 = N B cos θ − FB sin θ N 2 = (1.12974W ) cos15° − (0.07248W ) sin 15° N 2 = 1.07249W



FA = μs NA

www.elsolucionario.net PROBLEM 8.64 (Continued) ΣFx = 0: FB cos θ + N B sin θ + μs N 2 − P = 0 P = FB cos θ + N B sin θ + μ s N 2 P = (0.07248W ) cos 15° + (1.12974W ) sin 15° + 0.2(1.07249W ) P = 0.5769W P = 283 N




W = mg : P = 0.5769(50 kg)(9.81 m/s 2 )


www.elsolucionario.net PROBLEM 8.65 A 15° wedge is forced under a 50-kg pipe as shown. Knowing that the coefficient of static friction at both surfaces of the wedge is 0.20, determine the largest coefficient of static friction between the pipe and the vertical wall for which slipping will occur at A.

SOLUTION Free body: Pipe ΣM A = 0: N B r cos θ − μ B N B r − ( μ B N B sin θ )r − Wr = 0

NB =

W cos θ − μ B (1 + sin θ )

W cos15° − 0.2(1 + sin15°) N B = 1.4002W


NB =

ΣFx = 0: N A − N B sin θ − μ B N B cos θ = 0 N A = N B (sin θ + μ B cos θ ) = (1.4002W )(sin15° + 0.2 × cos15°) N A = 0.63293W ΣFy = 0: − FA − W + N B cos θ − μ B N B sin θ = 0 FA = N B (cos θ − μ B sin θ ) − W FA = (1.4002W )(cos15° − 0.2 × sin θ ) − W FA = 0.28001W

For slipping at A:

FA = μ A N A

μA =

FA 0.28001W = N A 0.63293W

μ A = 0.442 


www.elsolucionario.net PROBLEM 8.66* A 200-N block rests as shown on a wedge of negligible weight. The coefficient of static friction μs is the same at both surfaces of the wedge, and friction between the block and the vertical wall may be neglected. For P = 100 N, determine the value of μs for which motion is impending. (Hint: Solve the equation obtained by trial and error.)

SOLUTION Free body: Wedge

Law of sines:

R2 P = sin(90° − φs ) sin(15° + 2φs ) R2 = P

sin(90° − φs ) sin(15° + 2φs )

(1)

Free body: Block ΣFy = 0

Vertical component of R2 is 200 N Return to force triangle of wedge. Note P = 100 N 100 N = (200 N) tan φ + (200 N) tan(15° + φs ) 0.5 = tan φ + tan(15° + φs )

Solve by trial and error

φs = 6.292 μs = tan φs = tan 6.292°

μs = 0.1103 



Force triangle:

www.elsolucionario.net PROBLEM 8.67* Solve Problem 8.66 assuming that the rollers are removed and that μs is the coefficient of friction at all surfaces of contact. PROBLEM 8.66* A 200-N block rests as shown on a wedge of negligible weight. The coefficient of static friction μs is the same at both surfaces of the wedge, and friction between the block and the vertical wall may be neglected. For P = 100 N, determine the value of μs for which motion is impending. (Hint: Solve the equation obtained by trial and error.)

SOLUTION Free body: Wedge

Law of sines:

R2 P = sin (90° − φs ) sin (15° + 2φs ) R2 = P

sin (90° − φs ) sin (15° + 2φs )

(1)

Free body: Block (Rollers removed) Force triangle:

Law of sines:

R2 W = sin (90° + φs ) sin (75° − 2φs ) R2 = W

sin (90° + φs ) sin (75° − 2θ s )

(2)



Force triangle:


Equate R2 from Eq. (1) and Eq. (2): P

sin (90° − φs ) sin (90° + φs ) =W sin (15° + 2φs ) sin (75° − 2φs ) P = 100 lb W = 200 N sin (90° + φs )sin (15° + 2φs ) 0.5 = sin (75° − 2φs )sin (90° − φs )

Solve by trial and error:

φs = 5.784° μs = 0.1013 


μs = tan φs = tan 5.784°


www.elsolucionario.net PROBLEM 8.68 Derive the following formulas relating the load W and the force P exerted on the handle of the jack discussed in Section 8.6. (a) P = (Wr/a) tan (θ + φs), to raise the load; (b) P = (Wr/a) tan (φs − θ ), to lower the load if the screw is self-locking; (c) P = (Wr/a) tan (θ − φs), to hold the load if the screw is not self-locking.

SOLUTION FBD jack handle: See Section 8.6.

ΣM C = 0: aP − rQ = 0 or

P=

r Q a

FBD block on incline: Raising load

Q = W tan (θ + φs )

(b)

Lowering load if screw is self-locking (i.e., if φs > θ )

Q = W tan (φs − θ )

(c)

r P = W tan (θ + φs )  a

r P = W tan (φs − θ )  a

Holding load is screw is not self-locking (i.e., if φs < θ )

Q = W tan (θ − φs )

r P = W tan (θ − φs )  a



(a)

www.elsolucionario.net PROBLEM 8.69 The square-threaded worm gear shown has a mean radius of 2 in. and a lead of 0.5 in. The large gear is subjected to a constant clockwise couple of 9.6 kip · in. Knowing that the coefficient of static friction between the two gears is 0.12, determine the couple that must be applied to shaft AB in order to rotate the large gear counterclockwise. Neglect friction in the bearings at A, B, and C.

SOLUTION Free body: Large gear ΣM C = 0: W (16 in.) − 9.6 kip ⋅ in. = 0


W = 0.6 kip = 600 lb

Block-and-Incline analysis of worn gear

0.5 in. = 0.039789 2π (2 in.) θ = 2.28°

tan θ =

μs = 0.12, φs = tan −1 0.12 = 6.84° Q = (600 lb) tan 9.12° = 96.32 lb

Torque = Qr = (96.32 lb)(2 in.) = 192.6 lb ⋅ in. Torque = 16.05 lb ⋅ ft 


www.elsolucionario.net PROBLEM 8.70 In Problem 8.69, determine the couple that must be applied to shaft AB in order to rotate the large gear clockwise. PROBLEM 8.69 The square-threaded worm gear shown has a mean radius of 2 in. and a lead of 0.5 in. The large gear is subjected to a constant clockwise couple of 9.6 kip ⋅ in. Knowing that the coefficient of static friction between the two gears is 0.12, determine the couple that must be applied to shaft AB in order to rotate the large gear counterclockwise. Neglect friction in the bearings at A, B, and C.

SOLUTION Free body: Large gear

Block-and-incline analysis of worm gear From Prob. 8.69 we have θ = 2.28° and φs = 6.84°. Since θ < φs , gear is self-looking and a torque must be applied to it to rotate large gear clockwise.

φs − θ = 6.84° − 2.28° = 4.56°

Q = (600 lb) tan 4.56° = 47.85 lb Torque = Qr = (47.85 lb)(2 in.) = 95.70 lb ⋅ in. Torque = 7.98 lb ⋅ ft 



See solution to Prob. 8.69. We find W = 600 lb

www.elsolucionario.net PROBLEM 8.71 High-strength bolts are used in the construction of many steel structures. For a 24-mmnominal-diameter bolt the required minimum bolt tension is 210 kN. Assuming the coefficient of friction to be 0.40, determine the required couple that should be applied to the bolt and nut. The mean diameter of the thread is 22.6 mm, and the lead is 3 mm. Neglect friction between the nut and washer, and assume the bolt to be square-threaded.

SOLUTION


FBD block on incline:

3 mm (22.6 mm)π = 2.4195°

θ = tan −1

φs = tan −1 μ s = tan −1 0.40 φs = 21.801° Q = (210 kN) tan (21.801° + 2.4195°) Q = 94.468 kN d Torque = Q 2 22.6 mm = (94.468 kN) 2 = 1067.49 N ⋅ m

Torque = 1068 N ⋅ m 


www.elsolucionario.net PROBLEM 8.72 The position of the automobile jack shown is controlled by a screw ABC that is single-threaded at each end (right-handed thread at A, left-handed thread at C). Each thread has a pitch of 0.1 in. and a mean diameter of 0.375 in. If the coefficient of static friction is 0.15, determine the magnitude of the couple M that must be applied to raise the automobile.

SOLUTION Free body: Parts A, D, C, E Two-force members Joint D: FAD = FCD

Symmetry:


ΣFy = 0: 2 FCD sin 25° − 800 lb = 0 FCD = 946.5 lb

Joint C: FCE = FCD

Symmetry:

ΣFx = 0: 2 FCD cos 25° − FAC = 0 FAC = 2(946.5 lb) cos 25° FAC = 1715.6 lb

Block-and-incline analysis of one screw:

0.1 in. π (0.375 in.) θ = 4.852°

tan θ =

φs = tan −1 0.15 = 8.531° Q = (1715.6 lb) tan13.383° Q = 408.2 lb

But, we have two screws:

 0.375 in.  Torque = 2Qr = 2(408.2 lb)   2  

Torque = 153.1 lb ⋅ in. 


www.elsolucionario.net PROBLEM 8.73 For the jack of Problem 8.72, determine the magnitude of the couple M that must be applied to lower the automobile. PROBLEM 8.72 The position of the automobile jack shown is controlled by a screw ABC that is single-threaded at each end (right-handed thread at A, left-handed thread at C). Each thread has a pitch of 0.1 in. and a mean diameter of 0.375 in. If the coefficient of static friction is 0.15, determine the magnitude of the couple M that must be applied to raise the automobile.

SOLUTION Free body: Parts A, D, C, E Two-force members


Joint D: FAD = FCD

Symmetry:

ΣFy = 0: 2 FCD sin 25° − 800 lb = 0 FCD = 946.5 lb

Joint C: FCE = FCD

Symmetry:

ΣFx = 0: 2 FCD cos 25° − FAC = 0 FAC = 2(946.5 lb) cos 25°; FAC = 1715.6 lb

Block-and-incline analysis of one screw: tan θ =

0.1 in.

π (0.375 in.)

θ = 4.852° φs = tan −1 0.15 = 8.531°

Since φs > θ , the screw is self-locking Q = (1715.6 lb) tan 3.679° Q = 110.3 lb

For two screws:

1 Torque = 2(110.3 lb) (0.375 in.) 2

Torque = 41.4 lb ⋅ in. 


www.elsolucionario.net PROBLEM 8.74 In the gear-pulling assembly shown the square-threaded screw AB has a mean radius of 15 mm and a lead of 4 mm. Knowing that the coefficient of static friction is 0.10, determine the couple that must be applied to the screw in order to produce a force of 3 kN on the gear. Neglect friction at end A of the screw.

SOLUTION Block/Incline: 4 mm 30π mm = 2.4302°

φs = tan −1 μs = tan −1 (0.10) = 5.7106° Q = (3000 N) tan (8.1408°) = 429.14 N

Couple = rQ = (0.015 m)(429.14 N) = 6.4371 N ⋅ m

M = 6.44 N ⋅ m 



θ = tan −1

www.elsolucionario.net PROBLEM 8.75 The ends of two fixed rods A and B are each made in the form of a single-threaded screw of mean radius 6 mm and pitch 2 mm. Rod A has a right-handed thread and rod B has a left-handed thread. The coefficient of static friction between the rods and the threaded sleeve is 0.12. Determine the magnitude of the couple that must be applied to the sleeve in order to draw the rods closer together.

SOLUTION To draw rods together: Screw at A 2 mm 2π (6 mm)

θ = 3.037° φs = tan −1 0.12 = 6.843° Q = (2 kN) tan 9.88° = 348.3 N Torque at A = Qr = (348.3 N)(0.006 m) = 2.0898 N ⋅ m

Total torque = 4.18 N ⋅ m 

Same torque required at B



tan θ =

www.elsolucionario.net PROBLEM 8.76 Assuming that in Problem 8.75 a right-handed thread is used on both rods A and B, determine the magnitude of the couple that must be applied to the sleeve in order to rotate it. PROBLEM 8.75 The ends of two fixed rods A and B are each made in the form of a single-threaded screw of mean radius 6 mm and pitch 2 mm. Rod A has a right-handed thread and rod B has a left-handed thread. The coefficient of static friction between the rods and the threaded sleeve is 0.12. Determine the magnitude of the couple that must be applied to the sleeve in order to draw the rods closer together.

SOLUTION

Torque at A = 2.09 N ⋅ m Screw at B: Loosening

θ = 3.037° φs = 6.843° Q = (2 kN) tan 3.806° = 133.1 N

Torque at B = Qr = (133.1 N)(0.006 m) = 0.79860 N ⋅ m

Total torque = 2.0848 N ⋅ m + 0.79860 N ⋅ m

Total torque = 2.89 N ⋅ m 



From the solution to Problem 8.70,

www.elsolucionario.net PROBLEM 8.77 A lever of negligible weight is loosely fitted onto a 30-mm-radius fixed shaft as shown. Knowing that a force P of magnitude 275 N will just start the lever rotating clockwise, determine (a) the coefficient of static friction between the shaft and the lever, (b) the smallest force P for which the lever does not start rotating counterclockwise.

SOLUTION (a)

Impending motion W = (40 kg)(9.81 m/s 2 ) = 392.4 N

160 P − 100W P +W (160 mm)(275 N) − (100 mm)(392.4 N) rf = 275 N + 392.4 N r f = 7.132 mm rf =

r f = r sin φs = r μ s

μs = (b)

Impending motion

rf r

=

7.132 mm = 0.2377 30 mm

μs = 0.238 

r f = r sin φs = r μ s = (30 mm)(0.2377) r f = 7.132 mm

ΣM D = 0: P(160 + rf ) − W (100 − rf ) = 0

P =W

100 − r f 160 + rf

P = (392.4 N) P = 218.04 N

100 mm − 7.132 mm 160 mm + 7.132 mm

P = 218 N 



ΣM D = 0: P(160 − rf ) − W (100 + rf ) = 0

www.elsolucionario.net PROBLEM 8.78 A hot-metal ladle and its contents weigh 130 kips. Knowing that the coefficient of static friction between the hooks and the pinion is 0.30, determine the tension in cable AB required to start tipping the ladle.

SOLUTION


Free body: Ladle

sin φs ≈ tan φs = μ s = 0.30 rbearing = 8 in. rf = rbearing sin φs = (8 in.) (0.30) = 2.4 in.

R is tangent to friction circle at A. ΣM A = 0 : T (64 in. + rf ) − (130 kips)rf = 0 T (64 + 2.4) − (130)(2.4) = 0 T = 4.70 kips 


www.elsolucionario.net PROBLEM 8.79 The double pulley shown is attached to a 10-mm-radius shaft that fits loosely in a fixed bearing. Knowing that the coefficient of static friction between the shaft and the poorly lubricated bearing is 0.40, determine the magnitude of the force P required to start raising the load.

SOLUTION ΣM D = 0: P(45 − r f ) − W (90 + rf ) = 0 90 + rf 45 − rf

= (196.2 N) P = 449.82 N

90 mm + 4 mm 45 mm − 4 mm P = 450 N 



P =W

www.elsolucionario.net PROBLEM 8.80 The double pulley shown is attached to a 10-mm-radius shaft that fits loosely in a fixed bearing. Knowing that the coefficient of static friction between the shaft and the poorly lubricated bearing is 0.40, determine the magnitude of the force P required to start raising the load.

SOLUTION Find P required to start raising load

P =W

90 − r f 45 − rf

= (196.2 N) P = 411.54 N

90 mm − 4 mm 45 mm − 4 mm P = 412 N 



ΣM D = 0: P(45 − r f ) − W (90 − rf ) = 0

www.elsolucionario.net PROBLEM 8.81 The double pulley shown is attached to a 10-mm-radius shaft that fits loosely in a fixed bearing. Knowing that the coefficient of static friction between the shaft and the poorly lubricated bearing is 0.40, determine the magnitude of the smallest force P required to maintain equilibrium.

SOLUTION Find smallest P to maintain equilibrium ΣM D = 0: P(45 + rf ) − W (90 − rf ) = 0 90 − rf 45 + rf

= (196.2 N) P = 344.35 N

90 mm − 4 mm 45 mm + 4 mm P = 344 N 



P =W

www.elsolucionario.net PROBLEM 8.82 The double pulley shown is attached to a 10-mm-radius shaft that fits loosely in a fixed bearing. Knowing that the coefficient of static friction between the shaft and the poorly lubricated bearing is 0.40, determine the magnitude of the smallest force P required to maintain equilibrium.

SOLUTION

ΣM D = 0: P(45 + rf ) − W (90 + r f ) = 0 P =W

90 + rf 45 + rf

= (196.2 N) P = 376.38 N

90 mm + 4 mm 45 mm + 4 mm P = 376 N 



Find smallest P to maintain equilibrium

www.elsolucionario.net PROBLEM 8.83 The block and tackle shown are used to raise a 150-lb load. Each of the 3-in.diameter pulleys rotates on a 0.5-in.-diameter axle. Knowing that the coefficient of static friction is 0.20, determine the tension in each portion of the rope as the load is slowly raised.

SOLUTION Axle diameter = 0.5 in.  0.5 in.  r f = r sin φs ≈ μs r = 0.20   = 0.05 in.  2 

Pulley BC:

ΣM B = 0: TCD (3 in.) − (150 lb)(1.5 in. + rf ) = 0 1 TCD = (150 lb)(1.5 in. + 0.05 in.) 3 ΣFy = 0: TAB + 77.5 lb − 150 lb = 0

Pulley DE:

TCD = 77.5 lb  TAB = 72.5 lb 

ΣM B = 0: TCD (1.5 + rf ) − TEF (1.5 − rf ) = 0 TEF = TCD

1.5 + rf 1.5 − rf

= (77.5 lb)

1.5 in. + 0.05 in. 1.5 in. − 0.05 in.

TEF = 82.8 lb 



For each pulley:

www.elsolucionario.net PROBLEM 8.84 The block and tackle shown are used to lower a 150-lb load. Each of the 3-in.diameter pulleys rotates on a 0.5-in.-diameter axle. Knowing that the coefficient of static friction is 0.20, determine the tension in each portion of the rope as the load is slowly lowered.

SOLUTION

Pulley BC:

 0.5 in.  rf = r μs =   0.2 = 0.05 in.  2 

ΣM B = 0: TCD (3 in.) − (150 lb)(1.5 in. − rf ) = 0 TCD =

(150 lb)(1.5 in. − 0.05 in.) 3 in.

ΣFy = 0: TAB + 72.5 lb − 150 lb = 0

Pulley DE:

TCD = 72.5 lb  TAB = 77.5 lb 

TCD (1.5 in. − rf ) − TEF (1.5 in. + rf ) = 0 TEF = TCD

1.5 in. − rf 1.5 in. + r f

= (72.5 lb)

1.5 in. − 0.05 in. 1.5 in. + 0.05 in.

TEF = 67.8 lb 



For each pulley:

www.elsolucionario.net PROBLEM 8.85 A scooter is to be designed to roll down a 2 percent slope at a constant speed. Assuming that the coefficient of kinetic friction between the 25-mm-diameter axles and the bearings is 0.10, determine the required diameter of the wheels. Neglect the rolling resistance between the wheels and the ground.

tan θ =

2 = 0.02 100

Since a scooter rolls at constant speed, each wheel is in equilibrium. Thus, W and R must have a common line of action tangent to the friction circle. rf = μ k r = (0.10)(12.5 mm) = 1.25 mm rf OB 1.25 mm = = tan θ tan θ 0.02 = 62.5 mm

OA =

Diameter of wheel = 2(OA) = 125.0 mm 



SOLUTION

www.elsolucionario.net PROBLEM 8.86 The link arrangement shown is frequently used in highway bridge construction to allow for expansion due to changes in temperature. At each of the 60-mmdiameter pins A and B the coefficient of static friction is 0.20. Knowing that the vertical component of the force exerted by BC on the link is 200 kN, determine (a) the horizontal force that should be exerted on beam BC to just move the link, (b) the angle that the resulting force exerted by beam BC on the link will form with the vertical.

SOLUTION Bearing:

= 0.20(30 mm) = 6 mm

Resultant forces R must be tangent to friction circles at Points C and D. (a) Ry = Vertical component = 200 kN Rx = Ry tan θ = (200 kN) tan 1.375° = 4.80 kN Horizontal force = 4.80 kN 

(b) 6 mm 250 mm sin θ = 0.024 sin θ =

θ = 1.375° 





r = 30 mm rf = μs r

www.elsolucionario.net PROBLEM 8.87 A lever AB of negligible weight is loosely fitted onto a 2.5-in.-diameter fixed shaft. Knowing that the coefficient of static friction between the fixed shaft and the lever is 0.15, determine the force P required to start the lever rotating counterclockwise.

SOLUTION rf = μs r

ΣM D = 0: P(5 in. + rf ) − (50 lb) rf = 0 50(0.1875) 5.1875 = 1.807 lb

P=

P = 1.807 lb 



= 0.15(1.25 in.) = 0.1875 in.

www.elsolucionario.net PROBLEM 8.88 A lever AB of negligible weight is loosely fitted onto a 2.5-in.-diameter fixed shaft. Knowing that the coefficient of static friction between the fixed shaft and the lever is 0.15, determine the force P required to start the lever rotating counterclockwise.

SOLUTION rf = μ s r


= 0.15(1.25 in.) rf = 0.1875 in. 2 in. 5 in . γ = 21.801°

tan γ =

In ΔEOD: OD = (2 in.) 2 + (5 in.) 2 = 5.3852 in. rf OE sin θ = = OD OD 0.1875 in. = 5.3852 in. θ = 1.99531°

Force triangle: P = (50 lb) tan (γ + θ ) = (50 lb) tan 23.796° = 22.049 lb

P = 22.0 lb




www.elsolucionario.net PROBLEM 8.89 A lever AB of negligible weight is loosely fitted onto a 2.5-in.-diameter fixed shaft. Knowing that the coefficient of static friction between the fixed shaft and the lever is 0.15, determine the force P required to start the lever rotating clockwise.

SOLUTION rf = μs r

ΣM D = 0: P(5 in. − rf ) − (50 lb)r f = 0 50(0.1875) 5 − 0.1875 = 1.948 lb

P=

P = 1.948 lb 



= 0.15(1.25 in.) r f = 0.1875 in.

www.elsolucionario.net PROBLEM 8.90 A lever AB of negligible weight is loosely fitted onto a 2.5-in.-diameter fixed shaft. Knowing that the coefficient of static friction between the fixed shaft and the lever is 0.15, determine the force P required to start the lever rotating clockwise.

SOLUTION rf = μ s r


= 0.15(1.25 in.) = 0.1875 in. 5 in. tan β = 2 in.

β = 68.198° In ΔEOD: OD = (2 in.) 2 + (5 in.) 2 OD = 5.3852 in. OE 0.1875 in. sin θ = = OD 5.3852 in. θ = 1.99531°

Force triangle:

P=

50 50 lb = tan ( β + θ ) tan 70.193°

P = 18.01 lb




www.elsolucionario.net PROBLEM 8.91 A loaded railroad car has a mass of 30 Mg and is supported by eight 800-mm-diameter wheels with 125-mmdiameter axles. Knowing that the coefficients of friction are μ s = 0.020 and μk = 0.015, determine the horizontal force required (a) to start the car moving, (b) to keep the car moving at a constant speed. Neglect rolling resistance between the wheels and the track.

SOLUTION rf = μ r ; R = 400 mm rf R

=

P = W tan θ = W

μr R

μr R

62.5 mm P =Wμ 400 mm = 0.15625Wμ

For one wheel:

For eight wheels of railroad car:

(a)

To start motion:

1 W = (30 mg)(9.81 m/s 2 ) 8 1 = (294.3 kN) 8

1 ΣP = 8(0.15625) (294.3 kN) μ 8 = (45.984μ ) kN

μs = 0.020 ΣP = (45.984)(0.020) = 0.9197 kN

(b)

To maintain motion:

ΣP = 920 N 

μk = 0.015 ΣP = (45.984)(0.015) = 0.6897 kN

ΣP = 690 N 



sin θ = tan θ =

www.elsolucionario.net PROBLEM 8.92 Knowing that a couple of magnitude 30 N ⋅ m is required to start the vertical shaft rotating, determine the coefficient of static friction between the annular surfaces of contact.

SOLUTION

M=

So,

30 N ⋅ m =

R3 − R13 2 μ s N 22 3 R2 − R12 2 (0.06 m)3 − (0.025 m)3 μ s (4000 N) 3 (0.06 m) 2 − (0.025 m) 2

μs = 0.1670 



For annular contact regions, use Equation 8.8 with impending slipping:

www.elsolucionario.net PROBLEM 8.93 A 50-lb electric floor polisher is operated on a surface for which the coefficient of kinetic friction is 0.25. Assuming that the normal force per unit area between the disk and the floor is uniformly distributed, determine the magnitude Q of the horizontal forces required to prevent motion of the machine.

SOLUTION See Figure 8.12 and Eq. (8.9). Using: R = 9 in. P = 50 lb

μk = 0.25 2 2 μk PR = (0.25)(50 lb)(9 in.) 3 3 = 75 lb ⋅ in.

M =

ΣM y = 0 yields:

M = Q(20 in.) 75 lb ⋅ in. = Q(20 in.) Q = 3.75 lb 



and

www.elsolucionario.net PROBLEM 8.94* The frictional resistance of a thrust bearing decreases as the shaft and bearing surfaces wear out. It is generally assumed that the wear is directly proportional to the distance traveled by any given point of the shaft and thus to the distance r from the point to the axis of the shaft. Assuming, then, that the normal force per unit area is inversely proportional to r, show that the magnitude M of the couple required to overcome the frictional resistance of a worn-out end bearing (with contact over the full circular area) is equal to 75 percent of the value given by Eq. (8.9) for a new bearing.

SOLUTION Using Figure 8.12, we assume k ΔA: ΔA = r Δθ Δ r r ΔN =

P = ΣΔN

We write

P= ΔN =

or

2π

  0

R 0

k r Δθ Δr = k Δθ Δr r


ΔN =



P = dN

k Δθ Δr = 2π Rk ; k =

PΔθ Δr 2π R

P 2π R

PΔθ Δr 2π R R μ P 2πμk P R 2 1 k rdrdθ = ⋅ = μk PR 0 2π R 2π R 2 2

ΔM = r ΔF = r μ k ΔN = r μk M=

2π

  0

From Eq. (8.9) for a new bearing,

Thus,

M new =

2 μk PR 3

M = M new

1 2 2 3

=

3 4

M = 0.75M new 


www.elsolucionario.net PROBLEM 8.95* Assuming that bearings wear out as indicated in Problem 8.94, show that the magnitude M of the couple required to overcome the frictional resistance of a worn-out collar bearing is M = 12 μk P ( R1 + R2 )

where P = magnitude of the total axial force R1, R2 = inner and outer radii of collar

SOLUTION ΔN k = . ΔA r

As in the text,

ΔF = μΔN

ΔM = r ΔF

The total normal force P is P = lim ΣΔN = ΔA→ 0

P = 2π

Total couple:



R2 R1

ΔA→ 0



  

  0

k    dθ rdr r 

R2 R1

kdr = 2π k ( R2 − R1 ) or k =

M worn = lim ΣΔM =

M worn = 2πμ k

2π

R2 R1

2π



   0

R2 R1

(rdr ) = πμ k

(

rμ

k  rdr  dθ r 

R22

−

R12

)=

P 2π ( R2 − R1 )

(

πμ P R22 − R12 2π ( R2 − R1 )

) M worn =

1 μ P( R2 + R1 )  2



Let normal force on ΔA be ΔN , and

www.elsolucionario.net PROBLEM 8.96* Assuming that the pressure between the surfaces of contact is uniform, show that the magnitude M of the couple required to overcome frictional resistance for the conical bearing shown is M=

2 μ k P R23 − R13 3 sin θ R22 − R12

SOLUTION

So

ΔN = k ΔA

ΔN = k. ΔA ΔA = r Δs Δφ

Δs =

Δr sin θ


Let normal force on ΔA be ΔN and

where φ is the azimuthal angle around the symmetry axis of rotation. ΔFy = ΔN sin θ = kr Δr Δφ

Total vertical force:

P = lim ΣΔFy ΔA→0

P=

2π

  

  0

R2 R1

 krdr  dφ = 2π k 

(

P = π k R22 − R12

Friction force: Moment: Total couple:

)

or k =



π

R2

rdr

R1

(

P R22

− R12

)

ΔF = μΔN = μ k ΔA ΔM = r ΔF = r μ kr M = lim ΣΔM = ΔA→0

M = 2π

μk sin θ



R2 R1

Δr Δφ sin θ 2π 

R2

0

R1

   

r 2 dr =

μk 2  r dr  dφ sin θ 

2 πμ P 3 sin θ π R22 − R32

(

)

(R

3 2

− R33

)

M=

2 μ P R23 − R13  3 sin θ R22 − R12


www.elsolucionario.net PROBLEM 8.97 Solve Problem 8.93 assuming that the normal force per unit area between the disk and the floor varies linearly from a maximum at the center to zero at the circumference of the disk. PROBLEM 8.93 A 50-lb electric floor polisher is operated on a surface for which the coefficient of kinetic friction is 0.25. Assuming that the normal force per unit area between the disk and the floor is uniformly distributed, determine the magnitude Q of the horizontal forces required to prevent motion of the machine.

r ΔN  = k 1 −  . R ΔA  

Let normal force on ΔA be ΔN and

r r   ΔF = μΔN = μ k 1 −  ΔA = μ k 1 −  r Δr Δθ R R   P = lim ΣΔN = ΔA→0

P = 2π k





0

R 0

 r  k 1 −  rdr  dθ R  

 R 2 R3  r rdr k 1 2 π − = −     R   2 3R 

R 0

1 P = π kR 2 3

or k =

3P π R2

 r  r μ k 1 −  rdr  dθ 0 0 R   3  R r = 2πμ k  r 2 −  dr 0 R 

M = lim Σr ΔF = ΔA→0

2π

 

2π



 

R



 R3 R 4  1 3 = 2πμ k  −  = πμ kR 3 4 6 R   πμ 3P 3 1 R = μ PR = 6 π R2 2

where

μ = μk = 0.25 R = 9 in.



SOLUTION


P = W = 50 lb 1 (0.25)(50 lb)(9 in.) 2 = 56.250 lb ⋅ in.

M=

Finally

Q=

M 56.250 lb ⋅ in. = d 20 in.

Q = 2.81 lb 


Then


www.elsolucionario.net PROBLEM 8.98 Determine the horizontal force required to move a 2500-lb automobile with 23-in.-diameter tires along a horizontal road at a constant speed. Neglect all forms of friction except rolling resistance, and assume the coefficient of rolling resistance to be 0.05 in.

SOLUTION

r = 11.5 in. b = 0.05 in. b θ = sin −1 r b  P = W tan θ = W tan  sin −1  for each wheel, so for total r  0.05   P = 2500 lb tan  sin −1 11.5  

P = 10.87 lb 



FBD wheel:

www.elsolucionario.net PROBLEM 8.99 Knowing that a 6-in.-diameter disk rolls at a constant velocity down a 2 percent incline, determine the coefficient of rolling resistance between the disk and the incline.

SOLUTION

tan θ = slope = 0.02 b = r tan θ = (3 in.)(0.02)

b = 0.0600 in. 



FBD disk:

www.elsolucionario.net PROBLEM 8.100 A 900-kg machine base is rolled along a concrete floor using a series of steel pipes with outside diameters of 100 mm. Knowing that the coefficient of rolling resistance is 0.5 mm between the pipes and the base and 1.25 mm between the pipes and the concrete floor, determine the magnitude of the force P required to slowly move the base along the floor.

SOLUTION


FBD pipe:

W = mg = (900 kg)(9.81 m/s 2 ) = 8829.0 N .5 mm + 1.25 mm θ = sin −1 100 mm = 1.00273° P = W tan θ for each pipe, so also for total P = (8829.0 N) tan (1.00273°)

P = 154.4 N 


www.elsolucionario.net PROBLEM 8.101 Solve Problem 8.85 including the effect of a coefficient of rolling resistance of 1.75 mm. PROBLEM 8.85 A scooter is to be designed to roll down a 2 percent slope at a constant speed. Assuming that the coefficient of kinetic friction between the 25-mm-diameter axles and the bearings is 0.10, determine the required diameter of the wheels. Neglect the rolling resistance between the wheels and the ground.

SOLUTION Since the scooter rolls at a constant speed, each wheel is in equilibrium. Thus, W and R must have a common line of action tangent to the friction circle. a = Radius of wheel 2 = 0.02 100

Since b and r f are small compared to a, tan θ ≈

Data:

rf + b a

=

μk r + b a

= 0.02

μk = 0.10, b = 1.75 mm, r = 12.5 mm (0.10)(12.5 mm) + 1.75 mm = 0.02 a a = 150 mm

Diameter = 2a = 300 mm 



tan θ =

www.elsolucionario.net PROBLEM 8.102 Solve Problem 8.91 including the effect of a coefficient of rolling resistance of 0.5 mm. PROBLEM 8.91 A loaded railroad car has a mass of 30 Mg and is supported by eight 800-mm-diameter wheels with 125-mm-diameter axles. Knowing that the coefficients of friction are μ s = 0.020 and μk = 0.015, determine the horizontal force required (a) to start the car moving, (b) to keep the car moving at a constant speed. Neglect rolling resistance between the wheels and the track.

SOLUTION For one wheel: rf = μ r tan θ ≈ sin θ ≈

a

μr + b

a W W μr + b Q = tan θ = a 8 8

For eight wheels of car:

P =W

μr + b

a W = mg = (30 Mg)(9.81 m/s 2 ) = 294.3 kN a = 400 mm, r = 62.5 mm, b = 0.5 mm

(a)

To start motion:

μ = μs = 0.02 P = (294.3 kN)

(0.020)(62.5 mm) + 0.5 mm 400 mm P = 1.288 kN 

(b)

To maintain constant speed

μ = μk = 0.015 P = (294.3 kN)

(0.015)(62.5 mm) + 0.5 mm 400 mm P = 1.058 kN 



tan θ =

rf + b

www.elsolucionario.net PROBLEM 8.103 A 300-lb block is supported by a rope that is wrapped 1 12 times around a horizontal rod. Knowing that the coefficient of static friction between the rope and the rod is 0.15, determine the range of values of P for which equilibrium is maintained.

β = 1.5 turns = 3π rad For impending motion of W up, P = We μs β = (300 lb)e(0.15)3π = 1233.36 lb

For impending motion of W down, P = We− μs β = (300 lb)e− (0.15)3π = 72.971 lb 73.0 lb ≤ P ≤ 1233 lb 

For equilibrium,



SOLUTION

www.elsolucionario.net PROBLEM 8.104 A hawser is wrapped two full turns around a bollard. By exerting an 80-lb force on the free end of the hawser, a dockworker can resist a force of 5000 lb on the other end of the hawser. Determine (a) the coefficient of static friction between the hawser and the bollard, (b) the number of times the hawser should be wrapped around the bollard if a 20,000-lb force is to be resisted by the same 80-lb force.

SOLUTION

β = 2 turns = 2(2π ) = 4π T1 = 80 lb, ln

T2 = 5000 lb

T2 = μs β T1

μs = μs =

1

β

ln

T2 1 5000 lb ln = 80 lb T1 4π

1 4.1351 ln 62.5 = 4π 4π

μs = 0.329 

T1 = 80 lb, T2 = 20,000 lb, μs = 0.329

(b) ln

T2 = μs β T1

β= β=

1

μ

ln

T2 1 20, 000 lb = ln T1 0.329 80 lb

1 5.5215 = 16.783 ln(250) = 0.329 0.329

Number of turns =

16.783 2π

Number of turns = 2.67 



(a)

www.elsolucionario.net PROBLEM 8.105 A rope ABCD is looped over two pipes as shown. Knowing that the coefficient of static friction is 0.25, determine (a) the smallest value of the mass m for which equilibrium is possible, (b) the corresponding tension in portion BC of the rope.

SOLUTION We apply Eq. (8.14) to pipe B and pipe C. T2 = e μs β T1 T2 = WA , T1 = TBC

μs = 0.25, β =

2π 3

WA = e0.25(2π /3) = eπ /6 TBC

Pipe C:

(1)

T2 = TBC , T1 = WD , μ s = 0.25, β =

π 3

TBC = e0.025(π /3) = eπ /12 WD

(a)

(2)

Multiplying Eq. (1) by Eq. (2): WA = eπ /6 ⋅ eπ /12 = eπ /6 + π /12 = eπ /4 = 2.193 WD WA

WA W mA 50 kg g WD = m= D = = = g 2.193 2.193 2.193 2.193 m = 22.8 kg 

(b)

TBC =

From Eq. (1):

WA

eπ /6

=

(50 kg)(9.81 m/s 2 ) = 291 N  1.688



Pipe B:

(8.14)

www.elsolucionario.net PROBLEM 8.106 A rope ABCD is looped over two pipes as shown. Knowing that the coefficient of static friction is 0.25, determine (a) the largest value of the mass m for which equilibrium is possible, (b) the corresponding tension in portion BC of the rope.

SOLUTION See FB diagrams of Problem 8.105. We apply Eq. (8.14) to pipes B and C. T1 = WA , T2 = TBC , μ s = 0.25, β =

2π 3

T T2 = e μs β : BC = e0.25(2π /3) = eπ /6 T1 WA

Pipe C:

T1 = TBC , T2 = WD , μ s = 0.25, β =

(1)

π 3

T2 WD = e μs β : = e0.25(π /3) = eπ /12 T1 TBC

(a)

(2)

Multiply Eq. (1) by Eq. (2): WD = eπ /6 ⋅ eπ /12 = eπ /6 + π /12 = eπ /4 = 2.193 WA W0 = 2.193WA

(b)

From Eq. (1):

m = 2.193mA = 2.193(50 kg)

m = 109.7 kg 

TBC = WAeπ /6 = (50 kg)(9.81 m/s 2 )(1.688) = 828 N 



Pipe B:

www.elsolucionario.net PROBLEM 8.107 Knowing that the coefficient of static friction is 0.25 between the rope and the horizontal pipe and 0.20 between the rope and the vertical pipe, determine the range of values of P for which equilibrium is maintained.

SOLUTION Horizontal pipe μs = μh

For motion of P to be impending downward: P = e μh (π /2) , Q

Q = e μv π , S

S = e μh (π /2) 400 N

Multiply equation member by member: P Q S = e( μh + 2 μv + μh )(π /2) Q S 400

or:

P = e ( μ h + μ v )π 400 N

(1)

For motion of P to be impending upward, we find in a similar way P = e − ( μ h + μv ) π 400 N

(2)

Given data:

μh = 0.25, μv = 0.20

From (1):

P = (400 N)e0.45π = 1644 N

From (2):

P = (400 N)e −0.45π = 97.3 N 97.3 N ≤ P ≤ 1644 N 

Range for equilibrium:



Vertical pipe μs = μv

www.elsolucionario.net PROBLEM 8.108 Knowing that the coefficient of static friction is 0.30 between the rope and the horizontal pipe and that the smallest value of P for which equilibrium is maintained is 80 N, determine (a) the largest value of P for which equilibrium is maintained, (b) the coefficient of static friction between the rope and the vertical pipe.

SOLUTION Horizontal pipe: μs = μh

For motion of P to be impending downward: P = e μh (π /2) , Q

Q = e μv π , S

S = e μh (π /2) 400 N

Multiply equation member by member: P Q S = e( μh + 2 μv + μh )(π /2) Q S 400 P = e ( μ h + μ v )π 400 N

or:

(1)

For motion of P to be impending upward, we find in a similar way P = e − ( μ h + μv ) π 400 N

(2)

Setting P = Pmax in Eq. (1), P = 80 N in Eq. (2) and μh = 0.30 in both, we get Pmax 80 N = e(0.30 + μv )π (1′); = e− (0.30 + μv )π (2′) 400 N 400 N

(a)

Multiplying (1′) by (2′): Pmax 80 ⋅ = e0 = 1 400 400

(b)

From (2′) : (0.30 + μv )π = ln

Pmax = 2000 N 

400 = ln 5 = 1.60944 80 0.30 + μv = 0.512

μv = 0.212 



Vertical pipe: μs = μv

www.elsolucionario.net PROBLEM 8.109 A band brake is used to control the speed of a flywheel as shown. The coefficients of friction are μs = 0.30 and μk = 0.25. Determine the magnitude of the couple being applied to the flywheel, knowing that P = 45 N and that the flywheel is rotating counterclockwise at a constant speed.

SOLUTION


Free body: Cylinder

Since slipping of band relative to cylinder is clockwise, T1 and T2 are located as shown. From free body: Lever ABC ΣM C = 0: (45 N)(0.48 m) − T2 (0.12 m) = 0 T2 = 180 N

Free body: Lever ABC

From free body: Cylinder Using Eq. (8.14) with μk = 0.25 and β = 270° =

3π rad: 2

T2 = e μs β = e(0.25)(3π / 2) = e3π /8 T1 T1 =

e

T2 3π /8

=

180 N = 55.415 N 3.2482

ΣM D = 0: (55.415 N)(0.36 m) − (180 N)(0.36 m) + M = 0

M = 44.9 N ⋅ m




www.elsolucionario.net PROBLEM 8.110 The setup shown is used to measure the output of a small turbine. When the flywheel is at rest, the reading of each spring scale is 14 lb. If a 105-lb · in. couple must be applied to the flywheel to keep it rotating clockwise at a constant speed, determine (a) the reading of each scale at that time, (b) the coefficient of kinetic friction. Assume that the length of the belt does not change.

SOLUTION Since the length of the belt is constant, the spring in scale B will increase in length by δ and the spring in scale A will decrease by the same amount. Thus, the sum of the readings in scales A and B remains constant: TA + TB = 14 lb + 14 lb

TA + TB = 28 lb (1)

On the other hand, the sum of the moments of TA and TB about axle must be equal to moment of couple: (TB − TA )(9.375 in.) = 105 lb ⋅ in.,

TB − TA = 11.2 lb (2)

Solving (1) and (2) simultaneously TA = 8.40 lb;

(b)

Apply Eq. (8.13) with T2 = TB , ln

T2 = μs β : T1

TB = 19.60 lb  T1 = TA , β = 180° = π rad.

μsπ = ln

19.60 = 0.84730 8.40

μs = 0.270 



(a)

www.elsolucionario.net PROBLEM 8.111 The setup shown is used to measure the output of a small turbine. The coefficient of kinetic friction is 0.20 and the reading of each spring scale is 16 lb when the flywheel is at rest. Determine (a) the reading of each scale when the flywheel is rotating clockwise at a constant speed, (b) the couple that must be applied to the flywheel. Assume that the length of the belt does not change.

SOLUTION Since the length of the belt is constant, the spring in scale B will increase in length by δ and the spring in scale A will decrease by the same amount. Thus, the sum of the readings in scales A and B remains constant: TA + TB = 16 lb + 16 lb

TA + TB = 32 lb (1)

We now apply Eq. (8.14) with T1 = TA , T2 = TB , μk = 0.20, β = 180° = π rad. T2 T = e μk B : B = e0.20π = 1.87446 T1 TA TB = 1.87446 TA (2)

Substituting (2) into (1): TA + 1.87446TA = 32 lb TA = 11.1325 lb

TA = 11.13 lb 

TB = 20.868 lb

TB = 20.9 lb 

From (1): TB = 32 lb − 11.1325 lb

(b)

Couple applied to flywheel: M = (TB − TA )r = (20.868 lb − 11.1325 lb)(9.375 in.) M = 91.3 lb ⋅ in.





(a)

www.elsolucionario.net PROBLEM 8.112 A flat belt is used to transmit a couple from drum B to drum A. Knowing that the coefficient of static friction is 0.40 and that the allowable belt tension is 450 N, determine the largest couple that can be exerted on drum A.

SOLUTION FBD’s drums:

π


7π 6 6 π 5π β B = 180° − 30° = π − = 6 6

β A = 180° + 30° = π +

=

Since β B < β A , slipping will impend first on B (friction coefficients being equal) So

T2 = Tmax = T1e μs β B 450 N = T1e(0.4)5π /6

or T1 = 157.914 N

ΣM A = 0: M A + (0.12 m)(T1 − T2 ) = 0 M A = (0.12 m)(450 N − 157.914 N) = 35.05 N ⋅ m M A = 35.1 N ⋅ m 


www.elsolucionario.net PROBLEM 8.113 A flat belt is used to transmit a couple from pulley A to pulley B. The radius of each pulley is 60 mm, and a force of magnitude P = 900 N is applied as shown to the axle of pulley A. Knowing that the coefficient of static friction is 0.35, determine (a) the largest couple that can be transmitted, (b) the corresponding maximum value of the tension in the belt.

SOLUTION Drum A: T2 = e μsπ = e(0.35)π T1

β = 180° = π radians (a)

Torque:

ΣM A = 0: M − (675.15 N)(0.06 m) + (224.84 N)(0.06 m) M = 27.0 N ⋅ m 

(b)

ΣFx = 0: T1 + T2 − 900 N = 0 T1 + 3.0028 T1 − 900 N = 0 4.00282 T1 = 900 T1 = 224.841 N T2 = 3.0028(224.841 N) = 675.15 N Tmax = 675 N 



T2 = 3.0028 T1

www.elsolucionario.net PROBLEM 8.114 Solve Problem 8.113 assuming that the belt is looped around the pulleys in a figure eight. PROBLEM 8.113 A flat belt is used to transmit a couple from pulley A to pulley B. The radius of each pulley is 60 mm, and a force of magnitude P = 900 N is applied as shown to the axle of pulley A. Knowing that the coefficient of static friction is 0.35, determine (a) the largest couple that can be transmitted, (b) the corresponding maximum value of the tension in the belt.

SOLUTION

β = 240° = 240°

60 1 = 120 2 θ = 30°

π

sin θ =

4 = π 180° 3 T2 = e μs β = e0.35(4/3π ) T1 T2 = 4.3322 T1

(a)

Torque:

ΣM B = 0: M − (844.3 N)(0.06 m) + (194.9 N)(0.06 m) = 0 M = 39.0 N ⋅ m 

(b)

ΣFx = 0: (T1 + T2 ) cos30° − 900 N (T1 + 4.3322 T1 ) cos 30° = 900 T1 = 194.90 N T2 = 4.3322(194.90 N) = 844.3 N Tmax = 844 N 



Drum A:

www.elsolucionario.net PROBLEM 8.115 The speed of the brake drum shown is controlled by a belt attached to the control bar AD. A force P of magnitude 25 lb is applied to the control bar at A. Determine the magnitude of the couple being applied to the drum, knowing that the coefficient of kinetic friction between the belt and the drum is 0.25, that a = 4 in., and that the drum is rotating at a constant speed (a) counterclockwise, (b) clockwise.

SOLUTION (a)

Counterclockwise rotation Free body: Drum


r = 8 in. β = 180° = π radians T2 = e μk β = e0.25π = 2.1933 T1 T2 = 2.1933T1

Free body: Control bar ΣM C = 0: T1 (12 in.) − T2 (4 in.) − (25 lb)(28 in.) = 0 T1 (12) − 2.1933T1 (4) − 700 = 0 T1 = 216.93 lb T2 = 2.1933(216.93 lb) = 475.80 lb

Return to free body of drum ΣM E = 0: M + T1 (8 in.) − T2 (8 in.) = 0 M + (216.96 lb)(8 in.) − (475.80 lb)(8 in.) = 0 M = 2070.9 lb ⋅ in.

(b)

M = 2070 lb ⋅ in. 

Clockwise rotation r = 8 in. β = π rad T2 = e μk β = e0.25π = 2.1933 T1 T2 = 2.1933T1


www.elsolucionario.net PROBLEM 8.115 (Continued) Free body: Control rod ΣM C = 0: T2 (12 in.) − T1 (4 in.) − (25 lb)(28 in.) = 0 2.1933T1 (12) − T1 (4) − 700 = 0 T1 = 31.363 lb T2 = 2.1933(31.363 lb) T2 = 68.788 lb

Return to free body of drum ΣM E = 0: M + T1 (8 in.) − T2 (8 in.) = 0 M + (31.363 lb)(8 in.) − (68.788 lb)(8 in.) = 0 M = 299 lb ⋅ in. 


M = 299.4 lb ⋅ in.


www.elsolucionario.net PROBLEM 8.116 The speed of the brake drum shown is controlled by a belt attached to the control bar AD. Knowing that a = 4 in., determine the maximum value of the coefficient of static friction for which the brake is not self-locking when the drum rotates counterclockwise.

SOLUTION r = 8 in., β = 180° = π radians T2 = e μ s β = e μsπ T1


T2 = e μsπ T1

Free body: Control rod

ΣM C = 0: P(28 in.) − T1 (12 in.) + T2 (4 in.) = 0 28 P − 12T1 + e μπ T1 (4) = 0 P=0

For self-locking brake: 12 T1 = 4 T1e μsπ e μs π = 3 μ sπ = ln 3 = 1.0986

μs =

1.0986

π

= 0.3497

μs = 0.350 


www.elsolucionario.net PROBLEM 8.117 The speed of the brake drum shown is controlled by a belt attached to the control bar AD. Knowing that the coefficient of static friction is 0.30 and that the brake drum is rotating counterclockwise, determine the minimum value of a for which the brake is not self-locking.

SOLUTION r = 8 in., β = π radians

T2 = 2.5663T1

Free body: Control rod

b = 16 in. − a ΣM C = 0: P(16 in. + b) − T1b + T2 a = 0

For brake to be self-locking,

P=0 T2 a = T1b ; 2.5663T1a = T1 (16 − a)

2.5663a = 16 − a 3.5663a = 16

a = 4.49 in. 



T2 = e μs β = e0.30π = 2.5663 T1

www.elsolucionario.net PROBLEM 8.118 Bucket A and block C are connected by a cable that passes over drum B. Knowing that drum B rotates slowly counterclockwise and that the coefficients of friction at all surfaces are μs = 0.35 and μk = 0.25, determine the smallest combined mass m of the bucket and its contents for which block C will (a) remain at rest, (b) start moving up the incline, (c) continue moving up the incline at a constant speed.

SOLUTION Free body: Drum 2

μ π T2 =e 3 mg

(a)

(1)


T2 = mge 2 μπ /3

Smallest m for block C to remain at rest Cable slips on drum. Eq. (1) with μk = 0.25; T2 = mge2(0.25)π /3 = 1.6881mg Block C: At rest, motion impending ΣF = 0: N − mC g cos 30° N = mC g cos 30° F = μ s N = 0.35 mC g cos 30° mC = 100 kg ΣF = 0: T2 + F − mC g sin 30° = C 1.6881 mg + 0.35mC g cos 30° − mC g sin 30° = 0 1.6881m = 0.19689mC m = 0.11663mC = 0.11663(100 kg);

(b)

m = 11.66 kg 

Smallest m to start block moving up No slipping at both drum and block:

μs = 0.35

Eq. (1):

T2 = mge 2(0.35)π /3 = 2.0814 mg



Block C: Motion impending mC = 100 kg ΣF = 0: N − mg cos 30° N = mC g cos30° F = μs N = 0.35mC g cos 30° ΣF = 0: T2 − F − mC g sin 30° = 0 2.0814mg − 0.35mC g cos 30° − mC g sin 30° = 0 2.0814m = 0.80311mC m = 38.6 kg 

(c)

Smallest m to keep block moving up drum: No slipping: μs = 0.35 Eq. (1) with μs = 0.35 T2 = mg 2 μsπ /3 = mge2(0.35)π /3 T2 = 2.0814mg

Block C: Moving up plane, thus μk = 0.25 Motion up ΣF = 0: N − mC g cos 30° = 0 N = mC g cos 30° F = μk N = 0.25 mC g cos 30° ΣF = 0: T2 − F − mC g sin 30° = 0 2.0814mg − 0.25mC g cos 30° − mC g sin 30° = 0 2.0814m = 0.71651mC m = 0.34424mC = 0.34424 (100 kg) m = 34.4 kg 



m = 0.38585mC = 0.38585(100 kg)

www.elsolucionario.net PROBLEM 8.119 Solve Problem 8.118 assuming that drum B is frozen and cannot rotate. PROBLEM 8.118 Bucket A and block C are connected by a cable that passes over drum B. Knowing that drum B rotates slowly counterclockwise and that the coefficients of friction at all surfaces are μs = 0.35 and μk = 0.25, determine the smallest combined mass m of the bucket and its contents for which block C will (a) remain at rest, (b) start moving up the incline, (c) continue moving up the incline at a constant speed.

SOLUTION Block C remains at rest: Motion impends T2 = e μk β = e0.35(2π /3) mg T2 = 2.0814mg

Drum:

Block C:

Motion impends ΣF = 0: N − mC g cos 30° = 0 N = mC g cos30° F = μs N = 0.35mC g cos 30° ΣF = 0: T2 + F − mC g sin 30° = 0 2.0814mg + 0.35mC g cos 30° − mC g sin 30° = 0 2.0814mg = 0.19689mC m = 0.09459mC = 0.09459(100 kg)

(b)

Block C: Starts moving up

m = 9.46 kg 

μs = 0.35

Drum: Impending motion of cable T2 = e μs β T1 mg = e0.35(2/3π ) T1 mg 2.0814 = 0.48045mg

T1 =



(a)


Block C: Motion impends ΣF = 0: N − mC g cos 30° N = mC g cos30° F = μs N = 0.35mC g cos 30° ΣF = 0: T1 − F − mC g sin 30° = 0 0.48045mg − 0.35mC g cos 30° − 0.5mC g = 0 0.48045m = 0.80311mC m = 1.67158mC = 1.67158(100 kg)

(c)

m = 167.2 kg 

Smallest m to keep block moving Drum: Motion of cable

μk = 0.25


T2 = e μk β = e0.25(2/3π ) T1 mg = 1.6881 T1 T1 =

mg = 0.59238mg 1.6881

Block C: Block moves ΣF = 0: N − mC g cos 30° = 0 N = mC g cos30° F = μk N = 0.25mC g cos 30° ΣF = 0: T1 − F − mC g sin 30° = 0 0.59238mg − 0.25mC g cos 30° − 0.5mC g = 0 0.59238m = 0.71651mC m = 1.20954mC = 1.20954(100 kg)

m = 121.0 kg 


www.elsolucionario.net PROBLEM 8.120 A cable is placed around three parallel pipes. Knowing that the coefficients of friction are μs = 0.25 and μk = 0.20, determine (a) the smallest weight W for which equilibrium is maintained, (b) the largest weight W that can be raised if pipe B is slowly rotated counterclockwise while pipes A and C remain fixed.

SOLUTION

μ = μ s = 0.25 at all pipes.

TAC = e0.25π TBC

50 lb = e0.25π /2 TAB

TBC = e0.25π / 2 W

50 lb TAB TBC ⋅ ⋅ = eπ /8 ⋅ eπ /4 ⋅ eπ /8 = eπ /8+π / 4+π /8 = eπ / 2 = 4.8105 TAB TBC W 50 lb = 4.8015; W = 10.394 lb W

(b)

Pipe B rotated

β=

π 2

; μ = μk

50 lb = e0.2π /2 TAB

β = π ; μ = μs

TBC = e0.25π TAB

W = 10.39 lb 

β=

π 2

; μ = μk

TBC = e0.2π / 2 W

50 lb TAB TBC ⋅ ⋅ = eπ /10 ⋅ e −π /4 ⋅ eπ /10 TAB TBC W = eπ /10 −π /4 +π /10 = e −π / 20 = 0.85464 50 lb = 0.85464 W 50 lb W= = 58.504 lb 0.85464

W = 58.5 lb 



(a)

www.elsolucionario.net PROBLEM 8.121 A cable is placed around three parallel pipes. Two of the pipes are fixed and do not rotate; the third pipe is slowly rotated. Knowing that the coefficients of friction are μs = 0.25 and μk = 0.20, determine the largest weight W that can be raised (a) if only pipe A is rotated counterclockwise, (b) if only pipe C is rotated clockwise.

SOLUTION Pipe A rotates

β=

π 2

; μ = μs

β = π , μ = μk

TAB = e0.25π /2 50 lb

β=

π 2

; μ = μk

TBC = e0.2π /2 W

TAB = e0.2π TBC

TAB TBC W ⋅ ⋅ = eπ /8 ⋅ e−π /5 ⋅ e−π /10 50 lb TAB TBC = eπ (1/8−1/5−1/10) = e −7π / 40 = 0.57708 W = 0.57708; W = 28.854 lb 50 lb

(b)

Pipe C rotates

β=

π

; μ = μk

β = π ; μ = μk

50 lb = e0.2π / 2 TAB

TAB = e0.2π TBC

2

β=

π 2

W = 28.9 lb 

, μ = μs

W = e0.25π /2 TBC

50 lb TAB TBC ⋅ ⋅ = eπ /10 ⋅ eπ /5 ⋅ e −π /8 = e7π /40 = 0.57708 TAB TBC W 50 lb = 0.57708 W W = 28.854 lb

W = 28.9 lb 



(a)

www.elsolucionario.net PROBLEM 8.122 A cable is placed around three parallel pipes. Knowing that the coefficients of friction are μs = 0.25 and μk = 0.20, determine (a) the smallest weight W for which equilibrium is maintained, (b) the largest weight W that can be raised if pipe B is slowly rotated counterclockwise while pipes A and C remain fixed.

SOLUTION Smallest W for equilibrium

B = π , μ = μs

TAC = e0.25π 50 lb

TBC = e0.25π TAC

W = e0.25π TBC

TAC TBC W ⋅ ⋅ = eπ /4 ⋅ eπ / 4 ⋅ eπ / 4 = e3π / 4 = 10.551 50 lb TAC TBC W = 10.551; 50 lb

(b)

W = 4.739 lb

W = 4.74 lb 

Largest W which can be raised by pipe B rotated

β = π , μ = μk

β = π , μ = μk

β = π , μ = μs

50 lb = e0.2π TAC

TAC = e0.2π TBC

W = e0.25π TBC

50 lb TAC TBC ⋅ ⋅ = eπ /5 ⋅ eπ /5 ⋅ e−π /4 = eπ (1/5+1/5−1/4) TAC TBC W = e3π / 20 = 1.602 50 lb = 1.602; W

W=

50 lb = 31.21 lb 1.602

W = 31.2 lb 



(a)

www.elsolucionario.net PROBLEM 8.123 A cable is placed around three parallel pipes. Two of the pipes are fixed and do not rotate; the third pipe is slowly rotated. Knowing that the coefficients of friction are μs = 0.25 and μk = 0.20, determine the largest weight W that can be raised (a) if only pipe A is rotated counterclockwise, (b) if only pipe C is rotated clockwise.

SOLUTION Pipe A rotates

β = π , μ = μs

β = π , μ = μk

TAC = e0.25π 50 lb

TAC = e0.2π TBC

β = π , μ = μk

TBC = e0.2π W

TAC TBC W ⋅ ⋅ = eπ /4 ⋅ e−π /5 ⋅ e−π /5 50 lb TAC TBC = eπ (1/ 4−1/5−1/5) = e−3π / 20 = 0.62423 W = 0.62423; W = 31.21 lb 50 lb

(b)

Pipe C rotates

β = π , μ = μk

50 lb = e0.2π TAC

β = π , μ = μs

TBC = e0.25π TAC

W = 31.2 lb 

β = π , μ = μk

TBC = e0.2π W

50 lb TAC TBC ⋅ ⋅ = eπ /5 ⋅ e −π / 4 ⋅ eπ /5 = eπ (1/5−1/4 +1/5) = e3π / 20 TAC TBC W 50 lb = e3π /20 = 1.602 W 50 lb W= = 31.21 lb 1.602

W = 31.2 lb 



(a)

www.elsolucionario.net PROBLEM 8.124 A recording tape passes over the 20-mm-radius drive drum B and under the idler drum C. Knowing that the coefficients of friction between the tape and the drums are μs = 0.40 and μk = 0.30 and that drum C is free to rotate, determine the smallest allowable value of P if slipping of the tape on drum B is not to occur.

SOLUTION


FBD drive drum:

ΣM B = 0: r (TA − T ) − M = 0 TA − T =

Impending slipping:

M 300 N ⋅ mm = = 15.0000 N r 20 mm

TA = Te μs β = Te0.4π

So

T (e0.4π − 1) = 15.0000 N

or

T = 5.9676 N

If C is free to rotate, P = T

P = 5.97 N 


www.elsolucionario.net PROBLEM 8.125 Solve Problem 8.124 assuming that the idler drum C is frozen and cannot rotate. PROBLEM 8.124 A recording tape passes over the 20-mm-radius drive drum B and under the idler drum C. Knowing that the coefficients of friction between the tape and the drums are μs = 0.40 and μk = 0.30 and that drum C is free to rotate, determine the smallest allowable value of P if slipping of the tape on drum B is not to occur.

SOLUTION


FBD drive drum:

ΣM B = 0: r (TA − T ) − M = 0 TA − T =

Impending slipping:

M = 300 N ⋅ mm = 15.0000 N r

TA = Te μs β = Te0.4π

So

(e0.4π − 1)T = 15.000 N

or

T = 5.9676 N

If C is fixed, the tape must slip So

P = Te μk βC = (5.9676 N)e0.3π / 2 = 9.5600 N

P = 9.56 N 


www.elsolucionario.net PROBLEM 8.126 The strap wrench shown is used to grip the pipe firmly without marring the external surface of the pipe. Knowing that the coefficient of static friction is the same for all surfaces of contact, determine the smallest value of μs for which the wrench will be self-locking when a = 200 mm, r = 30 mm, and θ = 65°.

For wrench to be self-locking ( P = 0), the value of μs must prevent slipping of strap which is in contact with the pipe from Point A to Point B and must be large enough so that at Point A the strap tension can increase from zero to the minimum tension required to develop “belt friction” between strap and pipe. Free body: Wrench handle

Geometry

In ΔCDH:

On wrench handle

a tan θ a CD = sin θ DE = BH = CH − BC a DE = −r tan θ a AD = CD − CA = −r sin θ

CH =

ΣM D = 0: TB ( DE ) − F ( AD ) = 0 a −r TB AD sin θ = = a F DE −r tan θ

(1)



SOLUTION


Free body: Strap at Point A

ΣF = 0: T1 − 2 F = 0 T1 = 2 F

(2)

Pipe and strap

μs β = ln

Eq. (8.13):

μs =

1

β

T2 T1 ln

TB 2F

(3)

Return to free body of wrench handle ΣFx = 0: N sin θ + F cos θ − TB = 0 T N sin θ = B − cos θ F F

Since F = μ s N , we have or

1

μs

sin θ =

μs =

TB − cos θ F sin θ TB − cos θ F

(4)

(Note: For a given set of data, we seek the larger of the values of μs from Eqs. (3) and (4).) For

Eq. (1):

a = 200 mm, r = 30 mm, θ = 65° 200 mm − 30 mm TB sin 65 ° = 200 mm F − 30 mm tan 65° 190.676 mm = = 3.0141 63.262 mm

β = 2π − θ = 2π − 65°

π 180°

= 5.1487 radians



β = (2π − θ ) radians


Eq. (3):

1 3.0141 ln 5.1487 rad 2 0.41015 = 5.1487

μs =

= 0.0797

Eq. (4):



sin 65° 3.0141 − cos 65° 0.90631 = 2.1595

μs =

= 0.3497



μs = 0.350 


We choose the larger value:


www.elsolucionario.net PROBLEM 8.127 Solve Problem 8.126 assuming that θ = 75°. PROBLEM 8.126 The strap wrench shown is used to grip the pipe firmly without marring the external surface of the pipe. Knowing that the coefficient of static friction is the same for all surfaces of contact, determine the smallest value of μs for which the wrench will be self-locking when a = 200 mm, r = 30 mm, and θ = 65°.

For wrench to be self-locking ( P = 0), the value of μs must prevent slipping of strap which is in contact with the pipe from Point A to Point B and must be large enough so that at Point A the strap tension can increase from zero to the minimum tension required to develop “belt friction” between strap and pipe. Free body: Wrench handle

Geometry

In ΔCDH:

On wrench handle

a tan θ a CD = sin θ DE = BH = CH − BC a −r DE = tan θ a −r AD = CD − CA = sin θ

CH =

ΣM D = 0: TB ( DE ) − F ( AD ) = 0

a −r TB AD sin θ = = a F DE −r tan θ

(1)



SOLUTION


Free body: Strap at Point A

ΣF = 0: T1 − 2 F = 0

T1 = 2 F

(2)

Pipe and strap

μs β = ln

Eq. (8.13):

μs =

1

β

T2 T1 ln

TB 2F

(3)

Return to free body of wrench handle ΣFx = 0: N sin θ + F cos θ − TB = 0

T N sin θ = B − cos θ F F

Since F = μ s N , we have or

1

μs

sin θ =

μs =

TB − cos θ F sin θ TB − cos θ F

(4)

(Note: For a given set of data, we seek the larger of the values of μs from Eqs. (3) and (4).) For

Eq. (1):

a = 200 mm, r = 30 mm, θ = 75° 200 mm − 30 mm TB sin 75 ° = 200 mm F − 30 mm tan 75° 177.055 mm = = 7.5056 23.590 mm

β = 2π − θ = 2π − 75°

π 180°

= 4.9742



β = (2π − θ ) radians


Eq. (3):

1 7.5056 ln 4.9742 rad 2 1.3225 = 4.9742

μs =

= 0.2659

Eq. (4):



sin 75° 7.5056 − cos 75° 0.96953 = 7.2468

μs =

= 0.1333



μs = 0.266 


We choose the larger value:


www.elsolucionario.net PROBLEM 8.128 The 10-lb bar AE is suspended by a cable that passes over a 5-in.-radius drum. Vertical motion of end E of the bar is prevented by the two stops shown. Knowing that μs = 0.30 between the cable and the drum, determine (a) the largest counterclockwise couple M 0 that can be applied to the drum if slipping is not to occur, (b) the corresponding force exerted on end E of the bar.

