CHAPTER 32
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 N, a = 75.1∞ R = 179 N
75.1∞ 
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3
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.
75 N
P
20°
35°
Q
R
Rx = –P sin 20º + Q sin 35º = 0
fi P sin 20º = Q sin 35º
Q=
P sin 20∞
sin 20∞
= 75 N
sin 35∞
sin 35∞
Q = 44.7 N b
(b)
Magnitude of R:
ur
R = P cos 20∞ + Q cos 35∞
= 107.093 N
R = 107.1 N
R = 107.1 N 
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4
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
a = 50.2∞
b = 56.3∞
We measure:
(a)
Parallelogram law:
(b)
Triangle rule:
We measure:
R = 684 N,
g = 66.5º
R = 684 N
66.5º 
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5
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:
R = 3.30 kN, a = 66.6∞ R = 3.30 kN
66.6∞ 
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6
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 b
sin 60∞
=
1200 N 1500 N
sin b = 0.69282
b = 43.854∞
a + b + 60∞ = 180∞
a = 180∞ - 60∞ - 43.854∞
= 76.146∞
(b)
Law of sines:
Fbb¢
1500 N
=
sin 76.146∞ sin 60∞
a = 76.1∞ 
Fbb¢ = 1682 N 
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7
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 a
sin 60∞
=
600 N 1500 N
sin a = 0.34641
a = 20.268∞ (b)
a = 20.3∞ 
a + b + 60∞ = 180∞
b = 180∞ - 60∞ - 20.268∞
= 99.732∞
Faa¢
1500 N
=
sin 99.732∞ sin 60∞
Faa¢ = 1707 N 
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8
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 a sin 25∞
=
50 N
35 N
sin a = 0.60374
a = 37.138∞ (b)
a = 37.1∞ 
a + b + 25∞ = 180∞
b = 180∞ - 25∞ - 37.138∞
= 117.86∞
R
35 N =
sin 117.86 sin 25∞
R = 73.2 N 
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9
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
(a)
Parallelogram law:
We measure:
4 kN
30°
R = 5.4 kN a = 12º
R
a
R = 5.4 kN
12º 
R = 5.4 kN
12º 
25°
2 kN
(b)
Triangle rule:
We measure:
25°
R = 5.4 kN
2 kN
4 kN
30°
D
R
a = 12º
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10
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)
(b)
1600 N
P
=
sin 25∞ sin 75∞
P = 3660 N 
25∞ + b + 75∞ = 180∞
b = 180∞ - 25∞ - 75∞
= 80∞
1600 N
R
=
sin 25∞ sin 80∞
R = 3730 N 
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11
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 2 = (1600 N)2 + (2500 N)2 - 2(1600 N)(2500 N) cos 75∞
P = 2596 N
Using the law of sines:
sin a
sin 75∞
=
1600 N 2596 N
a = 36.5∞
P is directed 90∞ - 36.5∞ or 53.5∞ below the horizontal.
P = 2600 N
53.5∞ 
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12
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
Using the triangle rule and the law of sines:
(a)
(b)
b + 50∞ + 60∞ = 180∞
b = 180∞ - 50∞ - 60∞
= 70∞
2125 N
P
=
sin 70∞ sin 60∞
P = 1958 N 
2125 N
R
=
sin 70∞ sin 50∞
R = 1732 N 
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13
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
Using the triangle rule and the law of sines:
(a)
(a + 30∞) + 60∞ + b
b
b
sin (90∞ - a )
2125 N
= 180∞
= 180∞ - (a + 30∞) - 60∞
= 90∞ - a
sin 60∞
=
3000 N
90∞ - a = 37.8∞
(b)
3000 N
R
=
sin (52.2∞ + 30∞) sin 60∞
a = 52.2∞ 
R = 3430 N 
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14
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 = (50 N) sin 25∞ P = 21.1 N Ø 
(b)
R = (50 N) cos 25∞ R = 45.3 N 
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15
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 = (2125 N) cos 30∞ P = 1840 N Æ 
(b)
R = (2125 N) sin 30∞ R = 1062.5 N 
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16
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∞ + a = 180∞
a = 125∞
R2 = P 2 + Q 2 - 2 PQ cos a
R2 = (75 N )2 + (125 N)2
- 2(75 N)(125 N) cos125∞
R2 = 32004.56
R = 178.898 N
Using the law of sines:
sin 125∞
sin b
=
125 N 178.898 N
b = 34.915∞
70∞ + b = 104.915∞ R = 178.9 N
104.9º 
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17
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 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
2.5
3
a = 39.81∞
2
tan b =
3
b = 33.69∞
tan a =
Using the triangle rule:
Using the law of cosines:
Using the law of sines:
a + b + y = 180∞
39.81∞ + 33.69∞ + y = 180∞
y = 106.5∞
R2 = (600 N )2 + (200 N )2 - 2(600 N )(200 N ) cos106.5∞
R = 684.22 N
sin 106.5∞
sin g
=
200 N 684.22 N
g = 16.28∞
f = (90∞ - a ) + g
f = (90∞ - 39.81∞) + 16.28∞
f = 66.47∞
R = 684 N
66.5º 
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18
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:
Using the law of sines:
R2 = (2 kN )2 + (3 kN )2
- 2(2 kN )(3 kN ) cos 80∞
R = 3.304 kN
sin 80∞
sin g
=
2 kN 3.304 kN
g = 36.59∞
b + g + 80∞ = 180∞
g = 180∞ - 80∞ - 36.59∞
g = 63.41∞
f = 180∞ - b + 50∞
f = 66.59∞
R = 3.30 kN
66.6∞ 
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19
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
Then
g = 180∞ - ( 40∞ + 20∞)
= 120∞
R2 = (15 kN )2 + (10 kN )2
- 2(15 kN )(10 kN ) cos120∞
= 475 kN 2
R = 21.794 kN
and
10 kN 21.794 kN
=
sin 120∞
sin a
Ê 10 kN ˆ
sin a = Á
sin 120∞
Ë 21.794 kN ˜¯
= 0.39737
a = 23.414
Hence:
f = a + 50∞ = 73.414
R = 21.8 kN
73.4° 
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20
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
Then
g = 180∞ - ( 40∞ + 20∞)
= 120∞
R2 = (10 kN )2 + (15 kN )2
- 2(10 kN )(15 kN ) cos120∞
= 475 kN 2
R = 21.794 kN
and
15 kN 21.794 kN
=
sin 120∞
sin a
Ê 15 kN ˆ
sin a = Á
sin 120∞
Ë 21.794 kN ˜¯
= 0.59605
a = 36.588∞
Hence:
f = a + 50∞ = 86.588∞ R = 21.8 kN
86.6° b
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21
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
Then
b = 180∞ - (50∞ + 25∞)
= 105∞
R2 = (75 N )2 + (50 N )2
- 2(75 N )(50 N ) cos 105∞
R2 = 10066.1 N 2
R = 100.330 N
and
sin g
sin 105∞
=
75 N 100.330 N
sin g = 0.72206
g = 46.225∞
Hence:
g - 25∞ = 46.225∞ - 25∞ = 21.225∞
R = 100.3 N
21.2∞ 
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22
PROBLEM 2.21
Determine the x and y components of each of the forces shown.
Solution
Compute the following distances:
OA = (600)2 + (800)2
= 1000 mm
OB = (560)2 + (900)2
= 1060 mm
OC = ( 480)2 + (900)2
= 1020 mm
800-N Force:
424-N Force:
408-N Force:
Fx = +(800 N )
800
1000
Fx = +640 N 
Fy = +(800 N )
600
1000
Fy = +480 N 
Fx = -( 424 N )
560
1060
Fx = -224 N 
Fy = -( 424 N )
900
1060
Fy = -360 N 
Fx = +( 408 N )
480
1020
Fx = +192.0 N 
Fy = -( 408 N )
900
1020
Fy = -360 N 
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23
PROBLEM 2.22
Determine the x and y components of each of the forces shown.
Solution
Compute the following distances:
OA = (2100)2 + (2000)2
= 2900 mm
OB = (700)2 + (2400)2
= 2500 mm
OC = (2250)2 + (1200)2
= 2550 mm
145-N Force:
250-N Force:
255-N Force:
Fx = +(145 N )
2100
2900
Fx = +105 N 
Fy = +(145 N )
2000
2900
Fy = 100 N 
Fx = -(250 N )
700
2500
Fx = -70 N 
Fy = +(250 N )
2400
2500
Fy = 240 N 
Fx = +(255 N )
1200
2550
Fx = +120.0 N 
Fy = -(255 N )
2250 2550
Fy = -225 N 
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24
PROBLEM 2.23
Determine the x and y components of each of the forces shown.
Solution
200-N Force:
250-N Force:
300-N Force:
Fx = +(200 N ) cos 60∞ Fx = 100.0 N 
Fy = -(200 N )sin 60∞
Fy = -173.2 N 
Fx = -(250 N )sin 50∞
Fx = -191.5 N 
Fy = -(250 N ) cos 50∞
Fy = -160.7 N 
Fx = +(300 N ) cos 25∞ Fx = 272 N 
Fy = +(300 N )sin 25∞ Fy = 126.8 N 
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25
PROBLEM 2.24
Determine the x and y components of each of the forces shown.
Solution
80-N Force:
120-N Force:
150-N Force:
Fx = +(80 N ) cos 40∞ Fx = 61.3 N 
Fy = +(80 N )sin 40∞ Fy = 51.4 N 
Fx = +(120 N ) cos 70∞ Fx = 41.0 N 
Fy = +(120 N )sin 70∞ Fy = 112.8 N 
Fx = -(150 N ) cos 35∞
Fx = -122. 9 N 
Fy = +(150 N )sin 35∞ Fy = 86.0 N 
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26
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
P sin 35∞ = 1500 N
(a)
P=
(b)
Vertical component
1500 N
sin 35∞
P = 2620 N 
Pv = P cos 35∞
= (2615.17) cos 35∞ Pv = 2140 N 
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27
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 = P sin 20∞
P = 2193 N (b)
P = 2190 N 
PABC = P cos 20∞
= (2193 N ) cos 20∞ PABC = 2060 N 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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28
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)
Px
sin 38∞
120 N
=
sin 38∞
P=
= 194.91 N (b)
or
P = 194.9 N 
Px
tan 38∞
120 N
=
tan 38∞
Py =
= 153.59 N or
Py = 153.6 N 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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29
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)
P=
Py
cos 38∞
180 N
=
cos 38∞
= 228.4 N (b)
P = 228 N 
Px = Py tan 38∞
= (180 N ) tan 38∞
= 140.63 N Px = 140.6 N 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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30
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)
Px = P sin 55∞
Px
sin 55∞
1200 N
=
sin 55∞
= 1464.9 N P=
(b)
P = 1465 N 
Px = Py tan 55∞
Px
tan 55∞
1200 N
=
tan 55∞
= 840.2 N Py =
Py = 840 N Ø 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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31
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)
P=
Py
cos 55∞
1750 N
cos 55∞
= 3051 N =
(b)
P = 3050 N 
Px = P sin 55∞
= (3051 N )sin 55∞
= 2499.2 N
Px = 2500 N 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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32
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 = Rx i + Ry j
= (155 N)i + (115 N) j
tan a =
Ry
Rx
115
155
a = 36.573∞
=
R = (155)2 + (115)2
= 193.0 N
R = 193.0 N
36.6∞ 
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33
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
+741.0
+112.8
150 N
-122.9
+86.0
Rx = -20.6
Ry = +250.2
R = Rx i + Ry j
= ( -20.6 N )i + (250.2 N ) j
Ry
tan a =
Rx
250.2 N
20.6 N
tan a = 12.1456
a = 85.293∞
tan a =
R=
250.2 N
sin 85.293∞
R = 251 N
85.3∞ 
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34
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. (N)
y Comp. (N)
200 N
+100
-173.2
250 N
-191.5
-160.7
300 N
+272
+126.8
Rx = +180.5
Ry = -207.1
R = Rx i + Ry j
= +(180.5 N )i - (207.1 N ) j
Ry
tan a =
Rx
207.1 N
180.5 N
tan a = 1.147
a = 48.9∞
tan a =
R = (180.5)2 + ( -207.1)2 = 274.72 N R = 275 N
48.9∞ 
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35
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 N
+640
+480
424 N
-224
-360
408 N
+192
-360
Rx = +608
Ry = -240
R = Rx i + Ry j
= (608 N )i + ( -240 N ) j
tan a =
Ry
Rx
240
608
a = 21.541∞
=
240 N
sin(21.541∞)
= 653.65 N R=
R = 654 N
21.5∞ 
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36
PROBLEM 2.35
Knowing that a = 35°, determine the resultant of the three forces
shown.
