Vector Mechanics for Engineers Statics 7th - Cap 02

PROBLEM 2.1 Two forces are applied to an eye bolt fastened to a beam. Determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule.

SOLUTION (a)

(b)

We measure:

R

8.4 kN

D

19q R

1

8.4 kN

19q W

PROBLEM 2.2 The cable stays AB and AD help support pole AC. Knowing that the tension is 500 N in AB and 160 N 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

D

We measure:

51.3q, E

59q

(a)

(b)

We measure:

R

575 N, D

67q R

2

575 N

67q W

PROBLEM 2.3 Two forces P and Q are applied as shown at point A of a hook support. 15 lb and Q 25 lb, determine graphically the Knowing that P magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule.

SOLUTION (a)

(b)

We measure:

R

37 lb, D

76q R

3

37 lb

76q W

PROBLEM 2.4 Two forces P and Q are applied as shown at point A of a hook support. 45 lb and Q 15 lb, determine graphically the Knowing that P magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule.

SOLUTION (a)

(b)

We measure:

R

61.5 lb, D

86.5q R

4

61.5 lb

86.5q W

PROBLEM 2.5 Two control rods are attached at A to lever AB. Using trigonometry and knowing that the force in the left-hand rod is F1 120 N, determine (a) the required force F2 in the right-hand rod if the resultant R of the forces exerted by the rods on the lever is to be vertical, (b) the corresponding magnitude of R.

SOLUTION

Graphically, by the triangle law F2 # 108 N

We measure:

R # 77 N By trigonometry: Law of Sines F2 sin D

D

90q 28q

R sin 38q

62q, E

120 sin E

180q 62q 38q

80q

Then: F2 sin 62q

R sin 38q

120 N sin 80q or (a) F2 (b)

5

R

107.6 N W 75.0 N W

PROBLEM 2.6 Two control rods are attached at A to lever AB. Using trigonometry and 80 N, determine knowing that the force in the right-hand rod is F2 (a) the required force F1 in the left-hand rod if the resultant R of the forces exerted by the rods on the lever is to be vertical, (b) the corresponding magnitude of R.

SOLUTION

Using the Law of Sines F1 sin D

D

90q 10q

R sin 38q

80q, E

80 sin E

180q 80q 38q

62q

Then: F1 sin 80q

R sin 38q

80 N sin 62q or (a) F1 (b) R

6

89.2 N W 55.8 N W

PROBLEM 2.7 The 50-lb force is to be resolved into components along lines a-ac and b-bc. (a) Using trigonometry, determine the angle D knowing that the component along a-ac is 35 lb. (b) What is the corresponding value of the component along b-bc ?

SOLUTION

Using the triangle rule and the Law of Sines (a)

sin E 35 lb

sin 40q 50 lb

sin E

0.44995

E

26.74q

D E 40q

Then:

180q

D

113.3q W

(b) Using the Law of Sines:

Fbbc sin D

50 lb sin 40q

Fbbc

7

71.5 lb W

PROBLEM 2.8 The 50-lb force is to be resolved into components along lines a-ac and b-bc. (a) Using trigonometry, determine the angle D knowing that the component along b-bc is 30 lb. (b) What is the corresponding value of the component along a-ac ?

SOLUTION

Using the triangle rule and the Law of Sines (a)

sin D 30 lb

sin 40q 50 lb

sin D

0.3857

D (b)

D E 40q E Faac sin E Faac

22.7q W

180q 117.31q 50 lb sin 40q § sin E · 50 lb ¨ ¸ © sin 40q ¹ Faac

8

69.1 lb W

PROBLEM 2.9 To steady a sign as it is being lowered, two cables are attached to the sign at A. Using trigonometry and knowing that D 25q, determine (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.

SOLUTION

Using the triangle rule and the Law of Sines Have:

D

180q 35q 25q 120q

Then:

P sin 35q

R sin120q

360 N sin 25q or (a) P (b) R

9

489 N W 738 N W

PROBLEM 2.10 To steady a sign as it is being lowered, two cables are attached to the sign at A. Using trigonometry and knowing that the magnitude of P is 300 N, determine (a) the required angle D if the resultant R of the two forces applied at A is to be vertical, (b) the corresponding magnitude of R.

SOLUTION

Using the triangle rule and the Law of Sines (a) Have:

360 N sin D sin D

300 N sin 35q 0.68829

D E

(b)

43.5q W

180 35q 43.5q 101.5q

Then:

R sin101.5q

300 N sin 35q or R

10

513 N W

PROBLEM 2.11 Two forces are applied as shown to a hook support. Using trigonometry and knowing that the magnitude of P is 14 lb, determine (a) the required angle D 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 the Law of Sines

(a) Have:

20 lb sin D

14 lb sin 30q

sin D

0.71428

D 180q 30q 45.6q

E

(b)

45.6q W

104.4q Then:

R sin104.4q

14 lb sin 30q R

11

27.1 lb W

PROBLEM 2.12 For the hook support of Problem 2.3, using trigonometry and knowing that the magnitude of P is 25 lb, determine (a) the required magnitude of the force Q if the resultant R of the two forces applied at A is to be vertical, (b) the corresponding magnitude of R. Problem 2.3: Two forces P and Q are applied as shown at point A of a 15 lb and Q 25 lb, determine hook support. Knowing that P graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule.

SOLUTION Using the triangle rule and the Law of Sines

Q sin15q

(a) Have:

25 lb sin 30q Q

E

(b)

12.94 lb W

180q 15q 30q 135q R sin135q

Thus:

R

§ sin135q · 25 lb ¨ ¸ © sin 30q ¹

25 lb sin 30q 35.36 lb R

12

35.4 lb W

PROBLEM 2.13 For the hook support of Problem 2.11, determine, using 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. Problem 2.11: Two forces are applied as shown to a hook support. Using trigonometry and knowing that the magnitude of P is 14 lb, determine (a) the required angle D if the resultant R of the two forces applied to the support is to be horizontal, (b) the corresponding magnitude of R.

SOLUTION (a) The smallest force P will be perpendicular to R, that is, vertical

20 lb sin 30q

P

10 lb (b)

R

10 lb W

20 lb cos 30q 17.32 lb

13

P

R

17.32 lb W

PROBLEM 2.14 As shown in Figure P2.9, two cables are attached to a sign at A to steady the sign as it is being lowered. Using trigonometry, determine (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 We observe that force P is minimum when D is 90q, that is, P is horizontal

Then:

(a) P

360 N sin 35q or P

And:

(b) R

206 N

W

360 N cos 35q or R

14

295 N W

PROBLEM 2.15 For the hook support of Problem 2.11, determine, using trigonometry, the magnitude and direction of the resultant of the two forces applied to the support knowing that P 10 lb and D 40q.

Problem 2.11: Two forces are applied as shown to a hook support. Using trigonometry and knowing that the magnitude of P is 14 lb, determine (a) the required angle D 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 Law of Cosines

R2

10 lb 2 20 lb 2 2 10 lb 20 lb cos110q ª¬100 400 400 0.342 º¼ lb 2

636.8 lb 2 R

25.23 lb

Using now the Law of Sines 10 lb sin E

25.23 lb sin110q

sin E

§ 10 lb · ¨ ¸ sin110q © 25.23 lb ¹ 0.3724

So:

E

21.87q

Angle of inclination of R, I is then such that:

IE I Hence:

30q 8.13q

R

15

25.2 lb

8.13q W

PROBLEM 2.16 Solve Problem 2.1 using trigonometry

Problem 2.1: Two forces are applied to an eye bolt fastened to a beam. Determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule.

SOLUTION Using the force triangle, the Law of Cosines and the Law of Sines

180q 50q 25q

D

We have:

105q Then:

R2

4.5 kN 2 6 kN 2 2 4.5 kN 6 kN cos105q 70.226 kN 2

or

R

8.3801 kN

8.3801 kN sin105q

Now:

sin E

6 kN sin E

§ 6 kN · ¨ ¸ sin105q © 8.3801 kN ¹ 0.6916

E

43.756q

R

16

8.38 kN

18.76q W


Problem 2.2: The cable stays AB and AD help support pole AC. Knowing that the tension is 500 N in AB and 160 N 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 From the geometry of the problem: tan 1

2 2.5

38.66q

E

tan 1

1.5 2.5

30.96q

180q 38.66 30.96q

T

Now:

D

110.38

And, using the Law of Cosines: R2

500 N 2 160 N 2 2 500 N 160 N cos110.38q 331319 N 2 R

575.6 N

Using the Law of Sines: 160 N sin J sin J

575.6 N sin110.38q

§ 160 N · ¨ ¸ sin110.38q © 575.6 N ¹ 0.2606

J I

15.1q

90q D J

66.44q

R

17

576 N

66.4q W


Problem 2.3: Two forces P and Q are applied as shown at point A of a hook support. Knowing that P 15 lb and Q 25 lb, determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule.

SOLUTION Using the force triangle and the Laws of Cosines and Sines We have: 180q 15q 30q

J

135q Then:

R2

15 lb 2 25 lb 2 2 15 lb 25 lb cos135q 1380.3 lb 2

or

R

37.15 lb

and 25 lb sin E sin E

37.15 lb sin135q

§ 25 lb · ¨ ¸ sin135q © 37.15 lb ¹ 0.4758

E Then:

28.41q

D E 75q D

180q

76.59q

R

18

37.2 lb

76.6q W

PROBLEM 2.19 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 30 kN in member A and 20 kN in member B, determine, using 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

Then:

180q 45q 25q

J

We have: R2

110q

30 kN 2 20 kN 2 2 30 kN 20 kN cos110q 1710.4 kN 2

R

41.357 kN

and 20 kN sin D

41.357 kN sin110q

§ 20 kN · ¨ ¸ sin110q © 41.357 kN ¹

sin D

0.4544

D

27.028q

I

Hence:

D 45q

72.028q

R

19

41.4 kN

72.0q W

PROBLEM 2.20 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 20 kN in member A and 30 kN in member B, determine, using 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

Then:

180q 45q 25q

J

We have: R2

110q

30 kN 2 20 kN 2 2 30 kN 20 kN cos110q 1710.4 kN 2

R

41.357 kN

and 30 kN sin D

41.357 kN sin110q

§ 30 kN · ¨ ¸ sin110q © 41.357 kN ¹

sin D

0.6816

D Finally:

42.97q

I

D 45q

87.97q

R

20

41.4 kN

88.0q W

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

SOLUTION 20 kN Force: Fx

20 kN cos 40q,

Fx

15.32 kN W

Fy

20 kN sin 40q,

Fy

12.86 kN W

Fx

30 kN cos 70q,

Fy

30 kN sin 70q,

Fx

42 kN cos 20q,

Fx

39.5 kN W

Fy

42 kN sin 20q,

Fy

14.36 kN W

30 kN Force: 10.26 kN W

Fx Fy

28.2 kN W

42 kN Force:

21


SOLUTION 40 lb Force: Fx

40 lb sin 50q,

Fx

30.6 lb W

Fy

40 lb cos 50q,

Fy

25.7 lb W

Fx

60 lb cos 60q,

Fy

60 lb sin 60q,

Fx

80 lb cos 25q,

Fx

72.5 lb W

Fy

80 lb sin 25q,

Fy

33.8 lb W

60 lb Force: Fx Fy

30.0 lb W 52.0 lb W

80 lb Force:

22


SOLUTION We compute the following distances: OA

48 2 90 2

102 in.

OB

56 2 90 2

106 in.

OC

80 2 60 2

100 in.

Then: 204 lb Force:

Fx

102 lb

48 , 102

Fy

102 lb

90 , 102

Fx

212 lb

56 , 106

Fx

112.0 lb W

Fy

212 lb

90 , 106

Fy

180.0 lb W

Fx

400 lb

80 , 100

Fx

320 lb W

Fy

400 lb

60 , 100

Fy

240 lb W

Fx

Fy

48.0 lb W 90.0 lb W

212 lb Force:

400 lb Force:

23


SOLUTION We compute the following distances: OA

70 2 240 2

OB

210 2 200 2

290 mm

OC

120 2 225 2

255 mm

250 mm

500 N Force: Fx

§ 70 · 500 N ¨ ¸ © 250 ¹

Fy

§ 240 · 500 N ¨ ¸ © 250 ¹

Fy

480 N W

Fx

§ 210 · 435 N ¨ ¸ © 290 ¹

Fx

315 N W

Fy

§ 200 · 435 N ¨ ¸ © 290 ¹

Fy

300 N W

Fx

§ 120 · 510 N ¨ ¸ © 255 ¹

Fx

240 N W

Fy

§ 225 · 510 N ¨ ¸ © 255 ¹

140.0 N W

Fx

435 N Force:

510 N Force:

24

Fy

450 N W

PROBLEM 2.25 While emptying a wheelbarrow, a gardener exerts on each handle AB a force P directed along line CD. Knowing that P must have a 135-N horizontal component, determine (a) the magnitude of the force P, (b) its vertical component.

SOLUTION

(a)

P

Px cos 40q 135 N cos 40q

(b)

Py

Px tan 40q

or P

176.2 N W

or Py

113.3 N W

P sin 40q

135 N tan 40q

25

PROBLEM 2.26 Member BD exerts on member ABC a force P directed along line BD. Knowing that P must have a 960-N vertical component, determine (a) the magnitude of the force P, (b) its horizontal component.

SOLUTION

(a)

P

Py sin 35q 960 N sin 35q

(b)

Px

or P

1674 N W

or Px

1371 N W

Py tan 35q 960 N tan 35q

26

PROBLEM 2.27 Member CB of the vise shown exerts on block B a force P directed along line CB. Knowing that P must have a 260-lb horizontal component, determine (a) the magnitude of the force P, (b) its vertical component.