SOLUTION Drum: Slipping impends

μs = 0.30

TD = 2.5663TB

(a)

Free-body: Drum and bar ΣM C = 0: M 0 − E (8 in.) = 0 M 0 = (3.78649 lb)(8 in.) M 0 = 30.3 lb ⋅ in.

= 30.27 lb ⋅ in.

(b)

Bar AE:



ΣFy = 0: TB + TD − E − 10 lb = 0 TB + 2.5663TB − E − 10 lb = 0 3.5663TB − E − 10 lb = 0 E = 3.5663TB − 10 lb

(1)

ΣM D = 0: E (3 in.) − (10 lb)(5 in.) + TB (10 in.) = 0 (3.5663TB − 10 lb)(3 in.) − 50 lb ⋅ in. + TB (10 in.) = 0 20.699TB = 80 TB = 3.8649 lb

Eq. (1):

E = 3.5663(3.8649 lb) − 10 lb E = +3.78347 lb

E = 3.78 lb 



T2 T = e μβ : D = e0.30π = 2.5663 T1 TB

www.elsolucionario.net PROBLEM 8.129 Solve Problem 8.128 assuming that a clockwise couple M 0 is applied to the drum. PROBLEM 8.128 The 10-lb bar AE is suspended by a cable that passes over a 5-in.-radius drum. Vertical motion of end E of the bar is prevented by the two stops shown. Knowing that μs = 0.30 between the cable and the drum, determine (a) the largest counterclockwise couple M 0 that can be applied to the drum if slipping is not to occur, (b) the corresponding force exerted on end E of the bar.

SOLUTION

μs = 0.30 T2 = e μβ T1 TB = e0.30π = 2.5663 TD TB = 2.5663TD

(a)

Free body: Drum and bar ΣM C = 0: M 0 − E (8 in.) = 0 M 0 = (2.1538 lb)(8 in.) M 0 = 17.23 lb ⋅ in.

(b)



Bar AE: ΣFy = 0: TB + TD + E − 10 lb = 0 = 2.5663TD + TD + E − 10 lb E = −3.5663TD + 10 lb

(1)

ΣM B = 0: TD (10 in.) − (10 lb)(5 in.) + E (13 in.) = 0 TD (10 in.) − 50 lb ⋅ in. + ( − 3.5663TD + 10 lb)(13 in.) = 0 −36.362TD + 80 lb ⋅ in. = 0; TD = 2.200 lb E = −3.5633(2.200 lb) + 10 lb

Eq. (1):

E = + 2.1538 lb

E = 2.15 lb 



Drum: Slipping impends

www.elsolucionario.net PROBLEM 8.130 Prove that Eqs. (8.13) and (8.14) are valid for any shape of surface provided that the coefficient of friction is the same at all points of contact.

SOLUTION ΣFn = 0: ΔN − [T + (T + ΔT )]sin ΔN = (2T + ΔT ) sin ΣFt = 0: [(T + ΔT ) − T ]cos ΔF = ΔT cos

or

ΔT cos

So

So and

Δθ − ΔF = 0 2

Δθ 2

ΔF = μ s ΔN

Impending slipping:

In limit as

Δθ 2

Δθ

Δθ Δθ sin Δθ = μs 2T sin + μs ΔT 2 2 2

0: dT = μsTdθ



T2 T1

ln

dT = T



β 0

or

dT = μ s dθ T

μ s dθ

T2 = μs β T1

or T2 = T1e μs β 

(Note: Nothing above depends on the shape of the surface, except it is assumed to be a smooth curve.)



or

Δθ =0 2

www.elsolucionario.net PROBLEM 8.131 Complete the derivation of Eq. (8.15), which relates the tension in both parts of a V belt.

SOLUTION Small belt section: Side view:

End view:

ΔN α Δθ =0 sin − [T + (T + ΔT )]sin 2 2 2

ΣFx = 0: [(T + ΔT ) − T ]cos ΔF = μ s ΔN  ΔT cos

Impending slipping:

In limit as

Δθ

0: dT =

μ sTdθ α sin

so

or



T2 T1

2

μs dT = α T sin 2

ln

or



β 0


ΣFy = 0: 2

Δθ − ΔF = 0 2

Δθ 2T + ΔT Δθ = μs sin α 2 2 sin 2

μs dT = dθ α T sin 2

dθ

μβ T2 = s T1 sin α 2 T2 = T1e

or

μ s β / sin α2




www.elsolucionario.net PROBLEM 8.132 Solve Problem 8.112 assuming that the flat belt and drums are replaced by a V belt and V pulleys with α = 36°. (The angle α is as shown in Figure 8.15a.) PROBLEM 8.112 A flat belt is used to transmit a couple from drum B to drum A. Knowing that the coefficient of static friction is 0.40 and that the allowable belt tension is 450 N, determine the largest couple that can be exerted on drum A.

SOLUTION Since β is smaller for pulley B. The belt will slip first at B.


 π rad  5  = π rad  180°  6

β = 150° 

T2 μ β / sin α2 =e s T1

450 N (0.4) 5 π / sin18° =e 6 = e3.389 T1 450 N = 29.63; T1 = 15.187 N T1

Torque on pulley A:

ΣM B = 0: M − (Tmax − T1 )(0.12 m) = 0 M − (450 N − 15.187 N)(0.12 m) = 0 M = 52.18 N ⋅ m

M = 52.2 N ⋅ m 


www.elsolucionario.net PROBLEM 8.133 Solve Problem 8.113 assuming that the flat belt and pulleys are replaced by a V belt and V pulleys with α = 36°. (The angle α is as shown in Figure 8.15a.) PROBLEM 8.113 A flat belt is used to transmit a couple from pulley A to pulley B. The radius of each pulley is 60 mm, and a force of magnitude P = 900 N is applied as shown to the axle of pulley A. Knowing that the coefficient of static friction is 0.35, determine (a) the largest couple that can be transmitted, (b) the corresponding maximum value of the tension in the belt.

SOLUTION Pulley A:

β = π rad


T2 μ β / sin α2 =e s T1 T2 = e0.35π / sin18° T1 T2 = e3.558 = 35.1 T1 T2 = 35.1T1 ΣFx = 0: T1 + T2 − 900 N = 0 T1 + 35.1T1 − 900 N = 0 T1 = 24.93 N T2 = 35.1(24.93 N) = 875.03 N ΣM A = 0: M − T2 (0.06 m) + T1 (0.06 m) = 0 M − (875.03 N)(0.06 m) + (24.93 N)(0.06 m) = 0 M = 51.0 N ⋅ m  Tmax = T2

Tmax = 875 N 



SOLUTION Assume equilibrium: ΣFy = 0: N − (800 N) cos 25° + (200 N) sin 10° = 0 N = 690.3 N

ΣFx = 0: − F + (800 N) sin 25° − (200 N) cos 10° = 0 F = 141.13 N

F = 141.13 N

Maximum friction force: Fm = μs N = (0.20)(690.3 N) = 138.06 N

Since F > Fm ,

Block moves down



Friction force: F = μk N



= (0.15)(690.3 N) = 103.547 N 

F = 103.5 N





N = 690.3 N

www.elsolucionario.net PROBLEM 8.135 Three 4-kg packages A, B, and C are placed on a conveyor belt that is at rest. Between the belt and both packages A and C the coefficients of friction are μs = 0.30 and μk = 0.20; between package B and the belt the coefficients are μs = 0.10 and μk = 0.08. The packages are placed on the belt so that they are in contact with each other and at rest. Determine which, if any, of the packages will move and the friction force acting on each package.

SOLUTION Consider C by itself: Assume equilibrium ΣFy = 0: N C − W cos 15° = 0

ΣFx = 0: FC − W sin 15° = 0 FC = W sin 15° = 0.259W

But

Fm = μ s NC = 0.30(0.966W ) = 0.290W

Thus, FC < Fm

Package C does not move  FC = 0.259W = 0.259(4 kg)(9.81 m/s 2 ) = 10.16 N FC = 10.16 N



Consider B by itself: Assume equilibrium. We find, FB = 0.259W N B = 0.966W

But

Fm = μ s N B = 0.10(0.966W ) = 0.0966W

Thus, FB > Fm .

Package B would move if alone 



N C = W cos 15° = 0.966W


Consider A and B together: Assume equilibrium FA = FB = 0.259W N A = N B = 0.966W FA + FB = 2(0.259W ) = 0.518W ( FA ) m + ( FB ) m = 0.3 N A + 0.1N B = 0.386W

Thus,

FA + FB > ( FA ) m + ( FB ) m



FA = μk N A = 0.2(0.966)(4)(9.81) 





A and B move 

FA = 7.58 N



FB = 3.03 N




FB = μk N B = 0.08(0.966)(4)(9.81)


www.elsolucionario.net PROBLEM 8.136 The cylinder shown is of weight W and radius r. Express in terms W and r the magnitude of the largest couple M that can be applied to the cylinder if it is not to rotate, assuming the coefficient of static friction to be (a) zero at A and 0.30 at B, (b) 0.25 at A and 0.30 at B.

SOLUTION FBD cylinder: For maximum M, motion impends at both A and B FA = μ A N A FB = μ B N B ΣFx = 0: N A − FB = 0


N A = FB = μ B N B FA = μ A N A = μ A μ B N B ΣFy = 0: N B + FA − W = 0 N B (1 + μ A μ B ) = W 1

or

NB =

and

FB = μ B N B =

1 + μ A μB

W

μB W 1 + μ AμB μ A μB FA = μ A μ B N B = W 1 + μ A μB ΣM C = 0: M − r ( FA + FB ) = 0 M = Wr μ B

(a)

1 + μA 1 + μ A μB

For μ A = 0 and μ B = 0.30: M = 0.300Wr 

(b)

For μ A = 0.25 and μ B = 0.30: M = 0.349Wr 


www.elsolucionario.net PROBLEM 8.137 End A of a slender, uniform rod of length L and weight W bears on a surface as shown, while end B is supported by a cord BC. Knowing that the coefficients of friction are μs = 0.40 and μk = 0.30, determine (a) the largest value of θ for which motion is impending, (b) the corresponding value of the tension in the cord.

SOLUTION Free-body diagram

Motion impends: tan φs = 0.4

φs = 21.80° (a)

Since BG = GA, it follows that BD = DC and AD bisects ∠BAC

θ 2

θ 2

+ φs = 90°

+ 21.8° = 90°

θ = 136.4° 

(b)

Force triangle (right triangle): T = W cos 21.8° T = 0.928W 



Three-force body. Line of action of R must pass through D, where T and R intersect.

www.elsolucionario.net PROBLEM 8.138 A worker slowly moves a 50-kg crate to the left along a loading dock by applying a force P at corner B as shown. Knowing that the crate starts to tip about the edge E of the loading dock when a = 200 mm, determine (a) the coefficient of kinetic friction between the crate and the loading dock, (b) the corresponding magnitude P of the force.

SOLUTION Free body: Crate Three-force body. Reaction E must pass through K where P and W intersect. Geometry:

JK = 0.9 m + HK = 1.06077 m 0.4 m tan φs = = 0.37708 1.06077 m

μs = tan φs = 0.377 

φs = 20.66° Force triangle: W = (50 kg)(9.81 m/s) = 490.5 N

(b) Law of sines:

P 490.5 N = sin 20.66° sin 84.34° P = 173.91 N

P = 173.9 N 



HK = (0.6 m) tan15° = 0.16077 m

(a)

www.elsolucionario.net PROBLEM 8.139 A window sash weighing 10 lb is normally supported by two 5-lb sash weights. Knowing that the window remains open after one sash cord has broken, determine the smallest possible value of the coefficient of static friction. (Assume that the sash is slightly smaller than the frame and will bind only at Points A and D.)

SOLUTION T = 5 lb ΣFx = 0: N A − N D = 0 N A = ND

Impending motion:

FA = μ s N A FD = μ s N D ΣM D = 0: (18 in.)W − (27 in.) NA − (36 in.) FA = 0

W = 10 lb

3 N A + 2μs N A 2 2W NA = 3 + 4μs W=

ΣFy = 0: FA − W + T + FD = 0 FA + FD = W − T =

Now

W 2

FA + FD = μ s ( N A + N D ) = 2μs N A

Then

W 2W = 2μs 2 3 + 4μs

μs = 0.750 

or



FBD window:

www.elsolucionario.net PROBLEM 8.140 The slender rod AB of length l = 600 mm is attached to a collar at B and rests on a small wheel located at a horizontal distance a = 80 mm from the vertical rod on which the collar slides. Knowing that the coefficient of static friction between the collar and the vertical rod is 0.25 and neglecting the radius of the wheel, determine the range of values of P for which equilibrium is maintained when Q = 100 N and θ = 30°.


SOLUTION For motion of collar at B impending upward: F = μs N

ΣM B = 0: Ql sin θ −

Ca =0 sin θ

l C = Q   sin 2 θ a l ΣFx = 0: N = C cos θ = Q   sin 2 θ cos θ a

ΣFy = 0: P + Q − C sin θ − μ s N = 0 l l P + Q − Q   sin 3 θ − μ s Q   sin 2 θ cos θ = 0 a a l  P = Q  sin 2 θ (sin θ − μs cos θ ) − 1 a 

Substitute data:

(1)

 600 mm 2  P = (100 N)  sin 30°(sin 30° − 0.25cos 30°) − 1  80 mm 

P = −46.84 N (P is directed ) P = −46.8 N 



For motion of collar, impending downward: F = μs N

In Eq. (1) we substitute − μ s for μ s . l  P = Q  sin 2 θ (sin θ + μ s cos θ ) − 1 a   600 mm   P = (100 N)  sin 2 30°(sin 30° + 0.25cos θ ) − 1  80 mm  P = + 34.34 N  −46.8 N ≤ P ≤ 34.3 N 


For equilibrium:


www.elsolucionario.net PROBLEM 8.141 The machine part ABC is supported by a frictionless hinge at B and a 10° wedge at C. Knowing that the coefficient of static friction is 0.20 at both surfaces of the wedge, determine (a) the force P required to move the wedge to the left, (b) the components of the corresponding reaction at B.

SOLUTION

μs = 0.20 φs = tan −1 μs = tan −1 0.20 = 11.3099°

10° + 11.3099° = 21.3099° ΣM B = 0: ( RC cos 21.3099°)(10) − (120 lb)(8) = 0 RC = 103.045 lb

Free body: Wedge

Force triangle:



Free body: ABC


Law of sines: ( R = 103.045 lb) P = C sin 32.6198° sin 78.690°

(a) (b)

P = 56.6 lb



B x = 82.6 lb



Returning to free body of ABC: ΣFx = 0: Bx + 120 − (103.045) sin 21.3099° = 0 Bx = −82.552 lb ΣFy = 0: By + (103.045) cos 21.3099° = 0 B y = 96.0 lb 


By = −96.000 lb


www.elsolucionario.net PROBLEM 8.142 A conical wedge is placed between two horizontal plates that are then slowly moved toward each other. Indicate what will happen to the wedge (a) if μs = 0.20, (b) if μs = 0.30.

SOLUTION

(a)

φs = tan −1 μs = tan −1 0.2 = 11.31°. As θ decrease, the minimum angle at the contact approaches 12.5° > φs = 11.31°, so the wedge will slide up and out from the slot. 

(b)

φs = tan −1 μs = tan −1 0.3 = 16.70°. As θ decreases, the angle at one contact reaches 16.7°. (At this time the angle at the other contact is 25° − 16.7° = 8.3° < φs ). The wedge binds in the slot. 



As the plates are moved, the angle θ will decrease.

www.elsolucionario.net PROBLEM 8.143 In the machinist’s vise shown, the movable jaw D is rigidly attached to the tongue AB that fits loosely into the fixed body of the vise. The screw is single-threaded into the fixed base and has a mean diameter of 0.75 in. and a pitch of 0.25 in. The coefficient of static friction is 0.25 between the threads and also between the tongue and the body. Neglecting bearing friction between the screw and the movable head, determine the couple that must be applied to the handle in order to produce a clamping force of 1 kip.

SOLUTION Free body: Jaw D and tongue AB P is due to elastic forces in clamped object. ΣFy = 0: N H − N J = 0 N J = N H = N

For final tightening, FH = FJ = μ s N = 0.25 N ΣFx = 0: W − P − 2(0.25 N) = 0 N = 2(W − P)

(1)

ΣM H = 0: P(3.75) − W (2) − N (3) + (0.25 N)(1.25) = 0 3.75P − 2W − 2.6875 N = 0

Substitute Eq. (1) into Eq. (2):

(2)

3.75 P − 2W − 2.6875[2(W − P)] = 0 7.375W = 9.125P = 9.125(1 kip) W = 1.23729 kips

Block-and-incline analysis of screw: tan φs = μ s = 0.25

φs = 14.0362° 0.25 in. π (0.75 in.) θ = 6.0566° θ + φs = 20.093° tan θ =

Q = (1.23729 kips) tan 20.093° = 0.45261 kip  0.75 in.  T = Qr = (452.61 lb)    2 

T = 169.7 lb ⋅ in. 



W is force exerted by screw.

www.elsolucionario.net PROBLEM 8.144 A lever of negligible weight is loosely fitted onto a 75-mmdiameter fixed shaft. It is observed that the lever will just start rotating if a 3-kg mass is added at C. Determine the coefficient of static friction between the shaft and the lever.

SOLUTION ΣM O = 0: WC (150) − WD (100) − Rr f = 0

WD = (30 kg)(9.81 m/s 2 ) R = WC + WD = (53 kg)(9.81)

Thus, after dividing by 9.81, 23(150) − 30(100) − 53 rf = 0 rf = 8.49 mm

But

μs ≈

rf r

=

8.49 mm 37.5 mm

μs ≈ 0.226 



WC = (23 kg)(9.81 m/s 2 )

But

www.elsolucionario.net PROBLEM 8.145 In the pivoted motor mount shown the weight W of the 175-lb motor is used to maintain tension in the drive belt. Knowing that the coefficient of static friction between the flat belt and drums A and B is 0.40, and neglecting the weight of platform CD, determine the largest couple that can be transmitted to drum B when the drive drum A is rotating clockwise.

SOLUTION


FBD motor and mount:

Impending belt slip: cw rotation T2 = T1e μs β = T1e0.40π = 3.5136T1 ΣM D = 0: (12 in.)(175 lb) − (7 in.)T2 − (13 in.)T1 = 0 2100 lb = [(7 in.)(3.5136) + 13 in.]T1 T1 = 55.858 lb, T2 = 3.5136 T1 = 196.263 lb

FBD drum at B: ΣM B = 0: M B − (3 in.)(196.263 lb − 55.858 lb) = 0

r = 3 in. M B = 421 lb ⋅ in. 


www.elsolucionario.net PROBLEM 8.F1 Draw the free-body diagram needed to determine the smallest force P for which equilibrium of the 7.5-kg block is maintained.

SOLUTION

W = (7.5 kg)(9.81 m/s 2 ) = 73.575 N

Free body: Block Since we seek smallest P, motion impends downward, with Fm = μ s N = 0.45 N



r = 3 in.

www.elsolucionario.net PROBLEM 8.F2 Two blocks A and B are connected by a cable as shown. Knowing that the coefficient of static friction at all surfaces of contact is 0.30 and neglecting the friction of the pulleys, draw the free-body diagrams needed to determine the smallest force P required to move the blocks.

SOLUTION

Free body: Block A Motion impends to left, with

Free body: Block B

Motion impends to left, with FB = μ s N B = 0.30 N B



FA = μ S N A = 0.30 N A

www.elsolucionario.net PROBLEM 8.F3 The cylinder shown is of weight W and radius r, and the coefficient of static friction μs is the same at A and B. Draw the free-body diagram needed to determine the largest couple M that can be applied to the cylinder if it is not to rotate.

SOLUTION

Free body: Cylinder For maximum M, motion impends at both A and B, with

FB = μ S N B

Free body: Block B



FA = μ s N A

www.elsolucionario.net PROBLEM 8.F4 A uniform crate of mass 30 kg must be moved up along the 15° incline without tipping. Knowing that the force P is horizontal, draw the free-body diagram needed to determine the largest allowable coefficient of static friction between the crate and the incline, and the corresponding force P.

SOLUTION

W = (30 kg) (9.81 m/s 2 ) = 294.3 N

For impending tip, N must act at C. For impending slip,

Fm = μ max N



Free body: Crate



CHAPTER 9





www.elsolucionario.net PROBLEM 9.1 Determine by direct integration the moment of inertia of the shaded area with respect to the y axis.

SOLUTION

x  dI y = x 2 dA = x 2  h1 + (h 2 − h1 )  dx a  a h 2 − h1 3  I y =  h1 x 2 + x  dx 0 a   3 4 h 2 − h1 a a = h1 + a 3 4 3 3 ba h a = 1 + 2 12 4



Iy =

a3 (h1 + 3h2 )  12



x y = h1 + (h 2 − h1 ) ; dA = ydx a


SOLUTION y = kx1/3

For x = a :

b = ka1/3

Thus:

y=

b 1/3 x a1/3

dI y = x 2 dA = x 2 ydx b 1/3 b x dx = 1/3 x 7/3 dx 1/3 a a a b b  3  Iy = dI y = 1/3 x7/3 dx = 1/3  a10/3  0 10 a a  

dIy = x 2





Iy =

3 3 a b  10



k = b /a1/3


SOLUTION  x x2  y = 4h  − 2  a a  dA = ydx   dx 



a

 x4  a3 a3  x5  I y = 4h  − 2  = 4h  −  5   4 a 5a  0  4

1 I y = ha3  5



 x x2 dI y = x 2 dA = 4hx 2  − 2 a a a  x3 x4  I y = 4h  − 2  dx 0  a a 


SOLUTION y = kx 4

Thus:

b = ka 4 k=

b a4

y=

b 4 x a4

dA = (b − y )dx dIy = x 2 da = x 2 (b − y )dx b   = x 2  b − 4 x 4  dx a  



Iy = dI y =



a 0

1 3 1 3 b 6 2  bx − 4 x  dx = 3 a b − 7 a b a  

I y = 4a3b /21 



For x = a :

www.elsolucionario.net PROBLEM 9.5 Determine by direct integration the moment of inertia of the shaded area with respect to the x axis.

SOLUTION y = h1 + (h 2 − h1 )

1 = 12 =

1 3



dI x =

3

a 0

1 3 y dx 3

x  h1 + (h 2 − h1 ) a  dx   a

x  a    h1 + (h 2 − h1 ) a   h − h     2 1 0 4

2 2 a a (h 2 + h1 )( h 2 + h1 )(h 2 − h1 ) h 42 − h14 = ⋅ h 2 − h1 12(h 2 − h1 ) 12

(

)

Ix =

a 2 ( h1 + h22 )(h1 + h 2 )  12





I x = dI y =

x a


SOLUTION y=

b 1/3 x a1/3 3

1 3 1 b 1 b3  y dx =  1/3 x1/3  dx = xdx 3 3 a 3 a 



I x = dI x =



1 b3 a 2 b3 xdx = 3 a 3 a 2

a1 0

Ix =

1 3 ab  6



dI x =


SOLUTION  x x2  y = 4h  − 2  a a 



I x = dI x =

64h3 3



3

   dx  

a 0

x3 x4 x5 x 6  3 − 3 4 + 3 5 − 6 a a a a

64h3 1 x 4 3 x5 1 x6 1 x7 = − + − 3 4 a 3 5 a 4 2 a5 7 a 6 =

  dx 

a

0

64h  1 3 1 1  64h  3  a − + −  = a  3 3 4 5 2 7  420  3

3

Ix =

16 3 ah  105



1 1   x x2 dI x = y 3 dx =  4h  − 2 3 3   a a


SOLUTION See figure of solution of Problem 9.4. b 4 x a4

1 1 1 1 b3 12 dI x = b3 dx − y 3 dx = b3 dx − x dx 3 3 3 3 a12 a 1 1 b3 12  I x = dI x =  b3 − x  dx 0 3 3 a12  





1 1 b3 a13 = b3 a − 3 3 a12 13 1 1  =  −  ab3  3 39  12 3  13 1  ab =  −  ab3 = 39 39 39   I x = 4ab3 /13 



y=


SOLUTION

Then

Now

Then

y1 = y2 = b

y1 :

b = ka 2

y2 :

b = ma or m = y1 =

b 2 x a2

y2 =

b x a

or k =

(

b a2 b a


x1 = a,

At

)

1 3 y2 − y13 dx 3 b3  1  b3 =  3 x3 − 6 x6  dx 3 a a 

dI x =



I x = dI x = =



a 0

b3  1 3 1 6  x − 6 x  dx 3  a3 a  a

1 b  1 4  x − 6 x7   3 3  4a 7a 0 3

or

Ix =

1 3 ab  28



SOLUTION y = kx n

For x = a :

b = ka n

Thus:

b n x an b y = n xn a y=

1 3 1 b3 3 n y dx = x dx 3 3 a3n b 3 a 3n b3 a 3n +1 Ix = dI x = 3n x dx = 3n 0 3a 3a (3n + 1)

dIx =





Ix =

ab3  3(3n + 1)



k = b /a n


At x = a, y = b :

b = ke a/a

Then

y=

Now

dI x =

Then

or k =

b e

b x/a e = be x/a −1 e

1 3 1 y dx = (be x/a −1 )3 dx 3 3 1 3 3( x/a −1) = be dx 3



I x = dI x =



a 0

a

1 3 3( x/a −1) b3  a  be dx =  e3( x/a −1)  3 3 3 0

1 = ab3 (1 − e−3 ) 9

or

I x = 0.1056ab3 



SOLUTION


SOLUTION

Then

Now

Then

y1 = y2 = b

y1 :

b = ka 2

y2 :

b = ma or m = y1 =

b 2 x a2

y2 =

b x a

or k =

b a2 b a

 b b   dI y = x 2 dA = x 2 [( y2 − y1 )dx = x 2  x − 2 x 2  dx  a    a



I y = dI y =



a 0

1 1  b  x3 − 2 x 4  dx a a  a

1 1  = b  x 4 − 2 x5  4 a 5 a  0

or

Iy =

1 3 ab  20



x1 = a,

At


SOLUTION y = kx n

For x = a :

b = ka n

Thus:

y=

b n x an

dI y = x 2 dA = x 2 ydx b n b x dx = n x n + 2 dx n a a b a n+ 2 b a n+3 Iy = dI y = n x dx = n a 0 a n+3

dIy = x 2





Iy =

a 3b  n+3



k = b /a n


SOLUTION y = b:

b = kea/a

or

k=

b e

Then

y=

b x/a e = be x/a −1 e


At x = a,

dI y = x 2 dA = x 2 ( ydx)

Now

= x 2 (be x /a −1dx)



I y = dI y =

Then



a

bx 2 e x/a −1dx

0

Now use integration by parts with

Then



a 0

u = x2

dv = e x /a −1dx

du = 2 xdx

v = ae x /a −1 a

x 2 e x /a −1dx =  x 2 ae x /a −1  − 0

= a 3 − 2a



a



a 0

( ae x /a −1 )2 xdx

xe x /a −1dx

0

Using integration by parts with

Then

u=x

dv = e x/a −1dx

du = dx

v = ae x/a −1

  I y = b  a3 − 2a ( xae x/a −1 ) |0a −  

{ = b {a



a 0

 ( ae x/a −1 ) dx   

} ) }

= b a3 − 2a  a 2 − ( a 2 e x/a −1 ) |0a  3

− 2a  a 2 − ( a 2 − a 2 e−1

or

I y = 0.264a3b 


www.elsolucionario.net PROBLEM 9.15 Determine the moment of inertia and the radius of gyration of the shaded area shown with respect to the x axis.

SOLUTION

For

y2 = k2 x1/ 2

x = 0 and

y1 = y2 = b

b = k1a 2

b = k2 a1/ 2

k1 =

b a2

y1 =

b 2 x a2

k2 =

b a

1/ 2


y1 = k1 x 2

Thus, y2 =

b 1/ 2 x a 1/2

dA = ( y2 − y1 )dx A=



a 0

b 1/2 b 2   a1/ 2 x − a 2 x  dx  

2 ba 3/2 ba 3 − 3 a1/2 3a 2 1 A = ab 3 A=

1 3 1 y2 dx − y13 dx 3 3 3 1 b 1 b3 6 x3/ 2 dx − x dx = 3/ 2 3a 3 a6

dI x =



b3 a 3/ 2 b3 a 6 x dx − x dx 3a 3/2 0 3a 6 0 1  b3 a 5/2 b3 a 7  2 = 3/ 2 5 − 6 =  −  ab3 7 15 21 3a   ( 2 ) 3a



I x = dI x =

k x2

I = x = A

(

3 35

ab3 ab b

)



Ix =

3 3 ab  35

kx = b

9  35



SOLUTION x2 y2 + =1 a 2 b2 x = a 1−

y2 b2


dA = xdy dI x = y 2 dA = y 2 xdy





I x = dI x =

b −b

y = b sin θ

Set:

Ix = a

π /2

π

− /2

= ab3



b −b

y2 1 −

y2 dy b2

dy = b cos θ dθ b 2 sin 2 θ 1 − sin 2 θ

π /2

π

− /2

1 = ab3 4

xy 2 dy = a

b cos θ dθ

sin 2 θ cos 2 θ dθ = ab3

π /2

π

− /2

1 2 sin 2θ dθ 4 π /2

1 1 1   (1 − cos 4θ )dθ = ab3 θ − sin 4θ  −π /2 2 8 4   −π /2



π /2

 π  π  π 1 = ab3  −  −   = ab 2 8  2  2  8

1 I x = π ab3  8

π /2 y2 = 1 − sin 2 θ b cos θ dθ dy a 2 /2 π −b −b − b π /2 π /2 1 cos 2 θ dθ = ab (1 + cos 2θ ) dθ = ab −π /2 −π /2 2



A = dA =



+b

xdy = a



b





1 = ab θ + sin 2θ 2 Ix =

k x2 A



1−

k x2

π /2

= −π /2

ab  π  π   1  −  −   = π ab 2  2  2  2

3 I x 81 π ab 1 = = 1 = b2 A π ab 4 2

kx =

1 b  2


www.elsolucionario.net PROBLEM 9.17 Determine the moment of inertia and the radius of gyration of the shaded area shown with respect to the y axis.

SOLUTION See figure of solution on Problem 9.15. 1 A = ab 3

b b  b  x 2  1/2 x1/ 2 − 2 x 2  dx = 1/2 a a a 

Iy =



Iy =

b b7/2 b a5  2 1  3 ⋅ − ⋅ =  − a b a1/ 2 ( 72 ) a 2 5  7 5 

k y2

=

0

Iy A

( =

3 35

a 3b

)



a 0

x5/ 2 dx −

b a2



a

x 4 dx

0

Iy =

3 3 ab  35

ky = a

ab 3

9  35



a

dI y = x 2 dA = x 2 ( y2 − y1 )dx


SOLUTION x2 y2 + =1 a 2 b2 x2 a2

y = b 1−


dA = 2 ydx dI y = x 2 dA = 2 x 2 ydx



I y = dI y =

Set:



x = a sin θ I y = 2b



0

=



1 3 ab 2

0

2 x 2 ydx = 2b



a 0

x2 1 −

a 2 sin 2 θ 1 − sin 2 θ

π /2 0



sin 2 θ cos 2 θ dθ = 2a3b

π /2 1 0

a cos θ dθ

2

(1 − cos 4θ )dθ =



π /2 0

1 2 sin 2θ dθ 4

1 3 1 a b θ − sin 4θ 4 4

1 π  π = a 3b  − 0  = a 3b 4 2  8

From solution of Problem 9.16:

x2 dx a2

dx = a cos θ dθ

π /2

= 2 a 3b

a

π /2 0

1 I y = π a 3b  8

1 A = π ab 2

Thus: I y = k y2 A k y2 =

Iy A

=

1 π a 3b 8 1 π ab 2

=

1 2 a 4

ky =

1 a  2



SOLUTION y1 : b = ka 2

or k =

y2 : b = 2b − ca 2

b a2

or c =

b a2

 x2 y2 = b  2 − 2 a 

b 2 x a2

  

Then

y1 =

Now

  x2 dA = ( y2 − y1 )dx = b  2 − 2 a  

Then

Now

Then



A = dA =



 b 2 2b 2 2  − 2 x  dx = 2 (a − x )dx a a   a

2b 2 2b  1  4 (a − x 2 )dx = 2  a 2 x − x3  = ab 2 3 0 3 a a 

a 0

 1  1   x2 1 dI x =  y23 − y13  dx =  b  2 − 2 3  3    a 3 

3    b 2 3    −  2 x   dx      a 

=

1 b3 (8a 6 − 12a 4 x 2 + 6a 2 x 4 − x6 − x 6 )dx 3 a6

=

2 b3 (4a 6 − 6a 4 x 2 + 3a 2 x 4 − x 6 ) dx 3 a6



I x = dI x =



a 0

2 b3 (4a 6 − 6a 4 x 2 + 3a 2 x 4 − x 6 )dx 3 a6 a

2b  6 3 1  4a x − 2a 4 x 3 + a 6 x 5 − x 7  3 a 6  5 7 0 172 3 = ab 105

=

and

k x2 =

3

Ix = A


At x = a1, y1 = y2 = b :

172 ab3 105 4 ab 3

=

or

I x = 1.638ab3 

or

k x = 1.108b 

43 2 b 35



SOLUTION y1 = y2 = b :

y1 : b = ka 2

or k =

y2 : b = 2b − ca 2

Then

Now

b a2

or c =

b 2 x a2  x2 y2 = b  2 − 2 a 

b a2

y1 =

  

dA = ( y2 − y1 ) dx    x2  b = b  2 − 2  − 2 x 2  dx a  a    2b = 2 (a 2 − x 2 ) dx a

Then



A = dA =



a

0

2b 2 (a − x 2 )dx a2 a

2b  2 1  a x − x3  2  3 0 a  4 = ab 3 =

Now

 2b  dI y = x 2 dA = x 2  2 ( a 2 − x 2 ) dx  a 



At x = a,


Then



I y = dI y =



a

0

2b 2 2 x (a − x 2 ) dx a2 a

=

2b  1 2 3 1 5  a x − x  5 0 a 2  3

or

and

k y2 =

Iy A

=

4 3 ab 15 4 ab 3

=

Iy =

4 3 ab  15

1 2 a 5 ky =

a 5




or


www.elsolucionario.net PROBLEM 9.21 Determine the polar moment of inertia and the polar radius of gyration of the shaded area shown with respect to Point P.

x=

a a y 1 + = (a + y ) 2 2a 2

dA = x dy =

1 A= 2



a 0

dA =



1 I x = y 2 dA = 2



a 0



1 1 y2 (a + y )dy = ay + 2 2 2 a



a 0

1 1 Iy = 2 24

a

= 0

1 3 1 x dy = 3 3



a 0

1 1 (a + y )dy = 2 2

y2

0

1 y3 y4 = a + 2 3 4 1 Iy = 2

1 (a + y )dy 2



a 0

= 0

1  2 a2  3 2  a +  = a 2  2  4

A=

Ix =

7 4 a  12

Iy =

5 4 a  16

JO =

43 4 a  48

3

1   2 (a + y )  dy   a

(a + y )3 dy =

1 1 1  15 4 4 4 (a + y ) 4 = (2a) − a  = 96 a 24 4 96 0

1 5 4 Iy = a 2 32

From Eq. (9.4):

3 2 a  2

(ay 2 + y 3 )dy

11 1 4 1 7 4 + a = a 2  3 4  2 12

a 0



a

JO = I x + I y = J O = kO2 A

7 4 5 4  28 + 15  4 a + a = a 12 16  48 

kO2 =

JO = A

43 4 a 48 3 2 a 2

=

43 2 a 72

kO = a

43 72

kO = 0.773a 



SOLUTION


SOLUTION

First note

Now

y=

b x a

dA = ( y + 2b) dx b = ( x + 2a )dx a 1 3 1 3  dI x =  ytop − ybottom  dx 3 3   3  1  b  =  x  − (−2b)3  dx 3  a  

=

Then

1 b3 3 ( x + 8a3 ) dx 3 a3



I x = dI x =



1 b3 2 ( x + 8a3 ) dx − a 3 a3 a

a

Also Then

=

1 b3  1 4  x + 8a 3 x  3  3 a 4 −a

=

1 b3 3 a3

 1 4  1   16 3 3 4 3   (a) + 8a (a )  −  (−a ) + 8a (−a )   = ab  4  3  4

b  dI y = x 2 dA = x 2  ( x + 2a ) dx  a  



I y = dI y =



a −a

b 2 x ( x + 2a) dx a a

=

b 1 4 2 3 x + ax  a  4 3  −a

=

b  1 4 2 2 4 3 3 1 4 3    ( a) + a (a)  −  ( −a ) + a( −a )   = a b a  4 3 3  4  3



By observation


Now

JP = Ix + Iy =

16 3 4 3 ab + a b 3 3 JP =

and

k P2 =

4 ab(a 2 + 4b 2 )  3

4 ab(a 2 + 4b 2 ) JP 3 = A (2a)(3b) − 12 (2a)(2b)

1 = (a 2 + 4b 2 ) 3

kP =

a 2 + 4b2  3


or



SOLUTION At x = 2a,

y = 2a : 2a = k1 (2a) 2

y2 :

At x = 0, At x = 2a,

y = a:

or k1 =

a=c

y = 2a :

2a = a + k2 (2a )2

Then

y1 =

1 2 x 2a

Then

Now

k2 =

or

y2 = a + =

Now

1 2a

1 4a

1 2 x 4a

1 (4a 2 + x 2 ) 4a

1 2 1 dA = ( y2 − y1 )dx =  (4a 2 + x 2 ) − x dx 2a   4a 1 = (4a 2 − x 2 ) dx 4a



A = dA = 2



2a 0

2a

1 1  2 1  8 (4a 2 − x 2 )dx = 4a x − x 3  = a 2  4a 2a  3 0 3

3 3 1  1   1 1  1   dI x =  y23 − y13  dx =   (4a 2 + x 2 )  −  x 2   dx 3  3   4a 3   2a  

1 1 1  (64a 6 + 48a 4 x 2 + 12a 2 x 4 + x6 ) − 3 x6  dx  3 3  64a 8a  1 (64a 6 + 48a 4 x 2 + 12a 2 x 4 − 7 x 6 )dx = 3 192a =



y1 :


Then



I x = dI x = 2



2a 0

1 (64a 6 + 48a 4 x 2 + 12a 2 x 4 − 7 x 6 )dx 192a3

1 = 96a3

2a

12 2 5  6 4 3 7 64a x + 16a x + 5 a x − x   0

12 2  6 4 3 5 7 64a (2a) + 16a (2a ) + 5 a (2a) − (2a )    1 4 12  32 4 = a 128 + 128 + × 32 − 128  = a 96  5  15 =

Then

1  dI y = x 2 dA = x 2  (4a 2 − x 2 )dx   4a 



I y = dI y = 2 =



2a 0

2a

1 2 1 4 2 3 1 5 x (4a 2 − x 2 )dx = a x − x  4a 2a  3 5 0

1 4 2 1  32 4  1 1  32 4 a (2a )3 − (2a )5  = a  − = a  2a  3 5  2  3 5  15

Now

JP = I x + I y =

and

k P2 =

JP = A

32 4 32 4 a + a 15 15

64 4 a 15 8 2 a 3

=

8 2 a 5

or

JP =

64 4 a  15

or k P = 1.265a 



Also

1 96a3


SOLUTION The equation of the circle is x2 + y 2 = r 2

Now

x = r 2 − y2 dA = xdy = r 2 − y 2 dy





r

Then

A = dA = 2

Let

y = r sin θ ; dy = r cos θ dθ

Then

A=2

π /2

π

− /6

− r/2

r 2 − y 2 dy

r 2 − (r sin θ ) 2 r cos dθ π /2

 θ sin 2θ  r cos θ dθ = 2r  + =2 −π /6 4  −π /6 2



π /2

2

2

2

 π  − π sin − π3 = 2r 2  2 −  6 + 4  2  2

  3 2 π    = 2r  + 8    3

= 2.5274r 2

Now

dI x = y 2 dA = y 2

Then

I x = dI x = 2

Let Then

Now





( r

− r/2

r 2 − y 2 dy

)

y 2 r 2 − y 2 dy

y = r sin θ ; dy = r cos θ dθ π /2

Ix = 2

π

=2

π

− /6

π /2

− /6

(r sin θ )2 r 2 − (r sin θ ) 2 r cos θ dθ r 2 sin 2 θ (r cos θ )r cos θ dθ

sin 2θ = 2sin θ cos θ  sin 2 θ cos 2 θ =

1 sin 2θ 4



So that


Ix = 2



π /2

r 4  θ sin 4θ  1  r 4  sin 2 2θ  dθ =  − −π /6 2 2 8  −π /6 4  π /2

 π2  π6 sin − 23π  −  − 8  2  2 3 r4  π =  −   2  3 16 

=

Also

dI y =

1 3 1 x dy = 3 3





(

r 2 − y2

3

1 2 (r − y 2 )3/2 dy − r/ 2 3 r

I y = dI y = 2

Let

y = r sin θ ; dy = r cos θ dθ

Now Then

Iy =

2 3

Iy =

2 3

π /2

π

− /6

π /2

π

− /6

    

) dy

Then

Then

r4 2

[r 2 − (r sin θ ) 2 ]3/ 2 r cos θ dθ ( r 3 cos3 θ )r cos θ dθ

1 cos 4 θ = cos 2 θ (1 − sin 2 θ ) = cos 2 θ − sin 2 2θ 4 Iy = =

2 3

π /2

π

− /6

1   r 4  cos 2 θ − sin 2 2θ  dθ 4  

2 4  θ sin 2θ r  + 3  2 4

π /2

 1  θ sin 4θ   −  −  8   −π / 6  4 2

2 4   π2 1  π2    − π6 sin − π3 1  − π6 sin − 23π + −  − r  −   −  3   2 4  2    2 4 4  2 8  2 π π π 1  3  π 1  3  = r4  − + +  +   −   3  4 16 12 4  2  48 32  2   =

=

        

2 4π 9 3  r  +  3  4 64 



Then


Now

JP = Ix + Iy =

r4 2

π 3  2 4 π 9 3   −  + r  +  64   3 16  3  4

π 3 = r4  + = 1.15545r 4  3 16   

k P2 =

J P 1.15545r 4 = A 2.5274r 2

or

J P = 1.155r 4  k P = 0.676r 


and

or


www.elsolucionario.net PROBLEM 9.25 (a) Determine by direct integration the polar moment of inertia of the semiannular area shown with respect to Point O. (b) Using the result of part a, determine the moments of inertia of the given area with respect to the x and y axes.

SOLUTION (a)

By definition dJ O = r 2 dA = r 2 (π rdr ) = π r 3 dr



J O = dJ O =



R2

R1

π r 3 dr

R

1  2 =  r4   4  R1

or

(b)

JO =

π 4

(R

4 2

− R14 

)

(R

4 2

− R14 

First note that symmetry implies ( I x )1 = ( I y )1

( I x )2 = ( I y )2

Also have I x = ( I x )1 + ( I x ) 2

and

I y = ( I y )1 + ( I y ) 2 = ( I x )1 + ( I x )2 = I x

Now

JO = I x + I y

Ix = I y

= 2I x ∴ Ix = I y =

1 JO 2

or

Ix = Iy =

π 8

)



Then

www.elsolucionario.net PROBLEM 9.26 (a) Show that the polar radius of gyration kO of the semiannular area shown is approximately equal to the mean radius Rm = (R1 + R2)/2 for small values of the thickness t = R2 − R1. (b) Determine the percentage error introduced by using Rm in place of kO for the following values of 1 1 t/Rm: 1, 2 , and 10 .

SOLUTION From the solution to Problem 9.25 have JO =

Now

kO2 =

π 4

(R

4 2

JO = A

)

− R14 π

4

π

2

(R (R

(

4 2 2 2

− R14 − R12

) = 1 (R ) 2

2 2

− R12 R22

)( R

2 2

−

+ R12

)

=

1 2 R2 + R12 2

Now

Rm =

1 ( R1 + R2 ) 2

Then

1 R1 = Rm − t 2

)

R11

(1) t = R2 − R1 1 R2 = Rm + t 2

Substituting into Eq. (1), 1  1   1   Rm + t  +  Rm − t  2  2   2  2

kO2 = =

2

 

1 2 1 2 1 2 2  Rm + Rm t + t + Rm − Rm t + t  2 4 4 

1 = Rm2 + t 2 4

 1 2 2  kO = Rm + t 4 

  

If t << R1 , R2 Then t << Rm So that

kO2 ≈ Rm2

or kO ≈ Rm 



(a)


(b)

The percentage error, % error, is given by % error = =

Rm − kO × 100% kO Rm − Rm2 + 14 t 2 Rm2 + 14 t 2

  1 =   1 + 12 

( ) t Rm

2

× 100%

  − 1 × 100%   

  t 1   = 1: % error =  − 1 × 100% 2 Rm  1 + ( 12 + 1)    or

% error = −10.56% 

  t 1 1   = : % error =  − 1 × 100% 2 Rm 2  1 + ( 12 × 12 )    or

% error = −2.99% 

  t 1  1  = : % error =  − 1 × 100% 2 Rm 10  1 + ( 12 × 101 )    or % error = −0.1248% 



Then, for

www.elsolucionario.net PROBLEM 9.27 Determine the polar moment of inertia and the polar radius of gyration of the shaded area shown with respect to Point O.

SOLUTION At θ = π , R = 2a :

2a = a + k (π ) a

or

k=

Then

R=a+

π a

π

 θ θ = a 1 +  π 





A = dA =

Then

A=



π 0

π

a (1+θ /π )

0

0

 

rdr dθ =



π 0

a (1+θ /π )

1 2  2 r   0


dA = (dr )(r dθ ) = rdr dθ

Now

dθ

π

2 3 1 2 θ  1 π  θ   a  1 +  dθ = a 2   1 +   2  π 2  3  π   0

 π 3  7 1 = π a 2 1 +  − (1)3  = π a 2 6  π   6

Now Then

dJ O = r 2 dA = r 2 (rdr dθ )



J O = dJ O = =



π 0

π

a (1+θ /π )

0

0

 

a (1+θ /π )

1 4  4 r   0

r 3 dr dθ

dθ =



π 0

4

1 4 θ  a  1 +  dθ 4  π

π

5  π 5  1 π  θ   1 = a 4  1 +   = π a 4 1 +  − (1)5  4  5  π   20  π   0

and

kO2 =

JO = A

31 π a4 20 7 π a2 6

=

93 2 a 70

31 4 πa  20

or

JO =

or

kO = 1.153a 


www.elsolucionario.net PROBLEM 9.28 Determine the polar moment of inertia and the polar radius of gyration of the isosceles triangle shown with respect to Point O.

SOLUTION By observation:

y=

h

or

x=

b y 2h

x

Now

 b dA = xdy =   2h

and

dI x = y 2 dA =

Then



 y  dy 

b 3 y dy 2h

I x = dI x = 2



h 0

b 3 y dy 2h

b y4 = h 4

From above: Now

y=

= 0

1 3 bh 4

2h x b

2h   dA = (h − y )dx =  h − x dx b   =

and

h

dI y = x 2 dA = x 2

h (b − 2 x)dx b

h (b − 2 x)dx b



b 2


Then



I y = dI y = 2



b/2 0

h 2 x (b − 2 x)dx b b/2

h 1 1  = 2  bx3 − x 4  b 3 2 0

3 4 h b  b  1  b   1 = 2    −    = b3 h b  3  2  2  2   48

Now

JO = I x + I y =

and

kO2 =

bh 48

(12h 2 + b 2 ) 1 2

bh

=

1 (12h 2 + b 2 ) 24

or

or

JO =

bh (12h 2 + b 2 )  48

kO =

12h 2 + b 2  24


JO = A

1 3 1 3 bh + b h 4 48


www.elsolucionario.net PROBLEM 9.29* Using the polar moment of inertia of the isosceles triangle of Problem 9.28, show that the centroidal polar moment of inertia of a circular area of radius r is π r 4 /2. (Hint: As a circular area is divided into an increasing number of equal circular sectors, what is the approximate shape of each circular sector?) PROBLEM 9.28 Determine the polar moment of inertia and the polar radius of gyration of the isosceles triangle shown with respect to Point O.

SOLUTION First the circular area is divided into an increasing number of identical circular sectors. The sectors can be approximated by isosceles triangles. For a large number of sectors the approximate dimensions of one of the isosceles triangles are as shown.

JO =

(Δ J O )sector  =

Now

bh (12h 2 + b 2 ) 48

b = r Δθ

Then with

dJ O sector dθ

and h = r

1 (r Δθ )(r )[12r 2 + (r Δθ ) 2 ] 48 1 4 r Δθ (12 + Δθ 2 )  48

 Δ J O sector  1 4 2  = lim   = lim  r [12 + (Δθ ) ] 0 Δθ →0 Δ θ →  48   Δθ  =

Then


For an isosceles triangle (see Problem 9.28):



( J O )circle = dJ O sector =



2π 0

1 4 r 4

1 4 1 2π r dθ = r 4 [θ ]0 4 4

or

( J O )circle =

π 2

r 4 


www.elsolucionario.net PROBLEM 9.30* Prove that the centroidal polar moment of inertia of a given area A cannot be smaller than A2 /2π . (Hint: Compare the moment of inertia of the given area with the moment of inertia of a circle that has the same area and the same centroid.)

SOLUTION From the solution to sample Problem 9.2, the centroidal polar moment of inertia of a circular area is ( J C )cir =

π 2

r4

The area of the circle is

So that

[ J C ( A)]cir =

A2 2π

Two methods of solution will be presented. However, both methods depend upon the observation that as a given element of area dA is moved closer to some Point C. The value of J C will be decreased ( J C =  r 2 dA; as r decreases, so must J C ). Solution 1

Imagine taking the area A and drawing it into a thin strip of negligible width and of sufficient length so that its area is equal to A. To minimize the value of ( J C ) A, the area would have to be distributed as closely as possible about C. This is accomplished by winding the strip into a tightly wound roll with C as its center; any voids in the roll would place the corresponding area farther from C than is necessary, thus increasing the value of ( J C ) A. (The process is analogous to rewinding a length of tape back into a roll.) Since the shape of the roll is circular, with the centroid of its area at C, it follows that ( JC ) A ≥

A2 Q.E.D.  2π

  

where the equality applies when the original area is circular.



Acir = π r 2


Solution 2

Consider an area A with its centroid at Point C and a circular area of area A with its center (and centroid) at Point C. Without loss of generality, assume that A1 = A2

A3 = A4

It then follows that ( J C ) A = ( J C )cir + [ J C ( A1 ) − J C ( A2 ) + J C ( A3 ) − J C ( A4 )]

Now observe that J C ( A1 ) − J C ( A2 ) ≥ 0 J C ( A3 ) − J C ( A4 ) ≥ 0

( J C ) A ≥ ( J C )cir

or ( J C ) A ≥

A2 Q.E.D.  2π



since as a given area is moved farther away from C its polar moment of inertia with respect to C must increase.

www.elsolucionario.net PROBLEM 9.31 Determine the moment of inertia and the radius of gyration of the shaded area with respect to the x axis.

SOLUTION


First note that A = A1 + A2 + A3 = [(24)(6) + (8)(48) + (48)(6)] mm 2 = (144 + 384 + 288) mm 2 = 816 mm 2

Now

I x = ( I x )1 + ( I x )2 + ( I x )3

where 1 (24 mm)(6 mm)3 + (144 mm 2 )(27 mm)2 12 = (432 + 104,976) mm 4

( I x )1 =

= 105, 408 mm 4 1 (8 mm)(48 mm)3 = 73,728 mm 4 12 1 ( I x )3 = (48 mm)(6 mm)3 + (288 mm 2 )(27 mm) 2 12 = (864 + 209,952) mm 4 = 210,816 mm 4

( I x )2 =

Then

I x = (105, 408 + 73, 728 + 210,816) mm 4

= 389,952 mm 4

and

k x2 =

I x 389,952 mm 4 = A 816 mm 2

or

or

I x = 390 × 103 mm 4  k x = 21.9 mm 


www.elsolucionario.net PROBLEM 9.32 Determine the moment of inertia and the radius of gyration of the shaded area with respect to the x axis.

SOLUTION First note that


A = A1 − A2 − A3

= [(5)(6) − (4)(2) − (4)(1)] in 2 = (30 − 8 − 4) in 2 = 18 in 2

Now where

I x = ( I x )1 − ( I x ) 2 − ( I x )3

( I x )1 =

1 (5 in.)(6 in.)3 = 90 in 4 12

1 (4 in.)(2 in.)3 + (8 in 2 )(2 in.) 2 12 2 = 34 in 4 3

( I x )2 =

1 3  (4 in.)(1 in.)3 + (4 in 2 )  in.  12 2  1 = 9 in 4 3

2

( I x )3 =

Then

2 1  I x =  90 − 34 − 9  in 4 3 3 

and

k x2 =

I x 46.0 in 4 = A 18 in 4

or

I x = 46.0 in 4 

or k x = 1.599 in. 


www.elsolucionario.net PROBLEM 9.33 Determine the moment of inertia and the radius of gyration of the shaded area with respect to the y axis.

SOLUTION


First note that A = A1 + A2 + A3

= [(24 × 6) + (8)(48) + (48)(6)] mm 2 = (144 + 384 + 288) mm 2 = 816 mm 2

Now

I y = ( I y )1 + ( I y ) 2 + ( I y )3

where 1 (6 mm)(24 mm)3 = 6912 mm 4 12 1 ( I y ) 2 = (48 mm)(8 mm)3 = 2048 mm 4 12 1 ( I y )3 = (6 mm)(48 mm)3 = 55, 296 mm 4 12 ( I y )1 =

Then

and

I y = (6912 + 2048 + 55, 296) mm 4 = 64, 256 mm 4

k y2 =

Iy A

=

64, 256 mm 4 816 mm 2

or

I y = 64.3 × 103 mm 4 

or

k y = 8.87 mm 


www.elsolucionario.net PROBLEM 9.34 Determine the moment of inertia and the radius of gyration of the shaded area with respect to the y axis.

SOLUTION First note that


A = A1 − A2 − A3

= [(5)(6) − (4)(2) − (4)(1)] in 2 = (30 − 8 − 4) in 2 = 18 in 2

Now

I y = ( I y )1 − ( I y ) 2 − ( I y )3

where ( I y )1 =

1 (6 in.)(5 in.)3 = 62.5 in 4 12

( I y )2 =

1 2 (2 in.)(4 in.)3 = 10 in 4 12 3

( I y )3 =

1 1 (1 in.)(4 in.)3 = 5 in 4 12 3

Then

2 1  I y =  62.5 − 10 − 5  in 4 3 3 

and

k y2 =

Iy A

=

46.5 in 4 18 in 2

or

I y = 46.5 in 4 

or k y = 1.607 in. 


www.elsolucionario.net PROBLEM 9.35 Determine the moments of inertia of the shaded area shown with respect to the x and y axes when a = 20 mm.

SOLUTION By symmetry:

Ix = Iy

Square Semicircle 1:

Ix =

1 (80 mm) 4 = 3.413 × 106 mm 4 12

I AA′ =

π (20 mm) 4 = 62.83 × 103 mm 4 8

I AA′ = I x′ + Ad 2

π (20 mm) 2 (8.488 mm) 2 2 I x′ = 17.56 × 103 mm 4

62.83 × 103 mm 4 = I x′ +

I x = I x′ + A(40 mm − 8.488 mm) 2

= 17.56 × 103 mm 4 +

π (20 mm) 2 (31.512 mm) 2 2

I x = 641.5 × 103 mm 4



Given area = square − 4(semicircles)


Semicircle 2:

Same as 1:

I x = 641.5 × 103 mm 4

Semicircles 3 and 4 are equivalent to one 20-mm circle when computing Ix Ix =

π (20 mm)4 = 125.66 × 103 mm 4 4

Entire area = square − 4 (semicircles) I x = 3.413 × 106 mm 4 − 2(641.5 × 103 mm 4 ) − 125.66 × 10−3 mm 4

= 2.0047 × 106 mm 4


I x = I y = 2.00 × 106 mm 4 


www.elsolucionario.net PROBLEM 9.36 Determine the moments of inertia of the shaded area shown with respect to the x and y axes when a = 20 mm.

SOLUTION

where

I x = ( I x )1 + 2( I x ) 2

1 (40 mm)(40 mm)3 12 = 213.33 × 103 mm 4

( I x )1 =

2 π π  4 × 20   ( I x )2 =  (20 mm)4 − (20 mm) 2  mm   2  3π    8

+

 4 × 20   (20 mm)2  + 20  mm  2 3 π   

π

2

= 527.49 × 103 mm 4

Then

I x = [213.33 + 2(527.49)] × 103 mm 4

or Also where

I y = ( I y )1 + 2( I y )2 ( I y )1 = ( I y )2 =

Then

I x = 1.268 × 106 mm 4 

1 (40 mm)(40 mm)3 = 213.33 × 103 mm 4 12

π 8

(20 mm)4 = 62.83 × 103 mm 4

I y = [213.33 + 2(62.83)] × 103 mm 4

or

I y = 339 × 103 mm 4 



We have

www.elsolucionario.net PROBLEM 9.37 The shaded area is equal to 50 in 2 . Determine its centroidal moments of inertia I x and I y , knowing that I y = 2 I x and that the polar moment of inertia of the area about Point A is JA = 2250 in 4 .

SOLUTION Given:

A = 50 in 2

I y = 2 I x , J A = 2250 in 4 J A = J C + A(6 in.) 2

J C = 450 in 4 JC = I x + I y

with

450 in 4 = I x + 2 I x

I y = 2I x I x = 150.0 in 4  I y = 2 I x = 300 in 4 



2250 in 4 = J C + (50 in 2 )(6 in.) 2

www.elsolucionario.net PROBLEM 9.38 The polar moments of inertia of the shaded area with respect to Points A, B, and D are, respectively, JA = 2880 in 4, JB = 6720 in 4, and JD = 4560 in4. Determine the shaded area, its centroidal moment of inertia JC , and the distance d from C to D.

SOLUTION See figure at solution of Problem 9.39. J A = 2880 in 4 , J B = 6720 in 4 , J D = 4560 in 4 J B = J C + A(CB ) 2 ; 6720 in 4 = J C + A(62 + d 2 )

(1)

J D = J C + A(CD) 2 ; 4560 in 4 = J C + Ad 2

(2)

Eq. (1) subtracted by Eq. (2): J B − J D = 2160 in 4 = A(6) 2 J A = J C + A( AC ) 2 ; 2880 in 4 = J C + (60 in 2 )(6 in.) 2

Eq. (2):

4560 in 4 = 720 in 4 + (60 in 2 )d 2

A = 60.0 in 2  J C = 720 in 4  d = 8.00 in. 



Given:

www.elsolucionario.net PROBLEM 9.39 Determine the shaded area and its moment of inertia with respect to the centroidal axis parallel to AA′, knowing that d1 = 30 mm and d2 = 10 mm, and that the moments of inertia with respect to AA′ and BB′ are 4.1 × 106 mm4 and 6.9 × 106 mm4, respectively.

SOLUTION

(1)

I BB′ = 6.9 × 106 mm 4 = I + A(40 mm) 2 I BB′ − I AA′ = (6.9 − 4.1) × 106 = A(402 − 302 ) 2.8 × 106 = A(700)

Eq. (1):

4.1 × 106 = I + (4000)(30)2

A = 4000 mm 2  I = 500 × 103 mm 4 



I AA′ = 4.1 × 106 mm 4 = I + A(30 mm)2

www.elsolucionario.net PROBLEM 9.40 Knowing that the shaded area is equal to 7500 mm2 and that its moment of inertia with respect to AA′ is 31 × 106 mm4, determine its moment of inertia with respect to BB′, for d1 = 60 mm and d2 = 15 mm.

SOLUTION A = 7500 mm 2

I AA′ = I + Ad12 ; 31 × 106 mm 2 = I + (7500 mm 2 )(60 mm) 2 I = 4 × 106 mm 4 I BB′ = I + Ad 2 = 4 × 106 mm 4 + (7500 mm 2 )(60 mm + 15 mm) 2 = 4 × 106 + 7500(75) 2 = 46.188 × 106 mm 4 I BB′ = 46.2 × 106 mm 4 



Given:

www.elsolucionario.net PROBLEM 9.41 Determine the moments of inertia I x and I y of the area shown with respect to centroidal axes respectively parallel and perpendicular to side AB.

SOLUTION

Dimensions in mm

First locate centroid C of the area.

1

Then

A, mm 2

x , mm

108 × 60 = 6480

54

349,920

46

–59,616

2

1 − × 72 × 36 = −1296 2

Σ

5184

xA, mm3

290,304

X Σ A = Σ xA : X (5184 mm 2 ) = 290,304 mm3

or

X = 56.0 mm

Now

I x = ( I x )1 − ( I x )2

where


Symmetry implies Y = 30 mm.

1 (108 mm)(60 mm)3 = 1.944 × 106 mm 4 12 1  1  ( I x )2 = 2  (72 mm)(18 mm)3 +  × 72 mm × 18 mm  (6 mm) 2  2   36  ( I x )1 =

= 2(11, 664 + 23,328) mm 4 = 69.984 × 103 mm 4 [( I x ) 2 is obtained by dividing A2 into

Then

]

I x = (1.944 − 0.069984) × 106 mm 4

or

I x = 1.874 × 106 mm 4 



Also where

I y = ( I y )1 − ( I y ) 2

1 (60 mm)(108 mm)3 + (6480 mm 2 )[(56.54) mm]2 12 = (6, 298,560 + 25,920) mm 4 = 6.324 × 106 mm 4

( I y )1 =

1 (36 mm)(72 mm)3 + (1296 mm 2 )[(56 − 46) mm]2 36 = (373, 248 + 129, 600) mm 4 = 0.502 × 106 mm 4

( I y )2 =

Then

I y = (6.324 − 0.502)106 mm 4


or I y = 5.82 × 106 mm 4 



SOLUTION First locate C of the area.

A, mm 2

1

Then

y , mm

yA, mm3

14

14,112

36 × 28 = 1008

2

1 × 36 × 42 = 756 2

Σ

1764

where

Dimensions in mm

45,864

Y Σ A = Σ yA : Y (1764 mm 2 ) = 45,864 mm3

or Now

31,752

42

Y = 26.0 mm

I x = ( I x )1 + ( I x ) 2 1 (36 mm)(28 mm)3 + (1008 mm 2 )[(26 − 14) mm]2 12 = (65,856 + 145,152) mm 4 = 211.008 × 103 mm 4

( I x )1 =

1 (36 mm)(42 mm)3 + (756 mm 2 )[(42 − 26) mm]2 36 = (74,088 + 193,536) mm 4 = 267.624 × 103 mm 4

( I x )2 =

Then

I x = (211.008 + 267.624) × 103 mm 4

or

I x = 479 × 103 mm 4 



Symmetry implies X = 18 mm.


Also where

I y = ( I y )1 + ( I y ) 2 1 (28 mm)(36 mm)3 = 108.864 × 103 mm 4 12 1  1  ( I y ) 2 = 2  (42 mm)(18 mm)3 +  × 18 mm × 42 mm  (6 mm) 2  2   36  ( I y )1 =

= 2(6804 + 13, 608) mm 4 = 40.824 × 103 mm 4 [( I y )2 is obtained by dividing A2 into

I y = (108.864 + 40.824 × 103 mm 4

or I y = 149.7 × 103 mm 4 


Then

]




Then

A, in 2

x , in.

1

5 × 8 = 40

2.5

2

−2 × 5 = −10

1.9

Σ

30

where

4

100

160

4.3

–19

–43

81

117

X = 2.70 in. Y Σ A = Σ yA: Y (30 in 2 ) = 117 in 3 Y = 3.90 in.

or Now

yA, in 3

X Σ A = Σ xA: X (30 in 2 ) = 81 in 3

or and

xA, in 3

y , in.

