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, α = 75.1°
R = 179 N
75.1°
!
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3
PROBLEM 2.2
Two forces P and Q are applied as shown at Point A of a hook support. Knowing that
P = 60 lb and Q = 25 lb, 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 = 77.1 lb, α = 85.4°
R = 77.1 lb
85.4°
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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4
PROBLEM 2.3
The cable stays AB and AD help support pole AC. Knowing
that the tension is 120 lb in AB and 40 lb in AD, determine
graphically the magnitude and direction of the resultant of the
forces exerted by the stays at A using (a) the parallelogram
law, (b) the triangle rule.
SOLUTION
We measure:
(a)
Parallelogram law:
(b)
Triangle rule:
We measure:
α = 51.3°
β = 59.0°
γ = 67.0°
R = 139.1 lb,
R = 139.1 lb
67.0°
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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, α = 66.6°
R = 3.30 kN
66.6°
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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6
PROBLEM 2.5
The 300-lb force is to be resolved into components along lines a-a′ and b-b′.
(a) Determine the angle α by trigonometry knowing that the component
along line a-a′ is to be 240 lb. (b) What is the corresponding value of the
component along b-b′?
SOLUTION
(a)
Using the triangle rule and law of sines:
sin β
sin 60°
=
240 lb 300 lb
sin β = 0.69282
β = 43.854°
α + β + 60° = 180°
α = 180° − 60° − 43.854°
= 76.146°
(b)
Law of sines:
Fbb′
300 lb
=
sin 76.146° sin 60°
α = 76.1°
Fbb′ = 336 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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7
PROBLEM 2.6
The 300-lb force is to be resolved into components along lines a-a′ and b-b′.
(a) Determine the angle α by trigonometry knowing that the component along
line b-b′ is to be 120 lb. (b) What is the corresponding value of the component
along a-a′?
SOLUTION
Using the triangle rule and law of sines:
(a)
sin α
sin 60°
=
120 lb 300 lb
sin α = 0.34641
α = 20.268°
(b)
α = 20.3°
α + β + 60° = 180°
β = 180° − 60° − 20.268°
= 99.732°
Faa′
300 lb
=
sin 99.732° sin 60°
Faa′ = 341 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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 α if the resultant R of the two forces applied to the support is to
be horizontal, (b) the corresponding magnitude of R.
SOLUTION
Using the triangle rule and law of sines:
(a)
sin α sin 25°
=
50 N
35 N
sin α = 0.60374
α = 37.138°
(b)
α = 37.1°
α + β + 25° = 180°
β = 180° − 25° − 37.138°
= 117.86°
R
35 N
=
sin117.86 sin 25°
R = 73.2 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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9
PROBLEM 2.8
For the hook support of Problem 2.1, knowing that the magnitude of P is 75 N,
determine by trigonometry (a) the required magnitude of the force Q if the
resultant R of the two forces applied at A is to be vertical, (b) the corresponding
magnitude of R.
PROBLEM 2.1 Two forces P and Q are applied as shown at Point A of a hook
support. Knowing that 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 law of sines:
(a)
(b)
Q
75 N
=
sin 20° sin 35°
Q = 44.7 N
α + 20° + 35° = 180°
α = 180° − 20° − 35°
= 125°
R
75 N
=
sin125° sin 35°
R = 107.1 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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 α = 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° + β + 75° = 180°
β = 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:
Using the law of sines:
P 2 = (1600 N)2 + (2500 N)2 − 2(1600 N)(2500 N) cos 75°
P = 2596 N
sin α
sin 75°
=
1600 N 2596 N
α = 36.5°
P is directed 90° − 36.5° or 53.5° below the horizontal.
P = 2600 N
53.5°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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12
PROBLEM 2.11
A steel tank is to be positioned in an excavation. Knowing that
α = 20°, determine by trigonometry (a) the required magnitude
of the force P if the resultant R of the two forces applied at A is
to be vertical, (b) the corresponding magnitude of R.
SOLUTION
Using the triangle rule and the law of sines:
(a)
β + 50° + 60° = 180°
β = 180° − 50° − 60°
= 70°
(b)
425 lb
P
=
sin 70° sin 60°
P = 392 lb
425 lb
R
=
sin 70° sin 50°
R = 346 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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 500 lb, determine by trigonometry (a) the
required angle α if the resultant R of the two forces applied at A
is to be vertical, (b) the corresponding magnitude of R.
SOLUTION
Using the triangle rule and the law of sines:
(a)
(α + 30°) + 60° + β = 180°
β = 180° − (α + 30°) − 60°
β = 90° − α
sin (90° − α ) sin 60°
425 lb
(b)
=
500 lb
90° − α = 47.40°
α = 42.6°
R
500 lb
=
sin (42.6° + 30°) sin 60°
R = 551 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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°
(b)
R = (50 N) cos 25°
P = 21.1 N
R = 45.3 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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 α = 20°, determine by trigonometry
(a) the required magnitude of the force P if the resultant R
of the two forces applied at A is to be vertical, (b) the
corresponding magnitude of R.
SOLUTION
The smallest force P will be perpendicular to R.
(a)
P = (425 lb) cos 30°
(b)
R = (425 lb)sin 30°
P = 368 lb
R = 213 lb
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16
PROBLEM 2.15
Solve Problem 2.2 by trigonometry.
PROBLEM 2.2 Two forces P and Q are applied as shown at Point A of a hook
support. Knowing that P = 60 lb and Q = 25 lb, 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° + α = 180°
α = 125°
R 2 = P 2 + Q 2 − 2 PQ cos α
R 2 = (60 lb)2 + (25 lb) 2
− 2(60 lb)(25 lb) cos125°
R 2 = 3600 + 625 + 3000(0.5736)
R = 77.108 lb
Using the law of sines:
sin β
sin125°
=
25 lb 77.108 lb
β = 15.402°
70° + β = 85.402°
R = 77.1 lb
85.4°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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 120 lb in AB and
40 lb in AD, determine graphically the magnitude and
direction of the resultant of the forces exerted by the stays
at A using (a) the parallelogram law, (b) the triangle rule.
SOLUTION
8
10
α = 38.66°
6
tan β =
10
β = 30.96°
tan α =
Using the triangle rule:
Using the law of cosines:
Using the law of sines:
α + β + ψ = 180°
38.66° + 30.96° + ψ = 180°
ψ = 110.38°
R 2 = (120 lb)2 + (40 lb) 2 − 2(120 lb)(40 lb) cos110.38°
R = 139.08 lb
sin γ
sin110.38°
=
40 lb
139.08 lb
γ = 15.64°
φ = (90° − α ) + γ
φ = (90° − 38.66°) + 15.64°
φ = 66.98°
R = 139.1 lb
67.0°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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:
R 2 = (2 kN)2 + (3 kN)2
− 2(2 kN)(3 kN) cos80°
R = 3.304 kN
Using the law of sines:
sin γ
sin 80°
=
2 kN 3.304 kN
γ = 36.59°
β + γ + 80° = 180°
γ = 180° − 80° − 36.59°
γ = 63.41°
φ = 180° − β + 50°
φ = 66.59°
R = 3.30 kN
66.6°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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
γ = 180° − (40° + 20°)
= 120°
Then
R 2 = (15 kN) 2 + (10 kN)2
− 2(15 kN)(10 kN) cos120°
= 475 kN 2
R = 21.794 kN
and
Hence:
10 kN 21.794 kN
=
sin α
sin120°
10 kN !
sin α = "
# sin120°
$ 21.794 kN %
= 0.39737
α = 23.414
φ = α + 50° = 73.414
R = 21.8 kN
73.4°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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
γ = 180° − (40° + 20°)
= 120°
Then
R 2 = (10 kN) 2 + (15 kN)2
− 2(10 kN)(15 kN) cos120°
= 475 kN 2
R = 21.794 kN
and
Hence:
15 kN 21.794 kN
=
sin α
sin120°
15 kN !
sin α = "
# sin120°
$ 21.794 kN %
= 0.59605
α = 36.588°
φ = α + 50° = 86.588°
R = 21.8 kN
86.6°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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21
PROBLEM 2.20
For the hook support of Problem 2.7, knowing that P = 75 N and α = 50°,
determine by trigonometry the magnitude and direction of the resultant of
the two forces applied to the support.
PROBLEM 2.7 Two forces are applied as shown to a hook support.
Knowing that the magnitude of P is 35 N, determine by trigonometry (a) the
required angle α if the resultant R of the two forces applied to the support is
to be horizontal, (b) the corresponding magnitude of R.
SOLUTION
Using the force triangle and the laws of cosines and sines:
We have
β = 180° − (50° + 25°)
= 105°
Then
R 2 = (75 N) 2 + (50 N) 2
− 2(75 N)(50 N) cos 105°
2
R = 10066.1 N 2
R = 100.330 N
and
Hence:
sin γ
sin105°
=
75 N 100.330 N
sin γ = 0.72206
γ = 46.225°
γ − 25° = 46.225° − 25° = 21.225°
R = 100.3 N
21.2°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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 = (84) 2 + (80) 2
= 116 in.
OB = (28)2 + (96)2
= 100 in.
OC = (48)2 + (90)2
= 102 in.
29-lb Force:
50-lb Force:
51-lb Force:
Fx = + (29 lb)
84
116
Fx = +21.0 lb
Fy = +(29 lb)
80
116
Fy = +20.0 lb
Fx = −(50 lb)
28
100
Fx = −14.00 lb
Fy = +(50 lb)
96
100
Fy = + 48.0 lb
Fx = + (51 lb)
48
102
Fx = +24.0 lb
Fy = −(51 lb)
90
102
Fy = −45.0 lb
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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24
PROBLEM 2.23
Determine the x and y components of each of the forces shown.
SOLUTION
40-lb Force:
50-lb Force:
60-lb Force:
Fx = + (40 lb) cos 60°
Fx = 20.0 lb
Fy = −(40 lb)sin 60°
Fy = −34.6 lb
Fx = −(50 lb)sin 50°
Fx = −38.3 lb
Fy = −(50 lb) cos 50°
Fy = −32.1 lb
Fx = + (60 lb) cos 25°
Fx = 54.4 lb
Fy = +(60 lb)sin 25°
Fy = 25.4 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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 300-lb horizontal component, determine
(a) the magnitude of the force P, (b) its vertical component.
SOLUTION
P sin 35° = 300 lb
(a)
P=
(b)
Vertical component
300 lb
sin 35°
P = 523 lb
Pv = P cos 35°
= (523 lb) cos 35°
Pv = 428 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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
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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
or
Py = 153.6 N
Px
tan 38°
120 N
=
tan 38°
Py =
= 153.59 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
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 or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
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
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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31
PROBLEM 2.30
Cable AC exerts on beam AB a force P directed along line AC. Knowing that P
must have a 350-lb vertical component, determine (a) the magnitude of the
force P, (b) its horizontal component.
SOLUTION
(a)
P=
Py
cos 55°
350 lb
cos 55°
= 610.2 lb
=
(b)
P = 610 lb
Px = P sin 55°
= (610.2 lb) sin 55°
= 499.8 lb
Px = 500 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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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. (lb)
y Comp. (lb)
29 lb
+21.0
+20.0
50 lb
–14.00
+48.0
51 lb
+24.0
–45.0
Rx = +31.0
Ry = + 23.0
R = Rx i + R y j
= (31.0 lb) i + (23.0 lb) j
Ry
tan α =
Rx
23.0
31.0
α = 36.573°
23.0 lb
R=
sin (36.573°)
=
= 38.601 lb
R = 38.6 lb
36.6°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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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
+41.0
+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 α =
Rx
250.2 N
20.6 N
tan α = 12.1456
α = 85.293°
tan α =
R=
250.2 N
sin 85.293°
R = 251 N
85.3°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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. (lb)
y Comp. (lb)
40 lb
+20.00
–34.64
50 lb
–38.30
–32.14
60 lb
+54.38
+25.36
Rx = +36.08
Ry = −41.42
R = Rx i + Ry j
= ( +36.08 lb)i + (−41.42 lb) j
Ry
tan α =
Rx
41.42 lb
36.08 lb
tan α = 1.14800
α = 48.942°
tan α =
R=
41.42 lb
sin 48.942°
R = 54.9 lb
48.9°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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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 lb
+640
+480
424 lb
–224
–360
408 lb
+192
–360
Rx = +608
Ry = −240
R = Rx i + Ry j
= (608 lb)i + (−240 lb) j
tan α =
Ry
Rx
240
608
α = 21.541°
=
240 N
sin(21.541°)
= 653.65 N
R=
R = 654 N
21.5°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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36
PROBLEM 2.35
Knowing that α = 35°, determine the resultant of the three forces
shown.
SOLUTION
Fx = +(100 N) cos 35° = +81.915 N
100-N Force:
Fy = −(100 N)sin 35° = −57.358 N
Fx = +(150 N) cos 65° = +63.393 N
150-N Force:
Fy = −(150 N) sin 65° = −135.946 N
Fx = −(200 N) cos 35° = −163.830 N
200-N Force:
Fy = −(200 N)sin 35° = −114.715 N
Force
x Comp. (N)
y Comp. (N)
100 N
+81.915
−57.358
150 N
+63.393
−135.946
200 N
−163.830
−114.715
Rx = −18.522
Ry = −308.02
R = Rx i + Ry j
= (−18.522 N)i + (−308.02 N) j
tan α =
Ry
Rx
308.02
18.522
α = 86.559°
=
R=
308.02 N
sin 86.559
R = 309 N
86.6°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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37
!
PROBLEM 2.36
Knowing that the tension in cable BC is 725 N, determine the
resultant of the three forces exerted at Point B of beam AB.
SOLUTION
Cable BC Force:
500-N Force:
780-N Force:
and
840
= −525 N
1160
840
Fy = (725 N)
= 500 N
1160
Fx = −(725 N)
3
Fx = −(500 N) = −300 N
5
4
Fy = −(500 N) = −400 N
5
12
= 720 N
13
5
Fy = −(780 N) = −300 N
13
Fx = (780 N)
Rx = ΣFx = −105 N
R y = ΣFy = −200 N
R = (−105 N)2 + (−200 N) 2
= 225.89 N
Further:
tan α =
200
105
α = tan −1
200
105
= 62.3°
R = 226 N
Thus:
62.3°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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38
PROBLEM 2.37
Knowing that α = 40°, determine the resultant of the three forces
shown.
SOLUTION
60-lb Force:
Fx = (60 lb) cos 20° = 56.38 lb
Fy = (60 lb)sin 20° = 20.52 lb
80-lb Force:
Fx = (80 lb) cos 60° = 40.00 lb
Fy = (80 lb)sin 60° = 69.28 lb
120-lb Force:
Fx = (120 lb) cos 30° = 103.92 lb
Fy = −(120 lb)sin 30° = −60.00 lb
and
Rx = ΣFx = 200.30 lb
R y = ΣFy = 29.80 lb
R = (200.30 lb) 2 + (29.80 lb) 2
= 202.50 lb
Further:
tan α =
29.80
200.30
α = tan −1
29.80
200.30
R = 203 lb
= 8.46°
8.46°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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39
!
PROBLEM 2.38
Knowing that α = 75°, determine the resultant of the three forces
shown.
SOLUTION
60-lb Force:
Fx = (60 lb) cos 20° = 56.38 lb
Fy = (60 lb) sin 20° = 20.52 lb
80-lb Force:
Fx = (80 lb) cos 95 ° = −6.97 lb
Fy = (80 lb)sin 95° = 79.70 lb
120-lb Force:
Fx = (120 lb) cos 5 ° = 119.54 lb
Fy = (120 lb) sin 5° = 10.46 lb
Then
Rx = ΣFx = 168.95 lb
R y = ΣFy = 110.68 lb
and
R = (168.95 lb) 2 + (110.68 lb) 2
= 201.98 lb
110.68
168.95
tan α = 0.655
α = 33.23°
tan α =
R = 202 lb
33.2°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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40
PROBLEM 2.39
For the collar of Problem 2.35, determine (a) the required value of
α if the resultant of the three forces shown is to be vertical, (b) the
corresponding magnitude of the resultant.
SOLUTION
Rx = ΣFx
= (100 N) cos α + (150 N) cos (α + 30°) − (200 N) cos α
Rx = −(100 N) cos α + (150 N) cos (α + 30°)
(1)
Ry = ΣFy
= −(100 N) sin α − (150 N)sin (α + 30°) − (200 N)sin α
Ry = −(300 N) sin α − (150 N)sin (α + 30°)
(a)
(2)
For R to be vertical, we must have Rx = 0. We make Rx = 0 in Eq. (1):
−100 cos α + 150cos (α + 30°) = 0
−100cos α + 150 (cos α cos 30° − sin α sin 30°) = 0
29.904cos α = 75sin α
29.904
75
= 0.3988
α = 21.74°
tan α =
(b)
α = 21.7°
Substituting for α in Eq. (2):
Ry = −300sin 21.74° − 150sin 51.74°
= −228.9 N
R = | Ry | = 228.9 N
R = 229 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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41
PROBLEM 2.40
For the beam of Problem 2.36, determine (a) the required
tension in cable BC if the resultant of the three forces exerted at
Point B is to be vertical, (b) the corresponding magnitude of the
resultant.
SOLUTION
Rx = ΣFx = −
Rx = −
21
TBC + 420 N
29
Ry = ΣFy =
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
Substituting for TBC in Eq. (2):
20
(580 N) − 700 N
29
Ry = −300 N
Ry =
R = | Ry | = 300 N
R = 300 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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 = ΣFx = TAC sin10° + (50 lb) cos 35° + (75 lb) cos 60°
= TAC sin10° + 78.46 lb
(1)
Ry = ΣFy = (50 lb)sin 35° + (75 lb)sin 60° − TAC cos10°
Ry = 93.63 lb − TAC cos10°
(a)
(2)
Set Ry = 0 in Eq. (2):
93.63 lb − TAC cos10° = 0
TAC = 95.07 lb
(b)
TAC = 95.1 lb
Substituting for TAC in Eq. (1):
Rx = (95.07 lb) sin10° + 78.46 lb
= 94.97 lb
R = Rx
R = 95.0 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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43
!
PROBLEM 2.42
For the block of Problems 2.37 and 2.38, determine (a) the
required value of α if the resultant of the three forces shown is to
be parallel to the incline, (b) the corresponding magnitude of the
resultant.
SOLUTION
Select the x axis to be along a a′.
Then
Rx = ΣFx = (60 lb) + (80 lb) cos α + (120 lb)sin α
(1)
Ry = ΣFy = (80 lb)sin α − (120 lb) cos α
(2)
and
(a)
Set Ry = 0 in Eq. (2).
(80 lb) sin α − (120 lb) cos α = 0
Dividing each term by cos α gives:
(80 lb) tan α = 120 lb
120 lb
80 lb
α = 56.310°
tanα =
(b)
α = 56.3°
Substituting for α in Eq. (1) gives:
Rx = 60 lb + (80 lb) cos 56.31° + (120 lb)sin 56.31° = 204.22 lb
Rx = 204 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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44
PROBLEM 2.43
Two cables are tied together at C and are loaded as shown.
Knowing that α = 20°, determine the tension (a) in cable AC,
(b) in cable BC.
SOLUTION
Free-Body Diagram
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)
TBC =
1962 N
sin 50° = 1735.49 N
sin 60°
TBC = 1.735 kN
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
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)
TBC =
500 N
sin 40° = 326.35 N
sin 80°
TBC = 326 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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46
PROBLEM 2.45
Two cables are tied together at C and are loaded as shown.
Knowing that P = 500 N and α = 60°, determine the tension in
(a) in cable AC, (b) in cable BC.
SOLUTION
Free-Body Diagram
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)
TBC =
500 N
sin 75°
sin 70°
TBC = 514 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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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/s 2 )
= 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)
TBC =
(1962 N) sin 105°
sin 60°
TBC = 2190 N
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
48
PROBLEM 2.47
Knowing that α = 20°, determine the tension (a) in cable AC,
(b) in rope BC.
SOLUTION
Free-Body Diagram
Law of sines:
Force Triangle
TAC
T
1200 lb
= BC =
sin 110° sin 5° sin 65°
(a)
TAC =
1200 lb
sin 110°
sin 65°
TAC = 1244 lb
(b)
TBC =
1200 lb
sin 5°
sin 65°
TBC = 115.4 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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49
PROBLEM 2.48
Knowing that α = 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
300 lb
= BC =
sin 35° sin 50° sin 95°
(a)
FAC =
300 lb
sin 35°
sin 95°
FAC = 172.7 lb
(b)
TBC =
300 lb
sin 50°
sin 95°
TBC = 231 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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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 = 500 lb and Q = 650 lb, determine the magnitudes of
the forces exerted on the rods A and B.
SOLUTION
Free-Body Diagram
Resolving the forces into x- and y-directions:
R = P + Q + FA + FB = 0
Substituting components:
R = −(500 lb) j + [(650 lb) cos 50°]i
− [(650 lb) sin 50°] j
+ FB i − ( FA cos 50°)i + ( FA sin 50°) j = 0
In the y-direction (one unknown force)
−500 lb − (650 lb)sin 50° + FA sin 50° = 0
Thus,
FA =
500 lb + (650 lb) sin 50°
sin 50°
= 1302.70 lb
In the x-direction:
Thus,
FA = 1303 lb
(650 lb) cos 50° + FB − FA cos 50° = 0
FB = FA cos 50° − (650 lb) cos50°
= (1302.70 lb) cos 50° − (650 lb) cos 50°
= 419.55 lb
FB = 420 lb
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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 = 750 lb and FB = 400 lb, determine the magnitudes of
P and Q.
SOLUTION
Free-Body Diagram
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
− [(750 lb) cos 50°]i
+ [(750 lb)sin 50°] j + (400 lb)i
In the x-direction (one unknown force)
Q cos 50° − [(750 lb) cos 50°] + 400 lb = 0
Q=
(750 lb) cos 50° − 400 lb
cos 50°
= 127.710 lb
In the y-direction:
− P − Q sin 50° + (750 lb) sin 50° = 0
P = −Q sin 50° + (750 lb) sin 50°
= −(127.710 lb)sin 50° + (750 lb) sin 50°
= 476.70 lb
P = 477 lb; Q = 127.7 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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
ΣFx = 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
Σ Fy = 0: − FD +
With FA and FB as above:
FC = 6.40 kN
3
3
FB − FA = 0 !
5
5
3
3
FD = (16 kN) − (8 kN) !
5
5
FD = 4.80 kN
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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
3
3
ΣFy = 0: − FD − FA + FB = 0
5
5
3
FA
5
or
FB = FD +
With
FA = 5 kN, FD = 8 kN
FB =
5'
3
(
6 kN + (5 kN) *
3 )+
5
,
ΣFx = 0: − FC +
FB = 15.00 kN
4
4
FB − FA = 0
5
5
4
( FB − FA )
5
4
= (15 kN − 5 kN)
5
FC =
FC = 8.00 kN
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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54
PROBLEM 2.53
Two cables tied together at C are loaded as shown. Knowing that
Q = 60 lb, determine the tension (a) in cable AC, (b) in cable BC.
SOLUTION
ΣFy = 0: TCA − Q cos 30° = 0
With
Q = 60 lb
(a)
TCA = (60 lb)(0.866)
(b)
ΣFx = 0: P − TCB − Q sin 30° = 0
With
TCA = 52.0 lb
P = 75 lb
TCB = 75 lb − (60 lb)(0.50)
or TCB = 45.0 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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 60 lb in
either cable.
SOLUTION
Free-Body Diagram
ΣFx = 0: −TBC − Q cos 60° + 75 lb = 0
TBC = 75 lb − Q cos 60°
(1)
ΣFy = 0: TAC − Q sin 60° = 0
TAC = Q sin 60°
(2)
TAC # 60 lb:
Requirement
Q sin 60° # 60 lb
From Eq. (2):
Q # 69.3 lb
TBC # 60 lb:
Requirement
From Eq. (1):
75 lb − Q sin 60° # 60 lb
Q $ 30.0 lb
30.0 lb # Q # 69.3 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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56
PROBLEM 2.55
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 α = 30° and β = 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
ΣFx = 0: TACB cos 10° − TACB cos 30° − TCD cos 30° = 0
TCD = 0.137158TACB
(1)
ΣFy = 0: TACB sin 10° + TACB sin 30° + TCD sin 30° − 900 = 0
0.67365TACB + 0.5TCD = 900
(a)
Substitute (1) into (2):
0.67365 TACB + 0.5(0.137158 TACB ) = 900
TACB = 1212.56 N
(b)
From (1):
(2)
TCD = 0.137158(1212.56 N)
TACB = 1213 N
TCD = 166.3 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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 α = 25° and β = 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
ΣFx = 0: TACB cos 15° − TACB cos 25° − (80 N) cos 25° = 0
TACB = 1216.15 N
ΣFy = 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) TACB = 1216 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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 α.
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
sin β
sin (25° + 45°)
=
600 N
784.02 N
β = 46.0°
α = 46.0° + 25° = 71.0°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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59
PROBLEM 2.58
For the situation described in Figure P2.47, determine (a) the
value of α for which the tension in rope BC is as small as
possible, (b) the corresponding value of the tension.
PROBLEM 2.47 Knowing that α = 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,
α = 5°
= 5.00°
TBC = (1200 lb) sin 5°
TBC = 104.6 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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60
PROBLEM 2.59
For the structure and loading of Problem 2.48, determine (a) the value of α for
which the tension in cable BC is as small as possible, (b) the corresponding
value of the tension.
SOLUTION
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:
α = 90° − 30°
α = 60.0°
TBC = (300 lb)sin 50°
(b)
or
TBC = 229.81 lb
TBC = 230 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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61
PROBLEM 2.60
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)
ΣFy = 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:
BC
2.1 m
=
870 N 630 N
BC = 2.90 m
L = 2( BC )
= 5.80 m
L = 5.80 m
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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 α.
SOLUTION
Free-Body Diagram: C
Force Triangle
Force triangle is isosceles with
2 β = 180° − 85°
β = 47.5°
P = 2(800 N)cos 47.5° = 1081 N
(a)
P = 1081 N
Since P . 0, the solution is correct.
(b)
α = 180° − 50° − 47.5° = 82.5°
α = 82.5°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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 α.
SOLUTION
Free-Body Diagram
(a)
Law of cosines:
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.
P = 1294 N
(b)
Law of sines:
sin β
sin 85°
=
1200 N 1294 N
β = 67.5°
α = 180° − 50° − 67.5°
α = 62.5°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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64
!
PROBLEM 2.63
Collar A is connected as shown to a 50-lb load and can slide on
a frictionless horizontal rod. Determine the magnitude of the
force P required to maintain the equilibrium of the collar when
(a) x = 4.5 in., (b) x = 15 in.
SOLUTION
(a)
Free Body: Collar A
Force Triangle
P 50 lb
=
4.5 20.5
(b)
Free Body: Collar A
P = 10.98 lb
Force Triangle
P 50 lb
=
15
25
P = 30.0 lb
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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65
PROBLEM 2.64
Collar A is connected as shown to a 50-lb load and can slide on a
frictionless horizontal rod. Determine the distance x for which the
collar is in equilibrium when P = 48 lb.
SOLUTION
Free Body: Collar A
Force Triangle
N 2 = (50) 2 − (48) 2 = 196
N = 14.00 lb
Similar Triangles
x
48 lb
=
20 in. 14 lb
x = 68.6 in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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66
!
PROBLEM 2.65
A 160-kg load is supported by the rope-and-pulley arrangement shown.
Knowing that β = 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
ΣFx = 0: 2P sin 20° − P cos α = 0
and
cos α = 0.8452 or α = ± 46.840°
α = + 46.840
For
ΣFy = 0: 2P cos 20° + P sin 46.840° − 1569.60 N = 0
or
P = 602 N
46.8°
α = −46.840
For
ΣFy = 0: 2P cos 20° + P sin( −46.840°) − 1569.60 N = 0
or
P = 1365 N
46.8°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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67
PROBLEM 2.66
A 160-kg load is supported by the rope-and-pulley arrangement shown.
Knowing that α = 40°, determine (a) the angle β, (b) the magnitude of
the force P that must be exerted on the free end of the rope to maintain
equilibrium. (See the hint for Problem 2.65.)
SOLUTION
Free-Body Diagram: Pulley A
(a)
ΣFx = 0: 2 P sin sin β − P cos 40° = 0
1
cos 40°
2
β = 22.52°
sin β =
β = 22.5°
(b)
ΣFy = 0: P sin 40° + 2 P cos 22.52° − 1569.60 N = 0
P = 630 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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68
PROBLEM 2.67
A 600-lb 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)
ΣFy = 0: 2T − (600 lb) = 0
T=
1
(600 lb)
2
T = 300 lb
(b)
ΣFy = 0: 2T − (600 lb) = 0
T=
1
(600 lb)
2
T = 300 lb
(c)
ΣFy = 0: 3T − (600 lb) = 0
1
T = (600 lb)
3
T = 200 lb
(d)
ΣFy = 0: 3T − (600 lb) = 0
1
T = (600 lb)
3
T = 200 lb
(e)
ΣFy = 0: 4T − (600 lb) = 0
T=
1
(600 lb)
4
T = 150.0 lb
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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69
!
PROBLEM 2.68
Solve Parts b and d of Problem 2.67, assuming that
the free end of the rope is attached to the crate.
PROBLEM 2.67 A 600-lb crate is supported by
several rope-and-pulley arrangements as shown.
Determine for each arrangement the tension in the
rope. (See the hint for Problem 2.65.)
SOLUTION
Free-Body Diagram of Pulley and Crate
(b)
ΣFy = 0: 3T − (600 lb) = 0
1
T = (600 lb)
3
T = 200 lb
(d)
ΣFy = 0: 4T − (600 lb) = 0
T=
1
(600 lb)
4
T = 150.0 lb
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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70
!
PROBLEM 2.69
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
ΣFx = 0: TACB (cos 25° − cos 55°) − (750 N) cos 55° = 0
(a)
TACB = 1292.88 N
Hence:
TACB = 1293 N
ΣFy = 0: TACB (sin 25° + sin 55°) + (750 N) sin 55° − Q = 0
(b)
!
(1292.88 N)(sin 25° + sin 55°) + (750 N) sin 55° − Q = 0
Q = 2219.8 N
or
Q = 2220 N
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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71
PROBLEM 2.70
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
ΣFx = 0: TACB (cos 25° − cos 55°) − P cos 55° = 0
P = 0.58010TACB
or
(1)
ΣFy = 0: TACB (sin 25° + sin 55°) + P sin 55° − 1800 N = 0
1.24177TACB + 0.81915 P = 1800 N
or
(a)
(2)
Substitute Equation (1) into Equation (2):
1.24177TACB + 0.81915(0.58010TACB ) = 1800 N
TACB = 1048.37 N
Hence:
TACB = 1048 N
(b)
P = 0.58010(1048.37 N) = 608.16 N
Using (1),
P = 608 N
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72
PROBLEM 2.71
Determine (a) the x, y, and z components of the 750-N force, (b) the
angles θx, θy, and θz that the force forms with the coordinate axes.
SOLUTION
Fh = F sin 35°
= (750 N)sin 35°
Fh = 430.2 N
(a)
Fx = Fh cos 25°
= (430.2 N) cos 25°
Fx = +390 N,
(b)
Fy = F cos 35°
= (750 N) cos 35°
Fy = +614 N,
cos θ x =
cos θ y =
cos θ z =
Fx +390 N
=
750 N
F
Fy
F
=
+614 N
750 N
Fz +181.8 N
=
750 N
F
Fz = Fh sin 25°
= (430.2 N) sin 25°
Fz = +181.8 N
θ x = 58.7°
θ y = 35.0°
θ z = 76.0°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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73
!
PROBLEM 2.72
Determine (a) the x, y, and z components of the 900-N force, (b) the
angles θx, θy, and θz that the force forms with the coordinate axes.
SOLUTION
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 θ x =
cos θ y =
cos θ z =
Fx −130.1 N
=
900 N
F
Fy
Fz = Fh cos 20°
= (380.4 N) cos 20°
Fz = +357 N
θ x = 98.3°
+816 N
900 N
θ y = 25.0°
Fz +357 N
=
900 N
F
θ z = 66.6°
F
=
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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74
!
PROBLEM 2.73
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
θ x, θ y, and θ z 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 θ x =
110.3 N
= 287.97 N
sin 30° sin 50°
F = 288 N
Fx
110.3 N
=
= 0.38303
F 287.97 N
θ x = 67.5°
Fy = F cos 30° = 249.39
cos θ y =
Fy
F
=
249.39 N
= 0.86603
287.97 N
θ y = 30.0°
Fz = − F sin 30° cos 50°
= −(287.97 N)sin 30°cos 50°
= −92.552 N
cos θ z =
Fz −92.552 N
=
= −0.32139
F
287.97 N
θ z = 108.7°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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75
PROBLEM 2.74
A horizontal circular plate is suspended as shown from three wires that
are attached to a support at D and form 30° angles with the vertical.
Knowing that the z component of the force exerted by wire BD on the
plate is –32.14 N, determine (a) the tension in wire BD, (b) the angles
θx, θy, and θz that the force exerted at B forms with the coordinate axes.
SOLUTION
(a)
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
Fx = − F sin 30° cos 40°
= −(100.0 N)sin 30°cos 40°
= −38.302 N
cos θ x =
Fx 38.302 N
=
= −0.38302
100.0 N
F
θ x = 112.5°
Fy = F cos 30° = 86.603 N
cos θ y =
Fy
F
=
86.603 N
= 0.86603
100 N
θ y = 30.0°
Fz = −32.14 N
cos θ z =
Fz −32.14 N
=
= −0.32140
F
100 N
θ z = 108.7°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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 60 lb,
determine (a) the components of the force exerted by this wire
on the plate, (b) the angles θ x , θ y , and θ z that the force forms
with the coordinate axes.
SOLUTION
(a)
(b)
Fx = −(60 lb) sin 30°cos 60° = −15 lb
Fx = −15.00 lb
Fy = (60 lb) cos 30° = 51.96 lb
Fy = +52.0 lb
Fz = (60 lb) sin 30° sin 60° = 25.98 lb
Fz = +26.0 lb
cos θ x =
cos θ y =
cos θ z =
Fx −15.0 lb
=
= −0.25
60 lb
F
Fy
θ x = 104.5°
51.96 lb
= 0.866
60 lb
θ y = 30.0°
Fz 25.98 lb
=
= 0.433
F
60 lb
θ z = 64.3°
F
=
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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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 –20.0 lb, determine (a) the tension in wire CD, (b) the angles
θx, θy, and θz that the force exerted at C forms with the coordinate axes.
SOLUTION
(a)
Fx = − F sin 30° cos 60° = −20 lb (Given)
F=
(b)
cos θ x =
20 lb
= 80 lb
sin 30°cos 60°
F = 80.0 lb
Fx −20 lb
=
= − 0.25
80 lb
F
θ x = 104.5°
Fy = (80 lb) cos 30° = 69.282 lb
cos θ y =
Fy
F
=
69.282 lb
= 0.86615
80 lb
θ y = 30.0°
Fz = (80 lb)sin 30° sin 60° = 34.641 lb
cos θ z =
Fz 34.641
=
= 0.43301
80
F
θ z = 64.3°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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 120 lb, determine (a) the components of the force
exerted by this wire on the pole, (b) the angles θx, θy, and θz that the
force forms with the coordinate axes.
SOLUTION
(a)
Fx = (120 lb) cos 60° cos 20°
Fx = 56.382 lb
Fx = +56.4 lb
Fy = −(120 lb)sin 60°
Fy = −103.923 lb
Fy = −103.9 lb
Fz = −(120 lb) cos 60° sin 20°
Fz = −20.521 lb
(b)
cos θ x =
cos θ y =
cos θ z =
Fz = −20.5 lb
Fx 56.382 lb
=
F
120 lb
Fy
F
=
−103.923 lb
120 lb
Fz −20.52 lb
=
F
120 lb
θ x = 62.0°
θ y = 150.0°
θ z = 99.8°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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 85 lb, determine (a) the components of the force
exerted by this wire on the pole, (b) the angles θx, θy, and θz that the
force forms with the coordinate axes.
SOLUTION
(a)
Fx = (85 lb)sin 36° sin 48°
= 37.129 lb
Fx = 37.1 lb
Fy = −(85 lb) cos 36°
= −68.766 lb
Fy = −68.8 lb
Fz = (85 lb)sin 36° cos 48°
Fz = 33.4 lb
= 33.431 lb
(b)
cos θ x =
cos θ y =
cos θ z =
Fx 37.129 lb
=
F
85 lb
Fy
F
=
−68.766 lb
85 lb
Fz 33.431 lb
=
F
85 lb
θ x = 64.1°
θ y = 144.0°
θ z = 66.8°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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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
cos θ x =
cos θ y =
cos θ y =
Fx 320 N
=
F 570 N
Fy
F
=
F = 570 N
θ x = 55.8°
400 N
570 N
θ y = 45.4°
Fz −250 N
=
F
570 N
θ z = 116.0°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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81
PROBLEM 2.80
Determine the magnitude and direction of the force F = (240 N)i − (270 N)j + (680 N)k.
SOLUTION
F = Fx2 + Fy2 + Fz2
F = (240 N) 2 + (−270 N) 2 + (680 N)
cos θ x =
cos θ y =
cos θ z =
Fx 240 N
=
F 770 N
Fy
F
=
F = 770 N
θ x = 71.8°
−270 N
770 N
Fz 680 N
=
F 770 N
θ y = 110.5°
θ z = 28.0°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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82
PROBLEM 2.81
A force acts at the origin of a coordinate system in a direction defined by the angles θx = 70.9° and
θy = 144.9°. Knowing that the z component of the force is –52.0 lb, determine (a) the angle θz, (b) the other
components and the magnitude of the force.
SOLUTION
(a)
We have
(cos θ x ) 2 + (cos θ y )2 + (cos θ z ) 2 = 1
(cos θ y ) 2 = 1 − (cos θ y ) 2 − (cos θ z ) 2
Since Fz , 0 we must have cos θ z , 0
Thus, taking the negative square root, from above, we have:
cos θ z = − 1 − (cos 70.9°) 2 − (cos144.9°) 2 = 0.47282
(b)
θ z = 118.2°
Then:
F=
and
Fz
52.0 lb
=
= 109.978 lb
cos θ z 0.47282
Fx = F cos θ x = (109.978 lb) cos 70.9°
Fx = 36.0 lb
Fy = F cos θ y = (109.978 lb) cos144.9°
Fy = −90.0 lb
F = 110.0 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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 θy = 55° and θz = 45°.
Knowing that the x component of the force is − 500 lb, determine (a) the angle θx, (b) the other components
and the magnitude of the force.
SOLUTION
(a)
We have
(cos θ x ) 2 + (cos θ y )2 + (cos θ z ) 2 = 1
(cos θ y ) 2 = 1 − (cos θ y ) 2 − (cos θ z ) 2
Since Fx , 0 we must have cos θ x , 0
Thus, taking the negative square root, from above, we have:
cos θ x = − 1 − (cos 55) 2 − (cos 45) 2 = 0.41353
(b)
θ x = 114.4°
Then:
Fx
500 lb
=
= 1209.10 lb
cos θ x 0.41353
F = 1209 lb
Fy = F cos θ y = (1209.10 lb) cos 55°
Fy = 694 lb
Fz = F cos θ z = (1209.10 lb) cos 45°
Fz = 855 lb
F=
and
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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, θz = 151.2°,
and Fy < 0, determine (a) the components Fy and Fz, (b) the angles θx and θy.
SOLUTION
Fz = F cos θ 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
cos θ x =
cos θ y =
Fy = −62.0 lb
Fx
80 N
=
= 0.38095
F 210 N
Fy
F
=
61.929 N
= −0.29490
210 N
θ x = 67.6°
θ y = 107.2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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 θx = 32.5°, Fy = − 60 N,
and Fz > 0, determine (a) the components Fx and Fz, (b) the angles θy and θz.
SOLUTION
(a)
We have
Fx = F cos θ x = (230 N) cos 32.5°
Then:
Fx = −194.0 N
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
Fz = 108.036 N
Fy
−60 N
= − 0.26087
F 230 N
F 108.036 N
cos θ z = z =
= 0.46972
F
230 N
cos θ y =
=
θ y = 105.1°
θ z = 62.0°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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 525 lb, determine the
components of the force exerted by the wire on the bolt at B.
SOLUTION
!
BA = (20 ft)i + (100 ft) j − (25 ft)k
BA = (20 ft) 2 + (100 ft) 2 + (−25 ft)2
= 105 ft
F=F
BA
!
BA
=F
BA
525 lb
[(20 ft)i + (100 ft) j − (25 ft)k ]
=
105 ft
F = (100.0 lb)i + (500 lb) j − (125.0 lb)k
Fx = +100.0 lb, Fy = +500 lb, Fz = −125.0 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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 315 lb, determine the
components of the force exerted by the wire on the bolt at D.
SOLUTION
!
DA = (20 ft)i + (100 ft) j + (70 ft)k
DA = (20 ft) 2 + (100 ft) 2 + ( +70 ft) 2
= 126 ft
F=F
DA
!
DA
=F
DA
315 lb
[(20 ft)i + (100 ft) j + (74 ft)k ]
=
126 ft
F = (50 lb)i + (250 lb) j + (185 lb)k
Fx = +50 lb, Fy = +250 lb, Fz = +185.0 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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
!
DB = (480 mm)i − (510 mm) j + (320 mm)k
DB = (480 mm) 2 + (510 mm 2 ) + (320 mm) 2
= 770 mm
F=F
DB
!
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
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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
!
EB = (270 mm)i − (400 mm) j + (600 mm)k
EB = (270 mm) 2 + (400 mm) 2 + (600 mm) 2
= 770 mm
F=F
EB
!
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
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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
!
BA = −(900 mm)i + (600 mm) j + (360 mm)k
BA = (900 mm) 2 + (600 mm) 2 + (360 mm) 2
= 1140 mm
TBA = TBA
BA
!
BA
BA
1425 N
=
[ −(900 mm)i + (600 mm) j + (360 mm)k ]
1140 mm
= −(1125 N)i + (750 N) j + (450 N)k
= TBA
TBA
(TBA ) x = −1125 N, (TBA ) y = 750 N, (TBA ) z = 450 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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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
!
CA = −(900 mm)i + (600 mm) j − (920 mm)k
CA = (900 mm)2 + (600 mm)2 + (920 mm) 2
= 1420 mm
TCA = TCA λCA
!
CA
= TCA
CA
2130 N
TCA =
[−(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
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
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° sin15°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 − 0.21985]
= (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 θ x =
cos θ y =
cos θ z =
R = 515 N
Rx 174.367 N
=
= 0.33853
515.07 N
R
Ry
θ x = 70.2°
456.42 N
= 0.88613
515.07 N
θ y = 27.6°
Rz 163.011 N
=
= 0.31648
R
515.07 N
θ z = 71.5°
R
=
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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° sin15°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 θ x =
cos θ y =
cos θ z =
R = 515 N
Rx 91.532 N
=
= 0.177708
R 515.07 N
Ry
θ x = 79.8°
429.81 N
= 0.83447
515.07 N
θ y = 33.4°
Rz 268.66 N
=
= 0.52160
R 515.07 N
θ z = 58.6°
R
=
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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94
!
PROBLEM 2.93
Knowing that the tension is 425 lb in cable AB and 510 lb in
cable AC, determine the magnitude and direction of the resultant
of the forces exerted at A by the two cables.
SOLUTION
!
AB = (40 in.)i − (45 in.) j + (60 in.)k
AB = (40 in.) 2 + (45 in.) 2 + (60 in.) 2 = 85 in.
!
AC = (100 in.)i − (45 in.) j + (60 in.)k
AC = (100 in.) 2 + (45 in.) 2 + (60 in.) 2 = 125 in.
!
(40 in.)i − (45 in.) j + (60 in.)k !
AB
= (425 lb) "
TAB = TAB AB = TAB
#
85 in.
AB
$
%
TAB = (200 lb)i − (225 lb) j + (300 lb)k
!
AC
(100 in.)i − (45 in.) j + (60 in.)k !
= (510 lb) "
TAC = TAC AC = TAC
#
AC
125 in.
$
%
TAC = (408 lb)i − (183.6 lb) j + (244.8 lb)k
R = TAB + TAC = (608)i − (408.6 lb) j + (544.8 lb)k
Then:
and
R = 912.92 lb
R = 913 lb
cos θ x =
608 lb
= 0.66599
912.92 lb
cos θ y =
408.6 lb
= −0.44757
912.92 lb
θ y = 116.6°
cos θ z =
544.8 lb
= 0.59677
912.92 lb
θ z = 53.4°
θ x = 48.2°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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95
!
PROBLEM 2.94
Knowing that the tension is 510 lb in cable AB and 425 lb in cable
AC, determine the magnitude and direction of the resultant of the
forces exerted at A by the two cables.
SOLUTION
!
AB = (40 in.)i − (45 in.) j + (60 in.)k
AB = (40 in.) 2 + (45 in.) 2 + (60 in.) 2 = 85 in.
!
AC = (100 in.)i − (45 in.) j + (60 in.)k
AC = (100 in.) 2 + (45 in.) 2 + (60 in.) 2 = 125 in.
!
(40 in.)i − (45 in.) j + (60 in.)k !
AB
= (510 lb) "
TAB = TAB AB = TAB
#
85 in.
AB
$
%
TAB = (240 lb)i − (270 lb) j + (360 lb)k
!
AC
(100 in.)i − (45 in.) j + (60 in.)k !
= (425 lb) "
TAC = TAC AC = TAC
#
AC
125 in.
$
%
TAC = (340 lb)i − (153 lb) j + (204 lb)k
R = TAB + TAC = (580 lb)i − (423 lb) j + (564 lb)k
Then:
and
R = 912.92 lb
R = 913 lb
cos θ x =
580 lb
= 0.63532
912.92 lb
cos θ y =
−423 lb
= −0.46335
912.92 lb
cos θ z =
564 lb
= 0.61780
912.92 lb
θ x = 50.6°
θ y = 117.6°
θ z = 51.8°
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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
!
BD = −(480 mm)i + (510 mm) j − (320 mm)k
BD = (480 mm) 2 + (510 mm) 2 + (320 mm) 2 = 770 mm
!
BD
FBD = TBD 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
!
BE = −(270 mm)i + (400 mm) j − (600 mm)k
BE = (270 mm) 2 + (400 mm) 2 + (600 mm) 2 = 770 mm
!
BE
FBE = TBE 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
cos θ x =
−375 N
747.83 N
θ x = 120.1°
cos θ y =
455 N
747.83 N
θ y = 52.5°
cos θ z =
−460 N
747.83 N
θ z = 128.0°
!
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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 = ΣFx = +1125 + 1350 = +2475 N
Ry = ΣFy = −750 − 900 = −1650 N
Rz = ΣFz = −450 + 1380 = + 930 N
R = Rx2 + Ry2 + Rz2
= (+2475) 2 + (−1650) 2 + ( +930)2
= 3116.6 N
cos θ x =
cos θ y =
cos θ z =
R = 3120 N
Rx +2475
=
R 3116.6
Ry
θ x = 37.4°
−1650
3116.6
θ y = 122.0°
Rz
+ 930
=
R 3116.6
θ z = 72.6°
R
=
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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 150 lb and that the resultant of the forces exerted
at A by wires AC and AD must be contained in the xy plane,
determine (a) the tension in AD, (b) the magnitude and direction
of the resultant of the two forces.
SOLUTION
R = TAC + TAD
= (150 lb)(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)
(1)
Since Rz = 0, The coefficient of k must be zero.
(150 lb)( − cos 60° sin 20°) + TAD (sin 36° cos 48°) = 0
TAD = 65.220 lb
(b)
TAD = 65.2 lb
Substituting for TAD into Eq. (1) gives:
R = [(150 lb) cos 60° cos 20° + (65.220 lb) sin 36° sin 48°)]i
− [(150 lb) sin 60° + (65.220 lb) cos 36°]j + 0
R = (98.966 lb)i − (182.668 lb) j
R = (98.966 lb)2 + (182.668 lb) 2
= 207.76 lb
R = 208 lb
cos θ x =
98.966 lb
207.76 lb
θ x = 61.6°
cos θ y =
182.668 lb
207.76 lb
θ y = 151.6°
cos θ z = 0
θ z = 90.0°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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 125 lb and that the resultant of the forces exerted
at A by wires AC and AD must be contained in the xy plane,
determine (a) the tension in AC, (b) the magnitude and direction
of the resultant of the two forces.
SOLUTION
R = TAC + TAD
= TAC (cos 60° cos 20°i − sin 60° j − cos 60° sin 20°k )
+ (125 lb)(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°) + (125 lb)(sin 36° cos 48°) = 0
TAC = 287.49 lb
(b)
TAC = 287 lb
Substituting for TAC into Eq. (1) gives:
R = [(287.49 lb) cos 60° cos 20° + (125 lb) sin 36° sin 48°]i
− [(287.49 lb) sin 60° + (125 lb) cos 36°]j + 0
R = (189.677 lb)i − (350.10 lb) j
R = (189.677 lb)2 + (350.10 lb) 2
R = 398 lb
= 398.18 lb
cos θ x =
189.677 lb
398.18 lb
θ x = 61.6°
cos θ y =
350.10 lb
398.18 lb
θ y = 151.6°
cos θ z = 0
θ z = 90.0°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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
!
AB = − (4.20 m)i − (5.60 m) j
AB = 7.00 m
!
AC = (2.40 m)i − (5.60 m) j + (4.20 m)k AC = 7.40 m
!
AD = − (5.60 m) j − (3.30 m)k
AD = 6.50 m
!
AB
and
TAB = TAB AB = TAB
= (− 0.6i − 0.8 j)TAB
AB!
AC
TAC = TAC AC = TAC
= (0.32432 − 0.75676 j + 0.56757k )TAC
AC!
AD
TAD = TAD AD = TAD
= ( − 0.86154 j − 0.50769k )TAD
AD
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101
PROBLEM 2.99 (Continued)
Equilibrium condition
ΣF = 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
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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
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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
TAC =
Substituting into expressions for TAB and TAD gives:
TAB = 0.54053(371.69 N)
TAD = 1.11795(371.69 N)
TAB = 201 N, TAC = 372 N, TAD = 416 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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 750 lb.
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
!
AB = − (36 in.)i + (60 in.) j − (27 in.)k
AB = 75 in.
!
AC = (60 in.) j + (32 in.)k
AC = 68 in.
!
AD = (40 in.)i + (60 in.) j − (27 in.)k
AD = 77 in.
!
AB
TAB = TAB AB = TAB
and
AB
= (− 0.48i + 0.8 j − 0.36k )TAB
!
AC
TAC = TAC AC = TAC
AC
= (0.88235 j + 0.47059k )TAC
!
AD
TAD = TAD AD = TAD
AD
= (0.51948i + 0.77922 j − 0.35065k )TAD
Equilibrium Condition with
W = − Wj
ΣF = 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:
−0.48TAB + 0.51948TAD = 0
0.8TAB + 0.88235TAC + 0.77922TAD − W = 0
−0.36TAB + 0.47059TAC − 0.35065TAD = 0
Substituting TAB = 750 lb in Equations (1), (2), and (3) and solving the resulting set of equations, using
conventional algorithms for solving linear algebraic equations, gives:
TAC = 1090.1 lb
TAD = 693 lb
W = 2100 lb
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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 616 lb.
SOLUTION
See Problem 2.103 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3)
below:
− 0.48TAB + 0.51948TAD = 0
0.8TAB + 0.88235TAC + 0.77922TAD − W = 0
−0.36TAB + 0.47059TAC − 0.35065TAD = 0
Substituting TAD = 616 lb in Equations (1), (2), and (3) above, and solving the resulting set of equations using
conventional algorithms, gives:
TAB = 667.67 lb
TAC = 969.00 lb
W = 1868 lb
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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 544 lb.
SOLUTION
See Problem 2.103 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3)
below:
− 0.48TAB + 0.51948TAD = 0
0.8TAB + 0.88235TAC + 0.77922TAD − W = 0
− 0.36TAB + 0.47059TAC − 0.35065TAD = 0
Substituting TAC = 544 lb in Equations (1), (2), and (3) above, and solving the resulting set of equations using
conventional algorithms, gives:
TAB = 374.27 lb
TAD = 345.82 lb
W = 1049 lb
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109
PROBLEM 2.106
A 1600-lb 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:
−0.48TAB + 0.51948TAD = 0
0.8TAB + 0.88235TAC + 0.77922TAD − W = 0
−0.36TAB + 0.47059TAC − 0.35065TAD = 0
Substituting W = 1600 lb in Equations (1), (2), and (3) above, and solving the resulting set of equations using
conventional algorithms, gives:
TAB = 571 lb
TAC = 830 lb
TAD = 528 lb
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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
!
ΣFA = 0: TAB + TAC + TAD + P = 0 where P = Pi
!
AB = −(960 mm)i − (240 mm)j + (380 mm)k
AB = 1060 mm
!
AC = −(960 mm)i − (240 mm) j − (320 mm)k AC = 1040 mm !
!
AD = −(960 mm)i + (720 mm) j − (220 mm)k AD = 1220 mm
!
48 12
19 !
AB
= TAB " − i − j + k #
TAB = TAB AB = TAB
53
53
53 %
AB
$
!
12
3
4 !
AC
TAC = TAC AC = TAC
= TAC " − i − j − k #
13
13
13
AC
$
%
305 N
[( −960 mm)i + (720 mm) j − (220 mm)k ]
TAD = TAD AD =
1220 mm
= −(240 N)i + (180 N) j − (55 N)k
Substituting into ΣFA = 0, factoring i, j, k , and setting each coefficient equal to φ gives:
i: P =
48
12
TAB + TAC + 240 N
53
13
(1)
j:
12
3
TAB + TAC = 180 N
53
13
(2)
k:
19
4
TAB − TAC = 55 N
53
13
(3)
Solving the system of linear equations using conventional algorithms gives:
TAB = 446.71 N
TAC = 341.71 N
P = 960 N
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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
ΣFA = 0: TAB + TAC + Qj + (1200 N)i = 0
!
AB = −(960 mm)i − (240 mm)j + (380 mm)k AB = 1060 mm
!
AC = −(960 mm)i − (240 mm) j − (320 mm)k AC = 1040 mm
!
AB
48 12
19 !
TAB = TAB AB = TAB
= " − i − j + k # TAB
AB $ 53 53
53 %
!
AC
12
3
4 !
TAC = TAC AC = TAC
= " − i − j − k # TAC
AC $ 13 13 13 %
Substituting into ΣFA = 0, factoring i, j, k , and setting each coefficient equal to φ gives:
i: −
48
12
TAB − TAC + 1200 N = 0
53
13
(1)
j: −
12
3
TAB − TAC + Q = 0
53
13
(2)
k:
19
4
TAB − TAC = 0
53
13
(3)
Solving the resulting system of linear equations using conventional algorithms gives:
TAB = 605.71 N
TAC = 705.71 N
Q = 300.00 N
0 # Q , 300 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. !
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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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 630 lb, determine the vertical force P exerted by the
tower on the pin at A.
SOLUTION
Free Body A:
We write
ΣF = 0: TAB + TAC + TAD + Pj = 0
!
AB = −45i − 90 j + 30k AB = 105 ft
!
AC = 30i − 90 j + 65k AC = 115 ft
!
AD = 20i − 90 j − 60k AD = 110 ft
!
AB
TAB = TAB AB = TAB
AB
3
6
2 !
= " − i − j + k # TAB
7
7 %
$ 7
!
AC
TAC = TAC AC = TAC
AC
6
18
13 !
j + k # TAC
=" i−
23
23 %
$ 23
!
AD
TAD = TAD AD = TAD
AD
2
9
6 !
= " i − j − k # TAD
11
11
11
$
%
Substituting into the Eq. ΣF = 0 and factoring i, j, k :
3
6
2
!
" − TAB + TAC + TAD # i
7
23
11
$
%
6
18
9
!
+ " − TAB − TAC − TAD + P # j
23
11
$ 7
%
2
13
6
!
+ " TAB + TAC − TAD # k = 0
23
11
$7
%
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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113
PROBLEM 2.109 (Continued)
Setting the coefficients of i, j, k , equal to zero:
i:
3
6
2
− TAB + TAC + TAD = 0
7
23
11
(1)
j:
6
18
9
− TAB − TAC − TAD + P = 0
7
23
11
(2)
k:
2
13
6
TAB + TAC − TAD = 0
7
23
11
(3)
Set TAB = 630 lb in Eqs. (1) – (3):
6
2
TAC + TAD = 0
23
11
(1′)
18
9
TAC − TAD + P = 0
23
11
(2′)
13
6
TAC − TAD = 0
23
11
(3′)
−270 lb +
−540 lb −
180 lb +
Solving,
TAC = 467.42 lb TAD = 814.35 lb P = 1572.10 lb
P = 1572 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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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 920 lb, 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:
3
6
2
− TAB + TAC + TAD = 0
7
23
11
(1)
6
18
9
− TAB − TAC − TAD + P = 0
7
23
11
(2)
2
13
6
TAB + TAC − TAD = 0
7
23
11
(3)
Substituting for TAC = 920 lb in Equations (1), (2), and (3) above and solving the resulting set of equations
using conventional algorithms gives:
Solving,
3
2
− TAB + 240 lb + TAD = 0
7
11
(1′)
6
9
− TAB − 720 lb − TAD + P = 0
7
11
(2′)
2
6
TAB + 520 lb − TAD = 0
7
11
(3′)
TAB = 1240.00 lb
TAD = 1602.86 lb
P = 3094.3 lb
P = 3090 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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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:
ΣF = 0: TAB + TAC + TAD + Pj = 0
We have:
!
AB = −(320 mm)i − (480 mm)j + (360 mm)k AB = 680 mm
!
AC = (450 mm)i − (480 mm) j + (360 mm)k AC = 750 mm
!
AD = (250 mm)i − (480 mm) j − ( 360 mm ) k AD = 650 mm
Thus:
TAB = TAB
AB
TAC = TAC
AC
TAD = TAD
AD
!
AB
8
12
9 !
= − i − j + k TAB
AB "$ 17 17 17 #%
!
AC
= TAC
= ( 0.6i − 0.64 j + 0.48k ) TAC
AC
!
AD
5
9.6
7.2 !
= TAD
=" i−
j−
k TAD
AD $ 13
13
13 #%
= TAB
Dimensions in mm
Substituting into the Eq. ΣF = 0 and factoring i, j, k :
8
5
!
" − TAB + 0.6TAC + TAD # i
13
$ 17
%
12
9.6
!
TAD + P # j
+ " − TAB − 0.64TAC −
17
13
$
%
9
7.2
!
TAD # k = 0
+ " TAB + 0.48TAC −
13
$ 17
%
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116
PROBLEM 2.111 (Continued)
Setting the coefficient of i, j, k equal to zero:
i:
−
8
5
TAB + 0.6TAC + TAD = 0
17
13
(1)
j:
−
12
9.6
TAB − 0.64TAC −
TAD + P = 0
7
13
(2)
9
7.2
TAB + 0.48TAC −
TAD = 0
17
13
(3)
8
5
TAB + 36 N + TAD = 0
17
13
(1′)
9
7.2
TAB + 28.8 N −
TAD = 0
17
13
(3′)
k:
Making TAC = 60 N in (1) and (3):
−
Multiply (1′) by 9, (3′) by 8, and add:
554.4 N −
12.6
TAD = 0 TAD = 572.0 N
13
Substitute into (1′) and solve for TAB :
TAB =
17
5
!
36 + × 572 #
"
8 $
13
%
TAB = 544.0 N
Substitute for the tensions in Eq. (2) and solve for P:
12
9.6
(544 N) + 0.64(60 N) +
(572 N)
17
13
= 844.8 N
P=
Weight of plate = P = 845 N
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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
(1)
12
9.6
TAB + 0.64 TAC −
TAD + P = 0
17
13
(2)
9
7.2
TAB + 0.48TAC −
TAD = 0
17
13
(3)
−
−
Making TAD = 520 N in Eqs. (1) and (3):
8
TAB + 0.6TAC + 200 N = 0
17
(1′)
9
TAB + 0.48TAC − 288 N = 0
17
(3′)
−
Multiply (1′) by 9, (3′) by 8, and add:
9.24TAC − 504 N = 0 TAC = 54.5455 N
Substitute into (1′) and solve for TAB :
TAB =
17
(0.6 × 54.5455 + 200) TAB = 494.545 N
8
Substitute for the tensions in Eq. (2) and solve for P:
12
9.6
(494.545 N) + 0.64(54.5455 N) +
(520 N)
17
13
Weight of plate = P = 768 N
= 768.00 N
P=
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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 2100 lb.
SOLUTION
See Problem 2.109 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and
(3) below:
3
6
2
− TAB + TAC + TAD = 0
7
23
11
(1)
6
18
9
− TAB − TAC − TAD + P = 0
7
23
11
(2)
2
13
6
TAB + TAC − TAD = 0
7
23
11
(3)
Substituting for P = 2100 lb in Equations (1), (2), and (3) above and solving the resulting set of equations
using conventional algorithms gives:
3
6
2
− TAB + TAC + TAD = 0
7
23
11
(1′)
6
18
9
− TAB − TAC − TAD + 2100 lb = 0
7
23
11
(2′)
2
13
6
TAB + TAC − TAD = 0
7
23
11
(3′)
TAB = 841.55 lb
TAC = 624.38 lb
TAD = 1087.81 lb
TAB = 842 lb
TAC = 624 lb
TAD = 1088 lb
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119
!
PROBLEM 2.114
A horizontal circular plate weighing 60 lb 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
ΣFx = 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)
ΣFy = 0: −TAD (cos 30°) − TBD (cos 30°) − TCD (cos 30°) + 60 lb = 0
TAD + TBD + TCD = 69.282 lb
or
(2)
ΣFz = 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 = 29.5 lb
TBD = 10.25 lb
!
!
!
TCD = 29.5 lb
!
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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
(1)
12
9.6
TAB − 0.64TAC −
TAD + 792 N = 0
17
13
(2)
9
7.2
TAB + 0.48TAC −
TAD = 0
17
13
(3)
−
−
Solving Equations (1), (2), and (3) by conventional algorithms gives
TAB = 510.00 N
TAB = 510 N
TAC = 56.250 N
TAC = 56.2 N
TAD = 536.25 N
TAD = 536 N
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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
ΣFA = 0: TAB + TAC + TAD + P + Q = 0
Where
P = Pi and Q = Qj
!
AB = −(960 mm)i − (240 mm) j + (380 mm)k AB = 1060 mm
!
AC = −(960 mm)i − (240 mm) j − (320 mm)k AC = 1040 mm
!
AD = −(960 mm)i + (720 mm) j − (220 mm)k AD = 1220 mm
!
48 12
19 !
AB
TAB = TAB AB = TAB
= TAB " − i − j + k #
53
53
53 %
AB
$
!
12
3
4 !
AC
TAC = TAC AC = TAC
= TAC " − i − j − k #
13
13
13
AC
$
%
!
48
36
11 !
AD
= TAD " − i +
TAD = TAD AD = TAD
j− k#
61
61
61 %
AD
$
Substituting into ΣFA = 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
TAC = 1025 N
TAD = 915 N
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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
(1)
−
12
3
36
TAB − TAC + TAD + Q = 0
53
13
61
(2)
19
4
11
TAB − TAC − TAD = 0
53
13
61
(3)
Setting P = 2880 N and Q = 576 N gives:
−
48
12
48
TAB − TAC − TAD + 2880 N = 0
53
13
61
(1′)
12
3
36
TAB − TAC + TAD + 576 N = 0
53
13
61
(2′)
19
4
11
TAB − TAC − TAD = 0
53
13
61
(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
!
TAC = 1560 N
TAD = 183.0 N
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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
(1)
−
12
3
36
TAB − TAC + TAD + Q = 0
53
13
61
(2)
19
4
11
TAB − TAC − TAD = 0
53
13
61
(3)
Setting P = 2880 N and Q = −576 N gives:
−
48
12
48
TAB − TAC − TAD + 2880 N = 0
53
13
61
(1′)
12
3
36
TAB − TAC + TAD − 576 N = 0
53
13
61
(2′)
19
4
11
TAB − TAC − TAD = 0
53
13
61
(3′)
−
Solving the resulting set of equations using conventional algorithms gives:
TAB = 1249.29 N
TAC = 490.31 N
TAD = 1646.97 N
TAB = 1249 N
TAC = 490 N
TAD = 1647 N
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125
PROBLEM 2.119
Using two ropes and a roller chute, two workers are unloading a 200-lb
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, –20 in., 40 in.), B(–40 in.,
50 in., 0), and C(45 in., 40 in., 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
!
AB = (40 in.)i + (70 in.)j − (40 in.)k
and
AB = (40 in.) 2 + (70 in.) 2 + (40 in.)2
= 90 in.
!
AB
TAB = T AB = TAB
AB
TAB
[(−40 in.)i + (70 in.)j − (40 in.)k ]
=
90 in.
4
7
4 !
TAB = TAB " − i + j − k #
9
9 %
$ 9
!
AC = (45 in.)i + (60 in.)j − (40 in.)k
AC = (45 in.) 2 + (60 in.)2 + (40 in.)2 = 85 in.
!
AC
TAC = T AC = TAC
AC
TAC
[(45 in.)i + (60 in.)j − (40 in.)k ]
=
85 in.
9
12
8 !
TAC = TAC " i + j − k #
$ 17 17 17 %
Then:
ΣF = 0: N + TAB + TAC + W = 0
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126
PROBLEM 2.119 (Continued)
With W = 200 lb, and equating the factors of i, j, and k to zero, we obtain the linear algebraic equations:
i:
j:
k:
4
9
− TAB + TAC = 0
9
17
(1)
7
12
2
TAB + TAC +
− 200 lb = 0
9
17
5
(2)
4
8
1
− TAB − TAC +
N =0
9
17
5
(3)
Using conventional methods for solving linear algebraic equations we obtain:
TAB = 65.6 lb
TAC = 55.1 lb
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127
PROBLEM 2.120
Solve Problem 2.119 assuming that a third worker is exerting a force
P = −(40 lb)i on the counterweight.
PROBLEM 2.119 Using two ropes and a roller chute, two workers are
unloading a 200-lb 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, –20 in., 40 in.),
B(–40 in., 50 in., 0), and C(45 in., 40 in., 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.
4
7
4 !
TAB = TAB " − i + j − k #
9
9 %
$ 9
9
12
8 !
TAC = TAC " i + j − k #
17
17
17
$
%
Then:
Where
and
ΣFA = 0: N + TAB + TAC + P + W = 0
P = −(40 lb)i
W = (200 lb)j
Equating the factors of i, j, and k to zero, we obtain the linear equations:
4
9
i : − TAB + TAC − 40 lb = 0
9
17
j:
k:
2
7
12
N + TAB + TAC − 200 lb = 0
9
17
5
1
5
4
8
N − TAB − TAC = 0
9
17
Using conventional methods for solving linear algebraic equations we obtain
TAB = 24.8 lb
TAC = 96.4 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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
AB
"
AB
=T
AB
(−130 mm)i + (400 mm) j + (160 mm)k
=T
450 mm
13
40
16 !
=T "− i +
j+ k#
45
45 %
$ 45
TAC = T
AC
"
AC
=T
AC
( −150 mm)i + (400 mm) j + (−240 mm)k
=T
490 mm
15
40
24 !
= T "− i +
j− k#
49
49
49 %
$
ΣF = 0: TAB + TAC + Q + P + W = 0
Setting coefficients of i, j, k equal to zero:
i: −
13
15
T − T +P=0
45
49
0.59501T = P
(1)
j: +
40
40
T + T −W = 0
45
49
1.70521T = W
(2)
k: +
16
24
T − T +Q =0
45
49
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
0.134240(220.50 N) = Q
Q = 29.6 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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
Eq. (3):
0.134240(275.63 N) = Q
Q = 37.0 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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
!
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
AB
= TAB
!
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
!
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
!
AC TAC
TAC = T AC = TAC
=
[(0)i + (1.6 m) j + (1.2 m)k ]
AC 2 m
TAC = TAC (0.8 j + 0.6k )
!
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
!
T
AD
TAD = T AD = TAD
= AD [(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)
!
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
!
AE
TAE = T 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:
ΣF = 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
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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) W = 1202 N
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134
!
PROBLEM 2.125
Collars A and B are connected by a 25-in.-long wire and can slide
freely on frictionless rods. If a 60-lb force Q is applied to collar B as
shown, determine (a) the tension in the wire when x = 9 in., (b) the
corresponding magnitude of the force P required to maintain the
equilibrium of the system.
SOLUTION
Free Body Diagrams of Collars:
A:
B:
AB
ΣF = 0: Pi + N y j + N z k + TAB λ AB = 0
Collar A:
Substitute for
!
AB − xi − (20 in.) j + zk
=
=
AB
25 in.
AB
and set coefficient of i equal to zero:
P−
AB
and set coefficient of k equal to zero:
60 lb −
x = 9 in.
(a)
From Eq. (2):
(b)
(1)
ΣF = 0: (60 lb)k + N x′ i + N y′ j − TAB λ AB = 0
Collar B:
Substitute for
TAB x
=0
25 in.
From Eq. (1):
TAB z
=0
25 in.
(2)
(9 in.)2 + (20 in.) 2 + z 2 = (25 in.) 2
z = 12 in.
60 lb − TAB (12 in.)
25 in.
P=
(125.0 lb)(9 in.)
!
25 in.
TAB = 125.0 lb
P = 45.0 lb
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135
!
PROBLEM 2.126
Collars A and B are connected by a 25-in.-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 = 120 lb
and Q = 60 lb.
SOLUTION
See Problem 2.125 for the diagrams and analysis leading to Equations (1) and (2) below:
P=
TAB x
=0
25 in.
(1)
60 lb −
TAB z
=0
25 in.
(2)
For P = 120 lb, Eq. (1) yields
TAB x = (25 in.)(20 lb)
(1′)
From Eq. (2)
TAB z = (25 in.) (60 lb)
(2′)
x
=2
z
Dividing Eq. (1′) by (2′):
Now write
(3)
x 2 + z 2 + (20 in.) 2 = (25 in.) 2
(4)
Solving (3) and (4) simultaneously
4 z 2 + z 2 + 400 = 625
z 2 = 45
z = 6.708 in.
From Eq. (3)
x = 2 z = 2(6.708 in.)
= 13.416 in.
x = 13.42 in., z = 6.71 in.
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136
!
PROBLEM 2.127
The direction of the 75-lb forces may vary, but the angle between the forces is
always 50°. Determine the value of for which the resultant of the forces acting at A
is directed horizontally to the left.
SOLUTION
We must first replace the two 75-lb forces by their resultant R1 using the triangle rule.
R1 = 2(75 lb) cos 25°
= 135.946 lb
R1 = 135.946 lb
α + 25°
Next we consider the resultant R 2 of R1 and the 240-lb force where R 2 must be horizontal and directed to
the left. Using the triangle rule and law of sines,
sin (α + 25°) sin (30°)
=
240 lb
135.946
sin (α + 25°) = 0.88270
α + 25° = 61.970°
α = 36.970°
α = 37.0°
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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 40-lb vertical force.
