statics_13esi_solutions-manual_c02-180897.pdf

.

.

2–2.

If u = 60° and F = 450 N, determine the magnitude of the resultant force and its direction, measured counterclockwise counterclockwise from the positive x axis.

 y F

u

15 700 N

SOLUTION The parallelogram law of addition and the triangular rule are shown in Figs.a and b, respectively. Applying the law of consines to Fig. b, FR =

 2 700 700

2

+ 4502 - 2(700)(450) cos 45°

= 497.01 N = 497 N

Ans.

This yields sin a sin 45° = 700 497.01 Thus, the direction of angle f of Thus, positive x axis axis,, is

a = 95.19° FR

measured counterc counterclockwise lockwise from the

f = a + 60° = 95.19° + 60° = 155°

Ans.

 x

2–3.

 y

If the magnitude of the resultant force is to be 500 N, directed along the positive  y axis axis,, determ determine ine the magnitu magnitude de of force F and its direction u.

F

u

15 700 N

SOLUTION The parallelogram law of a ddition and the triangular rule are shown in Figs. a and b, respectively. Applying the law of cosines to Fig. b, F =

 2 500 500

2

+ 7002 - 2(500)(700) cos 105°

= 959.78 N = 960 N

Ans.

Applying the law of sines to Fig. b, and using using this this resul result, t, yiel yields ds sin (90° + u) sin 105° = 700 959.78 u = 45.2°

Ans.

 x

2–4.

Determine the magnitude of the resultant force FR = F1 + F2 and its direction, measured clockwise clockwise from the positive u axis. 70 u

30 45 F 2

SOLUTION FR =

 2 (300) (300) + (500) - 2(300)(500) cos 95° = 605.1 = 605 N 2

2

Ans.

605.1 500 = sin 95° sin u u = 55.40° f = 55.40°  + 30° = 85.4°

Ans.

500 N

F 1 v

300 N

2–5.

Resolv Reso lve e th the e fo forc rce e F1 in into to co comp mpon onen ents ts ac acti ting ng al alon ong g th the e u and v axe axess and det determ ermine ine the mag magnit nitude udess of the co compo mponen nents ts.. 70 u

30 45 F 2

SOLUTION F1u

sin 40°

=

300 sin 110°

F1u = 205N F1v

sin 30°

=

Ans.

300 sin 110°

F1v = 160N

Ans.

500 N

F 1 v

300 N

2–6.

actting along the  u and Resolve the f orce F2 intocomponents ac v axes and d etermine the  magnitudes of  the components. 

70

u 30 

45 F 2

SOLUTION F2u

sin 45° F2u F2v

sin 65° F2v

=

500 sin 70°

=

376  N

=

500 sin 70°

=

482  N

Ans.

Ans.



500  N

F 1 v



300  N

If FB = 2 kN and the resultant force acts along the positive u axis, de dete term rmine the magnitude of the resultant force and the angle u. 2–7.

 y

F  A  3 kN  x u 30

B

 A

u FB

If the resultant force is required to act along the positive u axis and have a magnitude of 5 kN, kN, determine the required magnitude of FB and its direction u .

2–8.

 y

F  A  3 kN  x u 30

B

 A

u FB

2–9. Resolve F1 into components along the u and v axes and determine the magnitudes of these components components..

v

F 1

F 2

SOLUTION

150 15 0N

30

30

Sine law: F1v

sin 30° F1u

sin 45°

105

=

250 sin 105°

F1v

=

129 N

Ans.

=

250 sin 105°

F1u

=

183N

Ans.

250 N

u

2–10. Resolve F2 into components along the u and v axes and determine the magnitudes of these components.

v

F 1

F 2

SOLUTION

150 15 0N

30

30

Sine law: F2v

sin 30° F2u

sin 75°

105

=

150 sin 75°

F2v

=

77.6 77 .6 N

Ans.

=

150 sin 75°

F2u

=

150 N

Ans.

250 N

u

.

.

.

.

.

2–13. The device is used for surgical replacement of the knee joint. If the force acting along the leg is 360 N, determine its components along the  x and  y ¿ axes.

 y¿

 y

10

 x x¿ ¿  x x

60

360 N

.

.

2–14. The device is used for surgical replacement of the knee joint. If the force acting acting along the leg leg is 360 N, N, determine its components along the  x ¿ and  y axes.

 y¿

 y

10

 x¿  x

60

360 N

.

.

2–15.

The plate is subjected to the two forces at  A and B as shown.If u = 60°, determ determine ine the magnitu magnitude de of the result resultant ant of these two forces and its direction measured clockwise from the horizontal.

F  A u

 A

SOLUTION  Parallelogram Law: The parallelogram law of addition is shown in Fig. a. Trigonometry: Using law of cosines (Fig. b), we ha have ve FR

=

 2 8

2

+ 62 - 2(8)(6) cos 100° 40

= 10 10.8 .80 0 kN = 10 10.8 .8 kN

Ans.

The angle u can be determined using law of sines (Fig. (Fig. b).

F B

sin u sin 100° = 6 10.80 sin u = 0.5470 u

= 33.16°

Thus,, the direction f of FR measured from the x axis is Thus f

= 33.16° - 30° = 3.16°

B

Ans.

6 kN

8 kN

2–16.

Determine the angle of u for connecting member  A to the plate so that the resultant force of F A and FB is dir direc ected ted horizontally horizon tally to the right. Also Also,, what is the magnitude magnitude of the resultant force?

F  A u

 A

SOLUTION  Parallelogram Law: The parallelogram law of addition is shown in Fig. a. Trigonometry: Using law of sines (F (Fig ig .b), we ha have ve

sin (90° - u) sin 50° = 6 8

40

B

sin (90° - u) = 0.5745 u

= 54.93° = 54.9°

F B

Ans.

From the triangle, f = 180° - (90° - 54.93°) - 50° = 94.93° . Thus, using law of   cosines, the magnitude of FR is FR

=

 2 8

2

+ 62 - 2(8)(6) cos 94.93°

= 10 10.4 .4 kN

Ans.

6 kN

8 kN

.

.

.

.

.

2–19.

Deter Det ermi min ne th the e ma magn gnit itu ude an and d di dirrec ecti tion on of th the e res esu ult ltan antt FR = F1 + F2 + F3 of th the e th thre ree e for orce cess by fi firs rstt fi fin ndi ding ng th the e resultant F ¿ = F1 + F2 and th then fo forming FR = F ¿ + F3.

 y F 1

30 N 5 3 4

F 3

50 N

 x 20

SOLUTION F¿

=

 2 (20) (20)

2

F 2

+ (30)2 - 2(20)(30) cos 73.13° = 30.85 N

30.85 30 = ; sin 73.13° sin (70° - u¿) FR

=

 2 (30.85) (30.85)

2

u¿

= 1.47°

+ (50)2 - 2(30.85)(50) cos 1.47° = 19.18 = 19 19.2 .2 N

19.18 30.85 = ; sin 1.47° sin u

u

= 2.37°

Ans.

Ans.

20 N

2–20.

Determ Dete rmin ine e the ma magn gnit itu ude an and d di dirrec ecti tio on of th the e res esu ult ltan antt FR = F1 + F2 + F3 of th the e thr hree ee for orcces by fi firrst fi fin ndi ding ng th the e resultant F ¿ = F2 + F3 and th then fo forming FR = F ¿ + F1.

 y F 1

30 N 5 3 4

F 3

50 N

 x 20

SOLUTION F

¿

=

 2 (20) (20)

2

20 sin u FR

¿

=

=

F 2

+ (50) - 2(20)(50) cos 70° = 47 47.0 .07 7N 2

47.07 ; sin 70°

 2 (47.07) (47.07)

2

u¿

+ (30)2 - 2(47.07)(30) cos 13.34° = 19.18 = 19 19.2 .2 N

19.18 30 = ; sin 13.34° sin f u

= 23.53°

f

Ans.

= 21.15°

= 23.53° - 21.15° = 2.37°

Ans.

20 N

2–21.

Two force forcess act on the screw eye. If F1 = 400 N and F2 = 600 N, de dete term rmin ine e the the an angl gle e u(0° … u … 180°) between them, so that the resultant force has a magnitude of FR = 800 N.

F1

u

SOLUTION The parallelogram law of addition and triangular rule are shown in Figs. a and b, respectively.Applying law of cosines to Fig. b, 800 =

 2 400 400

2

+ 6002 - 2(400)(600) cos (180° - u°)

8002 = 4002 + 6002 - 480000 cos (180° - u) cos (180° - u) = - 0.25 180° - u = 104.48 u = 75.52° = 75.5°

Ans.

F2

2–22.

Two forces F1 and F2 act on the screw screw eye. If their lines of  action are at an angle u apart and the magnitude of each force is F1 = F2 = F, determine the magnitude of the resultant force FR and the angle between FR and F1.

F1

u

SOLUTION F sin f

=

F sin (u - f)

sin (u - f) = sin f u

-f=f

f

=

FR

=

u

Ans.

2

 2 (F)

2

+ (F)2 - 2(F)(F) cos (18 (180° 0° - u)

Since cos (180° - u) = - cos u FR

Since cos

= FA

a 2 b = A 1 + 2cos u

F2

2 2 B 2 1 + cos

u

u

Then FR

= 2F cos

a2b u

Ans.

2–23.

Two force forcess act on the screw eye. If F = 600 N, determ determine ine u the magnitude of the resultant force force and the angle if the resultant force is directed vertically upward.

 y

F

500 N

30 u

 x

SOLUTION The parallelogram law of addition and triangular rule are shown in Figs. a and b respectively.Applying law of sines to Fig. b, sin u sin 30° = ;  sin u = 0.6 600 500

u = 36.87° = 36.9°

Ans.

Using the result of u , f = 180° - 30° - 36.87° = 113.13°

Again, applying law of sines using the result of f , FR

sin 113.13°

=

500 ; sin 30°

FR = 919.61 N = 920 N

Ans.

2–24.

Two forces are applied at the end of a screw eye in order to remove the post. post. Determi Determine ne the angle angle u 0° … u … 90° and the magnitude of force F so that the resultant force acting on the post is directed vertically upward and has a magnitude of 750 N.

1

 y

2

F 500 N

30

°

θ 

 x 

SOLUTION  Parallelogram Law: The parallelogram law of addition is shown in Fig. a. Trigonometry: Using law of sines si nes (Fig. b), we have have

sin f 750

=

sin 30° 500

sin f = 0.750 f

1

2

= 131.41° By observation, f 7 90°

Thus, u

= 180° - 30° - 131.41° = 18.59° = 18.6°

Ans.

F 500 = sin 18.59° sin 30° F

= 319 N

Ans.

2-25

.

Determine the magnitude and direction of the resultant force FR . Express the result in terms of  the magnitudes of the component F2 and resultant FR  and the angle  .

Solution:  F 1

2

2

2

  F R   F 2  2 FR F 2 cos  

0deg     cos    Since cos  180deg

,

 F 1 

2

2

F R   F 2  2  F R F 2 cos  

Ans.

2–26.

The beam is to be hoisted using two two chains. Determine the magnitudes of forces F A and FB acting on each chain in order to develop a resultant force of 600 N directed along the positive y axis axis.. Set u = 45°.

 y FB

F A

u

30

SOLUTION  x

FA

600 =  ; sin 45° sin 105° FB

sin 30°

=

600  ; sin 105°

FA

= 439N

Ans.

FB

= 31 311 1N

Ans.

2–27.

The beam is to be hoisted using two chains. chains. If the resultant force is to be 600 N directed along the positive  y axis, determine the magnitudes of forces F A and FB acting on each chain and the angle u of FB so that the magnitude of FB is a minimum. F A acts at 30° 30° from the the y axis, as shown.

 y FB

F A

u

30

SOLUTION  x

For minimum F B, req requir uire e u

= 60°

Ans.

FA

= 600 cos 30° = 520N

Ans.

FB

= 600 sin 30° = 30 300 0N

Ans.

2–28.

If the resultant force of the two tugboats is 3 kN, directed along the positive x axis, determine the required magnitude of force FB and its direction u.

 y

 A

F  A  2 kN

30

 x

u

C 

SOLUTION The parallelogram law of addition and the triangular rule are shown in Figs. a and b, respectively. Applying the law of cosines to Fig. b, FB =

 2 2

2

+ 32 - 2(2)(3)cos 30°

= 1.615kN = 1.61 kN

Ans.

Using this result and applying the law of sines to Fig. b, yields sin u sin 30° = 2 1.615

u = 38.3°

Ans.

FB

B

2–29.

If FB = 3 kN and u = 45°, determine the magnitude of the resultant force of the two tugboats and its direction measured clockwise from the positive x axis.

 y

 A

F  A  2 kN

30

 x

u

C 

SOLUTION The parallelogram law of addition and the triangular rule are shown in Figs. Figs. a and b, respectively. Applying the law of cosines to Fig. b, FR =

 2 2

2

+ 32 - 2(2)(3) cos 105°

= 4.013 kN = 4.01 kN

Ans.

Using this result and applying the law of sines to Fig. b, yields sin a sin 105° = 3 4.013

a = 46.22°

Thus,, the direction angle f of FR, measured clockwise from the positive x axis, is Thus f = a - 30° = 46.22° - 30° = 16.2°

Ans.

FB

B

2–30.

If the resultant force of the two tugboats is required to be directed direc ted toward towardss the positi positive ve x axis axis,, and FB is to be a minimum,, determ minimum determine ine the magnitu magnitude de of FR and FB and the angle u.

 y

 A

F  A  2 kN

30

 x

u

SOLUTION

C 

FB

For FB to be minimum, it has to be directed perpendicular to FR . Thus Thus,, u = 90°

Ans.

The parallelogram law of addition and triangular rule are shown in Figs. a and b, respectively. By applying simple trigonometry to Fig. b, FB = 2 sin 30° = 1 kN

Ans.

FR = 2 cos 30° = 1.73 kN

Ans.

B

2-31

.

If the tension in the cable is F1, determine the magnitude and direction of the resultant force acting on the pulley. This angle defines the same angle   of  of line AB line AB on  on the tailboard block. Given:  F 1  400 N  1  30 deg

Solution:  F  R 

2

2

F 1   F 1  2 F 1 F 1 cos 90 deg   1 

 F  R  400 N sin  90 deg     F  R

Ans.



   90 deg  asin

   60deg

sin   1   F 1

  F  R   sin   1    F 1   Ans.