SOLUTION

We note: CB exerts force P on B along CB, and the horizontal component of P is Px

260 lb.

Then: (a)

P sin 50q

Px

Px sin 50q

P

260 lb sin50q P

339.4 lb (b)

Px

Py tan 50q

Py

Px tan 50q

339 lb W

260 lb tan 50q 218.2 lb

27

Py

218 lb W

PROBLEM 2.28 Activator rod AB exerts on crank BCD a force P directed along line AB. Knowing that P must have a 25-lb component perpendicular to arm BC of the crank, determine (a) the magnitude of the force P, (b) its component along line BC.

SOLUTION

Using the x and y axes shown. (a)

Py Then:

P

25 lb Py sin 75q 25 lb sin 75q

(b)

Px

or P

25.9 lb W

or Px

6.70 lb W

Py tan 75q 25 lb tan 75q

28

PROBLEM 2.29 The guy wire BD exerts on the telephone pole AC a force P directed along BD. Knowing that P has a 450-N component along line AC, determine (a) the magnitude of the force P, (b) its component in a direction perpendicular to AC.

SOLUTION

Note that the force exerted by BD on the pole is directed along BD, and the component of P along AC is 450 N. Then: (a)

(b)

P

Px

450 N cos 35q

549.3 N P

549 N W

Px

315 N W

450 N tan 35q 315.1 N

29

PROBLEM 2.30 The guy wire BD exerts on the telephone pole AC a force P directed along BD. Knowing that P has a 200-N perpendicular to the pole AC, determine (a) the magnitude of the force P, (b) its component along line AC.

SOLUTION

(a)

P

Px sin 38q 200 N sin 38q 324.8 N

(b)

Py

or P

325 N W

or Py

256 N W

Px tan 38q 200 N tan 38q 255.98 N

30

PROBLEM 2.31 Determine the resultant of the three forces of Problem 2.24. Problem 2.24: Determine the x and y components of each of the forces shown.

SOLUTION

From Problem 2.24: F500

R

140 N i 480 N j

F425

315 N i 300 N j

F510

240 N i 450 N j

6F

415 N i 330 N j

Then:

D

R

tan 1

330 415

38.5q

415 N 2 330 N 2

Thus:

530.2 N

R

31

530 N

38.5q W


SOLUTION

From Problem 2.21:

R

F20

15.32 kN i 12.86 kN j

F30

10.26 kN i 28.2 kN j

F42

39.5 kN i 14.36 kN j

6F

34.44 kN i 55.42 kN j

Then:

D

R

tan 1

55.42 34.44

58.1q

55.42 kN 2 34.44 N 2

65.2 kN R

32

65.2 kN

58.2q W


SOLUTION The components of the forces were determined in 2.23. Force

x comp. (lb)

y comp. (lb)

40 lb

30.6

25.7

60 lb

30

51.96

80 lb

72.5 Rx

R

71.9

33.8 Ry

43.86

Rxi Ry j

71.9 lb i 43.86 lb j

R

tan D

43.86 71.9

D

31.38q

71.9 lb 2 43.86 lb 2 84.23 lb R

33

84.2 lb

31.4q W

PROBLEM 2.34 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 The components of the forces were determined in Problem 2.23.

F204

48.0 lb i 90.0 lb j

F212

112.0 lb i 180.0 lb j 320 lb i 240 lb j

F400 Thus

R

Rx R y

256 lb i 30.0 lb j

R Now:

30.0 256

tan D tan 1

D

30.0 256

6.68q

and

256 lb 2 30.0 lb 2

R

257.75 lb

R

34

258 lb

6.68q W

PROBLEM 2.35 Knowing that D shown.

35q, determine the resultant of the three forces

SOLUTION 300-N Force: Fx

300 N cos 20q

281.9 N

Fy

300 N sin 20q

102.6 N

Fx

400 N cos55q

229.4 N

Fy

400 N sin 55q

327.7 N

Fx

600 N cos 35q

491.5 N

600 N sin 35q

344.1 N

400-N Force:

600-N Force:

Fy

and 6Fx

Rx Ry R

6Fy

1002.8 N 86.2 N

1002.8 N 2 86.2 N 2

1006.5 N

Further: tan D

D

tan 1

86.2 1002.8 86.2 1002.8

4.91q

R

35

1007 N

4.91q W



SOLUTION 300-N Force: Fx

300 N cos 20q

281.9 N

Fy

300 N sin 20q

102.6 N

Fx

400 N cos85q

34.9 N

Fy

400 N sin 85q

398.5 N

Fx

600 N cos 5q

597.7 N

600 N sin 5q

52.3 N

400-N Force:

600-N Force:

Fy

and

R

Rx

6Fx

914.5 N

Ry

6Fy

448.8 N

914.5 N 2 448.8 N 2

1018.7 N

Further: tan D

D

tan 1

448.8 914.5 448.8 914.5

26.1q

R

36

1019 N

26.1q W

PROBLEM 2.37 Knowing that the tension in cable BC is 145 lb, determine the resultant of the three forces exerted at point B of beam AB.

SOLUTION Cable BC Force:

145 lb

84 116

105 lb

Fy

145 lb

80 116

100 lb

Fx

100 lb

3 5

60 lb

Fy

100 lb

4 5

80 lb

Fx

156 lb

12 13

144 lb

Fx

100-lb Force:

156-lb Force:

156 lb

Fy

5 13

60 lb

and Rx R

6Fx

21 lb,

6Fy

Ry

21 lb 2 40 lb 2

40 lb

45.177 lb

Further: 40 21

tan D

D Thus:

tan 1

40 21

62.3q

R

37

45.2 lb

62.3q W



SOLUTION The resultant force R has the x- and y-components:

140 lb cos 50q 60 lb cos85q 160 lb cos 50q

Rx

6Fx

Rx

7.6264 lb

Ry

6Fy

Ry

289.59 lb

and

140 lb sin 50q 60 lb sin 85q 160 lb sin 50q

Further: tan D

D Thus:

tan 1

290 7.6 290 7.6

88.5q

R

38

290 lb

88.5q W

PROBLEM 2.39 Determine (a) the required value of D if the resultant of the three forces shown is to be vertical, (b) the corresponding magnitude of the resultant.

SOLUTION For an arbitrary angle D , we have: 6Fx

Rx

140 lb cosD 60 lb cos D 35q 160 lb cosD

(a) So, for R to be vertical: Rx

6Fx

140 lb cosD 60 lb cos D 35q 160 lb cosD

0

Expanding, cos D 3 cos D cos 35q sin D sin 35q

0

Then: tan D

cos 35q sin 35q

1 3

or

D

§ cos 35q tan 1 ¨ ¨ sin 35q ©

1 3

· ¸¸ ¹

D

40.265q

40.3q W

(b) Now: R

Ry

6Fy

140 lb sin 40.265q 60 lb sin 75.265q 160 lb sin 40.265q R

39

R

252 lb W

PROBLEM 2.40 For the beam of Problem 2.37, determine (a) the required tension in cable BC if the resultant of the three forces exerted at point B is to be vertical, (b) the corresponding magnitude of the resultant.

Problem 2.37: Knowing that the tension in cable BC is 145 lb, determine the resultant of the three forces exerted at point B of beam AB.

SOLUTION We have: Rx

6Fx

or

84 12 3 TBC 156 lb 100 lb 116 13 5 Rx

0.724TBC 84 lb

and 80 5 4 TBC 156 lb 100 lb 116 13 5

Ry

6Fy

Ry

0.6897TBC 140 lb

(a) So, for R to be vertical, Rx

0.724TBC 84 lb

0 TBC

116.0 lb W

(b) Using TBC R

Ry

116.0 lb

0.6897 116.0 lb 140 lb

60 lb R

40

R

60.0 lb W

PROBLEM 2.41 Boom AB is held in the position shown by three cables. Knowing that the tensions in cables AC and AD are 4 kN and 5.2 kN, respectively, determine (a) the tension in cable AE if the resultant of the tensions exerted at point A of the boom must be directed along AB, (b) the corresponding magnitude of the resultant.

SOLUTION

Choose x-axis along bar AB. Then (a) Require

Ry

6Fy

0:

TAE

or

(b)

4 kN cos 25q 5.2 kN sin 35q TAE sin 65q

R

0

7.2909 kN TAE

7.29 kN W

R

9.03 kN W

6Fx 4 kN sin 25q 5.2 kN cos 35q 7.2909 kN cos 65q

9.03 kN

41

PROBLEM 2.42 For the block of Problems 2.35 and 2.36, determine (a) the required value of D of the resultant of the three forces shown is to be parallel to the incline, (b) the corresponding magnitude of the resultant. Problem 2.35: Knowing that D three forces shown.

35q, determine the resultant of the

Problem 2.36: Knowing that D three forces shown.

65q, determine the resultant of the

SOLUTION

Selecting the x axis along aac, we write

(a) Setting Ry

Rx

6Fx

300 N 400 N cos D 600 N sin D

(1)

Ry

6Fy

400 N sin D 600 N cosD

(2)

0 in Equation (2): tan D

Thus

600 400

D

1.5 56.3q W

(b) Substituting for D in Equation (1):

Rx

300 N 400 N cos 56.3q 600 N sin 56.3q

Rx

1021.1 N

R

42

Rx

1021 N W

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

From the geometry, we calculate the distances:

AC

16 in. 2 12 in. 2

20 in.

BC

20 in. 2 21 in. 2

29 in.

Then, from the Free Body Diagram of point C: 6Fx or and or

0: TBC

6Fy

0:

16 21 TAC TBC 20 29

0

29 4 u TAC 21 5 12 20 TAC TBC 600 lb 20 29

12 20 § 29 4 · TAC u TAC ¸ 600 lb ¨ 20 29 © 21 5 ¹

Hence:

TAC

0 0

440.56 lb

(a)

TAC

441 lb W

(b)

TBC

487 lb W

43

PROBLEM 2.44 Knowing that D rope BC.

25q, determine the tension (a) in cable AC, (b) in


Force Triangle

Law of Sines: TAC sin115q

TBC sin 5q

(a)

TAC

5 kN sin115q sin 60q

(b)

TBC

5 kN sin 5q sin 60q

5 kN sin 60q

5.23 kN

0.503 kN

44

TAC

TBC

5.23 kN W 0.503 kN W

PROBLEM 2.45 Knowing that D 50q and that boom AC exerts on pin C a force directed long line AC, determine (a) the magnitude of that force, (b) the tension in cable BC.


Force Triangle

Law of Sines: FAC sin 25q

TBC sin 60q

400 lb sin 95q

(a)

FAC

400 lb sin 25q sin 95q

169.69 lb

(b)

TBC

400 sin 60q sin 95q

347.73 lb

45

FAC

TBC

169.7 lb W 348 lb W

PROBLEM 2.46 Two cables are tied together at C and are loaded as shown. Knowing that D 30q, determine the tension (a) in cable AC, (b) in cable BC.


Force Triangle

Law of Sines: TAC sin 60q

TBC sin 55q

2943 N sin 65q

(a)

TAC

2943 N sin 60q sin 65q

2812.19 N

TAC

2.81 kN W

(b)

TBC

2943 N sin 55q sin 65q

2659.98 N

TBC

2.66 kN W

46

PROBLEM 2.47 A chairlift has been stopped in the position shown. Knowing that each chair weighs 300 N and that the skier in chair E weighs 890 N, determine that weight of the skier in chair F.

SOLUTION Free-Body Diagram Point B

In the free-body diagram of point B, the geometry gives:

T AB

tan 1

9.9 16.8

30.51q

T BC

tan 1

12 28.8

22.61q

Thus, in the force triangle, by the Law of Sines: Force Triangle

TBC sin 59.49q TBC

Free-Body Diagram Point C

1190 N sin 7.87q 7468.6 N

In the free-body diagram of point C (with W the sum of weights of chair and skier) the geometry gives:

T CD

tan 1

1.32 7.2

10.39q

Hence, in the force triangle, by the Law of Sines: Force Triangle

W sin12.23q

W Finally, the skier weight

7468.6 N sin100.39q 1608.5 N

1608.5 N 300 N

1308.5 N skier weight

47

1309 N W

PROBLEM 2.48 A chairlift has been stopped in the position shown. Knowing that each chair weighs 300 N and that the skier in chair F weighs 800 N, determine the weight of the skier in chair E.

SOLUTION Free-Body Diagram Point F

In the free-body diagram of point F, the geometry gives:

T EF

tan 1

12 28.8

22.62q

T DF

tan 1

1.32 7.2

10.39q

Thus, in the force triangle, by the Law of Sines: Force Triangle

TEF sin100.39q TBC

Free-Body Diagram Point E

1100 N sin12.23q

5107.5 N

In the free-body diagram of point E (with W the sum of weights of chair and skier) the geometry gives: tan 1

T AE

9.9 16.8

30.51q

Hence, in the force triangle, by the Law of Sines:

W sin 7.89q

Force Triangle

W Finally, the skier weight

5107.5 N sin 59.49q 813.8 N

813.8 N 300 N

513.8 N skier weight

48

514 N W

PROBLEM 2.49 Four wooden members are joined with metal plate connectors and are in equilibrium under the action of the four fences shown. Knowing that FA 510 lb and FB 480 lb, determine the magnitudes of the other two forces.


