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CHAPTER 3CHAPTER 2CHAPTER 2

3

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

PROBLEM 2.1

Two forces P and Q are applied as shown at Point A of a hook support. Knowing that P = 75 N and Q = 125 N, determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule.

SOLutiOn

(a) Parallelogram law:

(b) Triangle rule:

We measure: R = = ∞179 75 1 N, a . R = 179 N 75 1. ∞

4


PROBLEM 2.2

For the hook support of Problem 2.1, knowing that the magnitude of P is 75-N, 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.

SOLutiOn

(a) Since R is vertical, x component of R is zero.

R

QP

75 N

20° 35°

Rx = –P sin 20º + Q sin 35º = 0

fi P sin 20º = Q sin 35º

Q = ∞

∞= ∞

∞P

Nsin

sin

sin

sin

20

3575

20

35

Q = 44.7 N

(b) Magnitude of R:

R

N

R 107.1 N

ur= ∞ + ∞

==

P Qcos cos

.

20 35

107 093

R = 107.1 N

5


PROBLEM 2.3

The cable stays AB and AD help support pole AC. Knowing that the tension is 600 N in AB and 200 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: ab

= ∞= ∞

50 2

56 3

.

.


(b) Triangle rule:

We measure: R = 684 N, g = 66.5º R = 684 N 66.5º

6


PROBLEM 2.4

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

SOLutiOn


(b) Triangle rule:

We measure: R = = ∞3 30 66 6. .kN, a R = 3 30. kN 66 6. ∞

7


PROBLEM 2.5

The 1500-N force is to be resolved into components along lines a-a¢ and b-b¢. (a) Determine the angle a by trigonometry knowing that the component along line a-a¢ is to be 1200-N. (b) What is the corresponding value of the component along b-b¢?

SOLutiOn

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

sin sin

sin .

.

b

bb

a ba

1200

60

1500

0 69282

43 854

60 180

18

N N= ∞

== ∞

+ + ∞ = ∞= 00 60 43 854

76 146

∞ - ∞ - ∞= ∞

.

. a = ∞76 1.

(b) Law of sines: Fbb¢

∞=

∞sin .76 146

1500 N

sin60 F

bb¢ = 1682 N

8


PROBLEM 2.6

The 1500-N force is to be resolved into components along lines a-a¢ and b-b¢. (a) Determine the angle a by trigonometry knowing that the component along line b-b¢ is to be 600 N. (b) What is the corresponding value of the component along a-a¢?

SOLutiOn

Using the triangle rule and law of sines:

(a) sin sina600

60

1500N N= ∞

sin .

.

aa

== ∞

0 34641

20 268 a = ∞20 3.

(b) a bb

+ + ∞ = ∞= ∞ - ∞ - ∞= ∞

60 180

180 60 20 268

99 732

.

.

Faa¢

∞=

∞sin .99 732

1500 N

sin60 Faa¢ = 1707 N

9


PROBLEM 2.7

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

SOLutiOn

Using the triangle rule and law of sines:

(a) sin sin

sin .

a

a50

25

350 60374

N N=

∞

=

a = ∞37 138. a = ∞37 1.

(b) a bb

+ + ∞ = ∞= ∞ - ∞ - ∞= ∞

25 180

180 25 37 138

117 86

.

.

R

sin . sin117 86

35

25=

∞ N R = 73 2. N

10


PROBLEM 2.8

A disabled automobile is pulled by means of ropes subjected to the two forces as shown. Determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule.

SOLutiOn


We measure:

R = 5.4 kN a = 12º

R = 5.4 kN 12º

(b) Triangle rule:

We measure:

R = 5.4 kN a = 12º

R = 5.4 kN 12º

R4 kN

2 kN25°

30°a

2 kN

R

4 kN30°

25°

11


PROBLEM 2.9

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

SOLutiOn

Using the triangle rule and the law of sines:

(a) 1600

75

N

sin 25∞=

∞P

sin P = 3660 N

(b) 25 75 180

180 25 75

80

∞ + + ∞ = ∞= ∞ - ∞ - ∞= ∞

bb

1600

80

N

sin 25∞=

∞R

sin R = 3730 N

12


PROBLEM 2.10

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

SOLutiOn

Using the law of cosines: P

P

2 1600 2500 2 1600

2596

= + -=

( ( ( N) N) N)(2500 N)cos 75

N

2 2 ∞

Using the law of sines: sin sin

.

a

a

1600

75

2596

36 5

N N= ∞

= ∞

P is directed 90 36 5∞ ∞ ∞- . or 53.5 below the horizontal. P = 2600 N 53 5. ∞

13


PROBLEM 2.11

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

SOLutiOn


(a) bb

+ ∞ + ∞ = ∞= ∞ - ∞ - ∞= ∞

50 60 180

180 50 60

70

2125

70 60

N

sin sin∞=

∞P

P = 1958 N

(b) 2125

70 50

N

sin sin∞=

∞R R = 1732 N

14


PROBLEM 2.12

A steel tank is to be positioned in an excavation. Knowing that the magnitude of P is 3000 N, determine by trigonometry (a) the required angle a if the resultant R of the two forces applied at A is to be vertical, (b) the corresponding magnitude of R.

SOLutiOn


(a) ( )

( )

sin ( )

a bb ab a

a

+ ∞ + ∞ + = ∞= ∞ - + ∞ - ∞= ∞ -

∞ -

30 60 180

180 30 60

90

90

2125 N==

∞sin60

3000 N

90 37 8∞ - = ∞a . a = ∞52 2.

(b) R

sin ( . ) sin52 2 30

3000

60∞ + ∞=

∞N

R = 3430 N

15


PROBLEM 2.13

For the hook support of Problem 2.7, determine by trigonometry (a) the magnitude and direction of the smallest force P for which the resultant R of the two forces applied to the support is horizontal, (b) the corresponding magnitude of R.

SOLutiOn

The smallest force P will be perpendicular to R.

(a) P = ∞( sin50 25 N) P = 21 1. N Ø

(b) R = ∞( cos50 25 N) R = 45 3. N

16


PROBLEM 2.14

For the steel tank of Problem 2.11, determine by trigonometry (a) the magnitude and direction of the smallest force P for which the resultant R of the two forces applied at A is vertical, (b) the corresponding magnitude of R.

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

SOLutiOn

The smallest force P will be perpendicular to R.

(a) P = ∞( cos2125 30N) P = 1840 N Æ

(b) R = ∞( sin2125 30N) R = 1062.5 N

17


PROBLEM 2.15

Solve Problem 2.1 by trigonometry.

PROBLEM 2.1 Two forces P and Q are applied as shown at Point A of a hook support. Knowing that P = 75 N and Q = 125 N, determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule.

SOLutiOn

Using the triangle rule and the law of cosines:

20 35 180

125

2

75 125

2 7

2 2 2

2 2 2

∞ + ∞ + = ∞= ∞

= + -

= +-

aa

aR P Q PQ

R

cos

( ) (

(

N N)

55 125

32004 56

178 898

2

N)(125 N)

N

cos

.

.

∞

==

R

R

Using the law of sines: sin sin

.

.

b

b125

125

178 898

34 915

N N=

∞

= ∞

70 104 915∞ + = ∞b . R = 178.9 N 104.9º

18


PROBLEM 2.16


PROBLEM 2.3 The cable stays AB and AD help support pole AC. Knowing that the tension is 600 N in AB and 200 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

tan.

.

tan

.

a

a

b

b

=

= ∞

=

= ∞

2 5

339 81

2

333 69

Using the triangle rule: a b yyy

+ + = ∞∞ + ∞ + = ∞

= ∞

180

39 81 33 69 180

106 5

. .

.

Using the law of cosines: R

R

2 600 200 2 600 200 106 5

684 22

2 2= + - ∞=

( ) ( ) ( )( )cos .

.

N N N N

N

Using the law of sines: sin sin .g

200

106 5

N 684.22 N=

∞

gf a gff

= ∞= ∞ - += ∞ - ∞ + ∞= ∞

16 28

90

90 39 81 16 28

66 47

.

( )

( . ) .

. R = 684 N 66.5º

19


PROBLEM 2.17


PROBLEM 2.4 Two forces are applied at Point B of beam AB. Determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule.

SOLutiOn

Using the law of cosines: R

R

2 2 22 3

2 2 3 80

3 304

= +- ∞

=

( ) ( )

( )( )cos

.

kN kN

kN kN

kN

Using the law of sines: sin sing2

80

kN 3.304 kN=

∞

gb g

gg

b

= ∞+ + ∞ = ∞

= ∞ - ∞ - ∞= ∞= ∞ - + ∞

36 59

80 180

180 80 36 59

63 41

180 50

.

.

.

ff == ∞66 59. R = 3 30. kN 66 6. ∞

20


PROBLEM 2.18

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

SOLutiOn

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

We have g = ∞ - ∞ + ∞= ∞

180 40 20

120

( )

Then R

R

2 2 215 10

2 15 10 120

475

21 79

= +- ∞

==

( ) ( )

( )( )cos

.

kN kN

kN kN

kN2

44 kN

and 10 21 794

120

10120

kN kN

kN

21.794 kN

sin

.

sin

sin sin

a

a

=∞

= ÊËÁ

ˆ¯̃

∞

= 00 39737

23 414

.

.a =

Hence: f a= + ∞ =50 73 414. R = 21 8. kN 73.4°

21


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 10 kN in member A and 15 kN in member B, determine by trigonometry the magnitude and direction of the resultant of the forces applied to the bracket by members A and B.

SOLutiOn

Using the force triangle and the laws of cosines and sines

We have g = ∞ - ∞ + ∞= ∞

180 40 20

120

( )

Then R

R

2 2 210 15

2 10 15 120

475

21 79

= +- ∞

==

( ) ( )

( )( )cos

.

kN kN

kN kN

kN2

44 kN

and 15 21 794

120

15120

kN kN

kN

21.794 kN

sin

.

sin

sin sin

a

a

=∞

= ÊËÁ

ˆ¯̃

∞

= 00 59605

36 588

.

.a = ∞

Hence: f = + ∞ = ∞a 50 86 588. R = 21 8. kN 86.6°

22


PROBLEM 2.20

For the hook support of Problem 2.7, knowing that P = 75 N and a = 50 ,∞ determine by trigonometry the magnitude and direction of the resultant of the two forces applied to the support.

PROBLEM 2.7 Two forces are applied as shown to a hook support. Knowing that the magnitude of P is 35 N, determine by trigonometry (a) the required angle a if the resultant R of the two forces applied to the support is to be horizontal, (b) the corresponding magnitude of R.

SOLutiOn

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

We have b = ∞ - ∞ + ∞= ∞

180 50 25

105

( )

Then R

R

R

2 2 2

2 2

75 50

2 75 50 105

10066 1

100 330

= +- ∞

==

( ) ( )

( )( )cos

.

.

N N

N N

N

NN

and sin

N N

g

g

g

75

105

100 330

0 72206

46 225

= ∞

=

= ∞

sin

.

sin .

.

Hence: g - ∞ = ∞ - ∞ = ∞25 46 225 25 21 225. . R = 100 3. N 21 2. ∞

23


PROBLEM 2.21

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

SOLutiOn

Compute the following distances:

OA

OB

OC

= +=

= +=

= +

( ) ( )

( ) ( )

( )

600 800

1000

560 900

1060

480

2 2

2 2

2

mm

mm

(( )900

1020

2

= mm

800-N Force: Fx = +( )800800

1000 N Fx = +640 N

Fy = +( )800600

1000 N Fy = +480 N

424-N Force: Fx = -( )424560

1060 N Fx = -224 N

Fy = -( )424900

1060 N Fy = -360 N

408-N Force: Fx = +( )408480

1020 N Fx = +192 0. N

Fy = -( )408900

1020 N Fy = -360 N

24


PROBLEM 2.22


SOLutiOn

Compute the following distances:

OA

OB

OC

= +=

= +=

=

( ) ( )

( ) ( )

(

2100 2000

2900

700 2400

2500

225

2 2

2 2

mm

mm

00 1200

2550

2 2) ( )+= mm

145-N Force: Fx = +( )1452100

2900 N Fx = +105 N

Fy = +( )1452000

2900 N Fy = 100 N

250-N Force: Fx = -( )250700

2500 N Fx = -70 N

Fy = +( )2502400

2500 N Fy = 240 N

255-N Force: Fx = +( )2551200

2550 N Fx = +120 0. N

Fy = -( )2552250

2550 N Fy = -225 N

25


PROBLEM 2.23


SOLutiOn

200-N Force: Fx = + ∞( )cos200 60 N Fx = 100 0. N

Fy = - ∞( )sin200 60 N Fy = -173 2. N

250-N Force: Fx = - ∞( )sin250 50N Fx = -191 5. N

Fy = - ∞( )cos250 50 N Fy = -160 7. N

300-N Force: Fx = + ∞( )cos300 25 N Fx = 272 N

Fy = + ∞( )sin300 25 N Fy = 126 8. N

26


PROBLEM 2.24


SOLutiOn


Fy = + ∞( )sin80 40 N Fy = 51 4. N


Fy = + ∞( )sin120 70 N Fy = 112 8. N

150-N Force: Fx = - ∞( )cos150 35 N Fx = -122. 9 N

Fy = + ∞( )sin150 35 N Fy = 86.0 N

27


PROBLEM 2.25

Member BD exerts on member ABC a force P directed along line BD. Knowing that P must have a 1500 N horizontal component, determine (a) the magnitude of the force P, (b) its vertical component.

SOLutiOn

(a) P sin35 1500∞ = N

P =∞

1500

35

N

sin P = 2620 N

(b) Vertical component P Pv = ∞cos35

= ∞( . )cos2615 17 35 Pv = 2140 N

28


PROBLEM 2.26

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

SOLutiOn

(a) 750 20 N = ∞P sin

P = 2193 N P = 2190 N

(b) P PABC = ∞cos20

= ∞( )cos2193 20 N PABC = 2060 N

29


PROBLEM 2.27

The guy wire BD exerts on the telephone pole AC a force P directed along BD. Knowing that P must have a 120-N component perpendicular to the pole AC, determine (a) the magnitude of the force P, (b) its component along line AC.