SOLUTION
Triangle rule:
Law of cosines:
Law of sines:
P 2 = (30) 2 + (40) 2 − 2(30)(40) cos 25°
P = 18.0239 lb
sin α
sin 25°
=
30 lb 18.0239 lb
α = 44.703°
90° − α = 45.297°
P = 18.02 lb
45.3°
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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 240-lb 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°
=
240 lb
sin 40°
or P = 373 lb
=
240 lb
tan 40°
or Px = 286 lb
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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:
ΣFx = 0: −
12 ft
7.5 ft
TAC +
TBC = 0
12.5 ft
8.5 ft
TBC = 1.08800TAC
ΣFy = 0:
(a)
3.5 ft
4 ft
TAC +
TBC − 396 lb = 0
12 ft
8.5 ft
3.5 ft
4 ft
(1.08800TAC ) − 396 lb = 0
TAC +
12.5 ft
8.5 ft
(0.28000 + 0.51200)TAC = 396 lb
TAC = 500.0 lb
(b)
TBC = (1.08800)(500.0 lb)
TAC = 500 lb
TBC = 544 lb
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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)
ΣFx = 0: −
(b)
ΣFy = 0:
12
4
TAC + (360 N) = 0
13
5
TAC = 312 N
5
3
(312 N) + TBC + (360 N) − 480 N = 0
13
5
TBC = 480 N − 120 N − 216 N
TBC = 144 N
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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
ΣFx = 0: −
12
4
TAC + P = 0
13
5
TAC =
ΣFy = 0:
Substitute for TAC from (1):
13
P
15
(1)
5
3
TAC + TBC + P − 480 N = 0
13
5
3
! 5 "! 13 "
# 13 $# 15 $ P + TBC + 5 P − 480 N = 0
% &% &
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
0 , P , 514 N
Allowable range:
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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 θ x = 69.3° and
θ z = 57.9°. Knowing that the y component of the force is –174.0 lb, determine (a) the angle θ y , (b) the other
components and the magnitude of the force.
SOLUTION
(a)
To determine θ y , use the relation
cos 2 θ y + cos 2 θ y + cos 2 θ z = 1
or
cos 2 θ y = 1 − cos 2 θ x − cos 2 θ y
Since Fy , 0, we must have cos θ y , 0
cos θ y = − 1 − cos 2 69.3° − cos 2 57.9°
θ y = 140.3°
= − 0.76985
(b)
Fy
−174.0 lb
= 226.02 lb
−0.76985
F = 226 lb
Fx = F cos θ x = (226.02 lb) cos 69.3°
Fx = 79.9 lb
Fz = F cos θ z = (226.02 lb) cos 57.9°
Fz = 120.1 lb
F=
cos θ y
=
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143
!
PROBLEM 2.134
Cable AB is 65 ft long, and the tension in that cable is 3900 lb.
Determine (a) the x, y, and z components of the force exerted by
the cable on the anchor B, (b) the angles θ x , θ y , and θ z defining
the direction of that force.
SOLUTION
56 ft
65 ft
= 0.86154
cos θ y =
From triangle AOB:
θ y = 30.51°
Fx = − F sin θ y cos 20°
(a)
= −(3900 lb)sin 30.51° cos 20°
Fx = −1861 lb
Fy = + F cos θ y = (3900 lb)(0.86154)
!
Fz = + (3900 lb)sin 30.51° sin 20°
(b)
cos θ x =
From above:
Fx
1861 lb
=−
= − 0.4771
3900 lb
F
θ y = 30.51°
cos θ z =
Fz
677 lb
=+
= + 0.1736
3900 lb
F
Fy = +3360 lb
Fz = +677 lb
θ x = 118.5°
θ y = 30.5°
θ z = 80.0°
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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
!!!"
AB = −15.588i + 15 j + 12k
AB = 24.739 m
!!!"
AC = −15.588i + 18.60 j − 15k
AC = 28.530 m
!!!"
AB
TAB = TAB !AB = TAB
AB
−15.588i + 15 j + 12k
TAB = (10 kN)
24.739
TAB = (6.301 kN)i + (6.063 kN) j + (4.851 kN)k
!!!"
−15.588i + 18.60 j − 15k
AC
TAC = TAC !AC = TAC
(7.5 kN)
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
R = 15.13 kN
= 15.130 kN
cos θ x =
cos θ y =
cos θ z =
Rx −10.399 kN
=
= − 0.6873
15.130 kN
R
Ry
θ z = 133.4°
10.953 kN
= 0.7239
15.130 kN
θ y = 43.6°
Rz
0.908 kN
=
= 0.0600
R 15.130 kN
θ z = 86.6°
R
=
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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:
ΣF = 0: TAB + TAC + TAD + W = 0
!!!"
AB = (450 mm)i + (600 mm) j
AB = 750 mm
!!!"
AC = (600 mm) j − (320 mm)k
AC = 680 mm
!!!"
AD = (500 mm)i + (600 mm) j + (360 mm)k
AD = 860 mm
!!!"
AB ! 450
600 "
=#
TAB = TAB λ AB = TAB
i+
j TAB
AB % 750
750 $&
= (0.6i + 0.8 j)TAB
!!!"
AC ! 600
320 "
8 "
! 15
TAC = TAC λ AC = TAC
=#
j−
k $ TAC = # j − k $ TAC
680 &
AC % 680
% 17 17 &
!!!"
600
360 "
AD ! 500
= #−
TAD = TAD λ AD = TAD
i+
j+
k TAD
AD % 860
860
860 $&
30
18 "
! 25
TAD = # − i +
j + k $ TAD
43
43 &
% 43
Substitute into ΣF = 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 "
!
# 0.8 × 0.96899 + 17 × 0.88953 + 53 $ TAD − 1165 N = 0
%
&
2.2578TAD − 1165 N = 0
TAD = 516 N
From (1):
TAB = 0.96899(516 N)
TAB = 500 N
From (3):
TAC = 0.88953(516 N)
TAC = 459 N
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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
For both Problems 2.137 and 2.138:
Free Body Diagrams of Collars:
( AB) 2 = x 2 + y 2 + z 2
Here
(0.525 m) 2 = (.20 m) 2 + y 2 + z 2
y 2 + z 2 = 0.23563 m 2
or
Thus, why y given, z is determined,
Now
!AB
!!!"
AB
=
AB
1
(0.20i − yj + zk )m
0.525 m
= 0.38095i − 1.90476 yj + 1.90476 zk
=
Where y and z are in units of meters, m.
From the F.B. Diagram of collar A:
ΣF = 0: N x i + N z k + Pj + TAB λ AB = 0
Setting the j coefficient to zero gives:
P − (1.90476 y )TAB = 0
P = 341 N
With
TAB =
341 N
1.90476 y
Now, from the free body diagram of collar B:
ΣF = 0: N x i + N y j + Qk − TAB !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 ) =
y
(1.90476) 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 m 2 − (0.155 m) 2
z = 0.46 m
and
341 N
0.155(1.90476)
= 1155.00 N
TAB =
(a)
TAB = 1.155 kN
or
and
341 N(0.46 m)(0.866)
(0.155 m)
= (1012.00 N)
Q=
(b)
Q = 1.012 kN
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 m 2
341 N
TAB =
1.90476 y
341 N
Q=
z
y
With y = 275 mm = 0.275 m, we obtain:
z 2 = 0.23563 m 2 − (0.275 m) 2
z = 0.40 m
and
TAB =
(a)
341 N
= 651.00
(1.90476)(0.275 m)
TAB = 651 N
or
and
Q=
(b)
341 N(0.40 m)
(0.275 m)
Q = 496 N
or
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150
!
CHAPTER 3
PROBLEM 3.1
A foot valve for a pneumatic system is hinged at B. Knowing
that a = 28°, determine the moment of the 16-N force about
Point B by resolving the force into horizontal and vertical
components.
SOLUTION
Note that
and
θ = α − 20° = 28° − 20° = 8°
Fx = (16 N) cos8° = 15.8443 N
Fy = (16 N) sin 8° = 2.2268 N
Also
x = (0.17 m) cos 20° = 0.159748 m
y = (0.17 m) sin 20° = 0.058143 m
Noting that the direction of the moment of each force component about B is counterclockwise,
MB = xFy + yFx
= (0.159748 m)(2.2268 N)
+ (0.058143 m)(15.8443 N)
= 1.277 N ⋅ m
or M B = 1.277 N ⋅ m
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153
PROBLEM 3.2
A foot valve for a pneumatic system is hinged at B. Knowing
that a = 28°, determine the moment of the 16-N force about
Point B by resolving the force into components along ABC and
in a direction perpendicular to ABC.
SOLUTION
First resolve the 4-lb force into components P and Q, where
Q = (16 N) sin 28°
= 7.5115 N
Then
M B = rA/B Q
= (0.17 m)(7.5115 N)
or M B = 1.277 N ⋅ m
= 1.277 N ⋅ m
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154
PROBLEM 3.3
A 300-N force is applied at A as shown. Determine (a) the moment
of the 300-N force about D, (b) the smallest force applied at B that
creates the same moment about D.
SOLUTION
(a)
Fx = (300 N) cos 25°
= 271.89 N
Fy = (300 N) sin 25°
= 126.785 N
F = (271.89 N)i + (126.785 N) j
!
r = DA = −(0.1 m)i − (0.2 m) j
MD = r × F
M D = [−(0.1 m)i − (0.2 m) j] × [(271.89 N)i + (126.785 N) j]
= −(12.6785 N ⋅ m)k + (54.378 N ⋅ m)k
= (41.700 N ⋅ m)k
M D = 41.7 N ⋅ m
(b)
The smallest force Q at B must be perpendicular to
!
DB at 45°
!
M D = Q ( DB )
41.700 N ⋅ m = Q (0.28284 m)
Q = 147.4 N
45°
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155
PROBLEM 3.4
A 300-N force is applied at A as shown. Determine (a) the
moment of the 300-N force about D, (b) the magnitude and
sense of the horizontal force applied at C that creates the
same moment about D, (c) the smallest force applied at C that
creates the same moment about D.
SOLUTION
(a)
See Problem 3.3 for the figure and analysis leading to the determination of MD
M D = 41.7 N ⋅ m
(b)
Since C is horizontal C = C i
!
r = DC = (0.2 m)i − (0.125 m) j
M D = r × C i = C (0.125 m)k
41.7 N ⋅ m = (0.125 m)(C )
C = 333.60 N
(c)
C = 334 N
The smallest force C must be perpendicular to DC; thus, it forms α with the vertical
0.125 m
0.2 m
α = 32.0°
tan α =
M D = C ( DC ); DC = (0.2 m) 2 + (0.125 m) 2
= 0.23585 m
41.70 N ⋅ m = C (0.23585 m)
C = 176.8 N
58.0°
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156
PROBLEM 3.5
An 8-lb force P is applied to a shift lever. Determine the moment of P about B
when a is equal to 25°.
SOLUTION
First note
Px = (8 lb) cos 25°
= 7.2505 lb
Py = (8 lb)sin 25°
= 3.3809 lb
Noting that the direction of the moment of each force component about B is
clockwise, have
M B = − xPy − yPx
= −(8 in.)(3.3809 lb)
− (22 in.)(7.2505 lb)
= −186.6 lb ⋅ in.
or M B = 186.6 lb ⋅ in.
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157
PROBLEM 3.6
For the shift lever shown, determine the magnitude and the direction of the smallest
force P that has a 210-lb ⋅ in. clockwise moment about B.
SOLUTION
For P to be minimum, it must be perpendicular to the line joining Points A and B. Thus,
α =θ
8 in.
22 in.
= 19.98°
= tan −1
and
Where
M B = dPmin
d = rA/B
= (8 in.) 2 + (22 in.)2
= 23.409 in.
Then
210 lb ⋅ in.
23.409 in.
= 8.97 lb
Pmin =
Pmin = 8.97 lb
19.98°
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158
PROBLEM 3.7
An 11-lb force P is applied to a shift lever. The moment of P about B is clockwise
and has a magnitude of 250 lb ⋅ in. Determine the value of a.
SOLUTION
By definition
M B = rA/B P sin θ
where
θ = α + (90° − φ )
and
φ = tan −1
also
8 in.
= 19.9831°
22 in.
rA/B = (8 in.) 2 + (22 in.) 2
= 23.409 in.
Then
or
250 lb ⋅ in = (23.409 in.)(11 lb)
x sin(α + 90° − 19.9831°)
sin (α + 70.0169°) = 0.97088
or
α + 70.0169° = 76.1391°
and
α + 70.0169° = 103.861°
α = 6.12° 33.8°
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159
PROBLEM 3.8
It is known that a vertical force of 200 lb is required to remove the nail
at C from the board. As the nail first starts moving, determine (a) the
moment about B of the force exerted on the nail, (b) the magnitude of
the force P that creates the same moment about B if a = 10°, (c) the
smallest force P that creates the same moment about B.
SOLUTION
(a)
M B = rC/B FN
We have
= (4 in.)(200 lb)
= 800 lb ⋅ in.
or MB = 800 lb ⋅ in.
(b)
By definition
M B = rA/B P sin θ
θ = 10° + (180° − 70°)
= 120°
Then
800 lb ⋅ in. = (18 in.) × P sin120°
or P = 51.3 lb
(c)
For P to be minimum, it must be perpendicular to the line joining
Points A and B. Thus, P must be directed as shown.
Thus
M B = dPmin
d = rA/B
or
or
800 lb ⋅ in. = (18 in.)Pmin
Pmin = 44.4 lb
Pmin = 44.4 lb
20°
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160
PROBLEM 3.9
A winch puller AB is used to straighten a fence post. Knowing that the tension in cable BC is 1040 N and
length d is 1.90 m, determine the moment about D of the force exerted by the cable at C by resolving that
force into horizontal and vertical components applied (a) at Point C, (b) at Point E.
SOLUTION
(a)
Slope of line
EC =
12
(TAB )
13
12
= (1040 N)
13
= 960 N
Then
TABx =
and
TABy =
Then
0.875 m
5
=
1.90 m + 0.2 m 12
5
(1040 N)
13
= 400 N
M D = TABx (0.875 m) − TABy (0.2 m)
= (960 N)(0.875 m) − (400 N)(0.2 m)
= 760 N ⋅ m
(b)
We have
or M D = 760 N ⋅ m
M D = TABx ( y ) + TABx ( x)
= (960 N)(0) + (400 N)(1.90 m)
= 760 N ⋅ m
or M D = 760 N ⋅ m
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161
!
PROBLEM 3.10
It is known that a force with a moment of 960 N ⋅ m about D is required to straighten the fence post CD. If
d = 2.80 m, determine the tension that must be developed in the cable of winch puller AB to create the
required moment about Point D.
SOLUTION
Slope of line
EC =
0.875 m
7
=
2.80 m + 0.2 m 24
Then
TABx =
24
TAB
25
and
TABy =
7
TAB
25
We have
M D = TABx ( y ) + TABy ( x)
24
7
TAB (0) + TAB (2.80 m)
25
25
= 1224 N
960 N ⋅ m =
TAB
or TAB = 1224 N
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162
PROBLEM 3.11
It is known that a force with a moment of 960 N ⋅ m about D is required to straighten the fence post CD. If the
capacity of winch puller AB is 2400 N, determine the minimum value of distance d to create the specified
moment about Point D.
SOLUTION
The minimum value of d can be found based on the equation relating the moment of the force TAB about D:
M D = (TAB max ) y (d )
where
M D = 960 N ⋅ m
(TAB max ) y = TAB max sin θ = (2400 N) sin θ
Now
sin θ =
0.875m
(d + 0.20) 2 + (0.875) 2 m
0.875
960 N ⋅ m = 2400 N "
" (d + 0.20)2 + (0.875) 2
$
or
(d + 0.20) 2 + (0.875) 2 = 2.1875d
or
(d + 0.20) 2 + (0.875) 2 = 4.7852d 2
or
!
# (d )
#
%
3.7852d 2 − 0.40d − .8056 = 0
Using the quadratic equation, the minimum values of d are 0.51719 m and − .41151 m.
Since only the positive value applies here, d = 0.51719 m
or d = 517 mm
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163
PROBLEM 3.12
The tailgate of a car is supported by the hydraulic lift BC. If the
lift exerts a 125-lb force directed along its centerline on the ball
and socket at B, determine the moment of the force about A.
SOLUTION
First note
dCB = (12.0 in.) 2 + (2.33 in.) 2
= 12.2241 in.
Then
and
12.0 in.
12.2241 in.
2.33 in.
sin θ =
12.2241 in.
cos θ =
FCB = FCB cos θ i − FCB sin θ j
=
125 lb
[(12.0 in.) i − (2.33 in.) j]
12.2241 in.
Now
M A = rB/A × FCB
where
rB/A = (15.3 in.) i − (12.0 in. + 2.33 in.) j
= (15.3 in.) i − (14.33 in.) j
Then
M A = [(15.3 in.)i − (14.33 in.) j] ×
125 lb
(12.0i − 2.33j)
12.2241 in.
= (1393.87 lb ⋅ in.)k
= (116.156 lb ⋅ ft)k
or M A = 116.2 lb ⋅ ft
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164
PROBLEM 3.13
The tailgate of a car is supported by the hydraulic lift BC. If the lift
exerts a 125-lb force directed along its centerline on the ball and
socket at B, determine the moment of the force about A.
SOLUTION
First note
dCB = (17.2 in.) 2 + (7.62 in.) 2
= 18.8123 in.
Then
and
17.2 in.
18.8123 in.
7.62 in.
sin θ =
18.8123 in.
cos θ =
FCB = ( FCB cos θ )i − ( FCB sin θ ) j
=
125 lb
[(17.2 in.)i + (7.62 in.) j]
18.8123 in.
Now
M A = rB/A × FCB
where
rB/A = (20.5 in.)i − (4.38 in.) j
Then
MA = [(20.5 in.)i − (4.38 in.) j] ×
= (1538.53 lb ⋅ in.)k
= (128.2 lb ⋅ ft)k
125 lb
(17.2i − 7.62 j)
18.8123 in.
or M A = 128.2 lb ⋅ ft
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165
PROBLEM 3.14
A mechanic uses a piece of pipe AB as a lever when tightening an
alternator belt. When he pushes down at A, a force of 485 N is
exerted on the alternator at B. Determine the moment of that force
about bolt C if its line of action passes through O.
SOLUTION
We have
M C = rB/C × FB
Noting the direction of the moment of each force component about C is
clockwise.
M C = xFBy + yFBx
Where
and
x = 120 mm − 65 mm = 55 mm
y = 72 mm + 90 mm = 162 mm
FBx =
FBy =
65
2
(65) + (72) 2
72
2
(65) + (72) 2
(485 N) = 325 N
(485 N) = 360 N
M C = (55 mm)(360 N) + (162)(325 N)
= 72450 N ⋅ mm
= 72.450 N ⋅ m
or M C = 72.5 N ⋅ m
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166
PROBLEM 3.15
Form the vector products B × C and B′ × C, where B = B′, and use the results
obtained to prove the identity
sin a cos β =
1
1
sin ( a + β ) + sin ( a − β ).
2
2
SOLUTION
Note:
B = B(cos β i + sin β j)
B′ = B(cos β i − sin β j)
C = C (cos α i + sin α j)
By definition:
Now
| B × C | = BC sin (α − β )
(1)
| B′ × C | = BC sin (α + β )
(2)
B × C = B(cos β i + sin β j) × C (cos α i + sin α j)
= BC (cos β sin α − sin β cos α )k
and
(3)
B′ × C = B(cos β i − sin β j) × C (cos α i + sin α j)
= BC (cos β sin α + sin β cos α ) k
(4)
Equating the magnitudes of B × C from Equations (1) and (3) yields:
BC sin(α − β ) = BC (cos β sin α − sin β cos α )
(5)
Similarly, equating the magnitudes of B′ × C from Equations (2) and (4) yields:
BC sin(α + β ) = BC (cos β sin α + sin β cos α )
(6)
Adding Equations (5) and (6) gives:
sin(α − β ) + sin(α + β ) = 2cos β sin α
or sin α cos β =
1
1
sin(α + β ) + sin(α − β )
2
2
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167
PROBLEM 3.16
A line passes through the Points (20 m, 16 m) and (−1 m, −4 m). Determine the perpendicular distance d from
the line to the origin O of the system of coordinates.
SOLUTION
d AB = [20 m − ( −1 m)]2 + [16 m − ( −4 m)]2
= 29 m
Assume that a force F, or magnitude F(N), acts at Point A and is
directed from A to B.
Then,
Where
F = F λ AB
λ AB =
=
By definition
Where
Then
rB − rA
d AB
1
(21i + 20 j)
29
M O = | rA × F | = dF
rA = −(1 m)i − (4 m) j
M O = [ −(−1 m)i − (4 m) j] ×
F
[(21 m)i + (20 m) j]
29 m
F
[−(20)k + (84)k ]
29
& 64 '
= ( F )k N ⋅ m
* 29 +
=
Finally
& 64 '
( 29 F ) = d ( F )
*
+
64
d=
m
29
d = 2.21 m
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168
PROBLEM 3.17
The vectors P and Q are two adjacent sides of a parallelogram. Determine the area of the parallelogram when
(a) P = −7i + 3j − 3k and Q = 2i + 2j + 5k, (b) P = 6i − 5j − 2k and Q = −2i + 5j − k.
SOLUTION
(a)
We have
A = |P × Q|
where
P = −7i + 3j − 3k
Q = 2i + 2 j + 5k
Then
i j k
P × Q = −7 3 −3
2 2 5
= [(15 + 6)i + ( −6 + 35) j + ( −14 − 6)k ]
= (21)i + (29) j(−20)k
A = (20) 2 + (29) 2 + (−20) 2
(b)
We have
A = |P × Q|
where
P = 6i − 5 j − 2k
or A = 41.0
Q = −2i + 5 in. j − 1k
Then
i
j k
P × Q = 6 −5 −2
−2 5 −1
= [(5 + 10)i + (4 + 6) j + (30 − 10)k ]
= (15)i + (10) j + (20)k
A = (15) 2 + (10) 2 + (20) 2
or A = 26.9
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169
PROBLEM 3.18
A plane contains the vectors A and B. Determine the unit vector normal to the plane when A and B are equal
to, respectively, (a) i + 2j − 5k and 4i − 7j − 5k, (b) 3i − 3j + 2k and −2i + 6j − 4k.
SOLUTION
(a)
We have
where
=
A×B
|A × B|
A = 1i + 2 j − 5k
B = 4i − 7 j − 5k
Then
i j k
A × B = 1 +2 −5
4 −7 −5
= (−10 − 35)i + (20 + 5) j + (−7 − 8)k
= 15(3i − 1j − 1k )
and
|A × B | = 15 (−3)2 + (−1) 2 + (−1)2 = 15 11
=
(b)
We have
where
=
15(−3i − 1j − 1k )
15 11
=
or
1
11
(−3i − j − k )
A×B
|A × B|
A = 3i − 3 j + 2k
B = −2i + 6 j − 4k
Then
i
j k
A × B = 3 −3 2
−2 6 −4
= (12 − 12)i + (−4 + 12) j + (18 − 6)k
= (8 j + 12k )
and
|A × B| = 4 (2) 2 + (3) 2 = 4 13
=
4(2 j + 3k )
4 13
or
=
1
13
(2 j + 3k )
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170
PROBLEM 3.19
Determine the moment about the origin O of the force F = 4i + 5j − 3k that acts at a Point A. Assume that the
position vector of A is (a) r = 2i − 3j + 4k, (b) r = 2i + 2.5j − 1.5k, (c) r = 2i + 5j + 6k.
SOLUTION
(a)
i j k
M O = 2 −3 4
4 5 −3
= (9 − 20)i + (16 + 6) j + (10 + 12)k
(b)
i
j
k
M O = 2 2.5 −1.5
4
5 −3
= (−7.5 + 7.5)i + (−6 + 6) j + (10 − 10)k
(c)
M O = −11i + 22 j + 22k
MO = 0
i j k
MO = 2 5 6
4 5 −3
= (−15 − 30)i + (24 + 6) j + (10 − 20)k
M O = −45i + 30 j − 10k
Note: The answer to Part b could have been anticipated since the elements of the last two rows of the
determinant are proportional.
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171
PROBLEM 3.20
Determine the moment about the origin O of the force F = −2i + 3j + 5k that acts at a Point A. Assume that the
position vector of A is (a) r = i + j + k, (b) r = 2i + 3j − 5k, (c) r = −4i + 6j + 10k.
SOLUTION
(a)
i j k
MO = 1 1 1
−2 3 5
= (5 − 3)i + (−2 − 5) j + (3 + 2)k
(b)
i j k
MO = 2 3 − 5
−2 3 5
= (15 + 15)i + (10 − 10) j + (6 + 6)k
(c)
M O = 2i − 7 j + 5k
M O = 30i + 12k
i j k
M O = −4 6 10
−2 3 5
= (30 − 30)i + ( −20 + 20) j + (−12 + 12)k
MO = 0
Note: The answer to Part c could have been anticipated since the elements of the last two rows of the
determinant are proportional.
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172
PROBLEM 3.21
A 200-N force is applied as shown to the bracket ABC. Determine
the moment of the force about A.
SOLUTION
We have
M A = rC/A × FC
where
rC/A = (0.06 m)i + (0.075 m) j
FC = −(200 N) cos 30° j + (200 N)sin 30°k
Then
i
j
k
M A = 200 0.06
0.075
0
− cos 30° sin 30°
0
= 200[(0.075sin 30°)i − (0.06sin 30°) j − (0.06 cos 30°)k ]
or M A = (7.50 N ⋅ m)i − (6.00 N ⋅ m) j − (10.39 N ⋅ m)k
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173
PROBLEM 3.22
Before the trunk of a large tree is felled, cables AB and BC are
attached as shown. Knowing that the tensions in cables AB and BC
are 555 N and 660 N, respectively, determine the moment about O
of the resultant force exerted on the tree by the cables at B.
SOLUTION
We have
M O = rB/O × FB
where
rB/O = (7 m) j
FB = TAB + TBC
TAB =
=
TBC =
=
BATAB
−(0.75 m)i − (7 m) j + (6 m)k
(.75) 2 + (7) 2 + (6) 2 m
(555 N)
BC TBC
(4.25 m)i − (7 m) j + (1 m)k
(4.25) 2 + (7) 2 + (1) 2 m
(660 N)
FB = [−(45.00 N)i − (420.0 N) j + (360.0 N)k ]
+ [(340.0 N)i − (560.0 N) j + (80.00 N)k ]
= (295.0 N)i − (980.0 N) j + (440.0 N)k
and
i
j
k
MO = 0
7
0 N⋅m
295 980 440
= (3080 N ⋅ m)i − (2070 N ⋅ m)k
or M O = (3080 N ⋅ m)i − (2070 N ⋅ m)k
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174
PROBLEM 3.23
The 6-m boom AB has a fixed end A. A steel cable is stretched from
the free end B of the boom to a Point C located on the vertical wall.
If the tension in the cable is 2.5 kN, determine the moment about A of
the force exerted by the cable at B.
SOLUTION
First note
d BC = (−6)2 + (2.4) 2 + (−4) 2
= 7.6 m
2.5 kN
(−6i + 2.4 j − 4k )
7.6
Then
TBC =
We have
M A = rB/A × TBC
where
rB/A = (6 m)i
Then
M A = (6 m)i ×
2.5 kN
(−6i + 2.4 j − 4k )
7.6
or M A = (7.89 kN ⋅ m) j + (4.74 kN ⋅ m)k
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175
PROBLEM 3.24
A wooden board AB, which is used as a temporary prop to support
a small roof, exerts at Point A of the roof a 57-lb force directed
along BA. Determine the moment about C of that force.
SOLUTION
We have
M C = rA/C × FBA
where
rA/C = (48 in.)i − (6 in.) j + (36 in.)k
and
FBA =
BA FBA
−(5 in.)i + (90 in.) j − (30 in.)k !
# (57 lb)
="
2
2
2
"
#
(5)
(90)
(30)
in.
+
+
$
%
= −(3 lb)i + (54 lb) j − (18 lb)k
i
j k
M C = 48 6 36 lb ⋅ in.
3 54 18
= −(1836 lb ⋅ in.)i + (756 lb ⋅ in.) j + (2574 lb ⋅ in.)
or M C = −(153.0 lb ⋅ ft)i + (63.0 lb ⋅ ft) j + (215 lb ⋅ ft)k
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176
PROBLEM 3.25
The ramp ABCD is supported by cables at corners C and D.
The tension in each of the cables is 810 N. Determine the
moment about A of the force exerted by (a) the cable at D,
(b) the cable at C.
SOLUTION
(a)
We have
M A = rE/A × TDE
where
rE/A = (2.3 m) j
TDE =
=
DE TDE
(0.6 m)i + (3.3 m) j − (3 m)k
(0.6) 2 + (3.3)2 + (3)2 m
(810 N)
= (108 N)i + (594 N) j − (540 N)k
i
j
k
MA = 0
2.3
0 N⋅m
108 594 −540
= −(1242 N ⋅ m)i − (248.4 N ⋅ m)k
or M A = −(1242 N ⋅ m)i − (248 N ⋅ m)k
(b)
We have
M A = rG/A × TCG
where
rG/A = (2.7 m)i + (2.3 m) j
TCG =
=
CG TCG
−(.6 m)i + (3.3 m) j − (3 m)k
(.6) 2 + (3.3) 2 + (3) 2 m
(810 N)
= −(108 N)i + (594 N) j − (540 N)k
i
j
k
M A = 2.7 2.3
0 N⋅m
−108 594 −540
= −(1242 N ⋅ m)i + (1458 N ⋅ m) j + (1852 N ⋅ m)k
or M A = −(1242 N ⋅ m)i + (1458 N ⋅ m) j + (1852 N ⋅ m)k
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177
PROBLEM 3.26
A small boat hangs from two davits, one of which is shown in the
figure. The tension in line ABAD is 82 lb. Determine the moment
about C of the resultant force RA exerted on the davit at A.
SOLUTION
We have
R A = 2FAB + FAD
where
and
FAB = −(82 lb) j
!
AD
6i − 7.75 j − 3k
= (82 lb)
FAD = FAD
AD
10.25
FAD = (48 lb)i − (62 lb) j − (24 lb)k
Thus
R A = 2FAB + FAD = (48 lb)i − (226 lb) j − (24 lb)k
Also
rA/C = (7.75 ft) j + (3 ft)k
Using Eq. (3.21):
i
j
k
M C = 0 7.75
3
48 − 226 −24
= (492 lb ⋅ ft)i + (144 lb ⋅ ft) j − (372 lb ⋅ ft)k
M C = (492 lb ⋅ ft)i + (144.0 lb ⋅ ft) j − (372 lb ⋅ ft)k
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178
PROBLEM 3.27
In Problem 3.22, determine the perpendicular distance from
Point O to cable AB.
PROBLEM 3.22 Before the trunk of a large tree is felled,
cables AB and BC are attached as shown. Knowing that the
tensions in cables AB and BC are 555 N and 660 N,
respectively, determine the moment about O of the resultant
force exerted on the tree by the cables at B.
SOLUTION
We have
where
| M O | = TBA d
d = perpendicular distance from O to line AB.
Now
M O = rB/O × TBA
and
rB/O = (7 m) j
TBA =
=
BATAB
−(0.75 m)i − (7 m) j + (6 m)k
(0.75) 2 + (7) 2 + (6) 2 m
(555 N)
= −(45.0 N)i − (420 N) j + (360 N)k
i
j
k
MO = 0
7
0 N⋅m
−45 −420 360
= (2520.0 N ⋅ m)i + (315.00 N ⋅ m)k
and
| M O | = (2520.0) 2 + (315.00) 2
= 2539.6 N ⋅ m
2539.6 N ⋅ m = (555 N)d
or
d = 4.5759 m
or d = 4.58 m
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179
PROBLEM 3.28
In Problem 3.22, determine the perpendicular distance from
Point O to cable BC.
PROBLEM 3.22 Before the trunk of a large tree is felled,
cables AB and BC are attached as shown. Knowing that the
tensions in cables AB and BC are 555 N and 660 N,
respectively, determine the moment about O of the resultant
force exerted on the tree by the cables at B.
SOLUTION
We have
where
| M O | = TBC d
d = perpendicular distance from O to line BC.
M O = rB/O × TBC
rB/O = 7 mj
TBC =
=
BC TBC
(4.25 m)i − (7 m) j + (1 m)k
(4.25)2 + (7) 2 + (1) 2 m
(660 N)
= (340 N)i − (560 N) j + (80 N)k
i
j
k
MO = 0
7
0
340 −560 80
= (560 N ⋅ m)i − (2380 N ⋅ m)k
and
| M O | = (560)2 + (2380) 2
= 2445.0 N ⋅ m
2445.0 N ⋅ m = (660 N)d
d = 3.7045 m
or d = 3.70 m
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180
PROBLEM 3.29
In Problem 3.24, determine the perpendicular distance from Point D
to a line drawn through Points A and B.
PROBLEM 3.24 A wooden board AB, which is used as a temporary
prop to support a small roof, exerts at Point A of the roof a 57-lb
force directed along BA. Determine the moment about C of that
force.
SOLUTION
We have
where
| M D | = FBA d
d = perpendicular distance from D to line AB.
M D = rA/D × FBA
rA/D = −(6 in.) j + (36 in.)k
FBA =
=
BA FBA
(−(5 in.)i + (90 in.) j − (30 in.)k )
(5)2 + (90) 2 + (30) 2 in.
(57 lb)
= −(3 lb)i + (54 lb) j − (18 lb)k
i
j
k
M D = 0 −6 36 lb ⋅ in.
−3 54 −18
= −(1836.00 lb ⋅ in.)i − (108.000 lb ⋅ in.) j − (18.0000 lb ⋅ in.)k
and
| M D | = (1836.00) 2 + (108.000)2 + (18.0000)2
= 1839.26 lb ⋅ in.
1839.26 lb ⋅ in = (57 lb)d
d = 32.268 in.
or d = 32.3 in.
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181
PROBLEM 3.30
In Problem 3.24, determine the perpendicular distance from Point C to a
line drawn through Points A and B.
PROBLEM 3.24 A wooden board AB, which is used as a temporary
prop to support a small roof, exerts at Point A of the roof a 57-lb force
directed along BA. Determine the moment about C of that force.
SOLUTION
We have
where
| M C | = FBA d
d = perpendicular distance from C to line AB.
M C = rA/C × FBA
rA/C = (48 in.)i − (6 in.) j + (36 in.)k
FBA =
=
BA FBA
( −(5 in.)i + (90 in.) j − (30 in.)k )
(5) 2 + (90) 2 + (30) 2 in.
(57 lb)
= −(3 lb)i + (54 lb) j − (18 lb)k
i
j
k
M C = 48 −6 36 lb ⋅ in.
−3 54 −18
= −(1836.00lb ⋅ in.)i − (756.00 lb ⋅ in.) j + (2574.0 lb ⋅ in.)k
and
| M C | = (1836.00) 2 + (756.00) 2 + (2574.0)2
= 3250.8 lb ⋅ in.
3250.8 lb ⋅ in. = 57 lb
d = 57.032 in.
or d = 57.0 in.
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182
PROBLEM 3.31
In Problem 3.25, determine the perpendicular distance from
Point A to portion DE of cable DEF.
PROBLEM 3.25 The ramp ABCD is supported by cables at
corners C and D. The tension in each of the cables is 810 N.
Determine the moment about A of the force exerted by (a) the
cable at D, (b) the cable at C.
SOLUTION
We have
where
| M A | = TDE d
d = perpendicular distance from A to line DE.
M A = rE/A × TDE
rE/A = (2.3 m) j
TDE =
=
DE TDE
(0.6 m)i + (3.3 m) j − (3 m)k
(0.6) 2 + (3.3) 2 + (3) 2 m
(810 N)
= (108 N)i + (594 N) j − (540 N)k
i
j
k
MA = 0
2.3
0 N⋅m
108 594 540
= − (1242.00 N ⋅ m)i − (248.00 N ⋅ m)k
and
|M A | = (1242.00)2 + (248.00) 2
= 1266.52 N ⋅ m
1266.52 N ⋅ m = (810 N)d
d = 1.56360 m
or d = 1.564 m
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183
PROBLEM 3.32
In Problem 3.25, determine the perpendicular distance from
Point A to a line drawn through Points C and G.
PROBLEM 3.25 The ramp ABCD is supported by cables at
corners C and D. The tension in each of the cables is 810 N.
Determine the moment about A of the force exerted by (a) the
cable at D, (b) the cable at C.
SOLUTION
We have
where
|M A | = TCG d
d = perpendicular distance from A to line CG.
M A = rG/A × TCG
rG/A = (2.7 m)i + (2.3 m) j
TCG =
=
CG TCG
−(0.6 m) i + (3.3 m) j − (3 m) k
(0.6) 2 + (3.3) 2 + (3) 2 m
(810 N)
= −(108 N) i + (594 N) j − (540 N) k
i
j
k
0 N⋅m
M A = 2.7 2.3
−108 594 −540
= −(1242.00 N ⋅ m)i + (1458.00 N ⋅ m) j + (1852.00 N ⋅ m)k
and
|M A | = (1242.00) 2 + (1458.00) 2 + (1852.00)2
= 2664.3 N ⋅ m
2664.3 N ⋅ m = (810 N)d
d = 3.2893 m
or d = 3.29 m
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184
PROBLEM 3.33
In Problem 3.26, determine the perpendicular distance from
Point C to portion AD of the line ABAD.
PROBLEM 3.26 A small boat hangs from two davits, one of
which is shown in the figure. The tension in line ABAD is
82 lb. Determine the moment about C of the resultant force
RA exerted on the davit at A.
SOLUTION
First compute the moment about C of the force FDA exerted by the line on D:
From Problem 3.26:
FDA = − FAD
= −(48 lb) i + (62 lb) j + (24 lb)k
M C = rD/C × FDA
= + (6 ft)i × [−(48 lb)i + (62 lb) j + (24 lb)k ]
= −(144 lb ⋅ ft) j + (372 lb ⋅ ft)k
M C = (144) 2 + (372)2
= 398.90 lb ⋅ ft
Then
M C = FDA d
Since
FDA = 82 lb
d=
=
MC
FDA
398.90 lb ⋅ ft
82 lb
d = 4.86 ft
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185
PROBLEM 3.34
Determine the value of a that minimizes the
perpendicular distance from Point C to a section
of pipeline that passes through Points A and B.
SOLUTION
Assuming a force F acts along AB,
|M C | = |rA / C × F| = F ( d )
d = perpendicular distance from C to line AB
Where
F=
=
AB F
(24 ft) i + (24 ft) j − (28) k
(24) 2 + (24)2 + (18) 2 ft
F
F
(6) i + (6) j − (7) k
11
= (3 ft)i − (10 ft) j − (a − 10 ft)k
=
rA/C
i
j
k
F
M C = 3 −10 10a
11
6 6
−7
= [(10 + 6a )i + (81 − 6a) j + 78 k ]
|M C | = |rA/C × F 2 |
Since
or
F
11
|rA/C × F 2 | = ( dF ) 2
1
(10 + 6a) 2 + (81 − 6a) 2 + (78)2 = d 2
121
Setting
d
da
(d 2 ) = 0 to find a to minimize d
1
[2(6)(10 + 6a) + 2(−6)(81 − 6a)] = 0
121
Solving
a = 5.92 ft
or a = 5.92 ft
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186
PROBLEM 3.35
Given the vectors P = 3i − j + 2k , Q = 4i + 5 j − 3k , and S = −2i + 3j − k , compute the scalar products P ⋅ Q,
P ⋅ S, and Q ⋅ S.
SOLUTION
P ⋅ Q = (3i − 1j + 2k ) ⋅ (4i − 5 j − 3k )
= (3)(4) + (−1)(−5) + (2)(−3)
=1
or
P ⋅Q =1
or
P ⋅ S = −11
or
Q ⋅ S = 10
!
P ⋅ S = (3i − 1j + 2k ) ⋅ (−2i + 3j − 1k )
= (3)(−2) + (−1)(3) + (2)( −1)
= −11
Q ⋅ S = (4i − 5 j − 3k ) ⋅ ( −2i + 3j − 1k )
= (4)(−2) + (5)(3) + ( −3)(−1)
= 10
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187
PROBLEM 3.36
Form the scalar products B ⋅ C and B′ ⋅ C, where B = B ′, and use the
results obtained to prove the identity
cos a cos β =
1
1
cos (a + β ) + cos (a − β ).
2
2
SOLUTION
By definition
B ⋅ C = BC cos(α − β )
B = B [(cos β )i + (sin β ) j]
C = C [(cos α )i + (sin α ) j]
where
(B cos β )( C cos α ) + ( B sin β )(C sin α ) = BC cos (α − β )
cos β cos α + sin β sin α = cos(α − β )
or
(1)
B′ ⋅ C = BC cos (α + β )
By definition
B′ = [(cos β )i − (sin β ) j]
where
(B cos β ) (C cos α ) + (− B sin β )(C sin α ) = BC cos (α + β )
or
cos β cos α − sin β sin α = cos (α + β )
(2)
Adding Equations (1) and (2),
2 cos β cos α = cos (α − β ) + cos (α + β )
or cos α cos β =
1
1
cos (α + β ) + cos (α − β )
2
2
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188
PROBLEM 3.37
Section AB of a pipeline lies in the yz plane and forms an angle
of 37° with the z axis. Branch lines CD and EF join AB as
shown. Determine the angle formed by pipes AB and CD.
SOLUTION
First note
!
AB = AB (sin 37° j − cos 37°k )
CD = CD(− cos 40° cos 55° j + sin 40° j − cos 40° sin 55°k )
Now
!
!
AB ⋅ CD = ( AB)(CD ) cos θ
or
AB (sin 37° j − cos 37°k ) ⋅ CD (− cos 40° cos 55°i + sin 40° j − cos 40° sin 55°k )
= (AB)(CD) cos θ
or
cos θ = (sin 37°)(sin 40°) + (− cos 37°)(− cos 40° sin 55°)
= 0.88799
or θ = 27.4°
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189
PROBLEM 3.38
Section AB of a pipeline lies in the yz plane and forms an angle
of 37° with the z axis. Branch lines CD and EF join AB as
shown. Determine the angle formed by pipes AB and EF.
SOLUTION
First note
Now
or
!
AB = AB (sin 37° j − cos 37°k )
!
EF = EF (cos 32° cos 45°i + sin 32° j − cos 32° sin 45°k )
! !
AB ⋅ EF = ( AB )( EF ) cos θ
AB (sin 37° j − cos 37°k ) ⋅ EF (cos 32° cos 45° j + sin 32° j − cos 32° sin 45°k )
= ( AB )( EF ) cos θ
or
cos θ = (sin 37°)(sin 32°) + ( − cos 37°)( − cos 32° sin 45°)
= 0.79782
or θ = 37.1°
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190
PROBLEM 3.39
Consider the volleyball net shown.
Determine the angle formed by guy
wires AB and AC.
SOLUTION
First note
AB = (−6.5)2 + (−8)2 + (2) 2 = 10.5 ft
AC = (0) 2 + (−8) 2 + (6)2 = 10 ft
and
By definition
or
or
!
AB = −(6.5 ft)i − (8 ft) j + (2 ft)k
!
AC = −(8 ft) j + (6 ft)k
!
!
AB ⋅ AC = ( AB )( AC ) cos θ
(−6.5i − 8 j + 2k ) ⋅ (−8 j + 6k ) = (10.5)(10) cos θ
(−6.5)(0) + ( −8)( −8) + (2)(6) = 105cos θ
cos θ = 0.72381
or θ = 43.6°
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191
PROBLEM 3.40
Consider the volleyball net shown.
Determine the angle formed by guy
wires AC and AD.
SOLUTION
First note
AC = (0)2 + (−8) 2 + (6) 2
= 10 ft
AD = (4) 2 + ( −8) 2 + (1) 2
and
By definition
or
= 9 ft
!
AC = −(8 ft)j + (6 ft)k
!
AD = (4 ft) j − (8 ft) j + (1 ft)k
!
!
AC ⋅ AD = ( AC )( AD ) cos θ
(−8 j + 6k ) ⋅ (4i − 8 j + k ) = (10)(9) cos θ
(0)(4) + ( −8)( −8) + (6)(1) = 90 cos θ
or
cos θ = 0.777 78
or θ = 38.9°
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192
PROBLEM 3.41
Knowing that the tension in cable AC is 1260 N, determine
(a) the angle between cable AC and the boom AB, (b) the
projection on AB of the force exerted by cable AC at Point A.
SOLUTION
(a)
First note
AC = (−2.4) 2 + (0.8)2 + (1.2) 2
= 2.8 m
AB = (−2.4) 2 + (−1.8)2 + (0) 2
and
By definition
or
(b)
= 3.0 m
!
AC = −(2.4 m)i + (0.8 m) j + (1.2 m)k
!
AB = −(2.4 m)i − (1.8 m) j
! !
AC ⋅ AB = ( AC )( AB) cos θ
(−2.4i + 0.8 j + 1.2k ) ⋅ (−2.4i − 1.8 j) = (2.8)(30) × cos θ
or
(−2.4)(−2.4) + (0.8)(−1.8) + (1.2)(0) = 8.4cos θ
or
cos θ = 0.514 29
We have
or θ = 59.0°
(TAC ) AB = TAC ⋅ λ AB
= TAC cos θ
= (1260 N)(0.51429)
or (TAC ) AB = 648 N
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193
PROBLEM 3.42
Knowing that the tension in cable AD is 405 N, determine (a) the
angle between cable AD and the boom AB, (b) the projection on
AB of the force exerted by cable AD at Point A.
SOLUTION
(a)
First note
AD = (−2.4) 2 + (1.2) 2 + (−2.4) 2
= 3.6 m
AB = (−2.4) 2 + (−1.8) 2 + (0) 2
= 3.0 m
and
# = −(2.4 m)i + (1.2 m) j − (2.4 m)k
AD
AB = −(2.4 m)i − (1.8 m) j
# ⋅ AB = ( AD )( AB) cos θ
AD
By definition,
(−2.4i + 1.2 j − 2.4k ) ⋅ (−2.4i − 1.8 j) = (3.6)(3.0) cos θ
(−2.4)(−2.4) + (1.2)(−1.8) + (−2.4)(0) = 10.8cos θ
cos θ =
(b)
1
3
θ = 70.5°
(TAD ) AB = TAD ⋅ λ AB
= TAD cos θ
&1'
= (405 N) ( )
* 3+
(TAD ) AB = 135.0 N
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194
PROBLEM 3.43
Slider P can move along rod OA. An elastic cord PC is
attached to the slider and to the vertical member BC. Knowing
that the distance from O to P is 6 in. and that the tension in the
cord is 3 lb, determine (a) the angle between the elastic cord
and the rod OA, (b) the projection on OA of the force exerted
by cord PC at Point P.
SOLUTION
First note
OA = (12)2 + (12)2 + (−6) 2 = 18 in.
OA 1
= (12i + 12 j − 6k )
OA 18
1
= (2i + 2 j − k )
3
Then
λ OA =
Now
1
OP = 6 in. - OP = (OA)
3
The coordinates of Point P are (4 in., 4 in., −2 in.)
!
PC = (5 in.)i + (11 in.) j + (14 in.)k
so that
PC = (5) 2 + (11) 2 + (14)2 = 342 in.
and
(a)
We have
or
or
!
PC ⋅ λ OA = ( PC ) cos θ
1
(5i + 11j + 14k ) ⋅ (2i + 2 j − k ) = 342 cos θ
3
1
[(5)(2) + (11)(2) + (14)( −1)]
3 342
= 0.324 44
cos θ =
or θ = 71.1°
(b)
We have
(TPC ) OA = TPC ⋅ λ OA
= (TPC λ PC ) ⋅ λ OA
PC
⋅ λ OA
PC
= TPC cos θ
= TPC
= (3 lb)(0.324 44)
or (TPC )OA = 0.973 lb
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195
PROBLEM 3.44
Slider P can move along rod OA. An elastic cord PC is
attached to the slider and to the vertical member BC.
Determine the distance from O to P for which cord PC and
rod OA are perpendicular.
SOLUTION
First note
OA = (12)2 + (12)2 + (−6)2 = 18 in.
Then
λ OA =
OA 1
= (12i + 12 j − 6k )
OA 18
1
= (2i + 2 j − k )
3
Let the coordinates of Point P be (x in., y in., z in.). Then
!
PC = [(9 − x)in.]i + (15 − y )in.] j + [(12 − z )in.]k
Also,
and
!
d
OP = dOP λ OA = OP (2i + 2 j − k )
3
!
OP = ( x in.)i + ( y in.) j + ( z in.)k
2
2
1
x = dOP y = dOP z = dOP
3
3
3
The requirement that OA and PC be perpendicular implies that
!
λ OA ⋅ PC = 0
or
1
(2 j + 2 j − k ) ⋅ [(9 − x)i + (15 − y ) j + (12 − z )k ] = 0
3
or
2
2
&
'
&
'
& 1
'!
(2) ( 9 − dOP ) + (2) (15 − dOP ) + (−1) "12 − ( − dOP ) # = 0
3
3
*
+
*
+
* 3
+%
$
or dOP = 12.00 in.
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196
PROBLEM 3.45
Determine the volume of the parallelepiped of Fig. 3.25 when
(a) P = 4i − 3j + 2k, Q = −2i − 5j + k, and S = 7i + j − k,
(b) P = 5i − j + 6k, Q = 2i + 3j + k, and S = −3i − 2j + 4k.
SOLUTION
Volume of a parallelepiped is found using the mixed triple product.
(a)
Vol = P ⋅ (Q × S)
4 −3 2
= −2 −5 1 in.3
7 1
−1
= (20 − 21 − 4 + 70 + 6 − 4)
= 67
or Volume = 67.0
(b)
Vol = P ⋅ (Q × S)
5 −1 6
= 2 3 1 in.3
−3 −2 4
= (60 + 3 − 24 + 54 + 8 + 10)
= 111
or Volume = 111.0
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197
PROBLEM 3.46
Given the vectors P = 4i − 2 j + 3k , Q = 2i + 4 j − 5k , and S = S x i − j + 2k , determine the value of S x for which
the three vectors are coplanar.
SOLUTION
If P, Q, and S are coplanar, then P must be perpendicular to (Q × S).
P ⋅ (Q × S) = 0
(or, the volume of a parallelepiped defined by P, Q, and S is zero).
−2 3
4 −5 = 0
−1 2
Then
4
2
Sx
or
32 + 10S x − 6 − 20 + 8 − 12S x = 0
Sx = 7
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198
PROBLEM 3.47
The 0.61 × 1.00-m lid ABCD of a storage bin is hinged
along side AB and is held open by looping cord DEC over a
frictionless hook at E. If the tension in the cord is 66 N,
determine the moment about each of the coordinate axes of
the force exerted by the cord at D.
SOLUTION
First note
z = (0.61) 2 − (0.11)2
= 0.60 m
Then
d DE = (0.3) 2 + (0.6) 2 + (−0.6) 2
= 0.9 m
66 N
(0.3i + 0.6 j − 0.6k )
0.9
= 22[(1 N)i + (2 N) j − (2 N)k ]
and
TDE =
Now
M A = rD/A × TDE
where
rD/A = (0.11 m) j + (0.60 m)k
Then
i
j
k
M A = 22 0 0.11 0.60
−2
1
2
= 22[(−0.22 − 1.20)i + 0.60 j − 0.11k ]
= − (31.24 N ⋅ m)i + (13.20 N ⋅ m) j − (2.42 N ⋅ m)k
M x = −31.2 N ⋅ m, M y = 13.20 N ⋅ m, M z = −2.42 N ⋅ m
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199
PROBLEM 3.48
The 0.61 × 1.00-m lid ABCD of a storage bin is hinged
along side AB and is held open by looping cord DEC over a
frictionless hook at E. If the tension in the cord is 66 N,
determine the moment about each of the coordinate axes of
the force exerted by the cord at C.
SOLUTION
First note
z = (0.61) 2 − (0.11)2
= 0.60 m
Then
dCE = (−0.7) 2 + (0.6) 2 + (−0.6) 2
= 1.1 m
66 N
(−0.7i + 0.6 j − 0.6k )
1.1
= 6[−(7 N)i + (6 N) j − (6 N)k ]
and
TCE =
Now
M A = rE/A × TCE
where
rE/A = (0.3 m)i + (0.71 m) j
Then
i
j
k
M A = 6 0.3 0.71 0
−7
−6
6
= 6[ −4.26i + 1.8 j + (1.8 + 4.97)k ]
= − (25.56 N ⋅ m)i + (10.80 N ⋅ m) j + (40.62 N ⋅ m)k
M x = −25.6 N ⋅ m, M y = 10.80 N ⋅ m, M z = 40.6 N ⋅ m
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200
PROBLEM 3.49
To lift a heavy crate, a man uses a block and tackle attached to the
bottom of an I-beam at hook B. Knowing that the moments about the y
and the z axes of the force exerted at B by portion AB of the rope are,
respectively, 120 N ⋅ m and −460 N ⋅ m, determine the distance a.
SOLUTION
First note
!
BA = (2.2 m)i − (3.2 m) j − ( a m)k
Now
M D = rA/D × TBA
where
rA/D = (2.2 m)i + (1.6 m) j
Then
TBA =
TBA
(2.2i − 3.2 j − ak ) (N)
d BA
MD =
i
j
k
TBA
2.2 1.6
0
d BA
2.2 −3.2 − a
=
Thus
TBA
{−1.6a i + 2.2a j + [(2.2)(−3.2) − (1.6)(2.2)]k}
d BA
M y = 2.2
TBA
a
d BA
M z = −10.56
Then forming the ratio
TBA
d BA
(N ⋅ m)
(N ⋅ m)
My
Mz
T
2.2 dBA (N ⋅ m)
120 N ⋅ m
BA
=
−460 N ⋅ m −10.56 TdBA (N ⋅ m)
or a = 1.252 m
BA
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201
PROBLEM 3.50
To lift a heavy crate, a man uses a block and tackle attached to the
bottom of an I-beam at hook B. Knowing that the man applies a 195-N
force to end A of the rope and that the moment of that force about the y
axis is 132 N ⋅ m, determine the distance a.
SOLUTION
d BA = (2.2) 2 + (−3.2) 2 + (−a ) 2
First note
= 15.08 + a 2 m
195 N
(2.2i − 3.2 j − a k )
d BA
and
TBA =
Now
M y = j ⋅ (rA/D × TBA )
where
rA/0 = (2.2 m)i + (1.6 m) j
Then
0
1
0
195
My =
2.2 1.6
0
d BA
2.2 −3.2 − a
=
195
(2.2a ) (N ⋅ m)
d BA
Substituting for My and dBA
132 N ⋅ m =
or
195
15.08 + a 2
(2.2a )
0.30769 15.08 + a 2 = a
Squaring both sides of the equation
0.094675(15.08 + a 2 ) = a 2
or a = 1.256 m
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202
PROBLEM 3.51
A small boat hangs from two davits, one of which is shown in
the figure. It is known that the moment about the z axis of the
resultant force RA exerted on the davit at A must not exceed
279 lb ⋅ ft in absolute value. Determine the largest allowable
tension in line ABAD when x = 6 ft.
SOLUTION
First note
R A = 2TAB + TAD
Also note that only TAD will contribute to the moment about the z axis.
Now
AD = (6) 2 + (−7.75) 2 + (−3) 2
= 10.25 ft
!
AD
=T
AD
T
(6i − 7.75 j − 3k )
=
10.25
Then,
TAD
Now
M z = k ⋅ (rA/C × TAD )
where
rA/C = (7.75 ft) j + (3 ft)k
Then for Tmax ,
0
0
1
Tmax
279 =
0 7.75 3
10.25
6 −7.75 −3
=
Tmax
| − (1)(7.75)(6)|
10.25
or Tmax = 61.5 lb
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203
PROBLEM 3.52
For the davit of Problem 3.51, determine the largest allowable
distance x when the tension in line ABAD is 60 lb.
SOLUTION
From the solution of Problem 3.51, TAD is now
TAD = T
=
AD
AD
60 lb
x 2 + ( −7.75) 2 + (−3)2
( xi − 7.75 j − 3k )
Then M z = k ⋅ (rA / C × TAD ) becomes
279 =
279 =
60
x 2 + (−7.75) 2 + ( −3) 2
60
0
0
1
0 7.75 3
x −7.75 −3
| − (1)(7.75)( x) |
2
x + 69.0625
279 x 2 + 69.0625 = 465 x
0.6 x 2 + 69.0625 = x
Squaring both sides:
0.36 x 2 + 24.8625 = x 2
x 2 = 38.848
x = 6.23 ft
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204
PROBLEM 3.53
To loosen a frozen valve, a force F of magnitude 70 lb is
applied to the handle of the valve. Knowing that θ = 25°,
Mx = −61 lb ⋅ ft, and M z = −43 lb ⋅ ft, determine φ and d.
SOLUTION
We have
ΣM O : rA/O × F = M O
where
rA/O = −(4 in.)i + (11 in.) j − (d )k
F = F (cos θ cos φ i − sin θ j + cos θ sin φ k )
For
F = 70 lb, θ = 25°
F = (70 lb)[(0.90631cos φ )i − 0.42262 j + (0.90631sin φ )k ]
i
M O = (70 lb)
−4
−0.90631cos φ
j
k
11
−d
in.
−0.42262 0.90631sin φ
= (70 lb)[(9.9694sin φ − 0.42262d ) i + (−0.90631d cos φ + 3.6252sin φ ) j
+ (1.69048 − 9.9694cos φ )k ] in.
and
M x = (70 lb)(9.9694sin φ − 0.42262d )in. = −(61 lb ⋅ ft)(12 in./ft)
(1)
M y = (70 lb)(−0.90631d cos φ + 3.6252sin φ ) in.
(2)
M z = (70 lb)(1.69048 − 9.9694cos φ ) in. = −43 lb ⋅ ft(12 in./ft)
(3)
From Equation (3)
& 634.33 '
φ = cos −1 (
) = 24.636°
* 697.86 +
or
From Equation (1)
& 1022.90 '
d =(
) = 34.577 in.
* 29.583 +
or d = 34.6 in.
φ = 24.6°
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205
PROBLEM 3.54
When a force F is applied to the handle of the valve shown, its
moments about the x and z axes are, respectively, M x = −77 lb ⋅ ft
and M z = −81 lb ⋅ ft. For d = 27 in., determine the moment My of
F about the y axis.
SOLUTION
We have
ΣM O : rA/O × F = M O
Where
rA/O = −(4 in.)i + (11 in.) j − (27 in.)k
F = F (cos θ cos φ i − sin θ j + cos θ sin φ k )
MO = F
i
−4
cos θ cos φ
j
11
− sin θ
k
−27
lb ⋅ in.
cos θ sin φ
= F [(11cos θ sin φ − 27 sin θ )i
+ (−27 cos θ cos φ + 4 cos θ sin φ ) j
+ (4sin θ − 11cos θ cos φ )k ](lb ⋅ in.)
and
M x = F (11cos θ sin φ − 27sin θ )(lb ⋅ in.)
(1)
M y = F (−27 cos θ cos φ + 4cos θ sin φ ) (lb ⋅ in.)
(2)
M z = F (4sin θ − 11cos θ cos φ ) (lb ⋅ in.)
(3)
Now, Equation (1)
cos θ sin φ =
1 & Mx
'
+ 27sin θ )
11 (* F
+
(4)
and Equation (3)
cos θ cos φ =
M '
1&
4sin θ − z )
(
11 *
F +
(5)
Substituting Equations (4) and (5) into Equation (2),
.0
M '!
1&
1 &M
' ! /0
M y = F 1−27 " ( 4sin θ − z ) # + 4 " ( x + 27 sin θ ) # 2
F +%
+ % 04
03
$11 *
$11 * F
or
My =
1
(27 M z + 4 M x )
11
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206
PROBLEM 3.54 (Continued)
Noting that the ratios 27
and 114 are the ratios of lengths, have
11
27
4
(−81 lb ⋅ ft) + (−77 lb ⋅ ft)
11
11
= 226.82 lb ⋅ ft
My =
or M y = −227 lb ⋅ ft
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207
PROBLEM 3.55
The frame ACD is hinged at A and D and is supported by a
cable that passes through a ring at B and is attached to hooks
at G and H. Knowing that the tension in the cable is 450 N,
determine the moment about the diagonal AD of the force
exerted on the frame by portion BH of the cable.
SOLUTION
M AD = λ AD ⋅ (rB/A × TBH )
Where
and
1
λ AD = (4i − 3k )
5
rB/A = (0.5 m)i
d BH = (0.375)2 + (0.75) 2 + (−0.75) 2
= 1.125 m
Then
Finally
450 N
(0.375i + 0.75 j − 0.75k )
1.125
= (150 N)i + (300 N) j − (300 N)k
TBH =
MAD
4
0
−3
1
= 0.5 0
0
5
150 300 −300
1
= [(−3)(0.5)(300)]
5
or M AD = − 90.0 N ⋅ m
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208
PROBLEM 3.56
In Problem 3.55, determine the moment about the diagonal AD
of the force exerted on the frame by portion BG of the cable.
SOLUTION
M AD = λ AD ⋅ (rB/A × TBG )
Where
and
1
λ AD = (4i − 3k )
5
rB/A = (0.5 m) j
BG = (−0.5) 2 + (0.925)2 + (−0.4)2
= 1.125 m
Then
Finally
$
450 N
( −0.5i + 0.925 j − 0.4k )
T BG =
1.125
= −(200 N)i + (370 N) j − (160 N)k
MAD
4
0
−3
1
0.5
0
0
=
5
−200 370 −160
1
= [(−3)(0.5)(370)]
5
M AD = −111.0 N ⋅ m
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209
PROBLEM 3.57
The triangular plate ABC is supported by ball-and-socket joints
at B and D and is held in the position shown by cables AE and
CF. If the force exerted by cable AE at A is 55 N, determine the
moment of that force about the line joining Points D and B.
SOLUTION
First note
d AE = (0.9) 2 + (−0.6) 2 + (0.2) 2 = 1.1 m
Then
TAE =
Also
Then
Now
where
Then
55 N
(0.9i − 0.6 j + 0.2k )
1.1
= 5[(9 N)i − (6 N) j + (2 N)k ]
DB = (1.2) 2 + ( −0.35) 2 + (0)2
λ DB
= 1.25 m
!
DB
=
DB
1
(1.2i − 0.35 j)
=
1.25
1
=
(24i − 7 j)
25
M DB = λ DB ⋅ (rA/D × TAE )
TDA = −(0.1 m) j + (0.2 m)k
M DB
24 −7
0
1
=
(5) 0 −0.1 0.2
25
−6
9
2
1
= (−4.8 − 12.6 + 28.8)
5
or M DB = 2.28 N ⋅ m
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210
PROBLEM 3.58
The triangular plate ABC is supported by ball-and-socket joints
at B and D and is held in the position shown by cables AE and
CF. If the force exerted by cable CF at C is 33 N, determine the
moment of that force about the line joining Points D and B.
SOLUTION
First note
dCF = (0.6) 2 + (−0.9) 2 + (−0.2) 2 = 1.1 m
Then
TCF =
Also
DB = (1.2) 2 + ( −0.35) 2 + (0)2
33 N
(0.6i − 0.9 j + 0.2k )
1.1
= 3[(6 N)i − (9 N) j − (2 N)k ]
= 1.25 m
!
DB
=
DB
1
(1.2i − 0.35 j)
=
1.25
1
=
(24i − 7 j)
25
Then
λ DB
Now
M DB = λ DB ⋅ (rC/D × TCF )
where
rC/D = (0.2 m) j − (0.4 m)k
Then
M DB
24 −7
0
1
=
(3) 0 0.2 −0.4
25
6 −9 −2
=
3
(−9.6 + 16.8 − 86.4)
25
or M DB = −9.50 N ⋅ m
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211
PROBLEM 3.59
A regular tetrahedron has six edges of length a. A force P is
directed as shown along edge BC. Determine the moment of P
about edge OA.
SOLUTION
We have
M OA = λOA ⋅ (rC/O × P)
where
From triangle OBC
(OA) x =
a
2
(OA) z = (OA) x tan 30° =
Since
a& 1 '
a
(
)=
2* 3 + 2 3
(OA) 2 = (OA) 2x + (OA) 2y + (OAz )2
2
or
& a '
&a'
a 2 = ( ) + (OA) 2y + (
)
*2+
*2 3+
(OA) y = a 2 −
Then
and
rA/O =
OA
=
2
a2 a2
2
−
=a
4 12
3
2
a
a
i+a
j+
k
2
3
2 3
1
2
1
i+
j+
k
2
3
2 3
P = λBC P =
(a sin 30°)i − (a cos 30°)k
P
( P) = (i − 3k )
2
a
rC/O = ai
M OA
1
2
=
1
2
3
0
2 3
&P'
(a) ( )
0
*2+
1
0
− 3
=
1
aP & 2 '
aP
(( −
)) (1)(− 3) =
2 * 3+
2
M OA =
aP
2
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212
PROBLEM 3.60
A regular tetrahedron has six edges of length a. (a) Show that two
opposite edges, such as OA and BC, are perpendicular to each
other. (b) Use this property and the result obtained in Problem 3.59
to determine the perpendicular distance between edges OA and BC.
SOLUTION
(a)
For edge OA to be perpendicular to edge BC,
!
!
OA ⋅ BC = 0
where
From triangle OBC
(OA) x =
a
2
a& 1 '
a
(
)=
2* 3 + 2 3
! &a'
& a '
OA = ( ) i + (OA) y j + (
)k
*2+
*2 3+
!
BC = ( a sin 30°) i − (a cos 30°) k
(OA) z = (OA) x tan 30° =
and
=
Then
or
so that
(b)
Have M OA
a
a 3
a
i−
k = (i − 3 k )
2
2
2
& a ' !
a
a
" i + (OA) y j + (
) k # ⋅ (i − 3k ) = 0
2
*2 3+ %
$2
a2
a2
+ (OA) y (0) −
=0
4
4
!
!
OA ⋅ BC = 0
!
!
OA is perpendicular to BC .
!
!
= Pd , with P acting along BC and d the perpendicular distance from OA to BC .
From the results of Problem 3.57
M OA =
Pa
2
Pa
2
or d =
= Pd
a
2
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213
!
PROBLEM 3.61
A sign erected on uneven ground is
guyed by cables EF and EG. If the
force exerted by cable EF at E is 46 lb,
determine the moment of that force
about the line joining Points A and D.
SOLUTION
First note that BC = (48) 2 + (36) 2 = 60 in. and that
BE
BC
=
45
60
= 34 . The coordinates of Point E are
then ( 34 × 48, 96, 34 × 36 ) or (36 in., 96 in., 27 in.). Then
d EF = (−15) 2 + (−110) 2 + (30)2
= 115 in.
46 lb
( −15i − 110 j + 30k )
115
= 2[−(3 lb)i − (22 lb) j + (6 lb)k ]
Then
TEF =
Also
AD = (48)2 + ( −12)2 + (36)2
Then
Now
λ AD
= 12 26 in.
!
AD
=
AD
1
(48i − 12 j + 36k )
=
12 26
1
=
(4i − j + 3k )
26
M AD = λ AD ⋅ (rE/A × TEF )
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214
PROBLEM 3.61 (Continued)
where
Then
rE/A = (36 in.)i + (96 in.) j + (27 in.)k
M AD =
=
−1 3
4
(2) 36 96 27
26
−3 −22 6
1
2
26
(2304 + 81 − 2376 + 864 + 216 + 2376)
or M AD = 1359 lb ⋅ in.
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215
!
PROBLEM 3.62
A sign erected on uneven ground is
guyed by cables EF and EG. If the
force exerted by cable EG at E is 54 lb,
determine the moment of that force
about the line joining Points A and D.
SOLUTION
First note that BC = (48) 2 + (36) 2 = 60 in. and that
BE
BC
=
45
60
= 43 . The coordinates of Point E are
then ( 34 × 48, 96, 34 × 36 ) or (36 in., 96 in., 27 in.). Then
d EG = (11) 2 + (−88) 2 + (−44) 2
= 99 in.
Then
Also
54 lb
(11i − 88 j − 44k )
99
= 6[(1 lb)i − (8 lb) j − (4 lb)k ]
TEG =
AD = (48)2 + ( −12)2 + (36)2
= 12 26 in.
!
AD
=
AD
1
(48i − 12 j + 36k )
=
12 26
1
(4i − j + 3k )
=
26
Then
λ AD
Now
M AD = λ AD ⋅ (rE/A × TEG )
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216
PROBLEM 3.62 (Continued)
where
Then
rE/A = (36 in.)i + (96 in.) j + (27 in.)k
M AD =
=
4 −1 3
(6) 36 96 27
26
1 − 8 −4
1
6
26
(−1536 − 27 − 864 − 288 − 144 + 864)
or M AD = −2350 lb ⋅ in.
!
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217
!
PROBLEM 3.63
Two forces F1 and F2 in space have the same magnitude F. Prove that the moment of F1 about the line of
action of F2 is equal to the moment of F2 about the line of action of F1 .
SOLUTION
First note that
F1 = F1λ1
and F2 = F2 λ 2
Let M1 = moment of F2 about the line of action of M1 and M 2 = moment of F1 about the line of
action of M 2
Now, by definition
M1 = λ1 ⋅ (rB/A × F2 )
= λ1 ⋅ (rB/A × λ 2 ) F2
M 2 = λ2 ⋅ (rA/B × F1 )
= λ2 ⋅ (rA/B × λ1 ) F1
Since
F1 = F2 = F
and rA/B = −rB/A
M1 = λ1 ⋅ (rB/A × λ 2 ) F
M 2 = λ 2 ⋅ (−rB/A × λ1 ) F
Using Equation (3.39)
so that
λ1 ⋅ (rB/A × λ 2 ) = λ 2 ⋅ ( −rB/A × λ1 )
M 2 = λ 1⋅ (rB/A × λ 2 ) F !
M12 = M 21
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218
!
PROBLEM 3.64
In Problem 3.55, determine the perpendicular distance between
portion BH of the cable and the diagonal AD.
PROBLEM 3.55 The frame ACD is hinged at A and D and is
supported by a cable that passes through a ring at B and is
attached to hooks at G and H. Knowing that the tension in the
cable is 450 N, determine the moment about the diagonal AD
of the force exerted on the frame by portion BH of the cable.
SOLUTION
From the solution to Problem 3.55:
TBH = 450 N
TBH = (150 N)i + (300 N) j − (300 N)k
| M AD | = 90.0 N ⋅ m
1
λ AD = (4i − 3k )
5
Based on the discussion of Section 3.11, it! follows that only the perpendicular component of TBH will
contribute to the moment of TBH about line AD.
Now
(TBH )parallel = TBH ⋅ λ AD
1
= (150i + 300 j − 300k ) ⋅ (4i − 3k )
5
1
= [(150)(4) + (−300)(−3)]
5
= 300 N
Also
so that
TBH = (TBH ) parallel + (TBH )perpendicular
(TBH )perpendicular = (450) 2 − (300) 2 = 335.41 N
Since λ AD and (TBH )perpendicular are perpendicular, it follows that
M AD = d (TBH ) perpendicular
or
90.0 N ⋅ m = d (335.41 N)
d = 0.26833 m
d = 0.268 m
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219
!
PROBLEM 3.65
In Problem 3.56, determine the perpendicular distance between
portion BG of the cable and the diagonal AD.
PROBLEM 3.56 In Problem 3.55, determine the moment
about the diagonal AD of the force exerted on the frame by
portion BG of the cable.
SOLUTION
From the solution to Problem 3.56:
ΤBG = 450 N
TBG = −(200 N)i + (370 N) j − (160 N)k
| M AD | = 111 N ⋅ m
1
λ AD = (4i − 3k )
5
Based on the discussion of Section 3.11, it" follows that only the perpendicular component of TBG will
contribute to the moment of TBG about line AD.
Now
(TBG ) parallel = TBG ⋅ λ AD
1
= ( −200i + 370 j − 160k ) ⋅ (4i − 3k )
5
1
= [(−200)(4) + (−160)(−3)]
5
= −64 N
Also
so that
TBG = (TBG ) parallel + (TBG ) perpendicular
(TBG ) perpendicular = (450) 2 − ( −64) 2 = 445.43 N
Since λ AD and (TBG ) perpendicular are perpendicular, it follows that
M AD = d (TBG ) perpendicular
or
111 N ⋅ m = d (445.43 N)
d = 0.24920 m
d = 0.249 m
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220
PROBLEM 3.66
In Problem 3.57, determine the perpendicular distance between
cable AE and the line joining Points D and B.
PROBLEM 3.57 The triangular plate ABC is supported by
ball-and-socket joints at B and D and is held in the position
shown by cables AE and CF. If the force exerted by cable AE
at A is 55 N, determine the moment of that force about the line
joining Points D and B.
SOLUTION
From the solution to Problem 3.57
ΤAE = 55 N
TAE = 5[(9 N)i − (6 N) j + (2 N)k ]
| M DB | = 2.28 N ⋅ m
λ DB =
1
(24i − 7 j)
25
Based on the discussion of Section 3.11, it! follows that only the perpendicular component of TAE will
contribute to the moment of TAE about line DB.