2-32

.

Determine the magnitude of the resultant force and its direction, measured clockwise from the positive x positive  x axis.  axis. Given:  F 1  70 N  F 2  50 N  F 3  65 N    30 deg    45 deg Solution:

 

F  Rx =  F  x;

F  RX   F 1  F 2 cos     F 3 cos   



F  Ry =  F  y;

F  RY   F 2 sin     F 3 sin   

 F  R 

F  RX    F  RY 



   atan

2

2

   F  RY       F  RX   

 F  R  97.8 N

Ans.

   46.5 46.5 deg

Ans.

2–33.

Determine the magnitude of the resultant force and its direction, direc tion,measu measured red coun counterc terclock lockwise wise from the posit positive ive x axis.

 y F 3

750 N

45

 x

SOLUTION

3

5 4

+ FR = © Fx ; x

:

FRx

=

4 (850) - 625 sin 30° - 750 sin 45° = - 16 162. 2.8 8N 5

30

F 2

+ c FRy = © Fy ;

FRy FR

3 5

= - (850) - 625 cos 30° + 750 cos 45° = - 520 520.9 .9 N =

 2 ( - 162.8)

2

B

 - 520.9 - 162.8

+ ( - 520.9)2 = 546 N

R

f

= tan - 1

u

= 180° + 72.64° = 253°

Ans.

= 72.64° Ans.

625 N

F 1

850 N

2–34.

 y

60

Resolve F1 and F2 into their x and y components.

30 F 1  400 N

45

SOLUTION F1 = {400 sin 30°( + i) + 400 cos 30°( + j)} N

= {200i + 346 j} N

F 2  250 N

Ans.

F2 = {250 cos 45°( + i) + 250 sin 45°( - j)} N

= {177i -177 j} N

Ans.

 x

2–35.

Determine the magnitude of the resultant force and its direction measured counterclockwise from the positive x axis.

 y

60 30 F 1  400 N

SOLUTION  Rectangular Components: By referring to Fig. a, th the e x and y components of F1 and F2 can be written as

(F1)x = 400 sin 30° = 200 N

(F1)y = 400 cos 30° = 346.41 N

(F2)x = 250 cos 45° = 176.78 N

(F2)y = 250 sin 45° = 176.78 N

F 2  250 N

 Resultant Force: Summing the force components algebraically along the x and y axes axes,, we have :

+ © (F ) = © F ; R x x

(FR)x = 200 + 176.78 = 376.78 N

+ c © (FR)y = © Fy;

(FR)y = 346.41 - 176.78 = 169.63 N c

The magnitude of the resultant force FR is FR =

 2 (F )

2 R x

+ (FR)y2 =

 2 376.78 376.78

2

+ 169.632 = 413 N

Ans.

The direc direction tion angle u of FR , Fig Fig.. b , measured counterclockwise counterclockwise from the positive axiss, is axi u = tan-1

c (( )) d = tan a 169.63 b = 24.2° 376.78 FR FR

y

x

-1

45

Ans.

 x

2–36.

Resolve each force force acting acting on the gusset plate into its x and y components, and express each force as a Cartesian vector. vector.

 y F 3  650 N 3

5

F 2  750 N

4

45  x F 1  900 N

F1 = {900( + i)} = {900i} N

Ans.

F2 = {750 cos 45°( + i) + 750 sin 45°( + j)} N = {530i + 530 j} N

Ans.

F3 =

 e 650 a 45 b ( + ) + 650 a 35 b ( - ) f  N i

= {520 i - 390 j)} N

 j

Ans.

2–37.

Determine the magnitude of the resultant force acting on the plate and its direction, measured counterclockwise from the positive x axis.

 y F 3  650 N 3

5

F 2  750 N

4

SOLUTION 45

 Rectangular Components: By referring to Fig. a, th the e x and y components of F1, F2, and F3 can be written as

(F1)x = 900 N

(F1)y = 0

(F2)x = 750 cos 45° = 530.33 N (F3)x = 650

(F2)y = 750 sin 45° = 530.33 N

a 45 b = 520 N

(F3)y = 650

a 35 b = 390 N

 Resultant For  Resultant Force: ce: Summi Summing ng the force components algebraicall algebraically y along the x y axes axes,, we have

+ © (F ) = © F ; R x x

(FR)x = 900 + 530.33 + 520 = 1950.33 N

+ c © (FR)y = © Fy;

(FR)y = 530.33 - 390 = 140.33 N c

:

and

:

The magnitude of the resultant force FR is FR =

 2 (F )

2 R x

+ (FR)y2 =

 2 1950.33 1950.33

2

+ 140.332 = 1955 N = 1.96 kN Ans.

The direction angle u of FR , measured clockwise from the positive x axis, is u = tan-1

140.33 c (( )) d = tan a 1950.33 b = 4.12° FR FR

y

x

-1

Ans.

 x F 1  900 N

.

2–39.  y

Resolve each force acting on the support into its  x and  y components, and express each force as a Cartesian vector.

F 2  600 N

F 1  800 N

45 60  x 5

SOLUTION

F1 = {800 cos 60°( + i) + 800 sin 60°( + j)} N

= {400i + 693 j} N

Ans.

F2 = {600 sin 45°( - i) + 600 cos 45°( + j)} N

Ans.

= { - 424i + 424 j} N F3 =

 e 650 a 1213 b ( + ) + 650 a 135 b ( - ) f  N i

= {600i - 250 j} N

 j

Ans.

13 12

F 3  650 N

2–40.  y

Determine the magnitude of the resultant force and its direction u, measured counterclockwise from the positive x axis.

F 2  600 N

F 1  800 N

45 60  x 5

SOLUTION  Rectangular Components: By referring to Fig. a, th the e x and y components of F1, F2, and F3 can be written as

(F1)x = 800 cos 60° = 400 N

(F1)y = 800 sin 60° = 692.82 N

(F2)x = 600 sin 45° = 424.26 N

(F2)y = 600 cos 45° = 424.26 N

(F3)x = 650

a 1213 b = 600 N

(F3)y = 650

a 135 b = 250 N

 Resu ltan t Force Force:: Summing the force components algebraically along the x and y axes axes,, we have

+ © (F ) = © F ; R x x

(FR)x = 400 - 424.26 + 600 = 575.74 N

+ c © (FR)y = © Fy;

(FR)y = - 692.82 + 424.26 - 250 = 867.08 N c

:

:

The magnitude of the resultant force FR is FR =

 2 (F )

2 R x

+ (FR)y2 =

 2 575.74 575.74

2

+ 867.082 = 1041 N = 1.04 kN Ans.

The direction angle u of FR , Fig Fig.. b, measured counterclockw counterclockwise ise from the positive x axis axis,, is u = tan-1

c (( )) d = tan a 867.08 b = 56.4° 575.74 FR FR

y

x

-1

Ans.

13 12

F 3  650 N

2-41

.

Determine the magnitude of the resultant force and its direction, measured counterclockwise counterclockwise from the positive x positive x axis.  axis. Units Used: 3

kN  10  N

Given:  F 1  30 kN  F 2  26 kN    30 deg c  5 d   12

Solution:

  F   =  F  ;   Rx

 x

 

 F  Rx   F 1 sin    

  



F  Ry =  F  y;

F  Ry   F 1 cos    

   

 F  R 

2

F  Rx   F   Ry

   atan

  F  Ry    F  Rx 

    180 180 deg   

2

c 2

  F 

2

2

F  Rx  25 kN

c  d    d  2

  F 

2

2

F  Ry  2 kN

c  d   

F  R  25.1 25.1 kN

Ans.

   4.5 deg

   184.5 184.5 deg

Ans.

2–42.

Determine the magnitude and orientation u of FB so that the result resultant ant force is directed along the positive positive y axis and has a magnitude of 1500 N.

 y F B

F   A  = 700 N

30° A

 B

θ 

 x 

SOLUTION  Scalar Notation: Summing the force components algebraically, we have

+ FR = © Fx ; x

:

0 = 700 sin 30° - FB cos u FB cos u

+  c FRy = © Fy ;

= 350

(1)

1500 = 700 cos 30° + FB sin u FB sin u

= 893.8

(2)

Solving Eq. Eq. (1) and (2) yields yields u

= 68.6° FB = 960 N

Ans.

2–43.

Determine the Determine the magnitude magnitude and orientat orientation, ion, measur measured ed counterclockwise from the positive  y axis axis,, of the resultant resultant force acting on the brack bracket, et, if FB = 600 N and u = 20°.

 y FB

F  A

30

700 N

A

B u

 x

SOLUTION  Scalar Notation: Summing the force components algebraically, we have

+ FR = ©Fx ; x

FRx

:

= 700 sin 30° - 600 cos 20° = - 21 213. 3.8 8 N = 213 213.8 .8 N

+ c FRy = © Fy ;

FRy

;

= 700 cos 30° + 600 sin 20° = 81 811. 1.4 4 N  c

The magnitude of the resultant force FR is FR

=

 2 F

2 Rx

+ F2Ry =

 2 213.8 213.8

2

+ 811.42 = 839 N

Ans.

The direction angle u measured counterclockwise from the positive y axis is u

= tan - 1

 F Rx FRy

= tan - 1

¢ ≤ 213.8 811.4

= 14.8°

Ans.

2–44.

The magnitude of the resultant force acting on the bracket is to be 400 N. Determine the magnitude of F1 if f = 30° .

 y

u

F 2  650 N 4

5 3

45

 Rectangular Components: By referring to Fig. a, th the e x and y components of F1, F2, and F3 can be written as

(F1)x = F1 cos 30° = 0.8660F1

(F1)y = F1 sin 30° = 0.5F1

a 35 b = 390 N

(F2)y = 650

(F3)x = 500 cos 45° = 353.55 N

a 45 b = 520 N

(F3)y = 500 sin 45° = 353.55 N

 Resu ltan t Force Force:: Summing the force components algebraically along the x and y axes axes,, we have

+ © (F ) = © F R x x;

(FR)x = 0.8660F1 - 390 + 353.55

:

= 0.8660F1 - 36.45 + c © (FR)y = © Fy;

(FR)y = 0.5F1 + 520 - 353.55

= 0.5F1 + 166.45 Since the magnitude of the resultant force is FR = 400 N , we can write

 2 (F ) + (F ) 400 = 2 (0.8660 (0.8660F - 36.45) FR =

2 R x

2 R y

1

2

+ (0.5F1 + 166.45)2

F12 + 103.32F1 - 130967.17 = 0

Ans.

Solving, F1 = 314 N

f

45

SOLUTION

(F2)x = 650

F1

or

F1 = - 417 N

Ans.

The negative sign indicates that F1 = 417 N must act in the opposite sense to that shown in the figure.

F 3  500 N

 x

2–45.  y

If the resultant force acting on the bracket is to be directed along the positive u axis axis,, and the magnitu magnitude de of F1 is required to be minimum, determi determine ne the magnitudes magnitudes of the resultant force and F1.

u

F 2  650 N 4

5 3

45

 Rectangular Components: By referring to Figs. a and b, th the e  x and y components of  a nd FR can be written as F1, F2, F3, an

(F1)x = F1 cos f

(F1)y = F1 sin f

a 35 b = 390 N

(F2)y = 650

(F3)x = 500 cos 45° = 353.55 N

a 45 b = 520 N

(F3)y = 500 sin 45° = 353.55 N

(FR)x = FR cos 45° = 0.7071FR

(FR)y = FR sin 45° = 0.7071FR

 Resultant Force: Summing the force components algebraically along the x and axes,, we have y axes

+ © (F ) = © F ; R x x + c © (FR)y = © Fy;

:

0.7071FR = F1 cos f - 390 + 353.55

(1)

0.7071FR = F1 sin f + 520 - 353.55

(2)

Eliminating Eliminat ing FR from Eqs Eqs.. (1) and (2), yields F1 =

202.89 cos f - sin f

(3)

The first first derivative of Eq. (3) is dF1 df

=

sin f + cos f (cos f - sin f)2

(4)

The secon second d derivative of Eq. (3) is d2F1 2

df

For F1 to be mi ni m um,

=

dF1 df

2(sin f + cos f)2 (cos f - sin f)

3

+

1 cos f - sin f

(5)

Thus, f ro rom Eq. ( 4) 4) = 0 . Th sin f + cos f = 0 tan f = - 1 f = - 45°

Substituting Substi tuting f = - 45° into Eq. (5), yields d2F1 df2

= 0.7071 > 0

This shows that f = - 45° indeed produ produces ces minimum F1 . Thu Thus, s, from Eq. (3) F1 =

f

45

SOLUTION

(F2)x = 650

F1

202.89 = 143.47 N = 143 N cos ( - 45°) - sin ( - 45°)

Ans.

Substituting Substi tuting f = - 45° and F1 = 143.47 N into either Eq. (1) or Eq. (2), yields FR = 91.9 N

Ans.

F 3  500 N

 x

2–46.  y

If the magnitude of the resultant force a cting on the bracket is 600 N, direct directed ed along the positi positive ve u axis axis,, determ determine ine the magnitude of F and its direction f .

u

F 2  650 N 4

5 3

45

F1 f

45

SOLUTION  Rectangular Components: By referring to Figs. a and b, th the e x and y components of  a nd FR can be written as F1, F2, F3, an

(F1)x = F1 cos f (F2)x = 650

(F1)y = F1 sin f

a 35 b = 390 N

(F2)y = 650

(F3)x = 500 cos 45° = 353.55 N

a 45 b = 520 N

(F3)y = 500 cos 45° = 353.55 N

(FR)x = 600 cos 45° = 424.26 N

(FR)y = 600 sin 45° = 424.26 N

 Resultant Force: Summing the force components algebraically along the x and axes,, we have y axes

+ © (F ) = © F ; R x x

:

424.26 = F1 cos f - 390 + 353.55

(1)

F1 cos f = 460.71

+ c © (FR)y = © Fy;

424.26 = F1 sin f + 520 - 353.55

(2)

F1 sin f = 257.82

Solving Solvin g Eqs. Eqs. (1) and (2), (2), yields f = 29.2°

F1 = 528 N

Ans.

F 3  500 N

 x

2–47.

Determine the magnitude and direction u  of the resultant force FR. Express the result in terms of the magnitudes of  the components F1 and F2 and the angle f.