SOLutiOn

(a) PPx=

∞

=∞

sin

sin

38

120 N

38

= 194 91. N or NP = 194 9.

(b) PP

yx=

∞

=∞

tan

120 N

tan

38

38

= 153 59. N or NPy = 153 6.

30


PROBLEM 2.28

The guy wire BD exerts on the telephone pole AC a force P directed along BD. Knowing that P has a 180-N component along line AC, determine (a) the magnitude of the force P, (b) its component in a direction perpendicular to AC.

SOLutiOn

(a) PPy=

∞

=∞

cos38180 N

cos 38

= 228 4. N P = 228 N

(b) P Px y= ∞

= ∞

tan

( ) tan

38

180 38 N

= 140 63. N Px = 140 6. N

31


PROBLEM 2.29

Member CB of the vise shown exerts on block B a force P directed along line CB. Knowing that P must have a 1200- N horizontal component, determine (a) the magnitude of the force P, (b) its vertical component.

SOLutiOn

We note:

CB exerts force P on B along CB, and the horizontal component of P is Px = 1200 N:

Then

(a) P Px = ∞sin55

PPx=

∞

=∞

=

sin

sin

.

55

1200

55

1464 9

N

N P = 1465 N

(b) P Px y= ∞tan55

PP

yx=

∞

=∞

=

tan

tan

.

55

1200

55

840 2

N

N Py = 840 N Ø

32


PROBLEM 2.30

Cable AC exerts on beam AB a force P directed along line AC. Knowing that P must have a 1750-N vertical component, determine (a) the magnitude of the force P, (b) its horizontal component.

SOLutiOn

(a) PPy=

∞cos 55

=∞

=

1750

55

3051

N

N

cos

P = 3050 N

(b) P Px = ∞sin 55

= ∞

=

( )sin

.

3051 55

2499 2

N

N Px = 2500 N

33


PROBLEM 2.31

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

Components of the forces were determined in Problem 2.22:

Force x Comp. (N) y Comp. (N)

145 N +105 +100

250 N –70 +240

255 N +120 –225

Rx = +155 Ry = +115

R i j

i j

= +

= +

=

=

= ∞

=

R R

R

R

R

x y

y

x

( (

tan

.

( )

155 115

115

15536 573

155 2

N) N)

a

a

++=

( )

.

115

193 0

2

N R = 193.0 N 36 6. ∞

34


PROBLEM 2.32



SOLutiOn



80 N +61.3 +51.4

120 N +741.0 +112.8

150 N -122.9 +86.0

Rx = -20 6. Ry = +250 2.

R i j

i j

= +

= - +

=

=

R R

R

R

x y

y

x

( . ) ( . )

tan

tan.

20 6 250 2

250 2

N N

N

20.6 N

a

a

ttan .

.

aa

== ∞

12 1456

85 293

R =∞

250 2

85 293

.

sin .

N R = 251 N 85 3. ∞

35


PROBLEM 2.33



SOLutiOn


200 N +100 -173.2

250 N -191.5 -160.7

300 N +272 +126.8

Rx = +180 5. Ry = -207 1.

R i j

i j

= +

= + -

=

=

R R

R

R

x y

y

x

( . ) ( . )

tan

tan.

.

180 5 207 1

207 1

180 5

N N

N

a

aNN

tan .

.

aa

== ∞

1 147

48 9

R = + - =( . ) ( . ) .180 5 207 1 274 722 2 N R = 275 N 48 9. ∞

36


PROBLEM 2.34



SOLutiOn



800 N +640 +480

424 N -224 -360

408 N +192 -360

Rx = +608 Ry = -240

R i j

i j

= +

= + -

=

=

= ∞

R R

R

R

x y

y

x

( ) ( )

tan

.

608 240

240

60821 541

N N

a

a

R =

=

240

653 65

N

sin(21.541 )

N

∞. R = 654 N 21 5. ∞

37


PROBLEM 2.35

Knowing that a = 35°, determine the resultant of the three forces shown.

SOLutiOn

100-N Force: F

Fx

y

= + ∞ = += - ∞ = -

( )cos .

( )sin .

100 35 81 915

100 35 57 358

N N

N N

150-N Force: F

Fx

y

= + ∞ = += - ∞ = -

( cos .

( )sin .

150 65 63 393

150 65 135 946

N) N

N N

200-N Force: F

Fx

y

= - ∞ = -= - ∞ = -

( )cos .

( )sin .

200 35 163 830

200 35 114 715

N N

N N


100 N +81.915 -57.358

150 N +63.393 -135.946

200 N -163.830 -114.715

Rx = -18 522. Ry = -308 02.

R i j

i j

= +

= - + -

=

=

R R

R

R

x y

y

x

( . ) ( . )

tan

.

.

18 522 308 02

308 02

18 522

N N

a

a == ∞86 559.

R = 308 02

86 559

.

sin .

N R = 309 N 86 6. ∞

38


PROBLEM 2.36

Four forces act on bolt A as shown. Determine the resultant of the forces on the bolt.

SOLutiOn

The x and y components of each force are determined by trigonometry as shown and are entered in the table below.

Force Magnitude N x Component N y Component N

F1

150 +129.9 +75.0

F2

80 –27.4 +75.2

F3

110 0 –110.0

F4

100 +96.6 –25.9

Rx = +199 1. Ry = +14 3.

Thus the resultant R of the four forces is

R = Rxi+R

yj R = (199.1 N)i + (14.3 N)j

The magnitude and direction of the resultant may now be determined. From the triangle shown, we have

tan.

..a a= = = ∞

R

Ry

x

14 3

199 14 1

N

N

R = =14 3

199 6.

sin.

NN

a R = 199.6 N 4.1º

(F1 sin 30°)

(F1 cos 30°)

(F4 cos 15°) i– (F4 sin 15°) j

(F2 sin 20°) i

(F2 cos 20°) j

– F3j

Ra

Rx = (199.1 N) iRy = (14.3 N) j

15°30°

20°

F4=100 N

F2=150 NF1=80 N

y

Ax

F3=110 N

39


PROBLEM 2.37

Knowing that a = 40°, determine the resultant of the three forces shown.

SOLutiOn

300-N Force: F

Fx

y

= ∞ == ∞ =

( )cos .

( )sin .

300 20 281 91

300 20 102 61

N N

N N

400-N Force: F

Fx

y

= ∞ == ∞ =

( )cos .

( )sin .

400 60 200 0

400 60 346 41

N N

N N

600-N Force: F

Fx

y

= ∞ == - ∞ = -

( )cos .

( )sin .

600 30 519 62

600 30 300 0

N N

N N

and R F

R F

R

x x

y y

= == =

= +=

SS

1001 53

149 02

1001 53 149 02

1012 5

2 2

.

.

( . ) ( . )

.

N

N

66 N

Further: tan.

..a = =149 02

1001 530 14879

a == ∞

-tan .

.

1 0 14879

8 463 R = 1013 N 8 46. ∞

40


PROBLEM 2.38

Knowing that a = 75°, determine the resultant of the forces shown.

SOLutiOn

300-N Force: F

Fx

y

= ∞ == ∞ =

( ) cos .

( )sin .

300 20 281 91

300 20 102 61

N N

N N

400-N Force: F

Fx

y

= ∞ = -= ∞ =

( ) cos .

( )sin .

400 95 34 862

400 95 398 48

N N

N N

600-N Force: F

Fx

y

= ∞ == ∞ =

( ) cos .

( )sin .

600 5 597 72

600 5 52 293

N N

N N

Then R F

R Fx x

y y

= == =

SS

844 77

553 38

.

.

N

N

and R = +=

( . ) ( . )

.

844 77 553 38

1009 88

2 2

N

tan.

.tan .

.

a

aa

=

== ∞

553 38

844 770 655

33 23 R = 1010 N 33 2. ∞

41


PROBLEM 2.39

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

SOLutiOn

R F

R

x x

x

== + + ∞ -= -

S( )cos ( )cos( ) ( )cos

(

100 150 30 200

100

N N N

N

a a a))cos ( )cos( )a a+ + ∞150 30 N (1)

R F

R

y y

y

=

= - - + ∞ -= -

S

( )sin ( )sin ( ) ( )sin

(

100 150 30 200

300

N N N

a a aNN N)sin ( )sin ( )a a- + ∞150 30 (2)

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

- + + ∞ =

- + ∞ - ∞100 150 30

100 150 30 30

cos cos( )

cos (cos cos sin sin

a aa a a

0

))

. cos sin

==

0

29 904 75a a

tan.

.

.

a

a

=

== ∞

29 904

750 3988

21 74 a = ∞21 7.

(b) Substituting for a in Eq. (2):

Ry = - ∞ - ∞

= -

300 21 74 150 51 74

228 9

sin . sin .

. N

R Ry= =| | .228 9 N R = 229 N

42


PROBLEM 2.40

For the beam of Sample Problem 2.3, 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.

SOLutiOn

R F T

R T

x x BC

x BC

= = - + -

= - +

S 840

1160

12

13780

3

5500

21

29420

( ) ( ) N N

N (1)

R F T

R T

y y BC

y BC

= = - -

= -

S 800

1160

5

13780

4

5500

20

29700

( ) ( ) N N

N (2)

(a) For R to be vertical, we must have Rx = 0

Set Rx = 0 in Eq. (1) - + =21

29420TBC N 0 TBC = 580 N

(b) Substituting for TBC in Eq. (2):

R

R

y

y

= -

= -

20

29580 700

300

( ) N N

N

R Ry= =| | 300 N R = 300 N

43


PROBLEM 2.41

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

SOLutiOn

Using the x and y axes shown:

R F T

Tx x AC

AC

= = ∞ + ∞ + ∞= ∞ +

S sin ( )cos ( )cos

sin

10 250 35 375 60

10 392

N N

..29 N (1)

R F T

R T

y y AC

y A

= = ∞ + ∞ - ∞

= -

S ( )sin ( )sin cos

.

250 35 375 60 10

468 15

N N

N CC cos10∞ (2)

(a) Set Ry = 0 in Eq. (2):

468 15 10 0

475 37

. cos

.

N

N

- ∞ ==

T

TAC

AC TAC = 475 N

(b) Substituting for TAC in Eq. (1):

R

R R

x

x

= ∞ +==

( . )sin .475 37 10 392 29 N N

474.84 N

R = 475 N

44


PROBLEM 2.42

For the block of Problems 2.37, determine (a) the required value of a if the resultant of the three forces shown is to be parallel to the incline, (b) the corresponding magnitude of the resultant.

SOLutiOn

Select the x axis to be along a a ¢.

Then

R Fx x= = + +S ( ) ( )cos ( )sin300 400 600 N N Na a (1)

and

R Fy y= = -S ( )sin ( )cos400 600 N Na a (2)

(a) Set Ry = 0 in Eq. (2).

( )sin ( )cos400 600 0N Na a- =

Dividing each term by cosa gives:

( ) tan

.

400 600

600

40056 310

N N

tan N

N

a

a

a

=

=

= ∞ a = ∞56 3.

(b) Substituting for a in Eq. (1) gives:

Rx = + ∞ + ∞ =300 400 56 31 600 56 31 1021 11 N N N N( )cos . ( )sin . . Rx = 1021 N

45


PROBLEM 2.43

Two cables are tied together at C and are loaded as shown. Knowing that a = 20°, determine the tension (a) in cable AC, (b) in cable BC.

SOLutiOn

Free-Body Diagram Force Triangle

Law of sines: T TAC BC

sin sin sin70 50

1962

60∞=

∞=

∞ N

(a) TAC =∞

∞ =1962

6070 2128 9

N N

sinsin . TAC = 2 13. kN

(b) TBC =∞

∞ =1962

6050 1735 49

N N

sinsin . TBC = 1 735. kN

46


PROBLEM 2.44

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

SOLutiOn



sin sin sin60 40

500

80∞=

∞=

∞ N

(a) TAC =∞

∞ =500

8060 439 69

N N

sinsin . TAC = 440 N

(b) TBC =∞

∞ =500

8040 326 35

N N

sinsin . TBC = 326 N

47


PROBLEM 2.45

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

SOLutiOn



sin sin35 75

500

∞=

∞=

∞ N

sin 70

(a) TAC =∞

∞500

7035

N

sinsin TAC = 305 N

(b) TBC =∞

∞500

7075

N

sinsin TBC = 514 N

48


PROBLEM 2.46

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

SOLutiOn


W =

==

mg

kg m/s

N

2( )( . )200 9 81

1962


sin sin sin15 105

1962

60∞=

∞=

∞ N

(a) TAC =∞

∞( sin

sin

1962 15

60

N) TAC = 586 N

(b) TBC =∞

∞( sin

sin

1962 105

60

N) TBC = 2190 N

49


PROBLEM 2.47

Knowing that a = ∞20 , determine the tension (a) in cable AC, (b) in rope BC.

SOLutiOn



sin sin sin110 5

6000

65∞=

∞=

∞ N

(a) TAC =∞

∞6000

65110

N

sinsin TAC = 6220 N

(b) TBC =∞

∞6000

655

N

sinsin TBC = 577 N

50


PROBLEM 2.48

Knowing that a = ∞55 and that boom AC exerts on pin C a force directed along line AC, determine (a) the magnitude of that force, (b) the tension in cable BC.

SOLutiOn


Law of sines: F TAC BC

sin sin sin35 50

1500

95∞=

∞=

∞ N

(a) FAC =∞

∞1500

9535

N

sinsin FAC = 864 N

(b) TBC =∞

∞1500

9550

N

sinsin TBC = 1153 N

51


PROBLEM 2.49

Two forces P and Q are applied as shown to an aircraft connection. Knowing that the connection is in equilibrium and that P = 2500 N and Q = 3250 N, determine the magnitudes of the forces exerted on the rods A and B.