Now
(TAE )parallel = TAE ⋅ λ DB
= 5(9i − 6 j + 2k ) ⋅
1
(24i − 7 j)
25
1
= [(9)(24) + (−6)(−7)]
5
= 51.6 N
Also
so that
TAE = (TAE ) parallel + (TAE ) perpendicular
(TAE )perpendicular = (55) 2 + (51.6)2 = 19.0379 N
Since λ DB and (TAE )perpendicular are perpendicular, it follows that
M DB = d (TAE ) perpendicular
or
2.28 N ⋅ m = d (19.0379 N)
d = 0.119761
d = 0.1198 m
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221
PROBLEM 3.67
In Problem 3.58, determine the perpendicular distance
between cable CF and the line joining Points D and B.
PROBLEM 3.58 The triangular plate ABC is supported by
ball-and-socket joints at B and D and is held in the position
shown by cables AE and CF. If the force exerted by cable CF at
C is 33 N, determine the moment of that force about the line
joining Points D and B.
SOLUTION
From the solution to Problem 3.58
ΤCF = 33 N
TCF = 3[(6 N)i − (9 N) j − (2 N)k ]
| M DB | = 9.50 N ⋅ m
λ DB =
1
(24i − 7 j)
25
Based on the discussion of Section 3.11, it! follows that only the perpendicular component of TCF will
contribute to the moment of TCF about line DB.
Now
(TCF )parallel = TCF ⋅ λ DB
= 3(6i − 9 j − 2k ) ⋅
1
(24i − 7 j)
25
3
[(6)(24) + (−9)( −7)]
25
= 24.84 N
=
Also
so that
TCF = (TCF ) parallel + (TCF ) perpendicular
(TCF )perpendicular = (33) 2 − (24.84) 2
= 21.725 N
Since λ DB and (TCF )perpendicular are perpendicular, it follows that
| M DB | = d (TCF ) perpendicular
or
9.50 N ⋅ m = d × 21.725 N
or d = 0.437 m
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222
PROBLEM 3.68
In Problem 3.61, determine the perpendicular
distance between cable EF and the line joining
Points A and D.
PROBLEM 3.61 A sign erected on uneven
ground is guyed by cables EF and EG. If the
force exerted by cable EF at E is 46 lb,
determine the moment of that force about
the line joining Points A and D.
SOLUTION
From the solution to Problem 3.61
TEF = 46 lb
TEF = 2[−(3 lb)i − (22 lb) j + (6 lb)k ]
| M AD | = 1359 lb ⋅ in.
λ AD =
1
26
(4i − j + 3k )
Based on the discussion of Section 3.11, it! follows that only the perpendicular component of TEF will
contribute to the moment of TEF about line AD.
Now
(TEF ) parallel = TEF ⋅ λ AD
= 2(−3i − 22 j + 6k ) ⋅
=
1
26
(4i − j + 3k )
2
[(−3)(4) + (−22)(−1) + (6)(3)]
26
= 10.9825 lb
Also
so that
TEF = (TEF ) parallel + (TEF )perpendicular
(TEF ) perpendicular = (46) 2 − (10.9825) 2 = 44.670 lb
Since λ AD and (TEF ) perpendicular are perpendicular, it follows that
M AD = d (TEF ) perpendicular
or
1359 lb ⋅ in. = d × 44.670 lb
or d = 30.4 in.
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223
!
PROBLEM 3.69
In Problem 3.62, determine the perpendicular
distance between cable EG and the line joining
Points A and D.
PROBLEM 3.62 A sign erected on uneven
ground is guyed by cables EF and EG. If the
force exerted by cable EG at E is 54 lb,
determine the moment of that force about the
line joining Points A and D.
SOLUTION
From the solution to Problem 3.62
TEG = 54 lb
TEG = 6[(1 lb)i − (8 lb) j − (4 lb)k ]
| M AD | = 2350 lb ⋅ in.
1
λ AD =
26
(4i − j + 3k )
Based on the discussion of Section 3.11, it! follows that only the perpendicular component of TEG will
contribute to the moment of TEG about line AD.
Now
(TEG ) parallel = TEG ⋅ λ AD
= 6(i − 8 j − 4k ) ⋅
=
Thus,
6
26
1
26
(4i − j + 3k )
[(1)(4) + ( −8)( −1) + (−4)(3)] = 0
(TEG ) perpendicular = TEG = 54 lb
Since λ AD and (TEG ) perpendicular are perpendicular, it follows that
| M AD | = d (TEG ) perpendicular
or
2350 lb ⋅ in. = d × 54 lb
or d = 43.5 in.
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224
!
PROBLEM 3.70
Two parallel 60-N forces are applied to a lever as shown.
Determine the moment of the couple formed by the two forces
(a) by resolving each force into horizontal and vertical components
and adding the moments of the two resulting couples, (b) by using
the perpendicular distance between the two forces, (c) by summing
the moments of the two forces about Point A.
SOLUTION
(a)
We have
where
ΣM B : − d1C x + d 2 C y = M
d1 = (0.360 m) sin 55°
= 0.29489 m
d 2 = (0.360 m) sin 55°
= 0.20649 m
C x = (60 N) cos 20°
= 56.382 N
C y = (60 N)sin 20°
= 20.521 N
M = −(0.29489 m)(56.382 N)k + (0.20649 m)(20.521 N)k
= −(12.3893 N ⋅ m)k
(b)
We have
M = Fd (−k )
= 60 N[(0.360 m) sin(55° − 20°)]( −k )
= −(12.3893 N ⋅ m)k
(c)
We have
or M = 12.39 N ⋅ m
or M = 12.39 N ⋅ m
ΣM A : Σ(rA × F) = rB/A × FB + rC/A × FC = M
i
j
k
sin 55° 0
M = (0.520 m)(60 N) cos 55°
− cos 20° − sin 20° 0
i
j
k
+ (0.800 m)(60 N) cos 55° sin 55° 0
cos 20° sin 20° 0
= (17.8956 N ⋅ m − 30.285 N ⋅ m)k
= −(12.3892 N ⋅ m)k
or M = 12.39 N ⋅ m
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225
PROBLEM 3.71
A plate in the shape of a parallelogram is acted upon by two
couples. Determine (a) the moment of the couple formed by the
two 21-lb forces, (b) the perpendicular distance between the 12-lb
forces if the resultant of the two couples is zero, (c) the value of α
if the resultant couple is 72 lb ⋅ in. clockwise and d is 42 in.
SOLUTION
(a)
M1 = d1 F1
We have
d1 = 16 in.
where
F1 = 21 lb
M1 = (16 in.)(21 lb)
= 336 lb ⋅ in.
(b)
(c)
M1 + M 2 = 0
We have
or
336 lb ⋅ in. − d 2 (12 lb) = 0
d 2 = 28.0 in.
M total = M1 + M 2
We have
or
or M1 = 336 lb ⋅ in.
−72 lb ⋅ in. = 336 lb ⋅ in. − (42 in.)(sin α )(12 lb)
sin α = 0.80952
α = 54.049°
and
or α = 54.0°
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226
PROBLEM 3.72
A couple M of magnitude 18 N ⋅ m is applied to the
handle of a screwdriver to tighten a screw into a
block of wood. Determine the magnitudes of the
two smallest horizontal forces that are equivalent to
M if they are applied (a) at corners A and D, (b) at
corners B and C, (c) anywhere on the block.
SOLUTION
(a)
M = Pd
We have
or
18 N ⋅ m = P(.24 m)
P = 75.0 N
or Pmin = 75.0 N
d BC = ( BE ) 2 + ( EC ) 2
(b)
= (.24 m) 2 + (.08 m) 2
= 0.25298 m
M = Pd
We have
18 N ⋅ m = P(0.25298 m)
P = 71.152 N
or
P = 71.2 N
d AC = ( AD) 2 + ( DC ) 2
(c)
= (0.24 m) 2 + (0.32 m) 2
= 0.4 m
M = Pd AC
We have
18 N ⋅ m = P(0.4 m)
P = 45.0 N
or
P = 45.0 N
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227
PROBLEM 3.73
Four 1-in.-diameter pegs are attached to a board as shown. Two
strings are passed around the pegs and pulled with the forces
indicated. (a) Determine the resultant couple acting on the board.
(b) If only one string is used, around which pegs should it pass
and in what directions should it be pulled to create the same
couple with the minimum tension in the string? (c) What is the
value of that minimum tension?
SOLUTION
M = (35 lb)(7 in.) + (25 lb)(9 in.)
= 245 lb ⋅ in. + 225 lb ⋅ in.
(a)
M = 470 lb ⋅ in.
(b)
With only one string, pegs A and D, or B and C should be used. We have
6
8
tan θ =
θ = 36.9°
90° − θ = 53.1°
Direction of forces:
(c)
With pegs A and D:
θ = 53.1°
With pegs B and C:
θ = 53.1°
The distance between the centers of the two pegs is
82 + 62 = 10 in.
Therefore, the perpendicular distance d between the forces is
1 !
d = 10 in. + 2 " in. #
$2 %
= 11 in.
M = Fd
We must have
470 lb ⋅ in. = F (11 in.)
F = 42.7 lb
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228
PROBLEM 3.74
Four pegs of the same diameter are attached to a board as shown.
Two strings are passed around the pegs and pulled with the
forces indicated. Determine the diameter of the pegs knowing
that the resultant couple applied to the board is 485 lb·in.
counterclockwise.
SOLUTION
M = d AD FAD + d BC FBC
485 lb ⋅ in. = [(6 + d )in.](35 lb) + [(8 + d )in.](25 lb)
d = 1.250 in.
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229
PROBLEM 3.75
The shafts of an angle drive are acted upon by the two couples shown.
Replace the two couples with a single equivalent couple, specifying its
magnitude and the direction of its axis.
SOLUTION
Based on
where
M = M1 + M 2
M1 = −(8 lb ⋅ ft)j
M 2 = −(6 lb ⋅ ft)k
M = −(8 lb ⋅ ft)j − (6 lb ⋅ ft)k
and
|M| = (8) 2 + (6) 2 = 10 lb ⋅ ft
or M = 10.00 lb ⋅ ft
M
|M|
−(8 lb ⋅ ft)j − (6 lb ⋅ ft)k
=
10 lb ⋅ ft
= −0.8 j − 0.6k
=
or
M = |M| = (10 lb ⋅ ft)(−0.8 j − 0.6k )
cos θ x = 0
cos θ y = −0.8
θ x = 90°
θ y = 143.130°
cos θ z = −0.6
θ z = 126.870°
or θ x = 90.0° θ y = 143.1° θ z = 126.9°
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230
PROBLEM 3.76
If P = 0, replace the two remaining couples with a single
equivalent couple, specifying its magnitude and the direction
of its axis.
SOLUTION
We have
where
M = M1 + M 2
M1 = rG/C × F1
rG/C = −(0.3 m)i
F1 = (18 N)k
M1 = −(0.3 m)i × (18 N)k
= (5.4 N ⋅ m) j
Also,
M 2 = rD/F × F2
rD/F = −(.15 m)i + (.08 m) j
F2 = λED F2
=
(.15 m)i + (.08 m) j + (.17 m)k
(.15)2 + (.08) 2 + (.17) 2 m
(34 N)
= 141.421 N ⋅ m(.15i + .08j + .17k )
i
j
k
M 2 = 141.421 N ⋅ m −.15 .08 0
−.15 .08 .17
= 141.421(.0136i + 0.0255 j)N ⋅ m
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231
PROBLEM 3.76 (Continued)
and
M = [(5.4 N ⋅ m)j] + [141.421(.0136i + .0255 j) N ⋅ m]
= (1.92333 N ⋅ m)i + (9.0062 N ⋅ m)j
| M | = (M x )2 + (M y )2
= (1.92333)2 + (9.0062) 2
= 9.2093 N ⋅ m
λ=
or M = 9.21 N ⋅ m
M (1.92333 N ⋅ m)i + (9.0062 N ⋅ m) j
=
|M|
9.2093 N ⋅ m
= 0.20885 + 0.97795
cos θ x = 0.20885
θ x = 77.945°
or
θ x = 77.9°
or
θ y = 12.05°
or
θ z = 90.0°
cos θ y = 0.97795
θ y = 12.054°
cos θ z = 0.0
θ z = 90°
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232
PROBLEM 3.77
If P = 0, replace the two remaining couples with a single
equivalent couple, specifying its magnitude and the direction
of its axis.
SOLUTION
M = M1 + M 2 ; F1 = 16 lb, F2 = 40 lb
M1 = rC × F1 = (30 in.)i × [−(16 lb) j] = −(480 lb ⋅ in.)k
M 2 = rE/B × F2 ; rE/B = (15 in.)i − (5 in.) j
d DE = (0) 2 + (5) 2 + (10) 2 = 5 5 in.
F2 =
40 lb
5 5
(5 j − 10k )
= 8 5[(1 lb) j − (2 lb)k ]
i
j k
M 2 = 8 5 15 −5 0
0 1 −2
= 8 5[(10 lb ⋅ in.)i + (30 lb ⋅ in.) j + (15 lb ⋅ in.)k ]
M = −(480 lb ⋅ in.)k + 8 5[(10 lb ⋅ in.)i + (30 lb ⋅ in.) j + (15 lb ⋅ in.)k ]
= (178.885 lb ⋅ in.)i + (536.66 lb ⋅ in.) j − (211.67 lb ⋅ in.)k
M = (178.885) 2 + (536.66) 2 + (−211.67) 2
= 603.99 lb ⋅ in
M = 604 lb ⋅ in.
M
= 0.29617i + 0.88852 j − 0.35045k
M
cos θ x = 0.29617
λ axis =
cos θ y = 0.88852
θ x = 72.8° θ y = 27.3° θ z = 110.5°
cos θ z = −0.35045
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233
PROBLEM 3.78
If P = 20 lb, replace the three couples with a single
equivalent couple, specifying its magnitude and the
direction of its axis.
SOLUTION
From the solution to Problem. 3.77
16 lb force:
M1 = −(480 lb ⋅ in.)k
40 lb force:
M 2 = 8 5[(10 lb ⋅ in.)i + (30 lb ⋅ in.) j + (15 lb ⋅ in.)k ]
P = 20 lb
M 3 = rC × P
= (30 in.)i × (20 lb)k
= (600 lb ⋅ in.) j
M = M1 + M 2 + M 3
= −(480)k + 8 5 (10i + 30 j + 15k ) + 600 j
= (178.885 lb ⋅ in)i + (1136.66 lb ⋅ in.) j − (211.67 lb ⋅ in.)k
M = (178.885) 2 + (113.66) 2 + (211.67) 2
= 1169.96 lb ⋅ in
M = 1170 lb ⋅ in.
M
= 0.152898i + 0.97154 j − 0.180921k
M
cos θ x = 0.152898
λ axis =
cos θ y = 0.97154
θ x = 81.2° θ y = 13.70° θ z = 100.4°
cos θ z = −0.180921
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234
PROBLEM 3.79
If P = 20 N, replace the three couples with a single
equivalent couple, specifying its magnitude and the
direction of its axis.
SOLUTION
We have
M = M1 + M 2 + M 3
where
i
j k
M1 = rG/C × F1 = −0.3 0 0 N ⋅ m = (5.4 N ⋅ m)j
0
0 18
M 2 = rD/F
i
j
k
× F2 = −.15 .08 0 141.421 N ⋅ m
−.15 .08 .17
= 141.421(.0136i + .0255 j)N ⋅ m
(See Solution to Problem 3.76.)
i
j
k
M 3 = rC/A × F3 = 0.3 0 0.17 N ⋅ m
0 20
0
= −(3.4 N ⋅ m)i + (6 N ⋅ m)k
M = [(1.92333 − 3.4)i + (5.4 + 3.6062) j + (6)k ] N ⋅ m
= −(1.47667 N ⋅ m)i + (9.0062 N ⋅ m) j + (6 N ⋅ m)
| M | = M x2 + M y2 + M z2
= (1.47667) + (9.0062) + (6)2
or M = 10.92 N ⋅ m
= 10.9221 N ⋅ m
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235
PROBLEM 3.79 (Continued)
M −1.47667 + 9.0062 + 6
=
|M|
10.9221
= −0.135200i + 0.82459 j + 0.54934k
=
cos θ x = −0.135200 θ x = 97.770
or θ x = 97.8°
cos θ y = 0.82459
θ y = 34.453
or θ y = 34.5°
cos θ z = 0.54934
θ z = 56.678
or θ z = 56.7°
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236
PROBLEM 3.80
Shafts A and B connect the gear box to the wheel assemblies
of a tractor, and shaft C connects it to the engine. Shafts A and
B lie in the vertical yz plane, while shaft C is directed along
the x axis. Replace the couples applied to the shafts with a
single equivalent couple, specifying its magnitude and the
direction of its axis.
SOLUTION
Represent the given couples by the following couple vectors:
M A = −1600sin 20° j + 1600cos 20°k
= −(547.232 N ⋅ m) j + (1503.51 N ⋅ m)k
M B = 1200sin 20° j + 1200 cos 20°k
= (410.424 N ⋅ m) j + (1127.63 N ⋅ m)k
M C = −(1120 N ⋅ m)i
The single equivalent couple is
M = M A + M B + MC
= −(1120 N ⋅ m)i − (136.808 N ⋅ m) j + (2631.1 N ⋅ m)k
M = (1120) 2 + (136.808)2 + (2631.1)2
= 2862.8 N ⋅ m
−1120
cos θ x =
2862.8
−136.808
cos θ y =
2862.8
2631.1
cos θ z =
2862.8
M = 2860 N ⋅ m θ x = 113.0° θ y = 92.7° θ z = 23.2°
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237
PROBLEM 3.81
The tension in the cable attached to the end C of an
adjustable boom ABC is 560 lb. Replace the force
exerted by the cable at C with an equivalent forcecouple system (a) at A, (b) at B.
SOLUTION
(a)
Based on
ΣF : FA = T = 560 lb
FA = 560 lb
or
20°
ΣM A : M A = (T sin 50°)(d A )
= (560 lb)sin 50°(18 ft)
= 7721.7 lb ⋅ ft
M A = 7720 lb ⋅ ft
or
(b)
Based on
ΣF : FB = T = 560 lb
FB = 560 lb
or
20°
ΣM B : M B = (T sin 50°)(d B )
= (560 lb) sin 50°(10 ft)
= 4289.8 lb ⋅ ft
M B = 4290 lb ⋅ ft
or
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238
PROBLEM 3.82
A 160-lb force P is applied at Point A of a structural member.
Replace P with (a) an equivalent force-couple system at C,
(b) an equivalent system consisting of a vertical force at B
and a second force at D.
SOLUTION
(a)
Based on
ΣF : PC = P = 160 lb
or PC = 160 lb
60°
ΣM C : M C = − Px d cy + Py dCx
where
Px = (160 lb) cos 60°
= 80 lb
Py = (160 lb)sin 60°
dCx
= 138.564 lb
= 4 ft
dCy = 2.75 ft
M C = (80 lb)(2.75 ft) + (138.564 lb)(4 ft)
= 220 lb ⋅ ft + 554.26 lb ⋅ ft
= 334.26 lb ⋅ ft
(b)
Based on
or M C = 334 lb ⋅ ft
ΣFx : PDx = P cos 60°
= (160 lb) cos 60°
= 80 lb
ΣM D : ( P cos 60°)( d DA ) = PB (d DB )
[(160 lb) cos 60°](1.5 ft) = PB (6 ft)
PB = 20.0 lb
or PB = 20.0 lb
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239
PROBLEM 3.82 (Continued)
ΣFy : P sin 60° = PB + PDy
(160 lb)sin 60° = 20.0 lb + PDy
PDy = 118.564 lb
PD = ( PDx ) 2 + ( PDy ) 2
= (80) 2 + (118.564) 2
= 143.029 lb
PDy !
#
$ PDx %
118.564 !
= tan −1 "
#
$ 80 %
= 55.991°
θ = tan −1 "
or PD = 143.0 lb
56.0°
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240
PROBLEM 3.83
The 80-N horizontal force P acts on a bell crank as shown.
(a) Replace P with an equivalent force-couple system at B.
(b) Find the two vertical forces at C and D that are equivalent
to the couple found in Part a.
SOLUTION
(a)
ΣF : FB = F = 80 N
Based on
or FB = 80.0 N
ΣM : M B = Fd B
= 80 N (.05 m)
= 4.0000 N ⋅ m
M B = 4.00 N ⋅ m
or
(b)
If the two vertical forces are to be equivalent to MB, they must be
a couple. Further, the sense of the moment of this couple must be
counterclockwise.
Then, with FC and FD acting as shown,
ΣM : M D = FC d
4.0000 N ⋅ m = FC (.04 m)
FC = 100.000 N
or FC = 100.0 N
ΣFy : 0 = FD − FC
FD = 100.000 N
or FD = 100.0 N
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241
PROBLEM 3.84
A dirigible is tethered by a cable attached to its cabin at B. If
the tension in the cable is 1040 N, replace the force exerted by
the cable at B with an equivalent system formed by two
parallel forces applied at A and C.
SOLUTION
Require the equivalent forces acting at A and C be parallel and at an
angle of α with the vertical.
Then for equivalence,
ΣFx : (1040 N)sin 30° = FA sin α + FB sin α
(1)
ΣFy : −(1040 N) cos 30° = − FA cos α − FB cos α
(2)
Dividing Equation (1) by Equation (2),
( FA + FB ) sin α
(1040 N) sin 30°
=
−(1040 N) cos 30° −( FA + FB ) cos α
Simplifying yields α = 30°
Based on
ΣM C : [(1040 N) cos 30°](4 m) = ( FA cos 30°)(10.7 m)
FA = 388.79 N
FA = 389 N
or
60°
Based on
ΣM A : − [(1040 N) cos 30°](6.7 m) = ( FC cos 30°)(10.7 m)
FC = 651.21 N
FC = 651 N
or
60°
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242
PROBLEM 3.85
The force P has a magnitude of 250 N and is applied at the end C
of a 500-mm rod AC attached to a bracket at A and B. Assuming
α = 30° and β = 60°, replace P with (a) an equivalent force-couple
system at B, (b) an equivalent system formed by two parallel forces
applied at A and B.
SOLUTION
(a)
ΣF : F = P or F = 250 N
Equivalence requires
60°
ΣM B : M = −(0.3 m)(250 N) = −75 N ⋅ m
The equivalent force-couple system at B is
F = 250 N
(b)
M = 75.0 N ⋅ m
60°
Require
Equivalence then requires
ΣFx : 0 = FA cos φ + FB cos φ
FA = − FB
or cos φ = 0
ΣFy : − 250 = − FA sin φ − FB sin φ
Now if
FA = − FB & −250 = 0 reject
cos φ = 0
or
and
Also
φ = 90°
FA + FB = 250
ΣM B : − (0.3 m)(250 N) = (0.2m) FA
or
FA = −375 N
and
FB = 625 N
FA = 375 N
60°
FB = 625 N
60°
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243
PROBLEM 3.86
Solve Problem 3.85, assuming α = β = 25°.
SOLUTION
(a)
Equivalence requires
ΣF : FB = P or FB = 250 N
25.0°
ΣM B : M B = −(0.3 m)[(250 N)sin 50°] = −57.453 N ⋅ m
The equivalent force-couple system at B is
FB = 250 N
(b)
25.0°
M B = 57.5 N ⋅ m
Require
Equivalence requires
M B = d AE Q (0.3 m)[(250 N) sin 50°]
= [(0.2 m) sin 50°]Q
Q = 375 N
Adding the forces at B:
FA = 375 N
25.0°
FB = 625 N
25.0°
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244
PROBLEM 3.87
A force and a couple are applied as shown to the end of a
cantilever beam. (a) Replace this system with a single force F
applied at Point C, and determine the distance d from C to a
line drawn through Points D and E. (b) Solve Part a if the
directions of the two 360-N forces are reversed.
SOLUTION
(a)
We have
ΣF : F = (360 N) j − (360 N) j − (600 N)k
or F = −(600 N)k
ΣM D : (360 N)(0.15 m) = (600 N)(d )
and
d = 0.09 m
or d = 90.0 mm below ED
(b)
We have from Part a
F = −(600 N)k
ΣM D : −(360 N)(0.15 m) = −(600 N)( d )
and
d = 0.09 m
or d = 90.0 mm above ED
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245
PROBLEM 3.88
The shearing forces exerted on the cross section of a steel channel can
be represented by a 900-N vertical force and two 250-N horizontal
forces as shown. Replace this force and couple with a single force F
applied at Point C, and determine the distance x from C to line BD.
(Point C is defined as the shear center of the section.)
SOLUTION
Replace the 250-N forces with a couple and move the 900-N force to Point C such that its moment about H is
equal to the moment of the couple
M H = (0.18)(250 N)
= 45 N ⋅ m
Then
or
M H = x(900 N)
45 N ⋅ m = x(900 N)
x = 0.05 m
F = 900 N
x = 50.0 mm
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246
PROBLEM 3.89
While tapping a hole, a machinist applies the horizontal forces
shown to the handle of the tap wrench. Show that these forces
are equivalent to a single force, and specify, if possible, the
point of application of the single force on the handle.
SOLUTION
Since the forces at A and B are parallel, the force at B can be replaced with the sum of two forces with one of
the forces equal in magnitude to the force at A except with an opposite sense, resulting in a force-couple.
Have FB = 2.9 lb − 2.65 lb = 0.25 lb, where the 2.65 lb force be part of the couple. Combining the two
parallel forces,
M couple = (2.65 lb)[(3.2 in. + 2.8 in.) cos 25°]
= 14.4103 lb ⋅ in.
and
M couple = 14.4103 lb ⋅ in.
A single equivalent force will be located in the negative z-direction
Based on
ΣM B : −14.4103 lb ⋅ in. = [(.25 lb) cos 25°](a )
a = 63.600 in.
F′ = (.25 lb)(cos 25°i + sin 25°k )
F′ = (0.227 lb)i + (0.1057 lb)k and is applied on an extension of handle BD at a
distance of 63.6 in. to the right of B
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247
PROBLEM 3.90
Three control rods attached to a lever ABC exert on it the
forces shown. (a) Replace the three forces with an equivalent
force-couple system at B. (b) Determine the single force that
is equivalent to the force-couple system obtained in Part a,
and specify its point of application on the lever.
SOLUTION
(a)
First note that the two 20-lb forces form A couple. Then
F = 48 lb
θ
where
θ = 180° − (60° + 55°) = 65°
and
M = ΣM B
= (30 in.)(48 lb) cos55° − (70 in.)(20 lb) cos 20°
= −489.62 lb ⋅ in
The equivalent force-couple system at B is
F = 48.0 lb
(b)
65°
M = 490 lb ⋅ in.
The single equivalent force F ′ is equal to F. Further, since the sense of M is clockwise, F ′ must be applied
between A and B. For equivalence.
ΣM B : M = − aF ′ cos 55°
where a is the distance from B to the point of application of F′. Then
−489.62 lb ⋅ in. = −a (48.0 lb) cos 55°
or
a = 17.78 in.
F ′ = 48.0 lb
65.0°
and is applied to the lever 17.78 in.
To the left of pin B
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248
PROBLEM 3.91
A hexagonal plate is acted upon by the force P and the couple shown.
Determine the magnitude and the direction of the smallest force P for
which this system can be replaced with a single force at E.
SOLUTION
From the statement of the problem, it follows that ΣM E = 0 for the given force-couple system. Further,
for Pmin, must require that P be perpendicular to rB/E . Then
ΣM E : (0.2 sin 30° + 0.2)m × 300 N
+ (0.2 m)sin 30° × 300 N
− (0.4 m) Pmin = 0
or
Pmin = 300 N
Pmin = 300 N
30.0°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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249
PROBLEM 3.92
A rectangular plate is acted upon by the force and couple
shown. This system is to be replaced with a single equivalent
force. (a) For α = 40°, specify the magnitude and the line of
action of the equivalent force. (b) Specify the value of α if the line
of action of the equivalent force is to intersect line CD 300 mm
to the right of D.
SOLUTION
(a)
The given force-couple system (F, M) at B is
F = 48 N
and
M = ΣM B = (0.4 m)(15 N) cos 40° + (0.24 m)(15 N)sin 40°
or
M = 6.9103 N ⋅ m
The single equivalent force F′ is equal to F. Further for equivalence
ΣM B : M = dF ′
6.9103 N ⋅ m = d × 48 N
or
d = 0.14396 m
or
F ′ = 48 N
and the line of action of F′ intersects line AB 144 mm to the right of A.
(b)
Following the solution to Part a but with d = 0.1 m and α unknown, have
ΣM B : (0.4 m)(15 N) cos α + (0.24 m)(15 N) sin α
= (0.1 m)(48 N)
or
5cos α + 3sin α = 4
Rearranging and squaring
25 cos 2 α = (4 − 3 sin α )2
Using cos 2 α = 1 − sin 2 α and expanding
25(1 − sin 2 α ) = 16 − 24 sin α + 9 sin 2 α
or
Then
34 sin 2 α − 24 sin α − 9 = 0
24 ± (−24) 2 − 4(34)(−9)
2(34)
sin α = 0.97686 or sin α = −0.27098
sin α =
α = 77.7° or α = −15.72°
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250
PROBLEM 3.93
An eccentric, compressive 1220-N force P is applied to the end
of a cantilever beam. Replace P with an equivalent force-couple
system at G.
SOLUTION
We have
ΣF : − (1220 N)i = F
F = − (1220 N)i
Also, we have
ΣM G : rA/G × P = M
i
j
k
1220 0 −.1 −.06 N ⋅ m = M
−1 0
0
M = (1220 N ⋅ m)[(−0.06)(−1) j − ( −0.1)( −1)k ]
or M = (73.2 N ⋅ m) j − (122 N ⋅ m)k
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251
PROBLEM 3.94
To keep a door closed, a wooden stick is wedged between the
floor and the doorknob. The stick exerts at B a 175-N force
directed along line AB. Replace that force with an equivalent
force-couple system at C.
SOLUTION
We have
ΣF : PAB = FC
where
PAB =
=
AB PAB
(33 mm)i + (990 mm) j − (594 mm)k
(175 N)
1155.00 mm
or FC = (5.00 N)i + (150 N) j − (90.0 N)k
We have
ΣM C : rB/C × PAB = M C
i
j
k
M C = 5 0.683 −0.860 0 N ⋅ m
1
30
−18
= (5){(− 0.860)(−18)i − (0.683)(−18) j
+ [(0.683)(30) − (0.860)(1)]k}
or M C = (77.4 N ⋅ m)i + (61.5 N ⋅ m) j + (106.8 N ⋅ m)k
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252
PROBLEM 3.95
An antenna is guyed by three cables as shown. Knowing
that the tension in cable AB is 288 lb, replace the force
exerted at A by cable AB with an equivalent force-couple
system at the center O of the base of the antenna.
SOLUTION
We have
d AB = (−64)2 + (−128) 2 + (16) 2 = 144 ft
Then
TAB =
Now
288 lb
( −64i − 128 j + 16k )
144
= (32 lb)( −4i − 8 j + k )
M = M O = rA / O × TAB
= 128 j × 32(−4i − 8 j + k )
= (4096 lb ⋅ ft)i + (16,384 lb ⋅ ft)k
The equivalent force-couple system at O is
F = −(128.0 lb)i − (256 lb) j + (32.0 lb)k
M = (4.10 kip ⋅ ft)i + (16.38 kip ⋅ ft)k
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253
PROBLEM 3.96
An antenna is guyed by three cables as shown. Knowing
that the tension in cable AD is 270 lb, replace the force
exerted at A by cable AD with an equivalent force-couple
system at the center O of the base of the antenna.
SOLUTION
We have
d AD = ( −64) 2 + (−128)2 + (−128) 2
= 192 ft
Then
Now
270 lb
(−64i − 128 j + 128k )
192
= (90 lb)(−i − 2 j − 2k )
TAD =
M = M O = rA/O × TAD
= 128 j × 90(−i − 2 j − 2k )
= −(23, 040 lb ⋅ ft)i + (11,520 lb ⋅ ft)k
The equivalent force-couple system at O is
F = −(90.0 lb)i − (180.0 lb) j − (180.0 lb)k
M = −(23.0 kip ⋅ ft)i + (11.52 kip ⋅ ft)k
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254
PROBLEM 3.97
Replace the 150-N force with an equivalent force-couple
system at A.
SOLUTION
Equivalence requires
ΣF : F = (150 N)(− cos 35° j − sin 35°k )
= −(122.873 N) j − (86.036 N)k
ΣMA : M = rD/A × F
where
Then
rD/A = (0.18 m)i − (0.12 m) j + (0.1 m)k
i
j
k
0.1
N⋅m
−0.12
M = 0.18
0
−122.873 −86.036
= [( −0.12)(−86.036) − (0.1)(−122.873)]i
+ [−(0.18)(−86.036)]j
+ [(0.18)(−122.873)]k
= (22.6 N ⋅ m)i + (15.49 N ⋅ m) j − (22.1 N ⋅ m)k
The equivalent force-couple system at A is
F = −(122.9 N) j − (86.0 N)k
M = (22.6 N ⋅ m)i + (15.49 N ⋅ m) j − (22.1 N ⋅ m)k
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255
!
PROBLEM 3.98
A 77-N force F1 and a 31-N ⋅ m couple M1 are
applied to corner E of the bent plate shown. If
F1 and M1 are to be replaced with an equivalent
force-couple system (F2, M2) at corner B and if
(M2)z = 0, determine (a) the distance d, (b) F2
and M2.
SOLUTION
(a)
ΣM Bz : M 2 z = 0
We have
k ⋅ (rH /B × F1 ) + M 1z = 0
where
(1)
rH /B = (0.31 m)i − (0.0233) j
F1 =
EH F1
(0.06 m)i + (0.06 m) j − (0.07 m)k
(77 N)
0.11 m
= (42 N)i + (42 N) j − (49 N)k
=
M1z = k ⋅ M1
M1 =
=
EJ M 1
− di + (0.03 m) j − (0.07 m)k
d 2 + 0.0058 m
(31 N ⋅ m)
Then from Equation (1),
0
0
1
( −0.07 m)(31 N ⋅ m)
0.31 −0.0233 0 +
=0
2
0.0058
+
d
42
42
−49
Solving for d, Equation (1) reduces to
(13.0200 + 0.9786) −
From which
2.17 N ⋅ m
d 2 + 0.0058
d = 0.1350 m
=0
or d = 135.0 mm
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256
PROBLEM 3.98 (Continued)
(b)
F2 = F1 = (42i + 42 j − 49k )N
or F2 = (42 N)i + (42 N) j − (49 N)k
M 2 = rH /B × F1 + M1
i
j
k
(0.1350)i + 0.03j − 0.07k
(31 N ⋅ m)
= 0.31 −0.0233 0 +
0.155000
−49
42
42
= (1.14170i + 15.1900 j + 13.9986k ) N ⋅ m
+ (−27.000i + 6.0000 j − 14.0000k ) N ⋅ m
M 2 = − (25.858 N ⋅ m)i + (21.190 N ⋅ m) j
or M 2 = − (25.9 N ⋅ m)i + (21.2 N ⋅ m) j
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257
PROBLEM 3.99
A 46-lb force F and a 2120-lb ⋅ in. couple M are
applied to corner A of the block shown. Replace
the given force-couple system with an equivalent
force-couple system at corner H.
SOLUTION
We have
Then
Also
d AJ = (18) 2 + (−14) 2 + (−3) 2 = 23 in.
46 lb
(18i − 14 j − 3k )
23
= (36 lb)i − (28 lb) j − (6 lb)k
F=
d AC = (−45) 2 + (0)2 + (−28)2 = 53 in.
2120 lb ⋅ in.
( −45i − 28k )
53
= −(1800 lb ⋅ in.)i − (1120 lb ⋅ in.)k
Then
M=
Now
M ′ = M + rA/H × F
where
Then
rA/H = (45 in.)i + (14 in.) j
i
j
k
0
M ′ = (−1800i − 1120k ) + 45 14
36 −28 −6
= (−1800i − 1120k ) + {[(14)(−6)]i + [−(45)( −6)]j + [(45)(−28) − (14)(36)]k}
= (−1800 − 84)i + (270) j + (−1120 − 1764)k
= −(1884 lb ⋅ in.)i + (270 lb ⋅ in.)j − (2884 lb ⋅ in.)k
= −(157 lb ⋅ ft)i + (22.5 lb ⋅ ft) j − (240 lb ⋅ ft)k
F′ = (36.0 lb)i − (28.0 lb) j − (6.00 lb)k
The equivalent force-couple system at H is
M ′ = −(157 lb ⋅ ft)i + (22.5 lb ⋅ ft) j − (240 lb ⋅ ft)k
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258
PROBLEM 3.100
The handpiece for a miniature industrial grinder weighs 0.6 lb, and
its center of gravity is located on the y axis. The head of the
handpiece is offset in the xz plane in such a way that line BC forms
an angle of 25° with the x direction. Show that the weight of the
handpiece and the two couples M1 and M2 can be replaced with a
single equivalent force. Further, assuming that M1 = 0.68 lb ⋅ in.
and M2 = 0.65 lb ⋅ in., determine (a) the magnitude and the direction
of the equivalent force, (b) the point where its line of action
intersects the xz plane.
SOLUTION
First assume that the given force W and couples M1 and M2 act at the origin.
Now
W = − Wj
and
M = M1 + M 2
= − ( M 2 cos 25°)i + ( M 1 − M 2 sin 25°)k
Note that since W and M are perpendicular, it follows that they can be replaced
with a single equivalent force.
F = W or F = − Wj = − (0.6 lb) j
or F = − (0.600 lb)j
(a)
We have
(b)
Assume that the line of action of F passes through Point P(x, 0, z). Then for equivalence
M = rP/0 × F
where
rP/0 = xi + zk
− ( M 2 cos 25°)i + ( M1 − M 2 sin 25°)k
i
j
= x 0
0 −W
Equating the i and k coefficients,
(b)
For
z=
k
z = (Wz )i − (Wx)k
0
− M z cos 25°
W
and
x = −"
$
M 1 − M 2 sin 25° !
#
W
%
W = 0.6 lb M1 = 0.68 lb ⋅ in. M 2 = 0.65 lb ⋅ in.
x=
0.68 − 0.65sin 25°
= 0.67550 in.
− 0.6
or
z=
− 0.65cos 25°
= − 0.98183 in.
0.6
or z = − 0.982 in.
x = 0.675 in.
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259
PROBLEM 3.101
A 4-m-long beam is subjected to a variety of loadings. (a) Replace each loading with an equivalent forcecouple system at end A of the beam. (b) Which of the loadings are equivalent?
SOLUTION
(a)
(a)
We have
ΣFy : −400 N − 200 N = Ra
or
and
R a = 600 N
ΣM A : 1800 N ⋅ m − (200 N)(4 m) = M a
or M a = 1000 N ⋅ m
(b)
We have
ΣFy : − 600 N = Rb
or
and
ΣM A : − 900 N ⋅ m = M b
or
(c)
We have
M b = 900 N ⋅ m
ΣFy : 300 N − 900 N = Rc
or
and
R b = 600 N
R c = 600 N
ΣM A : 4500 N ⋅ m − (900 N)(4 m) = M c
or
M c = 900 N ⋅ m
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260
PROBLEM 3.101 (Continued)
(d)
We have
and
(e)
We have
ΣFy : − 400 N + 800 N = Rd
or
R d = 400 N
or
M d = 900 N ⋅ m
ΣM A : (800 N)(4 m) − 2300 N ⋅ m = M d
ΣFy : − 400 N − 200 N = Re
or
and
ΣM A : 200 N ⋅ m + 400 N ⋅ m − (200 N)(4 m) = M e
or
( f ) We have
M e = 200 N ⋅ m
ΣFy : − 800 N + 200 N = R f
or
and
R e = 600 N
R f = 600 N
ΣM A : − 300 N ⋅ m + 300 N ⋅ m + (200 N)(4 m) = M f
or M f = 800 N ⋅ m
(g)
We have
ΣFy : − 200 N − 800 N = Rg
or
and
R g = 1000 N
ΣM A : 200 N ⋅ m + 4000 N ⋅ m − (800 N)(4 m) = M g
or M g = 1000 N ⋅ m
(h)
We have
ΣFy : − 300 N − 300 N = Rh
or
and
ΣM A : 2400 N ⋅ m − 300 N ⋅ m − (300 N)(4 m) = M h
or
(b)
R h = 600 N
M h = 900 N ⋅ m
!
Therefore, loadings (c) and (h) are equivalent.
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261
PROBLEM 3.102
A 4-m-long beam is loaded as shown. Determine the loading of
Problem 3.101 which is equivalent to this loading.
SOLUTION
We have
and
ΣFy : − 200 N − 400 N = R
or R = 600 N
ΣM A : −400 N ⋅ m + 2800 N ⋅ m − (400 N)(4 m) = M
M = 800 N ⋅ m
or
Equivalent to case ( f ) of Problem 3.101
Problem 3.101 Equivalent force-couples at A
Case
R
(a)
600 N
1000 N ⋅ m
(b)
600 N
900 N ⋅ m
(c)
600 N
900 N ⋅ m
(d)
400 N
900 N ⋅ m
(e)
600 N
200 N ⋅ m
(f )
600 N
800 N ⋅ m
(g)
1000 N
1000 N ⋅ m
(h)
600 N
900 N ⋅ m
M
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262
!
PROBLEM 3.103
Determine the single equivalent force and the distance from Point A to its line of action for the beam and
loading of (a) Problem 3.101b, (b) Problem 3.101d, (c) Problem 3.101e.
PROBLEM 3.101 A 4-m-long beam is subjected to a variety of loadings. (a) Replace each loading with an
equivalent force-couple system at end A of the beam. (b) Which of the loadings are equivalent?
SOLUTION
(a)
For equivalent single force at distance d from A
ΣFy : −600 N = R
We have
or R = 600 N
and
ΣM C : (600 N)(d ) − 900 N ⋅ m = 0
or d = 1.500 m
(b)
ΣFy : − 400 N + 800 N = R
We have
or R = 400 N
and
ΣM C : (400 N)( d ) + (800 N)(4 − d )
− 2300 N ⋅ m = 0
(c)
d = 2.25 m
or
R = 600 N
ΣFy : − 400 N − 200 N = R
We have
and
or
ΣM C : 200 N ⋅ m + (400 N)(d )
− (200 N)(4 − d ) + 400 N ⋅ m = 0
or d = 0.333 m
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263
PROBLEM 3.104
Five separate force-couple systems act at the corners of a piece of sheet metal, which has been bent into the
shape shown. Determine which of these systems is equivalent to a force F = (10 lb)i and a couple of moment
M = (15 lb ⋅ ft)j + (15 lb ⋅ ft)k located at the origin.
SOLUTION
First note that the force-couple system at F cannot be equivalent because of the direction of the force [The
force of the other four systems is (10 lb)i]. Next move each of the systems to the origin O; the forces remain
unchanged.
A: M A = ΣM O = (5 lb ⋅ ft) j + (15 lb ⋅ ft)k + (2 ft)k × (10 lb)i
= (25 lb ⋅ ft) j + (15 lb ⋅ ft)k
D : M D = ΣM O = −(5 lb ⋅ ft) j + (25 lb ⋅ ft)k
+ [(4.5 ft) j + (1 ft) j + (2 ft)k ] × 10 lb)i
= (15 lb ⋅ ft)i + (15 lb ⋅ ft)k
G : M G = ΣM O = (15 lb ⋅ ft)i + (15 lb ⋅ ft) j
I : M I = ΣM I = (15 lb ⋅ ft) j − (5 lb ⋅ ft)k
+ [(4.5 ft)i + (1 ft) j] × (10 lb) j
= (15 lb ⋅ ft) j − (15 lb ⋅ ft)k
The equivalent force-couple system is the system at corner D.
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264
!
PROBLEM 3.105
The weights of two children sitting at ends A and B of a seesaw
are 84 lb and 64 lb, respectively. Where should a third child sit
so that the resultant of the weights of the three children will
pass through C if she weighs (a) 60 lb, (b) 52 lb.
SOLUTION
(a)
For the resultant weight to act at C,
Then
ΣM C = 0 WC = 60 lb
(84 lb)(6 ft) − 60 lb(d ) − 64 lb(6 ft) = 0
d = 2.00 ft to the right of C
(b)
For the resultant weight to act at C,
Then
ΣM C = 0 WC = 52 lb
(84 lb)(6 ft) − 52 lb(d ) − 64 lb(6 ft) = 0
d = 2.31 ft to the right of C
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265
!
PROBLEM 3.106
Three stage lights are mounted on a pipe as shown.
The lights at A and B each weigh 4.1 lb, while the one
at C weighs 3.5 lb. (a) If d = 25 in., determine the
distance from D to the line of action of the resultant
of the weights of the three lights. (b) Determine the
value of d so that the resultant of the weights passes
through the midpoint of the pipe.
SOLUTION
For equivalence
ΣFy : − 4.1 − 4.1 − 3.5 = − R or R = 11.7 lb
ΣFD : − (10 in.)(4.1 lb) − (44 in.)(4.1 lb)
−[(4.4 + d )in.](3.5 lb) = −( L in.)(11.7 lb)
375.4 + 3.5d = 11.7 L (d , L in in.)
or
d = 25 in.
(a)
We have
375.4 + 3.5(25) = 11.7 L or
L = 39.6 in.
The resultant passes through a Point 39.6 in. to the right of D.
L = 42 in.
(b)
We have
375.4 + 3.5d = 11.7(42)
or d = 33.1 in.
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266
PROBLEM 3.107
A beam supports three loads of given magnitude and a fourth load whose magnitude is a function of position.
If b = 1.5 m and the loads are to be replaced with a single equivalent force, determine (a) the value of a so that
the distance from support A to the line of action of the equivalent force is maximum, (b) the magnitude of the
equivalent force and its point of application on the beam.
SOLUTION
For equivalence
ΣFy : −1300 + 400
a
− 400 − 600 = − R
b
a!
R = " 2300 − 400 # N
b
$
%
or
ΣM A :
a
a!
400 # − a (400) − (a + b)(600) = − LR
"
b%
2$
1000a + 600b − 200
L=
or
(1)
2300 − 400
a2
b
a
b
10a + 9 −
Then with
b = 1.5 m L =
4 2
a
3
(2)
8
23 − a
3
Where a, L are in m
(a)
Find value of a to maximize L
dL "$
=
da
8 !
8 !
4 ! 8!
10 − a #" 23 − a # − "10a + 9 − a 2 #" − #
3 %$
3 % $
3 %$ 3 %
8 !
" 23 − 3 a #
$
%
2
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267
PROBLEM 3.107 (Continued)
230 −
or
16a 2 − 276a + 1143 = 0
276 ± (−276) 2 − 4(16)(1143)
2(16)
Then
a=
or
a = 10.3435 m and a = 6.9065 m
Since
(b)
184
80
64 2 80
32
a− a+
a + a + 24 − a 2 = 0
3
3
9
3
9
or
AB = 9 m, a must be less than 9 m
a = 6.91 m
6.9065
1.5
Using Eq. (1)
R = 2300 − 400
and using Eq. (2)
4
10(6.9065) + 9 − (6.9065)2
3
= 3.16 m
L=
8
23 − (6.9065)
3
or R = 458 N
R is applied 3.16 m to the right of A.
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268
!
!
PROBLEM 3.108
Gear C is rigidly attached to arm AB. If the forces and couple
shown can be reduced to a single equivalent force at A,
determine the equivalent force and the magnitude of the
couple M.
SOLUTION
We have
For equivalence
ΣFx : −18sin 30° + 25cos 40° = Rx
or
Rx = 10.1511 lb
ΣFy : −18cos 30° − 40 − 25sin 40° = Ry
or
Then
Ry = −71.658 lb
R = (10.1511) 2 + (71.658) 2
= 72.416
tanθ =
or
Also
71.658
10.1511
θ = 81.9°
R = 72.4 lb
81.9°
ΣM A : M − (22 in.)(18 lb)sin 35° − (32 in.)(40 lb) cos 25°
− (48 in.)(25 lb) sin 65° = 0
M = 2474.8 lb ⋅ in.
or M = 206 lb ⋅ ft
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269
PROBLEM 3.109
A couple of magnitude M = 54 lb ⋅ in. and the three forces shown are
applied to an angle bracket. (a) Find the resultant of this system of
forces. (b) Locate the points where the line of action of the resultant
intersects line AB and line BC.
SOLUTION
(a)
We have
ΣF : R = (−10 j) + (30 cos 60°)i
+ 30 sin 60° j + (−45i )
= −(30 lb)i + (15.9808 lb) j
or R = 34.0 lb
(b)
28.0°
First reduce the given forces and couple to an equivalent force-couple system (R , M B ) at B.
We have
ΣM B : M B = (54 lb ⋅ in) + (12 in.)(10 lb) − (8 in.)(45 lb)
= −186 lb ⋅ in.
Then with R at D
or
and with R at E
or
ΣM B : −186 lb ⋅ in = a(15.9808 lb)
a = 11.64 in.
ΣM B : −186 lb ⋅ in = C (30 lb)
C = 6.2 in.
The line of action of R intersects line AB 11.64 in. to the left of B and intersects line BC 6.20 in.
below B.
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270
PROBLEM 3.110
A couple M and the three forces shown are applied to an angle bracket.
Find the moment of the couple if the line of action of the resultant of the
force system is to pass through (a) Point A, (b) Point B, (c) Point C.
SOLUTION
In each case, must have M iR = 0
(a)
M AB = ΣM A = M + (12 in.)[(30 lb) sin 60°] − (8 in.)(45 lb) = 0
M = +48.231 lb ⋅ in.
(b)
M = 48.2 lb ⋅ in.
M BR = ΣM B = M + (12 in.)(10 lb) − (8 in.)(45 lb) = 0
M = +240 lb ⋅ in.
(c)
M = 240 lb ⋅ in.
M CR = ΣM C = M + (12 in.)(10 lb) − (8 in.)[(30 lb) cos 60°] = 0
M =0
M=0
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271
PROBLEM 3.111
Four forces act on a 700 × 375-mm plate as shown. (a) Find
the resultant of these forces. (b) Locate the two points where
the line of action of the resultant intersects the edge of the
plate.
SOLUTION
(a)
R = ΣF
= (−400 N + 160 N − 760 N)i
+ (600 N + 300 N + 300 N) j
= −(1000 N)i + (1200 N) j
R = (1000 N) 2 + (1200 N)2
= 1562.09 N
1200 N !
tan θ = " −
#
$ 1000 N %
= −1.20000
θ = −50.194°
(b)
R = 1562 N
50.2°
M CR = Σr × F
= (0.5 m)i × (300 N + 300 N) j
= (300 N ⋅ m)k
(300 N ⋅ m)k = xi × (1200 N) j
x = 0.25000 m
x = 250 mm
(300 N ⋅ m) = yj × ( −1000 N)i
y = 0.30000 m
y = 300 mm
Intersection 250 mm to right of C and 300 mm above C
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272
PROBLEM 3.112
Solve Problem 3.111, assuming that the 760-N force is directed
to the right.
PROBLEM 3.111 Four forces act on a 700 × 375-mm plate as
shown. (a) Find the resultant of these forces. (b) Locate the two
points where the line of action of the resultant intersects the edge
of the plate.
SOLUTION
R = ΣF
(a)
= ( −400N + 160 N + 760 N)i
+ (600 N + 300 N + 300 N) j
= (520 N)i + (1200 N) j
R = (520 N) 2 + (1200 N) 2 = 1307.82 N
1200 N !
tan θ = "
# = 2.3077
$ 520 N %
θ = 66.5714°
R = 1308 N
66.6°
M CR = Σr × F
(b)
= (0.5 m)i × (300 N + 300 N) j
= (300 N ⋅ m)k
(300 N ⋅ m)k = xi × (1200 N) j
x = 0.25000 m
or
x = 0.250 mm
(300 N ⋅ m)k = [ x′i + (0.375 m) j] × [(520 N)i + (1200 N) j]
= (1200 x′ − 195)k
x′ = 0.41250 m
or
x′ = 412.5 mm
Intersection 412 mm to the right of A and 250 mm to the right of C
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273
PROBLEM 3.113
A truss supports the loading shown. Determine the equivalent
force acting on the truss and the point of intersection of its
line of action with a line drawn through Points A and G.
SOLUTION
We have
R = ΣF
R = (240 lb)(cos 70°i − sin 70° j) − (160 lb) j
+ (300 lb)(− cos 40°i − sin 40° j) − (180 lb) j
R = −(147.728 lb)i − (758.36 lb) j
R = Rx2 + Ry2
= (147.728) 2 + (758.36) 2
= 772.62 lb
Ry !
#
$ Rx %
−758.36 !
= tan −1 "
#
$ −147.728 %
= 78.977°
θ = tan −1 "
or R = 773 lb
We have
ΣM A = dRy
where
ΣM A = −[240 lb cos 70°](6 ft) − [240 lbsin 70°](4 ft)
79.0°
− (160 lb)(12 ft) + [300 lb cos 40°](6 ft)
− [300 lb sin 40°](20 ft) − (180 lb)(8 ft)
= −7232.5 lb ⋅ ft
−7232.5 lb ⋅ ft
−758.36 lb
= 9.5370 ft
d=
or d = 9.54 ft to the right of A
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274
PROBLEM 3.114
Pulleys A and B are mounted on bracket CDEF. The tension
on each side of the two belts is as shown. Replace the four
forces with a single equivalent force, and determine where its
line of action intersects the bottom edge of the bracket.
SOLUTION
Equivalent force-couple at A due to belts on pulley A
We have
ΣF : − 120 lb − 160 lb = RA
R A = 280 lb
We have
ΣM A :
−40 lb(2 in.) = M A
M A = 80 lb ⋅ in.
Equivalent force-couple at B due to belts on pulley B
We have
ΣF: (210 lb + 150 lb)
25° = R B
R B = 360 lb
We have
25°
ΣM B : − 60 lb(1.5 in.) = M B
M B = 90 lb ⋅ in.
Equivalent force-couple at F
We have
ΣF: R F = ( − 280 lb) j + (360 lb)(cos 25°i + sin 25° j)
= (326.27 lb)i − (127.857 lb) j
R = RF
2
2
= RFx
+ RFy
= (326.27)2 + (127.857)2
= 350.43 lb
RFy !
#
$ RFx %
−127.857 !
= tan −1 "
#
$ 326.27 %
= −21.399°
θ = tan −1 "
or R F = R = 350 lb
21.4°
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275
PROBLEM 3.114 (Continued)
We have
ΣM F : M F = − (280 lb)(6 in.) − 80 lb ⋅ in.
− [(360 lb) cos 25°](1.0 in.)
+ [(360 lb) sin 25°](12 in.) − 90 lb ⋅ in.
M F = − (350.56 lb ⋅ in.)k
To determine where a single resultant force will intersect line FE,
M F = dRy
d=
MF
Ry
−350.56 lb ⋅ in.
−127 ⋅ 857 lb
= 2.7418 in.
=
or d = 2.74 in.
!
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276
PROBLEM 3.115
A machine component is subjected to the forces and couples
shown. The component is to be held in place by a single rivet
that can resist a force but not a couple. For P = 0, determine the
location of the rivet hole if it is to be located (a) on line FG,
(b) on line GH.
SOLUTION
We have
First replace the applied forces and couples with an equivalent force-couple system at G.
ΣFx : 200cos 15° − 120 cos 70° + P = Rx
Thus
Rx = (152.142 + P) N
or
ΣFy : − 200sin 15° − 120sin 70° − 80 = Ry
Ry = −244.53 N
or
ΣM G : − (0.47 m)(200 N) cos15° + (0.05 m)(200 N)sin15°
+ (0.47 m)(120 N) cos 70° − (0.19 m)(120 N)sin 70°
− (0.13 m)( P N) − (0.59 m)(80 N) + 42 N ⋅ m
+ 40 N ⋅ m = M G
M G = −(55.544 + 0.13P) N ⋅ m
or
(1)
Setting P = 0 in Eq. (1):
Now with R at I
ΣM G : − 55.544 N ⋅ m = − a(244.53 N)
a = 0.227 m
or
and with R at J
ΣM G : − 55.544 N ⋅ m = −b(152.142 N)
b = 0.365 m
or
(a)
The rivet hole is 0.365 m above G.
(b)
The rivet hole is 0.227 m to the right of G.
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277
PROBLEM 3.116
Solve Problem 3.115, assuming that P = 60 N.
PROBLEM 3.115 A machine component is subjected to the
forces and couples shown. The component is to be held in
place by a single rivet that can resist a force but not a couple.
For P = 0, determine the location of the rivet hole if it is to
be located (a) on line FG, (b) on line GH.
SOLUTION
See the solution to Problem 3.115 leading to the development of Equation (1)
M G = −(55.544 + 0.13P) N ⋅ m
Rx = (152.142 + P) N
and
P = 60 N
For
We have
Rx = (152.142 + 60)
= 212.14 N
M G = −[55.544 + 0.13(60)]
= −63.344 N ⋅ m
Then with R at I
ΣM G : −63.344 N ⋅ m = −a(244.53 N)
a = 0.259 m
or
and with R at J
ΣM G : −63.344 N ⋅ m = −b(212.14 N)
b = 0.299 m
or
(a)
The rivet hole is 0.299 m above G.
(b)
The rivet hole is 0.259 m to the right of G.
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278
PROBLEM 3.117
A 32-lb motor is mounted on the floor. Find the resultant of the
weight and the forces exerted on the belt, and determine where
the line of action of the resultant intersects the floor.
SOLUTION
We have
ΣF : (60 lb)i − (32 lb) j + (140 lb)(cos 30°i + sin 30° j) = R
R = (181.244 lb)i + (38.0 lb) j
or R = 185.2 lb
We have
11.84°
ΣM O : ΣM O = xRy
− [(140 lb) cos 30°][(4 + 2 cos 30°)in.] − [(140 lb) sin 30°][(2 in.)sin 30°]
− (60 lb)(2 in.) = x(38.0 lb)
x=
and
1
(− 694.97 − 70.0 − 120) in.
38.0
x = −23.289 in.
Or, resultant intersects the base (x axis) 23.3 in. to the left of
the vertical centerline (y axis) of the motor.
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279
PROBLEM 3.118
As follower AB rolls along the surface of member C, it exerts
a constant force F perpendicular to the surface. (a) Replace F
with an equivalent force-couple system at the Point D
obtained by drawing the perpendicular from the point of
contact to the x axis. (b) For a = 1 m and b = 2 m, determine
the value of x for which the moment of the equivalent forcecouple system at D is maximum.
SOLUTION
(a)
The slope of any tangent to the surface of member C is
dy d (
x2
=
*b ""1 − 2
dx dx *, $ a
! ) −2b
## + = 2 x
% -+ a
Since the force F is perpendicular to the surface,
dy !
tan α = − " #
$ dx %
−1
=
a2 1 !
2b "$ x #%
For equivalence
ΣF : F = R
ΣM D : ( F cos α )( y A ) = M D
where
cos α =
2bx
(a 2 ) 2 + (2bx)2
x2 !
y A = b ""1 − 2 ##
$ a %
x3 !
2 Fb 2 " x − 2 #
a %
$
MD =
4
2 2
a + 4b x
Therefore, the equivalent force-couple system at D is
R=F
a2 !
tan −1 ""
##
$ 2bx %
x3 !
2 Fb2 " x − 2 #
a %
$
M=
a 4 + 4b 2 x 2
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280
PROBLEM 3.118 (Continued)
(b)
To maximize M, the value of x must satisfy
dM
=0
dx
a = 1 m, b = 2 m
where, for
M=
8F ( x − x3 )
1 + 16 x 2
(1
)
1 + 16 x 2 (1 − 3x 2 ) − ( x − x3 ) * (32 x)(1 + 16 x 2 ) −1/ 2 +
dM
2
,
- =0
= 8F
2
dx
(1 + 16 x )
(1 + 16 x 2 )(1 − 3x 2 ) − 16 x( x − x3 ) = 0
32 x 4 + 3x 2 − 1 = 0
or
x2 =
−3 ± 9 − 4(32)(−1)
= 0.136011 m 2
2(32)
Using the positive value of x2
x = 0.36880 m
and − 0.22976 m 2
or x = 369 mm
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281
!
PROBLEM 3.119
Four forces are applied to the machine component
ABDE as shown. Replace these forces by an equivalent
force-couple system at A.
SOLUTION
R = −(50 N) j − (300 N)i − (120 N)i − (250 N)k
R = −(420 N)i − (50 N)j − (250 N)k
rB = (0.2 m)i
rD = (0.2 m)i + (0.16 m)k
rE = (0.2 m)i − (0.1 m) j + (0.16 m)k
M RA = rB × [−(300 N)i − (50 N) j]
+ rD × (−250 N)k + r × ( − 120 N)i
i
j
k
i
j
k
= 0.2 m
0
0 + 0.2 m 0 0.16 m
−300 N −50 N 0
0
0 −250 N
i
j
k
+ 0.2 m −0.1 m 0.16 m
−120 N
0
0
= −(10 N ⋅ m)k + (50 N ⋅ m) j − (19.2 N ⋅ m) j − (12 N ⋅ m)k
Force-couple system at A is
R = −(420 N)i − (50 N) j − (250 N)k M RA = (30.8 N ⋅ m) j − (220 N ⋅ m)k
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282
PROBLEM 3.120
Two 150-mm-diameter pulleys are mounted
on line shaft AD. The belts at B and C lie in
vertical planes parallel to the yz plane. Replace
the belt forces shown with an equivalent forcecouple system at A.
SOLUTION
Equivalent force-couple at each pulley
Pulley B
R B = (145 N)(− cos 20° j + sin 20°k ) − 215 Nj
= − (351.26 N) j + (49.593 N)k
M B = − (215 N − 145 N)(0.075 m)i
= − (5.25 N ⋅ m)i
Pulley C
R C = (155 N + 240 N)(− sin10° j − cos10°k )
= − (68.591 N) j − (389.00 N)k
M C = (240 N − 155 N)(0.075 m)i
= (6.3750 N ⋅ m)i
Then
R = R B + R C = − (419.85 N) j − (339.41)k
or R = (420 N) j − (339 N)k
M A = M B + M C + rB/ A × R B + rC/ A × R C
i
j
k
0
0 N⋅m
= − (5.25 N ⋅ m)i + (6.3750 N ⋅ m)i + 0.225
0
−351.26 49.593
i
j
k
+ 0.45
0
0
N⋅m
0
−68.591 −389.00
= (1.12500 N ⋅ m)i + (163.892 N ⋅ m) j − (109.899 N ⋅ m)k
or M A = (1.125 N ⋅ m)i + (163.9 N ⋅ m) j − (109.9 N ⋅ m)k
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283
!
PROBLEM 3.121
While using a pencil sharpener, a student applies the forces and
couple shown. (a) Determine the forces exerted at B and C knowing
that these forces and the couple are equivalent to a force-couple
system at A consisting of the force R = (2.6 lb)i + Ry j − (0.7 lb)k
and the couple M RA = M x i + (1.0 lb · ft)j − (0.72 lb · ft)k. (b) Find
the corresponding values of Ry and M x .
SOLUTION
(a)
From the statement of the problem, equivalence requires
ΣF : B + C = R
or
ΣFx : Bx + C x = 2.6 lb
(1)
ΣFy : − C y = R y
(2)
ΣFz : − C z = −0.7 lb or C z = 0.7 lb
and
ΣM A : (rB/A × B + M B ) + rC/A × C = M AR
or
1.75 !
ΣM x : (1 lb ⋅ ft) + "
ft # (C y ) = M x
$ 12
%
(3)
3.75 !
1.75 !
3.5 !
ΣM y : "
ft # ( Bx ) + "
ft # (C x ) + "
ft # (0.7 lb) = 1 lb ⋅ ft
$ 12 %
$ 12 %
$ 12 %
or
Using Eq. (1)
3.75Bx + 1.75C x = 9.55
3.75Bx + 1.75(2.6 Bx ) = 9.55
or
Bx = 2.5 lb
and
C x = 0.1 lb
3.5 !
ΣM z : − "
ft # (C y ) = −0.72 lb ⋅ ft
$ 12 %
C y = 2.4686 lb
or
B = (2.5 lb)i C = (0.1000 lb)i − (2.47 lb) j − (0.700 lb)k
(b)
Eq. (2) &
Using Eq. (3)
Ry = −2.47 lb
1.75 !
1+ "
# (2.4686) = M x
$ 12 %
or M x = 1.360 lb ⋅ ft
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284
PROBLEM 3.122
A mechanic uses a crowfoot wrench to loosen a bolt at C. The
mechanic holds the socket wrench handle at Points A and B
and applies forces at these points. Knowing that these forces
are equivalent to a force-couple system at C consisting of the
force C = (8 lb)i + (4 lb)k and the couple M C = (360 lb · in.)i,
determine the forces applied at A and at B when Az = 2 lb.
SOLUTION
We have
ΣF :
or
Fx : Ax + Bx = 8 lb
A+B=C
Bx = −( Ax + 8 lb)
(1)
ΣFy : Ay + By = 0
Ay = − By
or
(2)
ΣFz : 2 lb + Bz = 4 lb
Bz = 2 lb
or
(3)
ΣM C : rB/C × B + rA/C × A = M C
We have
i
j
8
Bx
0
By
k
i
j
2 + 8
2
Ax
0
Ay
k
8 lb ⋅ in. = (360 lb ⋅ in.)i
2
(2 By − 8 Ay )i + (2 Bx − 16 + 8 Ax − 16) j
or
+ (8By + 8 Ay )k = (360 lb ⋅ in.)i
From
i-coefficient
j-coefficient
k-coefficient
2 By − 8 Ay = 360 lb ⋅ in.
−2 Bx + 8 Ax = 32 lb ⋅ in.
8 By + 8 Ay = 0
(4)
(5)
(6)
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285
PROBLEM 3.122 (Continued)
From Equations (2) and (4):
2 By − 8(− By ) = 360
By = 36 lb
From Equations (1) and (5):
Ay = 36 lb
2(− Ax − 8) + 8 Ax = 32
Ax = 1.6 lb
From Equation (1):
Bx = −(1.6 + 8) = −9.6 lb
A = (1.600 lb)i − (36.0 lb) j + (2.00 lb)k
B = −(9.60 lb)i + (36.0 lb) j + (2.00 lb)k
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286
PROBLEM 3.123
As an adjustable brace BC is used to bring a wall into plumb, the
force-couple system shown is exerted on the wall. Replace this
force-couple system with an equivalent force-couple system at A
if R = 21.2 lb and M = 13.25 lb · ft.
SOLUTION
We have
where
ΣF : R = R A = RλBC
BC
=
RA =
or
We have
where
(42 in.)i − (96 in.) j − (16 in.)k
106 in.
21.2 lb
(42i − 96 j − 16k )
106
R A = (8.40 lb)i − (19.20 lb) j − (3.20 lb)k
ΣM A : rC/A × R + M = M A
rC/A = (42 in.)i + (48 in.)k =
1
(42i + 48k )ft
12
= (3.5 ft)i + (4.0 ft)k
R = (8.40 lb)i − (19.50 lb) j − (3.20 lb)k
M = −λBC M
−42i + 96 j + 16k
(13.25 lb ⋅ ft)
106
= −(5.25 lb ⋅ ft)i + (12 lb ⋅ ft) j + (2lb ⋅ ft)k
=
Then
i
j
k
3.5
0
4.0 lb ⋅ ft + (−5.25i + 12 j + 2k )lb ⋅ ft = M A
8.40 −19.20 −3.20
M A = (71.55 lb ⋅ ft)i + (56.80 lb ⋅ ft)j − (65.20 lb ⋅ ft)k
or M A = (71.6 lb ⋅ ft)i + (56.8 lb ⋅ ft)j − (65.2 lb ⋅ ft)k
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287
PROBLEM 3.124
A mechanic replaces a car’s exhaust system by firmly clamping the catalytic converter FG to its mounting
brackets H and I and then loosely assembling the mufflers and the exhaust pipes. To position the tailpipe AB,
he pushes in and up at A while pulling down at B. (a) Replace the given force system with an equivalent
force-couple system at D. (b) Determine whether pipe CD tends to rotate clockwise or counterclockwise
relative to muffler DE, as viewed by the mechanic.
SOLUTION
(a)
Equivalence requires
ΣF : R = A + B
= (100 N)(cos 30° j − sin 30° k ) − (115 N) j
= −(28.4 N) j − (50 N)k
and
where
ΣM D : M D = rA/D × FA + rB/D × FB
rA/D = −(0.48 m)i − (0.225 m) j + (1.12 m)k
rB/D = −(0.38 m)i + (0.82 m)k
Then
i
j
k
i
j
k
M D = 100 −0.48 −0.225
1.12
+ 115 −0.38 0 0.82
0 cos 30° − sin 30°
0 −1
0
= 100[(0.225sin 30° − 1.12cos 30°)i + (−0.48sin 30°) j
+ (−0.48cos 30°)k ] + 115[(0.82)i + (0.38)k ]
= 8.56i − 24.0 j + 2.13k
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288
PROBLEM 3.124 (Continued)
The equivalent force-couple system at D is
R = −(28.4 N) j − (50.0 N)k
M D = (8.56 N ⋅ m)i − (24.0 N ⋅ m) j + (2.13 N ⋅ m)k
(b)
Since ( M D ) z is positive, pipe CD will tend to rotate counterclockwise relative to muffler DE.
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289
PROBLEM 3.125
For the exhaust system of Problem 3.124, (a) replace the given force system with an equivalent force-couple
system at F, where the exhaust pipe is connected to the catalytic converter, (b) determine whether pipe EF
tends to rotate clockwise or counterclockwise, as viewed by the mechanic.
SOLUTION
(a)
Equivalence requires
ΣF : R = A + B
= (100 N)(cos 30° j − sin 30° k ) − (115 N) j
= −(28.4 N) j − (50 N)k
and
where
M F : M F = rA/F × A + rB/F × B
rA/F = −(0.48 m)i − (0.345 m) j + (2.10 m)k
rB/F = −(0.38 m)i − (0.12 m) j + (1.80 m)k
Then
i
j
k
i
j
k
M F = 100 −0.48 −0.345
2.10
+ 115 −0.38 0.12 1.80
0 cos 30° − sin 30°
0 −1
0
M F = 100[(0.345 sin 30° − 2.10 cos 30°)i + (−0.48 sin 30°) j
+ (−0.48 cos 30°)k ] + 115[(1.80)i + (0.38)k ]
= 42.4i − 24.0 j + 2.13k
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290
PROBLEM 3.125 (Continued)
The equivalent force-couple system at F is
R = −(28.4 N) j − (50 N)k
M F = (42.4 N ⋅ m)i − (24.0 N ⋅ m) j + (2.13 N ⋅ m)k
(b)
Since ( M F ) z is positive, pipe EF will tend to rotate counterclockwise relative to the mechanic.
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291
PROBLEM 3.126
The head-and-motor assembly of a radial drill press was originally
positioned with arm AB parallel to the z axis and the axis of the
chuck and bit parallel to the y axis. The assembly was then rotated
25° about the y axis and 20° about the centerline of the horizontal
arm AB, bringing it into the position shown. The drilling process
was started by switching on the motor and rotating the handle to
bring the bit into contact with the workpiece. Replace the force
and couple exerted by the drill press with an equivalent forcecouple system at the center O of the base of the vertical column.
SOLUTION
We have
R = F = (11 lb)[( sin 20° cos 25°)]i − (cos 20°) j − (sin 20° sin 25°)k ]
= (3.4097 lb)i − (10.3366 lb)j − (1.58998 lb)k
or
R = (3.41 lb)i − (10.34 lb) j − (1.590 lb)k
We have
M O = rB/O × F × M C
where
rB/O = [(14 in.) sin 25°]i + (15 in.) j + [(14 in.) cos 25°]k
= (5.9167 in.)i + (15 in.) j + (12.6883 in.)k
M C = (90 lb ⋅ in.)[(sin 20° cos 25°)i − (cos 20°) j − (sin 20° sin 25°)k ]
= (27.898 lb ⋅ in.)i − (84.572 lb ⋅ in.) j − (13.0090 lb ⋅ in.)k
i
j
k
15
12.6883 lb ⋅ in.