SOLUTION

F1

F 2R

=

F 21

+

F 22

- 2F1F2 cos (180° - f)

Since cos (180° - f) = - cos f, FR

=

 2 

F 21

f u

+

F 22

+ 2F1F2 cos f

Ans.

From the figure, tan u =

u

FR

F1 sin f F2

+ F1 cos f

¢

F1 sin f

= tan –1

F2

+ F1 cos f

≤

Ans.

F2

2–48.

If F1 = 600N and f = 30°, determine the magnitude of the resultant force acting on the eyebolt and its direction measured clockwise from the positive x axis.

y

F1

f

 x

SOLUTION  Rectangular Components: By referring to Fig. a, th the e  x and  y components of each force can be written as

60

(F1)x = 600 cos 30° = 519.62 N (F1)y = 600 sin 30° = 300 N (F2)x = 500 cos 60° = 250 N (F3)x = 450

5

(F2)y = 500 sin 60° = 433. 433.01 01 N

a 35 b = 270 N

(F3)y = 450

F 3

a 45 b = 363600 N

 Resultantt Force:  Resultan Force: Summing the force components algebraically along the x and y axes,

+ © (FR)x = ©Fx ;

:

(FR)x = 519.62 + 250 - 270 = 499 499.6 .62 2N

:

+ c © (FR)y = ©Fy ; (FR)y = 300 - 433.01 - 360 = - 493.01 N = 49 493. 3.01 01 N T The magnitude of the resultant force FR is FR

=

 2 (F )

2 R x

+ (FR)y2 =

 2 499.62 499.62

2

+ 493.012 = 70 701. 1.91 91 N = 702 N

Ans.

The direction angle u of FR, Fi Fig. g. b, measur measured ed clockwise clockwise from the x axis axis,, is u

= tan - 1

B R (FR)y (FR)x

= tan - 1

¢

493.01 499.62

≤

= 44.6°

4

3

Ans.

F 2 450 N

500 N

2–49.

If the magnitude of the resultant force acting on the eyebolt is 600 N and its direction measured clockwise from the positive  x axis is u = 30°, determine the magnitude of  F1 and the angle f .

 y

F1

f

SOLUTION  Rectangular Components: By referring to Figs. a and b, th the e x and y components of  F1, F2, F3, an and d FR can be written as

(F1)x = F1 cos f

(F1)y = F1 sin f

(F2)x = 500 cos 60° = 250 N

(F2)y = 500 sin 60° = 433 433.0 .01 1N

(F3)x = 450

a 35 b = 270 N

(F3)y = 450

(FR)x = 600 cos 30° = 519 519.6 .62 2N

5

+ © (FR)x = ©Fx ;

F 3

a 45 b = 360 N

(FR)y = 600 sin 30° = 300 N

519.62 = F1 cos f + 250 - 270 F1 cos f

= 539.62

(1)

+ c © (FR)y = © Fy ; - 300 = F1 sin f - 433.01 - 360 F1 sin f

= 493.01

(2)

Solving Solvin g Eqs. Eqs. (1) and (2), (2), yields f

= 42.4°

F1

= 731 N

4

3

 Resultant   Resultan t  Force:  Force: Summing the force components algebraically along the x and y axes, :

60

Ans.

F 2 450 N

500 N

2–50.

Determine the magnitude of F1 and its direction u so that the resultant force is directed vertically upward and has a magnitude of 800 N.

 y F1

u

600 N

400 N 5 3 4

30

SOLUTION

 x  A

 Scalar Notation: Summing the force components algebraically, we have

+ F = ©F ; Rx x

:

FRx

= 0 = F1 sin u + 400 cos 30° - 600 F1 sin u

+ c  FRy = © Fy ;

FRy

a 45 b

= 133.6

= 800 = F1 cos u + 400 sin 30° + 600 F1 cos u

= 240

(1)

a 53 b (2)

Solving Eqs. (1) and (2) yields u

= 29.1°

F1

= 275 N

Ans.

2–51.

Determine the magnitude and direction measured Determine counterclockwise from the positive  x axis of the resultant force of the three forces acting on the ring  A. Take F1 = 500 N and u = 20°.

 y F1

u

600 N

400 N 5 3 4

30

SOLUTION

 x  A

 Scalar Notation: Summing the force components algebraically, algebraically, we have

+ F = ©F ; Rx x

:

FRx

= 500 sin 20° + 400 cos 30° - 600 = 37 37.4 .42 2N

+ c FRy = ©Fy ;

FRy

a 45 b

:

= 500 cos 20° + 400 sin 30° + 600

a 35 b

= 1029.8 N c The magnitude of the resultant force FR is FR

=

 2 F

2 Rx

+ F 2Ry =

 2 37.42 37.42

2

+ 1029.82 = 1030.5 N = 1. 1.03 03 kN

Ans.

The direction angle u measured counterclockwise from positive x axis is u

= tan - 1

FRy FRx

= tan - 1

a 1029.8  b = 87.9° 37.42

Ans.

2–52.

Determine the magnitude of force F so that the resultant FR of the three three forces forces is as small as as possible possible.. What is the minimum magnitude of FR?

5 kN

SOLUTION 30 F

 Scalar Notation: Summing the force components algebraically, we have

+ FR = © Fx ; x

FRx

:

= 5 - 0.50F + c FRy = ©Fy ;

FRy

4 kN

= 5 - F sin 30° :

= F cos 30° - 4 = 0.8660F - 4  c

The magnitude of the resultant force FR is FR

 2 F + F = 2 (5 (5 - 0.50F) + (0.8660F - 4) = 2 F - 11.93F + 41 =

2 Rx

2 Ry

2

2

2

FR2

2FR

¢

(1)

= F2 - 11.93F + 41 dFR

= 2F - 11.93

dF

FR

d2FR 2

dF

+

dFR dF

*

dFR dF

(2)

≤

=1

In order to obtain the minimum resultant force FR, 2FR F

dFR dF

(3) dFR dF

= 0. Fr From Eq. (2 (2)

= 2F - 11.93 = 0

= 5. 5.96 964 4 kN = 5. 5.96 96 kN

Ans.

Substituting F = 5. 5.96 964 4 kN into Eq. (1), we have FR

=

 2 5.964 5.964

2

- 11.93(5.964) + 41

= 2.330 kN = 2. 2.33 33 kN Substituting FR = 2.330 kN with

B

dFR dF

(2.330) d2FR dF2

d2FR dF2

= 0 into Eq. (3), we have

R

+0 =1

= 0.429 7 0

Hence, F = 5.96 kN is indeed producing a minimum resultant force.

Ans.

2–53.

Determine the magnitude of force F so that the resultant force of the three forces is as small as possible. possible. What is the magnitude of the resultant force?

F

14 kN

30

45 8 kN

SOLUTION + F = ©F ; Rx x

FRz = 8 - F cos 45° - 14 cos 30°

:

= - 4.1244 - F cos 45° +  c FRy = © Fy ;

FRy = - F sin 45° + 14 sin 30°

= 7 - F sin 45° FR2 = ( - 4.1244 - F cos 45°)2 + (7 - F sin 45°)2

2FR

dFR dF

 coss 45°) + 2(7 - F sin 45°)( - sin 45°) = 0 = 2( - 4.1244 - F cos 45°)( -  co

F = 2.03 kN

From Fro m Eq. (1) (1);;

(1)

FR = 7.87 kN

Ans.  

Ans.

Also, from the figure require (FR)x¿ = 0 = © Fx¿;

F + 14 sin 15° - 8 cos 45° = 0 F = 2.03 kN

(FR)y¿ = © Fy¿ ;

Ans.

FR = 14 cos 15° - 8 sin 45° FR = 7.87 kN

Ans.

2–54.

Three for Three forces ces act on the bra bracke cket.Deter t.Determin mine e the mag magnit nitude ude and dirrec di ecti tion on u of F1 so th tha at th the e res esul ulta tan nt fo forrce isdi isdirrec ecte ted d al alon ong g th the e posi po siti tiv ve x ¿ ax axisan isand d ha hass a ma magn gnit itud ude e of 1 kN kN..

 y

450 45 0N

F 2

45

F 3

200 20 0N  x

SOLUTION 30

+ FRx = ©Fx ;

1000 cos 30° = 200 + 450 cos 45° + F1 cos(u + 30°)

+ c FRy = ©Fy ;

- 1000 sin 30° = 450 sin 45° - F1 sin(u + 30°)

:

F1 sin(u

+ 30°) = 818.198

F1 cos(u

+ 30°) = 347.827

u

+ 30° = 66.97°,

F1

= 88 889 9N

u

= 37.0°

u

F1

Ans. Ans.

 x¿

2–55.

If  F1 = 300 00 N  N and u = 20 20°°, dete etermin rmine e the mag agni nittud ude e and direct dir ectiion, meas easur ure ed co coun unte terrcloc ockwi kwise se fr fro om the x ¿ axi xis, s, of  the r  res esul ulta tan nt f orce of  f tthe thr hree ee f   f orce cess ac acttin ing g on the br  brac ack ket et..

 y

F 2   450 450 N  N

45

SOLUTION

F 3   200 200 N  N  x

30

+ F = ©F ; Rx x

FRx = 300 co coss 50° 50° + 200 + 450 cos 45 45°° = 711. 11.0 03 N

+ c FRy = © Fy ;

FRy = - 300 sin in 50  50°° + 450 sin in 45  45°° = 88 88.3 .38 8 N

:

FR = 2  (711. 11.0 03) + (88 88.3 .38 8) 2

f ¿ (angle fr fro om x axi xiss) = ta tan n-1

B

88.38 88.3 8 711. 11.0 03

2

= 717 N

u

F1

Ans.

R

f ¿ = 7.1 .10 0° f  (  (a angle fr fro om x ¿ axi xiss) = 30° + 7.1 .10 0° f = 37.1°

Ans.

 x¿

.

.

.

.

2–58.

If the magnitude of the resultant force acting on the bracket is to be 450 N directed along along the positive positive u axis axis,, determi determine ne the magnitude of F1 and its direction f.

 y u

F1

f 30

 x

SOLUTION

F 2

 Rectangular Components: By referr referring ing to Fig. Fig. a, th the e  x and  y components of F1, F2, F3, an and d FR can be written as

(F1)x = F1 sin f

(F1)y = F1 cos f

(F2)x = 200  N

( F2)y = 0

(F3)x = 260

¢ ≤ 5 13

= 100 N

(F3)y = 260

(FR)x = 450 cos 30° = 389.71 N

¢ ≤ 12 13

(FR)y = 450 sin 30° = 225 N

 Resultantt Force:  Resultan Force: Summing the force components algebraically along the x and  y axes,

+ © (F ) = © F ; 389.71 = F sin f + 200 + 100 R x x 1

:

F1 sin f = 89.71

(1)

+ c © (FR)y = © Fy; 225 = F1 cos f - 240 F1 cos f = 465

(2)

Solving Eqs Eqs.. (1) and (2), (2), yields f = 10.9°

474 4N F1 = 47

13

12 5

F 3

240 0N = 24

Ans.

200 N

260 N

2–59.

If the resultant force acting on the bracket is required to be a min minimum imum,, det determ ermine ine the mag magnitu nitudes des of F1 and the resultant result ant force force.. Set f = 30°.

 y u

F1

f 30

 x

SOLUTION

F 2

 Rectangular Components: By referring to Fig.  a, th the e  x and  y components of F1, F2, and F3 can be written as

(F1)x = F1 sin 30° = 0.5F1

(F1)y = F1 cos 30° = 0.8660F1

(F2)x = 200 N

(F2)y = 0

(F3)x = 260

a 135 b = 100 N

(F3)y = 260

 Resultant  Resu ltant Force: Summing the force components algebraically along the x and  y axes,

+ © (F ) = © F ; (F ) = 0.5F + 200 + 100 = 0.5F + 300 R x x R x 1 1

:

+ c © (FR)y = © Fy; (FR)y = 0.8660F1 - 240 The magnitude of the resultant force FR is FR = 2 (FR)x2 + (FR)y2

(0.5F1 + 300)2 + (0.8660F1 - 240)2 = 2 (0.5

= 2 F 21 - 115.69F1 + 147 600

(1)

Thus, FR2 = F 21 - 115.69F1 + 14 147 7 60 600 0

(2)

The first derivative derivative of Eq. (2) is 2FR For FR to be minimum,

dFR dF1

2FR

dFR dF1

= 2F1 - 115.69

(3)

Thus, fr f rom Eq. (3 ( 3) = 0 . Th

dFR dF1

= 2F1 - 115.69 = 0

F1 = 57 57.8 .846 46 N = 57.8 N

Ans.

from Eq. Eq. (1), FR = 2 (57.846) (57.846)2 - 115.69(57.846) + 147 600 = 380 N

13

12 5

F 3

a 1213 b = 242400 N

Ans.

200 N

260 N

2–60.

The stock mounted on the lathe is subjected to a force of  60 N. Determi Determine ne the coord coordinate inate direct direction ion angle b and express the force as a Cartesian vector. z

60 N



45

SOLUTION 1 =

 2 cos cos

2

b

a + cos2 b + cos2 g



60

1 = cos2 60° + cos2 b + cos2 45° cos b = ; 0.5

 y  x

b = 60°, 120°

Use b = 120°

Ans.

F = 60 N(cos 60°i + cos 120° j + cos 45°k)

= {30i - 30 j + 42.4k} N

Ans.

2–61. Determine the coordinate angle g for F2 and then express each force acting on the bracket as a Cartesian vector.

z

F 1  450 N

45 30 45

60

 x F 2  600 N

 y

2–62. Determine the magnitude and coordinate direction angles of the resultant force acting on the bracket.

z

F 1  450 N

45 30 45

60

 x F 2  600 N

 y

2–63.

The bolt is subjected to the force F, which has components components acting along the  x, y, z axes as shown.If shown. If the magnitude of F is 80 N, and a 60° and g 45°, determi determine ne the magnitu magnitudes des of its components. =

z

=

Fz

g

F F y

 y

a

SOLUTION

cos b

=

=

2 1 2 1

-

-

cos2 a

-

cos2 60°

F x

cos2g -

cos2 45°  x

b

=

120°

Fx

=

|80 cos 60°|

Fy

=

|80 cos 120°|

Fz

=

|80 cos 45°|

=

40 N 40 N

Ans.

56.6 56 .6 N

Ans.

=

=

Ans.

b

 2-64

.

Determine the magnitude and coordinate direction angles of the force F acting on the stake.

