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: 8.4 kNR

19

8.4 kNR 19

1

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

We measure: 51.3 , 59

(a)

(b)

We measure: 575 N, 67R

575 NR 67

2

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

(a)

(b)

We measure: 37 lb, 76R

37 lbR 76

3

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

SOLUTION

(a)

(b)

We measure: 61.5 lb, 86.5R

61.5 lbR 86.5

4

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

We measure: 2 108 NF

77 NR

By trigonometry: Law of Sines

2 120sin sin 38 sinF R

90 28 62 , 180 62 38 80

Then:

2 120 Nsin 62 sin 38 sin80

F R

or (a) 2 107.6 NF

(b) 75.0 NR

5

PROBLEM 2.6 Two control rods are attached at A to lever AB. Using trigonometry and knowing that the force in the right-hand rod is F2 80 N, determine (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

1 80sin sin 38 sin

F R

90 10 80 , 180 80 38 62

Then:

1 80 Nsin80 sin 38 sin 62

F R

or (a) 1 89.2 NF

(b) 55.8 NR

6

PROBLEM 2.7 The 50-lb force is to be resolved into components along lines -a a and

- .b b (a) Using trigonometry, determine the angle knowing that the component along -a a is 35 lb. (b) What is the corresponding value of the component along - ?b b

SOLUTION

Using the triangle rule and the Law of Sines

(a) sin sin 4035 lb 50 lb

sin 0.44995

26.74

Then: 40 180

113.3

(b) Using the Law of Sines:

50 lbsin sin 40

bbF

71.5 lbbbF

7

PROBLEM 2.8 The 50-lb force is to be resolved into components along lines -a a and

- .b b (a) Using trigonometry, determine the angle knowing that the component along -b b is 30 lb. (b) What is the corresponding value of the component along - ?a a

SOLUTION

Using the triangle rule and the Law of Sines

(a) sin sin 4030 lb 50 lb

sin 0.3857

22.7

(b) 40 180

117.31

50 lbsin sin 40

aaF

sin50 lbsin 40aaF

69.1 lbaaF

8

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 25 , 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: 180 35 25

120

Then: 360 Nsin 35 sin120 sin 25

P R

or (a) 489 NP

(b) 738 NR

9

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 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 300 Nsin sin 35

sin 0.68829

43.5

(b) 180 35 43.5

101.5

Then: 300 Nsin101.5 sin 35

R

or 513 NR

10

ÐÎÑÞÔÛÓ îòïï

Two forces are applied as shown to a hook support. Using trigonometry and knowing that the magnitude of P is 14 lb, determine (¿) the required angle if the resultant R of the two forces applied to the support is to be horizontal, (¾) the corresponding magnitude of R.

ÍÑÔËÌ×ÑÒ

Using the triangle rule and the Law of Sines

(¿) Have: 20 lb 14 lbsin sin 30

sin 0.71428

45.6

(¾) 180 30 45.6

104.4

Then: 14 lbsin104.4 sin 30

Î

27.1 lbÎ

11

ÐÎÑÞÔÛÓ îòïî

For the hook support of Problem 2.3, using trigonometry and knowing that the magnitude of P is 25 lb, determine (¿) the required magnitude of the force Q if the resultant R of the two forces applied at ß is to be vertical, (¾) the corresponding magnitude of R.

Problem 2.3: Two forces P and Q are applied as shown at point ß of a hook support. Knowing that Ð 15 lb and Ï 25 lb, determine graphically the magnitude and direction of their resultant using (¿) the parallelogram law, (¾) the triangle rule.

ÍÑÔËÌ×ÑÒ

Using the triangle rule and the Law of Sines

(¿) Have: 25 lbsin15 sin 30Ï

12.94 lbÏ

(¾) 180 15 30

135

Thus: 25 lbsin135 sin30Î

sin13525 lb 35.36 lbsin30

Î

35.4 lbÎ

12

ÐÎÑÞÔÛÓ îòïí

For the hook support of Problem 2.11, determine, using trigonometry, (¿) the magnitude and direction of the smallest force P for which the resultant R of the two forces applied to the support is horizontal, (¾) 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 (¿) the required angle if the resultant R of the two forces applied to the support is to be horizontal, (¾) the corresponding magnitude of R.

ÍÑÔËÌ×ÑÒ

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

20 lb sin30Ð

10 lb 10 lbP

(¾) 20 lb cos30Î

17.32 lb 17.32 lbÎ

13

ÐÎÑÞÔÛÓ îòïì

As shown in Figure P2.9, two cables are attached to a sign at ß to steady the sign as it is being lowered. Using trigonometry, determine (¿) the magnitude and direction of the smallest force P for which the resultant Rof the two forces applied at ß is vertical, (¾) the corresponding magnitude of R.

ÍÑÔËÌ×ÑÒ

We observe that force P is minimum when is 90 , that is, P is horizontal

Then: (¿) 360 N sin 35Ð

or 206 NP

And: (¾) 360 N cos35Î

or 295 NÎ

14

ÐÎÑÞÔÛÓ îòïë

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 Ð 10 lb and 40 .

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 (¿) the required angle if the resultant R of the two forces applied to the support is to be horizontal, (¾) the corresponding magnitude of R.

ÍÑÔËÌ×ÑÒ

Using the force triangle and the Law of Cosines

2 22 10 lb 20 lb 2 10 lb 20 lb cos110Î

2100 400 400 0.342 lb

2636.8 lb

25.23 lbÎ

Using now the Law of Sines

10 lb 25.23 lbsin sin110

10 lbsin sin11025.23 lb

0.3724

So: 21.87

Angle of inclination of Î, is then such that:

30

8.13

Hence: 25.2 lbR 8.13

15

ÐÎÑÞÔÛÓ îòïê

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 (¿) the parallelogram law, (¾) the triangle rule.

ÍÑÔËÌ×ÑÒ

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

We have: 180 50 25

105

Then: 2 22 4.5 kN 6 kN 2 4.5 kN 6 kN cos105Î

270.226 kN

or 8.3801 kNÎ

Now: 8.3801 kN 6 kNsin105 sin

6 kNsin sin1058.3801 kN

0.6916

43.756

8.38 kNR 18.76

16

ÐÎÑÞÔÛÓ îòïé

Solve Problem 2.2 using trigonometry

Problem 2.2: The cable stays ßÞ and ßÜ help support pole ßÝ. Knowing that the tension is 500 N in ßÞ and 160 N in ßÜ, determine graphically the magnitude and direction of the resultant of the forces exerted by the stays at ß using (¿) the parallelogram law, (¾) the triangle rule.

ÍÑÔËÌ×ÑÒ

From the geometry of the problem:

1 2tan 38.662.5

1 1.5tan 30.962.5

Now: 180 38.66 30.96 110.38

And, using the Law of Cosines:

2 22 500 N 160 N 2 500 N 160 N cos110.38Î

2331319 N

575.6 NÎ

Using the Law of Sines:

160 N 575.6 Nsin sin110.38

160 Nsin sin110.38575.6 N

0.2606

15.1

90 66.44

576 NR 66.4

17

ÐÎÑÞÔÛÓ îòïè

Solve Problem 2.3 using trigonometry

Problem 2.3: Two forces P and Q are applied as shown at point ß of a hook support. Knowing that Ð 15 lb and Ï 25 lb, determine graphically the magnitude and direction of their resultant using (¿) the parallelogram law, (¾) the triangle rule.

ÍÑÔËÌ×ÑÒ

Using the force triangle and the Laws of Cosines and Sines

We have:

180 15 30

135

Then: 2 22 15 lb 25 lb 2 15 lb 25 lb cos135Î

21380.3 lb

or 37.15 lbÎ

and

25 lb 37.15 lbsin sin135

25 lbsin sin13537.15 lb

0.4758

28.41

Then: 75 180

76.59

37.2 lbR 76.6

18

ÐÎÑÞÔÛÓ îòïç

Two structural members ß and Þ are bolted to a bracket as shown. Knowing that both members are in compression and that the force is 30 kN in member ß and 20 kN in member Þ, determine, using trigonometry, the magnitude and direction of the resultant of the forces applied to the bracket by members ß and Þ.

ÍÑÔËÌ×ÑÒ

Using the force triangle and the Laws of Cosines and Sines

We have: 180 45 25 110

Then: 2 22 30 kN 20 kN 2 30 kN 20 kN cos110Î

21710.4 kN

41.357 kNÎ

and

20 kN 41.357 kNsin sin110

20 kNsin sin11041.357 kN

0.4544

27.028

Hence: 45 72.028

41.4 kNR 72.0

19

ÐÎÑÞÔÛÓ îòîð

Two structural members ß and Þ are bolted to a bracket as shown. Knowing that both members are in compression and that the force is 20 kN in member ß and 30 kN in member Þ, determine, using trigonometry, the magnitude and direction of the resultant of the forces applied to the bracket by members ß and Þ.