SOLutiOn

Resolving the forces into x- and y-directions:

R P Q F F= + + + =A B 0

Substituting components: R j i

j

i

= - + ∞- ∞+ -

( ) [( )cos ]

[( )sin ]

( cos

2500 3250 50

3250 50

50

N N

N

F FB A ∞∞ + ∞ =) ( sin )i jFA 50 0

In the y-direction (one unknown force)

- - ∞ + ∞ =2500 3250 50 50 0N N( )sin sinFA

Thus, FA =+ ∞

∞2500 3250 50

50

N N( )sin

sin

= 6513 5. N FA = 6510 N

In the x-direction: ( )cos cos3250 50 50 0N ∞ + - ∞ =F FB A

Thus, F FB A= ∞ - ∞= ∞ - ∞

cos ( )cos

( . )cos ( )cos

50 3250 50

6513 5 50 3250 50

N

N N

= 2097 7. N FB = 2100 N

Free-Body Diagram

52


PROBLEM 2.50

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 = 3750 N and FB = 2000 N, determine the magnitudes of P and Q.

SOLutiOn

Resolving the forces into x- and y-directions:

R P Q F F= + + + =A B 0

Substituting components: R = - + ∞ - ∞- ∞+ ∞

P Q Qj i j

i

cos sin

[( cos ]

[( sin ]

50 50

3750 50

3750 50

N)

N) jj i+ (2000 N)

In the x-direction (one unknown force)

Q cos [( cos ]50 3750 50 2000 0∞ - ∞ + = N) N

Q =∞ -∞

=

( cos

cos

.

3750 50 2000

50

638 55

N) N

N

In the y-direction: - - ∞ + ∞ =P Q sin ( sin50 3750 50 0 N)

P Q= - ∞ + ∞= - ∞ + ∞

sin ( sin

( . sin ( sin

50 3750 50

638 55 50 3750 50

N)

N) N)

== 2383 5. N P Q= =2380 639 N N;

Free-Body Diagram

53


PROBLEM 2.51

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

SOLutiOn

Free-Body Diagram of Connection

SF F F Fx B C A= - - =03

5

3

50:

With F

FA

B

==

8

16

kN

kN

FC = -4

516

4

58( ( kN) kN) FC = 6 40. kN

S F F F Fy D B A= - + - =03

5

3

50:

With FA and F

B as above: FD = -3

516

3

58( ( kN) kN) FD = 4 80. kN

54


PROBLEM 2.52

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

SOLutiOn

Free-Body Diagram of Connection

SF F F Fy D A B= - - + =03

5

3

50:

or F F FB D A= + 3

5

With F FA D= =5 8kN kN,

FB = +ÈÎÍ

˘˚̇

5

36

3

55kN kN( ) FB = 15 00. kN

SF F F Fx C B A= - + - =04

5

4

50:

F F FC B A= -

= -

4

54

515 5

( )

( )kN kN FC = 8 00. kN

55


PROBLEM 2.53

Two cables tied together at C are loaded as shown. Knowing that Q = 300 N, determine the tension (a) in cable AC, (b) in cable BC.

SOLutiOn

SF T Qy CA= - ∞ =0 30 0: cos

With Q = 300 N

(a) TCA = ( )( . )300 0 866N TCA = 260 N

(b) SF P T Qx CB= - - ∞ =0 30 0: sin

With P = 375 N

TCB = -375 300 0 50N N( )( . ) or TCB = 225 N

56


PROBLEM 2.54

Two cables tied together at C are loaded as shown. Determine the range of values of Q for which the tension will not exceed 300-N in either cable.

SOLutiOn

Free-Body Diagram

SF T Qx BC= - - ∞ + =0 60 375 0: cos N

T QBC = - ∞375 60 N cos (1)

SF T Qy AC= - ∞ =0 60 0: sin

T QAC = ∞sin60 (2)

Requirement TAC 300 N

From Eq. (2): Q sin60 300∞ N

Q 346 41. N

Requirement TBC 300 N

From Eq. (1): 375 60 300 N N- ∞Q cos

Q 150 N 150 0 346. N N Q

57


PROBLEM 2.55

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

SOLutiOn

Free-Body Diagram

+ SF T T Tx ACB ACB CD= ∞ - ∞ - ∞ =0 10 30 30 0: cos cos cos

T TCD ACB= 0 137158. (1)

+SF T T Ty ACB ACB CD= ∞ + ∞ + ∞ - =0 10 30 30 900 0: sin sin sin

0 67365 0 5 900. .T TACB CD+ = (2)

(a) Substitute (1) into (2): 0 67365 0 5 0 137158 900. . ( . )T TACB ACB+ =

TACB = 1212 56. N TACB = 1213 N

(b) From (1): TCD = 0 137158 1212 56. ( . N) TCD = 166 3. N

58


PROBLEM 2.56

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

SOLutiOn

Free-Body Diagram

+ SF T Tx ACB ACB= ∞ - ∞ - ∞ =0 15 25 80 25 0: cos cos ( cos N)

TACB = 1216 15. N

+SFy = ∞ + ∞0 1216 15 15 1216 15 25: ( . sin ( . sin N) N)

+ ∞ - =

=( sin

.

80 25 0

862 54

N)

N

W

W (a) W = 863 N

(b) TACB = 1216 N

59


PROBLEM 2.57

For the cables of Problem 2.45, it is known that the maximum allowable tension is 600 N in cable AC and 750 N in cable BC. Determine (a) the maximum force P that can be applied at C, (b) the corresponding value of a.

SOLutiOn


(a) Law of cosines P2 2 2600 750 2 600 750 25 45= + - ∞ + ∞( ) ( ) ( )( )cos( )

P = 784 02. N P = 784 N

(b) Law of sines sin sin ( )

.

b600

25 45

784 02 N N=

∞ + ∞

b = ∞46 0. a = ∞ + ∞ = ∞46 0 25 71 0. .

60


PROBLEM 2.58

For the situation described in Figure P2.47, determine (a) the value of a for which the tension in rope BC is as small as possible, (b) the corresponding value of the tension.

PROBLEM 2.47 Knowing that a = ∞20 , determine the tension (a) in cable AC, (b) in rope BC.

SOLutiOn


To be smallest, TBC must be perpendicular to the direction of TAC .

(a) Thus, a = ∞5 a = ∞5 00.

(b) TBC = ∞( sin6000 5 N) TBC = 523 N

61


PROBLEM 2.59

For the structure and loading of Problem 2.48, determine (a) the value of a for which the tension in cable BC is as small as possible, (b) the corresponding value of the tension.

SOLutiOn

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

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

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

(a) We observe: a = ∞ - ∞90 30 a = ∞60 0.

(b) TBC = ∞( )sin1500 50N

or TBC = 1149 07. N TBC = 1149 N

62


PROBLEM 2.60

Knowing that portions AC and BC of cable ACB must be equal, determine the shortest length of cable that can be used to support the load shown if the tension in the cable is not to exceed 870 N.

SOLutiOn

Free-Body Diagram: C

(ForT = 725 N)

+SF Ty y= - =0 2 1200 0: N

T

T T T

T

T

y

x y

x

x

=

+ =

+ ==

600

600 870

630

2 2 2

2 2 2

N

N N

N

( ) ( )

By similar triangles: BC

BC

870

2 1

630

2 90

N

m

N

m

=

=

.

.

L BC==

2

5 80

( )

. m L = 5 80. m

63


PROBLEM 2.61

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

SOLutiOn

Free-Body Diagram: C Force Triangle

Force triangle is isosceles with2 180 85

47 5

bb

= ∞ - ∞= ∞.

(a) P = =2 800( N)cos 47.5 1081 N∞

Since P 0, the solution is correct. P = 1081 N

(b) a = ∞ - ∞ - ∞ = ∞180 50 47 5 82 5. . a = ∞82 5.

64


PROBLEM 2.62

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

SOLutiOn


(a) Law of cosines: P

P

2 21200 600 2 1200 85

1294

= + - ∞=

( ( ( cos N) N) N)(600 N)

N

2

Since P �1200 N, the solution is correct.

P = 1294 N b(b) Law of sines:

sin sin

.

.

b

ba

1200

85

129467 5

180 50 67 5

N N=

∞

= ∞= ∞ - ∞ - ∞ a = ∞62 5. b

65


PROBLEM 2.63

Collar A is connected as shown to a 250-N 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 = 125 mm, (b) x = 375 mm.

SOLutiOn

(a) Free Body: Collar A Force Triangle

P

P

125

250

515 39

60 634

=

=

N

N

.

.

P = 60 6. N

(b) Free Body: Collar A Force Triangle

P

375

250= N

625 P = 150 0. N

66


PROBLEM 2.64

Collar A is connected as shown to a 250 N load and can slide on a frictionless horizontal rod. Determine the distance x for which the collar is in equilibrium when P = 240 N.

SOLutiOn

Free Body: Collar A Force Triangle

N

N

2 2 2250 240 4900

70 0

= - ==

( ) ( )

. NSimilar Triangles

x

500

240

mm

N

70 N=

x = 1714 mm

67


PROBLEM 2.65

A 160-kg load is supported by the rope-and-pulley arrangement shown. Knowing that b = ∞20 , determine the magnitude and direction of the force P that must be exerted on the free end of the rope to maintain equilibrium. (Hint: The tension in the rope is the same on each side of a simple pulley. This can be proved by the methods of Chapter 4.)

SOLutiOn

Free-Body Diagram: Pulley A

+ SF P Px = ∞ - =0 20 0: 2 sin cosa

and cos . .

.

a aa

= = ± ∞= +

0 8452 46 840

46 840

or

For +SF P Py = ∞ + ∞ - =0 20 46 840 1569 60: 2 N 0cos sin . .

or P = 602 N 46 8. ∞

For a = -46 840.

+SF P Py = ∞ + - ∞ - =0 20 46 840 1569 60 0: 2 Ncos sin( . ) .

or P = 1365 N 46 8. ∞

68


SOLutiOn

Free-Body Diagram: Pulley A

(a) SF P Px = - ∞ =0 2 40 0: sin sin cosb

sin cos

.

b

b

= ∞

= ∞

1

240

22 52

b = ∞22 5.

(b) SF P Py = ∞ + ∞ - =0 40 2 22 52 1569 60 0: sin cos . . N

P = 630 N

PROBLEM 2.66

A 160-kg load is supported by the rope-and-pulley arrangement shown. Knowing that a = ∞40 , determine (a) the angle b, (b) the magnitude of the force P that must be exerted on the free end of the rope to maintain equilibrium. (See the hint for Problem 2.65.)

69


PROBLEM 2.67

A 3000 N crate is supported by several rope-and-pulley arrangements as shown. Determine for each arrangement the tension in the rope. (See the hint for Problem 2.65.)

SOLutiOn

Free-Body Diagram of Pulley

(a) +SF T

T

y = - =

=

0 3000 0

1

23000

: 2 N)

N)

(

(

T = 1500 N

(b) +SF T

T

y = - =

=

0 3000 0

1

23000

: 2 N)

N)

(

(

T = 1500 N

(c) +SF T

T

y = - =

=

0 3000 0

1

33000

: 3 N)

N)

(

(

T = 1000 N

(d) +SF T

T

y = - =

=

0 3000 0

1

33000

: 3 N)

N)

(

(

T = 1000 N

(e) +SF T

T

y = - =

=

0 3000 0

1

43000

: 4 N)

N)

(

(

T = 750 N

70


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 3000-N crate is supported by several rope-and-pulley arrangements as shown. Determine for each arrangement the tension in the rope. (See the hint for Problem 2.65.)

SOLutiOn

Free-Body Diagram of Pulley and Crate

(b) +SF T

T

y = - =

=

0 3000 0

1

33000

: 3 N)

N)

(

(

T = 1000 N

(d) +SF T

T

y = - =

=

0 3000

1

43000

: 4 N) 0

N)

(

(

T = 750 N

71


SOLutiOn

Free-Body Diagram: Pulley C

(a) + SF Tx ACB= ∞ - ∞ - ∞ =0 25 55 750 0: N 55(cos cos ) ( )cos

Hence: TACB = 1292 88. N

TACB = 1293 N

(b) +SF T Qy ACB= ∞ + ∞ + ∞ - =0 25 55 750 55 0

1292 88 2

: N)

N

(sin sin ) ( sin

( . )(sin 55 55 750 55 0∞ + ∞ + ∞ - =sin ) ( )sin N Q

or Q = 2219 8. N Q = 2220 N

PROBLEM 2.69

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

72


PROBLEM 2.70

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

SOLutiOn

Free-Body Diagram: Pulley C

+ SF T Px ACB= ∞ - ∞ - ∞ =0 25 55 55 0: (cos cos ) cos

or P TACB= 0 58010. (1)

+SF T Py ACB= ∞ + ∞ + ∞ - =0 25 55 55 1800: N 0(sin sin ) sin

or 1 24177 0 81915 1800. .T PACB + = N (2)

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

1 24177 0 81915 0 58010 1800. . ( . )T TACB ACB+ = N

Hence: TACB = 1048 37. N

TACB = 1048 N

(b) Using (1), P = =0 58010 1048 37 608 16. ( . ) . N N

P = 608 N

73


SOLutiOn

F F

F

h

h

= ∞= ∞=

sin

( sin

.

35

750 35

430 2

N)

N

(a)

F Fx h= ∞cos 25 F Fy = ∞cos35 F Fz h= ∞sin25

= ( .430 2 N)cos25∞ = (750 N)cos 35∞ = ∞( . )sin430 2 25 N

Fx = +390 N, Fy = +614 N, Fz = +181 8. N

(b) cos qxxF

F= = +390 N

750 N qx = ∞58 7.

cos q yyF

F= = +614

750

N

N q y = ∞35 0.

cos.q z

zF

F= = +181 8

750

N

N q z = ∞76 0.

PROBLEM 2.71

Determine (a) the x, y, and z components of the 750-N force, (b) the angles q

x, q

y, and q

z that the force forms with the coordinate axes.

74


PROBLEM 2.72

Determine (a) the x, y, and z components of the 900-N force, (b) the angles q

x, q

y, and q

z that the force forms with the coordinate axes.

SOLutiOn

F F

F

h

h

= ∞= ∞=

cos

( cos

.