Resolving the forces into x and y components: 6Fx

6Fy

0: FC 510 lb sin15q 480 lb cos15q

0: FD 510 lb cos15q 480 lb sin15q

49

0 or FC

332 lb W

or FD

368 lb W

0

PROBLEM 2.50 Four wooden members are joined with metal plate connectors and are in equilibrium under the action of the four fences shown. Knowing that FA 420 lb and FC 540 lb, determine the magnitudes of the other two forces.

SOLUTION

Resolving the forces into x and y components: 6Fx

0: FB cos15q 540 lb 420 lb cos15q

6Fy

0

or

0: FD 420 lb cos15q 671.6 lb sin15q

50

FB

671.6 lb

FB

672 lb W

or FD

232 lb W

0

PROBLEM 2.51 Two forces P and Q are applied as shown to an aircraft connection. 400 lb and Knowing that the connection is in equilibrium and the P Q 520 lb, determine the magnitudes of the forces exerted on the rods A and B.


Resolving the forces into x and y directions:

R

P Q FA FB

0

Substituting components: R

400 lb j ª¬ 520 lb cos 55q º¼ i ª¬ 520 lb sin 55qº¼ j

FBi FA cos 55q i FA sin 55q j

0

In the y-direction (one unknown force) 400 lb 520 lb sin 55q FA sin 55q

0

Thus, FA

400 lb 520 lb sin 55q sin 55q

1008.3 lb FA

1008 lb W

In the x-direction:

520 lb cos55q FB FA cos 55q

0

Thus, FB

FA cos 55q 520 lb cos 55q

1008.3 lb cos 55q 520 lb cos 55q 280.08 lb FB

51

280 lb W

PROBLEM 2.52 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 600 lb and FB 320 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

320 lb i ª¬ 600 lb cos 55qº¼ i ª¬ 600 lb sin 55qº¼ j Pi Q cos 55q i Q sin 55q j

0

In the x-direction (one unknown force) 320 lb 600 lb cos 55q Q cos 55q

0

Thus, Q

320 lb 600 lb cos 55q cos 55q

42.09 lb Q

42.1 lb W

P

457 lb W

In the y-direction:

600 lb sin 55q P Q sin 55q

0

Thus, P

600 lb sin 55q Q sin 55q

52

457.01 lb

PROBLEM 2.53 Two cables tied together at C are loaded as shown. Knowing that W 840 N, determine the tension (a) in cable AC, (b) in cable BC.


From geometry: The sides of the triangle with hypotenuse CB are in the ratio 8:15:17. The sides of the triangle with hypotenuse CA are in the ratio 3:4:5. Thus: 3 15 15 0: TCA TCB 680 N 5 17 17

6Fx

0

or 1 5 TCA TCB 5 17

200 N

(1)

and 6Fy

0:

4 8 8 TCA TCB 680 N 840 N 5 17 17

0

or 1 2 TCA TCB 5 17

290 N

(2)

Solving Equations (1) and (2) simultaneously: TCA

(a) (b)

TCB

53

750 N W 1190 N W

PROBLEM 2.54 Two cables tied together at C are loaded as shown. Determine the range of values of W for which the tension will not exceed 1050 N in either cable.


From geometry: The sides of the triangle with hypotenuse CB are in the ratio 8:15:17. The sides of the triangle with hypotenuse CA are in the ratio 3:4:5. Thus: 3 15 15 0: TCA TCB 680 N 5 17 17

6Fx

0

or 1 5 TCA TCB 5 17

(1)

200 N

and 6Fy

4 8 8 TCA TCB 680 N W 5 17 17

0:

0

or 1 2 TCA TCB 5 17

80 N

1 W 4

(2)

Then, from Equations (1) and (2) TCB

680 N

TCA

25 W 28

17 W 28

Now, with T d 1050 N TCA : TCA or

W

1050 N

25 W 28

1176 N

and TCB : TCB or

W

54

1050 N

609 N

680 N

17 W 28 ? 0 d W d 609 N W

PROBLEM 2.55 The cabin of an aerial tramway is suspended from a set of wheels that can roll freely on the support cable ACB and is being pulled at a constant 35q, that the speed by cable DE. Knowing that D 40q and E combined weight of the cabin, its support system, and its passengers is 24.8 kN, and assuming the tension in cable DF to be negligible, determine the tension (a) in the support cable ACB, (b) in the traction cable DE.

SOLUTION Note: In Problems 2.55 and 2.56 the cabin is considered as a particle. If considered as a rigid body (Chapter 4) it would be found that its center of gravity should be located to the left of the centerline for the line CD to be vertical. Now 6Fx

0: TACB cos 35q cos 40q TDE cos 40q

0

or 0.0531TACB 0.766TDE

0

(1)

and 0: TACB sin 40q sin 35q TDE sin 40q 24.8 kN

6Fy

0

or 0.0692TACB 0.643TDE

24.8 kN

(2)

From (1) TACB

14.426TDE

Then, from (2) 0.0692 14.426TDE 0.643TDE

24.8 kN

and

55

(b) TDE

15.1 kN W

(a) TACB

218 kN W

PROBLEM 2.56 The cabin of an aerial tramway is suspended from a set of wheels that can roll freely on the support cable ACB and is being pulled at a constant speed by cable DE. Knowing that D 42q and E 32q, that the tension in cable DE is 20 kN, and assuming the tension in cable DF to be negligible, determine (a) the combined weight of the cabin, its support system, and its passengers, (b) the tension in the support cable ACB.


First, consider the sum of forces in the x-direction because there is only one unknown force: 6Fx

0: TACB cos 32q cos 42q 20 kN cos 42q

0

or 0.1049TACB

14.863 kN (b) TACB

141.7 kN W

Now 6Fy

0: TACB sin 42q sin 32q 20 kN sin 42q W

0

or

141.7 kN 0.1392 20 kN 0.6691 W

0 (a) W

56

33.1 kN W

PROBLEM 2.57 A block of weight W is suspended from a 500-mm long cord and two springs of which the unstretched lengths are 450 mm. Knowing that the 1500 N/m and kAD 500 N/m, constants of the springs are kAB determine (a) the tension in the cord, (b) the weight of the block.

SOLUTION Free-Body Diagram At A

First note from geometry: The sides of the triangle with hypotenuse AD are in the ratio 8:15:17. The sides of the triangle with hypotenuse AB are in the ratio 3:4:5. The sides of the triangle with hypotenuse AC are in the ratio 7:24:25. Then: FAB

k AB LAB Lo

and LAB

0.44 m 2 0.33 m 2

0.55 m

So: FAB

1500 N/m 0.55 m 0.45 m 150 N

Similarly, FAD

k AD LAD Lo

Then:

0.66 m 2 0.32 m 2

LAD FAD

0.68 m

1500 N/m 0.68 m 0.45 m 115 N

(a) 6Fx

0:

4 7 15 150 N TAC 115 N 5 25 17

0

or TAC

57

66.18 N

TAC

66.2 N W

PROBLEM 2.57 CONTINUED (b) and 6Fy

0:

3 24 8 150 N 66.18 N 115 N W 5 25 17 or W

58

0 208 N W

PROBLEM 2.58 A load of weight 400 N is suspended from a spring and two cords which are attached to blocks of weights 3W and W as shown. Knowing that the constant of the spring is 800 N/m, determine (a) the value of W, (b) the unstretched length of the spring.

SOLUTION Free-Body Diagram At A

First note from geometry: The sides of the triangle with hypotenuse AD are in the ratio 12:35:37. The sides of the triangle with hypotenuse AC are in the ratio 3:4:5. The sides of the triangle with hypotenuse AB are also in the ratio 12:35:37. Then: 6Fx

0:

4 35 12 3W W Fs 5 37 37

0

or Fs

4.4833W

and 6Fy

0:

3 12 35 3W W Fs 400 N 5 37 37

0

Then: 3 12 35 3W W 4.4833W 400 N 5 37 37

0

or W

62.841 N

and Fs

281.74 N

or (a)

W

59

62.8 N W

PROBLEM 2.58 CONTINUED (b) Have spring force Fs

k LAB Lo

Where FAB

k AB LAB Lo

and LAB

0.360 m 2 1.050 m 2

1.110 m

So: 281.74 N

800 N/m 1.110 L0 m or L0

60

758 mm W

PROBLEM 2.59 For the cables and loading of Problem 2.46, determine (a) the value of D for which the tension in cable BC is as small as possible, (b) the corresponding value of the tension.

SOLUTION The smallest TBC is when TBC is perpendicular to the direction of TAC

Free-Body Diagram At C

Force Triangle

D

(a) (b)

TBC

55.0q W

2943 N sin 55q 2410.8 N TBC

61

2.41 kN W

PROBLEM 2.60 Knowing that portions AC and BC of cable ACB must be equal, determine the shortest length of cable which can be used to support the load shown if the tension in the cable is not to exceed 725 N.

SOLUTION Free-Body Diagram: C For T 725 N

6Fy

0: 2Ty 1000 N Ty

500 N

Tx2 Ty2 Tx2 500 N

Tx

0

2

T2

725 N 2

525 N

By similar triangles: BC 725 ? BC L

2 BC

1.5 m 525 2.07 m 4.14 m L

62

4.14 m W

PROBLEM 2.61 Two cables tied together at C are loaded as shown. Knowing that the maximum allowable tension in each cable is 200 lb, determine (a) the magnitude of the largest force P which may be applied at C, (b) the corresponding value of D.

SOLUTION Free-Body Diagram: C

Force Triangle

Force triangle is isoceles with 2E

180q 85q

E P

(a)

47.5q

2 200 lb cos 47.5q

Since P ! 0, the solution is correct.

(b)

D

270 lb

P

180q 55q 47.5q

63

77.5q

D

270 lb W 77.5q W

PROBLEM 2.62 Two cables tied together at C are loaded as shown. Knowing that the maximum allowable tension is 300 lb in cable AC and 150 lb in cable BC, determine (a) the magnitude of the largest force P which may be applied at C, (b) the corresponding value of D.

SOLUTION Free-Body Diagram: C

Force Triangle

(a) Law of Cosines: P2

P

300 lb 2 150 lb 2 2 300 lb 150 lb cos85q 323.5 lb

Since P ! 300 lb, our solution is correct.

P

324 lb W

(b) Law of Sines:

or

D

sin E 300

sin 85q 323.5q

sin E

0.9238

E

67.49q

180q 55q 67.49q

57.5q

D

64

57.5q W

PROBLEM 2.63 For the structure and loading of Problem 2.45, determine (a) the value of D for which the tension in cable BC is as small as possible, (b) the corresponding value of the tension.

SOLUTION TBC must be perpendicular to FAC to be as small as possible. Free-Body Diagram: C

(a) We observe: (b) or

Force Triangle is a right triangle

D TBC

400 lb sin 60q

TBC

346.4 lb

65

D

55q

TBC

55q W

346 lb W

PROBLEM 2.64 Boom AB is supported by cable BC and a hinge at A. Knowing that the boom exerts on pin B a force directed along the boom and that the tension in rope BD is 70 lb, determine (a) the value of D for which the tension in cable BC is as small as possible, (b) the corresponding value of the tension.

SOLUTION Free-Body Diagram: B

TBD FAB TBC

(a) Have:

0

where magnitude and direction of TBD are known, and the direction of FAB is known.

Then, in a force triangle:

D

By observation, TBC is minimum when (b) Have

TBC

90.0q W

70 lb sin 180q 70q 30q 68.93 lb TBC

66

68.9 lb W

PROBLEM 2.65 Collar A shown in Figure P2.65 and P2.66 can slide on a frictionless vertical rod and is attached as shown to a spring. The constant of the spring is 660 N/m, and the spring is unstretched when h 300 mm. Knowing that the system is in equilibrium when h 400 mm, determine the weight of the collar.

SOLUTION Free-Body Diagram: Collar A Fs

Have:

k LcAB LAB

where: LcAB

0.3 m 2 0.4 m 2

LAB

0.3 2 m

0.5 m Fs

Then:

660 N/m 0.5 0.3 2 m

49.986 N For the collar: 6Fy

0: W

4 49.986 N 5

0 or W

67

40.0 N W

PROBLEM 2.66 The 40-N collar A can slide on a frictionless vertical rod and is attached as shown to a spring. The spring is unstretched when h 300 mm. Knowing that the constant of the spring is 560 N/m, determine the value of h for which the system is in equilibrium.

SOLUTION 6Fy

Free-Body Diagram: Collar A

hFs

or

Fs

Now.. where

LcAB

Then:

h ª560 «¬

or

h

0: W

0.3 2 h2

k LcAB LAB LAB

m

0.09 h 2 0.3 2 º »¼

14h 1

0

40 0.09 h 2

0.3 2 h2

Fs

0.09 h 2

4.2 2h

0.3 2 m 40 0.09 h 2

hm

Solving numerically, h

68

415 mm W

PROBLEM 2.67 A 280-kg crate is supported by several rope-and-pulley arrangements as shown. Determine for each arrangement the tension in the rope. (Hint: The tension in the rope is the same on each side of a simple pulley. This can be proved by the methods of Chapter 4.)

SOLUTION Free-Body Diagram of pulley

6Fy

(a)

0: 2T 280 kg 9.81 m/s 2 T

0

1 2746.8 N 2 T

6Fy

(b)

0: 2T 280 kg 9.81 m/s 2 T

0

1 2746.8 N 2 T

6Fy

(c)

0: 3T 280 kg 9.81 m/s 2 T

(d)

6Fy

6Fy

(e)

69

0

T

916 N W

T

916 N W

T

687 N W

0

1 2746.8 N 3

0: 4T 280 kg 9.81 m/s 2 T

1373 N W

1 2746.8 N 3

0: 3T 280 kg 9.81 m/s 2 T

1373 N W

0

1 2746.8 N 4

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 280-kg crate is supported by several rope-and-pulley arrangements as shown. Determine for each arrangement the tension in the rope. (Hint: The tension in the rope is the same on each side of a simple pulley. This can be proved by the methods of Chapter 4.)