Solution
100-N Force:
Fx = +(100 N ) cos 35∞ = +81.915 N
Fy = -(100 N )sin 35∞ = -57.358 N
150-N Force:
Fx = +(150 N) cos 65∞ = +63.393 N
Fy = -(150 N )sin 65∞ = -135.946 N
200-N Force:
Fx = -(200 N ) cos 35∞ = -163.830 N
Fy = -(200 N )sin 35∞ = -114.715 N
Force
x Comp. (N)
100 N
+81.915
-57.358
150 N
+63.393
-135.946
200 N
-163.830
-114.715
Rx = -18.522
y Comp. (N)
Ry = -308.02
R = Rx i + Ry j
= ( -18.522 N )i + ( -308.02 N ) j
tan a =
Ry
Rx
308.02
18.522
a = 86.559∞
=
R=
308.02 N
sin 86.559
R = 309 N
86.6∞ 
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37
y
F1 =80 N
PROBLEM 2.36
20°
F2 = 150 N
Four forces act on bolt A as shown. Determine the resultant of the
forces on the bolt.
30°
A
15°
x
F4 = 100 N
F3 = 110 N
Solution
The x and y components of each force are determined by trigonometry as shown and are entered in the table
below.
(F2 cos 20°) j
Force Magnitude N x Component N
(F1 sin 30°)
(F1 cos 30°)
(F2 sin 20°) i
(F4 cos 15°) i
– (F4 sin 15°) j
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
– F3j
Thus the resultant R of the four forces is
R = (199.1 N)i + (14.3 N)j b
R = Rxi+Ryj
The magnitude and direction of the resultant may now be determined. From the triangle shown, we have
a
Ry = (14.3 N)j
R
Rx = (199.1 N) i
tan a =
Ry
Rx
=
R=
14.3 N
199.1 N
a = 4.1∞
14.3 N
= 199.6 N sin a
R = 199.6 N
4.1º b
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38
PROBLEM 2.37
Knowing that a = 40°, determine the resultant of the three forces
shown.
Solution
300-N Force:
Fx = (300 N ) cos 20∞ = 281.91 N
Fy = (300 N )sin 20∞ = 102.61 N
400-N Force:
Fx = ( 400 N ) cos 60∞ = 200.0 N
Fy = ( 400 N )sin 60∞ = 346.41 N
600-N Force:
Fx = (600 N ) cos 30∞ = 519.62 N
Fy = -(600 N )sin 30∞ = -300.0 N
and
Rx = SFx = 1001.53 N
Ry = SFy = 149.02 N
R = (1001.53)2 + (149.02)2
= 1012.56 N
Further:
tan a =
149.02
= 0.14879
1001.53
a = tan -1 0.14879
= 8.463∞
R = 1013 N
8.46∞ b
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39
PROBLEM 2.38
Knowing that a = 75°, determine the resultant of the forces shown.
Solution
300-N Force:
Fx = (300 N ) cos 20∞ = 281.91 N
Fy = (300 N )sin 20∞ = 102.61 N
400-N Force:
Fx = ( 400 N ) cos 95 ∞ = -34.862 N
Fy = ( 400 N )sin 95∞ = 398.48 N
600-N Force:
Fx = (600 N ) cos 5 ∞ = 597.72 N
Fy = (600 N )sin 5∞ = 52.293 N
Then
Rx = SFx = 844.77 N
Ry = SFy = 553.38 N
and
R = (844.77)2 + (553.38)2
= 1009.88 N
553.38
844.77
tan a = 0.655
a = 33.23∞ tan a =
R = 1010 N
33.2∞ b
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40
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
Rx = SFx
= (100 N ) cos a + (150 N ) cos (a + 30∞) - (200 N ) cos a
Rx = -(100 N ) cos a + (150 N ) cos (a + 30∞)
(1)
Ry = SFy
= -(100 N )sin a - (150 N )sin (a + 30∞) - (200 N )sin a
Ry = -(300 N )sin a - (150 N )sin (a + 30∞)
(a)
(2)
For R to be vertical, we must have Rx = 0. We make Rx = 0 in Eq. (1):
-100 cos a + 150 cos (a + 30∞) = 0
-100 cos a + 150 (cos a cos 30∞ - sin a sin 30∞) = 0
29.904 cos a = 75 sin a
29.904
75
= 0.3988
a = 21.74∞ tan a =
(b)
a = 21.7∞ b
Substituting for a in Eq. (2):
Ry = -300 sin 21.74∞ - 150 sin 51.74∞
= -228.9 N
R = | Ry | = 228.9 N R = 229 N b
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41
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
Rx = SFx = Rx = -
21
TBC + 420 N
29
Ry = SFy =
Ry =
(a)
(1)
800
5
4
TBC - (780 N ) - (500 N )
1160
13
5
20
TBC - 700 N
29
(2)
For R to be vertical, we must have Rx = 0
Set Rx = 0 in Eq. (1)
(b)
840
12
3
TBC + (780 N ) - (500 N )
1160
13
5
-
21
TBC + 420 N = 0
29
TBC = 580 N b
Substituting for TBC in Eq. (2):
20
(580 N ) - 700 N
29
Ry = -300 N
Ry =
R = | Ry | = 300 N R = 300 N b
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42
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:
Rx = SFx = TAC sin 10∞ + (250 N ) cos 35∞ + (375 N ) cos 60∞
= TAC sin 10∞ + 392.29 N
(1)
(2)
Ry = SFy = (250 N )sin 35∞ + (375 N )sin 60∞ - TAC cos10∞
Ry = 468.15 N - TAC cos10∞
(a)
Set Ry = 0 in Eq. (2):
468.15 N - TAC cos10∞ = 0
TAC = 475 N b
TAC = 475.37 N
(b)
Substituting for TAC in Eq. (1):
Rx = ( 475.37 N )sin 10∞ + 392.29 N
= 474.84 N
R = Rx
R = 475 N b
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43
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
Rx = SFx = (300 N ) + ( 400 N ) cos a + (600 N )sin a (1)
Ry = SFy = ( 400 N )sin a - (600 N ) cos a (2)
and
(a)
Set Ry = 0 in Eq. (2).
( 400 N )sin a - (600 N ) cos a = 0
Dividing each term by cosa gives:
( 400 N ) tan a = 600 N
600 N
tana =
400 N
a = 56.310∞ (b)
a = 56.3∞ b
Substituting for a in Eq. (1) gives:
Rx = 300 N + ( 400 N ) cos 56.31∞ + (600 N )sin 56.31∞ = 1021.11 N Rx = 1021 N b
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44
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
Law of sines:
Force Triangle
TAC
T
1962 N
= BC =
sin 70∞ sin 50∞ sin 60∞
(a)
TAC =
1962 N
sin 70∞ = 2128.9 N sin 60∞
TAC = 2.13 kN b
(b)
TBC =
1962 N
sin 50∞ = 1735.49 N sin 60∞
TBC = 1.735 kN b
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45
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
Law of sines:
Force Triangle
TAC
T
500 N
= BC =
sin 60∞ sin 40∞ sin 80∞
(a)
TAC =
500 N
sin 60∞ = 439.69 N sin 80∞
TAC = 440 N b
(b)
TBC =
500 N
sin 40∞ = 326.35 N sin 80∞
TBC = 326 N b
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46
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
Free-Body Diagram
Law of sines:
Force Triangle
TAC
T
500 N
= BC =
sin 35∞ sin 75∞ sin 70∞
(a)
TAC =
500 N
sin 35∞ sin 70∞
TAC = 305 N b
(b)
TBC =
500 N
sin 75∞ sin 70∞
TBC = 514 N b
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47
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
W = mg
= (200 kg)(9.81 m/s2 )
= 1962 N
Law of sines:
TAC
TBC
1962 N
=
=
sin 15∞ sin 105∞ sin 60∞
(a)
TAC =
(1962 N) sin 15∞
sin 60∞
TAC = 586 N b
(b)
TBC =
(1962 N) sin 105∞
sin 60∞
TBC = 2190 N b
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48
PROBLEM 2.47
Knowing that a = 20∞, determine the tension (a) in cable AC,
(b) in rope BC.
Solution
Free-Body Diagram
Law of sines:
Force Triangle
TAC
T
6000 N
= BC =
sin 110∞ sin 5∞ sin 65∞
(a)
TAC =
6000 N
sin 110∞ sin 65∞
(b)
TBC =
6000 N
sin 5∞ sin 65∞
TAC = 6220 N b
TBC = 577 N b
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49
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
Free-Body Diagram
Law of sines:
Force Triangle
FAC
T
1500 N
= BC =
sin 35∞ sin 50∞ sin 95∞
(a)
FAC =
1500 N
sin 35∞ sin 95∞
FAC = 864 N b
(b)
TBC =
1500 N
sin 50∞ sin 95∞
TBC = 1153 N b
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50
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
Free-Body Diagram
Resolving the forces into x- and y-directions:
R = P + Q + FA + FB = 0
Substituting components:
R = -(2500 N ) j + [(3250 N ) cos 50∞]i
- [(3250 N )sin 50∞]j
+ FB i - ( FA cos 50∞)i + ( FA sin 50∞) j = 0
In the y-direction (one unknown force)
-2500 N - (3250 N )sin 50∞ + FA sin 50∞ = 0
Thus,
FA =
2500 N + (3250 N )sin 50∞
sin 50∞
= 6513.5 N In the x-direction:
Thus,
FA = 6510 N b
(3250 N ) cos 50∞ + FB - FA cos 50∞ = 0
FB = FA cos 50∞ - (3250 N ) cos 50∞
= (6513.5 N ) cos 50∞ - (3250 N ) cos 50∞
= 2097.7 N FB = 2100 N b
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51
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.
