Solucionario desde capitulo 2 al 5

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3
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 125 N,Q = 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: 179 N, 75.1R α= = ° 179 N=R 75.1° W

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4
PROBLEM 2.2
Two forces P and Q are applied as shown at Point A of a hook support. Knowing that
P 60 lb= and 25 lb,Q = 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: 77.1lb, 85.4R α= = ° 77.1lb=R 85.4° W

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5
PROBLEM 2.3
The cable stays AB and AD help support pole AC. Knowing
that the tension is 120 lb in AB and 40 lb in AD, determine
graphically the magnitude and direction of the resultant of the
forces exerted by the stays at A using (a) the parallelogram
law, (b) the triangle rule.
SOLUTION
We measure: 51.3
59.0
α
β
= °
= °
(a) Parallelogram law:
(b) Triangle rule:
We measure: 139.1 lb,R = 67.0γ = ° 139.1lbR = 67.0° W

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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
(a) Parallelogram law:
(b) Triangle rule:
We measure: 3.30 kN, 66.6R α= = ° 3.30 kN=R 66.6° W

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7
PROBLEM 2.5
The 300-lb force is to be resolved into components along lines a-a′ and b-b′.
(a) Determine the angle α by trigonometry knowing that the component
along line a-a′ is to be 240 lb. (b) What is the corresponding value of the
component along b-b′?
SOLUTION
(a) Using the triangle rule and law of sines:
sin sin 60
240 lb 300 lb
sin 0.69282
43.854
60 180
180 60 43.854
76.146
β
β
β
α β
α
°
=
=
= °
+ + ° = °
= ° − ° − °
= ° 76.1α = ° W
(b) Law of sines:
300 lb
sin76.146 sin 60
bbF ′
=
° °
336 lbbbF ′ = W

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8
PROBLEM 2.6
The 300-lb force is to be resolved into components along lines a-a′ and b-b′.
(a) Determine the angle α by trigonometry knowing that the component along
line b-b′ is to be 120 lb. (b) What is the corresponding value of the component
along a-a′?
SOLUTION
Using the triangle rule and law of sines:
(a)
sin sin 60
120 lb 300 lb
α °
=
sin 0.34641
20.268
α
α
=
= ° 20.3α = ° W
(b) 60 180
180 60 20.268
99.732
α β
β
+ + ° = °
= ° − ° − °
= °
300 lb
sin99.732 sin 60
aaF ′
=
° °
341 lbaaF ′ = W

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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 α 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 25
50 N 35 N
sin 0.60374
α
α
°
=
=
37.138α = ° 37.1α = ° W
(b) 25 180
180 25 37.138
117.86
α β
β
+ + ° = °
= ° − ° − °
= °
35 N
sin117.86 sin 25
R
=
°
73.2 NR = W

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10
PROBLEM 2.8
For the hook support of Problem 2.1, knowing that the magnitude of P is 75 N,
determine by trigonometry (a) the required magnitude of the force Q if the
resultant R of the two forces applied at A is to be vertical, (b) the corresponding
magnitude of R.
PROBLEM 2.1 Two forces P and Q are applied as shown at Point A of a hook
support. Knowing that 75 NP = and 125 N,Q = determine graphically the
magnitude and direction of their resultant using (a) the parallelogram law, (b) the
triangle rule.
SOLUTION
Using the triangle rule and law of sines:
(a)
75 N
sin 20 sin35°
Q
=
°
44.7 NQ = W
(b) 20 35 180
180 20 35
125
α
α
+ ° + ° = °
= ° − ° − °
= °
75 N
sin125 sin35°
R
=
°
107.1 NR = W

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11
PROBLEM 2.9
A trolley that moves along a horizontal beam is acted upon by two
forces as shown. (a) Knowing that 25 ,α = ° 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 N
sin 25° sin 75
P
=
°
3660 NP = W
(b) 25 75 180
180 25 75
80
β
β
° + + ° = °
= ° − ° − °
= °
1600 N
sin 25° sin80
R
=
°
3730 NR = W

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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: 2 2 2
(1600 N) (2500 N) 2(1600 N)(2500 N)cos 75°
2596 N
P
P
= + −
=
Using the law of sines:
sin sin75
1600 N 2596 N
36.5
α
α
°
=
= °
P is directed 90 36.5 or 53.5°° − ° below the horizontal. 2600 N=P 53.5° W

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13
PROBLEM 2.11
A steel tank is to be positioned in an excavation. Knowing that
α = 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
Using the triangle rule and the law of sines:
(a) 50 60 180
180 50 60
70
β
β
+ ° + ° = °
= ° − ° − °
= °
425 lb
sin 70 sin 60
P
=
° °
392 lbP = W
(b)
425 lb
sin 70 sin50
R
=
° °
346 lbR = W

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14
PROBLEM 2.12
A steel tank is to be positioned in an excavation. Knowing that
the magnitude of P is 500 lb, determine by trigonometry (a) the
required angle α if the resultant R of the two forces applied at A
is to be vertical, (b) the corresponding magnitude of R.
SOLUTION
Using the triangle rule and the law of sines:
(a) ( 30 ) 60 180
180 ( 30 ) 60
90
sin (90 ) sin60
425 lb 500 lb
α β
β α
β α
α
+ ° + ° + = °
= ° − + ° − °
= ° −
° − °
=
90 47.40α° − = ° 42.6α = ° W
(b)
500 lb
sin (42.6 30 ) sin 60
R
=
° + ° °
551 lbR = W

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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) (50 N)sin 25P = ° 21.1 N=P W
(b) (50 N)cos 25R = ° 45.3 NR = W

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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 α = 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) (425 lb)cos30P = ° 368 lb=P W
(b) (425 lb)sin30R = ° 213 lbR = W

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17
PROBLEM 2.15
Solve Problem 2.2 by trigonometry.
PROBLEM 2.2 Two forces P and Q are applied as shown at Point A of a hook
support. Knowing that 60 lbP = and 25 lb,Q = 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:
2 2 2
2 2 2
2
20 35 180
125
2 cos
(60 lb) (25 lb)
2(60 lb)(25 lb)cos125
3600 625 3000(0.5736)
77.108 lb
R P Q PQ
R
R
R
α
α
α
° + ° + = °
= °
= + −
= +
− °
= + +
=
Using the law of sines:
sin sin125
25 lb 77.108 lb
15.402
β
β
°
=
= °
70 85.402β° + = ° 77.1lb=R 85.4° W

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18
PROBLEM 2.16
Solve Problem 2.3 by trigonometry.
PROBLEM 2.3 The cable stays AB and AD help support
pole AC. Knowing that the tension is 120 lb in AB and
40 lb 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
8
tan
10
38.66
6
tan
10
30.96
α
α
β
β
=
= °
=
= °
Using the triangle rule: 180
38.66 30.96 180
110.38
α β ψ
ψ
ψ
+ + = °
° + ° + = °
= °
Using the law of cosines: 2 22
(120 lb) (40 lb) 2(120 lb)(40 lb)cos110.38
139.08 lb
R
R
= + − °
=
Using the law of sines:
sin sin110.38
40 lb 139.08 lb
γ °
=
15.64
(90 )
(90 38.66 ) 15.64
66.98
γ
φ α γ
φ
φ
= °
= ° − +
= ° − ° + °
= ° 139.1 lb=R 67.0° W

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19
PROBLEM 2.17
Solve Problem 2.4 by trigonometry.
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: 2 2 2
(2 kN) (3 kN)
2(2 kN)(3 kN)cos80
3.304 kN
R
R
= +
− °
=
Using the law of sines:
sin sin80
2 kN 3.304 kN
γ °
=
36.59
80 180
180 80 36.59
63.41
180 50
66.59
γ
β γ
γ
γ
φ β
φ
= °
+ + ° = °
= ° − ° − °
= °
= ° − + °
= ° 3.30 kN=R 66.6° W