Given:  F h  40 N    70 deg c  3 d   4

Solution:

 F  F h

 

c  d   

 

2

d 

2

 

 F   50 N

Ans.

 F  x  F h cos   

F  y  F h sin   

F  z  

 F  x  13.7 N

F  y  37.6 N

F  z   30 N

   acos

  F  x    F  

   74.1 74.1 deg

   acos

  F  y    F  

   41.3 41.3 deg

   

   acos

c 2

  F 

2

c  d   

  F  z     F  

   53.1 53.1 deg

Ans.

2-6 5

.

Determine the magnitude and coordinate direction angles of the force F acting on the stake.

Given: F h := 40N

:= 50deg c := 3 d := 4 Solution:

2

F := F h⋅

c +d

2

d

F = 50 N

Ans.

F x := F h⋅ cos ( )

F y := Fh⋅ sin ( )

c

F z :=

2

c +d F x = 25.7 N

:= acos

F y = 30.6 N

⎛ Fx ⎞

:= acos

⎝  F  ⎠

= 59.1 deg

Ans.

F z = 30 N

⎛ Fy ⎞

:= acos

⎝  F  ⎠

= 52.2 deg

⋅F 2

Ans.

⎛ Fz ⎞ ⎝  F  ⎠

= 53.1 deg

Ans.

2–66.

Determine the magnitude and coordinate direction angles of  F1 = 60i - 50 j + 40k  N and F2 = - 40i - 85 j + 30k  N. Sketch each f orce on an x, y, z ref erence frame.

5

6

5

6

SOLUTION

60 i

F1

=

F1

=

a1

=

b1

=

g1

=

F2

= -

F2

=

a2

=

b2

=

g2

=

-

50 j

+

40 k

1 2  1 502  1402 60 b 46 9 cos a 87 7496 50 cos a b 25 87 7496 40 cos a b 62 9 87 7496 2  60

2

+

2

-

-1

=

.

-

-1

=

.

-1

40 i

85 j

-

+

=

-1

2

+

-

. 1

-1

-

. 1

cos-1

30 98.615

=

87.7 N

Ans. Ans.

°

Ans.

. °

Ans.

30 k

1 2  1 852  1302 40 cos a b 4 98 6 5 85 cos a b 50 98 6 5 2  - 40

87.7496

. °

1

=

.

2

+

2

-

+

2

=

98.615

=

98.6 N

Ans.

=

11 °

Ans.

=

1

°

Ans.

=

72.3°

Ans.

Express each force in Cartesian vector form.

2–67.

( F1 )

 x

( F1 ) ( F1 )

 y

=

0

( F2 )

x

=

2 kN

( F2 )

y

5 cos 60

=

5 cos 45

( F2 ) = 5 cos 60 Thus, F1 = 0( i ) + 2 ( − j) + 0 ( k ) = { −2 j} N F2  = 2.5 ( i )  + 3.54 ( j) + 2.5 ( k ) = {2.5i +   3.54 j + 2.5k } N  z

=

0

z 



=





z

=

2.5 kN

=

3.54 kN

=

F 2  2 kN

45

 y 60

2.5 kN

Ans.

Ans.

F 1  5 kN

60

 x

2–68. z

Express each f orce as a Cartesian vector.

30 30

 x

SOLUTION  Rectangular Components: By ref erring to Figs. a and b, the x, y, and z components of  F1 and F2 can be written as

(F1)x = 300 cos 30° = 259.8 N

(F2)x = 500 cos 45° sin 30° = 176.78 N

(F1)y = 0

(F2)y = 500 cos 45 ° cos 30° = 306.19 N

(F1)t = 300 sin 30° = 150 N

(F2)z = 500 sin 45° = 353.55 N

F1 = 259.81( + i) + 0 j + 150( - k) Ans.

F2 = 176.78( + i) + 306.19( + j) + 353.55( - k)

= 2{177i + 306 j - 354k} N

45

F 2  500 N

Thus, F1 and F2 can be written in Cartesian vector f orm as

= {260i - 150k} N

F 1  300 N

Ans.

y

2–69.

Determine t he magnitude a nd coo rdinate direction angles of  t  the resultant f orce acting on the hook.

z

30 30

 x

SOLUTION  Force Vectors: Vectors: By resolving F1 and F2 into their ir x  x, y, and z components ts,, as shown in Figs. a and b, respectively, F1 and F2 can be expessed in Cartesian vector f orm as

F 1  300 N

45

F1 = 300 cos 30°( + i) + 0 j + 300 sin 30°( - k) F 2  500 N

= {259.81i - 150k} N F2 = 500 cos 45°sin 30°( + i) + 500 cos 45° cos 30°( + j) + 500 sin 45°( - k)

= {176.78i - 306.19 j - 353.55k} N  Resultant Force:  Resultant Force: The resultant f orce acting on the hook can be obtained by vectorally adding F1 and F2.Thus,

FR = F1 + F2

= (259.81i - 150k) + (176.78i + 306.19 j - 353.55k) = {436.58i) + 306.19 j - 503.55k} N The magnitude of  FR is

 2 (F ) + (F ) (F ) (436.58) + (306.19) + ( - 503.55) =  2 (436

FR =

2

R x

R y

2

R z

2

2

2

2

= 733.43 N = 733 N

Ans.

The coordinate direction angles of  FR are ux = cos-1 uy = cos-1 uz = cos-1

58 c ( ) d = cos a 436 b = 53 5 733 43 9 c ( ) d = cos a 306 b = 65 3 733 43 c ( ) d = cos a -7335034355 b = 33 FR x FR

FR

y

-1

-1

FR

FR

FR

z

. .

. °

Ans.

.1 .

. °

Ans.

.

-1

.

1

°

Ans.

y

. . . .

2–71. If  t  the resultant f orce acting on the bracket is directed along the positive y axis, determine the magnitude of  t  the resultant f orce and the coordinate direction angles of  F so that b 6 90°.

z

g

F 

500 N

SOLUTION ir x  x, y, and z components  Force Vectors: By resolving F1 and F  in to their ts,, as shown in Figs. a and b, respectively, F1 and F can be expressed in Cartesian vector f orm as

F1

=

600 cos 30 ° sin 30°( + i)

=

{259.81i

+

450 j

600 cos 30° cos 30°( + j)

+

+

b a

600 sin 30°( - k)

30

300k} N

-

 y  x

F

500 cos ai

=

+

5000 cos b j 50

+

30

5000 cos gk 50 F 1

Since the resultant f orce FR i s directed towards the positive y axis, then FR

600 N

FR j

=

 Resultant Force: FR

=

F1

+

F

FR j

=

(259 .81i

FR j

=

(259 .81

+ +

450 j

300k)

-

500 cos a)i

+

+

(500 cos a i

(450

+

5000 cos b j 50

5000 cos b ) j 50

+

+

+

5000 cos g k) 50

(5000 cos g (50

-

300) k

Equating the i, j, and k components, 0

=

259.81

a

=

121.31°

FR

However, since cos2 a cos b

= ;

If w If  we substitute cos b

=

450

+

500 cos g

g

=

53.13°

-

121°

Ans.

5000 cos b 50

=

+

5000 cos a 50

=

0

cos2 b

+

2 1

=

+

=

300

-

53.1°

cos2 g

cos2 121.31°

(1)

1, a

=

-

Ans. =

121.31°, and g

cos2 53.13°

= ;

=

53.13°,

0.6083

0.6083 into Eq. (1), 2

FR

=

450

+

500(0.6083)

=

754 N

Ans. 2

and b

=

cos - 1 (0.6083)

=

52.5°

Ans.

2

2

2-72

.

Specify the magnitude  F 3 and directions  3,   3, and  3 of F3 so that the resultant force of the three forces is FR . Units Used: 3

kN  10  N Given:  F 1  12 kN

c  5

 F 2  10 kN

d   12

   30 deg

 0  FR  

9

   0 

kN

Solution: F 3x  1 kN

Initial Guesses:

Given

  F 3x   F 3y  

F 3y  1 kN

  F 3x    0   FR    F 3y    F 1 cos          sin       F 3z   Find  F 3x  F 3y  F 3z 

F3

  F 3z  

  3  

 d  

 F 2 2

2

c  d 

acos

 

F3

 

 

F3

 

  3 

0



  c  

  F 3x    F 3y    F 3z  

  3    3  

F 3z   1 kN

  9.2   F3

 1.4 kN

  2.2    

F3

 9.6 9.6 kN

Ans.

 15.5 

  3    98.4  deg   3    77.0 

Ans.

 2-73

.

Determine the magnitude and coordinate direction angles of  F3 so that the resultant of the three forces acts along the positive y positive  y axis  axis and has magnitude F  magnitude  F . Given:  F   600    F 1  180 N  F 2  300 N  1  30 deg  2  40 deg Solution: Initial guesses:    40 deg     50 deg

   50 deg F 3  45 N

Given  F  Rx =  F  x;

0   F 1  F 2 cos   1  sin   2   F 3 cos   

 F  Ry =  F  y;

F  F 2 cos   1  cos   2   F 3 cos   

 F  Rz  =  F  z ;

0   F 2 sin   1   F 3 cos    2

2

2

cos     cos     cos     1

  F 3                     

Find F 3         

F 3  428 N

88.3                20.6  deg      69.5   

Ans.

 2-74

.

Determine the magnitude and coordinate direction angles of F3 so that the resultant of the three forces is zero. Given:  F 1  180 N

 1  30 deg

 F 2  300 N

 2  40 deg

Solution: Initial guesses:

   40 deg

   50 deg

    50 deg

F 3  45 N

Given  F  Rx =  F  x;

0  F 1  F 2 cos   1  sin   2   F 3 cos   

 F  Ry =  F  y;

0  F 2 cos   1  cos   2   F 3 cos   

 F  Rz  =  F  z ;

0  F 2 sin   1   F 3 cos    2

2

2

cos     cos     cos     1

  F 3                 

Find  F 3         

F 3  250 N

      87.0         142.9  deg       53.1  

Ans.

2–75. z

Determine the coordinate direction angles of  f f  f orce F1. F 1  600 N

5 3

F 2  450 N

4

45 30

SOLUTION

 x

 Rectangular Components: By ref erring to Figs. a, the x, y, and z components of  F1 can be written as

(F1)x = 600

a 45 b  cos 30

°N

(F1)y = 600

a 45 b  s

in 30° N

(F1)z = 600

a 35 b

N

Thus, F1 expressed in Cartesian vector f orm can be written as

e 45 cos 30 ( + ) +

f

4 3 sin 30°( - j) +  ( + k) N 5 5 = 600[0.6928i - 0.4 j + 0.6k] N

F1 = 600

°

i

Theref ore, the unit vector f or F1 is given by

uF1 =

F1 F1

=

600(0.6928i - 0.4 j + 0.6k = 0.6928i - 0.4 j + 0.6k 600

The coordinate direction angles of  F1 are a = cos-1(uF1)x = cos-1(0.6928) = 46.1°

Ans.

b = cos-1(uF1)y = cos-1( - 0.4) = 114°

Ans.

g = cos-1(uF1)z = cos-1(0.6) = 53.1°

Ans.

 y

2–76. z

Determine t he magnitude a nd coordinate direction a ngles of  t  the resultant f orce acting on the eyebolt. F 1  600 N

5 3

F 2  450 N

4

30  x

SOLUTION ir x  x, y, and z components  Force Vectors: Vectors: By resolving F1 and F2 into their ts,, as shown in Figs. a and b, respectively, they are expressed in Cartesian vector f orm as

a 45 b cos 30 ( + ) + 600 a 45 b s = 54 5 69 - 240 + 360 6

F1 = 600

°

1 .

i

in 30°( - j) + 600

i

 j

a 35 b ( + ) k

k N

F2 = 0i + 450 cos 45 °( + j) + 450 sin 45°( + k)

 5

6

= 318.20 j + 318.20k N  Resultant Force: The resultant f orce acting on the eyebolt can be obtained by vectorally adding F1 and F2.Thus, FR = F 1 + F 2

= (415.69i - 240 j + 360k) + (318.20 j + 318.20k)

 5

6

= 415.69i + 78.20 j + 678.20k N The magnitude of  FR is given by

 3 (F ) + (F ) + (F ) =  3 (4 (415.69) + (78.20) + (678.20)

FR =

R x

2

R y

2

2

R z

2

2

2

= 799.29 N = 799 N

Ans.

The coordinate direction angles of  FR are a = cos-1

b = cos-1

g = cos-1

c ( ) d = cos a 47995 6929 b = 58 7 78 20 c ( ) d = cos a 799 b = 84 4 29 20 c ( ) d = cos a 678 b = 32 0 799 29 FR

x

-1

y

-1

FR

FR

FR

FR

FR

1 . . .

.

z

-1

. .

. °

Ans.

. °

Ans.

. °

Ans.

45  y

2–77. The cables attached to the screw  eye a re subjected to the three f orces shown. Express each f orce in Ca rtesian vector f orm and determine the magnitude and coordinate direction angles of  t  the resultant f orce.

 z F 1 = 350 N

40

°

F 3 = 250 N

60

°

120

°

45

°

 y 60

°

SOLUTION Cartesian Vector Notation: F1

5

350 sin 40° j

=

.  j

 j

=

F2

F3

45

°

6

F 2 = 100 N

 5224 98 268 2 6  5225 268 6 005cos 45 cos 60 cos 20 6  570 7 50 0 50 0 6  570 7 50 0 50 0 6 2505cos 60 cos 35 cos 60 6  5 25 0 76 78 25 0 6  5 25 77 25 6

=

k  N

+

°i

1

=

.1 k  N

+

=

. 1i

=

. i

.  j

+

.  j

+

°i

= =

1

. i

=

1

i

. k  N

-

+

° j

1

.  j

1  j

°k  N

1

+

Ans.

. k  N

-

+

1

-

-

° j

+

+

1

Ans.

°k  N

+

. k  N

k  N

1

Ans.

 Resultant Force: FR

=

F1

+

F2

 5170 7  5 95 7

. 1

= =

1

. 1i

+

F3

+

125.0 i

+

98.20 j

+

+

2  1224 98 .

+

50.0

-

2  1268 2

 j 176.78  j

6

+

.1

-

50.0

+

26

125.0 k  N

343.12k  N

The magnitude of  t  the resultant f orce is FR

=

2 F2Rx

=

2 195.712

=

407.03 N

+

F2Ry +

=

+

F2Rz

98.202

+

343.122

407 N

Ans.