M O = 5.9167
3.4097 −10.3366 1.58998
+ (27.898 − 84.572 − 13.0090) lb ⋅ in.
= (135.202 lb ⋅ in.)i − (31.901 lb ⋅ in.) j − (125.313 lb ⋅ in.)k
or M O = (135.2 lb ⋅ in.)i − (31.9 lb ⋅ in.) j − (125.3 lb ⋅ in.)k
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292
PROBLEM 3.127
Three children are standing on a 5 × 5-m raft. If the weights
of the children at Points A, B, and C are 375 N, 260 N, and
400 N, respectively, determine the magnitude and the point
of application of the resultant of the three weights.
SOLUTION
We have
ΣF : FA + FB + FC = R
−(375 N) j − (260 N) j − (400 N) j = R
−(1035 N) j = R
or
We have
R = 1035 N
ΣM x : FA ( z A ) + FB ( z B ) + FC ( zC ) = R ( z D )
(375 N)(3 m) + (260 N)(0.5 m) + (400 N)(4.75 m) = (1035 N)(z D )
z D = 3.0483 m
We have
or
z D = 3.05 m
ΣM z : FA ( x A ) + FB ( xB ) + FC ( xC ) = R ( xD )
375 N(1 m) + (260 N)(1.5 m) + (400 N)(4.75 m) = (1035 N)( xD )
xD = 2.5749 m
or
xD = 2.57 m
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293
PROBLEM 3.128
Three children are standing on a 5 × 5-m raft. The weights
of the children at Points A, B, and C are 375 N, 260 N, and
400 N, respectively. If a fourth child of weight 425 N
climbs onto the raft, determine where she should stand if
the other children remain in the positions shown and the
line of action of the resultant of the four weights is to pass
through the center of the raft.
SOLUTION
We have
ΣF : FA + FB + FC = R
−(375 N) j − (260 N) j − (400 N) j − (425 N) j = R
R = −(1460 N) j
We have
ΣM x : FA ( z A ) + FB ( z B ) + FC ( zC ) + FD ( z D ) = R( z H )
(375 N)(3 m) + (260 N)(0.5 m) + (400 N)(4.75 m)
+ (425 N)(z D ) = (1460 N)(2.5 m)
z D = 1.16471 m
We have
or
z D = 1.165 m
ΣM z : FA ( x A ) + FB ( xB ) + FC ( xC ) + FD ( xD ) = R ( xH )
(375 N)(1 m) + (260 N)(1.5 m) + (400 N)(4.75 m)
+ (425 N)(xD ) = (1460 N)(2.5 m)
xD = 2.3235 m
or
xD = 2.32 m
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294
PROBLEM 3.129
Four signs are mounted on a frame spanning a highway,
and the magnitudes of the horizontal wind forces acting on
the signs are as shown. Determine the magnitude and the
point of application of the resultant of the four wind forces
when a = 1 ft and b = 12 ft.
SOLUTION
We have
Assume that the resultant R is applied at Point P whose coordinates are (x, y, 0).
Equivalence then requires
ΣFz : − 105 − 90 − 160 − 50 = − R
or R = 405 lb
ΣM x : (5 ft)(105 lb) − (1 ft)(90 lb) + (3 ft)(160 lb)
+ (5.5 ft)(50 lb) = − y (405 lb)
or
y = −2.94 ft
ΣM y : (5.5 ft)(105 lb) + (12 ft)(90 lb) + (14.5 ft)(160 lb)
+ (22.5 ft)(50 lb) = − x(405 lb)
x = 12.60 ft
or
R acts 12.60 ft to the right of member AB and 2.94 ft below member BC.
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295
PROBLEM 3.130
Four signs are mounted on a frame spanning a highway,
and the magnitudes of the horizontal wind forces acting on
the signs are as shown. Determine a and b so that the point
of application of the resultant of the four forces is at G.
SOLUTION
Since R acts at G, equivalence then requires that ΣM G of the applied system of forces also be zero. Then
at
G : ΣM x : − (a + 3) ft × (90 lb) + (2 ft)(105 lb)
+ (2.5 ft)(50 lb) = 0
or a = 0.722 ft
!
ΣM y : − (9 ft)(105 ft) − (14.5 − b) ft × (90 lb)
!
+ (8 ft)(50 lb) = 0
or b = 20.6 ft
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296
!
PROBLEM 3.131*
A group of students loads a 2 × 3.3-m flatbed trailer with two
0.66 × 0.66 × 0.66-m boxes and one 0.66 × 0.66 × 1.2-m box.
Each of the boxes at the rear of the trailer is positioned so that
it is aligned with both the back and a side of the trailer.
Determine the smallest load the students should place in a
second 0.66 × 0.66 × 1.2-m box and where on the trailer they
should secure it, without any part of the box overhanging the
sides of the trailer, if each box is uniformly loaded and the line
of action of the resultant of the weights of the four boxes is to
pass through the point of intersection of the centerlines of the
trailer and the axle. (Hint: Keep in mind that the box may be
placed either on its side or on its end.)
SOLUTION
For the smallest weight on the trailer so that the resultant force of the four weights acts over the axle at the
intersection with the center line of the trailer, the added 0.66 × 0.66 × 1.2-m box should be placed adjacent to
one of the edges of the trailer with the 0.66 × 0.66-m side on the bottom. The edges to be considered are
based on the location of the resultant for the three given weights.
We have
ΣF : − (224 N) j − (392 N) j − (176 N) j = R
R = −(792 N) j
We have
ΣM z : − (224 N)(0.33 m) − (392 N)(1.67 m) − (176 N)(1.67 m) = ( −792 N)( x)
xR = 1.29101 m
We have
ΣM x : (224 N)(0.33 m) + (392 N)(0.6 m) + (176 N)(2.0 m) = (792 N)( z )
z R = 0.83475 m
From the statement of the problem, it is known that the resultant of R from the original loading and the
lightest load W passes through G, the point of intersection of the two center lines. Thus, ΣM G = 0.
Further, since the lightest load W is to be as small as possible, the fourth box should be placed as far from G
as possible without the box overhanging the trailer. These two requirements imply
(0.33 m # x # 1 m)(1.5 m # z # 2.97 m)
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297
PROBLEM 3.131* (Continued)
xL = 0.33 m
With
at
G : ΣM Z : (1 − 0.33) m × WL − (1.29101 − 1) m × (792 N) = 0
WL = 344.00 N
or
Now must check if this is physically possible,
at
G : ΣM x : ( Z L − 1.5)m × 344 N) − (1.5 − 0.83475)m × (792 N) = 0
Z L = 3.032 m
or
which is not acceptable.
Z L = 2.97 m:
With
at
G : ΣM x : (2.97 − 1.5)m × WL − (1.5 − 0.83475)m × (792 N) = 0
WL = 358.42 N
or
Now check if this is physically possible
at
G : ΣM z : (1 − X L )m × (358.42 N) − (1.29101 − 1)m × (792 N) = 0
or
X L = 0.357 m ok!
WL = 358 N
The minimum weight of the fourth box is
And it is placed on end (A 0.66 × 0.66-m side down) along side AB with the center of the box 0.357 m
from side AD.
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298
PROBLEM 3.132*
Solve Problem 3.131 if the students want to place as much
weight as possible in the fourth box and at least one side of
the box must coincide with a side of the trailer.
PROBLEM 3.131* A group of students loads a 2 × 3.3-m
flatbed trailer with two 0.66 × 0.66 × 0.66-m boxes and one
0.66 × 0.66 × 1.2-m box. Each of the boxes at the rear of the
trailer is positioned so that it is aligned with both the back
and a side of the trailer. Determine the smallest load the
students should place in a second 0.66 × 0.66 × 1.2-m box
and where on the trailer they should secure it, without any
part of the box overhanging the sides of the trailer, if each
box is uniformly loaded and the line of action of the
resultant of the weights of the four boxes is to pass through
the point of intersection of the centerlines of the trailer and
the axle. (Hint: Keep in mind that the box may be placed
either on its side or on its end.)
SOLUTION
First replace the three known loads with a single equivalent force R applied at coordinate ( X R , 0, Z R )
Equivalence requires
ΣFy : − 224 − 392 − 176 = − R
or
R = 792 N
ΣM x : (0.33 m)(224 N) + (0.6 m)(392 N)
+ (2 m)(176 N) = z R (792 N)
or
z R = 0.83475 m
ΣM z : − (0.33 m)(224 N) − (1.67 m)(392 N)
− (1.67 m)(176 N) = xR (792 N)
or
xR = 1.29101 m
From the statement of the problem, it is known that the resultant of R and the heaviest loads WH passes
through G, the point of intersection of the two center lines. Thus,
ΣM G = 0
Further, since WH is to be as large as possible, the fourth box should be placed as close to G as possible while
keeping one of the sides of the box coincident with a side of the trailer. Thus, the two limiting cases are
xH = 0.6 m or
z H = 2.7 m
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299
PROBLEM 3.132* (Continued)
Now consider these two possibilities
With xH = 0.6 m:
at
G : ΣM z : (1 − 0.6)m × WH − (1.29101 − 1)m × (792 N) = 0
WH = 576.20 N
or
Checking if this is physically possible
at
or
G : ΣM x : ( z H − 1.5)m × (576.20 N) − (1.5 − 0.83475)m × (792 N) = 0
z H = 2.414 m
which is acceptable.
With z H = 2.7 m
at
or
G : ΣM x : (2.7 − 1.5) WH − (1.5 − 0.83475)m × (792 N) = 0
WH = 439 N
Since this is less than the first case, the maximum weight of the fourth box is
WH = 576 N
and it is placed with a 0.66 × 1.2-m side down, a 0.66-m edge along side AD, and the center 2.41 m
from side DC.
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300
PROBLEM 3.133
Three forces of the same magnitude P act on a cube of side a as
shown. Replace the three forces by an equivalent wrench and
determine (a) the magnitude and direction of the resultant force
R, (b) the pitch of the wrench, (c) the axis of the wrench.
SOLUTION
Force-couple system at O:
R = Pi + Pj + Pk = P (i + j + k )
M OR = aj × Pi + ak × Pj + ai × Pk
= − Pak − Pai − Paj
M OR = − Pa (i + j + k )
Since R and M OR have the same direction, they form a wrench with M1 = M OR . Thus, the axis of the wrench
is the diagonal OA. We note that
cos θ x = cos θ y = cos θ z =
a
a 3
=
1
3
R = P 3 θ x = θ y = θ z = 54.7°
M1 = M OR = − Pa 3
Pitch = p =
(a)
(b)
(c)
M 1 − Pa 3
=
= −a
R
P 3
R = P 3 θ x = θ y = θ z = 54.7°
–a
Axis of the wrench is diagonal OA
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301
PROBLEM 3.134*
A piece of sheet metal is bent into the shape shown and is acted
upon by three forces. If the forces have the same magnitude P,
replace them with an equivalent wrench and determine (a) the
magnitude and the direction of the resultant force R, (b) the pitch
of the wrench, (c) the axis of the wrench.
SOLUTION
(
)
First reduce the given forces to an equivalent force-couple system R, M OR at the origin.
We have
ΣF : − Pj + Pj + Pk = R
R = Pk
or
"5 # !
ΣM O : − (aP) j + $ −( aP)i + & aP ' k % = M OR
(2 ) +
*
5 #
"
M OR = aP & −i − j + k '
2 )
(
or
(a)
Then for the wrench
R=P
and
axis
=
R
=k
R
cos θ x = 0 cos θ y = 0 cos θ z = 1
or
(b)
θ x = 90° θ y = 90° θ z = 0°
Now
M1 = λ axis ⋅ M OR
5 #
"
= k ⋅ aP & −i − j + k '
2 )
(
5
= aP
2
Then
P=
M1 25 aP
=
R
P
or P =
5
a
2
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302
PROBLEM 3.134* (Continued)
(c)
The components of the wrench are (R , M1 ), where M1 = M1 λ axis , and the axis of the wrench is
assumed to intersect the xy plane at Point Q whose coordinates are (x, y, 0). Thus require
M z = rQ × R R
Where
M z = M O × M1
Then
5 # 5
"
aP & −i − j + k ' − aPk = ( xi + yj) + Pk
2 ) 2
(
Equating coefficients
i : − aP = yP
or
j: − aP = − xP or
y = −a
x=a
The axis of the wrench is parallel to the z axis and intersects the xy plane at
x = a, y = −a.
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303
PROBLEM 3.135*
The forces and couples shown are applied to two screws as a piece
of sheet metal is fastened to a block of wood. Reduce the forces and
the couples to an equivalent wrench and determine (a) the resultant
force R, (b) the pitch of the wrench, (c) the point where the axis of
the wrench intersects the xz plane.
SOLUTION
First, reduce the given force system to a force-couple system.
We have
ΣF : − (20 N)i − (15 N) j = R
We have
ΣM O : Σ(rO × F ) + ΣM C = M OR
R = 25 N
M OR = −20 N(0.1 m)j − (4 N ⋅ m)i − (1 N ⋅ m)j
= −(4 N ⋅ m)i − (3 N ⋅ m) j
R = −(20.0 N)i − (15.0 N)j
(a)
(b)
We have
R
R
= (−0.8i − 0.6 j) ⋅ [−(4 N ⋅ m)]i − (3 N ⋅ m)j]
M1 = λR ⋅ M OR
λ=
= 5 N⋅m
Pitch
p=
M1 5 N ⋅ m
=
= 0.200 m
R
25 N
or p = 0.200 m
(c)
From above note that
M1 = M OR
Therefore, the axis of the wrench goes through the origin. The line of action of the wrench lies in the xy
plane with a slope of
y=
3
x
4
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304
PROBLEM 3.136*
The forces and couples shown are applied to two screws as a piece
of sheet metal is fastened to a block of wood. Reduce the forces and
the couples to an equivalent wrench and determine (a) the resultant
force R, (b) the pitch of the wrench, (c) the point where the axis of
the wrench intersects the xz plane.
SOLUTION
First, reduce the given force system to a force-couple at the origin.
We have
ΣF : − (10 lb) j − (11 lb) j = R
R = − (21 lb) j
We have
ΣM O : Σ(rO × F ) + ΣM C = M OR
M OR
i
j
k
i
j
k
= 0 0 20 lb ⋅ in. + 0 0 −15 lb ⋅ in. − (12 lb ⋅ in) j
0 −10 0
0 −11 0
= (35 lb ⋅ in.)i − (12 lb ⋅ in.) j
R = − (21 lb) j
(a)
(b)
We have
or R = − (21.0 lb) j
R
R
= (− j) ⋅ [(35 lb ⋅ in.)i − (12 lb ⋅ in.) j]
M1 =
R
⋅ M OR
R
=
= 12 lb ⋅ in. and M1 = −(12 lb ⋅ in.) j
and pitch
p=
M 1 12 lb ⋅ in.
=
= 0.57143 in.
R
21 lb
or
p = 0.571 in.
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305
PROBLEM 3.136* (Continued)
(c)
We have
M OR = M1 + M 2
M 2 = M OR − M1 = (35 lb ⋅ in.)i
Require
M 2 = rQ/O × R
(35 lb ⋅ in.)i = ( xi + zk ) × [ −(21 lb) j]
35i = −(21x)k + (21z )i
From i:
From k:
35 = 21z
z = 1.66667 in.
0 = − 21x
z=0
The axis of the wrench is parallel to the y axis and intersects the xz plane at x = 0, z = 1.667 in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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306
!
PROBLEM 3.137*
Two bolts at A and B are tightened by applying the forces
and couples shown. Replace the two wrenches with a
single equivalent wrench and determine (a) the resultant
R, (b) the pitch of the single equivalent wrench, (c) the
point where the axis of the wrench intersects the xz plane.
SOLUTION
First, reduce the given force system to a force-couple at the origin.
We have
ΣF : − (84 N) j − (80 N)k = R
R = 116 N
and
ΣM O : Σ(rO × F) + ΣM C = M OR
i
j k
i
j k
0.6 0 .1 + 0.4 0.3 0 + ( −30 j − 32k )N ⋅ m = M OR
0 84 0
0
0 80
M OR = − (15.6 N ⋅ m)i + (2 N ⋅ m) j − (82.4 N ⋅ m)k
R = − (84.0 N) j − (80.0 N)k
(a)
(b)
We have
M1 =
R
⋅ M OR
R
=
R
R
−84 j − 80k
⋅ [− (15.6 N ⋅ m)i + (2 N ⋅ m) j − (82.4 N ⋅ m)k ]
116
= 55.379 N ⋅ m
=−
and
Then pitch
M1 = M1λR = − (40.102 N ⋅ m) j − (38.192 N ⋅ m)k
p=
M 1 55.379 N ⋅ m
=
= 0.47741 m
R
116 N
or p = 0.477 m
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307
PROBLEM 3.137* (Continued)
(c)
We have
M OR = M1 + M 2
M 2 = M OR − M1 = [(−15.6i + 2 j − 82.4k ) − (40.102 j − 38.192k )] N ⋅ m
= − (15.6 N ⋅ m)i + (42.102 N ⋅ m) j − (44.208 N ⋅ m)k
Require
M 2 = rQ/O × R
(−15.6i + 42.102 j − 44.208k ) = ( xi + zk ) × (84 j − 80k )
= (84 z )i + (80 x) j − (84 x)k
From i:
or
From k:
or
−15.6 = 84 z
z = − 0.185714 m
z = − 0.1857 m
−44.208 = −84 x
x = 0.52629 m
x = 0.526 m
The axis of the wrench intersects the xz plane at
x = 0.526 m
y = 0 z = − 0.1857 m
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308
!
PROBLEM 3.138*
Two bolts at A and B are tightened by applying the forces and
couples shown. Replace the two wrenches with a single equivalent
wrench and determine (a) the resultant R, (b) the pitch of the
single equivalent wrench, (c) the point where the axis of the
wrench intersects the xz plane.
SOLUTION
First, reduce the given force system to a force-couple at the origin at B.
(a)
We have
15 #
" 8
ΣF : − (26.4 lb)k − (17 lb) & i + j ' = R
17
17
(
)
R = − (8.00 lb)i − (15.00 lb) j − (26.4 lb)k
and
We have
R = 31.4 lb
ΣM B : rA/B × FA + M A + M B = M RB
M RB
i
j
k
15 #
" 8
0 − 220k − 238 & i + j ' = 264i − 220k − 14(8i + 15 j)
= 0 −10
( 17 17 )
0 0 − 26.4
M RB = (152 lb ⋅ in.)i − (210 lb ⋅ in.)j − (220 lb ⋅ in.)k
(b)
We have
R
R
−8.00i − 15.00 j − 26.4k
=
⋅ [(152 lb ⋅ in.)i − (210 lb ⋅ in.) j − (220 lb ⋅ in.)k ]
31.4
= 246.56 lb ⋅ in.
M1 =
R
⋅ M OR
R
=
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309
PROBLEM 3.138* (Continued)
and
Then pitch
(c)
We have
M1 = M1λR = − (62.818 lb ⋅ in.)i − (117.783 lb ⋅ in.) j − (207.30 lb ⋅ in.)k
p=
M 1 246.56 lb ⋅ in.
=
= 7.8522 in.
31.4 lb
R
or p = 7.85 in.
M RB = M1 + M 2
M 2 = M RB − M1 = (152i − 210 j − 220k ) − ( − 62.818i − 117.783j − 207.30k )
= (214.82 lb ⋅ in.)i − (92.217 lb ⋅ in.) j − (12.7000 lb ⋅ in.)k
Require
M 2 = rQ/B × R
i
j
k
214.82i − 92.217 j − 12.7000k = x
0
z
−8 −15 −26.4
= (15 z )i − (8 z ) j + (26.4 x) j − (15 x)k
From i:
214.82 = 15 z
From k:
−12.7000 = −15 x
The axis of the wrench intersects the xz plane at
z = 14.3213 in.
x = 0.84667 in.
x = 0.847 in. y = 0 z = 14.32 in.
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310
!
PROBLEM 3.139*
Two ropes attached at A and B are used to move the trunk of a
fallen tree. Replace the forces exerted by the ropes with an
equivalent wrench and determine (a) the resultant force R,
(b) the pitch of the wrench, (c) the point where the axis of the
wrench intersects the yz plane.
SOLUTION
(a)
(
)
First replace the given forces with an equivalent force-couple system R, M OR at the origin.
We have
d AC = (6) 2 + (2) 2 + (9) 2 = 11 m
d BD = (14) 2 + (2)2 + (5) 2 = 15 m
Then
1650 N
= (6i + 2 j + 9k )
11
= (900 N)i + (300 N) j + (1350 N)k
TAC =
and
1500 N
= (14i + 2 j + 5k )
15
= (1400 N)i + (200 N) j + (500 N)k
TBD =
Equivalence then requires
ΣF : R = TAC + TBD
= (900i + 300 j + 1350k )
+(1400i + 200 j + 500k )
= (2300 N)i + (500 N) j + (1850 N)k
ΣM O : M OR = rA × TAC + rB × TBD
= (12 m)k × [(900 N)i + (300 N)j + (1350 N)k ]
+ (9 m)i × [(1400 N)i + (200 N)j + (500 N)k ]
= −(3600)i + (10800 − 4500) j + (1800)k
= −(3600 N ⋅ m)i + (6300 N ⋅ m)j + (1800 N ⋅ m)k
The components of the wrench are (R , M1 ), where
R = (2300 N)i + (500 N) j + (1850 N)k
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311
PROBLEM 3.139* (Continued)
!
(b)
We have
R = 100 (23)2 + (5) 2 + (18.5) 2 = 2993.7 N
Let
λ axis =
Then
1
R
(23i + 5 j + 18.5k )
=
R 29.937
M1 = λ axis ⋅ M OR
1
(23i + 5 j + 18.5k ) ⋅ (−3600i + 6300 j + 1800k )
29.937
1
[(23)( −36) + (5)(63) + (18.5)(18)]
=
0.29937
= −601.26 N ⋅ m
=
Finally
P=
M1 −601.26 N ⋅ m
=
R
2993.7 N
or P = − 0.201 m
(c)
We have
M1 = M 1
axis
= (−601.26 N ⋅ m) ×
1
(23i + 5 j + 18.5k )
29.937
or
M1 = −(461.93 N ⋅ m)i − (100.421 N ⋅ m) j − (371.56 N ⋅ m)k
Now
M 2 = M OR − M1
= (−3600i + 6300 j + 1800k )
− ( −461.93i − 100.421j − 371.56k )
= − (3138.1 N ⋅ m)i + (6400.4 N ⋅ m)j + (2171.6 N ⋅ m)k
For equivalence
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312
PROBLEM 3.139* (Continued)
Thus require
M 2 = rP × R
r = ( yj + zk )
Substituting
i
j
k
−3138.1i + 6400.4 j + 2171.6k = 0
y
z
2300 500 1850
Equating coefficients
j : 6400.4 = 2300 z
or
k : 2171.6 = −2300 y or
The axis of the wrench intersects the yz plane at
z = 2.78 m
y = − 0.944 m
y = −0.944 m
z = 2.78 m
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313
!
PROBLEM 3.140*
A flagpole is guyed by three cables. If the tensions in the
cables have the same magnitude P, replace the forces exerted
on the pole with an equivalent wrench and determine (a) the
resultant force R, (b) the pitch of the wrench, (c) the point
where the axis of the wrench intersects the xz plane.
SOLUTION
(a)
First reduce the given force system to a force-couple at the origin.
We have
ΣF : P
BA
+P
DC
+P
DE
=R
3 # "3
4 # " −9
4
12 # !
"4
R = P $& j − k ' + & i − j ' + & i − j + k ' %
5 ) (5
5 ) ( 25
5
25 ) +
*( 5
R=
R=
We have
3P
(2i − 20 j − k )
25
3P
27 5
(2) 2 + (20) 2 + (1)2 =
P
25
25
ΣM : Σ(rO × P) = M OR
3P #
4P #
4P
12 P #
" −4 P
" 3P
" −9 P
j−
k ' + (20a) j × &
i−
j ' + (20a) j × &
i−
j+
k ' = M OR
(24a) j × &
5
5
5
5
25
5
25
(
)
(
)
(
)
M OR =
24 Pa
( −i − k )
5
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314
PROBLEM 3.140* (Continued)
(b)
We have
M1 = λR ⋅ M OR
where
λR =
Then
M1 =
and pitch
R 3P
25
1
=
=
(2i − 20 j − k )
(2i − 20 j − k )
R 25
27 5 P 9 5
p=
1
9 5
(2i − 20 j − k ) ⋅
M 1 −8Pa " 25 # −8a
=
&
'=
R 15 5 ( 27 5 P ) 81
M1 = M1λR =
(c)
Then
M 2 = M OR − M1 =
24 Pa
−8 Pa
(−i − k ) =
5
15 5
or p = − 0.0988a
−8 Pa " 1 #
8Pa
(−2i + 20 j + k )
&
' (2i − 20 j − k ) =
675
15 5 ( 9 5 )
24 Pa
8Pa
8 Pa
( −i − k ) −
(−2i + 20 j + k ) =
(−430i − 20 j − 406k )
5
675
675
M 2 = rQ/O × R
Require
" 8Pa #
" 3P #
& 675 ' (−403i − 20 j − 406k ) = ( xi + zk ) × & 25 ' (2i − 20 j − k )
(
)
(
)
" 3P #
=&
' [20 zi + ( x + 2 z ) j − 20 xk ]
( 25 )
From i:
8(− 403)
Pa
" 3P #
= 20 z &
' z = −1.99012a
675
( 25 )
From k:
8(−406)
Pa
" 3P #
= −20 x &
' x = 2.0049a
675
( 25 )
The axis of the wrench intersects the xz plane at
x = 2.00a, z = −1.990a
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315
!
PROBLEM 3.141*
Determine whether the force-and-couple system shown can be
reduced to a single equivalent force R. If it can, determine R
and the point where the line of action of R intersects the yz
plane. If it cannot be so reduced, replace the given system with
an equivalent wrench and determine its resultant, its pitch, and
the point where its axis intersects the yz plane.
SOLUTION
First, reduce the given force system to a force-couple at the origin.
We have
ΣF : FA + FG = R
(40 mm)i + (60 mm) j − (120 mm)k !
R = (50 N)k + 70 N $
%
140 mm
*
+
= (20 N)i + (30 N) j − (10 N)k
and
We have
R = 37.417 N
ΣM O : Σ(rO × F) + ΣM C = M OR
M OR = [(0.12 m) j × (50 N)k ] + {(0.16 m)i × [(20 N)i + (30 N) j − (60 N)k ]}
+ (10 N ⋅ m) $
*
(160 mm)i − (120 mm) j !
%
200 mm
+
(40 mm)i − (120 mm) j + (60 mm)k !
+ (14 N ⋅ m) $
%
140 mm
*
+
R
M 0 = (18 N ⋅ m)i − (8.4 N ⋅ m) j + (10.8 N ⋅ m)k
To be able to reduce the original forces and couples to a single equivalent force, R and M must be
perpendicular. Thus, R ⋅ M = 0.
Substituting
?
(20i + 30 j − 10k ) ⋅ (18i − 8.4 j + 10.8k ) = 0
?
or
(20)(18) + (30)(−8.4) + (−10)(10.8) = 0
or
0=0
R and M are perpendicular so that the given system can be reduced to the single equivalent force
R = (20.0 N)i + (30.0 N) j − (10.00 N)k
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316
PROBLEM 3.141* (Continued)
!
Then for equivalence
Thus require
M OR = rp × R rp = yj + zk
Substituting
i
j
k
18i − 8.4 j + 10.8k = 0 y
z
20 30 −10
Equating coefficients
j: − 8.4 = 20 z
k:
or
z = −0.42 m
10.8 = −20 y or
y = −0.54 m
The line of action of R intersects the yz plane at
x=0
y = −0.540 m z = −0.420 m
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317
!
PROBLEM 3.142*
Determine whether the force-and-couple system shown can be
reduced to a single equivalent force R. If it can, determine R
and the point where the line of action of R intersects the yz
plane. If it cannot be so reduced, replace the given system with
an equivalent wrench and determine its resultant, its pitch, and
the point where its axis intersects the yz plane.
SOLUTION
First determine the resultant of the forces at D. We have
d DA = (−12) 2 + (9) 2 + (8) 2 = 17 in.
d ED = (−6) 2 + (0)2 + (−8)2 = 10 in.
Then
34 lb
= (−12i + 9 j + 8k )
17
= −(24 lb)i + (18 lb) j + (16 lb)k
FDA =
and
30 lb
= (−6i − 8k )
10
= −(18 lb)i − (24 lb)k
FED =
Then
ΣF : R = FDA + FED
= (−24i + 18 j + 16k + ( −18i − 24k )
= −(42 lb)i + (18 lb)j − (8 lb)k
For the applied couple
d AK = ( −6) 2 + (−6) 2 + (18) 2 = 6 11 in.
Then
M=
160 lb ⋅ in.
( −6i − 6 j + 18k )
6 11
160
=
[−(1 lb ⋅ in.)i − (1 lb ⋅ in.)j + (3 lb ⋅ in.)k ]
11
To be able to reduce the original forces and couple to a single equivalent force, R and M
must be perpendicular. Thus
?
R ⋅ M =0
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318
PROBLEM 3.142* (Continued)
Substituting
(−42i + 18 j − 8k ) ⋅
160
or
11
160
11
?
(−i − j + 3k ) = 0
?
[(−42)(−1) + (18)(−1) + (−8)(3)] = 0
0 =0
or
R and M are perpendicular so that the given system can be reduced to the single equivalent force
R = −(42.0 lb)i + (18.00 lb) j − (8.00 lb)k
Then for equivalence
M = rP/D × R
Thus require
where
rP/D = −(12 in.)i + [( y − 3)in.] j + ( z in.)k
Substituting
i
j
k
( −i − j + 3k ) = −12 ( y − 3) z
11
−42
−8
18
= [( y − 3)( −8) − ( z )(18)]i
160
+ [( z )(−42) − (−12)(−8)]j
+ [( −12)(18) − ( y − 3)(−42)]k
Equating coefficients
j: −
k:
160
= − 42 z − 96
or
11
480
= −216 + 42( y − 3) or
11
The line of action of R intersects the yz plane at
x=0
z = −1.137 in.
y = 11.59 in.
y = 11.59 in. z = −1.137 in.
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319
PROBLEM 3.143*
Replace the wrench shown with an equivalent system consisting of two
forces perpendicular to the y axis and applied respectively at A and B.
SOLUTION
Express the forces at A and B as
A = Ax i + Az k
B = Bx i + Bz k
Then, for equivalence to the given force system
ΣFx : Ax + Bx = 0
(1)
ΣFz : Az + Bz = R
(2)
ΣM x : Az ( a) + Bz ( a + b) = 0
(3)
ΣM z : − Ax (a) − Bx (a + b) = M
(4)
Bx = − Ax
From Equation (1),
Substitute into Equation (4)
− Ax ( a) + Ax ( a + b) = M
M
M
and Bx = −
Ax =
b
b
From Equation (2),
Bz = R − Az
and Equation (3),
Az a + ( R − Az )(a + b) = 0
" a#
Az = R &1 + '
( b)
and
a#
"
Bz = R − R &1 + '
( b)
a
Bz = − R
b
"M
A=&
( b
Then
a#
#
"
' i + R &1 + b ' k
)
(
)
"M
B = −&
( b
# "a #
'i − & b R'k
) (
)
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320
!
PROBLEM 3.144*
Show that, in general, a wrench can be replaced with two forces chosen in such a way that one force passes
through a given point while the other force lies in a given plane.
SOLUTION
First, choose a coordinate system so that the xy plane coincides with the given plane. Also, position the
coordinate system so that the line of action of the wrench passes through the origin as shown in Figure a.
Since the orientation of the plane and the components (R, M) of the wrench are known, it follows that the
scalar components of R and M are known relative to the shown coordinate system.
A force system to be shown as equivalent is illustrated in Figure b. Let A be the force passing through the
given Point P and B be the force that lies in the given plane. Let b be the x-axis intercept of B.
The known components of the wrench can be expressed as
R = Rx i + Ry j + Rz k
and M = M x i + M y j + M z k
while the unknown forces A and B can be expressed as
A = Ax i + Ay j + Az k and B = Bx i + Bz k
Since the position vector of Point P is given, it follows that the scalar components (x, y, z) of the
position vector rP are also known.
Then, for equivalence of the two systems
ΣFx : Rx = Ax + Bx
(1)
ΣFy : Ry = Ay
(2)
ΣFz : Rz = Az + Bz
(3)
ΣM x : M x = yAz − zAy
(4)
ΣM y : M y = zAx − xAz − bBz
(5)
ΣM z : M z = xAy − yAx
(6)
Based on the above six independent equations for the six unknowns ( Ax , Ay , Az , Bx , By , Bz , b), there
exists a unique solution for A and B.
Ay = Ry
From Equation (2)
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321
PROBLEM 3.144* (Continued)
Equation (6)
"1#
Ax = & ' ( xRy − M z )
( y)
Equation (1)
"1#
Bx = Rx − & ' ( xRy − M z )
( y)
Equation (4)
"1#
Az = & ' ( M x + zRy )
( y)
Equation (3)
"1#
Bz = Rz − & ' ( M x + zRy )
( y)
b=
Equation (5)
( xM x + yM y + zM z )
( M x − yRz + zRy )
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322
!
!
PROBLEM 3.145*
Show that a wrench can be replaced with two perpendicular forces, one of which is applied at a given point.
SOLUTION
First, observe that it is always possible to construct a line perpendicular to a given line so that the constructed
line also passes through a given point. Thus, it is possible to align one of the coordinate axes of a rectangular
coordinate system with the axis of the wrench while one of the other axes passes through the given point.
See Figures a and b.
We have
R = Rj and M = Mj
and are known.
The unknown forces A and B can be expressed as
A = Ax i + Ay j + Az k
and B = Bx i + By j + Bz k
The distance a is known. It is assumed that force B intersects the xz plane at (x, 0, z). The for equivalence
ΣFx :
0 = Ax + Bx
(1)
ΣFy :
R = Ay + By
(2)
ΣFz :
0 = Az + Bz
(3)
0 = − zBy
(4)
ΣM x :
ΣM y : M = − aAz − xBz + zBx
ΣM z :
(5)
0 = aAy + xBy
(6)
Since A and B are made perpendicular,
A ⋅ B = 0 or
There are eight unknowns:
Ax Bx + Ay B y + Az Bz = 0
(7)
Ax , Ay , Az , Bx , By , Bz , x, z
But only seven independent equations. Therefore, there exists an infinite number of solutions.
0 = − zBy
Next consider Equation (4):
If By = 0, Equation (7) becomes
Ax Bx + Az Bz = 0
Ax2 + Az2 = 0
Using Equations (1) and (3) this equation becomes
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323
PROBLEM 3.145* (Continued)
Since the components of A must be real, a nontrivial solution is not possible. Thus, it is required that By ≠ 0,
so that from Equation (4), z = 0.
To obtain one possible solution, arbitrarily let Ax = 0.
(Note: Setting Ay , Az , or Bz equal to zero results in unacceptable solutions.)
The defining equations then become
0 = Bx
(1)′
R = Ay + By
(2)
0 = Az + Bz
(3)
M = − aAz − xBz
(5)′
0 = aAy + xBy
(6)
Ay By + Az Bz = 0
(7)′
Then Equation (2) can be written
Ay = R − By
Equation (3) can be written
Bz = − Az
Equation (6) can be written
x=−
aAy
By
Substituting into Equation (5)′,
"
R − By
M = − aAz − & − a
&
By
(
M
Az = −
By
aR
or
#
' ( − Az )
'
)
(8)
Substituting into Equation (7)′,
" M
#" M
#
( R − By ) By + & −
By '&
By ' = 0
( aR )( aR )
or
By =
a 2 R3
a2 R2 + M 2
Then from Equations (2), (8), and (3)
a2 R2
RM 2
=
a2 R2 + M 2 a2 R2 + M 2
#
M " a 2 R3
aR 2 M
Az = −
=
−
&& 2 2
'
aR ( a R + M 2 ')
a2 R2 + M 2
Ay = R −
Bz =
aR 2 M
a2 R2 + M 2
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324
PROBLEM 3.145* (Continued)
In summary
A=
RM
( Mj − aRk )
a R2 + M 2
B=
aR 2
(aRj + Mk )
a2 R2 + M 2
2
Which shows that it is possible to replace a wrench with two perpendicular forces, one of which is applied at
a given point.
Lastly, if R . 0 and M . 0, it follows from the equations found for A and B that Ay . 0 and By . 0.
From Equation (6), x , 0 (assuming a . 0). Then, as a consequence of letting Ax = 0, force A lies in a
plane parallel to the yz plane and to the right of the origin, while force B lies in a plane parallel to the yz plane
but to the left to the origin, as shown in the figure below.
!
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325
PROBLEM 3.146*
Show that a wrench can be replaced with two forces, one of which has a prescribed line of action.
SOLUTION
First, choose a rectangular coordinate system where one axis coincides with the axis of the wrench and
another axis intersects the prescribed line of action (AA′). Note that it has been assumed that the line of action
of force B intersects the xz plane at Point P(x, 0, z). Denoting the known direction of line AA′ by
λ A = λx i + λ y j + λz k
it follows that force A can be expressed as
A = Aλ A = A(λx i + λ y j + λz k )
Force B can be expressed as
B = Bx i + By j + Bz k
Next, observe that since the axis of the wrench and the prescribed line of action AA′ are known, it follows that
the distance a can be determined. In the following solution, it is assumed that a is known.
Then, for equivalence
ΣFx : 0 = Aλx + Bx
(1)
ΣFy : R = Aλ y + By
(2)
ΣFz : 0 = Aλz + Bz
(3)
ΣM x : 0 = − zBy
(4)
ΣM y : M = − aAλz + zBx − xBz
(5)
ΣM x : 0 = − aAλ y + xBy
(6)
Since there are six unknowns (A, Bx, By, Bz, x, z) and six independent equations, it will be possible to
obtain a solution.
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326
PROBLEM 3.146* (Continued)
Case 1: Let z = 0 to satisfy Equation (4)
Now Equation (2)
Equation (3)
Equation (6)
Aλ y = R − By
Bz = − Aλz
x=−
aAλ y
By
" a
= −&
& By
(
#
' ( R − By )
'
)
Substitution into Equation (5)
" a
M = − aAλz − $ − &
$* &( By
A=−
!
#
' ( R − By )(− Aλz ) %
'
%+
)
1 "M #
B
λz &( aR ') y
Substitution into Equation (2)
R=−
By =
Then
1 "M #
B λ + By
λz &( aR ') y y
λz aR 2
λz aR − λ y M
MR
R
=
λz aR − λ y M λ − aR λ
y
z
M
λx MR
Bx = − Aλx =
λz aR − λ y M
A=−
Bz = − Aλz =
λz MR
λz aR − λ y M
A=
In summary
B=
and
R
λz aR − λ y M
P
λA
aR
λy −
λz
M
(λx Mi + λz aRj + λz M k )
"
R #
x = a &1 −
'
&
By ')
(
" λz aR − λ y M
= a $1 − R &
& λ aR 2
$*
z
(
#!
'' %
) %+
or x =
λy M
λz R
Note that for this case, the lines of action of both A and B intersect the x axis.
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327
PROBLEM 3.146* (Continued)
Case 2: Let By = 0 to satisfy Equation (4)
Now Equation (2)
A=
R
λy
Equation (1)
"λ
Bx = − R & x
& λy
(
#
'
'
)
Equation (3)
"λ
Bz = − R & z
& λy
(
#
'
'
)
Equation (6)
aAλ y = 0
which requires a = 0
Substitution into Equation (5)
"λ
M = z $− R & x
$* &( λ y
#!
"λ
'% − x $− R & z
'%
$* &( λ y
)+
#!
"M #
' % or λz x − λx z = & ' λ y
'%
( R)
)+
This last expression is the equation for the line of action of force B.
In summary
" R
A=&
& λy
(
#
' λA
'
)
" R
B=&
& λy
(
#
' ( −λ x i − λx k )
'
)
Assuming that λx , λ y , λz . 0, the equivalent force system is as shown below.
Note that the component of A in the xz plane is parallel to B.!
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328
PROBLEM 3.147
A crate of mass 80 kg is held in the position shown. Determine
(a) the moment produced by the weight W of the crate about E,
(b) the smallest force applied at B that creates a moment of
equal magnitude and opposite sense about E.
SOLUTION
(a)
By definition
We have
W = mg = 80 kg(9.81 m/s 2 ) = 784.8 N
ΣM E : M E = (784.8 N)(0.25 m)
M E = 196.2 N ⋅ m
(b)
For the force at B to be the smallest, resulting in a moment (ME) about E, the line of action of force FB
must be perpendicular to the line connecting E to B. The sense of FB must be such that the force
produces a counterclockwise moment about E.
Note:
We have
d = (0.85 m) 2 + (0.5 m) 2 = 0.98615 m
ΣM E : 196.2 N ⋅ m = FB (0.98615 m)
FB = 198.954 N
and
" 0.85 m #
' = 59.534°
( 0.5 m )
θ = tan −1 &
FB = 199.0 N
or
59.5°
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329
!
PROBLEM 3.148
It is known that the connecting rod AB exerts on the crank BC a 1.5-kN force
directed down and to the left along the centerline of AB. Determine the moment
of the force about C.
SOLUTION
Using (a)
M C = y1 ( FAB ) x + x1 ( FAB ) y
" 7
#
" 24
#
= (0.028 m) & × 1500 N ' + (0.021 m) & × 1500 N '
( 25
)
( 25
)
= 42 N ⋅ m
or M C = 42.0 N ⋅ m
Using (b)
M C = y2 ( FAB ) x
" 7
#
= (0.1 m) & × 1500 N ' = 42 N ⋅ m
( 25
)
or M C = 42.0 N ⋅ m
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330
!
PROBLEM 3.149
A 6-ft-long fishing rod AB is securely anchored in the sand of a beach. After a fish takes the bait, the resulting
force in the line is 6 lb. Determine the moment about A of the force exerted by the line at B.
SOLUTION
We have
Then
Txz = (6 lb) cos 8° = 5.9416 lb
Tx = Txz sin 30° = 2.9708 lb
Ty = TBC sin 8° = − 0.83504 lb
Tz = Txz cos 30° = −5.1456 lb
Now
M A = rB/A × TBC
where
rB/A = (6sin 45°) j − (6cos 45°)k
=
6 ft
2
(j − k)
i
j
k
−1
0
1
2
2.9708 −0.83504 −5.1456
6
6
6
=
(−5.1456 − 0.83504)i −
(2.9708) j −
(2.9708)k
2
2
2
6
Then
MA =
or
M A = −(25.4 lb ⋅ ft)i − (12.60 lb ⋅ ft) j − (12.60 lb ⋅ ft)k
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331
!
PROBLEM 3.150
Ropes AB and BC are two of the ropes used to
support a tent. The two ropes are attached to a
stake at B. If the tension in rope AB is 540 N,
determine (a) the angle between rope AB and
the stake, (b) the projection on the stake of the
force exerted by rope AB at Point B.
SOLUTION
First note
BA = (−3) 2 + (3)2 + (−1.5) 2 = 4.5 m
BD = (−0.08) 2 + (0.38)2 + (0.16) 2 = 0.42 m
λ BD
(a)
TBA
(−3i + 3j − 1.5k )
4.5
T
= BA (−2i + 2 j − k )
3
BD
1
=
=
( −0.08i + 0.38 j + 0.16k )
BD 0.42
1
= (−4i + 19 j + 8k )
21
TBA =
Then
We have
or
or
TBA ⋅ λ BD = TBA cos θ
TBA
1
( −2i + 2 j − k ) ⋅ (−4i + 19 j + 8k ) = TBA cos θ
3
21
1
[(−2)( −4) + (2)(19) + (−1)(8)]
63
= 0.60317
cos θ =
or θ = 52.9°
(b)
We have
(TBA ) BD = TBA ⋅ λ BD
= TBA cos θ
= (540 N)(0.60317)
or (TBA ) BD = 326 N
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332
!
PROBLEM 3.151
A farmer uses cables and winch pullers B and E to plumb
one side of a small barn. If it is known that the sum of the
moments about the x axis of the forces exerted by the
cables on the barn at Points A and D is equal to 4728 lb ⋅ ft,
determine the magnitude of TDE when TAB = 255 lb.
SOLUTION
The moment about the x axis due to the two cable forces can be found using the z components of each
force acting at their intersection with the xy-plane (A and D). The x components of the forces are parallel to
the x axis, and the y components of the forces intersect the x axis. Therefore, neither the x or y components
produce a moment about the x axis.
We have
ΣM x : (TAB ) z ( y A ) + (TDE ) z ( yD ) = M x
where
(TAB ) z = k ⋅ TAB
= k ⋅ (TAB λ AB )
" − i − 12 j + 12k # !
= k ⋅ $ 255 lb &
'%
17
(
)+
*
= 180 lb
(TDE ) z = k ⋅ TDE
= k ⋅ (TDE λDE )
" 1.5i − 14 j + 12k # !
= k ⋅ $TDE &
'%
18.5
(
)+
*
= 0.64865TDE
y A = 12 ft
yD = 14 ft
M x = 4728 lb ⋅ ft
(180 lb)(12 ft) + (0.64865TDE )(14 ft) = 4728 lb ⋅ ft
and
TDE = 282.79 lb
or TDE = 283 lb
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333
!
PROBLEM 3.152
Solve Problem 3.151 when the tension in cable AB is 306 lb.
PROBLEM 3.151 A farmer uses cables and winch pullers B
and E to plumb one side of a small barn. If it is known that the
sum of the moments about the x axis of the forces exerted by
the cables on the barn at Points A and D is equal to 4728 lb ⋅ ft,
determine the magnitude of TDE when TAB = 255 lb.
SOLUTION
The moment about the x axis due to the two cable forces can be found using the z components of each force
acting at the intersection with the xy plane (A and D). The x components of the forces are parallel to the x axis,
and the y components of the forces intersect the x axis. Therefore, neither the x or y components produce a
moment about the x axis.
We have
ΣM x : (TAB ) z ( y A ) + (TDE ) z ( yD ) = M x
Where
(TAB ) z = k ⋅ TAB
= k ⋅ (TAB
AB )
" − i − 12 j + 12k # !
= k ⋅ $306 lb &
'%
17
(
)+
*
= 216 lb
(TDE ) z = k ⋅ TDE
= k ⋅ (TDE
DE )
" 1.5i − 14 j + 12k # !
= k ⋅ $TDE &
'%
18.5
(
)+
*
= 0.64865TDE
y A = 12 ft
yD = 14 ft
M x = 4728 lb ⋅ ft
(216 lb)(12 ft) + (0.64865TDE )(14 ft) = 4728 lb ⋅ ft
and
TDE = 235.21 lb
or TDE = 235 lb
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334
!
PROBLEM 3.153
A wiring harness is made by routing either two or
three wires around 2-in.-diameter pegs mounted on a
sheet of plywood. If the force in each wire is 3 lb,
determine the resultant couple acting on the plywood
when a = 18 in. and (a) only wires AB and CD are in
place, (b) all three wires are in place.
SOLUTION
In general, M = ΣdF, where d is the perpendicular distance between the lines of action of the two forces
acting on a given wire.
(a)
We have
M = d AB FAB + dCD FCD
4
"
#
= (2 + 24) in. × 3 lb + & 2 + × 28 ' in. × 3 lb
5
(
)
or M = 151.2 lb ⋅ in.
(b)
We have
M = [d AB FAB + dCD FCD ] + d EF FEF
= 151.2 lb ⋅ in. − 28 in. × 3 lb
or M = 67.2 lb ⋅ in.
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335
!
PROBLEM 3.154
A worker tries to move a rock by applying a 360-N force
to a steel bar as shown. (a) Replace that force with an
equivalent force-couple system at D. (b) Two workers
attempt to move the same rock by applying a vertical force
at A and another force at D. Determine these two forces if
they are to be equivalent to the single force of Part a.
SOLUTION
(a)
(a)
We have
(b)
ΣF : 360 N(− sin 40°i − cos 40° j) = −(231.40 N)i − (275.78 N) j = F
or F = 360 N
We have
where
50°
ΣM D : rB/D × R = M
rB/D = −[(0.65 m) cos 30°]i + [(0.65 m)sin 30°]j
= −(0.56292 m)i + (0.32500 m) j
i
j
k
M = −0.56292 0.32500 0 N ⋅ m
−231.40 −275.78 0
= [155.240 + 75.206)N ⋅ m]k
= (230.45 N ⋅ m)k
(b)
We have
where
or M = 230 N ⋅ m
ΣM D : M = rA/D × FA
rB/D = −[(1.05 m) cos 30°]i + [(1.05 m)sin 30°]j
= −(0.90933 m)i + (0.52500 m) j
i
j
k
FA = −0.90933 0.52500 0 N ⋅ m
0
−1
= [230.45 N ⋅ m]k
0
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336
PROBLEM 3.154 (Continued)
(0.90933FA )k = 230.45k
or
FA = 253.42 N
We have
or FA = 253 N
ΣF : F = FA + FD
−(231.40 N)i − (275.78 N) j = −(253.42 N) j + FD (− cos θ i − sin θ j)
From
i : 231.40 N = FD cos θ
(1)
j: 22.36 N = FD sin θ
(2)
Equation (2) divided by Equation (1)
tan θ = 0.096629
θ = 5.5193° or θ = 5.52°
Substitution into Equation (1)
FD =
231.40
= 232.48 N
cos 5.5193°
or FD = 232 N
5.52°
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337
PROBLEM 3.155
A 110-N force acting in a vertical plane parallel to the yz plane
is applied to the 220-mm-long horizontal handle AB of a
socket wrench. Replace the force with an equivalent forcecouple system at the origin O of the coordinate system.
SOLUTION
We have
ΣF : PB = F
where
PB = 110 N[− (sin15°) j + (cos15°)k ]
= −(28.470 N) j + (106.252 N)k
or F = −(28.5 N) j + (106.3 N)k
We have
where
ΣM O : rB/O × PB = M O
rB/O = [(0.22 cos 35°)i + (0.15) j − (0.22sin 35°)k ]m
= (0.180213 m)i + (0.15 m) j − (0.126187 m)k
i
j
k
0.180213 0.15 0.126187 N ⋅ m = M O
0
106.3
−28.5
M O = [(12.3487)i − (19.1566) j − (5.1361)k ]N ⋅ m
or M O = (12.35 N ⋅ m)i − (19.16 N ⋅ m)j − (5.13 N ⋅ m)k
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338
PROBLEM 3.156
Four ropes are attached to a crate and exert the forces shown.
If the forces are to be replaced with a single equivalent force
applied at a point on line AB, determine (a) the equivalent
force and the distance from A to the point of application of
the force when α = 30°, (b) the value of α so that the single
equivalent force is applied at Point B.
SOLUTION
We have
(a)
For equivalence
ΣFx : −100 cos 30° + 400 cos 65° + 90 cos 65° = Rx
or
Rx = 120.480 lb
ΣFy : 100 sin α + 160 + 400 sin 65° + 90 sin 65° = R y
or
R y = (604.09 + 100sin α ) lb
With α = 30°
R y = 654.09 lb
Then
R = (120.480) 2 + (654.09) 2
(1)
= 665 lb
Also
654.09
120.480
or θ = 79.6°
tan θ =
ΣM A : (46 in.)(160 lb) + (66 in.)(400 lb) sin 65°
+ (26 in.)(400 lb) cos 65° + (66 in.)(90 lb) sin 65°
+ (36 in.)(90 lb) cos 65° = d (654.09 lb)
or
ΣM A = 42, 435 lb ⋅ in. and d = 64.9 in.
R = 665 lb
79.6°
and R is applied 64.9 in. To the right of A.
(b)
We have d = 66 in.
Then
or
Using Eq. (1)
ΣM A : 42, 435 lb ⋅ in = (66 in.) Ry
Ry = 642.95 lb
642.95 = 604.09 + 100sin α
or α = 22.9°
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339
PROBLEM 3.157
A blade held in a brace is used to tighten a screw at A.
(a) Determine the forces exerted at B and C, knowing
that these forces are equivalent to a force-couple
system at A consisting of R = −(30 N)i + Ry j + Rz k
and M RA = − (12 N · m)i. (b) Find the corresponding
values of Ry and Rz . (c) What is the orientation of the
slot in the head of the screw for which the blade is least
likely to slip when the brace is in the position shown?
SOLUTION
(a)
Equivalence requires
or
Equating the i coefficients
Also
or
Equating coefficients
ΣF : R = B + C
−(30 N)i + Ry j + Rz k = − Bk + (−Cx i + C y j + C z k )
i : − 30 N = −Cx
or Cx = 30 N
ΣM A : M RA = rB/A × B + rC/A × C
−(12 N ⋅ m)i = [(0.2 m)i + (0.15 m)j] × (− B)k
+(0.4 m)i × [−(30 N)i + C y j + Cz k ]
i : − 12 N ⋅ m = −(0.15 m) B
k : 0 = (0.4 m)C y
or B = 80 N
or C y = 0
j: 0 = (0.2 m)(80 N) − (0.4 m)C z or Cz = 40 N
B = −(80.0 N)k C = −(30.0 N)i + (40.0 N)k
(b)
Now we have for the equivalence of forces
−(30 N)i + Ry j + Rz k = −(80 N)k + [(−30 N)i + (40 N)k ]
Equating coefficients
j: R y = 0
Ry = 0
k : Rz = −80 + 40
(c)
or
!
Rz = −40.0 N
First note that R = −(30 N)i − (40 N)k. Thus, the screw is best able to resist the lateral force Rz
when the slot in the head of the screw is vertical.
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340
PROBLEM 3.158
A concrete foundation mat in the shape of a regular hexagon of side
12 ft supports four column loads as shown. Determine the magnitudes
of the additional loads that must be applied at B and F if the resultant
of all six loads is to pass through the center of the mat.
SOLUTION
From the statement of the problem it can be concluded that the six applied loads are equivalent to the resultant
R at O. It then follows that
ΣM O = 0 or ΣM x = 0 ΣM z = 0
For the applied loads.
Then
ΣM x = 0: (6 3 ft) FB + (6 3 ft)(10 kips) − (6 3 ft)(20 kips)
− (6 3 ft) FF = 0
FB − FF = 10
or
(1)
ΣM z = 0: (12 ft)(15 kips) + (6 ft) FB − (6 ft)(10 kips)
− (12 ft)(30 kips) − (6 ft)(20 kips) + (6 ft) FF = 0
FB + FF = 60
or
(2)
Then (1) + (2) -
FB = 35.0 kips
and
FF = 25.0 kips
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341
CHAPTER 4
PROBLEM 4.1
A 2100-lb tractor is used to lift 900 lb of gravel. Determine
the reaction at each of the two (a) rear wheels A, (b) front
wheels B.
SOLUTION
(a)
Rear wheels
ΣM B = 0: + (2100 lb)(40 in.) − (900 lb)(50 in.) + 2 A(60 in.) = 0
A = +325 lb
(b)
Front wheels
ΣM A : − (2100 lb)(20 in.) − (900 lb)(110 in.) − 2 B(60 in.) = 0
B = +1175 lb
Check:
A = 325 lb
B = 1175 lb
+ΣFy = 0: 2 A + 2 B − 2100 lb − 900 lb = 0
2(325 lb) + 2(1175 lb) − 2100 lb − 900 = 0
0 = 0 (Checks) !
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345
PROBLEM 4.2
A gardener uses a 60-N wheelbarrow to transport a 250-N bag of
fertilizer. What force must she exert on each handle?
SOLUTION
Free-Body Diagram:
!
ΣM A = 0: (2 F )(1 m) − (60 N)(0.15 m) − (250 N)(0.3 m) = 0
F = 42.0 N
!
!
!
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346
PROBLEM 4.3
The gardener of Problem 4.2 wishes to transport a second 250-N
bag of fertilizer at the same time as the first one. Determine the
maximum allowable horizontal distance from the axle A of the
wheelbarrow to the center of gravity of the second bag if she can
hold only 75 N with each arm.
PROBLEM 4.2 A gardener uses a 60-N wheelbarrow to transport
a 250-N bag of fertilizer. What force must she exert on each handle?
SOLUTION
Free-Body Diagram:
!
ΣMA = 0: 2(75 N)(1 m) − (60 N)(0.15 m)
− (250 N)(0.3 m) − (250 N) x = 0
x = 0.264 m
!
!
!
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347
PROBLEM 4.4
For the beam and loading shown, determine (a) the reaction at A,
(b) the tension in cable BC.
SOLUTION
Free-Body Diagram:
(a)
Reaction at A:
ΣFx = 0: Ax = 0
ΣMB = 0: (15 lb)(28 in.) + (20 lb)(22 in.) + (35 lb)(14 in.)
+ (20 lb)(6 in.) − Ay (6 in.) = 0
Ay = +245 lb
(b)
Tension in BC
A = 245 lb
ΣM A = 0: (15 lb)(22 in.) + (20 lb)(16 in.) + (35 lb)(8 in.)
− (15 lb)(6 in.) − FBC (6 in.) = 0
FBC = +140.0 lb
Check:
FBC = 140.0 lb
ΣFy = 0: − 15 lb − 20 lb = 35 lb − 20 lb + A − FBC = 0
−105 lb + 245 lb − 140.0 = 0
0 = 0 (Checks) !
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348
PROBLEM 4.5
Two crates, each of mass 350 kg, are placed as shown in the
bed of a 1400-kg pickup truck. Determine the reactions at each
of the two (a) rear wheels A, (b) front wheels B.
SOLUTION
Free-Body Diagram:
W = (350 kg)(9.81 m/s2 ) = 3.434 kN
Wt = (1400 kg)(9.81 m/s 2 ) = 13.734 kN
(a)
Rear wheels
ΣM B = 0: W (1.7 m + 2.05 m) + W (2.05 m) + Wt (1.2 m) − 2 A(3 m) = 0
(3.434 kN)(3.75 m) + (3.434 kN)(2.05 m)
+ (13.734 kN)(1.2 m) − 2 A(3 m) = 0
A = +6.0663 kN
(b)
Front wheels
A = 6.07 kN
ΣFy = 0: − W − W − Wt + 2 A + 2 B = 0
−3.434 kN − 3.434 kN − 13.734 kN + 2(6.0663 kN) + 2B = 0
B = + 4.2347 kN
B = 4.23 kN
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349
PROBLEM 4.6
Solve Problem 4.5, assuming that crate D is removed and that
the position of crate C is unchanged.
PROBLEM 4.5 Two crates, each of mass 350 kg, are placed
as shown in the bed of a 1400-kg pickup truck. Determine the
reactions at each of the two (a) rear wheels A, (b) front wheels B.
SOLUTION
Free-Body Diagram:
W = (350 kg)(9.81 m/s2 ) = 3.434 kN
Wt = (1400 kg)(9.81 m/s 2 ) = 13.734 kN
(a)
Rear wheels
ΣM B = 0: W (1.7 m + 2.05 m) + Wt (1.2 m) − 2 A(3 m) = 0
(3.434 kN)(3.75 m) + (13.734 kN)(1.2 m) − 2 A(3 m) = 0
A = + 4.893 kN
(b)
Front wheels
A = 4.89 kN
ΣM y = 0: − W − Wt + 2 A + 2 B = 0
−3.434 kN − 13.734 kN + 2(4.893 kN) + 2B = 0
B = +3.691 kN
B = 3.69 kN
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350
PROBLEM 4.7
A T-shaped bracket supports the four loads shown. Determine the
reactions at A and B (a) if a = 10 in., (b) if a = 7 in.
SOLUTION
Free-Body Diagram:
ΣFx = 0: Bx = 0
ΣM B = 0: (40 lb)(6 in.) − (30 lb)a − (10 lb)(a + 8 in.) + (12 in.) A = 0
(40a − 160)
12
A=
(1)
ΣM A = 0: − (40 lb)(6 in.) − (50 lb)(12 in.) − (30 lb)(a + 12 in.)
− (10 lb)(a + 20 in.) + (12 in.) B y = 0
By =
Bx = 0 B =
Since
(a)
(b)
(1400 + 40a)
12
(1400 + 40a)
12
(2)
For a = 10 in.
Eq. (1):
A=
(40 × 10 − 160)
= +20.0 lb
12
Eq. (2):
B=
(1400 + 40 × 10)
= +150.0 lb
12
B = 150.0 lb
Eq. (1):
A=
(40 × 7 − 160)
= +10.00 lb
12
A = 10.00 lb
Eq. (2):
B=
(1400 + 40 × 7)
= +140.0 lb
12
B = 140.0 lb
A = 20.0 lb
For a = 7 in.
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351
PROBLEM 4.8
For the bracket and loading of Problem 4.7, determine the smallest
distance a if the bracket is not to move.
PROBLEM 4.7 A T-shaped bracket supports the four loads shown.
Determine the reactions at A and B (a) if a = 10 in., (b) if a = 7 in.
SOLUTION
Free-Body Diagram:
For no motion, reaction at A must be downward or zero; smallest distance a for no motion
corresponds to A = 0.
ΣM B = 0: (40 lb)(6 in.) − (30 lb)a − (10 lb)(a + 8 in.) + (12 in.) A = 0
A=
(40a − 160)
12
A = 0: (40a − 160) = 0
a = 4.00 in.
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352
!
PROBLEM 4.9
The maximum allowable value of each of the reactions is 180 N.
Neglecting the weight of the beam, determine the range of the
distance d for which the beam is safe.
SOLUTION
ΣFx = 0: Bx = 0
B = By
ΣM A = 0: (50 N) d − (100 N)(0.45 m − d ) − (150 N)(0.9 m − d ) + B (0.9 m − d ) = 0
50d − 45 + 100d − 135 + 150d + 0.9 B − Bd
d=
180 N ⋅ m − (0.9 m) B
300 A − B
(1)
ΣM B = 0: (50 N)(0.9 m) − A(0.9 m − d ) + (100 N)(0.45 m) = 0
45 − 0.9 A + Ad + 45 = 0
(0.9 m) A − 90 N ⋅ m
A
(2)
d$
180 − (0.9)180 18
=
= 0.15 m
300 − 180
120
d $ 150.0 mm "
d#
(0.9)180 − 90 72
=
= 0.40 m
180
180
d=
Since B # 180 N, Eq. (1) yields.
Since A # 180 N, Eq. (2) yields.
Range:
d # 400 mm "
!
150.0 mm # d # 400 mm
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353
PROBLEM 4.10
Solve Problem 4.9 if the 50-N load is replaced by an 80-N load.
PROBLEM 4.9 The maximum allowable value of each of the
reactions is 180 N. Neglecting the weight of the beam, determine
the range of the distance d for which the beam is safe.
SOLUTION
ΣFx = 0: Bx = 0
B = By
ΣM A = 0: (80 N)d − (100 N)(0.45 m − d ) − (150 N)(0.9 m − d ) + B(0.9 m − d ) = 0
80d − 45 + 100d − 135 + 150d + 0.9 B − Bd = 0
d=
180 N ⋅ m − 0.9 B
330N − B
(1)
ΣM B = 0: (80 N)(0.9 m) − A(0.9 m − d ) + (100 N)(0.45 m) = 0
d=
0.9 A − 117
A
(2)
Since B # 180 N, Eq. (1) yields.
d $ (180 − 0.9 × 180)/(330 − 180) =
18
= 0.12 m
150
d = 120.0 mm "
Since A # 180 N, Eq. (2) yields.
d # (0.9 × 180 − 112)/180 =
Range:
45
= 0.25 m
180
d = 250 mm "
120.0 mm # d # 250 mm
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354
PROBLEM 4.11
For the beam of Sample Problem 4.2, determine the range of
values of P for which the beam will be safe, knowing that the
maximum allowable value of each of the reactions is 30 kips and
that the reaction at A must be directed upward.
SOLUTION
ΣFx = 0: Bx = 0
B = By
ΣM A = 0: − P(3 ft) + B (9 ft) − (6 kips)(11 ft) − (6 kips)(13 ft) = 0
P = 3B − 48 kips
(1)
ΣM B = 0: − A(9 ft) + P (6 ft) − (6 kips)(2 ft) − (6 kips)(4 ft) = 0
P = 1.5 A + 6 kips
(2)
Since B # 30 kips, Eq. (1) yields.
P # (3)(30 kips) − 48 kips
P # 42.0 kips "
Since 0 # A # 30 kips. Eq. (2) yields.
0 + 6 kips # P # (1.5)(30 kips)1.6 kips
6.00 kips # P # 51.0 kips
"
Range of values of P for which beam will be safe:
6.00 kips # P # 42.0 kips
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355
PROBLEM 4.12
The 10-m beam AB rests upon, but is not attached to, supports at
C and D. Neglecting the weight of the beam, determine the range
of values of P for which the beam will remain in equilibrium.
SOLUTION
Free-Body Diagram:
ΣM C = 0: P(2 m) − (4 kN)(3 m) − (20 kN)(8 m) + D(6 m) = 0
P = 86 kN − 3D
(1)
ΣM D = 0: P(8 m) + (4 kN)(3 m) − (20 kN)(2 m) − C (6 m) = 0
P = 3.5 kN + 0.75C
(2)
For no motion C $ 0 and D $ 0
For C $ 0 from (2) P # 3.50 kN
For D $ 0 from (1) P # 86.0 kN
Range of P for no motion:
3.50 kN # P # 86.0 kN
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356
PROBLEM 4.13
The maximum allowable value of each of the reactions is 50 kN, and
each reaction must be directed upward. Neglecting the weight of the
beam, determine the range of values of P for which the beam is safe.
SOLUTION
Free-Body Diagram:
ΣM C = 0: P(2 m) − (4 kN)(3 m) − (20 kN)(8 m) + D(6 m) = 0
P = 86 kN − 3D
(1)
ΣM D = 0: P(8 m) + (4 kN)(3 m) − (20 kN)(2 m) − C (6 m) = 0
P = 3.5 kN + 0.75C
(2)
For C $ 0, from (2):
P $ 3.50 kN
"
For D $ 0, from (1):
P # 86.0 kN
"
P # 3.5 kN + 0.75(50 kN)
P # 41.0 kN
"
P $ 86 kN − 3(50 kN)
P $ −64.0 kN
"
For C # 50 kN, from (2):
For D # 50 kN, from (1):
Comparing the four criteria, we find
3.50 kN # P # 41.0 kN
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357
PROBLEM 4.14
For the beam and loading shown, determine the range of the
distance a for which the reaction at B does not exceed 100 lb
downward or 200 lb upward.
SOLUTION
Assume B is positive when directed
Sketch showing distance from D to forces.
ΣM D = 0: (300 lb)(8 in. − a ) − (300 lb)(a − 2 in.) − (50 lb)(4 in.) + 16 B = 0
−600a + 2800 + 16B = 0
(2800 + 16B)
600
(1)
[2800 + 16(−100)] 1200
=
= 2 in.
600
600
a $ 2.00 in. "
a=
For B = 100 lb = −100 lb, Eq. (1) yields:
a$
For B = 200 = +200 lb, Eq. (1) yields:
a#
Required range:
[2800 + 16(200)] 6000
=
= 10 in.
600
600
a # 10.00 in. "
2.00 in. # a # 10.00 in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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358
PROBLEM 4.15
Two links AB and DE are connected by a bell crank as
shown. Knowing that the tension in link AB is 720 N,
determine (a) the tension in link DE, (b) the reaction at C.
SOLUTION
Free-Body Diagram:
ΣM C = 0: FAB (100 mm) − FDE (120 mm) = 0
FDE =
(a)
For
(1)
FAB = 720 N
FDE =
(b)
5
FAB
6
5
(720 N)
6
FDE = 600 N
3
ΣFx = 0: − (720 N) + C x = 0
5
C x = +432 N
4
ΣFy = 0: − (720 N) + C y − 600 N = 0
5
C y = +1176 N
C = 1252.84 N
α = 69.829°
C = 1253 N
69.8°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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359
PROBLEM 4.16
Two links AB and DE are connected by a bell crank as
shown. Determine the maximum force that may be safely
exerted by link AB on the bell crank if the maximum
allowable value for the reaction at C is 1600 N.
SOLUTION
See solution to Problem 4.15 for F. B. D. and derivation of Eq. (1)
FDE =
5
FAB
6
(1)
3
3
ΣFx = 0: − FAB + C x = 0 C x = FAB
5
5
ΣFy = 0: −
4
FAB + C y − FDE = 0
5
4
5
− FAB + C y − FAB = 0
5
6
49
Cy =
FAB
30
C = C x2 + C y2
1
(49) 2 + (18) 2 FAB
30
C = 1.74005FAB
=
For
C = 1600 N, 1600 N = 1.74005FAB
FAB = 920 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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360
PROBLEM 4.17
The required tension in cable AB is 200 lb. Determine (a) the vertical
force P that must be applied to the pedal, (b) the corresponding
reaction at C.
SOLUTION
Free-Body Diagram:
BC = 7 in.
(a)
ΣM C = 0: P(15 in.) − (200 lb)(6.062 in.) = 0
P = 80.83 lb
(b)
P = 80.8 lb
ΣFy = 0: C x − 200 lb = 0
C x = 200 lb
ΣFy = 0: C y − P = 0 C y − 80.83 lb = 0
C y = 80.83 lb
α = 22.0°
C = 215.7 lb
C = 216 lb
22.0°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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361
PROBLEM 4.18
Determine the maximum tension that can be developed in cable AB if
the maximum allowable value of the reaction at C is 250 lb.
SOLUTION
Free-Body Diagram:
BC = 7 in.
ΣM C = 0: P (15 in.) − T (6.062 in.) = 0
P = 0.40415T
ΣFy = 0: − P + C y = 0 − 0.40415 P + C y = 0
C y = 0.40415T
ΣFx = 0: −T + C x = 0
Cx = T
C = C x2 + C y2 = T 2 + (0.40415T ) 2
C = 1.0786T
For
C = 250 lb
250 lb = 1.0786T
T = 232 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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362
PROBLEM 4.19
The bracket BCD is hinged at C and attached to a control
cable at B. For the loading shown, determine (a) the tension
in the cable, (b) the reaction at C.
SOLUTION
At B:
Ty
Tx
=
0.18 m
0.24 m
3
Ty = Tx
4
(a)
(1)
ΣM C = 0: Tx (0.18 m) − (240 N)(0.4 m) − (240 N)(0.8 m) = 0
Tx = +1600 N
Eq. (1)
Ty =
3
(1600 N) = 1200 N
4
T = Tx2 + Ty2 = 16002 + 12002 = 2000 N
(b)
T = 2.00 kN
ΣFx = 0: Cx − Tx = 0
Cx − 1600 N = 0 C x = +1600 N
C x = 1600 N
ΣFy = 0: C y − Ty − 240 N − 240 N = 0
C y − 1200 N − 480 N = 0
C y = +1680 N
C y = 1680 N
α = 46.4°
C = 2320 N
C = 2.32 kN
46.4°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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363
PROBLEM 4.20
Solve Problem 4.19, assuming that a = 0.32 m.
PROBLEM 4.19 The bracket BCD is hinged at C and
attached to a control cable at B. For the loading shown,
determine (a) the tension in the cable, (b) the reaction at C.
SOLUTION
At B:
Ty
Tx
=
0.32 m
0.24 m
4
Ty = Tx
3
ΣM C = 0: Tx (0.32 m) − (240 N)(0.4 m) − (240 N)(0.8 m) = 0
Tx = 900 N
Eq. (1)
Ty =
4
(900 N) = 1200 N
3
T = Tx2 + Ty2 = 9002 + 12002 = 1500 N
T = 1.500 kN
ΣFx = 0: C x − Tx = 0
C x − 900 N = 0 C x = +900 N
C x = 900 N
ΣFy = 0: C y − Ty − 240 N − 240 N = 0
C y − 1200 N − 480 N = 0
C y = +1680 N
C y = 1680 N
α = 61.8°
C = 1906 N
C = 1.906 kN
61.8°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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364
PROBLEM 4.21
Determine the reactions at A and B when (a) h = 0,
(b) h = 200 mm.