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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 180 (40 20 )
120
γ = ° − ° + °
= °
Then 2 2 2
2
(15 kN) (10 kN)
2(15 kN)(10 kN)cos120
475 kN
21.794 kN
R
R
= +
− °
=
=
and
10 kN 21.794 kN
sin sin120
10 kN
sin sin120
21.794 kN
0.39737
23.414
α
α
α
=
°
§ ·
= °¨ ¸
© ¹
=
=
Hence: 50 73.414φ α= + ° = 21.8 kN=R 73.4° W

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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 180 (40 20 )
120
γ = ° − ° + °
= °
Then 2 2 2
2
(10 kN) (15 kN)
2(10 kN)(15 kN)cos120
475 kN
21.794 kN
R
R
= +
− °
=
=
and
15 kN 21.794 kN
sin sin120
15 kN
sin sin120
21.794 kN
0.59605
36.588
α
α
α
=
°
§ ·
= °¨ ¸
© ¹
=
= °
Hence: 50 86.588φ α= + ° = ° 21.8 kN=R 86.6° W

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22
PROBLEM 2.20
For the hook support of Problem 2.7, knowing that 75P = N and α 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 α 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 180 (50 25 )
105
β = ° − ° + °
= °
Then 2 2 2
2 2
(75 N) (50 N)
2(75 N)(50 N)cos105
10066.1 N
100.330 N
R
R
R
= +
− °
=
=
and
sin sin105
75 N 100.330 N
sin 0.72206
46.225
γ
γ
γ
°
=
=
= °
Hence: 25 46.225 25 21.225γ − ° = ° − ° = ° 100.3 N=R 21.2° W

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23
PROBLEM 2.21
Determine the x and y components of each of the forces shown.
SOLUTION
Compute the following distances:
2 2
2 2
2 2
(600) (800)
1000 mm
(560) (900)
1060 mm
(480) (900)
1020 mm
OA
OB
OC
= +
=
= +
=
= +
=
800-N Force:
800
(800 N)
1000
xF = + 640 NxF = + W
600
(800 N)
1000
yF = + 480 NyF = + W
424-N Force:
560
(424 N)
1060
xF = − 224 NxF = − W
900
(424 N)
1060
yF = − 360 NyF = − W
408-N Force:
480
(408 N)
1020
xF = + 192.0 NxF = + W
900
(408 N)
1020
yF = − 360 NyF = − W

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24
PROBLEM 2.22
Determine the x and y components of each of the forces shown.
SOLUTION
Compute the following distances:
2 2
2 2
2 2
(84) (80)
116 in.
(28) (96)
100 in.
(48) (90)
102 in.
OA
OB
OC
= +
=
= +
=
= +
=
29-lb Force:
84
(29 lb)
116
xF = + 21.0 lbxF = + W
80
(29 lb)
116
yF = + 20.0 lbyF = + W
50-lb Force:
28
(50 lb)
100
xF = − 14.00 lbxF = − W
96
(50 lb)
100
yF = + 48.0 lbyF = + W
51-lb Force:
48
(51 lb)
102
xF = + 24.0 lbxF = + W
90
(51 lb)
102
yF = − 45.0 lbyF = − W

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25
PROBLEM 2.23
Determine the x and y components of each of the forces shown.
SOLUTION
40-lb Force: (40 lb)cos60xF = + ° 20.0 lbxF = W
(40 lb)sin60yF = − ° 34.6 lbyF = − W
50-lb Force: (50 lb)sin50xF = − ° 38.3 lbxF = − W
(50 lb)cos50yF = − ° 32.1 lbyF = − W
60-lb Force: (60 lb)cos25xF = + ° 54.4 lbxF = W
(60 lb)sin 25yF = + ° 25.4 lbyF = W

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26
PROBLEM 2.24
Determine the x and y components of each of the forces shown.
SOLUTION
80-N Force: (80 N)cos40xF = + ° 61.3 NxF = W
(80 N)sin 40yF = + ° 51.4 NyF = W
120-N Force: (120 N)cos70xF = + ° 41.0 NxF = W
(120 N)sin 70yF = + ° 112.8 NyF = W
150-N Force: (150 N)cos35xF = − ° 122. 9 NxF = − W
(150 N)sin35yF = + ° 86.0 NyF = W

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27
PROBLEM 2.25
Member BD exerts on member ABC a force P directed along line BD.
Knowing that P must have a 300-lb horizontal component, determine
(a) the magnitude of the force P, (b) its vertical component.
SOLUTION
(a) sin35 300 lbP ° =
300 lb
sin35
P =
°
523 lbP = W
(b) Vertical component cos35vP P= °
(523 lb)cos35= ° 428 lbvP = W

26. 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.
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 N sin 20P= °
2193 NP = 2190 NP = W
(b) cos20ABCP P= °
(2193 N)cos20= ° 2060 NABCP = W

27. 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.
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)
sin38
120 N
sin38
xP
P =
°
=
°
194.91 N= or 194.9 NP = W
(b)
tan38
120 N
tan38
x
y
P
P =
°
=
°
153.59 N= or 153.6 NyP = W

28. 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.
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)
cos38
180 N
cos 38°
yP
P =
°
=
228.4 N= 228 NP = W
(b) tan38
(180 N)tan38
x yP P= °
= °
140.63 N= 140.6 NxP = W

29. 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.
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 1200 N:xP =
Then
(a) sin55xP P= °
sin55
1200 N
sin55
1464.9 N
xP
P =
°
=
°
= 1465 NP = W
(b) tan55x yP P= °
tan55
1200 N
tan55
840.2 N
x
y
P
P =
°
=
°
= 840 Ny =P W

30. 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.
32
PROBLEM 2.30
Cable AC exerts on beam AB a force P directed along line AC. Knowing that P
must have a 350-lb vertical component, determine (a) the magnitude of the
force P, (b) its horizontal component.
SOLUTION
(a)
cos 55
yP
P =
°
350 lb
cos 55
610.2 lb
=
°
= 610 lbP = W
(b) sin 55xP P= °
(610.2 lb)sin 55
499.8 lb
= °
= 500 lbxP = W

31. 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.
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. (lb) y Comp. (lb)
29 lb +21.0 +20.0
50 lb –14.00 +48.0
51 lb +24.0 –45.0
31.0xR = + 23.0yR = +
(31.0 lb) (23.0 lb)
tan
23.0
31.0
36.573
23.0 lb
sin (36.573 )
x y
y
x
R R
R
R
R
α
α
= +
= +
=
=
= °
=
°
i j
i j
R
38.601 lb= 38.6 lb=R 36.6° W

32. 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.
34
PROBLEM 2.32
Determine the resultant of the three forces of Problem 2.24.
PROBLEM 2.24 Determine the x and y components of each of the
forces shown.
SOLUTION
Components of the forces were determined in Problem 2.24:
Force x Comp. (N) y Comp. (N)
80 N +61.3 +51.4
120 N +41.0 +112.8
150 N –122.9 +86.0
20.6xR = − 250.2yR = +
( 20.6 N) (250.2 N)
tan
250.2 N
tan
20.6 N
tan 12.1456
85.293
x y
y
x
R R
R
R
α
α
α
α
= +
= − +
=
=
=
= °
R i j
i j
250.2 N
sin85.293
R =
°
251 N=R 85.3° W

33. 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.
35
PROBLEM 2.33
Determine the resultant of the three forces of Problem 2.23.
PROBLEM 2.23 Determine the x and y components of each of
the forces shown.
SOLUTION
Force x Comp. (lb) y Comp. (lb)
40 lb +20.00 –34.64
50 lb –38.30 –32.14
60 lb +54.38 +25.36
36.08xR = + 41.42yR = −
( 36.08 lb) ( 41.42 lb)
tan
41.42 lb
tan
36.08 lb
tan 1.14800
48.942
x y
y
x
R R
R
R
α
α
α
α
= +
= + + −
=
=
=
= °
R i j
i j
41.42 lb
sin 48.942
R =
°
54.9 lb=R 48.9° W