The coordinate direction angles are

cos a

=

cos b

=

cos g

=

°

 x 

cos 40°k  N

+

60

FRx FR FRy FR FRz FR

=

195.71 407.03

a

=

61.3°

Ans.

=

98.20 407.03

b

=

76.0°

Ans.

=

343.12 407.03

g

=

32.5°

Ans.

2–78. Three f orces act on the ring. If  t  the resultant f orce FR  h as a magnitude and direction as shown, determine the magnitude and the coordinate direction angles of  f f  f orce F3.

z

F3

F R

110 N

F 2

120 N

SOLUTION Cartesian Vector Notation: FR

F1

F 1

=

120{cos 45°sin 30°i

+

=

{42.43i

+

73.48 j

84.85k} N

=

80

b

+

3  k  N 5

4  i 5

+

r

F2

=

{ - 110k} N

F3

=

{F3x i

cos 45°cos 30° j

{64.0i

=

+

F3z k} N

sin 45°k} N

=

F1

{42.43i

+

+

F2

+

73.48 j

48.0k} N

84.85k}

=

E A 64.0

F3 +

+

F3 x B i

+

F3 y j

+

A 48.0

-

110

+

F3 z B k F

Equating i, j and k components, we have 64.0

48.0

F3 x

+

=

110

-

42.43

F3 z

+

84.85

=

21.57 N

F3x

= -

F3 y

=

73.48 N

F3 z

=

146.85 N

The magnitude of  f f  f orce F3 i s F3

=

2 F 23 x + F 23 y

+

F 23 z

=

2 ( - 21.57)2

+

73.482

=

165.62 N

=

+

146.852

166 N

Ans.

The coordinate direction angles f or F3 are

cos a

=

cos b

=

cos g

=

F3 x F3 F3 y F3 =

=

=

F3 z F3

21.57 165.62

a

=

97.5°

Ans.

73.48 165.62

b

=

63.7°

Ans.

g

=

27.5°

Ans.

-

=

146.85 165.62

45

 y 4

30

 Resultant Force: FR

5 3

 x

F3y j

+

+

+

80  N

2–79.

Determine the coordinate direction angles of  F1 and FR.

z

F3

F R

110 N

F 2

120 N

SOLUTION Unit Vector of  F1 and F R : uF1

=

uR

=

cos 45° sin 30°i

=

0.3536i

+

80  N

5 3

4  i 5

+

3  k 5

F 1 =

0.8i

+

+

0.6124 j

 y 4

30

0.6k

cos 45° cos 30° j +

45

+

0.7071k

sin 45°k  x

hus, the coordinate direction angles F1 and FR are

cos aF1

=

0.8

aF1

=

36.9°

Ans.

cos b F1

=

0

b F1

=

90.0°

Ans.

cos gF1

=

0.6

gF1

=

53.1°

Ans.

cos aR

=

0.3536

aR

=

69.3°

Ans.

cos b R

=

0.6124

bR

=

52.2°

Ans.

cos gR

=

0.7071

gR

45.0°

Ans.

=

2–80. The mast is subjected to the three f orces shown. Determine the coordinate direction angles a1, b 1, g1 of  F1 so that the resultant f orce acting on the mast is

z F1

 5350 6

FR =

i N.

g1 a1 b1

F 3  300 N  y  x

. . .

F 2  200 N

The mast is subjected to the three forces shown. Determine the coordinate direction angles a1, b 1, g1 of  F so that the resultant force acting on the mast is zero. 1 2–81.

z F1

g1 a1 b1

F 3  300 N  y  x

. . .

F 2  200 N

the magnitude and coordinate 2–82. Determine direction angles of F2 so that the resultant of the two forces acts along the positive  x axis and has a magnitude of 500 N.

z

F2

g2 b2 a2

 y 

60 

15

 x F 1  180 N

. . . .

2–83. Determine the magnitude and coordinate direction angles of F2 so that the resultant of the two forces is zero.

z

F2

g2 b2 a2

 y 

60 

15

 x F 1  180 N

. . . .

2–84.

The pole is subjected to the force F, which has component componentss acting along the  x, y, z axes as shown. If the magnitude of F is 3 kN, b = 30°, a nd nd g = 75°, d et etermine the magnitudes of   its three components.

z

Fz

F

g

SOLUTION

b 2

cos a 2

cos a a

=

+ +

2

cos b

+

2

cos 30°

2

cos g

=

 y

1

2

+

cos 75°

=

1.28 1. 28 kN

=

F y

a

1

F x

64.67°

Fx

=

3 cos 64.6 64.67° 7°

Fy

=

3 co coss 30° 30°

=

2.60 kN

Ans.

Fz

=

3 co coss 75° 75°

=

0.77 0. 776 6 kN

Ans.

Ans.

 x

2–85.

The pole is subjected to the force F which has components Fx = 1.5 kN and Fz = 1.25 kN. If b = 75°, de deter termin mine e the magnitudes of F and F y.

z

Fz

F

g

SOLUTION

b 2

cos a

a 1.5 b F

F Fy

= =

+

2

cos b

2 +

2

cos g

+

cos2 75°

+

=

a

2

F

2.02 2. 02 kN 2.02 2.0 2 cos 75° 75°

 y

1

 a 1.25 b

=

1

F x

Ans. =

0.52 0. 523 3 kN

F y

Ans.

 x

.

2–87.

Determine t he lengths o f  wires  AD, BD, and CD. The ring at D  i s  midway between  A and B.

z

C 

D

 A

0.5  m

SOLUTION 2  m

0.5  m

D

a 2 2 0, 0 2 2,

rAD

+

=

+

(1

-

2)i

= - 1i +

 1. 5

(1

-

(1

-

0)i

=

1i

-

1 j

=

(1

-

0)i

=

1i

+

1 j

rAD

=

2 (-1)2

rBD

=

2 12

+

( - 1)2

rCD

=

2 12

+

12

rCD

+

+

+

0) j

-

 m

=

(1

+

(1

-

2  m

D(1, 1, 1) m

-

1.5  m

1.5)k  x

2) j

+

(1

-

0.5)k

0) j

+

(1

-

2)k

0.5k

+

-

b

0.5k

=

rBD

0.5

+

2

+

1 j

B

(1

-

1k 12

+

+ +

( - 0.5)2 0.52

( - 1)2

=

=

=

1.50  m

1.50 m

1.73 m

Ans. Ans. Ans.

 y

2–88.  y

Determine the length of  member AB of  the truss by first establishing a Cartesian position  vector from A to B and then determining its magnitude.

B  A

1.5 m 1.2 m

SOLUTION rAB

40

1.5 = (1.1) = - 0.80)i + (1.5 - 1.2) j tan 40°

 x O

 2 (2(2.09)

2

+ (0.3)2 = 2.11 m

0.8 m 0.3 m

rAB = {2.09i + 0.3 j} m rAB =

C 

Ans.

2–89.

 5

6

z

If  F = 350i - 250 j - 450k N and ca ble AB is 9 m long, determine the x, y, z coordinates of  f p point A.

 A

F

B

SOLUTION

 x

 Position  Posi tion Vector: The position vector rAB, directed from point A to point B, is given by rAB = [0 - ( - x)]i + (0 - y) j + (0 - z)k

= xi - y j - zk Unit Vector: Knowing the magnitude of  rAB is 9 m, the unit vector f or rAB is given by uAB =

rAB rAB

=

xi - y j - zk

9

The unit vector f or f orce F is

uF =

F = F

350i - 250 j - 450k 3

3502 + ( - 250)2 + ( - 450)2

= 0.5623i - 0.4016 j - 0.7229k

Since f orce F is also directed from point A to point B, then uAB = uF xi - y j - zk

9

z

= 0.5623i - 0.4016 j - 0.7229k

Equating the i, j, and k components, x = 0.5623 9

x = 5.06 m

Ans.

-y = - 0.4016 9

y = 3.61 m

Ans.

-z = 0.7229 9

z = 6.51 m

Ans.

x

 y

 y

2–90. z

Express FB and FC in Cartesian vector f orm.

C 

B

2m 3.5 m

F B  600 N

SOLUTION

0.5 m

 Force Vectors: Vectors:  Th e unit vectors uB and uC of  FB and FC must be determined first. From Fig. a uB =

rB rB

=

3 ( - 1.5 - 0.5)

2

rC rC

=

1.5 m  x

+ [ - 2.5 - ( - 1.5)]2 + (2 - 0)2

1 2 2 i -  j + k 3 3 3

 j + (3.5 - 0)k ( - 1.5 - 0.5)i + [0.5 - ( - 1.5)] j

3 ( - 1.5 - 0.5)

2

= -

+ [0.5 - ( - 1.5)]2 + (3.5 - 0)2

4 4 7 i +  j + k 9 9 9

Thus, the f orce vectors FB and FC are given by

a 4 = 450 a 9

2 2 1 FB = FB uB = 600 - i -  j + k 3 3 3 FC = FC uC

4 7 i +  j + k 9 9

b = 5- 400 - 200 b = 5 - 200 + 200

6 50 6

i

 j + 400k N

Ans.

i

 j + 3

Ans.

k N

F C   450 N

1.5 m

1m

 j + (2 - 0)k ( - 1.5 - 0.5)i + [ - 2.5 - ( - 1.5)] j

= uC =

 A

0.5 m  y

2–91. z

Determine the magnitude and coo rdinate direction angles of  t  the resultant f orce acting at A.

C 

B

2m 3.5 m

F B  600 N

SOLUTION

0.5 m  A

 Force Vectors: Vectors:  The unit vectors uB and uC of  FB and FC must be determined first. From Fig. a rB

uB =

rB

=

3 ( - 1.5 - 0.5)

2

= rC

uC =

rC

=

1.5 m  x

+ [ - 2.5 - ( - 1.5)]2 + (2 - 0)2

1 2 2 i -  j + k 3 3 3

 j + (3.5 - 0)k ( - 1.5 - 0.5)i + [0.5 - ( - 1.5)] j

3 ( - 1.5 - 0.5)

2

= -

+ [0.5 - ( - 1.5)]2 + (3.5 - 0)2

4 4 7 i +  j + k 9 9 9

Thus, the f orce vectors FB and FC are given by

a 4 = 450 a 9

2 2 1 FB = FB uB = 600 - i -  j + k 3 3 3 FC = FC uC

4 7 i +  j + k 9 9

b = 5- 400 - 200 b = 5- 200 + 200

6 50 6

i

 j + 400k N

i

 j + 3

k N

 Resultant Force: FR = FB + FC = ( - 400i - 200 j + 400k) + ( - 200i + 200 j + 350k)

 5- 600 + 750 6

=

i

k N

The magnitude of  FR is

 3 (F ) + (F ) + (F ) = 3 ( - 600) + 0 + 750 = 960.47 N = 960 N

FR =

R x

2

R y

2

2

2

R z

2

2

The coordinate direction angles of  FR are a = cos-1

c ( ) d = cos a 960- 60047 b = 29 c ( ) d = cos a 9600 47 b = 90 c ( ) d = cos a 96076047 b = 8 7

b = cos-1

g = cos-1

FR

x

-1

FR

FR

.

y

-1

FR

FR

FR

.

z

-1

.

1

°

°

3 . °

1.5 m

1m

 j + (2 - 0)k ( - 1.5 - 0.5)i + [ - 2.5 - ( - 1.5)] j

Ans.

Ans.

Ans.

F C   450 N

0. 5 m  y

2–92. If  FB = 560 N and FC = 700 N,  d etermine the magnitude and coordinate direction angles of  t  the resultant f orce acting on the flag pole.

z

 A 6m

FB FC 

SOLUTION  Force Vectors: Vectors:  The unit vectors uB and uC of  FB and FC must be determined first. From Fig. a rB

uB =

rB

rC

uC =

rC

(2 - 0)i + ( - 3 - 0) j + (0 - 6)k 2

2

2

=

2

2

3m 3m 2m

 x

2

Thus, the f orce vectors FB and FC are given by

a 27 = 700 a 7

FB = FB uB = 560

FC = FC uC

3

i -

6 3  j - k 7 7

2 6 i +  j - k 7 7

b = 5 60 - 240 - 480 6 b = 5 00 + 200 - 600 6 1

i

 j

k N

3

i

 j

k N

 Resultant Force: FR = FB + FC = (160i - 240 j - 480k) + (300i + 200 j - 600k)

 5

6

= 460i - 40 j + 1080k N The magnitude of  FR is

 3 (F ) (460) =  3 (460)

FR =

R x

2

2

+ (FR)y 2 + (FR)z 2 + ( - 40)2 + ( - 1080)2 = 1174.56 N = 1.17 kN

Ans.

The coordinate direction angles of  FR are a = cos-1

c ( ) d = cos a 460 b = 66 9 74 56 c ( ) d = cos a -744056 b = 92 0 c ( ) d = cos a -7408056 b = 57

b = cos-1

g = cos-1

FR

x

-1

FR

FR

y

FR

z

.

11

.

-1

FR

FR

11

1

-1

11

.

1

B

3 2 6 i -  j - k 7 7 7

3 (2(2 - 0) + ( - 3 - 0) + (0 - 6) (3 - 0)i + (2 - 0) j + (0 - 6)k 3 2 6 = = i +  j - k 3 (3 - 0) + (2 - 0) + (0 - 6) 7 7 7 =

2m

. °

Ans.

. °

Ans.

°

Ans.

C 

 y

2–93. z

If FB = 700 N, and FC = 560 N, determ determine ine the magnitu magnitude de and coordinate direction angles of the resultant force acting on the flag pole.

 A 6m

FB FC 

SOLUTION  Force Vectors: Vectors: The unit vectors uB and uC of FB and FC must be determined first. From Fig. a rB

uB =

rB

rC

uC =

rC

=

(2 - 0)i + ( - 3 - 0) j + (0 - 6)k

3 (2(2 - 0)

2

=

+ ( - 3 - 0)2 + (0 - 6)2

(3 - 0)i + (2 - 0) j + (0 - 6)k

3 (3(3 - 0)

2

=

+ (2 - 0)2 + (0 - 6)2

=

a 27 3 = 560 a 7

3 6  j - k 7 7

i -

FC = FC uC

2 6 i +  j - k 7 7

3m 3m 2m

 x

3 2 6 i +  j - k 7 7 7

b = 5200 - 300 - 600 6 N b = 5240 + 160 - 480 6 N i

 j

k

i

 j

k

 Resultant Force: FR = FB + FC = (200i - 300 j - 600k) + (240i + 160 j - 480k)

 5

6

= 440i - 140 j - 1080k  N The magnitude of FR is

 3 (F ) (440) = 3 (440)

FR =

R x

2

2

+ (FR)y 2 + (FR)z 2 + ( - 140)2 + ( - 1080)2 = 1174.56 N = 1.17 kN

Ans.