I x = ( I x )1 − ( I x )2

1 (5 in.)(8 in.)3 + (40 in 2 )[(4 − 3.9) in.]2 12 = (213.33 + 0.4) in 4 = 213.73 in 4

( I x )1 =

1 (2 in.)(5 in.)3 + (10 in 2 )[(4.3 − 3.9) in.]2 12 = (20.83 + 1.60) = 22.43 in 4

( I x )2 =



SOLUTION


Then

I x = (213.73 − 22.43) in 4

Also

I y = ( I y )1 − ( I y ) 2

where

or I x = 191.3 in 4 

1 (8 in.)(5 in.)3 + (40 in 2 )[(2.7 − 2.5) in.]2 12 = (83.333 + 1.6) in 4 = 84.933 in 4

( I y )1 =

1 (5 in.)(2 in.)3 + (10 in 2 )[(2.7 − 1.9) in.]2 12 = (3.333 + 6.4) in 4 = 9.733 in 4

( I y )2 =

I y = (84.933 − 9.733) in 4

or

I y = 75.2 in 4 


Then



First locate centroid C of the area. A, in 2

or and or

y , in.

xA, in 3

yA, in 3

1

3.6 × 0.5 = 1.8

1.8

0.25

3.24

0.45

2

0.5 × 3.8 = 1.9

0.25

2.4

0.475

4.56

3

1.3 × 1 = 1.3

0.65

4.8

0.845

6.24

4.560

11.25

Σ

Then

x , in.

5.0 X Σ A = Σ xA: X (5 in 2 ) = 4.560 in 3 X = 0.912 in. Y Σ A = Σ yA: Y (5 in 2 ) = 11.25 in 3 Y = 2.25 in.



SOLUTION


Now where

I x = ( I x )1 + ( I x )2 + ( I x )3 1 (3.6 in.)(0.5 in.)3 + (1.8 in 2 )[(2.25 − 0.25) in.]2 12 = (0.0375 + 7.20) in 4 = 7.2375 in 4

( I x )1 =

1 (0.5 in.)(3.8 in.)3 + (1.9 in 2 )[(2.4 − 2.25) in.]2 12 = (2.2863 + 0.0428)in 4 = 2.3291 in 4

( I x )2 =

1 (1.3 in.)(1 in.)3 + (1.3 in 2 )[(4.8 − 2.25 in.)]2 12 = (0.1083 + 8.4533) in 4 = 8.5616 in 4

( I x )3 =

I x = (7.2375 + 2.3291 + 8.5616) in 4 = 18.1282 in 4 I x = 18.13 in 4 

or Also where

I y = ( I y )1 + ( I y ) 2 + ( I y )3 1 (0.5 in.)(3.6 in.)3 + (1.8 in 2 )[(1.8 − 0.912) in.]2 12 = (1.9440 + 1.4194) in 4 = 3.3634 in 4

( I y )1 =

1 (3.8 in.)(0.5 in.)3 + (1.9 in 2 )[(0.912 − 0.25) in.]2 12 = (0.0396 + 0.8327) in 4 = 0.8723 in 4

( I y )2 =

1 (1 in.)(1.3 in.)3 + (1.3 in 2 )[(0.912 − 0.65) in.]2 12 = (0.1831 + 0.0892) in 4 = 0.2723 in 4

( I y )3 =

Then

I y = (3.3634 + 0.8723 + 0.2723) in 4 = 4.5080 in 4

or

I y = 4.51 in 4 



Then

www.elsolucionario.net PROBLEM 9.45 Determine the polar moment of inertia of the area shown with respect to (a) Point O, (b) the centroid of the area.

Area, in2

x , in.

(4.5) 2 = 15.904

1.9099

30.375

π

1.2732

–9.00

Section 1

2

π 4

−

4

(3) 2 = −7.069

Σ

Then

8.835

xA, in 3

21.375

X A = Σ xA: X (8.835 in 2 ) = 21.375 in 3 X = 2.419 in.

or Then


SOLUTION

JO =

π 8

(4.5 in.)4 −

π 8

(3 in.) 4 = 129.22 in 4

J O = 129.2 in 4 

OC = 2 X = 2(2.419 in.) = 3.421 in. J O = J C + A(OC ) 2 : 129.22 in 4 = J C + (8.835 in.)(3.421 in.) 2

J C = 25.8 in 4 



SOLUTION

Area, in2

Section 1

π 4

(4) 2 = 4π

2

1 (4)(4) = 8 2

3

1 (4)(4) = 8 2

Σ

28.566

x , in.

xA, in 3

16 3π

21.333

−

4 3

4 3


Determination of centroid C of entire section:

−10.6667

10.6667 21.333

X Σ A = Σ xA: X (28.566 in 2 ) = 21.333 in 3 X = 0.74680 in.; by symmetry, Y = X = 0.74680 in.

(a)

(b)

For section :

Ix =

11 4 1 π r  = π (4) 4 = 50.265 in 4 4  4  16

For sections ,:

Ix =

1 3 1 bh = (4)(4)3 = 21.333 in 4 12 12

For total area,

I x = 50.265 + 2(21.333) = 92.931 in 4

By symmetry,

I y = I x ; J O = I x + I y = 185.862 in 4

Parallel axis theorem:

J C = J O − A(OC ) 2 = J O − A( X 2 + Y 2 ) J C = 185.862 − (28.566)(0.746802 + 0.746802 )

J O = 185.9 in 4 

J C = 154.0 in 4 



SOLUTION

Determination of centroid C of entire section Y Σ A = Σ yA Y (4000 mm 2 ) = 122.67 × 103 mm3 Y = 30.667 mm

Area mm2

(a)

1

1 (160)(80) = 6400 2

2

1 − (80)(60) = −2400 2

Σ

4000

y , mm

yA, mm3

80 3

170.67 × 103

20

−48 × 103 122.67 × 103

Polar moment of inertia J O :

Section :

Ix =

1 (160)(80)3 = 6.8267 × 106 mm 4 12

For Iy consider the following two triangles 1  I y = 2  (80)(80)3  = 6.8267 × 106 mm 4 12  J O = I x + I y = (6.8267 + 6.8267)106 = 13.653 × 106 mm 4 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1456


Symmetry: X = 0

Dimensions in mm


(b)

Section :

Ix =

1 (80)(60)3 = 1.44 × 106 mm 4 12

For Iy consider the following two triangles 1  I y = 2  (60)(40)3  = 0.640 × 106 mm 4 12   J O = I x + I y = (1.44 + 0.640)106 = 2.08 × 106 mm 4

Entire section

J O = ( J O )1 − ( J O ) 2 = 13.653 × 106 − 2.08 × 106 J O = 11.573 × 106 mm 4

(c)

J O = 11.57 × 106 mm 4 

Polar moment of inertia J O of intire area

11.573 × 106 = J C + (4000)(30.667) 2 J C = 7.811 × 106 mm 4 J C = 7.81 × 106 mm 4 



J O = J C + AY 2


SOLUTION

A, mm 2

1

2 Σ

π 2 −

(54)(36) = 3053.6

48

π

24

2

π

(18) 2 = −508.9

π

y , mm

yA, mm3

= 15.2789

46,656

= 7.6394

−3888 42,768

2544.7

Y Σ A = Σ y A: Y (2544.7 mm 2 ) = 42, 768 mm3

Then

Y = 16.8067 mm 

or J O = ( J O )1 − ( J O ) 2

(a)

=

π

(54 mm)(36 mm)[(54 mm)2 + (36 mm)2 ] −

8 = (3.2155 × 106 − 0.0824 × 106 ) mm 4

π 4

(18 mm) 4

= 3.1331 × 106 mm 4

or

J O = 3.13 × 106 mm 4 

J O = J C + A(Y )2

(b) or

J C = 3.1331 × 106 mm 4 − (2544.7 mm 2 )(16.8067 mm) 2

or

J C = 2.41 × 106 mm 4 




www.elsolucionario.net PROBLEM 9.49 Two channels and two plates are used to form the column section shown. For b = 200 mm, determine the moments of inertia and the radii of gyration of the combined section with respect to the centroidal x and y axes.


SOLUTION

From Figure 9.13B For C250 × 22.8 A = 2890 mm 2 I x = 28.0 × 106 mm 4 I y = 0.945 × 106 mm 4

Total area

A = 2[2890 mm 4 + (10 mm)(375 mm)] = 13.28 × 103 mm 2

Given b = 200 mm:

1  I x = 2[28.0 × 106 mm 4 ] + 2  (375 mm)(10 mm)3 + (375 mm)(10 mm)(132 mm) 2  12   = 186.743 × 106 k x2 =

I x = 186.7 × 106 mm 4 

I x 186.743 × 106 = = 14.0620 × 103 mm 2 A 13.28 × 103

k x = 118.6 mm 



Channel

Plate

1  I y = Σ( I y + Ad 2 ) = 2[ I y + Ad 2 ] + 2  (10 mm)(375 mm)3  12  2    6 4 2  200 mm = 2 0.945 × 10 mm + (2890 mm )  + 16.10 mm   + 87.891 × 106 mm 4 2     = 2[0.945 × 106 + 38.955 × 106 ] + 87.891 × 106 = 167.691 × 106 mm 4 Iy A

=

167.691 × 106 = 12.6273 × 103 13.28 × 103

k y = 112.4 mm 


k y2 =

I y = 167.7 × 106 mm 4 


www.elsolucionario.net PROBLEM 9.50 1

Two L6 × 4 × 2 -in. angles are welded together to form the section shown. Determine the moments of inertia and the radii of gyration of the combined section with respect to the centroidal x and y axes.


SOLUTION

From Figure 9.13A From Figure 9.13A:

Area = A = 4.75 in2 I x′ = 17.3 in 4 I y ′ = 6.22 in 4 I x = 2[ I x′ + AyO2 ] = 2[17.3 in 4 + (4.75 in 2 )(1.02 in.) 2 ] = 44.484 in 4 I x = 44.5 in 4 

Total area = 2(4.75) = 9.50 in2 k x2 =

Ix 44.484 in 4 = = 4.6825 in 2 2 area 9.50 in

k x = 2.16 in. 

I y = 2[ I y′ + AxO2 ] = 2[6.22 in 4 + (4.75 in 2 )(1.269 in.) 2 ] = 27.738 in 4

I y = 27.7 in 4 

Total area = 2(4.75) = 9.50 in2 k y2 =

Iy area

=

27.738 in 4 = 2.9198 9.50 in 2

k y = 1.709 in 


www.elsolucionario.net PROBLEM 9.51 Two channels are welded to a rolled W section as shown. Determine the moments of inertia and the radii of gyration of the combined section with respect to the centroidal x and y axes.

SOLUTION W section:

A = 9.12 in 2 I x = 110 in 4

Channel:

A = 3.37 in 2 I x = 1.31 in 4 I y = 32.5 in 4 Atotal = AW + 2Achan = 9.12 + 2(3.37) = 15.86 in 2

Now

where

I x = ( I x ) W + 2( I x )chan ( I x )chan = I xchan + Ad 2 = 1.31 in 4 + (3.37 in 2 )(4.572 in.) 2 = 71.754 in 4

Then

I x = (110 + 2 × 71.754) in 4 = 253.51 in 4

and

k x2 =

Also

I y = ( I y ) W + 2( I y )chan

Ix 253.51 in 4 = Atotal 15.86 in 2

= (37.1 + 2 × 32.5) in 4 = 102.1 in 4

and

k y2 =

Iy Atotal

=

102.1 in 4 15.86 in 2

I x = 254 in 4 

k x = 4.00 in. 

I y = 102.1 in 4  k y = 2.54 in. 



I y = 37.1 in 4

www.elsolucionario.net PROBLEM 9.52 Two 20-mm steel plates are welded to a rolled S section as shown. Determine the moments of inertia and the radii of gyration of the combined section with respect to the centroidal x and y axes.

SOLUTION

I x = 90.3 × 106 mm 4 I y = 3.88 × 106 mm 4

Atotal = AS + 2 Aplate

Note:

= 6010 mm 2 + 2(160 mm)(20 mm) = 12, 410 mm 2 I x = ( I x )S + 2( I x )plate

Now

( I x ) plate = I xplate + Ad 2

where

1 (160 mm)(20 mm)3 + (3200 mm 2 )[(152.5 + 10) mm]2 12 = 84.6067 × 106 mm 4 =

I x = (90.3 + 2 × 84.6067) × 106 mm 4

Then

= 259.5134 × 106 mm 4 k x2 =

and Also

Ix 259.5134 × 106 mm 4 = Atotal 12410 mm 2

or I x = 260 × 106 mm 4  or

k x = 144.6 mm 

I y = ( I y )S + 2( I y ) plate

1  = 3.88 × 106 mm 4 + 2  (20 mm)(160 mm)3  12  

or I y = 17.53 × 106 mm 4 

= 17.5333 × 106 mm 4

and

k y2 =

Iy Atotal

=

17.5333 × 106 mm 4 12, 410 mm 2

or

k y = 37.6 mm 



A = 6010 mm 2

S section:

www.elsolucionario.net PROBLEM 9.53 A channel and a plate are welded together as shown to form a section that is symmetrical with respect to the y axis. Determine the moments of inertia of the combined section with respect to its centroidal x and y axes.

From Figure 9.13B

B = 3.37 in 2

For C8 × 11.5:

I y = 1.31 in 4

(Note change of axes)

I y = 32.5 in 4


SOLUTION

Location of centroid Y Σ A = Σ y A:

Y [3.7 in 2 + (12 in.)(0.5 in.)] = (3.37 in 2 )(1.938 in.) Y = 0.69702 in.

Moment of inertia with respect to x axis. 1 (12 in.)(0.5 in.)3 + (12 in.)(0.5 in.)(0.69702 in.) 2 12 = 0.125 + 2.9150 = 3.0400 in 4

Plate:

I x = I x + AY 2 =

Channel:

I x = I x′ + AY02 = 1.31 in 4 + (3.37 in.)(1.938 in. − 0.69702 in.) 2 = 1.31 + 5.1899 = 6.4999 in 4

Entire section:

I x = 3.0400 in 4 + 6.4999 in 4 = 9.5399 in 4

I x = 9.54 in 4 

Moment of inertia with respect to y axis. 1 (0.5 in.)(12 in.)3 = 72 in 4 12

Plate:

Iy =

Channel:

I y = 32.5 in 4

Entire section:

I y = 72 in 4 + 32.5 in 4 = 104.5 in 4

I y = 104.5 in 4 


www.elsolucionario.net PROBLEM 9.54 The strength of the rolled W section shown is increased by welding a channel to its upper flange. Determine the moments of inertia of the combined section with respect to its centroidal x and y axes.

SOLUTION A = 14, 400 mm 2

W section:

I x = 554 × 106 mm 4

A = 2890 mm 2

Channel:

I x = 0.945 × 106 mm 4 I y = 28.0 × 106 mm 4

First locate centroid C of the section.

W Section Channel

Σ

Then

A, mm2

y , mm

yA, mm3

14,400

−231

−33,26,400

2,890

49.9

1,44,211 −31,82,189

17,290

Y Σ A = Σ y A: Y (17, 290 mm 2 ) = −3,182,189 mm3

Y = −184.047 mm

or Now where

I x = ( I x ) W + ( I x )C ( I x ) W = I x + Ad 2 = 554 × 106 mm 4 + (14, 400 mm 2 )(231 − 184.047) 2 mm 2 = 585.75 × 106 mm 4 ( I x )C = I x − Ad 2 = 0.945 × 106 mm 4 + (2,890 mm 2 )(49.9 + 184.047)2 mm 2 = 159.12 × 106 mm 4



I y = 63.3 × 106 mm 4


Then

also

I x = (585.75 + 159.12) × 106 mm 4

or

I x = 745 × 106 mm 4 

or

I y = 91.3 × 106 mm 4 

I y = ( I y ) W + ( I y )C


= (63.3 + 28.0) × 106 mm 4


www.elsolucionario.net PROBLEM 9.55 Two L76 × 76 × 6.4-mm angles are welded to a C250 × 22.8 channel. Determine the moments of inertia of the combined section with respect to centroidal axes respectively parallel and perpendicular to the web of the channel.

SOLUTION A = 929 mm 2

Angle:

I x = I y = 0.512 × 106 mm4 A = 2890 mm 2

Channel:

I x = 0.945 × 106 mm 4


I y = 28.0 × 106 mm 4

First locate centroid C of the section

Then

A, mm 2

y , mm

yA, mm3

Angle

2(929) = 1858

21.2

39,389.6

Channel

2890

−16.1

−46,529

Σ

4748

−7139.4

Y Σ A = Σ y A: Y (4748 mm 2 ) = −7139.4 mm3

Y = −1.50366 mm

or Now where

I x = 2( I x ) L + ( I x )C ( I x ) L = I x + Ad 2 = 0.512 × 106 mm 4 + (929 mm 2 )[(21.2 + 1.50366) mm]2 = 0.990859 × 106 mm 4 ( I x )C = I x + Ad 2 = 0.949 × 106 mm 4 + (2890 mm 2 )[(16.1 − 1.50366) mm]2 = 1.56472 × 106 mm 4

Then

I x = [2(0.990859) + 1.56472 × 106 ] mm 4

or

I x = 3.55 × 106 mm 4 



Also where

I y = 2( I y ) L + ( I y )C

( I y ) L = I y + Ad 2 = 0.512 × 106 mm 4 + (929 mm 2 )[(127 − 21.2) mm]2 = 10.9109 × 106 mm 4 ( I y )C = I y

Then

I y = [2(10.9109) + 28.0] × 106 mm 4 I y = 49.8 × 106 mm4 


or


www.elsolucionario.net PROBLEM 9.56 Two L4 × 4 × 12 -in. angles are welded to a steel plate as shown. Determine the moments of inertia of the combined section with respect to centroidal axes respectively parallel and perpendicular to the plate.

1 For 4 × 4 × -in. angle: 2

A = 3.75 in 2 , I x = I y = 5.52 in 4 YA = Σ y A Y (12.5 in 2 ) = 33.85 in 3 Y = 2.708 in.

Area, in2

y in.

yA, in 3

Plate

(0.5)(10) = 5

5

25

Two angles

2(3.75) = 7.5

1.18

8.85

Section

Σ

33.85

12.5



SOLUTION


Entire section: 1  I x = Σ( I x′ + Ad 2 ) =  (0.5)(10)3 + (0.5)(10)(2.292) 2  + 2[5.52 + (3.75)(1.528) 2 ]  12 

= 41.667 + 26.266 + 1604 + 17.511 = 96.48 in 4 Iy =

I x = 96.5 in 4 

1 (10)(0.5)3 + 2[5.52 + (3.75)(1.43) 2 ] 12

I y = 26.5 in 4 


= 0.104 + 11.04 + 15.367 = 26.51 in 4


www.elsolucionario.net PROBLEM 9.57 The panel shown forms the end of a trough that is filled with water to the line AA′. Referring to section 9.2, determine the depth of the point of application of the resultant of the hydrostatic forces acting on the panel (the center of pressure).

SOLUTION From section 9.2:





R = γ ydA, M AA′ = γ y 2 dA

Let yP = distance of center of pressure from AA′. We must have:

 

2 M AA′ γ y dA I AA′ = = RyP = M AA′ : yP = R γ ydA y A

For triangular panel: I AA′ =

1 3 ah 12

yP =

I AA′ = yA

1 y= h 3 1 12

ah3

( h) ( 1 3

1 2

ah )

A=

1 ah 2 yP =

1 h  2



(1)


SOLUTION See solution of Problem 9.57 for derivation of Eq. (1): yP =

I AA′ yA

(1)

1 1 ( a − b)h3 + bh3 12 3 1 3 1 3 = ah + bh 2 4 1 1 1 1  1 yA = y1 A1 + y2 A2 = h  (a − b)h  + h(bh) = ah 2 + bh 2 3 2 6 3  2

I AA′ =

yP =

I AA′ = yA

1 ah3 + 14 bh3 12 1 ah 2 + 13 bh 2 6

yP =

a + 3b h  2a + 4b



Divide trapezoid as shown:


SOLUTION Using the equation developed on page 475 of the text have:

For a semi ellipse:

Then

I AA′ yA

π

I AA′ =

ab3 8 4b y= 3π

yP =

π

8

4b 3π

ab3

× π2 ab

A=

π 2

ab

or

or

yP =

3π b  16



yP =

www.elsolucionario.net PROBLEM 9.60* The panel shown forms the end of a trough that is filled with water to the line AA′. Referring to section 9.2, determine the depth of the point of application of the resultant of the hydrostatic forces acting on the panel (the center of pressure).

SOLUTION Using the equation developed on page 491 of the text:

Now By observation

I AA′ yA

y=

2 h 5

A=

4 ah 3

1 dI AA′ = (h − y )3 dx 3

y=


For a parabola:

yP =

h 2 x a2 3

So that

Then

dI AA′

1 h 1 h3 2  =  h − 2 x 2  dx = (a − x 2 )3 dx 6 3 3a a  1 h 6 = (a − 3a 4 x 2 + 3a 2 x 4 − x 6 )dx 3 a6

I AA′ = 2



h3 6 (a − 3a 4 x 2 + 3a 2 x 4 − x 6 )dx 3 a6

a1 0

a

2 h3  6 3 1  a x − a 4 x3 + a 2 x5 − x7  3 a 6  5 7 0 32 3 ah = 105 =

Finally,

yp =

32 105 2 5

ah3

h × 43 ah

or

yP =

4 h  7


www.elsolucionario.net PROBLEM 9.61 A vertical trapezoidal gate that is used as an automatic valve is held shut by two springs attached to hinges located along edge AB. Knowing that each spring exerts a couple of magnitude 1470 N · m, determine the depth d of water for which the gate will open.

SOLUTION

R = γ yA

yP =

I SS ′ yA

where R is the resultant of the hydrostatic forces acting on the gate and yP is the depth to the point of application of R. Now 1  1  yA = Σ yA = [(h + 0.17) m]  × 1.2 m × 0.51 m  + [(h + 0.34) m]  × 0.84 × 0.51 m  2  2  3 = (0.5202h + 0.124848) m

Recalling that γ = py, we have R = (103 kg/m3 × 9.81 m/s 2 )(0.5202h + 0.124848) m3 = 5103.162(h + 0.24) N

Also where

I SS ′ = ( I SS ′ )1 + ( I SS ′ ) 2

( I SS ′ )1 = I x + Ad 2 =

1 1  (1.2 m)(0.51 m)3 +  × 1.2 m × 0.51 m  [(h + 0.17) m]2 36 2 

= [0.0044217 + 0.306(h + 0.17)2 ] m 4 = (0.306h 2 + 0.10404h + 0.0132651) m 4 ( I SS ′ ) 2 = I X 2 + Ad 2 =

1 1  (0.84 m)(0.51 m)3 +  × 0.84 m × 0.51 m  [(h + 0.34) m]2 36 2  

= [0.0030952 + 0.2142( h + 0.34) 2 ] m 4 = (0.2142h 2 + 0.145656h + 0.0278567) m 4



From section 9.2:


I SS ′ = ( I SS ′ )1 + ( I SS ′ ) 2

Then

= (0.5202h 2 + 0.249696h + 0.0411218) m 4

yP =

and

=

(0.5202h 2 + 0.244696h + 0.0411218) m 4 (0.5202h + 0.124848) m3 h 2 + 0.48h + 0.07905 m h + 0.24

For the gate to open, require that ΣM AB : M open = ( yP − h) R

or

 h2 + 0.48h + 0.07905  2940 N ⋅ m =  − h  m × 5103.162(h + 0.24) N h + 0.24   5103.162(0.24h + 0.07905) = 2940

or

h = 2.0711 m

Then

d = (2.0711 + 0.79) m

or

d = 2.86 m 



Substituting

www.elsolucionario.net PROBLEM 9.62 The cover for a 0.5-m-diameter access hole in a water storage tank is attached to the tank with four equally spaced bolts as shown. Determine the additional force on each bolt due to the water pressure when the center of the cover is located 1.4 m below the water surface.

SOLUTION From section 9.2: R = γ yA

yP =

I AA′ yA

where R is the resultant of the hydrostatic forces acting on the cover


and yP is the depth to the point of application of R. Recalling that γ = p ⋅ y, we have R = (103 kg/m3 × 9.81 m/s 2 )(1.4 m)[π (0.25 m) 2 ] = 2696.67 N

π

(0.25 m) 4 + [π (0.25 m) 2 ](1.4 m) 2 4 = 0.387913 m 4

I AA′ = I x + Ay 2 =

Also

yP =

Then

0.387913 m 4 = 1.41116 m (1.4 m)[π (0.25 m) 2 ]

Now note that symmetry implies FA = FB

FC = FD

Next consider the free-body of the cover. We have

ΣM CD = 0: [2(0.32 m) sin 45°](2 FA ) − [0.32 sin 45° − (1.41116 − 1.4)] m × (2696.67 N) = 0

or Then or

FA = 640.92 N ΣFz = 0: 2(640.92 N) + 2 FC − 2696.67 N = 0 FC = 707.42 N FA = FB = 641 N  FC = FD = 707 N 

and


www.elsolucionario.net PROBLEM 9.63* Determine the x coordinate of the centroid of the volume shown. (Hint: The height y of the volume is proportional to the x coordinate; consider an analogy between this height and the water pressure on a submerged surface.)

SOLUTION

Now where

Then



h x a



x dV = xEL dV h  xEL = x dV = ydA =  x  dA a 

x=

 x ( xdA) =  x dA = ( I )  xdA  xdA ( xA) 2

h a

z A

h a

A

where ( I z ) A and ( x A ) A pertain to area.

OABC: ( I z ) A is the moment of inertia of the area with respect to the z axis, x A is the x coordinate of the centroid of the area, and A is the area of OABC. Then ( I z ) A = ( I z ) A1 + ( I z ) A2 1 1 = (b)(a)3 + (b)( a)3 3 12 5 3 = ab 12

and

( xA) A = Σ xA  a    a  1  =   ( a × b)  +   × a × b     2    3  2 2 = a 2b 3



y=

First note that


Finally,

x=

5 3 ab 12 2 2 a b 3

or

5 x = a  8

Analogy with hydrostatic pressure on a submerged plate: Recalling that P = γ y, it follows that the following analogies can be established. Height y ~ P

Recalling that

It can then be concluded that

yP =

Mx R

 dM  =   dF   

 

x ~ yP



dV = ydA ~ pdA = dF xdV = x( ydA) ~ ydF = dM

www.elsolucionario.net PROBLEM 9.64* Determine the x coordinate of the centroid of the volume shown; this volume was obtained by intersecting an elliptic cylinder with an oblique plane. (Hint: The height y of the volume is proportional to the x coordinate; consider an analogy between this height and the water pressure on a submerged surface.)

SOLUTION Following the “Hint,” it can be shown that (see solution to Problem 9.63) x=

(I z ) A x ( xA) A

( I z ) A = I z + Ad 2 =

π

(39 mm)(64 mm)3 + [π (64 mm)(39 mm)](64 mm) 2

4 = 12.779520π × 106 mm 4

( xA) A = (64 mm)[π (64 mm)(39 mm)] = 0.159744π × 106 mm3

Finally

x=

12.779520π × 106 mm 4 0.159744π × 106 mm 4

or

x = 80.0 mm 



where ( I z ) A and ( xA) A are the moment of inertia and the first moment of the area, respectively, of the elliptical area of the base of the volume with respect to the z axis. Then

www.elsolucionario.net PROBLEM 9.65* Show that the system of hydrostatic forces acting on a submerged plane area A can be reduced to a force P at the centroid C of the area and two couples. The force P is perpendicular to the area and is of magnitude P = γ Ay sin θ , where γ is the specific weight of the liquid, and the couples are M x′ = (γ I x′ sin θ )i and M y′ = (γ I x′y ′ sin θ ) j, where I x′y′ =  x′y ′ dA (see section 9.8). Note that the couples are independent of the depth at which the area is submerged.

SOLUTION The pressure p at an arbitrary depth ( y sin θ ) is p = γ ( y sin θ ) dF = (γ y sin θ )dA

Equivalence of the force P and the system of infinitesimal forces dF requires







ΣF : P = dF = γ y sin θ dA = γ sin θ y dA or

P = γ Ay sin θ 

Equivalence of the force and couple (P, M x′ + M y′ ) and the system of infinitesimal hydrostatic forces requires



ΣM x : − yP − M x′ = (− ydF ) Now







− ydF = − y (γ y sin θ )dA = −γ sin θ y 2 dA = −(γ sin θ ) I x

Then or

− yP − M x′ = −(γ sin θ ) I x M x′ = (γ sin θ ) I x − y (γ Ay sin θ ) = γ sin θ ( I x − Ay 2 )

or

M x′ = γ I x′ sin θ 



ΣM y : xP + M y ′ = x dF Now

 x dF =  x(γ y sin θ )dA = γ sin θ  xy dA = (γ sin θ ) I xy

(Equation 9.12)



so that the hydrostatic force dF exerted on an infinitesimal area dA is


Then or

xP + M y′ = (γ sin θ ) I xy M y′ = (γ sin θ ) I xy − x (γ Ay sin θ ) = γ sin θ ( I xy − Ax y ) M y′ = γ I x′y ′ sin θ 


or


www.elsolucionario.net PROBLEM 9.66* Show that the resultant of the hydrostatic forces acting on a submerged plane area A is a force P perpendicular to the area and of magnitude P = γ Ay sin θ = pA, where γ is the specific weight of the liquid and p is the pressure at the centroid C of the area. Show that P is applied at a Point CP, called the center of pressure, whose coordinates are xP = I xy /Ay and y P = I x /Ay , where I xy =  xydA (see section 9.8). Show also that the difference of ordinates y P − y is equal to k x2′ / y and thus depends upon the depth at which the area is submerged.

SOLUTION The pressure P at an arbitrary depth ( y sin θ ) is

so that the hydrostatic force dP exerted on an infinitesimal area dA is dP = (γ y sin θ )dA

The magnitude P of the resultant force acting on the plane area is then







P = dP = γ y sin θ dA = γ sin θ ydA = γ sin θ ( yA)

Now

p = γ y sin θ

P = pA 

Next observe that the resultant P is equivalent to the system of infinitesimal forces d P . Equivalence then requires



ΣM x : − yP P = − ydP Now

 ydP =  y(γ y sin θ )dA = γ sin θ  y dA 2

= (γ sin θ ) I x

Then or

y P P = (γ sin θ ) I x

yP =

(γ sin θ ) I x γ sin θ ( yA)

or

yP =

Ix  Ay



P = γ ( y sin θ )




ΣM y : xP P = xdP

 xdP =  x(γ y sin θ )dA = γ sin θ  xydA = (γ sin θ ) I xy

Then or

(Equation 9.12)

xP P = (γ sin θ ) I xy

xP =

(γ sin θ ) I xy

γ sin θ ( yA) or

Now

I x = I x′ + A y 2

From above

I x = ( A y ) yP

By definition

I x′ = k x′ A

Substituting

( Ay ) yP = k x2′ A + Ay 2

xP =

I xy Ay



2

yP − y =

Rearranging yields

k x2′  y

Although k x′ is not a function of the depth of the area (it depends only on the shape of A), y is dependent on the depth. ( y P − y ) = f (depth) 



Now

www.elsolucionario.net PROBLEM 9.67 Determine by direct integration the product of inertia of the given area with respect to the x and y axes.

SOLUTION y = a 1− =

x2 4a 2

1 4a 2 − x 2 2

We have

dI xy = dI x′y′ + xEL yEL dA

where

dI x′y′ = 0 (symmetry) yEL =

xEL = x

1 1 4a 2 − x 2 y= 2 4

dA = ydx =

Then


First note

1 4a 2 − x 2 dx 2



I xy = dI xy =



2a 0

1 = 8 =



a4 8

1  1  4a 2 − x 2  4a 2 − x 2  dx x 4  2  2a

0

2a

1 1  (4a x − x )dx =  2a 2 x 2 − x 4  8 4 0 2

3

1 4  2  2(2) − 4 (2)    or

I xy =

1 4 a  2



SOLUTION dI xy = dI x′y′ + xEL yEL dA

But

dI x′y′ = 0 (by symmetry) xEL = x

yEL =

1 y 2

y h = x b

y=



I xy = dI xy = =

1 h2 2 b2



b 0

h x b



b 0

1 1  x  y  ydx = 2 2 

x3 dx =



b 0

2

h  x  x  dx b 

1 h2 b4 2 b2 4 1 I xy = b 2 h 2  8



dA = yd y


SOLUTION

For

x = a, b = ka 3

Thus,

y=

or k =

b a3

dI xy = dI x′y ′ + xEL yEL dA, But

with

xEL = x



y 2

yEL =

I xy = dI xy =



a

0

dI x′y′ = 0 (by symmetry)

dA = ydx

1  y x   ydx = 2 2  



a 0

2

 b  x  3 x3  dx a  

a

=

1 b 2  x8    2 a6  8 0

I xy = a 2b2 /16 



b 3 x a3


SOLUTION  x x2 y = 4h  − 2 a a

  

dI xy = dI x′y′ + xEL yEL dA dI x′y′ = 0 (by symmetry) xEL = x

yEL =



I xy = dI xy = = 8h

2





a 0

1 y 2

1 x 2

dA = ydx

1  y  ydx = 2 



 x x2 16h 2 x  − 2 0 a a a

a 0

  dx 

4 x3 x 4 x5  2 x5 x6 2 x − +  2 − 2 3 + 4  dx = 8h a a  4 a 2 5a 3 6 a 4 a

a

0

1 2 1  15 − 24 + 10  = 8h 2 a 2  − +  = 8 h 2 a 2   4 5 6 60     I xy =

2 2 2 a h  15

Check: Apply parallel axis theorem to area see Figure 5.8: Area =

2 ah 3

by symmetry I x′y′ = 0  a  2  2  2 I xy = I x′y′ + xc yc A = 0 +    h   ah  = a 2 h 2  2   5   3  15



But

www.elsolucionario.net PROBLEM 9.71 Using the parallel-axis theorem, determine the product of inertia of the area shown with respect to the centroidal x and y axes.

SOLUTION Dimensions in mm

I xy = ( I xy )1 + ( I xy ) 2 + ( I xy )3

We have

( I xy ) 2 = 0

and for the other rectangles I xy = I x′y′ + x yA

I x′y′ = 0 (symmetry)

where

I xy = ( x yA)1 + ( x yA)3

Thus A, mm 2

x , mm

y , mm

x yA, mm 4

1

10 × 80 = 800

−55

20

−880,000

3

10 × 80 = 800

55

−20

−880,000

Σ

−1,760,000 I xy = −1.760 × 106 mm 4 



Now symmetry implies


SOLUTION

For rectangle I xy = 0, by symmetry For each triangle:

Compare these triangles with the triangle of sample Problem 9.6, where I x′y′ = −

1 2 2 b h . 72

For orientation of axes of this problem, I x′y′ = +

1 2 2 1 (60 mm)2 (40 mm)2 = +80 × 103 mm 4 b h = 72 72



Given area = Rectangle − (Two triangles)


Area, mm 2

x , mm

y , mm

x yA, mm 4

1

1 (60)(40) = 1200 2

−40

+26.67

−1.280 × 106

2

1 (60)(40) = 1200 2

+40

−26.67

−1.280 × 106

Σ x y A = −2.56 × 106 mm 4

For two triangles:

I xy = Σ( I x′y′ + x y A) = +2(80 × 103 ) − 2.56 × 106 = −2.40 × 106 mm 4 I xy = +2.40 × 106 mm 4 


Since we must subtract triangles,



We have

I xy = ( I xy )1 + ( I xy ) 2

For each semicircle

I xy = I x′y′ + x yA

and


Thus

I xy = Σ x yA

A, in 2 1 2

π 2

π 2

x , in.

(6)2 = 18π

−3

(6)2 = 18π

3

Σ

y , in. −

8

π

8

π

xyA

432 432 864

I xy = 864 in 4 



SOLUTION



SOLUTION

I xy = ( I xy )1 + ( I xy ) 2

We have For each rectangle

I xy = I x′y′ + x yA I x′y′ = 0 (symmetry)

and

I xy = Σ x yA

Thus

A, in 2

x , in.

y , in.

x yA, in 4

1

3 × 0.25 = 0.75

−0.520

0.362

−0.141180

2

0.25 × 1.75 = 0.4375

0.855

−0.638

−0.238652 −0.379832

Σ

I xy = −0.380 in 4 





SOLUTION I xy = ( I xy )1 + ( I xy ) 2 + ( I xy )3

We have

( I xy )1 = 0

Now symmetry implies


where


Thus

I xy = ( x yA) 2 + ( x yA)3

A, mm 2

X , mm

y , mm

x yA, mm 4

2

8 × 32 = 256

–46

–20

235,520

3

8 × 32 = 256

46

20

235,520

Σ

471,040 I xy = 471 × 103 mm 4 



and for the other rectangles


I xy = ( I xy )1 + ( I xy ) 2 + ( I xy )3

We have Now, symmetry implies

( I xy )1 = 0


and for each triangle

1 where, using the results of Sample Problem 9.6, I x′y′ = − 72 b2 h2 for both triangles. Note that the sign of I x′y′

is unchanged because the angles of rotation are 0° and 180° for triangles 2 and 3, respectively. Now A, in 2

x , in.

y , in.

x yA, in 4

2

1 (9)(15) = 67.5 2

–9

7

–4252.5

3

1 (9)(15) = 67.5 2

9

–7

–4252.5

Σ

Then

–8505  1  I xy = 2  − (9 in.)2 (15 in.)2  − 8505 in 4 72   = −9011.25 in 4

or

I xy = −9010 in 4 



SOLUTION


We have

I xy = ( I xy )1 + ( I xy ) 2 + ( I xy )3

For each rectangle


and



SOLUTION

I xy = Σ x yA

Thus

A, in 2

x , in.

y , in.

1

3.6 × 0.5 = 1.8

0.888

–2.00

2

0.5 × 3.8 = 1.9

–0.662

0.15

–0.18867

3

1.3 × 1.0 = 1.3

–0.262

2.55

–0.86853

Σ

x yA, in 4

–3.196

–4.25320 I xy = −4.25 in 4 



We have

I xy = ( I xy )1 + ( I xy ) 2

For each rectangle


and


Thus

I xy = Σ x yA

A, mm 2

1 2

76 × 12.7 = 965.2

12.7 × (127 − 12.7) = 1451.61

x yA, mm 4

x , mm

y , mm

–19.1

–37.85

697,777

12.55

25.65

467,284

Σ

1,165,061 I xy = 1.165 × 106 mm4 



SOLUTION

www.elsolucionario.net PROBLEM 9.79 Determine for the quarter ellipse of Problem 9.67 the moments of inertia and the product of inertia with respect to new axes obtained by rotating the x and y axes about O (a) through 45° counterclockwise, (b) through 30° clockwise.

SOLUTION Ix = = Iy = =

From Problem 9.67: First note

π 16

π 8

a4

π 16

π 2

I xy =

(2a )( a)3

(2a )3 (a)

a4

1 4 a 2

1 1π π  5 (I x + I y ) =  a4 + a 4  = π a4 2 2 8 2  16 1 1π π  3 (I x − I y ) =  a4 − a4  = − π a4 2 2 8 2  16

Now use Equations (9.18), (9.19), and (9.20). Equation (9.18):

I x′ = =

Equation (9.19):

I y′ = =

Equation (9.20):

I x′y′ =

1 1 ( I x + I y ) + ( I x − I y ) cos 2θ − I xy sin 2θ 2 2 5 3 1 π a 4 − π a 4 cos 2θ − a 4 sin 2θ 16 16 2 1 1 ( I x + I y ) − ( I x − I y ) cos 2θ + I xy sin 2θ 2 2 5 3 1 π a 4 + π a 4 cos 2θ + a 4 sin 2θ 16 16 2 1 ( I x − I y )sin 2θ + I xy cos 2θ 2

=−

3 1 π a 4 sin 2θ + a 4 cos 2θ 16 2



From Figure 9.12:


(a)

θ = +45°:

I x′ =

I y′ =

5 3 1 π a 4 − π a 4 cos 90° − a 4 sin 90° 16 16 2

or

I x′ = 0.482a 4 

or

I y′ = 1.482a 4 

5 3 1 π + π a 4 cos 90° + a 4 16 16 2

I x′y′ = −

3 1 π a 4 sin 90° + a 4 cos 90° 16 2

(b)

θ = −30°:

I x′ =

I y′ =

5 3 1 π a 4 − π a 4 cos( −60°) − a 4 sin( −60°) 16 16 2

or

I x′ = 1.120a 4 

or

I y′ = 0.843a 4 

or

I x′y′ = 0.760a 4 

5 3 1 π a 4 + π a 4 cos(−60°) + a 4 sin( −60°) 16 16 2

I x′y′ = −

3 1 π a 4 sin(−60°) + a 4 cos(−60°) 16 2



or I x′y′ = −0.589a 4 

www.elsolucionario.net PROBLEM 9.80 Determine the moments of inertia and the product of inertia of the area of Problem 9.72 with respect to new centroidal axes obtained by rotating the x and y axes 30° counterclockwise.

SOLUTION I xy = 2.40 × 106 mm4

Now, with two rectangles and two triangles:

1  1  I x = 2  (60 mm)(40 mm)3  + 2  (60 mm)(40 mm)3  3  12  I x = 3.20 × 106 mm 4 1  1  I y = 2  (40 mm)(60 mm)3  + 2  (40 mm)(60 mm)3  = 7.20 × 106 mm 4 3  12 

Now

1 (I x + I y ) = 2 1 (I x − I y ) = 2

1 (3.20 + 7.20) × 106 = 5.20 × 106 mm 4 2 1 (3.20 − 7.20) × 106 = −2.00 × 106 mm 4 2

Using Eqs. (9.18), (9.19), (9.20): Eq. (9.18):

1 1 ( I x + I y ) + ( I x − I y ) cos 2θ − I xy sin 2θ 2 2 = [5.20 + (−2.00) cos 60° − (2.40)sin 60°] × 106 mm 4

I x′ =

or

I x′ = 2.12 × 106 mm 4 



From Problem 9.72:


Eq. (9.19):

1 1 ( I x + I y ) − ( I x − I y ) cos 2θ + I xy sin 2θ 2 2 = [5.20 − ( −2.00) cos 60° + (2.40) sin 60°] × 106 mm 4

I y′ =

or Eq. (9.20):

I y′ = 8.28 × 106 mm 4 

1 ( I x − I y )sin 2θ + I xy sin 2θ 2 = [(−2.00)sin 60° + (2.40) cos 60°] × 106 mm 4

I x′y′ =

I x′y′ = −0.532 × 106 mm 4 


or


www.elsolucionario.net PROBLEM 9.81 Determine the moments of inertia and the product of inertia of the area of Problem 9.73 with respect to new centroidal axes obtained by rotating the x and y axes 60° counterclockwise.

SOLUTION From Problem 9.73: I xy = 864 in 4

( I x )1 = ( I x ) 2 =

π

(6 in.) 4 8 = 162π in 4

where

Then

I x = 2(162π in 4 ) = 324π in 4

Also

I y = ( I y )1 + ( I y )2 ( I y )1 = ( I y ) 2 =

where

π  (6 in.)4 +  (6 in.) 2  (3 in.) 2 = 324π in 4 8 2 

I y = 2(324π in 4 ) = 648π in 4

Then Now

π

1 (I x + I y ) = 2 1 (I x − I y ) = 2

1 (324π + 648π ) = 486π in 4 2 1 (324π − 648π ) = −162π in 4 2

Using Eqs. (9.18), (9.19), and (9.20): Eq. (9.18):

1 1 ( I x + I y ) + ( I x − I y ) cos 2θ − I xy sin 2θ 2 2 = [486π + (−162π ) cos120° − 864 sin120°] in 4

I x′ =

or

I x′ = 1033 in 4 



I x = ( I x )1 + ( I x )2

Now


Eq. (9.19):

or

I y′ = 2020 in 4 

or

I x′y′ = −873 in 4 

1 ( I x − I y ) sin 2θ + I xy cos 2θ 2 = [(−162π )sin120° + 864 cos120°] in 4

I x′y′ =


Eq. (9.20):

1 1 ( I x + I y ) − ( I x − I y ) cos 2θ + I xy sin 2θ 2 2 = [486π − (−162π ) cos120° + 864sin120°] in 4

I y′ =


www.elsolucionario.net PROBLEM 9.82 Determine the moments of inertia and the product of inertia of the area of Problem 9.75 with respect to new centroidal axes obtained by rotating the x and y axes 45° clockwise.

SOLUTION From Problem 9.75: I xy = 471,040 mm 4

where

I x = ( I x )1 + ( I x )2 + ( I x )3 1 (100 mm)(8 mm)3 12 = 4266.67 mm 4

( I x )1 =

1 (8 mm)(32 mm)3 + [(8 mm)(32 mm)](20 mm) 2 12 = 124, 245.33

( I x ) 2 = ( I x )3 =

Then

I x = [4266.67 + 2(124, 245.33)] mm 4 = 252, 757 mm 4

Also

I y = ( I y )1 + ( I y ) 2 + ( I y )3

where

Then Now

1 (8 mm)(100 mm)3 = 666, 666.7 mm 4 12 1 ( I y ) 2 = ( I y )3 = (32 mm)(8 mm)3 + [(8 mm)(32 mm)](46 mm) 2 12 = 543,061.3 mm 4 ( I y )1 =

I y = [666,666.7 + 2(543,061.3)] mm 4 = 1,752,789 mm 4 1 (I x + I y ) = 2 1 (I x − I y ) = 2

1 (252, 757 + 1, 752, 789) mm 4 = 1,002, 773 mm 4 2 1 (252, 757 − 1,752,789) mm 4 = −750, 016 mm 4 2

Using Eqs. (9.18), (9.19), and (9.20): Eq. (9.18):

1 1 ( I x + I y ) + ( I x − I y ) cos 2θ − I xy sin 2θ 2 2 = [1, 002,773 + (−750,016) cos(−90°) − 471, 040sin( −90°)]

I x′ =

or



I x′ = 1.474 × 106 mm 4 



Now


Eq. (9.19):

1 1 ( I x + I y ) − ( I x − I y ) cos 2θ + I xy sin 2θ 2 2 = [1, 002, 773 − (−750, 016) cos(−90°) + 471, 040sin(−90°)]

I y′ =

or Eq. (9.20):

I y′ = 0.532 × 106 mm4 

1 ( I x − I y ) sin 2θ + I xy cos 2θ 2 = [(−750, 016)sin(−90°) + 471, 040 cos( −90°)]

I x′y′ =

I x′y ′ = 0.750 × 106 mm 4 


or


www.elsolucionario.net PROBLEM 9.83 Determine the moments of inertia and the product of inertia of the L3 × 2 × 14 -in. angle cross section of Problem 9.74 with respect to new centroidal axes obtained by rotating the x and y axes 30° clockwise.

SOLUTION From Figure 9.13: I x = 0.390 in 4 I y = 1.09 in 4

I xy = −0.37983 in 4 Now

1 (I x + I y ) = 2 1 (I x − I y ) = 2

1 (0.390 + 1.09) in 4 = 0.740 in 4 2 1 (0.390 − 1.09) in 4 = −0.350 in 4 2

Using Eqs. (9.18), (9.19), and (9.20): Eq. (9.18):

1 1 ( I x + I y ) + ( I x − I y ) cos 2θ − I xy sin 2θ 2 2 = [0.740 + ( −0.350) cos( −60°) − (−0.37983)sin(−60°)]

I x′ =

or Eq. (9.19):

1 1 ( I x + I y ) − ( I x − I y ) cos 2θ + I xy sin 2θ 2 2 = [0.740 − (−0.350) cos(−60°) + (−0.37983) sin( −60°)]

I y′ =

or Eq. (9.20):

I x′ = 0.236 in 4 

I y′ = 1.244 in 4 

1 ( I x − I y )sin 2θ + I xy cos 2θ 2 = [(−0.350) sin(−60°) + (−0.37983) cos(−60°)]

I x′y′ =

or

I x′y′ = 0.1132 in 4 



From Problem 9.74:

www.elsolucionario.net PROBLEM 9.84 Determine the moments of inertia and the product of inertia of the L127 ×76 × 12.7-mm angle cross section of Problem 9.78 with respect to new centroidal axes obtained by rotating the x and y axes 45° counterclockwise.

SOLUTION From Figure 9.13: I x = 3.93 × 106 mm 4

From Problem 9.78: I xy = 1.165061 × 106 mm4 Now

1 1 ( I x + I y ) = (3.93 + 1.06) × 106 mm 4 2 2 = 2.495 × 106 mm 4 1 1 ( I x − I y ) = (3.93 − 1.06) × 106 mm 4 = 1.435 × 106 mm 4 2 2

Using Eqs. (9.18), (9.19), and (9.20): Eq. (9.18):

1 1 ( I x + I y ) + ( I x − I y ) cos 2θ − I xy sin 2θ 2 2 = [2.495 + 1.435cos 90° − 1.165061sin 90°] × 106 mm 4

I x′ =

or Eq. (9.19):

1 1 ( I x + I y ) − ( I x − I y ) cos 2θ + I xy sin 2θ 2 2 = [2.495 − 1.435cos 90° + 1.165061sin 90°] × 106 mm 4

I y′ =

or Eq. (9.20):

I x′ = 1.330 × 106 mm 4 

I y′ = 3.66 × 106 mm4 

1 ( I x − I y ) sin 2θ + I xy cos 2θ 2 = [(1.435sin 90° + 1.165061cos90°] × 106 mm 4

I x′y′ =

or

I x′y′ = 1.435 × 106 mm4 



I y = 1.06 × 106 mm 4

www.elsolucionario.net PROBLEM 9.85 For the quarter ellipse of Problem 9.67, determine the orientation of the principal axes at the origin and the corresponding values of the moments of inertia.

SOLUTION Ix =

Problem 9.67:

I xy =

Now Eq. (9.25):

π 8

Iy =

a4

a4

2

1 4 a 2

tan2θ m = −

2 I xy Ix − I y

=−

2 π

8

=

Then

π

(

1 2

a4

)

a − π2 a 4 4

8 = 0.84883 3π

2θ m = 40.326° and

220.326°

or Also Eq. (9.27):

I max, min = =

Ix + Iy 2

θ m = 20.2° and 110.2° 

2

 Ix − I y  2 ±   + I xy 2  

1π 4 π 4  a + a  2  8 2  2

1  π π  1  ±   a4 − a4  +  a4  2  2  2 8

2

= (0.981,748 ± 0.772, 644) a 4

or and

I max = 1.754a 4  I min = 0.209a 4 

By inspection, the a axis corresponds to I min and the b axis corresponds to I max .





From Problem 9.79:

www.elsolucionario.net PROBLEM 9.86 For the area indicated, determine the orientation of the principal axes at the origin and the corresponding values of the moments of inertia. Area of Problem 9.72.


I x = 3.20 × 106 mm4 I y = 7.20 × 106 mm 4

Now Eq. (9.25): Then

I xy = 2.40 × 106 mm 4 tan2θ m = −

2 I xy Ix − Iy

=−

2(2.40 × 106 ) = 1.200 (3.20 − 7.20) × 106

2θ m = 50.194° and 230.194°

or Ix + I y

 Ix − Iy ±   2 

θ m = 25.1° and 115.1° 

2

 2  + I xy  

Also Eq. (9.27):

I max,min =

Then

2  3.20 + 7.20   3.20 − 7.20  2 I max,min =  ±  + (2.40) × 106 mm 4    2 2    

2

= (5.20 ± 3.1241) × 106 mm 4

or and

I max = 8.32 × 106 mm 4  I min = 2.08 × 106 mm 4 






From Problem 9.72:


SOLUTION I x = 324π in 4

Problem 9.73:

I xy = 864 in 4

Now Eq. (9.25):

tan2θ m = −

2 I xy Ix − I y

I y = 648π in 4

=−

2(864) 324π − 648π

= 1.69765

Then

2θ m = 59.500° and 239.500°

θ m = 29.7° and 119.7° 

or

Also Eq. (9.27):

I max, min =

Ix + I y 2

 Ix − Iy ±   2 

2

  + I xy2   2

Then

I max, min

324π + 648π  324π − 648π  2 = ±   + 864 2 2   = (1526.81 ± 1002.75) in 4

or

I max = 2530 in 4 

and

I min = 524 in 4 






From Problem 9.81:



I x = 252, 757 mm 4

I y = 1,752,789 mm 4

Now Eq. (9.25):

I xy = 471,040 mm 4 tan 2θ m = −

2 I xy Ix − Iy

2(471, 040) 252, 757 − 1, 752, 789 = 0.62804 =−

Then

2θ m = 32.130° and 212.130°

or Also Eq. (9.27):

I max, min =

Then

I max, min =

Ix − I y 2

 Ix − Iy ±   2 

θ m = 16.07° and 106.1° 

2

  + I xy2   2

252,757 + 1, 752,789  252, 757 − 1, 752,789  2 ±   + 471040 2 2  

= (1, 002, 773 ± 885,665) mm 4

or

I max = 1.888 × 106 mm 4 

and

I min = 0.1171 × 106 mm 4 




Problem 9.75:

www.elsolucionario.net PROBLEM 9.89 For the angle cross section indicated, determine the orientation of the principal axes at the origin and the corresponding values of the moments of inertia. The L3 × 2 × 14 -in. angle cross section of Problem 9.74.

SOLUTION I x = 0.390 in 4

Problem 9.74:

I xy = −0.37983 in 4

Now Eq. (9.25):

tan 2θ m = −

2 I xy Ix − I y

I y = 1.09 in 4

=−

2(−0.37983) 0.390 − 1.09

= −1.08523

Then

2θ m = −47.341° and 132.659°

or

Also Eq. (9.27):

I max, min =

Then

I max, min =

Ix + I y 2

 Ix − I y ±   2 

θ m = −23.7° and 66.3° 

2

  + I xy2   2

0.390 + 1.09  0.390 − 1.09  2 ±   + ( −0.37983) 2 2  

= (0.740 ± 0.51650)2 in 4

or and

I max = 1.257 in 4  I min = 0.224 in 4 






From Problem 9.83:

www.elsolucionario.net PROBLEM 9.90 For the angle cross section indicated, determine the orientation of the principal axes at the origin and the corresponding values of the moments of inertia. The L127 × 76 × 12.7-mm angle cross section of Problem 9.78.


I x = 3.93 × 106 mm 4

Problem 9.78: Now Eq. (9.25):

Then

I xy = 1.165061 × 106 mm 4 tan 2θ m = −

2(1.165061 × 106 ) Ix − Iy (3.93 − 1.06) × 106 = −0.81189 2 I xy

=−

2θ m = −39.073° and 140.927°

or θ m = −19.54° and 70.5°  Also Eq. (9.27):

Then

I max, min =

I max, min

Ix + I y 2

 Ix − I y ±   2 

2

  + I xy2  

2  3.93 + 1.06   3.93 − 1.06   = ±  + 1.1650612  × 106 mm4    2 2    

= (2.495 ± 1.84840) × 106 mm 4

or

I max = 4.34 × 106 mm 4 

and

I min = 0.647 × 106 mm 4 

By inspection, the a axis corresponds to I max and the b axis corresponds to I min.



I y = 1.06 × 106 mm 4

www.elsolucionario.net PROBLEM 9.91 Using Mohr’s circle, determine for the quarter ellipse of Problem 9.67 the moments of inertia and the product of inertia with respect to new axes obtained by rotating the x and y axes about O (a) through 45° counterclockwise, (b) through 30° clockwise.

SOLUTION π 8

Iy =

π 2

a4

1 4 a 2

I xy =

Problem 9.67:

a4

The Mohr’s circle is defined by the diameter XY, where 1  1  π π X  a 4 , a 4  and Y  a 4, − a 4  2  2  8 2

Now

I ave =

1 1π π  5 ( I x + I y ) =  a 4 + a 4  = π a 4 = 0.98175a 4 2 2 8 2  16 2

and

2

2  Ix − I y   1  π 4 π 4   1 4  2 R=   + I xy =   a − a   +  a  2   2  2  8  2 

= 0.77264a 4

The Mohr’s circle is then drawn as shown. tan 2θ m = −

=−

2 I xy Ix − I y

2 π 8

(

1 2

a4

)

a − π2 a 4 4

= 0.84883

or

2θ m = 40.326°



Ix =

From Problem 9.79:


α = 90° − 40.326°

Then

= 49.674°

β = 180° − (40.326° + 60°) = 79.674°

(a)

θ = +45°:

I x′ = I ave − R cos α = 0.98175a 4 − 0.77264a 4 cos 49.674°

or

I x′ = 0.482a 4 

I y′ = I ave + R cos α = 0.98175a 4 + 0.77264a 4 cos 49.674° or

I y′ = 1.482a 4 

or

I x′y′ = −0.589a 4 

(b)

θ = −30°:

I x′ = I ave + R cos β = 0.98175a 4 + 0.77264a 4 cos 79.674°

or

I x′ = 1.120a 4 

I y′ = I ave − R cos β = 0.98175a 4 − 0.77264a 4 cos 79.674° or

I y′ = 0.843a 4 

or

I x′y′ = 0.760a 4 

I x′y′ = R sin β = 0.77264a 4 sin 79.674°



I x′y′ = − R sin α = −0.77264a 4 sin 49.674°

www.elsolucionario.net PROBLEM 9.92 Using Mohr’s circle, determine the moments of inertia and the product of inertia of the area of Problem 9.72 with respect to new centroidal axes obtained by rotating the x and y axes 30° counterclockwise.

SOLUTION I x = 3.20 × 106 mm4

From Problem 9.80:

I y = 7.20 × 106 mm4

The Mohr’s circle is defined by the diameter XY, where X (3.20 × 106 , 2.40 × 106 ) and Y (7.20 × 106, − 2.40 × 106 ). Now

and

I ave =

1 1 ( I x + I y ) = (3.20 + 7.20) × 106 = 5.20 × 106 mm 4 2 2

2 2  1  1      2 R =  ( I x − I y )  + I xy =   (3.20 − 7.20)  + (2.40)2  × 106 mm4 2     2 

= 3.1241 × 106 mm4


2 I xy Ix − Iy

2(2.40 × 106 ) (3.20 − 7.20) × 106 = 1.200 =−

or Then

2θ m = 50.1944°

α = 60° − 50.1944° = 9.8056°



I xy = 2.40 × 106 mm4

From Problem 9.72:


Then

I x′ = I ave − R cos α = (5.20 − 3.1241cos 9.8056°) × 106

or

I x′ = 2.12 × 106 mm4 

or

I y′ = 8.28 × 106 mm 4 

I y ′ = I ave + R cos α = (5.20 + 3.1241cos9.8056°) × 106

I x′y′ = − R sin α = −(3.1241 × 106 )sin 9.8056°


or I x′y′ = −0.532 × 106 mm 4 


www.elsolucionario.net PROBLEM 9.93 Using Mohr’s circle, determine the moments of inertia and the product of inertia of the area of Problem 9.73 with respect to new centroidal axes obtained by rotating the x and y axes 60° counterclockwise.

SOLUTION I x = 324π in 4

From Problem 9.81:

I y = 648π in 4

The Mohr’s circle is defined by the diameter XY, where X (324π , 864) and Y (648π , − 864). Now

I ave =

1 1 ( I x + I y ) = (324π + 648π ) = 1526.81 in 4 2 2 2

and

1 1  R =  ( I x − I y ) 2 + I xy2 =  (324π + 648π )  + 8642 2 2 

= 1002.75 in 4


2 I xy Ix − I y

2(864) 324π − 648π = 1.69765 =−

or Then

2θ m = 59.500°

α = 120° − 59.500° = 60.500°



I xy = 864 in 4

Problem 9.73:


Then



I x′ = I ave − R cos α = 1526.81 − 1002.75cos 60.500° or

I x′ = 1033 in 4 

or

I y′ = 2020 in 4 

or

I x′y ′ = −873 in 4 

I y = I ave + R cos α = 1526.81 + 1002.75cos 60.500° 


I x′y′ = − R sin α = −1002.75sin 60.500°


www.elsolucionario.net PROBLEM 9.94 Using Mohr’s circle, determine the moments of inertia and the product of inertia of the area of Problem 9.75 with respect to new centroidal axes obtained by rotating the x and y axes 45° clockwise.


I x = 252, 757 mm 4

I y = 1, 752, 789 mm 4 I xy = 471, 040 mm 4

The Mohr’s circle is defined by the diameter XY, where X (252,757; 471,040) and Y (1,752,789; −471, 040). Now

1 1 ( I x + I y ) = (252, 757 + 1, 752,789) 2 2 = 1, 002, 773 mm 4

I ave =

2

and

1  R =  ( I x − I y )  + I xy2 2  1  =  (252, 757 − 1, 752,789)2  + 471, 0402 2   = 885,665 mm 4


2 I xy Ix − Iy

2(471, 040) 252, 757 − 1, 752, 789 = 0.62804 =−

or Then

2θ m = 32.130°

α = 180° − (32.130 + 90°) = 57.870°



Problem 9.75:


Then

I x′ = I ave + R cos α = 1,002, 773 + 885,665cos57.870° I x′ = 1.474 × 106 mm 4 

or I y′ = I ave − R cos α = 1, 002,773 − 885,665cos 57.870°

or

I y′ = 0.532 × 106 mm4 

or

I x′y ′ = 0.750 × 106 mm 4 


I x′y′ = R sin α = 885, 665sin 57.870°


www.elsolucionario.net PROBLEM 9.95 Using Mohr’s circle, determine the moments of inertia and the product of inertia of the L3 × 2 × 14 -in. angle cross section of Problem 9.74 with respect to new centroidal axes obtained by rotating the x and y axes 30° clockwise.


I x = 0.390 in 4

I y = 1.09 in 4

I xy = −0.37983 in 4 The Mohr’s circle is defined by the diameter XY, where X (0.390, − 0.37983) and Y (1.09, 0.37983). Now

1 (I x + I y ) 2 1 = (0.390 + 1.09) 2 = 0.740 in 4

I ave =

2

1  R =  ( I x − I y )  + I xy2 2 

and

2

1  =  (0.390 − 1.09)  + (−0.37983)2 2 

= 0.51650 in 4


2 I xy Ix − I y

2(−0.37983) 0.390 − 1.09 = −1.08523 =−

or Then

2θ m = −47.341°

α = 60° − 47.341° = 12.659°



Problem 9.74:


Then

I x′ = I ave − R cos α = 0.740 − 0.51650 cos 12.659° or

I x′ = 0.236 in 4 

I y ′ = I ave + R cos α = 0.740 + 0.51650 cos 12.659°

or

I y = 1.244 in 4 

or

I x′y′ = 0.1132 in 4 


I x′y′ = R sin α = 0.51650 sin 12.659°


www.elsolucionario.net PROBLEM 9.96 Using Mohr’s circle, determine the moments of inertia and the product of inertia of the L127 × 76 × 12.7-mm angle cross section of Problem 9.78 with respect to new centroidal axes obtained by rotating the x and y axes 45° counterclockwise.


I x = 3.93 × 106 mm 4

Problem 9.78:

I xy = 1.165061 × 106 mm 4

The Mohr’s circle is defined by the diameter XY, where X (3.93 × 106 , 1.165061 × 106 ), Y (1.06 × 106 , −1.165061 × 106 ). Now

I ave =

1 1 ( I x + I y ) = (3.93 + 1.06) × 106 = 2.495 × 106 mm 4 2 2 2

and

1  R =  ( I x − I y )  + I xy2 2   2  1      =   (3.93 − 1.06)  + 1.1650612  × 106 mm 4    2 

= 1.84840 × 106 mm 4


2 I xy Ix − Iy

2(1.165061 × 106 ) (3.93 − 1.06) × 106 = −0.81189 =−

or

2θ m = −39.073°



I y = 1.06 × 106 mm 4


Then

α = 180° − (39.073° + 90°) = 50.927°

Then

I x′ = I ave − R cos α = (2.495 − 1.84840 cos 50.927°) × 106

or

I x′ = 1.330 × 106 mm 4 

I y′ = I ave + R cos α = (2.495 + 1.84840 cos50.927°) × 106 or

I y′ = 3.66 × 106 mm4 

or

I x′y′ = 1.435 × 106 mm4 


I x′y′ = R sin α = (1.84840 × 106 )sin 50.927°


www.elsolucionario.net PROBLEM 9.97 For the quarter ellipse of Problem 9.67, use Mohr’s circle to determine the orientation of the principal axes at the origin and the corresponding values of the moments of inertia.