34. 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.
36
PROBLEM 2.34
Determine the resultant of the three forces of Problem 2.21.
PROBLEM 2.21 Determine the x and y components of each of
the forces shown.
SOLUTION
Components of the forces were determined in Problem 2.21:
Force x Comp. (N) y Comp. (N)
800 lb +640 +480
424 lb –224 –360
408 lb +192 –360
608xR = + 240yR = −
(608 lb) ( 240 lb)
tan
240
608
21.541
x y
y
x
R R
R
R
α
α
= +
= + −
=
=
= °
R i j
i j
240 N
sin(21.541°)
653.65 N
R =
= 654 N=R 21.5° W

35. 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.
37
PROBLEM 2.35
Knowing that α = 35°, determine the resultant of the three forces
shown.
SOLUTION
100-N Force: (100 N)cos35 81.915 N
(100 N)sin35 57.358 N
x
y
F
F
= + ° = +
= − ° = −
150-N Force: (150 N)cos65 63.393 N
(150 N)sin 65 135.946 N
x
y
F
F
= + ° = +
= − ° = −
200-N Force: (200 N)cos35 163.830 N
(200 N)sin35 114.715 N
x
y
F
F
= − ° = −
= − ° = −
Force x Comp. (N) y Comp. (N)
100 N +81.915 −57.358
150 N +63.393 −135.946
200 N −163.830 −114.715
18.522xR = − 308.02yR = −
( 18.522 N) ( 308.02 N)
tan
308.02
18.522
86.559
x y
y
x
R R
R
R
α
α
= +
= − + −
=
=
= °
R i j
i j
308.02 N
sin86.559
R = 309 N=R 86.6° W

36. 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.
38
PROBLEM 2.36
Knowing that the tension in cable BC is 725 N, determine the
resultant of the three forces exerted at Point B of beam AB.
SOLUTION
Cable BC Force:
840
(725 N) 525 N
1160
840
(725 N) 500 N
1160
x
y
F
F
= − = −
= =
500-N Force:
3
(500 N) 300 N
5
4
(500 N) 400 N
5
x
y
F
F
= − = −
= − = −
780-N Force:
12
(780 N) 720 N
13
5
(780 N) 300 N
13
x
y
F
F
= =
= − = −
and
2 2
105 N
200 N
( 105 N) ( 200 N)
225.89 N
x x
y y
R F
R F
R
= Σ = −
= Σ = −
= − + −
=
Further:
1
200
tan
105
200
tan
105
62.3
α
α −
=
=
= °
Thus: 226 N=R 62.3° W

37. 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.
39
PROBLEM 2.37
Knowing that α = 40°, determine the resultant of the three forces
shown.
SOLUTION
60-lb Force: (60 lb)cos20 56.38 lb
(60 lb)sin 20 20.52 lb
x
y
F
F
= ° =
= ° =
80-lb Force: (80 lb)cos60 40.00 lb
(80 lb)sin 60 69.28 lb
x
y
F
F
= ° =
= ° =
120-lb Force: (120 lb)cos30 103.92 lb
(120 lb)sin30 60.00 lb
x
y
F
F
= ° =
= − ° = −
and
2 2
200.30 lb
29.80 lb
(200.30 lb) (29.80 lb)
202.50 lb
x x
y y
R F
R F
R
= Σ =
= Σ =
= +
=
Further:
29.80
tan
200.30
α =
1 29.80
tan
200.30
8.46
α −
=
= ° 203 lb=R 8.46° W

38. 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.
40
PROBLEM 2.38
Knowing that α = 75°, determine the resultant of the three forces
shown.
SOLUTION
60-lb Force: (60 lb)cos 20 56.38 lb
(60 lb)sin 20 20.52 lb
x
y
F
F
= ° =
= ° =
80-lb Force: (80 lb)cos 95 6.97 lb
(80 lb)sin 95 79.70 lb
x
y
F
F
= ° = −
= ° =
120-lb Force: (120 lb)cos 5 119.54 lb
(120 lb)sin 5 10.46 lb
x
y
F
F
= ° =
= ° =
Then 168.95 lb
110.68 lb
x x
y y
R F
R F
= Σ =
= Σ =
and 2 2
(168.95 lb) (110.68 lb)
201.98 lb
R = +
=
110.68
tan
168.95
tan 0.655
33.23
α
α
α
=
=
= ° 202 lb=R 33.2° W

39. 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.
41
PROBLEM 2.39
For the collar of Problem 2.35, determine (a) the required value of
α if the resultant of the three forces shown is to be vertical, (b) the
corresponding magnitude of the resultant.
SOLUTION
(100 N)cos (150 N)cos( 30 ) (200 N)cos
(100 N)cos (150 N)cos( 30 )
x x
x
R F
R
α α α
α α
= Σ
= + + ° −
= − + + ° (1)
(100 N)sin (150 N)sin( 30 ) (200 N)sin
(300 N)sin (150 N)sin( 30 )
y y
y
R F
R
α α α
α α
= Σ
= − − + ° −
= − − + ° (2)
(a) For R to be vertical, we must have 0.xR = We make 0xR = in Eq. (1):
100cos 150cos( 30 ) 0
100cos 150(cos cos 30 sin sin 30 ) 0
29.904cos 75sin
α α
α α α
α α
− + + ° =
− + ° − ° =
=
29.904
tan
75
0.3988
21.74
α
α
=
=
= ° 21.7α = ° W
(b) Substituting for α in Eq. (2):
300sin 21.74 150sin51.74
228.9 N
yR = − ° − °
= −
| | 228.9 NyR R= = 229 NR = W

40. 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.
42
PROBLEM 2.40
For the beam of Problem 2.36, 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
840 12 3
(780 N) (500 N)
1160 13 5
21
420 N
29
x x BC
x BC
R F T
R T
= Σ = − + −
= − + (1)
800 5 4
(780 N) (500 N)
1160 13 5
20
700 N
29
y y BC
y BC
R F T
R T
= Σ = − −
= − (2)
(a) For R to be vertical, we must have 0xR =
Set 0xR = in Eq. (1)
21
420 N 0
29
BCT− + = 580 NBCT = W
(b) Substituting for BCT in Eq. (2):
20
(580 N) 700 N
29
300 N
y
y
R
R
= −
= −
| | 300 NyR R= = 300 NR = W

41. 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.
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:
sin10 (50 lb)cos35 (75 lb)cos60
sin10 78.46 lb
x x AC
AC
R F T
T
= Σ = ° + ° + °
= ° + (1)
(50 lb)sin35 (75 lb)sin 60 cos10
93.63 lb cos10
y y AC
y AC
R F T
R T
= Σ = ° + ° − °
= − ° (2)
(a) Set 0yR = in Eq. (2):
93.63 lb cos10 0
95.07 lb
AC
AC
T
T
− ° =
= 95.1 lbACT = W
(b) Substituting for ACT in Eq. (1):
(95.07 lb)sin10 78.46 lb
94.97 lb
x
x
R
R R
= ° +
=
= 95.0 lbR = W

42. 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.
44
PROBLEM 2.42
For the block of Problems 2.37 and 2.38, determine (a) the
required value of α 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
(60 lb) (80 lb)cos (120 lb)sinx xR F α α= Σ = + + (1)
and
(80 lb)sin (120 lb)cosy yR F α α= Σ = − (2)
(a) Set 0yR = in Eq. (2).
(80 lb)sin (120 lb)cos 0α α− =
Dividing each term by cosα gives:
(80 lb)tan 120 lb
120 lb
tan
80 lb
56.310
α
α
α
=
=
= ° 56.3α = ° W
(b) Substituting for α in Eq. (1) gives:
60 lb (80 lb)cos56.31 (120lb)sin56.31 204.22 lbxR = + ° + ° = 204 lbxR = W

43. 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.
45
PROBLEM 2.43
Two cables are tied together at C and are loaded as shown.
Knowing that α = 20°, determine the tension (a) in cable AC,
(b) in cable BC.
SOLUTION
Free-Body Diagram Force Triangle
Law of sines:
1962 N
sin70 sin50 sin 60
AC BCT T
= =
° ° °
(a)
1962 N
sin 70 2128.9 N
sin60
ACT = ° =
°
2.13 kNACT = W
(b)
1962 N
sin50 1735.49 N
sin60
BCT = ° =
°
1.735 kNBCT = W