SOLUTION Free-Body Diagram of pulley and crate (b) 6Fy

0: 3T 280 kg 9.81 m/s 2 T

0

1 2746.8 N 3 T

916 N W

(d)

6Fy

0: 4T 280 kg 9.81 m/s 2 T

0

1 2746.8 N 4 T

70

687 N W

PROBLEM 2.69 A 350-lb load is supported by the rope-and-pulley arrangement shown. Knowing that E 25q, determine the magnitude and direction of the force P which should 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 Chapter 4.)

SOLUTION Free-Body Diagram: Pulley A

6Fx

0: 2P sin 25q P cos D

0

and cos D

0.8452

D

For 6Fy

or

D

r32.3q

32.3q

0: 2P cos 25q P sin 32.3q 350 lb or P

D

For 6Fy

32.3q W

149.1 lb

32.3q

0: 2P cos 25q P sin 32.3q 350 lb or P

71

0

274 lb

0 32.3q W

PROBLEM 2.70 A 350-lb load is supported by the rope-and-pulley arrangement shown. Knowing that D 35q, determine (a) the angle E, (b) the magnitude of the force P which should 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 Chapter 4.)

SOLUTION Free-Body Diagram: Pulley A

6Fx

0: 2 P sin E P cos 25q

0

1 cos 25q 2

or E

Hence: sin E

(a)

0: 2P cos E P sin 35q 350 lb

6Fy

(b)

24.2q W 0

Hence: 2P cos 24.2q P sin 35q 350 lb or

P

72

145.97 lb

0 P

146.0 lb W

PROBLEM 2.71 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 800 N, over the pulley A and supports a load P. Knowing that P determine (a) the tension in cable ACB, (b) the magnitude of load Q.

SOLUTION Free-Body Diagram: Pulley C

6Fx

(a)

0: TACB cos 30q cos 50q 800 N cos 50q

TACB

Hence

0

2303.5 N TACB

6Fy

(b)

0: TACB sin 30q sin 50q 800 N sin 50q Q

2303.5 N sin 30q sin 50q 800 N sin 50q Q or

Q

73

3529.2 N

2.30 kN W

0

0 Q

3.53 kN W

PROBLEM 2.72 A 2000-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 the cable ACB, (b) the magnitude of load P.

SOLUTION Free-Body Diagram: Pulley C

6Fx

0: TACB cos 30q cos 50q P cos 50q

or

P 6Fy

0.3473TACB

(1)

0: TACB sin 30q sin 50q P sin 50q 2000 N 1.266TACB 0.766P

or

0

0

2000 N

(2)

(a) Substitute Equation (1) into Equation (2): 1.266TACB 0.766 0.3473TACB Hence:

TACB

2000 N

1305.5 N TACB

1306 N W

(b) Using (1) P

0.3473 1306 N

453.57 N

P

74

454 N W

PROBLEM 2.73 Determine (a) the x, y, and z components of the 200-lb force, (b) the angles Tx, Ty, and Tz that the force forms with the coordinate axes.

SOLUTION (a)

Fx

Fy

Fz

200 lb cos 30q cos 25q

200 lb sin 30q

200 lb cos 30q sin 25q

156.98 lb Fx

157.0 lb W

Fy

100.0 lb W

100.0 lb

73.1996 lb Fz

73.2 lb W

(b)

cosT x

156.98 200

or T x

38.3q W

cosT y

100.0 200

or T y

60.0q W

cosT z

75

73.1996 200

or T z

111.5q W

PROBLEM 2.74 Determine (a) the x, y, and z components of the 420-lb force, (b) the angles Tx, Ty, and Tz that the force forms with the coordinate axes.

SOLUTION (a)


Fx

134.985 lb 135.0 lb W

Fx Fy

Fz

(b)

420 lb cos 20q

394.67 lb

420 lb sin 20q cos 70q

cosT x

Fy

395 lb W

Fz

49.1 lb W

49.131 lb

134.985 420

Tx cosT y

cosT z

108.7q W

394.67 420

Ty

20.0q W

Tz

83.3q W

49.131 420

76

PROBLEM 2.75 To stabilize a tree partially uprooted in a storm, cables AB and AC are attached to the upper trunk of the tree and then are fastened to steel rods anchored in the ground. Knowing that the tension in cable AB is 4.2 kN, determine (a) the components of the force exerted by this cable on the tree, (b) the angles Tx, Ty, and Tz that the force forms with axes at A which are parallel to the coordinate axes.

SOLUTION

(a)

(b)

Fx

Fy

Fz

4.2 kN sin 50q cos 40q

4.2 kN cos 50q

Fx

2.46 kN W

Fy

2.70 kN W

Fz

2.07 kN W

2.6997 kN

4.2 kN sin 50q sin 40q

cosT x

2.4647 kN

2.0681 kN

2.4647 4.2

Tx

77

54.1q W

PROBLEM 2.75 CONTINUED

cosT y

2.7 4.2

Ty cosT z

130.0q W

2.0681 4.0

Tz

78

60.5q W

PROBLEM 2.76 To stabilize a tree partially uprooted in a storm, cables AB and AC are attached to the upper trunk of the tree and then are fastened to steel rods anchored in the ground. Knowing that the tension in cable AC is 3.6 kN, determine (a) the components of the force exerted by this cable on the tree, (b) the angles Tx, Ty, and Tz that the force forms with axes at A which are parallel to the coordinate axes.

SOLUTION

(a)

Fx

3.6 kN cos 45q sin 25q

1.0758 kN

Fx Fy

Fz

(b)

3.6 kN sin 45q

3.6 kN cos 45q cos 25q

cosT x

1.076 kN W

2.546 kN

Fy

2.55 kN W

Fz

2.31 kN W

2.3071 kN

1.0758 3.6

Tx

79

107.4q W

PROBLEM 2.76 CONTINUED cosT y

2.546 3.6

Ty cosT z

135.0q W

2.3071 3.6

Tz

80

50.1q W

PROBLEM 2.77 A horizontal circular plate is suspended as shown from three wires which are attached to a support at D and form 30q angles with the vertical. Knowing that the x component of the force exerted by wire AD on the plate is 220.6 N, determine (a) the tension in wire AD, (b) the angles Tx, Ty, and Tz that the force exerted at A forms with the coordinate axes.

SOLUTION (a)

Fx

F sin 30q sin 50q

220.6 N sin30q sin50q

F

(b)

cosT x

Fy cosT y

220.6 N (Given)

Fz

Fx F

220.6 575.95

F cos 30q Fy F

575.95 N 576 N W

Tx

67.5q W

Ty

30.0q W

0.3830

498.79 N

498.79 575.95

0.86605

F sin 30q cos 50q

575.95 N sin 30q cos 50q

185.107 N cosT z

F

185.107 575.95

Fz F

0.32139

Tz

81

108.7q W

PROBLEM 2.78 A horizontal circular plate is suspended as shown from three wires which are attached to a support at D and form 30q angles with the vertical. Knowing that the z component of the force exerted by wire BD on the plate is –64.28 N, determine (a) the tension in wire BD, (b) the angles Tx, Ty, and Tz that the force exerted at B forms with the coordinate axes.

SOLUTION (a)

Fz

F sin 30q sin 40q F

64.28 N (Given)

64.28 N sin30q sin40q

(b)

200.0 N

200.0 N sin 30q cos 40q

76.604 N

Fy

cosT y

cosT z

200 N W

Tx

112.5q W

F sin 30q cos 40q

Fx

cosT x

F

76.604 200.0

Fx F

F cos 30q Fy F

0.38302

173.2 N

173.2 200

0.866

Fz

64.28 N

Fz F

64.28 200

0.3214

82

Ty

Tz

30.0q W

108.7q W

PROBLEM 2.79 A horizontal circular plate is suspended as shown from three wires which are attached to a support at D and form 30q angles with the vertical. Knowing that the tension in wire CD is 120 lb, determine (a) the components of the force exerted by this wire on the plate, (b) the angles Tx, Ty, and Tz that the force forms with the coordinate axes.

SOLUTION (a)

120 lb sin 30q cos 60q

Fx

30 lb 30.0 lb W

Fx

120 lb cos 30q

Fy

103.92 lb 103.9 lb W

Fy Fz


51.96 lb 52.0 lb W

Fz (b)

cosT x

30.0 120

Fx F

0.25

Tx cosT y

cosT z

Fy F

Fz F

83

103.92 120

51.96 120

104.5q W

0.866

Ty

30.0q W

Tz

64.3q W

0.433

PROBLEM 2.80 A horizontal circular plate is suspended as shown from three wires which are attached to a support at D and form 30q angles with the vertical. Knowing that the x component of the forces exerted by wire CD on the plate is –40 lb, determine (a) the tension in wire CD, (b) the angles Tx, Ty, and Tz that the force exerted at C forms with the coordinate axes.

SOLUTION (a)

F sin 30q cos 60q

Fx

40 lb (Given)

40 lb sin30q cos60q

F

160 lb 160.0 lb W

F cosT x

(b)

Fx F

40 160

0.25

Tx Fy

160 lb cos 30q

cosT y

Fz

Fy F

103.92 lb

103.92 160

0.866

160 lb sin 30q sin 60q cosT z

Fz F

104.5q W

69.282 160

84

Ty

30.0q W

Tz

64.3q W

69.282 lb 0.433

PROBLEM 2.81 Determine the magnitude and F 800 lb i 260 lb j 320 lb k.

direction

of

the

force

SOLUTION F

Fx2 Fy2 Fz2

800 lb 2 260 lb 2 320 lb 2

F

900 lb W

cosT x

Fx F

800 900

0.8889

Tx

27.3q W

cosT y

Fy

260 900

0.2889

Ty

73.2q W

0.3555

Tz

cosT z

F Fz F

85

320 900

110.8q W

PROBLEM 2.82 Determine the magnitude and direction F 400 N i 1200 N j 300 N k.

of

the

force

SOLUTION F

Fx2 Fy2 Fz2

cosT x

cosT y cosT z

400 N 2 1200 N 2 300 N 2 Fx F Fy F Fz F

400 1300 1200 1300 300 1300

0.30769

0.92307

0.23076

86

F

Tx Ty Tz

1300 N W 72.1q W

157.4q W 76.7q W

PROBLEM 2.83 A force acts at the origin of a coordinate system in a direction defined by the angles Tx 64.5q and Tz 55.9q. Knowing that the y component of the force is –200 N, determine (a) the angle Ty, (b) the other components and the magnitude of the force.

SOLUTION (a) We have

cosT x 2 cosT y

2

cosT z

2

1 cosT y

2

1 cosT y

2 cosT z 2

Since Fy 0 we must have cosT y 0

Thus, taking the negative square root, from above, we have: cosT y

1 cos 64.5q cos 55.9q 2

2

0.70735

Ty

135.0q W

(b) Then:

and

Fy

200 N 0.70735

F

cosT y

Fx

F cosT x

282.73 N cos 64.5q

Fx

121.7 N W

Fz

F cosT z

282.73 N cos 55.9q

Fy

158.5 N W

282.73 N

F

87

283 N W

PROBLEM 2.84 A force acts at the origin of a coordinate system in a direction defined by the angles Tx 75.4q and Ty 132.6q. Knowing that the z component of the force is –60 N, determine (a) the angle Tz, (b) the other components and the magnitude of the force.

SOLUTION (a) We have

cosT x 2 cosT y

2

cosT z

2

1 cosT y

2

1 cosT y

2 cosT z 2

Since Fz 0 we must have cosT z 0

Thus, taking the negative square root, from above, we have: cosT z

1 cos 75.4q cos132.6q 2

2

0.69159

Tz

133.8q W

F

86.8 N W

Fx

21.9 N W

(b) Then:

and

F

Fz cosT z

60 N 0.69159

86.757 N

Fx

F cosT x

86.8 N cos 75.4q

Fy

F cosT y

86.8 N cos132.6q

88

Fy

58.8 N W

PROBLEM 2.85 A force F of magnitude 400 N acts at the origin of a coordinate system. Knowing that Tx 28.5q, Fy –80 N, and Fz ! 0, determine (a) the components Fx and Fz, (b) the angles Ty and Tz.

SOLUTION (a) Have

Fx

400 N cos 28.5q

F cosT x

Fx

351.5 N W

Fz

173.3 N W

Then: Fx2 Fy2 Fz2

F2

400 N 2

So:

352.5 N 2 80 N 2 Fz2

Hence: Fz

400 N 2 351.5 N 2 80 N 2

(b)

cosT y cosT z

Fy F Fz F

89

80 400 173.3 400

0.20

0.43325

Ty Tz

101.5q W 64.3q W

PROBLEM 2.86 A force F of magnitude 600 lb acts at the origin of a coordinate system. Knowing that Fx 200 lb, Tz 136.8q, Fy 0, determine (a) the components Fy and Fz, (b) the angles Tx and Ty.

SOLUTION (a)

F cosT z

Fz

600 lb cos136.8q 437.4 lb

Fz

437 lb W

Fy

359 lb W

Then: Fx2 Fy2 Fz2

F2

So:

600 lb 2 200 lb 2 Fy

Hence:

Fy

2

437.4 lb

2

600 lb 2 200 lb 2 437.4 lb 2

358.7 lb (b)

Fx F

cosT x cosT y

Fy F

200 600

358.7 600

0.333

0.59783

90

Tx Ty

70.5q W

126.7q W

PROBLEM 2.87 A transmission tower is held by three guy wires anchored by bolts at B, C, and D. If the tension in wire AB is 2100 N, determine the components of the force exerted by the wire on the bolt at B.