Ix =

Problem 9.67:

I xy =

π 8

Iy =

a4

π 2

a4

1 4 a 2

1  1  π π X  a 4 , a 4  and Y  a 4 , − a 4  8 2 2 2    

Now

I ave =

1 1π π  ( I x + I y ) =  a 4 + a 4  = 0.98175a 4 2 2 8 2 

and 2

1  2 R =  ( I x − I y )  + I xy 2  2

1  π π   1  =   a4 − a 4  +  a4  2   2  2 8

2

= 0.77264a 4

The Mohr’s circle is then drawn as shown. tan 2θ m = − =−

2 I xy Ix − I y 2 π

8

(

1 2

a4 π

)

a − 2 a4 4

= 0.84883

or and

2θ m = 40.326°

θ m = 20.2°



The Mohr’s circle is defined by the diameter XY, where


The principal axes are obtained by rotating the xy axes through 20.2° counterclockwise 



about O.



I max, min = I ave ± R = 0.98175a 4 ± 0.77264a 4 or

I max = 1.754 a 4 

and

I min = 0.209 a 4 

From the Mohr’s circle it is seen that the a axis corresponds to I min and the b axis corresponds to I max . 



Now

www.elsolucionario.net PROBLEM 9.98 Using Mohr’s circle, determine for the area indicated the orientation of the principal centroidal axes and the corresponding values of the moments of inertia. Area of Problem 9.72.


I x = 3.20 × 106 mm 4

I y = 7.20 × 106 mm4 I xy = 2.40 × 106 mm4

The Mohr’s circle is defined by the diameter XY, where X (3.20 × 106 , 2.40 × 106 ) and Y (7.20 × 106 , − 2.40 × 106 ). 1 (I x + I y ) 2 1 = (3.20 + 7.20) × 106 mm 4 2 = 5.20 × 106 mm 4

Now

I ave =

and

1  R =  ( I x − I y )  + I xy2 2  

2

2  1      =   (3.20 − 7.20)  + (2.40) 2  × 106 mm 4    2 

= 3.1241 × 106 mm 4

The Mohr’s circle is then drawn as shown. tan 2θ m = − =−

2 I xy Ix − I y 2(2.40 × 106 ) = 1.200 (3.20 − 7.20) × 106

or

2θ m = 50.194°

and

θ m = 25.097°



From Problem 9.72:


The principal axes are obtained by rotating the xy axes through 25.1° counterclockwise  about C.

I max, min = I ave ± R = (5.20 + 3.1241) × 106

or

I max = 8.32 × 106 mm 4 

and

I min = 2.08 × 106 mm 4 




Now


SOLUTION

I xy = −9011.25 in 4 Now where

I x = ( I x )1 + ( I x )2 + ( I x )3 ( I x )1 =

1 (24 in.)(4 in.)3 = 128 in 4 12

( I x ) 2 = ( I x )3 =

1 1  (9 in.)(15 in.)3 +  (9 in.)(15 in.)  (7 in.)2 36 2  

= 4151.25 in 4

Then

I x = [128 + 2(4151.25)] in 4 = 8430.5 in 4

Also where

I y = ( I y )1 + ( I y ) 2 + ( I y )3 ( I y )1 =

1 (4 in.)(24 in.)3 = 4608 in 4 12

( I y ) 2 = ( I y )3 =

1 1  (15 in.)(9 in.)3 +  (9 in.)(15 in.)  (9 in.) 2 36 2 

= 5771.25 in 4

Then

I y = [4608 + 2(5771.25)] in 4 = 16150.5 in 4



From Problem 9.76:


The Mohr’s circle is defined by the diameter XY, where X (8430.5, − 9011.25) and Y (16150.5, 9011.25). Now

I ave =

1 1 ( I x + I y ) = (8430.5 + 16150.5) = 12290.5 in 4 2 2 2

and

1  R =  ( I x − I y )  + I xy2 2  2

1  =  (8430.5 − 16150.5)  + (−9011.25) 2 2  

= 9803.17 in 4

The Mohr’s circle is then drawn as shown. 2 I xy Ix − I y

2( −9011.25) 8430.5 − 16150.5 = −2.33452 =−

or and

2θ m = −66.812°

θ m = −33.4° The principal axes are obtained by rotating the xy axes through 33.4° clockwise 



Now

about C.

I max, min = I ave ± R = 12290.5 ± 9803.17

or

I max = 22.1 × 103 in 4 

and

I min = 2490 in 4 




tan 2θ m = −

www.elsolucionario.net PROBLEM 9.100 Using Mohr’s circle, determine for the area indicated the orientation of the principal centroidal axes and the corresponding values of the moments of inertia. Area of Problem 9.73


I x = 324π in 4

Problem 9.73:

I xy = 864 in 4

I y = 648π in 4

The Mohr’s circle is defined by the diameter XY, where X (324π , 864) and Y (648π , − 864).

and

I ave =

1 1 ( I x + I y ) = (324π + 648π ) = 1526.81 in 4 2 2


Now

1  2 R =  ( I x − I y ) 2  + I xy 2   2

1  =  (324π − 648π )  + 8642 2  

= 1002.75 in 4


2 I xy Ix − I y

2(864) 324π − 648π 1.69765 = =−

or

2θ m = 59.4998°

and

θ m = 29.7° The principal axes are obtained by rotating the xy axes through 29.7° counterclockwise  about C.



Now

I max,min = I ave ± R = 1526.81 ± 1002.75

or

I max = 2530 in 4 

and

I min = 524 in 4 





SOLUTION I x = 0.390 in 4

From Problem 9.83:

I y = 1.09 in 4

I xy = −0.37983 in 4

Problem 9.74:

Now

I ave =


The Mohr’s circle is defined by the diameter XY, where X (0.390, − 0.37983) and Y (1.09, 0.37983). 1 1 ( I x + I y ) = (0.390 + 1.09) = 0.740 in 4 2 2 2

and

1  R =  ( I x − I y )  + I xy2 2  2

1  =  (0.390 − 1.09)  + (−0.37983) 2 2 

= 0.51650 in 4


2 I xy Ix − Iy

2( −0.37983) 0.390 − 1.09 = −1.08523 =−

Then and

2θ m = −47.341°

θ m = −23.7° The principal axes are obtained by rotating the xy axes through 23.7° clockwise  about C.



Now

I max, min = I ave ± R = 0.740 ± 0.51650

or

I max = 1.257 in 4 

and

I min = 0.224 in 4 




www.elsolucionario.net PROBLEM 9.102 Using Mohr’s circle, determine for the area indicated the orientation of the principal centroidal axes and the corresponding values of the moments of inertia. Area of Problem 9.77 (The moments of inertia I x and I y of the area of Problem 9.102 were determined in Problem 9.44).


I x = 18.1282 in 4

Problem 9.77:

I xy = −4.25320 in 4

The Mohr’s circle is defined by the diameter XY, where X (18.1282, − 4.25320) and Y (4.5080, 4.25320). Now

I ave =

1 1 ( I x + I y ) = (18.1282 + 4.5080) = 11.3181 in 4 2 2 2

1  R =  ( I x − I y )  + I xy2 2 

and

2

1  =  (18.1282 − 4.5080)  + ( −4.25320)2 2 

= 8.02915 in 4


2 I xy Ix − Iy

2(−4.25320) 18.1282 − 4.5080 = 0.62454 =−

or and

2θ m = 31.986°

θ m = 15.99° The principal axes are obtained by rotating the xy axes through 15.99° counterclockwise  about C.



I y = 4.5080 in 4


Now

I max, min = I ave ± R = 11.3181 ± 8.02915 I max = 19.35 in 4 

or

I min = 3.29 in 4 

and


From the Mohr’s circle it is seen that the a axis corresponds to I max and the b axis corresponds to I min . 


www.elsolucionario.net PROBLEM 9.103 The moments and product of inertia of an L4 × 3 × 14 -in. angle cross section with respect to two rectangular axes x and y through C are, respectively, I x = 1.33 in4, I y = 2.75 in4, and I xy < 0, with the minimum value of the moment of inertia of the area with respect to any axis through C being I min = 0.692 in4. Using Mohr’s circle, determine (a) the product of inertia I xy of the area, (b) the orientation of the principal axes, (c) the value of I max .

SOLUTION (Note: A review of a table of rolled-steel shapes reveals that the given values of I x and I y are obtained when the 4-in. leg of the angle is parallel to the x axis. Further, for I xy < 0, the angle must be oriented as shown.) 1 1 ( I x + I y ) = (1.33 + 2.75) = 2.040 in 4 2 2

I ave =

and

I min = I ave − R or

R = 2.040 − 0.692

= 1.348 in 4

Using I ave and R, the Mohr’s circle is then drawn as shown; note that for the diameter XY, X (1.33, I xy ) and Y (2.75, |I xy |). 2

(a)

We have

1  R 2 =  ( I x − I y )  + I xy2 2  

or

1  I xy2 = 1.3482 −  (1.33 − 2.75)  2  

2

Solving for I xy and taking the negative root (since I xy < 0) yields I xy = −1.14586 in 4 . I xy = −1.146 in 4 

(b)

We have

tan 2θ m = −

2 I xy Ix − Iy

=−

2(−1.14586) 1.33 − 2.75

= −1.61389

or

2θ m = −58.217°

θ m = −29.1°

The principal axes are obtained by rotating the xy axes through 29.1° clockwise  about C. (c)

We have

I max = I ave + R = 2.040 + 1.348

or

I max = 3.39 in 4 



Now

www.elsolucionario.net PROBLEM 9.104 Using Mohr’s circle, determine for the cross section of the rolledsteel angle shown the orientation of the principal centroidal axes and the corresponding values of the moments of inertia. (Properties of the cross sections are given in Figure 9.13.)

SOLUTION I x = 0.162 × 106 mm 4

From Figure 9.13B:

We have

I xy = ( I xy )1 + ( I xy ) 2

For each rectangle


and

I x′y′ = 0

I xy = Σ x yA

(symmetry)

A, mm 2

x , mm

y , mm

x yA, mm 4

1

76 × 6.4 = 486.4

−13.1

9.2

−58620.93

2

6.4 × (51 − 6.4) = 285.44

21.7

–16.3

–100962.98

Σ

–159583.91 I xy = −159584 mm 4

The Mohr’s circle is defined by the diameter XY where X (0.162 × 106 , − 0.159584 × 106 ) and Y (0.454 × 106 , 0.159584 × 106 ) Now

1 1 ( I x + I y ) = (0.162 + 0.454) × 106 2 2 6 = 0.3080 × 10 mm 4

I ave =



I y = 0.454 × 106 mm 4


and

2 2  1  1      2 R =  ( I x − I y )  + I xy =   (0.162 − 0.454)  + (−0.159584) 2  × 106 2  2   

= 0.21629 × 106 mm 4


2 I xy Ix − Iy

2(−0.159584 × 106 ) (0.162 − 0.454) × 106 = −1.09304 =−

and

2θ m = −47.545

θ m = −23.8° The principal axes are obtained by rotating the xy axes through 23.8° clockwise  About C.

Now

I max, min = I ave ± R = (0.3080 ± 0.21629) × 106 I max = 0.524 × 106 mm 4 

or

I min = 0.0917 × 106 mm 4 

and




or

www.elsolucionario.net PROBLEM 9.105 Using Mohr’s circle, determine for the cross section of the rolledsteel angle shown the orientation of the principal centroidal axes and the corresponding values of the moments of inertia. (Properties of the cross sections are given in Figure 9.13.)

I x = 3.93 × 106 mm 4

From Figure 9.13B:

I y = 1.06 × 106 mm 4 I xy = −1.165061 × 106 mm4

Problem 9.7B:

(Note that the figure of Problem 9.105 is obtained by replacing x with –x in the figure of Problem 9.78; thus the change in sign of I xy .) The Mohr’s circle is defined by the diameter XY, where X (3.93 × 106 , − 1.165061 × 106 ) and Y (1.06 × 106 , 1.165061 × 106 ). Now

1 (I x + I y ) 2 1 = (3.93 + 1.06) × 106 2 = 2.495 × 106 mm 4

I ave =

2

and

1  R =  ( I x − I y )  + I xy2 2  2  1      =   (3.93 − 1.06)  + (−1.165061) 2  × 106 2    

= 1.84840 × 106 mm 4



SOLUTION



2 I xy Ix − I y

2(−1.165061 × 106 ) (3.93 − 1.06) × 106 = 0.81189 =−

or and

2θ m = 39.073°

θ m = 19.54° The principal axes are obtained by rotating the xy axes through 19.54° counterclockwise 

Now

I max, min = I ave ± R = (2.495 ± 1.84840) × 106 or

I max = 4.34 × 106 mm 4 

and

I min = 0.647 × 106 mm 4 

From the Mohr’s circle it is seen that the a axis corresponds to I max and the b axis corresponds to I min .



about C.

www.elsolucionario.net PROBLEM 9.106* For a given area the moments of inertia with respect to two rectangular centroidal x and y axes are I x = 1200 in4 and I y = 300 in4, respectively. Knowing that after rotating the x and y axes about the centroid 30° counterclockwise, the moment of inertia relative to the rotated x axis is 1450 in4, use Mohr’s circle to determine (a) the orientation of the principal axes, (b) the principal centroidal moments of inertia.

SOLUTION We have

I ave =

1 1 ( I x + I y ) = (1200 + 300) = 750 in.4 2 2

Now observe that I x > I ave , I x′ > I x , and 2θ = +60°. This is possible only if I xy < 0. Therefore, assume I xy < 0 and (for convenience) I x′y′ > 0. Mohr’s circle is then drawn as shown.

Now using ΔABD:

2θ m + α = 60°

R=

I x − I ave 1200 − 750 = cos 2θ m cos 2θ m =

Using ΔAEF:

R=

450 cos 2θ m

(in 4 )

I x′ − I ave 1450 − 750 = cos α cos α 700 (in 4 ) = cos α

Then

450 700 = cos 2θ m cos α

or

9cos (60° − 2θ m ) = 14 cos 2θ m

Expanding:

9(cos 60° cos 2θ m + sin 60° sin 2θ m ) = 14 cos 2θ m tan 2θ m =

or


We have

α = 60° − 2θ m

14 − 9 cos 60° = 1.21885 9 sin 60°

2θ m = 50.633° and θ m = 25.3°

or

(Note: 2θ m < 60° implies assumption I x′y′ > 0 is correct.) Finally, (a)

R=

450 = 709.46 in 4 cos 50.633°

From the Mohr’s circle it is seen that the principal axes are obtained by rotating the given centroidal x and y axes through θ m about the centroid C or 25.3° counterclockwise 



(b)

We have

I max, min = I ave ± R = 750 ± 709.46

or

I max = 1459 in 4 

and

I min = 40.5 in 4 




www.elsolucionario.net PROBLEM 9.107 It is known that for a given area I y = 48 × 106 mm4 and I xy = –20 × 106 mm4, where the x and y axes are rectangular centroidal axes. If the axis corresponding to the maximum product of inertia is obtained by rotating the x axis 67.5° counterclockwise about C, use Mohr’s circle to determine (a) the moment of inertia I x of the area, (b) the principal centroidal moments of inertia.

SOLUTION

From the Mohr’s circle we have 2θ m = 2(67.5°) − 90° = 45°

(a)

From the Mohr’s circle we have Ix = Iy + 2

(b)

We have

and Now

| I xy | tan 2θ m

= 48 × 106 + 2

20 × 106 tan 45°

or

I x = 88.0 × 106 mm 4 

or

I max = 96.3 × 106 mm 4 

and

I min = 39.7 × 106 mm 4 

1 1 ( I x + I y ) = (88.0 + 48) × 106 2 2 = 68.0 × 106 mm 4

I ave =

R=

| I xy | sin 2θ m

=

20 × 106 = 28.284 × 106 mm 4 sin 45°

I max, min = I ave ± R = (68.0 ± 28.284) × 106



First assume I x > I y and then draw the Mohr’s circle as shown. (Note: Assuming I x < I y is not consistent with the requirement that the axis corresponding to ( I xy ) max is obtained after rotating the x axis through 67.5° CCW.)

www.elsolucionario.net PROBLEM 9.108 Using Mohr’s circle, show that for any regular polygon (such as a pentagon) (a) the moment of inertia with respect to every axis through the centroid is the same, (b) the product of inertia with respect to every pair of rectangular axes through the centroid is zero.

SOLUTION

Because the y axis is an axis of symmetry, it follows that I xy = 0. Since I xy = 0, the x and y axes must be principal axes. Assuming I x = I max and I y = I min , the Mohr’s circle is then drawn as shown.

Now rotate the coordinate axes through an angle α as shown; the resulting moments of inertia, I x′ and I y′ , and product of inertia, I x′y′ , are indicated on the Mohr’s circle. However, the x′ axis is an axis of symmetry, which implies I x′y′ = 0. For this to be possible on the Mohr’s circle, the radius R must be equal to zero (thus, the circle degenerates into a point). With R = 0, it immediately follows that (a) (b)

I x = I y = I x′ = I y ′ = I ave (for all moments of inertia with respect to an axis through C)  I xy = I x′y′ = 0 (for all products of inertia with respect to all pairs of rectangular axes with origin at C) 



Consider the regular pentagon shown, with centroidal axes x and y.

www.elsolucionario.net PROBLEM 9.109 Using Mohr’s circle, prove that the expression I x′ I y′ − I x2′y′ is independent of the orientation of the x′ and y′ axes, where Ix′, Iy′, and Ix′y′ represent the moments and product of inertia, respectively, of a given area with respect to a pair of rectangular axes x′ and y′ through a given Point O. Also show that the given expression is equal to the square of the length of the tangent drawn from the origin of the coordinate system to Mohr’s circle.

SOLUTION

From the Mohr’s circle I x′ = I ave + R cos 2θ I y′ = I ave − R cos 2θ I x′y′ = R sin 2θ

Then, forming the expression

I x′ I y′ − I x2′y′

I x′ I y′ − I x2′y′ = ( I ave + R cos 2θ )( I ave − R cos 2θ ) − ( R sin 2θ ) 2

(

)

2 = I ave − R 2 cos 2 2θ − ( R 2 sin 2 2θ ) 2 = I ave − R2

which is a constant

I x′ I y′ − I x2′y′ is independent of the orientation of the coordinate axes Q.E.D. 

Shown is a Mohr’s circle, with line OA, of length L, the required tangent.

Noting that

OAC is a right angle, it follows that 2 L2 = I ave − R2

or

L2 = I x′ I y′ − I x2′y′ Q.E.D. 



First observe that for a given area A and origin O of a rectangular coordinate system, the values of I ave and R are the same for all orientations of the coordinate axes. Shown below is a Mohr’s circle, with the moments of inertia, I x′ and I y′ , and the product of inertia, I x′y′ , having been computed for an arbitrary orientation of the x′y ′ axes.

www.elsolucionario.net PROBLEM 9.110 Using the invariance property established in the preceding problem, express the product of inertia Ixy of an area A with respect to a pair of rectangular axes through O in terms of the moments of inertia Ix and Iy of A and the principal moments of inertia Imin and Imax of A about O. Use the formula obtained to calculate the product of inertia Ixy of the L3 × 2 × 14 -in. angle cross section shown in Figure 9.13A, knowing that its maximum moment of inertia is 1.257 in4.

SOLUTION

Case 1:

I x′ = I x , I y ′ = I y , I x′y′ = I xy

Case 2:

I x′ = I max , I y′ = I min , I x′y′ = 0

The invariance property then requires 2 I x I y − I xy = I max I min

From Figure 9.13A:

or I xy = ± I x I y − I max I min 

I x = 1.09 in 4 I y = 0.390 in 4

Using Eq. (9.21): Substituting or Then

I x + I y = I max + I min 1.09 + 0.390 = 1.257 + I min I min = 0.223 in 4 I xy = (1.09)(0.390) − (1.257)(0.223) = ±0.381 in 4

The two roots correspond to the following two orientations of the cross section. I xy = −0.381 in 4 

For

I xy = 0.381 in 4 

and for



Consider the following two sets of moments and products of inertia, which correspond to two different orientations of the coordinate axes whole origin is at Point O.

www.elsolucionario.net PROBLEM 9.111 A thin plate of mass m is cut in the shape of an equilateral triangle of side a. Determine the mass moment of inertia of the plate with respect to (a) the centroidal axes AA′ and BB′, (b) the centroidal axis CC′ that is perpendicular to the plate.

SOLUTION Area =

1  3  3 2 a a = a 2  2  4

I mass = ρ t I area =

(a)

Axis AA′:

m I area A

 1  3   a 3  3 4 I AA′,area = 2   a   = a   12  2   2   96 I AA′, mass

 3 4 1 m = I AA′, area = 3 a 2  a = ma 2  96  24 A 4  

I BB′, area

1  3  3 4 = a a = a   36  2  96

m



3

Axis BB″:

I BB′, mass =

(b)

Eq. (9.38):

m I BB′, A

area =

(we check that I AA = I BB )

 3 4 1 a2  a = ma 2   4  96  24

m 3

 1  I CC ′ = IAA′ + I BB′ = 2  ma 2  24  



I CC ′ =

1 ma 2  12



Mass = m = ρV = ρ tA

www.elsolucionario.net PROBLEM 9.112 The elliptical ring shown was cut from a thin, uniform plate. Denoting the mass of the ring by m, determine its mass moment of inertia with respect to (a) the centroidal axis BB′, (b) the centroidal axis CC′ that is perpendicular to the plane of the ring.

SOLUTION Mass = m = ρV = ρ tA = ρ t × η[(2a)(2b) − ab] = 3π ρ tab

I mass = ρ tI area

Also

=

(a)

Using Figure 9.12,

Then

m I area 3π ab

I BB′,area =

π

[(2a)(2b)3 − ab3 ]

4 15 = π ab3 4

I BB′, mass =

m 15 × π ab3 3π ab 4

or (b)

I BB′ =

5 2 mb  4

Using Figure 9.12 and symmetry, we can conclude that I AA′, mass =

Now

5 2 ma 4

I CC ′, mass = I AA′, =

mass + I BB′, mass

5 2 5 2 ma + mb 4 4

or

I CC ′ =

5 m(a 2 + b 2 )  4



First note

www.elsolucionario.net PROBLEM 9.113 A thin semicircular plate has a radius a and a mass m. Determine the mass moment of inertia of the plate with respect to (a) the centroidal axis BB′, (b) the centroidal axis CC′ that is perpendicular to the plate.

SOLUTION mass = m = ρ tA I mass = ρ t I area =

m I area A

I AA′,area = I DD′,area =

1π 4 1 4  a  = πa 2 4  8

I AA′, mass = I DD′,mass =

m m 1 4 1 2 I AA′,area = π a  = ma 1 π a2  8 A   4 2

I BB′ = I DD′ − m( AC ) 2 =

(a)

1 2  4a  ma − m   4  3π 

= (0.25 − 0.1801) ma 2

(b)

Eq. (9.38):

I CC ′ = I AA′ + I BB′ =

2

I BB′ = 0.0699ma 2 

1 ma 2 + 0.0699ma 2 4 I CC ′ = 0.320ma 2 



1 A = π a2 2

Area:

www.elsolucionario.net PROBLEM 9.114 The quarter ring shown has a mass m and was cut from a thin, uniform plate. Knowing that r1 = 34 r2, determine the mass moment of inertia of the quarter ring with respect to (a) the axis AA′, (b) the centroidal axis CC′ that is perpendicular to the plane of the quarter ring.

SOLUTION mass = m = ρV = ρ tA = ρt

(r 4

2 2

− r12

)

I mass = ρ t I area

Also

=

m

π 4

(a)

π

Using Figure 9.12,

I AA′,area =

Then

I AA′, mass =

=

π 16

(

r22

(r

4 2

− r12

)

− r14

)

m

π

(r 4

− r12

)

m 2 r2 + r12 4

)

2 2

(

I area

×

π

(r 16

m 3  =  r22 +  r2  4  4 

(b)


First note

4 2

2

 

− r14

)

or

I AA′ =

25 2 mr2  64

Symmetry implies I BB′, mass = I AA′,mass

Then

I DD′ = I AA′ + I BB′  25  = 2  mr22  64   25 2 = mr2 32



Now locate centroid C. X Σ A = Σx A

or

π  4r  π  4r  π  π X  r22 − r12  = 2  r22  − 1  r12  4 4  3π  4  3π  4   4 r23 − r13 3π r22 − r12

or

X=

Now

r =X 2 3 3 4 2 r2 − ( 4 r2 ) = 3π r 2 − ( 3 r )2 2 4 2

=

Finally, or

37 2 r2 21π

I DD′ = I CC ′ + mr 2

 37 2  25 2 mr2 = I CC ′ + m  r  21π 2  32  

2

or

I CC ′ = 0.1522mr22 



3

www.elsolucionario.net PROBLEM 9.115 A piece of thin, uniform sheet metal is cut to form the machine component shown. Denoting the mass of the component by m, determine its mass moment of inertia with respect to (a) the x axis, (b) the y axis.

SOLUTION First note

mass = m = ρV = ρ tA

I mass = ρ tI area

Also

=

(a)


 1  a  = ρ t (2a )( a) − (2a)    2  2   3 = ρ ta 2 2

Now

Then

2m I area 3a 2

I x,area = ( I x )1,area − 2( I x ) 2,area =

 1 a  1 ( a)(2a)3 − 2    (a)3  12 12 2    

=

7 4 a 12

I x, mass =

2m 7 4 × a 3a 2 12

or (b)

We have

Ix =

7 ma 2  18

I z ,area = ( I z )1,area − 2( I z )2,area 3 1 1 a  = (2a)(a )3 − 2  (a)    3  2   12

=

Then

31 4 a 48

2m 31 4 × a 3a 2 48 31 2 = ma 72

I z ,mass =



Finally,

I y ,mass = I x ,mass + I z ,mass 7 31 2 ma 2 + ma 18 72 59 2 ma = 72 =

I y = 0.819ma 2 


or


www.elsolucionario.net PROBLEM 9.116 A piece of thin, uniform sheet metal is cut to form the machine component shown. Denoting the mass of the component by m, determine its mass moment of inertia with respect to (a) the axis AA′, (b) the axis BB′, where the AA′ and BB′ axes are parallel to the x axis and lie in a plane parallel to and at a distance a above the xz plane.

SOLUTION First note that the x axis is a centroidal axis so that I = I x ,mass + md 2

I x, mass =

(a)

(b)

We have

We have

I AA′, mass =

I BB′, mass =

7 ma 2 18 7 ma 2 + m(a) 2 18

or

I AA′ = 1.389ma 2 

or

I BB′ = 2.39ma 2 

7 ma 2 + m(a 2) 2 18



and that from the solution to Problem 9.115,

www.elsolucionario.net PROBLEM 9.117 A thin plate of mass m was cut in the shape of a parallelogram as shown. Determine the mass moment of inertia of the plate with respect to (a) the x axis, (b) the axis BB′, which is perpendicular to the plate.

SOLUTION mass = m = ρ t A

(a)

m I area A

Consider parallelogram as made of horizontal strips and slide strips to form a square since distance from each strip to x axis is unchanged.

1 I x,area = a 4 3 I x, mass =

(b)

m m I x,area = 2 A a

1 4 3a   

1 I x = ma 2  3

For centroidal axis y′: 1  1 I y′,area = 2  a 4  = a 4 12  6 m m 1  1 I y′,mass = I y′,area = 2  a 4  = ma 2 A a 6  6 1 7 I y′′ = I y′ + ma 2 = ma 2 + ma 2 = ma 2 6 6

For axis BB′ ⊥ to plate, Eq. (9.38): 1 7 I BB′ = I x + I y′′ = ma 2 + ma 2 3 6

I BB′ =

3 2 ma  2



I mass = ρ t I area =

www.elsolucionario.net PROBLEM 9.118 A thin plate of mass m was cut in the shape of a parallelogram as shown. Determine the mass moment of inertia of the plate with respect to (a) the y axis, (b) the axis AA′, which is perpendicular to the plate.

SOLUTION See sketch of solution of Problem 9.117. From part b of solution of Problem 9.117: I y′ =

1 2 ma 6

I y = I y′ + ma 2 =

(b)

1 2 ma + ma 2 6

Iy =

7 ma 2  6

I AA′ =

1 ma 2  2

From solution of Problem 9.115: I y′ =

Eq. (9.38):

1 2 ma 6

1 and I x = ma 2 3

I AA′ = I y′ + I x =

1 2 1 2 ma + ma 6 3



(a)

www.elsolucionario.net PROBLEM 9.119 Determine by direct integration the mass moment of inertia with respect to the z axis of the right circular cylinder shown, assuming that it has a uniform density and a mass m.

SOLUTION m = ρV = ρπ a 2 L

For the element shown:

dm = ρπ a 2 dx m = dx L

and

dI z = dI z + x 2 dm =

Then

1 2 a dm + x 2 dm 4



I z = dI z =



L

0

1 2  2  m  a + x  dx  4  L  L

=

m 1 2 1  a x + x3  L  4 3 0

Iz =

1 m(3a 2 + 4 L2 )  12



For the cylinder:

www.elsolucionario.net PROBLEM 9.120 The area shown is revolved about the x axis to form a homogeneous solid of revolution of mass m. Using direct integration, express the mass moment of inertia of the solid with respect to the x axis in terms of m and h.

SOLUTION We have

y=

2h − h x+h a

so that

r=

h ( x + a) a

dm = ρπ r 2 dx dI x =


For the element shown: 1 2 r dm 2

2

h  = ρπ  ( x + a)  dx a  h2 1 h2 2 ( x a ) dx [( x + a)3 ]0a + = ρπ 0 3 a2 a2 1 h2 7 = ρπ 2 (8a 3 − a3 ) = ρπ ah 2 3 3 a



Then

m = dm =

Now

I x = dI x =





a



1 2 1 r ( ρπ r 2 dx) = ρπ 2 2

ρπ



ah

0

4

  a ( x + a )  dx  

1 1 h4 1 h4 5 a [( x a ) ] (32a5 − a 5 ) + = ρπ × ρπ 0 4 4 2 5a 10 a 31 = ρπ ah 4 10 =

From above: Then

3 7

ρπ ah 2 = m Ix =

31  3  2 93 2 m h = mh 10  7  70

or

I x = 1.329 mh 2 


www.elsolucionario.net PROBLEM 9.121 The area shown is revolved about the x axis to form a homogeneous solid of revolution of mass m. Determine by direct integration the mass moment of inertia of the solid with respect to (a) the x axis, (b) the y axis. Express your answers in terms of m and the dimensions of the solid.

SOLUTION We have at

(a, h): h =

k a

k = ah

or


For the element shown: r=4 dm = ρπ r 2 dx 2

 ah  = ρπ   dx  x 



m = dm =

Then



3a a

2

 ah   dx  x 

ρπ 

3a

 1 = ρπ a h  −   x a 2 2

 1  1  2 = ρπ a 2 h2  − −  −   = ρπ ah 2  3a  a   3

(a)

For the element:

Then

dI x =

1 2 1 r dm = ρπ r 4 dx 2 2



I x = dI x =



3a a

3a

4

1 1  ah   1 1 ρπ   dx = ρπ a 4 h 4  − 3  2 2  x   3 x a

 1 3  1 3  1 26 1 = − ρπ a 4 h 4   −    = × ρπ ah4 6 3 6 27 a a       =

1 2 13 13 2 mh × ρπ ah 2 × h 2 = 6 3 9 54

or

I x = 0.241 mh 2 



(b)

For the element:

dI y = dI y + x 2 dm =

Then

1 2 r dm + x 2 dm 4



I y = dI y =



3a  1  a

2 2  ah   ah  2    + x  ρπ   dx  x   4  x  

= ρπ a h

2 2



3a

3a  a

2 2   1 a 2 h2 2 2 1 a h + = − + x ρπ 1 dx a h    4 3  4 x   12 x a

2 2   1 a 2 h2    3    1 a h =  m  a − + − − + a a 3  3 3  2    12 (3a)   12 (a )  

 1 26 h 2  3  13 2  ma  × h + 3a 2  + 2a  = m  2  108   12 27 a 

or

I y = m(3a 2 + 0.1204h 2 ) 



=

www.elsolucionario.net PROBLEM 9.122 Determine by direct integration the mass moment of inertia with respect to the x axis of the pyramid shown, assuming that it has a uniform density and a mass m.

For element shown:

ab  y  y  dm = ρ dV = ρ  a  b  dy = ρ 2 y 2 dy h  h  h  2

dI x′ =

Parallel-axis theorem

1  y 1 b 2 2  ab  2 1 ab3 4 ρ ρ b dm y y dy y dy = = 12  h  12 h 2  h 2  12 h 4

1 y where d 2 = y 2 +  b  2 h  1 ab3 1 b 2 2  ab 2 dI x = ρ 4 y 4 dy +  y 2 + y ρ y dy 12 h 4 h2  h2 

2

dI x = dI x′ + d 2 dm

 ρ ab3 ab  =  + ρ 2  y 4 dy 4 h  3 h  ρ ab3 ab  + ρ 2  I x = dI x =  4 h  3 h





h

0

y 4 dy =

ρ ab3 h 15

+

ρ abh3 5



SOLUTION


1 ρ abh 3

For pyramid,

m = ρv =

Thus:

2 3h 2  1  b I x =  ρ abh   +  5  3  5

1 m(b 2 + 3h 2 )  5


Ix =


www.elsolucionario.net PROBLEM 9.123 Determine by direct integration the mass moment of inertia with respect to the x axis of the pyramid shown, assuming that it has a uniform density and a mass m.

SOLUTION See figure of solution of Problem 9.122 for element shown

For thin plate:


 y  y  dm = ρ  b  a  dy  h  h  ab dm = ρ 2 y 2 dy h dI y = dI x′′ + dI z′′ 2

2

1 y  1 y  1 dI y =  b  dm +  a  dm = 2 (b 2 + a 2 ) y 2 dm 3 h  3 h  3h 1  ab  ρ ab = 2 (a 2 + b2 ) y 2  ρ 2 y 2 dy  = 4 (a 2 + b 2 ) y 4 dy 3h  h  3h h ρ ab ρ ab 2 I y = dI y = 4 (a 2 + b 2 ) y 4 dy = (a + b 2 ) h 0 15 3h



For pyramid,

m = ρv =



1 ρ abh 3

1 1 I y =  ρ abh  (a 2 + b 2 ) 3 5 1 I y = m(a 2 + b 2 )  5


www.elsolucionario.net PROBLEM 9.124 Determine by direct integration the mass moment of inertia with respect to the y axis of the paraboloid shown, assuming that it has a uniform density and a mass m.

SOLUTION at x = h, r = a; a 2 = kh; k = Thus: r 2 =

a2 x h

dm = ρπ r 2 dx = ρπ m=



h

0

a2 h

dm = ρπ

a2 xdy h

a2 h

dI y = dI y′ + mx 2 =



h

0

xdx; m =

1 ρπ a 2 h 2

 1 a2  1 2 r dm + x 2 dm =  x + x 2  dm 4 4 h 

 1 a2 a2 a2 x + x 2  ρπ xdy = ρπ  0 0 4 h h h   a 2  1 a 2 h3 h 4  = ρπ +   h  4 h 3 4 

Iy =



a

Iy =



a

Iy =

Recall: m =

1 ρπ a 2 h; 2



h

 1 a2 2 x + x3  dx  0 4 h  

1 ρπ a 2 h(a 2 + 3h 2 )  12

1  1  I y =  ρπ a 2 h   (a 2 + 3h 2 ) 2   6  Iy =

1 m( a 2 + 3h 2 )  6



r 2 = y 2 + z 2 = kx:

www.elsolucionario.net PROBLEM 9.125 A thin rectangular plate of mass m is welded to a vertical shaft AB as shown. Knowing that the plate forms an angle θ with the y axis, determine by direct integration the mass moment of inertia of the plate with respect to (a) the y axis, (b) the z axis.


SOLUTION

Projection on yz plane

Mass of plate: m = ρ tab (a)

For element shown:

dm = ρ btdv dI y = dI y′ + (v sin θ ) 2 dm =

1 2 b dm + v 2 sin 2 θ dm 12

 1  =  b 2 + v 2 sin 2 θ  ρ btdv  12  a 1  I y = dI y = ρ bt  b 2 + v 2 sin 2 θ  dv 0  12 





1 2 v3 b v + sin θ = ρ bt 12 3

a

0

 1  a3 = ρ bt  ab 2 + sin 2 θ  3  12 

1 = ( ρ tab) (b 2 + 4a 2 sin 2 θ ) 12 Iy =

1 m(b 2 + 4a 2 sin 2 θ )  12



(b)

dI z = dI z′ + (v cos θ ) 2 dm =

1 2 b dm + v 2 cos 2 θ dm 12

 1  =  b 2 + v 2 cos 2 θ  ρ btdv  12  a 1  I z = dI z = ρ bt  b 2 + v 2 cos 2 θ  dv 0  12 





1 2 v3 = ρ bt b v + cos 2 θ 12 3

a

0

 1  a3 = ρ bt  ab 2 + cos 2 θ  3  12 

1 = ( ρ tab) (b 2 + 4a 2 cos 2 θ ) 12 1 m(b 2 + 4a 2 cos 2 θ )  12


Iz =


www.elsolucionario.net PROBLEM 9.126* A thin steel wire is bent into the shape shown. Denoting the mass per unit length of the wire by m′, determine by direct integration the mass moment of inertia of the wire with respect to each of the coordinate axes.

SOLUTION dy = − x −1/3 (a 2/3 − x 2/3 )1/2 dx

First note

2

 dy  1 +   = 1 + x −2/3 (a 2/3 − x 2/3 )  dx  a =   x


Then

2/3

2

 dy  dm = m′dL = m′ 1 +   dx  dx 

For the element shown:

1/3

a = m′    x





a

Then

m = dm =

Now

I x = y 2 dm =



0

m′

dx

a a1/3 3 3 dx = m′a1/3  x 2/3  = m′a 1/3 0 2 2 x

 a1/3  (a 2/3 − x 2/3 )3  m′ 1/3 dx  0  x  a  a2  = m′a1/3  1/3 − 3a 4/3 x1/3 + 3a 2/3 x − x5/3  dx 0 x  



a



a

9 3 3 3  = m′a1/3  a 2 x 2/3 − a 4/3 x 4/3 + a 2/3 x 2 − x8/3  2 4 2 8  0 3 9 3 3 3   = m′a3  − + −  = m′a3  2 4 2 8 8

or Symmetry implies  

Ix =

1 2 ma  4

Iy =

1 ma 2  4



Alternative solution:



 a1/3  x 2  m′ 1/3 dx  = m′a1/3 0  x  a 3 3 = m′a1/3 ×  x8/3  = m′a3 0 8 8

I y = x 2 dm =

=

Also



a



a

x5/3 dx

0

1 2 ma 4



I z = ( x 2 + y 2 ) dm = I y + I x Iz =

1 2 ma  2


or


www.elsolucionario.net PROBLEM 9.127 Shown is the cross section of an idler roller. Determine its mass moment of inertia and its radius of gyration with respect to the axis AA′. (The specific weight of bronze is 0.310 lb/in3; of aluminum, 0.100 lb/in3; and of neoprene, 0.0452 lb/in3.)

SOLUTION First note for the cylindrical ring shown that

π

(d 4

2 2

)

− d12 =

π 4

(

ρ t d 22 − d12

)

and, using Figure 9.28, that 2

2

1  d 2  1  d1  m2 − m1   2  2  2  2 π  π   1   =  ρ t × d 22  d 22 −  ρ t × d12  d12  8  4  4   

I AA′ =

1π  =  ρ t  d 24 − d14 8 4  1π  =  ρ t  d 22 − d12 8 4  1 = m d12 + d 22 8

(

)

(

)( d

(

2 2

+ d12

)

)

Now treat the roller as three concentric rings and, working from the bronze outward, we have 2 2  1   3   1   2 2 3  13 m= × ft/s (0.310 lb/in )  in.    −    in 4 32.2  16   8   4    2 2   1   3   2 3  11 + (0.100 lb/in )  in.    −    in  16   2   8   2 2   11   1   1   + (0.0452 lb/in 3 )  in.  1  −    in 2   16   8   2    = (479.96 + 183.41 + 769.80) × 10−6 lb ⋅ s 2 /ft

π

= 1.4332 × 10−3 lb ⋅ s 2 /ft



m = ρV = ρ t ×


and

 1  2  3  2   3  2  1  2  1  I AA′ = (479.96)   +    + (183.41)   +    8  4   8    8   2     1 2  1 2   1 ft 2 + (769.80) 1  +     × 10−6 lb ⋅ s 2 /ft × in 2 × 144 in 2  8   2    = (84.628 + 62.191 + 1012.78) × 10−9 lb ⋅ ft ⋅ s 2

= 1.15960 × 10−6 lb ⋅ ft ⋅ s 2

or

I AA′ = 1.160 × 10−6 lb ⋅ ft ⋅ s 2 

2 k AA ′ =

Then

k AA′ = 28.445 × 10−3 ft

or

k AA′ = 0.341 in. 



I AA′ 1.15960 × 10−6 lb ⋅ ft ⋅ s 2 = = 809.09 × 10−6 ft 2 m 1.4332 × 10−3 lb ⋅ s 2 /ft

Now

www.elsolucionario.net PROBLEM 9.128 Shown is the cross section of a molded flat-belt pulley. Determine its mass moment of inertia and its radius of gyration with respect to the axis AA′. (The density of brass is 8650 kg/m3 and the density of the fiber-reinforced polycarbonate used is 1250 kg/m3.)

SOLUTION

m = ρV = ρ t ×

π 4

(d

2 2

− d12

)

and, using Figure 9.28, that 2

2

1  d2  1 d  m2 − m1  1  2  2  2  2 π  π   1   =  ρ t × d 22  d 22 −  ρ t × d12  d12  8  4  4    1π  =  ρ t  d 24 − d14 8 4  1π  =  ρ t  d 22 − d12 d 22 + d12 8 4  1 = m d12 + d 22 8 Now treat the pulley as four concentric rings and, working from the brass outward, we have I AA′ =

(

)

(

)(

(

m=

π

)

)

{

8650 kg/m3 × (0.0175 m) × (0.0112 − 0.0052 ) m 2 4 + 1250 kg/m3 [(0.0175 m) × (0.0172 − 0.0112 ) m 2 + (0.002 m) × (0.0222 − 0.017 2 ) m 2 + (0.0095 m) × (0.0282 − 0.0222 ) m 2 ]}

= (11.4134 + 2.8863 + 0.38288 + 2.7980) × 10−3 kg = 17.4806 × 10−3 kg



First note for the cylindrical ring shown that


and

1 I AA′ = [(11.4134)(0.0052 + 0.0112 ) + (2.8863)(0.0112 + 0.017 2 ) 8 + (0.38288)(0.017 2 + 0.0222 ) + (2.7980)(0.0222 + 0.0282 )] × 10−3 kg × m 2 = (208.29 + 147.92 + 37.00 + 443.48) × 10−9 kg ⋅ m 2 = 836.69 × 10−9 kg ⋅ m 2

or Now

2 k AA ′ =

I AA′ = 837 × 10−9 kg ⋅ m 2 

I AA′ 836.69 × 10−9 kg ⋅ m 2 = = 47.864 × 10−6 m 2 −3 m 17.4806 × 10 kg k AA′ = 6.92 mm 


or


www.elsolucionario.net PROBLEM 9.129 The machine part shown is formed by machining a conical surface into a circular cylinder. For b = 12 h, determine the mass moment of inertia and the radius of gyration of the machine part with respect to the y axis.

Mass:

mcyl = ρπ a 2 h 1 mcyl a 2 2 1 = ρπ a 4 h 2

I y : I cyl =

mcone =


SOLUTION

1 h 1 ρπ a 2 = ρπ a 2 h 3 2 6

3 mcone a 2 10 1 ρπ a 4 h = 20

I cone =

For entire machine part: 1 5 ρπ a 2 h = ρπ a 2 h 6 6 1 1 9 = ρπ a 4 h − ρπ a 4 h = ρπ a 4 h 2 20 20

m = mcyl − mcone = ρπ a 2 h − I y = I cyl − I cone

or

5  6  9  I y =  ρπ a 2 h    a 2 6   5  20 

Then

k y2 =

I 27 2 = a m 50

Iy =

27 2 ma  50

k y = 0.735a 


www.elsolucionario.net PROBLEM 9.130 Given the dimensions and the mass m of the thin conical shell shown, determine the mass moment of inertia and the radius of gyration of the shell with respect to the x axis. (Hint: Assume that the shell was formed by removing a cone with a circular base of radius a from a cone with a circular base of radius a + t, where t is the thickness of the wall. In the resulting expressions, neglect terms containing t2, t3, etc. Do not forget to account for the difference in the heights of the two cones.)

First note

h h = a+t a

or

h′ =

h (a + t ) a

For a cone of height H whose base has a radius r, have Ix =

3 mr 2 10

m = ρV = ρ ×

Then

Ix = =

π 3

r2H

3 π 2  2  ρr H  r 10  3 

π 10

ρr4H

Now following the hint have

π

ρ[(a + t ) 2 h′ − a 2 h] 3 b π   = ρ  (a + t ) 2 = (a + t ) − a 2 h  3  a 

mshell = mouter − minner =

=

Neglecting the t2 and t3 terms obtain

  π t t   ρ a 2 h 1 +  − 1 = ρ a 2 h 1 + 3 +  − 1 3 a 3 a      3

π

mshell = πρ aht



SOLUTION


Also

( I x )shell = ( I x )outer − ( I x )inner = =

π 10

π 10

ρ [(a + t ) 4 h′ − a 4 h]  

 

h a

ρ ( a + t )4 × (a + t ) − a 4 h 

5   π t t   = ρ a h 1 +  − 1 = ρ a 4 h 1 + 5 +  − 1 10 a    a   10

π

4

Neglecting t2 and higher order terms, obtain

Now

k x2 =

π 2

ρ a3 ht

2 I x 12 ma = m m

or or

Ix =

1 2 ma  2

kx =

a 2





( I x )shell =

www.elsolucionario.net PROBLEM 9.131 A square hole is centered in and extends through the aluminum machine component shown. Determine (a) the value of a for which the mass moment of inertia of the component with respect to the axis AA′, which bisects the top surface of the hole, is maximum, (b) the corresponding values of the mass moment of inertia and the radius of gyration with respect to the axis AA′. (The specific weight of aluminum is 0.100 lb/in3.)

SOLUTION First note

m1 = ρV1 = ρ b 2 L

and

m2 = ρV2 = ρ a 2 L

Using Figure 9.28 and the parallel-axis theorem, we have I AA′ = ( I AA′ )1 − ( I AA′ ) 2 2 1 a  =  m1 (b 2 + b 2 ) + m1     2   12 2 1 a  −  m2 (a 2 + a 2 ) + m2     2   12 1  1  5  = ( ρ b 2 L)  b 2 + a 2  − ( ρ a 2 L)  a 2  6 4 12     ρL 4 2 2 4 = (2b + 3b a − 5a ) 12

Then

dI AA′ ρ L = (6b 2 a − 20a 3 ) = 0 da 12

or

a=0 d 2 I AA′

Also

da

=

ρL 12

d 2 I AA′

Now for a = 0, and for a = b

2

da 2 3 , 10

a=b

and

d 2 I AA′ da 2

3 10

(6b 2 − 60a 2 ) =

1 ρ L(b2 − 10a 2 ) 2

>0 <0



(a)


( I AA′ )max occurs when

Then

a=b

3 10

a = (4.2 in.)

3 10

or From part a: ( I AA′ ) max

ρ L 

2 4   3  3   49 2b + 3b  b = − 5 b = ρ Lb 4  10   10   240 12       

=

4

2

 1 ft  49 γ AL 4 49 0.100 lb/in 3 Lb = × × (15 in.)(4.2 in.) 4   2 240 g 240 32.2 ft/s  12 in. 

or Now where

2 k AA ′ =

2

(I AA′ )max = 20.6 × 10−3 lb ⋅ ft ⋅ s 2 

( I AA′ ) max m

m = m1 − m2 = ρ L(b 2 − a 2 ) 2   3   2  = ρL b − b  10        7 = ρ Lb2 10

Then

2 k AA ′ =

49 ρ Lb 4 240 7 ρ Lb 2 10

=

7 2 7 b = (4.2 in.)2 24 24

or

k AA′ = 2.27 in. 



(b)

a = 2.30 in. 

www.elsolucionario.net PROBLEM 9.132 The cups and the arms of an anemometer are fabricated from a material of density ρ. Knowing that the mass moment of inertia of a thin, hemispherical shell of mass m and thickness t with respect to its centroidal axis GG′ is 5ma 2 /12, determine (a) the mass moment of inertia of the anemometer with respect to the axis AA′, (b) the ratio of a to l for which the centroidal moment of inertia of the cups is equal to 1 percent of the moment of inertia of the cups with respect to the axis AA′.

SOLUTION First note and

marm = ρVarm = ρ ×

π 4

d 2l

dmcup = ρ dVcup = ρ[(2π a cos θ )(t )(adθ )]

Then



mcup = dmcup =



π /2 0

2πρ a 2 t cos θ dθ

= 2πρ a 2 t[sin θ ]π0 /2 = 2πρ a 2 t

Now

( I AA′ )anem = ( I AA′ )cups + ( I AA′ )arms

Using the parallel-axis theorem and assuming the arms are slender rods, we have 2    ( I AA′ )anem = 3 ( I GG ′ )cup + mcup d AG  + 3  I arm + marm d AGarm  2 2  1  5  a    l  = 3  mcup a 2 + mcup (l + a) 2 +     + 3  marm l 2 + marm    12  2     2     2 

5  = 3mcup  a 2 + 2la + l 2  + marm l 2 3  5  π  = 3(2πρ a 2 t )  a 2 + 2la + l 2  +  ρ d 2 l  (l 2 ) 3  4 

or

  5 a2 a  d 2l  ( I AA′ )anem = πρ l 2 6a 2 t  2 + 2 + 1 +   l 3 l  4  



(a)


(b)

We have

or

( I GG ′ )cup ( I AA′ )cup

= 0.01

5 5  mcup a 2 = 0.01mcup  a 2 + 2la + l 2  (from part a) 12 3  

Then

5  5ζ 2 = 0.12  ζ 2 + 2ζ + 1 3 

or

40ζ 2 − 2ζ − 1 = 0 2 ± (−2) 2 − 4(40)(−1) 2(40)

Then

ζ =

or

ζ = 0.1851 and ζ = − 0.1351 a = 0.1851  l



a Now let ζ = . l

www.elsolucionario.net PROBLEM 9.133 After a period of use, one of the blades of a shredder has been worn to the shape shown and is of mass 0.18 kg. Knowing that the mass moments of inertia of the blade with respect to the AA′ and BB′ axes are 0.320 g ⋅ m 2 and 0.680 g ⋅ m 2 , respectively, determine (a) the location of the centroidal axis GG′, (b) the radius of gyration with respect to axis GG′.

SOLUTION (a)

d B = (0.08 − d A ) m

We have and, using the parallel axis

I AA′ = I GG′ + md A2

Then

(

I BB′ − I AA′ = m d B2 − d A2

)

= m (0.08 − d A ) 2 − d A2  = m(0.0064 − 0.16d A )

Substituting:

(0.68 − 0.32) × 10−3 kg ⋅ m 2 = 0.18 kg(0.0064 − 0.16d A ) m 2

or

d A = 27.5 mm 

to the right of A (b)

We have

I AA′ = I GG′ + md A2

or

I GG ′ = 0.32 × 10−3 kg ⋅ m 2 − 0.18 kg ⋅ (0.0275 m)2 = 0.183875 × 10−3 kg ⋅ m 2

Then

2 kGG ′ =

I GG′ 0.183875 × 10−3 kg ⋅ m 2 = = 1.02153 × 10−3 m 2 m 0.18 kg

or

kGG ′ = 32.0 mm 



I BB′ = I GG′ + md B2


SOLUTION First note that the given shape can be formed adding a small cone to a cylinder and then removing a larger cone as indicated.

h h + 2.4 = 0.4 1.2

Now

or h = 1.2 in.

The weight of the body is given by W = mg = g (m1 + m2 − m3 ) = ρ g (V1 + V2 − V3 )

or

0.9 lb = ρ × 32.2 ft/s 2

π π   3  1 ft  2 2 2 π (0.8) (2.4) + 3 (0.2) (1.2) − 3 (0.6) (3.6)  in ×  12 in.     

3

= ρ × 32.2 ft/s 2 (2.79253 + 0.02909 − 0.78540) × 10−3 ft 3

or Then

ρ = 13.7266 lb ⋅ s 2 /ft 4 m1 = (13.7266)(2.79253 × 10 −3 ) = 0.038332 lb ⋅ s 2 /ft m2 = (13.7266)(0.02909 × 10 −3 ) = 0.000399 lb ⋅ s 2 /ft m3 = (13.7266)(0.78540 × 10−3 ) = 0.010781 lb ⋅ s 2 /ft



Determine the mass moment of inertia of the 0.9-lb machine component shown with respect to the axis AA′.


Finally, using Figure 9.28, we have I AA′ = ( I AA′ )1 + ( I AA′ ) 2 − ( I AA′ )3 =

1 3 3 m1a12 + m2 a22 − m3 a32 2 10 10

3 3 1   1 ft  =  (0.038332)(0.8) 2 + (0.000399)(0.2) 2 − (0.010781)(0.6)2  (lb ⋅ s 2 /ft) × in 2 ×   10 10 2   12 in. 

2

= (85.1822 + 0.0333 − 8.0858) × 10−6 lb ⋅ ft ⋅ s 2 I AA′ = 77.1 × 10−6 lb ⋅ ft ⋅ s 2 


or


www.elsolucionario.net To the instructor: The following formulas for the mass moment of inertia of thin plates and a half cylindrical shell are derived at this time for use in the solutions of Problems 9.135 through 9.140. Thin rectangular plate ( I x )m = ( I x′ ) m + md 2 =

 b  2  h  2  1 m(b 2 + h 2 ) + m   +    12  2   2  

1 = m(b2 + h 2 ) 3 ( I y ) m = ( I y′ ) m + md 2 2

1 1 b = mb 2 + m   = mb 2 12 3 2

2

=

1 1 h mh 2 + m   = mh 2 12 2 3  

Thin triangular plate We have

1  m = ρV = ρ  bht  2  1 3 bh 36

and

I z ,area =

Then

I z ,mass = ρ tI z ,area = ρt ×

1 3 bh 36

=

1 mh 2 18

Similarly,

I y ,mass =

1 mb2 18

Now

I x, mass = I y ,mass + I z ,mass =

1 m(b2 + h 2 ) 18

Thin semicircular plate We have and

π  m = ρV = ρ  a 2 t  2   I y ,area = I z ,area =

π 8

a4



I z = ( I z′ ) m + md 2

www.elsolucionario.net (Continued)

I y ,mass = I z ,mass = ρ tI y ,area = ρt × =

π 8

a4

1 2 ma 4 1 ma 2 2

Now


Also

I x, mass = I x′,mass + my 2

or

 1 16 I x′,mass = m  − 2  2 9π

 2 a 

and

I z ,mass = I z′,mass + my 2

or

 1 16 I z ′,mass = m  − 2  4 9π

 2 a 

y=z =

4a 3π

Thin Quarter-Circular Plate We have

π  m = ρV = ρ  a 2 t  4  

π

and

I y ,area = I z ,area =

Then

I y ,mass = I z ,mass = ρ tI y ,area

16

a4

= ρt ×

π

16 1 = ma 2 4

a4

1 ma 2 2

Now


Also

I x, mass = I x′,mass + m( y 2 + z 2 )

or

 1 32 I x′,mass = m  − 2  2 9π

and

I y ,mass = I y′,mass + mz 2

or

 1 16 I y′,mass = m  − 2  4 9π

 2 a 

 2 a  



Then

www.elsolucionario.net PROBLEM 9.135 A 2-mm thick piece of sheet steel is cut and bent into the machine component shown. Knowing that the density of steel is 7850 kg/m3, determine the mass moment of inertia of the component with respect to each of the coordinate axes.

SOLUTION First compute the mass of each component. We have m = ρSTV = ρST t A m1 = (7850 kg/m3 )(0.002 m)(0.76 × 0.48) m 2 = 5.72736 kg 1  m2 = (7850 kg/m3 )(0.002 m)  × 0.76 × 0.48  m 2 2   = 2.86368 kg π  m3 = (7850 kg/m3 )(0.002 m)  × 0.482  m 2 4  = 2.84101 kg

Using Figure 9.28 for component 1 and the equations derived above for components 2 and 3, we have I x = ( I x )1 + ( I x )2 + ( I x )3 2 1  0.48   2 =  (5.72736 kg)(0.48 m) + (5.72736 kg)  m   2   12 2 1  0.48    1  +  (2.86368 kg)(0.48 m) 2 + (2.86368 kg)  m   +  (2.84101 kg)(0.48 m) 2  18 3 2      

= [(0.109965 + 0.329896) + (0.036655 + 0.073310) + (0.327284)] kg ⋅ m 2 = (0.439861 + 0.109965 + 0.327284) kg ⋅ m 2

or

I x = 0.877 kg ⋅ m 2 

        PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1587


Then


I y = ( I y )1 + ( I y )2 + ( I y )3  1  0.76 2  0.48 2  2  =  (5.72736 kg)(0.762 + 0.482 ) m 2 + (5.72736 kg)   +  m  12  2   2     2 1  0.76    1  2 +  (2.86368 kg)(0.76 m) + (2.86368 kg)  m   +  (2.84101 kg)(0.48 m) 2   3    4  18

= [(0.385642 + 1.156927) + (0.091892 + 0.183785) + (0.163642)] kg ⋅ m 2 = (1.542590 + 0.275677 + 0.163642) kg ⋅ m 2

or

I y = 1.982 kg ⋅ m 2 

2 1  0.76   =  (5.72736 kg)(0.76 m) 2 + (5.72736 kg)  m   2   12  1  0.76 2  0.48 2  2  +  (2.86368 kg)(0.762 + 0.482 ) m 2 + (2.86368 kg)   +  m  18   3   3   1  +  (2.84101 kg)(0.48 m) 2  4 

I z = [(0.275677 + 0.827031) + (0.128548 + 0.257095) + (0.163642)] kg ⋅ m 2 = (1.102708 + 0.385643 + 0.163642) kg ⋅ m 2

or

I z = 1.652 kg ⋅ m 2 



I z = ( I z )1 + ( I z ) 2 + ( I z )3


SOLUTION First compute the mass of each component. We have m = ρSTV = ρST tA

m1 = (7850 kg/m3 )(0.002 m)(0.35 × 0.39) m 2 = 2.14305 kg π  m2 = (7850 kg/m3 )(0.002 m)  × 0.1952  m 2 = 0.93775 kg 2  1  m3 = (7850 kg/m3 )(0.002 m)  × 0.39 × 0.15  m 2 = 0.45923 kg 2 

Using Figure 9.28 for component 1 and the equations derived above for components 2 and 3, we have I x = ( I x )1 + ( I x )2 + ( I x )3 2 1  0.39   =  (2.14305 kg)(0.39 m) 2 + (2.14305 kg)  m   2   12  1 16    4 × 0.195 2  2 +  − 2  (0.93775 kg)(0.195 m) + (0.93775 kg)  + (0.195) 2  m 2   2 9π    3π   

 1  0.39 2  0.15 2  2  2 2 2 +  (0.45923 kg)[(0.39) + (0.15) ] m + (0.45923 kg)   +  m   3   3   18  = [(0.027163 + 0.081489) + (0.011406 + 0.042081) + (0.004455 + 0.008909)] kg ⋅ m 2 = (0.108652 + 0.053487 + 0.013364) kg ⋅ m 2 = 0.175503 kg ⋅ m 2

or

I x = 175.5 × 10−3 kg ⋅ m 2 



Then


I y = ( I y )1 + ( I y )2 + ( I y )3  1  0.35 2  0.39 2  2  =  (2.14305 kg)[(0.35) 2 + (0.39) 2 ] m 2 + (2.14305 kg)   +   m  12  2   2     1  +  (0.93775 kg)(0.195 m)2 + (0.93775 kg)(0.195 m) 2  4  2  1   0.39   2  +  (0.45923 kg)(0.39 m) 2 + (0.45923 kg) (0.35) 2 +    m   3   18   = [(0.049040 + 0.147120) + (0.008914 + 0.035658) + (0.003880 + 0.064017)] kg ⋅ m 2

= 0.308629 kg ⋅ m 2

or

I y = 309 × 10−3 kg ⋅ m 2 

I z = ( I z )1 + ( I z ) 2 + ( I z )3 2 1  0.35   m  =  (2.14305 kg)(0.35 m) 2 + (2.14305 kg)   2   12

1  +  (0.93775 kg)(0.195 m) 2  4  2  1   0.15   2  2 2 +  (0.45923 kg)(0.15 m) + (0.45923 kg) (0.35) +    m  18  3     

= [(0.021877 + 0.065631) + 0.008914) + (0.000574 + 0.057404)] kg ⋅ m 2 = (0.087508 + 0.008914 + 0.057978) kg ⋅ m 2 = 0.154400 kg ⋅ m 2

or

I z = 154.4 × 10−3 kg ⋅ m 2 



= (0.196160 + 0.044572 + 0.067897) kg ⋅ m 2

www.elsolucionario.net PROBLEM 9.137 A subassembly for a model airplane is fabricated from three pieces of 1.5-mm plywood. Neglecting the mass of the adhesive used to assemble the three pieces, determine the mass moment of inertia of the subassembly with respect to each of the coordinate axes. (The density of the plywood is 780 kg/m3 .)

SOLUTION First compute the mass of each component. We have m = ρV = ρ tA

1  m1 = m2 = (780 kg/m3 )(0.0015 m)  × 0.3 × 0.12  m 2 2  = 21.0600 × 10−3 kg π  m3 = (780 kg/m3 )(0.0015 m)  × 0.122  m 2 = 13.2324 × 10−3 kg 4  

Using the equations derived above and the parallel-axis theorem, we have ( I x )1 = ( I x ) 2 I x = ( I x )1 + ( I x ) 2 + ( I x )3 2 1  0.12   m  = 2  (21.0600 × 10−3 kg)(0.12 m) 2 + (21.0600 × 10−3 kg)   3   18 1  +  (13.2324 × 10−3 kg)(0.12 m) 2  2 

= [2(16.8480 + 33.6960) + (95.2733)] × 10−6 kg ⋅ m 2 = [2(50.5440) + (95.2733)] × 10−6 kg ⋅ m 2

or

I x = 196.4 × 10−6 kg ⋅ m 2 

        PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1591


Then

www.elsolucionario.net PROBLEM 9.137 (Continued)  

I y = ( I y )1 + ( I y )2 + ( I y )3



2 1  0.3   −3 2 m  =  (21.0600 × 10 kg)(0.3 m) + (21.0600 kg)   3   18



1 +  (21.0600 × 10−3 kg)[(0.3)2 + (0.12) 2 ] m 2  18  0.3 2  0.12 2  2  + (21.0600 × 10 kg)   +  m    3   3   −3

 

1  +  (13.2324 × 10−3 kg)(0.12 m) 2   4  = [(105.300 + 210.600) + (122.148 + 244.296) 



= (315.900 + 366.444 + 47.637) × 10−6 kg ⋅ m 2 

or

I y = 730 × 10−6 kg ⋅ m 2 

Symmetry implies I y = I z I z = 730 × 10−6 kg ⋅ m 2 



+ (47.637)] × 10−6 kg ⋅ m 2

www.elsolucionario.net PROBLEM 9.138 The cover for an electronic device is formed from sheet aluminum that is 0.05 in. thick. Determine the mass moment of inertia of the cover with respect to each of the coordinate axes. (The specific weight of aluminum is 0.100 lb/in 3 .)

SOLUTION First compute the mass of each component. We have

Then

γ g

tA

m1 =

0.100 lb/in 3 × 0.05 in. × (3 × 2.4) in 2 = 1.11801 × 10−3 lb ⋅ s 2 /ft 2 32.2 ft/s

m2 =

0.100 lb/in 3 × 0.05 in. × (3 × 6.2) in 2 = 2.88820 × 10−3 lb ⋅ s 2 /ft 32.2 ft/s 2

m3 = m4 =

0.100 lb/in 3 × 0.05 in. × (2.4 × 6.2) in 2 = 2.31056 × 10 −3 lb ⋅ s 2 /ft 32.2 ft/s 2

Using Figure 9.28 and the parallel-axis theorem, we have ( I x )3 = ( I x ) 4 I x = ( I x )1 + ( I x )2 + ( I x )3 + ( I x ) 4

1 =  (1.11801 × 10−3 lb ⋅ s 2 /ft)(2.4 in.) 2 12 2 2  2.4    1 ft  + (1.11801 × 10−3 lb ⋅s 2 /ft)  in.      2    12 in.  1 +  (2.88820 × 10−3 lb ⋅ s 2 /ft)(6.2 in.) 2 12 2 2  6.2    1 ft  in.    + (2.88820 × 10−3 lb ⋅ s 2 /ft)    2    12 in.  1 + 2  (2.31056 × 10−3 lb ⋅ s 2 /ft) (2.4)2 + (6.2)2  in 2 12  2.4 2  6.2 2  2   1 ft 2 + (2.31056 × 10 lb ⋅ s /ft)   +   in  12 in.   2   2      −3

2



m = ρV =

www.elsolucionario.net PROBLEM 9.138 (Continued) = [(3.7267 + 11.1801) + (64.2491 + 192.7472) + 2(59.1011 + 177.3034)] × 10−6 lb ⋅ ft ⋅ s 2 = [14.9068 + 256.9963 + 2(236.4045)] × 10−6 lb ⋅ ft ⋅ s 2

or

I x = 745 × 10−6 lb ⋅ ft ⋅ s 2 

I y = ( I y )1 + ( I y ) 2 + ( I y )3 + ( I y ) 4

1 =  (1.11801 × 10−3 lb ⋅ s 2 /ft)(3 in.) 2 12 2 2 3    1 ft  + (1.11801 × 10−3 lb ⋅ s 2 /ft)  in.     2    12 in. 