65

900 65

380 4

N)

N

(a)

F Fx h= ∞sin 20 F Fy = ∞sin65 F Fz h= ∞cos 20

= ( .380 4 N)sin 20∞ = ( )sin900 N 65∞ = ∞( . )cos380 4 20 N

Fx = -130 1. N, Fy = +816 N, Fz = +357 N

(b) cos.qx

xF

F= = -130 1 N

900 N qx = ∞98 3.

cos q yyF

F= = +816

900

N

N q y = ∞25 0.

cos q zzF

F= = +357

900

N

N q z = ∞66 6.

75


PROBLEM 2.73

A horizontal circular plate is suspended as shown from three wires that are attached to a support at D and form 30° angles with the vertical. Knowing that the x component of the force exerted by wire AD on the plate is 110.3 N, determine (a) the tension in wire AD, (b) the angles q

x, q

y, and q

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

SOLutiOn

(a) F Fx = ∞ ∞ =sin sin .30 50 110 3 N (Given)

F = =110 3287 97

..

N

sin 30 sin 50 N

∞ ∞ F = 288 N

(b) cos.

.qxxF

F= = =110 3

0 38303 N

287.97 N qx = ∞67 5.

F F

F

F

y

yy

= ∞ =

= = =

cos .

cos.

..

30 249 39

249 39

287 970 86603q N

N q y = ∞30 0.

F Fz = - ∞ ∞= -= -

sin cos

( .

.

30 50

287 97

92 552

N)sin 30 cos 50

N

∞ ∞

cos.

..q z

zF

F= = - = -92 552

287 970 32139

N

N q z = ∞108 7.

76


PROBLEM 2.74

A horizontal circular plate is suspended as shown from three wires that are attached to a support at D and form 30° angles with the vertical. Knowing that the z component of the force exerted by wire BD on the plate is –32.14 N, determine (a) the tension in wire BD, (b) the angles q

x, q

y, and q

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

SOLutiOn

(a) F Fz = - ∞ ∞ =sin sin . ( )30 40 32 14 N Given

F =∞ ∞

=32 14

30 40100 0

.

sin sin. N F = 100 0. N

(b) F Fx = - ∞ ∞= -= -

sin cos

( .

.

30 40

100 0

38 302

N) sin 30 cos 40

N

∞ ∞

cos.

..qx

xF

F= = = -

38 302

100 00 38302

N

N qx = ∞112 5.

F Fy = ∞ =cos .30 86 603 N

cos.q y

yF

F= = =86 603

100

N

N0.86603 q y = ∞30 0.

Fz = -32 14. N

cos.

.q zzF

F= = - = -32 14

1000 32140

N

N q z = ∞108 7.

77


PROBLEM 2.75

A horizontal circular plate is suspended as shown from three wires that are attached to a support at D and form 30° angles with the vertical. Knowing that the tension in wire CD is 300 N, determine (a) the components of the force exerted by this wire on the plate, (b) the angles q qx y, , and q z that the force forms with the coordinate axes.

SOLutiOn

(a) Fx = - = -( )sin300 75 N 30 cos 60 N∞ ∞ Fx = -75 0. N

Fy = ∞ =( )cos .300 30 259 81 N N Fy = +260 N

Fz = ∞ ∞ =( )sin sin .300 30 60 129 9 N N Fz = +129 9. N

(b) cos .qxxF

F= = - = -75

3000 25

N

N qx = ∞104 5.

cos.

.q yyF

F= = =259 81

3000 866

N

N q y = ∞30 0.

cos.

.q zzF

F= = =129 9

3000 433

N

N q z = ∞64 3.

78


PROBLEM 2.76

A horizontal circular plate is suspended as shown from three wires that are attached to a support at D and form 30° angles with the vertical. Knowing that the x component of the force exerted by wire CD on the plate is –100 N, determine (a) the tension in wire CD, (b) the angles q

x, q

y, and q

z that the force exerted at C forms with the coordinate axes.

SOLutiOn

(a) F Fx = - ∞ ∞ = -sin cos30 60 100 N (Given)

F =∞

=100

60400

N

sin 30N

∞cos F = 400 N

(b) cos .qxxF

F= =

-= -

1000 25

N

400 N qx = ∞104 5.

F

F

F

y

yy

= ∞ =

= = =

( )cos .

cos.

.

400 30 346 41

346 410 866

N N

N

400 Nq q y = ∞30 0.

F

F

F

z

zz

= ∞ ∞ =

= = =

( )sin sin .

cos.

.

400 30 60 173 21

173 21

4000 43301

N N

N

Nq q z = ∞64 3.

79


PROBLEM 2.77

The end of the coaxial cable AE is attached to the pole AB, which is strengthened by the guy wires AC and AD. Knowing that the tension in wire AC is 600 N, determine (a) the components of the force exerted by this wire on the pole, (b) the angles q

x, q

y, and q

z that the

force forms with the coordinate axes.

SOLutiOn

(a) Fx = ∞ ∞( cos cos600 60 20 N)

Fx = 281 91. N Fx = +282 N

F

F

y

y

= - ∞

= -

( sin

.

600 60

519 62

N)

N Fy = -520 N

F

Fz

z

= - ∞ ∞= -

( cos sin

.

600 60 20

102 61

N)

N Fz = -102 6. N

(b) cos.qx

xF

F= = 281 91 N

600 N qx = ∞62 0.

cos.q y

yF

F= = -519 62 N

600 N q y = ∞150 0.

cos.q z

zF

F= = -102 61 N

600 N q z = ∞99 8.

80


PROBLEM 2.78

The end of the coaxial cable AE is attached to the pole AB, which is strengthened by the guy wires AC and AD. Knowing that the tension in wire AD is 425 N, determine (a) the components of the force exerted by this wire on the pole, (b) the angles q

x, q

y, and q

z that the

force forms with the coordinate axes.

SOLutiOn

(a) Fx = ∞ ∞( sin sin425 36 48 N)

= 185 64. N Fx = 185 6. N

Fy = - ∞

= -

( cos

.

425 36

343 83

N)

N Fy = -344 N

Fz = ∞ ∞=

( sin cos

.

425 36 48

167 15

N)

N Fz = 167 2. N

(b) cos.qx

xF

F= = 185 64 N

425 N qx = ∞64 1.

cos.q y

yF

F= = -343 83 N

425 N q y = ∞144 0.

cos.q z

zF

F= = 167 15 N

425 N q z = ∞66 8.

81


PROBLEM 2.79

Determine the magnitude and direction of the force F = (320 N)i + (400 N)j = (250 N)k.

SOLutiOn

F F F F

F

x y z= + +

= + + -

2 2 2

2320 400 250( ( ( N) N) N)2 2 F = 570 N

cos qxxF

F= = 320

570

N

N qx = ∞55 8.

cos q yyF

F= = 400 N

570 N q y = ∞45 4.

cos q zzF

F= = -250 N

570 N q z = ∞116 0.

82


PROBLEM 2.80

Determine the magnitude and direction of the force F = (240 N)i 2 (270 N)j + (680 N)k.

SOLutiOn

F F F F

F

x y z= + +

= + - +

2 2 2

2240 270 680( ( ( N) N) N)2 2 F = 770 N

cos qxxF

F= = 240

770

N

N qx = ∞71 8.

cos q yyF

F= = -270 N

770 N q y = ∞110 5.

cos q zzF

F= = 680 N

770 N q z = ∞28 0.

83


PROBLEM 2.81

A force acts at the origin of a coordinate system in a direction defined by the angles qx = 70.9° and

qy = 144.9°. Knowing that the z component of the force is –260 N, determine (a) the angle q

z, (b) the other

components and the magnitude of the force.

SOLutiOn

(a) We have

(cos ) (cos ) (cos ) (cos ) (cos ) (cos )q q q q q qx y z y y z2 2 2 2 2 21 1+ + = = - -fi

Since Fz 0 we must have cosq z 0

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

cos (cos . ) (cos . ) .q z = - - ∞ - ∞ =1 70 9 144 9 0 472822 2 q z = ∞118 2.

(b) Then:

FFz

z

= = =cos

.q

260549 89

N

0.47282N

and F Fx x= = ∞cos ( . )cos .q 549 89 70 9N Fx = 179 9. N

F Fy y= = ∞cos ( . )cos .q 549 89 144 9N Fy = -450 N

F = 550 N

84


PROBLEM 2.82

A force acts at the origin of a coordinate system in a direction defined by the angles qy = 55° and q

z = 45°.

Knowing that the x component of the force is 22500 N, determine (a) the angle qx, (b) the other components

and the magnitude of the force.

SOLutiOn

(a) We have

(cos ) (cos ) (cos ) (cos ) (cos ) (cos )q q q q q qx y z y y z2 2 2 2 2 21 1+ + = = - -fi

Since Fx 0 we must have cosqx 0

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

cos (cos ) (cos ) .qx = - - - =1 55 45 0 413532 2 qx = ∞114 4.

(b) Then:

FFx

x

= = =cos

.q

25006045 5

N

0.41353N F = 6050 N

and F Fy y= = ∞cos ( . )cosq 6045 5 55N Fy = 3470 N

F Fz z= = ∞cos ( . )cosq 6045 5 45N Fz = 4270 N

85


PROBLEM 2.83

A force F of magnitude 210 N acts at the origin of a coordinate system. Knowing that Fx = 80 N, q

z = 151.2°,

and Fy < 0, determine (a) the components F

y and F

z, (b) the angles q

x and q

y.

SOLutiOn

(a) F Fz z= = ∞cos ( )cos .q 210 151 2N

= -184 024. N Fz = -184 0. N

Then: F F F Fx y z2 2 2 2= + +

So: ( ) ( ) ( ) ( . )210 80 184 0242 2 2 2N N N= + +Fy

Hence: Fy = - - -( ) ( ) ( . )210 80 184 0242 2 2N N N

= -61 929. N Fy = -62 0. N

(b) cos .qxxF

F= = =

800 38095

N

210 N qx = ∞67 6.

cos.

.q yyF

F= = = -

61 929

2100 29490

N

N q y = 107 2.

86


PROBLEM 2.84

A force F of magnitude 230 N acts at the origin of a coordinate system. Knowing that qx = 32.5°, F

y = 2 60 N,

and Fz > 0, determine (a) the components F

x and F

z, (b) the angles q

y and q

z.

SOLutiOn

(a) We have

F Fx x= = ∞cos ( )cos .q 230 32 5N Fx = -194 0. N

Then: Fx = 193 980. N

F F F Fx y z2 2 2 2= + +

So: ( ) ( . ) ( )230 193 980 602 2 2 2N N N= + - + Fz

Hence: Fz = + - - -( ) ( . ) ( )230 193 980 602 2 2N N N Fz = 108 0. N

(b) Fz = 108 036. N

cos .q yyF

F= =

-= -

600 26087

N

230 N q y = ∞105 1.

cos.

.q zzF

F= = =

108 036

2300 46972

N

N q z = ∞62 0.

87


PROBLEM 2.85

A transmission tower is held by three guy wires anchored by bolts at B, C, and D. If the tension in wire AB is 2625 N, determine the components of the force exerted by the wire on the bolt at B.

SOLutiOn

BA

BA

F BA

u ruu= + -

= + +

=

=

=

( ) ( ) ( )

( ) ( ) ( )

.

7 34 8

7 34 8

35 623

2 2 2

m m m

m

i j k

F l

FFBA

BA

u ruu

= + -

= +

26257 34 8

515 82 2

N

35.623 mm m m

N

[( ) ( ) ( ) ]

( . ) (

i j k

F i 5505 4 589 51. ) ( . )N Nj k-

F F Fx y z= + = + = -516 2510 590N, N, N

88


PROBLEM 2.86

A transmission tower is held by three guy wires anchored by bolts at B, C, and D. If the tension in wire AD is 1575 N, determine the components of the force exerted by the wire on the bolt at D.

SOLutiOn

DA

DA

F D

u ruu= + +

= + +

=

=

( ) ( ) ( )

( ) ( ) ( )

.

7 34 25

7 34 25

42 778

2 2 2

m m m

m

i j k

F l AA

FDA

DA=

= + +

=

u ruu

15757 34 25

257 73

N

42.778 mm m m

N

[( ) ( ) ( ) ]

( . )

i j k

F i ++ +( . ) ( . )1251 81 920 45N Nj k

F F Fx y z= + = + = +258 1252 920N, N, N

89


PROBLEM 2.87

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

SOLutiOn

DB

DB

u ruu= - +

= +

( ( ) ( )

( ( )

480 510 320

480 510

mm) mm mm

mm) mm2 2

i j k

++

=

=

=

= -

(

[( (

320

770

385480

mm)

mm

N

770 mm mm)

2

F F

FDB

DB

DBlu ruu

i 5510 320

240 255 160

mm) mm)

N) N N

j k+

= - +

( ]

( ( ) ( )i j k

F F Fx y z= + = - = +240 255 160 0 N, N N, .

90


PROBLEM 2.88

For the frame and cable of Problem 2.87, determine the components of the force exerted by the cable on the support at E.

PROBLEM 2.87 A frame ABC is supported in part by cable DBE that passes through a frictionless ring at B. Knowing that the tension in the cable is 385 N, determine the components of the force exerted by the cable on the support at D.

SOLutiOn

EB

EB

u ruu= - +

= +

( ( ) ( )

( ( )

270 400 600

270 400 2

mm) mm mm

mm) mm2

i j k

++

=

=

=

= -

(

[( (

600

770

385270

mm)

mm

N

770 mm mm)

2

F F

FEB

EB

EBlu ruu

i 4400 600

135 200 300

mm) mm)

N) N N

j k+

= - +

( ]

( ( ) ( )F i j k

F F Fx y z= + = - = +135 0 200 300. , N, N N

91


PROBLEM 2.89

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

SOLutiOn

BA

BA

u ruu= - + +

= +

( ( ) ( )

( ( )

900 600 360

900 6002

mm) mm mm

mm) mm

i j k

22 2360

1140

1425

+

=

=

=

=

( mm)

mm

N

1140 m

T

T

BA BA BA

BA

BA

T

TBA

BA

lu ruu

mm mm mm mm

N) N

[ ( ) ( ) ( ) ]

( ( ) (

- + +

= - + +

900 600 360

1125 750 45

i j k

i j 00 N)k

( ) ( ) , ( )T T TBA x BA y BA z= - = =1125 750 450N, N N

92


PROBLEM 2.90

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

SOLutiOn

CA

CA

u ruu= - + -

= +

( ( ) ( )

( ( )

900 600 920

900 6002

mm) mm mm

mm) mm

i j k

22 2920

1420

2130

+

=

=

=

=

( mm)

mm

N

1420 m

T

T

CA CA CA

CA

CA

T

TCA

CA

lu ruu

mm mm mm mm

N) N

[ ( ) ( ) ( ) ]

( ( ) (

- + -

= - + -

900 600 920

1350 900 13

i j k

i j 880 N)k

( ) ( ) , ( )T T TCA x CA y CA z= - = = -1350 900 1380N, N N

93


PROBLEM 2.91

Find the magnitude and direction of the resultant of the two forces shown knowing that P = 300 N and Q = 400 N.