SOLUTION
Free-Body Diagram:
ΣM A = 0: ( B cos 60°)(0.5 m) − ( B sin 60°)h − (150 N)(0.25 m) = 0
37.5
B=
0.25 − 0.866h
(a)
(1)
When h = 0:
B=
Eq. (1):
37.5
= 150 N
0.25
B = 150.0 N
30.0°
ΣFy = 0: Ax − B sin 60° = 0
Ax = (150)sin 60° = 129.9 N
A x = 129.9 N
ΣFy = 0: Ay − 150 + B cos 60° = 0
Ay = 150 − (150) cos 60° = 75 N
A y = 75 N
α = 30°
A = 150.0 N
(b)
A = 150.0 N
30.0°
When h = 200 mm = 0.2 m
Eq. (1):
B=
37.5
= 488.3 N
0.25 − 0.866(0.2)
B = 488 N
30.0°
ΣFx = 0: Ax − B sin 60° = 0
Ax = (488.3) sin 60° = 422.88 N
A x = 422.88 N
ΣFy = 0: Ay − 150 + B cos 60° = 0
Ay = 150 − (488.3) cos 60° = −94.15 N
A y = 94.15 N
α = 12.55°
A = 433.2 N
A = 433 N
12.6°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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365
PROBLEM 4.22
For the frame and loading shown, determine the reactions at A and E
when (a) α = 30°, (b) α = 45°.
SOLUTION
Free-Body Diagram:
ΣM A = 0: ( E sin α )(8 in.) + ( E cos α )(5 in.)
− (20 lb)(10 in.) − (20 lb)(3 in.) = 0
(a)
When α = 30°:
E=
260
8sin α + 5cos α
E=
260
= 31.212 lb
8sin 30° + 5cos 30°
E = 31.2 lb
60.0°
ΣFx = 0: Ax − 20 lb + (31.212 lb) sin 30° = 0
Ax = +4.394 lb
A x = 4.394 lb
ΣFy = 0: Ay − 20° + (31.212 lb) cos 30° = 0
Ay = −7.03 lb
A y = 7.03 lb
A = 8.29 lb
58.0°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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366
PROBLEM 4.22 (Continued)
(b)
When α = 45° :
E=
260
= 28.28 lb
8sin 45° + 5cos α
E = 28.3 lb
45.0°
ΣFx = 0: Ax − 20 lb + (28.28 lb)sin 45° = 0
Ax = 0
Ax = 0
ΣFy = 0: Ay − 20 lb + (28.28 lb) cos 45° = 0
Ay = 0
Ay = 0
A=0
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367
PROBLEM 4.23
For each of the plates and loadings shown, determine the reactions at A and B.
SOLUTION
(a)
Free-Body Diagram:
ΣM A = 0: B(20 in.) − (50 lb)(4 in.) − (40 lb)(10 in.) = 0
B = +30 lb
B = 30.0 lb
ΣFx = 0: Ax + 40 lb = 0
Ax = −40 lb
A x = 40.0 lb
ΣFy = 0: Ay + B − 50 lb = 0
Ay + 30 lb − 50 lb = 0
Ay = +20 lb
A y = 20.0 lb
α = 26.56°
A = 44.72 lb
A = 44.7 lb
26.6°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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368
PROBLEM 4.23 (Continued)
(b)
Free-Body Diagram:
ΣM A = 0: ( B cos 30°)(20 in.) − (40 lb)(10 in.) − (50 lb)(4 in.) = 0
B = 34.64 lb
B = 34.6 lb
60.0°
ΣFx = 0: Ax − B sin 30° + 40 lb
Ax − (34.64 lb) sin 30° + 40 lb = 0
Ax = −22.68 lb
A x = 22.68 lb
ΣFy = 0: Ay + B cos 30° − 50 lb = 0
Ay + (34.64 lb) cos 30° − 50 lb = 0
Ay = +20 lb
A y = 20.0 lb
α = 41.4°
A = 30.24 lb
A = 30.2 lb
41.4°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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369
PROBLEM 4.24
For each of the plates and loadings shown, determine the reactions at A and B.
SOLUTION
(a)
Free-Body Diagram:
ΣM B = 0: A(20 in.) + (50 lb)(16 in.) − (40 lb)(10 in.) = 0
A = +20 lb
A = 20.0 lb
ΣFx = 0: 40 lb + Bx = 0
Bx = −40 lb
B x = 40 lb
ΣFy = 0: A + By − 50 lb = 0
20 lb + By − 50 lb = 0
By = +30 lb
α = 36.87°
B = 50 lb
B y = 30 lb
B = 50.0 lb
36.9°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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370
PROBLEM 4.24 (Continued)
(b)
ΣM A = 0: − ( A cos 30°)(20 in.) − (40 lb)(10 in.) + (50 lb)(16 in.) = 0
A = 23.09 lb
A = 23.1 lb
60.0°
ΣFx = 0: A sin 30° + 40 lb + Bx = 0
(23.09 lb) sin 30° + 40 lb + 8 x = 0
Bx = −51.55 lb
B x = 51.55 lb
ΣFy = 0: A cos 30° + By − 50 lb = 0
(23.09 lb) cos 30° + By − 50 lb = 0
By = +30 lb
B y = 30 lb
α = 30.2°
B = 59.64 lb
B = 59.6 lb
30.2°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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371
PROBLEM 4.25
Determine the reactions at A and B when (a) α = 0, (b) α = 90°,
(c) α = 30°.
SOLUTION
(a)
α =0
ΣM A = 0: B(20 in.) − 750 lb ⋅ in. = 0
B = 37.5 lb
ΣFx = 0: Ax = 0
ΣFy = 0: Ay + 37.5 lb = 0
Ay = −37.5 lb
B = 37.5 lb
A = 37.5 lb
(b)
α = 90°
ΣM A = 0: B (12 in.) − 750 lb ⋅ in. = 0
B = 62.5 lb
ΣFA = 0: Ax − 62.5 lb = 0,
Ax = 62.5 lb
ΣFy = 0: Ay = 0
A = 62.5 lb
(c)
B = 62.5 lb
α = 30°
ΣM A = 0: ( B cos 30°)(20 in.) + ( B sin 30°)(12 in.) − 750 lb ⋅ in. = 0
B = 32.16 lb
ΣFx = 0: Ax − (32.16 lb) sin 30° = 0
Ax = 16.08 lb
ΣFy = 0: Ay + (32.16 lb) cos 30° = 0
Ay = −27.85 lb
A = 32.16 lb α = 60.0°
A = 32.2 lb
60.0°
B = 32.2 lb
60.0°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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372
PROBLEM 4.26
A rod AB hinged at A and attached at B to cable BD supports
the loads shown. Knowing that d = 200 mm, determine
(a) the tension in cable BD, (b) the reaction at A.
SOLUTION
Free-Body Diagram:
(a)
Move T along BD until it acts at Point D.
ΣM A = 0: (T sin 45°)(0.2 m) + (90 N)(0.1 m) + (90 N)(0.2 m) = 0
T = 190.92 N
(b)
T = 190.9 N
ΣFx = 0: Ax − (190.92 N) cos 45° = 0
Ax = +135 N
A x = 135.0 N
ΣFy = 0: Ay − 90 N − 90 N + (190.92 N) sin 45° = 0
Ay = +45 N
A y = 45.0 N
A = 142.3 N
18.43°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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373
PROBLEM 4.27
A rod AB hinged at A and attached at B to cable BD supports
the loads shown. Knowing that d = 150 mm, determine (a) the
tension in cable BD, (b) the reaction at A.
SOLUTION
Free-Body Diagram:
tan α =
(a)
10
; α = 33.69°
15
Move T along BD until it acts at Point D.
ΣM A = 0: (T sin 33.69°)(0.15 m) − (90 N)(0.1 m) − (90 N)(0.2 m) = 0
T = 324.5 N
(b)
T = 324 N
ΣFx = 0: Ax − (324.99 N) cos 33.69° = 0
Ax = +270 N
A x = 270 N
ΣFy = 0: Ay − 90 N − 90 N + (324.5 N)sin 33.69° = 0
Ay = 0
Ay = 0
A = 270 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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374
PROBLEM 4.28
A lever AB is hinged at C and attached to a control cable at A. If
the lever is subjected to a 500-N horizontal force at B, determine
(a) the tension in the cable, (b) the reaction at C.
SOLUTION
Triangle ACD is isosceles with
C = 90° + 30° = 120°
A=
D=
1
(180° − 120°) = 30°
2
Thus DA forms angle of 60° with horizontal.
(a)
We resolve FAD into components along AB and perpendicular to AB.
ΣM C = 0: ( FAD sin 30°)(250 mm) − (500 N)(100 mm) = 0
(b)
FAD = 400 N
ΣFx = 0: − (400 N) cos 60° + C x − 500 N = 0
C x = +300 N
ΣFy = 0: − (400 N) sin 60° + C y = 0
C y = +346.4 N
C = 458 N
49.1°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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375
PROBLEM 4.29
A force P of magnitude 280 lb is applied to member ABCD, which is
supported by a frictionless pin at A and by the cable CED. Since the cable
passes over a small pulley at E, the tension may be assumed to be the
same in portions CE and ED of the cable. For the case when a = 3 in.,
determine (a) the tension in the cable, (b) the reaction at A.
SOLUTION
Free-Body Diagram:
(a)
ΣM A = 0: − (280 lb)(8 in.)
7
T (12 in.)
25
24
− T (8 in.) = 0
25
T (12 in.) −
(12 − 11.04)T = 840
(b)
ΣFx = 0:
T = 875 lb
7
(875 lb) + 875 lb + Ax = 0
25
Ax = −1120
ΣFy = 0: Ay − 280 lb −
Ay = +1120
A x = 1120 lb
24
(875 lb) = 0
25
A y = 1120 lb
A = 1584 lb
45.0°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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376
PROBLEM 4.30
Neglecting friction, determine the tension in cable ABD and the reaction
at support C.
SOLUTION
Free-Body Diagram:
ΣM C = 0: T (0.25 m) − T (0.1 m) − (120 N)(0.1 m) = 0
T = 80.0 N
ΣFx = 0: C x − 80 N = 0 C x = +80 N
C x = 80.0 N
ΣFy = 0: C y − 120 N + 80 N = 0 C y = +40 N
C y = 40.0 N
C = 89.4 N
26.6°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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377
PROBLEM 4.31
Rod ABC is bent in the shape of an arc of circle of radius R. Knowing
the θ = 30°, determine the reaction (a) at B, (b) at C.
SOLUTION
Free-Body Diagram:
ΣM D = 0: C x ( R) − P( R) = 0
Cx = + P
ΣFx = 0: C x − B sin θ = 0
P − B sin θ = 0
B = P/sin θ
B=
P
sin θ
θ
ΣFy = 0: C y + B cos θ − P = 0
C y + ( P/sin θ ) cos θ − P = 0
1 !
Cy = P 1 −
!
tan θ #
"
For θ = 30°:
(a)
(b)
B = P/sin 30° = 2 P
B = 2P
60.0°
Cx = + P Cx = P
C y = P(1 − 1/tan 30°) = − 0.732/P
C y = 0.7321P
C = 1.239P
36.2°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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378
!
PROBLEM 4.32
Rod ABC is bent in the shape of an arc of circle of radius R. Knowing the
θ = 60°, determine the reaction (a) at B, (b) at C.
SOLUTION
See the solution to Problem 4.31 for the free-body diagram and analysis leading to the following expressions:
Cx = + P
1 %
$
Cy = P 1 −
tan θ !#
"
P
B=
sin θ
For θ = 60°:
(a)
(b)
B = P/sin 60° = 1.1547 P
B = 1.155P
30.0°
Cx = + P Cx = P
C y = P(1 − 1/tan 60°) = + 0.4226 P
C y = 0.4226 P
C = 1.086P
22.9°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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379
!
PROBLEM 4.33
Neglecting friction, determine the tension in cable ABD and the reaction
at C when θ = 60°.
SOLUTION
ΣM C = 0: T (2a + a cos θ ) − Ta + Pa = 0
T=
P
1 + cos θ
(1)
ΣFx = 0: C x − T sin θ = 0
C x = T sin θ =
P sin θ
1 + cos θ
ΣFy = 0: C y + T + T cos θ − P = 0
C y = P − T (1 + cos θ ) = P − P
1 + cos θ
1 + cos θ
Cy = 0
C y = 0, C = C x
Since
C=P
sin θ
1 + cos θ
(2)
For θ = 60°:
P
P
=
1 + cos 60° 1 + 12
Eq. (1):
T=
Eq. (2):
C=P
sin 60°
0.866
=P
1 + cos 60°
1 + 12
T=
2
P
3
C = 0.577P
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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380
!
PROBLEM 4.34
Neglecting friction, determine the tension in cable ABD and the reaction
at C when θ = 45°.
SOLUTION
Free-Body Diagram:
Equilibrium for bracket:
ΣM C = 0: − T (a ) − P(a ) + (T sin 45°)(2a sin 45°)
+ (T cos 45°)(a + 2a cos 45°) = 0
T = 0.58579
or T = 0.586 P
ΣFx = 0: C x + (0.58579 P) sin 45° = 0
C x = 0.41422 P
ΣFy = 0: C y + 0.58579 P − P + (0.58579 P) cos 45° = 0
Cy = 0
or C = 0.414P
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381
!
PROBLEM 4.35
A light rod AD is supported by frictionless pegs at B and C and rests
against a frictionless wall at A. A vertical 120-lb force is applied
at D. Determine the reactions at A, B, and C.
SOLUTION
Free-Body Diagram:
ΣFx = 0: A cos 30° − (120 lb) cos 60° = 0
A = 69.28 lb
A = 69.3 lb
ΣM B = 0: C (8 in.) − (120 lb)(16 in.) cos 30°
+ (69.28 lb)(8 in.)sin 30° = 0
C = 173.2 lb
C = 173.2 lb
60.0°
B = 34.6 lb
60.0°
ΣM C = 0: B(8 in.) − (120 lb)(8 in.) cos 30°
+ (69.28 lb)(16 in.) sin 30° = 0
B = 34.6 lb
Check:
ΣFy = 0: 173.2 − 34.6 − (69.28)sin 30° − (120)sin 60° = 0
0 = 0(check) !
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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382
PROBLEM 4.36
A light bar AD is suspended from a cable BE and supports a 50-lb
block at C. The ends A and D of the bar are in contact with
frictionless vertical walls. Determine the tension in cable BE and
the reactions at A and D.
SOLUTION
Free-Body Diagram:
ΣFx = 0:
A= D
ΣFy = 0:
TBE = 50.0 lb
We note that the forces shown form two couples.
ΣM = 0: A(8 in.) − (50 lb)(3 in.) = 0
A = 18.75 lb
A = 18.75 lb
D = 18.75 lb
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383
!
PROBLEM 4.37
Bar AC supports two 400-N loads as shown. Rollers at A and C
rest against frictionless surfaces and a cable BD is attached at B.
Determine (a) the tension in cable BD, (b) the reaction at A, (c) the
reaction at C.
SOLUTION
Similar triangles: ABE and ACD
AE BE
=
;
AD CD
(a)
ΣM A = 0: Tx (0.25 m) −
BE
0.15 m
=
; BE = 0.075 m
0.5 m 0.25 m
$ 0.075 %
Tx ! (0.5 m) − (400 N)(0.1 m) − (400 N)(0.4 m) = 0
" 0.35 #
Tx = 1400 N
0.075
(1400 N)
0.35
= 300 N
Ty =
T = 1432 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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384
PROBLEM 4.37 (Continued)
(b)
ΣFy = 0: A − 300 N − 400 N − 400 N = 0
A = + 1100 N
(c)
A = 1100 N
ΣFx = 0: −C + 1400 N = 0
C = + 1400 N
C = 1400 N
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385
!
PROBLEM 4.38
Determine the tension in each cable and the reaction at D.
SOLUTION
tan α =
0.08 m
0.2 m
α = 21.80°
tan β =
0.08 m
0.1 m
β = 38.66°
ΣM B = 0: (600 N)(0.1 m) − (TCF sin 38.66°)(0.1 m) = 0
TCF = 960.47 N
TCF = 96.0 N
ΣM C = 0: (600 N)(0.2 m) − (TBE sin 21.80°)(0.1 m) = 0
TBE = 3231.1 N
TBE = 3230 N
ΣFy = 0: TBE cos α + TCF cos β − D = 0
(3231.1 N) cos 21.80° + (960.47 N) cos 38.66° − D = 0
D = 3750.03 N
D = 3750 N
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386
!
PROBLEM 4.39
Two slots have been cut in plate DEF, and the plate has
been placed so that the slots fit two fixed, frictionless pins A
and B. Knowing that P = 15 lb, determine (a) the force each
pin exerts on the plate, (b) the reaction at F.
SOLUTION
Free-Body Diagram:
ΣFx = 0: 15 lb − B sin 30° = 0
B = 30.0 lb
60.0°
ΣM A = 0: − (30 lb)(4 in.) + B sin 30°(3 in.) + B cos 30°(11 in.) − F (13 in.) = 0
−120 lb ⋅ in. + (30 lb) sin 30°(3 in.) + (30 lb) cos 30°(11 in.) − F (13 in.) = 0
F = + 16.2145 lb
F = 16.21 lb
ΣFy = 0: A − 30 lb + B cos 30° − F = 0
A − 30 lb + (30 lb) cos 30° − 16.2145 lb = 0
A = + 20.23 lb
A = 20.2 lb
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387
!
PROBLEM 4.40
For the plate of Problem 4.39 the reaction at F must be
directed downward, and its maximum allowable value is
20 lb. Neglecting friction at the pins, determine the required
range of values of P.
PROBLEM 4.39 Two slots have been cut in plate DEF,
and the plate has been placed so that the slots fit two fixed,
frictionless pins A and B. Knowing that P = 15 lb, determine
(a) the force each pin exerts on the plate, (b) the reaction
at F.
SOLUTION
Free-Body Diagram:
ΣFx = 0: P − B sin 30° = 0
B = 2P
60°
ΣM A = 0: − (30 lb)(4 in.) + B sin 30°(3 in.) + B cos 30°(11 in.) − F (13 in.) = 0
−120 lb ⋅ in.+ 2 P sin 30°(3 in.) + 2 P cos 30°(11 in.) − F (13 in.) = 0
−120 + 3P + 19.0525P − 13F = 0
P=
13E + 120
22.0525
For F = 0:
P=
13(0) + 120
= 5.442 lb
22.0525
For P = 20 lb:
P=
13(20) + 120
= 17.232 lb
22.0525
For
0 # F # 20 lb:
(1)
5.44 lb # P # 17.231 lb
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388
!
PROBLEM 4.41
Bar AD is attached at A and C to collars that can move freely
on the rods shown. If the cord BE is vertical (α = 0), determine
the tension in the cord and the reactions at A and C.
SOLUTION
Free-Body Diagram:
ΣFy = 0: − T cos 30° + (80 N) cos 30° = 0
T = 80 N
T = 80.0 N
ΣM C = 0: ( A sin 30°)(0.4 m) − (80 N)(0.2 m) − (80 N)(0.2 m) = 0
A = + 160 N
A = 160.0 N
30.0°
C = 160.0 N
30.0°
ΣM A = 0: (80 N)(0.2 m) − (80 N)(0.6 m) + (C sin 30°)(0.4 m) = 0
C = + 160 N
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389
!
PROBLEM 4.42
Solve Problem 4.41 if the cord BE is parallel to the rods
(α = 30°).
PROBLEM 4.41 Bar AD is attached at A and C to
collars that can move freely on the rods shown. If the
cord BE is vertical (α = 0), determine the tension in the
cord and the reactions at A and C.
SOLUTION
Free-Body Diagram:
ΣFy = 0: − T + (80 N) cos 30° = 0
T = 69.282 N
T = 69.3 N
ΣM C = 0: − (69.282 N) cos 30°(0.2 m)
− (80 N)(0.2 m) + ( A sin 30°)(0.4 m) = 0
A = + 140.000 N
A = 140.0 N
30.0°
C = 180.0 N
30.0°
ΣM A = 0: + (69.282 N) cos 30°(0.2 m)
− (80 N)(0.6 m) + (C sin 30°)(0.4 m) = 0
C = + 180.000 N
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390
!
PROBLEM 4.43
An 8-kg mass can be supported in the three different ways shown. Knowing that the pulleys have a 100-mm
radius, determine the reaction at A in each case.
SOLUTION
!
W = mg = (8 kg)(9.81m/s 2 ) = 78.48 N
!
(a)
ΣFx = 0: Ax = 0
ΣFy = 0: Ay − W = 0
!
A y = 78.48 N
ΣM A = 0: M A − W (1.6 m) = 0
!
M A = + (78.48 N)(1.6 m)
!
M A = 125.56 N ⋅ m
A = 78.5 N
!
(b)
!
M A = 125.6 N ⋅ m
ΣFx = 0: Ax − W = 0
A x = 78.48
ΣFy = 0: Ay − W = 0
A y = 78.48
A = (78.48 N) 2 = 110.99 N
!
45°
ΣM A = 0: M A − W (1.6 m) = 0
!
M A = + (78.48 N)(1.6 m)
!
!
A = 111.0 N
!
(c)
45°
M A = 125.56 N ⋅ m
M A = 125.6 N ⋅ m
ΣFx = 0: Ax = 0
ΣFy = 0: Ay − 2W = 0
!
!
Ay = 2W = 2(78.48 N) = 156.96 N
ΣM A = 0: M A − 2W (1.6 m) = 0
M A = + 2(78.48 N)(1.6 m)
A = 157.0 N
M A = 125.1 N ⋅ m
M A = 125 N ⋅ m
!
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391
PROBLEM 4.44
A tension of 5 lb is maintained in a tape as it passes through
the support system shown. Knowing that the radius of each
pulley is 0.4 in., determine the reaction at C.
SOLUTION
From f.b.d. of system
ΣFx = 0: C x + (5 lb) = 0
Cx = −5 lb
ΣFy = 0: C y − (5 lb) = 0
C y = 5 lb
Then
C = ( Cx ) 2 + (C y ) 2
= (5) 2 + (5) 2
= 7.0711 lb
and
θ = tan −1
$ +5 %
! = −45°
" −5 #
or C = 7.07 lb
45.0°
ΣM C = 0: M C + (5 lb)(6.4 in.) + (5 lb)(2.2 in.) = 0
M C = − 43.0 lb ⋅ in
or
M C = 43.0 lb ⋅ in.
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392
!
PROBLEM 4.45
Solve Problem 4.44, assuming that 0.6-in.-radius pulleys
are used.
PROBLEM 4.44 A tension of 5 lb is maintained in a tape
as it passes through the support system shown. Knowing
that the radius of each pulley is 0.4 in., determine the
reaction at C.
SOLUTION
From f.b.d. of system
ΣFx = 0: C x + (5 lb) = 0
Cx = −5 lb
ΣFy = 0: C y − (5 lb) = 0
C y = 5 lb
Then
C = (Cx ) 2 + (C y ) 2
= (5) 2 + (5) 2
= 7.0711 lb
and
θ = tan −1
$ 5 %
! = −45.0°
" −5 #
or C = 7.07 lb
45.0°
ΣM C = 0: M C + (5 lb)(6.6 in.) + (5 lb)(2.4 in.) = 0
M C = − 45.0 lb ⋅ in.
or
M C = 45.0 lb ⋅ in.
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393
!
PROBLEM 4.46
A 6-m telephone pole weighing 1600 N is used to support the ends of
two wires. The wires form the angles shown with the horizontal and
the tensions in the wires are, respectively, T1 = 600 N and T2 = 375 N.
Determine the reaction at the fixed end A.
SOLUTION
Free-Body Diagram:
ΣFx = 0: Ax + (375 N) cos 20° − (600 N) cos10° = 0
Ax = +238.50 N
ΣFy = 0: Ay − 1600 N − (600 N)sin10° − (375 N) sin 20° = 0
Ay = +1832.45 N
A = 238.502 + 1832.452
1832.45
θ = tan −1
238.50
A = 1848 N
82.6°
ΣM A = 0: M A + (600 N) cos10°(6 m) − (375 N) cos 20°(6 m) = 0
M A = −1431.00 N ⋅ m
M A = 1431 N ⋅ m
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394
PROBLEM 4.47
Beam AD carries the two 40-lb loads shown. The beam is held by a
fixed support at D and by the cable BE that is attached to the
counterweight W. Determine the reaction at D when (a) W = 100 lb,
(b) W = 90 lb.
SOLUTION
W = 100 lb
(a)
From f.b.d. of beam AD
ΣFx = 0: Dx = 0
ΣFy = 0: D y − 40 lb − 40 lb + 100 lb = 0
Dy = −20.0 lb
or
D = 20.0 lb
ΣM D = 0: M D − (100 lb)(5 ft) + (40 lb)(8 ft)
+ (40 lb)(4 ft) = 0
M D = 20.0 lb ⋅ ft
or M D = 20.0 lb ⋅ ft
W = 90 lb
(b)
From f.b.d. of beam AD
ΣFx = 0: Dx = 0
ΣFy = 0: D y + 90 lb − 40 lb − 40 lb = 0
Dy = −10.00 lb
or
D = 10.00 lb
ΣM D = 0: M D − (90 lb)(5 ft) + (40 lb)(8 ft)
+ (40 lb)(4 ft) = 0
M D = −30.0 lb ⋅ ft
or M D = −30.0 lb ⋅ ft
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395
PROBLEM 4.48
For the beam and loading shown, determine the range of values of W
for which the magnitude of the couple at D does not exceed 40 lb ⋅ ft.
SOLUTION
For Wmin ,
From f.b.d. of beam AD
M D = − 40 lb ⋅ ft
ΣM D = 0: (40 lb)(8 ft) − Wmin (5 ft)
+ (40 lb)(4 ft) − 40 lb ⋅ ft = 0
Wmin = 88.0 lb
For Wmax ,
From f.b.d. of beam AD
M D = 40 lb ⋅ ft
ΣM D = 0: (40 lb)(8 ft) − Wmax (5 ft)
+ (40 lb)(4 ft) + 40 lb ⋅ ft = 0
Wmax = 104.0 lb
or 88.0 lb # W # 104.0 lb
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396
PROBLEM 4.49
Knowing that the tension in wire BD is 1300 N, determine the
reaction at the fixed support C of the frame shown.
SOLUTION
T = 1300 N
5
Tx = T
13
= 500 N
12
Ty = T
13
= 1200 N
ΣM x = 0: C x − 450 N + 500 N = 0
C x = −50 N
ΣFy = 0: C y − 750 N − 1200 N = 0 C y = +1950 N
C x = 50 N
C y = 1950 N
C = 1951 N
88.5°
ΣM C = 0: M C + (750 N)(0.5 m) + (4.50 N)(0.4 m)
− (1200 N)(0.4 m) = 0
M C = −75.0 N ⋅ m
M C = 75.0 N ⋅ m
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397
PROBLEM 4.50
Determine the range of allowable values of the tension in wire
BD if the magnitude of the couple at the fixed support C is not to
exceed 100 N · m.
SOLUTION
ΣM C = 0: (750 N)(0.5 m) + (450 N)(0.4 m) −
−
$ 5 %
T ! (0.6 m)
" 13 #
$ 12 %
T ! (0.15 m) + M C = 0
" 13 #
375 N ⋅ m + 180 N ⋅ m −
T=
$ 4.8 %
m !T + M C = 0
" 13 #
13
(555 + M C )
4.8
For
M C = −100 N ⋅ m: T =
13
(555 − 100) = 1232 N
4.8
For
M C = +100 N ⋅ m: T =
13
(555 + 100) = 1774 N
4.8
For
|M C | # 100 N ⋅ m:
1.232 kN # T # 1.774 kN
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398
PROBLEM 4.51
A vertical load P is applied at end B of rod BC. (a) Neglecting the weight of the
rod, express the angle θ corresponding to the equilibrium position in terms of
P, l, and the counterweight W. (b) Determine the value of θ corresponding to
equilibrium if P = 2W .
SOLUTION
(a)
Triangle ABC is isosceles.
We have
CD = ( BC ) cos
θ
2
= l cos
θ
2
θ%
$
ΣM C = 0: W l cos ! − P(l sin θ ) = 0
2#
"
Setting
sin θ = 2sin
θ
θ
θ
W − 2 P sin
(b)
For P = 2W :
sin
θ
2
θ
2
or
θ
θ
cos : Wl cos − 2 Pl sin cos = 0
2
2
2
2
2
θ
2
=
θ
2
=0
θ = 2sin −1 $
W %
!
2
" P#
W
W
=
= 0.25
2 P 4W
θ = 29.0°
= 14.5°
= 165.5° θ = 331°(discard)
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399
PROBLEM 4.52
A slender rod AB, of weight W, is attached to blocks A and B,
which move freely in the guides shown. The blocks are connected
by an elastic cord that passes over a pulley at C. (a) Express the
tension in the cord in terms of W and θ. (b) Determine the value of
θ for which the tension in the cord is equal to 3W.
SOLUTION
(a)
From f.b.d. of rod AB
&$ 1 %
'
ΣM C = 0: T (l sin θ ) + W ( ! cos θ ) − T (l cos θ ) = 0
2
*" #
+
T=
W cos θ
2(cos θ − sin θ )
Dividing both numerator and denominator by cos θ,
T=
(b)
For T = 3W ,
or
3W =
W$
1
%
2 " 1 − tan θ !#
or T =
( W2 )
(1 − tan θ )
( W2 )
(1 − tan θ )
1
1 − tan θ =
6
θ = tan −1
$5%
! = 39.806°
"6#
or
θ = 39.8°
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400
PROBLEM 4.53
Rod AB is acted upon by a couple M and two forces, each of magnitude P.
(a) Derive an equation in θ, P, M, and l that must be satisfied when the rod
is in equilibrium. (b) Determine the value of θ corresponding to equilibrium
when M = 150 N · m, P = 200 N, and l = 600 mm.
SOLUTION
Free-Body Diagram:
(a)
From free-body diagram of rod AB
ΣM C = 0: P(l cos θ ) + P(l sin θ ) − M = 0
or sinθ + cosθ =
(b)
For
M
Pl
M = 150 lb ⋅ in., P = 20 lb, and l = 6 in.
150 lb ⋅ in.
5
sin θ + cos θ =
= = 1.25
(20 lb)(6 in.) 4
sin 2 θ + cos 2 θ = 1
Using identity
sin θ + (1 − sin 2 θ )1/2 = 1.25
(1 − sin 2 θ )1/2 = 1.25 − sin θ
1 − sin 2 θ = 1.5625 − 2.5sin θ + sin 2 θ
2sin 2 θ − 2.5sin θ + 0.5625 = 0
Using quadratic formula
sin θ =
=
or
−( −2.5) ± (625) − 4(2)(0.5625)
2(2)
2.5 ± 1.75
4
sin θ = 0.95572 and sin θ = 0.29428
θ = 72.886° and θ = 17.1144°
or θ = 17.11° and θ = 72.9°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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401
PROBLEM 4.54
Rod AB is attached to a collar at A and rests against a small roller at C.
(a) Neglecting the weight of rod AB, derive an equation in P, Q, a, l, and
θ that must be satisfied when the rod is in equilibrium. (b) Determine
the value of θ corresponding to equilibrium when P = 16 lb, Q = 12 lb,
l = 20 in., and a = 5 in.
SOLUTION
Free-Body Diagram:
ΣFy = 0: C cos θ − P − Q = 0
C=
ΣM A = 0: C
(a)
P+Q
cos θ
a
− Pl cos θ = 0
cos θ
P+Q
a
⋅
− Pl cos θ = 0
cos θ cos θ
(b)
For
cos3 θ =
a( P + Q)
Pl
P = 16 lb, Q = 12 lb, l = 20 in., and a = 5 in.
(5 in.)(16 lb + 12 lb)
(16 lb)(20 in.)
= 0.4375
cos3 θ =
cos θ = 0.75915
θ = 40.6°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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402
PROBLEM 4.55
A collar B of weight W can move freely along the vertical rod shown.
The constant of the spring is k, and the spring is unstretched when
θ = 0. (a) Derive an equation in θ, W, k, and l that must be satisfied
when the collar is in equilibrium. (b) Knowing that W = 300 N,
l = 500 mm, and k = 800 N/m, determine the value of θ corresponding
to equilibrium.
SOLUTION
First note
T = ks
Where
k = spring constant
s = elongation of spring
l
=
−l
cos θ
l
(1 − cos θ )
=
cos θ
kl
T=
(1 − cos θ )
cos θ
(a)
From f.b.d. of collar B
or
(b)
For
ΣFy = 0: T sin θ − W = 0
kl
(1 − cos θ )sin θ − W = 0
cos θ
or tan θ − sin θ =
W
kl
W = 3 lb
l = 6 in.
k = 8 lb/ft
6 in.
= 0.5 ft
l=
12 in./ft
tan θ − sin θ =
Solving numerically,
3 lb
= 0.75
(8 lb/ft)(0.5 ft)
θ = 57.957°
or
θ = 58.0°
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403
PROBLEM 4.56
A vertical load P is applied at end B of rod BC. The constant of the
spring is k, and the spring is unstretched when θ = 90°. (a) Neglecting
the weight of the rod, express the angle θ corresponding to equilibrium
in terms of P, k, and l. (b) Determine the value of θ corresponding to
equilibrium when P = 14 kl.
SOLUTION
First note
T = tension in spring = ks
where
s = elongation of spring
= ( AB)θ − ( AB)θ = 90°
$θ %
$ 90° %
! − 2l sin 2 !
2
" #
"
#
& $ θ % $ 1 %'
= 2l (sin ! −
!)
* " 2 # " 2 #+
= 2l sin
& $ θ % $ 1 %'
T = 2kl (sin
!)
!−
* " 2 # " 2 #+
(a)
From f.b.d. of rod BC
(1)
&
$ θ %'
ΣM C = 0: T (l cos ! ) − P(l sin θ ) = 0
" 2 #+
*
Substituting T from Equation (1)
& $ θ % $ 1 %' &
$ θ %'
2kl (sin ! −
! ) (l cos ! ) − P ( l sin θ ) = 0
" 2 #+
* " 2 # " 2 #+ *
& $ θ % $ 1 %'
&
$θ %
$θ %
$ θ %'
2kl 2 (sin ! −
! ) cos ! − Pl ( 2sin ! cos ! ) = 0
2
2
2
" #
" #
" 2 #+
*
* " # " 2 #+
Factoring out
2l cos
$θ %
! , leaves
"2#
& $ θ % $ 1 %'
$θ %
kl (sin
! ) − P sin ! = 0
!−
2
2
"2#
#+
* " # "
or
sin
$ θ % 1 $ kl %
!=
!
2 " kl − P #
"2#
&
'
)
* 2(kl − P) +
θ = 2sin −1 (
kl
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404
PROBLEM 4.56 (Continued)
(b)
P=
kl
4
&
kl
'
kl )
kl
−
2
(* (
4 ))
+
&
'
kl $ 4 %
= 2sin −1 (
!)
* 2 " 3kl # +
θ = 2sin −1 (
= 2sin −1
$ 4 %
!
"3 2 #
= 2sin −1 (0.94281)
= 141.058°
or θ = 141.1°
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405
PROBLEM 4.57
Solve Sample Problem 4.56, assuming that the spring is unstretched
when θ = 90°.
SOLUTION
First note
T = tension in spring = ks
where
s = deformation of spring
= rβ
F = kr β
From f.b.d. of assembly
or
ΣM 0 = 0: W (l cos β ) − F (r ) = 0
Wl cos β − kr 2 β = 0
cos β =
For
kr 2
β
Wl
k = 250 lb/in.
r = 3 in.
l = 8 in.
W = 400 lb
cos β =
or
(250 lb/in.)(3 in.)2
β
(400 lb)(8 in.)
cos β = 0.703125β
Solving numerically,
β = 0.89245 rad
or
β = 51.134°
Then
θ = 90° + 51.134° = 141.134°
or θ = 141.1°
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406
PROBLEM 4.58
A slender rod AB, of weight W, is attached to blocks A and B that
move freely in the guides shown. The constant of the spring is k,
and the spring is unstretched when θ = 0. (a) Neglecting the weight
of the blocks, derive an equation in W, k, l, and θ that must be
satisfied when the rod is in equilibrium. (b) Determine the value
of θ when W = 75 lb, l = 30 in., and k = 3 lb/in.
SOLUTION
Free-Body Diagram:
Spring force:
Fs = ks = k (l − l cos θ ) = kl (1 − cos θ )
ΣM D = 0: Fs (l sin θ ) − W
(a)
$l
%
cos θ ! = 0
"2
#
kl (1 − cos θ )l sin θ −
kl (1 − cos θ ) tan θ −
(b)
For given values of
W
l cos θ = 0
2
W
=0
2
or (1 − cos θ ) tan θ =
W
2kl
W = 75 lb
l = 30 in.
k = 3 lb/in.
(1 − cos θ ) tan θ = tan θ − sin θ
75 lb
=
2(3 lb/in.)(30 in.)
= 0.41667
Solving numerically:
θ = 49.710°
or
θ = 49.7°
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407
PROBLEM 4.59
Eight identical 500 × 750-mm rectangular plates, each of mass m = 40 kg, are held in a vertical plane as
shown. All connections consist of frictionless pins, rollers, or short links. In each case, determine whether
(a) the plate is completely, partially, or improperly constrained, (b) the reactions are statically determinate or
indeterminate, (c) the equilibrium of the plate is maintained in the position shown. Also, wherever possible,
compute the reactions.
SOLUTION
1.
Three non-concurrent, non-parallel reactions
(a)
Plate: completely constrained
(b)
Reactions: determinate
(c)
Equilibrium maintained
A = C = 196.2 N
2.
Three non-concurrent, non-parallel reactions
(a)
Plate: completely constrained
(b)
Reactions: determinate
(c)
Equilibrium maintained
B = 0, C = D = 196.2 N
3.
Four non-concurrent, non-parallel reactions
(a)
Plate: completely constrained
(b)
Reactions: indeterminate
(c)
Equilibrium maintained
A x = 294 N
,
D x = 294 N
( A y + D y = 392 N )
4.
Three concurrent reactions (through D)
(a)
Plate: improperly constrained
(b)
Reactions: indeterminate
(c)
No equilibrium
(ΣM D ≠ 0)
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408
PROBLEM 4.59 (Continued)
5.
Two reactions
(a)
Plate: partial constraint
(b)
Reactions: determinate
(c)
Equilibrium maintained
C = R = 196.2 N
6.
Three non-concurrent, non-parallel reactions
(a)
Plate: completely constrained
(b)
Reactions: determinate
(c)
Equilibrium maintained
B = 294 N
7.
8.
, D = 491 N
53.1°
Two reactions
(a)
Plate: improperly constrained
(b)
Reactions determined by dynamics
(c)
No equilibrium
(ΣFy ≠ 0)
For non-concurrent, non-parallel reactions
(a)
Plate: completely constrained
(b)
Reactions: indeterminate
(c)
Equilibrium maintained
B = D y = 196.2 N
(C + D x = 0)
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409
PROBLEM 4.60
The bracket ABC can be supported in the eight different ways shown. All connections consist of smooth pins,
rollers, or short links. For each case, answer the questions listed in Problem 4.59, and, wherever possible,
compute the reactions, assuming that the magnitude of the force P is 100 lb.
SOLUTION
1.
Three non-concurrent, non-parallel reactions
(a)
Bracket: complete constraint
(b)
Reactions: determinate
(c)
Equilibrium maintained
A = 120.2 lb
2.
3.
56.3°, B = 66.7 lb
Four concurrent, reactions (through A)
(a)
Bracket: improper constraint
(b)
Reactions: indeterminate
(c)
No equilibrium
(ΣM A ≠ 0)
Two reactions
(a)
Bracket: partial constraint
(b)
Reactions: determinate
(c)
Equilibrium maintained
A = 50 lb , C = 50 lb
4.
Three non-concurrent, non-parallel reactions
(a)
Bracket: complete constraint
(b)
Reactions: determinate
(c)
Equilibrium maintained
A = 50 lb , B = 83.3 lb
36.9°, C = 66.7 lb
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410
PROBLEM 4.60 (Continued)
5.
6.
Four non-concurrent, non-parallel reactions
(a)
Bracket: complete constraint
(b)
Reactions: indeterminate
(c)
Equilibrium maintained
(ΣM C = 0) A y = 50 lb
Four non-concurrent, non-parallel reactions
(a)
Bracket: complete constraint
(b)
Reactions: indeterminate
(c)
Equilibrium maintained
A x = 66.7 lb , B = 66.7 lb
( A y + B y = 100 lb )
7.
Three non-concurrent, non-parallel reactions
(a)
Bracket: complete constraint
(b)
Reactions: determinate
(c)
Equilibrium maintained
A = C = 50 lb
8.
Three concurrent, reactions (through A)
(a)
Bracket: improper constraint
(b)
Reactions: indeterminate
(c)
No equilibrium
(ΣM A ≠ 0)
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411
PROBLEM 4.61
Determine the reactions at A and B when a = 180 mm.
SOLUTION
Reaction at B must pass through D where A and 300-N load intersect.
∆BCD :
Free-Body Diagram:
(Three-force member)
240
180
β = 53.13°
tan β =
Force triangle
A = (300 N) tan 53.13°
= 400 N
300 N
cos 53.13°
= 500 N
!
A = 400 N
B=
B = 500 N
53.1°
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412
!
PROBLEM 4.62
For the bracket and loading shown, determine the range of values of the
distance a for which the magnitude of the reaction at B does not exceed
600 N.
SOLUTION
Reaction at B must pass through D where A and 300-N load intersect.
Free-Body Diagram:
(Three-force member)
a=
240 mm
tan β
(1)
Force Triangle
(with B = 600 N)
300 N
= 0.5
600 N
β = 60.0°
cos β =
Eq. (1)
240 mm
tan 60.0°
= 138.56 mm
a=
For B # 600 N a $ 138.6 mm
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413
!
PROBLEM 4.63
Using the method of Section 4.7, solve Problem 4.17.
PROBLEM 4.17 The required tension in cable AB is 200 lb.
Determine (a) the vertical force P that must be applied to the
pedal, (b) the corresponding reaction at C.
SOLUTION
Free-Body Diagram:
(Three-Force body)
Reaction at C must pass through E, where D and 200-lb force intersect.
6.062 in.
15 in.
β = 22.005°
tan β =
Force triangle
(a)
P = (200 lb)tan 22.005°
P = 80.83 lb
(b)
C=
P = 80.8 lb
200 lb
= 215.7 lb
cos 22.005°
C = 216 lb
22.0°
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414
!
PROBLEM 4.64
Using the method of Section 4.7, solve Problem 4.18.
PROBLEM 4.18 Determine the maximum tension that can
be developed in cable AB if the maximum allowable value of
the reaction at C is 250 lb.
SOLUTION
Free-Body Diagram:
(Three -Force body)
Reaction at C must pass through E, where D and the force T intersect.
6.062 in.
15 in.
β = 22.005°
tan β =
Force triangle
T = (250 lb) cos 22.005°
T = 231.8 lb
!
T = 232 lb
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415
!
PROBLEM 4.65
The spanner shown is used to rotate a shaft. A pin fits in a
hole at A, while a flat, frictionless surface rests against the
shaft at B. If a 60-lb force P is exerted on the spanner at D,
find the reactions at A and B.
SOLUTION
Free-Body Diagram:
(Three-Force body)
The line of action of A must pass through D, where B and P intersect.
3sin 50°
3cos 50° + 15
= 0.135756
α = 7.7310°
tan α =
Force triangle
60 lb
sin 7.7310°
= 446.02 lb
60 lb
B=
tan 7.7310°
= 441.97 lb
A=
A = 446 lb
7.73°
B = 442 lb
!
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416
PROBLEM 4.66
Determine the reactions at B and D when b = 60 mm.
SOLUTION
Since CD is a two-force member, the line of action of reaction at D must pass through Points C and D.
Free-Body Diagram:
(Three-Force body)
Reaction at B must pass through E, where the reaction at D and 80-N force intersect.
220 mm
250 mm
β = 41.348°
tan β =
Force triangle
Law of sines
80 N
B
D
=
=
sin 3.652° sin 45° sin131.348°
B = 888.0 N
D = 942.8 N
! B = 888 N
41.3°
D = 943 N
45.0°
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417
PROBLEM 4.67
Determine the reactions at B and D when b = 120 mm.
SOLUTION
Since CD is a two-force member, line of action of reaction at D must pass through C and D
.
Free-Body Diagram:
(Three-Force body)
Reaction at B must pass through E, where the reaction at D and 80-N force intersect.
280 mm
250 mm
β = 48.24°
tan β =
Force triangle
Law of sines
80 N
B
D
=
=
sin 3.24° sin135° sin 41.76°
B = 1000.9 N
D = 942.8 N !
B = 1001 N
48.2° D = 943 N
45.0°
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418
PROBLEM 4.68
Determine the reactions at B and C when a = 1.5 in.
SOLUTION
Since CD is a two-force member, the force it exerts on member ABD is directed along DC.
Free-Body Diagram of ABD: (Three-Force member)
The reaction at B must pass through E, where D and the 50-lb load intersect.
Triangle CFD:
Triangle EAD:
Triangle EGB:
Force triangle
3
= 0.6
5
α = 30.964°
tan α =
AE = 10 tan α = 6 in.
GE = AE − AG = 6 − 1.5 = 4.5 in.
GB
3
=
GE 4.5
β = 33.690°
tan β =
B
D
50 lb
=
=
sin120.964° sin 33.690° sin 25.346°
B = 100.155 lb
D = 64.789 lb
!
B = 100.2 lb
56.3°
C = D = 64.8 lb
31.0°
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419
!
PROBLEM 4.69
A 50-kg crate is attached to the trolley-beam system shown.
Knowing that a = 1.5 m, determine (a) the tension in cable CD,
(b) the reaction at B.
SOLUTION
Free-Body Diagram:
Three-Force body: W and TCD intersect at E.
0.7497 m
1.5 m
β = 26.56°
tan β =
Force triangle 3 forces intersect at E.
W = (50 kg) 9.81 m/s 2
= 490.5 N
Law of sines
TCD
490.5 N
B
=
=
sin 61.56° sin 63.44° sin 55°
TCD = 498.9 N
B = 456.9 N
TCD = 499 N
(a)
B = 457 N
(b)
26.6°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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420
!
PROBLEM 4.70
Solve Problem 4.69, assuming that a = 3 m.
PROBLEM 4.69 A 50-kg crate is attached to the trolley-beam
system shown. Knowing that a = 1.5 m, determine (a) the tension
in cable CD, (b) the reaction at B.
SOLUTION
W and TCD intersect at E
Free-Body Diagram:
Three-Force body:
AE 0.301 m
=
AB
3m
β = 5.722°
tan β =
Force Triangle (Three forces intersect at E.)
W = (50 kg) 9.81 m/s 2
= 490.5 N
Law of sines
TCD
490.5 N
B
=
=
sin 29.278° sin 95.722° sin 55°
TCD = 997.99 N
B = 821.59 N
TCD = 998 N
(a)
B = 822 N
(b)
5.72°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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421
PROBLEM 4.71
One end of rod AB rests in the corner A and the other end is attached to
cord BD. If the rod supports a 40-lb load at its midpoint C, find the
reaction at A and the tension in the cord.
SOLUTION
Free-Body Diagram: (Three-Force body)
The line of action of reaction at A must pass through E, where T and the 40-lb load intersect.
EF 23
=
AF 12
α = 62.447°
tan α =
5
EH
=
DH 12
β = 22.620°
tan β =
Force triangle
A
T
40 lb
=
=
sin 67.380° sin 27.553° sin 85.067°
A = 37.1 lb
62.4°
T = 18.57 lb
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422
PROBLEM 4.72
Determine the reactions at A and D when β = 30°.
SOLUTION
From f.b.d. of frame ABCD
ΣM D = 0: − A(0.18 m) + [(150 N) sin 30°](0.10 m)
+ [(150 N) cos 30°](0.28 m) = 0
A = 243.74 N
or A = 244 N
!
ΣFx = 0: (243.74 N) + (150 N) sin 30° + Dx = 0
Dx = −318.74 N
ΣFy = 0: D y − (150 N) cos 30° = 0
Dy = 129.904 N
Then
D = ( Dx ) 2 + Dx2
= (318.74) 2 + (129.904) 2
= 344.19 N
and
Dy !
#
$ Dx %
129.904 !
= tan −1 "
#
$ −318.74 %
θ = tan −1 "
= −22.174°
or D = 344 N
22.2°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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423
PROBLEM 4.73
Determine the reactions at A and D when β = 60°.
SOLUTION
From f.b.d. of frame ABCD
ΣM D = 0: − A(0.18 m) + [(150 N) sin 60°](0.10 m)
+ [(150 N) cos 60°](0.28 m) = 0
A = 188.835 N
or A = 188.8 N
!
ΣFx = 0: (188.835 N) + (150 N) sin 60° + Dx = 0
Dx = −318.74 N
ΣFy = 0: D y − (150 N) cos 60° = 0
Dy = 75.0 N
Then
D = ( Dx ) 2 + ( Dy ) 2
= (318.74) 2 + (75.0) 2
= 327.44 N
and
Dy !
#
$ Dx %
θ = tan −1 "
75.0 !
= tan −1 "
#
$ −318.74 %
= −13.2409°
or D = 327 N
13.24°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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424
PROBLEM 4.74
A 40-lb roller, of diameter 8 in., which is to be used on a tile floor, is resting
directly on the subflooring as shown. Knowing that the thickness of each tile
is 0.3 in., determine the force P required to move the roller onto the tiles if
the roller is (a) pushed to the left, (b) pulled to the right.
SOLUTION
See solution to Problem 4.73 for free-body diagram and analysis leading to the following equations:
T=
P
1 + cos θ
C=P
(1)
sin θ
1 + cos θ
(2)
For θ = 45°
P
P
=
1 + cos 45° 1.7071
Eq. (1):
T=
Eq. (2):
C=P
sin 45°
0.7071
=P
1 + cos 45°
1.7071
T = 0.586 P
C = 0.444P
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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425
PROBLEM 4.75
Member ABC is supported by a pin and bracket at B and by an
inextensible cord attached at A and C and passing over a frictionless
pulley at D. The tension may be assumed to be the same in portions AD
and CD of the cord. For the loading shown and neglecting the size of
the pulley, determine the tension in the cord and the reaction at B.
SOLUTION
Free-Body Diagram:
Reaction at B must pass through D.
7 in.
12 in.
α = 30.256°
7 in.
tan β =
24 in.
β = 16.26°
tan α =
Force triangle
Law of sines
T
T − 72 lb
B
=
=
sin 59.744° sin13.996° sin106.26
T (sin13.996°) = (T − 72 lb)(sin 59.744°)
T (0.24185) = (T − 72)(0.86378)
T = 100.00 lb
sin 106.26°
sin 59.744
= 111.14 lb
T = 100.0 lb
B = (100 lb)
B = 111.1 lb
30.3°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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426
PROBLEM 4.76
Member ABC is supported by a pin and bracket at B and by an
inextensible cord attached at A and C and passing over a frictionless
pulley at D. The tension may be assumed to be the same in portions
AD and CD of the cord. For the loading shown and neglecting the
size of the pulley, determine the tension in the cord and the reaction
at B.
SOLUTION
Free-Body Diagram:
Force triangle
Reaction at B must pass through D.
tan α =
120
; α = 36.9°
160
T T − 75 lb B
=
=
4
3
5
3T = 4T − 300; T = 300 lb
5
5
B = T = (300 lb) = 375 lb
4
4
B = 375 lb
36.9°!
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427
PROBLEM 4.77
Rod AB is supported by a pin and bracket at A and rests against a frictionless
peg at C. Determine the reactions at A and C when a 170-N vertical force is
applied at B.
SOLUTION
Free-Body Diagram:
(Three-Force body)
The reaction at A must pass through D where C and 170-N force intersect.
160 mm
300 mm
α = 28.07°
tan α =
We note that triangle ABD is isosceles (since AC = BC) and, therefore
CAD = α = 28.07°
Also, since CD ⊥ CB, reaction C forms angle α = 28.07° with horizontal.
Force triangle
We note that A forms angle 2α with vertical. Thus A and C form angle
180° − (90° − α ) − 2α = 90° − α
Force triangle is isosceles and we have
A = 170 N
C = 2(170 N)sin α
= 160.0 N
A = 170.0 N
33.9°
C = 160.0 N
28.1°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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428
!
PROBLEM 4.78
Solve Problem 4.77, assuming that the 170-N force applied at B is horizontal
and directed to the left.
PROBLEM 4.77 Rod AB is supported by a pin and bracket at A and rests
against a frictionless peg at C. Determine the reactions at A and C when a
170-N vertical force is applied at B.
SOLUTION
Free-Body Diagram: (Three-Force body)
The reaction at A must pass through D, where C and the 170-N force intersect.
160 mm
300 mm
α = 28.07°
tan α =
We note that triangle ADB is isosceles (since AC = BC). Therefore
Also
A=
B = 90° − α .
ADB = 2α
Force triangle
The angle between A and C must be 2α − α = α
α = 28.07°
Thus, force triangle is isosceles and
A = 170.0 N
C = 2(170 N) cos α = 300 N
A = 170.0 N
56.1°
C = 300 N
28.1°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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429
!
PROBLEM 4.79
Using the method of Section 4.7, solve Problem 4.21.
PROBLEM 4.21 Determine the reactions at A and B when
(a) h = 0, (b) h = 200 mm.
SOLUTION
!
!
Free-Body Diagram:
(a)
h=0
Reaction A must pass through C where 150-N weight and
B interect.
Force triangle is equilateral
(b)
A = 150.0 N
30.0°
B = 150.0 N
30.0°
h = 200 mm
55.662
250
β = 12.552°
!
tan β =
!
Force triangle
Law of sines
A
B
150 N
=
=
sin17.448° sin 60° sin102.552°
A = 433.247 N
B = 488.31 N
!
!
A = 433 N
!
B = 488 N
12.55°
30.0°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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430
!
PROBLEM 4.80
Using the method of Section 4.7, solve Problem 4.28.
PROBLEM 4.28 A lever AB is hinged at C and attached to a control
cable at A. If the lever is subjected to a 500-N horizontal force at B,
determine (a) the tension in the cable, (b) the reaction at C.
SOLUTION
Free-Body Diagram:
(Three-Force body)
Reaction at C must pass through E, where FAD and 500-N force intersect.
Since AC = CD = 250 mm, triangle ACD is isosceles.
We have
C = 90° + 30° = 120°
and
A=
D=
1
(180° − 120°) = 30°
2
Dimensions in mm
On the other hand, from triangle BCF:
CF = ( BC )sin 30° = 200 sin 30° = 100 mm
FD = CD − CF = 250 − 100 = 150 mm
From triangle EFD, and since
D = 30° :
EF = ( FD ) tan 30° = 150 tan 30° = 86.60 mm
From triangle EFC:
100 mm
CF
=
EF 86.60 mm
α = 49.11°
tan α =
Force triangle
Law of sines
FAD
C
500 N
=
=
sin 49.11° sin 60° sin 70.89°
FAD = 400 N, C = 458 N
FAD = 400 N
(a)
C = 458 N
(b)
49.1°
!
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431
PROBLEM 4.81
Knowing that θ = 30°, determine the reaction (a) at B, (b) at C.
SOLUTION
Reaction at C must pass through D where force P and reaction at B intersect.
In ∆ CDE:
Free-Body Diagram:
(Three-Force body)
( 3 − 1) R
R
= 3 −1
β = 36.2°
tan β =
Force triangle
Law of sines
P
B
C
=
=
sin 23.8° sin126.2° sin 30°
B = 2.00 P
C = 1.239 P
(a)
B = 2P
60.0°
(b)
C = 1.239P
36.2°
!
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432
PROBLEM 4.82
Knowing that θ = 60°, determine the reaction (a) at B, (b) at C.
SOLUTION
Reaction at C must pass through D where force P and reaction at B intersect.
In ∆CDE:
tan β =
R−
=1−
R
3
Free-Body Diagram:
(Three-Force body)
R
1
3
β = 22.9°
Force triangle
Law of sines
P
B
C
=
=
sin 52.9° sin 67.1° sin 60°
B = 1.155P
C = 1.086 P
(a)
B = 1.155P
30.0°
(b)
C = 1.086P
22.9°
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433
!
PROBLEM 4.83
Rod AB is bent into the shape of an arc of circle and is lodged between two
pegs D and E. It supports a load P at end B. Neglecting friction and the
weight of the rod, determine the distance c corresponding to equilibrium
when a = 20 mm and R = 100 mm.
SOLUTION
Free-Body Diagram:
yED = xED = a,
Since
Slope of ED is
slope of HC is
45°
45°
Also
DE = 2 a
and
a
1!
DH = HE = " # DE =
2
$2%
For triangles DHC and EHC
sin β =
a
2
R
=
a
2R
Now
c = R sin(45° − β )
For
a = 20 mm and R = 100 mm
sin β =
20 mm
2(100 mm)
= 0.141421
β = 8.1301°
and
c = (100 mm) sin(45° − 8.1301°)
= 60.00 mm
or c = 60.0 mm
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434
!
PROBLEM 4.84
A slender rod of length L is attached to collars that can slide freely along the
guides shown. Knowing that the rod is in equilibrium, derive an expression
for the angle θ in terms of the angle β.
SOLUTION
As shown in the free-body diagram of the slender rod AB, the three forces intersect at C. From the force
geometry
Free-Body Diagram:
tan β =
xGB
y AB
where
y AB = L cos θ
and
xGB =
tan β =
1
L sin θ
2
1
2
L sin θ
L cos θ
1
= tan θ
2
or tan θ = 2 tan β
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435
!
PROBLEM 4.85
An 8-kg slender rod of length L is attached to collars that can slide
freely along the guides shown. Knowing that the rod is in equilibrium
and that β = 30°, determine (a) the angle θ that the rod forms with the
vertical, (b) the reactions at A and B.
SOLUTION
(a)
As shown in the free-body diagram of the slender rod AB, the three forces intersect at C. From the
geometry of the forces
Free-Body Diagram:
tan β =
xCB
yBC
where
xCB =
1
L sin θ
2
and
yBC = L cos θ
tan β =
1
tan θ
2
or
tan θ = 2 tan β
For
β = 30°
tan θ = 2 tan 30°
= 1.15470
θ = 49.107°
or
θ = 49.1°
!
W = mg = (8 kg)(9.81 m/s2 ) = 78.480 N
(b)
From force triangle
A = W tan β
= (78.480 N) tan 30°
= 45.310 N
and
B=
or
W
78.480 N
=
= 90.621 N
cos β
cos 30°
A = 45.3 N
or B = 90.6 N
!
60.0°
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436
!
PROBLEM 4.86
A slender uniform rod of length L is held in equilibrium as shown, with one
end against a frictionless wall and the other end attached to a cord of length S.
Derive an expression for the distance h in terms of L and S. Show that this
position of equilibrium does not exist if S . 2L.
SOLUTION
!
From the f.b.d. of the three-force member AB, forces must intersect at D.
Since the force T intersects Point D, directly above G,
yBE = h
For triangle ACE:
S 2 = ( AE ) 2 + (2h) 2
(1)
For triangle ABE:
L2 = ( AE ) 2 + (h) 2
(2)
Subtracting Equation (2) from Equation (1)
S 2 − L2 = 3h 2
!
(3)
!
h=
S 2 − L2
3
!
or
!
As length S increases relative to length L, angle θ increases until
rod AB is vertical. At this vertical position:
!
h+L=S
or h = S − L
Therefore, for all positions of AB
h$S − L
or
S 2 − L2
$S −L
3
or
S 2 − L2 $ 3( S − L) 2
(4)
= 3( S 2 − 2 SL + L2 )
!
= 3S 2 − 6 SL + 3L2
or
0 $ 2S 2 − 6SL + 4 L2
and
0 $ S 2 − 3SL + 2 L2 = ( S − L)( S − 2 L)
!
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you are using it without permission.
437
PROBLEM 4.86 (Continued)
For
S−L=0 S =L
Minimum value of S is L
For
S − 2L = 0 S = 2L
Maximum value of S is 2L
Therefore, equilibrium does not exist if S . 2 L
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438
!
PROBLEM 4.87
A slender uniform rod of length L = 20 in. is held in equilibrium as shown,
with one end against a frictionless wall and the other end attached to a cord
of length S = 30 in. Knowing that the weight of the rod is 10 lb, determine
(a) the distance h, (b) the tension in the cord, (c) the reaction at B.
SOLUTION
From the f.b.d. of the three-force member AB, forces must intersect at D.
Since the force T intersects Point D, directly above G,
yBE = h
For triangle ACE:
S 2 = ( AE ) 2 + (2h) 2
(1)
For triangle ABE:
L2 = ( AE ) 2 + (h) 2
(2)
Subtracting Equation (2) from Equation (1)
S 2 − L2 = 3h 2
!
!
h=
or
!
(a)
S 2 − L2
3
L = 20 in. and S = 30 in.
For
(30) 2 − (20) 2
3
= 12.9099 in.
h=
(b)
!
W = 10 lb
We have
2h !
#
$ s %
θ = sin −1 "
and
!
or h = 12.91 in.
& 2(12.9099) '
= sin −1 (
)
30
*
+
θ = 59.391°
!
!
!
!
!
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439
PROBLEM 4.87 (Continued)
From the force triangle
W
sin θ
10 lb
=
sin 59.391°
= 11.6190 lb
T=
(c)
or
T = 11.62 lb
W
tan θ
10 lb
=
tan 59.391°
B=
or B = 5.92 lb
= 5.9161 lb
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440
!
PROBLEM 4.88
A uniform rod AB of length 2R rests inside a hemispherical bowl
of radius R as shown. Neglecting friction, determine the angle θ
corresponding to equilibrium.
SOLUTION
Based on the f.b.d., the uniform rod AB is a three-force body. Point E is the point of intersection of the three
forces. Since force A passes through O, the center of the circle, and since force C is perpendicular to the rod,
triangle ACE is a right triangle inscribed in the circle. Thus, E is a point on the circle.
Note that the angle α of triangle DOA is the central angle corresponding to the inscribed angle θ of
triangle DCA.
α = 2θ
The horizontal projections of AE , ( x AE ), and AG , ( x AG ), are equal.
x AE = x AG = x A
or
( AE ) cos 2θ = ( AG ) cos θ
and
(2 R) cos 2θ = R cos θ
cos 2θ = 2 cos 2 θ − 1
Now
then
or
4 cos 2 θ − 2 = cos θ
4 cos 2 θ − cos θ − 2 = 0
Applying the quadratic equation
cos θ = 0.84307 and cos θ = − 0.59307
θ = 32.534° and θ = 126.375°(Discard)
or θ = 32.5°
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441
!
PROBLEM 4.89
A slender rod of length L and weight W is attached to a collar
at A and is fitted with a small wheel at B. Knowing that the
wheel rolls freely along a cylindrical surface of radius R, and
neglecting friction, derive an equation in θ, L, and R that must
be satisfied when the rod is in equilibrium.
SOLUTION
Free-Body Diagram (Three-Force body)
Reaction B must pass through D where B and W intersect.
Note that ∆ABC and ∆BGD are similar.
AC = AE = L cos θ
In ∆ABC:
(CE ) 2 + ( BE )2 = ( BC )2
(2 L cos θ ) 2 + ( L sin θ )2 = R 2
2
R!
2
2
" # = 4cos θ + sin θ
$L%
2
R!
2
2
" # = 4cos θ + 1 − cos θ
$L%
2
R!
2
" # = 3cos θ + 1
$L%
2
'
1&
cos 2 θ = (" R !# − 1)
3 *$ L %
+
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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442
!
PROBLEM 4.90
Knowing that for the rod of Problem 4.89, L = 15 in., R = 20 in., and
W = 10 lb, determine (a) the angle θ corresponding to equilibrium,
(b) the reactions at A and B.
SOLUTION
See the solution to Problem 4.89 for free-body diagram and analysis leading to the following equation
2
'
1&
cos 2 θ = (" R !# − 1)
3 *$ L %
+
For L = 15 in., R = 20 in., and W = 10 lb.
2
'
1 & 20 in. !
cos 2 θ = ("
# − 1) ; θ = 59.39°
3 ($ 15 in. %
)
*
+
(a)
In ∆ABC:
θ = 59.4°
BE
L sin θ
1
=
= tan θ
CE 2 L cos θ 2
1
tan α = tan 59.39° = 0.8452
2
α = 40.2°
tan α =
Force triangle
A = W tan α = (10 lb) tan 40.2° = 8.45 lb
W
(10 lb)
=
= 13.09 lb
B=
cos α cos 40.2°
A = 8.45 lb
(b)
B = 13.09 lb
(c)
49.8°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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443
!
PROBLEM 4.91
A 4 × 8-ft sheet of plywood weighing 34 lb has been temporarily
placed among three pipe supports. The lower edge of the sheet
rests on small collars at A and B and its upper edge leans against
pipe C. Neglecting friction at all surfaces, determine the reactions
at A, B, and C.
SOLUTION
rG/B =
3.75 1.3919
i+
j+k
2
2
We have 5 unknowns and 6 Eqs. of equilibrium.
Plywood sheet is free to move in z direction, but equilibrium is maintained (ΣFz = 0).
ΣM B = 0: rA/B × ( Ax i + Ay j) + rC/B × ( −Ci ) + rG/B × ( − wj) = 0
i
0
Ax
j
0
Ay
k
i
j
k
i
j
k
4 + 3.75 1.3919 2 + 1.875 0.696 1 = 0
0
0
0
0
−C
−34 0
−4 Ay i + 4 Ax j − 2Cj + 1.3919Ck + 34i − 63.75k = 0
Equating coefficients of unit vectors to zero:
i:
− 4 Ay + 34 = 0
j:
− 2C + 4 Ax = 0
k : 1.3919C − 63.75 = 0
Ay = 8.5 lb
Ax =
1
1
C = (45.80) = 22.9 lb
2
2
C = 45.80 lb
C = 45.8 lb
ΣFx = 0:
Ax + Bx − C = 0:
Bx = 45.8 − 22.9 = 22.9 lb
ΣFy = 0:
Ay + By − W = 0:
By = 34 − 8.5 = 25.5 lb
A = (22.9 lb)i + (8.5 lb) j B = (22.9 lb)i + (25.5 lb) j C = −(45.8 lb)i
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444
PROBLEM 4.92
Two tape spools are attached to an axle supported by
bearings at A and D. The radius of spool B is 30 mm and
the radius of spool C is 40 mm. Knowing that TB = 80 N
and that the system rotates at a constant rate, determine the
reactions at A and D. Assume that the bearing at A does
not exert any axial thrust and neglect the weights of the
spools and axle.
SOLUTION
Dimensions in mm
We have six unknowns and six Eqs. of equilibrium.
ΣM A = 0: (90i + 30k ) × (−80 j) + (210i + 40 j) × (−TC k ) + (300i ) × ( Dx i + Dy j + Dz k ) = 0
−7200k + 2400i + 210TC j − 40TC i + 300 D y k − 300 Dz j = 0
Equate coefficients of unit vectors to zero:
i:
j:
2400 − 40TC = 0
TC = 60 N
210TC − 300 Dz = 0 (210)(60) − 300 Dz = 0
Dz = 42 N
k : −7200 + 300 Dy = 0
Dy = 24 N
ΣFx = 0:
Dx = 0
ΣFy = 0:
Ay + D y − 80 N = 0
Ay = 80 − 24 = 56 N
ΣFz = 0: Az + Dz − 60 N = 0
Az = 60 − 42 = 18 N
A = (56.0 N) j + (18.00 N)k
D = (24.0 N) j + (42.0 N)k
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445
PROBLEM 4.93
Solve Problem 4.92, assuming that the spool C is replaced
by a spool of radius 50 mm.
PROBLEM 4.92 Two tape spools are attached to an axle
supported by bearings at A and D. The radius of spool B
is 30 mm and the radius of spool C is 40 mm. Knowing
that TB = 80 N and that the system rotates at a constant rate,
determine the reactions at A and D. Assume that the bearing
at A does not exert any axial thrust and neglect the weights
of the spools and axle.
SOLUTION
Dimensions in mm
We have six unknowns and six Eqs. of equilibrium.
ΣM A = 0: (90i + 30k ) × (−80 j) + (210i + 50 j) × ( −TC k ) + (300i) × ( Dx i + D y j + Dz k ) = 0
−7200k + 2400i + 210TC j − 50TC i + 300 Dy k − 300 Dz j = 0
Equate coefficients of unit vectors to zero:
i:
j:
2400 − 50TC = 0
TC = 48 N
210TC − 300 Dz = 0 (210)(48) − 300 Dz = 0
Dz = 33.6 N
k : − 7200 + 300 Dy = 0
ΣFx = 0:
Dy = 24 N
Dx = 0
ΣFy = 0: Ay + Dy − 80 N = 0
Ay = 80 − 24 = 56 N
ΣFz = 0:
Az = 48 − 33.6 = 14.4 N
Az + Dz − 48 = 0
A = (56.0 N) j + (14.40 N)k
D = (24.0 N) j + (33.6 N)k
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446
PROBLEM 4.94
Two transmission belts pass over sheaves welded to an
axle supported by bearings at B and D. The sheave at A
has a radius of 2.5 in., and the sheave at C has a radius
of 2 in. Knowing that the system rotates at a constant
rate, determine (a) the tension T, (b) the reactions at B
and D. Assume that the bearing at D does not exert any
axial thrust and neglect the weights of the sheaves and
axle.
SOLUTION
Assume moment reactions at the bearing supports are zero. From f.b.d. of shaft
(a)
(b)
ΣM x -axis = 0: (24 lb − 18 lb)(5 in.) + (30 lb − T )(4 in.) = 0
T = 37.5 lb
ΣFx = 0: Bx = 0
ΣM D ( z -axis) = 0: (30 lb + 37.5 lb)(6 in.) − By (12 in.) = 0
By = 33.75 lb
ΣM D ( y -axis) = 0: (24 lb + 18 lb)(20 in.) + Bz (12 in.) = 0
Bz = −70.0 lb
or B = (33.8 lb) j − (70.0 lb)k
ΣM B ( z -axis) = 0: − (30 lb + 37.5 lb)(6 in.) + D y (12 in.) = 0
Dy = 33.75 lb
ΣM B ( y -axis) = 0: (24 lb + 18 lb)(8 in.) + Dz (12 in.) = 0
Dz = −28.0 lb !
or D = (33.8 lb) j − (28.0 lb)k
!
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447
PROBLEM 4.95
A 200-mm lever and a 240-mm-diameter pulley are
welded to the axle BE that is supported by bearings at C
and D. If a 720-N vertical load is applied at A when the
lever is horizontal, determine (a) the tension in the cord,
(b) the reactions at C and D. Assume that the bearing
at D does not exert any axial thrust.
SOLUTION
Dimensions in mm
We have six unknowns and six Eqs. of equilibrium–OK
ΣM C = 0: (−120k ) × ( Dx i + Dy j) + (120 j − 160k ) × Ti + (80k − 200i ) × (−720 j) = 0
−120 Dx j + 120 D y i − 120Tk − 160Tj + 57.6 × 103 i + 144 × 103 k = 0
Equating to zero the coefficients of the unit vectors:
k:
−120T + 144 × 103 = 0
i:
120 Dy + 57.6 × 103 = 0
j: − 120 Dx − 160(1200 N) = 0
(b)
ΣFx = 0:
Cx + Dx + T = 0
ΣFy = 0:
C y + Dy − 720 = 0
ΣFz = 0:
Cz = 0
(a) T = 1200 N
Dy = −480 N
Dx = −1600 N
Cx = 1600 − 1200 = 400 N
C y = 480 + 720 = 1200 N
C = (400 N)i + (1200 N) j D = −(1600 N)i − (480 N) j
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448
PROBLEM 4.96
Solve Problem 4.95, assuming that the axle has
been rotated clockwise in its bearings by 30° and
that the 720-N load remains vertical.
PROBLEM 4.95 A 200-mm lever and a 240-mmdiameter pulley are welded to the axle BE that is
supported by bearings at C and D. If a 720-N
vertical load is applied at A when the lever is
horizontal, determine (a) the tension in the cord,
(b) the reactions at C and D. Assume that the
bearing at D does not exert any axial thrust.