 3 2  6.2 2  2   1 ft 2 + (2.88820 × 10−3 lb ⋅ s 2 /ft)   +    in     2   2     12 in.  1 +  (2.31056 × 10−3 lb ⋅ s 2 /ft)(6.2 in.)2 12 2 2  6.2    1 ft  + (2.31056 × 10−3 lb ⋅ s 2 /ft)  in.      2    12 in.  1 +  (2.31056 × 10−3 lb ⋅ s 2 /ft)(6.2 in.) 2 12 2 2   6.2   2   1 ft  + (2.31056 × 10−3 lb ⋅ s 2 /ft)  (3)2 +  in       2      12 in.  = [(5.8230 + 17.4689) + (79.2918 + 237.8754) + (51.3993 + 154.1978)

+ (51.3993 + 298.6078)] × 10−6 lb ⋅ ft ⋅ s 2 = (23.2919 + 317.1672 + 205.5971 + 350.0071) × 10−6 lb ⋅ ft ⋅ s 2

or

I y = 896 × 10−6 lb ⋅ ft ⋅ s 2 



1 +  (2.88820 × 10−3 lb ⋅ s 2 /ft)[(3)2 + (6.2)2 ] in 2 12


I z = ( I z )1 + ( I z ) 2 + ( I z )3 + ( I z ) 4

1 =  (1.11801 × 10−3 lb ⋅ s 2 /ft)[(3) 2 + (2.4) 2 ] in 2 12  3 2  2.4 2  2   1 ft 2 + (1.11801 × 10−3 lb ⋅ s 2 /ft)   +    in      12 in.   2   2   1 +  (2.88820 × 10 −3 lb ⋅ s 2 /ft)(3 in.) 2 12 2 2 3    1 ft  + (2.88820 × 10 lb ⋅ s /ft)  in.     2    12 in.  −3

1 +  (2.31056 × 10−3 lb ⋅ s 2 /ft)(2.4 in.) 2 12 2 2  2.4    1 ft  + (2.31056 × 10−3 lb ⋅ s 2 / ft)  in.      2    12 in.  1 +  (2.31056 × 10−3 lb ⋅ s 2 /ft)(2.4 in.) 2 12 2   2.4  + (2.31056 × 10−3 lb ⋅ s 2 /ft)  (3)2 +    2   = [(9.5497 + 28.6490) + (15.0427 + 45.1281)

 2   1 ft 2  in       12 in. 

+ (7.7019 + 23.1056) + (7.7019 + 167.5156)] × 10−6 lb ⋅ ft ⋅ s 2 = (38.1987 + 60.1708 + 30.8075 + 175.2175) × 10−6 lb ⋅ ft ⋅ s 2

or

I z = 304 × 10−6 lb ⋅ ft ⋅ s 2 



2

www.elsolucionario.net PROBLEM 9.139 A framing anchor is formed of 0.05-in.-thick galvanized steel. Determine the mass moment of inertia of the anchor with respect to each of the coordinate axes. (The specific weight of galvanized steel is 470 lb/ft 3 .)

SOLUTION

m = ρV =

γ G.S. g

tA

Then 470 lb/ft 3  1 ft  × 0.05 in. × (2.25 × 3.5) in 2 ×   32.2 ft/s 2  12 in.  = 3325.97 × 10−6 lb ⋅ s 2 /ft

3

m1 =

m2 =

470 lb/ft 3  1 ft  × 0.05 in. × (2.25 × 1) in 2 ×   2 32.2 ft/s  12 in. 

3

= 950.28 × 10−6 lb ⋅ s 2 /ft 470 lb/ft 3 1   1 ft  × 0.05 in. ×  × 2 × 4.75  in 2 ×   2 32.2 ft/s 2    12 in.  = 2006.14 × 10−6 lb ⋅ s 2 /ft

3

m3 =

Using Figure 9.28 for components 1 and 2 and the equations derived above for component 3, we have I x = ( I x )1 + ( I x )2 + ( I x )3 1 =  (3325.97 × 10−6 lb ⋅ s 2 /ft)(3.5 in.) 2 12  3.5 2    1 ft  in    + (3325.97 × 10−6 lb ⋅ s 2 /ft)    2    12 in.  1 +  (950.28 × 10−6 lb ⋅ s 2 /ft)(1 in.) 2 12

2

2   1 ft 2  1  + (950.28 × 10−6 lb ⋅ s 2 /ft) (3.5) 2 +    in 2     2      12 in. 



First compute the mass of each component. We have


1 +  (2006.14 × 10−6 lb ⋅ s 2 /ft)[(4.75) 2 + (2) 2 ] in 2 18 2 2   1 ft 2  2  1   + (2006.14 × 10−6 lb ⋅ s 2 /ft)  × 4.75  +  × 2   in 2      3     12 in.   3

= [(23.578 + 70.735) + (0.550 + 82.490) + (20.559 + 145.894)] × 10−6 lb ⋅ ft ⋅ s 2 = (94.313 + 83.040 + 166.453) × 10−6 lb ⋅ ft ⋅ s 2

or

I x = 344 × 10−6 lb ⋅ ft ⋅ s 2 

1 =  (3325.97 × 10−6 lb ⋅ s 2 /ft)(2.25 in.)2 12 2 2  2.25    1 ft  + (3325.97 × 10−6 lb ⋅ s2 /ft)  in.      2    12 in. 

1 +  (950.28 × 10 −6 lb ⋅ s 2 /ft)[(2.25) 2 + (1) 2 ] in 2 12  2.25 2  1 2  2   1 ft 2 + (950.28 × 10−6 lb ⋅ s 2 /ft)   +    in     2   2     12 in.  1 +  (2006.14 × 10−6 lb ⋅ s 2 /ft)(2 in.)2 18 2   1 ft 2  1   + (2006.14 × 10−6 lb ⋅ s 2 /ft)  (2.25)2 +  × 2   in 2     3     12 in.   = [(9.744 + 29.232) + (3.334 + 10.002)

+ (3.096 + 76.720)] × 10−6 lb ⋅ ft ⋅ s 2 = (38.976 + 13.336 + 79.816) × 10−6 lb ⋅ ft ⋅ s 2

or

I y = 132.1 × 10−6 lb ⋅ ft ⋅ s 2 



I y = ( I y )1 + ( I y ) 2 + ( I y )3


I z = ( I z )1 + ( I z ) 2 + ( I z )3 1 =  (3325.97 × 10 −6 lb ⋅ s 2 /ft)[(2.25) 2 + (3.5)2 ] in 2 12  2.25 2  3.5 2  2   1 ft 2 + (3325.97 × 10−6 lb ⋅ s 2 /ft)   +   in      12 in.   2   2  

1 +  (950.28 × 10−6 lb ⋅ s2 /ft)(2.25 in.) 2 12

2   1 ft 2  2   + (2006.14 × 10−6 lb ⋅ s 2 /ft) (2.25)2 +  × 4.75   in 2    3      12 in.  = [(33.322 + 99.967) + (2.784 + 89.192)

+ (17.463 + 210.231)] × 10−6 lb ⋅ ft ⋅ s 2 = (133.289 + 91.976 + 227.694) × 10−6 lb ⋅ ft ⋅ s 2

or

I z = 453 × 10−6 lb ⋅ ft ⋅ s 2 



 2.25 2  2   1 ft 2 2 + (950.28 × 10−6 lb ⋅ s 2 /ft)  + (3.5)  in      2     12 in.  1 +  (2006.14 × 10−6 lb ⋅ s 2 /ft)(4.75 in.) 2 18

www.elsolucionario.net PROBLEM 9.140* A farmer constructs a trough by welding a rectangular piece of 2-mm-thick sheet steel to half of a steel drum. Knowing that the density of steel is 7850 kg/m3 and that the thickness of the walls of the drum is 1.8 mm, determine the mass moment of inertia of the trough with respect to each of the coordinate axes. Neglect the mass of the welds.


Then = 2.76948 kg

m2 = (7850 kg/m3 )(0.0018 m)(π × 0.285 × 0.84) m 2 = 10.62713 kg

π  m3 = m4 = (7850 kg/m3 )(0.0018 m)  × 0.2852  m 2 2  = 1.80282 kg

Using Figure 9.28 for component 1 and the equations derived above for components 2 through 4, we have ( I x )3 = ( I x ) 4 I x = ( I x )1 + ( I x ) 2 + ( I x )3 + ( I x ) 4 2 1 0.21  2   m  =  (2.76948 kg)(0.21 m) 2 + (2.76948 kg)  0.285 − 2   12  1  + [(10.62713 kg)(0.285 m)2 ] + 2  (1.80282 kg)(0.285 m) 2  2  

= [(0.01018 + 0.08973) + (0.86319) + 2(0.07322)] kg ⋅ m 2 = [(0.09991 + 0.86319 + 2(0.07322)] kg ⋅ m 2

or

I x = 1.110 kg ⋅ m 2 

   



m1 = (7850 kg/m3 )(0.002 m)(0.84 × 0.21) m 2

www.elsolucionario.net PROBLEM 9.140* (Continued) I y = ( I y )1 + ( I y )2 + ( I y )3 + ( I y ) 4 1 =  (2.76948 kg)[(0.84) 2 + (0.21) 2 ] m 2 12 2  0.84 2  0.21   2 0.285 + (2.76948 kg)  + − m   2    2  

  

2  1  0.84   2 2 2 m  +  (10.62713 kg)[(0.84) + 6(0.285) ] m + (10.62713 kg)   2   12

1  +  (1.80282 kg)(0.285 m) 2 + (1.80282 kg)(0.84 m) 2  4  = [(0.17302 + 0.57827) + (1.05647 + 1.87463) + (0.03661) + (0.03661 + 1.27207)] kg ⋅ m 2 = (0.75129 + 2.93110 + 0.03661 + 1.30868) kg ⋅ m 2

or

I y = 5.03 kg ⋅ m 2 

I z = ( I z )1 + ( I z ) 2 + ( I z )3 + ( I z ) 4 2 1  0.84   =  (2.76948 kg)(0.84 m) 2 + (2.76948 kg)  m   2   12 2  1  0.84   2 2 2 m  +  (10.62713 kg)[(0.84) + 6(0.285) ] m + (10.62713 kg)   2   12 1  +  (1.80282 kg)(0.285 m) 2  4 

1  +  (1.80282 kg)(0.285 m)2 + (1.80282 kg)(0.84 m)2  4   = [(0.16285 + 0.48854) + (1.05647 + 1.87463) + (0.03661) + (0.03661 + 1.27207)] kg ⋅ m 2 = (0.65139 + 2.93110 + 0.03661 + 1.30868) kg ⋅ m 2

or

I z = 4.93 kg ⋅ m 2 



1  +  (1.80282 kg)(0.285 m) 2  4 

www.elsolucionario.net PROBLEM 9.141 The machine element shown is fabricated from steel. Determine the mass moment of inertia of the assembly with respect to (a) the x axis, (b) the y axis, (c) the z axis. (The density of steel is 7850 kg/m3 .)

SOLUTION First compute the mass of each component. We have m = ρSTV

m1 = (7850 kg/m3 )(π (0.08 m) 2 (0.04 m)] = 6.31334 kg

m2 = (7850 kg/m3 )[π (0.02 m) 2 (0.06 m)] = 0.59188 kg m3 = (7850 kg/m3 )[π (0.02 m)2 (0.04 m)] = 0.39458 kg

Using Figure 9.28 and the parallel-axis theorem, we have (a)

I x = ( I x )1 + ( I x )2 − ( I x )3 1  =  (6.31334 kg)[3(0.08)2 + (0.04) 2 ] m 2 + (6.31334 kg)(0.02 m) 2  12   1  +  (0.59188 kg)[3(0.02) 2 + (0.06)2 ] m 2 + (0.59188 kg)(0.03 m) 2  12  1  −  (0.39458 kg)[3(0.02)2 + (0.04) 2 ] m 2 + (0.39458 kg)(0.02 m) 2  12  = [(10.94312 + 2.52534) + (0.23675 + 0.53269) − (0.09207 + 0.15783)] × 10−3 kg ⋅ m 2 = (13.46846 + 0.76944 − 0.24990) × 10−3 kg ⋅ m 2 = 13.98800 × 10−3 kg ⋅ m 2

or I x = 13.99 × 10−3 kg ⋅ m 2 



Then


(b)

I y = ( I y )1 + ( I y )2 − ( I y )3 1  =  (6.31334 kg)(0.08 m) 2  2  1  +  (0.59188 kg)(0.02 m) 2 + (0.59188 kg)(0.04 m) 2  2  1  −  (0.39458 kg)(0.02 m 2 ) + (0.39458 kg)(0.04 m) 2  2   = [(20.20269) + (0.11838 + 0.94701) − (0.07892 + 0.63133)] × 10−3 kg ⋅ m 2 = (20.20269 + 1.06539 − 0.71025) × 10−3 kg ⋅ m 2

or (c)

I y = 20.6 × 10−3 kg ⋅ m 2 

I z = ( I z )1 + ( I z ) 2 − ( I z )3 1  =  (6.31334 kg)[3(0.08)2 + (0.04) 2 ] m 2 + (6.31334 kg)(0.02 m) 2  12   1  +  (0.59188 kg)[3(0.02) 2 + (0.06)2 ] m 2 + (0.59188 kg)[(0.04) 2 + (0.03) 2 ] m 2  12   1  −  (0.39458 kg)[3(0.02) 2 + (0.04) 2 ] m 2 + (0.03958 kg)[(0.04)2 + (0.02) 2 ] m 2  12   = [(10.94312 + 2.52534) + (0.23675 + 1.47970) − (0.09207 + 0.78916)] × 10−3 kg ⋅ m 2 = (13.46846 + 1.71645 − 0.88123) × 10−3 kg ⋅ m 2 = 14.30368 × 10−3 kg ⋅ m 2

or

I z = 14.30 × 10−3 kg ⋅ m 2 



= 20.55783 × 10−3 kg ⋅ m 2

www.elsolucionario.net To the Instructor: The following formulas for the mass of inertia of a semicylinder are derived at this time for use in the solutions of Problems 9.142 through 9.145. From Figure 9.28: Cylinder

( I x ) cyl =

1 mcyl a 2 2

( I y )cyl = ( I z )cyl =

1 mcyl (3a 2 + L2 ) 12

Symmetry and the definition of the mass moment of inertia ( I =  r 2 dm) imply

( I x )sc =

and However,

1 ( I )cylinder 2 11  mcyl a 2  2  2 

( I y )sc = ( I z )sc =

msc =

11  mcyl (3a 2 + L2 )   2 12 

1 mcyl 2 1 msc a 2 2

Thus,

( I x )sc =

and

( I y )sc = ( I z )sc =

1 msc (3a 2 + L2 ) 12

Also, using the parallel axis theorem find  1 16 I x′ = msc  − 2  2 9π

 2 a 

 1 16 I z′ = msc  − 2  4 9π

 2 1 2  a + 12 L   

where x′ and z ′ are centroidal axes.



( I )semicylinder =

www.elsolucionario.net PROBLEM 9.142 Determine the mass moments of inertia and the radii of gyration of the steel machine element shown with respect to the x and y axes. (The density of steel is 7850 kg/m3.)

First compute the mass of each component. We have m = ρSTV

Then

m1 = (7850 kg/m3 )(0.24 × 0.04 × 0.14) m3 = 10.5504 kg

π  m2 = m3 = (7850 kg/m3 )  (0.07)2 × 0.04  m3 = 2.41683 kg 2 

m4 = m5 = (7850 kg/m3 )[π (0.044) 2 × (0.04)] m3 = 1.90979 kg

Using Figure 9.28 for components 1, 4, and 5 and the equations derived above (before the solution to Problem 9.144) for a semicylinder, we have I x = ( I x )1 + ( I x )2 + ( I x )3 + ( I x ) 4 − ( I x )5

where ( I x ) 2 = ( I x )3

1  1  =  (10.5504 kg)(0.042 + 0.142 ) m 2  + 2  (2.41683 kg)[3(0.07 m) 2 + (0.04 m) 2 ] 12 12     1  +  (1.90979 kg)[3(0.044 m) 2 + (0.04 m) 2 ] + (1.90979 kg)(0.04 m) 2  12  1  −  (1.90979 kg)[3(0.044 m)2 + (0.04 m)2 ]  12  = [(0.0186390) + 2(0.0032829) + (0.0011790 + 0.0030557) − (0.0011790)] kg ⋅ m 2 = 0.0282605 kg ⋅ m 2

or

I x = 28.3 × 10−3 kg ⋅ m 2 



SOLUTION


I y = ( I y )1 + ( I y )2 + ( I y )3 + ( I y ) 4 − ( I y )5 ( I y ) 2 = ( I y )3

where Then

(I y )4 = |( I y )5 |

1  I y =  (10.5504 kg)(0.242 + 0.142 ) m 2  12     1 16 + 2 (2.41683 kg)  − 2  2 9π 

2 4 × 0.07  2    2 + + (0.07 m ) (2.41683 kg) 0.12 m    3π    

= [(0.0678742) + 2(0.0037881 + 0.0541678)] kg ⋅ m 2 = 0.1837860 kg ⋅ m 2

or m = m1 + m2 + m3 + m4 − m5 where m2 = m3 , m4 = |m5 | = (10.5504 + 2 × 2.41683) kg = 15.38406 kg

Then

and

k x2 =

k y2 =

I x 0.0282605 kg ⋅ m 2 = m 15.38406 kg

Iy m

=

or

k x = 42.9 mm 

or

k y = 109.3 mm 

0.1837860 kg ⋅ m 2 15.38406 kg



Also

I y = 183.8 × 10−3 kg ⋅ m 2 


SOLUTION First compute the mass of each component. We have m = ρSTV =

Then

m1 =

γ ST g

V

490 lb/ft 3  1 ft  × (2.7 × 3.7 × 9) in 3 ×   2 32.2 ft/s  12 in. 

3

= 791.780 × 10−3 lb ⋅ s 2 /ft 490 lb/ft 3  π   1 ft  m2 = ×  × 1.352 × 1.4  in 3 ×   2 32.2 ft/s  2   12 in. 

3

= 35.295 × 10−3 lb ⋅ s 2 /ft

m3 =

490 lb/ft 3  1 ft  × (π × 0.62 × 1.2) in 3 ×   2 32.2 ft/s  12 in. 

3

= 11.952 × 10 −3 lb ⋅ s 2 /ft 3

490 lb/ft 3  1 ft  m4 = × (0.9 × 0.6 × 9) in 3 ×  = 42.799 × 10−3 lb ⋅ s 2 /ft  2 32.2 ft/s  12 in. 

The mass moments of inertia are now computed using Figure 9.28 (components 1, 3, and 4) and the equations derived above (component 2).



Determine the mass moment of inertia of the steel machine element shown with respect to the y axis. (The specific weight of steel is 490 lb/ft 3 .)


Find: I y We have

I y = ( I y )1 + ( I y ) 2 + ( I y )3 − ( I y ) 4 1 =  (791.780 × 10−3 lb ⋅ s 2 /ft)[(2.7)2 + (9)2 ] in 2 12  2.7 2  9 2  2   1 ft 2 + (791.780 × 10 lb ⋅ s /ft)   +    in  ×  12 in.    2   2     1 +  (35.295 × 10 −3 lb ⋅ s 2 /ft)[3(1.35) 2 + (1.4) 2 ] in 2 12 −3

2 2  1.4   2   1 ft   + (35.295 × 10−3 lb ⋅ s 2 /ft) (1.35) 2 +  9 − × in      2      12 in.   1 +  (11.952 × 10−3 lb ⋅ s 2 /ft)[3(0.6) 2 + (1.2) 2 ] in 2 12 2 2  1.2   2   1 ft  + (11.952 × 10 lb ⋅ s /ft)[(1.35) +  9 + in  ×   2     12 in.  1 −  (42.799 × 10−3 lb ⋅ s 2 /ft)[(0.9)2 + (9) 2 ] in 2 12

−3

2

2

2   1 ft 2  9  + (42.799 × 10−3 lb ⋅ s2 /ft)  (1.35)2 +    in 2  ×    2      12 in.  = [(40.4550 + 121.3650) + (0.1517 + 17.3319)

+ (0.0174 + 7.8005) − (2.0263 + 6.5603)] × 10−3 lb ⋅ ft ⋅ s 2 = (161.8200 + 17.4836 + 7.8179 − 8.5866) × 10−3 lb ⋅ ft ⋅ s 2

or

I y = 0.1785 lb ⋅ ft ⋅ s 2 



2



γ ST g

V

Then m1 =

490 lb/ft 3  1 ft  × (2.7 × 3.7 × 9) in 3 ×   2 32.2 ft/s  12 in. 

3

= 791.780 × 10−3 lb ⋅ s 2 /ft

m2 =

490 lb/ft 3  π   1 ft  ×  × 1.352 × 1.4  in 3 ×   2 32.2 ft/s  2   12 in. 

3

= 35.295 × 10−3 lb ⋅ s 2 /ft

m3 =

490 lb/ft 3  1 ft  × (π × 0.62 × 1.2) in 3 ×   2 32.2 ft/s  12 in. 

3

= 11.952 × 10 −3 lb ⋅ s 2 /ft 3

m4 =

490 lb/ft 3  1 ft  −3 2 × (0.9 × 0.6 × 9) in 3 ×   = 42.799 × 10 lb ⋅ s /ft 2 32.2 ft/s  12 in. 

The mass moments of inertia are now computed using Figure 9.28 (components 1, 3, and 4) and the equations derived above (component 2).



Determine the mass moment of inertia of the steel machine element shown with respect to the z axis. (The specific weight of steel is 490 lb/ft 3 .)


Find: I z We have

I z = ( I z )1 + ( I z ) 2 + ( I z )3 − ( I z ) 4  1 =  (791.780 × 10 −3 lb ⋅ s 2 /ft)[(2.7) 2 + (3.7) 2 ] in 2  12  2.7 2  3.7 2  2   1 ft 2 + (791.780 lb ⋅ s 2 /ft)   +   in  ×  12 in.    2   2      2  (1.35 in.) 

2 2  4 × 1.35   2   1 ft   in + (35.295 × 10−3 lb ⋅ s 2 /ft) (1.35)2 +  3.7 + ×      3π       12 in.  1 +  (11.952 × 10−3 lb ⋅ s 2 /ft)(0.6 in.) 2 12

  1 ft  + (11.952 × 10−3 lb ⋅ s 2 /ft)[(1.35)2 + (3.7)2 ] in 2  ×     12 in. 

2

 1 −  (42.799 × 10−3 lb ⋅ s 2 /ft)[(0.9)2 + (0.6)2 ] in 2  12 2   0.6  2 + (42.799 × 10 lb ⋅ s /ft) (1.35) +    2   I z = [(9.6132 + 28.8395) + (0.1429 + 4.9219)

−3

2

 2   1 ft 2  in  ×      12 in. 

+ (0.0149 + 1.2875) − (0.0290 + 0.5684)] × 10−3 lb ⋅ ft ⋅ s 2 = (38.4527 + 5.0648 + 1.3024 − 0.5974) × 10−3 lb ⋅ ft ⋅ s 2

or

I z = 0.0442 lb ⋅ ft ⋅ s 2 



  1 16 +  (35.295 × 10−3 lb ⋅ s 2 /ft)  − 2  2 9π 

www.elsolucionario.net PROBLEM 9.145 Determine the mass moment of inertia of the steel fixture shown with respect to (a) the x axis, (b) the y axis, (c) the z axis. (The density of steel is 7850 kg/m3.)


Then

m1 = 7850 kg/m3 × (0.08 × 0.05 × 0.160) m3 = 5.02400 kg

m2 = 7850 kg/m3 × (0.08 × 0.038 × 0.07) m3 = 1.67048 kg π  m3 = 7850 kg/m3 ×  × 0.0242 × 0.04  m3 = 0.28410 kg 2  

Using Figure 9.28 for components 1 and 2 and the equations derived above for component 3, we have (a)

I x = ( I x )1 − ( I x ) 2 − ( I x )3  1  0.05 2  0.16 2  2  =  (5.02400 kg)[(0.05)2 + (0.16) 2 ] m 2 + (5.02400 kg)   +  m   2   2   12  2 2  0.038   0.07   2   1 2 2 2 −  (1.67048 kg)[(0.038) + (0.07) ] m + (1.67048 kg)  0.05 − + 0.05 + m  12 2   2     

  1 16 − (0.28410 kg)  − 2  4 9π 

1  2 2 2  (0.024) + (0.04)  m 12  

2 2  4 × 0.024   0.04   2  0.16 + (0.28410 kg)  0.05 − + − m    3π 2      

= [(11.7645 + 35.2936) − (0.8831 + 13.6745) − (0.0493 + 6.0187)] × 10−3 kg ⋅ m 2 = (47.0581 − 14.5576 − 6.0680) × 10−3 kg ⋅ m 2 = 26.4325 × 10−3 kg ⋅ m 2

or

I x = 26.4 × 10−3 kg ⋅ m 2 



m = ρSTV


(b)

I y = ( I y )1 − ( I y ) 2 − ( I y )3  1  0.08 2  0.16 2  2  2 2 2 =  (5.02400 kg)[(0.08) + (0.16) ] m + (5.02400 kg)   +  m  12  2   2     2  1  0.08 2  0.07   2  0.05 −  (1.67048 kg)[(0.08) 2 + (0.07) 2 ] m 2 + (1.67048 kg)  + + m    2    2   12  2  0.08 2  0.04   2   1 0.16 −  (0.28410 kg)[3(0.024)2 + (0.04) 2 ] m 2 + (0.28410 kg)  + − m    2   12   2  

= [(13.3973 + 40.1920) − (1.5730 + 14.7420) − (0.0788 + 6.0229)] × 10−3 kg ⋅ m 2 = (53.5893 − 16.3150 − 6.1017) × 10−3 kg ⋅ m 2

(c)

or

I y = 31.2 × 10−3 kg ⋅ m 2 

I z = ( I z )1 − ( I z ) 2 − ( I z )3  1  0.08 2  0.05 2  2  2 2 2 =  (5.02400 kg)[(0.08) + (0.05) ] m + (5.02400 kg)   +  m  12  2   2     2  1  0.08 2  0.038   2  0.05 −  (1.67048 kg)[(0.08) 2 + (0.038) 2 ] m 2 + (1.67048 kg)  + − m    2   12   2  

  1 16 − (0.28410 kg)  − 2  2 9π 

2  0.08 2  4 × 0.024   2   2 (0.024 m) (0.28410 kg) 0.05 + + −       m  3π     2   

= [(3.7261 + 11.1784) − (1.0919 + 4.2781) − (0.0523 + 0.9049)] × 10−3 kg ⋅ m 2 = (14.9045 − 5.3700 − 0.9572) × 10−3 kg ⋅ m 2 = 8.5773 × 10−3 kg ⋅ m 2

or

I z = 8.58 × 10−3 kg ⋅ m 2 



= 31.1726 × 10−3 kg ⋅ m 2

www.elsolucionario.net To the instructor: The following formulas for the mass moment of inertia of wires are derived or summarized at this time for use in the solutions of Problems 9.146 through 9.148. Slender Rod Ix = 0

I y′ = I z′ =

1 mL2 (Figure 9.28) 12

1 I y = I z = mL2 (Sample Problem 9.9) 3

Circle I y = r 2 dm = ma 2

Now

Iy = Ix + Iz

And symmetry implies

Ix = Iz Ix = Iz =

1 2 ma 2

Semicircle Following the above arguments for a circle, We have Ix = Iz =

1 2 ma 2

I y = ma 2

Using the parallel-axis theorem I z = I z ′ + mx 2

or

1 4 I z′ = m  − 2 2 π

x=

2a

π

 2 a  





We have

www.elsolucionario.net PROBLEM 9.146 Aluminum wire with a weight per unit length of 0.033 lb/ft is used to form the circle and the straight members of the figure shown. Determine the mass moment of inertia of the assembly with respect to each of the coordinate axes.

SOLUTION First compute the mass of each component. We have m=

1 1 ft × 0.033 lb/ft × (2π × 16 in.) × 2 12 in. 32.2 ft/s 2 −3 = 8.5857 × 10 lb ⋅ s /ft 1 1 ft m2 = m3 = m4 = m5 = × 0.033 lb/ft × 8 in. × 12 in. 32.2 ft/s 2 = 0.6832 lb ⋅ s 2 /ft

m1 =


Then

W 1 W  L = g g  L AL

Using the equations given above and the parallel-axis theorem, we have I x = ( I x )1 + ( I x )2 + ( I x )3 + ( I x ) 4 + ( I x )5 1   1 ft  =  (8.5857 × 10−3 lb ⋅ s 2 /ft)(16 in.) 2  ×   2    12 in.  1   1 ft  +  (0.6832 lb ⋅ s 2 /ft)(8 in.) 2  ×   3   12 in. 

2

 1 ft  + [0 + (0.6832 lb ⋅ s /ft)(8 in.) ] ×    12 in.  2

2

2

2

1   1 ft  +  (0.6832 lb ⋅ s 2 /ft)(8 in.)2 + (0.6832 lb ⋅ s 2 /ft)(42 + 162 ) in 2  ×   12   12 in.  1   1 ft  +  (0.6832 lb ⋅ s 2 /ft)(8 in.)2 + (0.6832 lb ⋅ s 2 /ft)(82 + 122 ) in 2  ×   12   12 in. 

2

2

= [(7.6315) + (0.1012) + (0.3036) + (0.0253 + 1.2905) + (0.0253 + 0.9868)] × 10−3 lb ⋅ ft ⋅ s 2 = (7.6315 + 0.1012 + 0.3036) + 1.3158 + 1.0121) × 10−3 lb ⋅ ft ⋅ s 2 = 10.3642 × 10−3 lb ⋅ ft ⋅ s 2

or

I x = 10.36 × 10−3 lb ⋅ ft ⋅ s 2 



( I y )2 = ( I y )4 ,

( I y )3 = ( I y )5

I y = ( I y )1 + ( I y ) 2 + ( I y )3 + ( I y ) 4 + ( I y )5  1 ft  = [(8.5857 × 10 lb ⋅ s /ft)(16 in.) ] ×    12 in.  −3

2

2

2

 1 ft  +2[0 + (0.6832 × 10−3 lb ⋅ s 2 /ft)(16 in.) 2 ] ×    12 in. 

2

1   1 ft  +2  (0.6832 × 10 −3 lb ⋅ s 2 /ft)(8 in.) 2 + (0.6832 lb ⋅ s 2 /ft)(12 in.) 2  ×   12   12 in. 

2

= [(15.2635) + 2(1.2146) + 2(0.0253 + 0.6832)] × 10−3 lb ⋅ ft ⋅ s 2 = [15.2635 + 2(1.2146) + 2(0.7085)] × 10−3 lb ⋅ ft ⋅ s 2

Symmetry implies

or

I y = 19.11 × 10−3 lb ⋅ ft ⋅ s 2  I z = 10.36 × 10−3 lb ⋅ ft ⋅ s 2 

Ix = Iz



= 19.1097 × 10−3 lb ⋅ ft ⋅ s 2

www.elsolucionario.net PROBLEM 9.147 The figure shown is formed of 18 -in.-diameter steel wire. Knowing that the specific weight of the steel is 490 lb/ft3, determine the mass moment of inertia of the wire with respect to each of the coordinate axes.


γ ST g

AL


Then 2 3 490 lb/ft 3  π  1    1 ft  m1 = m2 = ×   in.   × (π × 18 in.) ×   32.2 ft/s 2  4  8    12 in. 

= 6.1112 × 10−3 lb ⋅ s 2 /ft 2 3 490 lb/ft 3  π  1    1 ft  m3 = m4 = =   in.   × 18 in. ×   32.2 ft/s 2  4  8    12 in. 

= 1.9453 × 10−3 lb ⋅ s 2 /ft

Using the equations given above and the parallel-axis theorem, we have ( I x )3 = ( I x ) 4 I x = ( I x )1 + ( I x )2 + ( I x )3 + ( I x ) 4 1   1 ft  =  (6.1112 × 10−3 lb ⋅ s 2 /ft)(18 in.) 2  ×   2   12 in. 

2

1   1 ft  +  (6.1112 × 10−3 lb ⋅ s 2 /ft)(18 in.) 2 + (6.1112 × 10−3 lb ⋅ s 2 /ft)(18 in.) 2  ×   2   12 in. 

2

1   1 ft  + 2  (1.9453 × 10−3 lb ⋅ s 2 /ft)(18 in.) 2 + (6.1112 × 10−3 lb ⋅ s 2 /ft)(92 + 182 ) in 2  ×   12   12 in. 

2

= [(6.8751) + (6.8751 + 13.1502) + 2(0.3647 + 5.4712)] × 10 −3 lb ⋅ ft ⋅ s 2 = [6.8751 + 20.6252 + 2(5.8359)] × 10−3 lb ⋅ ft ⋅ s 2 = 39.1721 × 10−3 lb ⋅ ft ⋅ s 2

or

I x = 0.0392 lb ⋅ ft ⋅ s 2 

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( I y )1 = ( I y ) 2 , ( I y )3 = ( I y )4 I y = ( I y )1 + ( I y ) 2 + ( I y )3 + ( I y ) 4  1 ft  = 2[(6.1112 × 10 lb ⋅ s /ft)(18 in.) ] ×    12 in.  −3

2

2

2

 1 ft  + 2[(0 + 1.9453 × 10 −3 lb ⋅ s 2 /ft)(18 in.) 2 ] ×    12 in. 

2

= [2(13.7502) + 2(4.3769)] × 10−3 lb ⋅ ft ⋅ s 2 = 36.2542 × 10−3 lb ⋅ ft ⋅ s 2

or

I y = 0.0363 lb ⋅ ft ⋅ s 2 

or

I z = 0.0304 lb ⋅ ft ⋅ s 2 

( I z )3 = ( I z ) 4 I z = ( I z )1 + ( I z ) 2 + ( I z )3 + ( I z ) 4

 1 4 + (6.1112 × 10−3 lb ⋅ s 2 /ft)  − 2 2 π 

2

 2  (18 in.) 

 2 × 18 2  2   1 ft 2 2 + (6.1112 × 10−3 lb ⋅ s 2 /ft)  + (18)  in  ×     π     12 in.  1   1 ft  + 2  (1.9453 × 10−3 lb ⋅ s 2 /ft)(18 in.)2  ×   3   12 in. 

2

= [(6.8751) + (1.3024 + 19.3229) + 2(1.4590)] × 10−3 lb ⋅ ft ⋅ s 2 = [6.8751 + 20.6253 + 2(1.4590)] × 10−3 lb ⋅ ft ⋅ s 2 = 30.4184 × 10−3 lb ⋅ ft ⋅ s 2



1   1 ft  =  (6.1112 × 10 −3 lb ⋅ s 2 /ft)(18 in.) 2  ×   2   12 in. 

www.elsolucionario.net PROBLEM 9.148 A homogeneous wire with a mass per unit length of 0.056 kg/m is used to form the figure shown. Determine the mass moment of inertia of the wire with respect to each of the coordinate axes.

SOLUTION

m = (m/L) L = 0.056 kg/m × 1.2 m = 0.0672 kg

Using the equations given above and the parallel-axis theorem, we have I x = ( I x )1 + ( I x ) 2 + ( I x )3 + ( I x ) 4 + ( I x )5 + ( I x )6

Now Then

( I x )1 = ( I x )3 = ( I x ) 4 = ( I x )6

and ( I x )2 = ( I x )5

1  I x = 4  (0.0672 kg)(1.2 m) 2  + 2[0 + (0.0672 kg)(1.2 m) 2 ] 3  

= [4(0.03226) + 2(0.09677)] kg ⋅ m 2 = 0.32258 kg ⋅ m 2

or

I x = 0.323 kg ⋅ m 2 

I y = ( I y )1 + ( I y ) 2 + ( I y )3 + ( I y ) 4 + ( I y )5 + ( I y )6

Now Then

( I y )1 = 0, ( I y ) 2 = ( I y )6 , and ( I y )4 = ( I y )5 1  I y = 2  (0.0672 kg)(1.2 m) 2  + [0 + (0.0672 kg)(1.2 m) 2 ] 3  1   + 2  (0.0672 kg)(1.2 m) 2 + (0.0672 kg)(1.22 + 0.62 ) m 2  12 

= [2(0.03226) + (0.09677) + 2(0.00806 + 0.12096)] kg ⋅ m 2 = [2(0.03226) + (0.09677) + 2(0.12902)] kg ⋅ m 2 = 0.41933 kg ⋅ m 2

Symmetry implies

or

I y = Iz

I y = 0.419 kg ⋅ m 2  I z = 0.419 kg ⋅ m 2 



First compute the mass m of each component. We have

www.elsolucionario.net PROBLEM 9.149 Determine the mass products of inertia Ixy, Iyz, and Izx of the steel fixture shown. (The density of steel is 7850 kg/m3.)


m1 = 7850 kg/m3 × (0.08 × 0.05 × 0.16) m3 = 5.02400 kg

Then

m2 = 7850 kg/m3 × (0.08 × 0.038 × 0.07) m3 = 1.67048 kg π  m3 = 7850 kg/m3 ×  × 0.0242 × 0.04  m3 = 0.28410 kg 2 

Now observe that the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , of each component are zero because of symmetry. Now 0.038   m = 0.031 m y2 =  0.05 − 2   4 × 0.024   y3 =  0.05 −  m = 0.039814 m 3π   and then mx y , kg ⋅ m 2

my z , kg ⋅ m 2

mz x , kg ⋅ m 2

0.08

5.0240 × 10−3

10.0480 × 10−3

16.0768 × 10−3

0.031

0.085

2.0714 × 10−3

4.4017 × 10−3

5.6796 × 10−3

0.039814

0.14

0.4524 × 10−3

1.5836 × 10−3

1.5910 × 10−3

m, kg

x, m

y, m

1

5.02400

0.04

0.025

2

1.67048

0.04

3

0.28410

0.04

z, m

Finally, I xy = ( I xy )1 − ( I xy ) 2 − ( I xy )3 = [(0 + 5.0240) − (0 + 2.0714) − (0 + 0.4524)] × 10−3 = 2.5002 × 10−3 kg ⋅ m 2

or

I xy = 2.50 × 10−3 kg ⋅ m 2 



m = ρV


I yz = ( I yz )1 − ( I yz ) 2 − ( I yz )3 = [(0 + 10.0480) − (0 + 4.4017) − (0 + 1.5836)] × 10−3 = 4.0627 × 10−3 kg ⋅ m 2

or

I yz = 4.06 × 10−3 kg ⋅ m 2 

I zx = ( I zx )1 − ( I zx ) 2 − ( I zx )3 = [(0 + 16.0768) − (0 + 5.6796) − (0 + 1.5910)] × 10−3 = 8.8062 × 10−3 kg ⋅ m 2 I zx = 8.81 × 10−3 kg ⋅ m 2 


or


www.elsolucionario.net PROBLEM 9.150 Determine the mass products of inertia Ixy, Iyz, and Izx of the steel machine element shown. (The density of steel is 7850 kg/m3.)

SOLUTION 

Since the machine element is symmetrical with respect to the xy plane, Iyz = Izx = 0 Also,

I xy = 0

For the cylinder m = ρV = (7850 kg/m3 )π (0.022 m) 2 (0.02 m) = 0.2387 kg ′ = I x′y′ + mx y = 0 + (0.2387 kg)(0.06 m)(0.02 m) = +286.4 × 10−6 kg ⋅ m 2 I xy I xy = 286 × 10−6 kg ⋅ m 2 



for all components except the cylinder, since these are symmetrical with respect to the xz plane.

www.elsolucionario.net PROBLEM 9.151 Determine the mass products of inertia Ixy, Iyz, and Izx of the cast aluminum machine component shown. (The specific weight of aluminum is 0.100 lb/in3)


Then

m1 =

γ AL g

V

0.100 lb/in 3 × (5.9 × 1.8 × 2.2) in 3 2 32.2 ft/s

= 72.5590 × 10−3 lb ⋅ s 2 /ft m2 =

0.100 lb/in 3  π  ×  (1.1) 2 × 1.8 in 3 32.2 ft/s 2 2 

= 10.6248 × 10−3 lb ⋅ s 2 /ft m3 =

0.100 lb/in 3 × (1.4 × 1.1 × 1.2) in 3 2 32.2 ft/s

= 5.7391 × 10−3 lb ⋅ s2 /ft

Now observe that the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , of each component are zero because of symmetry. Now 4 × 1.1   x2 = −  5.9 + in. 3π   = − 6.36685 in. 1.1   y3 = 1.8 − in. 2   = 1.25 in.



m = ρ ALV =


and then m, lb ⋅ s 2 /ft

1

72.5590 × 10−3

2

10.6248 × 10−3

3

5.7391 × 10−3

Finally,

x , ft

y , ft

z , ft

mx y , lb ⋅ ft ⋅ s 2

my z , lb ⋅ ft ⋅ s 2

mz x , lb ⋅ ft ⋅ s 2

2.95 12

0.9 12

1.1 12

−1.33781 × 10−3

0.49884 × 10−3

−1.63510 × 10−3

6.36685 12

0.9 12

1.1 12

− 0.42279 × 10−3

0.07305 × 10−3

− 0.51674 × 10−3

0.7 12

1.25 12

1.3 12

− 0.03487 × 10−3

0.06476 × 10−3

− 0.03627 × 10−3

− −

−

I xy = ( I xy )1 + ( I xy ) 2 − ( I xy )3

or

I xy = −1.726 × 10−3 lb ⋅ ft ⋅ s 2 

I yz = ( I yz )1 + ( I yz ) 2 − ( I yz )3 = [(0 + 0.49884) + (0 + 0.07305) − (0 + 0.06476)] × 10−3

or

I yz = 0.507 × 10−3 lb ⋅ ft ⋅ s 2 

I zx = ( I zx )1 + ( I zx )2 − ( I zx )3 = [(0 − 1.63510) + (0 − 0.51674) − (0 − 0.3627)] × 10 −3

or

I zx = −2.12 × 10 −3 lb ⋅ ft ⋅ s 2 



= [(0 − 1.33781) + (0 − 0.42279) − (0 − 0.03487)] × 10−3

www.elsolucionario.net PROBLEM 9.152 Determine the mass products of inertia Ixy, Iyz, and Izx of the cast aluminum machine component shown. (The specific weight of aluminum is 0.100 lb/in3)


Then

m1 =

γ AL g

V

0.100 lb/in 3 × (7.8 × 0.8 × 1.6) in 3 32.2 ft/s 2

= 31.0062 × 10 −3 lb ⋅ s 2 /ft m2 =

0.100 lb/in 3 × (2.4 × 0.8 × 2) in 3 2 32.2 ft/s

= 11.9255 × 10−3 lb ⋅ s 2 /ft m3 =

0.100 lb/in 3  π  ×  (0.8) 2 × 0.8 in 3 = 2.4977 lb ⋅ s 2 /ft 2 32.2 ft/s 2 

m4 =

0.100 lb/in 3 × [π (0.55) 2 × 0.6] in 3 = 1.7708 lb ⋅ s 2 /ft 32.2 ft/s 2

Now observe that the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , of each component are zero because of symmetry. Now 4 × 0.8   x3 = −  7.8 + in. = −8.13953 in. 3π  



m = ρ ALV =

www.elsolucionario.net PROBLEM 9.152 (Continued) and then x , ft

y , ft

z , ft

mx y , lb ⋅ ft ⋅ s 2

my z , lb ⋅ ft ⋅ s 2

mz x , lb ⋅ ft ⋅ s 2

1

31.0062 × 10−3

−

3.9 12

0.4 12

−

0.8 12

− 335.901 × 10−6

− 68.903 × 10−6

671.801 × 10−6

2

11.9255 × 10−3

−

1.2 12

0.4 12

−

2.6 12

− 39.752 × 10−6

−86.129 × 10−6

258.386 × 10−6

3

2.4977 × 10−3

8.13953 12

0.4 12

−

0.8 12

− 56.473 × 10−6

− 5.550 × 10−6

112.945 × 10−6

4

1.7708 × 10−3

7.8 12

1.1 12

−

0.8 12

−105.511 × 10−6

−10.822 × 10−6

76.735 × 10−6

− 537.637 × 10−6

−171.404 × 10−6

1119.867 × 10−6

−

−

Σ

Then

I xy I yz

0 = Σ( I x′y′ + mx y ) 0 = Σ( I y ′z ′ + m y z )

0 I zx = Σ( I z ′x′ + mz x )

or

I xy = −538 × 10 −6 lb ⋅ ft ⋅ s 2 

or

I yz = −171.4 × 10−6 lb ⋅ ft ⋅ s 2 

or

I zx = 1120 × 10−6 lb ⋅ ft ⋅ s 2 



m, lb ⋅ s 2 /ft

www.elsolucionario.net PROBLEM 9.153 A section of sheet steel 2 mm thick is cut and bent into the machine component shown. Knowing that the density of steel is 7850 kg/m3, determine the mass products of inertia Ixy, Iyz, and Izx of the component.

First compute the mass of each component. We have Then

m = ρSTV = ρST tA

m1 = (7850 kg/m3 )[(0.002)(0.4)(0.45)]m3 = 2.8260 kg  1  m2 = 7850 kg/m3 (0.002)  × 0.45 × 0.18   m3 2   = 0.63585 kg

Now observe that ( I x′y′ )1 = ( I y′z ′ )1 = ( I z′x′ )1 = 0 ( I y′z′ ) 2 = ( I z′x′ ) 2 = 0

From Sample Problem 9.6:

( I x′y′ )2,area = −

1 2 2 b2 h2 72

Then

1  1  ( I x′y′ ) 2 = ρST t ( I x′y′ ) 2,area = ρST t  − b22 h22  = − m2 b2 h2 36  72 

Also

x1 = y1 = z2 = 0

0.45   x2 =  − 0.225 + m = − 0.075 m 3  



SOLUTION


Finally,

 1 I xy = Σ( I xy + mx y ) = (0 + 0) +  − (0.63585 kg)(0.45 m)(0.18 m)  36  0.18 m   + (0.63585 kg)(− 0.075 m)    3  = (−1.43066 × 10−3 − 2.8613 × 10−3 ) kg ⋅ m 2

and

or

I xy = − 4.29 × 10−3 kg ⋅ m 2 

or

I yz = 0 

or

I zx = 0 

I yz = Σ( I y′z′ + m y z ) = (0 + 0) + (0 + 0) = 0


I zx = Σ( I z ′x′ + m z x ) = (0 + 0) + (0 + 0) = 0



m1 = ρV = 7850 kg/m3 (0.120 m)(0.200 m)(0.002 m) = 0.3768 kg m2 = ρV = 7850 kg/m3 (0.120 m)(0.100 m)(0.002 m) = 0.1884 kg

For each panel the centroidal product of inertia is zero with respect to each pair of coordinate axes. m, kg

x, m

y, m

z, m

mxy

myz

mz x



0.3768

0.1

−0.06

0.1

−2.261 × 10−3

−2.261 × 10−3

+3.768 × 10−3



0.1884

0.2

−0.06

0.05

−2.261 × 10−3

−0.565 × 10−3

+1.884 × 10−3

−4.522 × 10−3

−2.826 × 10−3

+5.652 × 10−3



I xy = −4.52 × 10−3 kg ⋅ m 2  I yz = −2.83 × 10−3 kg ⋅ m 2  I zx = +5.65 × 10−3 kg ⋅ m 2 



SOLUTION



Then

m1 = (7850 kg/m3 )(0.002 m)(0.35 × 0.39) m 2 = 2.14305 kg π  m2 = (7850 kg/m3 )(0.002 m)  × 0.1952  m 2 2  = 0.93775 kg 1  m3 = (7850 kg/m3 )(0.002 m)  × 0.39 × 0.15  m2 2   = 0.45923 kg

Now observe that because of symmetry the centroidal products of inertia of components 1 and 2 are zero and ( I x′y′ )3 = ( I z′x′ )3 = 0

Also

( I y′z ′ )3,mass = ρST t ( I y′z′ )3,area

Using the results of Sample Problem 9.6 and noting that the orientation of the axes corresponds to a 90° rotation, we have ( I y′z ′ )3,area =

Then Also

1 2 2 b3 h3 72

 1  1 ( I y′z ′ )3 = ρST t  b32 h32  = m3b3 h3  72  36 y1 = x2 = 0 y2 =

4 × 0.195 m = 0.082761 m 3π



m = ρSTV = ρST tA


Finally, I xy = Σ( I x′y′ + mx y )   0.15   m  = (0 + 0) + (0 + 0) + 0 + (0.45923 kg)(0.35 m)  − 3    = − 8.0365 × 10−3 kg ⋅ m 2

or I xy = − 8.04 × 10−3 kg ⋅ m 2 

I yz = Σ( I y′z′ + my z ) = (0 + 0) + [0 + (0.93775 kg)(0.082761 m)(0.195 m)] 1  0.15  0.39   +  (0.45923 kg)(0.39 m)(0.15 m) + (0.45923 kg)  − m  m  3   3   36 = (15.1338 − 2.2388) × 10−3 kg ⋅ m 2 = 12.8950 × 10−3 kg ⋅ m 2

or I yz = 12.90 × 10−3 kg ⋅ m 2 

I zx = Σ( I z ′x′ + mz x ) = [0 + (2.14305 kg)(0.175 m)(0.195 m)] + (0 + 0)    0.39  + 0 + (0.45923 kg)  m  (0.35 m)   3    = (73.1316 + 20.8950) × 10−3 kg ⋅ m 2 = 94.0266 × 10−3 kg ⋅ m 2

or I zx = 94.0 × 10−3 kg ⋅ m 2 



= [(15.1338) + (0.7462 − 2.9850)] × 10−3 kg ⋅ m 2



1  m1 = m3 = (7850 kg/m3 )(0.002 m)  × 0.225 × 0.135  m 2 = 0.23844 kg 2   π   m2 = (7850 kg/m3 )(0.002 m)  × 0.2252  m 2 = 0.62424 kg 4  π  m4 = (7850 kg/m3 )(0.002 m)  × 0.1352  m 2 = 0.22473 kg 4  

Now observe that the following centroidal products of inertia are zero because of symmetry.

Also Now

( I x′y′ )1 = ( I y′z′ )1 = 0

( I x′y′ ) 2 = ( I y′z′ )2 = 0

( I x′y′ )3 = ( I z′x′ )3 = 0

( I x′y′ ) 4 = ( I z′x′ ) = 0

y1 = y2 = 0

x3 = x4 = 0

I xy = Σ( I x′y′ + mx y ) I xy = 0 

so that Using the results of Sample Problem 9.6, we have I uv ,area =

Now

1 2 2 b h 24

I uv ,mass = ρST t I uv ,area  1  = ρST t  b2 h 2  24   1 = mbh 12

Thus,

( I zx )1 =

1 m1b1h1 12



Then


( I yz )3 = −

While

1 m3b3 h3 12

because of a 90° rotation of the coordinate axes. To determine I uv for a quarter circle, we have

dI uv = d I u ′v′

0 + uEL vEL dm 1 1 2 v= a − u2 2 2

uEL = u vEL =

Where

dm = ρST t dA = ρST tv du = ρST t a 2 − u 2 du



a

0

1 2  (u )  a − u 2  ρST t a 2 − u 2 du 2  

(

)

=

1 ρST t 2

=

1 1  1 1 1 ρST t  a 2u 2 − u 4  = ρST t a 4 = ma 4 2 4 0 8 2π 2



a

0

u (a 2 − u 2 ) du a

( I zx ) 2 = −

Thus

1 m2 a22 2π

because of a 90° rotation of the coordinate axes. Also ( I yz ) 4 =

Finally, I yz

1 m4 a42 2π

0 0 = Σ( I yz ) = [( I y′z′ ) + m1 y1 z1 ] + [( I y′z′ ) + m2 y2 z2 ] + ( I yz )3 + ( I yz ) 4  1   1  =  − (0.23844 kg)(0.225 m)(0.135 m)  +  (0.22473 kg)(0.135 m) 2  π 12 2    

= (− 0.60355 + 0.65185) × 10−3 kg ⋅ m 2

or

I yz = 48.3 × 10−6 kg ⋅ m 2 

0 I zx = Σ( I zx ) = ( I zx )1 + ( I zx ) 2 + [( I z ′x′ )3 + m3 z3 x3 ] + [( I z′x′ ) 4 + m4 z4 x4 ] 0

1   1  =  (0.23844 kg)(0.225 m)(0.135 m)  +  − (0.62424 kg)(0.225 m)2  12   2π 

= (0.60355 − 5.02964) × 10−3 kg ⋅ m 2

or

I zx = − 4.43 × 10−3 kg ⋅ m 2 





I uv = dI uv =

Then

www.elsolucionario.net PROBLEM 9.157 The figure shown is formed of 1.5-mm-diameter aluminum wire. Knowing that the density of aluminum is 2800 kg/m3, determine the mass products of inertia Ixy, Iyz, and Izx of the wire figure.

First compute the mass of each component. We have m = ρ ALV = ρ AL AL

Then π  m1 = m4 = (2800 kg/m3 )  (0.0015 m) 2  (0.25 m) 4  

= 1.23700 × 10−3 kg π  m2 = m5 = (2800 kg/m3 )  (0.0015 m)2  (0.18 m) 4 

= 0.89064 × 10−3 kg π  m3 = m6 = (2800 kg/m3 )  (0.0015 m) 2  (0.3 m) 4 

= 1.48440 × 10−3 kg



SOLUTION


m, kg

x, m

y, m

z, m

mx y , kg ⋅ m 2

my z , kg ⋅ m 2

mz x , kg ⋅ m 2

1

1.23700 × 10−3

0.18

0.125

0

27.8325 × 10−6

0

0

2

0.89064 × 10−3

0.09

0.25

0

20.0394 × 10−6

0

0

3

1.48440 × 10−3

0

0.25

0.15

0

55.6650 × 10−6

0

4

1.23700 × 10−3

0

0.125

0.3

0

46.3875 × 10−6

0

5

0.89064 × 10−3

0.09

0

0.3

0

0

24.0473 × 10−6

6

1.48440 × 10−3

0.18

0

0.15

0

0

40.0788 × 10−6

47.8719 × 10−6

102.0525 × 10−6

64.1261 × 10−6

Σ

Then

0 I xy = Σ( I x′y′ + mx y ) 0 I yz = Σ( I y′z′ + my z ) 0 I zx = Σ( I z′x′ + mz x )

or

I xy = 47.9 × 10−6 kg ⋅ m 2 

or

I yz = 102.1 × 10−6 kg ⋅ m 2 

or

I zx = 64.1 × 10−6 kg ⋅ m 2 



Now observe that the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , of each component are zero because of symmetry.

www.elsolucionario.net PROBLEM 9.158 Thin aluminum wire of uniform diameter is used to form the figure shown. Denoting by m′ the mass per unit length of the wire, determine the mass products of inertia Ixy, Iyz, and Izx of the wire figure.


Then

π  π m1 = m5 = m′  R1  = m′R1 2  2 m2 = m4 = m′( R2 − R1 ) π  π m3 = m′  R2  = m′R2 2  2

Now observe that because of symmetry the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , of components 2 and 4 are zero and ( I x′y′ )1 = ( I z′x′ )1 = 0

( I x′y′ )3 = ( I y′z′ )3 = 0

( I y′z ′ )5 = ( I z′x′ )5 = 0

Also

x1 = x2 = 0

y2 = y3 = y4 = 0

z 4 = z5 = 0

Using the parallel-axis theorem [Equations (9.47)], it follows that I xy = I yz = I zx for components 2 and 4. To determine I uv for one quarter of a circular arc, we have dI uv = uvdm where and

u = a cos θ

v = a sin θ

dm = ρ dV = ρ [ A( adθ )]

where A is the cross-sectional area of the wire. Now π  π  m = m′  a  = ρ A  a  2  2 

so that and

dm = m′adθ dI uv = (a cos θ )(a sin θ )(m′adθ ) = m′a3 sin θ cos θ dθ



m m =   L = m′L L




I uv = dI uv =

Then



π /2 0

m′a3 sin θ cos θ dθ π /2

1  = m′a3  sin 2 θ  2 0

=

1 m′a3 2

1 m′R13 2

Thus,

( I yz )1 =

and

1 ( I zx )3 = − m′R23 2

1 ( I xy )5 = − m′R13 2

because of 90° rotations of the coordinate axes. Finally, 0

or 0 0 I yz = Σ( I yz ) = ( I yz )1 + [( I y′z′ )3 + m3 y3 z3 ] + [( I y′z′ )5 + m5 y5 z5 ] or 0

0 I zx = Σ( I zx ) = [( I z′x′ )1 + m1 z1 x1 ] + ( I zx )3 + [( I z ′x′ )5 + m5 z5 x5 ] or

1 I xy = − m′R13  2

I yz =

1 m′R13  2

1 I zx = − m′R23  2



I xy

0 = Σ( I xy ) = [( I x′y ′ )1 + m1 x1 y1 ] + [( I x′y′ )3 + m3 x3 y3 ] + ( I xy )5

www.elsolucionario.net PROBLEM 9.159 Brass wire with a weight per unit length w is used to form the figure shown. Determine the mass products of inertia Ixy, Iyz, and Izx of the wire figure.

SOLUTION

m=

W 1 = wL g g

W w (2π × a ) = 2π a g g w w m2 = (a) = a g g w w m3 = (2a) = 2 a g g w w 3  m4 =  2π × a  = 3π a g g 2  m1 =

Then

Now observe that the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , of each component are zero because of symmetry. m 1 2

2π

w a g

3

2

4

3π

Σ

w a g

w a g w a g

x

y

z

2a

a

−a

2a

1 a 2

0

w 3 a g

0

2a

0

a

0

0

2a

3 − a 2

2a

mx y w 3 a g

4π

− 9π

w 3 a g

w (1 − 5π )a 3 g

my z

mz x

w 3 a g

−2π

− 4π

w 3 a g

0 4

− 9π

w 3 a g

−11π

w 3 a g

w 3 a g

12π 4

w 3 a g

w (1 + 2π )a3 g





Then

0 + mx y )

or

0 = Σ( I y′z′ + my z )

or

I xy = Σ( I x′y′ I yz

I xy =

I yz = −11π

0 or

I zx = 4

w 3 a  g

w 3 a (1 + 2π )  g


I zx = Σ( I z′x′ + mz x )

w 3 a (1 − 5π )  g


www.elsolucionario.net PROBLEM 9.160 Brass wire with a weight per unit length w is used to form the figure shown. Determine the mass products of inertia Ixy, Iyz, and Izx of the wire figure.

SOLUTION

m=

W 1 = wL g g

W 3  3 w π × a  = π a g 2  2 g w w m2 = (3a ) = 3 a g g w w m3 = (π × a ) = π a g g m1 =

Then

Now observe that the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , of each component are zero because of symmetry. m 1

3 w π a 2 g

2

3

w a g

3

π

w a g

x −

y

z

2a

1 a 2

0

1 a 2

2a

(a )

−a

a

23  a π  2 

2

π

mx y w 3 a g

−9

0 −2

w 3 a g

−11

Σ

w 3 a g

my z 3 w 3 π a 2 g 3

mz x −

9w 3 a 4g

w 3 a g

−π

w 3 a g

wπ  + 3  a3  g2 

0 2 −

w 3 a g

1w 3 a 4g





Then

I xy

0 = Σ( I x′y′ + mx y )

I xy = −11

or

0 I yz = Σ( I y′z′ + my z )

or

0 + mz x )

or

1w 3 a (π + 6)  2g

I zx = −

1w 3 a  4g


I zx = Σ( I z′x′

I yz =

w 3 a  g


www.elsolucionario.net PROBLEM 9.161 Complete the derivation of Eqs. (9.47), which express the parallel-axis theorem for mass products of inertia.

SOLUTION We have and Consider







I xy = xydm I yz = yzdm I zx = zxdm x = x′ + x

y = y′ + y

(9.45)

z = z′ + z

(9.31)



I xy = xydm

Substituting for x and for y

 =  x′y ′dm + y  x′dm + x  y ′dm + x y  dm

By definition and



I x′y′ = x′y ′dm

 x′dm = mx ′  y′dm = my ′

However, the origin of the primed coordinate system coincides with the mass center G, so that x′ = y′ = 0 I xy = I x′y′ + mx y Q.E.D. 

The expressions for I yz and I zx are obtained in a similar manner.



I xy = ( x′ + x )( y ′ + y ) dm

www.elsolucionario.net PROBLEM 9.162 For the homogeneous tetrahedron of mass m shown, (a) determine by direct integration the mass product of inertia Izx, (b) deduce Iyz and Ixy from the result obtained in part a.

SOLUTION First divide the tetrahedron into a series of thin vertical slices of thickness dz as shown. a z  z + a = a 1 −  c c  

Now

x=−

and

b z  y = − z + b = b 1 −  c  c


(a)

The mass dm of the slab is 2

z 1  1  dm = ρ dV = ρ  xydz  = ρ ab 1 −  dz 2  2  c

Then



m = dm =



c 0

2

1 z  ρ ab 1 −  dz c 2   c

3  c  z  1 1 = ρ ab  − 1 −   = ρ abc 2 3 6 c     0

Now where

dI zx = dI z′x′ + zEL xEL dm dI z′x′ = 0 (symmetry)

and

zEL = z

xEL =

Then

I zx = dI zx =



=



1 1  z x = a 1 −  3 3  c c 0

2  1  z   1 z  z  a 1 −    ρ ab 1 −  dz   c  3  c    2 

1 2 ρa b 6



c

z2 z3 z4   z − 3 + 3 2 − 3  dz 0 c c c   c

m 1 z3 3 z 4 1 z5  = a  z2 − + −  c 2 c 4 c 2 5 c3  0

or

I zx =

1 mac  20



(b)

Because of the symmetry of the body, I xy and I yz can be deduced by considering the circular permutation of ( x, y, z ) and (a, b, c) . Thus,



I xy =

1 mab  20

I yz =

1 mbc  20

Alternative solution for part a:

Now

x=−

a y  y + a = a 1 −  b b 

and

z=−

c y  y + c = c 1 −  b  b


First divide the tetrahedron into a series of thin horizontal slices of thickness dy as shown.

The mass dm of the slab is 2

y 1  1  dm = ρ dV = ρ  xzdy  = ρ ac 1 −  dy 2  2  b

Now where and dI zx,area =

Then

dI zx = ρ tdI zx ,area t = dy

1 2 2 x z from the results of Sample Problem 9.6. 24 2 2  1   y    y    dIzx = ρ (dy )   a 1 −   c 1 −    24  b     b      4

4

1 y 1m  y  = ac 1 −  dy ρ a 2 c 2 1 −  dy = 24 b 4b  b 

Finally,



I zx = dI zx =



b 0

4

1m  y ac 1 −  dy 4 b  b b

5 1 m  b  y  ac  − 1 −   = 4 b  5  b   0

or

I zx =

1 mac  20



Alternative solution for part a: The equation of the included face of the tetrahedron is x y z + + =1 a b c x z  y = b 1 − −   a c

so that

For an infinitesimal element of sides dx, dy, and dz : dm = ρ dV = ρ dydxdz z  x = a 1 −  c  

From part a



I zx = zxdm =

c

a (1− z/c )

0

0



=ρ

= ρb = ρb =

b (1− x/a − z/c ) 0

c

a (1− z/c )

0

0



= ρb



  

c 0 c 0 c 0


Now zx( ρ dydxdz )

zx b (1 − ax −

z c

) dxdz a (1− z/c )

1 1 x3 1 z 2  z  x2 − x  − 3 a 2 c 0 2

dz

2 3 2 1  1 z z 1 z 2 z   z  a 2 1 −  − a3 1 −  − a 1 −   dz  2  c  3a  c  2 c  c   3

1 2  z a z 1 −  dz 6  c

1 2 ρa b 6



c

z2 z3 z4  3 3 z − + −  dz  0 c c 2 c3   c

m 1 z3 3 z4 1 z5  = a  z2 − + −  c 2 c 4 c 2 5 c3  0

or I zx =

1 mac  20


www.elsolucionario.net PROBLEM 9.163 The homogeneous circular cone shown has a mass m. Determine the mass moment of inertia of the cone with respect to the line joining the origin O and Point A.