44. 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.
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
Free-Body Diagram Force Triangle
Law of sines:
500 N
sin60 sin 40 sin80
AC BCT T
= =
° ° °
(a)
500 N
sin 60 439.69 N
sin80
ACT = ° =
°
440 NACT = W
(b)
500 N
sin 40 326.35 N
sin80
BCT = ° =
°
326 NBCT = W

45. 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.
47
PROBLEM 2.45
Two cables are tied together at C and are loaded as shown.
Knowing that P = 500 N and α = 60°, determine the tension in
(a) in cable AC, (b) in cable BC.
SOLUTION
Free-Body Diagram Force Triangle
Law of sines:
500 N
sin35 sin 75 sin70°
AC BCT T
= =
° °
(a)
500 N
sin35
sin 70
ACT = °
°
305 NACT = W
(b)
500 N
sin 75
sin 70
BCT = °
°
514 NBCT = W

46. 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.
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
Free-Body Diagram Force Triangle
2
mg
(200 kg)(9.81m/s )
1962 N
W =
=
=
Law of sines:
1962 N
sin 15 sin 105 sin 60
AC BCT T
= =
° ° °
(a)
(1962 N) sin 15
sin 60
ACT
°
=
°
586 NACT = W
(b)
(1962 N)sin 105
sin 60
BCT
°
=
°
2190 NBCT = W

47. 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.
49
PROBLEM 2.47
Knowing that 20 ,α = ° determine the tension (a) in cable AC,
(b) in rope BC.
SOLUTION
Free-Body Diagram Force Triangle
Law of sines:
1200 lb
sin 110 sin 5 sin 65
AC BCT T
= =
° ° °
(a)
1200 lb
sin 110
sin 65
ACT = °
°
1244 lbACT = W
(b)
1200 lb
sin 5
sin 65
BCT = °
°
115.4 lbBCT = W

48. 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.
50
PROBLEM 2.48
Knowing that 55α = ° 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
Free-Body Diagram Force Triangle
Law of sines:
300 lb
sin 35 sin 50 sin 95
AC BCF T
= =
° ° °
(a)
300 lb
sin 35
sin 95
ACF = °
°
172.7 lbACF = W
(b)
300 lb
sin 50
sin 95
BCT = °
°
231 lbBCT = W

49. 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.
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 500P = lb and 650Q = lb, determine the magnitudes of
the forces exerted on the rods A and B.
SOLUTION
Free-Body Diagram
Resolving the forces into x- and y-directions:
0A B= + + + =R P Q F F
Substituting components: (500 lb) [(650 lb)cos50 ]
[(650 lb)sin50 ]
( cos50 ) ( sin50 ) 0B A AF F F
= − + °
− °
+ − ° + ° =
R j i
j
i i j
In the y-direction (one unknown force)
500 lb (650 lb)sin50 sin50 0AF− − ° + ° =
Thus,
500 lb (650 lb)sin50
sin50
AF
+ °
=
°
1302.70 lb= 1303 lbAF = W
In the x-direction: (650 lb)cos50 cos50 0B AF F° + − ° =
Thus, cos50 (650 lb)cos50
(1302.70 lb)cos50 (650 lb)cos50
B AF F= ° − °
= ° − °
419.55 lb= 420 lbBF = W

50. 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
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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 750AF = lb and 400BF = lb, determine the magnitudes of
P and Q.
SOLUTION
Free-Body Diagram
Resolving the forces into x- and y-directions:
0A B= + + + =R P Q F F
Substituting components: cos 50 sin 50
[(750 lb)cos 50 ]
[(750 lb)sin 50 ] (400 lb)
P Q Q= − + ° − °
− °
+ ° +
R j i j
i
j i
In the x-direction (one unknown force)
cos 50 [(750 lb)cos 50 ] 400 lb 0Q ° − ° + =
(750 lb)cos 50 400 lb
cos 50
127.710 lb
Q
° −
=
°
=
In the y-direction: sin 50 (750 lb)sin 50 0P Q− − ° + ° =
sin 50 (750 lb)sin 50
(127.710 lb)sin 50 (750 lb)sin 50
476.70 lb
P Q= − ° + °
= − ° + °
= 477 lb; 127.7 lbP Q= = W

51. 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
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53
PROBLEM 2.51
A welded connection is in equilibrium under the action of the
four forces shown. Knowing that 8AF = kN and 16BF = kN,
determine the magnitudes of the other two forces.
SOLUTION
Free-Body Diagram of Connection
3 3
0: 0
5 5
x B C AF F F FΣ = − − =
With 8 kN
16 kN
A
B
F
F
=
=
4 4
(16 kN) (8 kN)
5 5
CF = − 6.40 kNCF = W
3 3
0: 0
5 5
y D B AF F F FΣ = − + − =
With FA and FB as above:
3 3
(16 kN) (8 kN)
5 5
DF = − 4.80 kNDF = W

52. 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
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you are using it without permission.
54
PROBLEM 2.52
A welded connection is in equilibrium under the action of the four
forces shown. Knowing that 5AF = kN and 6DF = kN, determine
the magnitudes of the other two forces.
SOLUTION
Free-Body Diagram of Connection
3 3
0: 0
5 5
y D A BF F F FΣ = − − + =
or
3
5
B D AF F F= +
With 5 kN, 8 kNA DF F= =
5 3
6 kN (5 kN)
3 5
BF
ª º
= +« »
¬ ¼
15.00 kNBF = W
4 4
0: 0
5 5
x C B AF F F FΣ = − + − =
4
( )
5
4
(15 kN 5 kN)
5
C B AF F F= −
= − 8.00 kNCF = W

53. 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
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you are using it without permission.
55
PROBLEM 2.53
Two cables tied together at C are loaded as shown. Knowing that
Q = 60 lb, determine the tension (a) in cable AC, (b) in cable BC.
SOLUTION
0: cos30 0y CAF T QΣ = − ° =
With 60 lbQ =
(a) (60 lb)(0.866)CAT = 52.0 lbCAT = W
(b) 0: sin30 0x CBF P T QΣ = − − ° =
With 75 lbP =
75 lb (60 lb)(0.50)CBT = − or 45.0 lbCBT = W

54. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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 60 lb in
either cable.
SOLUTION
Free-Body Diagram 0: cos60 75 lb 0x BCF T QΣ = − − ° + =
75 lb cos60BCT Q= − ° (1)
0: sin 60 0y ACF T QΣ = − ° =
sin 60ACT Q= ° (2)
Requirement 60 lb:ACT Յ
From Eq. (2): sin60 60 lbQ °Յ
69.3 lbQ Յ
Requirement 60 lb:BCT Յ
From Eq. (1): 75 lb sin 60 60 lbQ− °Յ
30.0 lbQ Ն 30.0 lb 69.3 lbQՅ Յ W

55. 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
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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 30α = ° and 10β = ° 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
0: cos 10 cos 30 cos 30 0x ACB ACB CDF T T TΣ = ° − ° − ° =
0.137158CD ACBT T= (1)
0: sin 10 sin 30 sin 30 900 0y ACB ACB CDF T T TΣ = ° + ° + ° − =
0.67365 0.5 900ACB CDT T+ = (2)
(a) Substitute (1) into (2): 0.67365 0.5(0.137158 ) 900ACB ACBT T+ =
1212.56 NACBT = 1213 NACBT = W
(b) From (1): 0.137158(1212.56 N)CDT = 166.3 NCDT = W

56. 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
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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 25α = ° and 15β = ° 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
0: cos 15 cos 25 (80 N)cos 25 0x ACB ACBF T TΣ = ° − ° − ° =
1216.15 NACBT =
0: (1216.15 N)sin 15 (1216.15 N)sin 25yFΣ = ° + °
(80 N)sin 25 0
862.54 N
W
W
+ ° − =
=
(a) 863 NW = W
(b) 1216 NACBT = W

57. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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 α.
SOLUTION
Free-Body Diagram Force Triangle
(a) Law of cosines 2 2 2
(600) (750) 2(600)(750)cos(25 45 )P = + − ° + °
784.02 NP = 784 NP = W
(b) Law of sines
sin sin (25 45 )
600 N 784.02 N
β ° + °
=
46.0β = ° 46.0 25 71.0α = ° + ° = ° W

58. 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
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60
PROBLEM 2.58
For the situation described in Figure P2.47, determine (a) the
value of α for which the tension in rope BC is as small as
possible, (b) the corresponding value of the tension.
PROBLEM 2.47 Knowing that 20 ,α = ° determine the tension
(a) in cable AC, (b) in rope BC.
SOLUTION
Free-Body Diagram Force Triangle
To be smallest, BCT must be perpendicular to the direction of .ACT
(a) Thus, 5α = ° 5.00Į = ° W
(b) (1200 lb)sin 5BCT = ° 104.6 lbBCT = W

59. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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61
PROBLEM 2.59
For the structure and loading of Problem 2.48, determine (a) the value of α for
which the tension in cable BC is as small as possible, (b) the corresponding
value of the tension.
SOLUTION
BCT must be perpendicular to ACF to be as small as possible.
Free-Body Diagram: C Force Triangle is a right triangle
To be a minimum, BCT must be perpendicular to .ACF
(a) We observe: 90 30α = ° − ° 60.0α = ° W
(b) (300 lb)sin 50BCT = °
or 229.81lbBCT = 230 lbBCT = W

60. 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
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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
(For 725 N)T =
0: 2 1200 N 0y yF TΣ = − =
2 2 2
2 2 2
600 N
(600 N) (870 N)
630 N
y
x y
x
x
T
T T T
T
T
=
+ =
+ =
=
By similar triangles:
2.1 m
870 N 630 N
2.90 m
BC
BC
=
=
2( )
5.80 m
L BC=
= 5.80 mL = W

61. 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
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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 α.
SOLUTION
Free-Body Diagram: C Force Triangle
Force triangle is isosceles with
2 180 85
47.5
β
β
= ° − °
= °
(a) 2(800 N)cos 47.5° 1081 NP = =
Since 0,P Ͼ the solution is correct. 1081 NP = W
(b) 180 50 47.5 82.5α = ° − ° − ° = ° 82.5α = ° W

62. 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
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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 α.
SOLUTION
Free-Body Diagram Force Triangle
(a) Law of cosines: 2 2 2
(1200 N) (600 N) 2(1200 N)(600 N)cos 85
1294 N
P
P
= + − °
=
Since 1200 N,P Ͼ the solution is correct.
1294 NP = W
(b) Law of sines:
sin sin 85
1200 N 1294 N
67.5
180 50 67.5
β
β
α
°
=
= °
= ° − ° − ° 62.5α = ° W

63. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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65
PROBLEM 2.63
Collar A is connected as shown to a 50-lb load and can slide on
a frictionless horizontal rod. Determine the magnitude of the
force P required to maintain the equilibrium of the collar when
(a) 4.5 in.,x = (b) 15 in.x =
SOLUTION
(a) Free Body: Collar A Force Triangle
50 lb
4.5 20.5
P
= 10.98 lbP = W
(b) Free Body: Collar A Force Triangle
50 lb
15 25
P
= 30.0 lbP = W

64. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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66
PROBLEM 2.64
Collar A is connected as shown to a 50-lb load and can slide on a
frictionless horizontal rod. Determine the distance x for which the
collar is in equilibrium when P = 48 lb.
SOLUTION
Free Body: Collar A Force Triangle
2 2 2
(50) (48) 196
14.00 lb
N
N
= − =
=
Similar Triangles
48 lb
20 in. 14 lb
x
=
68.6 in.x = W

65. 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
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67
PROBLEM 2.65
A 160-kg load is supported by the rope-and-pulley arrangement shown.
Knowing that 20 ,β = ° 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
0: 2 sin 20 cos 0xF P P αΣ = ° − =
and cos 0.8452 or 46.840
46.840
α α
α
= = ± °
= +
For 0: 2 cos20 sin 46.840 1569.60 N 0yF P PΣ = ° + ° − =
or 602 N=P 46.8° W
For 46.840α = −
0: 2 cos20 sin( 46.840 ) 1569.60 N 0yF P PΣ = ° + − ° − =
or 1365 N=P 46.8° W

66. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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68
PROBLEM 2.66
A 160-kg load is supported by the rope-and-pulley arrangement shown.
Knowing that 40 ,α = ° determine (a) the angle β, (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.)
SOLUTION
Free-Body Diagram: Pulley A
(a) 0: 2 sin sin cos 40 0xF P PβΣ = − ° =
1
sin cos 40
2
22.52
β
β
= °
= °
22.5β = ° W
(b) 0: sin 40 2 cos 22.52 1569.60 N 0yF P PΣ = ° + ° − =
630 NP = W

67. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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69
PROBLEM 2.67
A 600-lb 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) 0: 2 (600 lb) 0
1
(600 lb)
2
yF T
T
Σ = − =
=
300 lbT = W
(b) 0: 2 (600 lb) 0
1
(600 lb)
2
yF T
T
Σ = − =
=
300 lbT = W
(c) 0: 3 (600 lb) 0
1
(600 lb)
3
yF T
T
Σ = − =
=
200 lbT = W
(d) 0: 3 (600 lb) 0
1
(600 lb)
3
yF T
T
Σ = − =
=
200 lbT = W
(e) 0: 4 (600 lb) 0
1
(600 lb)
4
yF T
T
Σ = − =
=
150.0 lbT = W

68. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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 600-lb 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) 0: 3 (600 lb) 0
1
(600 lb)
3
yF T
T
Σ = − =
=
200 lbT = W
(d) 0: 4 (600 lb) 0
1
(600 lb)
4
yF T
T
Σ = − =
=
150.0 lbT = W

69. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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71
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 750 N,P = determine
(a) the tension in cable ACB, (b) the magnitude of load Q.
SOLUTION
Free-Body Diagram: Pulley C
(a) 0: (cos25 cos55 ) (750 N)cos55° 0x ACBF TΣ = ° − ° − =
Hence: 1292.88 NACBT =
1293 NACBT = W
(b) 0: (sin 25 sin55 ) (750 N)sin55 0
(1292.88 N)(sin 25 sin55 ) (750 N)sin55 0
y ACBF T Q
Q
Σ = ° + ° + ° − =
° + ° + ° − =
or 2219.8 NQ = 2220 NQ = W

70. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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
0: (cos25 cos55 ) cos55 0x ACBF T PΣ = ° − ° − ° =
or 0.58010 ACBP T= (1)
0: (sin 25 sin55 ) sin55 1800 N 0y ACBF T PΣ = ° + ° + ° − =
or 1.24177 0.81915 1800 NACBT P+ = (2)
(a) Substitute Equation (1) into Equation (2):
1.24177 0.81915(0.58010 ) 1800 NACB ACBT T+ =
Hence: 1048.37 NACBT =
1048 NACBT = W
(b) Using (1), 0.58010(1048.37 N) 608.16 NP = =
608 NP = W

71. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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73
PROBLEM 2.71
Determine (a) the x, y, and z components of the 750-N force, (b) the
angles θx, θy, and θz that the force forms with the coordinate axes.
SOLUTION
sin 35
(750 N)sin 35
430.2 N
h
h
F F
F
= °
= °
=
(a)
cos 25x hF F= ° cos35yF F= ° sin 25z hF F= °
(430.2 N)cos25°= (750 N)cos 35°= (430.2 N)sin 25= °
390 N,xF = + 614 N,yF = + 181.8 NzF = + W
(b)
390 N
cos
750 N
x
x
F
F
θ
+
= = 58.7xθ = ° W
614 N
cos
750 N
y
y
F
F
θ
+
= = 35.0yθ = ° W
181.8 N
cos
750 N
z
z
F
F
θ
+
= = 76.0zθ = ° W

72. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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74
PROBLEM 2.72
Determine (a) the x, y, and z components of the 900-N force, (b) the
angles θx, θy, and θz that the force forms with the coordinate axes.
SOLUTION
cos 65
(900 N)cos 65
380.4 N
h
h
F F
F
= °
= °
=
(a)
sin 20x hF F= ° sin 65yF F= ° cos 20z hF F= °
(380.4 N)sin 20°= (900 N)sin 65°= (380.4 N)cos20= °
130.1 N,xF = − 816 N,yF = + 357 NzF = + W
(b)
130.1 N
cos
900 N
x
x
F
F
θ
−
= = 98.3xθ = ° W
816 N
cos
900 N
y
y
F
F
θ
+
= = 25.0yθ = ° W
357 N
cos
900 N
z
z
F
F
θ
+
= = 66.6zθ = ° W

73. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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
θx, θy, and θz that the force exerted at A forms with the coordinate axes.
SOLUTION
(a) sin30 sin50 110.3 N (Given)xF F= ° ° =
110.3 N
287.97 N
sin 30° sin 50°
F = = 288 NF = W
(b)
110.3 N
cos 0.38303
287.97 N
x
x
F
F
θ = = = 67.5xθ = ° W
cos30 249.39
249.39 N
cos 0.86603
287.97 N
y
y
y
F F
F
F
θ
= ° =
= = = 30.0yθ = ° W
sin30 cos50
(287.97 N)sin 30°cos 50°
92.552 N
zF F= − ° °
= −
= −
92.552 N
cos 0.32139
287.97 N
z
z
F
F
θ
−
= = = − 108.7zθ = ° W

74. 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
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you are using it without permission.
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
θx, θy, and θz that the force exerted at B forms with the coordinate axes.
SOLUTION
(a) sin30 sin 40 32.14 N (Given)zF F= − ° ° =
32.14
100.0 N
sin30 sin 40
F = =
° °
100.0 NF = W
(b) sin30 cos40
(100.0 N)sin 30°cos 40°
38.302 N
xF F= − ° °
= −
= −
38.302 N
cos 0.38302
100.0 N
x
x
F
F
θ = = = − 112.5xθ = ° W
cos30 86.603 NyF F= ° =
86.603 N
cos 0.86603
100 N
y
y
F
F
θ = = = 30.0yθ = ° W
32.14 NzF = −
32.14 N
cos 0.32140
100 N
z
z
F
F
θ
−
= = = − 108.7zθ = ° W

75. 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.
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 60 lb,
determine (a) the components of the force exerted by this wire
on the plate, (b) the angles , ,x yθ θ and zθ that the force forms
with the coordinate axes.
SOLUTION
(a) (60 lb)sin30°cos 60° 15 lbxF = − = − 15.00 lbxF = − W
(60 lb)cos30 51.96 lbyF = ° = 52.0 lbyF = + W
(60 lb)sin30 sin 60 25.98 lbzF = ° ° = 26.0 lbzF = + W
(b)
15.0 lb
cos 0.25
60 lb
x
x
F
F
θ
−
= = = − 104.5xθ = ° W
51.96 lb
cos 0.866
60 lb
y
y
F
F
θ = = = 30.0yθ = ° W
25.98 lb
cos 0.433
60 lb
z
z
F
F
θ = = = 64.3zθ = ° W

76. 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.
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 –20.0 lb, determine (a) the tension in wire CD, (b) the angles
θx, θy, and θz that the force exerted at C forms with the coordinate axes.
SOLUTION
(a) sin30 cos60 20 lb (Given)xF F= − ° ° = −
20 lb
80 lb
sin 30°cos60
F = =
°
80.0 lbF = W
(b)
20 lb
cos 0.25
80 lb
x
x
F
F
θ
−
= = = − 104.5xθ = ° W
(80 lb)cos30 69.282 lb
69.282 lb
cos 0.86615
80 lb
y
y
y
F
F
F
θ
= ° =
= = = 30.0yθ = ° W
(80 lb)sin30 sin 60 34.641lb
34.641
cos 0.43301
80
z
z
z
F
F
F
θ
= ° ° =
= = = 64.3zθ = ° W

77. 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.
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 120 lb, determine (a) the components of the force
exerted by this wire on the pole, (b) the angles θx, θy, and θz that the
force forms with the coordinate axes.
SOLUTION
(a) (120 lb)cos 60 cos 20xF = ° °
56.382 lbxF = 56.4 lbxF = + W
(120 lb)sin 60
103.923 lb
y
y
F
F
= − °
= − 103.9 lbyF = − W
(120 lb)cos 60 sin 20
20.521 lb
z
z
F
F
= − ° °
= − 20.5 lbzF = − W
(b)
56.382 lb
cos
120 lb
x
x
F
F
θ = = 62.0xθ = ° W
103.923 lb
cos
120 lb
y
y
F
F
θ
−
= = 150.0yθ = ° W
20.52 lb
cos
120 lb
z
z
F
F
θ
−
= = 99.8zθ = ° W

78. 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,
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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 85 lb, determine (a) the components of the force
exerted by this wire on the pole, (b) the angles θx, θy, and θz that the
force forms with the coordinate axes.
SOLUTION
(a) (85 lb)sin 36 sin 48xF = ° °
37.129 lb= 37.1 lbxF = W
(85 lb)cos 36
68.766 lb
yF = − °
= − 68.8 lbyF = − W
(85 lb)sin 36 cos 48
33.431 lb
zF = ° °
= 33.4 lbzF = W
(b)
37.129 lb
cos
85 lb
x
x
F
F
θ = = 64.1xθ = ° W
68.766 lb
cos
85 lb
y
y
F
F
θ
−
= = 144.0yθ = ° W
33.431 lb
cos
85 lb
z
z
F
F
θ = = 66.8zθ = ° W

79. 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.
81
PROBLEM 2.79
Determine the magnitude and direction of the force F = (320 N)i + (400 N)j − (250 N)k.
SOLUTION
2 2 2
2 2 2
(320 N) (400 N) ( 250 N)
x y zF F F F
F
= + +
= + + − 570 NF = W
320 N
cos
570 N
x
x
F
F
θ = = 55.8xθ = ° W
400 N
cos
570 N
y
y
F
F
θ = = 45.4yθ = ° W
250 N
cos
570 N
z
y
F
F
θ
−
= = 116.0zθ = ° W

80. 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.
82
PROBLEM 2.80
Determine the magnitude and direction of the force F = (240 N)i − (270 N)j + (680 N)k.
SOLUTION
2 2 2
2 2
(240 N) ( 270 N) (680 N)
x y zF F F F
F
= + +
= + − + 770 NF = W
240 N
cos
770 N
x
x
F
F
θ = = 71.8xθ = ° W
270 N
cos
770 N
y
y
F
F
θ
−
= = 110.5yθ = ° W
680 N
cos
770 N
z
z
F
F
θ = = 28.0zθ = ° W

81. 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.
83
PROBLEM 2.81
A force acts at the origin of a coordinate system in a direction defined by the angles θx = 70.9° and
θy = 144.9°. Knowing that the z component of the force is –52.0 lb, determine (a) the angle θz, (b) the other
components and the magnitude of the force.
SOLUTION
(a) We have
2 2 2 2 2 2
(cos ) (cos ) (cos ) 1 (cos ) 1 (cos ) (cos )x y z y y zθ θ θ θ θ θ+ + = Ÿ = − −
Since 0zF Ͻ we must have cos 0zθ Ͻ
Thus, taking the negative square root, from above, we have:
2 2
cos 1 (cos70.9 ) (cos144.9 ) 0.47282zθ = − − ° − ° = 118.2zθ = ° W
(b) Then:
52.0 lb
109.978 lb
cos 0.47282
z
z
F
F
θ
= = =
and cos (109.978 lb)cos70.9x xF F θ= = ° 36.0 lbxF = W
cos (109.978 lb)cos144.9y yF F θ= = ° 90.0 lbyF = − W
110.0 lbF = W