ÍÑÔËÌ×ÑÒ

Using the force triangle and the Laws of Cosines and Sines

We have: 180 45 25 110

Then: 2 22 30 kN 20 kN 2 30 kN 20 kN cos110Î

21710.4 kN

41.357 kNÎ

and

30 kN 41.357 kNsin sin110

30 kNsin sin11041.357 kN

0.6816

42.97

Finally: 45 87.97

41.4 kNR 88.0

20

ÐÎÑÞÔÛÓ îòîï

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

ÍÑÔËÌ×ÑÒ

20 kN Force:

20 kN cos 40 ,xF 15.32 kNxF

20 kN sin 40 ,yF 12.86 kNyF

30 kN Force:

30 kN cos70 ,xF 10.26 kNxF

30 kN sin 70 ,yF 28.2 kNyF

42 kN Force:

42 kN cos20 ,xF 39.5 kNxF

42 kN sin 20 ,yF 14.36 kNyF

21

ÐÎÑÞÔÛÓ îòîî

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

ÍÑÔËÌ×ÑÒ

40 lb Force:

40 lb sin 50 ,xF 30.6 lbxF

40 lb cos50 ,yF 25.7 lbyF

60 lb Force:

60 lb cos60 ,xF 30.0 lbxF

60 lb sin 60 ,yF 52.0 lbyF

80 lb Force:

80 lb cos 25 ,xF 72.5 lbxF

80 lb sin 25 ,yF 33.8 lbyF

22

ÐÎÑÞÔÛÓ îòîí

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

ÍÑÔËÌ×ÑÒ

We compute the following distances:

2 2

2 2

2 2

48 90 102 in.

56 90 106 in.

80 60 100 in.

OA

OB

OC

Then:

204 lb Force:

48102 lb ,102xF 48.0 lbxF

90102 lb ,102yF 90.0 lbyF

212 lb Force:

56212 lb ,106xF 112.0 lbxF

90212 lb ,106yF 180.0 lbyF

400 lb Force:

80400 lb ,100xF 320 lbxF

60400 lb ,100yF 240 lbyF

23

ÐÎÑÞÔÛÓ îòîì

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

ÍÑÔËÌ×ÑÒ

We compute the following distances:

2 270 240 250 mmOA

2 2210 200 290 mmOB

2 2120 225 255 mmOC

500 N Force:

70500 N250xF 140.0 NxF

240500 N250yF 480 NyF

435 N Force:

210435 N290xF 315 NxF

200435 N290yF 300 NyF

510 N Force:

120510 N255xF 240 NxF

225510 N255yF 450 NyF

24

ÐÎÑÞÔÛÓ îòîë

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.

ÍÑÔËÌ×ÑÒ

(a)cos40

xPP

135 Ncos40

or 176.2 NP

(b) tan 40 sin 40y xP P P

135 N tan 40

or 113.3 NyP

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)sin35

yPP

960 Nsin35

or 1674 NP

(b)tan 35

yx

PP

960 Ntan 35

or 1371 NxP

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 260 lb.xP

Then:

(a) sin 50xP P

sin50xPP

260 lbsin50

339.4 lb 339 lbP

(b) tan 50x yP P

tan 50x

yPP

260 lbtan 50

218.2 lb 218 lbyP

27

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) 25 lbyP

Then: sin 75

yPP

25 lbsin 75

or 25.9 lbP

(b)tan 75

yx

PP

25 lbtan 75

or 6.70 lbxP

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 ACis 450 N.

Then:

(a) 450 N 549.3 Ncos35

P

549 NP

(b) 450 N tan 35xP

315.1 N

315 NxP

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)sin38

xPP

200 Nsin38

324.8 N or 325 NP

(b)tan 38

xy

PP

200 Ntan 38

255.98 N

or 256 NyP

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:

500 140 N 480 NF i j

425 315 N 300 NF i j

510 240 N 450 NF i j

415 N 330 NR F i j

Then:

1 330tan 38.5415

2 2415 N 330 N 530.2 NR

Thus: 530 NR 38.5

31

PROBLEM 2.32 Determine the resultant of the three forces of Problem 2.21.

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

SOLUTION

From Problem 2.21:

20 15.32 kN 12.86 kNF i j

30 10.26 kN 28.2 kNF i j

42 39.5 kN 14.36 kNF i j

34.44 kN 55.42 kNR F i j

Then:

1 55.42tan 58.134.44

2 255.42 kN 34.44 N 65.2 kNR

65.2 kNR 58.2

32

PROBLEM 2.33 Determine the resultant of the three forces of Problem 2.22.

Problem 2.22: Determine the x and y components of each of the forces shown.

SOLUTION

The components of the forces were determined in 2.23.

x yR RR i j

71.9 lb 43.86 lbi j

43.86tan71.9

31.38

2 271.9 lb 43.86 lbR

84.23 lb

84.2 lbR 31.4

Force comp. (lb)x comp. (lb)y

40 lb 30.6 25.7

60 lb 30 51.9680 lb 72.5 33.8

71.9xR 43.86yR

33

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.

204 48.0 lb 90.0 lbF i j

212 112.0 lb 180.0 lbF i j

400 320 lb 240 lbF i j

Thus

x yR R R

256 lb 30.0 lbR i j

Now:

30.0tan256

1 30.0tan 6.68256

and

2 2256 lb 30.0 lbR

257.75 lb

258 lbR 6.68

34

PROBLEM 2.35 Knowing that 35 , determine the resultant of the three forces shown.

SOLUTION

300-N Force:

300 N cos20 281.9 NxF

300 N sin 20 102.6 NyF

400-N Force:

400 N cos55 229.4 NxF

400 N sin 55 327.7 NyF

600-N Force:

600 N cos35 491.5 NxF

600 N sin 35 344.1 NyF

and

1002.8 Nx xR F

86.2 Ny yR F

2 21002.8 N 86.2 N 1006.5 NR

Further:

86.2tan1002.8

1 86.2tan 4.911002.8

1007 NR 4.91

35

PROBLEM 2.36 Knowing that 65 , determine the resultant of the three forces shown.

SOLUTION

300-N Force:

300 N cos20 281.9 NxF

300 N sin 20 102.6 NyF

400-N Force:

400 N cos85 34.9 NxF

400 N sin85 398.5 NyF

600-N Force:

600 N cos5 597.7 NxF

600 N sin 5 52.3 NyF

and

914.5 Nx xR F

448.8 Ny yR F

2 2914.5 N 448.8 N 1018.7 NR

Further:

448.8tan914.5

1 448.8tan 26.1914.5

1019 NR 26.1

36

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:

84145 lb 105 lb116xF

80145 lb 100 lb116yF

100-lb Force:

3100 lb 60 lb5xF

4100 lb 80 lb5yF

156-lb Force:

12156 lb 144 lb13xF

5156 lb 60 lb13yF

and

21 lb, 40 lbx x y yR F R F

2 221 lb 40 lb 45.177 lbR

Further:

40tan21

1 40tan 62.321

Thus: 45.2 lbR 62.3

37

PROBLEM 2.38 Knowing that 50 , determine the resultant of the three forces shown.

SOLUTION

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

140 lb cos50 60 lb cos85 160 lb cos50x xR F

7.6264 lbxR

and

140 lb sin 50 60 lb sin85 160 lb sin 50y yR F

289.59 lbyR

Further:

290tan7.6

1 290tan 88.57.6

Thus: 290 lbR 88.5

38

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

SOLUTION

For an arbitrary angle , we have:

140 lb cos 60 lb cos 35 160 lb cosx xR F

(a) So, for R to be vertical:

140 lb cos 60 lb cos 35 160 lb cos 0x xR F

Expanding,

cos 3 cos cos35 sin sin 35 0

Then:

13cos35

tansin35

or 1

1 3cos35tan 40.265

sin35 40.3

(b) Now:

140 lb sin 40.265 60 lb sin 75.265 160 lb sin 40.265y yR R F

252 lbR R

39

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:

84 12 3156 lb 100 lb116 13 5x x BCR F T

or 0.724 84 lbx BCR T

and

80 5 4156 lb 100 lb116 13 5y y BCR F T

0.6897 140 lby BCR T

(a) So, for R to be vertical,

0.724 84 lb 0x BCR T

116.0 lbBCT

(b) Using

116.0 lbBCT

0.6897 116.0 lb 140 lb 60 lbyR R

60.0 lbR R

40

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

0: 4 kN cos25 5.2 kN sin 35 sin 65 0y y AER F T

or 7.2909 kNAET

7.29 kNAET

(b) xR F

4 kN sin 25 5.2 kN cos35 7.2909 kN cos65

9.03 kN

9.03 kNR

41

PROBLEM 2.42 For the block of Problems 2.35 and 2.36, determine (a) the required value of 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 35 , determine the resultant of the three forces shown.

Problem 2.36: Knowing that 65 , determine the resultant of the three forces shown.

SOLUTION

Selecting the x axis along ,aa we write

300 N 400 N cos 600 N sinx xR F (1)

400 N sin 600 N cosy yR F (2)

(a) Setting 0yR in Equation (2):

Thus 600tan 1.5400

56.3

(b) Substituting for in Equation (1):

300 N 400 N cos56.3 600 N sin56.3xR

1021.1 NxR

1021 NxR R

42

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:

2 216 in. 12 in. 20 in.AC

2 220 in. 21 in. 29 in.BC

Then, from the Free Body Diagram of point C:

16 210: 020 29x AC BCF T T

or 29 421 5BC ACT T

and 12 200: 600 lb 020 29y AC BCF T T

or 12 20 29 4 600 lb 020 29 21 5AC ACT T

Hence: 440.56 lbACT

(a) 441 lbACT

(b) 487 lbBCT

43

PROBLEM 2.44 Knowing that 25 , determine the tension (a) in cable AC, (b) in rope BC.