SOLUTION
Dimensions in mm
We have six unknowns and six Eqs. of equilibrium.
ΣM C = 0: (−120k ) × ( Dx i + Dy j) + (120 j − 160k ) × T i + (80k − 173.21i ) × (−720 j) = 0
−120 Dx j + 120 D y i −120T k −160T j + 57.6 × 103 i + 124.71 × 103 k = 0
Equating to zero the coefficients of the unit vectors:
k : − 120T + 124.71 × 103 = 0
i:
T = 1039 N
120 Dy + 57.6 × 103 = 0 Dy = −480 N
j: − 120 Dx − 160(1039.2)
(b)
T = 1039.2 N
ΣFx = 0:
Cx + Dx + T = 0
ΣFy = 0:
C y + Dy − 720 = 0
ΣFz = 0:
Cz = 0
Dx = −1385.6 N
Cx = 1385.6 − 1039.2 = 346.4
C y = 480 + 720 = 1200 N
C = (346 N)i + (1200 N) j D = −(1386 N)i − (480 N) j
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449
PROBLEM 4.97
An opening in a floor is covered by a 1 × 1.2-m sheet of
plywood of mass 18 kg. The sheet is hinged at A and B and
is maintained in a position slightly above the floor by a small
block C. Determine the vertical component of the reaction
(a) at A, (b) at B, (c) at C.
SOLUTION
rB/A = 0.6i
rC/A = 0.8i + 1.05k
rG/A = 0.3i + 0.6k
W = mg = (18 kg)9.81
W = 176.58 N
ΣM A = 0: rB/A × Bj + rC/A × Cj + rG/A × (−Wj) = 0
(0.6i ) × Bj + (0.8i + 1.05k ) × Cj + (0.3i + 0.6k ) × ( −Wj) = 0
0.6 Bk + 0.8Ck − 1.05Ci − 0.3Wk + 0.6Wi = 0
Equate coefficients of unit vectors of zero:
0.6 !
i : 1.05C + 0.6W = 0 C = "
#176.58 N = 100.90 N
1.05
$
%
k : 0.6 B + 0.8C − 0.3W = 0
0.6 B + 0.8(100.90 N) − 0.3(176.58 N) = 0 B = −46.24 N
ΣFy = 0: A + B + C − W = 0
A − 46.24 N + 100.90 N + 176.58 N = 0
A = 121.92 N
(a ) A = 121.9 N (b) B = −46.2 N (c) C = 100.9 N
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450
PROBLEM 4.98
Solve Problem 4.97, assuming that the small block C
is moved and placed under edge DE at a point 0.15 m
from corner E.
PROBLEM 4.97 An opening in a floor is covered by
a 1 × 1.2-m sheet of plywood of mass 18 kg. The sheet
is hinged at A and B and is maintained in a position
slightly above the floor by a small block C. Determine
the vertical component of the reaction (a) at A, (b) at B,
(c) at C.
SOLUTION
rB/A = 0.6i
rC/A = 0.65i + 1.2k
rG/A = 0.3i + 0.6k
W = mg = (18 kg) 9.81 m/s 2
W = 176.58 N
ΣM A = 0: rB/A × Bj + rC/A × Cj + rG/A × (−Wj) = 0
0.6i × Bj + (0.65i + 1.2k ) × Cj + (0.3i + 0.6k ) × (−Wj) = 0
0.6 Bk + 0.65Ck − 1.2Ci − 0.3Wk + 0.6Wi = 0
Equate coefficients of unit vectors to zero:
0.6 !
i : −1.2C + 0.6W = 0 C = "
#176.58 N = 88.29 N
$ 1.2 %
k : 0.6 B + 0.65C − 0.3W = 0
0.6 B + 0.65(88.29 N) − 0.3(176.58 N) = 0 B = −7.36 N
ΣFy = 0: A + B + C − W = 0
A − 7.36 N + 88.29 N − 176.58 N = 0
A = 95.648 N
(a ) A = 95.6 N (b) − 7.36 N (c) 88.3 N
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451
PROBLEM 4.99
The rectangular plate shown weighs 80 lb and is supported by three
vertical wires. Determine the tension in each wire.
SOLUTION
Free-Body Diagram:
ΣM B = 0: rA/B × TA j + rC/B × TC j + rG/B × (−80 lb) j = 0
(60 in.)k × TA j + [(60 in.)i + (15 in.)k ] × TC j + [(30 in.)i + (30 in.)k ] × ( −80 lb) j = 0
−60TAi + 60TC k − 15TC i − 2400k + 2400i = 0
Equating to zero the coefficients of the unit vectors:
i:
60TA − 15(40) + 2400 = 0
TA = 30.0 lb
k:
60TC − 2400 = 0
TC = 40.0 lb
ΣFy = 0:
TA + TB + TC − 80 lb = 0
30 lb + TB + 40 lb − 80 lb = 0
TB = 10.00 lb
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452
PROBLEM 4.100
The rectangular plate shown weighs 80 lb and is supported by three
vertical wires. Determine the weight and location of the lightest block
that should be placed on the plate if the tensions in the three wires are
to be equal.
SOLUTION
Free-Body Diagram:
Let −Wb j be the weight of the block and x and z the block’s coordinates.
Since tensions in wires are equal, let
TA = TB = TC = T
ΣM 0 = 0: (rA × Tj) + (rB × Tj) + (rC × Tj) + rG × (−Wj) + ( xi + zk ) × ( −Wb j) = 0
or,
(75 k ) × Tj + (15 k ) × Tj + (60i + 30k ) × Tj + (30i + 45k ) × (−Wj) + ( xi + zk ) × (−Wb j) = 0
or,
−75Ti − 15T i + 60T k − 30T i − 30W k + 45W i − Wb × k + Wb zi = 0
Equate coefficients of unit vectors to zero:
i:
−120T + 45W + Wb z = 0
(1)
k:
60T − 30W − Wb x = 0
(2)
ΣFy = 0:
3T − W − Wb = 0
(3)
Eq. (1) + 40 Eq. (3):
5W + ( z − 40)Wb = 0
(4)
Eq. (2) – 20 Eq. (3):
−10W − ( x − 20)Wb = 0
(5)
Also,
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453
PROBLEM 4.100 (Continued)
Solving (4) and (5) for Wb /W and recalling of 0 # x # 60 in., 0 # z # 90 in.,
(4):
Wb
5
5
=
$
= 0.125
W 40 − z 40 − 0
(5):
Wb
10
10
=
$
= 0.5
W 20 − x 20 − 0
Thus, (Wb ) min = 0.5W = 0.5(80) = 40 lb
(Wb ) min = 40.0 lb
Making Wb = 0.5W in (4) and (5):
5W + ( z − 40)(0.5W ) = 0
z = 30.0 in.
−10W − ( x − 20)(0.5W ) = 0
x = 0 in.
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454
PROBLEM 4.101
Two steel pipes AB and BC, each having a mass per unit
length of 8 kg/m, are welded together at B and supported by
three wires. Knowing that a = 0.4 m, determine the tension
in each wire.
SOLUTION
W1 = 0.6m′g
W2 = 1.2m′g
ΣM D = 0: rA/D × TA j + rE/D × (−W1 j) + rF/D × (−W2 j) + rC/D × TC j = 0
(−0.4i + 0.6k ) × TA j + (−0.4i + 0.3k ) × (−W1 j) + 0.2i × (−W2 j) + 0.8i × TC j = 0
−0.4TAk − 0.6TA i + 0.4W1k + 0.3W1i − 0.2W2 k + 0.8TC k = 0
Equate coefficients of unit vectors to zero:
1
1
i : − 0.6TA + 0.3W1 = 0; TA = W1 = 0.6m′g = 0.3m′g
2
2
k : − 0.4TA + 0.4W1 − 0.2W2 + 0.8TC = 0
−0.4(0.3m′g ) + 0.4(0.6m′g ) − 0.2(1.2m′g ) + 0.8TC = 0
TC =
(0.12 − 0.24 − 0.24)m′g
= 0.15m′g
0.8
ΣFy = 0: TA + TC + TD − W1 − W2 = 0
0.3m′g + 0.15m′g + TD − 0.6m′g − 1.2m′g = 0
TD = 1.35m′g
2
m′g = (8 kg/m)(9.81m/s ) = 78.48 N/m
TA = 0.3m′g = 0.3 × 78.45
TA = 23.5 N
TB = 0.15m′g = 0.15 × 78.45
TB = 11.77 N
TC = 1.35m′g = 1.35 × 78.45
TC = 105.9 N
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455
PROBLEM 4.102
For the pipe assembly of Problem 4.101, determine (a) the
largest permissible value of a if the assembly is not to tip,
(b) the corresponding tension in each wire.
SOLUTION
W1 = 0.6m′g
W2 = 1.2m′g
ΣM D = 0: rA/D × TA j + rE/D × (−W1 j) + rF/D × (−W2 j) + rC/D × TC j = 0
(− ai + 0.6k ) × TA j + (− ai + 0.3k ) × (−W1 j) + (0.6 − a)i × (−W2 j) + (1.2 − a)i × TC j = 0
−TA ak − 0.6TA i + W1ak + 0.3W1i − W2 (0.6 − a)k + TC (1.2 − a )k = 0
Equate coefficients of unit vectors to zero:
1
1
i : − 0.6TA + 0.3W1 = 0; TA = W1 = 0.6m′g = 0.3m′g
2
2
k : − TA a + W1a − W2 (0.6 − a ) + TC (1.2 − a) = 0
−0.3m′ga + 0.6m′ga − 1.2m′g (0.6 − a ) + TC (1.2 − a) = 0
TC =
(a)
0.3a − 0.6a + 1.2(0.6 − a)
1.2 − a
For Max a and no tipping, TC = 0
−0.3a + 1.2(0.6 − a) = 0
−0.3a + 0.72 − 1.2a = 0
1.5a = 0.72
a = 0.480 m
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456
PROBLEM 4.102 (Continued)
(b)
Reactions:
m′g = (8 kg/m) 9.81 m/s 2 = 78.48 N/m
TA = 0.3m′g = 0.3 × 78.48 = 23.544 N
TA = 23.5 N
ΣFy = 0: TA + TC + TD − W1 − W2 = 0
TA + 0 + TD − 0.6m′g − 1.2m′g = 0
TD = 1.8m′g − TA = 1.8 × 78.48 − 23.544 = 117.72
TD = 117.7 N
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457
PROBLEM 4.103
The 24-lb square plate shown is supported by three vertical
wires. Determine (a) the tension in each wire when a = 10 in.,
(b) the value of a for which the tension in each wire is 8 lb.
SOLUTION
rB/A = ai + 30k
rC/A = 30i + ak
rG/A = 15i + 15k
By symmetry: B = C
ΣM A = 0: rB/A × Bj + rC × Cj + rG/A × (−Wj) = 0
(ai + 30k ) × Bj + (30i + ak ) × Bj + (15i + 15k ) × (−Wj) = 0
Bak − 30 Bi + 30 Bk − Bai − 15Wk + 15Wi = 0
Equate coefficient of unit vector i to zero:
i : − 30 B − Ba + 15W = 0
B=
15W
30 + a
C=B=
15W
30 + a
(1)
ΣFy = 0: A + B + C − W = 0
& 15W '
A+ 2(
) − W = 0;
* 30 + a +
(a)
For
a = 10 in.
Eq. (1)
C=B=
Eq. (2)
A=
A=
aW
30 + a
(2)
15(24 lb)
= 9.00 lb
30 + 10
10(24 lb)
= 6.00 lb
30 + 10
A = 6.00 lb B = C = 9.00 lb
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458
!
PROBLEM 4.103 (Continued)
(b)
For tension in each wire = 8 lb
Eq. (1)
8 lb =
15(24 lb)
30 + a
30 in. + a = 45
a = 15.00 in.
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459
PROBLEM 4.104
The table shown weighs 30 lb and has a diameter of 4 ft. It is supported by
three legs equally spaced around the edge. A vertical load P of magnitude
100 lb is applied to the top of the table at D. Determine the maximum
value of a if the table is not to tip over. Show, on a sketch, the area of the
table over which P can act without tipping the table.
SOLUTION
r = 2 ft b = r sin 30° = 1 ft
We shall sum moments about AB.
(b + r )C + (a − b) P − bW = 0
(1 + 2)C + (a − 1)100 − (1)30 = 0
1
C = [30 − (a − 1)100]
3
If table is not to tip, C $ 0
[30 − ( a − 1)100] $ 0
30 $ (a − 1)100
a − 1 # 0.3 a # 1.3 ft a = 1.300 ft
Only ⊥ distance from P to AB matters. Same condition must be satisfied for each leg. P must be located
in shaded area for no tipping
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460
PROBLEM 4.105
A 10-ft boom is acted upon by the 840-lb force shown. Determine the
tension in each cable and the reaction at the ball-and-socket joint at A.
SOLUTION
We have five unknowns and six Eqs. of equilibrium but equilibrium is maintained (ΣM x = 0).
Free-Body Diagram:
!
BD = (−6 ft)i + (7 ft) j + (6 ft)k BD = 11 ft
BE = (−6 ft)i + (7 ft) j − (6 ft)k BE = 11 ft
!
BD TBD
TBD = TBD
(−6i + 7 j + 6k )
=
BD 11
!
BE TBE
TBE = TBE
(−6i + 7 j − 6k )
=
BE 11
ΣM A = 0: rB × TBD + rB × TBE + rC × ( −840 j) = 0
6i ×
TBD
T
(−6i + 7 j + 6k ) + 6i × BE (−6i + 7 j − 6k ) + 10i × (−840 j) = 0
11
11
42
36
42
36
TBD k − TBD j + TBE k + TBE j − 8400k
11
11
11
11
Equate coefficients of unit vectors to zero.
i: −
k:
36
36
TBD + TBE = 0 TBE = TBD
11
11
42
42
TBD + TBE − 8400 = 0
11
11
42
!
2 " TBD # = 8400
$ 11
%
TBD = 1100 lb
TBE = 1100 lb
!
!
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461
!
PROBLEM 4.105 (Continued)
ΣFx = 0: Ax −
6
6
(1100 lb) − (1100 lb) = 0
11
11
Ax = 1200 lb
ΣFy = 0: Ay +
7
7
(1100 lb) + (1100 lb) − 840 lb = 0
11
11
Ay = −560 lb
ΣFz = 0: Az +
6
6
(1100 lb) − (1100 lb) = 0
11
11
Az = 0
A = (1200 lb)i − (560 lb) j
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462
PROBLEM 4.106
A 2.4-m boom is held by a ball-and-socket joint at C and by two
cables AD and AE. Determine the tension in each cable and the
reaction at C.
SOLUTION
Free-Body Diagram:
Five Unknowns and six Eqs. of equilibrium. Equilibrium is maintained (ΣMAC = 0).
rB = 1.2k
TAD
TAE
rA = 2.4k
!
AD = −0.8i + 0.6 j − 2.4k
AD = 2.6 m
!
AE = 0.8i + 1.2 j − 2.4k
AE = 2.8 m
!
AD TAD
=
=
(−0.8i + 0.6 j − 2.4k )
AD 2.6
!
AE TAE
=
=
(0.8i + 1.2 j − 2.4k )
AE 2.8
ΣM C = 0: rA × TAD + rA × TAE + rB × (−3 kN) j = 0
i
j
k
i
j
k
TAD
T
0
0
2.4
0
2.4 AE + 1.2k × (−3.6 kN) j = 0
+ 0
2.6
2.8
−0.8 0.6 −2.4
0.8 1.2 −2.4
Equate coefficients of unit vectors to zero.
i : − 0.55385 TAD − 1.02857 TAE + 4.32 = 0
(1)
j : − 0.73846 TAD + 0.68671 TAE = 0
TAD = 0.92857 TAE
Eq. (1):
(2)
−0.55385(0.92857) TAE − 1.02857 TAE + 4.32 = 0
1.54286 TAE = 4.32
TAE = 2.800 kN
TAE = 2.80 kN
!
!
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463
PROBLEM 4.106 (Continued)
Eq. (2):
TAD = 0.92857(2.80) = 2.600 kN
TAD = 2.60 kN
0.8
0.8
(2.6 kN) +
(2.8 kN) = 0
2.6
2.8
0.6
1.2
(2.6 kN) +
(2.8 kN) − (3.6 kN) = 0
ΣFy = 0: C y +
2.6
2.8
2.4
2.4
(2.6 kN) −
(2.8 kN) = 0
ΣFz = 0: C z −
2.6
2.8
ΣFx = 0: C x −
!
Cx = 0
C y = 1.800 kN
C z = 4.80 kN
C = (1.800 kN) j + (4.80 kN)k
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464
PROBLEM 4.107
Solve Problem 4.106, assuming that the 3.6-kN load is applied at
Point A.
PROBLEM 4.106 A 2.4-m boom is held by a ball-and-socket joint
at C and by two cables AD and AE. Determine the tension in each
cable and the reaction at C.
SOLUTION
Free-Body Diagram:
Five unknowns and six Eqs. of equilibrium. Equilibrium is maintained (ΣMAC = 0).
!
AD = −0.8i + 0.6 j − 2.4k
AD = 2.6 m
!
AE = 0.8i + 1.2 j − 2.4k
AE = 2.8 m
!
AD TAD
(−0.8i + 0.6 j − 2.4k )
TAD =
=
AD 2.6
!
AE TAE
(0.8i + 1.2 j − 2.4k )
TAE =
=
AE 2.8
ΣM C = 0: rA × TAD + rA × TAE + rA × (−3.6 kN) j
Factor rA :
or:
Coefficient of i:
rA × (TAD + TAE − (3.6 kN) j)
TAD + TAE − (3 kN) j = 0
−
(Forces concurrent at A)
TAD
T
(0.8) + AE (0.8) = 0
2.6
2.8
TAD =
Coefficient of j:
2.6
TAE
2.8
(1)
TAD
T
(0.6) + AE (1.2) − 3.6 kN = 0
2.6
2.8
2.6
0.6 ! 1.2
TAE "
TAE − 3.6 kN = 0
#+
2.8
$ 2.6 % 2.8
TAE "
$
0.6 + 1.2 !
# = 3.6 kN
2.8 %
TAE = 5.600 kN
TAE = 5.60 kN
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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465
PROBLEM 4.107 (Continued)
Eq. (1):
TAD =
2.6
(5.6) = 5.200 kN
2.8
0.8
0.8
(5.2 kN) +
(5.6 kN) = 0;
2.6
2.8
0.6
1.2
(5.2 kN) +
(5.6 kN) − 3.6 kN = 0
ΣFy = 0: C y +
2.6
2.8
2.4
2.4
(5.2 kN) −
(5.6 kN) = 0
ΣFz = 0: Cz −
2.6
2.8
ΣFx = 0: C x −
TAD = 5.20 kN
!
C = (9.60 kN)k
!
Cx = 0
Cy = 0
Cz = 9.60 kN
Note: Since forces and reaction are concurrent at A, we could have used the methods of Chapter 2.
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466
PROBLEM 4.108
A 600-lb crate hangs from a cable that passes over a
pulley B and is attached to a support at H. The 200-lb
boom AB is supported by a ball-and-socket joint at A
and by two cables DE and DF. The center of gravity
of the boom is located at G. Determine (a) the tension
in cables DE and DF, (b) the reaction at A.
SOLUTION
Free-Body Diagram:
WC = 600 lb
WG = 200 lb
We have five unknowns (TDE , TDF , Ax , Ay , Az ) and five equilibrium equations. The boom is free to spin about
the AB axis, but equilibrium is maintained, since ΣM AB = 0.
"
We have
BH = (30 ft)i − (22.5 ft) j
BH = 37.5 ft
"
8.8
DE = (13.8 ft)i −
(22.5 ft) j + (6.6 ft)k
12
= (13.8 ft)i − (16.5 ft) j + (6.6 ft)k
DE = 22.5 ft
"
DF = (13.8 ft)i − (16.5 ft) j − (6.6 ft)k
DF = 22.5 ft
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467
PROBLEM 4.108 (Continued)
TBH = TBH
Thus:
TDE = TDE
TDF = TDF
(a)
"
BH
30i − 22.5 j
= (600 lb)
= (480 lb)i − (360 lb) j
37.5
BH
"
DE TDE
(13.8i − 16.5 j + 6.6k )
=
DE 22.5
"
DF TDE
(13.8i − 16.5 j − 6.6k )
=
DF 22.5
ΣM A = 0: (rJ × WC ) + (rK × WG ) + (rH × TBH ) + (rE × TDE ) + (rF × TDF ) = 0
− (12i ) × (−600 j) − (6i ) × (−200 j) + (18i) × (480i − 360 j)
i
j
k
i
j
k
TDE
TDF
5
0
6.6 +
5
0
+
−6.6 = 0
22.5
22.5
13.8 −16.5 6.6
13.8 −16.5 −6.6
7200k + 1200k − 6480k + 4.84(TDE − TDF )i
or,
+
58.08
82.5
(TDE − TDF ) j −
(TDE + TDF )k = 0
22.5
22.5
Equating to zero the coefficients of the unit vectors:
i or j:
TDE − TDF = 0
k : 7200 + 1200 − 6480 −
TDE = TDF *
82.5
(2TDE ) = 0
22.5
TDE = 261.82 lb
TDE = TDF = 262 lb
(b)
13.8 !
ΣFx = 0: Ax + 480 + 2 "
# (261.82) = 0
$ 22.5 %
16.5 !
ΣFy = 0: Ay − 600 − 200 − 360 − 2 "
# (261.82) = 0
$ 22.5 %
ΣFz = 0: Az = 0
Ax = −801.17 lb
Ay = 1544.00 lb
A = −(801 lb)i + (1544 lb) j
*Remark: The fact that TDE = TDF could have been noted at the outset from the symmetry of structure with
respect to xy plane.
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468
PROBLEM 4.109
A 3-m pole is supported by a ball-and-socket joint at A and
by the cables CD and CE. Knowing that the 5-kN force acts
vertically downward (φ = 0), determine (a) the tension in
cables CD and CE, (b) the reaction at A.
SOLUTION
Free-Body Diagram:
By symmetry with xy plane
TCD = TCE = T
!
CD = −3i + 1.5 j + 1.2k
CD = 3.562 m
TCD
TCE
rB/A = 2i
"
−3i + 1.5 j + 1.2k
CD
=T
=T
CD
3.562
−3i + 1.5 j − 1.2k
=T
3.562
rC/A = 3i
ΣMA = 0: rC/A × TCD + rC/A × TCE + rB/A × (−5 kN) j = 0
i
j
k
i
j
k
i j k
T
T
+ 3 0
+ 2 0 0 =0
3 0
0
0
3.562
3.562
−3 1.5 1.2
−3 1.5 −1.2
0 −5 0
Coefficient of k:
T '
&
2 (3 × 1.5 ×
) − 10 = 0 T = 3.958 kN
3.562
*
+
ΣF = 0: A + TCD + TCE − 5 j = 0
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469
PROBLEM 4.109 (Continued)
Coefficient of k:
Az = 0
Coefficient of i:
Ax − 2[3.958 × 3/3.562] = 0
Ax = 6.67 kN
Coefficient of j:
Ay + 2[3.958 × 1.5/3.562] − 5 = 0
Ay = 1.667 kN
(a)
TCD = TCE = 3.96 kN
A = (6.67 kN)i + (1.667 kN) j
(b)
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470
PROBLEM 4.110
A 3-m pole is supported by a ball-and-socket joint at A and
by the cables CD and CE. Knowing that the line of action
of the 5-kN force forms an angle φ = 30° with the vertical
xy plane, determine (a) the tension in cables CD and CE,
(b) the reaction at A.
SOLUTION
Free-Body Diagram:
Five unknowns and six Eqs. of equilibrium but equilibrium is
maintained (ΣM AC = 0)
rB/A = 2i
rC/A = 3i
Load at B.
= −(5cos 30) j + (5sin 30)k
= −4.33j + 2.5k
!
CD = −3i + 1.5 j + 1.2k
CD = 3.562 m
!
CD
T
( −3i + 1.5 j + 1.2k )
TCD = TCD
=
CD 3.562
Similarly,
TCE =
T
(−3i + 1.5 j − 1.2k )
3.562
ΣMA = 0: rC/A × TCD + rC/A × TCE + rB/A × ( −4.33j + 2.5k ) = 0
i
j
k
i
j
k
i
j
k
TCD
TCE
+ 3
+2
3 0
0
0
0
0
0 =0
3.562
3.562
−3 1.5 1.2
−3 1.5 −1.2
0 −4.33 2.5
Equate coefficients of unit vectors to zero.
j: − 3.6
TCD
T
+ 3.6 CE − 5 = 0
3.562
3.562
−3.6TCD + 3.6TCE − 17.810 = 0
(1)
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471
PROBLEM 4.110 (Continued)
k : 4.5
TCD
T
+ 4.5 CE − 8.66 = 0
3.562
3.562
4.5TCD + 4.5TCE = 30.846
(2) + 1.25(1):
Eq. (1):
(2)
9TCE − 53.11 = 0 TCE = 5.901 kN
−3.6TCD + 3.6(5.901) − 17.810 = 0
TCD = 0.954 kN
ΣF = 0: A + TCD + TCE − 4.33j + 2.5k = 0
i : Ax +
0.954
5.901
(−3) +
(−3) = 0
3.562
3.562
Ax = 5.77 kN
j: Ay +
0.954
5.901
(1.5) +
(1.5) − 4.33 = 0
3.562
3.562
Ay = 1.443 kN
k : Az +
0.954
5.901
(1.2) +
( −1.2) + 2.5 = 0
3.562
3.562
Az = −0.833 kN
Answers:
(a)
TCD = 0.954 kN
TCE = 5.90 kN
A = (5.77 kN)i + (1.443 kN) j − (0.833 kN)k
(b)
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472
PROBLEM 4.111
A 48-in. boom is held by a ball-and-socket joint at C and
by two cables BF and DAE; cable DAE passes around a
frictionless pulley at A. For the loading shown, determine
the tension in each cable and the reaction at C.
SOLUTION
Free-Body Diagram:
Five unknowns and six Eqs. of equilibrium but equilibrium is
maintained (ΣMAC = 0).
T = Tension in both parts of cable DAE.
rB = 30k
rA = 48k
!
AD = −20i − 48k
AD = 52 in.
!
AE = 20 j − 48k
AE = 52 in.
!
BF = 16i − 30k
BF = 34 in.
!
AD T
T
TAD = T
= (−20i − 48k ) = ( −5i − 12k )
AD 52
13
!
AE T
T
TAE = T
= (20 j − 48k ) = (5 j − 12k )
AE 52
13
!
T
BF TBF
TBF = TBF
=
(16i − 30k ) = BF (8i − 15k )
BF
34
17
ΣM C = 0: rA × TAD + rA × TAE + rB × TBF + rB × (−320 lb) j = 0
i j k
i j k
i j k
T
T
T
+ 0 0 48
+ 0 0 30 BF + (30k ) × (−320 j) = 0
0 0 48
13
13
17
−5 0 −12
0 5 −12
8 0 −15
Coefficient of i:
−
240
T + 9600 = 0
13
T = 520 lb
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473
PROBLEM 4.111 (Continued)
Coefficient of j:
−
240
240
T+
TBD = 0
13
17
TBD =
17
17
T = (520) TBD = 680 lb
13
13
ΣF = 0: TAD + TAE + TBF − 320 j + C = 0
Coefficient of i:
−
20
8
(520) + (680) + Cx = 0
52
17
−200 + 320 + Cx = 0
Coefficient of j:
20
(520) − 320 + C y = 0
52
200 − 320 + C y = 0
Coefficient of k:
Cx = −120 lb
−
C y = 120 lb
48
48
30
(520) − (520) − (680) + Cz = 0
52
52
34
−480 − 480 − 600 + Cz = 0
Cz = 1560 lb
Answers: TDAE = T
TDAE = 520 lb
!
TBD = 680 lb
!
C = −(120.0 lb)i + (120.0 lb) j + (1560 lb)k
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474
PROBLEM 4.112
Solve Problem 4.111, assuming that the 320-lb load is
applied at A.
PROBLEM 4.111 A 48-in. boom is held by a ball-andsocket joint at C and by two cables BF and DAE; cable
DAE passes around a frictionless pulley at A. For the
loading shown, determine the tension in each cable and
the reaction at C.
SOLUTION
Free-Body Diagram:
Five unknowns and six Eqs. of equilibrium but equilibrium is
maintained (ΣMAC = 0).
T = tension in both parts of cable DAE.
rB = 30k
rA = 48k
!
AD = −20i − 48k
AD = 52 in.
!
AE = 20 j − 48k
AE = 52 in.
!
BF = 16i − 30k
BF = 34 in.
!
AD T
T
TAD = T
= (−20i − 48k ) = ( −5i − 12k )
AD 52
13
!
AE T
T
TAE = T
= (20 j − 48k ) = (5 j − 12k )
AE 52
13
!
T
BF TBF
TBF = TBF
=
(16i − 30k ) = BF (8i − 15k )
BF
34
17
ΣM C = 0: rA × TAD + rA × TAE + rB × TBF + rA × ( −320 lb) j = 0
i j k
i j k
i j k
T
T
T
+ 0 0 48
+ 0 0 30 BF + 48k × (−320 j) = 0
0 0 48
13
13
17
−5 0 −12
0 5 −12
8 0 −15
Coefficient of i:
−
240
T + 15360 = 0
13
T = 832 lb
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475
PROBLEM 4.112 (Continued)
Coefficient of j:
−
240
240
T+
TBD = 0
13
17
TBD =
17
17
T = (832)
13
13
TBD = 1088 lb
ΣF = 0: TAD + TAE + TBF − 320 j + C = 0
−
Coefficient of i:
20
8
(832) + (1088) + C x = 0
52
17
−320 + 512 + Cx = 0
20
(832) − 320 + C y = 0
52
Coefficient of j:
320 − 320 + C y = 0
Coefficient of k:
−
Cy = 0
48
48
30
(832) − (852) − (1088) + Cz = 0
52
52
34
−768 − 768 − 960 + Cz = 0
Answers:
Cx = −192 lb
TDAE = T
Cz = 2496 lb
TDAE = 832 lb
!
TBD = 1088 lb
!
C = −(192.0 lb)i + (2496 lb)k
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476
PROBLEM 4.113
A 20-kg cover for a roof opening is hinged at corners A and B.
The roof forms an angle of 30° with the horizontal, and the
cover is maintained in a horizontal position by the brace CE.
Determine (a) the magnitude of the force exerted by the brace,
(b) the reactions at the hinges. Assume that the hinge at A does
not exert any axial thrust.
SOLUTION
Force exerted by CD
F = F (sin 75°)i + F (cos 75°) j
F = F (0.2588i + 0.9659 j)
W = mg = 20 kg(9.81 m/s 2 ) = 196.2 N
rA/B = 0.6k
rC/B = 0.9i + 0.6k
rG/B = 0.45i + 0.3k
F = F (0.2588i + 0.9659 j)
ΣM B = 0: rG/B × (−196.2 j) + rC/B × F + rA/B × A = 0
i
j
k
i
j
k
i
0.45
0
0.3 + 0.9
0
0.6 F + 0
−196.2 0
Ax
0
0.2588 +0.9659 0
j
0
Ay
k
0.6 = 0
0
Coefficient of i :
+58.86 − 0.5796 F − 0.6 Ay = 0
(1)
Coefficient of j:
+0.1553F + 0.6 Ax = 0
(2)
Coefficient of k:
−88.29 + 0.8693F = 0: F = 101.56 N
Eq. (2):
+58.86 − 0.5796(101.56) − 0.6 Ay = 0
Eq. (3):
+0.1553(101.56) + 0.6 Ax = 0
Ay = 0
Ax = −26.29 N
F = 101.6 N
A = −(26.3 N)i
ΣF : A + B + F − Wj = 0
Coefficient of i:
26.29 + Bx + 0.2588(101.56) = 0
Bx = 0
Coefficient of j:
By + 0.9659(101.56) − 196.2 = 0 By = 98.1 N
Bz = 0
Coefficient of k:
B = (98.1 N) j
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477
PROBLEM 4.114
The bent rod ABEF is supported by bearings at C and D and by wire
AH. Knowing that portion AB of the rod is 250 mm long, determine
(a) the tension in wire AH, (b) the reactions at C and D. Assume that
the bearing at D does not exert any axial thrust.
SOLUTION
∆ABH is equilateral
Free-Body Diagram:
Dimensions in mm
rH/C = −50i + 250 j
rD/C = 300i
rF/C = 350i + 250k
T = T (sin 30°) j − T (cos 30°)k = T (0.5 j − 0.866k )
ΣM C = 0: rH/C × T + rD × D + rF/C × (−400 j) = 0
i
j
k
i
j
−50 250
0 T + 300 0
0
0.5 −0.866
0 Dy
Coefficient i:
k
i
j
k
0 + 350
0
250 = 0
Dz
0 −400 0
−216.5T + 100 × 103 = 0
T = 461.9 N
Coefficient of j:
T = 462 N
−43.3T − 300 Dz = 0
−43.3(461.9) − 300 Dz = 0
Dz = −66.67 N
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478
PROBLEM 4.114 (Continued)
Coefficient of k:
−25T + 300 D y − 140 × 103 = 0
−25(461.9) + 300 Dy − 140 × 103 = 0
D y = 505.1 N
D = (505 N) j − (66.7 N)k
ΣF = 0: C + D + T − 400 j = 0
Coefficient i:
Cx = 0
Cx = 0
Coefficient j:
C y + (461.9)0.5 + 505.1 − 400 = 0 C y = −336 N
Coefficient k:
C z − (461.9)0.866 − 66.67 = 0
C z = 467 N
C = −(336 N) j + (467 N)k
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479
PROBLEM 4.115
A 100-kg uniform rectangular plate is supported in the position
shown by hinges A and B and by cable DCE that passes over a
frictionless hook at C. Assuming that the tension is the same in
both parts of the cable, determine (a) the tension in the cable,
(b) the reactions at A and B. Assume that the hinge at B does
not exert any axial thrust.
SOLUTION
rB/A (960 − 180)i = 780i
Dimensions in mm
960
! 450
− 90 # i +
rG/A = "
k
2
$ 2
%
= 390i + 225k
rC/A = 600i + 450k
T = Tension in cable DCE
!
CD = −690i + 675 j − 450k
!
CE = 270i + 675 j − 450k
CD = 1065 mm
CE = 855 mm
T
(−690i + 675 j − 450k )
1065
T
(270i + 675 j − 450k )
TCE =
855
W = −mgi = −(100 kg)(9.81 m/s 2 ) j = −(981 N) j
TCD =
ΣM A = 0: rC/A × TCD + rC/A × TCE + rG/A × (−Wj) + rB/A × B = 0
i
j
k
i
j
k
T
T
600
0
450
450
+ 600 0
+
1065
855
270 675 −450
−690 675 −450
i
+ 390
0
j
0
−981
k
i
225 + 780
0
0
j
0
k
0 =0
By
Bz
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480
PROBLEM 4.115 (Continued)
Coefficient of i:
−(450)(675)
T
T
− (450)(675)
+ 220.725 × 103 = 0
1065
855
T = 344.6 N
Coefficient of j:
(−690 × 450 + 600 × 450)
T = 345 N
344.6
344.6
+ (270 × 450 + 600 × 450)
− 780 Bz = 0
1065
855
Bz = 185.49 N
Coefficient of k:
(600)(675)
344.6
344.6
+ (600)(675)
− 382.59 × 103 + 780 By
1065
855
By = 113.2N
B = (113.2 N) j + (185.5 N)k
ΣF = 0: A + B + TCD + TCE + W = 0
690
270
(344.6) +
(344.6) = 0
1065
855
Coefficient of i:
Ax −
Ax = 114.4 N
Coefficient of j:
Ay + 113.2 +
675
675
(344.6) +
(344.6) − 981 = 0
1065
855
Ay = 377 N
Coefficient of k:
Az + 185.5 −
450
450
(344.6) −
(344.6) = 0
1065
855
Az = 141.5 N
A = (114.4 N)i + (377 N) j + (144.5 N)k
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481
!
PROBLEM 4.116
Solve Problem 4.115, assuming that cable DCE is replaced
by a cable attached to Point E and hook C.
PROBLEM 4.115 A 100-kg uniform rectangular plate is
supported in the position shown by hinges A and B and by
cable DCE that passes over a frictionless hook at C.
Assuming that the tension is the same in both parts of the
cable, determine (a) the tension in the cable, (b) the
reactions at A and B. Assume that the hinge at B does not
exert any axial thrust.
SOLUTION
See solution to Problem 4.115 for free-body diagram and analysis leading to the following:
CD = 1065 mm
CE = 855 mm
T
(−690i + 675 j − 450k )
1065
T
(270i + 675 j − 450k )
TCE =
855
W = −mgi = −(100 kg)(9.81 m/s 2 ) j = −(981 N)j
TCD =
Now:
ΣM A = 0: rC/A × TCE + rG/A × ( −Wj) + rB/A × B = 0
i
j
k
i
j
k
i
j
T
+ 390
600 0
450
0
225 + 780 0
855
270 675 −450
0 −981 0
0 By
Coefficient of i:
−(450)(675)
k
0 =0
Bz
T
+ 220.725 × 103 = 0
855
T = 621.3 N
Coefficient of j:
Coefficient of k:
(270 × 450 + 600 × 450)
(600)(675)
T = 621 N
621.3
− 980 Bz = 0 Bz = 364.7 N
855
621.3
− 382.59 × 103 + 780 B y = 0 B y = 113.2 N
855
B = (113.2 N)j + (365 N)k
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482
PROBLEM 4.116 (Continued)
ΣF = 0: A + B + TCE + W = 0
270
(621.3) = 0
855
Coefficient of i:
Ax +
Coefficient of j:
Ay + 113.2 +
675
(621.3) − 981 = 0
855
Ay = 377.3 N
Coefficient of k:
Az + 364.7 −
450
(621.3) = 0
855
Az = −37.7 N !
!
Ax = −196.2 N
A = −(196.2 N)i + (377 N)j − (37.7 N)k
!
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483
!
PROBLEM 4.117
The rectangular plate shown weighs 75 lb and is held in the
position shown by hinges at A and B and by cable EF. Assuming
that the hinge at B does not exert any axial thrust, determine
(a) the tension in the cable, (b) the reactions at A and B.
SOLUTION
rB/A = (38 − 8)i = 30i
rE/A = (30 − 4)i + 20k
rG/A
= 26i + 20k
38
= i + 10k
2
= 19i + 10k
!
EF = 8i + 25 j − 20k
EF = 33 in.
!
AE T
T =T
= (8i + 25 j − 20k )
AE 33
ΣM A = 0: rE/A × T + rG/A × (−75 j) + rB/A × B = 0
i
j
k
i
j
k
i
j
T
+ 19 0 10 + 30 0
26 0 20
33
8 25 −20
0 −75 0
0 By
−(25)(20)
Coefficient of i:
Coefficient of j:
Coefficient of k:
(160 + 520)
(26)(25)
T
+ 750 = 0:
33
k
0 =0
Bz
T = 49.5 lb
49.5
− 30 Bz = 0: Bz = 34 lb
33
49.5
− 1425 + 30 By = 0: By = 15 lb
33
B = (15 lb)j + (34 lb)k
!
!
!
!
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484
!
PROBLEM 4.117 (Continued)
!
ΣF = 0: A + B + T − (75 lb)j = 0 !
Ax +
Coefficient of i:
Coefficient of j:
Coefficient of k:
Ay + 15 +
8
(49.5) = 0
33
25
(49.5) − 75 = 0
33
Az + 34 −
20
(49.5) = 0
33
Ax = −12.00 lb
Ay = 22.5 lb
Az = −4.00 lb !
A = −(12.00 lb)i + (22.5 lb)j − (4.00 lb)k
!
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485
!
PROBLEM 4.118
Solve Problem 4.117, assuming that cable EF is replaced by a
cable attached at points E and H.
PROBLEM 4.117 The rectangular plate shown weighs 75 lb and
is held in the position shown by hinges at A and B and by cable
EF. Assuming that the hinge at B does not exert any axial thrust,
determine (a) the tension in the cable, (b) the reactions at A and B.
SOLUTION
rB/A = (38 − 8)i = 30i
rE/A = (30 − 4)i + 20k
= 26i + 20k
38
i + 10k
2
= 19i + 10k
rG/A =
!
EH = −30i + 12 j − 20k
EH = 38 in.
!
EH T
T=T
= (−30i + 12 j − 20k )
EH 38
ΣM A = 0: rE/A × T + rG/A × (−75 j) + rB/A × B = 0
i
j
k
i
j
k
i
j
T
26 0 20
+ 19 0 10 + 30 0
38
0 −75 0
0 By
−30 12 −20
−(12)(20)
Coefficient of i:
Coefficient of j:
Coefficient of k:
(−600 + 520)
(26)(12)
T
+ 750 = 0
38
T = 118.75
k
0 =0
Bz
T = 118.8lb
118.75
− 30 Bz = 0 Bz = −8.33lb
38
118.75
− 1425 + 30 By = 0 By = 15.00 lb
38
B = (15.00lb)j − (8.33 lb)k
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486
PROBLEM 4.118 (Continued)
ΣF = 0:
Coefficient of j:
Coefficient of k:
30
(118.75) = 0
38
Ax = 93.75 lb
12
(118.75) − 75 = 0
38
Ay = 22.5 lb
Ax −
Coefficient of i:
Ay + 15 +
A + B + T − (75 lb)j = 0
Az − 8.33 −
20
(118.75) = 0
38
Az = 70.83 lb
A = (93.8 lb)i + (22.5 lb)j + (70.8 lb)k
!
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487
!
PROBLEM 4.119
Solve Problem 4.114, assuming that the bearing at D is removed
and that the bearing at C can exert couples about axes parallel to
the y and z axes.
PROBLEM 4.114 The bent rod ABEF is supported by bearings
at C and D and by wire AH. Knowing that portion AB of the rod
is 250 mm long, determine (a) the tension in wire AH, (b) the
reactions at C and D. Assume that the bearing at D does not exert
any axial thrust.
SOLUTION
Free-Body Diagram:
∆ABH is Equilateral
Dimensions in mm
rH/C = −50i + 250 j
rF/C = 350i + 250k
T = T (sin 30°) j − T (cos 30°)k = T (0.5 j − 0.866k )
ΣM C = 0: rF/C × (−400 j) + rH/C × T + ( M C ) y j + ( M C ) z k = 0
i
j
k
i
j
k
350
0
250 + −50 250
0 T + (M C ) y j + (M C ) z k = 0
0 −400 0
0
0.5 −0.866
Coefficient of i:
Coefficient of j:
+100 × 103 − 216.5T = 0 T = 461.9 N
T = 462 N
−43.3(461.9) + ( M C ) y = 0
( M C ) y = 20 × 103 N ⋅ mm
(M C ) y = 20.0 N ⋅ m
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488
PROBLEM 4.119 (Continued)
Coefficient of k:
−140 × 103 − 25(461.9) + ( M C ) z = 0
( M C ) z = 151.54 × 103 N ⋅ mm
(M C ) z = 151.5 N ⋅ m
ΣF = 0: C + T − 400 j = 0
M C = (20.0 N ⋅ m)j + (151.5 N ⋅ m)k
Coefficient of i:
Coefficient of j:
Coefficient of k:
Cx = 0
C y + 0.5(461.9) − 400 = 0 C y = 169.1 N
C z − 0.866(461.9) = 0 C z = 400 N
C = (169.1 N)j + (400 N)k
!
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489
!
PROBLEM 4.120
Solve Problem 4.117, assuming that the hinge at B is removed and
that the hinge at A can exert couples about axes parallel to the y
and z axes.
PROBLEM 4.117 The rectangular plate shown weighs 75 lb and
is held in the position shown by hinges at A and B and by cable
EF. Assuming that the hinge at B does not exert any axial thrust,
determine (a) the tension in the cable, (b) the reactions at A and B.
SOLUTION
rE/A = (30 − 4)i + 20k = 26i + 20k
rG/A = (0.5 × 38)i + 10k = 19i + 10k
!
AE = 8i + 25 j − 20k
AE = 33 in.
!
AE T
= (8i + 25 j − 20k )
T =T
AE 33
ΣM A = 0: rE/A × T + rG/A × (−75 j) + ( M A ) y j + ( M A ) z k = 0
i
j
k
i
j
k
T
26 0 20
+ 19 0 10 + ( M A ) y j + ( M A ) z k = 0
33
8 25 −20
0 −75 0
−(20)(25)
Coefficient of i:
Coefficient of j:
Coefficient of k:
(160 + 520)
(26)(25)
T
+ 750 = 0
33
T = 49.5 lb
49.5
+ ( M A ) y = 0 ( M A ) y = −1020 lb ⋅ in.
33
49.5
− 1425 + ( M A ) z = 0
33
( M A ) z = 450 lb ⋅ in.
ΣF = 0: A + T − 75 j = 0
Ax +
Coefficient of i:
8
(49.5) = 0
33
M A = −(1020 lb ⋅ in)j + (450 lb ⋅ in.)k
Ax = 12.00 lb
Coefficient of j:
Ay +
25
(49.5) − 75 = 0
33
Ay = 37.5 lb
Coefficient of k:
Az −
20
(49.5)
33
Az = 30.0 lb
A = −(12.00 lb)i + (37.5 lb)j + (30.0 lb)k
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490
PROBLEM 4.121
The assembly shown is used to control the tension T in a tape that
passes around a frictionless spool at E. Collar C is welded to rods ABC
and CDE. It can rotate about shaft FG but its motion along the shaft is
prevented by a washer S. For the loading shown, determine (a) the
tension T in the tape, (b) the reaction at C.
SOLUTION
Free-Body Diagram:
rA/C = 4.2 j + 2k
rE/C = 1.6i − 2.4 j
ΣM C = 0: rA/C × (−6 j) + rE/C × T (i + k ) + ( M C ) y j + ( M C ) z k = 0
(4.2 j + 2k ) × (−6 j) + (1.6i − 2.4 j) × T (i + k ) + ( M C ) y j + ( M C ) z k = 0
Coefficient of i:
Coefficient of j:
Coefficient of k:
12 − 2.4T = 0
T = 5 lb
−1.6(5 lb) + (M C ) y = 0 ( M C ) y = 8 lb ⋅ in.
2.4(5 lb) + ( M C ) z = 0 ( M C ) z = −12 lb ⋅ in.
M C = (8 lb ⋅ in.)j − (12 lb ⋅ in.)k
ΣF = 0: C x i + C y j + C z k − (6 lb)j + (5 lb)i + (5 lb)k = 0
Equate coefficients of unit vectors to zero.
C x = −5 lb C y = 6 lb C z = −5 lb
C = −(5.00 lb)i + (6.00 lb)j − (5.00 lb)k
!
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491
!
PROBLEM 4.122
The assembly shown is welded to collar A that fits on the vertical
pin shown. The pin can exert couples about the x and z axes but
does not prevent motion about or along the y axis. For the loading
shown, determine the tension in each cable and the reaction at A.
SOLUTION
Free-Body Diagram:
First note
TCF =
CF TCF
−(0.08 m)i + (0.06 m) j
=
(0.08) 2 + (0.06)2 m
TCF
= TCF (−0.8i + 0.6 j)
TDE =
DE TDE
=
(0.12 m)j − (0.09 m)k
(0.12) 2 + (0.09)2 m
TDE
= TDE (0.8 j − 0.6k )
(a)
From f.b.d. of assembly
ΣFy = 0: 0.6TCF + 0.8TDE − 480 N = 0
0.6TCF + 0.8TDE = 480 N
or
(1)
ΣM y = 0: − (0.8TCF )(0.135 m) + (0.6TDE )(0.08 m) = 0
TDE = 2.25TCF
or
(2)
Substituting Equation (2) into Equation (1)
0.6TCF + 0.8[(2.25)TCF ] = 480 N
TCF = 200.00 N
TCF = 200 N
or
!
and from Equation (2)
TDE = 2.25(200.00 N) = 450.00
TDE = 450 N
or
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492
!
PROBLEM 4.122 (Continued)
(b)
From f.b.d. of assembly
ΣFz = 0: Az − (0.6)(450.00 N) = 0
Az = 270.00 N
ΣFx = 0: Ax − (0.8)(200.00 N) = 0
Ax = 160.000 N
or A = (160.0 N)i + (270 N)k
ΣM x = 0:
MAx + (480 N)(0.135 m) − [(200.00 N)(0.6)](0.135 m)
− [(450 N)(0.8)](0.09 m) = 0
M Ax = −16.2000 N ⋅ m
ΣM z = 0:
MAz − (480 N)(0.08 m) + [(200.00 N)(0.6)](0.08 m)
+ [(450 N)(0.8)](0.08 m) = 0
M Az = 0
or M A = −(16.20 N ⋅ m)i
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493
PROBLEM 4.123
The rigid L-shaped member ABC is supported by a
ball-and-socket joint at A and by three cables. If a
450-lb load is applied at F, determine the tension
in each cable.
SOLUTION
Free-Body Diagram:
In this problem:
We have
Thus
a = 21 in.
!
CD = (24 in.)j − (32 in.)k CD = 40 in.
!
BD = −(42 in.)i + (24 in.)j − (32 in.)k BD = 58 in.
!
BE = (42 in.)i − (32 in.)k BE = 52.802 in.
TCD
TBD
TBE
!
CD
= TCD
= TCD (0.6 j − 0.8k )
CD
!
BD
= TBD
= TBD (−0.72414i + 0.41379 j − 0.55172k )
BD
!
BE
= TBE
= TBE (0.79542i − 0.60604k )
BE
ΣM A = 0: (rC × TCD ) + (rB × TBD ) + (rB × TBE ) + (rW × W) = 0
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494
PROBLEM 4.123 (Continued)
rC = −(42 in.)i + (32 in.)k
Noting that
rB = (32 in.)k
rW = − ai + (32 in.)k
and using determinants, we write
i
j
k
i
j
k
32 TCD +
0
0
32
−42 0
TBD
0 0.6 −0.8
−0.72414 0.41379 −0.55172
i
j
k
i
j
k
0
0
32
0
32 = 0
+
TBE + −a
0.79542 0 −0.60604
0 −450 0
Equating to zero the coefficients of the unit vectors:
i: −19.2TCD − 13.241TBD + 14400 = 0
(1)
j: − 33.6TCD − 23.172TBD + 25.453TBE = 0
(2)
k:
−25.2TCD + 450a = 0
(3)
Recalling that a = 21 in., Eq. (3) yields
TCD =
From (1):
450(21)
= 375 lb
25.2
−19.2(375) − 13.241TBD + 14400 = 0
TBD = 543.77 lb
From (2):
TCD = 375 lb
TBD = 544 lb
−33.6(375) − 23.172(543.77) + 25.453TBE = 0
TBE = 990.07 lb !
!
TBE = 990 lb
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495
!
PROBLEM 4.124
Solve Problem 4.123, assuming that the 450-lb
load is applied at C.
PROBLEM 4.123 The rigid L-shaped member
ABC is supported by a ball-and-socket joint at A
and by three cables. If a 450-lb load is applied
at F, determine the tension in each cable.
SOLUTION
See solution of Problem 4.123 for free-body diagram and derivation of Eqs. (1), (2), and (3):
−19.2TCD − 13.241TBD + 14400 = 0
(1)
−33.6TCD − 23.172TBD + 25.453TBE = 0
(2)
−25.2TCD + 450a = 0
(3)
In this problem, the 450-lb load is applied at C and we have a = 42 in. Carrying into (3) and solving for TCD ,
TCD = 750 lb
From (1):
From (2):
−19.2(750) − 13.241TBD + 14400 = 0
−33.6(750) − 0 + 25.453TBE = 0 !
TCD = 750 lb
TBD = 0
TBE = 990 lb
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496
!
PROBLEM 4.125
Frame ABCD is supported by a ball-and-socket joint at A and
by three cables. For a = 150 mm, determine the tension in
each cable and the reaction at A.
SOLUTION
First note
TDG = λ DG TDG =
−(0.48 m)i + (0.14 m)j
(0.48) 2 + (0.14) 2 m
TDG
−0.48i + 0.14 j
TDG
0.50
T
= DG (24i + 7 j)
25
=
TBE = λ BE TBE =
−(0.48 m)i + (0.2 m)k
(0.48) 2 + (0.2) 2 m
TBE
−0.48i + 0.2k
TBE
0.52
T
= BE (−12 j + 5k )
13
=
From f.b.d. of frame ABCD
7
!
ΣM x = 0: " TDG # (0.3 m) − (350 N)(0.15 m) = 0
25
$
%
TDG = 625 N
or
24
5
!
!
ΣM y = 0: " × 625 N # (0.3 m) − " TBE # (0.48 m) = 0
$ 25
%
$ 13
%
TBE = 975 N
or
7
!
ΣM z = 0: TCF (0.14 m) + " × 625 N # (0.48 m) − (350 N)(0.48 m) = 0
$ 25
%
TCF = 600 N
or
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497
PROBLEM 4.125 (Continued)
ΣFx = 0: Ax + TCF + (TBE ) x + (TDG ) x = 0
12
24
!
!
Ax − 600 N − " × 975 N # − " × 625 N # = 0
$ 13
% $ 25
%
Ax = 2100 N
ΣFy = 0: Ay + (TDG ) y − 350 N = 0
7
!
Ay + " × 625 N # − 350 N = 0
$ 25
%
Ay = 175.0 N
ΣFz = 0: Az + (TBE ) z = 0
5
!
Az + " × 975 N # = 0
$ 13
%
Az = −375 N
A = (2100 N)i + (175.0 N) j − (375 N)k
Therefore
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498
PROBLEM 4.126
Frame ABCD is supported by a ball-and-socket joint
at A and by three cables. Knowing that the 350-N
load is applied at D (a = 300 mm), determine the
tension in each cable and the reaction at A.
SOLUTION
First note
TDG = λ DG TDG =
−(0.48 m)i + (0.14 m)j
(0.48) 2 + (0.14)2 m
TDG
−0.48i + 0.14 j
TDG
0.50
T
= DG (24i + 7 j)
25
=
TBE = λ BE TBE =
−(0.48 m)i + (0.2 m)k
(0.48) 2 + (0.2) 2 m
TBE
−0.48i + 0.2k
TBE
0.52
T
= BE (−12i + 5k )
13
=
From f.b.d. of frame ABCD
7
!
ΣM x = 0: " TDG # (0.3 m) − (350 N)(0.3 m) = 0
$ 25
%
TDG = 1250 N
or
24
5
!
!
ΣM y = 0: " × 1250 N # (0.3 m) − " TBE # (0.48 m) = 0
25
13
$
%
$
%
TBE = 1950 N
or
7
!
ΣM z = 0: TCF (0.14 m) + " × 1250 N # (0.48 m) − (350 N)(0.48 m) = 0
$ 25
%
TCF = 0
or
!
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499
!
PROBLEM 4.126 (Continued)
ΣFx = 0: Ax + TCF + (TBE ) x + (TDG ) x = 0
12
24
!
!
Ax + 0 − " × 1950 N # − " × 1250 N # = 0 !
$ 13
% $ 25
%
Ax = 3000 N !
!
ΣFy = 0: Ay + (TDG ) y − 350 N = 0
7
!
Ay + " × 1250 N # − 350 N = 0
$ 25
%
Ay = 0
ΣFz = 0: Az + (TBE ) z = 0
5
!
Az + " × 1950 N # = 0
$ 13
%
Az = −750 N
A = (3000 N)i − (750 N)k
Therefore
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500
PROBLEM 4.127
Three rods are welded together to form a “corner” that is supported
by three eyebolts. Neglecting friction, determine the reactions at
A, B, and C when P = 240 lb, a = 12 in., b = 8 in., and c = 10 in.
SOLUTION
From f.b.d. of weldment
ΣM O = 0: rA/O × A + rB/O × B + rC/O × C = 0
i
12
0
j
0
Ay
k
i
0 + 0
Az Bx
j k
i
8 0 + 0
0 Bz
Cx
j k
0 10 = 0
Cy 0
(−12 Az j + 12 Ay k ) + (8 Bz i − 8 Bx k ) + (−10 C y i + 10 C x j) = 0
From i-coefficient
Bz = 1.25C y
or
j-coefficient
k-coefficient
or!
(3)
ΣF = 0: A + B + C − P = 0 !
( Bx + C x )i + ( Ay + C y − 240 lb) j + ( Az + Bz )k = 0 !
From i-coefficient
Bx + C x = 0
C x = − Bx
or
j-coefficient
or
(2)
12 Ay − 8 Bx = 0
Bx = 1.5 Ay
or
(1)
−12 Az + 10 C x = 0
C x = 1.2 Az
or
!
8 Bz − 10 C y = 0
(4)
Ay + C y − 240 lb = 0
Ay + C y = 240 lb
(5)
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501
PROBLEM 4.127 (Continued)
k-coefficient
Az + Bz = 0
Az = − Bz
or
(6)
Substituting C x from Equation (4) into Equation (2)
− Bz = 1.2 Az
(7)
Using Equations (1), (6), and (7)
Cy =
Bz
− Az
1 Bx ! Bx
=
=
=
1.25 1.25 1.25 "$ 1.2 #% 1.5
(8)
From Equations (3) and (8)
Cy =
1.5 Ay
1.5
or C y = Ay
and substituting into Equation (5)
2 Ay = 240 lb
Ay = C y = 120 lb
(9)
Using Equation (1) and Equation (9)
Bz = 1.25(120 lb) = 150.0 lb
Using Equation (3) and Equation (9)
Bx = 1.5(120 lb) = 180.0 lb
From Equation (4)
C x = −180.0 lb
From Equation (6)
Az = −150.0 lb
Therefore
A = (120.0 lb) j − (150.0 lb)k
!
!
B = (180.0 lb)i + (150.0 lb)k
!
!
C = −(180.0 lb)i + (120.0 lb) j
!
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502
PROBLEM 4.128
Solve Problem 4.127, assuming that the force P is removed and is
replaced by a couple M = +(600 lb ⋅ in.)j acting at B.
PROBLEM 4.127 Three rods are welded together to form a “corner”
that is supported by three eyebolts. Neglecting friction, determine the
reactions at A, B, and C when P = 240 lb, a = 12 in., b = 8 in., and
c = 10 in.
SOLUTION
From f.b.d. of weldment
ΣM O = 0: rA/O × A + rB/O × B + rC/O × C + M = 0
i
12
0
j
0
Ay
k
i
0 + 0
Az Bx
j k
i
8 0 + 0
0 Bz
Cx
j k
0 10 + (600 lb ⋅ in.) j = 0
Cy 0
(−12 Az j + 12 Ay k ) + (8 Bz j − 8 Bx k ) + ( −10C y i + 10C x j) + (600 lb ⋅ in.) j = 0
From i-coefficient
8 Bz − 10 C y = 0
C y = 0.8Bz
or
j-coefficient
(1)
−12 Az + 10 C x + 600 = 0
C x = 1.2 Az − 60
or
k-coefficient
(2)
12 Ay − 8 Bx = 0
Bx = 1.5 Ay
or
(3)
!
ΣF = 0: A + B + C = 0 !
!
( Bx + C x )i + ( Ay + C y ) j + ( Az + Bz )k = 0 !
From i-coefficient
C x = − Bx
(4)
j-coefficient
C y = − Ay
(5)
k-coefficient
Az = − Bz
(6)
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503
PROBLEM 4.128 (Continued)
Substituting C x from Equation (4) into Equation (2)
B !
Az = 50 − " x #
$ 1.2 %
(7)
2!
C y = 0.8 Bz = − 0.8 Az = " # Bx − 40
$3%
(8)
Using Equations (1), (6), and (7)
From Equations (3) and (8)
C y = Ay − 40
Substituting into Equation (5)
2 Ay = 40
Ay = 20.0 lb
From Equation (5)
C y = −20.0 lb
Equation (1)
Bz = −25.0 lb
Equation (3)
Bx = 30.0 lb
Equation (4)
C x = −30.0 lb
Equation (6)
Az = 25.0 lb
A = (20.0 lb) j + (25.0 lb)k
!
!
! B = (30.0 lb)i − (25.0 lb)k
!
!
C = − (30.0 lb)i − (20.0 lb) j
!
Therefore
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504
PROBLEM 4.129
In order to clean the clogged drainpipe AE, a plumber has
disconnected both ends of the pipe and inserted a power
snake through the opening at A. The cutting head of the
snake is connected by a heavy cable to an electric motor that
rotates at a constant speed as the plumber forces the cable
into the pipe. The forces exerted by the plumber and
the motor on the end of the cable can be represented by
the wrench F = −(48 N)k , M = −(90 N ⋅ m)k. Determine the
additional reactions at B, C, and D caused by the cleaning
operation. Assume that the reaction at each support consists
of two force components perpendicular to the pipe.
SOLUTION
From f.b.d. of pipe assembly ABCD
ΣFx = 0: Bx = 0
ΣM D ( x -axis) = 0: (48 N)(2.5 m) − Bz (2 m) = 0
Bz = 60.0 N
and B = (60.0 N)k
ΣM D ( z -axis) = 0: C y (3 m) − 90 N ⋅ m = 0
C y = 30.0 N
ΣM D ( y -axis) = 0: −C z (3 m) − (60.0 N)(4 m) + (48 N)(4 m) = 0
C z = −16.00 N
and C = (30.0 N) j − (16.00 N)k
ΣFy = 0: Dy + 30.0 = 0
Dy = −30.0 N
ΣFz = 0: Dz − 16.00 N + 60.0 N − 48 N = 0
Dz = 4.00 N
and D = − (30.0 N) j + (4.00 N)k
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505
!
PROBLEM 4.130
Solve Problem 4.129, assuming that the plumber exerts a force
F = −(48 N)k and that the motor is turned off (M = 0).
PROBLEM 4.129 In order to clean the clogged drainpipe AE, a
plumber has disconnected both ends of the pipe and inserted a
power snake through the opening at A. The cutting head of the snake
is connected by a heavy cable to an electric motor that rotates at a
constant speed as the plumber forces the cable into the pipe. The
forces exerted by the plumber and the motor on the end of the cable
can be represented by the wrench F = −(48 N)k , M = −(90 N ⋅ m)k.
Determine the additional reactions at B, C, and D caused by the
cleaning operation. Assume that the reaction at each support consists
of two force components perpendicular to the pipe.
SOLUTION
From f.b.d. of pipe assembly ABCD
ΣFx = 0: Bx = 0
ΣM D ( x -axis) = 0: (48 N)(2.5 m) − Bz (2 m) = 0
Bz = 60.0 N
and B = (60.0 N)k
ΣM D ( z -axis) = 0: C y (3 m) − Bx (2 m) = 0
Cy = 0
ΣM D ( y -axis) = 0: C z (3 m) − (60.0 N)(4 m) + (48 N)(4 m) = 0
C z = −16.00 N
and C = − (16.00 N)k
ΣFy = 0: Dy + C y = 0
Dy = 0
ΣFz = 0: Dz + Bz + C z − F = 0
Dz + 60.0 N − 16.00 N − 48 N = 0
Dz = 4.00 N
and D = (4.00 N)k
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506
!
PROBLEM 4.131
The assembly shown consists of an 80-mm rod
AF that is welded to a cross consisting of four
200-mm arms. The assembly is supported by a
ball-and-socket joint at F and by three short
links, each of which forms an angle of 45° with
the vertical. For the loading shown, determine
(a) the tension in each link, (b) the reaction at F.
SOLUTION
rE/F = −200 i + 80 j
TB = TB (i − j) / 2
rB/F = 80 j − 200k
TC = TC (− j + k ) / 2 rC/F = 200i + 80 j
TD = TD (− i + j) / 2
rD/E = 80 j + 200k
ΣM F = 0: rB/F × TB + rC/F × TC + rD/F × TD + rE/F × (− Pj) = 0
i j
k
i
j k
i
j
k
i
j k
Tc
TB
TD
0 80 −200
+ 200 80 0
+ 0 80 200
+ −200 80 0 = 0
2
2
2
1 −1
0
0 −1 1
−1 −1 0
0
−P 0
Equate coefficients of unit vectors to zero and multiply each equation by 2.
i : − 200 TB + 80 TC + 200 TD = 0
(1)
j: − 200 TB − 200 TC − 200 TD = 0
(2)
k : − 80 TB − 200 TC + 80 TD + 200 2 P = 0
80
(2):
200
− 80 TB − 80 TC − 80 TD = 0
Eqs. (3) + (4):
−160TB − 280TC + 200 2 P = 0
Eqs. (1) + (2):
−400TB − 120TC = 0
TB = −
(3)
(4)
(5)
120
TC − 0.3TC
400
(6)
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507
PROBLEM 4.131 (Continued)
Eqs. (6)
−160( −0.3TC ) − 280TC + 200 2 P = 0
(5):
−232TC + 200 2 P = 0
TC = 1.2191P
TC = 1.219 P
From Eq. (6):
TB = −0.3(1.2191P) = − 0.36574 = P
From Eq. (2):
− 200(− 0.3657 P) − 200(1.2191P) − 200Tθ D = 0
TB = −0.366 P
TD = − 0.8534 P
ΣF = 0:
TD = − 0.853P
F + TB + TC + TD − Pj = 0
i : Fx +
(− 0.36574 P)
2
−
Fx = − 0.3448P
j: Fy −
k : Fz +
=0
2
Fx = − 0.345P
(− 0.36574 P)
2
Fy = P
( − 0.8534 P)
−
(1.2191P)
2
−
(− 0.8534 P)
2
− 200 = 0
Fy = P
(1.2191P)
2
=0
Fz = − 0.8620 P Fz = − 0.862 P
F = − 0.345 Pi + Pj − 0.862 Pk
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508
!
PROBLEM 4.132
The uniform 10-kg rod AB is supported by a ball-and-socket joint at
A and by the cord CG that is attached to the midpoint G of the rod.
Knowing that the rod leans against a frictionless vertical wall at B,
determine (a) the tension in the cord, (b) the reactions at A and B.
SOLUTION
Free-Body Diagram:
Five unknowns and six Eqs. of equilibrium. But equilibrium is
maintained (ΣMAB = 0)
W = mg
= (10 kg) 9.81m/s 2
W = 98.1 N
!
GC = − 300i + 200 j − 225k GC = 425 mm
!
GC
T
=
T =T
(− 300i + 200 j − 225k )
GC 425
rB/ A = − 600i + 400 j + 150 mm
rG/ A = − 300i + 200 j + 75 mm
ΣMA = 0: rB/ A × B + rG/ A × T + rG/ A × (− Wj) = 0
i
j
k
i
j
k
i
j
k
T
− 600 400 150 + − 300 200
75
+ − 300 200 75
425
B
0
0
− 300 200 − 225
0
− 98.1 0
Coefficient of i : (−105.88 − 35.29)T + 7357.5 = 0
T = 52.12 N
T = 52.1 N
!
!
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509
!
PROBLEM 4.132 (Continued)
Coefficient of j : 150 B − (300 × 75 + 300 × 225)
52.12
=0
425
B = 73.58 N
B = (73.6 N)i
ΣF = 0: A + B + T − Wj = 0
Coefficient of i : Ax + 73.58 − 52.15
300
=0
425
Ax = 36.8 N
Coefficient of j: Ay + 52.15
200
− 98.1 = 0
425
Ay = 73.6 N
Coefficient of k : Az − 52.15
225
=0
425
Az = 27.6 N
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510
!
PROBLEM 4.133
The bent rod ABDE is supported by ball-and-socket joints at A
and E and by the cable DF. If a 60-lb load is applied at C as
shown, determine the tension in the cable.
SOLUTION
Free-Body Diagram:
!
DF = −16i + 11j − 8k
DF = 21 in.
!
DE T
= (−16i + 11j − 8k )
T=T
DF 21
rD/E = 16 i
rC/E = 16i − 14k
!
EA 7i − 24k
=
EA =
25
EA
ΣM EA = 0:
EA
⋅ (rB/E × T) +
EA
⋅ (rC/E ⋅ (− 60 j)) = 0
7
0 −24
7
0 −24
T
1
+ 16 0 −14
=0
16 0
0
21 × 25
25
−16 11 −8
0 −60 0
−
−7 × 14 × 60 + 24 × 16 × 60
24 × 16 × 11
=0
T+
21 × 25
25
201.14 T + 17,160 = 0
T = 85.314 lb
T = 85.3 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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511
!
PROBLEM 4.134
Solve Problem 4.133, assuming that cable DF is replaced by a
cable connecting B and F.
SOLUTION
Free-Body Diagram:
rB/ A = 9i
rC/ A = 9i + 10k
!
BF = −16i + 11j + 16k
BF = 25.16 in.
!
BF
T
(−16i + 11j + 16k )
=
T =T
BF 25.16
!
AE 7i − 24k
=
AE =
AE
25
ΣMAE = 0:
AF
⋅ (rB/ A × T) +
AE
⋅ (rC/ A ⋅ (− 60 j)) = 0
7
0 −24
7 0 −24
1
T
9
0
0
+9 0
10
=0
25 × 25.16
25
−16 11 16
0 −60 0
−
24 × 9 × 11
24 × 9 × 60 + 7 × 10 × 60
T+
25 × 25.16
25
94.436 T − 17,160 = 0
T = 181.7 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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512
!
PROBLEM 4.135
The 50-kg plate ABCD is supported by hinges along edge AB
and by wire CE. Knowing that the plate is uniform, determine
the tension in the wire.
SOLUTION
Free-Body Diagram:
W = mg = (50 kg)(9.81 m/s 2 )
W = 490.50 N
!
CE = − 240i + 600 j − 400k
CE = 760 mm
!
CE
T
(− 240i + 600 j − 400k )
T =T
=
CE 760
!
AB 480i − 200 j 1
=
=
= (12i − 5 j)
AB
520
13
AB
ΣMAB = 0:
AB
⋅ (rE/ A × T ) +
AB
⋅ (rG/ A × − Wj) = 0
rE/ A = 240i + 400 j; rG/ A = 240i − 100 j + 200k
−5
−5
12
0
12
0
T
1
+ 240 −100 200
=0
240 400
0
13 × 20
13
− 240 600 − 400
−W
0
0
(−12 × 400 × 400 − 5 × 240 × 400)
T
+ 12 × 200W = 0
760
T = 0.76W = 0.76(490.50 N)
T = 373 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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513
!
PROBLEM 4.136
Solve Problem 4.135, assuming that wire CE is replaced by a
wire connecting E and D.
PROBLEM 4.135 The 50-kg plate ABCD is supported by
hinges along edge AB and by wire CE. Knowing that the plate
is uniform, determine the tension in the wire.
SOLUTION
Free-Body Diagram:
Dimensions in mm
W = mg = (50 kg)(9.81 m/s 2 )
W = 490.50 N
!
DE = − 240i + 400 j − 400k
DE = 614.5 mm
!
DE
T
(240i + 400 j − 400k )
T =T
=
DE 614.5
!
AB 480i − 200 j 1
=
= (12i − 5 j)
AB =
520
13
AB
rE/ A = 240i + 400 j; rG/ A = 240i − 100 j + 200k
12 −5
0
12
5
0
T
1
+ 240 −100 200
=0
240 400
0
13 × 614.5
13
−W
240 400 − 400
0
0
(−12 × 400 × 400 − 5 × 240 × 400)
T
+ 12 × 200 × W = 0
614.5
T = 0.6145W = 0.6145(490.50 N)
T = 301 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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514
!
PROBLEM 4.137
Two rectangular plates are welded together to form the assembly
shown. The assembly is supported by ball-and-socket joints at B
and D and by a ball on a horizontal surface at C. For the loading
shown, determine the reaction at C.
SOLUTION
λ BD =
First note
−(6 in.)i − (9 in.) j + (12 in.)k
(6) 2 + (9) 2 + (12)2 in.
1
(− 6i − 9 j + 12k )
16.1555
= − (6 in.)i
=
rA/B
P = (80 lb)k
rC/D = (8 in.)i
C = (C ) j
From the f.b.d. of the plates
ΣM BD = 0: λ BD ⋅ (rA/B × P ) + λ BD ⋅ ( rC/D × C ) = 0
−6 −9 12
−6 −9 12
& 6(80) '
& C (8) '
−1 0 0 (
+ 1 0 0 (
=0
)
*16.1555 +
* 16.1555 )+
0 0 1
0 1 0
( −9)(6)(80) + (12)(8)C = 0
C = 45.0 lb
or C = (45.0 lb) j
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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515
!