Free-Body Diagram
Solution
Resolving the forces into x- and y-directions:
R = P + Q + FA + FB = 0
Substituting components:
R = - Pj + Q cos 50∞i - Q sin 50∞ j
- [(3750 N) cos 50∞]i
+ [(3750 N) sin 50∞]j + (2000 N)i
In the x-direction (one unknown force)
Q cos 50∞ - [(3750 N) cos 50∞] + 2000 N = 0
(3750 N) cos 50∞ - 2000 N
cos 50∞
= 638.55 N
Q=
In the y-direction:
- P - Q sin 50∞ + (3750 N) sin 50∞ = 0
P = -Q sin 50∞ + (3750 N) sin 50∞
= -(638.55 N) sin 50∞ + (3750 N) sin 50∞
= 2383.5 N
P = 2380 N; Q = 639 N b
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52
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
SFx = 0 :
With
3
3
FB - FC - FA = 0
5
5
FA = 8 kN
FB = 16 kN
FC =
4
4
(16 kN) - (8 kN) 5
5
FC = 6.40 kN 
3
3
S Fy = 0 : - FD + FB - FA = 0
5
5
With FA and FB as above:
3
3
FD = (16 kN) - (8 kN)
5
5
FD = 4.80 kN b
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53
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
or
With
3
3
SFy = 0 : - FD - FA + FB = 0
5
5
3
FB = FD + FA
5
FA = 5 kN, FD = 8 kN
5È
3
˘
FB = Í6 kN + (5 kN ) ˙ 3Î
5
˚
4
4
SFx = 0 : - FC + FB - FA = 0
5
5
4
FC = ( FB - FA )
5
4
= (15 kN - 5 kN ) 5
FB = 15.00 kN b
FC = 8.00 kN b
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54
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
SFy = 0 : TCA - Q cos 30∞ = 0
With
Q = 300 N
(a)
TCA = (300 N )(0.866) (b)
SFx = 0 : P - TCB - Q sin 30∞ = 0
With
TCA = 260 N b
P = 375 N
TCB = 375 N - (300 N )(0.50) or TCB = 225 N b
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55
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
SFx = 0 : -TBC - Q cos 60∞ + 375 N = 0
TBC = 375 N - Q cos 60∞ (1)
SFy = 0 : TAC - Q sin 60∞ = 0
TAC = Q sin 60∞ Requirement
(2)
TAC  300 N
Q sin 60∞  300 N
From Eq. (2):
Q  346.41 N
Requirement
TBC  300 N
375 N - Q cos 60∞  300 N
From Eq. (1):
Q 150 N 150.0 N  Q  346 N b
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56
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
+
SFx = 0 : TACB cos 10∞ - TACB cos 30∞ - TCD cos 30∞ = 0
TCD = 0.137158TACB +
SFy = 0 : TACB sin 10∞ + TACB sin 30∞ + TCD sin 30∞ - 900 = 0
0.67365TACB + 0.5TCD = 900 (a)
Substitute (1) into (2):
From (1):
(2)
0.67365 TACB + 0.5(0.137158 TACB ) = 900
TACB = 1212.56 N (b)
(1)
TCD = 0.137158 (1212.56 N) TACB = 1213 N b
TCD = 166.3 N b
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57
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
+
SFx = 0 : TACB cos 15∞ - TACB cos 25∞ - (80 N) cos 25∞ = 0
TACB = 1216.15 N
+
SFy = 0 : (1216.15 N) sin 15∞ + (1216.15 N) sin 25∞
+ (80 N) sin 25∞ - W = 0
W = 862.54 N
(a)
W = 863 N b
(b) TACB = 1216 N b
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58
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
Free-Body Diagram
(a)
Law of cosines
Force Triangle
P 2 = (600)2 + (750)2 - 2(600)(750) cos (25∞ + 45∞)
P = 784.02 N (b)
Law of sines
P = 784 N b
sin b
sin (25∞ + 45∞)
=
600 N
784.02 N
b = 46.0∞ a = 46.0∞ + 25∞ = 71.0∞ b
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59
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
Free-Body Diagram
Force Triangle
To be smallest, TBC must be perpendicular to the direction of TAC .
(a)
(b)
Thus,
a = 5∞ TBC = (6000 N) sin 5∞
a = 5.00∞
b
TBC = 523 N b
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60
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 = (1500 N )sin 50∞
(b)
or
TBC = 1149.07 N TBC = 1149 N b
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61
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 T = 725 N)
+
SFy = 0 : 2Ty - 1200 N = 0
Ty = 600 N
Tx2 + Ty2 = T 2
Tx2 + (600 N )2 = (870 N )2
Tx = 630 N
By similar triangles:
2.1 m
BC
=
870 N 630 N
BC = 2.90 m
L = 2( BC )
= 5.80 m L = 5.80 m b
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62
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 with
2b = 180∞ - 85∞
b = 47.5∞
P = 2(800 N)cos 47.5∞ = 1081 N
(a)
P = 1081 N b
Since P  0, the solution is correct.
(b)
a = 180∞ - 50∞ - 47.5∞ = 82.5∞ a = 82.5∞ b
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63
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
Free-Body Diagram
Force Triangle
P 2 = (1200 N)2 + (600 N)2 - 2(1200 N)(600 N) cos 85∞
P = 1294 N
Since P 1200 N, the solution is correct.
(a)
Law of cosines:
(b)
Law of sines:
P = 1294 N b
sin b
sin 85∞
=
1200 N 1294 N
b = 67.5∞
a = 180∞ - 50∞ - 67.5∞ a = 62.5∞ b
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64
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
250 N
=
125 515.39
P = 60.6 N b
P = 60.634 N
(b)
Free Body: Collar A
Force Triangle
P
250 N
=
375
625
P = 150.0 N b
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65
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
Similar Triangles
Force Triangle
N 2 = (250)2 - (240)2 = 4900
N = 70.0 N
x
240 N
=
500 mm 70 N
x = 1714 mm b
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66
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
SFx = 0 : 2P sin 20∞ - P cosa = 0
+
and
For
cos a = 0.8452 or a = ± 46.840∞
a = + 46.840
+
SFy = 0 : 2P cos 20∞ + P sin 46.840∞ - 1569.60 N = 0
or P = 602 N
46.8∞ b
a = -46.840
For
+
SFy = 0 : 2P cos 20∞ + P sin( -46.840∞) - 1569.60 N = 0
or P = 1365 N
46.8∞ b
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67
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.)
Solution
Free-Body Diagram: Pulley A
(a)
SFx = 0 : 2 P sin sin b - P cos 40∞ = 0
1
sin b = cos 40∞
2
b = 22.52∞
b = 22.5∞ b
(b)
SFy = 0 : P sin 40∞ + 2 P cos 22.52∞ - 1569.60 N = 0
P = 630 N b
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68
PROBLEM 2.67
A 3000 N crate is supported by several rope-andpulley 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)
+
SFy = 0 : 2T - (3000 N) = 0
1
T = (3000 N)
2
T = 1500 N b
(b)
+
SFy = 0 : 2T - (3000 N) = 0
1
T = (3000 N)
2
T = 1500 N b
(c)
+
SFy = 0 : 3T - (3000 N) = 0
1
T = (3000 N)
3
T = 1000 N b
(d)
+
SFy = 0 : 3T - (3000 N) = 0
1
T = (3000 N)
3
T = 1000 N b
(e)
+
SFy = 0 : 4T - (3000 N) = 0
T=
1
(3000 N)
4
T = 750 N b
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69
PROBLEM 2.68
Solve Parts b and d of Problem 2.67, assuming that
the free end of the rope is attached to the crate.
PROBLEM 2.67 A 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)
+
SFy = 0 : 3T - (3000 N) = 0
1
T = (3000 N)
3
T = 1000 N b
(d)
+
SFy = 0 : 4T - (3000 N) = 0
T=
1
(3000 N)
4
T = 750 N b
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70
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.
Solution
Free-Body Diagram: Pulley C
(a)
+
SFx = 0 : TACB (cos 25∞ - cos 55∞) - (750 N ) cos 55∞ = 0
TACB = 1292.88 N
Hence:
TACB = 1293 N b
(b)
+
SFy = 0 : TACB (sin 25∞ + sin 55∞) + (750 N) sin 55∞ - Q = 0
(1292.88 N )(sin 25∞ + sin 55∞) + (750 N )sin 55∞ - Q = 0
Q = 2219.8 N or
Q = 2220 N b
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71
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
+
SFx = 0 : TACB (cos 25∞ - cos 55∞) - P cos 55∞ = 0
P = 0.58010TACB or
+
SFy = 0 : TACB (sin 25∞ + sin 55∞) + P sin 55∞ - 1800 N = 0
1.24177TACB + 0.81915P = 1800 N (2)
or
(a)
(1)
Substitute Equation (1) into Equation (2):
1.24177TACB + 0.81915(0.58010TACB ) = 1800 N
TACB = 1048.37 N
Hence:
TACB = 1048 N b
(b)
P = 0.58010(1048.37 N ) = 608.16 N
Using (1),
P = 608 N b
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72
PROBLEM 2.71
Determine (a) the x, y, and z components of the 750-N force, (b) the angles
qx, qy, and qz that the force forms with the coordinate axes.
Solution
Fh = F sin 35∞
= (750 N) sin 35∞
Fh = 430.2 N
(a)
Fx = Fh cos 25∞
Fy = F cos 35∞
= ( 430.2 N) cos 25∞
Fx = +390 N,
(b)
Fz = Fh sin 25∞
= (750 N) cos 35∞
Fy = +614 N,
cos q x =
cos q y =
cos q z =
Fz = +181.8 N Fx +390 N
=
F
750 N
Fy
F
=
= ( 430.2 N )sin 25∞
+614 N
750 N
Fz +181.8 N
=
750 N
F
b
q x = 58.7∞ b
q y = 35.0∞ b
q z = 76.0∞ b
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73
PROBLEM 2.72
Determine (a) the x, y, and z components of the 900-N force, (b) the angles
qx, qy, and qz that the force forms with the coordinate axes.
Solution
Fh = F cos 65∞
= (900 N) cos 65∞
Fh = 380.4 N
(a)
Fx = Fh sin 20∞
= (380.4 N)sin 20∞
Fx = -130.1 N,
(b)
Fy = F sin 65∞
= (900 N )sin 65∞
Fy = +816 N,
cos q x =
cos q y =
cos q z =
Fx -130.1 N
=
F
900 N
Fy
Fz = Fh cos 20∞
= (380.4 N ) cos 20∞
Fz = +357 N b
q x = 98.3∞ b
+816 N
900 N
q y = 25.0∞ b
Fz +357 N
=
900 N
F
q z = 66.6∞ b
F
=
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74
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
qx, qy, and qz that the force exerted at A forms with the coordinate axes.
Solution
(a)
Fx = F sin 30∞ sin 50∞ = 110.3 N (Given)
F=
(b)
cos q x =
110.3 N
= 287.97 N sin 30∞ sin 50∞
F = 288 N b
Fx
110.3 N
=
= 0.38303 F 287.97 N
q x = 67.5∞ b
Fy = F cos 30∞ = 249.39
cos q y =
Fy
F
=
249.39 N
= 0.86603 287.97 N
q y = 30.0∞ b
Fz = - F sin 30∞ cos 50∞
= -(287.97 N) sin 30∞cos 50∞
= -92.552 N
cos q z =
Fz -92.552 N
=
= -0.32139 F
287.97 N
q z = 108.7∞ b
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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75
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
qx, qy, and qz that the force exerted at B forms with the coordinate axes.