SOLUTION

Free-Body Diagram Force Triangle

Law of Sines:

5 kNsin115 sin5 sin 60

AC BCT T

(a) 5 kN sin115 5.23 kNsin 60ACT 5.23 kNACT

(b) 5 kN sin 5 0.503 kNsin 60BCT 0.503 kNBCT

44

PROBLEM 2.45 Knowing that 50 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.

SOLUTION

Free-Body Diagram Force Triangle

Law of Sines:

400 lbsin 25 sin 60 sin 95

AC BCF T

(a) 400 lb sin 25 169.69 lbsin95ACF 169.7 lbACF

(b) 400 sin 60 347.73 lbsin95BCT 348 lbBCT

45

PROBLEM 2.46 Two cables are tied together at C and are loaded as shown. Knowing that

30 , determine the tension (a) in cable AC, (b) in cable BC.

SOLUTION

Free-Body Diagram Force Triangle

Law of Sines:

2943 Nsin 60 sin 55 sin 65

AC BCT T

(a) 2943 N sin 60 2812.19 Nsin 65ACT 2.81 kNACT

(b) 2943 N sin 55 2659.98 Nsin 65BCT 2.66 kNBCT

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

Force Triangle

Free-Body Diagram Point C

Force Triangle

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

1 9.9tan 30.5116.8AB

1 12tan 22.6128.8BC

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

1190 Nsin59.49 sin 7.87

BCT

7468.6 NBCT

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

1 1.32tan 10.397.2CD

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

7468.6 Nsin12.23 sin100.39

W

1608.5 NW

Finally, the skier weight 1608.5 N 300 N 1308.5 N

skier weight 1309 N

47

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

Force Triangle

Free-Body Diagram Point E

Force Triangle

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

1 12tan 22.6228.8EF

1 1.32tan 10.397.2DF

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

1100 Nsin100.39 sin12.23

EFT

5107.5 NBCT

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

1 9.9tan 30.5116.8AE

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

5107.5 Nsin 7.89 sin 59.49

W

813.8 NW

Finally, the skier weight 813.8 N 300 N 513.8 N

skier weight 514 N

48

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.

SOLUTION

Free-Body Diagram

Resolving the forces into x and y components:

0: 510 lb sin15 480 lb cos15 0x CF F

or 332 lbCF

0: 510 lb cos15 480 lb sin15 0y DF F

or 368 lbDF

49

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:

0: cos15 540 lb 420 lb cos15 0 or 671.6 lbx B BF F F

672 lbBF

0: 420 lb cos15 671.6 lb sin15 0y DF F

or 232 lbDF

50

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

SOLUTION

Free-Body Diagram Resolving the forces into x and y directions:

0A BR P Q F F

Substituting components:

400 lb 520 lb cos55 520 lb sin 55R j i j

cos55 sin55 0B A AF F Fi i j

In the y-direction (one unknown force)

400 lb 520 lb sin55 sin 55 0AF

Thus,

400 lb 520 lb sin 551008.3 lb

sin55AF

1008 lbAF

In the x-direction:

520 lb cos55 cos55 0B AF F

Thus,

cos55 520 lb cos55B AF F

1008.3 lb cos55 520 lb cos55

280.08 lb

280 lbBF

51

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.

SOLUTION

Free-Body Diagram Resolving the forces into x and y directions:

0A BR P Q F F

Substituting components:

320 lb 600 lb cos55 600 lb sin55R i i j

cos55 sin 55 0P Q Qi i j

In the x-direction (one unknown force)

320 lb 600 lb cos55 cos55 0Q

Thus,

320 lb 600 lb cos5542.09 lb

cos55Q

42.1 lbQ

In the y-direction:

600 lb sin 55 sin 55 0P Q

Thus,

600 lb sin 55 sin 55 457.01 lbP Q

457 lbP

52

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.

SOLUTION

Free-Body Diagram 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 15 150: 680 N 05 17 17x CA CBF T T

or

1 5 200 N5 17CA CBT T (1)

and

4 8 80: 680 N 840 N 05 17 17y CA CBF T T

or

1 2 290 N5 17CA CBT T (2)

Solving Equations (1) and (2) simultaneously:

(a) 750 NCAT

(b) 1190 NCBT

53

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.

SOLUTION

Free-Body Diagram 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 15 150: 680 N 05 17 17x CA CBF T T

or

1 5 200 N5 17CA CBT T (1)

and

4 8 80: 680 N 05 17 17y CA CBF T T W

or

1 2 180 N5 17 4CA CBT T W (2)

Then, from Equations (1) and (2)

17680 N28

2528

CB

CA

T W

T W

Now, with 1050 NT

25: 1050 N28CA CAT T W

or 1176 NW

and

17: 1050 N 680 N28CB CBT T W

or 609 NW 0 609 NW

54

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 speed by cable DE. Knowing that 40 and 35 , that the 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

0: cos35 cos 40 cos 40 0x ACB DEF T T

or

0.0531 0.766 0ACB DET T (1)

and

0: sin 40 sin 35 sin 40 24.8 kN 0y ACB DEF T T

or

0.0692 0.643 24.8 kNACB DET T (2)

From (1)

14.426ACB DET T

Then, from (2)

0.0692 14.426 0.643 24.8 kNDE DET T

and

(b) 15.1 kNDET

(a) 218 kNACBT

55

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 42 and 32 , 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.

SOLUTION

Free-Body Diagram

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

0: cos32 cos 42 20 kN cos42 0x ACBF T

or

0.1049 14.863 kNACBT

(b) 141.7 kNACBT

Now

0: sin 42 sin 32 20 kN sin 42 0y ACBF T W

or

141.7 kN 0.1392 20 kN 0.6691 0W

(a) 33.1 kNW

56

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 constants of the springs are kAB 1500 N/m and kAD 500 N/m, 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:

AB AB AB oF k L L

and

2 20.44 m 0.33 m 0.55 mABL

So:

1500 N/m 0.55 m 0.45 mABF

150 N

Similarly,

AD AD AD oF k L L

Then:

2 20.66 m 0.32 m 0.68 mADL

1500 N/m 0.68 m 0.45 mADF

115 N

(a)

4 7 150: 150 N 115 N 05 25 17x ACF T

or

66.18 NACT 66.2 NACT

57

PROBLEM 2.57 CONTINUED

(b) and

3 24 80: 150 N 66.18 N 115 N 05 25 17yF W

or 208 NW

58

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:

4 35 120: 3 05 37 37x sF W W F

or

4.4833sF W

and

3 12 350: 3 400 N 05 37 37y sF W W F

Then:

3 12 353 4.4833 400 N 05 37 37

W W W

or

62.841 NW

and

281.74 NsF

or

(a) 62.8 NW

59

PROBLEM 2.58 CONTINUED

(b) Have spring force

s AB oF k L L

Where

AB AB AB oF k L L

and

2 20.360 m 1.050 m 1.110 mABL

So:

0281.74 N 800 N/m 1.110 mL

or 0 758 mmL

60

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

SOLUTION

The smallest BCT is when BCT is perpendicular to the direction of ACT

Free-Body Diagram At C Force Triangle

(a) 55.0

(b) 2943 N sin 55BCT

2410.8 N

2.41 kNBCT

61

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: CFor 725 NT 0: 2 1000 N 0y yF T

500 NyT

2 2 2x yT T T

2 22 500 N 725 NxT

525 NxT

By similar triangles:

1.5 m725 525BC

2.07 mBC

2 4.14 mL BC

4.14 mL

62

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 .

SOLUTION

Free-Body Diagram: C Force Triangle

Force triangle is isoceles with

2 180 85

47.5

(a) 2 200 lb cos 47.5 270 lbP

Since 0,P the solution is correct. 270 lbP

(b) 180 55 47.5 77.5 77.5

63

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 .

SOLUTION

Free-Body Diagram: C Force Triangle

(a) Law of Cosines:

2 22 300 lb 150 lb 2 300 lb 150 lb cos85P

323.5 lbP

Since 300 lb,P our solution is correct. 324 lbP

(b) Law of Sines:

sin sin 85300 323.5

sin 0.9238

or 67.49

180 55 67.49 57.5

57.5

64

PROBLEM 2.63 For the structure and loading of Problem 2.45, determine (a) the value of

for which the tension in cable BC is as small as possible, (b) the corresponding value of the tension.

SOLUTION

BCT must be perpendicular to ACF to be as small as possible.

Free-Body Diagram: C Force Triangle is a right triangle

(a) We observe: 55 55

(b) 400 lb sin 60BCT

or 346.4 lbBCT 346 lbBCT

65

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 for which the tension in cable BC is as small as possible, (b) the corresponding value of the tension.

SOLUTION

Free-Body Diagram: B (a) Have: 0BD AB BCT F T

where magnitude and direction of BDT are known, and the direction of ABF is known.