The coordinate direction angles of FR are a = cos-1

440 c ( ) d = cos a 1174.56 b = 68.0° - 140 c ( ) d = cos a 1174.56 b = 96.8° - 1080 c ( ) d = cos a 1174.56 b = 157°

b = cos-1

g = cos-1

FR

x

-1

Ans.

y

-1

Ans.

z

-1

Ans.

FR

FR

FR

FR

FR

B

2 3 6 i -  j - k 7 7 7

Thus,, the force vectors FB and FC are given by Thus FB = FB uB = 700

2m

C 

 y

2–94.

The tower is held in place by three cables. cables. If the force of  each cable cable acting on the tower is shown, shown, determ determine ine the magnitude and coordinate direction angles a,  b ,  g of the resultant force. Take x = 1 5 m , y = 2 0 m .

z

D 600 N 400 N

8 00 00 N

SOLUTION FDA

=

FDB

=

a 6 800 a 25.06 16 600 a 34

15 400 i 34.66 -

-

24  k N 34.66

+

4  j 25.06

-

24  k 25.06

18  j 34

FR

=

FDA

=

{263.92i

=

2 (263.92)2

+

=

1490.83 1490. 83 N

1.49 1. 49 kN

+

FDB

+

-

=

-

=

cos-1

b

=

cos-1

g

=

cos-1

 x

1466.71k} N

( 40.8 6)2

+

( - 1466.71)2 Ans.

=

79.8°

Ans.

=

88.4°

Ans.

a 11466.71 b 490 49 0.83  -

B 6m

 x

 y  A

263.92 a 1490.83 b 40.86 a 1490.83 b

a

18 m

FDC

40.86j

+

4m

O

C 

b

=

-

16 m

24  k N 34

FDC

FR

i

i

24 m

b bN

+

20  j 34.66

=

169.7°

Ans.

 y

2–95. z

At a given instant, the position of a plane at A and a train at B are measured relative to a radar antenna at O . Determine the distance d between A and B at this instant. To solve the problem, proble m, formul formulate ate a positio position n vector vector,, direc directed ted from A to B , and then determine its magnitude.

 A

5 km 60 35

SOLUTION

O

 Position Vector: The coordinates of points  A and B are

 x

A( - 5 cos 60° cos 35°, - 5 cos 60° sin 35°, 5 sin 60°) km

B

= A( - 2.048, - 1.434, 4.330) km B(2 cos 25° sin 40°, 2 cos 25° cos 40°, - 2 sin 25°) km

= B(1.165, 1.389, - 0.845) km The position vector r AB can be established from the coordinates of points A and B. rAB = {[1.165 - ( - 2.048)]i + [1.389 - ( - 1.434)] j + ( - 0.845 - 4.330)k} km

= {3.213i + 2.822 j - 5.175)k} km The distance between points  A and B is d = rAB =

 2 3.213 3.213

2

+ 2.8222 + ( - 5.175)2 = 6.71 km

 y

40 25 2 km

Ans.

2–96. Two cables are used to secure the overhang boom in position and support the 1500-N load . If the resultant force is directed along the boom from po int  A towards O, determine the magnitudes of the resultant force and forces FB and FC . Set x = 3 m and z = 2 m.

z  x

2m

B 3m

C 

z

FB

6 m FC 

 A

 x

 y

1500 N

Two cables are used to secure the overhang boom in position and support the 1500-N load. If the resultant force is directed along the boom from point  A towards O, determine the values of  x and z for the coordinates of point magnitude de of of the resulta resultant nt force force.. Set C  and the magnitu FB = 1610 N and FC = 2400 N.

2–97.

z  x

2m

B 3m

C 

z

FB

6 m FC 

 A

 x

 y

1500 N

.

.

2–99.

Determine the magnitude and coordinate direction angles of  the resultant force acting at point  A.

z

1.5 m

 A F 1 F 2

150 N

200 N

4m

SOLUTION rAC

=

{3i

|rAC|

=

2 32

F2

=

200

-

0.5 j

-

4k} m

( - 0.5)2

+

a3

0.5 j - 4k 5.02494

i

-

rAB

=

(3 cos 60° i

rAB

=

(1.5i

|rAB|

=

(1.5)2 2 (1.5)

F1

=

150

FR

=

F1

FR

=

(157.4124) 2 2 (157.4124)

a

=

cos-1

b

=

cos-1

g

=

cos-1

+

(1.5

4.0981 j

a 1.5

+

+

i

F2

+ +

=

+

=

5.02494

=

(119.4044 i

-

19.9007 j

=

3 si sin n 60° 60°)) j

-

2m -

159.2059 k)

4k)

4k) +

-

(157.4124 i

( - 4)2

4k

+

b

=

=

5.9198

(38.0079 i

83.9389 j

-

+

103.8396 j

-

101.3545 k)

260.5607 k)

(83.9389) 2

+

( - 260.5604) 2

=

60.100°

=

60.1°

Ans.

=

74.585°

=

74.6°

Ans.

260.5607 a 315.7786  b

=

145.60°

=

146°

=

B

C 

 x

+

4.0981 j 5.9198

a 157.4124 b 315.7786 83.9389 a 315.7786 b  -

b

(4.0981) 2

+

2 25.25 25.25

( - 4)2

+

60

3m

315.7786

=

316 N

Ans.

Ans.

 y

3   m 

2–100.

The guy wires are used to support the telephone pole. Represent the force in each wire in Cartesian vector form. Neglect the diameter of the pole.

z

SOLUTION B

Unit Vector: =

{( - 1

rAC

=

2 ( - 1)2

=

-

rAC

=

rAC

0)i

-

+

(4

+

42

+

1i

+

0) j

+

(0

( - 4)2

=

5.74 5. 7455 m

-

4 j - 4k 5.745

= -

4)k} m

-

=

{ - 1i

4 j

+

-

2m

0.1741 i

+

0.6963 j

-

rBD

=

{(2

rBD

=

2 2 2

=

-

rBD rBD

0) i +

=

+

(-3

( - 3)2 2i

-

+

-

0) j

(0

+

( - 5.5)2

3 j - 5.5k 6.576

=

5.5)k} m

-

0.6963 k

{2i

-

3 j

-

5.5k} m

-

0.4562 j

-

0.8363 k

 Force Vector: FA

FB

=

=

FA uAC

FBuBD

=

250{ - 0.1741 i

=

{ - 43.52i

=

{ - 43.5i

=

175{0.3041 i

=

{53.22 i

=

{53.2i

+

+

-

-

0.6963 j

+

174.08 j

174 j +

-

-

0.6963 k} N

174.08 k} N

174k} N

0.4562 j

79.83 j

79.8 j

-

-

-

-

Ans.

0.8363 k} N

146.36 k} N

146k} N

Ans.

250 N

4m D

3m

6.57 6. 5766 m

0.3041 i

=

=

F  A

4k} m

 x

uBD

1755 N 17

F B

rAC

uAC

1.5 m  A

4m C 

1m

 y

2 –101

.

The force acting on the man, caused by his pulling on the anchor cord, is F. If the length of  the cord is L is  L,, determine the coordinates A coordinates A(( x, y, y , z ) of the anchor. Given:

  40   F



20

   50 

 N

 L  25 m

Solution:

r

 L

F F

  14.9   r



7.5

   18.6 

m

Ans.

2–102.

Each of  the f our f orces acting at E has a magnitude of  28 kN. Express each f orce as a Ca rtesian vector and determine the resultant f orce.

z

E FEC 

FEA FEB

FED

D

SOLUTION FEA FEA FEB

=

28

a 64 1

 i

=

{12i

=

6  i 28 14

-

8 j

a

4

-

14 -

+

=

{12i

FEC

=

28

FEC

=

{ - 12i

FED

=

28

FED

=

{ - 12i

FR

=

FEA

=

{ - 96k } kN

a a

8 j

6 i 14

-

-

12 k 14

-

12 k 14

+

8 j

-

6 i 14

-

4  j 14

8 j

-

24k} kN

FEB

+

FEC

+

-

12  k 14

24k } kN -

12  k 14

+

 x

b Ans.

b Ans.

FED Ans.

C  6  m

4  m 4  m

Ans.

4  j 14

 -

 A

b

24k} kN -

12  m

Ans.

+

 -

b

24k} kN

4  j 14

FEB

+

 j

B

6  m

 y

2–103

.

The cord exerts a force F  on the hook. If the cord is length L length L,, determine the location x,y location x,y of  of the  point of attachment B attachment B,, and the height z  height z  of  of the hook.

Units Used: Given:

 12  F 

9  N

 8  L  4 m a  1 m

Solution :

x  1 m

Initial guesses

 x  a  Given

y

   z    

=

L

F F

y  1 m

z  1 m

 x  y

   z 

 x   Find ( x  y  z)

y

   z 

 3.82  

2.12

 1.88 

m

Ans.

2–104

.

 F on the hook. If the cord length  L, the  L,  the dist ance z, The cord exerts a force of magnitude  F on ance z, and the  x component  x component of the for ce F  y of the point of attachment B ce F  x are given, determine the the location x, location x, y of attachment B of   of  the cord to the ground. Given:

F  30kN L  4 m z  2 m Fx  25kN a  1 m Solution :

Guesses x  1 m y  1 m

Given Fx

=

xa L

F

 x   Find ( x  y)  y 

L

2

=

2

2

2

( x  a)  y  z

 x    4.33  m  y   0.94 

Ans.

. . . .

2–106

.

The chandelier is supported by three chains which are concurrent concurrent at point O. If the force in each chain has magnitude F, express each force as a C artesian vector and determine the magnitude and coordinate direction angles of the resultant force. Given: F

 300N

a

 1.8 m

b

 1.2m

1

 120deg

2

 120deg

Solution:

3

 360deg  1   2

 b sin 

  rOA   b cos  1   

rOB 

1

FA  F

 

a

 b sin 

1

 2 

 b cos 

1

 2 

a

 

 

FB  F

  0   rOC 

b

   a 

FR  FA  FB  FC

FC  F

rOA rOA

rOB rOB

rOC rOC

  144.12   FA



83.21

   249.62 

N

Ans.

N

Ans.

N

Ans.

 144.12  FB



83.21

   249.62      

FC



FR 

0.00

 

166.41

   249.62   748.85 N

Ans.

      FR    acos     FR      

      90.00       90.00   deg      180.00 

Ans.

2–107

.

The chandelier is supported by three chains which are concurrent concurrent at point O. If the resultant force at O has magnitude F R  and is dir ected along the negative z  negative  z  axis,  axis, determine the force in each chain assuming FA = FB = FC = F. Given: a

 1.8m

b

 1.2 m

FR  650N Solution: a

2

2

b

F



F

 260.4 N

3a

FR 

Ans.

2–108. z

Determine the magnitude and coordinate direction angles of the resultant force. Set FB = 630 N , FC = 520 N and FD = 750 N, and x = 3 m and z = 3.5 m.

3m

B

4.5 m

2m C 

FC 

FB  A

SOLUTION

4m

 Force Vectors: The un unit it ve vect ctor orss uB, uC, an and d uD of FB, FC, an and d FD mu must st be de dete term rmin ined ed first. From Fig. Fig. a, uB =

uC =

uD =

rB rB

rC rC

rD rD

( - 3 - 0)i + (0 - 6) j + (4.5 - 2.5)k

3 ( - 3 - 0)

=

2

+ (0 - 6)2 + (4.5 - 2.5)2

(2 - 0)i + (0 - 6) j + (4 - 2.5)k

3 (2(2 - 0)

=

2

= -

+ (0 - 6)2 + (4 - 2.5)2

=

3 (0(0 - 3)

2

3 6 2 i -  j + k 7 7 7

+ (0 - 6)2 + ( - 3.5 - 2.5)2

=

1 2 2 i -  j -  k 3 3 3

Thu huss, th thee fo forc rcee ve vect ctor orss FB, FC, an and d FD ar aree gi give ven n by

a 37 - 67 + 27 b = 5 - 270 - 540 + 180 6 N 4 12 3 = 520 a + b = 5160 - 480 + 20 6 N 13 13 13 1 2 2 b = 5250 - 500 - 500 6 N = 750 a 3 3 3

FB = FB uB = 630 -

FC = FC uC

i

 j

i

FD = FD uD

k

 j

i

 j

i

k

k

 j

i

i

 j

 j

k

1

k

k

 Resultant Force: FR = FB + FC + FD = ( - 270i - 540 j + 180k) + (160 i - 480 j + 120k) + (250i - 500 j - 500k)

= [140i - 1520 j - 200k] N

The magnitude of FR is

 3 (F ) + (F ) =  3 140 140 + ( - 1520)

FR =

R x

2

R y

2

2

+ (FR)z 2

2

+ ( - 200) 2 = 1539.48 N = 1.54 kN

Ans.

The coordinat coordinatee direction angles of FR are a = cos-1

140 c ( ) d = cos a 1539.48 b = 84.8° - 1520 c ( ) d = cos a 1539.48 b = 171° - 200 c ( ) d = cos a 1539.48 b = 97.5°

b = cos-1

g = cos-1

FR

x

-1

Ans.

y

-1

Ans.

z

-1

Ans.

FR

FR

FR

FR

FR

FD

z

6m D

4 12 3 i  j + k 13 13 13

(3 - 0)i + (0 - 6) j + ( - 3.5 - 2.5)k

=

 x

O

 x

2.5 m  y

2–109. z

If the magnitude of the resultant force is 1300 N and acts along the axis of the strut, directed from point A towards O , determine the magnitudes of the three forces acting on the strut. Set x = 0 and z = 5.5 m .