Solution
(a)
Fz = - F sin 30∞ sin 40∞ = 32.14 N (Given )
F=
(b)
32.14
= 100.0 N sin 30∞ sin 40∞
F = 100.0 N b
Fx = - F sin 30∞ cos 40∞
= -(100.0 N) sin 30∞cos 40∞
= -38.302 N
cos q x =
Fx 38.302 N
=
= -0.38302 F
100.0 N
q x = 112.5∞ b
Fy = F cos 30∞ = 86.603 N
Fy
86.603 N
=
= 0.86603 F
100 N
Fz = -32.14 N
cos q y =
cos q z =
Fz -32.14 N
=
= -0.32140 F
100 N
q y = 30.0∞ b
q z = 108.7∞ b
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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76
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 x , q y , and q z that the force forms with the coordinate axes.
Solution
(a)
Fx = -(300 N )sin 30∞cos 60∞ = -75 N Fx = -75.0 N b
Fy = (300 N ) cos 30∞ = 259.81 N Fy = +260 N b
Fz = (300 N )sin 30∞ sin 60∞ = 129.9 N (b)
cos q x =
cos q y =
cos q z =
Fx -75 N
=
= -0.25 F 300 N
Fy
F
=
259.81 N
= 0.866 300 N
Fz 129.9 N
=
= 0.433 F
300 N
Fz = +129.9 N b
q x = 104.5∞ b
q y = 30.0∞ b
q z = 64.3∞ b
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77
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
qx, qy, and qz that the force exerted at C forms with the coordinate axes.
Solution
(a)
Fx = - F sin 30∞ cos 60∞ = -100 N (Given)
F=
(b)
cos q x =
100 N
= 400 N sin 30∞cos 60∞
F = 400 N b
Fx -100 N
=
= - 0.25 F
400 N
q x = 104.5∞ b
Fy = ( 400 N ) cos 30∞ = 346.41 N
cos q y =
Fy
F
=
346.41 N
= 0.866
400 N
Fz = ( 400 N )sin 30∞ sin 60∞ = 173.21 N
F 173.21 N
cos q z = z =
= 0.43301
400 N
F
q y = 30.0∞ b
q z = 64.3∞ b
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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78
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 qx, qy, and qz that the
force forms with the coordinate axes.
Solution
(a)
Fx = (600 N) cos 60∞ cos 20∞
Fx = 281.91 N Fx = +282 N b
Fy = -(600 N) sin 60∞
Fy = -519.62 N
Fz = -(600 N) cos 60∞ sin 20∞
Fz = -102.61 N
(b)
Fx 281.91 N
=
F
600 N
Fy -519.62 N
=
cos q y =
F
600 N
F
-102.61 N
cos q z = z =
F
600 N
cos q x =
Fy = -520 N b
Fz = -102.6 N b
q x = 62.0∞ b
q y = 150.0∞ b
q z = 99.8∞ b
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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79
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 qx, qy, and qz that the
force forms with the coordinate axes.
Solution
(a)
Fx = ( 425 N) sin 36∞ sin 48∞
= 185.64 N Fx = 185.6 N b
Fy = -( 425 N) cos 36∞
= -343.83 N
Fz = ( 425 N) sin 36∞ cos 48∞
= 167.15 N
(b)
Fx 185.64 N
=
F
425 N
Fy -343.83 N
=
cos q y =
F
425 N
F 167.15 N
cos q z = z =
F
425 N
cos q x =
Fy = -344 N b
Fz = 167.2 N b
q x = 64.1∞ b
q y = 144.0∞ b
q z = 66.8∞ b
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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80
PROBLEM 2.79
Determine the magnitude and direction of the force F = (320 N)i + (400 N)j = (250 N)k.
Solution
F = Fx2 + Fy2 + Fz2
F = (320 N)2 + ( 400 N)2 + ( -250 N)2 Fx 320 N
=
F 570 N
Fy 400 N
=
cos q y =
F 570 N
F
-250 N
cos q z = z =
570 N
F
cos q x =
F = 570 N b
q x = 55.8∞ b
q y = 45.4∞ b
q z = 116.0∞ b
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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81
PROBLEM 2.80
Determine the magnitude and direction of the force F = (240 N)i 2 (270 N)j + (680 N)k.
Solution
F = Fx2 + Fy2 + Fz2
F = (240 N)2 + ( -270 N)2 + (680 N)2 cos q x =
cos q y =
cos q z =
Fx 240 N
=
F 770 N
Fy
F
=
F = 770 N b
q x = 71.8∞ b
-270 N
770 N
Fz 680 N
=
F 770 N
q y = 110.5∞ b
q z = 28.0∞ b
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82
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 qz, (b) the other
components and the magnitude of the force.
Solution
(a)
We have
(cos q x )2 + (cos q y )2 + (cos q z )2 = 1 fi (cos q y )2 = 1 - (cos q y )2 - (cos q z )2
Since Fz  0 we must have cosq z  0
Thus, taking the negative square root, from above, we have:
cos q z = - 1 - (cos 70.9∞)2 - (cos144.9∞)2 = 0.47282 (b)
q z = 118.2∞ b
Then:
F=
and
Fz
260 N
=
= 549.89 N
cos q z 0.47282
Fx = F cos q x = (549.89 N ) cos 70.9∞ Fx = 179.9 N b
Fy = F cos q y = (549.89 N ) cos144.9∞ Fy = -450 N b
F = 550 N b
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83
PROBLEM 2.82
A force acts at the origin of a coordinate system in a direction defined by the angles qy = 55° and qz = 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 q x )2 + (cos q y )2 + (cos q z )2 = 1 fi (cos q y )2 = 1 - (cos q y )2 - (cos q z )2
Since Fx  0 we must have cosq x  0
Thus, taking the negative square root, from above, we have:
cos q x = - 1 - (cos 55)2 - (cos 45)2 = 0.41353 (b)
q x = 114.4∞ b
Then:
Fx
2500 N
=
= 6045.5 N cos q x 0.41353
Fy = F cos q y = (6045.5 N ) cos 55∞ Fy = 3470 N b
Fz = F cos q z = (6045.5 N ) cos 45∞ Fz = 4270 N b
F=
and
F = 6050 N b
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84
PROBLEM 2.83
A force F of magnitude 210 N acts at the origin of a coordinate system. Knowing that Fx = 80 N, qz = 151.2°,
and Fy < 0, determine (a) the components Fy and Fz, (b) the angles qx and qy.
Solution
Fz = F cos q z = (210 N ) cos151.2∞
(a)
= -184.024 N Then:
So:
Hence:
F 2 = Fx2 + Fy2 + Fz2
(210 N )2 = (80 N )2 + ( Fy )2 + (184.024 N )2
Fy = - (210 N )2 - (80 N )2 - (184.024 N )2
= -61.929 N (b)
Fz = -184.0 N b
Fy = -62.0 N b
Fx
80 N
=
= 0.38095 F 210 N
Fy 61.929 N
=
= -0.29490 cos q y =
F
210 N
cos q x =
q x = 67.6∞ b
q y = 107.2 b
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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85
PROBLEM 2.84
A force F of magnitude 230 N acts at the origin of a coordinate system. Knowing that qx = 32.5°, Fy = 2 60 N,
and Fz > 0, determine (a) the components Fx and Fz, (b) the angles qy and qz.
Solution
(a)
We have
Fx = F cos q x = (230 N ) cos 32.5∞ Then:
Fx = -194.0 N b
Fx = 193.980 N
F 2 = Fx2 + Fy2 + Fz2
So:
Hence:
(b)
(230 N )2 = (193.980 N )2 + ( -60 N )2 + Fz2
Fz = + (230 N )2 - (193.980 N )2 - ( -60 N )2 Fz = 108.0 N b
Fz = 108.036 N
Fy
-60 N
= - 0.26087 F 230 N
F 108.036 N
cos q z = z =
= 0.46972 F
230 N
cos q y =
=
q y = 105.1∞ b
q z = 62.0∞ b
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86
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
uuur
BA = (7 m)i + (34 m) j - (8 m)k
BA = (7)2 + (34)2 + (8)2
= 35.623 m
F = F l BA
uuur
BA
=F
BA
=
2625 N
[(7 m)i + (34 m) j - (8 m)k ]
35.623 m
F = (515.82 N )i + (2505.4 N ) j - (589.51 N )k
Fx = +516 N, Fy = +2510 N, Fz = -590 N b
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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87
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
uuur
DA = (7 m)i + (34 m) j + (25 m)k
DA = (7)2 + (34)2 + (25)2
= 42.778 m
F = F l DA
uuur
DA
=F
DA
=
1575 N
[(7 m)i + (34 m) j + (25 m)k ]
42.778 m
F = (257.73 N )i + (1251.81 N ) j + (920.45 N )k
Fx = +258 N, Fy = +1252 N, Fz = +920 N b
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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88
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
uuur
DB = ( 480 mm)i - (510 mm) j + (320 mm)k
DB = ( 480 mm)2 + (510 mm2 ) + (320 mm)2
= 770 mm
F = F l DB
uuur
DB
=F
DB
=
385 N
[( 480 mm)i - (510 mm)j + (320 mm)k ]
770 mm
= (240 N)i - (255 N ) j + (160 N )k
Fx = +240 N, Fy = -255 N, Fz = +160.0 N b
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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89
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
uuur
EB = (270 mm)i - ( 400 mm) j + (600 mm)k
EB = (270 mm)2 + ( 400 mm)2 + (600 mm)2
= 770 mm
F = F l EB
uuur
EB
=F
EB
=
385 N
[(270 mm)i - ( 400 mm)j + (600 mm)k ]
770 mm
F = (135 N)i - (200 N ) j + (300 N )k
Fx = +135.0 N, Fy = -200 N, Fz = +300 N b
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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90
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
uuur
BA = -(900 mm)i + (600 mm) j + (360 mm)k
BA = (900 mm)2 + (600 mm)2 + (360 mm)2
= 1140 mm
TBA = TBA l BA
uuur
BA
= TBA
BA
TBA =
1425 N
[-(900 mm)i + (600 mm) j + (360 mm)k ]
1140 mm
= -(1125 N)i + (750 N ) j + ( 450 N)k
(TBA ) x = -1125 N, (TBA ) y = 750 N, (TBA ) z = 450 N b
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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91
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
uuur
CA = -(900 mm)i + (600 mm) j - (920 mm)k
CA = (900 mm)2 + (600 mm)2 + (920 mm)2
= 1420 mm
TCA = TCA lCA
uuur
CA
= TCA
CA
TCA =
2130 N
[-(900 mm)i + (600 mm) j - (920 mm)k ]