SOLUTION 9 3  dOA =  a  + (−3a) 2 + (3a) 2 = a 2 2 

Then

λOA =

1 3  1 ai − 3aj + 3ak  = (i − 2 j + 2k )  a2  3

9 2

For a rectangular coordinate system with origin at Point A and axes aligned with the given x, y , z axes, we have (using Figure 9.28) 3 1  I x = I z = m  a 2 + (3a) 2  5 4  =

Also, symmetry implies

Iy =

3 ma 2 10

111 2 ma 20

I xy = I yz = I zx = 0

With the mass products of inertia equal to zero, Equation (9.46) reduces to I OA = I x λx2 + I y λ y2 + I z λz2 2

2

=

111 2  1  3  2  111 2  2  ma   + ma 2  −  + ma   20 3 10 20    3 3

=

193 2 ma 60

or

2

I OA = 3.22ma 2 



2

First note that

www.elsolucionario.net PROBLEM 9.164 The homogeneous circular cylinder shown has a mass m. Determine the mass moment of inertia of the cylinder with respect to the line joining the origin O and Point A that is located on the perimeter of the top surface of the cylinder.

SOLUTION Iy =

From Figure 9.28:

1 ma 2 2

2

Ix = Iz =

Symmetry implies

1 1 h m(3a 2 + h 2 ) + m   = m(3a 2 + 4h 2 ) 12 2 12  


For convenience, let Point A lie in the yz plane. Then

λOA =

1 h2 + a 2

( hj + ak )

With the mass products of inertia equal to zero, Equation (9.46) reduces to 0 I OA = I x λx2 + I y λ y2 + I z λz2  1 h = ma 2   2 2 2  h +a

2

  1 a  + m(3a 2 + 4h2 )   12  h2 + a 2  

or

   

2

I OA =

1 10h 2 + 3a 2 ma 2  12 h2 + a 2

Note: For Point A located at an arbitrary point on the perimeter of the top surface, λOA is given by

λOA =

1 h + a2 2

( a cos φ i + hj + a sin φ k )

which results in the same expression for I OA . 



and using the parallel-axis theorem

www.elsolucionario.net PROBLEM 9.165 Shown is the machine element of Problem 9.141. Determine its mass moment of inertia with respect to the line joining the origin O and Point A.


m = ρSTV =

0.284 lb/in 3 V = (0.008819 lb ⋅ s2 /ft ⋅ in 3 )V 32.2 ft/s 2

Then m1 = (7850 kg/m3 )[π (0.08 m)2 (0.04 m)] = 6.31334 kg m2 = (7850 kg/m3 )[π (0.02 m) 2 (0.06 m)] = 0.59188 kg m3 = (7850 kg/m3 )[π (0.02 m) 2 (0.04 m)] = 0.39458 kg

Symmetry implies and Now

I yz = I zx = 0

( I xy )1 = 0

( I x′y′ ) 2 = ( I x′y′ )3 = 0

I xy = Σ( I x′y′ + mx y ) = m2 x2 y2 − m3 x3 y3 = [0.59188 kg (0.04 m)(0.03 m)] − [0.39458 kg (−0.04 m)(−0.02 m)] = (0.71026 − 0.31566) × 10−3 kg ⋅ m 2 = 0.39460 × 10−3 kg ⋅ m 2



SOLUTION


From the solution to Problem 9.141, we have I x = 13.98800 × 10−3 kg ⋅ m 2 I y = 20.55783 × 10−3 kg ⋅ m 2 I z = 14.30368 × 10−3 kg ⋅ m 2

Substituting into Eq. (9.46)

λ OA =

1 13

(2i + 3j)

0 0 0 I OA = I x λx2 + I y λ y2 + I z λz2 − 2 I xy λx λ y − 2 I yz λ y λz − 2 I zx λz λx   2   3  = (13.98800)   + (20.55783)     13   13   2  2   2 −3 − 2(0.39460)     × 10 kg ⋅ m  13  13   2

2

= (4.30400 + 14.23234 − 0.36425) × 10−3 kg ⋅ m 2 or

I OA = 18.17 × 10−3 kg ⋅ m 2 



By observation

www.elsolucionario.net PROBLEM 9.166 Determine the mass moment of inertia of the steel fixture of Problems 9.145 and 9.149 with respect to the axis through the origin that forms equal angles with the x, y, and z axes.

SOLUTION From the solutions to Problems 9.145 and 9.149, we have I x = 26.4325 × 10−3 kg ⋅ m 2


Problem 9.145:

I y = 31.1726 × 10−3 kg ⋅ m 2 I z = 8.5773 × 10−3 kg ⋅ m 2

Problem 9.149:

I xy = 2.5002 × 10−3 kg ⋅ m 2 I yz = 4.0627 × 10−3 kg ⋅ m 2 I zx = 8.8062 × 10−3 kg ⋅ m 2

From the problem statement it follows that

λx = λ y = λz Now or

λx2 + λ y2 + λz2 = 1  3λx2 = 1 λx = λ y = λz =

1 3

Substituting into Eq. (9.46) I OL = I x λx2 + I y λ y2 + I z λz2 − 2 I xy λx λ y − 2 I yz λ y λz − 2 I zx λz λx

Noting that We have

λx2 = λ y2 = λz2 = λx λ y = λ y λz = λz λx =

1 3

1 I OL = [26.4325 + 31.1726 + 8.5773 3 − 2(2.5002 + 4.0627 + 8.8062)] × 10−3 kg ⋅ m 2 or

I OL = 11.81 × 10−3 kg ⋅ m 2 


www.elsolucionario.net PROBLEM 9.167 The thin bent plate shown is of uniform density and weight W. Determine its mass moment of inertia with respect to the line joining the origin O and Point A.

SOLUTION First note that and that

m1 = m2 =

λOA =

1 3

1W 2 g

(i + j + k )

I x = ( I x )1 + ( I x )2  1  1 W  2 1 W  a 2  =  a +    2 g  2    12  2 g   1  1 W +  12  2 g =

 2 1W 2  (a + a ) + 2 g 

 a 2  a 2     +      2   2   

1 W  1 1  2  1 1  2  1 W 2 a  +  a +  +  a  = 2 g  12 4  6 2  2 g

I y = ( I y )1 + ( I y )2  1  1 W =  12 2 g  

 2 1W 2  (a + a ) + 2 g 

 a 2  a 2     +      2   2   

 2  a 2   (a) +      2     1 W  1 1  2  1 5  2  W 2 = a +  + a  = a  + 2 g  6 2   12 4   g  1  1 W  2 1 W +  a + 2 g 12  2 g 

I z = ( I z )1 + ( I z ) 2  1 1W  2 1W =  a + 2 g 12  2 g 

a   2

2

 

2  1  1 W  2 1 W  2  a    +   (a ) +     a + 2 g   2    12  2 g  1 W  1 1  2  1 5  2  5 W 2 = + a +  + a  = a  2 g  12 4   12 4   6 g



Using Figure 9.28 and the parallel-axis theorem, we have


Now observe that the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , of both components are zero because of symmetry. Also, y1 = 0 0 1W a 1W 2 I xy = Σ( I x′y ′ + mx y ) = m2 x2 y2 = (a)   = a Then 2 g 2 4 g 0 1 W  a  a  1 W 2 I yz = Σ( I y′z′ + my z ) = m2 y2 z2 = a = 2 g  2   2  8 g 0 I zx = Σ( I z ′x′ + mz x ) = m1z1 x1 + m2 z2 x2 =

1 W  a  a  1 W  a  3W 2 (a ) = a +       2 g  2  2  2 g  2  8 g

Substituting into Equation (9.46)

Noting that

λx2 = λ y2 = λz2 = λx λ y = λ y λz = λz λx =

1 3

We have  1 W 2 1 W 2 3 W 2  1 1 W 2 W 2 5 W 2 I OA =  a + a + a − 2 a + a + a  3 2 g g 6 g 8 g 8 g 4 g  1 14  3  W =  − 2   a 2 3 6  4  g

or

I OA =

5 W 2 a  18 g



I OA = I x λx2 + I y λ y2 + I z λz2 − 2 I xy λx λ y − 2 I yz λ y λz − 2 I zx λz λx

www.elsolucionario.net PROBLEM 9.168 A piece of sheet steel of thickness t and specific weight γ is cut and bent into the machine component shown. Determine the mass moment of inertia of the component with respect to the joining the origin O and Point A.

SOLUTION

1

λ OA =

6

(i + 2 j + k )

Next compute the mass of each component. We have m = ρV =

Then

m1 = m2 =

γ g

γ g

(tA)

t (2a × 2a) = 4

γt g

a2

γ π

 π γt 2 t  × a2  = a g 2  2 g

Using Figure 9.28 for component 1 and the equations derived above (following the solution to Problem 9.134) for a semicircular plate for component 2, we have I x = ( I x )1 + ( I x )2

γt  1  γ t   =   4 a 2  [(2a) 2 + (2a )2 ] + 4 a 2 (a 2 + a 2 )  g 12  g    1  π γ t 2  2 π γ t 2  a a + a [(2a) 2 + (a) 2 ] +  2 g  4  2 g   π γt 2  1 2   a2  + 2  a2 + a  + 5  a2 3 2 g 4     γt 4 = 18.91335 a g =4

γt g



First note that


I y = ( I y )1 + ( I y ) 2  1  γt   γt =   4 a 2  (2a )2 + 4 a 2 ( a 2 )  g 12  g   2  π γ t  1 16    π γ t 2  4a  2 2 a  − 2 a + a  + + (a) 2    2 g  2 9π  2 g  3π     16 π γ t 2  1 16 1   a 2  + 1 a 2 + a  − 2 + 2 + 1 a 2 g 3  2 g 9π  2 9π  γt = 7.68953 a 4 g =4

γt

 1  γt   γt =   4 a 2  (2a) 2 + 4 a 2 ( a 2 )  g 12  g   2  π γ t  1 16    π γ t 2  4a  2 2 + + (2a) 2   a  − 2 a + a   2 g 2 g  4 9π   3π    

π γ t 2  1 16 16 1   a 2  + 1 a 2 + a  − 2 + 2 + 4  a2 g 2 g  4 9π 9π 3   γt = 12.00922 a 4 g =4

γt

Now observe that the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , of both components are zero because of symmetry. Also x1 = 0. 0 π γ t 2  4a  I xy = Σ( I x′y′ + mx y ) = m2 x2 y2 = a  Then  (2a) 2 g  3π  γt = 1.33333 a 4 g I yz = Σ( I y′z′ =4

γt g

0 + my z ) = m1 y1 z1 + m2 y2 z2

a 2 ( a)(a) +

= 7.14159

γt g

π γt 2 g

a 2 (2a)(a)

a4

0 π γ t 2  4a  I zx = Σ( I z ′x′ + mz x ) = m2 z2 x2 = a (a)   2 g  3π  γt = 0.66667 a 4 g PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1652


I z = ( I z )1 + ( I z ) 2


Substituting into Eq. (9.46) I OA = I x λx2 + I y λ y2 + I z λz2 − 2 I xy λx λ y − 2 I yz λ y λz − 2 I zx λz λx = 18.91335

2

 1  γt 4  2  a4   + 7.68953 a   g g  6  6

γt

2

2

  1  γ t 4   1  2  + 12.00922 a   − 2 1.33333 a     g  6 g    6  6   γ t   2  2   γ t 4   1  1  − 2  7.14159 a 4     − 2  0.66667 a     g g    6  6     6  6  γt = (3.15223 + 5.12635 + 2.00154 − 0.88889 − 4.76106 − 0.22222) a 4 g 4

or

I OA = 4.41

γt g

a 4 



γt

www.elsolucionario.net PROBLEM 9.169 Determine the mass moment of inertia of the machine component of Problems 9.136 and 9.155 with respect to the axis through the origin characterized by the unit vector λ = (− 4i + 8j + k )/9.

SOLUTION From the solutions to Problems 9.136 and 9.155. We have I x = 175.503 × 10−3 kg ⋅ m 2 I y = 308.629 × 10−3 kg ⋅ m 2 I z = 154.400 × 10−3 kg ⋅ m 2

Problem 9.155:

I xy = −8.0365 × 10−3 kg ⋅ m 2 I yz = 12.8950 × 10−3 kg ⋅ m 2 I zx = 94.0266 × 10−3 kg ⋅ m 2

Substituting into Eq. (9.46) I OL = I x λx2 + I y λ y2 + I z λz2 − 2 I xy λx λ y − 2 I yz λ y λz − 2 I zx λz λx 2 2 2   4 8 1 = 175.503  −  + 308.629   + 154.400    9 9 9 

 4  8   8  1  − 2(−8.0365)  −   − 2(12.8950)     9  9   9  9   1  4   − 2(94.0266)   −   × 10−3 kg ⋅ m 2  9  9   = (34.6673 + 243.855 + 1.906 − 6.350 − 2.547 + 9.287) × 10−3 kg ⋅ m 2

or

I OL = 281 × 10−3 kg ⋅ m 2 



Problem 9.136:

www.elsolucionario.net PROBLEM 9.170 For the wire figure of Problem 9.148, determine the mass moment of inertia of the figure with respect to the axis through the origin characterized by the unit vector λ = (−3i − 6j + 2k )/7.

SOLUTION

m m =   L = 0.056 kg/m × 1.2 m L = 0.0672 kg

Now observe that the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , for each component are zero because of symmetry. Also x1 = x6 = 0

Then

y4 = y5 = y6 = 0

z1 = z2 = z3 = 0

0 I xy = Σ( I x′y ′ + mx y ) = m2 x2 y2 + m3 x3 y3 = (0.0672 kg)(0.6 m)(1.2 m) + (0.0672 kg)(1.2 m)(0.6 m) = 0.096768 kg ⋅ m 2

0 I yz = Σ( I y′z′ + my z ) = 0 0 I zx = Σ( I z ′x′ + mz x ) = m4 z4 x4 + m5 z5 x5 = (0.0672 kg)(0.6 m)(1.2 m) + (0.0672 kg)(1.2 m)(0.6 m) = 0.096768 kg ⋅ m 2 From the solution to Problem 9.148, we have I x = 0.32258 kg ⋅ m 2 I y = I z = 0.41933 kg ⋅ m 2






0 I OL = I x λx2 + I y λ y2 + I z λz2 − 2 I xy λx λ y − 2 I yz λ y λz − 2 I zx λz λx 2 2 2   3  6 2 = 0.32258  −  + 0.41933  −  + 0.41933    7  7 7 

 3  6   2  3   − 2(0.096768)  −  −  − 2(0.096768)   −   kg ⋅ m 2 7 7     7  7   I OL = (0.059249 + 0.30808 + 0.034231 − 0.071095 + 0.023698) kg ⋅ m 2 I OL = 0.354 kg ⋅ m 2 


or




γ ST g

AL

m1 = m2 =

3 2  1 ft  490 lb/ft 3  π  1   × × × × π in. ( 18 in.)       32.2 ft/s 2  4  8    12 in. 

= 6.1112 × 10−3 lb ⋅ s 2 /ft m3 = m4 =

3 2  1 ft  490 lb/ft 3  π  1   × × × in. 18 in.       32.2 ft/s 2  4  8    12 in. 

= 1.9453 lb ⋅ s 2 /ft

Now observe that the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , for each component are zero because of symmetry. Also

x3 = x4 = 0

y1 = 0

Then

0 I xy = Σ( I x′y′ + mx y ) = m2 x2 y2 = (6.1112 × 10

−3

z1 = z2 = 0

 1 ft   2 × 18  lb ⋅ s /ft)  − in.  (18 in.)   π    12 in. 

2

2

= −8.75480 × 10−3 lb ⋅ ft ⋅ s 2

0 I yz = Σ( I y ′z′ + my z ) = m3 y3 z3 + m4 y4 z4 Now

m3 = m4 ,

y3 = y4 ,

z4 = − z3

0 I zx = Σ( I z′x′ + mz x ) or

I yz = 0

I zx = 0



Then


From the solution to Problem 9.147, we have I x = 39.1721 × 10 −3 lb ⋅ ft ⋅ s 2 I y = 36.2542 × 10−3 lb ⋅ ft ⋅ s 2 I z = 30.4184 × 10−3 lb ⋅ ft ⋅ s 2


0 0 I OL = I x λx2 + I y λ y2 + I z λz2 − 2 I xy λx λ y − 2 I yz λ y λz − 2 I zx λz λx

= (7.19488 + 26.6357 + 2.48313 + 6.43210) × 10−3 lb ⋅ ft ⋅ s 2

or

I OL = 0.0427 lb ⋅ ft ⋅ s 2 



2 2 2   3  6 2 = 39.1721 −  + 36.2542  −  + 30.4184    7  7 7   3  6   − 2(−8.75480)  −  −   × 10−3 lb ⋅ ft ⋅ s 2  7  7  


SOLUTION First compute the mass of each component. We have m=

W 1 = (W/L) AL L g g

m1 =

1 1 ft (0.033 lb/ft)(2π × 16 in.) × 2 12 in. 32.2 ft/s

= 8.5857 × 10 −3 lb ⋅ s 2 /ft m2 = m3 = m4 = m5 =

1 1 ft (0.033 lb/ft)(8 in.) × 2 12 in. 32.2 ft/s

= 0.6832 × 10−3 lb ⋅ ft/s 2

Now observe that the centroidal products of inertia, I x′y′ , I y′z′ , and I z′x′ , of each component are zero because of symmetry. Also

Then

x1 = x4 = x5 = 0 y1 = 0 z1 = z2 = z3 = 0 0 I xy = Σ( I x′y ′ + mx y ) = m2 x2 y2 + m3 x3 y3

 16  4  = 0.6832 × 10 −3 lb ⋅ s 2 /ft  ft  ft   12  12   12  8  + 0.6832 × 10−3 lb ⋅ s 2 /ft  ft  ft   12  12  = (0.30364 + 0.45547) × 10−3 lb ⋅ ft ⋅ s 2 = 0.75911 × 10−3 lb ⋅ ft ⋅ s 2

Symmetry implies

I yz = I xy

I yz = 0.75911 × 10−3 lb ⋅ ft ⋅ s 2

0 I zx = Σ( I z ′x′ + mz x ) = 0



Then


From the solution to Problem 9.146, we have I x = I z = 10.3642 × 10−3 lb ⋅ ft ⋅ s 2 I y = 19.1097 × 10−3 lb ⋅ ft ⋅ s 2

Substituting into Eq. (9.46) I OL =

I x λx2

+

I y λ y2

+

I z λz2

0 − 2 I xy λx λ y − 2 I yz λ y λz − 2 I zx λz λx

2 2 2   3  6 2 = 10.3642  −  + 19.1097  −  + 10.3642    7  7 7 

 3  6   6  2   − 2(0.75911)  −  −  − 2(0.75911)  −    × 10−3 lb ⋅ ft ⋅ s 2  7  7   7  7  

or

I OL = 16.61 × 10−3 lb ⋅ ft ⋅ s 2 



= (1.90663 + 14.03978 + 0.84606 − 0.55771 + 0.37181) × 10−3 lb ⋅ ft ⋅ s 2

www.elsolucionario.net PROBLEM 9.173 For the homogeneous circular cylinder shown, of radius a and length L, determine the value of the ratio a/L for which the ellipsoid of inertia of the cylinder is a sphere when computed (a) at the centroid of the cylinder, (b) at Point A.

SOLUTION (a)

From Figure 9.28: 1 2 ma 2 1 I y = I z = m(3a 2 + L2 ) 12

Now observe that symmetry implies


Using Eq. (9.48), the equation of the ellipsoid of inertia is then I x x 2 + I y y 2 + I z z 2 = 1:

1 1 1 ma 2 x 2 + m(3a 2 + L2 ) y 2 + m(3a 2 + L2 ) = 1 2 12 12

For the ellipsoid to be a sphere, the coefficients must be equal. Therefore 1 1 ma 2 = m(3a 2 + L2 ) 2 12

(b)

or

a 1 =  L 3

Using Figure 9.28 and the parallel-axis theorem, we have I x′ =

1 ma 2 2 2

I y′ = I z′ =

1 7 2 L 1 m(3a 2 + L2 ) + m   = m  a 2 + L  12 48  4 4

Now observe that symmetry implies I x′y′ = I y′z′ = I z′x′ = 0

From Part a it then immediately follows that 1 7 2 1 ma 2 = m  a 2 + L 2 48  4

or

a 7 =  L 12



Ix =

www.elsolucionario.net PROBLEM 9.174 For the rectangular prism shown, determine the values of the ratios b/a and c/a so that the ellipsoid of inertia of the prism is a sphere when computed (a) at Point A, (b) at Point B.

SOLUTION Using Figure 9.28 and the parallel-axis theorem, we have at Point A 1 I x′ = m(b 2 + c 2 ) 12 1 a m( a 2 + c 2 ) + m   12  2 1 = m(4a 2 + c 2 ) 12

2


(a)

I y′ =

2

I z′ =

1 1 a m(a 2 + b 2 ) + m   = m(4a 2 + b 2 ) 12 2 12  


Using Eq. (9.48), the equation of the ellipsoid of inertia is then I x′ x 2 + I y′ y 2 + I z′ z 2 = 1:

1 1 1 m(b 2 + c 2 ) x 2 + m(4a 2 + c 2 ) y 2 + m(4a 2 + b 2 ) z 2 = 1 12 12 12

For the ellipsoid to be a sphere, the coefficients must be equal. Therefore 1 1 m(b 2 + c 2 ) = m(4a 2 + c 2 ) 12 12 1 = m(4a 2 + b 2 ) 12

Then

b 2 + c 2 = 4a 2 + c 2

or

b =2  a

and

b 2 + c 2 = 4a 2 + b 2

or

c =2  a



(b)

Using Figure 9.28 and the parallel-axis theorem, we have at Point B 2

I x′′ =

1 1 c m(b 2 + c 2 ) + m   = m(b 2 + 4c 2 ) 12 12 2 2

1 1 c m ( a 2 + c 2 ) + m   = m ( a 2 + 4c 2 ) 12  2  12 1 I z′′ = m(a 2 + b 2 ) 12

I y′′ =

Now observe that symmetry implies I x′′y′′ = I y′′z′′ = I z′′x′′ = 0

1 1 1 m(b 2 + 4c 2 ) = m(a 2 + 4c 2 ) = m(a 2 + b 2 ) 12 12 12

Then

b 2 + 4c 2 = a 2 + 4c 2

or

and

b 2 + 4c 2 = a 2 + b 2

or

b =1  a c 1 =  a 2



From part a it then immediately follows that

www.elsolucionario.net PROBLEM 9.175 For the right circular cone of Sample Problem 9.11, determine the value of the ratio a/h for which the ellipsoid of inertia of the cone is a sphere when computed (a) at the apex of the cone, (b) at the center of the base of the cone.

SOLUTION (a)

From Sample Problem 9.11, we have at the apex A 3 ma 2 10 3 1  I y = I z = m  a 2 + h2  5 4  Ix =


Using Eq. (9.48), the equation of the ellipsoid of inertia is then I x x 2 + I y y 2 + I z z 2 = 1:

3 3 1 3 1   ma 2 x 2 + m  a 2 + h 2  y 2 + m  a 2 + h 2  z 2 = 1 10 5 4 5 4  

For the ellipsoid to be a sphere, the coefficients must be equal. Therefore, 3 3 1  ma 2 = m  a 2 + h 2  10 5 4 

(b)

or

a =2  h

From Sample Problem 9.11, we have

and at the centroid C

I x′ =

3 ma 2 10

I y′′ =

3  2 1 2 m a + h  20  4  2

Then

3  2 1 2 1 h I y′ = I z′ = m a + h  + m  = m(3a 2 + 2h 2 ) 20  4  20 4


From Part a it then immediately follows that 3 1 ma 2 = m(3a 2 + 2h 2 ) 10 20

or

a = h

2  3



Now observe that symmetry implies

www.elsolucionario.net PROBLEM 9.176 Given an arbitrary body and three rectangular axes x, y, and z, prove that the mass moment of inertia of the body with respect to any one of the three axes cannot be larger than the sum of the mass moments of inertia of the body with respect to the other two axes. That is, prove that the inequality I x ≤ I y + I z and the two similar inequalities are satisfied. Further, prove that I y ≥ 12 I x if the body is a homogeneous solid of revolution, where x is the axis of revolution and y is a transverse axis.

SOLUTION Iy + Iz ≥ Ix

To prove

 =  (x

I y = ( z 2 + x 2 )dm

By definition

Iz

2

+ y 2 ) dm





I y + I z = ( z 2 + x 2 ) dm + ( x 2 + y 2 ) dm

Then





= ( y 2 + z 2 )dm + 2 x 2 dm

Now

(y

2

+ z 2 ) dm = I x

 x dm ≥ 0 2

and

I y + Iz ≥ Ix

Q.E.D.

The proofs of the other two inequalities follow similar steps. (ii)

If the x axis is the axis of revolution, then I y = Iz

and from part (i) or

Iy + Iz ≥ Ix

2I y ≥ I x Iy ≥

or

1 Ix 2

Q.E.D. 



(i)

www.elsolucionario.net PROBLEM 9.177 Consider a cube of mass m and side a. (a) Show that the ellipsoid of inertia at the center of the cube is a sphere, and use this property to determine the moment of inertia of the cube with respect to one of its diagonals. (b) Show that the ellipsoid of inertia at one of the corners of the cube is an ellipsoid of revolution, and determine the principal moments of inertia of the cube at that point.

SOLUTION (a)

At the center of the cube have (using Figure 9.28) Ix = I y = Iz =

1 1 m(a 2 + a 2 ) = ma 2 12 6

Now observe that symmetry implies I xy = I yz = I zx = 0

1 2 2 1 2 2 1 2 2  6 ma  x +  6 ma  y +  6 ma  z = 1      

or

x2 + y2 + z 2 =

6 (= R 2 )  2 ma

which is the equation of a sphere. Since the ellipsoid of inertia is a sphere, the moment of inertia with respect to any axis OL through the center O of the cube must always  1  1 . be the same  R = I OL = ma 2   6 I OL   (b)

The above sketch of the cube is the view seen if the line of sight is along the diagonal that passes through corner A. For a rectangular coordinate system at A and with one of the coordinate axes aligned with the diagonal, an ellipsoid of inertia at A could be constructed. If the cube is then rotated 120° about the diagonal, the mass distribution will remain unchanged. Thus, the ellipsoid will also remain unchanged after it is rotated. As noted at the end of Section 9.17, this is possible only if the ellipsoid is an ellipsoid of revolution, where the diagonal is both the axis of revolution and a principal axis. 1 It then follows that I x′ = I OL = ma 2  6 In addition, for an ellipsoid of revolution, the two transverse principal moments of inertia are equal and any axis perpendicular to the axis of revolution is a principal axis. Then, applying the parallel-axis theorem between the center of the cube and corner A for any perpendicular axis  3  1 I y′ = I z′ = ma 2 + m  a  2  6  

2

or

I y′ = I z′ =

11 2 ma  12

(Note: Part b can also be solved using the method of Section 9.18.) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1666


Using Equation (9.48), the equation of the ellipsoid of inertia is


First note that at corner A 2 2 ma 3 1 = ma 2 4

Ix = I y = Iz = I xy = I yz = I zx

Substituting into Equation (9.56) yields k 3 − 2ma 2 k 2 +

55 2 6 121 3 9 m a k− m a =0 48 864

For which the roots are 1 2 ma 6  11 k2 = k3 = ma 2 12


k1 =


www.elsolucionario.net PROBLEM 9.178 Given a homogeneous body of mass m and of arbitrary shape and three rectangular axes x, y, and z with origin at O, prove that the sum Ix + Iy + Iz of the mass moments of inertia of the body cannot be smaller than the similar sum computed for a sphere of the same mass and the same material centered at O. Further, using the result of Problem 9.176, prove that if the body is a solid of revolution, where x is the axis of revolution, its mass moment of inertia Iy about a transverse axis y cannot be smaller than 3ma2/10, where a is the radius of the sphere of the same mass and the same material.

SOLUTION (i)

Using Equation (9.30), we have







I x + I y + I z = ( y 2 + z 2 )dm + ( z 2 + x 2 ) dm + ( x 2 + y 2 )dm





= 2 r 2 dm

where r is the distance from the origin O to the element of mass dm. Now assume that the given body can be formed by adding and subtracting appropriate volumes V1 and V2 from a sphere of mass m and radius a which is centered at O; it then follows that m1 = m2 (mbody = msphere = m)

Then ( I x + I y + I z ) body = ( I x + I y + I z )sphere + ( I x + I y + I z )V1 − ( I x + I y + I z )V2

or

( I x + I y + I z ) body = ( I x + I y + I z )sphere + 2



m1

r 2 dm − 2



r 2 dm

m2

Now, m1 = m2 and r1 ≥ r2 for all elements of mass dm in volumes 1 and 2.



m1

r 2 dm −



m2

r 2 dm ≥ 0 ( I x + I y + I z ) body ≥ ( I x + I y + I z )sphere

so that

Q.E.D.



= 2 ( x 2 + y 2 + z 2 )dm


(ii)

First note from Figure 9.28 that for a sphere Ix = Iy = Iz =

Thus

( I x + I y + I z )sphere =

2 2 ma 5

6 2 ma 5

For a solid of revolution, where the x axis is the axis of revolution, we have I y = Iz

Then, using the results of part (i)

Iy ≥

From Problem 9.178 we have or

6 2 ma 5 1 Ix 2

(2 I y − I x )body ≥ 0

Adding the last two inequalities yields (4 I y ) body ≥

6 2 ma 5 ( I y ) body ≥

or

3 ma 2 10

Q.E.D.



( I x + 2 I y ) body ≥

www.elsolucionario.net PROBLEM 9.179* The homogeneous circular cylinder shown has a mass m, and the diameter OB of its top surface forms 45° angles with the x and z axes. (a) Determine the principal mass moments of inertia of the cylinder at the origin O. (b) Compute the angles that the principal axes of inertia at O form with the coordinate axes. (c) Sketch the cylinder, and show the orientation of the principal axes of inertia relative to the x, y, and z axes.

SOLUTION (a)

First compute the moments of inertia using Figure 9.28 and the parallel-axis theorem.  a 2  a 2  13 1 2 m(3a 2 + a 2 ) + m   +    = ma 12 2 12  2     1 3 I y = ma 2 + m(a) 2 = ma 2 2 2


Ix = Iz =

Next observe that the centroidal products of inertia are zero because of symmetry. Then 0  a  a  1 I xy = I x′y′ + mx y = m  ma 2  −  = − 2 2  2  2  0 1  a  a  I yz = I y′z ′ + my z = m  −   ma 2 =− 2 2  2  2  0  a  a  1 2 I zx = I z′x′ + mz x = m    = ma  2  2  2 Substituting into Equation (9.56)  13 3 13  K 3 −  + +  ma 2 K 2  12 2 12  2

2

 13 3   3 13   13 13   1   1  1 +  ×  +  ×  +  ×  −  −  − −  −   12 2   2 12   12 12   2 2   2 2   2  2

 13 3 13   13   1   3  1  −  × × −    −  −     12 2 12   12   2 2   2  2 

2

 (ma 2 ) 2 K 

2

2

 1  1  1   1   13   2 3 −   −  − 2 −  −     ( ma ) = 0  12   2 2   2 2  2 2   2  



Simplifying and letting K = ma 2ζ yields

ζ3−

11 2 565 95 ζ + ζ− =0 3 144 96

Solving yields

ζ 1 = 0.363383

ζ2 =

19 12

ζ 3 = 1.71995

The principal moments of inertia are then K1 = 0.363ma 2  K 2 = 1.583ma 2 

(b)

To determine the direction cosines λx , λ y , λz of each principal axis, we use two of the equations of Equations (9.54) and (9.57). Thus, ( I x − K )λx − I xy λ y − I zx λz = 0

(9.54a)

− I zx λx − I yz λ y + ( I z − K )λz = 0

(9.54c)

λx2 + λ y2 + λz2 = 1

(9.57)

(Note: Since I xy = I yz , Equations (9.54a) and (9.54c) were chosen to simplify the “elimination” of λ y during the solution process.) Substituting for the moments and products of inertia in Equations (9.54a) and (9.54c)   1  13 2  1  ma 2  λ y −  ma 2  λz = 0  12 ma − K  λx −  −   2   2 2    1 1   13  −  ma 2  λx −  − ma 2  λ y +  ma 2 − K  λz = 0 2   12   2 2 

or

and

 13  12 − ζ 

1 1  λ y − λz = 0  λx + 2 2 2 

1 1  13 − λx + λy +  − ζ 2 2 2  12

  λz = 0 

(i)

(ii)

Observe that these equations will be identical, so that one will need to be replaced, if 13 1 −ζ = − 12 2

ζ =

or

19 12



K 3 = 1.720ma 2 


Thus, a third independent equation will be needed when the direction cosines associated with K 2 are determined. Then for K1 and K 3 Eq. (i) through Eq. (ii):

 13  1  13  1   − ζ −  −   λx +  − −  − ζ  2  12  2  12

   λz = 0 

λz = λx

or Substituting into Eq. (i):

 13  12 − ζ 

1 1  λ y − λx = 0  λx + 2 2 2   

λy = 2 2  ζ −

or

7  λx 12 

Substituting into Equation (9.57): 



7 

2

or

2  7    2 + 8  ζ −   λx2 = 1 12    

(iii)

K1: Substituting the value of ζ 1 into Eq. (iii): 2  7    2 + 8  0.363383 −   (λx )12 = 1 12    

or and then

(λx )1 = (λz )1 = 0.647249 7   (λ y )1 = 2 2  0.363383 −  (0.647249) 12   = −0.402662 (θ x )1 = (θ z )1 = 49.7°

(θ y )1 = 113.7° 

K 3 : Substituting the value of ζ 3 into Eq. (iii): 2  7     2 + 8 1.71995 −   (λx )32 = 1 12    

or and then

(λx )3 = (λz )3 = 0.284726 7   (λ y )3 = 2 2 1.71995 −  (0.284726) 12   = 0.915348 (θ x )3 = (θ z )3 = 73.5°

(θ y )3 = 23.7° 





λx2 +  2 2  ζ −  λx  + (λx ) 2 = 1 12    

www.elsolucionario.net PROBLEM 9.179* (Continued)  K2: For this case, the set of equations to be solved consists of Equations (9.54a), (9.54b), and (9.57). Now − I xy λx + ( I y − K )λ y − I yz λz = 0

(9.54b)

Substituting for the moments and products of inertia.     1 1 3  − − ma 2  λx +  ma 2 − K  λ y −  − ma 2  λz = 0 2   2 2   2 2  3

1



1

λx +  − ξ  λ y + λz = 0 2 2 2 2 2 

or

(iv)

Substituting the value of ξ 2 into Eqs. (i) and (iv):

or

− (λx ) 2 +

and

1 2

(λx ) 2 −

(λ y ) 2 − (λ z ) 2 = 0

2 (λ y ) 2 + (λz ) 2 = 0 6

Adding yields

(λ y ) 2 = 0

and then

(λ y ) 2 = −(λx ) 2

Substituting into Equation (9.57) (λx ) 22

or

+

(λx ) 2 =


1 1  13 19  (λ y ) 2 − (λz ) 2 = 0  12 − 12  (λx ) 2 + 2 2 2   1 1  3 19  (λx ) 2 +  −  (λ y ) 2 + (λ z ) 2 = 0 2 2 2 2  2 12 

(λ y )22 1 2

0 + (−λx ) 22 = 1 and (λz ) 2 = −

1 2

(θ x ) 2 = 45.0° (θ y )2 = 90.0° (θ z ) 2 = 135.0° 

(c)

Principal axes 1 and 3 lie in the vertical plane of symmetry passing through Points O and B. Principal axis 2 lies in the xz plane.


www.elsolucionario.net PROBLEM 9.180 For the component described in Problem 9.165, determine (a) the principal mass moments of inertia at the origin, (b) the principal axes of inertia at the origin. Sketch the body and show the orientation of the principal axes of inertia relative to the x, y, and z axes.

SOLUTION From the solutions to Problems 9.141 and 9.165 we have I x = 13.98800 × 10−3 kg ⋅ m 2

Problem 9.141:

I y = 20.55783 × 10−3 kg ⋅ m 2 I z = 14.30368 × 10−3 kg ⋅ m 2 I yz = I zx = 0

Problem 9.165:

I xy = 0.39460 × 10−3 kg ⋅ m 2

Eq. (9.55) then becomes

Thus Substituting:

Ix − K

− I xy

0

− I xy

Iy − K

0

0

0

Iz − K

2 = 0 or ( I x − K )( I y − K )( I z − K ) − ( I z − K ) I xy =0

Iz − K = 0

2 I x I y − ( I x + I y ) K + K 2 − I xy =0

K1 = 14.30368 × 10−3 kg ⋅ m 2

or

K1 = 14.30 × 10−3 kg ⋅ m 2 

−3 −3 −3 2 −3 2 and (13.98800 × 10 )(20.55783 × 10 ) − (13.98800 + 20.55783)(10 ) K + K − (0.39460 × 10 ) = 0

K 2 − (34.54583 × 10−3 ) K + 287.4072 × 10−6 = 0

or Solving yields

and

K 2 = 13.96438 × 10−3 kg ⋅ m 2

K 3 = 20.58145 × 10−3 kg ⋅ m 2

or

K 2 = 13.96 × 10−3 kg ⋅ m 2 

or

K 3 = 20.6 × 10−3 kg ⋅ m 2 



(a)


(b)

To determine the direction cosines λx , λ y , λz of each principal axis, use two of the equations of Eq. (9.54) and Eq. (9.57). Then K1:

Begin with Eqs. (9.54a) and (9.54b) with I yz = I zx = 0. ( I x − K1 )(λx )1 − I xy (λ y )1 = 0 − I xy (λx )1 + ( I y − K1 )(λ y )1 = 0

Substituting: [(13.98800 − 14.30368) × 10 −3 ](λx )1 − (0.39460 × 10 −3 )(λ y )1 = 0 −(0.39460 × 10−3 )(λx )1 + [(20.55783 − 14.30368) × 10−3 ](λ y )1 = 0

Then using Eq. (9.57)

0 0 (λx )12 + (λ y )12 + (λz )12 = 1 (λz )1 = 1

or

(θ x )1 = 90.0° (θ y )1 = 90.0° (θ z )1 = 0° 

K2: Begin with Eqs. (9.54b) and (9.54c) with I yz = I zx = 0. − I xy (λx ) 2 + ( I y − K 2 )(λ y ) 2 = 0

(i)

( I z − K 2 )(λz ) 2 = 0 I z ≠ K 2  (λz ) 2 = 0

Now Substituting into Eq. (i):

−(0.39460 × 10−3 )(λx ) z + [(20.55783 − 13.96438) × 10−3 ](λ y ) 2 = 0

or

(λ y ) 2 = 0.059847(λx ) 2

Using Eq. (9.57):

(λx ) 22

or

(λx ) 2 = 0.998214

and

(λ y ) 2 = 0.059740

+ [0.05984](λx ) 2 ] + 2

(λz ) 22

(θ x ) 2 = 3.4°

0 =1

(θ y ) 2 = 86.6°

(θ z ) 2 = 90.0° 

K3: Begin with Eqs. (9.54b) and (9.54c) with I yz = I zx = 0. − I xy (λx )3 + ( I y − K 3 )(λ y )3 = 0

(ii)

( I z − K 3 )(λz )3 = 0



(λx )1 = (λ y )1 = 0

Adding yields


I z ≠ K 3  (λz )3 = 0

Now Substituting into Eq. (ii):

−(0.39460 × 10−3 )(λx )3 + [(20.55783 − 20.58145) × 10−3 ](λ y )3 = 0

Using Eq. (9.57):

(λ y )3 = −16.70618(λx )3

0 (λx )32 + [−16.70618(λx )3 ]2 + (λz )32 = 1

or

(λx )3 = −0.059751 (axes right-handed set  "_")

and

(λ y )3 = 0.998211 (θ x )3 = 93.4°

(θ y )3 = 3.43°

(θ z )3 = 90.0° 

Note: Since the principal axes are orthogonal and (θ z ) 2 = (θ z )3 = 90°, it follows that |(λx ) 2 | = |(λ y )3 |

|(λ y ) 2 | = | (λz )3 |

The differences in the above values are due to round-off errors. (c)

Principal axis 1 coincides with the z axis, while principal axes 2 and 3 lie in the xy plane.



or

www.elsolucionario.net PROBLEM 9.181* For the component described in Problems 9.145 and 9.149, determine (a) the principal mass moments of inertia at the origin, (b) the principal axes of inertia at the origin. Sketch the body and show the orientation of the principal axes of inertia relative to the x, y, and z axes.

SOLUTION From the solutions to Problems 9.145 and 9.149, we have Problem 9.145:

I x = 26.4325 × 10−3 kg ⋅ m 2

Problem 9.149:

I xy = 2.5002 × 10−3 kg ⋅ m 2

I y = 31.1726 × 10−3 kg ⋅ m 2

I yz = 4.0627 × 10−3 kg ⋅ m 2

I z = 8.5773 × 10−3 kg ⋅ m 2

I zx = 8.8062 × 10−3 kg ⋅ m 2

Substituting into Eq. (9.56): K 3 − [(26.4325 + 31.1726 + 8.5773)(10−3 )]K 2 + [(26.4325)(31.1726) + (31.1726)(8.5773) + (8.5773)(26.4325) − (2.5002) 2 − (4.0627)2 − (8.8062)2 ](10−6 ) K − [(26.4325)(31.1726)(8.5773) − (26.4325)(4.0627)2 − (31.1726)(8.8062)2 − (8.5773)(2.5002) 2 − 2(2.5002)(4.0627)(8.8062)](10−9 ) = 0

or

K 3 − (66.1824 × 10−3 ) K 2 + (1217.76 × 10−6 ) K − (3981.23 × 10−9 ) = 0

Solving yields



K1 = 4.1443 × 10−3 kg ⋅ m 2

or

K1 = 4.14 × 10−3 kg ⋅ m 2 

K 2 = 29.7840 × 10−3 kg ⋅ m 2

or

K 2 = 29.8 × 10−3 kg ⋅ m 2 

K3 = 32.2541 × 10−3 kg ⋅ m 2 

or

K3 = 32.3 × 10−3 kg ⋅ m 2 



(a)


(b)

To determine the direction cosines λx , λ y , λz of each principal axis, use two of the equations of Eqs. (9.54) and Eq. (9.57). Then K1 : Begin with Eqs. (9.54a) and (9.54b).

( I x − K1 )(λx )1 − I xy (λ y )1 − I zx (λz )1 = 0 I xy (λx )1 + ( I y − K1 )(λ y )1 − I yz (λz )1 = 0

Substituting: [(26.4325 − 4.1443)(10−3 )](λx )1 − (2.5002 × 10−3 )(λ y )1 − (8.8062 × 10−3 )(λz )1 = 0 −(2.5002 × 10 −3 )(λx )1 + [(31.1726 − 4.1443)(10 −3 )](λ y )1 − (4.0627 × 10−3 )(λz )1 = 0

Simplifying −0.0925(λx )1 + (λ y )1 − 0.1503(λz )1 = 0

Adding and solving for (λz )1: (λz )1 = 2.4022(λx )1 (λ y )1 = [8.9146 − 3.5222(2.4022)](λx )1

and then

= 0.45357(λx )1

Now substitute into Eq. (9.57): (λx )12 + [0.45357(λx )1 ]2 + [2.4022(λx )1 ]2 = 1 (λx )1 = 0.37861

or and

(λ y )1 = 0.17173

(λz )1 = 0.90950 (θ x )1 = 67.8° (θ y )1 = 80.1° (θ z )1 = 24.6° 

K 2 : Begin with Eqs. (9.54a) and (9.54b). ( I x − K 2 )(λx )2 − I xy (λ y ) 2 − I zx (λz ) 2 = 0 − I xy (λx ) 2 + ( I y − K 2 )(λ y ) 2 − I yz (λz ) 2 = 0

Substituting: [(26.4325 − 29.7840)(10−3 )](λx )2 − (2.5002 × 10−3 )(λ y )2 − (8.8062 × 10−3 )(λz )2 = 0 −(2.5002 × 10 −3 )(λx ) 2 + [(31.1726 − 29.7840)(10−3 )](λ y ) 2 − (4.0627 × 10−3 )( λz ) 2 = 0



8.9146(λx )1 − (λ y )1 − 3.5222(λz )1 = 0


Simplifying −1.3405(λx )2 − (λ y ) 2 − 3.5222(λz ) 2 = 0 −1.8005(λx )2 + (λ y ) 2 − 2.9258(λz ) 2 = 0

Adding and solving for (λz ) 2 : (λz )2 = −0.48713(λx ) 2 (λ y ) 2 = [−1.3405 − 3.5222(−0.48713)](λx )2

and then

= 0.37527(λx ) 2

Now substitute into Eq. (9.57): (λx ) 22 + [0.37527(λx ) 2 ]2 + [−0.48713(λx ) 2 ]2 = 1

(λ y ) 2 = 0.31967

and

(λz )2 = −0.41496 (θ x )1 = 31.6° (θ y ) 2 = 71.4° (θ z ) 2 = 114.5° 

K3 :

Begin with Eqs. (9.54a) and (9.54b). ( I x − K3 )(λx )3 − I xy (λ y )3 − I zx (λz )3 = 0 − I xy (λx )3 + ( I y − K3 )(λ y )3 − I yz (λz )3 = 0

Substituting: [(26.4325 − 32.2541)(10−3 )](λx )3 − (2.5002 × 10−3 )(λ y )3 − (8.8062 × 10−3 )(λz )3 = 0 −(2.5002 × 10−3 )(λx )3 + [(31.1726 − 32.2541)(10−3 )](λ y )3 − (4.0627 × 10−3 )(λz )3 = 0

Simplifying −2.3285(λx )3 − (λ y )3 − 3.5222(λz )3 = 0 2.3118(λx )3 + (λ y )3 + 3.7565(λz )3 = 0

Adding and solving for (λz )3 : (λz )3 = 0.071276(λx )3

and then

(λ y )3 = [ −2.3285 − 3.5222(0.071276)](λx )3 = −2.5795(λx )3



(λx ) 2 = 0.85184

or


Now substitute into Eq. (9.57): (λx )32 + [−2.5795(λx )3 ]2 + [0.071276(λx )3 ]2 = 1 (λx )3 = 0.36134

or and

(i)

(λ y )3 = 0.93208

(λz )3 = 0.025755 (θ x )3 = 68.8° (θ y )3 = 158.8° (θ z )3 = 88.5° 

Note: Principal axis 3 has been labeled so that the principal axes form a right-handed set. To obtain the direction cosines corresponding to the labeled axis, the negative root of Eq. (i) must be chosen; that is, (λx )3 = −0.36134.

(θ x )3 = 111.2° (θ y )3 = 21.2° (θ z )3 = 91.5° 

Then



(c)

www.elsolucionario.net PROBLEM 9.182* For the component described in Problem 9.167, determine (a) the principal mass moments of inertia at the origin, (b) the principal axes of inertia at the origin. Sketch the body and show the orientation of the principal axes of inertia relative to the x, y, and z axes.

SOLUTION From the solution of Problem 9.167, we have 1W 2 a 2 g W I y = a2 g 5W 2 Iz = a 6 g Ix =

1W 2 a 4 g 1W 2 I yz = a 8 g 3W 2 I zx = a 8 g

I xy =

Substituting into Eq. (9.56):  1 5   W  K 3 −  + 1 +   a 2   K 2 6  g   2 2 2 2 2  1   5   5  1   1   1   3    W  +   (1) + (1)   +    −   −   −     a 2  K  6   6  2   4   8   8    g  2   3 2 2 2  1   5   1  1   3   5  1   1  1  3    W 2  −   (1)   −    − (1)   −    − 2       a  = 0  8   6  4   4  8  8    g  2   6   2  8  

Simplifying and letting K =

W 2 a K yields g K 3 − 2.33333K 2 + 1.53125K − 0.192708 = 0

Solving yields K1 = 0.163917 K 2 = 1.05402 K3 = 1.11539

The principal moments of inertia are then

K1 = 0.1639

W 2 a  g

K 2 = 1.054

W 2 a  g

K3 = 1.115

W 2 a  g



(a)


(b)

To determine the direction cosines λx , λ y , λz of each principal axis, use two of the equations of Eq. (9.54) and Eq. (9.57). Then K1 :

Begin with Eqs. (9.54a) and (9.54b). ( I x − K1 )(λx )1 − I xy (λ y )1 − I zx (λz )1 = 0 − I xy (λx )1 + ( I y − K 2 )(λ y )1 − I yz (λz )1 = 0

Substituting  1 1W 2 3W 2   W 2  a  (λ y )1 −  a  (λz )1 = 0  − 0.163917   a   (λx )1 −  2 g 4 g     8 g  

Simplifying 1.34433(λx )1 − (λ y )1 − 1.5(λz )1 = 0 −0.299013(λx )1 + (λ y )1 − 0.149507(λz )1 = 0

Adding and solving for (λz )1: (λz )1 = 0.633715(λx )1 (λ y )1 = [1.34433 − 1.5(0.633715)](λx )1

and then

= 0.393758(λx )1

Now substitute into Eq. (9.57): (λx )12 + [0.393758(λx )1 ]2 + (0.633715(λx )1 ]2 = 1 (λx )1 = 0.801504 

or and

(λ y )1 = 0.315599

(λz )1 = 0.507925 (θ x )1 = 36.7° (θ y )1 = 71.6° (θ z )1 = 59.5° 

K2 :

Begin with Eqs. (9.54a) and (9.54b): ( I x − k2 )(λx ) 2 − I xy (λ y ) 2 − I zx (λz ) 2 = 0 − I xy (λx ) 2 + ( I y − k2 )(λ y )2 − I yz (λz ) 2 = 0



 1W 2  W  1W 2 − a  (λx )1 + (1 − 0.163917)  a 2   (λ y )1 −  a  (λz )1 = 0 4 g   g  8 g  


Substituting  1 1W 2 3W 2   W 2  a  (λ y ) 2 −  a  (λz ) 2 = 0  − 1.05402   a   (λx ) 2 −   g  4 g  8 g   2  1W 2  W  1W 2 − a  (λx )2 + (1 − 1.05402)  a 2   (λ y ) 2 −  a  (λz ) 2 = 0 4 g   g  8 g  

Simplifying −2.21608(λx ) 2 − (λ y ) 2 − 1.5(λz ) 2 = 0 4.62792(λx ) 2 + (λ y ) 2 + 2.31396(λz )2 = 0

Adding and solving for (λz ) 2

(λ y ) 2 = [−2.21608 − 1.5(−2.96309)](λx ) 2

and then

= 2.22856(λx ) 2

Now substitute into Eq. (9.57): (λx ) 22 + [2.22856(λx ) 2 ]2 + [−2.96309(λx ) 2 ]2 = 1 (λx ) 2 = 0.260410

or and

(λ y ) 2 = 0.580339

(λz ) 2 = −0.771618 (θ x ) 2 = 74.9° (θ y ) 2 = 54.5° (θ z )2 = 140.5° 

 K3 : Begin with Eqs. (9.54a) and (9.54b):

( I x − K3 )(λx )3 − I xy (λ y )3 − I zx (λz )3 = 0 − I xy (λx )3 + ( I y − K3 )(λ y )3 − I yz (λz )3 = 0

Substituting  1 1W 2 3W 2   W 2  a  (λ y )3 −  a  (λz )3 = 0  − 1.11539   a   (λx )3 −   g  4 g  8 g   2  1W 2  W  1W 2 − a  (λx )3 + (1 − 1.11539)  a 2   (λ y )3 −  a  (λ z )3 = 0 4 g   g  8 g  

Simplifying −2.46156(λx )3 − (λ y )3 − 1.5(λz )3 = 0 2.16657(λx )3 + (λ y )3 + 1.08328(λz )3 = 0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1683


(λz )2 = −2.96309(λx ) 2


Adding and solving for (λz )3 (λz )3 = −0.707885(λx )3

and then

(λ y )3 = [ −2.46156 − 1.5(−0.707885)](λx )3 = −1.39973(λx )3

Now substitute into Eq. (9.57): (λx )32 + [ −1.39973(λx )3 ]2 + [−0.707885(λx )3 ]2 = 1

or

(λx )3 = 0.537577

and

(λ y )3 = −0.752463

(i)

(λz )3 = −0.380543

(c)


(θ x )3 = 122.5° (θ y )3 = 41.2° (θ z )3 = 67.6° 

Then



(θ x )3 = 57.5° (θ y )3 = 138.8° (θ z )3 = 112.4° 

www.elsolucionario.net PROBLEM 9.183* For the component described in Problem 9.168, determine (a) the principal mass moments of inertia at the origin, (b) the principal axes of inertia at the origin. Sketch the body and show the orientation of the principal axes of inertia relative to the x, y, and z axes.

SOLUTION From the solution to Problem 9.168, we have I x = 18.91335

γt

I xy = 1.33333

a4 g γt I y = 7.68953 a 4 g γt I z = 12.00922 a 4 g

I yz I zx

γt

a4 g γt = 7.14159 a 4 g γt = 0.66667 a 4 g

Substituting into Eq. (9.56):   γ t  K 3 − (18.91335 + 7.68953 + 12.00922)  a 4   K 2 g   + [(18.91335)(7.68953) + (7.68953)(12.00922) + (12.00922)(18.91335) 2

 γt  − (1.33333) 2 − (7.14159) 2 + (0.66667)2 ]  a 4  K g 

− [(18.91335)(7.68953)(12.00922) − (18.91335)(7.14159) 2 − (7.68953)(0.66667)2 − (12.00922)(1.33333) 2 3

γt  − 2(1.33333)(7.14159)(0.66667)]  a 4  = 0 g 

Simplifying and letting K=

γt g

a4 K

yields

K 3 − 38.61210 K 2 + 411.69009 K − 744.47027 = 0



(a)


Solving yields K1 = 2.25890 K 2 = 17.27274 K3 = 19.08046 K1 = 2.26

The principal moments of inertia are then

K 2 = 17.27 K 3 = 19.08

(b)

γt g

γt g

γt g

a4  a4  a4 

To determine the direction cosines λx , λ y , λz of each principal axis, use two of the equations of Eq. (9.54) and Eq. (9.57). Then

( I x − K1 )(λx )1 − I xy (λ y )1 − I zx (λz )1 = 0 − I xy (λx )1 + ( I y − K 2 )(λ y )1 − I yz (λz )1 = 0

Substituting   γ t 4   γt 4  γt 4   (18.91335 − 2.25890)  a   (λx )1 − 1.33333 a  (λ y )1 −  0.66667 a  (λz )1 = 0 a  g   g        γ t   γt  γt  − 1.33333 a 4  (λx )1 + (7.68953 − 2.25890)  a 4   (λ y )1 −  7.14159 a 4  (λz )1 = 0 g  g    g   

Simplifying 12.49087(λx )1 − (λ y )1 − 0.5(λz )1 = 0 −0.24552(λx )1 + (λ y )1 − 1.31506(λz )1 = 0

Adding and solving for (λz )1 (λz )1 = 6.74653(λx )1

and then

(λ y )1 = [12.49087 − (0.5)(6.74653)](λx )1 = 9.11761(λx )1

Now substitute into Eq. (9.57):

(λx )12 + [9.11761(λx )1 ]2 + [6.74653(λx )1 ]2 = 1

or

(λx )1 = 0.087825

and

(λ y )1 = 0.80075

(λz )1 = 0.59251 (θ x )1 = 85.0° (θ y )1 = 36.8° (θ z )1 = 53.7° 



K1 : Begin with Eqs. (9.54a) and (9.54b):


K 2 : Begin with Eqs. (9.54a) and (9.54b): ( I x − K 2 )(λx )2 − I xy (λ y ) 2 − I zx (λz ) 2 = 0 − I xy (λx ) 2 + ( I y − K 2 )(λ y ) 2 − I yz (λz ) 2 = 0

Substituting γt    γt  γt  [(18.91335 − 17.27274)  a 4  (λx )2 −  1.33333 a 4  (λ y ) 2 −  0.66667 a 4  (λz ) 2 = 0 g g g         γ t   γt  γt  − 1.33333 a 4  (λx ) 2 + (7.68953 − 17.27274)  a 4   (λ y )2 −  7.14159 a 4  (λz ) 2 = 0 g  g   g   

Simplifying 0.13913(λx )2 + (λ y ) 2 + 0.74522(λz ) 2 = 0

Adding and solving for (λz ) 2 (λz )2 = −5.58515(λx )2

and then

(λ y ) 2 = [1.23046 − (0.5)(−5.58515)](λx )2 = 4.02304 (λx ) 2

Now substitute into Eq. (9.57): (λx ) 22 + [4.02304(λx ) 2 ]2 + [−5.58515(λx ) 2 ]2 = 1

or

(λx ) 2 = 0.14377

and

(λ y ) 2 = 0.57839

(λz )2 = −0.80298 (θ x ) 2 = 81.7° (θ y ) 2 = 54.7° (θ z ) 2 = 143.4° 

K3 :

Begin with Eqs. (9.54a) and (9.54b): ( I x − K3 )(λx )3 − I xy (λ y )3 − I zx (λz )3 = 0 − I xy (λx )3 + ( I y − K 3 )(λ y )3 − I yz (λ y )3 = 0

Substituting γt    γt  γt  [(18.91335 − 19.08046)  a 4  (λx )3 − 1.33333 a 4  (λ y )3 −  0.66667 a 4  (λz )3 = 0 g  g  g       γ t   γt  γt  − 1.33333 a 4  (λx )3 + (7.68953 − 19.08046)  a 4   (λ y )3 −  7.14159 a 4  (λz )3 = 0 g  g  g     PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1687


1.23046(λx )2 − (λ y )2 − 0.5(λz ) 2 = 0


Simplifying −0.12533(λx )3 − (λ y )3 − 0.5(λz )3 = 0 0.11705(λx )3 + (λ y )3 + 0.62695(λz )3 = 0

Adding and solving for (λz )3 (λz )3 = 0.06522(λx )3

and then

(λ y )3 = [ −0.12533 − (0.5)(0.06522)](λx )3 = −0.15794(λx )3


or

(λx )3 = 0.98571

and

(λ y )3 = −0.15568

(i)

(λz )3 = 0.06429 (θ x )3 = 9.7° (θ y )3 = 99.0° (θ z )3 = 86.3° 

(c)


(θ x )3 = 170.3° (θ y )3 = 81.0° (θ z )3 = 93.7° 

Then



(λx )32 + [−0.15794(λx )3 ]2 + [0.06522(λx ]3 ]2 = 1

www.elsolucionario.net PROBLEM 9.184* For the component described in Problems 9.148 and 9.170, determine (a) the principal mass moments of inertia at the origin, (b) the principal axes of inertia at the origin. Sketch the body and show the orientation of the principal axes of inertia relative to the x, y, and z axes.

SOLUTION From the solutions to Problems 9.148 and 9.170. We have I x = 0.32258 kg ⋅ m 2

I y = I z = 0.41933 kg ⋅ m 2

I xy = I zx = 0.096768 kg ⋅ m 2

I yz = 0

Substituting into Eq. (9.56) and using Iy = Iz

I xy = I zx

I yz = 0

K 3 − [0.32258 + 2(0.41933)]K 2 + [2(0.32258)(0.41933) + (0.41933)2 − 2(0.096768)2 ]K − [(0.32258)(0.41933) 2 − 2(0.41933)(0.096768)2 ] = 0

Simplifying K 3 − 1.16124 K 2 + 0.42764 K − 0.048869 = 0

Solving yields

(b)

K1 = 0.22583 kg ⋅ m 2

or

K1 = 0.226 kg ⋅ m 2 

K 2 = 0.41920 kg ⋅ m 2

or

K 2 = 0.419 kg ⋅ m 2 

K3 = 0.51621 kg ⋅ m 2

or

K3 = 0.516 kg ⋅ m 2 

To determine the direction cosines λx , λ y , λz of each principal axis, use two of the equations of Eqs. (9.54) and (9.57). Then K1 : Begin with Eqs. (9.54b) and (9.54c): 0 − I xy (λx )1 + ( I y − K1 )(λ y )1 − I yz (λz )3 = 0 0 − I zx (λx )1 − I yz (λ y )1 + ( I z − K1 )(λz )3 = 0



(a)


Substituting −(0.096768)(λx )1 + (0.41933 − 0.22583)(λ y )1 = 0 −(0.096768)(λx )1 + (0.41933 − 0.22583)(λz )1 = 0

Simplifying yields (λ y )1 = (λz )1 = 0.50009(λx )1

Now substitute into Eq. (9.54): (λx ) 2 + 2[0.50009(λx )1 ]2 = 1

or

(λx )1 = 0.81645

and

(λ y )1 = (λz )1 = 0.40830

K2 :

Begin with Eqs. (9.54a) and (9.54b) ( I x − K 2 )(λx ) 2 − I xy (λ y ) 2 − I zx (λz ) 2 = 0 0 − I xy (λx ) 2 + ( I y − K 2 )(λ y ) 2 − I yz (λz ) 2 = 0

Substituting (0.32258 − 0.41920)(λx ) 2 − (0.096768)(λ y ) 2 − (0.096768)(λz ) 2 = 0

(i)

−(0.96768)(λx ) 2 + (0.41933 − 0.41920)(λ y ) 2 = 0

(ii)

Eq. (ii)  (λx ) 2 = 0 and then Eq. (i)  (λz )2 = −(λ y )2 Now substitute into Eq. (9.57): (λx ) 22

0 + (λ y ) 22 + [−(λ y ) 2 ]2 = 1 1

or

(λ y ) 2 =

and

(λz ) 2 = −

2 1 2

(θ x ) 2 = 90.0° (θ y ) 2 = 45.0° (θ z ) 2 = 135.0° 



(θ x )1 = 35.3° (θ y )1 = (θ z )1 = 65.9° 


K3 :

Begin with Eqs. (9.54b) and (9.54c):

0 I xy (λx )3 + ( I y − K3 )(λ y )3 + I yz (λz )3 = 0

0 − I zx (λx )3 − I yz (λ y )3 + ( I z − K3 )(λz )3 = 0 Substituting −(0.096768)(λx )3 + (0.41933 − 0.51621)(λ y )3 = 0 −(0.096768)(λx )3 + (0.41933 − 0.51621)(λz )3 = 0

Simplifying yields (λ y )3 = (λz )3 = − (λx )3

(λx )32 + 2[−(λx )3 ]2 = 1

(i)

1

or

(λx )3 =

and

(λ y )3 = (λz ) = −

3 1 3 (θ x )3 = 54.7° (θ y )3 = (θ z )3 = 125.3° 

(c)

Note: Principal axis 3 has been labeled so that the principal axes form a right-handed set. To obtain the direction cosines corresponding to the labeled axis, the negative root of Eq. (i) must be chosen; That is,

(λx )3 = −

1 3 (θ x )3 = 125.3°

Then

(θ y )3 = (θ z )3 = 54.7° 




www.elsolucionario.net PROBLEM 9.185 Determine by direct integration the moments of inertia of the shaded area with respect to the x and y axes.