PROBLEM 4.138
Two 2 × 4-ft plywood panels, each of weight 12 lb, are nailed
together as shown. The panels are supported by ball-and-socket
joints at A and F and by the wire BH. Determine (a) the location
of H in the xy plane if the tension in the wire is to be minimum,
(b) the corresponding minimum tension.
SOLUTION
Free-Body Diagram:
!
AF = 4i − 2 j − 4k
AF
rG1/A
AF = 6 ft
1
= (2i − j − 2k )
3
= 2i − j
rG2 /A = 4i − j − 2k
rB/A = 4i
ΣM AF = 0:
AF
⋅ (rG/A × ( −12 j) +
AF
⋅ (rG 2/A × (−12 j)) +
2 −1 −2
2 −1 −2
1
1
+ 4 −1 −2 +
2 −1 0
3
3
0 −12 0
0 −12 0
1
1
(2 × 2 × 12) + (−2 × 2 × 12 + 2 × 4 × 12) +
3
3
AF
AF
⋅ (rB/A × T ) = 0
AF
AF
⋅ (rB/A × T) = −32 or T ⋅ (
⋅ (rB/A × T) = 0
⋅ (rB/A × T) = 0
A/F
× rB/A ) = −32
(1)
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516
PROBLEM 4.138 (Continued)
Projection of T on (
AF
× rB/A ) is constant. Thus, Tmin is parallel to
AF
Corresponding unit vector is
1
5
1
1
× rB/A = (2i − j − 2k ) × 4i = (−8 j + 4k )
3
3
(−2 j + k )
Tmin = T ( −2 j + k )
Eq. (1):
1
(2)
5
T
&1
'
(−2 j + k ) ⋅ ( (2i − j − 2k ) × 4i ) = −32
3
5
*
+
T
1
(−2 j + k ) ⋅ ( −8 j + 4k ) = −32
3
5
T
(16 + 4) = −32
3 5
T =−
3 5(32)
= 4.8 5
20
T = 10.7331 lb
Eq. (2)
Tmin = T ( −2 j + k )
1
5
= 4.8 5( −2 j + k )
1
5
Tmin = −(9.6 lb)j + (4.8 lb k )
Since Tmin has no i component, wire BH is parallel to the yz plane, and x = 4 ft.
(a)
(b)
x = 4.00 ft;
y = 8.00 ft
Tmin = 10.73 lb
!
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517
!
PROBLEM 4.139
Solve Problem 4.138, subject to the restriction that H must lie on
the y axis.
PROBLEM 4.138 Two 2 × 4-ft plywood panels, each of weight 12 lb,
are nailed together as shown. The panels are supported by ball-andsocket joints at A and F and by the wire BH. Determine (a) the
location of H in the xy plane if the tension in the wire is to be
minimum, (b) the corresponding minimum tension.
SOLUTION
Free-Body Diagram:
!
AF = 4i − 2 j − 4k
AF
rG1/A
1
= (2i − j − 2k )
3
= 2i − j
rG2 /A = 4i − j − 2k
rB/A = 4i
ΣMAF = 0:
AF
⋅ (rG/A × (−12 j) +
AF
⋅ (rG2 /A × (−12 j)) +
2 −1 2
2 −1 −2
1
1
2 −1 0 + 4 −1 −2 +
3
3
0 −12 0
0 −12 0
1
1
(2 × 2 × 12) + (−2 × 2 × 12 + 2 × 4 × 12) +
3
3
AF
AF
AF
AF
⋅ (rB/A × T ) = 0
⋅ (rB/A × T) = 0
⋅ (rB/A × T) = 0
⋅ (rB/A × T) = −32
!
BH = −4i + yj − 4k
BH = (32 + y 2 )1/2
!
BH
−4i + yj − 4k
=T
T=T
BH
(32 + y 2 )1/ 2
(1)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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518
PROBLEM 4.139 (Continued)
Eq. (1):
AF
2 −1 −2
T
⋅ (rB/A × T ) = 4 0 0
= −32
3(32 + y 2 )1/2
−4 y −4
(−16 − 8 y )T = −3 × 32(32 + y 2 )1/2
T = 96
(32 + y 2 )1/2
8 y + 16
(2)
(8y +16) 12 (32 + y 2 ) −1/ 2 (2 y ) + (32 + y 2 )1/2 (8)
dT
= 0: 96
dy
(8 y + 16) 2
Numerator = 0:
(8 y + 16) y = (32 + y 2 )8
8 y 2 + 16 y = 32 × 8 + 8 y 2
Eq. (2):
T = 96
(32 + 162 )1/ 2
= 11.3137 lb
8 × 16 + 16
y = 16.00 ft
Tmin = 11.31 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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519
!
PROBLEM 4.140
The pipe ACDE is supported by ball-and-socket joints at A and E and
by the wire DF. Determine the tension in the wire when a 640-N load is
applied at B as shown.
SOLUTION
Free-Body Diagram:
Dimensions in mm
!
AE = 480i + 160 j − 240k
AE = 560 mm
!
AE 480i + 160 j − 240k
=
=
AE
AE
560
6i + 2 j − 3k
AE =
7
rB/A = 200i
rD/A = 480i + 160 j
!
DF = −480i + 330 j − 240k ;
DF = 630 mm
!
DF
−480i + 330 j − 240k
−16i + 11j − 8k
= TDF
= TDF
TDF = TDF
630
21
DF
ΣM AE =
AE
⋅ (rD/A × TDF ) +
AE
⋅ (rB/A × (−600 j)) = 0
6
2 −3
6
2
−3
TDE
1
480 160 0
0
0 =0
+ 200
21 × 7
7
0 −640 0
−16 11 −8
3 × 200 × 640
−6 × 160 × 8 + 2 × 480 × 8 − 3 × 480 × 11 − 3 × 160 × 16
TDF +
=0
21 × 7
7
−1120TDF + 384 × 103 = 0
TDF = 342.86 N
TDF = 343 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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520
!
PROBLEM 4.141
Solve Problem 4.140, assuming that wire DF is replaced by a wire
connecting C and F.
PROBLEM 4.140 The pipe ACDE is supported by ball-and-socket
joints at A and E and by the wire DF. Determine the tension in the wire
when a 640-N load is applied at B as shown.
SOLUTION
Free-Body Diagram:
Dimensions in mm
!
AE = 480i + 160 j − 240k
AE = 560 mm
!
AE 480i + 160 j − 240k
=
=
AE
560
AE
6i + 2 j − 3k
AE =
7
rB/A = 200i
rC/A = 480i
!
CF = −480i + 490 j − 240k ; CF = 726.70 mm
!
CE −480i + 490 j − 240k
TCF = TCF
=
CF
726.70
ΣMAE = 0:
AE
⋅ (rC/A × TCF ) +
AE
⋅ (rB/A × (−600 j)) = 0
6
2
6
2
−3
−3
TCF
1
480
0
0
0
0 =0
+ 200
726.7 × 7
7
0 −640 0
−480 +490 −240
2 × 480 × 240 − 3 × 480 × 490
3 × 200 × 640
TCF +
=0
726.7 × 7
7
−653.91TCF + 384 × 103 = 0
TCF = 587 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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521
!
PROBLEM 4.142
A hand truck is used to move two kegs, each of mass 40 kg.
Neglecting the mass of the hand truck, determine (a) the vertical
force P that should be applied to the handle to maintain equilibrium
when α = 35°, (b) the corresponding reaction at each of the two
wheels.
SOLUTION
Free-Body Diagram:
W = mg = (40 kg)(9.81 m/s 2 ) = 392.40 N
a1 = (300 mm)sinα − (80 mm)cosα
a2 = (430 mm)cosα − (300 mm)sinα
b = (930 mm)cosα
From free-body diagram of hand truck
Dimensions in mm
ΣM B = 0: P(b) − W ( a2 ) + W (a1 ) = 0
(1)
ΣFy = 0: P − 2W + 2 B = 0
(2)
α = 35°
For
a1 = 300sin 35° − 80 cos 35° = 106.541 mm
a2 = 430 cos 35° − 300sin 35° = 180.162 mm
b = 930cos 35° = 761.81 mm
(a)
From Equation (1)
P(761.81 mm) − 392.40 N(180.162 mm) + 392.40 N(106.54 mm) = 0
P = 37.921 N
(b)
or P = 37.9 N
From Equation (2)
37.921 N − 2(392.40 N) + 2 B = 0
or B = 373 N
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522
PROBLEM 4.143
Determine the reactions at A and C when (a) α = 0,
(b) α = 30°.
SOLUTION
(a)
α = 0°
From f.b.d. of member ABC
ΣM C = 0: (300 N)(0.2 m) + (300 N)(0.4 m) − A(0.8 m) = 0
A = 225 N
or
A = 225 N
ΣFy = 0: C y + 225 N = 0
C y = −225 N or C y = 225 N
ΣFx = 0: 300 N + 300 N + C x = 0
C x = −600 N or C x = 600 N
Then
C = Cx2 + C y2 = (600) 2 + (225) 2 = 640.80 N
and
θ = tan −1 "
Cy !
−1 −225 !
# = tan "
# = 20.556°
$ −600 %
$ Cx %
or
(b)
C = 641 N
20.6°
α = 30°
From f.b.d. of member ABC
ΣM C = 0: (300 N)(0.2 m) + (300 N)(0.4 m) − ( A cos 30°)(0.8 m)
+ ( A sin 30°)(20 in.) = 0
A = 365.24 N
or
A = 365 N
60.0°
ΣFx = 0: 300 N + 300 N + (365.24 N) sin 30° + C x = 0
Cx = −782.62
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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523
PROBLEM 4.143 (Continued)
ΣFy = 0: C y + (365.24 N) cos 30° = 0
C y = −316.31 N or C y = 316 N
Then
C = Cx2 + C y2 = (782.62) 2 + (316.31) 2 = 884.12 N
and
θ = tan −1 "
Cy !
−1 −316.31 !
# = tan "
# = 22.007°
$ −782.62 %
$ Cx %
C = 884 N
or
22.0°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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524
PROBLEM 4.144
A lever AB is hinged at C and attached to a control cable at A. If the
lever is subjected to a 75-lb vertical force at B, determine (a) the
tension in the cable, (b) the reaction at C.
SOLUTION
Free-Body Diagram:
Geometry:
x AC = (10 in.) cos 20° = 9.3969 in.
y AC = (10 in.) sin 20° = 3.4202 in.
, yDA = 12 in. − 3.4202 in. = 8.5798 in.
yDA !
−1 8.5798 !
# = tan "
# = 42.397°
$ 9.3969 %
$ x AC %
α = tan −1 "
β = 90° − 20° − 42.397° = 27.603°
Equilibrium for lever:
ΣM C = 0: TAD cos 27.603°(10 in.) − (75 lb)[(15 in.)cos 20°] = 0
(a)
TAD = 119.293 lb
TAD = 119.3 lb
ΣFx = 0: C x + (119.293 lb) cos 42.397° = 0
(b)
Cx = −88.097 lb
ΣFy = 0: C y − 75 lb − (119.293 lb) sin 42.397° = 0
C y = 155.435
Thus:
C = Cx2 + C y2 = (−88.097) 2 + (155.435) 2 = 178.665 lb
and
θ = tan −1
Cy
Cx
= tan −1
155.435
= 60.456°
88.097
C = 178.7 lb
60.5°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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525
PROBLEM 4.145
Neglecting friction and the radius of the pulley, determine
(a) the tension in cable ADB, (b) the reaction at C.
SOLUTION
Free-Body Diagram:
Dimensions in mm
Geometry:
Distance
AD = (0.36) 2 + (0.150) 2 = 0.39 m
Distance
BD = (0.2)2 + (0.15)2 = 0.25 m
Equilibrium for beam:
ΣM C = 0:
(a)
0.15 !
0.15 !
T # (0.36 m) − "
T # (0.2 m) = 0
(120 N)(0.28 m) − "
$ 0.39 %
$ 0.25 %
T = 130.000 N
or
T = 130.0 N
0.36 !
0.2 !
ΣFx = 0: Cx + "
(130.000 N) + "
#
# (130.000 N) = 0
$ 0.39 %
$ 0.25 %
(b)
Cx = − 224.00 N
0.15 !
0.15 !
ΣFy = 0: C y + "
(130.00 N) + "
#
# (130.00 N) − 120 N = 0
$ 0.39 %
$ 0.25 %
C y = − 8.0000 N
2
Thus:
C = Cx2 + C y2 = (−224) 2 + ( − 8 ) = 224.14 N
and
θ = tan −1
Cy
Cx
= tan −1
8
= 2.0454°
224
C = 224 N
2.05°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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526
!
PROBLEM 4.146
The T-shaped bracket shown is supported by a small wheel at E and pegs at C
and D. Neglecting the effect of friction, determine the reactions at C, D, and E
when θ = 30°.
SOLUTION
Free-Body Diagram:
ΣFy = 0: E cos 30° − 20 − 40 = 0
E=
60 lb
= 69.282 lb
cos 30°
E = 69.3 lb
60.0°
ΣM D = 0: (20 lb)(4 in.) − (40 lb)(4 in.)
− C (3 in.) + E sin 30°(3 in.) = 0
−80 − 3C + 69.282(0.5)(3) = 0
C = 7.9743 lb
C = 7.97 lb
ΣFx = 0: E sin 30° + C − D = 0
(69.282 lb)(0.5) + 7.9743 lb − D = 0
D = 42.615 lb
D = 42.6 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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527
!
PROBLEM 4.147
The T-shaped bracket shown is supported by a small wheel at E and pegs
at C and D. Neglecting the effect of friction, determine (a) the smallest
value of θ for which the equilibrium of the bracket is maintained, (b) the
corresponding reactions at C, D, and E.
SOLUTION
Free-Body Diagram:
ΣFy = 0: E cos θ − 20 − 40 = 0
E=
60
cos θ
(1)
ΣM D = 0: (20 lb)(4 in.) − (40 lb)(4 in.) − C (3 in.)
!
60
+"
sin θ # 3 in. = 0
θ
cos
$
%
1
C = (180 tan θ − 80)
3
(a)
For C = 0,
180 tan θ = 80
tan θ =
Eq. (1)
E=
4
θ = 23.962°
9
θ = 24.0°
60
= 65.659
cos 23.962°
ΣFx = 0: −D + C + E sin θ = 0
D = (65.659) sin 23.962 = 26.666 lb
(b)
C = 0 D = 26.7 lb
E = 65.71 lb
66.0°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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528
PROBLEM 4.148
For the frame and loading shown, determine the reactions at A and C.
SOLUTION
Since member AB is acted upon by two forces, A and B, they must be colinear, have the same magnitude, and
be opposite in direction for AB to be in equilibrium. The force B acting at B of member BCD will be equal in
magnitude but opposite in direction to force B acting on member AB. Member BCD is a three-force body with
member forces intersecting at E. The f.b.d.’s of members AB and BCD illustrate the above conditions. The
force triangle for member BCD is also shown. The angle β is found from the member dimensions:
6 in. !
# = 30.964°
10
$ in. %
β = tan −1 "
Applying of the law of sines to the force triangle for member BCD,
30 lb
B
C
=
=
sin(45° − β ) sin β sin135°
or
30 lb
B
C
=
=
sin14.036° sin 30.964° sin135°
A= B=
(30 lb)sin 30.964°
= 63.641 lb
sin14.036°
or
and
C=
A = 63.6 lb
45.0°
C = 87.5 lb
59.0°
(30 lb) sin135°
= 87.466 lb
sin14.036°
or
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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529
PROBLEM 4.149
Determine the reactions at A and B when β = 50°.
SOLUTION
Free-Body Diagram: (Three-force body)
Reaction A must pass through Point D where 100-N force
and B intersect
In right ∆ BCD
α = 90° − 75° = 15°
BD = 250 tan 75° = 933.01 mm
Dimensions in mm
In right ∆ ABD
tan γ =
150 mm
AB
=
BD 933.01 mm
γ = 9.13
Force Triangle
Law of sines
100 N
A
B
=
=
sin 9.13° sin15° sin155.87°
A = 163.1 N; B = 257.6 N
A = 163.1 N
74.1° B = 258 N
65.0°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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530
PROBLEM 4.150
The 6-m pole ABC is acted upon by a 455-N force as shown. The
pole is held by a ball-and-socket joint at A and by two cables BD
and BE. For a = 3 m, determine the tension in each cable and the
reaction at A.
SOLUTION
Free-Body Diagram:
Five unknowns and six Eqs. of equilibrium, but equilibrium is maintained
(ΣM AC = 0)
rB = 3j
rC = 6 j
!
CF = −3i − 6 j + 2k CF = 7 m
!
BD = 1.5i − 3j − 3k BD = 4.5 m
!
BE = 1.5i − 3j + 3k BE = 4.5 m
!
CF P
P=P
= (−3i − 6 j + 2k )
CE 7
!
T
BD TBD
(1.5i − 3j − 3k ) = BD (i − 2 j − 2k )
TBD = TBD
=
BD 4.5
3
!
BE TBD
TBE = TBE =
=
(i − 2 j + 2k )
BE
3
ΣM A = 0: rB × TBD + rB × TBE + rC × P = 0
i j k
i j k
i
j k
TBD
TBE
P
+ 0 3 0
+ 0 6 0
=0
0 3 0
3
3
7
−3 −6 2
1 −2 −2
1 −2 2
Coefficient of i:
−2TBD + 2TBE +
12
P=0
7
(1)
Coefficient of k:
−TBD − TBF +
18
P=0
7
(2)
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531
PROBLEM 4.150 (Continued)
Eq. (1) + 2 Eq. (2):
Eq. (2):
− 4TBD +
−
48
12
P = 0 TBD = P
7
7
12
18
6
P − TBE + P = 0 TBE = P
7
7
7
P = 445 N TBD =
Since
TBE =
12
(455)
7
6
(455)
7
TBD = 780 N
TBE = 390 N
ΣF = 0: TBD + TBE + P + A = 0
780 390 455
(3) + Ax = 0
+
−
3
3
7
Coefficient of i:
260 + 130 − 195 + Ax = 0
Coefficient of j:
−
780
390
455
(2) −
(2) −
(6) + Ay = 0
3
3
7
−520 − 260 − 390 + Ay = 0
Coefficient of k:
−
Ax = 195.0 N
Ay = 1170 N
780
390
455
(2) +
(2) +
(2) + Az = 0
3
3
7
−520 + 260 + 130 + Az = 0
Az = +130.0 N
A = −(195.0 N)i + (1170 N) j + (130.0 N)k
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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532
!
PROBLEM 4.151
Solve Problem 4.150 for a = 1.5 m.
PROBLEM 4.150 The 6-m pole ABC is acted upon by a 455-N
force as shown. The pole is held by a ball-and-socket joint at
A and by two cables BD and BE. For a = 3 m, determine the
tension in each cable and the reaction at A.
SOLUTION
Free-Body Diagram:
Five unknowns and six Eqs. of equilibrium but equilibrium is maintained
(ΣM AC = 0)
rB = 3j
ΣM A = 0: rB × TBD
rC = 6 j
!
CF = −1.5i − 6 j + 2k CF = 6.5 m
!
BD = 1.5i − 3j − 3k BD = 4.5 m
!
BE = 1.5i − 3j + 3k BE = 4.5 m
!
CF
P
P
( −1.5i − 6 j + 2k ) = (−3i − 12 j + 4k )
P=P
=
CE 6.5
13
!
T
BD TBD
(1.5i − 3j − 3k ) = BD (i − 2 j − 2k )
TBD = TBD
=
BD 4.5
3
!
BE TBD
TBE = TBE =
=
(i − 2 j + 2k )
BE
3
+ rB × TBE + rC × P = 0
i j k
i j k
i
j
k
TBD
TBE
P
+ 0 3 0
+ 0
=0
0 3 0
6
0
3
3
13
−3 −12 + 4
1 −2 −2
1 −2 2
Coefficient of i:
−2TBD + 2TBE +
24
P=0
13
(1)
Coefficient of k:
−TBD − TBE +
18
P=0
13
(2)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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533
PROBLEM 4.151 (Continued)
Eq. (1) + 2 Eq. (2):
Eq (2):
−4TBD +
−
60
15
P = 0 TBD = P
13
13
15
18
3
P − TBE + P = 0 TBE = P
13
13
13
P = 445 N TBD =
Since
TBE =
15
(455)
13
3
(455)
13
TBD = 525 N
!
TBE = 105.0 N
!
ΣF = 0: TBD + TBE + P + A = 0
525 105 455
(3) + Ax = 0
+
−
3
3
13
Coefficient of i:
175 + 35 − 105 + Ax = 0
Coefficient of j:
−
525
105
455
(2) −
(2) −
(12) + Ay = 0
3
3
13
−350 − 70 − 420 + Ay = 0
Coefficient of k:
Ax = 105.0 N
−
Ay = 840 N
525
105
455
(2) +
(2) +
(4) + Az = 0
3
3
13
−350 + 70 + 140 + Az = 0
Az = 140.0 N
A = −(105.0 N)i + (840 N) j + (140.0 N)k
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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534
PROBLEM 4.152
The rigid L-shaped member ABF is supported by a ball-and-socket
joint at A and by three cables. For the loading shown, determine the
tension in each cable and the reaction at A.
SOLUTION
Free-Body Diagram:
rB/A = 12i
rF/A = 12 j − 8k
rD/A = 12i − 16k
rE/A = 12i − 24k
rF/A = 12i − 32k
!
BG = −12i + 9k
BG = 15 in.
BG = −0.8i + 0.6k
!
DH = −12i + 16 j; DH = 20 in.; λDH = −0.6i + 0.8 j
!
FJ = −12i + 16 j; FJ = 20 in.; λFJ = −0.6i + 0.8 j
ΣM A = 0: rB/A × TBG λBG + rDH × TDH λDH + rF/A × TFJ λFJ
+rF/A × (−24 j) + rE/A × ( −24 j) = 0
i
j k
i
j
k
i
j
k
12 0 0 TBG + 12
0 −16 TDH + 12
0 −32 TFJ
−0.8 0 0.6
−0.6 0.8 0
−0.6 0.8 0
i
j
k
i
j
k
+ 12 0 −8 + 12 0 −24 = 0
0 −24 0
0 −24 0
Coefficient of i:
+12.8TDH + 25.6TFJ − 192 − 576 = 0
(1)
Coefficient of k:
+9.6TDH + 9.6TFJ − 288 − 288 = 0
(2)
3
4
Eq. (1) − Eq. (2):
9.6TFJ = 0
TFJ = 0
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535
PROBLEM 4.152 (Continued)
Eq. (1):
12.8TDH − 268 = 0
Coefficient of j:
−7.2TBG + (16 × 0.6)(60.0 lb) = 0
ΣF = 0:
TDH = 60 lb
TBG = 80.0 lb
A + TBG λ BG + TDH λ DH + TFJ − 24 j − 24 j = 0
Coefficient of i:
Ax + (80)( −0.8) + (60.0)(−0.6) = 0
Ax = 100.0 lb
Coefficient of j:
Ay + (60.0)(0.8) − 24 − 24 = 0
Ay = 0
Coefficient of k:
Az + (80.0)(+0.6) = 0
Az = −48.0 lb
A = (100.0 lb)i − (48.0 lb) j
Note: The value Ay = 0
Can be confirmed by considering ΣM BF = 0
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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536
PROBLEM 4.153
A force P is applied to a bent rod ABC, which may be supported in four different ways as shown. In each case,
if possible, determine the reactions at the supports.
SOLUTION
(a)
ΣM A = 0: − Pa + (C sin 45°)2a + (cos 45°)a = 0
3
2
C
=P C=
P
3
2
2 ! 1
ΣFx = 0: Ax − "
P
" 3 ## 2
$
%
C = 0.471P
Ax =
2 ! 1
ΣFy = 0: Ay − P + "
P
" 3 ## 2
$
%
P
3
Ay =
2P
3
A = 0.745P
(b)
45°
63.4°
ΣM C = 0: +Pa − ( A cos 30°)2a + ( A sin 30°)a = 0
A(1.732 − 0.5) = P
A = 0.812 P
A = 0.812P
60.0°
ΣFx = 0: (0.812 P)sin 30° + Cx = 0 C x = −0.406 P
ΣFy = 0: (0.812 P) cos 30° − P + C y = 0 C y = −0.297 P
C = 0.503P
36.2°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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537
PROBLEM 4.153 (Continued)
ΣM C = 0: + Pa − ( A cos 30°)2a + ( A sin 30°)a = 0
(c)
A(1.732 + 0.5) = P
A = 0.448P
A = 0.448P
ΣFx = 0: − (0.448P) sin 30° + Cx = 0
60.0°
Cx = 0.224 P
ΣFy = 0: (0.448 P) cos 30° − P + C y = 0 C y = 0.612 P
C = 0.652P
69.9°
!
(d)
Force T exerted by wire and reactions A and C all intersect at Point D.
ΣM D = 0: Pa = 0
Equilibrium not maintained
Rod is improperly constrained
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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538
!
CHAPTER 5
PROBLEM 5.1
Locate the centroid of the plane area shown.
SOLUTION
Dimensions in mm
Then
A, mm 2
x , mm
y , mm
x A, mm3
y A, mm3
1
6300
105
15
0.66150 × 106
0.094500 × 106
2
9000
225
150
2.0250 × 106
1.35000 × 106
Σ
15300
2.6865 × 106
1.44450 × 106
X=
Σ x A 2.6865 × 106
=
15300
ΣA
Y =
Σ y A 1.44450 × 106
=
ΣA
15300
X = 175.6 mm
Y = 94.4 mm
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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541
PROBLEM 5.2
Locate the centroid of the plane area shown.
SOLUTION
Dimensions in mm
Then
A, mm 2
x , mm
y , mm
x A, mm3
y A, mm3
1
1200
10
30
12000
36000
2
540
30
36
16200
19440
Σ
1740
28200
55440
X=
Σ x A 28200
=
1740
ΣA
X = 16.21 mm
Y =
Σ y A 55440
=
1740
ΣA
Y = 31.9 mm
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
542
PROBLEM 5.3
Locate the centroid of the plane area shown.
SOLUTION
Dimensions in in.
Then
A, in.2
x , in.
y , in.
x A, in.3
y A, in.3
1
1
× 12 × 15 = 90
2
8
5
720
450
2
21 × 15 = 315
22.5
7.5
7087.5
2362.5
Σ
405.00
7807.5
2812.5
X=
Σ x A 7807.5
=
Σ A 405.00
X = 19.28 in.
Y =
Σ y A 2812.5
=
Σ A 405.00
Y = 6.94 in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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543
PROBLEM 5.4
Locate the centroid of the plane area shown.
SOLUTION
Then
A, in.2
x , in.
y , in.
x A, in.3
y A, in.3
1
1
(12)(6) = 36
2
4
4
144
144
2
(6)(3) = 18
9
7.5
162
135
Σ
54
306
279
XA = Σ xA
X (54) = 306
X = 5.67 in.
YA = Σ yA
Y (54) = 279
Y = 5.17 in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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544
PROBLEM 5.5
Locate the centroid of the plane area shown.
SOLUTION
Then
A, in.2
x , in.
y , in.
x A, in.3
y A, in.3
1
14 × 20 = 280
7
10
1960
2800
2
−π (4) 2 = −16π
6
12
–301.59
–603.19
Σ
229.73
1658.41
2196.8
X=
Σ xA 1658.41
=
ΣA
229.73
X = 7.22 in.
Y =
Σ y A 2196.8
=
Σ A 229.73
Y = 9.56 in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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545
PROBLEM 5.6
Locate the centroid of the plane area shown.
SOLUTION
A, mm 2
x , mm
y , mm
x A, mm3
y A, mm3
1
1
(120)(75) = 4500
2
80
25
360 × 103
112.5 × 103
2
(75)(75) = 5625
157.5
37.5
885.94 × 103
210.94 × 103
163.169
43.169
−720.86 × 103
−190.716 × 103
525.08 × 103
132.724 × 103
3
Σ
Then
−
π
4
(75) 2 = −4417.9
5707.1
XA = Σx A
X (5707.1) = 525.08 × 103
X = 92.0 mm
YA = Σ y A
Y (5707.1) = 132.724 × 103
Y = 23.3 mm
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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546
PROBLEM 5.7
Locate the centroid of the plane area shown.
SOLUTION
A, in.2
x , in.
y , in.
x A, in.3
y A, in.3
(38)2 = 2268.2
0
16.1277
0
36581
2
−20 × 16 = −320
−10
8
3200
−2560
Σ
1948.23
3200
34021
1
Then
π
2
X=
Σ xA
3200
=
Σ A 1948.23
X = 1.643 in.
Y =
Σ y A 34021
=
Σ A 1948.23
Y = 17.46 in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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547
PROBLEM 5.8
Locate the centroid of the plane area shown.
SOLUTION
1
2
Σ
Then
−
A, in.2
x , in.
y , in.
x A, in.3
y A, in.3
30 × 50 = 1500
15
25
22500
37500
23.634
30
–8353.0
–10602.9
14147.0
26.897
π
2
(15) 2 = 353.43
1146.57
X=
Σ x A 14147.0
=
Σ A 1146.57
X = 12.34 in.
Y =
Σ y A 26897
=
Σ A 1146.57
Y = 23.5 in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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548
PROBLEM 5.9
Locate the centroid of the plane area shown.
SOLUTION
1
Σ
Then
x , mm
y , mm
x A, mm3
y A, mm3
(60)(120) = 7200
–30
60
−216 × 103
432 × 103
(60) 2 = 2827.4
25.465
95.435
72.000 × 103
269.83 × 103
(60) 2 = −2827.4
–25.465
25.465
72.000 × 103
−72.000 × 103
−72.000 × 103
629.83 × 103
π
2
3
A, mm 2
4
−
π
4
7200
XA = Σ x A
X (7200) = −72.000 × 103
YA = Σ y A
Y (7200) = 629.83 × 103
X = −10.00 mm
Y = 87.5 mm
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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549
PROBLEM 5.10
Locate the centroid of the plane area shown.
SOLUTION
Dimensions in mm
A, mm 2
1
Then
π
2
× 47 × 26 = 1919.51
2
1
× 94 × 70 = 3290
2
Σ
5209.5
x , mm
y , mm
x A, mm3
y A, mm3
0
11.0347
0
21181
−15.6667
−23.333
−51543
−76766
−51543
−55584
X=
Σ x A −51543
=
ΣA
5209.5
X = −9.89 mm
Y =
Σ y A −55584
=
ΣA
5209.5
Y = −10.67 mm
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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550
PROBLEM 5.11
Locate the centroid of the plane area shown.
SOLUTION
X =0
First note that symmetry implies
A, in.2
1
2
−
π (8) 2
2
π (12) 2
2
yA, in.3
= −100.531
3.3953
–341.33
= 226.19
5.0930
1151.99
125.659
Σ
Then
y , in.
Y =
!
810.66
Σ y A 810.66 in.3
=
Σ A 125.66 in.2
or Y = 6.45 in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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551
PROBLEM 5.12
Locate the centroid of the plane area shown.
SOLUTION
Then
A, mm 2
x , mm
y , mm
xA, mm3
yA, mm3
1
(15)(80) = 1200
40
7.5
48 × 103
9 × 103
2
1
(50)(80) = 1333.33
3
60
30
80 × 103
40 × 103
Σ
2533.3
128 × 103
49 × 103
X A = Σ xA
X (2533.3) = 128 × 103
X = 50.5 mm
YA = Σ yA
Y (2533.3) = 49 × 103
Y = 19.34 mm
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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552
PROBLEM 5.13
Locate the centroid of the plane area shown.
SOLUTION
1
2
Σ
Then
A, mm 2
x , mm
y , mm
x A, mm3
y A, mm3
1
× 30 × 20 = 200
3
9
15
1800
3000
12.7324
32.7324
9000.0
23137
10800
26137
π
4
(30)2 = 706.86
906.86
X=
Σ x A 10800
=
Σ A 906.86
X = 11.91 mm
Y =
Σ y A 26137
=
Σ A 906.86
Y = 28.8 mm
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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553
PROBLEM 5.14
Locate the centroid of the plane area shown.
SOLUTION
Dimensions in in.
Then
A, in.2
x , in.
y , in.
x A, in 3
y A, in.3
1
2
800
× (20)(20) =
3
3
12
7.5
3200
2000
2
−1
−400
(20)(20) =
3
3
15
6.0
–2000
–800
Σ
400
3
1200
1200
X=
Y =
Σ x A 1200
=
400 !
ΣA
"
#
$ 3 %
Σ yA 1200
=
400 !
ΣA
"
#
$ 3 %
X = 9.00 in.
Y = 9.00 in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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554
PROBLEM 5.15
Locate the centroid of the plane area shown.
SOLUTION
Then
A, mm 2
x , mm
y , mm
x A, mm3
y A, mm3
1
2
(75)(120) = 6000
3
28.125
48
168750
288000
2
1
− (75)(60) = −2250
2
25
20
–56250
–45000
Σ
3750
112500
243000
X ΣA = ΣxA
X (3750 mm 2 ) = 112500 mm3
and
or X = 30.0 mm
Y ΣA = Σ yA
Y (3750 mm 2 ) = 243000
or Y = 64.8 mm
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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555
PROBLEM 5.16
Determine the y coordinate of the centroid of the shaded area in
terms of r1, r2, and α.
SOLUTION
First, determine the location of the centroid.
y2 =
From Figure 5.8A:
Similarly
=
2
cos α
r2 π
3 ( 2 −α )
y1 =
2 cos α
r1
3 ( π2 − α )
Σ yA =
Then
π
2 sin ( 2 − α )
r2 π
3
( 2 −α )
=
π
!
A2 = " − α # r22
$2
%
π
!
A1 = " − α # r12
2
$
%
2
cos α & π
! ' 2 cos α & π
! 2'
r2
− α # r22 ) − r1 π
(
(" − α # r1 )
3 ( π2 − α ) *"$ 2
3
2
α
−
% +
% +
( 2 ) *$
2 3 3
r2 − r1 cos α
3
(
)
π
π
!
!
ΣA = " − α # r22 − " − α # r12
$2
%
$2
%
and
π
!
= " − α # r22 − r12
2
$
%
(
)
Y ΣA = Σ yA
Now
& π
' 2
!
Y (" − α # r22 − r12 ) = r23 − r13 cos α
2
%
*$
+ 3
(
)
(
)
Y =
2 r23 − r13
"
3 "$ r22 − r12
! 2 cos α !
## "
#
% $ π − 2α %
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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556
PROBLEM 5.17
Show that as r1 approaches r2, the location of the centroid
approaches that for an arc of circle of radius (r1 + r2 )/2.
SOLUTION
First, determine the location of the centroid.
y2 =
From Figure 5.8A:
Similarly
=
2
cos α
r2 π
3 ( 2 −α )
y1 =
2 cos α
r1
3 ( π2 − α )
Σ yA =
Then
π
2 sin ( 2 − α )
r2 π
3
( 2 −α )
=
π
!
A2 = " − α # r22
$2
%
π
!
A1 = " − α # r12
2
$
%
2
cos α & π
! ' 2 cos α & π
! 2'
r2
− α # r22 ) − r1 π
(
(" − α # r1 )
3 ( π2 − α ) *"$ 2
3
2
−
α
% +
% +
( 2 ) *$
2 3 3
r2 − r1 cos α
3
(
)
π
π
!
!
ΣA = " − α # r22 − " − α # r12
$2
%
$2
%
and
π
!
= " − α # r22 − r12
2
$
%
(
)
Y ΣA = Σ yA
Now
& π
' 2
!
Y (" − α # r22 − r12 ) = r23 − r13 cos α
2
%
*$
+ 3
(
)
Y =
(
)
2 r23 − r13
"
3 "$ r22 − r12
! 2cos α !
## "
#
% $ π − 2α %
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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557
PROBLEM 5.17 (Continued)
1
(r1 + r2 ) is
2
Using Figure 5.8B, Y of an arc of radius
Y =
=
Now
r23 − r13
r22
−
r12
=
=
sin( π − α )
1
(r1 + r2 ) π 2
2
( 2 −α)
1
cos α
(r1 + r2 ) π
2
( 2 −α)
(
(r2 − r1 ) r22 + r1r2 + r12
(1)
)
(r2 − r1 )(r2 + r1 )
r22 + r1r2 + r12
r2 + r1
r2 = r + ∆
Let
r1 = r − ∆
r=
Then
and
In the limit as ∆
r23 − r13
r22
−
r12
1
(r1 + r2 )
2
=
(r + ∆) 2 + (r + ∆)(r − ∆)( r − ∆) 2
(r + ∆ ) + (r − ∆ )
=
3r 2 + ∆ 2
2r
0 (i.e., r1 = r2 ), then
r23 − r13
r22 − r12
So that
=
3
r
2
=
3 1
× ( r1 + r2 )
2 2
Y =
2 3
cos α
× (r1 + r2 ) π
−α
3 4
2
or Y = ( r1 + r2 )
cos α
π − 2α
Which agrees with Equation (1).
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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558
PROBLEM 5.18
For the area shown, determine the ratio a/b for which x = y .
SOLUTION
Then
A
x
y
xA
yA
1
2
ab
3
3
a
8
3
b
5
a 2b
4
2ab2
5
2
1
− ab
2
1
a
3
2
b
3
Σ
1
ab
6
−
a 2b
6
a 2b
12
−
ab 2
3
ab 2
15
X Σ A = Σ xA
1 ! a 2b
X " ab # =
$ 6 % 12
or
1
a
2
Y ΣA = Σ y A
X=
1 ! ab 2
Y " ab # =
$ 6 % 15
2
b
5
or
Y =
Now
X =Y ,
1
2
a= b
2
5
or
a 4
=
b 5
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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559
PROBLEM 5.19
For the semiannular area of Problem 5.11, determine the ratio r2/r1 so
that y = 3r1/4.
SOLUTION
A
−
1
Σ
2
Let
4r1
3π
2
− r13
3
r22
4r2
3π
2 3
r2
3
(r
2
2
− r12
2 3 3
r2 − r1
3
)
(
)
Y Σ A = Σ yA
Then
or
r12
2
2
π
YA
π
π
2
Y
3
2
π
r1 × r22 − r12 = r23 − r13
4
2
3
2
3
'
r !
9π & r2 !
(" # − 1) = " 2 # − 1
16 ($ r1 %
) $ r1 %
*
+
(
)
p=
(
)
r2
r1
9π
[( p + 1)( p − 1)] = ( p − 1)( p 2 + p + 1)
16
or
16 p 2 + (16 − 9π ) p + (16 − 9π ) = 0
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560
PROBLEM 5.19 (Continued)
−(16 − 9π ) ± (16 − 9π ) 2 − 4(16)(16 − 9π )
2(16)
Then
p=
or
p = −0.5726
p = 1.3397
r2
= 1.340
r1
Taking the positive root
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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561
PROBLEM 5.20
A composite beam is constructed by bolting four
plates to four 60 × 60 × 12-mm angles as shown.
The bolts are equally spaced along the beam, and
the beam supports a vertical load. As proved
in mechanics of materials, the shearing forces
exerted on the bolts at A and B are proportional to
the first moments with respect to the centroidal x
axis of the red shaded areas shown, respectively,
in Parts a and b of the figure. Knowing that the
force exerted on the bolt at A is 280 N, determine
the force exerted on the bolt at B.
SOLUTION
From the problem statement: F is proportional to Qx .
(Qx ) B
FA
(Qx ) A
Therefore:
FA
FB
, or
=
(Qx ) A (Qx ) B
For the first moments:
12 !
(Qx ) A = " 225 + # (300 × 12)
2%
$
FB =
= 831600 mm3
12 !
(Qx ) B = (Qx ) A + 2 " 225 − # (48 × 12) + 2(225 − 30)(12 × 60)
2%
$
= 1364688 mm3
Then
FB =
1364688
(280 N)
831600
or FB = 459 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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562
PROBLEM 5.21
The horizontal x axis is drawn through the centroid C of the area shown, and
it divides the area into two component areas A1 and A2 . Determine the first
moment of each component area with respect to the x axis, and explain the
results obtained.
SOLUTION
Note that
Then
and
Qx = Σ y A
5 ! 1
!
(Qx )1 = " in. #" × 6 × 5 # in.2
$ 3 %$ 2
%
or
(Qx )1 = 25.0 in.3
2
! 1
!
(Qx ) 2 = " − × 2.5 in. #" × 9 × 2.5 # in.2
$ 3
%$ 2
%
1
! 1
!
+ " − × 2.5 in. #" × 6 × 2.5 # in.2
3
2
$
%$
%
or (Qx ) 2 = −25.0 in.3
Now
Qx = (Qx )1 + (Qx )2 = 0
This result is expected since x is a centroidal axis (thus y = 0)
and
Qx = Σ yA = Y ΣA( y = 0 , Qx = 0) !
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563
PROBLEM 5.22
The horizontal x axis is drawn through the centroid C of the area shown,
and it divides the area into two component areas A1 and A2 . Determine
the first moment of each component area with respect to the x axis, and
explain the results obtained.
SOLUTION
First determine the location of the centroid C. We have
Then
A, in.2
y ′, in.
y ′A, in.3
I
1
!
2 " × 2 × 1.5 # = 3
$2
%
0.5
1.5
II
1.5 × 5.5 = 8.25
2.75
22.6875
III
4.5 × 2 = 9
6.5
58.5
Σ
20.25
82.6875
Y ′ Σ A = Σ y ′A
Y ′(20.25) = 82.6875
or
Y ′ = 4.0833 in.
Now
Qx = Σ y1 A
Then
&1
'
(Qx )1 = ( (5.5 − 4.0833)in.) [(1.5)(5.5 − 4.0833)]in.2
*2
+
+ [(6.5 − 4.0833)in.][(4.5)(2)]in.2
and
or
(Qx )1 = 23.3 in.3
&1
'
(Qx ) 2 = − ( (4.0833 in.) ) [(1.5)(4.0833)]in.2
*2
+
& 1
'
!
− [(4.0833 − 0.5)in.] × 2 (" × 2 × 1.5 # in.2 )
%
*$ 2
+
or (Qx ) 2 = −23.3 in.3
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you are using it without permission.
564
PROBLEM 5.22 (Continued)
Now
Qx = (Qx )1 + (Qx )2 = 0
This result is expected since x is a centroidal axis (thus Y = 0)
and
Qx = Σ y A = Y Σ A (Y = 0 , Qx = 0) !
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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565
PROBLEM 5.23
The first moment of the shaded area with respect to the x axis is denoted by Qx.
(a) Express Qx in terms of b, c, and the distance y from the base of the shaded
area to the x axis. (b) For what value of y is Ox maximum, and what is that
maximum value?
SOLUTION
Shaded area:
A = b (c − y )
Qx = yA
=
Qx =
(a)
(b)
For Qmax :
For y = 0:
1
(c + y )[b(c − y )]
2
1
b (c 2 − y 2 )
2
dQ
= 0 or
dy
(Qx ) =
1
b(−2 y ) = 0
2
1 2
bc
2
y=0
(Qx ) =
1 2
bc
2
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
566
PROBLEM 5.24
A thin, homogeneous wire is bent to form the perimeter of the figure
indicated. Locate the center of gravity of the wire figure thus formed.
SOLUTION
Perimeter of Figure 5.1
Dimensions in mm
L
x
y
xL, mm 2
yL, mm 2
I
30
0
15
0
0.45 × 103
II
210
105
30
22.05 × 103
6.3 × 103
III
270
210
165
56.7 × 103
44.55 × 103
IV
30
225
300
6.75 × 103
9 × 103
V
300
240
150
72 × 103
45 × 103
VII
240
120
0
28.8 × 103
0
Σ
1080
186.3 × 103
105.3 × 103
X ΣL = Σ x L
X (1080 mm) = 186.3 × 103 mm 2
X = 172.5 mm
Y ΣL = Σ y L
Y (1080 mm) = 105.3 × 103 mm 2
Y = 97.5 mm
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
567
PROBLEM 5.25
A thin, homogeneous wire is bent to form the perimeter of the figure
indicated. Locate the center of gravity of the wire figure thus formed.
SOLUTION
First note that because wire is homogeneous, its center of gravity will coincide with the centroid of the
corresponding line.
Then
L, mm
x , mm
y , mm
xL, mm 2
yL, mm 2
1
20
10
0
200
0
2
24
20
12
480
288
3
30
35
24
1050
720
4
46.861
35
42
1640.14
1968.16
5
20
10
60
200
1200
6
60
0
30
0
1800
Σ
200.86
3570.1
5976.2
X ΣL = Σ x L
X (200.86) = 3570.1
X = 17.77 mm
Y ΣL = Σ y L
Y (200.86) = 5976.2
Y = 29.8 mm
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
568
!
PROBLEM 5.26
A thin, homogeneous wire is bent to form the perimeter of the figure
indicated. Locate the center of gravity of the wire figure thus formed.
SOLUTION
First note that because the wire is homogeneous, its center of gravity will coincide with the centroid of the
corresponding line.
Then
L, in.
x , in.
y , in.
xL, in.2
yL, in.2
1
33
16.5
0
544.5
0
2
15
33
7.5
495
112.5
3
21
22.5
15
472.5
315
4
122 + 152 = 19.2093
6
7.5
115.256
144.070
Σ
88.209
1627.26
571.57
X ΣL = Σx L
X (88.209) = 1627.26
and
or X = 18.45 in.
Y ΣL = Σ y L
Y (88.209) = 571.57
or
Y = 6.48 in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
569
!
PROBLEM 5.27
A thin, homogeneous wire is bent to form the perimeter of the
figure indicated. Locate the center of gravity of the wire figure
thus formed.
SOLUTION
First note that because the wire is homogeneous, its center of gravity will coincide with the centroid of the
corresponding line.
Y6 =
Then
2
π
(38 in.)
L, in.
x , in.
y , in.
xL, in.2
yL, in.2
1
18
–29
0
–522
0
2
16
–20
8
–320
128
3
20
–10
16
–200
320
4
16
0
8
0
128
5
38
19
0
722
0
6
π (38) = 119.381
0
24.192
0
2888.1
Σ
227.38
–320
3464.1
X=
Σx L
−320
=
ΣL
227.38
X = −1.407 in.
Y =
Σ y L 3464.1
=
ΣL
227.38
Y = 15.23 in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
570
PROBLEM 5.28
A uniform circular rod of weight 8 lb and radius 10 in. is attached to a pin at C and
to the cable AB. Determine (a) the tension in the cable, (b) the reaction at C.
SOLUTION
For quarter circle
(a)
r=
2r
π
2r !
ΣM C = 0: W " # − Tr = 0
$π %
2!
2!
T = W " # = (8 lb) " #
$π %
$π %
(b)
T = 5.09 lb
ΣFx = 0: T − C x = 0 5.09 lb − Cx = 0
C x = 5.09 lb
ΣFy = 0: C y − W = 0
C y = 8 lb
C y − 8 lb = 0
C = 9.48 lb
57.5°
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
571
PROBLEM 5.29
Member ABCDE is a component of a mobile and is formed from a single
piece of aluminum tubing. Knowing that the member is supported at C and
that l = 2 m, determine the distance d so that portion BCD of the member
is horizontal.
SOLUTION
First note that for equilibrium, the center of gravity of the component must lie on a vertical line through C.
Further, because the tubing is uniform, the center of gravity of the component will coincide with the centroid
of the corresponding line. Thus, X = 0
So that
Σx L = 0
Then
0.75
!
cos 55° # m × (0.75 m)
−" d −
2
$
%
+ (0.75 − d )m × (1.5 m)
&
1
!'
+ ((1.5 − d )m − " × 2 m × cos 55° # ) × (2 m) = 0
$2
%+
*
or
&1
'
(0.75 + 1.5 + 2)d = ( (0.75) 2 − 2 ) cos 55° + (0.75)(1.5) + 3
2
*
+
or d = 0.739 m
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
572
PROBLEM 5.30
Member ABCDE is a component of a mobile and is formed from a
single piece of aluminum tubing. Knowing that the member is
supported at C and that d is 0.50 m, determine the length l of arm DE
so that this portion of the member is horizontal.
SOLUTION
First note that for equilibrium, the center of gravity of the component must lie on a vertical line through C.
Further, because the tubing is uniform, the center of gravity of the component will coincide with the centroid
of the corresponding line. Thus,
X =0
So that
or
Σx L = 0
0.75
!
−"
sin 20° + 0.5sin 35° # m × (0.75 m)
$ 2
%
+ (0.25 m × sin 35°) × (1.5 m)
l!
+ "1.0 × sin 35° − # m × (l m) = 0
2
$
%
or
+ ( sin 35° − 2l ) l = 0
−0.096193
!!!!!!!!"
!!!!!!"
( xL)DE
( xL) AB +( xL)BD
The equation implies that the center of gravity of DE must be to the right of C.
Then
l 2 − 1.14715l + 0.192386 = 0
or
l=
or
l = 0.204 m
1.14715 ± (−1.14715)2 − 4(0.192386)
2
or l = 0.943 m
Note that sin 35° − 12 l . 0 for both values of l so both values are acceptable.!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
573
PROBLEM 5.31
The homogeneous wire ABC is bent into a semicircular arc and a straight section
as shown and is attached to a hinge at A. Determine the value of θ for which the
wire is in equilibrium for the indicated position.
SOLUTION
First note that for equilibrium, the center of gravity of the wire must lie on a vertical line through A. Further,
because the wire is homogeneous, its center of gravity will coincide with the centroid of the corresponding
line. Thus,
X =0
So that
Σx L = 0
Then
1
2r
!
!
" − 2 r cos θ # (r ) + " π − r cos θ # (π r ) = 0
$
%
$
%
or
cos θ =
4
1 + 2π
= 0.54921
or θ = 56.7°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
574
!
PROBLEM 5.32
Determine the distance h for which the centroid of the shaded
area is as far above line BB as possible when (a) k = 0.10,
(b) k = 0.80.
SOLUTION
A
y
yA
1
1
ba
2
1
a
3
1 2
a b
6
2
1
− (kb)h
2
1
h
3
1
− kbh 2
6
Σ
b
(a − kh)
2
b 2
( a − kh 2 )
6
Y Σ A = Σ yA
Then
&b
' b
Y ( ( a − kh) ) = (a 2 − kh 2 )
*2
+ 6
Y =
or
(1)
dY 1 −2kh(a − kh) − (a 2 − kh 2 )(− k )
=
=0
dh 3
( a − kh) 2
and
or
a 2 − kh 2
3(a − kh)
2h(a − kh) − a 2 + kh 2 = 0
(2)
Simplifying Eq. (2) yields
kh 2 − 2ah + a 2 = 0
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
575
PROBLEM 5.32 (Continued)
Then
2a ± (−2a) 2 − 4( k )(a 2 )
2k
a
= &*1 ± 1 − k '+
k
h=
Note that only the negative root is acceptable since h , a. Then
(a)
k = 0.10
h=
(b)
a &
1 − 1 − 0.10 '
+
0.10 *
or h = 0.513a
k = 0.80
h=
a &
1 − 1 − 0.80 '
+
0.80 *
or h = 0.691a
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
576
PROBLEM 5.33
Knowing that the distance h has been selected to maximize the
distance y from line BB to the centroid of the shaded area,
show that y = 2h/3.
SOLUTION
See solution to Problem 5.32 for analysis leading to the following equations:
Y =
a 2 − kh 2
3(a − kh)
(1)
2h(a − kh) − a 2 + kh 2 = 0
(2)
Rearranging Eq. (2) (which defines the value of h which maximizes Y ) yields
a 2 − kh 2 = 2h(a − kh)
Then substituting into Eq. (1) (which defines Y )
Y =
1
× 2h(a − kh)
3( a − kh)
or Y =
2
h
3
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
577
PROBLEM 5.34
Determine by direct integration the centroid of the area shown. Express your
answer in terms of a and h.
SOLUTION
y h
=
x a
h
y= x
a
xEL = x
1
yEL = y
2
dA = ydx
A=
-x
-
EL dA
-
a
0
ydx =
-
= xydx =
yEL dA =
-
a
0
1
"2
$
-
a
0
a
0
1
h !
" a x # dx = 2 ah
$
%
a
h !
h & x3 '
1
x " x # dx = ( ) = ha 2
a * 3 +0 3
$a %
1
!
y # ydx =
2
%
-
a
0
2
h !
1 h2
" a x # dx = 2 2
b
$
%
a
& x3 '
1 2
( ) = h a
3
* +0 6
1 ! 1
xA = xEL dA: x " ah # = ha 2
$2 % 3
x=
1 ! 1
yA = yEL dA: y " ah # = h 2 a
$2 % 6
1
y= h
3
-
-
2
a
3
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
578
PROBLEM 5.35
Determine by direct integration the centroid of the area shown. Express
your answer in terms of a and h.
SOLUTION
y1 : h = ka 2
At (a, h)
k=
or
h
a2
y2 : h = ma
m=
or
xEL = x
Now
yEL =
1
( y1 + y2 )
2
h '
&h
dA = ( y2 − y1 )dx = ( x − 2 x 2 ) dx
a
a
*
+
h
= 2 (ax − x 2 ) dx
a
and
-
A = dA =
Then
and
h
a
-
-
a
0
a
1 '
1
h
h &a
(ax − x 2 )dx = 2 ( x 2 − x3 ) = ah
3 +0 6
a2
a *2
a
h &a
1 '
1
&h
x 2 (ax − x 2 ) dx = 2 ( x3 − x 4 ) = a 2 h
0 (
4 + 0 12
a *3
*a
1
1 2
( y1 + y2 )[( y2 − y1 ) dx] =
yEL dA =
y2 − y12 dx
2
2
xEL dA =
-
a
-
=
1
2
- (
-
a
0
)
h 2 2 h2 4 !
"" 2 x − 4 x ##dx
a
$a
%
a
1 h2 & a 2 3 1 5 '
=
( x − x )
2 a4 * 3
5 +0
1
= ah 2
15
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
579
PROBLEM 5.35 (Continued)
1 ! 1
xA = xEL dA: x " ah # = a 2 h
$ 6 % 12
x=
1
a
2
1 ! 1
yA = yEL dA: y " ah # = ah 2
$ 6 % 15
y=
2
h
5
-
-
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
580
!
PROBLEM 5.36
Determine by direct integration the centroid of the area shown. Express your
answer in terms of a and h.
SOLUTION
For the element (EL) shown
At
x = a,
Then
x=
xEL
yEL
h
a3
a 1/3
y
h1/3
a 1/3
y dy
h1/3
1
1 a 1/3
= x=
y
2
2 h1/ 3
=y
-
A = dA =
Then
Hence
or k =
dA = xdy =
Now
and
y = h : h = ka3
-
xEL dA =
-
yEL dA =
-
h
0
h
0
-
h
0
a 1/3
3 a
y dy =
y 4/3
1/3
4 h1/3
h
( )
h
=
0
3
ah
4
h
1 a 1/3 a 1/3 ! 1 a 3 5/3 !
3
y " 1/3 y dy # =
y # = a2h
2/3 "
2 h1/3
2
5
10
h
h
$
%
$
%0
h
a
a 3
3
!
!
y " 1/3 y1/3 dy # = 1/3 " y 7/3 # = ah 2
$h
% h $7
%0 7
-
3 ! 3
x " ah # = a 2 h
$ 4 % 10
x=
2
a
5
!
-
3 ! 3
y " ah # = ah2
$4 % 7
y=
4
h
7
!
xA = xEL dA:
yA = yEL dA:
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
581
PROBLEM 5.37
Determine by direct integration the centroid of the area shown.
SOLUTION
y=
For the element (EL) shown
b 2
a − x2
a
dA = (b − y ) dx
and
xEL
)
(
b
a − a 2 − x 2 dx
a
=x
=
1
( y + b)
2
b
a + a 2 − x2
=
2a
yEL =
(
-
A = dA =
Then
-
To integrate, let
x = a sin θ :
Then
A=
=
-
π /2
0
a
0
)
)
(
b
a − a 2 − x 2 dx
a
a 2 − x 2 = a cos θ , dx = a cos θ dθ
b
(a − a cos θ )(a cos θ dθ )
a
b& 2
2θ
2 θ
( a sin θ − a " + sin
a*
2
4
$
π /2
!'
#)
%+ 0
π!
= ab "1 − #
4%
$
and
-x
EL dA
=
-
a
0
&b
'
x ( a − a 2 − x 2 dx )
a
*
+
)
(
π /2
b& a
1
!'
= (" x 2 + (a 2 − x 2 )3/2 # )
a *$ 2
3
%+ 0
=
1 3
ab
6
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
582
PROBLEM 5.37 (Continued)
-y
EL dA
=
-
a
0
b
&b
'
a + a 2 − x 2 ( a − a 2 − x 2 dx )
2a
a
*
+
b2
= 2
2a
=
) (
(
)
a
-
a
0
b 2 x3 !
( x ) dx = 2 "" ##
2a $ 3 % 0
2
1 2
ab
6
-
&
π !' 1
x ( ab "1 − # ) = a 2 b
4 %+ 6
* $
or x =
2a
3(4 − π )
-
&
π !' 1
y ( ab "1 − # ) = ab 2
4 %+ 6
* $
or y =
2b
3(4 − π )
xA = xEL dA:
yA = yEL dA:
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
583
!
PROBLEM 5.38
Determine by direct integration the centroid of the area shown.
SOLUTION
x =0
First note that symmetry implies
For the element (EL) shown
yEL =
2r
(Figure 5.8B)
π
dA = π rdr
-
A = dA =
Then
and
-y
EL dA
=
-
r2
r1
-
r2
r1
r2
π r d r = π ""
$ 2
r2
!
π 2 2
r2 − r1
## =
% r1 2
(
r2
1 !
2
(π rdr ) = 2 " r 3 # = r23 − r13
π
$3 %r 3
2r
(
)
)
1
So
&π
' 2
yA = yEL dA: y ( r22 − r12 ) = r23 − r13
2
*
+ 3
-
(
)
(
)
or y =
4 r23 − r13
3π r22 − r12
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
584
PROBLEM 5.39
Determine by direct integration the centroid of the area shown.
SOLUTION
y =0
First note that symmetry implies
xEL
-
A = dA =
Then
-
= a[ x]0a −
and
-
xEL dA =
1
a( adφ )
2
2
= a cos φ
3
dA =
dA = adx
xEL = x
-
a
0
a
0
adx −
α
1
- α 2 a dφ
2
−
a2 α
[φ ]α = a 2 (1 − α )
2
x(adx) −
α
2
1
!
- α 3 a cos φ "$ 2 a dφ #%
2
−
a
& x2 '
1
= a ( ) − a3 [sin φ ]α−α
2
* +0 3
1 2
!
= a3 " − sin α #
$2 3
%
1 2
!
xA = xEL dA: x [a 2 (1 − α )] = a3 " − sin α #
$2 3
%
-
or x =
3 − 4sin α
a
6(1 − α )
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
585
!
PROBLEM 5.40
Determine by direct integration the centroid of the area shown. Express
your answer in terms of a and b.
SOLUTION
x = 0,
At
y =b
b = k (0 − a) 2
y=
Now
xEL = x
-
A = dA =
Then
and
b
( x − a) 2 dx
a2
-
a
0
a
1
b
b &
3'
2
(
)
−
=
−
= ab
x
a
dx
x
a
(
)
2
2 *
+
0
3
3a
a
&b
' b a 3
x ( 2 ( x − a) 2 dx ) = 2
x − 2ax 2 + a 2 x dx
0 *a
0
+ a
4
2
2
b x
a 2! 1 2
= 2 ""
− ax3 +
x ## = a b
2
a $ 4 3
% 12
-
xEL dA =
-
yEL dA =
=
Hence
y
b
= 2 ( x − a)2
2 2a
dA = ydx =
and
b
a2
b
( x − a )2
a2
Then
yEL =
or k =
-
a
-
a
0
- (
)
a
2
b
&b
' b &1
'
( x − a) 2 ( 2 ( x − a) 2 dx ) = 4 ( ( x − a)5 )
2
5
2a
*a
+ 2a *
+0
1 2
ab
10
1 ! 1
xA = xEL dA: x " ab # = a 2 b
$ 3 % 12
-
1 ! 1
yA = yEL dA: y " ab # = ab 2
$ 3 % 10
-
x=
y=
1
a
4
3
b
10
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
586
PROBLEM 5.41
Determine by direct integration the centroid of the area shown. Express
your answer in terms of a and b.
SOLUTION
b 2
x
a2
b
y2 = k2 x 4 but b = k2 a 4 y2 = 4 x 4
a
b
x4 !
dA = ( y2 − y1 )dx = 2 "" x 2 − 2 ## dx
a $
a %
xEL = x
y1 = k1 x 2
yEL =
=
but b = k1a 2
y1 =
1
( y1 + y2 )
2
b
2a 2
x4
2
"" x + 2
a
$
,
A = dA =
b
a2
,
a
0
!
##
%
x4
2
"" x − 2
a
$
!
## dx
%
a
b & x3
x5 '
= 2 ( − 2)
a * 3 5a + 0
2
= ba
15
,x
EL dA
!
## dx
%
×
b
x4
2
x
−
2 "
a "$
a2
b
a2
,
x5 !
3
"" x − 2 ## dx
0
a %
$
=
b
a2
& x4
x6 '
−
(
2)
* 4 6a + 0
=
1 2
a b
12
=
,
=
a
0
a
a
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
587
PROBLEM 5.41 (Continued)
,
yEL dA =
=
,
a
0
b
x4 ! b
x4 !
2
2
+
−
x
x
"
#
"
# dx
2a 2 "$
a 2 #% a 2 "$
a 2 #%
b2
2a 4
,
a
0
x8
4
"" x − 4
a
$
!
## dx
%
a
2 2
b 2 & x5
x9 '
= 4 ( − 4) =
ab
2a * 5 9a + 0 45
2 ! 1
xA = xEL dA: x " ba # = a 2b
15
$
% 12
5
x= a
8
2
! 2 2
yA = yEL dA: y " ba # =
ab !
15
$
% 45
1
y= b
3
,
!
,
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
588
!
PROBLEM 5.42
Determine by direct integration the centroid of the area shown.
SOLUTION
xEL = x
We have
!
##
%
2 !
x x
dA = ydx = a ""1 − + 2 ## dx
L L %
$
yEL =
1
a
x x2
y = ""1 − + 2
2
2$
L L
,
A = dA =
Then
,
2L
0
2L
&
x x2 !
x2
x3 '
a ""1 − + 2 ## dx = a ( x −
+ 2)
L L %
2 L 3L + 0
$
*
8
= aL
3
and
,
xEL dA =
=
,
yEL dA =
=
,
2L
0
&
x x2
x ( a ""1 − + 2
L L
(* $
2L
! '
& x 2 x3
x4 '
+ 2)
## dx ) = a ( −
% )+
* 2 3L 4 L + 0
10 2
aL
3
,
2L
0
a2
2
a
x x2 ! &
x x2 ! '
""1 − + 2 ## ( a ""1 − + 2 ## dx )
2$
L L % *( $
L L % +)
,
EL
0
x
x2
x3 x 4 !
""1 − 2 + 3 2 − 2 3 + 4 ## dx
L
L
L
L %
$
2L
a2 &
x 2 x3
x4
x5 '
=
+ 2 − 3 + 4)
(x −
2 *
L L
2L 5L + 0
11
= a2 L
5
Hence
8 ! 10
xA = xEL dA: x " aL # = aL2
$3 % 3
,
1 ! 11
yA = yEL dA: y " a # = a 2
$8 % 5
,
x=
y=
5
L
4
33
a
40
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589
!
PROBLEM 5.43
Determine by direct integration the centroid of the area shown. Express
your answer in terms of a and b.
SOLUTION
x = a, y = b : a = kb2
For y2 at
b
Then
y2 =
Now
xEL = x
and for
a
0# x# :
2
yEL =
a
y2 b x1/ 2
=
2 2 a
x1/2
a
dx
1
a
b x 1 x1/2 !
# x # a : yEL = ( y1 + y2 ) = "" − +
#
2
2
2$ a 2
a #%
x1/2
dA = ( y2 − y1 )dx = b ""
$
Then
a
b2
x1/2
dA = y2 dx = b
For
or k =
,
A = dA =
,
a/2
0
b
a
x1/2
a
−
dx +
,
x 1!
+ # dx
a 2 #%
x1/ 2 x 1 !
b ""
− + ## dx
a/2
$ a a 2%
a
a
a/ 2
& 2 x3/2 x 2 1 '
b & 2 3/2 '
=
+
−
+ x)
x
b
(
)
a (* 3
+0
* 3 a 2a 2 + a/2
=
3/ 2
3/ 2
a! '
2 b & a!
(" # + (a)3/ 2 − " # )
3 a *($ 2 %
$ 2 % +)
2
a! ' 1&
a ! ' /.
/- 1 & 2
(a ) − $ % ! + (a) − $ % ! #
+ b "−
& 2 ' !) 2 (
& 2 ' ) *,
*+ 2a (
13
ab
=
24
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590
PROBLEM 5.43 (Continued)
and
1
xEL dA =
1
a/2
0
/ x1/ 2 0
x $$ b
dx %% +
a
&
'
1
- / x1/ 2 x 1 0 .
x b $$
− + %% ! dx
a/2
( & a a 2 ' )!
a
a
a/ 2
- 2 x5/2 x3 x 4 .
b - 2 5/2 .
=
−
+ !
x ! +b
a (5
)0
( 5 a 3a 4 ) a/2
5/2
5/2
2 b -/ a 0
/a0 .
5/2
+
−
(
)
a
$ %
$ % !
5 a (& 2 '
& 2 ' )!
3
2
*2 1 - 3 / a 0 . 1 - 2 / a 0 . *3
+ b "−
( a) − $ % ! + (a) − $ % ! #
3a (
& 2 ' !) 4 (
& 2 ' !) ,*
+*
71 2
a b
=
240
=
1
yEL dA =
1
a/2
0
+
1
b x1/ 2 - x1/ 2 .
b
dx !
2 a (
a )
b / x 1 x1/2 0 - / x1/ 2 x 1 0 .
− + % dx !
$ − +
% b$
a/ 2 2 $ a
2
a %' ( $& a a 2 %' )!
&
a
a
a/ 2
3
b2 - 1 2 .
b 2 -/ x 2 1 / x 1 0 0 .
$
x ! +
=
− $ − % %!
2a ( 2 ) 0
2 $& 2a 3a & a 2 ' %' !
(
) a/2
=
b -/ a 0
/a0
2
$ % + (a ) − $ %
4 a (& 2 '
&2'
=
11 2
ab
48
2
Hence
2.
b2 / a 1 0
!−
$ − %
!) 6a & 2 2 '
/ 13 0 71 2
xA = xEL dA: x $ ab % =
a b
& 24 ' 240
1
/ 13 0 11 2
yA = yEL dA: y $ ab % =
ab
& 24 ' 48
1
3
x=
17
a = 0.546a
130
y=
11
b = 0.423b
26
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you are using it without permission.
591
!
PROBLEM 5.44
Determine by direct integration the centroid of the area shown. Express
your answer in terms of a and b.
SOLUTION
x = a,
For y1 at
y = 2b 2b = ka 2
or k =
2b
a2
2b 2
x
a2
Then
y1 =
By observation
b
x0
/
y2 = − ( x + 2b) = b $ 2 − %
a
a'
&
Now
xEL = x
and for 0 # x # a :
yEL =
1
b
y1 = 2 x 2
2
a
For a # x # 2a :
yEL =
1
b/
x0
x0
/
y2 = $ 2 − % and dA = y2 dx = b $ 2 − % dx
2
2&
a'
a'
&
1
A = dA =
Then
1
a
0
and dA = y1dx =
2b 2
x dx +
a2
1
2a
a
2b 2
x dx
a2
x0
/
b $ 2 − % dx
a'
&
2a
a
2
- a/
2b - x3 .
7
x0 .
= 2
! + b − $ 2 − % ! = ab
6
a ' )!
a ( 3 )0
( 2&
0
and
1x
EL dA
a
/ 2b
0
x $ 2 x 2 dx % +
&a
'
- /
x0 .
x b $ 2 − % dx !
a
' )
( &
1
=
- 2 x3 .
2b - x 4 .
! +b x − !
2
3a ) 0
a ( 4 )0
(
0
1
2a
=
a
a
2a
1 2
1 - 2
2
3
(2a ) − (a )3 .) #
a b + b " -((2a) 2 − (a) 2 .) +
(
2
3a
+
,
7 2
= a b
6
=
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592
PROBLEM 5.44 (Continued)
1y
EL dA
=
1
a
0
b 2 - 2b 2 .
x
x dx ! +
a2 ( a2
)
1
2a
0
b/
x 0- /
x0 .
$ 2 − % b $ 2 − % dx !
2&
a '( &
a' )
2a
a
3
2b 2 - x5 .
b2 - a /
x0 .
= 4
− $2 − % !
! +
a ' )!
a ( 5 )0 2 ( 3 &
a
17 2
=
ab
30
Hence
/7 0 7
xA = xEL dA: x $ ab % = a 2b
&6 ' 6
1
/ 7 0 17 2
yA = yEL dA: y $ ab % =
ab
& 6 ' 30
1
x =a
y=
17
b
35
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593
!
PROBLEM 5.45
A homogeneous wire is bent into the shape shown. Determine by direct
integration the x coordinate of its centroid.
SOLUTION
First note that because the wire is homogeneous, its center of gravity coincides with the centroid of the
corresponding line
xEL = a cos3 θ
Now
and dL = dx 2 + dy 2
x = a cos3 θ : dx = −3a cos 2 θ sin θ dθ
Where
y = a sin 3 θ : dy = 3a sin 2 θ cos θ dθ
dL = [(−3a cos 2 θ sin θ dθ )2 + (3a sin 2 θ cos θ dθ )2 ]1/ 2
Then
= 3a cos θ sin θ (cos 2 θ + sin 2 θ )1/ 2 dθ
= 3a cos θ sin θ dθ
1
L = dL =
and
1x
EL dL
=
3
a
2
=
1
π /2
0
1
π /2
0
π /2
3a cos θ sin θ dθ = 3a
a cos3θ (3a cos θ sin θ dθ )
π /2
- 1
.
= 3a 2 − cos5 θ !
( 5
)0
Hence
-1 2 .
sin θ !
(2
)0
3
= a2
5
/3 0 3
xL = xEL dL : x $ a % = a 2
&2 ' 5
1
x=
2
a
5
Alternative Solution
/ x0
x = a cos3 θ 4 cos 2 θ = $ %
&a'
/ y0
y = a sin 3 θ 4 sin 2 θ = $ %
&a'
/x0
$a%
& '
2/3
/ y0
+$ %
&a'
2/3
2/3
2/3
= 1 or
y = (a 2/3 − x 2/3 )3/2
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594
PROBLEM 5.45 (Continued)
dy
= (a 2/3 − x 2/3 )1/ 2 (− x −1/3 )
dx
Then
Now
xEL = x
and
/ dy 0
dL = 1 + $ %
& dx '
2
{
dx = 1 + -(( a 2/3 − x 2/3 )1/2 (− x −1/3 ) .)
1
L = dL =
Then
and
Hence
1
xEL dL =
1
a
0
1
a
0
2
1/2
}
dx
a
a1/3
3
- 3 2/3 .
dx = a1/3
x ! = a
1/3
x
(2
)0 2
a
/ a1/3 0
3
-3
.
x $$ 1/3 dx %% = a1/3 x5/3 ! = a 2
(5
)0 5
&x
'
/3 0 3
xL = xEL dL : x $ a % = a 2
&2 ' 5
1
x=
2
a
5
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you are using it without permission.
595
PROBLEM 5.46
A homogeneous wire is bent into the shape shown. Determine by direct
integration the x coordinate of its centroid.