SOLutiOn

P i j k

i

= - ∞ ∞ + ∞ + ∞ ∞

= -

( )[ cos sin sin cos cos ]

( . )

300 30 15 30 30 15

67 243

N

N ++ +

= ∞ ∞ + ∞ - ∞

( ) ( . )

( )[cos cos sin cos s

150 250 95

400 50 20 50 50

N N

N

j k

Q i j iin ]

( )[ . . . ]

( . ) (

20

400 0 60402 0 76604 0 21985

241 61 30

∞

= + -

= +

k

i j

i

N

N

k

66 42 87 939

174 367 456 42 163 011

. ) ( . )

( . ) ( . ) ( . )

N N

N N N

j k

R P Q

i j

-

= +

= + + kk

R = + +

=

( . ) ( . ) ( . )

.

174 367 456 42 163 011

515 07

2 2 2N N N

N R = 515 N

cos.

..qx

xR

R= = =

174 367

515 070 33853

N

N qx = ∞70 2.

cos.

.q yyR

R= = =

456 420 88613

N

515.07 N q y = ∞27 6.

cos.

.q zzR

R= = =

163 0110 31648

N

515.07 N q z = ∞71 5.

94


PROBLEM 2.92

Find the magnitude and direction of the resultant of the two forces shown knowing that P = 400 N and Q = 300 N.

SOLutiOn

P i j k

i

= - ∞ ∞ + ∞ + ∞ ∞

= -

( )[ cos sin sin cos cos ]

( . )

400 30 15 30 30 15

89 678

N

N ++ +

= ∞ ∞ + ∞ - ∞

( ) ( . )

( )[cos cos sin cos s

200 334 61

300 50 20 50 50

N N

N

j k

Q i j iin ]

( . ) ( . ( .

( . )

20

181 21 229 81 65 954

91 532

∞

= + -

= +

= +

k

i j k

R P Q

i

N N) N)

N (( . ) ( . )

( . ) ( . ) ( . )

429 81 268 66

91 532 429 81 268 66

5

2 2 2

N N

N N N

j k+

= + +

=

R

115 07. N R = 515 N

cos.

..qx

xR

R= = =

91 532

515 070 177708

N

N qx = ∞79 8.

cos.

.q yyR

R= = =

429 810 83447

N

515.07 N q y = ∞33 4.

cos.

.q zzR

R= = =

268 660 52160

N

515.07 N q z = ∞58 6.

95


PROBLEM 2.93

Knowing that the tension is 2125 N in cable AB and 2550 N in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables.

SOLutiOn

AB

AB

u ruu= - +

= + +

( ) ( ) ( )

( ) ( ) (

1000 1125 1500

1000 1125 152 2

mm mm mmi j k

000 2125

2500 1125 1500

2500

2

2

)

( ) ( ) ( )

( )

=

= - +

=

mm

mm mm mmAC

AC

u ruui j k

++ + =

= = =

( ) ( )

( )(

1125 1500 3125

2125100

2 2 mm

NTAB AB AB ABT TAB

ABl

u ruu00 1125 1500

1000 1125

mm mm mm

2125 mm

N N

) ( ) ( )

( ) ( )

i j k

T i

- +È

ÎÍ

˘

˚˙

= -AB jj k

Ti

+

= = =-

( )

( )( ) (

1500

25502500 1125

N

Nmm mm

AC AC AC ACT TAC

ACl

u ruu)) ( )

( ) ( ) ( )

j k

T i j k

R T

+È

ÎÍ

˘

˚˙

= - +

=

1500

3125

2040 918 1224

mm

mm

N N NAC

ABB AC+ = - + =T i j k( ) ( ) ( ) .3040 2043 2724 4564 61N N N N

Then: R = 4564 6. N R = 4560 N

and cos .qx = =3040

0 66599N

4564.61 N qx = ∞48 2.

cos.

.q y = = -2043

4564 610 44757

N

N q y = ∞116 6.

cos.

.q z = =2724

4564 610 59677

N

N q z = ∞53 4.

96


PROBLEM 2.94

Knowing that the tension is 2550 N in cable AB and 2125 N in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables.

SOLutiOn

AB

AB

u ruu= - +

= + +

( ) ( ) ( )

( ) ( ) (

1000 1125 1500

1005 1125 152 2

mm mm mmi j k

000 2125

2500 1125 1500

2500

2

2

)

( ) ( ) ( )

( )

=

= - +

=

mm

mm mm mmAC

AC

u ruui j k

++ + =

= = =

( ) ( )

( )(

1125 1500 3125

2550100

2 2 mm

NTAB AB AB ABT TAB

ABl

u ruu00 1125 1500

1200 1350

mm mm mm

2125 mm

N N

) ( ) ( )

( ) ( )

i j k

T i

- +È

ÎÍ

˘

˚˙

= -AB jj k

Ti

+

= = =-

( )

( )( ) (

1800

21252500 1125

N

Nmm mm

AC AC AC ACT TAC

ACl

u ruu)) ( )

( ) ( ) ( )

j k

T i j k

R T

+È

ÎÍ

˘

˚˙

= - +

=

1500

3125

1700 765 1020

mm

mm

N N NAC

ABB AC+ = - +T i j k( ) ( ) ( )2900 2115 2820N N N

Then: R = 4564 6. N R = 4560 N

and cos .qx = =2900

0 63532N

4564.6 qx = ∞50 6.

cos.

.q y =-

= -2115

4564 60 46335

N

N q y = ∞117 6.

cos.

.q z = =2820

4564 60 61780

N

N q z = ∞51 8.

97


PROBLEM 2.95

For the frame of Problem 2.87, determine the magnitude and direction of the resultant of the forces exerted by the cable at B knowing that the tension in the cable is 385 N.

PROBLEM 2.87 A frame ABC is supported in part by cable DBE that passes through a frictionless ring at B. Knowing that the tension in the cable is 385 N, determine the components of the force exerted by the cable on the support at D.

SOLutiOn

BD

BD

u ruu= - + -

= +

( ) ( ) ( )

( ) ( )

480 510 320

480 5102

mm mm mm

mm mm

i j k

22 2320 770+ =( ) mm mm

F

i

BD BD BD BDT TBD

BD= =

= - +

lu ruu

( )

( )[ ( ) ( )

385

770480 510

N

mm mm mm jj k

i j k

-

= - + -

( ) ]

( ) ( ) ( )

320

240 255 160

mm

N N N

BE

BE

u ruu= - + -

= +

( ) ( ) ( )

( ) ( )

270 400 600

270 4002

mm mm mm

mm mm

i j k

22 2600 770+ =( ) mm mm

F

i

BE BE BE BET TBE

BE= =

= - +

lu ruu

( )

( )[ ( ) ( )

385

770270 400

N

mm mm mm jj k

i j k

-

= - + -

( ) ]

( ) ( ) ( )

600

135 200 300

mm

N N N

R F F i j k= + = - + -BD BE ( ) ( ) ( )375 455 460 N N N

R = + + =( ) ( ( ) .375 455 460 747 832 2 N N) N N2 R = 748 N

cos.

qx = -375

747 83

N

N qx = ∞120 1.

cos q y = 455 N

747.83 N q y = ∞52 5.

cos.

q z = -460

747 83

N

N q z = ∞128 0.

98


PROBLEM 2.96

For the cables of Problem 2.89, knowing that the tension is 1425 N in cable AB and 2130 N in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables.

SOLutiOn

T T

T T

AB BA

AB x AB y

= -

= + =

( )

( ) ( )

use results of Problem 2.89

N1125 -- = -

= -

750 450 N N

use results of Problem 2.90

( )

( )

(

T

T T

AB z

AC CA

TT T TAC x AC y AC z) ( ) ( )= + = - = +1350 900 1380 N N N

Resultant: R F

R F

R F

x x

y y

z z

= = + + = +

= = - - = -

= = -

S

S

S

1125 1350 2475

750 900 1650

45

N

N

00 1380 930

2475 1650 930

2 2 2

2 2 2

+ = +

= + +

= + + - + +

N

R R R Rx y z

( ) ( ) ( )

= 3116 6. N R = 3120 N

cos.

qxxR

R= = +2475

3116 6 qx = ∞37 4.

cos.

q yyR

R= = -1650

3116 6 q y = ∞122 0.

cos.

q zzR

R= =

+930

3116 6 q z = ∞72 6.

99


PROBLEM 2.97

The end of the coaxial cable AE is attached to the pole AB, which is strengthened by the guy wires AC and AD. Knowing that the tension in AC is 750 N and that the resultant of the forces exerted at A by wires AC and AD must be contained in the xy plane, determine (a) the tension in AD, (b) the magnitude and direction of the resultant of the two forces.

SOLutiOn

R T T

i j k

= += ∞ ∞ - ∞ - ∞ ∞

+

AC AD

ADT

( )(cos cos sin cos sin )750 60 20 60 60 20 N

((sin sin cos sin cos )36 48 36 36 48∞ ∞ - ∞ + ∞ ∞i j k (1)

(a) Since Rz = 0, The coefficient of k must be zero.

( )( cos sin ) (sin cos )750 60 20 36 48 0 N

- ∞ ∞ + ∞ ∞ =TAD

NTAD = 326 1. TAD = 326 N

(b) Substituting for TAD into Eq. (1) gives:

R i= ∞ ∞ + ∞ ∞

-[( ) cos cos ( . )sin sin )]

[( )si

750 60 20 326 1 36 48

750

N N

N nn ( . )cos ]60 326 1 36 0∞ + ∞ + N j

R i j= -

= +=

( . ) ( . )

( . ) ( . )

.

494 83 913 34

494 83 913 34

1038 77

2 2

N N

N

R

R = 1039 N

cos.qx = 494 83 N

1038.77 N qx = ∞61 6.

cos.q y = -913 34 N

1038.76 N q y = ∞151 6.

cos q z = 0 q z = ∞90 0.

100


PROBLEM 2.98

The end of the coaxial cable AE is attached to the pole AB, which is strengthened by the guy wires AC and AD. Knowing that the tension in AD is 625 N and that the resultant of the forces exerted at A by wires AC and AD must be contained in the xy plane, determine (a) the tension in AC, (b) the magnitude and direction of the resultant of the two forces.

SOLutiOn

R T T

i j k

= += ∞ ∞ - ∞ - ∞ ∞

+

AC AD

ACT (cos cos sin cos sin )

( )

60 20 60 60 20

625 N ((sin sin cos sin cos )36 48 36 36 48∞ ∞ - ∞ + ∞ ∞i j k (1)

(a) Since Rz = 0, The coefficient of k must be zero.

TAC ( cos sin ) ( )(sin cos )- ∞ ∞ + ∞ ∞ =60 20 625 36 48 0 N

NTAC = 1437 43. TAC = 1437 N

(b) Substituting for TAC into Eq. (1) gives:

R i= ∞ ∞ + ∞ ∞

-[( . ) cos cos ( )sin sin ]

[( .

1437 43 60 20 625 36 48

1437 43

N N

N N)sin ( )cos ]60 625 36 0∞ + ∞ +j

R i j= -

= +=

( . ) ( . )

( . ) ( . )

.

948 377 1750 49

948 377 1750 49

1990 89

2 2

N N

R

N R = 1991 N

cos.qx = 948 377

1990.89 qx = ∞61 6.

cos.q y = -1750 49 N

1990.89 q y = ∞151 5.

cos q z = 0 q z = ∞90 0.

101


PROBLEM 2.99

Three cables are used to tether a balloon as shown. Determine the vertical force P exerted by the balloon at A knowing that the tension in cable AB is 259 N.

SOLutiOn

The forces applied at A are: T T T PAB AC AD, , , and

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

AB AB

AC

u ruuu ruu

= - - =

= - +

( . ) ( . ) .

( . ) ( . )

4 20 5 60 7 00

2 40 5 60

m m m

m m

i j

i j (( . ) .

( . ) ( . ) .

4 20 7 40

5 60 3 30 6 50

m m

m m m

k

j k

AC

AD AD

=

= - - =u ruu

and T i j

T

AB AB AB AB AB

AC AC AC AC

T TAB

ABT

T TAC

= = = - -

= =

l

l

u ruu

u ruu( . . )0 6 0 8

AACT

T TAD

AD

AC

AD AD AD AD

= - +

= = =

( . . . )

(

0 32432 0 75676 0 56757j k

T lu ruu

-- -0 86154 0 50769. . )j k TAD

102


PROBLEM 2.99 (Continued)

Equilibrium condition SF PAB AC AD= + + + =0 0: T T T j

Substituting the expressions obtained for T T TAB AC AD, , and and factoring i, j, and k:

( . . ) ( . . . )

(

- + + - - - ++

0 6 0 32432 0 8 0 75676 0 86154T T T T T PAB AC AB AC ADi j

00 56757 0 50769 0. . )T TAC AD- =k

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

- + =0 6 0 32432 0. .T TAB AC (1)

- - - + =0 8 0 75676 0 86154 0. . .T T T PAB AC AD (2)

0 56757 0 50769 0. .T TAC AD- = (3)

Setting TAB = 259 N in (1) and (2), and solving the resulting set of equations gives

T

TAC

AD

==

479 15

535 66

.

.