82. 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.
84
PROBLEM 2.82
A force acts at the origin of a coordinate system in a direction defined by the angles θy = 55° and θz = 45°.
Knowing that the x component of the force is − 500 lb, determine (a) the angle θx, (b) the other components
and the magnitude of the force.
SOLUTION
(a) We have
2 2 2 2 2 2
(cos ) (cos ) (cos ) 1 (cos ) 1 (cos ) (cos )x y z y y zθ θ θ θ θ θ+ + = Ÿ = − −
Since 0xF Ͻ we must have cos 0xθ Ͻ
Thus, taking the negative square root, from above, we have:
2 2
cos 1 (cos55) (cos45) 0.41353xθ = − − − = 114.4xθ = ° W
(b) Then:
500 lb
1209.10 lb
cos 0.41353
x
x
F
F
θ
= = = 1209 lbF = W
and cos (1209.10 lb)cos55y yF F θ= = ° 694 lbyF = W
cos (1209.10 lb)cos45z zF F θ= = ° 855 lbzF = W

83. 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.
85
PROBLEM 2.83
A force F of magnitude 210 N acts at the origin of a coordinate system. Knowing that Fx = 80 N, θz = 151.2°,
and Fy < 0, determine (a) the components Fy and Fz, (b) the angles θx and θy.
SOLUTION
(a) cos (210 N)cos151.2z zF F θ= = °
184.024 N= − 184.0 NzF = − W
Then: 2 2 2 2
x y zF F F F= + +
So: 2 2 2 2
(210 N) (80 N) ( ) (184.024 N)yF= + +
Hence: 2 2 2
(210 N) (80 N) (184.024 N)yF = − − −
61.929 N= − 62.0 lbyF = − W
(b)
80 N
cos 0.38095
210 N
x
x
F
F
θ = = = 67.6xθ = ° W
61.929 N
cos 0.29490
210 N
y
y
F
F
θ = = = − 107.2yθ = W

84. 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.
86
PROBLEM 2.84
A force F of magnitude 230 N acts at the origin of a coordinate system. Knowing that θx = 32.5°, Fy = − 60 N,
and Fz > 0, determine (a) the components Fx and Fz, (b) the angles θy and θz.
SOLUTION
(a) We have
cos (230 N)cos32.5x xF F θ= = ° 194.0 NxF = − W
Then: 193.980 NxF =
2 2 2 2
x y zF F F F= + +
So: 2 2 2 2
(230 N) (193.980 N) ( 60 N) zF= + − +
Hence: 2 2 2
(230 N) (193.980 N) ( 60 N)zF = + − − − 108.0 NzF = W
(b) 108.036 NzF =
60 N
cos 0.26087
230 N
y
y
F
F
θ
−
= = = − 105.1yθ = ° W
108.036 N
cos 0.46972
230 N
z
z
F
F
θ = = = 62.0zθ = ° W

85. 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.
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 525 lb, determine the
components of the force exerted by the wire on the bolt at B.
SOLUTION
2 2 2
(20 ft) (100 ft) (25 ft)
(20 ft) (100 ft) ( 25 ft)
105 ft
525 lb
[(20 ft) (100 ft) (25 ft) ]
105 ft
(100.0 lb) (500 lb) (125.0 lb)
BA
BA
BA
F
BA
F
BA
= + −
= + + −
=
=
=
= + −
= + −
i j k
F Ȝ
i j k
F i j k
JJJG
JJJG
100.0 lb, 500 lb, 125.0 lbx y zF F F= + = + = − W

86. 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.
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 315 lb, determine the
components of the force exerted by the wire on the bolt at D.
SOLUTION
2 2 2
(20 ft) (100 ft) (70 ft)
(20 ft) (100 ft) ( 70 ft)
126 ft
315 lb
[(20 ft) (100 ft) (74 ft) ]
126 ft
(50 lb) (250 lb) (185 lb)
DA
DA
DA
F
DA
F
DA
= + +
= + + +
=
=
=
= + +
= + +
i j k
F Ȝ
i j k
F i j k
JJJG
JJJG
50 lb, 250 lb, 185.0 lbx y zF F F= + = + = + W

87. 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.
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
2 2 2
(480 mm) (510 mm) (320 mm)
(480 mm) (510 mm ) (320 mm)
770 mm
385 N
[(480 mm) (510 mm) (320 mm) ]
770 mm
(240 N) (255 N) (160 N)
DB
DB
DB
F
DB
F
DB
= − +
= + +
=
=
=
= − +
= − +
i j k
F Ȝ
i j k
i j k
JJJG
JJJG
240 N, 255 N, 160.0 Nx y zF F F= + = − = + W

88. 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.
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
2 2 2
(270 mm) (400 mm) (600 mm)
(270 mm) (400 mm) (600 mm)
770 mm
385 N
[(270 mm) (400 mm) (600 mm) ]
770 mm
(135 N) (200 N) (300 N)
EB
EB
EB
F
EB
F
EB
= − +
= + +
=
=
=
= − +
= − +
i j k
F Ȝ
i j k
F i j k
JJJG
JJJG
135.0 N, 200 N, 300 Nx y zF F F= + = − = + W

89. 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.
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
2 2 2
(900 mm) (600 mm) (360 mm)
(900 mm) (600 mm) (360 mm)
1140 mm
1425 N
[ (900 mm) (600 mm) (360 mm) ]
1140 mm
(1125 N) (750 N) (450 N)
BA BA BA
BA
BA
BA
BA
T
BA
T
BA
= − + +
= + +
=
=
=
= − + +
= − + +
i j k
T Ȝ
T i j k
i j k
JJJG
JJJG
( ) 1125 N, ( ) 750 N, ( ) 450 NBA x BA y BA zT T T= − = = W

90. 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.
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
2 2 2
(900 mm) (600 mm) (920 mm)
(900 mm) (600 mm) (920 mm)
1420 mm
2130 N
[ (900 mm) (600 mm) (920 mm) ]
1420 mm
(1350 N) (900 N) (1380 N)
CA CA CA
CA
CA
CA
CA
T
CA
T
CA
λ
= − + −
= + +
=
=
=
= − + −
= − + −
i j k
T
T i j k
i j k
JJJG
JJJG
( ) 1350 N, ( ) 900 N, ( ) 1380 NCA x CA y CA zT T T= − = = − W

91. 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.
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
(300 N)[ cos30 sin15 sin30 cos30 cos15 ]
(67.243 N) (150 N) (250.95 N)
(400 N)[cos50 cos20 sin50 cos50 sin 20 ]
(400 N)[0.60402 0.76604 0.21985]
(241.61 N) (306.42 N) (87.939 N)
(174.
= − ° ° + ° + ° °
= − + +
= ° ° + ° − ° °
= + −
= + −
= +
=
P i j k
i j k
Q i j k
i j
i j k
R P Q
2 2 2
367 N) (456.42 N) (163.011 N)
(174.367 N) (456.42 N) (163.011 N)
515.07 N
R
+ +
= + +
=
i j k
515 NR = W
174.367 N
cos 0.33853
515.07 N
x
x
R
R
θ = = = 70.2xθ = ° W
456.42 N
cos 0.88613
515.07 N
y
y
R
R
θ = = = 27.6yθ = ° W
163.011 N
cos 0.31648
515.07 N
z
z
R
R
θ = = = 71.5zθ = ° W

92. 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.
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
(400 N)[ cos30 sin15 sin30 cos30 cos15 ]
(89.678 N) (200 N) (334.61 N)
(300 N)[cos50 cos20 sin50 cos50 sin 20 ]
(181.21 N) (229.81 N) (65.954 N)
(91.532 N) (429.81 N) (268.66 N)
(91.5R
= − ° ° + ° + ° °
= − + +
= ° ° + ° − ° °
= + −
= +
= + +
=
P i j k
i j k
Q i j k
i j k
R P Q
i j k
2 2 2
32 N) (429.81 N) (268.66 N)
515.07 N
+ +
= 515 NR = W
91.532 N
cos 0.177708
515.07 N
x
x
R
R
θ = = = 79.8xθ = ° W
429.81 N
cos 0.83447
515.07 N
y
y
R
R
θ = = = 33.4yθ = ° W
268.66 N
cos 0.52160
515.07 N
z
z
R
R
θ = = = 58.6zθ = ° W