SOLUTION JJJG BA

BA

F

4 m 2 20 m 2 5 m 2 JJJG BA F BA

F O BA F

4 m i 20 m j 5 m k 21 m

2100 N ª 4 m i 20 m j 5 m k º¼ 21 m ¬

400 N i 2000 N j 500 N k Fx

91

400 N, Fy

2000 N, Fz

500 N W

PROBLEM 2.88 A transmission tower is held by three guy wires anchored by bolts at B, C, and D. If the tension in wire AD is 1260 N, determine the components of the force exerted by the wire on the bolt at D.

SOLUTION JJJG DA

4 m 2 20 m 2 14.8 m 2

DA

F

4 m i 20 m j 14.8 m k

JJJG DA F DA

F O DA F

25.2 m

1260 N ª 4 m i 20 m j 14.8 m k º¼ 25.2 m ¬

200 N i 1000 N j 740 N k Fx

92

200 N, Fy

1000 N, Fz

740 N W

PROBLEM 2.89 A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AB is 204 lb, determine the components of the force exerted on the plate at B.

SOLUTION JJJG BA

BA

F

32 in. i 48 in. j 36 in. k

32 in. 2 48 in. 2 36 in. 2 F O BA

JJJG BA F BA F

68 in.

204 lb ª 32 in. i 48 in. j 36 in. k º¼ 68 in. ¬

96 lb i 144 lb j 108 lb k Fx

93

96.0 lb, Fy

144.0 lb, Fz

108.0 lb W

PROBLEM 2.90 A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AD is 195 lb, determine the components of the force exerted on the plate at D.

SOLUTION JJJG DA

DA

F

25 in. i 48 in. j 36 in. k

25 in. 2 48 in. 2 36 in. 2

F O DA

JJJG DA F DA F

65 in.


75 lb i 144 lb j 108 lb k Fx

94

75.0 lb, Fy

144.0 lb, Fz

108.0 lb W

PROBLEM 2.91 A steel rod is bent into a semicircular ring of radius 0.96 m and is supported in part by cables BD and BE which are attached to the ring at B. Knowing that the tension in cable BD is 220 N, determine the components of this force exerted by the cable on the support at D.

SOLUTION JJJG DB DB

TDB

T O DB

0.96 m i 1.12 m j 0.96 m k

0.96 m 2 1.12 m 2 0.96 m 2 JJJG DB T DB TDB

1.76 m

220 N ª 0.96 m i 1.12 m j 0.96 m k º¼ 1.76 m ¬

120 N i 140 N j 120 N k TDB x

95

120.0 N, TDB y

140.0 N, TDB z

120.0 N W

PROBLEM 2.92 A steel rod is bent into a semicircular ring of radius 0.96 m and is supported in part by cables BD and BE which are attached to the ring at B. Knowing that the tension in cable BE is 250 N, determine the components of this force exerted by the cable on the support at E.

SOLUTION JJJG EB

TEB

0.96 m i 1.20 m j 1.28 m k

EB

0.96 m 2 1.20 m 2 1.28 m 2

T O EB

JJJG EB T EB

TEB

2.00 m

250 N ª 0.96 m i 1.20 m j 1.28 m k º¼ 2.00 m ¬

120 N i 150 N j 160 N k TEB x

120.0 N, TEB y

96

150.0 N, TEB z

160.0 N W

PROBLEM 2.93 Find the magnitude and direction of the resultant of the two forces shown knowing that P 500 N and Q 600 N.

SOLUTION P

500 lb > cos 30q sin15qi sin 30qj cos 30q cos15qk @ 500 lb > 0.2241i 0.50 j 0.8365k @ 112.05 lb i 250 lb j 418.25 lb k

Q

600 lb >cos 40q cos 20qi sin 40qj cos 40q sin 20qk @ 600 lb >0.71985i 0.64278j 0.26201k @ 431.91 lb i 385.67 lb j 157.206 lb k

R

R

PQ

319.86 lb i 635.67 lb j 261.04 lb k

319.86 lb 2 635.67 lb 2 261.04 lb 2

757.98 lb R

cosT x

cosT y

cosT z

Rx R

Ry R

Rz R

97

319.86 lb 757.98 lb

635.67 lb 757.98 lb

261.04 lb 757.98 lb

758 lb W

0.42199

Tx

65.0q W

Ty

33.0q W

Tz

69.9q W

0.83864

0.34439

PROBLEM 2.94 Find the magnitude and direction of the resultant of the two forces shown knowing that P 600 N and Q 400 N.

SOLUTION Using the results from 2.93:

600 lb > 0.2241i 0.50 j 0.8365k @

P

134.46 lb i 300 lb j 501.9 lb k Q

400 lb >0.71985i 0.64278j 0.26201k @ 287.94 lb i 257.11 lb j 104.804 lb k

R R

PQ

153.48 lb i 557.11 lb j 397.10 lb k

153.48 lb 2 557.11 lb 2 397.10 lb 2

701.15 lb R

cosT x

cosT y

cosT z

Rx R

Ry R

Rz R

153.48 lb 701.15 lb

557.11 lb 701.15 lb

397.10 lb 701.15 lb

98

701 lb W

0.21890

Tx

77.4q W

Ty

37.4q W

Tz

55.5q W

0.79457

0.56637

PROBLEM 2.95 Knowing that the tension is 850 N in cable AB and 1020 N in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables.

SOLUTION JJJG AB

AB

400 mm i 450 mm j 600 mm k

400 mm 2 450 mm 2 600 mm 2 JJJG AC

AC

TAB

TABO AB

1000 mm i 450 mm j 600 mm k

1000 mm 2 450 mm 2 600 mm 2 TAB

JJJG AB AB

TAC O AC

TAC

JJJG AC AC

TAC

R

ª 400 mm i 450 mm j 600 mm k º » 850 mm ¬ ¼

850 N «

ª 1000 mm i 450 mm j 600 mm k º » 1250 mm ¬ ¼

1020 N «

816 N i 367.2 N j 489.6 N k

TAB TAC

Then: and

1250 mm

400 N i 450 N j 600 N k

TAB TAC

850 mm

1216 N i 817.2 N j 1089.6 N k R

cosT x cosT y cosT z

1825.8 N

1216 1825.8

0.66601

817.2 1825.8

0.44758

1089.6 1825.8

0.59678

99

R

Tx Ty Tz

1826 N W 48.2q W 116.6q W 53.4q W

PROBLEM 2.96 Assuming that in Problem 2.95 the tension is 1020 N in cable AB and 850 N in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables.

SOLUTION JJJG AB

AB

400 mm 2 450 mm 2 600 mm 2 JJJG AC

TABO AB

TAB

JJJG AB AB

TAB TAC

TAC O AC

TAC

ª 400 mm i 450 mm j 600 mm k º » 850 mm ¬ ¼

480 N i 540 N j 720 N k ª 1000 mm i 450 mm j 600 mm k º » 1250 mm ¬ ¼

850 N «

680 N i 306 N j 408 N k

TAB TAC

Then: and

1250 mm

1020 N «

JJJG AC AC

TAC R

850 mm

1000 mm i 450 mm j 600 mm k

1000 mm 2 450 mm 2 600 mm 2

AC

TAB

400 mm i 450 mm j 600 mm k

1160 N i 846 N j 1128 N k R

1825.8 N

cosT x

1160 1825.8

0.6353

cosT y

846 1825.8

0.4634

cosT z

1128 1825.8

0.6178

100

R

Tx Ty Tz

1826 N W 50.6q W 117.6q W 51.8q W

PROBLEM 2.97 For the semicircular ring of Problem 2.91, determine the magnitude and direction of the resultant of the forces exerted by the cables at B knowing that the tensions in cables BD and BE are 220 N and 250 N, respectively.

SOLUTION For the solutions to Problems 2.91 and 2.92, we have TBD

120 N i 140 N j 120 N k

TBE

120 N i 150 N j 160 N k

RB

TBD TBE

Then:

240 N i 290 N j 40 N k and

R

cosT x

cosT y

cosT z

378.55 N

240 378.55

290 378.55

40 378.55

101

RB

379 N W

Tx

129.3q W

0.6340

0.7661

Ty

40.0q W

Tz

96.1q W

0.1057

PROBLEM 2.98 To stabilize a tree partially uprooted in a storm, cables AB and AC are attached to the upper trunk of the tree and then are fastened to steel rods anchored in the ground. Knowing that the tension in AB is 920 lb and that the resultant of the forces exerted at A by cables AB and AC lies in the yz plane, determine (a) the tension in AC, (b) the magnitude and direction of the resultant of the two forces.

SOLUTION Have TAB TAC

920 lb sin 50q cos 40qi cos 50qj sin 50q sin 40q j TAC cos 45q sin 25qi sin 45q j cos 45q cos 25q j

(a) TAB TAC

RA

RA x ?

RA x

6Fx

0

920 lb sin 50q cos 40q TAC cos 45q sin 25q

0:

0

or TAC

1806.60 lb

TAC

1807 lb W

(b)

RA y

6Fy : 920 lb cos 50q 1806.60 lb sin 45q

RA y

RA z

6Fz :

1868.82 lb

920 lb sin 50q sin 40q 1806.60 lb cos 45q cos 25q RA z

? RA

1610.78 lb

1868.82 lb j 1610.78 lb k

Then: RA

2467.2 lb

RA

102

2.47 kips W

PROBLEM 2.98 CONTINUED and cosT x cosT y cosT z

0 2467.2

1868.82 2467.2 1610.78 2467.2

103

0

0.7560 0.65288

Tx Ty Tz

90.0q W 139.2q W 49.2q W

PROBLEM 2.99 To stabilize a tree partially uprooted in a storm, cables AB and AC are attached to the upper trunk of the tree and then are fastened to steel rods anchored in the ground. Knowing that the tension in AC is 850 lb and that the resultant of the forces exerted at A by cables AB and AC lies in the yz plane, determine (a) the tension in AB, (b) the magnitude and direction of the resultant of the two forces.

SOLUTION Have TAB sin 50q cos 40qi cos 50q j sin 50q sin 40q j

TAB

850 lb cos 45q sin 25qi sin 45qj cos 45q cos 25qj

TAC (a)

RA x ?

RA x

0

0: TAB sin 50q cos 40q 850 lb cos 45q sin 25q

6Fx

TAB

0

432.86 lb

TAB

433 lb W

(b)

RA y

6Fy : 432.86 lb cos 50q 850 lb sin 45q

RA y

RA z

6Fz :

879.28 lb

432.86 lb sin 50q sin 40q 850 lb cos 45q cos 25q RA z

879.28 lb j 757.87 lb k

? RA RA

1160.82 lb

cosT x cosT y cosT z

757.87 lb

RA

0 1160.82

879.28 1160.82 757.87 1160.82

0

0.75746 0.65287

104

1.161 kips W

Tx Ty Tz

90.0q W 139.2q W 49.2q W

PROBLEM 2.100 For the plate of Problem 2.89, determine the tension in cables AB and AD knowing that the tension if cable AC is 27 lb and that the resultant of the forces exerted by the three cables at A must be vertical.

SOLUTION With:

JJJG AC

45 in. 2 48 in. 2 36 in. 2

AC

TAC

45 in. i 48 in. j 36 in. k

TAC O AC

TAC

JJJG AB

32 in. i 48 in. j 36 in. k

32 in. 2 48 in. 2 36 in. 2

AB

TAB

TABO AB

TAB

and

JJJG AB AB

JJJG AD

TAB ª 32 in. i 48 in. j 36 in. k º¼ 68 in. ¬

25 in. i 48 in. j 36 in. k

25 in. 2 48 in. 2 36 in. 2

AD TADO AD

TAD

68 in.

TAB 0.4706i 0.7059 j 0.5294k

TAB

TAD


16.2 lb i 17.28 lb j 12.96 k

TAC and

JJJG AC AC

75 in.

TAD

JJJG AD AD

65 in.

TAD ª 25 in. i 48 in. j 36 in. k º¼ 65 in. ¬

TAD 0.3846i 0.7385 j 0.5538k

105

PROBLEM 2.100 CONTINUED Now R

TAB TAD TAD TAB 0.4706i 0.7059 j 0.5294k ª¬16.2 lb i 17.28 lb j 12.96 k º¼

TAD 0.3846i 0.7385 j 0.5538k Since R must be vertical, the i and k components of this sum must be zero. Hence:

0.4706TAB 0.3846TAD 16.2 lb

0

(1)

0.5294TAB 0.5538TAD 12.96 lb

0

(2)

Solving (1) and (2), we obtain: TAB

244.79 lb,

TAD

106

257.41 lb TAB

245 lb W

TAD

257 lb W

PROBLEM 2.101 The support assembly shown is bolted in place at B, C, and D and supports a downward force P at A. Knowing that the forces in members AB, AC, and AD are directed along the respective members and that the force in member AB is 146 N, determine the magnitude of P.