1420 mm
= -(1350 N)i + (900 N ) j - (1380 N)k
(TCA ) x = -1350 N, (TCA ) y = 900 N, (TCA ) z = -1380 N b
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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92
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 = (300 N )[- cos 30∞ sin 15∞i + sin 30∞ j + cos 30∞ cos15∞k ]
= - (67.243 N )i + (150 N ) j + (250.95 N )k
Q = ( 400 N )[cos 50∞ cos 20∞i + sin 50∞ j - cos 50∞ sin 20∞k ]
= ( 400 N )[0.60402i + 0.76604 j - 0.21985k ]
= (241.61 N )i + (306.42 N ) j - (87.939 N )k
R =P+Q
= (174.367 N )i + ( 456.42 N ) j + (163.011 N )k
R = (174.367 N )2 + ( 456.42 N )2 + (163.011 N )2
= 515.07 N
cos q x =
cos q y =
cos q z =
Rx 174.367 N
=
= 0.33853 R
515.07 N
Ry
R = 515 N b
q x = 70.2∞ b
456.42 N
= 0.88613 515.07 N
q y = 27.6∞ b
Rz 163.011 N
=
= 0.31648 R
515.07 N
q z = 71.5∞ b
R
=
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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93
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 = ( 400 N )[- cos 30∞ sin 15∞i + sin 30∞ j + cos 30∞ cos15∞k ]
= - (89.678 N )i + (200 N ) j + (334.61 N )k
Q = (300 N )[cos 50∞ cos 20∞i + sin 50∞ j - cos 50∞ sin 20∞k ]
= (181.21 N )i + (229.81 N)j - (65.954 N)k
R =P+Q
= (91.532 N )i + ( 429.81 N ) j + (268.66 N )k
R = (91.532 N )2 + ( 429.81 N )2 + (268.66 N )2
= 515.07 N
cos q x =
cos q y =
cos q z =
Rx 91.532 N
=
= 0.177708 R 515.07 N
Ry
R = 515 N b
q x = 79.8∞ b
429.81 N
= 0.83447 515.07 N
q y = 33.4∞ b
Rz 268.66 N
=
= 0.52160 R 515.07 N
q z = 58.6∞ b
R
=
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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94
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
uuur
AB = (1000 mm)i - (1125 mm) j + (1500 mm)k
AB = (1000)2 + (1125)2 + (1500)2 = 2125 mm
uuur
AC = (2500 mm)i - (1125 mm) j + (1500 mm)k
AC = (2500)2 + (1125)2 + (1500)2 = 3125 mm
uuur
È (1000 mm)i - (1125 mm) j + (1500 mm)k ˘
AB
TAB = TAB l AB = TAB
= (2125 N ) Í
˙
AB
2125 mm
Î
˚
TAB = (1000 N )i - (1125 N ) j + (1500 N )k
uuur
È (2500 mm)i - (1125 mm) j + (1500 mm)k ˘
AC
TAC = TAC l AC = TAC
= (2550 N ) Í
˙
AC
3125 mm
Î
˚
TAC = (2040 N )i - (918 N ) j + (1224 N )k
R = TABB + TAC = (3040 N )i - (2043 N ) j + (2724 N )k = 4564.61 N
Then:
and
R = 4564.6 N R = 4560 N b
cos q x =
3040 N
= 0.66599 4564.61 N
cos q y =
2043 N
= -0.44757 4564.61 N
cos q z =
2724 N
= 0.59677 4564.61 N
q x = 48.2∞ b
q y = 116.6∞ b
q z = 53.4∞ b
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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95
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
uuur
AB = (1000 mm)i - (1125 mm) j + (1500 mm)k
AB = (1005)2 + (1125)2 + (1500)2 = 2125 mm
uuur
AC = (2500 mm)i - (1125 mm) j + (1500 mm)k
AC = (2500)2 + (1125)2 + (1500)2 = 3125 mm
uuur
È (1000 mm)i - (1125 mm) j + (1500 mm)k ˘
AB
TAB = TAB l AB = TAB
= (2550 N ) Í
˙
AB
2125 mm
Î
˚
TAB = (1200 N )i - (1350 N ) j + (1800 N )k
uuur
È (2500 mm)i - (1125 mm) j + (1500 mm)k ˘
AC
TAC = TAC l AC = TAC
= (2125 N ) Í
˙
AC
3125 mm
Î
˚
TAC = (1700 N )i - (765 N ) j + (1020 N )k
R = TABB + TAC = (2900 N )i - (2115 N ) j + (2820 N )k
Then:
and
R = 4564.6 N R = 4560 N b
cos q x =
2900 N
= 0.63532 4564.6
cos q y =
-2115 N
= -0.46335 4564.6 N
cos q z =
2820 N
= 0.61780 4564.6 N
q x = 50.6∞ b
q y = 117.6∞ b
q z = 51.8∞ b
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96
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
uuur
BD = -( 480 mm)i + (510 mm) j - (320 mm)k
BD = ( 480 mm)2 + (510 mm)2 + (320 mm)2 = 770 mm
uuur
BD
FBD = TBD l BD = TBD
BD
(385 N )
[-( 480 mm)i + (510 mm) j - (320 mm)k ]
=
(770 mm)
= -(240 N )i + (255 N ) j - (160 N )k
uuur
BE = -(270 mm)i + ( 400 mm) j - (600 mm)k
BE = (270 mm)2 + ( 400 mm)2 + (600 mm)2 = 770 mm
uuur
BE
FBE = TBE l BE = TBE
BE
(385 N )
[-(270 mm)i + ( 400 mm) j - (600 mm)k ]
=
(770 mm)
= -(135 N )i + (200 N ) j - (300 N )k
R = FBD + FBE = -(375 N )i + ( 455 N ) j - ( 460 N )k
R = (375 N )2 + ( 455 N)2 + ( 460 N )2 = 747.83 N R = 748 N b
cos q x =
-375 N
747.83 N
q x = 120.1∞ b
cos q y =
455 N
747.83 N
q y = 52.5∞ b
cos q z =
-460 N
747.83 N
q z = 128.0∞ b
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97
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
TAB = -TBA
( use results of Problem 2.89)
(TAB ) x = +1125 N (TAB ) y = -750 N (TAB ) z = - 450 N
TAC = -TCA
( use results of Problem 2.90)
(TAC ) x = +1350 N (TAC ) y = -900 N (TAC ) z = +1380 N
Resultant:
Rx = SFx = +1125 + 1350 = +2475 N
Ry = SFy = -750 - 900 = -1650 N
Rz = SFz = -450 + 1380 = + 930 N
R = Rx2 + Ry2 + Rz2
= ( +2475)2 + ( -1650)2 + ( +930)2
= 3116.6 N cos q x =
cos q y =
cos q z =
R = 3120 N b
Rx +2475
=
R 3116.6
Ry
q x = 37.4∞ b
-1650
3116.6
q y = 122.0∞ b
Rz
+ 930
=
R 3116.6
q z = 72.6∞ b
R
=
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98
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 = TAC + TAD
= (750 N )(cos 60∞ cos 20∞i - sin 60∞ j - cos 60∞ sin 20∞k )
+ TAD (sin 36∞ sin 48∞i - cos 36∞ j + sin 36∞ cos 48∞k ) (a)
Since Rz = 0, The coefficient of k must be zero.
(750 N )( - cos 60∞ sin 20∞) + TAD (sin 36∞ cos 48∞) = 0
TAD = 326.1 N (b)
(1)
TAD = 326 N b
Substituting for TAD into Eq. (1) gives:
R = [(750 N ) cos 60∞ cos 20∞ + (326.1 N )sin 36∞ sin 48∞)]i
- [(750 N )sin 60∞ + (326.1 N ) cos 36∞]j + 0
R = ( 494.83 N )i - (913.34 N ) j
R = ( 494.83)2 + (913.34)2
= 1038.77 N
494.83 N
1038.77 N
-913.34 N
cos q y =
1038.76 N
cos q z = 0 cos q x =
R = 1039 N b
q x = 61.6∞ b
q y = 151.6∞ b
q z = 90.0∞ b
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99
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 = TAC + TAD
= TAC (cos 60∞ cos 20∞i - sin 60∞ j - cos 60∞ sin 20∞k )
+ (625 N )(sin 36∞ sin 48∞i - cos 36∞ j + sin 36∞ cos 48∞k ) (a)
(1)
Since Rz = 0, The coefficient of k must be zero.
TAC ( - cos 60∞ sin 20∞) + (625 N )(sin 36∞ cos 48∞) = 0
TAC = 1437.43 N (b)
TAC = 1437 N b
Substituting for TAC into Eq. (1) gives:
R = [(1437.43 N ) cos 60∞ cos 20∞ + (625 N )sin 36∞ sin 48∞]i
- [(1437.43 N )sin 60∞ + (625 N ) cos 36∞]j + 0
R = (948.377 N )i - (1750.49 N ) j
R = (948.377)2 + (1750.49)2
= 1990.89 N
cos q x =
948.377
1990.89
R = 1991 N b
q x = 61.6∞ b
-1750.49 N
1990.89
cos q z = 0 cos q y =
q y = 151.5∞ b
q z = 90.0∞ b
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100
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:
TAB , TAC , TAD , and P
where P = Pj . To express the other forces in terms of the unit vectors i, j, k, we write
uuur
AB = - ( 4.20 m)i - (5.60 m) j
AB = 7.00 m
uuur
AC = (2.40 m)i - (5.60 m) j + ( 4.20 m)k AC = 7.40 m
uuur
AD = - (5.60 m) j - (3.30 m)k
AD = 6.50 m
uuur
AB
and
TAB = TAB l AB = TAB
= ( - 0.6i - 0.8 j)TAB
AB
uuur
AC
TAC = TAC l AC = TAC
= (0.32432 - 0.75676 j + 0.56757k )TAC
AC
uuur
AD
TAD = TAD l AD = TAD
= ( - 0.86154 j - 0.50769k )TAD
AD
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101
Problem 2.99 (Continued)
Equilibrium condition
SF = 0 : TAB + TAC + TAD + Pj = 0
Substituting the expressions obtained for TAB , TAC , and TAD and factoring i, j, and k:
( - 0.6TAB + 0.32432TAC )i + ( -0.8TAB - 0.75676TAC - 0.86154TAD + P ) j
+ (0.56757TAC - 0.50769TAD )k = 0
Equating to zero the coefficients of i, j, k:
- 0.6TAB + 0.32432TAC = 0 (1)
- 0.8TAB - 0.75676TAC - 0.86154TAD + P = 0 (2)
0.56757TAC - 0.50769TAD = 0 (3)
Setting TAB = 259 N in (1) and (2), and solving the resulting set of equations gives
TAC = 479.15 N
TAD = 535.66 N P = 1031 N ≠ b
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102
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:
- 0.6TAB + 0.32432TAC = 0 (1)
- 0.8TAB - 0.75676TAC - 0.86154TAD + P = 0 (2)
0.56757TAC - 0.50769TAD = 0 (3)
Substituting TAC = 444 N in Equations (1), (2), and (3) above, and solving the resulting set of
equations using conventional algorithms gives
TAB = 240 N
TAD = 496.36 N P = 956 N ≠ b
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103
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).
- 0.6TAB + 0.32432TAC = 0 (1)
- 0.8TAB - 0.75676TAC - 0.86154TAD + P = 0 (2)
0.56757TAC - 0.50769TAD = 0 (3)
Substituting TAD = 481 N in Equations (1), (2), and (3) above, and solving the resulting set of equations using
conventional algorithms gives
TAC = 430.26 N
TAB = 232.57 N P = 926 N ≠ b
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104
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).