3m

B

4.5 m

2m C 

FC 

FB  A

4m

SOLUTION  Force Vectors: Vectors: The unit vectors uB, uC, uD, an and uFR of FB, FC, FD, an and FR must be determined first. From Fig. Fig. a, uB =

uC =

uD =

uFR =

rB rB

rC rC

rD rD

=

=

( - 3 - 0)i + (0 - 6) j + (4.5 - 2.5)k

3 ( - 3 - 0)

2

2

rAO

+ (0 - 6)2 + (4 - 2.5)2

=

3 6 2 i -  j + k 7 7 7

4 12 3 i  j +  k 13 13 13

3 

(0 - 0)i + (0 - 6) j + (0 - 2.5)k

3 (0(0 - 0)

2

+ (0 - 6)2 + (0 - 2.5)2

= -

12 5  j +  k 13 13

Thu huss, th thee fo forc rcee vect ctor orss FB, FC, FD, and F  R are give ven n by 3 7

FB = FB uB = - FB i -

FC = FC uC =

6 2 FB j + FB k 7 7

4 12 3 FC i FC j + F k 13 13 13 C 3 5

FD = FD uD = - FD j -

a

FR = FR uR = 1300 -

4 F k 5 D

12 5  j k 13 13

b = [ - 1200 - 500 ] N  j

k

 Resultant Force: FR = FB + FC + FD

 a - 37 3 =  a 7

- 1200 j - 500k =

FBi -

- 1200 j - 500k

FB +

6 2 FB j + FBk 7 7

b + a 134

b  a - 67

4 F i + 13 C

FB -

FCi -

12 3 FC j + F k 13 13 C

12 3 FC - FD j 13 5

b + a 27

b + a - 35

FB +

(1) (2) (3)

Solving Eqs qs.. (1) (1),, (2 (2), ), and (3), (3), yi yield eldss FC = 442 N

FB = 318 N

FD = 866 N

FD j -

4 F k 5 D

b

3 4 FC - FD k 13 5

Equati quating ng the i, j, and k comp component onents, s,

3 4 0 = - FB + FC 7 13 6 12 3 - 1200 = - FB - FC - FD j 7 13 5 2 3 4 FC - FD - 500 = FB + 7 13 5

6m D

3 4 = -  j +  k 5 5 (0 - 0)2 + (0 - 6)2 + ( - 5.5 - 2.5)2 (0

=

FD

z

 x

(0 - 0)i + (0 - 6) j + ( - 5.5 - 2.5)k

=

rAO

+ (0 - 6)2 + (4.5 - 2.5)2

(2 - 0)i + (0 - 6) j + (4 - 2.5)k

3 (2(2 - 0)

= -

O

Ans.

b

 x

2.5 m  y

2 –110

.

The positions of point A point  A on the building and point B point  B on the antenna have been measured relative to the electronic distance meter (EDM) at O. Determine the distance between A between A and and B  B.. Hint:  B;; then determine its magnitude. Formulate a position vector directed from A from  A to to B Given: a  460 m b  653 m    60 deg     55 deg    30 deg    40 deg Solution:

rOA

 a cos    sin      a cos    cos       a sin     

rOB

 b cos    sin        b cos    cos       b sin       148.2 

rAB

 rOB  rOA

rAB

 492.4 m

    239.2  

rAB

 567.2 67.2 m

Ans.

2–111. The cylindrical plate is subjected to the three cable forces which are concurrent at point D. Express each each force force which the cables exert on the plate as a Cartesian vector, and determine the magnitude and coordinate direction angles of the resultant force.

z D

F C   5 kN

3m

F B  8 kN C 

B

45

30

 y

 A

0.75 m

F  A  6 kN

 x

.

.

. .

. .

.

2–112.

Given the three vectors A, B, and D, show th that A # (B + D) = (A # B) + (A # D).

SOLUTION Since the component of (B D, th then en

D) is equal to the sum of the components of B and

+

A # (B

+

D)

A y j

+

A zk) # [(Bx

=

A#B

+

A#D

(QED)

Also, A # (B

+

D)

=

(A x i

=

A x (Bx

+

Dx)

=

(A xBx

+

A yBy

=

(A # B)

+

+

+

A y (By +

(A # D)

A zBz)

+ +

+

Dx)i

Dy)

+

(By

+

Dy) j

A z (Bz

+

Dz)

(A xDx

+

+

A yDy

+

+

(Bz

+

Dz)k]

A zDz)

(QED)

2–113.

Determine the angle u between the edges of the sheetmetal bracket.

z

400 mm

250 mm

 x

SOLUTION r1

=

{400i

r2

=

{50i

r1 # r2 u

=

+

+

250k} mm ;

300 j} mm ;

(400) (50)

=

cos-1

=

cos-1

u

+

0(300)

+

r1

=

471.70 mm

r2

=

304.14 mm

250(0)

=

300 mm 50 mm

 y

20 000

r1 # r2

¢ ≤ ¢ r1 r2

20 000 (471.70) (304.14)

≤

=

82.0°

Ans.

2–114.

Determine the angle u between the sides of the triangular plate.

 z

3m  B

4m

SOLUTION

 53

 A

6 2 132  142  1 12 5 09 0990 90 m  52 3 6 m 2 122  132 3 6056 m 0 4122  1 12132 5 5 cos cos 15 0990213 60562

rAC

=

rAC

=

rAB

=

rAB

=

=

u

=

+

4 j

2

 j

+

=

-1

-

+

2

+

2

+

-

=

.

+

-

rAC # rAB rACrAB

74.219°

=

y

1 k m

2

1m

74.2°

=

3m

.

=

5m

 k

+

2

rAC # rAB u

 i

θ 

1m

 x 

=

-1

.

.

Ans.

C 

2–115.

Determine the length of side BC  of the triangular plate. Solve the problem by f inding the magnitude of rBC ; then check the result by f irst f inding , r  AB, and r  AC  and then using the cosine law.

z

3m

B

SOLUTION rBC

=

{3 i

2 j

rBC

=

2 (3) (3)2

+

+

-

4m

4 k} m

(2)2

+

( - 4)2

 A =

5.39 m

1m

Ans.

u

 y 1m

Also, rAC

=

{3 i

rAC

=

(3)2 2 (3)

rAB

=

{2 j

rAB

=

2 (2) (2)2

rAC # rAB

=

a

+

+

0

4 j +

-

1 k} m

(4)2

=

cos-1

u

=

74.219°

( - 1)2

5.09 0990 90 m

=

 x

3 k} m +

(3)2

=

3.6056 m

+

4(2)

+

( - 1)(3)

rAC # rAB

u

+

5m

rAC rAB

rBC

=

2 (5 (5.0990)2

rBC

=

5.39 m

b

+

=

cos-1

=

5

5 (5.0990)(3.6056)

(3.6056)2 - 2(5.0990)(3.6056) cos 74.219° Ans.

3m

C 

 2–116

.

   between the tails of Determine the angle   between of the two vectors. Given: r 1  9 m r 2  6 m    60 deg     45 deg    120 120 deg    30 deg    40 deg Solution: Determine the two position vectors and use the dot  product to find the angle

r1v

  sin    cos       r 1  sin   sin      cos     

   acos

 

r1v r2v

  r1v

 

r2v  

r2v

 r 2

 cos      cos      cos     

   109.4 109.4 deg

Ans.

2 –117

.

Determine the magnitude of the projected component of r1 along r2, and the projection of r2 along r1. Given: r 1  9 m r 2  6 m    60 deg     45 deg    120 120 deg deg    30 deg    40 deg Solution: Write the vectors and unit vectors

r1v

r2v

u1

 r 1

  sin   cos       sin   sin      cos    

 r 2

 cos      cos      cos    



r1v r1v

  5.01   r1v

 2.89 m

  6.89      

r2v

u2



r2v r2v

3

 

 4.24 m

   3    

  0.557   u1

 0.321

    0.766  

  u2

0.5

 

 0.707

    0.5  

The magnitude of the projection of r1 along r2.

r1v u 2

 2.99 m

Ans.

The magnitude of the projection of r2 along r1.

r2v u 1

 1.99 m

Ans.

2–118.

Determine the pro jection of the force F along the pole.

 z

F = {2 i + 4 j  + 10k} kN

O

 y

SOLUTION 2m

 12

Pro j F

=

F # ua

Pro j F

=

0.66 667 7 kN

=

 i

+

4 j

+

2a

2  i 10 k # 3

+

2  j 3

-

1  k 3

b Ans.

2m  x 

1m

2 –119

.

Determine the angle   between the two cords. Given: a  3m b  2m c  6m d   3 m e  4m

Solution:

 b  rAC

 a m

   c 

  0   rAB

 d  m

    e  

   acos

   

rAC r AB rAC

 

rAB  

   64.6 64.6 deg

Ans.

2–120. Two forces act on the hook. Determine the angle u between them.Also, what are the projections of F1 and F2 along the  y axis?

z

F 1  600 N

45

60

120 u

.

 y  x

. .

F2  {120i + 90 j – 80k}N

Determine the 2–12 2–121. Two forces act on the hook. Determine magnitude of the projection of F2 along F 1.

z

F 1  600 N

45

60

120 u

 y  x

.

F2  {120i + 90 j – 80k}N

2–122. Determine the magnitude of the projected component of force F AB acting along the z axis.



 A

F  AC   3 kN F  AB

9m

 3.5 kN D

4.5 m O

Unit Vector: The unit vector u AB must be determined first. From Fig. a, u AB =

r AB r  AB

 =

(4.5 – 0)i + (– ( –3 – 0 )j + (0 ( 0 – 9)k 2

2

(4.5 4.5 – 0) + (– (–3 – 0) + (0 ( 0 – 9)

 = 2

3 7

i –

2 7

 j –

6 7

k

3m  x

Thus, the force vector F AB is given by

 3 2 6   F AB = F ABu AB = 3.5  i – j – k   = {1.5 i – 1 j – 3k} kN 7 7    7 Vector Dot Product: The projected component of F AB along the z axis is

(F  AB)z = F AB · k = (1.5i – 1 j – 3k) · k = –3 kN The negative sign indicates that (F  AB)z is directed towards the negative z axis.  axi s. Thus (F  AB)z = 3 kN

Ans.

A(0, 0, 9) m

B(4.5, –3, 0) m

3m

B

3 m C 

30

 y

2–123. Determine the magnitude of the projected component of force F AC acting along the z axis.



 A

F  AC   3 kN F  AB

9m

 3.5 kN D

4.5 m O

3m

B

3m  x

Unit Vector: The unit vector u AC  must be determined first. From Fig. a, u AC  =

r AC  r  AC 

 =

(3 sin 3 0° 0° – 0)i + ( 3 co cos 30° – 0)j + ( 0 – 9 9))k

 = 0.1581 i + 0.2739 j – 0.9487 k

(3 s in in 3 0° 0° – 0)2 + (3 ( 3 co c os 3 0° 0° – 0)2 + (0 – 9 )2

Thus, the force vector F AC  is given by F AC  = F AC u AC  = 3(0.1581 i + 0.2739 j – 0.9487k) = {0.4743i + 0.8217 j – 2.8461k} kN Vector Dot Product: The projected component of F AC  along the z axis is

(F  AC )z = F AC  · k = (0.4743 i + 0.8217 j – 2.8461k) · k = –2.8461 kN The negative sign indicates that (F  AC )z is directed towards the negative z axi s. Thus (F  AC )z = 2.846 kN

Ans.

A(0, 0, 9) m

C(3 sin 30°, 3 cos 30°, 0) m

3 m C 

30

 y

2–124. Determine the projection of force F 400 N acting along line  AC of the pipe assembly. assembly. Express the result as a Cartesian vector.

z

=

F   400 N B

45

C  30

 A 3m

 x

4m

 y

N

N

2–125. Determine the magnitudes of the components of  force F 400 N acting parallel and perpendicular perpendicul ar to segment BC  of the pipe assembly.

z

=

F   400 N B

45

C  30

 A 3m

 x

N

4m

 y

2–126. Cable OA is used to support column Determine the angle u it makes with beam OC .

OB

.

z D 30

O f

C 

4m

u 8m

 x 8m

B

.

 A

 y

2–127. Cable OA is used to support column Determine the angle f it makes with beam OD.

OB.

z D 30

O f

C 

4m

u 8m

 x 8m

B

.

 A

 y

2–128

.

  and   between OA of the pole and each cable, AB  AC . Determine the angles  and  between the axis OA of cable,  AB and  and AC  Given:  F 1  50 N  F 2  35 N a  1m b  3m c  2m d   5 m e  4m  f   6 m  g   4 m Solution:

  0   rAO

  g 

    f  

rAB

 

 

   acos

   acos

rAO rAB

  rAO

  c  

  e  

rAB  

  rAO rAC     rAO rAC  



a

    f  

rAC

 ab

    f     

   52.4 52.4 deg deg

Ans.

   68.2 68.2 deg deg

Ans.

2 –129

.

The two cables exert the forces shown on the pole. Determine the magnitude of the  projected component component of each force acting along the axis OA of OA of the pole. Given:  F 1  50 N  F 2  35 N a  1m b  3m c  2m d   5 m e  4m  f   6 m  g   4 m Solution:

  c  

  e   rAB



a

    f  

F1v

 F 1

F2v

 F 2

rAC rAC rAB rAB

rAC

 ab

     f   

  0   rAO

  g 

    f  

uAO



rAO rAO

F 1AO  F1v u AO

F 1AO  18.5 N

Ans.

F 2AO  F2v u AO

F 2AO  21.3 N

Ans.

2–130.

Determine the angle u between the pipe segments BA and BC .

z

1.5 m

 A

2m

2m

1m B

 x

SOLUTION

 y

F  {30i  45 j  50k}  N

 Position Vectors: The position vectors rBA and rBC must be determined first. From Fig. a,

2m C 

rBA = (0 - 1.5)i + (0 - 2)j + (0 - 0)k = { - 1.5i - 2j} m rBC = (3.5 - 1.5)i + (3 - 2) j + (- 2 - 0)k = {2i + 1j - 2k} m

The magnitude of rBA and rBC are

 3 (- 1.5 ) + (- 2) = 2.5 m =  3 2 + 1 + (- 2) = 3 m 2

rBA = rBC

2

2

2

B(1.5, 2, 0) m

2

Vector Dot Product:

rBA # rBC = (- 1. 1.5 5 i - 2j) # (2i + 1j - 2k)

C(3.5, 3, –2) m

= (- 1.5)(2) + ( - 2)(1) + 0( - 2) = - 5 m2 Thus, u = cos-1

a

rBA # rBC  rBA rBC 

-5 b = cos c 2.5(3) d = 132° -1

Ans.

2–131.