Then, in a force triangle:

By observation, BCT is minimum when 90.0

(b) Have 70 lb sin 180 70 30BCT

68.93 lb

68.9 lbBCT

66

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 Have: s AB ABF k L L

where:

2 20.3 m 0.4 m 0.3 2 mAB ABL L

0.5 m

Then: 660 N/m 0.5 0.3 2 msF

49.986 N

For the collar:

40: 49.986 N 05yF W

or 40.0 NW

67

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

Free-Body Diagram: Collar A 2 20: 0

0.3y s

hF W Fh

or 240 0.09shF h

Now.. s AB ABF k L L

where 2 20.3 m 0.3 2 mAB ABL h L

Then: 2 2560 0.09 0.3 2 40 0.09h h h

or 214 1 0.09 4.2 2 mh h h h

Solving numerically,

415 mmh

68

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

(a)

(b)

(c)

(d)

(e)

20: 2 280 kg 9.81 m/s 0yF T

1 2746.8 N2

T

1373 NT

20: 2 280 kg 9.81 m/s 0yF T

1 2746.8 N2

T

1373 NT

20: 3 280 kg 9.81 m/s 0yF T

1 2746.8 N3

T

916 NT

20: 3 280 kg 9.81 m/s 0yF T

1 2746.8 N3

T

916 NT

20: 4 280 kg 9.81 m/s 0yF T

1 2746.8 N4

T

687 NT

69

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)

(d)

20: 3 280 kg 9.81 m/s 0yF T

1 2746.8 N3

T

916 NT

20: 4 280 kg 9.81 m/s 0yF T

1 2746.8 N4

T

687 NT

70

PROBLEM 2.69 A 350-lb load is supported by the rope-and-pulley arrangement shown. Knowing that 25 , 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 0: 2 sin 25 cos 0xF P P

and

cos 0.8452 or 32.3

For 32.3

0: 2 cos 25 sin 32.3 350 lb 0yF P P

or 149.1 lbP 32.3

For 32.3

0: 2 cos 25 sin 32.3 350 lb 0yF P P

or 274 lbP 32.3

71

PROBLEM 2.70 A 350-lb load is supported by the rope-and-pulley arrangement shown. Knowing that 35 , determine (a) the angle , (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 0: 2 sin cos25 0xF P P

Hence:

(a) 1sin cos252

or 24.2

(b) 0: 2 cos sin 35 350 lb 0yF P P

Hence:

2 cos 24.2 sin 35 350 lb 0P P

or 145.97 lbP 146.0 lbP

72

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

SOLUTION

Free-Body Diagram: Pulley C

(a) 0: cos30 cos50 800 N cos50 0x ACBF T

Hence 2303.5 NACBT

2.30 kNACBT

(b) 0: sin 30 sin50 800 N sin 50 0y ACBF T Q

2303.5 N sin 30 sin 50 800 N sin 50 0Q

or 3529.2 NQ 3.53 kNQ

73

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

0: cos30 cos50 cos50 0x ACBF T P

or 0.3473 ACBP T (1)

0: sin 30 sin 50 sin 50 2000 N 0y ACBF T P

or 1.266 0.766 2000 NACBT P (2)

(a) Substitute Equation (1) into Equation (2):

1.266 0.766 0.3473 2000 NACB ACBT T

Hence: 1305.5 NACBT

1306 NACBT

(b) Using (1)

0.3473 1306 N 453.57 NP

454 NP

74

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

SOLUTION

(a) 200 lb cos30 cos 25 156.98 lbxF

157.0 lbxF

200 lb sin30 100.0 lbyF

100.0 lbyF

200 lb cos30 sin 25 73.1996 lbzF

73.2 lbzF

(b) 156.98cos200x or 38.3x

100.0cos200y or 60.0y

73.1996cos200z or 111.5z

75

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

SOLUTION

(a) 420 lb sin 20 sin 70 134.985 lbxF

135.0 lbxF

420 lb cos20 394.67 lbyF

395 lbyF

420 lb sin 20 cos70 49.131 lbzF

49.1 lbzF

(b) 134.985cos420x

108.7x

394.67cos420y

20.0y

49.131cos420z

83.3z

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 x, y, and z that the force forms with axes at A which are parallel to the coordinate axes.

SOLUTION

(a) 4.2 kN sin 50 cos 40 2.4647 kNxF

2.46 kNxF

4.2 kN cos50 2.6997 kNyF

2.70 kNyF

4.2 kN sin 50 sin 40 2.0681 kNzF

2.07 kNzF

(b) 2.4647cos4.2x

54.1x

77

PROBLEM 2.75 CONTINUED

2.7cos4.2y

130.0y

2.0681cos4.0z

60.5z

78

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 x, y, and z that the force forms with axes at A which are parallel to the coordinate axes.

SOLUTION

(a) 3.6 kN cos 45 sin 25 1.0758 kNxF

1.076 kNxF

3.6 kN sin 45 2.546 kNyF

2.55 kNyF

3.6 kN cos 45 cos25 2.3071 kNzF

2.31 kNzF

(b) 1.0758cos3.6x

107.4x

79

PROBLEM 2.76 CONTINUED

2.546cos3.6y

135.0y

2.3071cos3.6z

50.1z

80

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

y, and z that the force exerted at A forms with the coordinate axes.

SOLUTION

(a) sin 30 sin 50 220.6 NxF F (Given)

220.6 N 575.95 Nsin30 sin50

F

576 NF

(b) 220.6cos 0.3830575.95

xx

FF

67.5x

cos30 498.79 NyF F

498.79cos 0.86605575.95

yy

FF

30.0y

sin30 cos50zF F

575.95 N sin 30 cos50

185.107 N

185.107cos 0.32139575.95

zz

FF

108.7z

81

PROBLEM 2.78 A horizontal circular plate is suspended as shown from three wires which are attached to a support at D and form 30 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 x,

y, and z that the force exerted at B forms with the coordinate axes.

SOLUTION

(a) sin 30 sin 40 64.28 NzF F (Given)

64.28 N 200.0 Nsin30 sin40

F 200 NF

(b) sin 30 cos 40xF F

200.0 N sin 30 cos 40

76.604 N

76.604cos 0.38302200.0

xx

FF

112.5x

cos30 173.2 NyF F

173.2cos 0.866200

yy

FF

30.0y

64.28 NzF

64.28cos 0.3214200

zz

FF

108.7z

82

PROBLEM 2.79 A horizontal circular plate is suspended as shown from three wires which are attached to a support at D and form 30 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

x, y, and z that the force forms with the coordinate axes.

SOLUTION

(a) 120 lb sin 30 cos60 30 lbxF

30.0 lbxF

120 lb cos30 103.92 lbyF

103.9 lbyF

120 lb sin30 sin 60 51.96 lbzF

52.0 lbzF

(b) 30.0cos 0.25120

xx

FF

104.5x

103.92cos 0.866120

yy

FF

30.0y

51.96cos 0.433120

zz

FF

64.3z

83

PROBLEM 2.80 A horizontal circular plate is suspended as shown from three wires which are attached to a support at D and form 30 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 x, y,and z that the force exerted at C forms with the coordinate axes.

SOLUTION

(a) sin 30 cos60 40 lbxF F (Given)

40 lb 160 lbsin30 cos60

F

160.0 lbF

(b) 40cos 0.25160

xx

FF

104.5x

160 lb cos30 103.92 lbyF

103.92cos 0.866160

yy

FF

30.0y

160 lb sin 30 sin 60 69.282 lbzF

69.282cos 0.433160

zz

FF

64.3z

84

PROBLEM 2.81 Determine the magnitude and direction of the force

800 lb 260 lb 320 lb .F i j k

SOLUTION

2 2 22 2 2 800 lb 260 lb 320 lbx y zF F F F 900 lbF

800cos 0.8889900

xx

FF

27.3x

260cos 0.2889900

yy

FF

73.2y

320cos 0.3555900

zz

FF

110.8z

85

PROBLEM 2.82 Determine the magnitude and direction of the force

400 N 1200 N 300 N .F i j k

SOLUTION

2 2 22 2 2 400 N 1200 N 300 Nx y zF F F F 1300 NF

400cos 0.307691300

xx

FF

72.1x

1200cos 0.923071300

yy

FF

157.4y

300cos 0.230761300

zz

FF

76.7z

86

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

SOLUTION

(a) We have

2 2 22 2 2cos cos cos 1 cos 1 cos cosx y z y y z

Since 0yF we must have cos 0y

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

2 2cos 1 cos64.5 cos55.9 0.70735y 135.0y

(b) Then:

200 N 282.73 Ncos 0.70735

y

y

FF

and cos 282.73 N cos64.5x xF F 121.7 NxF

cos 282.73 N cos55.9z zF F 158.5 NyF

283 NF

87

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

SOLUTION

(a) We have

2 2 22 2 2cos cos cos 1 cos 1 cos cosx y z y y z

Since 0zF we must have cos 0z

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

2 2cos 1 cos75.4 cos132.6 0.69159z 133.8z

(b) Then:

60 N 86.757 Ncos 0.69159

z

z

FF 86.8 NF

and cos 86.8 N cos75.4x xF F 21.9 NxF

cos 86.8 N cos132.6y yF F 58.8 NyF

88

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

SOLUTION

(a) Have

cos 400 N cos28.5x xF F 351.5 NxF

Then:

2 2 2 2x y zF F F F

So: 2 2 2 2400 N 352.5 N 80 N zF

Hence:

2 2 2400 N 351.5 N 80 NzF 173.3 NzF

(b)

80cos 0.20400

yy

FF

101.5y

173.3cos 0.43325400

zz

FF

64.3z

89

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

SOLUTION

(a) cos 600 lb cos136.8z zF F

437.4 lb 437 lbzF

Then:

2 2 2 2x y zF F F F

So: 22 2 2600 lb 200 lb 437.4 lbyF

Hence: 2 2 2600 lb 200 lb 437.4 lbyF

358.7 lb 359 lbyF

(b)

200cos 0.333600

xx

FF

70.5x

358.7cos 0.59783600

yy

FF

126.7y

90

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

4 m 20 m 5 mBA i j k

2 2 24 m 20 m 5 m 21 mBA

2100 N 4 m 20 m 5 m21 mBA

BAF FBA

F i j k

400 N 2000 N 500 NF i j k

400 N, 2000 N, 500 Nx y zF F F

91

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

4 m 20 m 14.8 mDA i j k

2 2 24 m 20 m 14.8 m 25.2 mDA

1260 N 4 m 20 m 14.8 m25.2 mDA

DAF FDA

F i j k

200 N 1000 N 740 NF i j k

200 N, 1000 N, 740 Nx y zF F F

92

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

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

2 2 232 in. 48 in. 36 in. 68 in.BA

204 lb 32 in. 48 in. 36 in.68 in.BA

BAF FBA

F i j k

96 lb 144 lb 108 lbF i j k

96.0 lb, 144.0 lb, 108.0 lbx y zF F F

93

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

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

2 2 225 in. 48 in. 36 in. 65 in.DA

195 lb 25 in. 48 in. 36 in.65 in.DA

DAF FDA

F i j k

75 lb 144 lb 108 lbF i j k

75.0 lb, 144.0 lb, 108.0 lbx y zF F F

94

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

0.96 m 1.12 m 0.96 mDB i j k

2 2 20.96 m 1.12 m 0.96 m 1.76 mDB

220 N 0.96 m 1.12 m 0.96 m1.76 mDB DB

DBT TDB

T i j k

120 N 140 N 120 NDBT i j k

120.0 N, 140.0 N, 120.0 NDB DB DBx y zT T T

95

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

0.96 m 1.20 m 1.28 mEB i j k

2 2 20.96 m 1.20 m 1.28 m 2.00 mEB

250 N 0.96 m 1.20 m 1.28 m2.00 mEB EB

EBT TEB

T i j k

120 N 150 N 160 NEBT i j k

120.0 N, 150.0 N, 160.0 NEB EB EBx y zT T T

96

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

SOLUTION

500 lb cos30 sin15 sin30 cos30 cos15P i j k

500 lb 0.2241 0.50 0.8365i j k

112.05 lb 250 lb 418.25 lbi j k

600 lb cos40 cos20 sin 40 cos40 sin 20Q i j k

600 lb 0.71985 0.64278 0.26201i j k

431.91 lb 385.67 lb 157.206 lbi j k

319.86 lb 635.67 lb 261.04 lbR P Q i j k

2 2 2319.86 lb 635.67 lb 261.04 lb 757.98 lbR

758 lbR

319.86 lbcos 0.42199757.98 lb

xx

RR

65.0x

635.67 lbcos 0.83864757.98 lb

yy

RR

33.0y

261.04 lbcos 0.34439757.98 lb

zz

RR

69.9z

97

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 0.2241 0.50 0.8365P i j k

134.46 lb 300 lb 501.9 lbi j k

400 lb 0.71985 0.64278 0.26201Q i j k

287.94 lb 257.11 lb 104.804 lbi j k

153.48 lb 557.11 lb 397.10 lbR P Q i j k

2 2 2153.48 lb 557.11 lb 397.10 lb 701.15 lbR

701 lbR

153.48 lbcos 0.21890701.15 lb

xx

RR

77.4x

557.11 lbcos 0.79457701.15 lb

yy

RR

37.4y

397.10 lbcos 0.56637701.15 lb

zz

RR

55.5z

98

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

400 mm 450 mm 600 mmAB i j k

2 2 2400 mm 450 mm 600 mm 850 mmAB

1000 mm 450 mm 600 mmAC i j k

2 2 21000 mm 450 mm 600 mm 1250 mmAC

400 mm 450 mm 600 mm850 N

850 mmAB AB ABABABT TAB

i j kT

400 N 450 N 600 NAB

T i j k

1000 mm 450 mm 600 mm1020 N

1250 mmAC AC ACACACT TAC

i j kT

816 N 367.2 N 489.6 NAC

T i j k

1216 N 817.2 N 1089.6 NAB ACR T T i j k

Then: 1825.8 NR 1826 NR

and 1216cos 0.666011825.8x 48.2x

817.2cos 0.447581825.8y 116.6y

1089.6cos 0.596781825.8z 53.4z

99

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

400 mm 450 mm 600 mmAB i j k

2 2 2400 mm 450 mm 600 mm 850 mmAB

1000 mm 450 mm 600 mmAC i j k

2 2 21000 mm 450 mm 600 mm 1250 mmAC

400 mm 450 mm 600 mm1020 N

850 mmAB AB AB ABABT TAB

i j kT

480 N 540 N 720 NABT i j k

1000 mm 450 mm 600 mm850 N

1250 mmAC AC AC ACACT TAC

i j kT

680 N 306 N 408 NACT i j k

1160 N 846 N 1128 NAB ACR T T i j k

Then: 1825.8 NR 1826 NR

and 1160cos 0.63531825.8x 50.6x

846cos 0.46341825.8y 117.6y

1128cos 0.61781825.8z 51.8z

100

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

120 N 140 N 120 NBDT i j k

120 N 150 N 160 NBET i j k

Then:

B BD BER T T

240 N 290 N 40 Ni j k

and 378.55 NR 379 NBR

240cos 0.6340378.55x

129.3x

290cos 0.7661378.55y

40.0y

40cos 0.1057378.55z

96.1z

101

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 yzplane, determine (a) the tension in AC, (b) the magnitude and direction of the resultant of the two forces.

SOLUTION Have

920 lb sin 50 cos40 cos50 sin50 sin 40ABT i j j

cos45 sin 25 sin 45 cos45 cos25AC ACTT i j j

(a)

A AB ACR T T

0A xR

0: 920 lb sin 50 cos 40 cos 45 sin 25 0A x ACxR F T

or

1806.60 lbACT 1807 lbACT

(b)

: 920 lb cos50 1806.60 lb sin 45A yyR F

1868.82 lbA yR

: 920 lb sin 50 sin 40 1806.60 lb cos45 cos25A zzR F

1610.78 lbA zR

1868.82 lb 1610.78 lbAR j k

Then:

2467.2 lbAR 2.47 kipsAR

102

PROBLEM 2.98 CONTINUED

and

0cos 02467.2x 90.0x

1868.82cos 0.75602467.2y 139.2y

1610.78cos 0.652882467.2z 49.2z

103

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 yzplane, determine (a) the tension in AB, (b) the magnitude and direction of the resultant of the two forces.

SOLUTION Have

sin 50 cos40 cos50 sin 50 sin 40AB ABTT i j j

850 lb cos45 sin 25 sin 45 cos45 cos 25ACT i j j

(a)

0A xR

0: sin 50 cos 40 850 lb cos 45 sin 25 0A x ABxR F T

432.86 lbABT 433 lbABT

(b)

: 432.86 lb cos50 850 lb sin 45A yyR F

879.28 lbA yR

: 432.86 lb sin 50 sin 40 850 lb cos45 cos25A zzR F

757.87 lbA zR

879.28 lb 757.87 lbAR j k

1160.82 lbAR 1.161 kipsAR

0cos 01160.82x 90.0x

879.28cos 0.757461160.82y 139.2y

757.87cos 0.652871160.82z 49.2z

104

PROBLEM 2.100 For the plate of Problem 2.89, determine the tension in cables AB and ADknowing 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:

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

2 2 245 in. 48 in. 36 in. 75 in.AC

27 lb 45 in. 48 in. 36 in.75 in.AC AC AC AC

ACT TAC

T i j k

16.2 lb 17.28 lb 12.96ACT i j k

and

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

2 2 232 in. 48 in. 36 in. 68 in.AB

32 in. 48 in. 36 in.68 in.

ABAB AB AB AB

AB TT TAB

T i j k

0.4706 0.7059 0.5294AB ABTT i j k

and

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

2 2 225 in. 48 in. 36 in. 65 in.AD

25 in. 48 in. 36 in.65 in.

ADAD AD AD AD

AD TT TAD

T i j k

0.3846 0.7385 0.5538AD ADTT i j k

105

PROBLEM 2.100 CONTINUED

Now

AB AD ADR T T T

0.4706 0.7059 0.5294 16.2 lb 17.28 lb 12.96ABT i j k i j k

0.3846 0.7385 0.5538ADT i j k

Since R must be vertical, the i and k components of this sum must be zero.

