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: 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

2 120

sin sin 38 sin

F R

90 28 62 , 180 62 38 80

Then:

2 120 N

sin 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


SOLUTION

Using the Law of Sines

1 80

sin sin 38 sin

F R

90 10 80 , 180 80 38 62

Then:

1 80 N

sin80 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 40

35 lb 50 lb

sin 0.44995

26.74

Then: 40 180

113.3

(b) Using the Law of Sines:

50 lb

sin 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


(a)sin sin 40

30 lb 50 lb

sin 0.3857

22.7

(b) 40 180

117.31

50 lb

sin sin 40

aaF

sin50 lb

sin 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


Have: 180 35 25

120

Then:360 N

sin 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


(a) Have: 360 N 300 N

sin sin 35

sin 0.68829

43.5

(b) 180 35 43.5

101.5

Then: 300 N

sin101.5 sin 35

R

or 513 NR

10

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 if the resultant R of the two forces applied to the support is to be

horizontal, (b) the corresponding magnitude of R.

SOLUTION


(a) Have: 20 lb 14 lb

sin sin 30

sin 0.71428

45.6

(b) 180 30 45.6

104.4

Then: 14 lb

sin104.4 sin 30

R

27.1 lbR

11

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

hook support. Knowing that P 15 lb and Q 25 lb, determine



SOLUTION


(a) Have: 25 lb

sin15 sin 30

Q

12.94 lbQ

(b) 180 15 30

135

Thus: 25 lb

sin135 sin 30

R

sin13525 lb 35.36 lb

sin 30R

35.4 lbR

12

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 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 30P

10 lb 10 lbP

(b) 20 lb cos30R

17.32 lb 17.32 lbR

13

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 is 90 , that is, P is horizontal

Then: (a) 360 N sin 35P

or 206 NP

And: (b) 360 N cos35R

or 295 NR

14

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 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

(a) the required angle 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

2 22 10 lb 20 lb 2 10 lb 20 lb cos110R

2100 400 400 0.342 lb

2636.8 lb

25.23 lbR

Using now the Law of Sines

10 lb 25.23 lb

sin sin110

10 lbsin sin110

25.23 lb

0.3724

So: 21.87

Angle of inclination of R, is then such that:

30

8.13

Hence: 25.2 lbR 8.13

15

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

We have: 180 50 25

105

Then:2 22 4.5 kN 6 kN 2 4.5 kN 6 kN cos105R

270.226 kN

or 8.3801 kNR

Now: 8.3801 kN 6 kN

sin105 sin

6 kNsin sin105

8.3801 kN

0.6916

43.756

8.38 kNR 18.76

16

PROBLEM 2.17


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:

1 2tan 38.66

2.5

1 1.5tan 30.96

2.5

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 cos110.38R

2331319 N

575.6 NR

Using the Law of Sines:

160 N 575.6 N

sin sin110.38

160 Nsin sin110.38

575.6 N

0.2606

15.1

90 66.44

576 NR 66.4

17

PROBLEM 2.18


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



SOLUTION

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 cos135R

21380.3 lb

or 37.15 lbR

and

25 lb 37.15 lb

sin sin135

25 lbsin sin135

37.15 lb

0.4758

28.41

Then: 75 180

76.59

37.2 lbR 76.6

18

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


We have: 180 45 25 110

Then:2 22 30 kN 20 kN 2 30 kN 20 kN cos110R

21710.4 kN

41.357 kNR

and

20 kN 41.357 kN

sin sin110

20 kNsin sin110

41.357 kN

0.4544

27.028

Hence: 45 72.028

41.4 kNR 72.0

19

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


We have: 180 45 25 110

Then:2 22 30 kN 20 kN 2 30 kN 20 kN cos110R

21710.4 kN

41.357 kNR

and

30 kN 41.357 kN

sin sin110

30 kNsin sin110

41.357 kN

0.6816

42.97

Finally: 45 87.97

41.4 kNR 88.0

20

PROBLEM 2.21

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

SOLUTION

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


42 kN Force:

42 kN cos 20 ,xF 39.5 kNxF


21

PROBLEM 2.22


SOLUTION

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

PROBLEM 2.23


SOLUTION

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

PROBLEM 2.24


SOLUTION

We compute the following distances:

2 270 240 250 mmOA

2 2210 200 290 mmOB

2 2120 225 255 mmOC

500 N Force:

70500 N

250xF 140.0 NxF

240500 N

250yF 480 NyF

435 N Force:

210435 N

290xF 315 NxF

200435 N

290yF 300 NyF

510 N Force:

120510 N

255xF 240 NxF

225510 N

255yF 450 NyF

24

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)cos 40

xPP

135 N

cos 40

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)sin 35

yPP

960 N

sin 35

or 1674 NP

(b)tan 35

yx

PP

960 N

tan 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

sin 50

xPP

260 lb

sin50

339.4 lb 339 lbP

(b) tan 50x yP P

tan 50

xy

PP

260 lb

tan 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 lb

sin 75

or 25.9 lbP

(b)tan 75

yx

PP

25 lb

tan 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)sin 38

xPP

200 N

sin 38

324.8 N or 325 NP

(b)tan 38

xy

PP

200 N

tan 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 38.5

415

2 2415 N 330 N 530.2 NR

Thus: 530 NR 38.5

31

PROBLEM 2.32



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 58.1

34.44

2 255.42 kN 34.44 N 65.2 kNR

65.2 kNR 58.2

32

PROBLEM 2.33



shown.

SOLUTION

The components of the forces were determined in 2.23.

x yR RR i j

71.9 lb 43.86 lbi j

43.86tan

71.9

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 51.96

80 lb 72.5 33.8

71.9xR 43.86yR

33

PROBLEM 2.34



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:

30.0tan

256

1 30.0tan 6.68

256

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 cos 20 281.9 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:

86.2tan

1002.8

1 86.2tan 4.91

1002.8

1007 NR 4.91

35

PROBLEM 2.36


shown.

SOLUTION

300-N Force:

300 N cos 20 281.9 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:

448.8tan

914.5

1 448.8tan 26.1

914.5

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 lb

116xF

80145 lb 100 lb

116yF

100-lb Force:

3100 lb 60 lb

5xF

4100 lb 80 lb

5yF

156-lb Force:

12156 lb 144 lb

13xF

5156 lb 60 lb

13yF

and

21 lb, 40 lbx x y yR F R F

2 221 lb 40 lb 45.177 lbR

Further:

40tan

21

1 40tan 62.3

21

Thus: 45.2 lbR 62.3

37

PROBLEM 2.38


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:

290tan

7.6

1 290tan 88.5

7.6

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:

13

cos35tan

sin 35

or

11 3

cos35tan 40.265

sin 35 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 lb

116 13 5x x BCR F T

or 0.724 84 lbx BCR T

and

80 5 4156 lb 100 lb

116 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 cos 25 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 600

tan 1.5400

56.3

(b) Substituting for in Equation (1):

300 N 400 N cos56.3 600 N sin 56.3xR

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: 0

20 29x AC BCF T T

or29 4

21 5BC ACT T

and12 20

0: 600 lb 020 29

y AC BCF T T

or12 20 29 4

600 lb 020 29 21 5

AC 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 kN

sin115 sin 5 sin 60

AC BCT T

(a)5 kN

sin115 5.23 kNsin 60

ACT 5.23 kNACT

(b)5 kN

sin 5 0.503 kNsin 60

BCT 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


Law of Sines:

400 lb

sin 25 sin 60 sin 95

AC BCF T

(a)400 lb

sin 25 169.69 lbsin 95

ACF 169.7 lbACF

(b)400

sin 60 347.73 lbsin 95

BCT 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


Law of Sines:

2943 N

sin 60 sin 55 sin 65

AC BCT T

(a)2943 N

sin 60 2812.19 Nsin 65

ACT 2.81 kNACT

(b)2943 N

sin 55 2659.98 Nsin 65

BCT 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 30.51

16.8AB

1 12tan 22.61

28.8BC

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

1190 N

sin 59.49 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 10.39

7.2CD

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

7468.6 N

sin12.23 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 22.62

28.8EF

1 1.32tan 10.39

7.2DF

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

1100 N

sin100.39 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 30.51

16.8AE

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

5107.5 N

sin 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 sin 55 0B A AF F Fi i j

In the y-direction (one unknown force)

400 lb 520 lb sin 55 sin 55 0AF

Thus,

400 lb 520 lb sin 551008.3 lb

sin 55AF

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 sin 55R 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 0