SOLUTION At x = a, Then

y = a:

a=

k a

y=

a2 x

or k = a 2

Now

=

Then

1 a6 dx 3 x3



I x = dI x =



2a

a6 1  1 1 dx = a 6  − 3 3x 3  2 x 2  a

2a 1

a

1  1 1  = − a6  −  2 6  (2a) (a) 2 

Now

dI y = x 2

or

1 I x = a4  8

dA = x 2 ( y dx)

 a2  = x 2  dx  = a 2 x dx  x 

Then



I y = dI y =



2a a

2a

a2 1  a 2 x dx = a 2  x 2  = [(2a) 2 − (a) 2 ] 2  2 a or

Iy =

3 4 a  2



3

1 1  a2  dI x = y 3 dx =   dx 3 3 x 


SOLUTION

Then

Now

Then

y1 = y2 = b :

b a3 b y2 : b = ma or m = a y1 : b = ka3

or k =


At x = a,

b 3 x a3 b y2 = x a y1 =

b b  dA = ( y2 − y1 )dx =  x − 3 x3  dx a a  



A = dA = 2



a 0

b 1 3  x − 2 x  dx a a  a

1 1 b 1  = 2  x 2 − 2 x 4  = ab a 2 4a 0 2

Now

Then

 b b   dI y = x 2 dA = x 2  x − 3 x3  dx  a    a



I y = dI y = 2



a 0

b 2 1  x  x − 2 x3  dx a  a  a

b 1 1 1 6 x = 2  x4 − a 4 6 a 2  0

and

k y2 =

Iy A

=

1 3 ab 6 1 ab 2

1 = a2 3

or

or

Iy =

1 3 ab  6

ky =

a  3



SOLUTION At x = 2a,

y = 0:

0 = c sin k (2a) 2ak = π

y2 :

At x = a,

y = h:

At x = a,

y = 2h :

At x = 2a,

y = 0:

or k =

h = c sin

π 2a

π 2a

( a) or

c=h

2h = ma − b 0 = m(2a) + b 2h , b = 4h a

Solving yields

m=−

Then

y1 = h sin

Now

π   2h dA = ( y2 − y1 )dx =  (− x + 2a) − h sin x dx a 2 a  

Then



π 2a

A = dA =



x

2a a

2h x + 4h a 2h = ( − x + 2a ) a

y2 = −

π  2 h  (− x + 2a) − sin x dx a a  2  2a

2a π   1 cos x = h  − ( − x + 2a ) 2 + 2a  a π  a

 2 a  1  2  = h  −  + (− a + 2a ) 2  = ah 1 −   π  π  a  = 0.36338ah



y1 :


We have

I x and k x 3 3 1  1   2h π   1   dI x =  y2 − y1  dx =   (− x + 2a)  −  h sin x  dx 3  3   a 2a   3  

=

Then Now Then

h3  8 π  ( − x + 2a )3 − sin 3 x  3 3 a 2a 



I x = dI x =



2a a

h3 3

8  3 3 π  a3 (− x + 2a) − sin 2a x  dx  

sin 3 θ = sin θ (1 − cos 2 θ ) = sin θ − sin θ cos 2 θ 2a 

8 π π π   3 x − sin x cos 2 x  dx  3 (− x + 2a) −  sin 2a 2a 2a    a

Ix =

h3 3



=

h3 3

2a π 2a  2  4 3 π  − a3 (− x + 2a) + π cos 2a x − 3π cos 2a x   a

=

 h3  2a 2a  2 + + 3 ( − a + 2a ) 4   −  3  π 3π  a 

=

2 3 2  ah 1 −  3  3π 

a

2a

I x = 0.52520ah3

and

k x2 =

I x 0.52520ah3 = 0.36338ah A

or

I x = 0.525ah3 

or

k x = 1.202h 



Find:


SOLUTION First locate C of the area:

Then

A, mm 2

y , mm

yA, mm3

1

12 × 22 = 264

11

2904

2

1 (24)(18) = 216 2

28

6048

Σ

480


Symmetry implies X = 12 mm.

8952

Y ΣA = Σ yA: Y (480 mm 2 ) = 8952 mm3 Y = 18.65 mm

Now where

I x = ( I x )1 + ( I x ) 2 1 (12 mm)(22 mm)3 + (264 mm 2 )[(18.65 − 11) mm]2 12 = 26,098 mm 4

( I x )1 =

1 (24 mm)(18 mm)3 + (216 mm 2 )[(28 − 18.65) mm]2 36 = 22,771 mm 4

( I x )2 =

Then

I x = (26.098 + 22.771) × 103 mm 4

or

I x = 48.9 × 103 mm 4 



Also where

I y = ( I y )1 + ( I y ) 2

1 (22 mm)(12 mm)3 = 3168 mm 4 12 1  1  ( I y ) 2 = 2  (18 mm)(12 mm)3 +  × 18 mm × 12 mm  (4 mm)2  2   36  ( I y )1 =

= 5184 mm 4 [( I y )2 is obtained by dividing A2 into

I y = (3168 + 5184) mm 4

or I y = 8.35 × 103 mm 4 


Then

]



SOLUTION

Determination of centroid C of entire section. Area, mm 2

Section

π

1

4

(100) 2 = 7.854 × 103

y , mm

yA, mm3

42.44

333.3 × 103

2

(50)(100) = 5 × 103

50

250 × 103

3

(100)(50) = 5 × 103

–25

−125 × 103

Σ

17.854 × 103

458.3 × 103

Y ΣA = Σ yA: Y (17.854 × 103 mm 2 ) = 458.3 × 103 mm3 Y = 25.67 mm

Distance O to C: (a)

X = Y = 25.67 mm

OC = 2Y = 2(25.67) = 36.30 mm

π

Section 1:

JO =

Section 2:

JO = J + A(OD) 2 =

8

(100) 4 = 39.27 × 106 mm 4  50 2  100 2  1 (50)(100)[502 + 1002 ] + (50)(100)   +    12  2   2  

JO = 5.208 × 106 + 15.625 × 106 = 20.83 × 106 mm 4



Symmetry: X = Y

Dimensions in mm


Section 3: Same as Section 2;

J O = 20.83 × 106 mm 4

Entire section: J O = 39.27 × 106 + 2(20.83 × 106 ) JO = 80.9 × 106 mm 4 

= 80.94 × 106

(b)

Recall that,

OC = 36.30 mm and

A = 17.854 × 103 mm 2

J O = J C + A(OC ) 2 80.94 × 106 mm 4 = J C + (17.854 × 103 mm 2 )(36.30 mm) 2 J C = 57.4 × 106 mm 4 


J C = 57.41 × 106 mm 4


www.elsolucionario.net PROBLEM 9.190 Two L5 × 3 × 12 -in. angles are welded to a 12 -in. steel plate. Determine the distance b and the centroidal moments of inertia I x and I y of the combined section, knowing that I y = 4 I x .

SOLUTION A = 3.75in 2

Angle:

I x = 9.43 in 4 I y = 2.55 in 4

Area, in2

y , in.

yA, in 3

Angle

2(3.75) = 7.50

1.74

13.05

Plate

(10)(0.5) = 5

−0.25

−1.25

Σ

Then

12.50

11.80

Y Σ A = Σ yA: Y (12.50 in 2 ) = 11.80 in 3 Y = 0.944 in.

or Now where


First locate centroid C of the section.

I x = 2( I x )angle + ( I x )plate ( I x )angle = I x + Ad 2 = 9.43 in 4 + (3.75 in 2 )[(1.74 − 0.944) in.]2 = 11.8061 in 4

1 (10 in.)(0.5 in.)3 + (5 in 2 )[(0.25 + 0.944) in.]2 12 = 7.2323 in 4

( I x ) plate = I x + Ad 2 =

Then

I x = [2(11.8061) + 7.2323] in 4 = 30.8445 in 4

or

I x = 30.8 in 4 



We have

I y = 4 I x = 4(30.8445 in 4 ) = 123.378 in 4

or Now where

I y = 2( I y )angle + ( I y )plate ( I y )angle = I y + Ad 2 = 2.55 in 4 + (3.75 in 2 )[(b − 0.746) in.]2 ( I y ) plate =

Then

I y = 123.4 in 4 

1 (0.5 in.)(10 in.)3 = 41.6667 in 4 12

123.378 in 4 = 2[2.55 + 3.75(b − 0.746) 2 ] in 4 + 41.6667 in 4 b = 3.94 in. 


or


www.elsolucionario.net PROBLEM 9.191 Using the parallel-axis theorem, determine the product of inertia of the L5 × 3 × 12 -in. angle cross section shown with respect to the centroidal x and y axes.

SOLUTION


We have I xy = ( I xy )1 + ( I xy ) 2

For each rectangle: I xy = I x′y′ + x yA I x′y′ = 0 (symmetry)

and

I xy = Σ x yA

Thus

A, in 2

1

3×

2

4.5 ×

1 = 1.5 2 1 = 2.25 2

x , in.

y , in.

x yA, in 4

0.754

−1.49

−1.68519

− 0.496

1.01

−1.12716 − 2.81235

Σ

I xy = − 2.81 in 4 


www.elsolucionario.net PROBLEM 9.192 For the L5 × 3 × 12 -in. angle cross section shown, use Mohr’s circle to determine (a) the moments of inertia and the product of inertia with respect to new centroidal axes obtained by rotating the x and y axes 30° clockwise, (b) the orientation of the principal axes through the centroid and the corresponding values of the moments of inertia.

SOLUTION

I x = 9.43 in 4

I y = 2.55 in 4

From the solution to Problem 9.191 I xy = −2.81235 in 4

The Mohr’s circle is defined by the diameter XY, where X (9.43 − 2.81235),

Now

I ave =

Y (2.55, 2.81235)

1 1 ( I x + I y ) = (9.43 + 2.55) = 5.99 in 4 2 2 2

and

2

1  1  R =  ( I x − I y )  + I xy2 =  (9.43 − 2.55)  + (−2.81235) 2 2 2     = 4.4433 in 4


2 I xy Ix − Iy

2( −2.81235) 9.43 − 2.55 = 0.81754 =−

or and Then

2θ m = 39.267°

θ m = 19.6335° α = 180° − (39.267° + 60°) = 80.733°



From Figure 9.13a:


(a)

We have

I x′ = I ave − R cos α = 5.99 − 4.4433cos80.733°

or

I x′ = 5.27 in 4 

or

I y′ = 6.71 in 4 

or

I x′y′ = − 4.39 in 4 

I y′ = I ave + R cos α = 5.99 + 4.4433cos80.733°

I x′y′ = − R sin α = −4.4433sin 80.733°

(b)

First observe that the principal axes are obtained by rotating the xy axes through 19.63° counterclockwise 

Now

I max, min = I ave ± R = 5.99 ± 4.4433

or

I max = 10.43 in 4  I min = 1.547 in 4 




about C.

www.elsolucionario.net PROBLEM 9.193 A thin plate of mass m has the trapezoidal shape shown. Determine the mass moment of inertia of the plate with respect to (a) the x axis, (b) the y axis.

SOLUTION First note

mass = m = ρV = ρ tA

I mass = ρ tI area =

Also (a)

Now

m I area 3a 2

I x,area = ( I x )1,area + ( I x ) 2,area 1 1 = (2a)(a)3 + (2a )(a )3 3 12 5 4 = a 6

Then (b)

I x, mass =

m 5 4 × a 3a 2 6

or

I x , mass =

5 ma 2  18

We have I z ,area = ( I z )1,area + ( I z )2, area 2 1 1 1  1    =  (a )(2a)3  +  ( a)(2a)3 + (2a)(a )  2a + × 2a   = 10a 4 2 3 3   36   

m 10 × 10a 4 = ma 2 2 3 3a

Then

I x, mass =

Finally,

I y ,mass = I x ,mass + I z ,mass =

5 10 ma 2 + ma 2 18 3

=

65 2 ma 18

or

I y , mass = 3.61ma 2 



1   = ρ t (2a )( a) + (2a)(a)  = 3ρ ta 2 2  

www.elsolucionario.net PROBLEM 9.194 A thin plate of mass m has the trapezoidal shape shown. Determine the mass moment of inertia of the plate with respect to (a) the centroidal axis CC′ that is perpendicular to the plate, (b) the axis AA′ that is parallel to the x axis and is located at a distance 1.5a from the plate.

SOLUTION First locate the centroid C. 1   X Σ A = Σ xA: X (2a 2 + a 2 ) = a (2a 2 ) +  2a + × 2a  ( a 2 ) 3   14 a 9


X=

or

1  1  Z Σ A = Σ zA: Z (2a 2 + a 2 ) =  a  (2a 2 ) +  a  ( a 2 ) 2  3 

Z=

or (a)

We have

4 a 9

I y , mass = I CC ′, mass + m( X 2 + Z 2 )

From the solution to Problem 9.117: I y ,mass =

Then

I cc′,mass =

65 2 ma 18  14 2  4 2  65 2 ma − m  a  +  a   18  9   9  

or (b)

We have

I cc′ = 0.994ma 2 

I x, mass = I BB′,mass + m( Z ) 2

and

I AA′, mass = I BB′, mass + m(1.5a) 2

Then

2  4   I AA′, mass = I x, mass + m (1.5a) 2 −  a    9   



From the solution to Problem 9.193: I x, mass =

Then

I AA′, mass =

5 ma 2 18 2  5 4   ma 2 + m (1.5a) 2 −  a   18  9   

I AA′ = 2.33ma 2 


or



SOLUTION First compute the mass of each component. We have m = ρSTV = ρST tA m1 = (7850 kg/m3 )(0.002 m)(0.3 m) 2 = 1.413 kg m2 = m3 = (7850 kg/m3 )(0.002 m) × (0.15 × 0.12) m 2 = 0.2826 kg

Using Figure 9.28 and the parallel-axis theorem, we have I x = ( I x )1 + 2( I x )2 1  =  (1.413 kg)(0.3 m) 2  12  1  + 2  (0.2826 kg)(0.12 m) 2 + (0.2828 kg)(0.152 + 0.06 2 ) m 2  12  = [(0.0105975) + 2(0.0003391 + 0.0073759)] kg ⋅ m 2 = [(0.0105975) + 2(0.0077150)] kg ⋅ m 2

or

I x = 26.0 × 10−3 kg ⋅ m 2 

I y = ( I y )1 + 2( I y )2 1  =  (1.413 kg)(0.32 + 0.32 ) m 2  12   1  + 2  (0.2826 kg)(0.15 m) 2 + (0.2826 kg)(0.0752 + 0.152 ) m 2  12  = [(0.0211950) + 2(0.0005299 + 0.0079481)] kg ⋅ m 2 = [(0.0211950) + 2(0.0084780)] kg ⋅ m 2

or

I y = 38.2 × 10−3 kg ⋅ m 2 



Then


I z = ( I z )1 + 2( I z ) 2 1  =  (1.413 kg)(0.3 m) 2  12   1   + 2  (0.2826 kg)(0.152 + 0.122 ) m 2 + (0.2826 kg)(0.0752 + 0.062 ) m 2  12   = [(0.0105975) + 2(0.0008690 + 0.0026070)] kg ⋅ m 2 = [(0.0105975) + 2(0.0034760)] kg ⋅ m 2 I z = 17.55 × 10−3 kg ⋅ m 2 


or


www.elsolucionario.net PROBLEM 9.196 Determine the mass moment of inertia and the radius of gyration of the steel machine element shown with respect to the x axis. (The density of steel is 7850 kg/m3 .)

SOLUTION

m = ρSTV

Then

m1 = (7850 kg/m3 )(0.09 × 0.03 × 0.15) m3 = 3.17925 kg π  m2 = (7850 kg/m3 )  (0.045) 2 × 0.04  m3 2  = 0.998791 kg m3 = (7850 kg/m3 )[π (0.038) 2 × 0.03] m3 = 1.06834 kg

Using Figure 9.28 for components 1 and 3 and the equation derived above (before the solution to Problem 9.142) for a semicylinder, we have I x = ( I x )1 + ( I x )2 − ( I x )3 1  =  (3.17925 kg)(0.032 + 0.152 ) m 2 + (3.17925 kg)(0.075 m) 2  12     1 16 + (0.998791 kg)  − 2  4 9π 

1  2 2  (0.045 m) + 12 (0.04 m)   

2   4 × 0.045   2  2 + (0.998791 kg) (0.13) +  + 0.015   m   3π    

1  −  (1.06834 kg)[3(0.038 m) 2 + (0.03 m)2 ] + (1.06834 kg)(0.06 m)2  12 





= [(0.0061995 + 0.0178833) + (0.0002745 + 0.0180409) − (0.0004658 + 0.0038460)] kg ⋅ m 2 = (0.0240828 + 0.0183154 − 0.0043118) kg ⋅ m 2 = 0.0380864 kg ⋅ m 2

or Now

and

I x = 38.1 × 10−3 kg ⋅ m 2 

m = m1 + m2 − m3 = (3.17925 + 0.998791 − 1.06834) kg = 3.10970 kg k x2 =

I x 0.0380864 kg ⋅ m 2 = m 3.10970 kg k x = 110.7 mm 


or




CHAPTER 10





www.elsolucionario.net PROBLEM 10.1 Determine the vertical force P that must be applied at G to maintain the equilibrium of the linkage.


SOLUTION

Assume δ y A :

δ yC =

0.04 4 δ yA = y A 0.09 9

δ yG =

0.12 m 4  2 δ yD = 1.5  δ y A  = δ y A 0.08 m 9  3

δφ =

,

4 9

δ yD = δ yC = δ yA

δ yD

4 4 50 = δ y A /0.08 = δ yA = δ yA 0.08 9 0.72 9

Virtual Work:

δ U = 0:

(80 N)δ y A + (18 N ⋅ m)δφ + Pδ yG = 0  50  2  80δ y A + 18  δ y A  + P  δ y A  = 0 9 3     2 80 + 100 + P = 0 3 P = −270 N

P = 270 N 


www.elsolucionario.net PROBLEM 10.2 Determine the vertical force P that must be applied at G to maintain the equilibrium of the linkage.


SOLUTION

Assuming δ y A it follows

δ yC =

12 δ y A = 1.5δ y A 8

δ yE = δ yC = 1.5δ y A δ yD =

18 δ y A = 3(1.5δ y A ) = 4.5δ y A 6

δ yG =

10 10 δ y A = (1.5δ y A ) = 2.5δ y A 6 6

Then, by virtual work

δ U = 0: (300 lb)δ y A − (100 lb)δ yD + Pδ yG = 0 300δ y A − 100(4.5δ y A ) + P(2.5δ y A ) = 0 300 − 450 + 2.5 P = 0 P = 60.0 lb

P = 60.0 lb 


www.elsolucionario.net PROBLEM 10.3 Determine the couple M that must be applied to member DEFG to maintain the equilibrium of the linkage.


SOLUTION

Assume δ y A :

δ yC = δφ =

0.04 4 δ yA = y A 0.09 9

δ yC 0.08

,

4 9

δ yD = δ yC = δ yA

4 4 50 = δ y A /0.08 = δ yA = δ yA 9 0.72 9

Virtual Work:

δ U = 0:

(80 N)δ y A + (18 N ⋅ m)δφ + M δφ = 0  50   50  80δ y A + 18  δ y A  + M  δ y A  = 0  9   9  50 80 + 100 + M =0 9

M = −32.4 N ⋅ m

M = 32.4 N ⋅ m




www.elsolucionario.net PROBLEM 10.4 Determine the couple M that must be applied to member DEFG to maintain the equilibrium of the linkage.

Assume δ y A :

δ yC =

12 δ y A = 1.5δ y A , 8

δ yD =

18 δ yE = 3(1.5δ y A ) = 4.5δ y A 6

δφ =

δ yE 6

=


SOLUTION

δ yE = δ yC = 1.5δ y A

1.5δ y A 1 = δ yA 6 4

Virtual Work:

δ U = 0:

(300 lb)δ y A − (100 lb)δ yD + M δφ = 0 1  300δ y A − 100(1.5δ y A ) + M  δ y A  = 0 4  1 300 − 450 + M = 0 4 M = + 600 lb ⋅ in.

M = 600 lb ⋅ in.




www.elsolucionario.net PROBLEM 10.5 Determine the force P required to maintain the equilibrium of the linkage shown. All members are of the same length and the wheels at A and B roll freely on the horizontal rod.

SOLUTION

yD = 2 yC

δ yD = 2δ yC

yF = 3 yC

δ yF = 3δ yC

yG = yH = 4 yC

δ yG = δ yH = 4δ yC Virtual Work:

δ U = (400 N)δ yC + (100 N)δ yD − Pδ yF + (75 N)δ yG + (150 N)δ yH = 0 400δ yC + 100(2δ yC ) − P(3δ yC ) + (75 + 150)(4δ yC ) = 0 3P = 400 + 200 + 900

P = +500 N

P = 500 N 



Using yC as independent variable:

www.elsolucionario.net PROBLEM 10.6 Solve Problem 10.5 assuming that the vertical force P is applied at Point E. PROBLEM 10.5 Determine the force P required to maintain the equilibrium of the linkage shown. All members are of the same length and the wheels at A and B roll freely on the horizontal rod.

SOLUTION

yD = yE = 2 yC

δ yD = δ yE = 2δ yC yG = yH = 4 yC

δ yG = δ y A = 4δ yC

Virtual Work:

δ U = (400 N)δ yC + (100 N)δ yD − Pδ yE + (75 N)δ yG + (150 N)δ yH = 0 400δ yC + 100(2δ yC ) − P(2δ yC ) + (75 + 150)(4δ yC ) = 0 2 P = 400 + 200 + 900

P = +750 N

P = 750 N 



Using yC as independent variable:

www.elsolucionario.net PROBLEM 10.7 The two-bar linkage shown is supported by a pin and bracket at B and a collar at D that slides freely on a vertical rod. Determine the force P required to maintain the equilibrium of the linkage.


SOLUTION

Assume δ y A :

δ yC =

8 in. 1 δ y A ; δ yC = δ y A 16 in. 2

Since bar CD move in translation

δ yE = δ yF = δ yC or Virtual Work:

δ yE = δ y F =

1 yA 2

δ U = 0: − Pδ y A + (100 lb)δ yE + (150 lb)δ yF = 0 1  1  − Pδ y A + 100  δ y A  + 150  δ y A  = 0 2  2  P = 125 lb P = 125.0 lb 



SOLUTION From sketch

δ y A = 4δ yC

Thus (a)

y A = 4 yC

Virtual Work:

δ U = 0: Pδ y A − Fδ yC = 0 P(4δ yC ) − F δ yC = 0 P= F = 60 lb: P =

1 F 4

1 (60 lb) = 15 lb 4 P = 15.00 lb 

(b)

Free body: Corkscrew ΣFy = 0 R + P − F = 0; R + 15 lb − 60 lb = 0

On corkscrew:

R = 45 lb

,

On bottle:

R = 45.0 lb 



Knowing that the maximum friction force exerted by the bottle on the cork is 60 lb, determine (a) the force P that must be applied to the corkscrew to open the bottle, (b) the maximum force exerted by the base of the corkscrew on the top of the bottle.

www.elsolucionario.net PROBLEM 10.9 Rod AD is acted upon by a vertical force P at end A, and by two equal and opposite horizontal forces of magnitude Q at points B and C. Derive an expression for the magnitude Q of the horizontal forces required for equilibrium.

SOLUTION xC = a sin θ

δ xC = a cos θδθ xB = 2a sin θ δ xB = 2a cos θδθ y A = 3a cos θ δ y A = −3a sin θδθ Virtual Work: We note that P tends to increase yA and −Q tends to increase xB, while Q tends to decrease xC. Therefore

δ U = Pδ y A + Qδ xB − Qδ xC = 0 = P( −3a sin θδθ ) + Q (2a cos θδθ ) − Q(a cos θδθ ) = 0 Q cos θ = 3P sin θ ′ Q = 3P tan θ 



We have

www.elsolucionario.net PROBLEM 10.10 The slender rod AB is attached to a collar A and rests on a small wheel at C. Neglecting the radius of the wheel and the effect of friction, derive an expression for the magnitude of the force Q required to maintain the equilibrium of the rod.

SOLUTION


For Δ AA′C : A′C = a tan θ y A = −( A′C ) = − a tan θ

δ yA = −

a δθ cos 2 θ

For ΔCC ′B : BC ′ = l sin θ − A′C = l sin θ − a tan θ yB = BC ′ = l sin θ − a tan θ

δ yB = l cos θδθ − Virtual Work:

a δθ cos 2 θ

δ U = 0: Qδ y A − Pδ yB = 0 a  a   δθ − P  l cos θ − − Q− 2  cos 2 θ  cos θ   a   a   Q = P  l cos θ −  2  cos 2 θ   cos θ  

  δθ = 0 

l  Q = P  cos3 θ − 1  a 


www.elsolucionario.net PROBLEM 10.11 The slender rod AB is attached to a collar A and rests on a small wheel at C. Neglecting the radius of the wheel and the effect of friction, derive an expression for the magnitude of the force Q required to maintain the equilibrium of the rod.

SOLUTION For Δ AA′C :

δ yA = −


A′C = a tan θ y A = −( A′C ) = − a tan θ a δθ cos 2 θ

For Δ BB′C : B′C = l sin θ − A′C = l sin θ − a tan θ B′C l sin θ − a tan θ = BB′ = tan θ tan θ xB = BB′ = l cos θ − a

δ xB = −l sin θδθ Virtual Work:

δ U = 0: Pδ xB − Qδ y A = 0 a   P(−l sin θδθ ) − Q  − δθ  = 0 2 cos θ  

or

Pl sin θ cos 2 θ = Qa

l Q = P sin θ cos 2 θ  a


www.elsolucionario.net PROBLEM 10.12 Knowing that the line of action of the force Q passes through Point C, derive an expression for the magnitude of Q required to maintain equilibrium.

We have

y A = 2l cos θ ; δ y A = −2l sin θδθ

θ

θ

CD = 2l sin ; δ (CD ) = l cos δθ 2 2

Virtual Work:

δ U = 0: − Pδ y A − Qδ (CD) = 0 θ   − P (−2l sin θδθ ) − Q  l cos δθ  = 0 2  

Q = 2P

sin θ  cos(θ /2)



SOLUTION

www.elsolucionario.net PROBLEM 10.13 Solve Problem 10.12 assuming that the force P applied at Point A acts horizontally to the left. PROBLEM 10.12 Knowing that the line of action of the force Q passes through Point C, derive an expression for the magnitude of Q required to maintain equilibrium.

We have

x A = 2l cos θ ; δ x A = 2l cos θδθ

θ

θ

CD = 2l sin ; δ (CD ) = l cos δθ 2 2

Virtual Work:

δ U = 0: Pδ x A − Qδ (CD) = 0 θ   P (2l cos θδθ ) − Q  l cos δθ  = 0 2  

Q = 2P

cos θ  cos(θ /2)



SOLUTION

www.elsolucionario.net PROBLEM 10.14 The mechanism shown is acted upon by the force P; derive an expression for the magnitude of the force Q required to maintain equilibrium.

SOLUTION Virtual Work: x A = 2l sin θ

δ x A = 2l cos θδθ and

yF = 3l cos θ

δ yF = −3l sin θδθ Virtual Work:

δ U = 0: Qδ x A + Pδ yF = 0 Q(2l cos θδθ ) + P(−3l sin θδθ ) = 0

Q=

3 P tan θ  2



We have

www.elsolucionario.net PROBLEM 10.15 Derive an expression for the magnitude of the couple M required to maintain the equilibrium of the linkage shown.

We have

xB = l cos θ

δ xB = −l sin θδθ

(1)

yC = l sin θ

δ yC = l cos θδθ

Now

1 2

δ xB = lδθ

Substituting from Equation (1) −l sin θδθ =

or Virtual Work:

1 lδφ 2

δφ = −2sin θδθ δ U = 0:

M δϕ + Pδ yC = 0

M (−2sin θδθ ) + P(l cos θδθ ) = 0 M=

or

1 cos θ Pl 2 sin θ

M=

Pl  2 tan θ



SOLUTION


SOLUTION We have

δ xB = l cos θδθ y A = l cos θ δ y A = −l sin θδθ Virtual Work:

δ U = 0: M δθ − Pδ xB + Pδ y A = 0 M δθ − P(l cos θδθ ) + P(−l sin θδθ ) = 0 M = Pl (sin θ + cos θ ) 



xB = l sin θ

www.elsolucionario.net PROBLEM 10.17 A uniform rod AB of length l and weight W is suspended from two cords AC and BC of equal length. Derive an expression for the magnitude of the couple M required to maintain equilibrium of the rod in the position shown.

yG = h cos θ =

1 l tan α cos θ 2

1 2

δ yG = − l tan α sin θδθ Virtual Work:

δ U = Wδ yG + M δθ = 0  1  W  − l tan α sin θδθ  + M δθ = 0  2  1 M = Wl tan α sin θ 2





SOLUTION

www.elsolucionario.net PROBLEM 10.18 Collar B can slide along rod AC and is attached by a pin to a block that can slide in the vertical slot shown. Derive an expression for the magnitude of the couple M required to maintain equilibrium.

SOLUTION R tan(90° − θ ) − Rδθ δ yB = cos 2 (90° − θ ) − Rδθ δ yB = 2 sin θ

Virtual Work:

δ U = 0:

δ U = − M δθ − Pδ yB = 0 − M δθ + PR

1 δθ = 0 sin 2 θ M=

PR sin 2 θ

M = PR csc2 θ 



yB =

www.elsolucionario.net PROBLEM 10.19 For the linkage shown, determine the couple M required for equilibrium when l = 1.8 ft, Q = 40 lb, and θ = 65°.

δC =

1 2

lδφ

cos θ

Virtual Work:

δU = 0 :

M δφ − Qδ C = 0

1 l  M δφ − Q   δφ = 0  2 cos θ 

M=

Data:

M=

1 Ql 2 cos θ

1 (40 lb)(1.8 ft) = 85.18 lb ⋅ ft 2 cos 65°

M = 85.2 lb ⋅ ft





SOLUTION

www.elsolucionario.net PROBLEM 10.20 For the linkage shown, determine the force Q required for equilibrium when l = 18 in., M = 600 lb ⋅ in., and θ = 70°.

δC =

1 lδφ 2 cos θ

Virtual Work:

δ U = 0:

M δφ − Qδ C = 0

1 l M δφ − Q   2 cos θ

  δφ = 0  Q=

Data:

Q=

2 M cos θ l

2(600 lb ⋅ in.)cos70° = 22.801 lb 18 in.

Q = 22.8 lb

70.0° 



SOLUTION

www.elsolucionario.net PROBLEM 10.21 A 4-kN force P is applied as shown to the piston of the engine system. Knowing that AB = 50 mm and BC = 200 mm, determine the couple M required to maintain the equilibrium of the system when (a) θ = 30°, (b) θ = 150°.

SOLUTION

Law of sines sin φ sin θ = AB BC sin φ =

AB sin θ BC

(1)

Now xC = AB cos θ + BC cos φ

δ xC = − AB sin θδθ − BC sin φδφ Now, from Equation (1)

cos φδφ =

δφ =

or

(2)

AB cos θδθ BC AB cos θ δθ BC cos φ

(3)

From Equation (2)  AB cos θ  δθ   BC cos φ 

δ xC = − AB sin θδθ − BC sin φ  or Then

δ xC = − δ xC = −

AB (sin θ cos φ + sin φ cos θ )δθ cos φ

AB sin(θ + φ ) δθ cos φ



Analysis of the geometry:


Virtual Work:

δ U = 0: − Pδ xC − M δθ = 0  AB sin(θ + φ )  δθ  − M δθ = 0 − P − cos φ   M = AB

Thus,

sin(θ + φ ) P cos φ

(4)

P = 4 kN, θ = 30°

Eq. (1):

Eq. (4):

sin φ =

50 mm sin 30° 200 mm

M = (0.05 m)

φ = 7.181°

sin (30°+7.181°) (4 kN) cos 7.818°

M = 121.8 N ⋅ m



M = 78.2 N ⋅ m



P = 4 kN, θ = 150°

(b) Eq. (1): Eq. (4):

sin φ =

50 mm sin160° 200 mm

M = (0.05 m)

φ = 7.181°

sin (150°+7.181°) (4 kN) cos 7.181°



(a)

www.elsolucionario.net PROBLEM 10.22 A couple M of magnitude 100 N ⋅ m is applied as shown to the crank of the engine system. Knowing that AB = 50 mm and BC = 200 mm, determine the force P required to maintain the equilibrium of the system when (a) θ = 60°, (b) θ = 120°.

SOLUTION

Law of sines sin φ sin θ = AB BC sin φ =

AB sin θ BC

(1)

Now xC = AB cos θ + BC cos φ

δ xC = − AB sin θδθ − BC sin φδφ Now, from Equation (1)

cos φδφ =

δφ =

or

(2)

AB cos θδθ BC AB cos θ δθ BC cos φ

(3)

From Equation (2)  AB cos θ  δ xC = − AB sin θδθ − BC sin φ  δθ   BC cos φ 

or Then

δ xC = − δ xC = −

AB (sin θ cos φ + sin φ cos θ )δθ cos φ

AB sin(θ + φ ) δθ cos φ



Analysis of the geometry:


Virtual Work:

δ U = 0: − Pδ xC − M δθ = 0  AB sin(θ + φ )  − P − δθ  − M δθ = 0 cos φ   M = AB

Thus,

sin(θ + φ ) P cos φ

(4)

M = 100 N ⋅ m, θ = 60°

Eq. (1): Eq. (4):

sin φ =

50 mm sin 60° 200 mm

100 N ⋅ m = (0.05 m)

φ = 12.504°

sin (60° + 12.504°) P cos 12.504°

P = 2047 N

P = 2.05 kN



P = 2.65 kN



M = 100 N ⋅ m, θ = 120°

(b) Eq. (1): Eq. (4):

sin φ =

50 mm sin120° φ = 12.504° 200 mm

100 N ⋅ m = (0.05 m)

sin (120° + 12.504°) P cos12.504°

P = 2649 N



(a)

www.elsolucionario.net PROBLEM 10.23 A slender rod of length l is attached to a collar at B and rests on a portion of a circular cylinder of radius r. Neglecting the effect of friction, determine the value of θ corresponding to the equilibrium position of the mechanism when l = 200 mm, r = 60 mm, P = 40 N, and Q = 80 N.

SOLUTION Geometry

OC = r OC r = cos θ = OB xB

r cos θ r sin θ δ xB = δθ cos 2 θ y A = l cos θ

δ y A = −l sin θδθ Virtual Work:

δ U = 0: P( −δ y A ) − Qδ xB = 0 Pl sin θδθ − Q

r sin θ δθ = 0 cos 2 θ cos 2 θ =

Qr Pl

(1)

Then, with l = 200 mm, r = 60 mm, P = 40 N, and Q = 80 N cos 2 θ =

or

(80 N)(60 mm) = 0.6 (40 N)(200 mm)

θ = 39.231°

θ = 39.2° 



xB =

www.elsolucionario.net PROBLEM 10.24 A slender rod of length l is attached to a collar at B and rests on a portion of a circular cylinder of radius r. Neglecting the effect of friction, determine the value of θ corresponding to the equilibrium position of the mechanism when l = 14 in., r = 5 in., P = 75 lb, and Q = 150 lb.

SOLUTION Geometry

OC = r OC r = cos θ = OB xB r cos θ r sin θ δ xB = δθ cos 2 θ y A = l cos θ ; δ y A = −l sin θδθ

Virtual Work:

δ U = 0: P( −δ y A ) − Qδ xB = 0 Pl sin θδθ − Q

r sin θ δθ = 0 cos 2 θ cos 2 θ =

Qr Pl

(1)

Then, with l = 14 in., r = 5 in., P = 75 lb, and Q = 150 lb Eq. (1) becomes:

cos 2 θ =

(150 lb)(5 in.) = 0.71429 (75 lb)(14 in.)

θ = 32.3° 



xB =

www.elsolucionario.net PROBLEM 10.25 Determine the value of θ corresponding to the equilibrium position of the rod of Problem 10.10 when l = 30 in., a = 5 in., P = 25 lb, and Q = 40 lb. PROBLEM 10.10 The slender rod AB is attached to a collar A and rests on a small wheel at C. Neglecting the radius of the wheel and the effect of friction, derive an expression for the magnitude of the force Q required to maintain the equilibrium of the rod.

For Δ AA′C :

A′C = a tan θ y A = −( A′C ) = − a tan θ

δ yA = −

a δθ cos 2 θ

For ΔCC ′B : BC ′ = l sin θ − A′C = l sin θ − a tan θ yB = BC ′ = l sin θ − a tan θ

δ yB = l cos θδθ −

a δθ cos 2 θ

Virtual Work:

δ U = 0: − Qδ y A − Pδ yB = 0 a  a    −Q − δθ − P  l cos θ −  δθ = 0 2  cos 2 θ   cos θ    a Q 2  cos θ

or

a     = P  l cos θ −  cos 2 θ   

l  Q = P  cos3 θ − 1 a 



SOLUTION


with

l = 30 in., a = 5 in., P = 25 lb, and Q = 40 lb  30 in. 3  (40 lb) = (25 lb)  cos θ − 1 5 in.  

cos3 θ = 0.4333

θ = 40.8° 


or


www.elsolucionario.net PROBLEM 10.26 Determine the values of θ corresponding to the equilibrium position of the rod of Problem 10.11 when l = 600 mm, a = 100 mm, P = 50 N, and Q = 90 N. PROBLEM 10.11 The slender rod AB is attached to a collar A and rests on a small wheel at C. Neglecting the radius of the wheel and the effect of friction, derive an expression for the magnitude of the force Q required to maintain the equilibrium of the rod.

SOLUTION For Δ AA′C :

A′C = a tan θ

δ yA = −


y A = −( A′C ) = − a tan θ a δθ cos 2 θ

For ΔBB′C : B′C = l sin θ − A′C = l sin θ − a tan θ BB′ =

B′C l sin θ − a tan θ = tan θ tan θ

xB = BB′ = l cos θ − a

δ xB = −l sin θδθ Virtual Work:

δ U = 0: Pδ xB − Qδ y A = 0 a   P(−l sin θδθ ) − Q  − δθ  = 0 2  cos θ  Pl sin θ cos 2 θ = Qa

or

l Q = P sin θ cos 2 θ a



with

l = 600 mm, a = 100 mm, P = 50 N, and Q = 90 N 90 N = (50 N)

or

600 mm sin θ cos 2 θ 100 mm

sin θ cos 2 θ = 0.300

θ = 19.81° and 51.9° 


Solving numerically


www.elsolucionario.net PROBLEM 10.27 Determine the value of θ corresponding to the equilibrium position of the mechanism of Problem 10.12 when P = 80 N and Q = 100 N. PROBLEM 10.12 Knowing that the line of action of the force Q passes through Point C, derive an expression for the magnitude of Q required to maintain equilibrium.

SOLUTION From geometry y A = 2l cos θ , δ y A = −2l sin θδθ

θ

θ


CD = 2l sin , δ (CD ) = l cos δθ 2 2

Virtual Work:

δ U = 0:

− Pδ y A − Qδ (CD) = 0

θ   − P(−2l sin θδθ ) − Q  l cos δθ  = 0 2   Q = 2P

or

sin θ cos ( θ2 )

P = 80 N, Q = 100 N

with

(100 N) = 2(80 N)

sinθ cos ( θ2 )

sin θ = 0.625 cos ( θ2 )

or

2sin ( θ2 ) cos ( θ2 ) cos ( θ2 )

= 0.625

θ = 36.42°

θ = 36.4° 

(Additional solutions discarded as not applicable are θ = ±180°)


www.elsolucionario.net PROBLEM 10.28 Determine the value of θ corresponding to the equilibrium position of the mechanism of Prob. 10.14 when P = 270 N and Q = 960 N.

SOLUTION Virtual Work: yF = 3l cos θ , δ yF = −3l sin θδθ

δ U = 0: δ U = 0:

Qδ x A + Pδ yF = 0 Qδθ x A + Pδ yF = 0

Q(2l cos θδθ ) + P(−3l sin θδθ ) = 0 Q=

Data:

3 P tan θ 2

P = 270 N, Q = 960 N (960 N) =

3 (270 N) tan θ 2

θ = 67.1° 



x A = 2l sin θ , δ x A = 2l cos θδθ

www.elsolucionario.net PROBLEM 10.29 A load W of magnitude 600 N is applied to the linkage at B. The constant of the spring is k = 2.5 kN/m, and the spring is unstretched when AB and BC are horizontal. Neglecting the weight of the linkage and knowing that l = 300 mm, determine the value of θ corresponding to equilibrium.

SOLUTION

δ xC = −2l sin θδθ yB = l sin θ δ yB = l cos θδθ


xC = 2l cos θ

F = ks = k (2l − xC ) = 2kl (1 − cos θ )

Virtual Work:

δ U = 0: F δ xC + W δ yB = 0 2kl (1 − cos θ )( −2l sin θδθ ) + W (l cos θδθ ) = 0 4kl 2 (1 − cos θ )sin θ = Wl cos θ

or Given:

(1 − cos θ ) tan θ =

l = 0.3 m, W = 600 N, k = 2500 N/m 600 N 4(2500 N/m)(0.3 m)

Then


or

(1 − cos θ ) tan θ = 0.2

Solving numerically

W 4kl

θ = 40.22°

θ = 40.2° 


www.elsolucionario.net PROBLEM 10.30 A vertical load W is applied to the linkage at B. The constant of the spring is k, and the spring is unstretched when AB and BC are horizontal. Neglecting the weight of the linkage, derive an equation in θ, W, l, and k that must be satisfied when the linkage is in equilibrium.

xC = 2l cos θ

δ xC = −2l sin θδθ yB = l sin θ δ yB = l cos θδθ F = ks = k (2l − xC ) = 2kl (1 − cos θ )

Virtual Work:

δ U = 0: F δ xC + W δ yB = 0 2kl (1 − cos θ )( −2l sin θδθ ) + W (l cos θδθ ) = 0 4kl 2 (1 − cos θ )sin θ = Wl cos θ

or


W 4kl (1 − cos θ ) tan θ =

From above

W  4kl



SOLUTION

www.elsolucionario.net PROBLEM 10.31 Two bars AD and DG are connected by a pin at D and by a spring AG. Knowing that the spring is 12 in. long when unstretched and that the constant of the spring is 125 lb/in., determine the value of x corresponding to equilibrium when a 900-lb load is applied at E as shown.

x x x 1 + = δ yE = δ x 3 6 2 2 F = ks = (125 lb/in.)( x − 12 in.)

yE =

Virtual Work:

δ U = 0: Fδ x + W δ yE = 0 1  −(125)( x − 12)δ x + (900)  δ x  = 0 2  −125 x + 1500 + 450 = 0

x = 15.60 in. 



SOLUTION

www.elsolucionario.net PROBLEM 10.32 Solve Problem 10.31 assuming that the 900-lb vertical force is applied at C instead of E. PROBLEM 10.31 Two bars AD and DG are connected by a pin at D and by a spring AG. Knowing that the spring is 12 in. long when unstretched and that the constant of the spring is 125 lb/in., determine the value of x corresponding to equilibrium when a 900-lb load is applied at E as shown.

1 1 x δ yC = δ x 6 6 F = ks = (125 lb/in.)( x − 12 in.)

yC =

Virtual Work:

1 

 

δ U = − Fδ x + W δ yC = 125( x − 12)δ x + 900  δ x  = 0 6 −125 x + 1500 + 150 = 0

x = 13.20 in. 



SOLUTION

www.elsolucionario.net PROBLEM 10.33 Two 5-kg bars AB and BC are connected by a pin at B and by a spring DE. Knowing that the spring is 150 mm long when unstretched and that the constant of the spring is 1 kN/m, determine the value of x corresponding to equilibrium.

SOLUTION

During the virtual displacement, Points D and E move apart a distance δ ( DE ) = 23 δ x and the total work done

by the forces exerted at D and E is − F ( 23 δ x ) 2



1



3



δ U = 0: − F  δ x  + 49.05 N  δ x  + 49.05 N  δ x  = 0 3  4  4  F = 73.575 N

For F = 73.575 N, elongation of spring is F 73.575 N = = 73.575 × 10−3 m k 1000 N/m = 73.575 mm

Since undeformed length of spring is 150 mm, total length is DE =

2 x = 150 mm + 73.575 mm 3

x = 355 mm 



Wbar = (5 kg)(9.81 m/s 2 ) = 49.05 N

First note:



Rod ABC is attached to blocks A and B that can move freely in the guides shown. The constant of the spring attached at A is k = 3 kN/m, and the spring is unstretched when the rod is vertical. For the loading shown, determine the value of θ corresponding to equilibrium.

SOLUTION xC = (0.4 m)sin θ

δ xC = 0.4cos θδθ y A = (0.2 m) cos θ

δ y A = −0.2sin θδθ Spring: Unstretched length = 0.2 m F = k (0.2 m − y A ) = k (0.2 − 0.2 cos θ ) = (300 N/m)(0.2)(1 − cos θ ) F = 600(1 − cos θ )

Virtual Work:

δ U = 0: (150 N)δ xC + Fδ y A = 0 150(0.4cos θδθ ) + 600(1 − cos θ )(−0.2sin θδθ ) = 0 150(0.4) 1 1 = : = (1 − cos θ ) tan θ 600(0.2) 2 2

θ = 57.2° 



www.elsolucionario.net PROBLEM 10.35 A vertical force P of magnitude 150 N is applied to end E of cable CDE, which passes over a small pulley D and is attached to the mechanism at C. The constant of the spring is k = 4 kN/m, and the spring is unstretched when θ = 0. Neglecting the weight of the mechanism and the radius of the pulley, determine the value of θ corresponding to equilibrium.

SOLUTION

< CBD = 90° − θ


l = BC = 0.2 m r = 0.1 m

θ  v = 2l sin  45° −  2  θ  δ v = −l cos  45° −  δθ 2   s = rθ δ s = rδθ F = ks = krθ

Virtual Work:

δ U = 0: − Pδ v − Fδ s = 0  θ   − P  −l cos  45° −   δθ − krθ (rδθ ) = 0 2    Pl θ = 2 cos ( 45° − θ2 ) kr Pl (150 N)(0.2 m) = = 0.75 2 kr (4000 N/m)(0.1 m) 2 0.75 =

θ

cos ( 45° − θ2 )

θ = 0.67623 rad = 38.745°

θ = 38.7° 


www.elsolucionario.net PROBLEM 10.36 A horizontal force P of magnitude 40 lb is applied to the mechanism at C. The constant of the spring is k = 9 lb/in., and the spring is unstretched when θ = 0. Neglecting the weight of the mechanism, determine the value of θ corresponding to equilibrium.

SOLUTION


s = rθ δ s = rδθ

Spring is unstretched at θ = 0° FSP = ks = krθ xC = l sin θ

δ xC = l cos θδθ Virtual Work:

δ U = 0: Pδ xC − FSPδ s = 0 P(l cos θδθ ) − krθ (rδθ ) = 0 Pl θ = 2 cos θ kr

or Thus

or

(40 lb)(12 in.) θ = (9 lb/in.)(5 in.) 2 cos θ

θ cos θ

= 2.1333

θ = 1.054 rad = 60.39°

θ = 60.4° 


www.elsolucionario.net PROBLEM 10.37 Knowing that the constant of spring CD is k and that the spring is unstretched when rod ABC is horizontal, determine the value of θ corresponding to equilibrium for the data indicated. P = 300 N, l = 400 mm, k = 5 kN/m.

SOLUTION y A = l sin θ

Spring:

v = CD

Unstretched when

θ =0

so that

v0 = 2l

For θ :

 90° + θ  v = 2l sin    2   

δ v = l cos  45° +

θ

 δθ 2

Stretched length:

θ  s = v − v0 = 2l sin  45° +  − 2l 2 

Then

  θ  F = ks = kl  2sin  45° +  − 2  2   

Virtual Work:

δ U = 0: Pδ y A − F δ v = 0   θ θ   Pl cos θδθ − kl  2sin  45° +  − 2  l cos  45° +  δθ = 0 2 2    

or

P 1 = kl cos θ =

1 cos θ

 θ  θ θ     2sin  45° +  cos  45° +  − 2 cos  45° +   2 2 2       θ  θ θ     2sin  45° +  cos  45° +  cos θ − 2 cos  45° +   2 2 2     

=1− 2

cos ( 45° θ2 ) cos θ



δ y A = l cos θδθ


Now, with

P = 300 N, l = 400 mm, and k = 5 kN/m cos ( 45° + (300 N) =1− 2 (5000 N/m)(0.4 m) cos θ

or

cos ( 45° + θ2 ) cos θ

θ

2

)

= 0.60104

θ = 22.6° 


Solving numerically


www.elsolucionario.net PROBLEM 10.38 Knowing that the constant of spring CD is k and that the spring is unstretched when rod ABC is horizontal, determine the value of θ corresponding to equilibrium for the data indicated. P = 75 lb, l = 15 in., k = 20 lb/in.

SOLUTION From the analysis of Problem 10.37, we have

with

θ

2

)

P = 75 lb, l = 15 in. and k = 20 lb/in. cos ( 45° + θ2 ) (75 lb) =1− 2 (20 lb/in.)(15 in.) cos θ

or

cos ( 45° + θ2 ) cos θ

= 0.53033

θ = 51.1° 

Solving numerically



cos ( 45° + P =1− 2 kl cos θ

www.elsolucionario.net PROBLEM 10.39 The lever AB is attached to the horizontal shaft BC that passes through a bearing and is welded to a fixed support at C. The torsional spring constant of the shaft BC is K; that is, a couple of magnitude K is required to rotate end B through 1 rad. Knowing that the shaft is untwisted when AB is horizontal, determine the value of θ corresponding to the position of equilibrium when P = 100 N, l = 250 mm, and K = 12.5 N · m/rad.

SOLUTION y A = l sin θ

We have

δ y A = l cos θδθ Virtual Work:

Pl cos θδθ − Kθδθ = 0

or with

θ cos θ

Pl K

(1)

P = 100 N, l = 250 mm and K = 12.5 N ⋅ m/rad

θ cos θ

or

=

θ cos θ

=

(100 N)(0.250 m) 12.5 N ⋅ m/rad

= 2.0000

θ = 59.0° 

Solving numerically



δ U = 0: Pδ y A − M δθ = 0

www.elsolucionario.net PROBLEM 10.40 Solve Problem 10.39 assuming that P = 350 N, l = 250 mm, and K = 12.5 N · m/rad. Obtain answers in each of the following quadrants: 0 < θ < 90°, 270° < θ < 360°, 360° < θ < 450°. PROBLEM 10.39 The lever AB is attached to the horizontal shaft BC that passes through a bearing and is welded to a fixed support at C. The torsional spring constant of the shaft BC is K; that is, a couple of magnitude K is required to rotate end B through 1 rad. Knowing that the shaft is untwisted when AB is horizontal, determine the value of θ corresponding to the position of equilibrium when P = 100 N, l = 250 mm, and K = 12.5 N · m/rad.

SOLUTION

P = 350 N, l = 250 mm and K = 12.5 N ⋅ m/rad

We have or

θ cos θ

θ cos θ

=

(350 N)(0.250 m) 12.5 N ⋅ m/rad

= 7 or θ = 7 cos θ

(1)

The solutions to this equation can be shown graphically using any appropriate graphing tool, such as Maple, with the command: plot ({theta, 7 * cos(theta)}, t = 0.5 * Pi/2); Thus, we plot y = θ and y = 7 cos θ in the range 0 ≤θ ≤

5π 2



Using Equation (1) of Problem 10.39 and


We observe that there are three points of intersection, which implies that Equation (1) has three roots in the specified range of θ . π  0 ≤ θ ≤ 90°   ; 2

θ = 1.37333 rad, θ = 78.69°

θ = 78.7° 



 3π  ≤ θ ≤ 2π  ; θ = 5.65222 rad, θ = 323.85°  270 ≤ θ ≤ 360°  2  

θ = 324° 



5π   360 ≤ θ ≤ 450°  2π ≤ θ ≤ ; 2  

θ = 379° 


θ = 6.61597 rad, θ = 379.07 


www.elsolucionario.net PROBLEM 10.41 The position of boom ABC is controlled by the hydraulic cylinder BD. For the loading shown, determine the force exerted by the hydraulic cylinder on pin B when θ = 65°.

SOLUTION We have

δ y A = −72sin θδθ

( BD )2 = ( BC )2 + (CD ) 2 − 2( BC )(CD ) cos(180° − θ ) = (45)2 + (24) 2 + 2(45)(24) cos θ ( BD )2 = 2601 + 2160 cos θ

Differentiating:

(1)

2( BD )δ ( BD ) = −2160sin θδθ

δ ( BD) = −

1080 sin θδθ BD

(2)

Virtual Work: Noting that P tends to decrease yA and FBD tends to increase BD, write

δ U = − Pδ y A + FBDδ ( BD) = 0  1080  sin θδθ  = 0 − P(−72sin θδθ ) + FBD  − BD  

FBD =

or, since

1 ( BD) P 15

P = 600 lb: FBD =

600 ( BD) = (40 lb)( BD ) 15

(3)

Making θ = 65° in Eq. (1), we have ( BD )2 = 2601 + 2160 cos 65° = 3513.9 BD = 59.278

Carrying into Eq. (3):

FBD = (40 lb)(59.278) = 2371.1 lb

FBD = 2370 lb 



yA = (72 in.) cos θ

www.elsolucionario.net PROBLEM 10.42 The position of boom ABC is controlled by the hydraulic cylinder BD. For the loading shown, (a) express the force exerted by the hydraulic cylinder on pin B as a function of the length BD, (b) determine the smallest possible value of the angle θ if the maximum force that the cylinder can exert on pin B is 2.5 kips.

SOLUTION (a)

See solution of Problem 10.41 for the derivation of Eq. (3):

(b)

For ( FBD )max = 2.5 kips = 2500 lb, we have 2500 lb = (40 lb)( BD) BD = 62.5

Carrying this value into Eq. (1) of Problem 10.41, write ( BD ) 2 = 2601 + 2160 cos θ (62.5) 2 = 2601 + 2160 cos θ cos θ = 0.60428

θ = 52.8° 



FBD = (40 lb)( BD) 

www.elsolucionario.net PROBLEM 10.43 The position of member ABC is controlled by the hydraulic cylinder CD. For the loading shown, determine the force exerted by the hydraulic cylinder on pin C when θ = 55°.

y A = (0.8 m)sin θ

δ y A = 0.8cos θδθ CD 2 = BC 2 + BD 2 − 2( BC )( BD ) cos(90° − θ ) CD 2 = 0.52 + 1.52 − 2(0.5)(1.5)sin θ CD 2 = 2.5 − 1.5sin θ 2(CD )(δ CD ) = −1.5cos θδθ

Virtual Work:

(1)

δ CD = −

3cos θ δθ 4CD

δ U = 0: − (10 kN)δ y A − FCDδ CD = 0  3cos θ  δθ  = 0 −10(0.8cos θδθ ) − FCD  − 4 CD  

FCD =

32 CD 3

(2)

For θ = 55°: Eq. (1): Eq. (2):

CD 2 = 2.5 − 1.5sin 55° = 1.2713; CD = 1.1275 m FCD =

32 32 CD = (1.1275) = 12.027 kN 3 3

FCD = 12.03 kN





SOLUTION

www.elsolucionario.net PROBLEM 10.44 The position of member ABC is controlled by the hydraulic cylinder CD. Determine the angle θ knowing that the hydraulic cylinder exerts a 15-kN force on pin C.

y A = (0.8 m)sin θ

δ y A = 0.8cos θδθ CD 2 = BC 2 + BD 2 − 2( BC )( BD ) cos(90° − θ ) CD 2 = 0.52 + 1.52 − 2(0.5)(1.5)sin θ CD 2 = 2.5 − 1.5sin θ 2(CD )(δ CD ) = −1.5cos θδθ ;

Virtual Work:

(1)

δ CD = −

3cos θ δθ 4CD

δ U = 0: − (10 kN)δ y A − FCDδ CD = 0  3cos θ  δθ  = 0 −10(0.8cos θδθ ) − FCD  −  4CD 

FCD =

32 CD 3

(2)

For FCD = 15 kN: Eq. (2): Eq. (1):

15 kN =

32 CD : 3

CD =

(1.40625) 2 = 2.5 − 1.5sin θ :

45 = 1.40625 m 32

sin θ = 0.34831

θ = 20.38°

θ = 20.4° 



SOLUTION

www.elsolucionario.net PROBLEM 10.45 The telescoping arm ABC is used to provide an elevated platform for construction workers. The workers and the platform together weigh 500 lb and their combined center of gravity is located directly above C. For the position when θ = 20°, determine the force exerted on pin B by the single hydraulic cylinder BD.

In Δ ADE :

From the geometry:


SOLUTION

AE 2.7 ft = DE 1.5 ft α = 60.945° 2.7 ft AD = = 3.0887 m sin 60.945°

tan α =

yC = (15 ft) sin θ

δ yC = (15 ft) cos θδθ Then, in triangle BAD:

Angle BAD = α + θ

Law of cosines: BD 2 = AB 2 + AD 2 − 2( AB )( AD ) cos(α + θ )

or

BD 2 = (7.2 ft)2 + (3.0887 ft) 2 − 2(7.2 ft)(3.0887 ft) cos(α + θ ) BD 2 = 61.380 ft 2 − (44.477 cos(α + θ )) ft 2

(1)



And then 2( BD )(δ BD) = (44.477 sin(α + θ ))δθ 44.477sin(α + θ ) δ BD = δθ 2( BD)

Virtual Work:

δ U = 0: − Pδ yC + FBDδ BD = 0  (44.477 ft 2 )sin(α + θ )  δθ  = 0 −(500 lb)(15 ft) cos θδθ + FBD  2( BD )  

Substituting

or

  cos θ FBD = 337.25 BD  lb/ft + sin( α θ )  

(2)

Now, with θ = 20° and α = 60.945° BD 2 = 61.380 − 44.477 cos(60.945° + 20°) BD 2 = 54.380 BD = 7.3743 ft

Equation (2):   cos 20° FBD = 337.25 (7.3743 ft)  lb/ft sin(60.945° + 20°)  

or

FBD = 2366 lb

FBD = 2370 lb





Equation (1):

www.elsolucionario.net PROBLEM 10.46 Solve Problem 10.45 assuming that the workers are lowered to a point near the ground so that θ = −20°. PROBLEM 10.45 The telescoping arm ABC is used to provide an elevated platform for construction workers. The workers and the platform together weigh 500 lb and their combined center of gravity is located directly above C. For the position when θ = 20°, determine the force exerted on pin B by the single hydraulic cylinder BD.

SOLUTION

Equation (1):

BD 2 = 61.380 − 44.477 cos(60.945° − 20°) BD 2 = 27.785 BD = 5.2711 ft

Equation (2):

FBD = 337.25

cos(−20°) (5.2711) sin(60.945° − 20°)

FBD = 2549 lb

or FBD = 2550 lb





Using the figure and analysis of Problem 10.45, including Equations (1) and (2), and with θ = −20°, we have

www.elsolucionario.net PROBLEM 10.47 Denoting by μs the coefficient of static friction between collar C and the vertical rod, derive an expression for the magnitude of the largest couple M for which equilibrium is maintained in the position shown. Explain what happens if μs ≥ tan θ .

SOLUTION

xB = l cos θ

δ xB = −l sin θδθ

(1)

and yC = l sin θ

δ yC = l cos θδθ

(2)

Member AB: We have 1 2

δ xB = lδφ Substituting from Equation (1), −l sin θδθ =

1 lδφ 2

δφ = −2sin θδθ

or

(3)

Free body of rod BC For M max, motion of collar C impends upward ΣM B = 0: Nl sin θ − ( P + μs N )(l cos θ ) = 0 N tan θ − μ s N = P N=

 P tan θ − μ s



Member BC: We have


Virtual Work:

δ U = 0: M δφ + ( P + μs N )δ yC = 0 M (−2sin θδθ ) + ( P + μs N )l cos θδθ = 0 M max =

P + μ s tan θP− μ ( P + μs N ) s l= l 2 tan θ 2 tan θ

or M max =

Pl  2(tan θ − μs )


If μs = tan θ , M = ∞, system becomes self-locking.


www.elsolucionario.net PROBLEM 10.48 Knowing that the coefficient of static friction between collar C and the vertical rod is 0.40, determine the magnitude of the largest and smallest couple M for which equilibrium is maintained in the position shown, when θ = 35°, l = 600 mm, and P = 300 N.

SOLUTION

M max =

With

Pl 2(tan θ + μ s )

θ = 35°, l = 0.6 m, P = 300 N (300 N)(0.6 m) 2(tan 35° − 0.4) = 299.80 N ⋅ m

M max =

M max = 300 N ⋅ m 

For M min , motion of C impends downward and F acts upward. The equations of Problem 10.50 can still be used if we replace μs by − μs . Then M min =

Pl 2(tan θ + μ s )

Substituting, (300 N)(0.6 m) 2(tan 35° + 0.4) = 81.803 N ⋅ m

M min =

M min = 81.8 N ⋅ m 



From the analysis of Problem 10.50, we have

www.elsolucionario.net PROBLEM 10.49 A block of weight W is pulled up a plane forming an angle α with the horizontal by a force P directed along the plane. If μ is the coefficient of friction between the block and the plane, derive an expression for the mechanical efficiency of the system. Show that the mechanical efficiency cannot exceed 12 if the block is to remain in place when the force P is removed.

SOLUTION Input work = Pδ x Output work = (W sin α )δ x

Efficiency: W sin αδ x Pδ x

or η =

W sin α P

ΣFx = 0: P − F − W sin α = 0 or ΣFy = 0: N − W cos α = 0 or

(1) P = W sin α + F

(2)

N = W cos α

F = μ N = μW cos α

Equation (2):

P = W sin α + μW cos α = W (sin α + μ cos α )

Equation (1):

η=

W sin α W (sin α + μ cos α )

or η =

1 1 + μ cot α



If block is to remain in place when P = 0, we know (see Chapter 8) that φs ≥ α or, since

μ = tan φs,

μ cot α ≥ tan α cot α = 1

Multiply by cot α : Add 1 to each side:

μ ≥ tan α

1 + μ cot α ≥ 2

Recalling the expression for η , we find

η≥

1  2



η=

www.elsolucionario.net PROBLEM 10.50 Derive an expression for the mechanical efficiency of the jack discussed in Section 8.6. Show that if the jack is to be self-locking, the mechanical efficiency cannot exceed 12 .

SOLUTION

Q = W tan(θ + φs ) y = x tan θ so that δ y = δ x tan θ Input work = Qδ x = W tan(θ + φs )δ x Output work = W δ y = W (δ x) tan θ

Efficiency:

η=

W tan θδ x ; W tan(θ + φs )δ x

η=

tan θ  tan(θ + φs )

From Page 432, we know the jack is self-locking if

φ ≥θ Then so that

θ + φs ≥ 2θ tan(θ + φs ) ≥ tan 2θ 

From above

η=

tan θ tan(θ + φs )

It then follows that

η≤

tan θ tan 2θ

But

Then

tan 2θ =

η≤

2 tan θ 1 − tan 2 θ tan θ (1 − tan 2 θ ) 1 − tan 2 θ = 2 tan θ 2

η≤

1  2



Recall Figure 8.9a. Draw force triangle

www.elsolucionario.net PROBLEM 10.51 Denoting by μ s the coefficient of static friction between the block attached to rod ACE and the horizontal surface, derive expressions in terms of P, μ s, and θ for the largest and smallest magnitude of the force Q for which equilibrium is maintained.

SOLUTION

ΣM B = 0: − x A +

xA P P = 0 or A = 2 2

Then:

F = μ s A = μs

Now

x A = 2l sin θ

P 1 = μs P 2 2

δ x A = 2l cos θδθ and

yF = 3l cos θ

δ yF = −3l sin θδθ Virtual Work:

δ U = 0: (Qmax − F )δ x A + Pδ yF = 0 1    Qmax − 2 μ s P  (2l cos θδθ ) + P(−3l sin θδθ ) = 0  

or

Qmax =

3 1 P tan θ + μ s P 2 2 Qmax =

P (3 tan θ + μ s )  2

For Qmin, motion of A impends to the right and F acts to the left. We change μ s to − μ s and find Qmin =

P (3tan θ − μ s )  2



For the linkage:

www.elsolucionario.net PROBLEM 10.52 Knowing that the coefficient of static friction between the block attached to rod ACE and the horizontal surface is 0.15, determine the magnitude of the largest and smallest force Q for which equilibrium is maintained when θ = 30°, l = 0.2 m, and P = 40 N.

SOLUTION Using the results of Problem 10.48 with l = 0.2 m P = 40 N, and μs = 0.15

We have

P (3 tan θ + μ s ) 2 (40 N) = (3tan 30° + 0.15) 2 = 37.64 N

Qmax =

Qmax = 37.6 N 

and

P (3 tan θ − μ s ) 2 (40 N) = (3tan 30° − 0.15) 2 = 31.64 N

Qmin =

Qmin = 31.6 N 



θ = 30°

www.elsolucionario.net PROBLEM 10.53 Using the method of virtual work, determine the reaction at E.

SOLUTION

From similar triangles: 2.1 δ yE = 1.75δ yE 1.2 2.7 δ yC = δ yE = 2.25δ yE 1.2 0.5 0.5 (2.25δ yE ) = 1.25δ yE δ yB = δ yC = 0.9 0.9

δ yD =

Virtual Work:

δ U = (2 kN)δ yB + (3 kN)δ yD − Eδ yE = 0 2(1.25δ yE ) + 3(1.75δ yE ) − Eδ yE = 0 E = +7.75 kN

E = 7.75 kN 



We release the support at E and assume a virtual displacement δ yE for Point E.

www.elsolucionario.net PROBLEM 10.54 Using the method of virtual work, determine separately the force and couple representing the reaction at H.

SOLUTION


Force at H. We give a vertical virtual displacement δ yH to Point H, keeping member FH horizontal.

From the geometry of the diagram:

δ yF = δ yG = δ yH 0.9 0.9 δ yF = δ yH = 0.75δ yH 1.2 1.2 1.5 1.5 (0.75δ yH ) = 1.25δ yH δ yC = δ yD = 0.9 0.9 0.5 0.5 (1.25δ yH ) = 0.69444δ yH δ yB = δ yC = 0.9 0.9

δ yD =

Virtual Work:

δ U = (2 kN)δ yB + (3 kN)δ yD − (5 kN)δ yG + H δ yH = 0 2(0.69444δ yH ) + 3(0.75δ yH ) − 5δ yH + H δ yH = 0 H = +1.3611 kN H = 1.361 kN 

Couple at H. We rotate beam FH through δθ about Point H.