N

N P = 1031 N ≠

103


PROBLEM 2.100

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

SOLutiOn

See Problem 2.99 for the figure and the analysis leading to the linear algebraic Equations (1), (2),and (3) below:

- + =0 6 0 32432 0. .T TAB AC (1)

- - - + =0 8 0 75676 0 86154 0. . .T T T PAB AC AD (2)

0 56757 0 50769 0. .T TAC AD- = (3)

Substituting TAC = 444 N in Equations (1), (2), and (3) above, and solving the resulting set ofequations using conventional algorithms gives

T

TAB

AD

==

240

496 36

N

N. P = 956 N ≠

104


PROBLEM 2.101

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

SOLutiOn

See Problem 2.99 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3).

- + =0 6 0 32432 0. .T TAB AC (1)

- - - + =0 8 0 75676 0 86154 0. . .T T T PAB AC AD (2)

0 56757 0 50769 0. .T TAC AD- = (3)

Substituting TAD = 481 N in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms gives

T

TAC

AB

==

430 26

232 57

.

.

N

N P = 926 N ≠ b

105


PROBLEM 2.102

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

SOLutiOn

See Problem 2.99 for the figure and analysis leading to the linear algebraic Equations (1), (2), and (3).

- + =0 6 0 32432 0. .T TAB AC (1)

- - - + =0 8 0 75676 0 86154 0. . .T T T PAB AC AD (2)

0 56757 0 50769 0. .T TAC AD- = (3)

From Eq. (1) T TAB AC= 0 54053.

From Eq. (3) T TAD AC= 1 11795.

Substituting for TAB and TAD in terms of TAC into Eq. (2) gives:

- - - + =0 8 0 54053 0 75676 0 86154 1 11795 0. ( . ) . . ( . )T T T PAC AC AC

2 1523 800

800

2 1523371 69

. ;

..

T P P

T

AC

AC

= =

=

=

N

N

NSubstituting into expressions for TAB and TAD gives:

T

TAB

AD

==

0 54053 371 69

1 11795 371 69

. ( . )

. ( . )

N

N T T TAB AC AD= = =201 372 416 N N N, , b

106


PROBLEM 2.103

A crate is supported by three cables as shown. Determine the weight of the crate knowing that the tension in cable AB is 3750 N.

SOLutiOn

The forces applied at A are: T T T WAB AC AD, , and

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

AB

AB

AC

u ruu

u ruu

= - + -=

=

( ) ( ) ( )

(

900 1500 675

1875

1500

mm mm mm

mm

mm

i j k

)) ( )

( ) ( ) ( )

j k

i j k

+=

= + -

800

1700

1000 1500 675

mm

mm

mm mm mm

AC

AD

AD

u ruu

== 1925 mm

and T

i j k

T

AB AB AB AB

AB

AC AC AC A

T TAB

ABT

T T

= =

= - + -

= =

l

l

u ruu

( . . . )0 48 0 8 0 36

CC

AC

AD AD AD AD

AC

ACT

T TAD

AD

u ruu

u ruu= +

= =

=

( . . )

(

0 88235 0 47059

0

j k

T l

.. . . )51948 0 77922 0 35065i j k+ - TAD

Equilibrium Condition with W j= -W

SF WAB AC AD= + + - =0 0: T T T j

107



Substituting the expressions obtained for T T TAB AC AD, , and and factoring i, j, and k:

( . . ) ( . . . )

(

- + + + + -+

0 48 0 51948 0 8 0 88235 0 77922T T T T T WAB AD AB AC ADi j

-- + - =0 36 0 47059 0 35065 0. . . )T T TAB AC AD kEquating to zero the coefficients of i, j, k:

- + =+ + - =

-

0 48 0 51948 0

0 8 0 88235 0 77922 0

0 36

. .

. . .

.

T T

T T T WAB AD

AB AC AD

TT T TAB AC AD+ - =0 47059 0 35065 0. .

Substituting TAB = 3750 N in Equations (1), (2), and (3) and solving the resulting set of equations, usingconventional algorithms for solving linear algebraic equations, gives:

T

TAC

AD

==

5450 24

3465

. N

N

W = 10509 N W = 10510 N b

108


PROBLEM 2.104

A crate is supported by three cables as shown. Determine the weight of the crate knowing that the tension in cable AD is 3080 N.

SOLutiOn

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

- + =+ + - =

-

0 48 0 51948 0

0 8 0 88235 0 77922 0

0 36

. .

. . .

.

T T

T T T WAB AD

AB AC AD

TT T TAB AC AD+ - =0 47059 0 35065 0. .

Substituting TAD = 3080 N in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms, gives:

T

T

AB

AC

=

=

3333 33

4844 66

.

.

N

N

W = 9341.35 N W = 9340 N b

109


PROBLEM 2.105

A crate is supported by three cables as shown. Determine the weight of the crate knowing that the tension in cable AC is 2720 N.

SOLutiOn


- + =+ + - =

-

0 48 0 51948 0

0 8 0 88235 0 77922 0

0 36

. .

. . .

.

T T

T T T WAB AD

AB AC AD

TT T TAB AC AD+ - =0 47059 0 35065 0. .

Substituting TAC = 2720 N in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms, gives:

T

T

AB

AD

=

=

1871 35

1729 1

.

.

N

N

W = 5244.4 N W = 5240 N b

110


PROBLEM 2.106

A 8000-N crate is supported by three cables as shown. Determine the tension in each cable.

SOLutiOn


- + =+ + - =

-

0 48 0 51948 0

0 8 0 88235 0 77922 0

0 36

. .

. . .

.

T T

T T T WAB AD

AB AC AD

TT T TAB AC AD+ - =0 47059 0 35065 0. .

Substituting W = 8000 N in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms, gives:

TAB

= 2854.6 N TAB = 2850 N b

TAC

= 4149.2 N NTAC = 4150 b

TAD

= 2637.7 N TAD = 2640 N b

111


PROBLEM 2.107

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

SOLutiOn SF T T T PA AB AC AD= + + + =0 0: where P i= P

AB AB

AC

u ruuu ruu

= - - + =

= -

( ) ( ( )

(

960 240 380 1060

96

mm mm) mm mmi j k

00 240 320 1040

960 72

mm mm mm mm

mm

) ( ) ( )

( ) (

i j k

i

- - =

= - +

AC

ADu ruu

00 220 1220 mm mm mm) ( )j k- =AD

T

T

AB AB AB AB AB

AC AC

T TAB

ABT

T

= = = - - +ÊËÁ

ˆ¯̃

=

l

l

u ruu48

53

12

53

19

53i j k

AAC AC AC

AD AD AD

TAC

ACT

T

= = - - -ÊËÁ

ˆ¯̃

= =

u ruu12

13

3

13

4

13

305

i j k

T l N

12220 mm mm mm mm

N N

[( ) ( ) ( ) ]

( ) ( )

- + -

= - +

960 720 220

240 180

i j k

i j -- ( )55 N k

Substituting into SFA = 0, factoring i j k, , , and setting each coefficient equal to f gives:

i : P T TAB AC= + +48

53

12

13240 N (1)

j: 12

53

3

13180T TAB AC+ = N (2)

k: 19

53

4

1355T TAB AC- = N (3)

Solving the system of linear equations using conventional algorithms gives:

T

TAB

AC

==

446 71

341 71

.

.

N

N P = 960 N b

112


PROBLEM 2.108

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

SOLutiOn

We assume that TAD = 0 and write SF T T j iA AB AC Q= + + + =0 1200 0: ( ) N

AB AB

AC

u ruuu ruu

= - - + =

= -

( ) ( ( )

(

960 240 380 1060

96

mm mm) mm mmi j k

00 240 320 1040 mm mm mm mm) ( ) ( )i j k- - =AC

T

T

AB AB AB AB AB

AC AC

T TAB

ABT

T

= = = - - +ÊËÁ

ˆ¯̃

=

l

l

u ruu48

53

12

53

19

53i j k

AAC AC ACTAC

ACT= = - - -Ê

ËÁˆ¯̃

u ruu12

13

3

13

4

13i j k

Substituting into SFA = 0, factoring i j k, , , and setting each coefficient equal to f gives:

i : - - + =48

53

12

131200T TAB AC N 0 (1)

j: - - + =12

53

3

13T T QAB AC 0 (2)

k :19

53

4

13T TAB AC- = 0 (3)

Solving the resulting system of linear equations using conventional algorithms gives:

T

T

Q

AB

AC

===

605 71

705 71

300 00

.

.

.

N

N

N 0 300� �Q N b

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

113


PROBLEM 2.109

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 3150 N, determine the vertical force P exerted by the tower on the pin at A.

SOLutiOn

Free Body A:

SF T T T j= + + + =0 0: AB AC AD P

AB AB

AC AC

AD

u ruuu ruuu ru

= - - + =

= - + =

15 30 10 35

10 30 22 38 523

i j k

i j k

m

m.uu

= - - =7 30 20 36 729i j k AD . m

We write T

i j k

AB AB AB AB

AB

T TAB

AB

T

= =

= - - +ÊËÁ

ˆ¯̃

lu ruu

3

7

6

7

2

7

T

i j k

AC AC AC AC

AC

T TAC

ACT

= =

= - +( )

lu ruu

0.2596 0 7788 0 5711. .

T

i j k

AD AD AD AD

AD

T TAD

ADT

= =

= - -( )

lu ruu

0 19058 0 8168 0 5445. . .

Substituting into the equation SF = 0 and factoring i j k, : ,

- + +ÊËÁ

ˆ¯̃

+ - - -

3

70 2596 0 19058

6

70 7788 0 8168

T T T

T T

AB AC AD

AB AC

. .

. .

i

TT P

T T T

AD

AB AC AD

+ÊËÁ

ˆ¯̃

+ + -ÊËÁ

ˆ¯̃

=

j

k2

70 5711 0 5445 0. .

114



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

i: - + + =0 4285 0 2596 0 19058 0. . .T T TAB AC AD (1)

j: - - - + =0 8571 0 7788 0 8168 0. . .T T T PAB AC AD (2)

k : 0 2857 0 5711 0 5445 0. . .T T TAB AC AD+ - = (3)

Set TAB = 3150 N in Eqs. (1) – (3):

- + + =1350 0 2596 0 19058 0 N . .T TAC AD (1¢)

- - - + =2700 0 7788 0 8168 0 N . .T T PAC AD (2¢)

900 0 5711 0 5445 0 N + - =. .T TAC AD (3¢)

Solving, T T PAC AD= = =2252 48 4015 41. . N N 7734.02 N P = 7730 N b

115


PROBLEM 2.110

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 4600 N, determine the vertical force P exerted by the tower on the pin at A.

SOLutiOn


i: - + + =0 4285 0 2596 0 19058 0. . .T T TAB AC AD (1)

j: - - - + =0 8571 0 7788 0 8168 0. . .T T T PAB AC AD (2)

k : 0 2857 0 5711 0 5445 0. . .T T TAB AC AD+ - = (3)

Substituting for TAC = 4600 N in Equations (1), (2), and (3) above and solving the resulting set of equations using conventional algorithms gives:

- + + =0 4285 1194 16 0 19058 0. . .T TAB AD N (1¢)

- - - + =0 8571 3582 48 0 8168 0. . .T T PAB AD N (2¢)

0 2857 2627 06 0 5445 0. . .T TAB AD+ - = N (3¢)

Solving, 6434 22

8200 76

.

.

N

N

15795 63. N P = 15800 N b

116


PROBLEM 2.111

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

SOLutiOn

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

Free Body A:

SF PAB AC AD= + + + =0 0: T T T j

We have:

AB AB

AC

u ruuu ruu

= - - + =

=

( ) ( ( )

(

320 480 360 680

450

mm mm) mm mm

i j k

mmm mm mm mm

mm mm

) ( ) ( )

( ) (

i j k

i

- + =

= -

480 360 750

250 480

AC

ADu ruu

))j k- ( ) =360 650 mm mmAD

Thus:

T

T

AB AB AB AB AB

AC AC AC

T TAB

ABT

T

= = = - - +ÊËÁ

ˆ¯̃

=

l

l

u ruu8

17

12

17

9

17i j k

== = - +( )

= = =

TAC

ACT

T TAD

AD

AC AC

AD AD AD AD

u ruu

u ruu0 6 0 64 0 48. . .i j k

T l 55

13

9 6

13

7 2

13i j k- -Ê

ËÁˆ¯̃

. .TAD

Dimensions in mm

Substituting into the equation SF = 0 and factoring i j k, , :

- + +ÊËÁ

ˆ¯̃

+ - - - +ÊË

8

170 6

5

13

12

170 64

9 6

13

T T T

T T T P

AB AC AD

AB AC AD

.

..

i

ÁÁˆ¯̃

+ + -ÊËÁ

ˆ¯̃

=

j

k9

170 48

7 2

130T T TAB AC AD.

.

117



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

i: - + + =8

170 6

5

130T T TAB AC AD. (1)

j: - - - + =12

70 64

9 6

130T T T PAB AC AD.

. (2)

k : 9

170 48

7 2

130T T TAB AC AD+ - =.

. (3)

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

- + + =8

1736

5

130T TAB AD N (1¢)

9

1728 8

7 2

130T TAB AD+ - =.

. N (3¢)

Multiply (1¢) by 9, (3¢) by 8, and add:

554 412 6

130 572 0.

.. N N- = =T TAD AD

Substitute into (1¢) and solve for TAB :

T TAB AB= + ¥ÊËÁ

ˆ¯̃

=17

836

5

13572 544 0. N

Substitute for the tensions in Eq. (2) and solve for P:

P = + +

=

12

17544 0 64 60

9 6

13572

844 8

( ) . ( ).

( )

.

N N N

N Weight of plate = =P 845 N b

118


PROBLEM 2.112

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

SOLutiOn


- + + =8

170 6

5


- + - + =12

170 64

9 6

130T T T PAB AC AD.