5 17 17x CA CBF T T

or

1 5200 N

5 17CA CBT T (1)

and

4 8 80: 680 N 840 N 0

5 17 17y CA CBF T T

or

1 2290 N

5 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 0

5 17 17x CA CBF T T

or

1 5200 N

5 17CA CBT T (1)

and

4 8 80: 680 N 0

5 17 17y CA CBF T T W

or

1 2 180 N

5 17 4CA CBT T W (2)

Then, from Equations (1) and (2)

17680 N

28

25

28

CB

CA

T W

T W

Now, with 1050 NT

25: 1050 N

28CA CAT T W

or 1176 NW

and

17: 1050 N 680 N

28CB 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 cos 42 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 0

5 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 0

5 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 0

5 37 37x sF W W F

or

4.4833sF W

and

3 12 350: 3 400 N 0

5 37 37y sF W W F

Then:

3 12 353 4.4833 400 N 0

5 37 37W W W

or

62.841 NW

and

281.74 NsF

or

(a) 62.8 NW

59


(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: C

For 725 NT0: 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 m

725 525

BC

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 sin85

300 323.5

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 0

5yF 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 2

0: 0

0.3y s

hF W F

h

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

12746.8 N

2T

1373 NT

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

12746.8 N

2T

1373 NT

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

12746.8 N

3T

916 NT

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

12746.8 N

3T

916 NT

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

12746.8 N

4T

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

12746.8 N

3T

916 NT

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

12746.8 N

4T

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 cos 25 0xF P P

Hence:

(a)1

sin cos 252

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 sin 50 800 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 sin 30 100.0 lbyF

100.0 lbyF

200 lb cos30 sin 25 73.1996 lbzF

73.2 lbzF

(b)156.98

cos200

x or 38.3x

100.0cos

200y or 60.0y

73.1996cos

200z 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 cos 20 394.67 lbyF

395 lbyF

420 lb sin 20 cos70 49.131 lbzF

49.1 lbzF

(b)134.985

cos420

x

108.7x

394.67cos

420y

20.0y

49.131cos

420z

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)2.4647

cos4.2

x

54.1x

77


2.7cos

4.2y

130.0y

2.0681cos

4.0z

60.5z

78

PROBLEM 2.76



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 cos 25 2.3071 kNzF

2.31 kNzF

(b)1.0758

cos3.6

x

107.4x

79


2.546cos

3.6y

135.0y

2.3071cos

3.6z

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 N575.95 N

sin30 sin50F

576 NF

(b)220.6

cos 0.3830575.95

xx

F

F

67.5x

cos30 498.79 NyF F

498.79cos 0.86605

575.95

yy

F

F

30.0y

sin 30 cos50zF F

575.95 N sin 30 cos50

185.107 N

185.107cos 0.32139

575.95

zz

F

F

108.7z

81

PROBLEM 2.78



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 N200.0 N

sin30 sin40F 200 NF

(b) sin 30 cos 40xF F

200.0 N sin 30 cos 40

76.604 N

76.604cos 0.38302

200.0

xx

F

F 112.5x

cos30 173.2 NyF F

173.2cos 0.866

200

yy

F

F 30.0y

64.28 NzF

64.28cos 0.3214

200

zz

F

F 108.7z

82

PROBLEM 2.79



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 sin 30 sin 60 51.96 lbzF

52.0 lbzF

(b)30.0

cos 0.25120

xx

F

F

104.5x

103.92cos 0.866

120

yy

F

F

30.0y

51.96cos 0.433

120

zz

F

F

64.3z

83

PROBLEM 2.80



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 lb160 lb

sin30 cos60F

160.0 lbF

(b)40

cos 0.25160

xx

F

F

104.5x

160 lb cos30 103.92 lbyF

103.92cos 0.866

160

yy

F

F

30.0y

160 lb sin 30 sin 60 69.282 lbzF

69.282cos 0.433

160

zz

F

F

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 0.8889

900

xx

F

F27.3x

260cos 0.2889

900

yy

F

F 73.2y

320cos 0.3555

900

zz

F

F 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 0.30769

1300

xx

F

F 72.1x

1200cos 0.92307

1300

yy

F

F 157.4y

300cos 0.23076

1300

zz

F

F 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 N282.73 N

cos 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 N86.757 N

cos 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 cos 28.5x 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 0.20

400

yy

F

F 101.5y

173.3cos 0.43325

400

zz

F

F 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 2

600 lb 200 lb 437.4 lbyF

Hence: 2 2 2

600 lb 200 lb 437.4 lbyF

358.7 lb 359 lbyF

(b)

200cos 0.333

600

xx

F

F 70.5x

358.7cos 0.59783

600

yy

F

F 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 N4 m 20 m 5 m

21 mBA

BAF F

BAF 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 N4 m 20 m 14.8 m

25.2 mDA

DAF F

DAF 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 lb32 in. 48 in. 36 in.