SOLUTION Note that AB, AC, and AD are in compression. Have

and

d BA

220 mm 2 192 mm 2 0 2

d DA

192 mm 2 192 mm 2 96 mm 2

dCA

0 2 192 mm 2 144 mm 2 FBAO BA

FBA

292 mm 288 mm

240 mm

146 N ª 220 mm i 192 mm jº¼ 292 mm ¬

110 N i 96 N j FCA

FCAO CA

FCA ª192 mm j 144 mm k º¼ 240 mm ¬ FCA 0.80j 0.60k

FDA

FDAO DA

FDA ª192 mm i 192 mm j 96 mm k º¼ 288 mm ¬

FDA > 0.66667i 0.66667 j 0.33333k @ With At A:

i-component:

P

6F

Pj

0: FBA FCA FDA P

110 N 0.66667 FDA

0

or

0 FDA

j-component:

96 N 0.80 FCA 0.66667 165 N P

k-component:

0.60FCA 0.33333 165 N

Solving (2) for FCA and then using that result in (1), gives

107

165 N 0

(1)

0

(2)

P

279 N W

PROBLEM 2.102 The support assembly shown is bolted in place at B, C, and D and supports a downward force P at A. Knowing that the forces in members AB, AC, and AD are directed along the respective members and that P 200 N, determine the forces in the members.

SOLUTION With the results of 2.101: FBAO BA

FBA

FBA ª 220 mm i 192 mm jº¼ 292 mm ¬

FBA > 0.75342i 0.65753j@ N FCA

FCAO CA

FCA ª192 mm j 144 mm k º¼ 240 mm ¬

FCA 0.80 j 0.60k FDA

FDAO DA

FDA ª192 mm i 192 mm j 96 mm k º¼ 288 mm ¬

FDA > 0.66667i 0.66667 j 0.33333k @ With:

P

6F

At A:

200 N j

0: FBA FCA FDA P

0

Hence, equating the three (i, j, k) components to 0 gives three equations i-component:

0.75342 FBA 0.66667 FDA

j-component:

0.65735FBA 0.80FCA 0.66667 FDA 200 N

k-component:

0.60FCA 0.33333FDA

0

(1) 0

(2)

0

(3)

Solving (1), (2), and (3), gives FBA

104.5 N,

FCA

65.6 N,

FDA

118.1 N FBA FCA FDA

108

104.5 N W 65.6 N W 118.1 N W

PROBLEM 2.103 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 AB is 60 lb.

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 JJJG AB 12.6 ft i 16.8 ft j AB 21 ft JJJG AC 7.2 ft i 16.8 ft j 12.6 ft k AC 22.2 ft JJJG AD 16.8 ft j 9.9 ft k AD 19.5 ft JJJG AB and TAB TABO AB TAB 0.6i 0.8j TAB AB JJJG AC TAC TAC O AC TAC 0.3242i 0.75676 j 0.56757k TAC AC JJJG AD TAD TADO AD TAD 0.8615j 0.50769k TAD AD

109

PROBLEM 2.103 CONTINUED Equilibrium Condition 6F

0: TAB TAC TAD Pj

0

Substituting the expressions obtained for TAB , TAC , and TAD and factoring i, j, and k:

0.6TAB 0.3242TAC i 0.8TAB 0.75676TAC 0.56757TAC 0.50769TAD k

0.8615TAD P j

0

Equating to zero the coefficients of i, j, k: 0.6TAB 0.3242TAC

0

(1)

0.8TAB 0.75676TAC 0.8615TAD P 0.56757TAC 0.50769TAD

(2)

0

0

(3)

Setting TAB 60 lb in (1) and (2), and solving the resulting set of equations gives TAC TAD

111 lb 124.2 lb P

110

239 lb W

PROBLEM 2.104 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 100 lb.

SOLUTION See Problem 2.103 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: 0.6TAB 0.3242TAC

0

(1)

0.8TAB 0.75676TAC 0.8615TAD P 0.56757TAC 0.50769TAD

0

(2)

0

(3)

Substituting TAC 100 lb in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms gives TAB

54 lb

TAD

112 lb P

111

215 lb W

PROBLEM 2.105 The crate shown in Figure P2.105 and P2.108 is supported by three cables. Determine the weight of the crate knowing that the tension in cable AB is 3 kN.

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 JJJG AB 0.72 m i 1.2 m j 0.54 m k , AB 1.5 m JJJG AC 1.2 m j 0.64 m k , AC 1.36 m JJJG AD 0.8 m i 1.2 m j 0.54 m k , AD 1.54 m JJJG AB and TAB TABO AB TAB 0.48i 0.8j 0.36k TAB AB JJJG AC TAC TAC O AC TAC 0.88235j 0.47059k TAC AC JJJG AD TAD TADO AD TAD 0.51948i 0.77922 j 0.35065k TAD AD Equilibrium Condition with W 6F

Wj

0: TAB TAC TAD Wj

0

Substituting the expressions obtained for TAB , TAC , and TAD and factoring i, j, and k:

0.48TAB 0.51948TAD i 0.8TAB 0.88235TAC 0.36TAB 0.47059TAC 0.35065TAD k

112

0.77922TAD W j 0

PROBLEM 2.105 CONTINUED 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 3 kN in Equations (1), (2) and (3) and solving the resulting set of equations, using conventional algorithms for solving linear algebraic equations, gives TAC

4.3605 kN

TAD

2.7720 kN W

113

8.41 kN W

PROBLEM 2.106 For the crate of Problem 2.105, determine the weight of the crate knowing that the tension in cable AD is 2.8 kN. Problem 2.105: The crate shown in Figure P2.105 and P2.108 is supported by three cables. Determine the weight of the crate knowing that the tension in cable AB is 3 kN.

SOLUTION See Problem 2.105 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: 0.48TAB 0.51948TAD

0

0.8TAB 0.88235TAC 0.77922TAD W 0.36TAB 0.47059TAC 0.35065TAD

0 0

Substituting TAD 2.8 kN in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms, gives TAB

3.03 kN

TAC

4.40 kN W

114

8.49 kN W

PROBLEM 2.107 For the crate of Problem 2.105, determine the weight of the crate knowing that the tension in cable AC is 2.4 kN. Problem 2.105: The crate shown in Figure P2.105 and P2.108 is supported by three cables. Determine the weight of the crate knowing that the tension in cable AB is 3 kN.


0

0.8TAB 0.88235TAC 0.77922TAD W 0.36TAB 0.47059TAC 0.35065TAD

0 0

Substituting TAC 2.4 kN in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms, gives TAB

1.651 kN

TAD

1.526 kN W

115

4.63 kN W

PROBLEM 2.108 A 750-kg crate is supported by three cables as shown. Determine the tension in each cable.


0

0.8TAB 0.88235TAC 0.77922TAD W 0.36TAB 0.47059TAC 0.35065TAD Substituting W

750 kg 9.81 m/s2

0 0

7.36 kN in Equations (1), (2), and (3) above, and solving the

resulting set of equations using conventional algorithms, gives

116

TAB

2.63 kN W

TAC

3.82 kN W

TAD

2.43 kN W

PROBLEM 2.109 A force P is applied as shown to a uniform cone which is supported by three cords, where the lines of action of the cords pass through the vertex 0 and that the tension in cord BE is A of the cone. Knowing that P 0.2 lb, determine the weight W of the cone.

SOLUTION Note that because the line of action of each of the cords passes through the vertex A of the cone, the cords all have the same length, and the unit vectors lying along the cords are parallel to the unit vectors lying along the generators of the cone. Thus, for example, the unit vector along BE is identical to the unit vector along the generator AB. O AB

Hence: It follows that:

O BE

cos 45qi 8j sin 45qk 65

TBE

TBE O BE

§ cos 45qi 8 j sin 45qk · TBE ¨ ¸ 65 © ¹

TCF

TCF O CF

§ cos 30qi 8j sin 30qk · TCF ¨ ¸ 65 © ¹

TDG

TDG O DG

§ cos15qi 8 j sin15qk · TDG ¨ ¸ 65 © ¹

117

PROBLEM 2.109 CONTINUED 6F

At A:

0: TBE TCF TDG W P

0

Then, isolating the factors of i, j, and k, we obtain three algebraic equations: i:

or

TBE T T cos 45q CF cos 30q DG cos15q P 65 65 65

TBE cos 45q TCF cos 30q TDG cos15q P 65 j: TBE

TBE TCF TDG W

or k:

or With P

8 8 8 TCF TDG W 65 65 65

65 8

0

(1)

0

0

(2)

TBE T T sin 45q CF sin 30q DG sin15q 65 65 65

TBE sin 45q TCF sin 30q TDG sin15q 0 and the tension in cord BE

0

0

(3)

0

0.2 lb:

Solving the resulting Equations (1), (2), and (3) using conventional methods in Linear Algebra (elimination, matrix methods or iteration – with MATLAB or Maple, for example), we obtain: TCF

0.669 lb

TDG

0.746 lb W

118

1.603 lb W

PROBLEM 2.110 A force P is applied as shown to a uniform cone which is supported by three cords, where the lines of action of the cords pass through the vertex A of the cone. Knowing that the cone weighs 1.6 lb, determine the range of values of P for which cord CF is taut.

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

i : TBE cos 45q TCF cos 30q TDG cos15q 65 P j: TBE TCF TDG W

65 8

0

k : TBE sin 45q TCF sin 30q TDG sin15q

0

(1) (2)

0

(3)

With W 1.6 lb , the range of values of P for which the cord CF is taut can found by solving Equations (1), (2), and (3) for the tension TCF as a function of P and requiring it to be positive (! 0). Solving (1), (2), and (3) with unknown P, using conventional methods in Linear Algebra (elimination, matrix methods or iteration – with MATLAB or Maple, for example), we obtain: TCF Hence, for TCF ! 0 or

1.729P 0.668 lb 1.729 P 0.668 ! 0 P 0.386 lb ? 0 P 0.386 lb W

119

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 3.6 kN, determine the vertical force P exerted by the tower on the pin at A.

SOLUTION The force in each cable can be written as the product of the magnitude of the force and the unit vector along the cable. That is, with JJJG AC 18 m i 30 m j 5.4 m k

18 m 2 30 m 2 5.4 m 2

AC

TAC

T O AC

TAC

JJJG AB

and

T O AB

TAB

JJJG AB AB

JJJG AD

Finally

T O AD

TAB ª 6 m i 30 m j 7.5 m k º¼ 31.5 m ¬

6 m i 30 m j 22.2 m k

6 m 2 30 m 2 22.2 m 2 TAD TAD

31.5 m

TAB 0.1905i 0.9524 j 0.2381k

TAB

TAD

6 m i 30 m j 7.5 m k

6 m 2 30 m 2 7.5 m 2

AB

AD

TAC ª18 m i 30 m j 5.4 m k º¼ 35.4 m ¬

TAC 0.5085i 0.8475j 0.1525k

TAC

TAB

JJJG AC AC

35.4 m

JJJG AD AD

37.8 m

TAD ª 6 m i 30 m j 22.2 m k º¼ 37.8 m ¬

TAD 0.1587i 0.7937 j 0.5873k

120

PROBLEM 2.111 CONTINUED With P

Pj, at A:

6F

0: TAB TAC TAD Pj

0

Equating the factors of i, j, and k to zero, we obtain the linear algebraic equations: i : 0.1905TAB 0.5085TAC 0.1587TAD

0

j: 0.9524TAB 0.8475TAC 0.7937TAD P k : 0.2381TAB 0.1525TAC 0.5873TAD

0

(1) 0

(2) (3)

In Equations (1), (2) and (3), set TAB 3.6 kN, and, using conventional methods for solving Linear Algebraic Equations (MATLAB or Maple, for example), we obtain: TAC

1.963 kN

TAD

1.969 kN

P

121

6.66 kN W

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 2.6 kN, determine the vertical force P exerted by the tower on the pin at A.

SOLUTION Based on the results of Problem 2.111, particularly Equations (1), (2) and (3), we substitute TAC 2.6 kN and solve the three resulting linear equations using conventional tools for solving Linear Algebraic Equations (MATLAB or Maple, for example), to obtain TAB

4.77 kN

TAD

2.61 kN P

122

8.81 kN W

PROBLEM 2.113 A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AC is 15 lb, determine the weight of the plate.

SOLUTION The (vector) force in each cable can be written as the product of the (scalar) force and the unit vector along the cable. That is, with JJJG AB 32 in. i 48 in. j 36 in. k

32 in. 2 48 in. 2 36 in. 2

AB

TAB

T O AB

TAB

TAB

AC T O AC

45 in. i 48 in. j 36 in. k

45 in. 2 48 in. 2 36 in. 2 TAC

JJJG AC AC

TAC JJJG AD

Finally, AD

123

TAB ª 32 in. i 48 in. j 36 in. k º¼ 68 in. ¬

TAB 0.4706i 0.7059 j 0.5294k JJJG AC

and

TAC

JJJG AB AB

68 in.

75 in.

TAC ª 45 in. i 48 in. j 36 in. k º¼ 75 in. ¬

TAC 0.60i 0.64 j 0.48k

25 in. i 48 in. j 36 in. k

25 in. 2 48 in. 2 36 in. 2

65 in.

PROBLEM 2.113 CONTINUED TAD

T O AD

TAD

TAD With W

JJJG AD AD

TAD ª 25 in. i 48 in. j 36 in. k º¼ 65 in. ¬

TAD 0.3846i 0.7385 j 0.5538k

Wj, at A we have:

6F

0: TAB TAC TAD Wj

0

Equating the factors of i, j, and k to zero, we obtain the linear algebraic equations: i : 0.4706TAB 0.60TAC 0.3846TAD

0

j: 0.7059TAB 0.64TAC 0.7385TAD W k : 0.5294TAB 0.48TAC 0.5538TAD

(1) 0

(2) (3)

0

In Equations (1), (2) and (3), set TAC 15 lb, and, using conventional methods for solving Linear Algebraic Equations (MATLAB or Maple, for example), we obtain: TAB

136.0 lb

TAD

143.0 lb W

124

211 lb W

PROBLEM 2.114 A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AD is 120 lb, determine the weight of the plate.