. (2)

9

170 48

7 2


. (3)

Making TAD = 520 N in Eqs. (1) and (3):

- + + =8

170 6 200 0T TAB AC. N (1¢)

9

170 48 288 0T TAB AC+ - =. N (3¢)

Multiply (1¢) by 9, (3¢) by 8, and add:

9 24 504 0 54 5455. .T TAC AC- = = N N

Substitute into (1¢) and solve for TAB :

T TAB AB= ¥ + =17

80 6 54 5455 200 494 545( . . ) . N

Substitute for the tensions in Eq. (2) and solve for P:

P = + +

=

12

17494 545 0 64 54 5455

9 6

13520

768 00

( . ) . ( . ).

( )

.

N N N

N Weight of plate = =P 768 N b

119


PROBLEM 2.113

For the transmission tower of Problems 2.109 and 2.110, determine the tension in each guy wire knowing that the tower exerts on the pin at A an upward vertical force of 10.5 kN.

SOLutiOn


i: - + + =0 4285 0 2596 0 19058 0. . .T T TAB AC AD (1)

j: - - - + =0 8571 0 7788 0 8168 0. . .T T T PAB AC AD (2)

k : 0 2857 0 5711 0 5445 0. . .T T TAB AC AD+ - = (3)

Substituting for P = 10500 N in Equations (1), (2), and (3) above and solving the resulting set of equations using conventional algorithms gives:

- + + =0 4285 0 2596 0 19058 0. . .T T TAB AC AD (1¢)

- - - + =0 8571 0 7788 0 8168 10500 0. . .T T TAB AC AD N (2¢)

0 2857 0 5711 0 5445 0. . .T T TAB AC AD+ - = (3¢)

T

T

T

AB

AC

AD

===

4277 09

3057 81

5451 38

.

.

.

N

N

N TAB = 4280 N b

TAC = 3060 N b

TAD = 5450 N b

120


PROBLEM 2.114

A horizontal circular plate weighing 300 N is suspended as shown from three wires that are attached to a support at D and form 30° angles with the vertical. Determine the tension in each wire.

SOLutiOn

SFx = 0:

- ∞ ∞ + ∞ ∞ + ∞T T TAD BD CD(sin )(sin ) (sin )(cos ) (sin )(cos30 50 30 40 30 60∞∞ =) 0

Dividing through by sin30∞ and evaluating:

- + + =0 76604 0 76604 0 5 0. . .T T TAD BD CD (1)

SF T T Ty AD BD CD= - ∞ - ∞ - ∞ + =0 30 30 30 300: N 0(cos ) (cos ) (cos )

or T T TAD BD CD+ + = 346 41. N (2)

SF T T Tz AD BD CD= ∞ ∞ + ∞ ∞ - ∞ ∞ =0 30 50 30 40 30 60 0: sin cos sin sin sin sin

or 0 64279 0 64279 0 86603 0. . .T T TAD BD CD+ - = (3)

Solving Equations (1), (2), and (3) simultaneously:

T

T

T

AD

BD

CD

===

147 579

51 253

147 578

.

.

.

N

N

N TAD = 147 6. N b

TBD = 51 3. N b

TCD = 147 6. N b

121


PROBLEM 2.115

For the rectangular plate of Problems 2.111 and 2.112, determine the tension in each of the three cables knowing that the weight of the plate is 792 N.

SOLutiOn

See Problem 2.111 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below. Setting P = 792 N gives:

- + + =8

170 6

5


- - - + =12

170 64

9 6

13792 0T T TAB AC AD.

. N (2)

9

170 48

7 2


. (3)

Solving Equations (1), (2), and (3) by conventional algorithms gives

TAB = 510 00. N TAB = 510 N b

TAC = 56 250. N TAC = 56 2. N b

TAD = 536 25. N TAD = 536 N b

122


PROBLEM 2.116

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

SOLutiOn

SF T T T P QA AB AC AD= + + + + =0 0:

Where P i= P and Q j= Q

AB AB

AC

u ruuu ruu

= - - + =

= -

( ( ) ( )

(

960 240 380 1060

96

mm) mm mm mmi j k

00 240 320 1040

960 72

mm) mm mm mm

mm)

i j k

i

- - =

= - +

( ) ( )

( (

AC

ADu ruu

00 220 1220 mm mm mm) ( )j k- =AD

T

T

AB AB AB AB AB

AC AC

T TAB

ABT

T

= = = - - +ÊËÁ

ˆ¯̃

=

l

l

u ruu48

53

12

53

19

53i j k

AAC AC AC

AD AD AD AD

TAC

ACT

T TAD

= = - - -ÊËÁ

ˆ¯̃

= =

u ruu

u

12

13

3

13

4

13i j k

T lrruu

ADTAD= - + -Ê

ËÁˆ¯̃

48

61

36

61

11

61i j k

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

i : - - - + =48

53

12

13

48

612880 0T T TAB AC AD N (1)

j : - - + =12

53

3

13

36

610T T TAB AC AD (2)

k :19

53

4

13

11

610T T TAB AC AD- - = (3)

123



Solving the system of linear equations using conventional algorithms gives:

T

T

T

AB

AC

AD

===

1340 14

1025 12

915 03

.

.

.

N

N

N TAB = 1340 N b

TAC = 1025 N b

TAD = 915 N b

124


PROBLEM 2.117

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

SOLutiOn

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

- - - + =48

53

12

13

48

610T T T PAB AC AD (1)

- - + + =12

53

3

13

36

610T T T QAB AC AD (2)

19

53

4

13

11

610T T TAB AC AD- - = (3)

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

- - - + =48

53

12

13

48

612880 0T T TAB AC AD N (1¢)

- - + + =12

53

3

13

36

61576 0T T TAB AC AD N (2¢)

19

53

4

13

11

610T T TAB AC AD- - = (3¢)

Solving the resulting set of equations using conventional algorithms gives:

T

T

T

AB

AC

AD

===

1431 00

1560 00

183 010

.

.

.

N

N

N TAB = 1431 N b

TAC = 1560 N b

TAD = 183 0. N b

125


PROBLEM 2.118

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

SOLutiOn

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

- - - + =48

53

12

13

48

610T T T PAB AC AD (1)

- - + + =12

53

3

13

36

610T T T QAB AC AD (2)

19

53

4

13

11

610T T TAB AC AD- - = (3)

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

- - - + =48

53

12

13

48

612880 0T T TAB AC AD N (1¢)

- - + - =12

53

3

13

36

61576 0T T TAB AC AD N (2¢)

19

53

4

13

11

610T T TAB AC AD- - = (3¢)

Solving the resulting set of equations using conventional algorithms gives:

T

T

T

AB

AC

AD

===

1249 29

490 31

1646 97

.

.

.

N

N

N TAB = 1249 N b

TAC = 490 N b

TAD = 1647 N b

126


PROBLEM 2.119

Using two ropes and a roller chute, two workers are unloading a 1000-N 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, –500 mm, 1000 mm), B(–1000 mm, 1250 mm, 0), and C(1125 mm, 1000 mm, 0), and assuming that no friction exists between the counterweight and the chute, determine the tension in each rope. (Hint: Since there is no friction, the force exerted by the chute on the counterweight must be perpendicular to the chute.)

SOLutiOn

From the geometry of the chute:

N j k

j k

= +

= +

N

N5

2

0 8944 0 4472

( )

( . . )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

AB

AB

TAB AB

u ruu= - + -=

=

( ( (1000 1750 1000

2250

mm) mm) mm)

mm

i j k

T l ==

= - + -

TAB

ABT

AB

AB

u ruu

22501000 1750 1000

mm mm) mm) mm)[( ( ( ]i j k

TT i j kAB ABT= - + -ÊËÁ

ˆ¯̃

4

9

7

9

4

9

and AC

AC

u ruu= + -=

( ( (1125 1500 1000

2125

mm) mm) mm)

mm

i j k

T

i j

AC AC AC

AC

T TAC

ACT

= =

= + -

lu ruu

21251125 1500 100

mm mm) mm)[( ( ( 00

9

17

12

17

8

17

mm) ]k

T i j kAC ACT= + -ÊËÁ

ˆ¯̃

Then: SF N T T W= + + + =0 0: AB AC

127



With W = 1000 N, and equating the factors of i, j, and k to zero, we obtain the linear algebraic equations:

i : - + =4

9

9

170T TAB AC (1)

j: 7

9

12

17

2

51000T TAB AC+ + - =N 0 (2)

k : - - + =4

9

8

17

1

50T T NAB AC (3)

Using conventional methods for solving linear algebraic equations we obtain:

TAB

= 327.94 N TAB = 328 N b

TAC

= 275.30 N TAC = 275 N b

128


PROBLEM 2.120

Solve Problem 2.119 assuming that a third worker is exerting a force P = -( 200 N)i on the counterweight.

PROBLEM 2.119 Using two ropes and a roller chute, two workers are unloading a 1000-N 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, –500 mm, 1000 mm), B(–1000 mm, 1250 mm, 0), and C(1125 mm, 1000 mm, 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

See Problem 2.119 for the analysis leading to the vectors describing the tension in each rope.

T i j k

T i j k

AB AB

AC AC

T

T

= - + -ÊËÁ

ˆ¯̃

= + -ÊËÁ

ˆ¯̃

4

9

7

9

4

9

9

17

12

17

8

17

Then: SF N T T P WA AB AC= + + + + =0 0:

Where P i= -( 200 N)

and W j= (1000 N)

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

i : - + - =4

9

9

17200T TAB AC 0

j:2

05

7

9

12

171000N T TAB AC+ + - =

k :1

5N T TAB AC- - =4

9

8

170

Using conventional methods for solving linear algebraic equations we obtain

TAB

= 123.89 N TAB = 123 9. N b

TAC

= 481.78 N TAC = 482 N b

129


PROBLEM 2.121

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

SOLutiOn

Free Body A: T

i j k

AB ABT

TAB

AB

T

T

=

=

= - + +

=

lu vuu

( ) ( ) ( )130 400 160

450

mm mm mm

mm

-- + +ÊËÁ

ˆ¯̃

13

45

40

45

16

45i j k

T

i j k

AC ACT

TAC

AC

T

=

=

= - + + -

=

lu vuu

( ) ( ) ( )150 400 240

490

mm mm mm

mm

TT

F AB AC

- + -ÊËÁ

ˆ¯̃

= + + + + =

15

49

40

49

24

49

0 0

i j k

T T Q P WS :

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

i : .- - + = =13

45

15

490 0 59501T T P T P (1)

j: .+ + - = =40

45

40

490 1 70521T T W T W (2)

k : .+ - + = =16

45

24

490 0 134240T T Q T Q (3)

130



Data: W T T= = =376 1 70521 376 220 50 N N N. .

0 59501 220 50. ( . ) N = P P = 131 2. N b

0 134240 220 50. ( . ) N = Q Q = 29 6. N b

131


PROBLEM 2.122

For the system of Problem 2.121, determine W and Q knowing that P = 164 N.

PROBLEM 2.121 A container of weight W is suspended from ring A. Cable BAC passes through the ring and is attached to fixed supports at B and C. Two forces P i= P and Q k= Q are applied to the ring to maintain the container in the position shown. Knowing that W = 376 N, determine P and Q. (Hint: The tension is the same in both portions of cable BAC.)

SOLutiOn

Refer to Problem 2.121 for the figure and analysis resulting in Equations (1), (2), and (3) for P, W, and Q in terms of T below. Setting P = 164 N we have:

Eq. (1): 0 59501 164. T = N T = 275 63. N

Eq. (2): 1 70521 275 63. ( . ) N = W W = 470 N b

Eq. (3): 0 134240 275 63. ( . ) N = Q Q = 37 0. N b

132


PROBLEM 2.123

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 that passes over a pulley at B and through ring A and that 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

AB

AB

u ruu= - + +

= - + +

( . ) ( . ) ( )

( . ) ( . ) ( )

0 78 1 6 0

0 78 1 6 02 2 2

m m m

m m

i j k

==

= =

= - + +

1 78

1 780 78 1 6

.

.[ ( . ) ( . )

m

m m m

T

i j

AB AB AB

AB

T TAB

ABT

lu ruu

(( ) ]

( . . )

0

0 4382 0 8989 0

m k

T i j kAB ABT= - + +

and AC

AC

A

u ruu= + +

= + + =

( ) ( . ) ( . )

( ) ( . ) ( . )

0 1 6 1 2

0 1 6 1 2 22 2 2

i j k

T

m m

m m m m

CC AC ACAC

AC AC

T TAC

AC

T

T

= = = + +

=

lu ruu

20 1 6 1 2

0 m

m m[( ) ( . ) ( . ) ]

( .

i j k

T 88 0 6j k+ . )

and AD

AD

u ruu= + +

= + + =

( . ) ( . ) ( . )

( . ) ( . ) ( . )

1 3 1 6 0 4

1 3 1 6 0 42 2 2

m m m

m m m

i j k

22 1

2 11 3 1 6 0 4

.

.[( . ) ( . ) ( .

m

m m m mT i jAD AD AD

ADT TAD

AD

T= = = + +l

u ruu)) ]

( . . . )

k

T i kAD ADT= + +0 6190 0 7619 0 1905j

133



Finally, AE

AE

u ruu= - + -

= - + + -

( . ) ( . ) ( . )

( . ) ( . ) ( .

0 4 1 6 0 86

0 4 1 6 0 82 2

m m m

m m

i j k

66 1 86

1 860 4 1 6

2m m

m m m

) .

.[ ( . ) ( .

=

= =

= - +

T

i

AE AE AE

AE

T TAE

AET

lu ruu

)) ( . ) ]

( . . . )

j k

T i j k

-

= - + -

0 86

0 2151 0 8602 0 4624

m

AE AET

With the weight of the container W j= -W , at A we have:

SF T T T j= + + - =0 0: AB AC AD W

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

- + - =0 4382 0 6190 0 2151 0. . .T T TAB AD AE (1)

0 8989 0 8 0 7619 0 8602 0. . . .T T T T WAB AC AD AE+ + + - = (2)

0 6 0 1905 0 4624 0. . .T T TAC AD AE+ - = (3)

Knowing that W = 1000 N and that because of the pulley system at BT T PAB AD= = , where P is the externally applied (unknown) force, we can solve the system of linear Equations (1), (2) and (3) uniquely for P.