Hence:

0.4706 0.3846 16.2 lb 0AB ADT T (1)

0.5294 0.5538 12.96 lb 0AB ADT T (2)

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

244.79 lb, 257.41 lbAB ADT T

245 lbABT

257 lbADT

106

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

2 2 2220 mm 192 mm 0 292 mmBAd

2 2 2192 mm 192 mm 96 mm 288 mmDAd

2 2 20 192 mm 144 mm 240 mmCAd

and 146 N 220 mm 192 mm292 mmBA BA BAFF i j

110 N 96 Ni j

192 mm 144 mm240 mm

CACA CA CA

FFF j k

0.80 0.60CAF j k

192 mm 192 mm 96 mm288 mm

DADA DA DA

FFF i j k

0.66667 0.66667 0.33333DAF i j k

With PP j

At A: 0: 0BA CA DAF F F F P

i-component: 110 N 0.66667 0DAF or 165 NDAF

j-component: 96 N 0.80 0.66667 165 N 0CAF P (1)

k-component: 0.60 0.33333 165 N 0CAF (2)

Solving (2) for CAF and then using that result in (1), gives 279 NP

107

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:

220 mm 192 mm292 mm

BABA BA BA

FFF i j

0.75342 0.65753 NBAF i j

192 mm 144 mm240 mm

CACA CA CA

FFF j k

0.80 0.60CAF j k

192 mm 192 mm 96 mm288 mm

DADA DA DA

FFF i j k

0.66667 0.66667 0.33333DAF i j k

With: 200 NP j

At A: 0: 0BA CA DAF F F F P

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

i-component: 0.75342 0.66667 0BA DAF F (1)

j-component: 0.65735 0.80 0.66667 200 N 0BA CA DAF F F (2)

k-component: 0.60 0.33333 0CA DAF F (3)

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

DA104.5 N, 65.6 N, 118.1 NBA CAF F F

104.5 NBAF

65.6 NCAF

118.1 NDAF

108

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 ABis 60 lb.

SOLUTION

The forces applied at A are:

, , and AB AC ADT T T P

where PP j . To express the other forces in terms of the unit vectors i, j, k, we write

12.6 ft 16.8 ftAB i j 21 ftAB

7.2 ft 16.8 ft 12.6 ft 22.2 ftAC ACi j k

16.8 ft 9.9 ftAD j k 19.5 ftAD

and 0.6 0.8AB AB AB AB ABABT T TAB

T i j

0.3242 0.75676 0.56757AC AC AC AC ACACT T TAC

T i j k

0.8615 0.50769AD AD AD AD ADADT T TAD

T j k

109

PROBLEM 2.103 CONTINUED

Equilibrium Condition

0: 0AB AC ADF PT T T j

Substituting the expressions obtained for , , and AB AC ADT T T andfactoring i, j, and k:

0.6 0.3242 0.8 0.75676 0.8615AB AC AB AC ADT T T T T Pi j

0.56757 0.50769 0AC ADT T k

Equating to zero the coefficients of i, j, k:

0.6 0.3242 0AB ACT T (1)

0.8 0.75676 0.8615 0AB AC ADT T T P (2)

0.56757 0.50769 0AC ADT T (3)

Setting 60 lbABT in (1) and (2), and solving the resulting set of equations gives

111 lbACT

124.2 lbADT

239 lbP

110

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 ACis 100 lb.

SOLUTION

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

0.6 0.3242 0AB ACT T (1)

0.8 0.75676 0.8615 0AB AC ADT T T P (2)

0.56757 0.50769 0AC ADT T (3)

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

54 lbABT

112 lbADT

215 lbP

111

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:

, , and AB AC ADT T T P

where PP j . To express the other forces in terms of the unit vectors i, j, k, we write

0.72 m 1.2 m 0.54 m ,AB i j k 1.5 mAB

1.2 m 0.64 m ,AC j k 1.36 mAC

0.8 m 1.2 m 0.54 m ,AD i j k 1.54 mAD

and 0.48 0.8 0.36AB AB AB AB ABABT T TAB

T i j k

0.88235 0.47059AC AC AC AC ACACT T TAC

T j k

0.51948 0.77922 0.35065AD AD AD AD ADADT T TAD

T i j k

Equilibrium Condition with WW j

0: 0AB AC ADF WT T T j

Substituting the expressions obtained for , , and AB AC ADT T T andfactoring i, j, and k:

0.48 0.51948 0.8 0.88235 0.77922AB AD AB AC ADT T T T T Wi j

0.36 0.47059 0.35065 0AB AC ADT T T k

112

PROBLEM 2.105 CONTINUED

Equating to zero the coefficients of i, j, k:

0.48 0.51948 0AB ADT T

0.8 0.88235 0.77922 0AB AC ADT T T W

0.36 0.47059 0.35065 0AB AC ADT T T

Substituting 3 kNABT in Equations (1), (2) and (3) and solving the resulting set of equations, using conventional algorithms for solving linear algebraic equations, gives

4.3605 kNACT

2.7720 kNADT

8.41 kNW

113

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:

0.48 0.51948 0AB ADT T

0.8 0.88235 0.77922 0AB AC ADT T T W

0.36 0.47059 0.35065 0AB AC ADT T T

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

3.03 kNABT

4.40 kNACT

8.49 kNW

114

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.

SOLUTION

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

0.48 0.51948 0AB ADT T

0.8 0.88235 0.77922 0AB AC ADT T T W

0.36 0.47059 0.35065 0AB AC ADT T T

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

1.651 kNABT

1.526 kNADT

4.63 kNW

115

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

SOLUTION

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

0.48 0.51948 0AB ADT T

0.8 0.88235 0.77922 0AB AC ADT T T W

0.36 0.47059 0.35065 0AB AC ADT T T

Substituting 2750 kg 9.81 m/s 7.36 kNW in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms, gives

2.63 kNABT

3.82 kNACT

2.43 kNADT

116

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 A of the cone. Knowing that P 0 and that the tension in cord BE is 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.

Hence: cos45 8 sin 4565AB BE

i j k

It follows that: cos 45 8 sin 4565BE BE BE BET T i j kT

cos30 8 sin 3065CF CF CF CFT T i j kT

cos15 8 sin1565DG DG DG DGT T i j kT

117

PROBLEM 2.109 CONTINUED

At A: 0: 0BE CF DGF T T T W P

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

: cos 45 cos30 cos15 065 65 65BE CF DGT T T Pi

or cos45 cos30 cos15 65 0BE CF DGT T T P (1)

8 8 8: 065 65 65BE CF DGT T T Wj

or 65 08BE CF DGT T T W (2)

: sin 45 sin 30 sin15 065 65 65BE CF DGT T Tk

or sin 45 sin 30 sin15 0BE CF DGT T T (3)

With 0P and the tension in cord 0.2 lb:BE

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:

0.669 lbCFT

0.746 lbDGT

1.603 lbW

118

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:

: cos 45 cos30 cos15 65 0BE CF DGT T T Pi (1)

65: 08BE CF DGT T T Wj (2)

: sin 45 sin 30 sin15 0BE CF DGT T Tk (3)

With 1.6 lbW , 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 CFT 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:

1.729 0.668 lbCFT P

Hence, for 0CFT 1.729 0.668 0P

or 0.386 lbP

0 0.386 lbP

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

18 m 30 m 5.4 mAC i j k

2 2 218 m 30 m 5.4 m 35.4 mAC

18 m 30 m 5.4 m35.4 m

ACAC AC AC

AC TT TAC

T i j k

0.5085 0.8475 0.1525AC ACTT i j k

and 6 m 30 m 7.5 mAB i j k

2 2 26 m 30 m 7.5 m 31.5 mAB

6 m 30 m 7.5 m31.5 m

ABAB AB AB

AB TT TAB

T i j k

0.1905 0.9524 0.2381AB ABTT i j k

Finally 6 m 30 m 22.2 mAD i j k

2 2 26 m 30 m 22.2 m 37.8 mAD

6 m 30 m 22.2 m37.8 m

ADAD AD AD

AD TT TAD

T i j k

0.1587 0.7937 0.5873AD ADTT i j k

120

PROBLEM 2.111 CONTINUED

With , at :P AP j

0: 0AB AC AD PF T T T j

Equating the factors of i, j, and k to zero, we obtain the linear algebraic equations:

: 0.1905 0.5085 0.1587 0AB AC ADT T Ti (1)

: 0.9524 0.8475 0.7937 0AB AC ADT T T Pj (2)

: 0.2381 0.1525 0.5873 0AB AC ADT T Tk (3)

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

1.963 kNACT

1.969 kNADT

6.66 kNP

121

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 2.6 kNACTand solve the three resulting linear equations using conventional tools for solving Linear Algebraic Equations (MATLAB or Maple, for example), to obtain

4.77 kNABT

2.61 kNADT

8.81 kNP

122

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

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

2 2 232 in. 48 in. 36 in. 68 in.AB

32 in. 48 in. 36 in.68 in.