- 0.6TAB + 0.32432TAC = 0 (1)
- 0.8TAB - 0.75676TAC - 0.86154TAD + P = 0 (2)
0.56757TAC - 0.50769TAD = 0 (3)
From Eq. (1)
TAB = 0.54053TAC
From Eq. (3)
TAD = 1.11795TAC
Substituting for TAB and TAD in terms of TAC into Eq. (2) gives:
- 0.8(0.54053TAC ) - 0.75676TAC - 0.86154(1.11795TAC ) + P = 0
2.1523TAC = P ; P = 800 N
800 N
2.1523
= 371.69 N
gives:
TAC =
Substituting into expressions for TAB and TAD
TAB = 0.54053(371.69 N )
TAD = 1.11795(371.69 N )
TAB = 201 N, TAC = 372 N, TAD = 416 N b
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105
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:
TAB , TAC , TAD and W
where P = Pj . To express the other forces in terms of the unit vectors i, j, k, we write
uuur
AB = - (900 mm)i + (1500 mm) j - (675 mm)k
AB = 1875 mm
uuur
AC = (1500 mm) j + (800 mm)k
AC = 1700 mm
uuur
AD = (1000 mm)i + (1500 mm) j - (675 mm)k
AD = 1925 mm
uuur
AB
and
TAB = TAB l AB = TAB
AB
= ( - 0.48i + 0.8 j - 0.36k )TAB
uuur
AC
TAC = TAC l AC = TAC
AC
= (0.88235 j + 0.47059k )TAC
uuur
AD
TAD = TAD l AD = TAD
AD
= (0.51948i + 0.77922 j - 0.35065k )TAD
Equilibrium Condition with W = -Wj
SF = 0 : TAB + TAC + TAD - Wj = 0
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106
Problem 2.103 (Continued)
Substituting the expressions obtained for TAB , TAC , and TAD and factoring i, j, and k:
( - 0.48TAB + 0.51948TAD )i + (0.8TAB + 0.88235TAC + 0.77922TAD - W ) j
+ ( - 0.36TAB + 0.47059TAC - 0.35065TAD )k = 0
Equating to zero the coefficients of i, j, k:
0.8TAB
-0.48TAB + 0.51948TAD = 0
+ 0.88235TAC + 0.77922TAD - W = 0
- 0.36TAB + 0.47059TAC - 0.35065TAD = 0
Substituting TAB = 3750 N in Equations (1), (2), and (3) and solving the resulting set of equations, using
conventional algorithms for solving linear algebraic equations, gives:
TAC = 5450.24 N
TAD = 3465 N
W = 10509 N
W = 10510 N b
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107
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:
0.8TAB
- 0.48TAB + 0.51948TAD = 0
+ 0.88235TAC + 0.77922TAD - W = 0
- 0.36TAB + 0.47059TAC - 0.35065TAD = 0
Substituting TAD = 3080 N in Equations (1), (2), and (3) above, and solving the resulting set of equations
using conventional algorithms, gives:
TAB = 3333.33 N
TAC = 4844.66 N
W = 9341.35 N
W = 9340 N b
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108
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
See Problem 2.103 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3)
below:
0.8TAB
- 0.48TAB + 0.51948TAD = 0
+ 0.88235TAC + 0.77922TAD - W = 0
- 0.36TAB + 0.47059TAC - 0.35065TAD = 0
Substituting TAC = 2720 N in Equations (1), (2), and (3) above, and solving the resulting set of equations
using conventional algorithms, gives:
TAB = 1871.35 N
TAD = 1729.1 N
W = 5244.4 N
W = 5240 N b
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109
PROBLEM 2.106
A 8000-N crate is supported by three cables as shown. Determine
the tension in each cable.
Solution
See Problem 2.103 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3)
below:
0.8TAB
-0.48TAB + 0.51948TAD = 0
+ 0.88235TAC + 0.77922TAD - W = 0
-0.36TAB + 0.47059TAC - 0.35065TAD = 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
TAC = 4150 N b
TAD = 2637.7 N
TAD = 2640 N b
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110
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
SFA = 0 : TAB + TAC + TAD + P = 0 where P = Pi
uuur
AB = -(960 mm)i - (240 mm)j + (380 mm)k
AB = 1060 mm
uuur
AC = -(960 mm)i - (240 mm) j - (320 mm)k AC = 1040 mm
uuur
AD = -(960 mm)i + (720 mm) j - (220 mm)k AD = 1220 mm
uuur
AB
19 ˆ
Ê 48 12
TAB = TAB l AB = TAB
= TAB Á - i - j + k ˜
Ë 53 53
AB
53 ¯
uuur
3
4 ˆ
AC
Ê 12
= TAC Á - i - j - k ˜
TAC = TAC l AC = TAC
Ë
13 13 13 ¯
AC
305 N
TAD = TAD l AD =
[( -960 mm)i + (720 mm) j - (220 mm)k ]
12
220 mm
= -(240 N )i + (180 N ) j - (55 N )k
Substituting into SFA = 0, factoring i, j, k , and setting each coefficient equal to f gives:
48
12
TAB + TAC + 240 N 53
13
12
3
j:
TAB + TAC = 180 N 53
13
19
4
k:
TAB - TAC = 55 N 53
13
Solving the system of linear equations using conventional algorithms gives:
i: P =
TAB = 446.71 N
TAC = 341.71 N (1)
(2)
(3)
P = 960 N b
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111
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
SFA = 0 : TAB + TAC + Qj + (1200 N )i = 0
uuur
AB = -(960 mm)i - (240 mm)j + (380 mm)k AB = 1060 mm
uuur
AC = -(960 mm)i - (240 mm) j - (320 mm)k AC = 1040 mm
uuur
AB Ê 48 12
19 ˆ
TAB = TAB l AB = TAB
= Á - i - j + k ˜ TAB
AB Ë 53 53
53 ¯
uuur
3
4 ˆ
AC Ê 12
= Á - i - j - k ˜ TAC
TAC = TAC l AC = TAC
AC Ë 13 13 13 ¯
Substituting into SFA = 0, factoring i, j, k , and setting each coefficient equal to f gives:
48
12
TAB - TAC + 1200 N = 0 53
13
12
3
j: - TAB - TAC + Q = 0 53
13
19
4
k:
TAB - TAC = 0 53
13
Solving the resulting system of linear equations using conventional algorithms gives:
i: -
(1)
(2)
(3)
TAB = 605.71 N
TAC = 705.71 N
0 Q 300 N b
Q = 300.00 N 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.
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112
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:
We write
SF = 0 : TAB + TAC + TAD + Pj = 0
uuur
AB = -15i - 30 j + 10k AB = 35 m
uuur
AC = 10i - 30 j + 22k AC = 38.523 m
uuur
AD = 36.729 m
AD = 7i - 30 j - 20k
uuur
AB
TAB = TAB l AB = TAB
AB
2 ˆ
Ê 3 6
= Á - i - j + k ˜ TAB
Ë 7
7
7 ¯
uuur
AC
TAC = TAC l AC = TAC
AC
= (0.2596i - 0.7788 j + 0.5711k ) TAC
uuur
AD
TAD = TAD l AD = TAD
AD
= (0.19058i - 0.8168 j - 0.5445k ) TAD
Substituting into the equation SF = 0 and factoring i, j, k :
ˆ
Ê 3
ÁË - TAB + 0.2596TAC + 0.19058TAD ˜¯ i
7
ˆ
Ê 6
+ Á - TAB - 0.7788TAC - 0.8168TAD + P ˜ j
¯
Ë 7
ˆ
Ê2
+ Á TAB + 0.5711TAC - 0.5445TAD ˜ k = 0
¯
Ë7
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113
Problem 2.109 (Continued)
Setting the coefficients of i, j, k equal to zero:
i:
-0.4285TAB + 0.2596TAC + 0.19058TAD = 0 (1)
j:
-0.8571TAB - 0.7788TAC - 0.8168TAD + P = 0 (2)
k:
0.2857TAB + 0.5711TAC - 0.5445TAD = 0 (3)
Set TAB = 3150 N in Eqs. (1) – (3):
Solving,
-1350 N + 0.2596TAC + 0.19058TAD = 0 (1¢)
-2700 N - 0.7788TAC - 0.8168TAD + P = 0 (2¢)
900 N + 0.5711TAC - 0.5445TAD = 0 (3¢)
TAC = 2252.48 N TAD = 4015.41 N
P = 7734.02 N P = 7730 N b
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114
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
See Problem 2.109 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3)
below:
i:
-0.4285TAB + 0.2596TAC + 0.19058TAD = 0 (1)
j:
-0.8571TAB - 0.7788TAC - 0.8168TAD + P = 0 (2)
k:
0.2857TAB + 0.5711TAC - 0.5445TAD = 0 (3)
Substituting for TAC = 4600 N in Equations (1), (2), and (3) above and solving the resulting set of equations
using conventional algorithms gives:
Solving,
-0.4285TAB + 1194.16 N + 0.19058TAD = 0 (1¢)
-0.8571TAB - 3582.48 N - 0.8168TAD + P = 0 (2¢)
0.2857TAB + 2627.06 N - 0.5445TAD = 0 (3¢)
6434.22 N
8200.76 N
15795.63 N P = 15800 N b
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115
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 = 0 : TAB + TAC + TAD + Pj = 0
We have:
uuur
AB = -(320 mm)i - ( 480 mm)j + (360 mm)k AB = 680 mm
uuur
AC = ( 450 mm)i - ( 480 mm) j + (360 mm)k AC = 750 mm
uuur
AD = (250 mm)i - ( 480 mm) j - (360 mm ) k AD = 650 mm
Thus:
TAB
TAC
TAD
uuur
AB Ê 8
12
9 ˆ
= TAB l AB = TAB
= Á - i - j + k ˜ TAB
AB Ë 17 17 17 ¯
uuur
AC
= TAC l AC = TAC
= (0.6i - 0.64 j + 0.48k ) TAC
AC
uuur
AD Ê 5
9.6
7.2 ˆ
= TAD l AD = TAD
=Á ijk ˜ TAD
Ë
AD
13
13
13 ¯
Dimensions in mm
Substituting into the equation SF = 0 and factoring i, j, k :
5
ˆ
Ê 8
ÁË - TAB + 0.6TAC + TAD ˜¯ i
17
13
9.6
Ê 12
ˆ
+ Á - TAB - 0.64TAC TAD + P ˜ j
Ë 17
¯
13
7.2
ˆ
Ê9
TAD ˜ k = 0
+ Á TAB + 0.48TAC ¯
Ë 17
13
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116
Problem 2.111 (Continued)
Setting the coefficient of i, j, k equal to zero:
8
5
TAB + 0.6TAC + TAD = 0 17
13
12
9.6
j: - TAB - 0.64TAC TAD + P = 0 7
13
9
7.2
k:
TAB + 0.48TAC TAD = 0 17
13
= 60 N in (1) and (3):
-
i:
Making TAC
8
5
TAB + 36 N + TAD = 0 17
13
9
7.2
TAB + 28.8 N TAD = 0 17
13
Multiply (1¢) by 9, (3¢) by 8, and add:
-
554.4 N -
(1)
(2)
(3)
(1¢)
(3¢)
12.6
TAD = 0 TAD = 572.0 N
13
Substitute into (1¢) and solve for TAB :
17 Ê
5
ˆ
TAB = 544.0 N
ÁË 36 + ¥ 572˜¯
8
13
Substitute for the tensions in Eq. (2) and solve for P:
TAB =
12
9.6
(544 N ) + 0.64(60 N ) +
(572 N )
17
13
= 844.8 N
P=
Weight of plate = P = 845 N b
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117
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
See Problem 2.111 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and
(3) below:
8
5
TAB + 0.6TAC + TAD = 0 17
13
12
9.6
- TAB + 0.64 TAC TAD + P = 0 17
13
9
7.2
TAB + 0.48TAC TAD = 0 13
17
= 520 N in Eqs. (1) and (3):
-