From the geometry of the diagram:

δ yG = 1.2δθ

δ yF = 1.8δθ

0.9 0.9 δ yF = (1.8δθ ) = 1.35δθ 1.2 1.2 1.5 1.5 (1.35δθ ) = 2.25δθ δ yC = δ yD = 0.9 0.9 0.5 0.5 (2.25δθ ) = 1.25δθ δ yB = δ yC = 0.9 0.9

δ yD =

Virtual Work:

δ U = (2 kN)δ yB + (3 kN)δ yD − (5 kN)δ yG + M H δθ = 0 2(1.25δθ ) + 3(1.35δθ ) − 5(1.2δθ ) + M H δθ = 0 M H = 550 N ⋅ m




MH = −0.550 kN ⋅ m


www.elsolucionario.net PROBLEM 10.55 Referring to Problem 10.43 and using the value found for the force exerted by the hydraulic cylinder CD, determine the change in the length of CD required to raise the 10-kN load by 15 mm. PROBLEM 10.43 The position of member ABC is controlled by the hydraulic cylinder CD. For the loading shown, determine the force exerted by the hydraulic cylinder on pin C when θ = 55°.

Virtual Work: Assume both δ yA and δ CD increase

δ U = 0: − (10 kN)δ y A − FCDδ CD = 0 Substitute:

δ y A = 15 mm and FCD = 12.03 kN −(10 kN)(15 mm) − (12.03 kN)δ CD = 0

δ CD = −12.47 mm δ CD = 12.47 mm shorter 

The negative sign indicates that CD shortened



SOLUTION

www.elsolucionario.net PROBLEM 10.56 Referring to Problem 10.45 and using the value found for the force exerted by the hydraulic cylinder BD, determine the change in the length of BD required to raise the platform attached at C by 2.5 in. PROBLEM 10.45 The telescoping arm ABC is used to provide an elevated platform for construction workers. The workers and the platform together weigh 500 lb and their combined center of gravity is located directly above C. For the position when θ = 20°, determine the force exerted on pin B by the single hydraulic cylinder BD.


SOLUTION

Virtual Work: Assume both δ yC and δ BD increase

δ U = 0: − (500 lb)δ yC + FBDδ BD = 0 Substitute:

δ yC = 2.5 in. and FBD = 2370 lb −(500 lb)(2.5 in.) + (2370 lb)δ BD = 0

δ BD = +0.527 in. The positive sign indicates that cylinder BD increases in length

δ BD = 0.527 in. longer 


www.elsolucionario.net PROBLEM 10.57 Determine the vertical movement of joint D if the length of member BF is increased by 1.5 in. (Hint: Apply a vertical load at joint D, and, using the methods of Chapter 6, compute the force exerted by member BF on joints B and F. Then apply the method of virtual work for a virtual displacement resulting in the specified increase in length of member BF. This method should be used only for small changes in the lengths of members.)

SOLUTION Apply vertical load P at D.

E= ΣFy = 0:

P 3

3 P FBF − = 0 5 3

FBF =

5 P 9

Virtual Work: We remove member BF and replace it with forces FBF and −FBF at pins F and B, respectively.  Denoting the  virtual displacements of Points B and F as δ rB and δ rF , respectively, and noting that P and δ D have the same direction, we have Virtual Work:

δ U = 0:

Pδ D + FBF ⋅ δ rF + (−FBF ) ⋅ δ rB = 0 Pδ D + FBF δ rF cos θ F − FBF δ rB cos θ B = 0 Pδ D − FBF (δ rB cos θ B − δ rF cos θ F ) = 0

where (δ rB cos θ B − δ rF cos θ F ) = δ BF , which is the change in length of member BF. Thus, Pδ D − FBF δ BF = 0 5  Pδ D −  P  (1.5 in.) = 0 9  δ D = 0.833 in.

δ D = 0.833 in. 



ΣM H = 0: − P(40 ft) + E (120 ft) = 0

www.elsolucionario.net PROBLEM 10.58 Determine the horizontal movement of joint D if the length of member BF is increased by 1.5 in. (See the hint for Problem 10.57.)

SOLUTION Apply horizontal load P at D. ΣM H = 0: P(30 ft) − E y (120 ft) = 0

ΣFy = 0:

P 4

3 P FBF − = 0 5 4 FBF =

5 P 12

Denoting the We remove member BF and replace it with forces FBF and −FBF at pins F and B, respectively.   virtual displacements of Points B and F as δ rB and δ rF , respectively, and noting that P and δ D have the same direction, we have Virtual Work:

δ U = 0:

Pδ D + FBF ⋅ δ rF + (−FBF ) ⋅ δ rB = 0

Pδ D + FBF δ rF cos θ F − FBF δ rB cos θ B = 0 Pδ D − FBF (δ rB cos θ B − δ rF cos θ F ) = 0

where (δ rB cos θ B − δ rF cos θ F ) = δ BF , which is the change in length of member BF. Thus, Pδ D − FBF δ BF = 0 5  Pδ D −  P  (1.5 in.) = 0 9  δ D = 0.625 in.

δ D = 0.625 in.





Ey =

www.elsolucionario.net PROBLEM 10.59 Using the method of Section 10.8, solve Problem 10.29. PROBLEM 10.29 A load W of magnitude 600 N is applied to the linkage at B. The constant of the spring is k = 2.5 kN/m, and the spring is unstretched when AB and BC are horizontal. Neglecting the weight of the linkage and knowing that l = 300 mm, determine the value of θ corresponding to equilibrium.

W = 600 N l = 0.3 m, and k = 2500 N ⋅ m

We have

600 N 4(2500 N/m)(0.3 m) = 0.2

(l − cos θ )tan θ =

θ = 40.2° 

Solving numerically



SOLUTION

www.elsolucionario.net PROBLEM 10.60 Using the method of Section 10.8, solve Problem 10.30. PROBLEM 10.30 A vertical load W is applied to the linkage at B. The constant of the spring is k, and the spring is unstretched when AB and BC are horizontal. Neglecting the weight of the linkage, derive an equation in θ, W, l, and k that must be satisfied when the linkage is in equilibrium.

1 2 ks + WyB 2 1 V = k (2i − xC )2 + WyB 2 xC = 2l cos θ and yB = −l sin θ V=

Thus

1 k (2l − 2l cos θ ) 2 − Wl sin θ 2 = 2kl 2 (1 − cos θ ) 2 − Wl sin θ

V=

dV = 2kl 2 2(1 − cos θ )sin θ − Wl cos θ = 0 dθ

(1 − cos θ )tan θ =

W  4kl



SOLUTION

www.elsolucionario.net PROBLEM 10.61 Using the method of Section 10.8, solve Problem 10.31. PROBLEM 10.31 Two bars AD and DG are connected by a pin at D and by a spring AG. Knowing that the spring is 12 in. long when unstretched and that the constant of the spring is 125 lb/in., determine the value of x corresponding to equilibrium when a 900-lb load is applied at E as shown.

SOLUTION V=

s = x − 12 in.

x x − 3 6 x =− 2

and

yE = −

Thus

V=

1  x k ( x − 12)2 − W   2 2

dV 1 = k ( x − 12) − W = 0 dx 2 x = 12 +

W 900 lb = 12 in. + 2k 2(125 lb/in.) x = 15.60 in. 



But

1 2 ks + WyE 2

www.elsolucionario.net PROBLEM 10.62 Using the method of Section 10.8, solve Problem 10.32. PROBLEM 10.32 Solve Problem 10.31 assuming that the 900-lb vertical force is applied at C instead of E. PROBLEM 10.31 Two bars AD and DG are connected by a pin at D and by a spring AG. Knowing that the spring is 12 in. long when unstretched and that the constant of the spring is 125 lb/in., determine the value of x corresponding to equilibrium when a 900-lb load is applied at E as shown.

SOLUTION

But and Thus

1 2 ks + WyC 2

s = x − 12 in. yE = −

x 6

1 1 k ( x − 12) 2 − Wx 2 6 dV 1 = k ( x − 12) − W = 0 dx 6 W 900 lb x = 12 + = 12 in. + 6k 6(125 lb/in.) V=

x = 13.20 in. 



V=

www.elsolucionario.net PROBLEM 10.63 Using the method of Section 10.8, solve Problem 10.33. PROBLEM 10.33 Two 5-kg bars AB and BC are connected by a pin at B and by a spring DE. Knowing that the spring is 150 mm long when unstretched and that the constant of the spring is 1 kN/m, determine the value of x corresponding to equilibrium.

SOLUTION Wbar = (5 kg)(9.81 m/s 2 ) = 49.05 N


First note:

Since unstretched length of spring is 150 mm, or 0.15 m, we have 2 x − 0.15 3 1 3x x V = k Δ 2S − (49.05 N) − (49.05 N) 2 4 4

ΔS =

2

1 2  (1000 N/m)  x − 0.15  − 12.2625 x − 36.7875 x 2 3   dV 2 2 = 1000  x − 0.15  − 12.2625 − 36.7875 = 0 dx 3 3 V=

x = 0.335 m

x = 335 mm 


www.elsolucionario.net PROBLEM 10.64 Using the method of Section 10.8, solve Problem 10.35. PROBLEM 10.35 A vertical force P of magnitude 150 N is applied to end E of cable CDE, which passes over a small pulley D and is attached to the mechanism at C. The constant of the spring is k = 4 kN/m, and the spring is unstretched when θ = 0. Neglecting the weight of the mechanism and the radius of the pulley, determine the value of θ corresponding to equilibrium.

SOLUTION

θ 2


1 2 BC = BD = l

β = (90° − θ ) = 45° −

θ  CD = 2l sin β = 2l sin  45 −  2  For θ = 0 : (CD )0 = 2l sin 45 = 2 l yE = (CD )0 − CD

θ  = 2 l − 2l sin  45° −  2  Potential energy: 1 2 ks − PyE 2  θ  1  V = k (rθ ) 2 − P  2l − 2l sin  45° −   2 2    V=

θ  1  dV  = kr 2θ + 2 Pl cos  45° −  −  = 0 2  2  dθ  θ Pl = 2 kr cos ( 45° − θ2 ) Pl (150 N)(0.2 m) = = 0.75 2 kr (4000 N/m)(0.1 m) 2 0.75 =

θ

cos ( 45° − θ2 )

θ = 0.67623 rad


θ = 38.745°

θ = 38.7° 


www.elsolucionario.net PROBLEM 10.65 Using the method of Section 10.8, solve Problem 10.37. PROBLEM 10.37 and 10.38 Knowing that the constant of spring CD is k and that the spring is unstretched when rod ABC is horizontal, determine the value of θ corresponding to equilibrium for the data indicated. PROBLEM 10.37

P = 300 N, l = 400 mm, k = 5 kN/m.

SOLUTION Spring

 90° + θ  v = 2l sin    2 

θ  v = 2l sin  45° +  2 

Deflection of spring

v0 = 2l sin 45° = 2l


Unstretched (θ = 0)

θ  s = v − v0 = 2l sin  45° +  − 2l 2  2

  1 1 θ  V = ks 2 + PyA = kl 2  2sin  45° +  − 2  + P(−l sin θ ) 2 2 2      dV θ θ   = kl 2  2sin  45° +  − 2  cos  45° +  − Pl cos θ = 0 dθ 2 2      θ  θ θ     2sin  45° +  cos  45° +  − 2 cos  45° +   = 2 2 2        θ  cos θ − 2 cos  45° +  = 2 

Divide each member by cos θ

1− 2

cos ( 45° + θ2 ) cos θ

=

P cos θ kl P cos θ kl P kl

Then with P = 300 N, l = 0.4 m and k = 5000 N/m 1− 2

or

cos ( 45° + θ2 ) cos θ

cos ( 45° + θ2 ) cos θ

300 N (5000 N/m)(0.4 m) = 0.15

=

= 0.60104

θ = 22.6° 

Solving numerically


www.elsolucionario.net PROBLEM 10.66 Using the method of Section 10.8, solve Problem 10.38. PROBLEM 10.37 and 10.38 Knowing that the constant of spring CD is k and that the spring is unstretched when rod ABC is horizontal, determine the value of θ corresponding to equilibrium for the data indicated. P = 75 lb, l = 15 in., k = 20 lb/in.

PROBLEM 10.38

SOLUTION Using the results of Problem 10.65 with P = 75 lb, l = 15 in. and k = 20 lb/in., we have cos ( 45° + θ2 ) cos θ

=

P kl

75 lb (20 lb/in.)(15 in.) = 0.25 =

or Solving numerically

cos ( 45° + θ2 ) cos θ

= 0.53033

θ = 51.058°

θ = 51.1° 



1− 2

www.elsolucionario.net PROBLEM 10.67 Show that equilibrium is neutral in Problem 10.1. PROBLEM 10.1 Determine the vertical force P that must be applied at G to maintain the equilibrium of the linkage.

We have

y A = −u ,

2 4u yG = − u , β = 3 0.72

V = (80 N)y A + P( yG ) − (18 N ⋅ m)β 4u  2  V = 80( −u ) + P  − u  − (18) 3 0.72   dV 2 = −80 − P − 100 = 0 du 3 P = 270 N

P = 270 N 

Substituting P = 270 N in the expression for V, we have V = 0. Thus V is constant and equilibrium is neutral 



SOLUTION

www.elsolucionario.net PROBLEM 10.68 Show that equilibrium is neutral in Problem 10.7. PROBLEM 10.7 The two-bar linkage shown is supported by a pin and bracket at B and a collar at D that slides freely on a vertical rod. Determine the force P required to maintain the equilibrium of the linkage.


SOLUTION

y A = 2u yE = −u yF = −u V = PyA + (100 lb)yE + (150 lb)yF V = P(2u ) + (100 lb)(−u ) + (150 lb)( −u ) dV = 2 P − 100 − 150 = 0 du P = 125 lb

Now, substitute P = 125 lb in expression for V, making V = 0. Thus, V is constant and equilibrium is neutral. 


www.elsolucionario.net PROBLEM 10.69 Two uniform rods, each of mass m, are attached to gears of equal radii as shown. Determine the positions of equilibrium of the system and state in each case whether the equilibrium is stable, unstable, or neutral.

SOLUTION Potential energy

For equilibrium: or Thus


 l  l  V = W  − sin θ  + W  cos θ  W = mg  2  2  l = W (cos θ − sin θ ) 2 dV Wl = ( − sin θ − cos θ ) dθ 2 d 2V Wl = (sin θ − cos θ ) 2 dθ 2 dV = 0: sin θ = − cos θ dθ tan θ = −1

θ = −45.0° and θ = 135.0°

Stability: At θ = −45.0°:

d 2V Wl = [sin( −45°) − cos 45°] 2 dθ 2 2 2 Wl  = −  − <0 2  2 2 

θ = −45.0°, Unstable  At θ = 135.0°:

d 2V Wl = (sin135° − cos135°) 2 dθ 2 2 Wl  2 = +  >0 2  2 2 

θ = 135.0°, Stable  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1792

www.elsolucionario.net PROBLEM 10.70 Two uniform rods, AB and CD, are attached to gears of equal radii as shown. Knowing that WAB = 8 lb and WCD = 4 lb, determine the positions of equilibrium of the system and state in each case whether the equilibrium is stable, unstable, or neutral.

SOLUTION

 l  l  V = (3.5 kg × 9.81 m/s 2 )  − sin θ  + (1.75 kg × 9.81 m/s 2 )  cos θ  2 2     = (8.5838 N)l (−2sin θ + cos θ )

dV = (8.5838 N)l ( −2 cos θ − sin θ ) dθ d 2V = (8.5838 N)l (2sin θ − cos θ ) dθ 2

Equilibrium: or Thus

dV = 0: − 2 cos θ − sin θ = 0 dθ tan θ = −2

θ = −63.4° and 116.6°

Stability: At θ = −63.4°:

d 2V = (8.5838 N)l[2sin( −63.4°) − cos( −63.4°)] dθ 2 = (8.5838 N)l (−1.788 − 0.448) < 0

θ = −63.4°, Unstable  At θ = 116.6°:

d 2V = (8.5838 N)l[2sin(116.6°) − cos(116.6°)] dθ 2 = (8.5838 N)l (1.788 + 0.447) > 0

θ = 116.6°, Stable 



Potential energy

www.elsolucionario.net PROBLEM 10.71 Two uniform rods, each of mass m and length l, are attached to gears as shown. For the range 0 ≤ θ ≤ 180°, determine the positions of equilibrium of the system and state in each case whether the equilibrium is stable, unstable, or neutral.

Potential energy


SOLUTION

l  l  V = W  cos1.5θ  + W  cos θ  W = mg 2  2 

dV Wl Wl = ( −1.5sin1.5θ ) + (− sin θ ) dθ 2 2 Wl = − (1.5sin1.5θ + sin θ ) 2 2 d V Wl = − (2.25cos1.5θ + cos θ ) 2 2 dθ

For equilibrium

dV = 0: 1.5sin1.5θ + sin θ = 0 dθ

Solutions: One solution, by inspection, is θ = 0, and a second angle less than 180° can be found numerically:

θ = 2.4042 rad = 137.8° Now

d 2V Wl = − (2.25cos1.5θ + cos θ ) 2 2 dθ

At θ = 0:

d 2V Wl = − (2.25cos 0° + cos 0°) 2 2 dθ =−

Wl (3.25)(< 0) 2

θ = 0, Unstable 



At θ = 137.8°:

d 2V Wl = − [2.25cos(1.5 × 137.8°) + cos137.8°] 2 2 dθ Wl (2.75)( > 0) 2

θ = 137.8°, Stable 


=


www.elsolucionario.net PROBLEM 10.72 Two uniform rods, each of mass m and length l, are attached to drums that are connected by a belt as shown. Assuming that no slipping occurs between the belt and the drums, determine the positions of equilibrium of the system and state in each case whether the equilibrium is stable, unstable, or neutral.

SOLUTION


W = mg l  l  V = W  cos 2θ  − W  cos θ  2  2  dV l = W ( −2sin 2 + sin θ ) 2 dθ 2 1 d V = W ( −4 cos 2θ − cos θ ) 2 dθ 2

Equilibrium: or Solving

dV Wl = 0: (−2sin 2θ + sin θ ) = 0 dθ 2 sin θ (−4cos θ + 1) = 0

θ = 0, 75.5°, 180°, and 284°

Stability:

d 2V l = W (−4 cos 2θ − cos θ ) 2 dθ 2

At θ = 0:

d 2V l = W (−4 − 1) < 0 2 2 dθ

At θ = 75.5°:

d 2V l = W ( −4(−.874) − .25) > 0 2 2 dθ

At θ = 180°:

d 2V l = W (−4 + 1) < 0 2 2 dθ

At θ = 284°:

d 2V l = W ( −4(−.874) − .25) > 0 2 dθ 2

θ = 0, Unstable  θ = 75.5°, Stable  θ = 180.0°, Unstable  θ = 284°, Stable 


www.elsolucionario.net PROBLEM 10.73 Using the method of Section 10.8, solve Problem 10.39. Determine whether the equilibrium is stable, unstable, or neutral. (Hint: The potential energy corresponding to the couple exerted by a torsion spring is 12 Kθ 2 , where K is the torsional spring constant and θ is the angle of twist.) PROBLEM 10.39 The lever AB is attached to the horizontal shaft BC that passes through a bearing and is welded to a fixed support at C. The torsional spring constant of the shaft BC is K; that is, a couple of magnitude K is required to rotate end B through 1 rad. Knowing that the shaft is untwisted when AB is horizontal, determine the value of θ corresponding to the position of equilibrium when P = 100 N, l = 250 mm, and K = 12.5 N ⋅ m/rad.

Potential energy

V=


SOLUTION 1 Kθ 2 − Pl sin θ 2

dV = Kθ − Pl cos θ dθ d 2V = K + Pl sin θ dθ 2

Equilibrium: For

dV K = 0: cos θ = θ dθ Pl P = 100 N, l = 0.25 m., K = 12.5 N ⋅ m/rad 12.5 N ⋅ m/rad θ (100)(0.25 m) = 0.500θ

cos θ =

Solving numerically, we obtain

θ = 1.02967 rad = 59.000°

θ = 59.0° 

Stability d 2V = (12.5 N ⋅ m/rad) + (100 N)(0.25 m)sin 59.0° > 0 dθ 2

Stable 


www.elsolucionario.net PROBLEM 10.74 In Problem 10.40, determine whether each of the positions of equilibrium is stable, unstable, or neutral. (See hint for Problem 10.73.) PROBLEM 10.40 Solve Problem 10.39 assuming that P = 350 N, l = 250 mm, and K = 12.5 N ⋅ m/rad. Obtain answers in each of the following quadrants: 0 < θ < 90°, 270° < θ < 360°, 360° < θ < 450°.

SOLUTION Potential energy

V=

1 Kθ 2 − Pl sin θ 2

Equilibrium

dV K = 0: cos θ = θ dθ Pl

For

P = 350 N, l = 0.250 m and 12.5 N ⋅ m/rad cos θ = θ (350 N)(0.250 m)

or

cos θ =

K = 12.5 N ⋅ m/rad

θ 7

Solving numerically

θ = 1.37333 rad, 5.652 rad, and 6.616 rad

or

θ = 78.7°, 323.8°, 379.1° 

Stability at θ = 78.7°:

d 2V = (12.5 N ⋅ m/rad) + (350 N)(0.250 m) sin 78.7° dθ 2

θ = 78.7°, Stable 

= 98.304 > 0 At θ = 323.8°:

d 2V = (12.5 N ⋅ m/rad) + (350 N)(0.250 m) sin 323.8° dθ 2

= −39.178 N ⋅ m < 0

At At θ = 379.1°:


dV = Kθ − Pl cos θ dθ d 2V = K + Pl sin θ dθ 2

θ = 324°, Unstable 

d 2V = (12.5 N ⋅ m/rad) + (350 N)(0.250 m)sin 379.1° dθ 2

= 44.132 N ⋅ m > 0

θ = 379°, Stable 


www.elsolucionario.net PROBLEM 10.75 A load W of magnitude 100 lb is applied to the mechanism at C. Knowing that the spring is unstretched when θ = 15°, determine that value of θ corresponding to equilibrium and check that the equilibrium is stable.

SOLUTION We have

yC = l cos θ

π 1 k[ r (θ − θ0 )]2 + WyC θ0 = 15° = rad 2 12 1 2 = kr (θ − θ0 )2 + Wl cos θ 2

dV = kr 2 (θ − θ0 ) − Wl sin θ dθ

Equilibrium

dV = 0: kr 2 (θ − θ0 ) − wl sin θ = 0 dθ

with

W = 100 lb, R = 50 lb./in., l = 20 in., and r = 5 in.

(1)

π   (50 lb./in.)(25 in.2 )  θ −  − (100 lb)(20 in.)sin θ = 0 12   or Solving numerically

0.625θ − sin θ = 0.16362

θ = 1.8145 rad = 103.97° θ = 104.0° 

Stability

d 2V = kr 2 − Wl cos θ 2 dθ

or

= 1250 − 2000cos θ

For θ = 104.0°:

= 1734 in. ⋅ lb > 0

(2)

Stable 



V=

www.elsolucionario.net PROBLEM 10.76 A load W of magnitude 100 lb is applied to the mechanism at C. Knowing that the spring is unstretched when θ = 30°, determine that value of θ corresponding to equilibrium and check that the equilibrium is stable.

SOLUTION π  Using the solution of Problem 10.75, particularly Equation (1), with 15° replaced by 30°  rad  : 6 

π  kr 2  θ −  − Wl sin θ = 0 6  k = 50 lb/in., W = 100 lb, r = 5 in., and l = 20 in.

With

π  (50 lb/in.)(25 in.2 )  θ −  − (100 lb)(20 in.)sin θ = 0 6  or Solving numerically,

1250θ − 654.5 − 200sin θ = 0

θ = 1.9870 rad = 113.8° θ = 113.8° 

Stability: Equation (2), Problem 75: d 2V = kr 2 − Wl cos θ dθ 2

or

= 1250 − 2000cos θ

For θ = 113.8°:

= 2057 in. ⋅ lb > 0

Stable 



For equilibrium

www.elsolucionario.net PROBLEM 10.77 A slender rod AB, of weight W, is attached to two blocks A and B that can move freely in the guides shown. Knowing that the spring is unstretched when y = 0, determine the value of y corresponding to equilibrium when W = 80 N, l = 500 mm, and k = 600 N/m.

SOLUTION Deflection of spring = s, where

s = l2 + y2 − l ds y = 2 dy l − y2

1 2 y ks − W 2 2 dV ds 1 = ks − W dy dy 2 dV = k l2 + y2 − l dy V=

(

)


Potential energy:

1 − W 2 l +y y

2

2

  l  y − 1W = k 1 −  2 l 2 + y 2  

  dV l y= 1W = 0: 1 − 2 2  dy 2 k l + y  

Equilibrium

W = 80 N, l = 0.500 m, and k = 600 N/m

Now Then

or Solving numerically,

  0.500 m 1 −  y = 1 (80 N) 2 2  2 (600 N/m) (0.500) + y    1 − 0.500  0.25 + y 2 

  y = 0.066667   y = 0.357 m

y = 357 mm 


www.elsolucionario.net PROBLEM 10.78 A slender rod AB, of weight W, is attached to two blocks A and B that can move freely in the guides shown. Knowing that both springs are unstretched when y = 0, determine the value of y corresponding to equilibrium when W = 80 N, l = 550 mm, and k = 600 N/m.

SOLUTION Spring deflections S AD = l − l 2 − y 2 S BC = l 2 + y 2 − l 1 2 1 2 y −W kS AD + kS BC 2 2 2 2 1 1 V = k l − l 2 − y2 + k 2 2  dV y = k l − l 2 − y2   l 2 − y2 dy 

)

(

(

)

(

) − W 2y    + k ( l + y − l )     l 2 + y2 − l

2

2

2

 W − l 2 + y 2  2 y

    dV l l  y = W = 0:  − 1 +  1 −  l 2 − y 2   dy 2k l 2 + y 2     

Data: W = 80 N, l = 0.5 m, k = 600 N/m   0.5 0.5 80  y= − = 0.066667 2 2  2(1200)  (0.5) 2 − y 2 (0.5) y +  

y = 0.252 m


y = 252 mm 



V=

www.elsolucionario.net PROBLEM 10.79 A slender rod AB, of weight W, is attached to two blocks A and B that can move freely in the guides shown. The constant of the spring is k, and the spring is unstretched when AB is horizontal. Neglecting the weight of the blocks, derive an equation in θ, W, l, and k that must be satisfied when the rod is in equilibrium.

Elongation of spring:

Potential energy:

s = l sin θ + l cos θ − l s = l (sin θ + cos θ − 1) 1 2 1 ks − W sin θ W = mg 2 2 1 l = kl 2 (sin θ + cos θ − 1)2 − mg sin θ 2 2

V=

dV 1 = kl 2 (sin θ + cos θ − 1)(cos θ − sin θ ) − mgl cos θ dθ 2

Equilibrium:

(1)

dV mg = 0: (sin θ + cos θ − 1)(cos θ − sin θ ) − cos θ = 0 dθ 2kl mg   or cos θ (sin θ + cos θ − 1)(1 − tan θ ) − =0  2kl  



SOLUTION

www.elsolucionario.net PROBLEM 10.80 A slender rod AB, of weight W, is attached to two blocks A and B that can move freely in the guides shown. Knowing that the spring is unstretched when AB is horizontal, determine three values of θ corresponding to equilibrium when W = 300 lb, l = 16 in., and k = 75 lb/in. State in each case whether the equilibrium is stable, unstable, or neutral.

SOLUTION Using the results of Problem 10.79, particularly the condition of equilibrium mg   cos θ (sin θ + cos θ − 1)(1 − tan θ ) − =0 2kl  

W 300 lb = = 0.25 2kl (16 in.)(75 lb/in.) cos θ [ (sin θ + cos θ − 1)(1 − tan θ ) − 0.25] = 0

Thus:

cos θ = 0 and (sinθ + cos θ − 1)(1 − tan θ ) = 0.25

First equation yields θ = 90°. Solving the second equation by trial, we find θ = 9.39° and 34.16° Values of θ for equilibrium are

θ = 9.39°, 34.2°, and 90.0° Stability: we differentiate Eq. (1). d2y 1 = kl 2 [(cos θ − sin θ )(cos θ − sin θ ) + (sin θ + cos θ − 1)(− sin θ − cos θ )] + wl sin θ 2 2 ds W   = kl 2 cos 2 θ + sin 2 θ − 2 cos θ sin θ − sin 2 θ − cos 2 θ − 2 cos θ sin θ + sin θ + cos θ + sin θ  2kl     W  = kl 2 1 +  sin θ + cos θ − 2sin 2θ   2kl   d 2V = kl 2 (1.25sin θ + cos θ − 2sin 2θ ) 2 dθ

θ = 9.39° :

d 2V = kl 2 (1.25sin 9.4 + cos 9.4 − 2sin18.8) 2 dθ = kl 2 (+0.55) < 0

Stable 



Now, with W = 300 lb, l = 16 in., and k = 75 lb/in.


θ = 34.2° :

d 2v = kl 2 (1.25sin 34.2° + cos 34.3° − 2sin 68.4°) 2 dθ = kl 2 (−0.33) < 0

θ = 90.0° :

Unstable 

d 2V = kl 2 (1.25sin 90° + cos 90° − 2sin180°) 2 dθ

Stable 


= kl 2 (1.25) > 0


www.elsolucionario.net PROBLEM 10.81 A spring AB of constant k is attached to two identical gears as shown. Knowing that the spring is undeformed when θ = 0, determine two values of the angle θ corresponding to equilibrium when P = 30 lb, a = 4 in., b = 3 in., r = 6 in., and k = 5 lb/in. State in each case whether the equilibrium is stable, unstable, or neutral.

SOLUTION Elongation of spring 1 2 ks − Pbθ 2 1 = k (2a sin θ ) 2 − Pbθ 2

V=

dV = 4ka 2 sin θ cos θ − Pb dθ = 2ka 2 sin 2θ − Pb

Equilibrium

(1)

dV Pb = 0: sin 2θ = dθ 2ka 2 sin 2θ =

(30 lb)(3 in.) ; sin 2θ = 0.5625 2(5 lb/in.)(4 in.) 2 2θ = 34.229° and 145.771°

θ = 17.11° and 72.9°  Stability: We differentiate Eq. (1) d 2V = 4ka 2 cos 2θ dθ 2

θ = 17.11° :

d 2v = 4ka 2 cos 34.2° = 4ka 2 (0.83) > 0 dθ 2

Stable 

θ = 72.9°:

d 2v = 4ka 2 cos145.8° = 4ka 2 ( −0.83) < 0 2 dθ

Unstable 



s = 2(a sin θ ) = 2a sin θ

www.elsolucionario.net PROBLEM 10.82 A spring AB of constant k is attached to two identical gears as shown. Knowing that the spring is undeformed when θ = 0, and given that a = 60 mm, b = 45 mm, r = 90 mm, and k = 6 kN/m, determine (a) the range of values of P for which a position of equilibrium exists, (b) two values of θ corresponding to equilibrium if the value of P is equal to half the upper limit of the range found in part a.

SOLUTION Elongation of spring s = 2(a sin θ ) = 2a sin θ

1 2 1 ks − Pbθ = k (2a sin θ )2 − Pbθ 2 2 V = 2ka 2 sin 2 θ − Pbθ V=

Equilibrium dV dV = 0: = 4ka 2 sin θ cos θ − Pb dθ dθ = 2ka 2 sin 2θ − Pb = 0 sin 2θ =

Pmax ;

Pmax b 2ka 2

Pmax (0.045 m)

(a)

(b)

Pb ; For 2ka 2

2(6000 N/m)(0.06 m) 2

For P =

1 Pmax , 2

(1)

=1 =1

Pmax = 960 N 

1 sin 2θ = ; 2θ = 30° and 150° 2

θ = 15.00° and 75.0° 



Potential energy

www.elsolucionario.net PROBLEM 10.83 A slender rod AB is attached to two collars A and B that can move freely along the guide rods shown. Knowing that β = 30° and P = Q = 400 N, determine the value of the angle θ corresponding to equilibrium.

Law of Sines

yA L = sin(90° + β − θ ) sin(90 − β ) yA L = cos(θ − β ) cos β

or

yA = L

cos(θ − β ) cos β

From the figure:

yB = L

cos(θ − β ) − L cos θ cos β

Potential Energy:

 cos(θ − β )  cos(θ − β ) V = − PyB − Qy A = − P  L − L cos θ  − QL cos β cos β    sin(θ − β )  dV sin(θ − β ) = − PL  − + sin θ  + QL dθ cos β cos β  

= L( P + Q )

sin(θ − β ) − PL sin θ cos β



SOLUTION


dV sin(θ − β ) = 0: L( P + Q ) − PL sin θ = 0 dθ cos β

Equilibrium or

( P + Q) sin(θ − β ) = P sin θ cos β ( P + Q)(sin θ cos β − cos θ sin β ) = P sin θ cos β −( P + Q ) cos θ sin β + Q sin θ cos β = 0 −

P + Q sin β sin θ + =0 Q cos β cos θ

tan θ =

With

P+Q tan β Q

(2)

P = Q = 400 N, β = 30° tan θ =

800 N tan 30° = 1.1547 400 N

θ = 49.1° 



or

www.elsolucionario.net PROBLEM 10.84 A slender rod AB is attached to two collars A and B that can move freely along the guide rods shown. Knowing that β = 30°, P = 100 N, and Q = 25 N, determine the value of the angle θ corresponding to equilibrium.

SOLUTION

(100 N)(25 N) tan 30° (25 N) = 57.735 θ = 89.007°

tan θ =

θ = 89.0° 



Using Equation (2) of Problem 10.83, with P = 100 N, Q = 25 N, and β = 30°, we have

www.elsolucionario.net PROBLEM 10.85 Cart B, which weighs 75 kN, rolls along a sloping track that forms an angle β with the horizontal. The spring constant is 5 kN/m, and the spring is unstretched when x = 0. Determine the distance x corresponding to equilibrium for the angle β indicated. Angle β = 30°.

x = (4 m) tan θ

(1)

yB = x sin β = 4 tan θ sin β AC = (4 m) cos θ

For x = 0, Stretch of spring.

( AC )0 = 4 m s = AC − ( AC )0 = V=

4  1  − 4 = 4 − 1 cos θ  cos θ 

1 2 ks − (75 kN) yB 2 2

=

1  1  (5 kN/m) 16  − 1 − (75 kN)4 tan θ sin β 2  cos θ 

sin β dV  1  sin θ = 80  − 1 − 300 2 dθ cos 2 θ  cos θ  cos θ

Equilibrium Given: Eq. (2): Solve by trial and error: Eq. (1):

dV  1  = 0:  − 1 sin θ = 3.75sin β cos θ dθ  

(2)

β = 30°, sin θ = 0.5  1   cos θ − 1 sin θ = 3.75(0.5) = 1.875  

θ = 70.46° x = (4 m) tan 70.46°

x = 11.27 m 



SOLUTION

www.elsolucionario.net PROBLEM 10.86 Cart B, which weighs 75 kN, rolls along a sloping track that forms an angle β with the horizontal. The spring constant is 5 kN/m, and the spring is unstretched when x = 0. Determine the distance x corresponding to equilibrium for the angle β indicated. Angle β = 60°.

x = (4 m) tan θ

(1)

yB = x sin β = 4 tan θ sin β AC = (4 m) cos θ

For x = 0, Stretch of spring:

( AC )0 = 4 m s = AC − ( AC )0 = V=

4  1  − 4 = 4 − 1 cos θ  cos θ 

1 2 ks − (75 kN) yB 2 2

=

1  1  (5 kN/m) 16  − 1 − (75 kN)4 tan θ sin β 2  cos θ 

sin β dV  1  sin θ = 80  − 1 − 300 2 dθ cos 2 θ  cos θ  cos θ

Equilibrium Given: Eq. (2):

dV  1  = 0:  − 1 sin θ = 3.75sin β cos θ dθ  

(2)

β = 60°, sin θ = 0.86603  1   cos θ − 1 sin θ = 3.75(0.86603) = 3.2476  


θ = 76.67°

Eq. (1):

x = (4 m) tan 26.67°

x = 16.88 m 



SOLUTION

www.elsolucionario.net PROBLEM 10.87 Collar A can slide freely on the semicircular rod shown. Knowing that the constant of the spring is k and that the unstretched length of the spring is equal to the radius r, determine the value of θ corresponding to equilibrium when W = 50 lb, r = 9 in., and k = 15 lb/in.

SOLUTION Stretch of spring

s = AB − r s = 2(r cos θ ) − r s = r (2cos θ − 1)

Equilibrium

1 2 ks − Wr sin 2θ W = mg 2 1 V = kr 2 (2 cos θ − 1) 2 − Wr sin 2θ 2 dV = − kr 2 (2 cos θ − 1)2sin θ − 2Wr cos 2θ dθ V=


Potential energy:

dV = 0: − kr 2 (2 cos θ − 1) sin θ − Wr cos 2θ = 0 dθ (2cos θ − 1) sin θ W =− cos 2θ kr

Now Then Solving numerically,

W (50 lb) = = 0.37037 kr (15 lb/in.)(9 in.) (2cos θ − 1)sin θ = −0.37037 cos 2θ

θ = 0.95637 rad = 54.8°

θ = 54.8° 


www.elsolucionario.net PROBLEM 10.88 Collar A can slide freely on the semicircular rod shown. Knowing that the constant of the spring is k and that the unstretched length of the spring is equal to the radius r, determine the value of θ corresponding to equilibrium when W = 50 lb, r = 9 in., and k = 15 lb/in.

SOLUTION


Stretch of spring s = AB − r = 2( r cos θ ) − r s = r (2cos θ − 1) 1 V = ks 2 − Wr cos 2θ 2 1 2 = kr (2 cos θ − 1) 2 − Wr cos 2θ 2 dV = −kr 2 (2 cos θ − 1)2sin θ + 2Wr sin 2θ dθ

Equilibrium dV = 0: − kr 2 (2 cos θ − 1) sin θ + Wr sin 2θ = 0 dθ − kr 2 (2cos θ − 1) sin θ + Wr (2sin θ cos θ ) = 0 (2cos θ − 1) sin θ W = 2 cos θ kr

or Now Then

W (50 lb) = = 0.37037 kr (15 lb/in.)(9 in.) 2cos θ − 1 = 0.37037 2 cos θ

θ = 37.4° 

Solving


www.elsolucionario.net PROBLEM 10.89 Two bars AB and BC of negligible weight are attached to a single spring of constant k that is unstretched when the bars are horizontal. Determine the range of values of the magnitude P of two equal and opposite forces P and −P for which the equilibrium of the system is stable in the position shown.


SOLUTION

x = l cos θ y = l sin θ 1 2 ky 2 1 v = 2 Pl cos θ + kl 2 sin 2 θ 2

V = 2 Px +

dV = −2 Pl sin θ + kl 2 sin θ cos θ dθ 1 = −2 Pl sin θ + kl 2 sin 2θ 2 2 d V = −2 Pl cos θ + kl 2 cos 2θ 2 dθ

(1)

For equilibrium position θ = 0 to be stable d 2V = −2 Pl + kl 2 > 0 2 dθ

(2) P<

1 kl  2



Note: For P = 12 kl , we have

d 2V dθ 2

= 0 and we must determine which is the first derivative to be ≠ 0.

Differentiating Eq. (1): d 3V = +2 Pl sin θ − 2kl 2 sin 2θ = 0 for θ = 0 dθ 3 d 4V = 2 Pl cos θ − 4kl 2 cos 2θ = 2 Pl − 4kl 2 for θ = 0 4 dθ

But P = 12 kl. Thus

d 4V dθ 4

= kl 2 − 4kl 2 < 0 and we conclude that the equilibrium is unstable for P = 12 kl.


The sign < in Eq. (2) is thus correct.


www.elsolucionario.net PROBLEM 10.90 A vertical bar AD is attached to two springs of constant k and is in equilibrium in the position shown. Determine the range of values of the magnitude P of two equal and opposite vertical forces P and −P for which the equilibrium position is stable if (a) AB = CD, (b) AB = 2CD.

SOLUTION For both (a) and (b): Since P and − P are vertical, they form a couple of moment


M P = + Pl sin θ

The forces F and −F exerted by springs must, therefore, also form a couple, with moment M F = − Fa cos θ

We have

dU = M P dθ + M F dθ = ( Pl sin θ − Fa cos θ )dθ

but

Thus,

1  F = ks = k  a sin θ  2  1   dU =  Pl sin θ − ka 2 sin θ cos θ  dθ 2  

From Equation (10.19), page 580, we have dV = − dU = − Pl sin θ dθ +

or

1 2 ka sin 2θ dθ 4

dV 1 = − Pl sin θ + ka 2 sin 2θ dθ 4

and

d 2V 1 = − Pl cos θ + ka 2 cos 2θ 2 dθ 2

For θ = 0:

d 2V 1 = − Pl + ka 2 2 2 dθ

(1)



d 2V 1 > 0, − Pl + ka 2 > 0 2 2 dθ P<

or (for Parts a and b) Note: To check that equilibrium is unstable for P =

ka 2 2l

, we differentiate (1) twice:

d 3V = + Pl sin θ − ka 2 sin 2θ = 0, for dθ 3 d 4V = Pl cos θ − 2ka 2 cos 2θ dθ 4

For θ = 0 Thus, equilibrium is unstable when

ka 2  2l

θ = 0,

d 4V ka 2 2 = Pl − 2 ka = − 2ka 2 < 0 2 dθ 4 P=

ka 2  2l



For Stability:

www.elsolucionario.net PROBLEM 10.91 Rod AB is attached to a hinge at A and to two springs, each of constant k. If h = 25 in., d = 12 in., and W = 80 lb, determine the range of values of k for which the equilibrium of the rod is stable in the position shown. Each spring can act in either tension or compression.

SOLUTION xC = d sin θ

yB = h cos θ

Potential Energy:

1  V = 2  kxC2 + WyB  2  = kd 2 sin 2 θ + Wh cos θ

dV = 2kd 2 sin θ cos θ − Wh sin θ dθ = kd 2 sin 2θ − Wh sin θ

Then

d 2V = 2kd 2 cos 2θ − Wh cos θ dθ 2

and

(1)

For equilibrium position θ = 0 to be stable, we must have d 2V = 2kd 2 − Wh > 0 dθ 2 1 kd 2 > Wh 2

or Note: For kd 2 = 12 Wh, we have

d 2V dθ 2

(2)

= 0, so that we must determine which is the first derivative that is not

equal to zero. Differentiating Equation (1), we write d 3V = −4kd 2 sin 2θ + Wh sin θ = 0 dθ 3 d 4V = −8kd 2 cos 2θ + Wh cos θ 2 dθ

For θ = 0:

for θ = 0

d 4V = −8kd 2 + Wh dθ 4



We have


Since kd 2 = 12 Wh,

d 4V dθ 4

= −4Wh + Wh < 0, we conclude that the equilibrium is unstable for kd 2 = 12 Wh and

the > sign in Equation (2) is correct. With Equation (2) gives

k (12 in.) 2 >

1 (80 lb)(25 in.) 2

k > 6.944 lb/in.

k > 6.94 lb/in. 


or

W = 80 lb, h = 25 in., and d = 12 in.


www.elsolucionario.net PROBLEM 10.92 Rod AB is attached to a hinge at A and to two springs, each of constant k. If h = 45 in., k = 6 lb/in., and W = 60 lb, determine the smallest distance d for which the equilibrium of the rod is stable in the position shown. Each spring can act in either tension or compression.

SOLUTION Using Equation (2) of Problem 10.91 with

(6 lb/in.)d 2 >

or

1 (60 lb)(45 in.) 2

d 2 > 225 in.2 d > 15.0000 in.

smallest d = 15.00 in. 



h = 45 in., k = 6 lb/in., and W = 60 lb

www.elsolucionario.net PROBLEM 10.93 Two bars are attached to a single spring of constant k that is unstretched when the bars are vertical. Determine the range of values of P for which the equilibrium of the system is stable in the position shown.

SOLUTION s=

L 2L sin φ = sin θ 3 3


For small values of φ and θ

φ = 2θ 2L L  1 V = P  cos φ + cos θ  + ks 2 3 3  2 2

PL 1  2L  (cos 2θ + 2cos θ ) + k  sin θ  3 2  3  PL 2 (−2sin 2θ − 2sin θ ) + kL2 sin θ cos θ = 3 9 PL 2 (2sin 2θ + 2sin θ ) + kL2 sin 2θ =− 3 9 PL 4 (4 cos 2θ + 2 cos θ ) + kL2 cos 2θ =− 3 9

V= dV dθ

d 2V dθ 2

when θ = 0:

d 2V 6 PL 4 2 =− + kL 2 3 9 dθ

For stability:

d 2V 4 > 0, − 2 PL + kL2 > 0 2 9 dθ

P<

2 kL  9


www.elsolucionario.net PROBLEM 10.94 Two bars are attached to a single spring of constant k that is unstretched when the bars are vertical. Determine the range of values of P for which the equilibrium of the system is stable in the position shown.

SOLUTION a=

2L L sin θ = sin φ 3 3


For small values of φ and θ

φ = 2θ s=

L sin θ 3

L  2L  1 cos θ + cos φ  + ks 2 V = P 3  3  2 =

1 L PL  (2 cos θ + cos 2θ ) + k  sin θ  3 2 3 

2

dV PL kL2 (−2sin θ − 2sin 2θ ) + sin θ cos θ = dθ 3 9 dV 2 PL kL2 (sin θ + sin 2θ ) + sin 2θ =− dθ 3 18 d 2V 2 PL kL2 (cos 2 cos 2 ) cos 2θ θ θ = − + + 3 9 dθ 2

when θ = 0:

dV 2 kL2 = − 2 PL + 9 dθ 2

For stability:

d 2V kL2 > 0, − 2 + >0 PL 9 dθ 2

P<

1 kL  18


www.elsolucionario.net PROBLEM 10.95 The horizontal bar BEH is connected to three vertical bars. The collar at E can slide freely on bar DF. Determine the range of values of Q for which the equilibrium of the system is stable in the position shown when a = 24 in., b = 20 in., and P = 150 lb.

First note For small values of θ and φ : or

A = a sin θ = b sin φ aθ = bφ a b

φ= θ V = P( a + b) cos φ − 2Q (a + b) cos θ   a  = ( a + b)  P cos  θ  − 2Q cos θ  b     a  dV a  = ( a + b)  − P sin  θ  + 2Q sin θ  dθ b   b   a2  d 2V a  = ( a + b ) − 2 P cos  θ  + 2Q cos θ   2 dθ b   b 

when θ = 0:

 a2  d 2V ( a b ) = +  − 2 P + 2Q  2 dθ  b 



SOLUTION


Stability:

d 2V a2 > 0: − 2 P + 2Q > 0 2 dθ b b2 Q a2

(1)

a2 P 2b 2

(2)

P<2 Q>

or with

P = 150 lb, a = 24 in., and b = 20 in.

Equation (1):

Q>

(24 in.) 2 (150 lb) = 108.000 lb 2(20 in.) 2 Q > 108.0 lb 


For stability


www.elsolucionario.net PROBLEM 10.96 The horizontal bar BEH is connected to three vertical bars. The collar at E can slide freely on bar DF. Determine the range of values of P for which the equilibrium of the system is stable in the position shown when a = 150 mm, b = 200 mm, and Q = 45 N.

SOLUTION

Q = 45 N, a = 150 mm, and b = 200 mm

Equation (2)

(200 mm) 2 (45 N) (150 mm)2 = 160.000 N

P<2

P < 160.0 N 

For stability



Using Equation (2) of Problem 10.95 with

www.elsolucionario.net PROBLEM 10.97* Bars AB and BC, each of length l and of negligible weight, are attached to two springs, each of constant k. The springs are undeformed, and the system is in equilibrium when θ1 = θ2 = 0. Determine the range of values of P for which the equilibrium position is stable.

SOLUTION We have

xB = l sin θ yC = l cos θ1 + l cos θ 2 V = PyC +

or

V = Pl (cos θ1 + cos θ 2 ) +

1 2 1 2 kxB + kxC 2 2 1 2 kl sin 2 θ1 + (sin θ1 + sin θ 2 ) 2  2

For small values of θ1 and θ 2 : 1 1 sin θ1 ≈ θ1 , sin θ 2 ≈ θ 2 , cosθ1 ≈ 1 − θ12 , cos θ 2 ≈ 1 − θ 22 2 2

Then

and

 θ2 θ2 V = Pl 1 − 1 + 1 − 2  2 2 

 1 2 2 2  + kl θ1 + (θ1 + θ 2 )     2

∂V = − Plθ1 + kl 2 [θ1 + (θ1 + θ 2 )] ∂ θ1 ∂V = − Plθ 2 + kl 2 (θ1 + θ 2 ) ∂ θ2 ∂ 2V = − Pl + 2kl 2 ∂ θ12

∂ 2V = − Pl + kl 2 ∂ θ 22

∂ 2V = kl 2 ∂ θ1∂ θ 2 Stability: Conditions for stability (see Page 583).



xC = l sin θ1 + l sin θ 2


For

θ1 = θ 2 = 0:

∂V ∂V = = 0 (condition satisfied) ∂ θ1 ∂ θ 2

 ∂ 2V   ∂ θ1∂ θ 2

Substituting

2

 ∂ 2V ∂ 2V <0  − 2 2  ∂ θ1 ∂ θ 2

(kl 2 ) 2 − (− Pl + 2kl 2 )(− Pl + kl ) < 0 k 2 l 4 − P 2l 2 + 3Pkl 3 − 2k 2 l 4 < 0 P 2 − 3klP + k 2l 2 > 0 3− 5 kl or 2

P<

or

P < 0.382kl or

P>

3+ 5 kl 2

P > 2.62kl

∂ 2V > 0: − Pl + 2kl 2 > 0 ∂ θ12 or

P<

1 kl 2

∂ 2V > 0: − Pl + kl 2 > 0 ∂ θ 22 or

P < kl

Therefore, all conditions for stable equilibrium are satisfied when

0 ≤ P < 0.382kl 



Solving

www.elsolucionario.net PROBLEM 10.98* Solve Problem 10.97 knowing that l = 800 mm and k = 2.5 kN/m. PROBLEM 10.97* Bars AB and BC, each of length l and of negligible weight, are attached to two springs, each of constant k. The springs are undeformed, and the system is in equilibrium when θ1 = θ 2 = 0. Determine the range of values of P for which the equilibrium position is stable.

SOLUTION

l = 800 mm and k = 2.5 kN/m P < 0.382kl = 0.382(2500 N/m)(0.8 m) = 764 N

P < 764 N 



From the analysis of Problem 10.97 with

www.elsolucionario.net PROBLEM 10.99* Two rods of negligible weight are attached to drums of radius r that are connected by a belt and spring of constant k. Knowing that the spring is undeformed when the rods are vertical, determine the range of values of P for which the equilibrium position θ1 = θ 2 = 0 is stable.

Left end of spring moves from a to b. Right end of spring moves from a′ to b′. Elongation of spring s = a′b′ − ab = rθ1 − rθ 2 = r (θ1 − θ 2 ) 1 2 ks + pl cos θ1 − wl cos θ 2 2 1 = kr 2 (θ1 − θ 2 ) 2 + pl cos θ1 − wl cos θ 2 2

V=

∂v = kr 2 (θ1 − θ 2 ) − pl sin θ1 ∂ θ1 ∂v = − kr 2 (θ1 − θ 2 ) + wl sin θ 2 ∂ θ2 ∂ 2v = kr 2 − pl cos θ1 ∂ θ12 ∂ 2v = kr 2 + wl cos θ 2 ∂ θ 22 ∂ 2v = − kr 2 ∂ θ1∂ θ 2



SOLUTION


For

θ1 = θ 2 = 0:

∂ 2v ∂ 2v r 2v = kr 2 − pl , = + kr 2 + wl , = − kr 2 2 2 , ∂ θ ∂ θ ∂ θ1 ∂ θ2 2

Conditions for stability (see Page 583)  ∂ 2v   ∂ θ1∂ θ 2

2

 ∂ 2v ∂ 2v ⋅ <0  − 2 2  ∂ θ1 ∂ θ 2

(kr 2 ) 2 − (kr 2 − pl )(kr 2 + wl ) < 0 pl (kr 2 + wl ) − kr 2 wl < 0 wkr 2 kr 2   ; P< 2 l  kr + wl

  + w 

W kr l

2



∂ 2v kr 2 2 > : kr − pl > 0; P < l ∂ θ12



P<

We choose:

kr 2 l

   

   + W 

W kr l

2



P<

www.elsolucionario.net PROBLEM 10.100* Solve Problem 10.99 knowing that k = 20 lb/in., r = 3 in., l = 6 in., and (a) W = 15 lb, (b) W = 60 lb. PROBLEM 10.99* Two rods of negligible weight are attached to drums of radius r that are connected by a belt and spring of constant k. Knowing that the spring is undeformed when the rods are vertical, determine the range of values of P for which the equilibrium position θ1 = θ 2 = 0 is stable.

k = 20 lb/in. r = 3 in. l = 6 in. kr 2 (20 lb/in.)(3 in.) 2 = = 30 lb l 6 in.

(a)

W = 15 lb: P < (30 lb)

15 lb (30 lb) + (15 lb)

P < 10.00 lb 

(b)

W = 60 lb: P < (30 lb)

60 lb (30 lb) + (60 lb)

P < 20.0 lb 



SOLUTION

www.elsolucionario.net PROBLEM 10.101 Determine the horizontal force P that must be applied at A to maintain the equilibrium of the linkage.

SOLUTION Assume δθ


δ x A = 10δθ δ yC = 4δθ δ yD = δ yC = 4δθ δφ =

δ yD 6

2 = δθ 3

δ xG = 15δφ 2  = 15  δθ  3  = 10δθ

Virtual Work: We shall assume that a force P and a couple M are applied to member ABC as shown.

δ U = − Pδ x A − M δθ + 30δ yC + 40δ yD + 180δφ + 80δ xG = 0 2  − P(10δθ ) − M δθ + 30(4δθ ) + 40(4δθ ) + 180  δθ  + 80(10δθ ) = 0 3  −10 P − M + 120 + 160 + 120 + 800 = 0 (10 in.) P + M = 1200 lb ⋅ in.

Making M = 0 in Eq. (1):

P = +120.0 lb

(1) P = 120.0 lb




www.elsolucionario.net PROBLEM 10.102 Determine the couple M that must be applied to member ABC to maintain the equilibrium of the linkage.

SOLUTION Assume δθ


δ x A = 10δθ δ yC = 4δθ δ yD = δ yC = 4δθ δφ =

δ yD 6

2 = δθ 3

δ xG = 15δ φ 2  = 15  δθ  3  = 10δθ

Virtual Work: We shall assume that a force P and a couple M are applied to member ABC as shown.

δ U = − Pδ x A − M δθ + 30δ yC + 40δ yD + 180δφ + 80δ xG = 0 2  − P(10δθ ) − M δθ + 30(4δθ ) + 40(4δθ ) + 180  δθ  + 80(10δθ ) = 0 3  −10 P − M + 120 + 160 + 120 + 800 = 0 (10 in.) P + M = 1200 lb ⋅ in. P=0

Now from Eq. (1) for

(1) M = 1200 lb ⋅ in. 


www.elsolucionario.net PROBLEM 10.103 A spring of constant 15 kN/m connects Points C and F of the linkage shown. Neglecting the weight of the spring and linkage, determine the force in the spring and the vertical motion of Point G when a vertical downward 120-N force is applied (a) at Point C, (b) at Points C and H.

SOLUTION yG = 4 yC yF = 3 yC yE = 2 yC

δ yH = 4δ yC δ yF = 3δ yC δ yE = 2δ yC

For spring: Δ = yF − yC

Q = Force in spring (assumed in tension) Q = + k Δ = k ( yF − yC ) = k (3 yC − yC ) = 2kyC

(1)

C = 120 N, E = F = H = 0

(a) Virtual Work:

δ U = 0: −(120 N)δ yC + Qδ yC − Qδ yF = 0 −120δ yC + Qδ yC − Q (3δ yC ) = 0 Q = −60 N

Eq. (1):

Q = 2kyC , − 60 N = 2(15 kN/m) yC ,

Q = 60.0 N C 

yC = −2 mm

yG = 4 yC = 4(−2 mm) = −8 mm

At Point G:

y G = 8.00 mm 

C = H = 120 N, E = F = 0

(b) Virtual Work:

δ U = 0: − (120 N)δ yC − (120 N) yH + Qδ yC − Qδ yF = 0 −120δ yC − 120(4δ yC ) + Qδ yC − Q(3δ yC ) = 0 Q = −300 N

Eq. (1): At Point G:

Q = 2kyC

Q = 300 N C 

−300 N = 2(15 kN/m)yC , yC = −10 mm yG = 4 yC = 4(−10 mm) = − 40 mm

y G = 40.0 mm 



yH = 4 yC

www.elsolucionario.net PROBLEM 10.104 Derive an expression for the magnitude of the force Q required to maintain the equilibrium of the mechanism shown.

We have

xD = 2l cosθ

so that δ xD = −2l sin θδθ

δ A = 2lδθ δ B = lδθ Virtual Work:

δ U = 0: − Qδ xD − Pδ A − Pδ B = 0 −Q (−2l sin θδθ ) − P(2lδθ ) − P(lδθ ) = 0 2Ql sin θ − 3Pl = 0

Q=

3 P  2 sin θ



SOLUTION


yE = 3a sin θ

δ yE = 3a cos θδθ yF = 4a sin θ δ yF = 4a cos θδθ Virtual Work:

δ U = 0:

− M δθ + Pδ yE + Pδ yF = 0

− M δθ + P(3a cos θδθ ) + P(4a cos θδθ ) = 0

M = 7Pa cos θ 



SOLUTION

www.elsolucionario.net PROBLEM 10.106 Two rods AC and CE are connected by a pin at C and by a spring AE. The constant of the spring is k, and the spring is unstretched when θ = 30°. For the loading shown, derive an equation in P, θ, l, and k that must be satisfied when the system is in equilibrium.

SOLUTION

δ yE = −l sin θ δθ Spring: Unstretched length = 2l x = 2(2l sin θ ) = 4l sin θ

δ x = 4l cos θδθ F = k ( x − 2l ) F = k (4l sin θ − 2l )

Virtual Work:

δ U = 0:

Pδ yE − F δ x = 0 P(−l sin θ δθ ) − k (4l sin θ − 2l )(4l cos θ δθ ) = 0 − P sin θ − 8kl (2sin θ − 1) cos θ = 0

or

P cos θ = (1 − 2sin θ ) 8kl sin θ

P 1 − 2sin θ =  8kl tan θ



yE = l cos θ

www.elsolucionario.net PROBLEM 10.107 A force P of magnitude 240 N is applied to end E of cable CDE, which passes under pulley D and is attached to the mechanism at C. Neglecting the weight of the mechanism and the radius of the pulley, determine the value of θ corresponding to equilibrium. The constant of the spring is k = 4 kN/m, and the spring is unstretched when θ = 90°.

SOLUTION π  s = r  −θ  2  δ s = − rδθ

CD = 2l sin


π  F = ks = kr  − θ  2  

θ 2

θ 1  δ (CD) = 2l cos  δθ  22  θ = l cos δθ 2

Virtual Work: Since F tends to decrease s and P tends to decrease CD, we have

δ U = − F δ s − Pδ (CD ) = 0 θ  π   − kr  − θ  (− rδθ ) − P  l cos δθ  = 0 2 2     π

2

−θ

cos θ

2

Solving by trial and error:

=

pl (240 N)(0.3 m) = = 1.25 2 kr (4000 N/m)(0.12 m)2

θ = 0.33868 rad

θ = 19.40° 


www.elsolucionario.net PROBLEM 10.108 Two identical rods ABC and DBE are connected by a pin at B and by a spring CE. Knowing that the spring is 4 in. long when unstretched and that the constant of the spring is 8 lb/in., determine the distance x corresponding to equilibrium when a 24-lb load is applied at E as shown.


SOLUTION

Deformation of spring s = EC − 4 in. =

2x −4 3 2

x 1 1 2  2x  ks − (24 lb) = (8 lb/in.)  − 4  − 4 x 2 6 2  3  dV  2x 2 = 8 − 4 − 4 dx  3 3 V=

Equilibrium:

dV =0 dx

16  2 x  − 4 − 4 = 0  3 3  2x  3 − 4 = 4  3  16  2x 3 =4+ 3 4

x = 7.13 in. 


www.elsolucionario.net PROBLEM 10.109 Solve Problem 10.108 assuming that the 24-lb load is applied at C instead of E. PROBLEM 10.108 Two identical rods ABC and DBE are connected by a pin at B and by a spring CE. Knowing that the spring is 4 in. long when unstretched and that the constant of the spring is 8 lb/in., determine the distance x corresponding to equilibrium when a 24-lb load is applied at E as shown.


SOLUTION

Deformation of spring s = EC − 4 in. =

2x −4 3 2

1 2 5x 1  2x  ks − (24 lb) = (8 lb/in.)  − 4  − 20 x 2 6 2 3   dV  2x 2 = 8 − 4  − 20 dx  3 3 V=

Equilibrium:

dV =0 dx

16  2 x  − 4  − 20 = 0 3  3  2x  3 − 4 = 20   3  16  2x = 4 + 3.75 3

x = 11.63 in. 


www.elsolucionario.net PROBLEM 10.110 Two bars AB and BC are attached to a single spring of constant k that is unstretched when the bars are vertical. Determine the range of values of P for which the equilibrium of the system is stable in the position shown.

s = (l − a) sin θ V = P(2l cos θ ) + = 2 Pl cos θ +

1 2 ks 2

1 k (l − a) 2 sin 2 θ 2

dV = −2 Pl sin θ + k (l − a) 2 sin θ cos θ dθ 1 = −2 Pl sin θ + k (l − a )2 sin 2θ 2 2 d V = −2 Pl cos θ + k (l − a) 2 cos 2θ dθ 2

when

Stability:

θ = 0:

(1)

d 2V = −2 Pl + k (l − a) 2 dθ 2

d 2V > 0: − 2 Pl + k (l − a )2 > 0 2 dθ

To check whether equilibrium is unstable for P =

k (l − a )2 2l

P<

k (l − a )2  2l

, we differentiate

Eq. (1) twice: d 3V = 2 Pl sin θ − 2k (l − a) 2 sin 2θ = 0, For θ = 0 dθ 3 d 4V = 2 Pl cos θ − 4k (l − a) 2 cos 2θ dθ 4



SOLUTION


For G = 0 and

P=

k (l − a) 2 2l

d 4V = 2 Pl − 4k (l − a) 2 dθ 4 = k (l − a)3 − 4k (l − a )2 < 0 P=

k (l − a) 2  2l


Thus equilibrium is unstable for


www.elsolucionario.net PROBLEM 10.111 A homogeneous hemisphere of radius r is placed on an incline as shown. Assuming that friction is sufficient to prevent slipping between the hemisphere and the incline, determine the angle θ corresponding to equilibrium when β = 10°.

Detail 3 CG = r 8 V = W (− rθ sin β − (CG ) cos θ ) 3 dV   = W  − r sin β + r sin θ  8 dθ  

Equilibrium:

dV 3 = 0, − sin β + sin θ = 0 dθ 8 3 sin β = sin θ 8

For β = 10°

3 sin10° = sin θ 8

(1) sin θ = 0.46306, θ = 27.6° 



SOLUTION

www.elsolucionario.net PROBLEM 10.112 A homogeneous hemisphere of radius r is placed on an incline as shown. Assuming that friction is sufficient to prevent slipping between the hemisphere and the incline, determine (a) the largest angle β for which a position of equilibrium exists, (b) the angle θ corresponding to equilibrium when the angle β is equal to half the value found in part a.

Detail 3 CG = r 8 V = W (− rθ sin β − (CG ) cos θ ) 3 dV   = W  − r sin β + r sin θ  8 dθ   dV 3 = 0, − sin β + sin θ = 0 dθ 8

Equilibrium:

3 sin β = sin θ 8

(a)

For β max , θ = 90° Eq. (1)

(b)

(1)

3 3 sin β max = sin 90°, sin β max = = 22.02° 8 8

β max = 22.0° 

When β = 12 β max = 11.01° Eq. (1)

3 sin11.01° = sin θ ; sin θ = 0.5093 8

θ = 30.6° 

Note: We can also use ΔCGD and law of sines to derive Eq. (1). sin β sin θ CG 3 ; sin β = sin θ ; sin β = sin θ = CG CD CD 8



SOLUTION



ANSWER-NEW-Vector Mechanics for Engineers Statics (Ferdinand P. Beer)-ANSWER.pdf

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