Determine the angles u and f made between the axes OA of the flag flag pole and and AB and A respectivelyy, of each each cable. cable.  AC  C , respectivel

z

1.5 m 2m

B

4m

SOLUTION rA C rAB rA O

= = =

{ - 2i {1.5i

=

=

-

4 j 4 j

F C 

1k} m ;

+ +

rA C

3k} m;

rAB

3k} m;

rA O

= = =

F B

4.58 m

40 N

55 N

6m

u

f

 A

5.22 5. 22 m 5.00 m

O 3m

cos - 1 cos - 1

rAC # rAO f

-

{ - 4 j

rA B # rA O u

-

C 

=

(1.5)(0)

¢ ¢

rAB # rAO rAB rAO

( - 2)(0)

(3)( - 3)

=

7

rAC # rAO

4m

 x  y

≤

+

=

74.4°

( - 4)( - 4)

=

cos - 1

a

=

cos - 1

13 a 4.58(5.00) b

rAC rAO

+

≤

7 5.22(5.00)

=

( - 4)( - 4)

+

Ans. +

(1)( - 3)

=

13

b =

55.4°

Ans.

2–132.

The cables each exert a force of 400 N on the post. Determine the magnitude of the projected component of F1 along the line of action of F2.

z

F 1

400 N

35

SOLUTION 120

 Force Vector: uF1

=

sin 35° cos 20° i

-

sin 35° sin 20° j

+

cos 35°k

u

20

F1

=

F1uF1

=

0.5390 i

0.1962 j

=

400(0.5390 i  - 0.1962 j  + 0.8192 k) N

=

{215.59i

-

-

78.47 j

+

+

=

cos 45°i

 x

327.66 k} N

=

0.7071 i

cos 60° j

+ +

0.5 j

-

+

cos 120°k

0.5k

 Projected Component of  F  F1 Along Line of Action of  F  F2:

(F1)F2

=

F1 # uF2

=

(215.59 i

=

(215.59)(0.7071)

= -

-

78.47 j

+ +

327.66 k) # (0.7071 i ( - 78.47)(0.5)

+

+

0.5 j

-

0.5k)

(327.66)( - 0.5)

50.6 50 .6 N

Negative sign indicates that the force component ( F1)F 2 acts in the opposite sense of direction to that of uF 2. Thus the magnitude is (F  1 ) F2

=

50.6 50 .6 N

60

45

0.8192 k

Unit Vector: The unit vector along the line of action of F2 is uF2

 y

Ans.

F 2

400 N

2–133.

Determine the angle angle u between the two two cables cables attached attached to the post.

z

F 1

400 N

35

SOLUTION 120

Unit Vector: uF1

=

sin 35° cos 20° i - sin 35° sin 20° j + cos 35°k

u

20

uF2

=

0.5390 i

=

cos 45°i

=

0.7071 i

-

0.1962 j

cos 60° j

+ +

+

0.5 j

-

+

cos 120°k

 x

0.5k

determined first. =

(0.5390 i

=

0.5390(0.7071)

= -

-

0.1962 j +

+

0.8192 k) # (0.7071 i

( - 0.1962)(0.5)

+

+

0.5 j

-

0.5k)

0.8192( - 0.5)

0.1265

Then, u

=

cos-1 A uF1 # uF2 B

=

60

45

0.8192 k

The Angle Between Two Vectors u: The dot product product of two unit vectors must be uF1 # uF2

 y

cos-1( - 0.1265)

=

97.3°

Ans.

F 2

400 N

2 –134

.

Force F is applied to the handle of the wrench. Determine the angle   between  between the tail of the force and the handle  AB.  AB. Given: a  300 mm b  500 mm  F   80 N  1  30 deg  2  45 deg Solution:

 cos   1 sin  2  Fv

 F   cos   1 cos   2 

     acos

sin   1 

 Fv u ab     F   

  

  0   uab

 1

    0  

   127.8 127.8 deg

Ans.

2–135

.

Determine the projected component of the force F acting along the axis  AB of  AB of the pipe. Given:  F   80 N a  4m b  3m c  12 m d   2 m e  6m

Solution: Find the force and the unit vector 

  e   rA

 a  b

   d c    

  6   rA

 e  rAB

 b

    d  



7

   10 

m

Fv

 F 

 6  rAB

 3 m

    2  

uAB



rA rA

rAB rAB

 Now find the projection using using the Dot product. product.  F  AB  Fv u AB

F  AB  31.1 N

Ans.

 35.3  Fv

 41.2  N

   58.8    

 0.9  uAB

 0.4

  0.3    

2–136.

Determine the components of F that act along rod  AC and perpendicula perpen dicularr to it. Poi Point nt B is locate located d at the midpoint midpoint of  the rod.

z

 A

B

4m

4m

O

SOLUTION

( - 3i

4 j

=

rAB

=

rAD

=

rAB

+

rBD

rBD

=

rAD

-

rAB

=

(4i

=

{5.5i

rBD

=

2 (5.5) (5.5)2

F

=

600

rAC

-

=

2

6 j

+

+

3i

-

4 j

+

a b rBD

rBD

4 j 2

4k) -

+

4k),

-

4k

+

-

=

= -

( - 1.5i

2 ( - 3)2

1.5i

+

2 j

+

2 j

=

7.0887 m

-

+

-

42

+

( - 4)2

=

F # rAC rAC

F| |  = 99.1408

=

=

6m 4m

2k

D

2k)

+

( - 2)2

465.528 i

+

338.5659 j

-

169.2829 k

(465.528 i

+

338.5659 j

-

169.2829 k) # ( - 3i

+

4 j

-

4k)

2 41 41 99.1 N

Ans.

Component of F perpendicular to rAC is F F2

+

F2| | = F2

F2

=

6002

F

=

591.75

-

=

6002

99.1408 2 =

C 

3m

2 41 41 m

2k} m

(4)2 =

rAC

Component of F along rAC is F| | F| |  =

600 N

 x

rAC

+

F 

592 N

Ans.

 y

2–137.

Determine the components of F that act along rod  A  AC  C  and perpendic perpe ndicular ular to it. Po Point int B is loca located ted 3 m along along the the rod from end C .

z

 A

B

4m

4m

O

SOLUTION

4 j

3i

4k

rCA

=

6.403124

rCB

=

3  (r ) 6.403124 CA

rOB

=

rOC

+

rCB

= -

3i

+

4 j

= -

1.59444 i

+

4m

+

2.1259 j

=

(4i

rOB

+

rBD

rBD

=

rOD

-

rOB

=

5.5944 i

rBD

=

2 (5.5944) (5.5944) 2

F

=

600(

rAC

=

( - 3i

+

3.8741 j

)

+

-

+

6 j)

-

F # rAC

=

rAC

F| |  = 82.4351

=

+

1.874085 k

4k),

(475.568 i

rOB

-

1.874085 k

+

+

rAC

+

( - 1.874085) 2

329.326 j

Component of F along rAC is F| F| |  =

1.874085 j

-

1.874085 k

(3.8741) 2

475.568 i

=

4 j

+

1.40556 i

=

r  CB

+

=

rBD

6m

D

rOD

rBD

=

-

=

7.0582

159.311 k

2 41 41

|

329.326 j

-

159.311 k) # ( - 3i

+

4 j

-

4k)

2 41 41 82.4 N

Ans.

Component of F perpendicular to rAC is F F2

+

F2| |

F2

F2

=

6002

F

=

594N

=

-

C 

3m

=

+

600 N

 x

rCA

-

F 

=

6002

82.4351 2 Ans.

 y

2–138.

Determine the magnitudes of the projected components of  the force F = 300N acting along the x and  y axes.

30

F 

z

 A

30

300 mm

SOLUTION O 300 mm

 Force Vecto Vector: r: The force vector F must be determined first. From Fig. Fig. a, F

= =

300 sin 30°sin 30° i

[ - 75i

+

259.81 j

+

+

3000 cos 30° 30 30° j

+

300 sin 30°cos 30°cos 30°k

 x

129.90 k] N

Vector Dot Product: The magnitudes of the projected component of F along the  x and y axes are Fx

Fy

=

F#i

A - 75i

=

= -

75(1)

= -

75 N

=

F # j

= =

+

75(0)

+

259.81 j

259.81(0)

A - 75i

=

+

+

+

+

129.90 k B # i

129.90(0)

259.81 j

259.81(1)

+

+

129.90 k B # j

129.90(0)

260 N

The negative sign indicates that F x is directed towards the negative x axis axis.. Thus Fx

=

75 N,

Fy

=

260 N

Ans.

300 mm

y

300 N

2–139.

Determine the magnitude of the projected component of  the force F = 300 N acting along line OA.

30

F 

z

 A

30

300 mm

O 300 mm

SOLUTION

300 mm

 x

 Force and and Unit Unit Vector: Vector: The force vector F and unit vector uOA must be determined

first. Fro From m Fig. Fig. a F

=

( - 300 sin 30° sin sin 30°i

=

{ - 75i

uOA

=

rOA rOA

+

259.81 j

=

3000 cos 30° 30 30° j

+

300 sin 30° cos cos 30°k)

129.90 k} N

+

( - 0.45

+

0)i

-

2 ( - 0.45

-

+

0)2

(0.3 +

-

(0.3

0) j -

(0.2598

+

0)2

-

(0.2598

+

0)k -

0)2

= -

0.75i

+

0.5 j

+

0.4330 k

Vector Dot Product: The magnitude of the projected component of F along line OA is FOA

=

F # uOA

=

A - 75i

=

( - 75)( - 0.75)

=

242 N

+

259.81 j +

+

129.90 k B #

259.81(0.5)

A - 0.75i

+

+

0.5 j

+

0.4330 k B

129.90(0.4330) Ans.

y

300 N

2–140.

Determine the length of the connecting rod  AB by first formulating a Cartesian position vector from  A to  B and then determining its magnitude magnitude..

 y

30

B

 A

SOLUTION

12 5 mm

r AB  = [400 =

-

{462.5 i

( - 125 si sin 30°)]i -

(462.5)2 rAB  = 2 (462.5)

+

(0

-

125 co cos 30°) j 400mm

108.25 j} mm +

(108.25) 2

 x O

=

4755 mm 47

Ans.

2–141.

Determine the  x and  y components of  F1 and F2.

y

45

F 1    200 N

30

SOLUTION F1x = 200 sin 45° = 141 N

Ans.

F1y = 200 cos 45° = 141 N

Ans.

F2x = - 150 co coss 30° = - 130 N

Ans.

F2y = 150 sin 30° = 75 N

Ans.

F 2    1 50  N  x

2–142.

Determine the magnitude of  the resul tant f orce and its dir ection, measured counte rclockwi se from the positive  x axi s.

 y

45

F 1    200 N

30

SOLUTION F 2    1 50  N

FRx = - 150 co coss 30° + 200s in 45° = 11.518 N

+ R F Rx = © Fx;

 x Q + FRy = © Fy;

FRy = 150 sin 30° + 200 cos 45° = 216.421 N

FR = 2  (11.518)2 + (216.421)2 = 217 N u =

tan - 1

¢

216.421 11.518

≤

=

87.0°

Ans. Ans.

2–143. Resolve the 250-N force into components acting along the u and v axes and determ ine the magnitudes of  these components. u

20 250 N 40

v

. .

2–144 . Express F1 and F2 as Cartesian vectors.

 y

F 2  = 26 kN

13

12 5

 x 

SOLUTION F1 = - 30 sin 30° i - 30 cos 30°  j 30° j

 5 -

6 1 2

=

 j kN 15.0 i - 26.0 26.0 j

F2 = -

5 12 26  i + 26  j 13 13

=

1 2

 j kN - 10.0 i + 24.0 24.0 j

°

30

Ans. F 1  = 30 kN

Ans.

2–145. Determine the magnitude of the resultant force and its direction measured counterclockwise from the positive  x axis.

 y

F 2  = 26 kN

13

12 5

 x 

SOLUTION + F = ©F ; Rx x

:

+  c FRy = © Fy ;

1 2 1 2== + 2= = 2 1 - 2 + 1 a b= =

FRx = - 30 sin 30° 30° -

FRy FR f

30 cos 30° 30° 25

tan-1

2

1.981 25

5 26 = - 25 kN 13 12 26 13 1.981

2

°

30

1.981 1. 981 kN

25.1 25 .1 kN

F 1  = 30 kN

Ans.

4.53°

u = 180° + 4.53° = 185°

Ans.

2–146. Cable  AB exerts a force of 80 N on the end of the 3-m-long boom OA. Determine the magnitude of the projection of this force along the boom.

z

B

4m

O

 y 80 N 60

.

 A  x

.

3m

2–147.  y

Determine the magnitude and direction of the resultant FR = F1 + F2 + F3 of the three forces by first finding the resultant F ¿ = F1 + F3 and then forming FR = F ¿ + F2 . Specify its direction measured counterclockwise from the positive x axis.

F 2  75 N F 1  80 N F 3  50 N 30

SOLUTION F¿ =

 2 (80) (80)

2

45

 x

+ (50)2 - 2(80)(50) cos 105° = 104.7 N

sin f sin 105°  ; = 80 104.7 FR =

 2 (104.7) (104.7)

2

30

f = 47.54°

+ (75)2 - 2(104.7)(75) cos 162.46°

FR = 177.7 = 178 N

sin b sin 162.46°  ; = 104.7 177.7

Ans. b = 10.23°

u = 75° + 10.23° = 85.2°

Ans.

2–148.

If u = 60° and F = 20 kN, determine the magnitude of the resultant force and its direction measured clockwise from the positive x axis.

 y

50 kN

5

3

4

 x u

SOLUTION + F = © F  ; Rx x

:

+ c FRy = © Fy ;

2

1

FRx

FRy = 50

F

a b + 2 2 (40) - 20 cos 60° = 58.28 kN

4 = 50 5

1

40 kN

a 35 b - 2 1 2 (40) - 20 sin 60° = - 1515.6.600 kN

FR = 2 (58.28) (58.28)2 + ( - 15.60)2 = 60.3 kN f  = tan - 1

B R 15.60 58.28

= 15.0°

1

Ans. Ans.

2–149. Determine the design angle u (0° … u …  90°) for strut  AB so that the 400-N hor izontal force has a component of 500 N d irected from A towards C . What is the component of force act ing along member  AB? Take f = 40°.

400

N

A

u B

f

C 

N

.

N

N

. N