P = 378 N b

134


PROBLEM 2.124

Knowing that the tension in cable AC of the system described in Problem 2.123 is 150 N, determine (a) the magnitude of the force P, (b) the weight W of the container.

PROBLEM 2.123 A container of weight W issuspended from ring A, to which cables AC and AE are attached. A force P is applied to the end F of a third cable that passes over a pulley at B and through ring A and that 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

Here, as in Problem 2.123, 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

T T PAB AD= =

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

(a) P = 454 N b

(b) W = 1202 N b

135


PROBLEM 2.125

Collars A and B are connected by a 625-mm long wire and can slide freely on frictionless rods. If a 300-N force Q is applied to collar B as shown, determine (a) the tension in the wire when x = 225 mm,(b) the corresponding magnitude of the force P required to maintain the equilibrium of the system.

SOLutiOn

Free Body Diagrams of Collars: A: B:

lABAB

AB

x z= = - - +u ruu

i k( )500

625

mm

mm

j

Collar A: SF i j k= + + + =0 0: P N N Ty z AB ABl

Substitute for lAB and set coefficient of i equal to zero:

PT xAB- =

6250

mm (1)

Collar B: SF k i j= + ¢ + ¢ - =0 300 0: ( ) N N N Tx y AB ABl

Substitute for lAB and set coefficient of k equal to zero:

300625

0 N mm

- =T zAB (2)

(a) x z

z

= + + ==

225 225 500 625

300

2 2 2 2 mm mm mm mm

mm

( ) ( ) ( )

From Eq. (2): 300300

6250 N

mm

mm- =

TAB ( ) TAB = 625 N b

(b) From Eq. (1): P = ( )( )625 225

625

N mm

mm P = 225 N b

136


PROBLEM 2.126

Collars A and B are connected by a 625-mm long wire and can slide freely on frictionless rods. Determine the distances x and z for which the equilibrium of the system is maintained when P = 600 N and Q = 300 N.

SOLutiOn

See Problem 2.125 for the diagrams and analysis leading to Equations (1) and (2) below:

PT xAB=

625 mm (1)

300625

0 N mm

- =T zAB (2)

For P = 600 N, Eq. (1) yields T xAB = ( )( )625 600 mm N (1¢)

From Eq. (2) T zAB = ( )( )625 300 mm N (2¢)

Dividing Eq. (1¢) by (2¢): x

z= 2 (3)

Now write x z2 2 2 2500 625+ + =( ) ( ) mm mm (4)

Solving (3) and (4) simultaneously

4 250000 390625

167

167 71

2 2

2

z z

z

z

+ + =

== . mm

From Eq. (3) x z==

2

335 4. mm x z= =167 7 335. mm, mm b

137


PROBLEM 2.127

The direction of the 375-N forces may vary, but the angle between the forces is always 50°. Determine the value of a for which the resultant of the forces acting at A is directed horizontally to the left.

SOLutiOn

We must first replace the two 375 N forces by their resultant R1 using the triangle rule.

R1 2 375 25

679 73

= ∞

=

( )cos

.

N

N

R1 679 73= . N a + ∞25

Next we consider the resultant R2 of R1 and the 1200 N force where R2 must be horizontal and directed to the left. Using the triangle rule and law of sines,

sin ( ) sin ( )

.

sin ( ) .

a

aa

+ ∞=

∞

+ ∞ =+ ∞ =

25

1200

30

679 73

25 0 88270

25 61

N N

..

.

970

36 970

∞= ∞a a = ∞37 0. b

138


PROBLEM 2.128

A stake is being pulled out of the ground by means of two ropes as shown. Knowing the magnitude and direction of the force exerted on one rope, determine the magnitude and direction of the force P that should be exerted on the other rope if the resultant of these two forces is to be a 200-N vertical force.

SOLutiOn

Triangle rule:

Law of cosines: P

P

2 2 2150 200 2 150 200 25

90 1195

= + - ∞=

( ) ( ) ( )( )cos

. N

Law of sines: sin sin

.

.

a

a150

25

90 119544 703

N N= ∞

= ∞

90 45 297∞ - = ∞a . P = 90 1. N 45 3. ∞ b

139


PROBLEM 2.129

Member BD exerts on member ABC a force P directed along line BD. Knowing that P must have a 1200 N vertical component, determine (a) the magnitude of the force P, (b) its horizontal component.

SOLutiOn

(a) PPy=

∞=

∞=

sin sin.

35

12001866 9

N

40N or P = 1867 N b

(b) PP

xy=

∞=

∞=

tan tan.

40

12001430 1

N

40N or Px = 1430 N b

140


PROBLEM 2.130

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

SOLutiOn

Free Body Diagram at C:

SFx AC BCT T= - + =06

6 25

3 75

4 250:

.

.

.

m

m

m

m

T TBC AC= 1 08800.

SFy AC BCT T= + - =01 75

6 25

2

4 252000 0:

.

. .

m

m

m

m N

(a) 1 75

6 25

2

4 251 08800 2000

.

. .( . )

m

m

m

m N 0

(0.28000 0.5

T TAC AC+ - =

+ 11200) NTAC = 2000

TAC = 2525 25. N TAC = 2530 N b

(b) TBC ==

( . )( . )

.

1 08800 2525 25

2747 5

N

N TBC = 2750 N b

141


PROBLEM 2.131

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

SOLutiOn

Free Body: C

(a) SFx ACT= - + =012

13

4

5360 0: ( ) N TAC = 312 N b

(b) SFy BCT= + + - =05

13312

3

5360 480 0: ( ) ( ) N N N

TBC = - -480 120 216 N N N TBC = 144 N b

142


PROBLEM 2.132

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

SOLutiOn

Free Body: C

SFx ACT= - + =012

13

4

50: P

T PAC = 13

15 (1)

SFy AC BCT T P= + + - =05

13

3

5480 0: N

Substitute for TAC from (1): 5

13

13

15

3

5480 0Ê

ËÁˆ¯̃

ÊËÁ

ˆ¯̃

+ + - =P T PBC N

T PBC = -48014

15 N (2)

From (1), TAC � 0 requires P � 0.

From (2), TBC � 0 requires 14

15480 514 29P P� � N, N.

Allowable range: 0 514� �P N b

143


PROBLEM 2.133

A force acts at the origin of a coordinate system in a direction defined by the angles 69.3qx = ° and q z = ∞57 9. . Knowing that the y component of the force is –870 N, determine (a) the angle q y , (b) the other components and the magnitude of the force.

SOLutiOn

(a) To determine q y , use the relation

cos cos cos2 2 2 1q q qy y z+ + =

or cos cos cos2 2 21q q qy x y= - -

Since Fy � 0, we must have cosq y � 0

cos cos . cos .

.

q y = - - ∞ - ∞

= -

1 69 3 57 9

0 76985

2 2

q y = ∞140 3. b

(b) FFy

y

= = --

=cos .

.q

870

0 769851130 09

N N F = 1130 N b

F Fx x= = ∞cos ( . )cos .q 1130 09 69 3 N Fx = 399 N b

F Fz z= = ∞cos ( . )cos .q 1130 09 57 9 N Fz = 601 N b

144


PROBLEM 2.134

Cable AB is 32.5 m long, and the tension in that cable is 20 kN. Determine (a) the x, y, and z components of the force exerted by the cable on the anchor B, (b) the angles qx , q y , and q z defining the direction of that force.

SOLutiOn

From triangle AOB: cos.

.

.

q

q

y

y

=

== ∞

28

32 50 86154

30 51

m

m

(a) F Fx y= - ∞

= - ∞ ∞ =

sin cos

( sin . cos .

q 20

20000 30 51 20 9541 43 N) N

Fx = -9 54. kN b

F Fy y= + = =cos ( ( . )q 20000 0 86154 17231 N) N Fy = +17 23. kN b

Fz = +(20000 N)sin 30.51 sin 20 = 3472.8 N° ° Fz = +3 47. kN b

(b) cos.

.qxxF

F= = - = -9541 43

0 4771 N

20000 N qx = ∞118 5. b

From above: q y = ∞30 51. q y = ∞30 5. b

cos.

.q zzF

F= = + = +3472 8

200000 1736

N

N q z = ∞80 0. b

145


PROBLEM 2.135

In order to move a wrecked truck, two cables are attached at A and pulled by winches B and C as shown. Knowing that the tension is 10 kN in cable AB and 7.5 kN in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables.

SOLutiOn

AB

AB

AC

u ruu

u ruu

= - + +

=

= - + -

15 588 15 12

24 739

15 588 18 60 1

.

.

. .

i j k

i j

m

55

28 530

1015 588 15

k

T

Ti

AC

T TAB

ABAB AB AB AB

AB

=

= =

= - +

.

( ).

m

kN

lu ruu

jj k+12

24 739.

T i j k

T

AB

AC AC AC ACT TAC

= + +

= =

( . ) ( . ) ( . )6 301 6 063 4 851 kN kN kN

lu ruuu

AC

AC

( . ). .

.

( . ) ( .

7 515 588 18 60 15

28 530

4 098 4 8

kN

kN

- + -

= - +

i j k

T i 990 3 943

10 399 10 953

kN kN

kN kN

) ( . )

( . ) ( . ) (

j k

R T T i j

-

= + = - + +AB AC 00 908

10 399 10 953 0 908

15 130

2 2 2

. )

( . ) ( . ) ( . )

.

kN

kN

k

R = + +

= R = 15 13. kN b

cos.

..qx

xR

R= = - = -10 399

15 1300 6873

kN

kN q z = ∞133 4. b

cos.

..q y

yR

R= = =10 953

15 1300 7239

kN

kN q y = ∞43 6. b

cos.

..q z

zR

R= = =0 908

15 1300 0600

kN

kN q z = ∞86 6. b

146


PROBLEM 2.136

A container of weight W = 1165 N is supported by three cables as shown. Determine the tension in each cable.

SOLUTION

Free Body: A SF T T T W= + + + =0 0: AB AC AD

AB

AB

AC

u ruu

u ruu

= +=

= -

( ) ( )

( ) (

450 600

750

600 320

mm mm

mm

mm m

i j

j mm

mm

mm mm mm

mm

)

( ) ( ) ( )

k

i j k

AC

AD

AD

=

= + +=

680

500 600 360

860

u ruu

We have: TAB AB AB AB ABT TAB

ABT= = = +Ê

ËÁˆ¯̃

= +

lu ruu

450

750

600

750

0 6 0 8

i j

i j( . . ))TAB

T j k j kAC AC AC AC ACT TAC

AC= = = -Ê

ËÁˆ¯̃

= -lu ruu

600

680

320

680

15

17

8

17T ÊÊ

ËÁˆ¯̃

= = = - + +Ê

TAC

AD AD AD ADT TAD

ADT i j kl

u ruu500

860

600

860

360

860ËËÁˆ¯̃

= - + +ÊËÁ

ˆ¯̃

T

T

AD

AD ADT i j k25

43

30

43

18

43

Substitute into SF = 0, factor i, j, k, and set their coefficient equal to zero:

0 625

430 0 96899. .T T T TAB AD AB AD- = = (1)

0 815

17

30

431165. T T TAB AC AD+ + - = N 0 (2)

- + = =8

17

18

430 0 88953T T T TAC AD AC AD. (3)

147



Substitute for TAB and TAC from (1) and (3) into (2):

0 8 0 9689915

170 88953

30

531165 0. . .¥ + ¥ +Ê

ËÁˆ¯̃

- =TAD N

2 2578 1165 0. TAD - = N TAD = 516 N b

From (1): TAB = 0 96899 516. ( ) N TAB = 500 N b

From (3): TAC = 0 88953 516. ( ) N TAC = 459 N b

148


PROBLEM 2.137

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

SOLutiOn

For both Problems 2.137 and 2.138: Free Body Diagrams of Collars:

( )AB x y z2 2 2 2= + +

Here ( . (. )0 525 20 2 2 2 m) m2 = + +y z

or y z2 2 20 23563+ = . m

Thus, why y given, z is determined,

Now lABAB

AB

y z

y

=

= - +

= - +

u ruu

1

0 5250 20

0 38095 1 90476 1 9.

( . )

. . . m

mi j k

i j 00476zkWhere y and z are in units of meters, m.

From the F.B. Diagram of collar A: SF i k j= + + + =0 0: N N P Tx z AB ABl

Setting the j coefficient to zero gives: P y TAB- =( . )1 90476 0

With P

TyAB

=

=

341

341

1 90476

N

N

.

Now, from the free body diagram of collar B: S lF i j k= + + - =0 0: N N Q Tx y AB AB

Setting the k coefficient to zero gives: Q T zAB- =( . )1 90476 0

And using the above result for TAB we have Q T zy

zz

yAB= = =341

1 904761 90476

341N N

( . )( . )

( )( )

149



Then, from the specifications of the problem, y = 155 mm = 0 155. m

z

z

2 2 20 23563 0 155

0 46

= -=

. ( . )

.

m m

mand

(a) TAB =

=

341

0 155 1 90476

1155 00

N

N

. ( . )

.

or TAB = 1 155. kN b

and

(b) Q =

=

341 46 N(0. m)(0.866)

(0.155 m)

(1012.00 N)

or Q = 1 012. kN b

150


PROBLEM 2.138

Solve Problem 2.137 assuming that y = 275 mm.

PROBLEM 2.137 Collars A and B are connected by a 525-mm-long wire and can slide freely on frictionless rods. If a force P j= (341 N) is applied to collar A, determine (a) the tension in the wire when y = 155 mm, (b) the magnitude of the force Q required to maintain the equilibrium of the system.

SOLutiOn

From the analysis of Problem 2.137, particularly the results:

y z

Ty

Qy

z

AB

2 2 0 23563

341

1 90476

341

+ =

=

=

.

.

m

N

N

2

With y = =275 0 275 mm m,. we obtain:

z

z

2 2 20 23563 0 275

0 40

= -=

. ( . )

.

m m

mand

(a) TAB = =341

1 90476 0 275651 00

N

m( . )( . ).

or TAB = 651 N b

and

(b) Q =341

0 275

N(0.40 m)

m( . ) or Q = 496 N b

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