68 in.BA

BAF F

BAF 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


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 lb25 in. 48 in. 36 in.

65 in.DA

DAF F

DAF 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 N0.96 m 1.12 m 0.96 m

1.76 mDB DB

DBT T

DBT 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 N0.96 m 1.20 m 1.28 m

2.00 mEB EB

EBT T

EBT 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 sin 30 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 cos 40 cos 20 sin 40 cos 40 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 0.42199

757.98 lb

xx

R

R

65.0x

635.67 lbcos 0.83864

757.98 lb

yy

R

R

33.0y

261.04 lbcos 0.34439

757.98 lb

zz

R

R

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 0.21890

701.15 lb

xx

R

R

77.4x

557.11 lbcos 0.79457

701.15 lb

yy

R

R

37.4y

397.10 lbcos 0.56637

701.15 lb

zz

R

R

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 ABAB

ABT T

AB

i j kT

400 N 450 N 600 NABT i j k

1000 mm 450 mm 600 mm1020 N

1250 mmAC AC ACAC

ACT T

AC

i j kT

816 N 367.2 N 489.6 NACT i j k

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

Then: 1825.8 NR 1826 NR

and1216

cos 0.666011825.8

x 48.2x

817.2cos 0.44758

1825.8y 116.6y

1089.6cos 0.59678

1825.8z 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 AB

ABT T

AB

i j kT

480 N 540 N 720 NABT i j k

1000 mm 450 mm 600 mm850 N

1250 mmAC AC AC AC

ACT T

AC

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

and1160

cos 0.63531825.8

x 50.6x

846cos 0.4634

1825.8y 117.6y

1128cos 0.6178

1825.8z 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 0.6340

378.55x

129.3x

290cos 0.7661

378.55y

40.0y

40cos 0.1057

378.55z

96.1z

101

PROBLEM 2.98



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

920 lb sin 50 cos 40 cos50 sin 50 sin 40ABT i j j

cos 45 sin 25 sin 45 cos 45 cos 25AC 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 cos 45 cos 25A zzR F

1610.78 lbA zR

1868.82 lb 1610.78 lbAR j k

Then:

2467.2 lbAR 2.47 kipsAR

102


and

0cos 0

2467.2x 90.0x

1868.82cos 0.7560

2467.2y 139.2y

1610.78cos 0.65288

2467.2z 49.2z

103

PROBLEM 2.99



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

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

850 lb cos 45 sin 25 sin 45 cos 45 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 cos 45 cos 25A zzR F

757.87 lbA zR

879.28 lb 757.87 lbAR j k

1160.82 lbAR 1.161 kipsAR

0cos 0

1160.82x 90.0x

879.28cos 0.75746

1160.82y 139.2y

757.87cos 0.65287

1160.82z 49.2z

104

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:

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

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

27 lb45 in. 48 in. 36 in.

75 in.AC AC AC AC

ACT T

ACT 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 T

ABT 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 T

ADT i j k

0.3846 0.7385 0.5538AD ADTT i j k

105


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

and146 N

220 mm 192 mm292 mm

BA 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 AB

ABT T T

ABT i j

0.3242 0.75676 0.56757AC AC AC AC AC

ACT T T

ACT i j k

0.8615 0.50769AD AD AD AD AD

ADT T T

ADT j k

109


Equilibrium Condition

0: 0AB AC ADF PT T T j

Substituting the expressions obtained for , , and AB AC ADT T T and

factoring 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 AB

ABT T T

ABT i j k

0.88235 0.47059AC AC AC AC AC

ACT T T

ACT j k

0.51948 0.77922 0.35065AD AD AD AD AD

ADT T T

ADT 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 and

factoring 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


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


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


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


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: cos 45 8 sin 45

65AB BE

i j k

It follows that: cos 45 8 sin 45

65BE BE BE BET T

i j kT

cos30 8 sin 30

65CF CF CF CFT T

i j kT

cos15 8 sin15

65DG DG DG DGT T

i j kT

117


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 65

BE CF DGT T TPi

or cos 45 cos30 cos15 65 0BE CF DGT T T P (1)