Making TAD
8
TAB + 0.6TAC + 200 N = 0 17
9
TAB + 0.48TAC - 288 N = 0 17
Multiply (1¢) by 9, (3¢) by 8, and add:
-
(1)
(2)
(3)
(1¢)
(3¢)
9.24TAC - 504 N = 0 TAC = 54.5455 N
Substitute into (1¢) and solve for TAB :
17
(0.6 ¥ 54.5455 + 200) TAB = 494.545 N
8
Substitute for the tensions in Eq. (2) and solve for P:
TAB =
12
9.6
( 494.545 N ) + 0.64(54.5455 N ) +
(520 N )
17
13
= 768.00 N
P=
Weight of plate = P = 768 N b
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118
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
See Problem 2.109 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below:
i:
-0.4285TAB + 0.2596TAC + 0.19058TAD = 0 (1)
j:
-0.8571TAB - 0.7788TAC - 0.8168TAD + P = 0 (2)
k:
0.2857TAB + 0.5711TAC - 0.5445TAD = 0 (3)
Substituting for P = 10500 N in Equations (1), (2), and (3) above and solving the resulting set of equations
using conventional algorithms gives:
-0.4285TAB + 0.2596TAC + 0.19058TAD = 0 (1¢)
-0.8571TAB - 0.7788TAC - 0.8168TAD + 10500 N = 0 (2¢)
0.2857TAB + 0.5711TAC - 0.5445TAD = 0 (3¢)
TAB = 4277.09 N
TAC = 3057.81 N
TAD = 5451.38 N TAB = 4280 N b
TAC = 3060 N b
TAD = 5450 N b
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119
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:
-TAD (sin 30∞)(sin 50∞) + TBD (sin 30∞)(cos 40∞) + TCD (sin 30∞)(cos 60∞∞) = 0
Dividing through by sin 30∞ and evaluating:
-0.76604TAD + 0.76604TBD + 0.5TCD = 0 (1)
SFy = 0 : -TAD (cos 30∞) - TBD (cos 30∞) - TCD (cos 30∞) + 300 N = 0
or
TAD + TBD + TCD = 346.41 N (2)
SFz = 0 : TAD sin 30∞ cos 50∞ + TBD sin 30∞ sin 40∞ - TCD sin 30∞ sin 60∞ = 0
or
0.64279TAD + 0.64279TBD - 0.86603TCD = 0 (3)
Solving Equations (1), (2), and (3) simultaneously:
TAD = 147.579 N
TBD = 51.253 N
TCD = 147.578 N TAD = 147.6 N b
TBD = 51.3 N b
TCD = 147.6 N b
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120
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
5
TAB + 0.6TAC + TAD = 0 17
13
12
9.6
- TAB - 0.64TAC TAD + 792 N = 0 17
13
9
7.2
TAB + 0.48TAC TAD = 0 17
13
Solving Equations (1), (2), and (3) by conventional algorithms gives
-
(1)
(2)
(3)
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
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121
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
SFA = 0 : TAB + TAC + TAD + P + Q = 0
Where
P = Pi and Q = Qj
uuur
AB = -(960 mm)i - (240 mm) j + (380 mm)k AB = 1060 mm
uuur
AC = -(960 mm)i - (240 mm) j - (320 mm)k AC = 1040 mm
uuur
AD = -(960 mm)i + (720 mm) j - (220 mm)k AD = 1220 mm
uuur
AB
19 ˆ
Ê 48 12
TAB = TAB l AB = TAB
= TAB Á - i - j + k ˜
Ë 53 53
AB
53 ¯
uuur
3
4 ˆ
AC
Ê 12
= TAC Á - i - j - k ˜
TAC = TAC l AC = TAC
Ë 13 13 13 ¯
AC
uuur
AD
Ê 48 36 11 ˆ
j- k
TAD = TAD l AD = TAD
= TAD Á - i +
Ë 61 61 61 ˜¯
AD
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
12
48
TAB - TAC - TAD + 2880 N = 0 53
13
61
(1)
j: -
12
3
36
TAB - TAC + TAD = 0 53
13
61
(2)
k:
19
4
11
TAB - TAC - TAD = 0 53
13
61
(3)
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122
Problem 2.116 (Continued)
Solving the system of linear equations using conventional algorithms gives:
TAB = 1340.14 N
TAC = 1025.12 N
TAD = 915.03 N TAB = 1340 N b
TAC = 1025 N b
TAD = 915 N b
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123
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
12
48
TAB - TAC - TAD + P = 0 53
13
61
12
3
36
- TAB - TAC + TAD + Q = 0 53
13
61
19
4
11
TAB - TAC - TAD = 0 53
13
61
-
(1)
(2)
(3)
Setting P = 2880 N and Q = 576 N gives:
-
48
12
48
TAB - TAC - TAD + 2880 N = 0 53
13
61
12
3
36
- TAB - TAC + TAD + 576 N = 0 53
13
61
19
4
11
TAB - TAC - TAD = 0 53
13
61
(1¢)
(2¢)
(3¢)
Solving the resulting set of equations using conventional algorithms gives:
TAB = 1431.00 N
TAC = 1560.00 N
TAD = 183.010 N TAB = 1431 N b
TAC = 1560 N b
TAD = 183.0 N b
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124
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
12
48
TAB - TAC - TAD + P = 0 53
13
61
12
3
36
- TAB - TAC + TAD + Q = 0 53
13
61
19
4
11
TAB - TAC - TAD = 0 53
13
61
-
(1)
(2)
(3)
Setting P = 2880 N and Q = -576 N gives:
48
12
48
TAB - TAC - TAD + 2880 N = 0 53
13
61
12
3
36
- TAB - TAC + TAD - 576 N = 0 53
13
61
19
4
11
TAB - TAC - TAD = 0 53
13
61
Solving the resulting set of equations using conventional algorithms gives:
-
(1¢)
(2¢)
(3¢)
TAB = 1249.29 N
TAC = 490.31 N
TAD = 1646.97 N TAB = 1249 N b
TAC = 490 N b
TAD = 1647 N b
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125
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
N=
(2 j + k )
5
= N (0.8944 j + 0.4472k )
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
uuur
AB = ( -1000 mm)i + (1750 mm)j - (1000 mm)k
AB = 2250 mm
uuur
AB
TAB = T l AB = TAB
AB
TAB
[( -1000 mm)i + (1750 mm)j - (1000 mm)k ]
=
2250 mm
and
Then:
7
4 ˆ
Ê 4
TAB = TAB Á - i + j - k ˜
Ë 9
9
9 ¯
uuur
AC = (1125 mm)i + (1500 mm)j - (1000 mm)k
AC = 2125 mm
uuur
AC
TAC = T l AC = TAC
AC
TAC
=
[(1125 mm)i + (1500 mm)j - (1000 mm)k ]
2125 mm
12
8 ˆ
Ê9
TAC = TAC Á i + j - k ˜
Ë 17 17 17 ¯
SF = 0 : N + TAB + TAC + W = 0
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126
Problem 2.119 (Continued)
With W = 1000 N, and equating the factors of i, j, and k to zero, we obtain the linear algebraic equations:
4
9
- TAB + TAC = 0 9
17
7
12
2
j:
- 1000 N = 0 TAB + TAC +
9
17
5
4
8
1
k : - TAB - TAC +
N =0
9
17
5
Using conventional methods for solving linear algebraic equations we obtain:
i:
(1)
(2)
(3)
TAB = 327.94 N
TAB = 328 N b
TAC = 275.30 N
TAC = 275 N b
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127
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.
7
4 ˆ
Ê 4
TAB = TAB Á - i + j - k ˜
Ë 9
9
9 ¯
12
8 ˆ
Ê9
TAC = TAC Á i + j - k ˜
Ë 17 17 17 ¯
Then:
Where
and
SFA = 0 : N + TAB + TAC + P + W = 0
P = ( -200 N)i
W = (1000 N)j
Equating the factors of i, j, and k to zero, we obtain the linear equations:
4
9
i: - TAB + TAC - 200 = 0
9
17
2
7
12
j:
N + TAB + TAC - 1000 = 0
9
17
5
1
4
8
k:
N - TAB - TAC = 0
9
17
5
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
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128
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 = Pi 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:
TAB = T l AB
uuuv
AB
=T
AB
=T
( -130 mm)i + ( 400 mm) j + (160 mm)k
450 mm
40
16 ˆ
Ê 13
j + k˜
= T Á- i +
Ë 45
45
45 ¯
TAC = T l AC
uuuv
AC
=T
AC
( -150 mm)i + ( 400 mm) j + ( -240 mm)k
=T
490 mm
40
24 ˆ
Ê 15
j - k˜
= T Á- i +
Ë 49
49
49 ¯
SF = 0 : TAB + TAC + Q + P + W = 0
Setting coefficients of i, j, k equal to zero:
13
15
T - T +P=0
45
49
40
40
j: + T + T - W = 0
45
49
16
24
k: + T - T + Q = 0
45
49
i: -
0.59501T = P (1)
1.70521T = W (2)
0.134240 T = Q (3)
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129
Problem 2.121 (Continued)
Data:
W = 376 N 1.70521T = 376 N T = 220.50 N
0.59501(220.50 N ) = P P = 131.2 N b
0.134240(220.50 N ) = Q Q = 29.6 N b
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130
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 = Pi
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
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):
0.59501T = 164 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
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131
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
uuur
AB = -(0.78 m)i + (1.6 m) j + (0 m)k
AB = ( -0.78 m)2 + (1.6 m)2 + (0)2
= 1.78 m
TAB = T l AB = TAB
uuur
AB
AB
TAB
[-(0.78 m)i + (1.6 m) j + (0 m)k ]
1.78 m
= TAB ( -0.4382i + 0.8989 j + 0k )
=
and
TAB
uuur
AC = (0)i + (1.6 m) j + (1.2 m)k
and
AC = (0 m)2 + (1.6 m)2 + (1.2 m)2 = 2 m
uuur
AC TAC
TAC = T l AC = TAC
=
[(0)i + (1.6 m) j + (1.2 m)k ]
AC 2 m
TAC = TAC (0.8 j + 0.6k )
uuur
AD = (1.3 m)i + (1.6 m) j + (0.4 m)k
AD = (1.3 m)2 + (1.6 m)2 + (0.4 m)2 = 2.1 m
uuur
AD TAD
TAD = T l AD = TAD
[(1.3 m)i + (1.6 m) j + (0.4 m)k ]
=
AD 2.1 m
TAD = TAD (0.6190i + 0.7619 j + 0.1905k )
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132
Problem 2.123 (Continued)
uuur
AE = -(0.4 m)i + (1.6 m) j - (0.86 m)k
Finally,
AE = ( -0.4 m)2 + (1.6 m)2 + ( -0.86 m)2 = 1.86 m
uuur
AE
TAE = T l AE = TAE
AE
T
= AE [-(0.4 m)i + (1.6 m) j - (0.86 m)k ]
1.86 m
TAE = TAE ( -0.2151i + 0.8602 j - 0.4624k )
With the weight of the container
W = -W j, at A we have:
SF = 0 : TAB + TAC + TAD - Wj = 0
Equating the factors of i, j, and k to zero, we obtain the following linear algebraic equations:
-0.4382TAB + 0.6190TAD - 0.2151TAE = 0 (1)
0.8989TAB + 0.8TAC + 0.7619TAD + 0.8602TAE - W = 0 (2)
0.6TAC + 0.1905TAD - 0.4624TAE = 0 (3)
Knowing that W = 1000 N and that because of the pulley system at B TAB = TAD = P , where P is the externally
applied (unknown) force, we can solve the system of linear Equations (1), (2) and (3) uniquely for P.