SOLUTION Based on the results of Problem 2.111, particularly Equations (1), (2) and (3), we substitute TAD 120 lb and solve the three resulting linear equations using conventional tools for solving Linear Algebraic Equations (MATLAB or Maple, for example), to obtain TAC

12.59 lb

TAB

114.1 lb W

125

177.2 lb W

PROBLEM 2.115 A horizontal circular plate having a mass of 28 kg is suspended as shown from three wires which are attached to a support D and form 30q angles with the vertical. Determine the tension in each wire.

SOLUTION 6Fx

0: TAD sin 30q sin 50q TBD sin 30q cos 40q TCD sin 30q cos 60q

0

Dividing through by the factor sin 30q and evaluating the trigonometric functions gives 0.7660TAD 0.7660TBD 0.50TCD

0

(1)

Similarly, 6Fz

0: TAD sin 30q cos 50q TBD sin 30q sin 40q TCD sin 30q sin 60q

or

0

0.6428TAD 0.6428TBD 0.8660TCD

From (1)

TAD

0

(2)

TBD 0.6527TCD

Substituting this into (2): TBD

0.3573TCD

(3)

TCD

(4)

Using TAD from above: TAD Now, 6Fy

0: TAD cos 30q TBD cos 30q TCD cos 30q

28 kg 9.81 m/s 2 or

TAD TBD TCD

126

0

317.2 N

PROBLEM 2.115 CONTINUED Using (3) and (4), above:

TCD 0.3573TCD TCD Then:

317.2 N

TAD TBD

127

TCD

135.1 N W 46.9 N W 135.1 N W

PROBLEM 2.116 A transmission tower is held by three guy wires attached to a pin at A and anchored by bolts at B, C, and D. Knowing that the tower exerts on the pin at A an upward vertical force of 8 kN, determine the tension in each wire.

SOLUTION

From the solutions of 2.111 and 2.112:

Using P

TAB

0.5409 P

TAC

0.295P

TAD

0.2959 P

8 kN:

TAB

4.33 kN W

TAC

2.36 kN W

TAD

2.37 kN W

128

PROBLEM 2.117 For the rectangular plate of Problems 2.113 and 2.114, determine the tension in each of the three cables knowing that the weight of the plate is 180 lb.

SOLUTION

From the solutions of 2.113 and 2.114:

Using P

TAB

0.6440 P

TAC

0.0709 P

TAD

0.6771P

180 lb:

TAB

115.9 lb W

TAC

12.76 lb W

TAD

121.9 lb W

129

PROBLEM 2.118 For the cone of Problem 2.110, determine the range of values of P for which cord DG is taut if P is directed in the –x direction.

SOLUTION From the solutions to Problems 2.109 and 2.110, have

TBE TCF TDG

(2c)

0.2 65

TBE sin 45q TCF sin 30q TDG sin15q

0

TBE cos 45q TCF cos 30q TDG cos15q P 65

(3) 0

(1c )

Applying the method of elimination to obtain a desired result: Multiplying (2c) by sin 45q and adding the result to (3):

TCF sin 45q sin 30q TDG sin 45q sin15q or

TCF

0.2 65 sin 45q

0.9445 0.3714TDG

(4)

Multiplying (2c) by sin 30q and subtracting (3) from the result: TBE sin 30q sin 45q TDG sin 30q sin15q or

TBE

0.6679 0.6286TDG

130

0.2 65 sin 30q (5)

PROBLEM 2.118 CONTINUED Substituting (4) and (5) into (1c) : 1.2903 1.7321TDG P 65 ? TDG is taut for P

0

1.2903 lb 65

or 0 d P 0.1600 lb W

131

PROBLEM 2.119 A force P is applied as shown to a uniform cone which is supported by three cords, where the lines of action of the cords pass through the vertex 0, A of the cone. Knowing that the cone weighs 2.4 lb and that P determine the tension in each cord.

SOLUTION Note that because the line of action of each of the cords passes through the vertex A of the cone, the cords all have the same length, and the unit vectors lying along the cords are parallel to the unit vectors lying along the generators of the cone. Thus, for example, the unit vector along BE is identical to the unit vector along the generator AB. Hence: O AB

OBE

cos 45qi 8j sin 45qk 65

It follows that:

At A:

TBE

TBE O BE

§ cos 45qi 8 j sin 45qk · TBE ¨ ¸ 65 © ¹

TCF

TCF O CF

§ cos 30qi 8j sin 30qk · TCF ¨ ¸ 65 © ¹

TDG

TDG O DG

§ cos15qi 8 j sin15qk · TDG ¨ ¸ 65 © ¹

6F

0: TBE TCF TDG W P

132

0

PROBLEM 2.119 CONTINUED Then, isolating the factors if i, j, and k we obtain three algebraic equations: i:

TBE T T cos 45q CF cos 30q DG cos15q 65 65 65

TBE cos 45q TCF cos 30q TDG cos15q

or j: TBE

8 8 8 TCF TDG W 65 65 65

TBE TCF TDG

or k:

or

0

2.4 65 8

0

(1)

0

0.3 65

TBE T T sin 45q CF sin 30q DG sin15q P 65 65 65


(2) 0

P 65

(3)

With P 0, the tension in the cords can be found by solving the resulting Equations (1), (2), and (3) using conventional methods in Linear Algebra (elimination, matrix methods or iteration–with MATLAB or Maple, for example). We obtain

133

TBE

0.299 lb W

TCF

1.002 lb W

TDG

1.117 lb W

PROBLEM 2.120 A force P is applied as shown to a uniform cone which is supported by three cords, where the lines of action of the cords pass through the vertex 0.1 lb, A of the cone. Knowing that the cone weighs 2.4 lb and that P determine the tension in each cord.

SOLUTION See Problem 2.121 for the analysis leading to the linear algebraic Equations (1), (2), and (3) below: TBE cos 45q TCF cos 30q TDG cos15q TBE TCF TDG

0

(1)

0.3 65


(2) P 65

(3)

With P 0.1 lb, solving (1), (2), and (3), using conventional methods in Linear Algebra (elimination, matrix methods or iteration–with MATLAB or Maple, for example), we obtain

134

TBE

1.006 lb W

TCF

0.357 lb W

TDG

1.056 lb W

PROBLEM 2.121 Using two ropes and a roller chute, two workers are unloading a 200-kg cast-iron counterweight from a truck. Knowing that at the instant shown the counterweight is kept from moving and that the positions of points A, B, and C are, respectively, A(0, –0.5 m, 1 m), B(–0.6 m, 0.8 m, 0), and C(0.7 m, 0.9 m, 0), and assuming that no friction exists between the counterweight and the chute, determine the tension in each rope. (Hint: Since there is no friction, the force exerted by the chute on the counterweight must be perpendicular to the chute.)

SOLUTION From the geometry of the chute: N 2j k 5

N

N 0.8944 j 0.4472k

As in Problem 2.11, for example, the force in each rope can be written as the product of the magnitude of the force and the unit vector along the cable. Thus, with JJJG AB 0.6 m i 1.3 m j 1 m k

0.6 m 2 1.3 m 2 1 m 2

AB

TAB

T O AB

TAB

JJJG AC

and AC T O AC

TAC

JJJG AC AC

1.8574 m

TAC ª 0.7 m i 1.4 m j 1 m k º¼ 1.764 m ¬

TAC 0.3769i 0.7537 j 0.5384k

6F

135

0.7 m i 1.4 m j 1 m k

0.7 m 2 1.4 m 2 1 m 2

TAC Then:

TAB ª 0.6 m i 1.3 m j 1 m k º¼ 1.764 m ¬

TAB 0.3436i 0.7444 j 0.5726k

TAB

TAC

JJJG AB AB

1.764 m

0: N TAB TAC W

0

PROBLEM 2.121 CONTINUED With W 200 kg 9.81 m/s 1962 N, and equating the factors of i, j, and k to zero, we obtain the linear algebraic equations: i : 0.3436TAB 0.3769TAC

0

(1)

j: 0.7444TAB 0.7537TAC 0.8944 N 1962 k : 0.5726TAB 0.5384TAC 0.4472 N

0

(2) (3)

0

Using conventional methods for solving Linear Algebraic Equations (elimination, MATLAB or Maple, for example), we obtain

N

1311 N

136

TAB

551 N W

TAC

503 N W

PROBLEM 2.122 Solve Problem 2.121 assuming that a third worker is exerting a force P (180 N)i on the counterweight. Problem 2.121: Using two ropes and a roller chute, two workers are unloading a 200-kg cast-iron counterweight from a truck. Knowing that at the instant shown the counterweight is kept from moving and that the positions of points A, B, and C are, respectively, A(0, –0.5 m, 1 m), B(–0.6 m, 0.8 m, 0), and C(0.7 m, 0.9 m, 0), and assuming that no friction exists between the counterweight and the chute, determine the tension in each rope. (Hint: Since there is no friction, the force exerted by the chute on the counterweight must be perpendicular to the chute.)

SOLUTION From the geometry of the chute: N 2j k 5

N

N 0.8944 j 0.4472k

As in Problem 2.11, for example, the force in each rope can be written as the product of the magnitude of the force and the unit vector along the cable. Thus, with JJJG AB 0.6 m i 1.3 m j 1 m k

0.6 m 2 1.3 m 2 1 m 2

AB

TAB

T O AB

TAB

JJJG AC

and AC T O AC

TAC

6F

137

0.7 m i 1.4 m j 1 m k

0.7 m 2 1.4 m 2 1 m 2

TAC

Then:

TAB ª 0.6 m i 1.3 m j 1 m k º¼ 1.764 m ¬

TAB 0.3436i 0.7444 j 0.5726k

TAB

TAC

JJJG AB AB

1.764 m

JJJG AC AC

1.8574 m

TAC ª 0.7 m i 1.4 m j 1 m k º¼ 1.764 m ¬

TAC 0.3769i 0.7537 j 0.5384k 0: N TAB TAC P W

0

PROBLEM 2.122 CONTINUED Where and

P W

180 N i

ª 200 kg 9.81 m/s 2 º j ¬ ¼ 1962 N j

Equating the factors of i, j, and k to zero, we obtain the linear equations: i : 0.3436TAB 0.3769TAC 180

0

j: 0.8944 N 0.7444TAB 0.7537TAC 1962 k : 0.4472 N 0.5726TAB 0.5384TAC

0

0

Using conventional methods for solving Linear Algebraic Equations (elimination, MATLAB or Maple, for example), we obtain N

1302 N

138

TAB

306 N W

TAC

756 N W

PROBLEM 2.123 A piece of machinery of weight W is temporarily supported by cables AB, AC, and ADE. Cable ADE is attached to the ring at A, passes over the pulley at D and back through the ring, and is attached to the support at E. Knowing that W 320 lb, determine the tension in each cable. (Hint: The tension is the same in all portions of cable ADE.)

SOLUTION The (vector) force in each cable can be written as the product of the (scalar) force and the unit vector along the cable. That is, with JJJG AB 9 ft i 8 ft j 12 ft k AB

and

JJJG AB AB

T O AB

TAB

TAB 0.5294i 0.4706 j 0.7059k

JJJG AC

TAB

0 i 8 ft j 6 ft k 0 ft 2 8 ft 2 6 ft 2 JJJG AC AC

T O AC

TAC

TAC 0.8 j 0.6k

JJJG AD

4 ft i 8 ft j 1 ft k

TAC

4 ft 2 8 ft 2 1 ft 2 JJJG AD AD

10 ft

TAC ª 0 ft i 8 ft j 6 ft k º¼ 10 ft ¬

TAC

AD

17 ft

TAB ª 9 ft i 8 ft j 12 ft k º¼ 17 ft ¬

TAB

AC

and

9 ft 2 8 ft 2 12 ft 2

9 ft

TADE ª 4 ft i 8 ft j 1 ft k º¼ 9 ft ¬

TAD

T O AD

TAD

TADE 0.4444i 0.8889 j 0.1111k

TADE

139

PROBLEM 2.123 CONTINUED Finally,

JJJG AE

AE

8 ft i 8 ft j 4 ft k 8 ft 2 8 ft 2 4 ft 2 JJJG AE AE

12 ft

TADE ª 8 ft i 8 ft j 4 ft k º¼ 12 ft ¬

TAE

T O AE

TAE

TADE 0.6667i 0.6667 j 0.3333k

With the weight of the machinery, W 6F

TADE

W j, at A, we have: 0: TAB TAC 2TAD Wj

0

Equating the factors of i, j, and k to zero, we obtain the following linear algebraic equations: 0.5294TAB 2 0.4444TADE 0.6667TADE

0

(1)

0.4706TAB 0.8TAC 2 0.8889TADE 0.6667TADE W 0.7059TAB 0.6TAC 2 0.1111TADE 0.3333TADE

0

(2)

0

(3)

Knowing that W 320 lb, we can solve Equations (1), (2) and (3) using conventional methods for solving Linear Algebraic Equations (elimination, matrix methods via MATLAB or Maple, for example) to obtain TAB

46.5 lb W

TAC

34.2 lb W

TADE

140

110.8 lb W

PROBLEM 2.124 A piece of machinery of weight W is temporarily supported by cables AB, AC, and ADE. Cable ADE is attached to the ring at A, passes over the pulley at D and back through the ring, and is attached to the support at E. Knowing that the tension in cable AB is 68 lb, determine (a) the tension in AC, (b) the tension in ADE, (c) the weight W. (Hint: The tension is the same in all portions of cable ADE.)