ABAB AB AB

AB TT TAB

T i j k

0.4706 0.7059 0.5294AB ABTT i j k

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

2 2 245 in. 48 in. 36 in. 75 in.AC

45 in. 48 in. 36 in.75 in.

ACAC AC AC

AC TT TAC

T i j k

0.60 0.64 0.48AC ACTT i j k

Finally, 25 in. 48 in. 36 in.AD i j k

2 2 225 in. 48 in. 36 in. 65 in.AD

123

PROBLEM 2.113 CONTINUED

25 in. 48 in. 36 in.65 in.

ADAD AD AD

AD TT TAD

T i j k

0.3846 0.7385 0.5538AD ADTT i j k

With ,WW j at A we have:

0: 0AB AC AD WF T T T j

Equating the factors of i, j, and k to zero, we obtain the linear algebraic equations:

: 0.4706 0.60 0.3846 0AB AC ADT T Ti (1)

: 0.7059 0.64 0.7385 0AB AC ADT T T Wj (2)

: 0.5294 0.48 0.5538 0AB AC ADT T Tk (3)

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

136.0 lbABT

143.0 lbADT

211 lbW

124

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 120 lbADT and solve the three resulting linear equations using conventional tools for solving Linear Algebraic Equations (MATLAB or Maple, for example), to obtain

12.59 lbACT

114.1 lbABT

177.2 lbW

125

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 30 angles with the vertical. Determine the tension in each wire.

SOLUTION

0: sin 30 sin 50 sin 30 cos 40x AD BDF T T

sin 30 cos60 0CDT

Dividing through by the factor sin 30 and evaluating the trigonometric functions gives

0.7660 0.7660 0.50 0AD BD CDT T T (1)

Similarly,

0: sin 30 cos50 sin30 sin 40z AD BDF T T

sin 30 sin 60 0CDT

or 0.6428 0.6428 0.8660 0AD BD CDT T T (2)

From (1) 0.6527AD BD CDT T T

Substituting this into (2):

0.3573BD CDT T (3)

Using ADT from above:

AD CDT T (4)

Now,

0: cos30 cos30 cos30y AD BD CDF T T T

228 kg 9.81 m/s 0

or 317.2 NAD BD CDT T T

126

PROBLEM 2.115 CONTINUED

Using (3) and (4), above:

0.3573 317.2 NCD CD CDT T T

Then: 135.1 NADT

46.9 NBDT

135.1 NCDT

127

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 A of the cone. Knowing that the cone weighs 2.4 lb and that P 0, 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:

cos45 8 sin 4565AB BE

i j k

It follows that:

cos 45 8 sin 4565BE BE BE BET T i j kT

cos30 8 sin 3065CF CF CF CFT T i j kT

cos15 8 sin1565DG DG DG DGT T i j kT

At A: 0: 0BE CF DGF T T T W P

132

PROBLEM 2.119 CONTINUED

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

: cos 45 cos30 cos15 065 65 65BE CF DGT T Ti

or cos45 cos30 cos15 0BE CF DGT T T (1)

8 8 8: 065 65 65BE CF DGT T T Wj

or 2.4 65 0.3 658BE CF DGT T T (2)

: sin 45 sin 30 sin15 065 65 65BE CF DGT T T Pk

or sin 45 sin 30 sin15 65BE CF DGT T T P (3)

With 0,P 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

0.299 lbBET

1.002 lbCFT

1.117 lbDGT

133

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 A of the cone. Knowing that the cone weighs 2.4 lb and that P 0.1 lb, 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:

cos 45 cos30 cos15 0BE CF DGT T T (1)

0.3 65BE CF DGT T T (2)

sin 45 sin 30 sin15 65BE CF DGT T T P (3)

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

1.006 lbBET

0.357 lbCFT

1.056 lbDGT

134

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:

2 0.8944 0.44725

N NN j k j k

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

0.6 m 1.3 m 1 mAB i j k

2 2 20.6 m 1.3 m 1 m 1.764 mAB

0.6 m 1.3 m 1 m1.764 m

ABAB AB AB

AB TT TAB

T i j k

0.3436 0.7444 0.5726AB ABTT i j k

and 0.7 m 1.4 m 1 mAC i j k

2 2 20.7 m 1.4 m 1 m 1.8574 mAC

0.7 m 1.4 m 1 m1.764 m

ACAC AC AC

AC TT TAC

T i j k

0.3769 0.7537 0.5384AC ACTT i j k

Then: 0: 0AB ACF N T T W

135

PROBLEM 2.121 CONTINUED

With 200 kg 9.81 m/s 1962 N,W and equating the factors of i, j,and k to zero, we obtain the linear algebraic equations:

: 0.3436 0.3769 0AB ACT Ti (1)

: 0.7444 0.7537 0.8944 1962 0AB ACT T Nj (2)

: 0.5726 0.5384 0.4472 0AB ACT T Nk (3)

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

1311 NN

551 NABT

503 NACT

136

PROBLEM 2.122 Solve Problem 2.121 assuming that a third worker is exerting a force

(180 N)P 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:

2 0.8944 0.44725

N NN j k j k

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

0.6 m 1.3 m 1 mAB i j k

2 2 20.6 m 1.3 m 1 m 1.764 mAB

0.6 m 1.3 m 1 m1.764 m

ABAB AB AB

AB TT TAB

T i j k

0.3436 0.7444 0.5726AB ABTT i j k

and 0.7 m 1.4 m 1 mAC i j k

2 2 20.7 m 1.4 m 1 m 1.8574 mAC

0.7 m 1.4 m 1 m1.764 m

ACAC AC AC

AC TT TAC

T i j k

0.3769 0.7537 0.5384AC ACTT i j k

Then: 0: 0AB ACF N T T P W

137

PROBLEM 2.122 CONTINUED

Where 180 NP i

and 2200 kg 9.81 m/sW j

1962 N j

Equating the factors of i, j, and k to zero, we obtain the linear equations:

: 0.3436 0.3769 180 0AB ACT Ti

: 0.8944 0.7444 0.7537 1962 0AB ACN T Tj

: 0.4472 0.5726 0.5384 0AB ACN T Tk

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

1302 NN

306 NABT

756 NACT

138

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

9 ft 8 ft 12 ftAB i j k

2 2 29 ft 8 ft 12 ft 17 ftAB

9 ft 8 ft 12 ft17 ft

ABAB AB AB

AB TT TAB

T i j k

0.5294 0.4706 0.7059AB ABTT i j k

and

0 8 ft 6 ftAC i j k

2 2 20 ft 8 ft 6 ft 10 ftAC

0 ft 8 ft 6 ft10 ft

ACAC AC AC

AC TT TAC

T i j k

0.8 0.6AC ACTT j k

and

4 ft 8 ft 1 ftAD i j k

2 2 24 ft 8 ft 1 ft 9 ftAD

4 ft 8 ft 1 ft9 ftADE

AD AD ADEAD TT TAD

T i j k

0.4444 0.8889 0.1111AD ADETT i j k

139

PROBLEM 2.123 CONTINUED

Finally,

8 ft 8 ft 4 ftAE i j k

2 2 28 ft 8 ft 4 ft 12 ftAE

8 ft 8 ft 4 ft12 ft

ADEAE AE ADE

AE TT TAE

T i j k

0.6667 0.6667 0.3333AE ADETT i j k

With the weight of the machinery, ,WW j at A, we have:

0: 2 0AB AC AD WF T T T j

Equating the factors of , , and i j k to zero, we obtain the following linear algebraic equations:

0.5294 2 0.4444 0.6667 0AB ADE ADET T T (1)

0.4706 0.8 2 0.8889 0.6667 0AB AC ADE ADET T T T W (2)

0.7059 0.6 2 0.1111 0.3333 0AB AC ADE ADET T T T (3)

Knowing that 320 lb,W 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

46.5 lbABT

34.2 lbACT

110.8 lbADET

140

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:

0.5294 2 0.4444 0.6667 0AB ADE ADET T T (1)

0.4706 0.8 2 0.8889 0.6667 0AB AC ADE ADET T T T W (2)

0.7059 0.6 2 0.1111 0.3333 0AB AC ADE ADET T T T (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) 50.0 lbACT

(b) 162.0 lbAET

(c) 468 lbW

141

PROBLEM 2.128 Solve Problem 2.127 assuming 550y 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 (680 N)P j is applied at A, determine (a) the tension in the wire when 300y 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:

2 2 20.84 my z

680 NABT

y

680 NQ zy

With 550 mm 0.55 m,y we obtain:

22 20.84 m 0.55 m

0.733 m

z

z

and

(a) 680 N 1236.4 N0.55ABT

or 1.236 kNABT

and

(b) 1236 0.866 N 906 NQ

or 0.906 kNQ

147

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) sin 35 3001bP

300 lbsin35

P

523 lbP

(b) Vertical Component

cos35vP P

523 lb cos35

428 lbvP

148

PROBLEM 2.130 A container of weight W is suspended from ring A, to which cables ACand 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 W 1000 N, determine the magnitude 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