P = 378 N b
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133
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 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
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
TAB = TAD = P
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
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134
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:
B:
A:
l AB
uuur
AB - xi - (500 mm) j + zk
=
=
AB
625 mm
S F = 0 : Pi + N y j + N z k + TAB l AB = 0
Collar A:
Substitute for l AB and set coefficient of i equal to zero:
PCollar B:
TAB x
=0
625 mm
(1)
SF = 0 : (300 N )k + N x¢ i + N y¢ j - TAB l AB = 0
Substitute for l AB and set coefficient of k equal to zero:
300 N x = 225 mm
(a)
From Eq. (2):
(b)
From Eq. (1):
TAB z
=0
625 mm
(2)
(225 mm)2 + (500 mm)2 + z 2 = (625 mm)2
z = 300 mm
300 N P=
TAB (300 mm)
=0
625 mm
(625 N )(225 mm)
625 mm
TAB = 625 N b
P = 225 N b
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135
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:
TAB x
625 mm
T z
300 N - AB
=0
625 mm
TAB x = (625 mm)(600 N ) (1¢)
TAB z = (625 mm)(300 N ) (2¢)
P=
For P = 600 N, Eq. (1) yields
From Eq. (2)
x
=2
z
Dividing Eq. (1¢) by (2¢):
Now write
(1)
(2)
(3)
x 2 + z 2 + (500 mm)2 = (625 mm)2 (4)
Solving (3) and (4) simultaneously
4 z 2 + z 2 + 250000 = 390625
z 2 = 167
z = 167.71 mm
From Eq. (3)
x = 2z
= 335.4 mm
x = 167.7 mm, z = 335 mm b
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136
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 N ) cos 25∞
= 679.73 N
R1 = 679.73 N
a + 25∞
Next we consider the resultant R 2 of R1 and the 1200 N force where R 2 must be horizontal and directed to
the left. Using the triangle rule and law of sines,
sin (a + 25∞) sin (30∞)
=
1200 N
679.73 N
sin (a + 25∞) = 0.88270
a + 25∞ = 61.970∞
a = 36.970∞ a = 37.0∞ b
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137
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:
Law of sines:
P 2 = (150)2 + (200)2 - 2(150)(200) cos 25∞
P = 90.1195 N
sin a
sin 25∞
=
150 N 90.1195 N
a = 44.703∞
90∞ - a = 45.297∞ P = 90.1 N
45.3∞ b
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138
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)
P=
(b)
Px =
Py
sin 35 ∞
Py
tan 40∞
=
1200 N
= 1866.9 N sin 40∞
or P = 1867 N b
=
1200 N
= 1430.1 N tan 40∞
or Px = 1430 N b
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139
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 = 0 : -
6m
3.75 m
TAC +
TBC = 0
6.25 m
4.25 m
TBC = 1.08800TAC
SFy = 0 :
(a)
1.75 m
2m
TAC +
TBC - 2000 N = 0
6.25 m
4.25 m
1.75 m
2m
TAC +
(1.08800TAC ) - 2000 N = 0
6.25 m
4.25 m
(0.28000 + 0.51200)TAC = 2000 N
TAC = 2525.25 N (b)
TAC = 2530 N b
TBC = (1.08800)(2525.25 N )
= 2747.5 N
TBC = 2750 N b
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140
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)
(b)
12
4
TAC + (360 N ) = 0 13
5
5
3
SFy = 0 :
(312 N ) + TBC + (360 N ) - 480 N = 0
13
5
TBC = 480 N - 120 N - 216 N SFx = 0 : -
TAC = 312 N b
TBC = 144 N b
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141
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
12
4
TAC + P = 0
13
5
13
TAC = P 15
5
3
SFy = 0 :
TAC + TBC + P - 480 N = 0
13
5
SFx = 0 : -
Substitute for TAC from (1):
(1)
3
Ê 5 ˆ Ê 13 ˆ
ÁË ˜¯ ÁË ˜¯ P + TBC + P - 480 N = 0
13 15
5
TBC = 480 N -
14
P
15
(2)
From (1), TAC 0 requires P 0.
From (2), TBC 0 requires
14
P 480 N, P 514.29 N
15
Allowable range:
0 P 514 N b
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142
PROBLEM 2.133
A force acts at the origin of a coordinate system in a direction defined by the angles q x = 69.3° 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
cos2 q y + cos2 q y + cos2 q z = 1
or
cos2 q y = 1 - cos2 q x - cos2 q y
Since Fy 0, we must have cosq y 0
cos q y = - 1 - cos2 69.3∞ - cos2 57.9∞
= - 0.76985
(b)
Fy
q y = 140.3∞ b
-870 N
= 1130.09 N -0.76985
F = 1130 N b
Fx = F cos q x = (1130.09 N ) cos 69.3∞ Fx = 399 N b
Fz = F cos q z = (1130.09 N ) cos 57.9∞ Fz = 601 N b
F=
cos q y
=
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143
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 q x , q y , and q z defining the
direction of that force.
Solution
28 m
32.5 m
= 0.86154
q y = 30.51∞
cos q y =
From triangle AOB:
(a)
Fx = - F sin q y cos 20∞
= -(20000 N) sin 30.51∞ cos 20∞ = 9541.43 N
Fx = -9.54 kN b
Fy = + F cos q y = (20000 N)(0.86154) = 17231 N Fz = +(20000 N) sin 30.51° sin 20° = 3472.8 N (b)
cos q x =
From above:
cos q z =
Fx -9541.43 N
=
= - 0.4771 F
20000 N
q y = 30.51∞ Fz
3472.8 N
=+
= + 0.1736 F
20000 N
Fy = +17.23 kN b
Fz = +3.47 kN b
q x = 118.5∞ b
q y = 30.5∞ b
q z = 80.0∞ b
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144
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
uuur
AB = -15.588i + 15 j + 12k
AB = 24.739 m
uuur
AC = -15.588i + 18.60 j - 15k
AC = 28.530 m
TAB = TAB l AB = TAB
TAB = (10 kN )
uuur
AB
AB
-15.588i + 15 j + 12k
24.739
TAB = (6.301 kN )i + (6.063 kN ) j + ( 4.851 kN )k
uuuur
AC
-15.588i + 18.60 j - 15k
(7.5 kN )
TAC = TAC l AC = TAC
28.530
AC
TAC = -( 4.098 kN )i + ( 4.890 kN ) j - (3.943 kN )k
R = TAB + TAC = -(10.399 kN )i + (10.953 kN ) j + (0.908 kN )k
R = (10.399)2 + (10.953)2 + (0.908)2
= 15.130 kN
Rx -10.399 kN
=
= - 0.6873 R
15.130 kN
Ry 10.953 kN
=
= 0.7239 cos q y =
R 15.130 kN
R
0.908 kN
cos q z = z =
= 0.0600 R 15.130 kN
cos q x =
R = 15.13 kN b
q z = 133.4∞ b
q y = 43.6∞ b
q z = 86.6∞ b
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145
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
We have:
S F = 0 : TAB + TAC + TAD + W = 0
uuur
AB = ( 450 mm)i + (600 mm) j
AB = 750 mm
uuur
AC = (600 mm) j - (320 mm)k
AC = 680 mm
uuur
AD = (500 mm)i + (600 mm) j + (360 mm)k
AD = 860 mm
uuur
AB Ê 450
600 ˆ
TAB = TAB l AB = TAB
=Á
i+
j˜ TAB
Ë
AB
750
750 ¯
= (0.6i + 0.8 j)TAB
uuur
AC Ê 600
320 ˆ
8 ˆ
Ê 15
TAC = TAC l AC = TAC
=Á
jk ˜ TAC = Á j - k ˜ TAC
¯
Ë
Ë
AC
680
680
17 17 ¯
uuur
AD Ê 500
600
360 ˆ
= ÁTAD = TAD l AD = TAD
i+
j+
k ˜ TAD
AD Ë 860
860
860 ¯
18 ˆ
Ê 25 30
TAD = Á - i +
j + k ˜ TAD
Ë 43 43
43 ¯
Substitute into SF = 0, factor i, j, k, and set their coefficient equal to zero:
0.6TAB 0.8TAB +
-
25
TAD = 0
43
TAB = 0.96899TAD 15
30
TAC + TAD - 1165 N = 0 17
43
8
18
TAC + TAD = 0
17
43
TAC = 0.88953TAD (1)
(2)
(3)
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146
Problem 2.136 (Continued)
Substitute for TAB and TAC from (1) and (3) into (2):
15
30 ˆ
Ê
ÁË 0.8 ¥ 0.96899 + ¥ 0.88953 + ˜¯ TAD - 1165 N = 0
17
53
2.2578TAD - 1165 N = 0 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
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147
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 = (341 N)j 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
Free Body Diagrams of Collars:
For both Problems 2.137 and 2.138:
( AB)2 = x 2 + y 2 + z 2
Here
or
(0.525 m)2 = (.20 m)2 + y 2 + z 2
y 2 + z 2 = 0.23563 m2
Thus, why y given, z is determined,
Now
l AB
uuur
AB
=
AB
1
(0.20i - yj + zk ) m
0.525 m
= 0.38095i - 1.90476 yj + 1.90476zk
Where y and z are in units of meters, m.
=
From the F.B. Diagram of collar A:
SF = 0 : N x i + N z k + Pj + TAB l AB = 0
Setting the j coefficient to zero gives:
P - (1.90476 y )TAB = 0
With
P = 341 N
341 N
TAB =
1.90476 y
Now, from the free body diagram of collar B:
SF = 0 : N x i + N y j + Qk - TAB l AB = 0
Setting the k coefficient to zero gives:
Q - TAB (1.90476 z ) = 0
And using the above result for TAB we have
Q = TAB z =
341 N
(341 N )( z )
(1.90476 z ) =
(1.90476) y
y
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148
Problem 2.137 (Continued)
Then, from the specifications of the problem, y = 155 mm = 0.155 m
z 2 = 0.23563 m2 - (0.155 m)2
z = 0.46 m
and
341 N
0.155(1.90476)
= 1155.00 N
TAB =
(a)
TAB = 1.155 kN b
or
and
341 N(0.46 m)(0.866)
(0.155 m)
= (1012.00 N)
Q=
(b)
Q = 1.012 kN b
or
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149
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 = (341 N)j
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 2 + z 2 = 0.23563 m2
341 N
TAB =
1.90476 y
341 N
Q=
z
y
With y = 275 mm = 0.275 m, we obtain:
z 2 = 0.23563 m2 - (0.275 m)2
z = 0.40 m
and
TAB =
(a)
341 N
= 651.00
(1.90476)(0.275 m)
TAB = 651 N b
or
and
Q=
(b)
341 N(0.40 m)
(0.275 m)
Q = 496 N b
or
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150