93. 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.
95
PROBLEM 2.93
Knowing that the tension is 425 lb in cable AB and 510 lb in
cable AC, determine the magnitude and direction of the resultant
of the forces exerted at A by the two cables.
SOLUTION
2 2 2
2 2 2
(40 in.) (45 in.) (60 in.)
(40 in.) (45 in.) (60 in.) 85 in.
(100 in.) (45 in.) (60 in.)
(100 in.) (45 in.) (60 in.) 125 in.
(40 in.) (45 in.) (60 in.)
(425 lb)
85 in.
AB AB AB AB
AB
AB
AC
AC
AB
T T
AB
= − +
= + + =
= − +
= + + =
− +
= = =
i j k
i j k
i j k
T Ȝ
JJJG
JJJG
JJJG
(200 lb) (225 lb) (300 lb)
(100 in.) (45 in.) (60 in.)
(510 lb)
125 in.
(408 lb) (183.6 lb) (244.8 lb)
(608) (408.6 lb) (544.8 lb)
AB
AC AC AC AC
AC
AB AC
AC
T T
AC
ª º
« »
¬ ¼
= − +
ª º− +
= = = « »
¬ ¼
= − +
= + = − +
T i j k
i j k
T Ȝ
T i j k
R T T i j k
JJJG
Then: 912.92 lbR = 913 lbR = W
and
608 lb
cos 0.66599
912.92 lb
xθ = = 48.2xθ = ° W
408.6 lb
cos 0.44757
912.92 lb
yθ = = − 116.6yθ = ° W
544.8 lb
cos 0.59677
912.92 lb
zθ = = 53.4zθ = ° W

94. 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.
96
PROBLEM 2.94
Knowing that the tension is 510 lb in cable AB and 425 lb in cable
AC, determine the magnitude and direction of the resultant of the
forces exerted at A by the two cables.
SOLUTION
2 2 2
2 2 2
(40 in.) (45 in.) (60 in.)
(40 in.) (45 in.) (60 in.) 85 in.
(100 in.) (45 in.) (60 in.)
(100 in.) (45 in.) (60 in.) 125 in.
(40 in.) (45 in.) (60 in.)
(510 lb)
85 in.
AB AB AB AB
AB
AB
AC
AC
AB
T T
AB
= − +
= + + =
= − +
= + + =
− +
= = =
i j k
i j k
i j k
T Ȝ
JJJG
JJJG
JJJG
(240 lb) (270 lb) (360 lb)
(100 in.) (45 in.) (60 in.)
(425 lb)
125 in.
(340 lb) (153 lb) (204 lb)
(580 lb) (423 lb) (564 lb)
AB
AC AC AC AC
AC
AB AC
AC
T T
AC
ª º
« »
¬ ¼
= − +
ª º− +
= = = « »
¬ ¼
= − +
= + = − +
T i j k
i j k
T Ȝ
T i j k
R T T i j k
JJJG
Then: 912.92 lbR = 913 lbR = W
and
580 lb
cos 0.63532
912.92 lb
xθ = = 50.6xθ = ° W
423 lb
cos 0.46335
912.92 lb
yθ
−
= = − 117.6yθ = ° W
564 lb
cos 0.61780
912.92 lb
zθ = = 51.8zθ = ° W

95. 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.
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
2 2 2
(480 mm) (510 mm) (320 mm)
(480 mm) (510 mm) (320 mm) 770 mm
BD
BD
= − + −
= + + =
i j k
JJJG
(385 N)
[ (480 mm) (510 mm) (320 mm) ]
(770 mm)
(240 N) (255 N) (160 N)
BD BD BD BD
BD
T T
BD
= =
= − + −
= − + −
F Ȝ
i j k
i j k
JJJG
2 2 2
(270 mm) (400 mm) (600 mm)
(270 mm) (400 mm) (600 mm) 770 mm
BE
BE
= − + −
= + + =
i j k
JJJG
(385 N)
[ (270 mm) (400 mm) (600 mm) ]
(770 mm)
(135 N) (200 N) (300 N)
BE BE BE BE
BE
T T
BE
= =
= − + −
= − + −
F Ȝ
i j k
i j k
JJJG
(375 N) (455 N) (460 N)BD BE= + = − + −R F F i j k
2 2 2
(375 N) (455 N) (460 N) 747.83 NR = + + = 748 NR = W
375 N
cos
747.83 N
xθ
−
= 120.1xθ = ° W
455 N
cos
747.83 N
yθ = 52.5yθ = ° W
460 N
cos
747.83 N
zθ
−
= 128.0zθ = ° W

96. 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.
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
(use results of Problem 2.89)
( ) 1125 N ( ) 750 N ( ) 450 N
(use results of Problem 2.90)
( ) 1350 N ( ) 900 N ( ) 1380 N
AB BA
AB x AB y AB z
AC CA
AC x AC y AC z
T T
T T T
T T
T T T
= −
= + = − = −
= −
= + = − = +
Resultant:
2 2 2
2 2 2
1125 1350 2475 N
750 900 1650 N
450 1380 930 N
( 2475) ( 1650) ( 930)
x x
y y
z z
x y z
R F
R F
R F
R R R R
= Σ = + + = +
= Σ = − − = −
= Σ = − + = +
= + +
= + + − + +
3116.6 N= 3120 NR = W
2475
cos
3116.6
x
x
R
R
θ
+
= = 37.4xθ = ° W
1650
cos
3116.6
y
y
R
R
θ
−
= = 122.0yθ = ° W
930
cos
3116.6
z
z
R
R
θ
+
= = 72.6zθ = ° W

97. 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.
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 150 lb 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
(150 lb)(cos 60 cos 20 sin 60 cos 60 sin 20 )
(sin 36 sin 48 cos 36 sin 36 cos 48 )
AC AD
ADT
= +
= ° ° − ° − ° °
+ ° ° − ° + ° °
R T T
i j k
i j k (1)
(a) Since 0,zR = The coefficient of k must be zero.
(150 lb)( cos 60 sin 20 ) (sin 36 cos 48 ) 0
65.220 lb
AD
AD
T
T
− ° ° + ° ° =
= 65.2 lbADT = W
(b) Substituting for ADT into Eq. (1) gives:
[(150 lb) cos 60 cos 20 (65.220 lb)sin 36 sin 48 )]
[(150 lb)sin 60 (65.220 lb)cos 36 ] 0
= ° ° + ° °
− ° + ° +
R i
j
2 2
(98.966 lb) (182.668 lb)
(98.966 lb) (182.668 lb)
207.76 lb
R
= −
= +
=
R i j
208 lbR = W
98.966 lb
cos
207.76 lb
xθ = 61.6xθ = ° W
182.668 lb
cos
207.76 lb
yθ = 151.6yθ = ° W
cos 0zθ = 90.0zθ = ° W

98. 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.
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 125 lb 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
(cos 60 cos 20 sin 60 cos 60 sin 20 )
(125 lb)(sin 36 sin 48 cos 36 sin 36 cos 48 )
AC AD
ACT
= +
= ° ° − ° − ° °
+ ° ° − ° + ° °
R T T
i j k
i j k (1)
(a) Since 0,zR = The coefficient of k must be zero.
( cos 60 sin 20 ) (125 lb)(sin 36 cos 48 ) 0
287.49 lb
AC
AC
T
T
− ° ° + ° ° =
= 287 lbACT = W
(b) Substituting for ACT into Eq. (1) gives:
[(287.49 lb) cos 60 cos 20 (125 lb)sin 36 sin 48 ]
[(287.49 lb)sin 60 (125 lb)cos 36 ] 0
= ° ° + ° °
− ° + ° +
R i
j
2 2
(189.677 lb) (350.10 lb)
(189.677 lb) (350.10 lb)
398.18 lb
R
= −
= +
=
R i j
398 lbR = W
189.677 lb
cos
398.18 lb
xθ = 61.6xθ = ° W
350.10 lb
cos
398.18 lb
yθ = 151.6yθ = ° W
cos 0zθ = 90.0zθ = ° W