8 8 8: 0

65 65 65BE CF DGT T T Wj

or65

08

BE CF DGT T T W (2)

: sin 45 sin 30 sin15 065 65 65

BE 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 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: 0

8BE 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

AC

AC AC AC

AC TT T

ACT i j k

0.5085 0.8475 0.1525AC AC

TT 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

AB

AB AB AB

AB TT T

ABT i j k

0.1905 0.9524 0.2381AB AB

TT 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

AD

AD AD AD

AD TT T

ADT i j k

0.1587 0.7937 0.5873AD AD

TT i j k

120


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


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 kNACT

and 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


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 T

ABT 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 T

ACT 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


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

ADAD AD AD

AD TT T

ADT 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


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 sin 30 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


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 2.116


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)

cos 45 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


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 2.119



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:

cos 45 8 sin 45

65AB BE

i j k

It follows that:

cos 45 8 sin 45

65BE BE BE BET T

i j kT

cos30 8 sin 30

65CF CF CF CFT T

i j kT

cos15 8 sin15

65DG DG DG DGT T

i j kT

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

132


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

: cos 45 cos30 cos15 065 65 65

BE CF DGT T Ti

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

8 8 8: 0

65 65 65BE CF DGT T T Wj

or2.4

65 0.3 658

BE CF DGT T T (2)

: sin 45 sin 30 sin15 065 65 65

BE CF DGT T TPk

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 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

NNN 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 T

ABT 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 T

ACT i j k

0.3769 0.7537 0.5384AC ACTT i j k

Then: 0: 0AB ACF N T T W

135


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

NNN 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 T

ABT 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 T

ACT i j k

0.3769 0.7537 0.5384AC ACTT i j k

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

137


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 T

ABT 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 T

ACT 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 ft

ADEAD AD ADE

AD TT T

ADT i j k

0.4444 0.8889 0.1111AD ADETT i j k

139


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 T

AET 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 2.125

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 T

ABT 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 T

ACT i j k

0.3177 0.9351 0.1688AC ACTT i j k

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

142


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 N684.5 N

1.7532T

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 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

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 N456.3 N

0.3506T

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

Here2 2 2 21 m 0.40 m y z

or 2 2 20.84 my z

Thus, with y given, z is determined.

Now

10.40 m 0.4

1 mAB

ABy z y z

ABi 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


Setting the k 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 N0.30

ABT

or 2.27 kNABT

and

(b) 2266.7 0.866 1963.2 NQ

or 1.963 kNQ

146

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 z

y

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 N0.55

ABT

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 lb

sin 35P

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 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 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 T

ABT 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 T

ACT 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 T

ADT i j k

0.6190 0.7619 0.1905AD ADTT i j k

149


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 T

AET 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 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

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 2.132

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 3

0: 05 5

x B C AF F F F

With 8 kN, 16 kNA BF F

4 416 kN 8 kN

5 5CF

6.40 kNCF

3 30: 0

5 5y D B AF F F F

With AF and BF as above:

3 316 kN 8 kN

5 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 3

0: 05 5

y D A BF F F F

or3

5B D AF F F

With 5 kN, 8 kNA DF F

5 36 kN 5 kN

3 5BF

15.00 kNBF

4 40: 0

5 5x C B AF F F F

4

5C B AF F F

415 kN 5 kN

5

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 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

4.5

0: 50 lb 020.5

xF P

or 10.98 lbP

(b) Triangle Proportions

15

0: 50 lb 025

xF 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 0

ˆ400x

xF

x

or 248ˆ ˆ400

50x 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 N480 mm 510 mm 320 mm

770 mmDB

DBF F

DBF 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 T

BDT 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 T

BDT 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


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 375cos

747.83x or 120.1x

1 455cos

747.83y or 52.5y

1 460cos

747.83z 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


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

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Vector Mechanics for Engineers Statics 7th - Cap 02

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Transcript of Vector Mechanics for Engineers Statics 7th - Cap 02