SOLUTION See Problem 2.123 for the analysis leading to the linear algebraic Equations (1), (2), and (3), below: 0.5294TAB 2 0.4444TADE 0.6667TADE

0

(1)

0.4706TAB 0.8TAC 2 0.8889TADE 0.6667TADE W 0.7059TAB 0.6TAC 2 0.1111TADE 0.3333TADE

0

(2)

0

(3)

Knowing that the tension in cable AB is 68 lb, we can solve Equations (1), (2) and (3) using conventional methods for solving Linear Algebraic Equations (elimination, matrix methods via MATLAB or Maple, for example) to obtain (a) TAC

50.0 lb W

(b) TAE

162.0 lb W

(c)

141

W

468 lb W

PROBLEM 2.125 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 is the position shown. Knowing that W 1200 N, determine P and Q. (Hint: The tension is the same in both portions of cable BAC.)

SOLUTION The (vector) force in each cable can be written as the product of the (scalar) force and the unit vector along the cable. That is, with JJJG AB 0.48 m i 0.72 m j 0.16 m k AB

0.48 m 2 0.72 m 2 0.16 m 2 JJJG AB AB

TAB ª 0.48 m i 0.72 m j 0.16 m k º¼ 0.88 m ¬

TAB

T O AB

TAB

TAB 0.5455i 0.8182 j 0.1818k

TAB

0.88 m

and JJJG AC

AC

0.24 m i 0.72 m j 0.13 m k 0.24 m 2 0.72 m 2 0.13 m 2 JJJG AC AC

TAC ª 0.24 m i 0.72 m j 0.13 m k º¼ 0.77 m ¬

TAC

T O AC

TAC

TAC 0.3177i 0.9351j 0.1688k

At A:

TAC

6F

0: TAB TAC P Q W

142

0.77 m

0

PROBLEM 2.125 CONTINUED Noting that TAB TAC because of the ring A, we equate the factors of i, j, and k to zero to obtain the linear algebraic equations: i:

0.5455 0.3177 T

or

P j:

W k:

With W

W

Q

0

1.7532T

0.1818 0.1688 T

or

0

0.2338T

0.8182 0.9351 T

or

P

Q

0

0.356T

1200 N: T

1200 N 1.7532

684.5 N 160.0 N W

P Q

143

240 N W

PROBLEM 2.126 For the system of Problem 2.125, determine W and P knowing that Q 160 N.

Problem 2.125: 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 is the position shown. Knowing that W 1200 N, determine P and Q. (Hint: The tension is the same in both portions of cable BAC.)

SOLUTION Based on the results of Problem 2.125, particularly the three equations relating P, Q, W, and T we substitute Q 160 N to obtain T

160 N 0.3506

456.3 N W P

144

800 N W 107.0 N W

PROBLEM 2.127 Collars A and B are connected by a 1-m-long wire and can slide freely on frictionless rods. If a force P (680 N) j is applied at A, determine (a) the tension in the wire when y 300 mm, (b) the magnitude of the force Q required to maintain the equilibrium of the system.

SOLUTION Free-Body Diagrams of collars

For both Problems 2.127 and 2.128:

AB 2 1 m 2

Here

x2 y 2 z 2

0.40 m 2

y2 z2

or

y2 z2

0.84 m 2

Thus, with y given, z is determined. Now O AB

JJJG AB AB

1 0.40i yj zk m 1m

0.4i yk zk

Where y and z are in units of meters, m. From the F.B. Diagram of collar A: 6F

0: N xi N zk Pj TAB O AB

0

Setting the j coefficient to zero gives: P yTAB With P

0

680 N, TAB

680 N y

Now, from the free body diagram of collar B: 6F

145

0: N xi N y j Qk TABO AB

0

PROBLEM 2.127 CONTINUED Setting the k coefficient to zero gives: Q TAB z

0

And using the above result for TAB we have Q

TAB z

680 N z y

Then, from the specifications of the problem, y 0.84 m 2 0.3 m

z2

? z

300 mm

0.3 m

2

0.866 m

and (a)

TAB

680 N 0.30

2266.7 N

or

TAB

2.27 kN W

Q

1.963 kN W

and Q

(b) or

146

2266.7 0.866

1963.2 N

PROBLEM 2.128 Solve Problem 2.127 assuming y

550 mm.

Problem 2.127: Collars A and B are connected by a 1-m-long wire and can slide freely on frictionless rods. If a force P (680 N) j is applied at A, determine (a) the tension in the wire when y 300 mm, (b) the magnitude of the force Q required to maintain the equilibrium of the system.

SOLUTION From the analysis of Problem 2.127, particularly the results: y2 z2

With y

550 mm

0.84 m 2

TAB

680 N y

Q

680 N z y

0.55 m, we obtain: z2

0.84 m 2 0.55 m

? z

0.733 m

TAB

680 N 0.55

2

and (a)

1236.4 N

or

TAB

1.236 kN W

Q

0.906 kN W

and (b)

Q or

147

1236 0.866 N

906 N

PROBLEM 2.129 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.

SOLUTION

(a)

P sin 35q P

300 1b

300 lb sin 35q P

523 lb W

Pv

428 lb W

(b) Vertical Component Pv

P cos 35q

523 lb cos 35q

148

PROBLEM 2.130 A container of weight W is suspended from ring A, to which cables AC and AE are attached. A force P is applied to the end F of a third cable which passes over a pulley at B and through ring A and which is attached 1000 N, determine the magnitude to a support at D. Knowing that W of P. (Hint: The tension is the same in all portions of cable FBAD.)

SOLUTION The (vector) force in each cable can be written as the product of the (scalar) force and the unit vector along the cable. That is, with JJJG AB 0.78 m i 1.6 m j 0 m k

0.78 m 2 1.6 m 2 0 2

AB

JJJG AB AB

TAB ª 0.78 m i 1.6 m j 0 m k º¼ 1.78 m ¬

TAB

T O AB

TAB

TAB 0.4382i 0.8989 j 0k

and

TAB

JJJG AC

AC

0 i 1.6 m j 1.2 m k 0 m 2 1.6 m 2 1.2 m 2 JJJG AC AC

TAC

T O AC

TAC

TAC 0.8 j 0.6k

and

JJJG AD

AD

1.78 m

TAC

2m

TAC ª 0 i 1.6 m j 1.2 m k º¼ 2m¬

1.3 m i 1.6 m j 0.4 m k 1.3 m 2 1.6 m 2 0.4 m 2 JJJG AD AD

TAD ª1.3 m i 1.6 m j 0.4 m k º¼ 2.1 m ¬

TAD

T O AD

TAD

TAD 0.6190i 0.7619 j 0.1905k

TAD

2.1 m

149

PROBLEM 2.130 CONTINUED Finally,

JJJG AE

0.4 m i 1.6 m j 0.86 m k

0.4 m 2 1.6 m 2 0.86 m 2

AE

JJJG AE AE

TAE ª 0.4 m i 1.6 m j 0.86 m k º¼ 1.86 m ¬

TAE

T O AE

TAE

TAE 0.2151i 0.8602 j 0.4624k

TAE

With the weight of the container W

1.86 m

Wj, at A we have: 6F

0: TAB TAC TAD Wj

0

Equating the factors of i, j, and k to zero, we obtain the following linear algebraic equations: 0.4382TAB 0.6190TAD 0.2151TAE

0

0.8989TAB 0.8TAC 0.7619TAD 0.8602TAE W 0.6TAC 0.1905TAD 0.4624TAE

(1) 0

(2)

0

(3)

Knowing that W 1000 N and that because of the pulley system at B TAB TAD P, where P is the externally applied (unknown) force, we can solve the system of linear equations (1), (2) and (3) uniquely for P. P

150

378 N W

PROBLEM 2.131 A container of weight W is suspended from ring A, to which cables AC and AE are attached. A force P is applied to the end F of a third cable which passes over a pulley at B and through ring A and which is attached to a support at D. Knowing that the tension in cable AC is 150 N, determine (a) the magnitude of the force P, (b) the weight W of the container. (Hint: The tension is the same in all portions of cable FBAD.)

SOLUTION Here, as in Problem 2.130, the support of the container consists of the four cables AE, AC, AD, and AB, with the condition that the force in cables AB and AD is equal to the externally applied force P. Thus, with the condition

TAB

TAD

P

and using the linear algebraic equations of Problem 2.131 with TAC

150 N, we obtain (a)

P

(b) W

151

454 N W 1202 N W

PROBLEM 2.132 Two cables tied together at C are loaded as shown. Knowing that Q 60 lb, determine the tension (a) in cable AC, (b) in cable BC.

SOLUTION 6Fy

With (a)

0: TCA Q cos 30q Q

TCA

0

60 lb

60 lb 0.866 TCA

6Fx

(b) With

0: P TCB Q sin 30q P

TCB

0

75 lb

75 lb 60 lb 0.50 or TCB

152

52.0 lb W

45.0 lb W

PROBLEM 2.133 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.

SOLUTION 6Fx

Have

0: TCA Q cos 30q

or

TCA

0

0.8660 Q

TCA d 60 lb

Then for

0.8660Q 60 lb Q d 69.3 lb

or 6Fy

From or

0: TCB TCB

P Q sin 30q

75 lb 0.50Q TCB d 60 lb

For

75 lb 0.50Q d 60 lb 0.50Q t 15 lb

or

Q t 30 lb

Thus,

30.0 d Q d 69.3 lb W

Therefore,

153

PROBLEM 2.134 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 Free-Body Diagram of Connection

6Fx

3 3 FB FC FA 5 5

0: FA

With

16 kN

4 4 16 kN 8 kN 5 5

FC

6Fy

8 kN, FB

0

0: FD

3 3 FB FA 5 5

FC

6.40 kN W

FD

4.80 kN W

0

With FA and FB as above: FD

154

3 3 16 kN 8 kN 5 5

PROBLEM 2.135 A welded connection is in equilibrium under the action of the four forces shown. Knowing that FA 5 kN and FD 6 kN, determine the magnitudes of the other two forces.

SOLUTION Free-Body Diagram of Connection

6Fy

0: FD

FB

or With

FA

3 3 FA FB 5 5

FD

0

3 FA 5

5 kN, FD

8 kN

5ª 3 º 6 kN 5 kN » 3 «¬ 5 ¼

FB

FB 6Fx

0: FC

FC

4 4 FB FA 5 5

15.00 kN W

0

4 FB FA 5 4 15 kN 5 kN 5 FC

155

8.00 kN W

PROBLEM 2.136 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 Diagram of Collar

(a)

Triangle Proportions

6Fx

0: P

4.5 50 lb 20.5 or P

(b)

0 10.98 lb W


6Fx

0: P

15 50 lb 25 or P

156

0 30.0 lb W

PROBLEM 2.137 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.

SOLUTION Free-Body Diagram of Collar


Hence:

6Fx

xˆ

or

0: 48

50 xˆ 400 xˆ 2

0

48 400 xˆ 2 50

xˆ 2

0.92 lb 400 xˆ 2

xˆ 2

4737.7 in 2

xˆ

157

68.6 in. W

PROBLEM 2.138 A frame ABC is supported in part by cable DBE which 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 The force in cable DB can be written as the product of the magnitude of the force and the unit vector along the cable. That is, with JJJG DB 480 mm i 510 mm j 320 mm k

480 2 510 2 320 2

DB

F

F O DB

JJJG DB F DB F

770 mm

385 N ª 480 mm i 510 mm j 320 mm k º¼ 770 mm ¬

240 N i 255 N j 160 N k Fx

158

240 N, Fy

255 N, Fz

160.0 N W

PROBLEM 2.139 A frame ABC is supported in part by cable DBE which passes through a frictionless ring at B. 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.

SOLUTION The force in each cable can be written as the product of the magnitude of the force and the unit vector along the cable. That is, with JJJG BD 0.48 m i 0.51 m j 0.32 m k BD

TBD

T O BD

0.48 m 2 0.51 m 2 0.32 m 2 TBD

JJJG BD BD

0.77 m

TBD ª 0.48 m i 0.51 m j 0.32 m k º¼ 0.77 m ¬

TBD

TBD 0.6234i 0.6623j 0.4156k

JJJG BE

0.27 m i 0.40 m j 0.6 m k

and

BE

TBE

T O BE

0.27 m 2 0.40 m 2 0.6 m 2 TBE TBE

JJJG BD BD

TBE ª 0.26 m i 0.40 m j 0.6 m k º¼ 0.770 m ¬

TBE 0.3506i 0.5195 j 0.7792k

Now, because of the frictionless ring at B, TBE cables is F

0.770 m

TBD

385 N and the force on the support due to the two

385 N 0.6234i 0.6623j 0.4156k 0.3506i 0.5195j 0.7792k 375 N i 455 N j 460 N k

159

PROBLEM 2.139 CONTINUED The magnitude of the resultant is F

Fx2 Fy2 Fz2

375 N 2 455 N 2 460 N 2

747.83 N

or F

748 N W

The direction of this force is:

Tx

cos 1

375 747.83

Ty

cos 1

455 747.83

or T y

Tz

cos 1

460 747.83

or T z

or T x

160

120.1q W 52.5q W 128.0q W

PROBLEM 2.140 A steel tank is to be positioned in an excavation. Using trigonometry, determine (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 Force Triangle

(a) For minimum P it must be perpendicular to the vertical resultant R ? P

425 lb cos 30q or P

(b)

R

368 lb

425 lb sin 30q or R

161

W

213 lb W

Vector Mechanics for Engineers Statics 7th - Cap 02

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