0.78 m 1.6 m 0 mAB i j k

2 2 20.78 m 1.6 m 0 1.78 mAB

0.78 m 1.6 m 0 m1.78 m

ABAB AB AB

AB TT TAB

T i j k

0.4382 0.8989 0AB ABTT i j k

and

0 1.6 m 1.2 mAC i j k

2 2 20 m 1.6 m 1.2 m 2 mAC

0 1.6 m 1.2 m2 m

ACAC AC AC

AC TT TAC

T i j k

0.8 0.6AC ACTT j k

and

1.3 m 1.6 m 0.4 mAD i j k

2 2 21.3 m 1.6 m 0.4 m 2.1 mAD

1.3 m 1.6 m 0.4 m2.1 m

ADAD AD AD

AD TT TAD

T i j k

0.6190 0.7619 0.1905AD ADTT i j k

149

PROBLEM 2.130 CONTINUED

Finally,

0.4 m 1.6 m 0.86 mAE i j k

2 2 20.4 m 1.6 m 0.86 m 1.86 mAE

0.4 m 1.6 m 0.86 m1.86 m

AEAE AE AE

AE TT TAE

T i j k

0.2151 0.8602 0.4624AE AETT i j k

With the weight of the container ,WW j at A we have:

0: 0AB AC AD WF T T T j

Equating the factors of , , and i j k to zero, we obtain the following linear algebraic equations:

0.4382 0.6190 0.2151 0AB AD AET T T (1)

0.8989 0.8 0.7619 0.8602 0AB AC AD AET T T T W (2)

0.6 0.1905 0.4624 0AC AD AET T T (3)

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

378 NP

150

PROBLEM 2.131 A container of weight W is suspended from ring A, to which cables ACand 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

AB ADT T P

and using the linear algebraic equations of Problem 2.131 with 150 N,ACT we obtain

(a) 454 NP

(b) 1202 NW

151

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

PP i and QQ k are applied to the ring to maintain the container is the position shown. Knowing that 1200W 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

0.48 m 0.72 m 0.16 mAB i j k

2 2 20.48 m 0.72 m 0.16 m 0.88 mAB

0.48 m 0.72 m 0.16 m0.88 m

ABAB AB AB

AB TT TAB

T i j k

0.5455 0.8182 0.1818AB ABTT i j k

and

0.24 m 0.72 m 0.13 mAC i j k

2 2 20.24 m 0.72 m 0.13 m 0.77 mAC

0.24 m 0.72 m 0.13 m0.77 m

ACAC AC AC

AC TT TAC

T i j k

0.3177 0.9351 0.1688AC ACTT i j k

At A: 0: 0AB ACF T T P Q W

142

PROBLEM 2.125 CONTINUED

Noting that AB ACT T because of the ring A, we equate the factors of , , and i j k to zero to obtain the linear algebraic equations:

: 0.5455 0.3177 0T Pi

or 0.2338P T

: 0.8182 0.9351 0T Wj

or 1.7532W T

: 0.1818 0.1688 0T Qk

or 0.356Q T

With 1200 N:W

1200 N 684.5 N1.7532

T

160.0 NP

240 NQ

143

PROBLEM 2.126 For the system of Problem 2.125, determine W and P knowing that

160Q 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 Band C. Two forces PP i and QQ k are applied to the ring to maintain the container is the position shown. Knowing that 1200W 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 160 NQ to obtain

160 N 456.3 N0.3506

T

800 NW

107.0 NP

144

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 (680 N)P j is applied at A, determine (a) the tension in the wire when 300y 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: 2 2 2 2AB x y z

Here 2 2 2 21 m 0.40 m y z

or 2 2 20.84 my z

Thus, with y given, z is determined.

Now

1 0.40 m 0.41 mAB

AB y z y zAB

i j k i k k

Where y and z are in units of meters, m.

From the F.B. Diagram of collar A:

0: 0x z AB ABN N P TF i k j

Setting the jcoefficient to zero gives:

0ABP yT

With 680 N,P

680 NABT

y

Now, from the free body diagram of collar B:

0: 0x y AB ABN N Q TF i j k

145

PROBLEM 2.127 CONTINUED

Setting thek coefficient to zero gives:

0ABQ T z

And using the above result for ABT we have

680 NABQ T z z

y

Then, from the specifications of the problem, 300 mm 0.3 my

22 20.84 m 0.3 mz

0.866 mz

and

(a) 680 N 2266.7 N0.30ABT

or 2.27 kNABT

and

(b) 2266.7 0.866 1963.2 NQ

or 1.963 kNQ

146

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:

0.5409ABT P

0.295ACT P

0.2959ADT P

Using 8 kN:P

4.33 kNABT

2.36 kNACT

2.37 kNADT

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:

0.6440ABT P

0.0709ACT P

0.6771ADT P

Using 180 lb:P

115.9 lbABT

12.76 lbACT

121.9 lbADT

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

0.2 65BE CF DGT T T (2 )

sin 45 sin 30 sin15 0BE CF DGT T T (3)

cos45 cos30 cos15 65 0BE CF DGT T T P (1 )

Applying the method of elimination to obtain a desired result:

Multiplying (2 ) by sin 45 and adding the result to (3):

sin 45 sin 30 sin 45 sin15 0.2 65 sin 45CF DGT T

or 0.9445 0.3714CF DGT T (4)

Multiplying (2 ) by sin 30 and subtracting (3) from the result:

sin 30 sin 45 sin 30 sin15 0.2 65 sin 30BE DGT T

or 0.6679 0.6286BE DGT T (5)

130

PROBLEM 2.118 CONTINUED

Substituting (4) and (5) into (1 ) :

1.2903 1.7321 65 0DGT P

DGT is taut for 1.2903 lb65

P

or 0.16000 lbP

131

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

0: cos30 0y CAF T Q

With 60 lbQ

(a) 60 lb 0.866CAT

52.0 lbCAT

(b) 0: sin 30 0x CBF P T Q

With 75 lbP

75 lb 60 lb 0.50CBT

or 45.0 lbCBT

152

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

Have 0: cos30 0x CAF T Q

or 0.8660 QCAT

Then for 60 lbCAT

0.8660 60 lbQ

or 69.3 lbQ

From 0: sin 30y CBF T P Q

or 75 lb 0.50CBT Q

For 60 lbCBT

75 lb 0.50 60 lbQ

or 0.50 15 lbQ

Thus, 30 lbQ

Therefore, 30.0 69.3 lbQ

153

PROBLEM 2.134 A welded connection is in equilibrium under the action of the four forces shown. Knowing that 8 kNAF and 16 kN,BF determine the magnitudes of the other two forces.

SOLUTION

Free-Body Diagram of Connection

3 30: 05 5x B C AF F F F

With 8 kN, 16 kNA BF F

4 416 kN 8 kN5 5CF

6.40 kNCF

3 30: 05 5y D B AF F F F

With AF and BF as above:

3 316 kN 8 kN5 5DF

4.80 kNDF

154

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

SOLUTION

Free-Body Diagram of Connection

3 30: 05 5y D A BF F F F

or 35B D AF F F

With 5 kN, 8 kNA DF F

5 36 kN 5 kN3 5BF

15.00 kNBF

4 40: 05 5x C B AF F F F

45C B AF F F

4 15 kN 5 kN5

8.00 kNCF

155

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 Prequired 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

4.50: 50 lb 020.5xF P

or 10.98 lbP

(b) Triangle Proportions

150: 50 lb 025xF P

or 30.0 lbP

156

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

Triangle Proportions

Hence: 2

ˆ500: 48 0ˆ400

xxF

x

or 248ˆ ˆ40050

x x

2 2ˆ ˆ0.92 lb 400x x

2 2ˆ 4737.7 inx

ˆ 68.6 in.x

157

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

480 mm 510 mm 320 mmDB i j k

2 2 2480 510 320 770 mmDB

385 N 480 mm 510 mm 320 mm770 mmDB

DBF FDB

F i j k

240 N 255 N 160 NF i j k

240 N, 255 N, 160.0 Nx y zF F F

158

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

0.48 m 0.51 m 0.32 mBD i j k

2 2 20.48 m 0.51 m 0.32 m 0.77 mBD

0.48 m 0.51 m 0.32 m0.77 m

BDBD BD BD

BD TT TBD

T i j k

0.6234 0.6623 0.4156BD BDTT i j k

and

0.27 m 0.40 m 0.6 mBE i j k

2 2 20.27 m 0.40 m 0.6 m 0.770 mBE

0.26 m 0.40 m 0.6 m0.770 m

BEBE BE BE

BD TT TBD

T i j k

0.3506 0.5195 0.7792BE BETT i j k

Now, because of the frictionless ring at B, 385 NBE BDT T and the force on the support due to the two cables is

385 N 0.6234 0.6623 0.4156 0.3506 0.5195 0.7792F i j k i j k

375 N 455 N 460 Ni j k

159

PROBLEM 2.139 CONTINUED

The magnitude of the resultant is

2 2 22 2 2 375 N 455 N 460 N 747.83 Nx y zF F F F

or 748 NF

The direction of this force is:

1 375cos747.83x or 120.1x

1 455cos747.83y or 52.5y

1 460cos747.83z or 128.0z

160

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

425 lb cos30P

or 368 lbP

(b) 425 lb sin 30R

or 213 lbR

161