SOLUTION
First note that because the wire is homogeneous, its center of gravity coincides with the centroid of the
corresponding line
Now
xEL = r cos θ
Then
L = dL =
and
1
1
xEL dL =
7π / 4
1π
/4
and dL = rdθ
7π /4
1π
/4
3
rdθ = r[θ ]π7π/ 4/ 4 = π r
2
r cos θ ( rdθ )
= r 2 [sin θ ]π7π/4/ 4
/ 1
1 0
= r2 $ −
−
%
2
2'
&
= −r 2 2
Thus
/3 0
xL = xdL : x $ π r % = −r 2 2
&2 '
1
x =−
2 2
r
3π
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596
PROBLEM 5.47*
A homogeneous wire is bent into the shape shown. Determine by
direct integration the x coordinate of its centroid. Express your
answer in terms of a.
SOLUTION
First note that because the wire is homogeneous, its center of gravity will coincide with the centroid of the
corresponding line.
x = a,
We have at
y=a
a = ka 3/2
1
y=
Then
a
or k =
1
a
x3/ 2
dy
3 1/2
=
x
dx 2 a
and
Now
xEL = x
and
/ dy 0
dL = 1 + $ % dx
& dx '
2
1/2
2
- / 3
0 .
x1/2 % !
= 1+ $
' !)
( &2 a
1
4a + 9 x dx
=
2 a
1
L = dL =
Then
1
a
1
0
2 a
dx
4a + 9 x dx
a
1 -2 1
.
× (4a + 9 x)3/ 2 !
3
9
2 a(
)0
a
= [(13)3/2 − 8]
27
= 1.43971a
=
and
1x
EL dL
=
1
a
0
x
- 1
(2 a
.
4a + 9 x dx !
)
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you are using it without permission.
597
PROBLEM 5.47* (Continued)
Use integration by parts with
u=x
du = dx
Then
1
xEL dL =
dv = 4a + 9 x dx
2
v=
(4a + 9 x)3/ 2
27
a
1 2* 2
3/2 .
×
+
x
a
x
(4
9
)
"
! −
2 a *+ ( 27
)0
1
a
0
3*
2
(4a + 9 x)3/ 2 dx #
27
*,
a
=
(13)3/2 2
1 -2
.
a −
(4a + 9 x)5/2 !
27
45
27 a (
)0
=
a2 2
2
3
3/ 2
5/2
"(13) − [(13) − 32]#
27 +
45
,
= 0.78566a 2
1
xL = xEL dL : x (1.43971a ) = 0.78566a 2
or x = 0.546a
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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598
PROBLEM 5.48*
Determine by direct integration the centroid of the area shown.
SOLUTION
We have
xEL = x
1
πx
a
yEL = y = cos
2
2
2L
and
dA = ydx = a cos
1
A = dA =
Then
1
πx
dx
2L
L/2
a cos
0
πx
2L
dx
L/ 2
πx.
- 2L
sin
=a
2 L !) 0
(π
2
=
and
1x
EL dA
π
aL
πx 0
/
= x $ a cos
dx
2 L %'
&
1
u=x
Use integration by parts with
dv = cos
du = dx
Then
πx
1 x cos 2L dx =
1
2L
v=
× sin
2L
π
πx
−
πx
dx
2L
sin
πx
2L
2L
1π
πx
dx
2L
2L
π x 2L
πx0
2L /
cos
x sin
=
+
π $&
2L π
2 L %'
π
xEL dA = a
=a
2L -
π (
x sin
πx
2L
+
sin
2L
π
cos
π x.
L/ 2
2 L !) 0
2 L -/ L
2 0 2L .
L %−
+
$$
% π !
π (& 2 2 π
'
)!
= 0.106374aL2
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599
PROBLEM 5.48* (Continued)
Also
1y
EL
dA =
1
L/ 2
0
a2
=
2
πx/
πx 0
a
dx %
cos
$ a cos
2
2L &
2L '
-x
L/ 2
sin π x .
+ 2π L !
L
(2
)! 0
=
a −2 / L L 0
$ +
%
2 & 4 2π '
= 0.20458a 2L
/ 2 0
xA = xEL dA: x $
aL % = 0.106374aL2
$ π
%
&
'
or x = 0.236 L
/ 2 0
yA = yEL dA: y $
aL % = 0.20458a 2L
$ π
%
&
'
or y = 0.454a
1
1
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600
PROBLEM 5.49*
Determine by direct integration the centroid of the area shown.
SOLUTION
2
2
r cos θ = aeθ cos θ
3
3
2
2 θ
= r sin θ = ae sin θ
3
3
xEL =
We have
yEL
dA =
and
1
1
( r )(rdθ ) = a 2 e2θ dθ
2
2
1
A = dA =
Then
1
π
0
π
1 2 2θ
1 -1
.
a e dθ = a 2 e 2θ !
2
2 (2
)0
1 2 2π
a (e − 1)
4
= 133.623a 2
=
and
1x
EL dA =
1
π
0
2 θ
/1
0
ae cos θ $ a 2 e2θ dθ %
3
&2
'
1
= a3
3
1
π
0
e3θ cos θ dθ
To proceed, use integration by parts, with
u = e3θ
du = 3e3θ dθ
and
dv = cos θ dθ
Then
1e
3θ
1
then
du = 3e3θ dθ
dv = sin θ dθ , then
Then
1e
3θ
v = sin θ
cos θ dθ = e3θ sin θ − sin θ (3e3θ dθ )
u = e3θ
Now let
and
v = − cos θ
sin θ dθ = e3θ sin θ − 3 - −e3θ cos θ − (− cos θ )(3e3θ dθ ) .!
(
)
1
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601
PROBLEM 5.49* (Continued)
So that
1
e3θ cos θ dθ =
e3θ
(sin θ + 3cos θ )
10
π
1
.
1 - e3θ
(sin θ + 3cos θ ) !
xEL dA = a 3
3 ( 10
)0
1y
Also
EL dA
=
a3
(−3e3π − 3) = −1239.26a3
30
=
1
π
0
2 θ
/1
0
ae sin θ $ a 2 e 2θ dθ %
3
&2
'
1
= a3
3
1
π
0
e3θ sin θ dθ
Using integration by parts, as above, with
u = e3θ
and
1
dv = sin θ dθ
Then
1e
So that
1
3θ
du = 3e3θ dθ
and
v = − cos θ
1
sin θ dθ = −e3θ cos θ − (− cos θ )(3e3θ dθ )
e3θ sin θ dθ =
e3θ
( − cos θ + 3sin θ )
10
π
1
.
1 - e3θ
(− cos θ + 3sin θ ) !
yEL dA = a3
3 ( 10
)0
=
Hence
a 3 3π
(e + 1) = 413.09a3
30
1
or x = −9.27a
1
or
xA = xEL dA: x (133.623a 2 ) = −1239.26a 3
yA = yEL dA: y (133.623a 2 ) = 413.09a3
y = 3.09a
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
602
PROBLEM 5.50
Determine the centroid of the area shown when a = 2 in.
SOLUTION
xEL = x
We have
yEL =
/ 10
dA = ydx = $1 − % dx
x'
&
and
1
A = dA =
Then
and
1
1/ 10
y = $1 − %
2
2&
x'
=
1
a
EL dA =
1
a
1x
1y
EL dA
1
1
1
a/
1
1 0 dx
2
a
$1 − x % 2 = [ x − ln x]1 = (a − ln a − 1) in.
&
'
a
.
/ a2
-/ 1 0 . - x 2
10
− x ! = $$
− a + %% in.3
x $1 − % dx ! =
x' ) ( 2
2'
(&
)1 & 2
1 / 1 0 -/ 1 0 . 1
1 − % $1 − % dx ! =
2 $&
x ' (&
x' ) 2
1
a/
1
2 1 0
$ 1 − x + 2 % dx
x '
&
a
=
11.
1/
10
x − 2ln x − ! = $ a − 2ln a − % in.3
2(
x )1 2 &
a'
1
xA = xEL dA: x =
1
yA = yEL dA: y =
a2
2
−a+
1
2
in.
a − ln a − 1
a − 2ln a − 1a
2( a − ln a − 1)
in.
Find: x and y when a = 2 in.
We have
x=
and
y=
1
2
(2) 2 − 2 +
1
2
2 − ln 2 − 1
2 − 2ln 2 − 12
2(2 − ln 2 − 1)
or
x = 1.629 in.
or y = 0.1853 in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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603
PROBLEM 5.51
Determine the value of a for which the ratio x / y is 9.
SOLUTION
xEL = x
We have
yEL =
1
1/ 10
y = $1 − %
2
2&
x'
/ 10
dA = ydx = $1 − % dx
x'
&
and
1
A = dA =
Then
1
a/
1
1 0 dx
= [ x − ln x]1a
$1 − %
x' 2
&
= (a − ln a − 1) in.2
and
1
xEL dA =
1
a
1
a
.
-/ 1 0 . - x 2
− x!
x $1 − % dx ! =
x' ) ( 2
(&
)1
/ a2
10
= $$
− a + %% in.3
2'
& 2
1
yEL dA =
1
a
1
1 / 1 0 -/ 1 0 . 1
1 − % $1 − % dx ! =
x ' (&
x' ) 2
2 $&
1
a/
1
2 1 0
$ 1 − x + 2 % dx
x '
&
a
=
11.
x − 2ln x − !
2(
x )1
=
1/
10
a − 2ln a − % in.3
$
2&
a'
1
xA = xEL dA: x =
1
yA = yEL dA: y =
a2
2
−a+
1
2
in.
a − ln a − 1
a − 2ln a − 1a
2( a − ln a − 1)
in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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604
PROBLEM 5.51 (Continued)
Find: a so that
x
=9
y
x xA
=
=
y yA
We have
Then
or
1
2
1
2
a2 − a +
1
2
( a − 2 ln a − 1a )
1x
1y
EL dA
EL dA
=9
a 3 − 11a 2 + a + 18a ln a + 9 = 0
Using trial and error or numerical methods and ignoring the trivial solution a = 1 in., find
a = 1.901 in. and a = 3.74 in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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605
PROBLEM 5.52
Determine the volume and the surface area of the solid obtained
by rotating the area of Problem 5.1 about (a) the line x = 240 mm,
(b) the y axis.
PROBLEM 5.1 Locate the centroid of the plane area shown.
SOLUTION
From the solution to Problem 5.1 we have
A = 15.3 × 103 mm 2
Σ xA = 2.6865 × 106 mm3
Σ yA = 1.4445 × 106 mm3
Applying the theorems of Pappus-Guldinus we have
(a)
Rotation about the line
x = 240 mm
Volume = 2π (240 − x ) A
= 2π (240 A − Σ xA)
= 2π [240(15.3 × 103 ) − 2.6865 × 106 ]
Volume = 6.19 × 106 mm3
Area = 2π X line L = 2π Σ( xline ) L
= 2π ( x1 L1 + x3 L3 + x4 L4 + x5 L5 + x6 L6 )
Where x1 ,
, x6 are measured with respect to line x = 240 mm.
Area = 2π [(120)(240) + (15)(30) + (30)(270)
+ (135)(210) + (240)(30)]
(b)
Area = 458 × 103 mm 2
Rotation about the y axis
Volume = 2π X area A = 2π (Σ xA)
= 2π (2.6865 × 106 mm3 )
Volume = 16.88 × 106 mm3
Area = 2π X line L = 2π Σ( xline ) L
= 2π ( x1 L1 + x2 L2 + x3 L3 + x4 L4 + x5 L5 )
= 2π [(120)(240) + (240)(300)
+ (225)(30) + (210)(270) + (105)(210)]
Area = 1.171 × 106 mm 2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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606
!
PROBLEM 5.53
Determine the volume and the surface area of the solid obtained by rotating
the area of Problem 5.2 about (a) the line y = 60 mm, (b) the y axis.
PROBLEM 5.2 Locate the centroid of the plane area shown.
SOLUTION
From the solution to Problem 5.2 we have
A = 1740 mm 2
Σ xA = 28200 mm3
Σ yA = 55440 mm3
Applying the theorems of Pappus-Guldinus we have
(a)
y = 60 mm
Rotation about the line
Volume = 2π (60 − y ) A
= 2π (60 A − Σ yA)
= 2π [60(1740) − 55440]
Volume = 308 × 103 mm3
Area = 2π Yline
= 2π Σ( yline ) L
= 2π ( y1 L1 + y2 L2 + y3 L3 + y4 L4 + y6 L6 )
Where y1 ,
, y6 are measured with respect to line y = 60 mm.
Area = 2π -(60)(20) + (48)(24) + (36)(30) + (18) (30) 2 + (36) 2 + (30)(60) .
!)
(
Area = 38.2 × 103 mm 2
(b)
Rotation about the y axis
Volume = 2π X area A = 2π (Σ xA) = 2π (28200 mm3 )
Volume = 177.2 × 106 mm3
Area = 2π X line L = 2π Σ( xline ) L = 2π ( x1 L1 + x2 L2 + x3 L3 + x4 L4 + x5 L5 )
= 2π -(10)(20) + (20)(24) + (35)(30) + (35) (30) 2 + (36) 2 + (10)(20) .
!)
(
Area = 22.4 × 103 mm 2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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607
PROBLEM 5.54
Determine the volume and the surface area of the solid obtained by rotating the
area of Problem 5.8 about (a) the x axis, (b) the y axis.
PROBLEM 5.8 Locate the centroid of the plane area shown.
SOLUTION
From the solution to Problem 5.8 we have
A = 1146.57 in.2
ΣxA = 14147.0 in.3
ΣyA = 26897 in.3
Applying the theorems of Pappus-Guldinus we have
(a)
Rotation about the x axis:
Volume = 2π Yarea A = 2π Σ y A
= 2π (26897 in.3 )
or Volume = 169.0 × 103 in.3
Area = 2π Yline A
= 2π Σ( yline ) A
= 2π ( y2 L2 + y3 L3 + y4 L4 + y5 L5 + y6 L6 )
= 2π [(7.5)(15) + (30)(π × 15) + (47.5)(5)
+ (50)(30) + (25)(50)]
(b)
or Area = 28.4 × 103 in.2
Rotation about the y axis
Volume = 2π X area A = 2π Σ x A
= 2π (14147.0 in.3 )
or Volume = 88.9 × 103 in.3
Area = 2π X line L = 2π Σ( xline ) L
= 2π ( x1 L1 + x2 L2 + x3 L3 + x4 L4 + x5 L5 )
.
2 × 15 0
/
= 2π (15)(30) + (30)(15) + $ 30 −
(π × 15) + (30)(5) + (15)(30) !
%
π '
&
(
)
Area = 15.48 × 103 in.2
or
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608
PROBLEM 5.55
Determine the volume of the solid generated by rotating the parabolic
area shown about (a) the x axis, (b) the axis AA′.
SOLUTION
First, from Figure 5.8a we have
4
ah
3
2
y= h
5
A=
Applying the second theorem of Pappus-Guldinus we have
(a)
Rotation about the x axis:
Volume = 2π yA
/ 2 0/ 4 0
= 2π $ h %$ ah %
& 5 '& 3 '
(b)
or Volume =
16
π ah 2
15
or Volume =
16 2
πa h
3
Rotation about the line AA′:
Volume = 2π (2a) A
/4 0
= 2π (2a) $ ah %
&3 '
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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609
PROBLEM 5.56
Determine the volume and the surface area of the chain link
shown, which is made from a 6-mm-diameter bar, if R = 10 mm
and L = 30 mm.
SOLUTION
The area A and circumference C of the cross section of the bar are
A=
π
4
d 2 and C = π d .
Also, the semicircular ends of the link can be obtained by rotating the cross section through a horizontal
semicircular arc of radius R. Now, applying the theorems of Pappus-Guldinus, we have for the volume V:
V = 2(Vside ) + 2(Vend )
= 2( AL) + 2(π RA)
= 2( L + π R) A
or
V = 2[30 mm + π (10 mm)]
-π
.
(6 mm) 2 !
(4
)
= 3470 mm3
For the area A:
or V = 3470 mm3
A = 2( Aside ) + 2( Aend )
= 2(CL) + 2(π RC )
= 2( L + π R)C
or
A = 2[30 mm + π (10 mm)][π (6 mm)]
= 2320 mm 2
or A = 2320 mm 2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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610
PROBLEM 5.57
Verify that the expressions for the volumes of the first four shapes in Figure 5.21 on Page 253 are correct.
SOLUTION
Following the second theorem of Pappus-Guldinus, in each case a specific
generating area A will be rotated about the x axis to produce the given shape.
Values of y are from Figure 5.8a.
(1)
Hemisphere: the generating area is a quarter circle
We have
(2)
2
V = π a3
3
/ 4a 0/ π 0
V = 2π y A = 2π $
%$ ha %
& 3π '& 4 '
2
or V = π a 2 h
3
Paraboloid of revolution: the generating area is a quarter parabola
We have
(4)
or
Semiellipsoid of revolution: the generating area is a quarter ellipse
We have
(3)
/ 4a 0/ π 2 0
V = 2π y A = 2π $
%$ a %
& 3π '& 4 '
/ 3 0/ 2 0
V = 2π y A = 2π $ a %$ ah %
& 8 '& 3 '
1
or V = π a 2 h
2
Cone: the generating area is a triangle
We have
/ a 0/ 1 0
V = 2π y A = 2π $ %$ ha %
& 3 '& 2 '
1
or V = π a 2 h
3
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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611
PROBLEM 5.58
A 34 -in.-diameter hole is drilled in a piece of 1-in.-thick steel; the hole
is then countersunk as shown. Determine the volume of steel removed
during the countersinking process.
SOLUTION
The required volume can be generated by rotating the area shown about the y axis. Applying the second
theorem of Pappus-Guldinus, we have
V = 2π x A
= 2π
-3 1 / 1 0 . -1 1
1 .
+ $ % in.! × × in. × in.!
8
3
4
2
4
4 )
& ' ) (
(
V = 0.0900 in.3
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
612
PROBLEM 5.59
Determine the capacity, in liters, of the punch bowl
shown if R = 250 mm.
SOLUTION
The volume can be generated by rotating the triangle and circular sector shown about the y axis. Applying the
second theorem of Pappus-Guldinus and using Figure 5.8a, we have
V = 2π x A = 2π Σ x A
= 2π ( x1 A1 + x2 A2 )
-/ 1 1 0 / 1 1
3 0 / 2 R sin 30° 0 / π 2 0 .
= 2π $ × R % $ × R ×
R% + $
R %!
%
$
2 '% &$ 3 × π6 '% &$ 6
' !)
(& 3 2 ' & 2 2
/ R3
R3 0
= 2π $$
+
%%
& 16 3 2 3 '
3 3
π R3
8
3 3
=
π (0.25 m)3
8
= 0.031883 m3
=
Since
103 l = 1 m3
V = 0.031883 m3 ×
103 l
1 m3
V = 31.9 l
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
613
PROBLEM 5.60
Three different drive belt profiles are to be
studied. If at any given time each belt makes
contact with one-half of the circumference of
its pulley, determine the contact area between
the belt and the pulley for each design.
SOLUTION
SOLUTION
Applying the first theorem of Pappus-Guldinus, the contact area AC of a belt
is given by:
AC = π yL = π Σ yL
where the individual lengths are the lengths of the belt cross section that are
in contact with the pulley.
AC = π [2( y1 L1 ) + y2 L2 ]
(a)
0.125 0 . - 0.125 in. .
*2 -/
*3
= π "2 $ 3 −
+ [(3 − 0.125)in.](0.625 in.) #
in.!
%
!
2 ' ) ( cos 20° )
+* (&
,*
AC = 8.10 in.2
or
AC = π [2( y1 L1 )]
(b)
-/
0.375 0 . / 0.375 in. 0
in.!
= 2π $ 3 − 0.08 −
2 '% ) &$ cos 20° '%
(&
AC = 6.85 in.2
or
AC = π [2( y1 L1 )]
(c)
-/
2(0.25) 0 .
in. [π (0.25 in.)]
= π $3 −
π %' !)
(&
AC = 7.01 in.2
or
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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614
PROBLEM 5.61
The aluminum shade for the small high-intensity lamp shown has a uniform thickness of 1 mm. Knowing that
the density of aluminum is 2800 kg/m3, determine the mass of the shade.
SOLUTION
The mass of the lamp shade is given by
m = ρV = ρ At
Where A is the surface area and t is the thickness of the shade. The area can be generated by rotating the line
shown about the x axis. Applying the first theorem of Pappus Guldinus we have
A = 2π yL = 2π Σ yL
= 2π ( y1 L1 + y2 L2 + y3 L3 + y4 L4 )
or
A = 2π
-13 mm
/ 13 + 16 0
2
2
(13 mm) + $
% mm × (32 mm) + (3 mm)
& 2 '
( 2
/ 16 + 28 0
+$
mm × (8 mm) 2 + (12 mm) 2
%
& 2 '
/ 28 + 33 0
2
2.
+$
% mm × (28 mm) + (5 mm) !
2
&
'
)
= 2π (84.5 + 466.03 + 317.29 + 867.51)
= 10903.4 mm 2
Then
m = ρ At
= (2800 kg/m3 )(10.9034 × 10−3 m 2 )(0.001 m)
m = 0.0305 kg
or
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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615
!
PROBLEM 5.62
The escutcheon (a decorative plate placed on a pipe where the
pipe exits from a wall) shown is cast from brass. Knowing that
the density of brass is 8470 kg/m3, determine the mass of the
escutcheon.
SOLUTION
The mass of the escutcheon is given by m = (density)V , where V is the volume. V can be generated by
rotating the area A about the x-axis.
From the figure:
L1 = 752 − 12.52 = 73.9510 m
L2 =
37.5
= 76.8864 mm
tan 26°
a = L2 − L1 = 2.9324 mm
12.5
= 9.5941°
75
26° − 9.5941°
α=
= 8.2030° = 0.143168 rad
2
φ = sin −1
Area A can be obtained by combining the following four areas:
Applying the second theorem of Pappus-Guldinus and using Figure 5.8a, we have
V = 2π yA = 2π Σ yA
Seg.
A, mm 2
y , mm
y A , mm3
1
1
(76.886)(37.5) = 1441.61
2
1
(37.5) = 12.5
3
18020.1
2
−α (75) 2 = −805.32
2(75)sin α
sin (α + φ ) = 15.2303
3α
−12265.3
3
1
− (73.951)(12.5) = −462.19
2
1
(12.5) = 4.1667
3
−1925.81
4
−(2.9354)(12.5) = −36.693
1
(12.5) = 6.25
2
−229.33
Σ
3599.7
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
616
PROBLEM 5.62 (Continued)
Then
V = 2π Σ yA
= 2π (3599.7 mm3 )
= 22618 mm3
m = (density)V
= (8470 kg/m3 )(22.618 × 10−6 m3 )
= 0.191574 kg
or m = 0.1916 kg
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
617
PROBLEM 5.63
A manufacturer is planning to produce 20,000 wooden pegs
having the shape shown. Determine how many gallons of
paint should be ordered, knowing that each peg will be given
two coats of paint and that one gallon of paint covers 100 ft2.
SOLUTION
The number of gallons of paint needed is given by
/ 1 gallon 0
Number of gallons = (Number of pegs)(Surface area of 1 peg) $
(2 coats)
2 %
& 100 ft '
Number of gallons = 400 As
or
( As ! ft 2 )
where As is the surface area of one peg. As can be generated by rotating the line shown about the x axis.
Using the first theorem of Pappus-Guldinus and Figures 5.8b,
We have
R = 0.875 in.
sin 2α =
0.5
0.875
2α = 34.850° α = 17.425°
or
As = 2π Y L = 2π Σ y L
yL, in.2
1
0.25
y , in.
"
0.125
2
0.5
0.25
0.125
3
0.0625
0.25 + 0.3125
= 0.28125
2
0.0175781
4
3 − 0.875(1 − cos 34.850) − 0.1875 = 2.6556
0.3125
0.82988
L, in.
5
6
π
2
0.5 −
× 0.1875 = 0.29452
2 × 0.1875
π
0.875sin17.425°
2α (0.875)
α
0.03125
= 0.38063
0.112103
× sin17.425°
0.137314
Σ y L = 1.25312 in.2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
618
PROBLEM 5.63 (Continued)
Then
As = 2π (1.25312 in.2 ) ×
1 ft 2
144 in.2
= 0.054678 ft 2
Finally
Number of gallons = 400 × 0.054678
= 21.87 gallons
Order 22 gallons
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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619
!
PROBLEM 5.64
The wooden peg shown is turned from a dowel 1 in. in
diameter and 4 in. long. Determine the percentage of the
initial volume of the dowel that becomes waste.
SOLUTION
To obtain the solution it is first necessary to determine the volume of the peg. That volume can be generated
by rotating the area shown about the x axis.
The generating area is next divided into six components as indicated
sin 2α =
0.5
0.875
2α = 34.850°
or
α = 17.425°
Applying the second theorem of Pappus-Guldinus and then using Figure 5.8a, we have
VPEG = 2π YA = 2π Σ yA
A, in.2
yA, in.3
y , in.
"
0.125
0.015625
1
0.5 × 0.25 × 0.125
2
[3 − 0.875(1 − cos 34.850°) − 0.1875]
× (0.3125) = 0.82987
0.15625
0.129667
3
0.1875 × 0.5 × 0.9375
0.25
0.023438
4
−
π
4
(0.1875)2 = −0.027612
0.5 −
4 × 0.1875
= 0.42042
3π
−0.011609
5
α (0.875)2
2 × 0.875sin17.425°
× sin17.425°
3α
0.04005
6
1
− (0.875cos 34.850°)(0.5) = −0.179517
2
1
(0.5) = 0.166667
3
−0.029920
Σ y L = 0.167252 in.3
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
620
PROBLEM 5.64 (Continued)
Then
V peg = 2π (0.167252 in.3 )
= 1.05088 in.3
Now
Vdowel =
=
π
4
π
(diameter)2 (length)
(1 in.)2 (4in.)
4
= 3.14159 in.3
Then
% Waste =
=
Vwaste
× 100%
Vdowel
Vdowel − V peg
Vdowel
× 100%
/ 1.05088 0
= $1 −
% × 100%
& 3.14159 '
or % Waste = 66.5%
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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621
!
PROBLEM 5.65*
The shade for a wall-mounted light is formed from a thin sheet of
translucent plastic. Determine the surface area of the outside of
the shade, knowing that it has the parabolic cross section shown.
SOLUTION
First note that the required surface area A can be generated by rotating the parabolic cross section through π
radians about the y axis. Applying the first theorem of Pappus-Guldinus we have
A = π xL
Now at
x = 100 mm,
250 = k (100)
and
2
y = 250 mm
or k = 0.025 mm −1
xEL = x
2
/ dy 0
dL = 1 + $ % dx
& dx '
where
dy
= 2kx
dx
Then
dL = 1 + 4k 2 x 2 dx
We have
xL = xEL dL =
1
1
100
x
0
( 1 + 4k x dx )
2 2
100
-1 1
.
(1 + 4k 2 x 2 )3/ 2 !
2
( 3 4k
)0
1
1
[1 + 4(0.025)2 (100) 2 ]3/2 − (1)3/2
=
2
12 (0.025)
xL =
{
}
= 17543.3 mm 2
Finally
A = π (17543.3 mm 2 )
or A = 55.1 × 103 mm 2
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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622
PROBLEM 5.66
For the beam and loading shown, determine (a) the magnitude
and location of the resultant of the distributed load, (b) the
reactions at the beam supports.
SOLUTION
1
(150 lb/ft) (9 ft) = 675 lb
2
1
RII = (120 lb/ft) (9 ft) = 540 lb
2
R = RI + RII = 675 + 540 = 1215 lb
RI =
XR = ΣXR :
X (1215) = (3)675) + (6)(540)
X = 4.3333 ft
R = 1215 lb
(a)
(b)
Reactions:
X = 4.33 ft
Σ M A = 0: B (9 ft) − (1215 lb) (4.3333 ft) = 0
B = 585.00 lb
B = 585 lb
Σ Fy = 0: A + 585.00 lb − 1215 lb = 0
A = 630.00 lb
A = 630 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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623
PROBLEM 5.67
For the beam and loading shown, determine (a) the magnitude and
location of the resultant of the distributed load, (b) the reactions at the
beam supports.
SOLUTION
(a)
RI = (4 m) ( 200 N/m ) = 800 N
We have
RII =
Then
or
and
Σ Fy :
2
(4 m) (600 N/m) = 1600 N
3
− R = − RI − RII
R = 800 + 1600 = 2400 N
Σ M A : − X (2400) = −2(800) − 2.5(1600)
X=
or
7
m
3
R = 2400 N
(b)
Reactions
X = 2.33 m
Σ Fx = 0: Ax = 0
/7 0
Σ M A = 0: (4 m) By − $ m % (2400 N) = 0
&3 '
By = 1400 N
or
ΣFy = 0: Ay + 1400 N − 2400 N = 0
Ay = 1000 N
or
A = 1000 N
B = 1400 N
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624
PROBLEM 5.68
Determine the reactions at the beam supports for the given
loading.
SOLUTION
1
(4 kN/m) (6 m)
2
= 12 kN
RII = (2 kN/m) (10 m)
RI =
= 20 kN
ΣFy = 0: A − 12 kN − 20 kN = 0
A = 32.0 kN
ΣM A = 0: M A − (12 kN) (2 m) − (20 kN) (5 m) = 0
M A = 124.0 kN ⋅ m
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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625
PROBLEM 5.69
Determine the reactions at the beam supports for the given
loading.
SOLUTION
We have
Then
1
(3ft) ( 480 lb/ft ) = 720 lb
2
1
R II = (6 ft) ( 600 lb/ft ) = 1800 lb
2
R III = (2 ft) ( 600 lb/ft ) = 1200 lb
R I=
ΣFx = 0: Bx = 0
ΣM B = 0: (2 ft)(720 lb) − (4 ft) (1800 lb) + (6 ft)C y − (7 ft) (1200 lb) = 0
or
C y = 2360 lb
C = 2360 lb
ΣFy = 0: −720 lb + By − 1800 lb + 2360 lb − 1200 lb = 0
or
By = 1360 lb
B = 1360 lb
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626
PROBLEM 5.70
Determine the reactions at the beam supports for the given loading.
SOLUTION
R I = (200 lb/ft) (15 ft)
R I = 3000 lb
1
(200 lb/ft) (6 ft)
2
R II = 600 lb
R II =
ΣM A = 0: − (3000 lb) (1.5 ft) − (600 lb) (9 ft + 2 ft) + B(15 ft) = 0
B = 740 lb
B = 740 lb
ΣFy = 0: A + 740 lb − 3000 lb − 600 lb = 0
A = 2860 lb
A = 2860 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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627
PROBLEM 5.71
Determine the reactions at the beam supports for the given loading.
SOLUTION
First replace the given loading with the loading shown below. The two loadings are equivalent because both
are defined by a linear relation between load and distance and the values at the end points are the same.
We have
Then
1
(3.6 m)(2200 N/m) = 3960 N
2
RII = (3.6 m)(1200 N/m) = 4320 N
RI =
ΣFx = 0: Bx = 0
ΣM B = 0: − (3.6 m) Ay + (2.4 m)(3960 N)
−(1.8 m)(4320 N) = 0
or
Ay = 480 N
A = 480 N
ΣFy = 0: 480 N − 3960 N + 4320 + By = 0
By = −840 N
or
B = 840 N
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628
PROBLEM 5.72
Determine the reactions at the beam supports for the given
loading.
SOLUTION
We have
Then
1
RI = (12 ft)(200 lb/ft) = 800 lb
3
1
RII = (6 ft)(100 lb/ft) = 200 lb
3
ΣFx = 0: Ax = 0
ΣFy = 0: Ay − 800 lb − 200 lb = 0
or
Ay = 1000 lb
A = 1000 lb
ΣM A = 0: M A − (3 ft)(800 lb) − (16.5 ft)(200 lb) = 0
or
M A = 5700 lb ⋅ ft
M A = 5700 lb ⋅ ft
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629
PROBLEM 5.73
Determine the reactions at the beam supports for the given loading.
SOLUTION
First replace the given loading with the loading shown below. The two loadings are equivalent because both
are defined by a parabolic relation between load and distance and the values at the end points are the same.
We have
RI = (6 m)(300 N/m) = 1800 N
RII =
Then
2
(6 m)(1200 N/m) = 4800 N
3
ΣFx = 0: Ax = 0
ΣFy = 0: Ay + 1800 N − 4800 N = 0
or
Ay = 3000 N
A = 3000 N
15 !
ΣM A = 0: M A + (3 m)(1800 N) − " m # (4800 N) = 0
$ 4 %
or
M A = 12.6 kN ⋅ m
M A = 12.6 kN ⋅ m
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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630
PROBLEM 5.74
Determine (a) the distance a so that the vertical reactions at
supports A and B are equal, (b) the corresponding reactions at
the supports.
SOLUTION
(a)
We have
Then
1
( a m )(1800 N/m ) = 900a N
2
1
RII = &(( 4 − a ) m ') ( 600 N/m ) = 300 ( 4 − a ) N
2
RI =
ΣFy = 0: Ay − 900a − 300(4 − a ) + B y = 0
or
Ay + By = 1200 + 600a
Now
Ay = By * Ay = By = 600 + 300a (N)
Also
(1)
&
a! '
ΣM B = 0: − (4 m) Ay + +" 4 − # m , &(( 900a ) N ')
3% )
($
&1
'
+ + (4 − a ) m , (&300 ( 4 − a ) N )' = 0
(3
)
or
Ay = 400 + 700a − 50a 2
Equating Eqs. (1) and (2)
600 + 300a = 400 + 700a − 50a 2
or
a 2 − 8a + 4 = 0
(2)
8 ± (−8) 2 − 4(1)(4)
2
Then
a=
or
a = 0.53590 m
Now
a#4m*
a = 7.4641 m
a = 0.536 m
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631
PROBLEM 5.74 (Continued)
(b)
We have
Eq. (1)
ΣFx = 0: Ax = 0
Ay = By
= 600 + 300(0.53590)
= 761 N
A = B = 761 N
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632
PROBLEM 5.75
Determine (a) the distance a so that the reaction at support B is
minimum, (b) the corresponding reactions at the supports.
SOLUTION
(a)
We have
Then
or
Then
(b)
Eq. (1)
and
1
(a m)(1800 N/m) = 900a N
2
1
RII = [(4 − a )m](600 N/m) = 300(4 − a) N
2
RI =
a !
8+a !
ΣM A = 0: − " m # (900a N) − "
m # &(300 ( 4 − a ) N ') + ( 4 m ) By = 0
$3 %
$ 3
%
By = 50a 2 − 100a + 800
dBy
da
= 100a − 100 = 0
By = 50(1)2 − 100(1) + 800 = 750 N
(1)
or a = 1.000 m
B = 750 N
ΣFx = 0: Ax = 0
ΣFy = 0: Ay − 900(1)N − 300(4 − 1)N + 750 N = 0
or
Ay = 1050 N
A = 1050 N
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633
PROBLEM 5.76
Determine the reactions at the beam supports for the given loading
when ω 0 = 150 lb/ft.
SOLUTION
We have
Then
1
(18 ft)(450 lb/ft) = 4050 lb
2
1
RII = (18 ft)(150 lb/ft) = 1350 lb
2
RI =
ΣFx = 0: Cx = 0
ΣM B = 0: − (44,100 kip ⋅ ft) − (2 ft) − (4050 lb)
− (8 ft)(1350 lb) + (12 ft)C y = 0
C y = 5250 lb
or
C = 5250 lb
ΣFy = 0: By − 4050 lb − 1350 lb + 5250 lb = 0
By = 150 lb
or
B = 150.0 lb
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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634
PROBLEM 5.77
Determine (a) the distributed load ω 0 at the end D of the
beam ABCD for which the reaction at B is zero, (b) the
corresponding reaction at C.
SOLUTION
(a)
We have
Then
1
(18 ft)(450 lb/ft) = 4050 lb
2
1
RII = (18 ft)(ω0 lb/ft) = 9 ω0 lb
2
RI =
ΣM C = 0: − (44,100 lb ⋅ ft) + (10 ft)(4050 lb) + (4 ft)(9ω0 lb) = 0
ω0 = 100.0 lb/ft
or
ΣFx = 0: C x = 0
(b)
ΣFy = 0: − 4050 lb − (9 × 100) lb + C y = 0
C y = 4950 lb
or
C = 4950 lb
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635
PROBLEM 5.78
The beam AB supports two concentrated loads and rests on soil that
exerts a linearly distributed upward load as shown. Determine the
values of ω A and ω B corresponding to equilibrium.
SOLUTION
1
RI = ω A (1.8 m) = 0.9ω A
2
1
RII = ωB (1.8 m) = 0.9ωB
2
ΣM D = 0: (24 kN)(1.2 − a) − (30 kN)(0.3 m) − (0.9ω A )(0.6 m) = 0
For
(1)
a = 0.6 m: 24(1.2 − 0.6) − (30)(0.3) − 0.54 ωa = 0
14.4 − 9 − 0.54ω A = 0
ΣFy = 0: − 24 kN − 30 kN + 0.9(10 kN/m) + 0.9ωB = 0
ω A = 10.00 kN/m
ωB = 50.0 kN/m
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636
PROBLEM 5.79
For the beam and loading of Problem 5.78, determine (a) the distance a
for which ω A = 20 kN/m, (b) the corresponding value of ω B .
PROBLEM 5.78 The beam AB supports two concentrated loads and
rests on soil that exerts a linearly distributed upward load as shown.
Determine the values of ω A and ω B corresponding to equilibrium.
SOLUTION
We have
1
(1.8 m)(20 kN/m) = 18 kN
2
1
RII = (1.8 m)(ω B kN/m) = 0.9ωB kN
2
RI =
ΣM C = 0: (1.2 − a)m × 24 kN − 0.6 m × 18 kN − 0.3 m × 30 kN = 0
(a)
a = 0.375 m
or
ΣFy = 0: − 24 kN + 18 kN + (0.9ωB ) kN − 30 kN = 0
(b)
ω B = 40.0 kN/m
or
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637
PROBLEM 5.80
The cross section of a concrete dam is as shown. For a 1-m-wide
dam section determine (a) the resultant of the reaction forces
exerted by the ground on the base AB of the dam, (b) the point of
application of the resultant of Part a, (c) the resultant of the
pressure forces exerted by the water on the face BC of the dam.
SOLUTION
(a)
Consider free body made of dam and triangular section of water shown. (Thickness = 1 m)
p = (7.2 m)(103 kg/m3 )(9.81m/s 2 )
2
(4.8 m)(7.2 m)(1 m)(2.4 × 103 kg/m3 )(9.81 m/s 2 )
3
= 542.5 kN
1
W2 = (2.4 m)(7.2 m)(1 m)(2.4 × 103 kg/m3 )(9.81 m/s 2 )
2
= 203.4 kN
1
W3 = (2.4 m)(7.2 m)(1 m)(103 kg/m3 )(9.81 m/s 2 )
2
= 84.8 kN
1
1
P = Ap = (7.2 m)(1 m)(7.2 m)(103 kg/m3 )(9.81 m/s 2 )
2
2
= 254.3 kN
W1 =
ΣFx = 0: H − 254.3 kN = 0
H = 254 kN
ΣFy = 0: V − 542.5 − 203.4 − 84.8 = 0
V = 830.7 kN
V = 831 kN
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638
PROBLEM 5.80 (Continued)
5
x1 = (4.8 m) = 3 m
8
1
x2 = 4.8 + (2.4) = 5.6 m
3
2
x3 = 4.8 + (2.4) = 6.4 m
3
(b)
ΣM A = 0: xV − Σ xW + P(2.4 m) = 0
x(830.7 kN) − (3 m)(542.5 kN) − (5.6 m)(203.4 kN)
− (6.4 m)(84.8 kN) + (2.4 m)(254.3 kN) = 0
x(830.7) − 1627.5 − 1139.0 − 542.7 + 610.3 = 0
x(830.7) − 2698.9 = 0
x = 3.25 m (To right of A)
(c)
Resultant on face BC
Direct computation:
P = ρ gh = (103 kg/m3 )(9.81 m/s 2 )(7.2 m)
P = 70.63 kN/m 2
BC = (2.4) 2 + (7.2) 2
= 7.589 m
θ = 18.43°
1
PA
2
1
= (70.63 kN/m 2 )(7.589 m)(1 m)
2
R=
R = 268 kN
18.43°
− R = 268 kN
18.43°
R = 268 kN
18.43°
Alternate computation: Use free body of water section BCD.
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639
PROBLEM 5.81
The cross section of a concrete dam is as shown. For a 1-ft-wide
dam section determine (a) the resultant of the reaction forces
exerted by the ground on the base AB of the dam, (b) the point of
application of the resultant of Part a, (c) the resultant of the
pressure forces exerted by the water on the face BC of the dam.
SOLUTION
The free body shown consists of a 1-ft thick section of the dam and the quarter circular section of water above
the dam.
Note:
4 × 21 !
x1 = " 21 −
# ft
3π %
$
= 12.0873 ft
x2 = (21 + 4) ft = 25 ft
4 × 21 !
x4 = " 50 −
ft
3π #%
$
= 41.087 ft
For area 3 first note.
I
II
Then
−
a
x
r2
1
r
2
π
4
r2
(
r−
& 1 (21)(21) 2 + ( 21 − 4×21 ) − π × 212
2
3π
4
x3 = 29 ft + (
(
(21) 2 − π4 (21) 2
*
= (29 + 4.6907)ft = 33.691 ft
4r
3π
) ') ft
)
+
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640
PROBLEM 5.81 (Continued)
(a)
Now
W = γV
So that
&π
'
W1 = (150 lb/ft 3 ) ( (21 ft)2 (1 ft) ) = 51,954 lb
4
*
+
W2 = (150 lb/ft 3 )[(8 ft)(21 ft)(1 ft)] = 25, 200 lb
&
'
π
!
W3 = (150 lb/ft 3 ) (" 212 − × 212 # ft 2 × (1 ft) ) = 14,196 lb
4
%
*$
+
&π
'
W4 = (62.4 lb/ft 3 ) ( (21 ft) 2 (1 ft) ) = 21, 613 lb
*4
+
Also
Then
P=
1
1
Ap = [(21 ft)(1 ft)][(62.4 lb/ft 3 )(21 ft)] = 13,759 lb
2
2
ΣFx = 0: H − 13,759 lb = 0
or H = 13.76 kips
ΣFy = 0: V − 51,954 lb − 25,200 lb − 14,196 lb − 21,613 lb = 0
or
(b)
We have
V = 112,963 lb
V = 113.0 kips
ΣM A = 0: x(112,963 lb) − (12.0873 ft)(51,954 lb) − (25 ft)(25,200 lb)
−(33.691 ft)(14,196 lb) − (41.087 ft)(21,613 lb)
+(7 ft)(13,759 lb) = 0
or
112,963x − 627,980 − 630, 000 − 478, 280 − 888, 010 + 96,313 = 0
x = 22.4 ft
or
(c)
Consider water section BCD as the free body
We have
Then
ΣF = 0
− R = 25.6 kips
57.5°
or R = 25.6 kips
57.5°
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641
!
PROBLEM 5.82
The 3 × 4-m side AB of a tank is hinged at its bottom A and is held in place by a
thin rod BC. The maximum tensile force the rod can withstand without breaking
is 200 kN, and the design specifications require the force in the rod not to exceed
20 percent of this value. If the tank is slowly filled with water, determine the
maximum allowable depth of water d in the tank.
SOLUTION
Consider the free-body diagram of the side.
We have
Now
Where
P=
1
1
Ap = A( ρ gd )
2
2
ΣM A = 0: hT −
d
P=0
3
h=3m
Then for d max .
(3 m)(0.2 × 200 × 103 N) −
d max
3
&1
'
3
3
2
( 2 (4 m × d max ) × (10 kg/m × 9.81 m/s × d max ) ) = 0
*
+
3
120 N ⋅ m − 6.54d max
N/m 2 = 0
or
d max = 2.64 m
or
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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642
!
PROBLEM 5.83
The 3 × 4-m side of an open tank is hinged at its bottom A and is held in
place by a thin rod BC. The tank is to be filled with glycerine, whose density
is 1263 kg/m3. Determine the force T in the rod and the reactions at the hinge
after the tank is filled to a depth of 2.9 m.
SOLUTION
Consider the free-body diagram of the side.
We have
Then
1
1
Ap = A( ρ gd )
2
2
1
= [(2.9 m)(4 m)] [(1263 kg/m3 )(9.81 m/s 2 )(2.9 m)]
2
= 208.40 kN
P=
ΣFy = 0: Ay = 0
2.9 !
ΣM A = 0: (3 m)T − "
m # (208.4 kN) = 0
$ 3 %
T = 67.151 kN
or
T = 67.2 kN
ΣFx = 0: Ax + 208.40 kN − 67.151 kN = 0
Ax = −141.249 kN
or
A = 141.2 kN
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643
!
PROBLEM 5.84
The friction force between a 6 × 6-ft square sluice gate AB and its guides is equal
to 10 percent of the resultant of the pressure forces exerted by the water on the face
of the gate. Determine the initial force needed to lift the gate if it weighs 1000 lb.
SOLUTION
Consider the free-body diagram of the gate.
Now
1
1
ApI = [(6 × 6) ft 2 ][(62.4 lb/ft 3 )(9 ft)]
2
2
= 10,108.8 lb
PI =
1
1
ApII = [(6 × 6) ft 2 ][(62.4 lb/ft 3 )(15 ft)]
2
2
= 16848 lb
PII =
Then
F = 0.1P = 0.1( PI + PII )
= 0.1(10108.8 + 16848)lb
= 2695.7 lb
Finally
ΣFy = 0: T − 2695.7 lb − 1000 lb = 0
or T = 3.70 kips
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644
!
PROBLEM 5.85
A freshwater marsh is drained to the ocean through an automatic tide
gate that is 4 ft wide and 3 ft high. The gate is held by hinges located
along its top edge at A and bears on a sill at B. If the water level in the
marsh is h = 6 ft, determine the ocean level d for which the gate will
open. (Specific weight of salt water = 64 lb/ft 3 .)
SOLUTION
Since gate is 4 ft wide
Thus:
w = (4 ft)p = 4γ (depth)
w1 = 4γ (h − 3)
w2 = 4γ h
w1′ = 4γ ′(d − 3)
w2′ = 4γ ′d
1
(3 ft)(w1′ − w1 )
2
1
= (3 ft)[4γ ′(d − 3) − 4λ (h − 3)] = 6γ ′(d − 3) − 6γ (h − 3)
2
PI′ − PI =
1
(3 ft)(w2′ − w2 )
2
1
= (3 ft)[4γ ′d − 4γ h] = 6γ ′d − 6γ h
2
PII′ − PII =
ΣM A = 0: (3 ft)B − (1 ft)(PI′ − PI ) − (2 ft)(PII′ − PII ) = 0
1
2
B = ( PI′ − PI ) − ( PII′ − PII )
3
3
1
2
= [6γ ′(d − 3) − 6γ (h − 3)] − [6γ ′d − 6γ h]
3
3
= 2γ (d − 3) − 2γ (h − 3) + 4γ ′d − 4γ h
B = 6γ ′(d − 1) − 6γ ( h − 1)
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645
PROBLEM 5.85 (Continued)
With
0 = 6γ ′(d − 1) − 6γ ( h − 1)
B = 0 and h = 6 ft:
d −1 = 5
Data:
γ
γ′
γ ′ = 64 lb/ft 3
γ = 62.4 lb/ft 3
62.4 lb/ft 3
64 lb/ft 3
= 4.875 ft
d −1 = 5
d = 5.88 ft
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646
PROBLEM 5.86
The dam for a lake is designed to withstand the additional force caused
by silt that has settled on the lake bottom. Assuming that silt is
equivalent to a liquid of density ρ s = 1.76 × 103 kg/m3 and considering a
1-m-wide section of dam, determine the percentage increase in the
force acting on the dam face for a silt accumulation of depth 2 m.
SOLUTION
First, determine the force on the dam face without the silt.
We have
1
1
Apw = A( ρ gh)
2
2
1
= [(6.6 m)(1 m)][(103 kg/m3 )(9.81 m/s 2 )(6.6 m)]
2
= 213.66 kN
Pw =
Next, determine the force on the dam face with silt
We have
1
Pw′ = [(4.6 m)(1 m)][(103 kg/m3 )(9.81 m/s2 )(4.6 m)]
2
= 103.790 kN
( Ps ) I = [(2.0 m)(1 m)][(103 kg/m3 )(9.81 m/s 2 )(4.6 m)]
= 90.252 kN
1
( Ps ) II = [(2.0 m)(1 m)][(1.76 × 103 kg/m3 )(9.81 m/s 2 )(2.0 m)]
2
= 34.531 kN
Then
P′ = Pw′ + ( Ps ) I + ( Ps ) II = 228.57 kN
The percentage increase, % inc., is then given by
% inc. =
P′ − Pw
× 100%
Pw
(228.57 − 213.66)
× 100%
213.66
= 6.9874%
=
% inc. = 6.98%
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647
PROBLEM 5.87
The base of a dam for a lake is designed to resist up to 120 percent of
the horizontal force of the water. After construction, it is found that
silt (that is equivalent to a liquid of density ρ s = 1.76 × 103 kg/m3 ) is
settling on the lake bottom at the rate of 12 mm/year. Considering a
1-m-wide section of dam, determine the number of years until the
dam becomes unsafe.
SOLUTION
From Problem 5.86, the force on the dam face before the silt is deposited, is Pw = 213.66 kN. The maximum
allowable force Pallow on the dam is then:
Pallow = 1.2 Pw = (1.5)(213.66 kN) = 256.39 kN
Next determine the force P′ on the dam face after a depth d of silt has settled
We have
1
Pw′ = [(6.6 − d )m × (1 m)][(103 kg/m3 )(9.81 m/s 2 )(6.6 − d )m]
2
= 4.905(6.6 − d ) 2 kN
( Ps ) I = [d (1 m)][(103 kg/m3 )(9.81 m/s 2 )(6.6 − d )m]
= 9.81 (6.6d − d 2 )kN
1
( Ps ) II = [d (1 m)][(1.76 × 103 kg/m3 )(9.81 m/s 2 )(d )m]
2
= 8.6328d 2 kN
P′ = Pw′ + ( Ps ) I + ( Ps ) II = [4.905(43.560 − 13.2000d + d 2 )
+ 9.81(6.6d − d 2 ) + 8.6328d 2 ]kN
= [3.7278d 2 + 213.66]kN
Now required that P′ = Pallow to determine the maximum value of d.
(3.7278d 2 + 213.66)kN = 256.39 kN
or
Finally
d = 3.3856 m
3.3856 m = 12 × 10−3
m
×N
year
or N = 282 years
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648
PROBLEM 5.88
A 0.5 × 0.8-m gate AB is located at the bottom of a tank filled
with water. The gate is hinged along its top edge A and rests on a
frictionless stop at B. Determine the reactions at A and B when
cable BCD is slack.
SOLUTION
First consider the force of the water on the gate.
We have
P=
1
1
Ap = A( ρ gh)
2
2
1
PI = [(0.5 m)(0.8 m)] × [(103 kg/m3 )(9.81 m/s 2 )(0.45 m)]
2
= 882.9 N
so that
1
PII = [(0.5 m)(0.8 m)] × [(103 kg/m3 )(9.81 m/s 2 )(0.93 m)]
2
= 1824.66 N
Reactions at A and B when T = 0
We have
ΣM A = 0:
1
2
(0.8 m)(882.9 N) + (0.8 m)(1824.66 N) − (0.8 m)B = 0
3
3
B = 1510.74 N
or
or
B = 1511 N
53.1°
A = 1197 N
53.1°
ΣF = 0: A + 1510.74 N − 882.9 N − 1824.66 N = 0
or
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649
!
PROBLEM 5.89
A 0.5 × 0.8-m gate AB is located at the bottom of a tank filled with
water. The gate is hinged along its top edge A and rests on a frictionless
stop at B. Determine the minimum tension required in cable BCD to
open the gate.
SOLUTION
First consider the force of the water on the gate.
We have
so that
P=
1
1
Ap = A( ρ gh)
2
2
1
PI = [(0.5 m)(0.8 m)] × [(103 kg/m3 )(9.81 m/s 2 )(0.45 m)]
2
= 882.9 N
1
PII = [(0.5 m)(0.8 m)] × [(103 kg/m3 )(9.81 m/s 2 )(0.93 m)]
2
= 1824.66 N
T to open gate
First note that when the gate begins to open, the reaction at B
Then
or
ΣM A = 0:
0.
1
2
(0.8 m)(882.9 N) + (0.8 m)(1824.66 N)
3
3
8 !
− (0.45 + 0.27)m × " T # = 0
$ 17 %
235.44 + 973.152 − 0.33882 T = 0
T = 3570 N
or
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650
!
PROBLEM 5.90
A long trough is supported by a continuous hinge along its
lower edge and by a series of horizontal cables attached to
its upper edge. Determine the tension in each of the cables,
at a time when the trough is completely full of water.
SOLUTION
Consider free body consisting of 20-in. length of the trough and water
l = 20-in. length of free body
&π
'
W = γ v = γ ( r 2l )
4
*
+
PA = γ r
P=
1
1
1
PA rl = (γ r )rl = γ r 2l
2
2
2
1 !
ΣM A = 0: Tr − Wr − P " r # = 0
$3 %
π
! 4r
Tr − " γ r 2 l #"
$ 4
%$ 3π
1 2 ! 1 !
!
# − " 2 γ r l #" 3 r # = 0
% $
%$ %
1
1
1
T = γ r 2l + γ r 2l = γ r 2l
3
6
2
Data:
γ = 62.4 lb/ft 3 r =
Then
T=
24
20
ft = 2ft l =
ft
12
12
1
20 !
(62.4 lb/ft 3 )(2 ft) 2 " ft #
2
12
$
%
= 208.00 lb
T = 208 lb
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651
PROBLEM 5.91
A 4 × 2-ft gate is hinged at A and is held in position by rod CD.
End D rests against a spring whose constant is 828 lb/ft. The spring
is undeformed when the gate is vertical. Assuming that the force
exerted by rod CD on the gate remains horizontal, determine the
minimum depth of water d for which the bottom B of the gate will
move to the end of the cylindrical portion of the floor.
SOLUTION
First determine the forces exerted on the gate by the spring and the water when B is at the end of the
cylindrical portion of the floor
sin θ =
We have
2
θ = 30°
4
Then
xSP = (3 ft) tan 30°
and
FSP = kxSP
= 828 lb/ft × 3 ft × tan30°
= 1434.14 lb
Assume
d $ 4 ft
We have
P=
1
1
Ap = A(γ h)
2
2
1
PI = [(4 ft)(2 ft)] × [(62.4 lb/ft 3 )(d − 4)ft]
2
= 249.6( d − 4)lb
Then
1
PII = [(4 ft)(2 ft)] × [(62.4 lb/ft 3 )(d − 4 + 4cos 30°)]
2
= 249.6(d − 0.53590°)lb
W =0
For d min so that gate opens,
Using the above free-body diagrams of the gate, we have
4 !
8 !
ΣM A = 0: " ft # [249.6(d − 4)lb] + " ft # [249.6( d − 0.53590)lb]
3
$
%
$3 %
−(3 ft)(1434.14 lb) = 0
or
(332.8d − 1331.2) + (665.6d − 356.70) − 4302.4 = 0
d = 6.00 ft
or
d $ 4 ft , assumption correct
d = 6.00 ft
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652
!
PROBLEM 5.92
Solve Problem 5.91 if the gate weighs 1000 lb.
PROBLEM 5.91 A 4 × 2-ft gate is hinged at A and is held in
position by rod CD. End D rests against a spring whose constant is
828 lb/ft. The spring is undeformed when the gate is vertical.
Assuming that the force exerted by rod CD on the gate remains
horizontal, determine the minimum depth of water d for which the
bottom B of the gate will move to the end of the cylindrical portion
of the floor.
SOLUTION
First determine the forces exerted on the gate by the spring and the water when B is at the end of the
cylindrical portion of the floor
sin θ =
We have
2
θ = 30°
4
Then
xSP = (3 ft) tan 30°
and
FSP = kxSP = 828 lb/ft × 3 ft × tan30°
= 1434.14 lb
Assume
d $ 4 ft
We have
P=
1
1
Ap = A(γ h)
2
2
1
PI = [(4 ft)(2 ft)] × [(62.4 lb/ft 3 )(d − 4)ft]
2
= 249.6(d − 4)lb
1
PII = [(4 ft)(2 ft)] × [(62.4 lb/ft 3 )(d − 4 + 4cos 30°)]
2
= 249.6(d − 0.53590°)lb
Then
For d min so that gate opens,
W = 1000 lb
Using the above free-body diagrams of the gate, we have
4 !
8 !
ΣM A = 0: " ft # [249.6(d − 4) lb] + " ft # [249.6(d − 0.53590) lb]
3
$
%
$3 %
− (3 ft)(1434.14 lb) − (1 ft)(1000 lb) = 0
or
(332.8d − 1331.2) + (665.6d − 356.70) − 4302.4 − 1000 = 0
d = 7.00 ft
or
d $ 4 ft , assumption correct
d = 7.00 ft
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653
!
PROBLEM 5.93
A prismatically shaped gate placed at the end of a freshwater channel is
supported by a pin and bracket at A and rests on a frictionless support at B.
The pin is located at a distance h = 0.10 m below the center of gravity C
of the gate. Determine the depth of water d for which the gate will open.
SOLUTION
First note that when the gate is about to open (clockwise rotation is impending), By
0 and the line of
action of the resultant P of the pressure forces passes through the pin at A. In addition, if it is assumed that the
gate is homogeneous, then its center of gravity C coincides with the centroid of the triangular area. Then
a=
d
− (0.25 − h)
3
and
b=
2
8 d!
(0.4) − " #
3
15 $ 3 %
Now
a 8
=
b 15
so that
d
− (0.25 − h)
3
2
(0.4) − 158 d3
3
( )
=
8
15
Simplifying yields
289
70.6
d + 15h =
45
12
(1)
Alternative solution
Consider a free body consisting of a 1-m thick section of the gate and the triangular section BDE of water
above the gate.
Now
1
1
Ap′ = (d × 1 m)( ρ gd )
2
2
1
= ρ gd 2 (N)
2
1 8
!
W ′ = ρ gV = ρ g " × d × d × 1 m #
2
15
$
%
4
= ρ gd 2 (N)
15
P′ =
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654
PROBLEM 5.93 (Continued)
Then with By = 0 (as explained above), we have
&2
1 8 !' 4
! &d
' 1
!
ΣM A = 0: ( (0.4) − " d # ) " ρ gd 2 # − ( − (0.25 − h) ) " ρ gd 2 # = 0
3
3
15
15
3
2
$
%+ $
% *
+$
%
*
Simplifying yields
289
70.6
d + 15h =
45
12
as above.
Find d ,
Substituting into Eq. (1)
h = 0.10 m
289
70.6
d + 15(0.10) =
45
12
or d = 0.683 m
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655
!
PROBLEM 5.94
A prismatically shaped gate placed at the end of a freshwater channel
is supported by a pin and bracket at A and rests on a frictionless support
at B. Determine the distance h if the gate is to open when d = 0.75 m.
SOLUTION
First note that when the gate is about to open (clockwise rotation is impending), By
0 and the line of
action of the resultant P of the pressure forces passes through the pin at A. In addition, if it is assumed that the
gate is homogeneous, then its center of gravity C coincides with the centroid of the triangular area. Then
a=
d
− (0.25 − h)
3
and
b=
2
8 d!
(0.4) − " #
3
15 $ 3 %
Now
a 8
=
b 15
so that
d
− (0.25 − h)
3
2
(0.4) − 158 d3
3
( )
=
8
15
Simplifying yields
289
70.6
d + 15h =
45
12
(1)
Alternative solution
Consider a free body consisting of a 1-m thick section of the gate and the triangular section BDE of water
above the gate.
Now
1
1
Ap′ = (d × 1 m)( ρ gd )
2
2
1
2
(N)
= ρ gd
2
1 8
!
W ′ = ρ gV = ρ g " × d × d × 1 m #
2
15
$
%
4
2
(N)
= ρ gd
15
P′ =
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656
PROBLEM 5.94 (Continued)
Then with By = 0 (as explained above), we have
&2
1 8 !' 4
! &d
' 1
!
ΣM A = 0: ( (0.4) − " d # ) " ρ gd 2 # − ( − (0.25 − h) ) " ρ gd 2 # = 0
3 $ 15 % + $ 15
% *3
+$ 2
%
*3
Simplifying yields
289
70.6
d + 15h =
45
12
as above.
Find h,
Substituting into Eq. (1)
d = 0.75 m
289
70.6
(0.75) + 15h =
45
12
or h = 0.0711 m
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657
!
PROBLEM 5.95
A 55-gallon 23-in.-diameter drum is placed on its side to act as a
dam in a 30-in.-wide freshwater channel. Knowing that the drum is
anchored to the sides of the channel, determine the resultant of the
pressure forces acting on the drum.
SOLUTION
Consider the elements of water shown. The resultant of the weights of water above each section of the drum
and the resultants of the pressure forces acting on the vertical surfaces of the elements is equal to the resultant
hydrostatic force acting on the drum. Then
1
1
ApI = A(γ h)
2
2
1 & 30 !
23 ! ' &
23 ! '
= (" # ft × " # ft ) × ( (62.4 lb/ft 3 ) "
ft # )
2 *$ 12 %
$ 12 % + *
$ 12 % +
PI =
= 286.542 lb
1
1
ApII = A(γ h)
2
2
1 & 30 !
11.5 ! ' &
11.5 ! '
3
= (" # ft × "
ft # )
# ft ) × ( (62.4 lb/ft ) "
2 *$ 12 %
12
$
% + *
$ 12 % +
PII =
= 71.635 lb
& 11.5 !2 2 π 11.5 ! 2 2 ' 30 !
W1 = γ V1 = (62.4 lb/ft 3 ) ("
# ft − "
# ft ) " ft # = 30.746 lb
4 $ 12 %
(*$ 12 %
)+ $ 12 %
& 11.5 !2 2 π 11.5 !2 2 ' 30 !
W2 = γ V2 = (62.4 lb/ft 3 ) ("
# ft + "
# ft ) " ft # = 255.80 lb
4 $ 12 %
(*$ 12 %
)+ $ 12 %
& π 11.5 !2 2 ' 30 !
ft # = 112.525 lb
W3 = γ V3 = (62.4 lb/ft 3 ) ( "
# ft ) "
*( 4 $ 12 %
+) $ 12 %
Then
ΣFx : Rx = (286.542 − 71.635) lb = 214.91 lb
ΣFy : Ry = (−30.746 + 255.80 + 112.525) lb = 337.58 lb
Finally
R = Rx2 + Ry2 = 400.18 1b
tan θ =
Ry
Rx
θ = 57.5°
R = 400 lb
57.5°
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658
!
PROBLEM 5.96
Determine the location of the centroid of the composite body shown
when (a) h = 2b, (b) h = 2.5b.
SOLUTION
V
x
xV
Cylinder I
π a 2b
1
b
2
1 2 2
πa b
2
Cone II
1 2
πa h
3
b+
1
h
4
1 2
1 !
π a h"b + h#
3
4 %
$
1 !
V = π a2 " b + h #
3 %
$
1
1
1
!
Σ xV = π a 2 " b 2 + hb + h 2 #
3
12 %
$2
(a)
For h = 2b :
1
&
' 5
V = π a 2 (b + (2b) ) = π a 2 b
3
*
+ 3
1
1
&1
'
Σ xV = π a 2 ( b 2 + (2b)b + (2b) 2 )
2
3
12
*
+
1
2
1
3
&
'
= π a 2b 2 ( + + ) = π a 2 b2
* 2 3 3+ 2
5
! 3
XV = Σ xV : X " π a 2b # = π a 2 b 2
$3
% 2
X=
9
b
10
Centroid is
1
b
10
to left of base of cone
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
659
PROBLEM 5.96 (Continued)
(b)
For h = 2.5b :
1
&
'
V = π a 2 (b + (2.5b) ) = 1.8333π a 2 b
3
*
+
1
1
&1
'
Σ xV = π a 2 ( b 2 + (2.5b)b + (2.5b) 2 )
3
12
*2
+
= π a 2b 2 [0.5 + 0.8333 + 0.52083]
= 1.85416π a 2b 2
XV = Σ xV : X (1.8333π a 2 b) = 1.85416π a 2b 2
X = 1.01136b
Centroid is 0.01136b to right of base of cone
Note: Centroid is at base of cone for h = 6b = 2.449b !
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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660
PROBLEM 5.97
Determine the y coordinate of the centroid of the body shown.
SOLUTION
First note that the values of Y will be the same for the given body and the body shown below. Then
V
y
Cone
1 2
πa h
3
1
− h
4
2
Cylinder
a!
1
−π " # b = − π a 2b
4
$2%
1
− b
2
Σ
We have
Then
π
12
a 2 (4h − 3b)
yV
−
1
π a 2 h2
12
1 2 2
πa b
8
−
π
24
a 2 (2h 2 − 3b 2 )
Y ΣV = Σ yV
π
&π
'
Y ( a 2 (4h − 3b) ) = − a 2 (2h 2 − 3b 2 )
24
*12
+
or Y = −
2h 2 − 3b 2
2(4h − 3b)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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661
!
PROBLEM 5.98
Determine the z coordinate of the centroid of the body shown. (Hint: Use
the result of Sample Problem 5.13.)
SOLUTION
First note that the body can be formed by removing a “half-cylinder” from a “half-cone,” as shown.
V
1 2
πa h
6
Half-Cone
Half-Cylinder
Σ
−
π a!
2
−
π
b = − a 2b
2 "$ 2 #%
8
π
24
zV
z
−
a
π
1
− a3 h
6
*
4 a!
2a
=−
"
#
3π $ 2 %
3π
a 2 (4h − 3b)
1 3
ab
12
−
1 3
a (2h − b)
12
From Sample Problem 5.13
We have
Then
Z ΣV = Σ zV
1
&π
'
Z ( a 2 (4h − 3b) ) = − a3 (2h − b)
12
* 24
+
or Z = −
a 4h − 2b !
π "$ 4h − 3b #%
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
662
!
PROBLEM 5.99
The composite body shown is formed by removing a semiellipsoid of
revolution of semimajor axis h and semiminor axis a/2 from a
hemisphere of radius a. Determine (a) the y coordinate of the centroid
when h = a/2, (b) the ratio h/a for which y = −0.4a.
SOLUTION
V
y
yV
Hemisphere
2 3
πa
3
3
− a
8
1
− π a4
4
2
Semiellipsoid
2
a!
1
− π " # h = − π a2 h
3 $2%
6
3
− h
8
ΣV =
Then
π
6
Σ yV = −
+
1
π a 2 h2
16
a 2 (4a − h)
π
16
a 2 (4a 2 − h 2 )
Y ΣV = Σ yV
Now
So that
π
&π
'
Y ( a 2 (4a − h) ) = − a 2 (4a 2 − h 2 )
16
*6
+
2
h!
h! '
3 &
Y " 4 − # = − a (4 − " # )
a%
8 (* $ a % )+
$
or
Y = ? when h =
(a)
Substituting
(1)
a
2
h 1
= into Eq. (1)
a 2
2
1!
3 &
1! '
Y " 4 − # = − a (4 − " # )
2%
8 *( $ 2 % +)
$
or
Y =−
45
a
112
Y = −0.402a
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
663
PROBLEM 5.99 (Continued)
h
= ? when Y = −0.4a
a
(b)
Substituting into Eq. (1)
2
h!
h! '
3 &
(−0.4a) " 4 − # = − a ( 4 − " # )
a%
8 (* $ a % )+
$
2
or
Then
h!
h!
3 " # − 3.2 " # + 0.8 = 0
$a%
$a%
2
h 3.2 ± (−3.2) − 4(3)(0.8)
=
a
2(3)
=
3.2 ± 0.8
6
or
h 2
=
a 5
and
h 2
=
a 3
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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664
!
PROBLEM 5.100
For the stop bracket shown, locate the x coordinate of the center
of gravity.
SOLUTION
Assume that the bracket is homogeneous so that its center of gravity coincides with the centroid of the
volume.
Then
V , mm3
x , mm
xV , mm 4
1
(100)(88)(12) = 105600
50
5280000
2
(100)(12)(88) = 105600
50
5280000
3
1
(62)(51)(10) = 15810
2
39
616590
4
1
− (66)(45)(12) = −17820
2
Σ
209190
X =
34 +
2
(66) = 78
3
−1389960
9786600
Σ xV 9786600
mm
=
209190
ΣV
or
X = 46.8 mm
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
665
PROBLEM 5.101
For the stop bracket shown, locate the z coordinate of the center of
gravity.
SOLUTION
Assume that the bracket is homogeneous so that it center of gravity coincides with the centroid of the volume.
V , mm3
Then
z , mm
zV , mm 4
6
633600
1
(100)(88)(12) = 105600
2
(100)(12)(88) = 105600
12 +
1
(88) = 56
2
5913600
3
1
(62)(51)(10) = 15810
2
1
12 + (51) = 29
3
458490
4
1
− (66)(45)(12) = −17820
2
Σ
209190
Z =
55 +
2
(45) = 85
3
−1514700
5491000
Σ zV 5491000
mm
=
209190
ΣV
or Z = 26.2 mm
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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666
!
PROBLEM 5.102
For the machine element shown, locate the y coordinate of the
center of gravity.
SOLUTION
First assume that the machine element is homogeneous so that its center of gravity will coincide with the
centroid of the corresponding volume.
V , mm3
x , mm
y , mm
xV, mm 4
yV, mm 4
I
(100)(18)(90) = 162000
50
9
8100000
1458000
II
(16)(60)(50) = 48000
92
48
4416000
2304000
III
π (12) 2 (10) = 4523.9
105
54
475010
244290
IV
−π (13) 2 (18) = −9556.7
28
9
–267590
–86010
Σ
204967.2
12723420
3920280
We have
Y ΣV = Σ yV
Y (204967.2 mm3 ) = 3920280 mm 4
or Y = 19.13 mm
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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667
!
PROBLEM 5.103
For the machine element shown, locate the y coordinate of the
center of gravity.
SOLUTION
For half cylindrical hole:
r = 1.25 in.
4(1.25)
yIII = 2 −
3π
= 1.470 in.
For half cylindrical plate:
r = 2 in.
4(2)
zIV = 7 +
= 7.85 in.
3π
V, in.3
y , in.
z , in.
yV , in.4
z V , in.4
I
Rectangular plate
(7)(4)(0.75) = 21.0
–0.375
3.5
–7.875
73.50
II
Rectangular plate
(4)(2)(1) = 8.0
1.0
2
8.000
16.00
III
–(Half cylinder)
(1.25) 2 (1) = 2.454
1.470
2
–3.607
–4.908
IV
Half cylinder
(2) 2 (0.75) = 4.712
–0.375
–7.85
–1.767
36.99
V
–(Cylinder)
−π (1.25) 2 (0.75) = −3.682
–0.375
7
1.381
–25.77
Σ
27.58
–3.868
95.81
−
π
2
π
2
Y ΣV = Σ yV
Y (27.58 in.3 ) = −3.868 in.4
Y = −0.1403 in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
668
!
PROBLEM 5.104
For the machine element shown, locate the z coordinate of the
center of gravity.
SOLUTION
For half cylindrical hole:
r = 1.25 in.
4(1.25)
yIII = 2 −
3π
= 1.470 in.
For half cylindrical plate:
r = 2 in.
4(2)
zIV = 7 +
= 7.85 in.
3π
V, in.3
Now
y , in.
z , in.
yV , in.4
z V , in.4
I
Rectangular plate
(7)(4)(0.75) = 21.0
–0.375
3.5
–7.875
73.50
II
Rectangular plate
(4)(2)(1) = 8.0
1.0
2
8.000
16.00
III
–(Half cylinder)
(1.25) 2 (1) = 2.454
1.470
2
–3.607
–4.908
IV
Half cylinder
(2) 2 (0.75) = 4.712
–0.375
–7.85
–1.767
36.99
V
–(Cylinder)
−π (1.25) 2 (0.75) = −3.682
–0.375
7
1.381
–25.77
Σ
27.58
–3.868
95.81
−
π
2
π
2
Z ΣV = zV
Z (27.58 in.3 ) = 95.81 in.4
Z = 3.47 in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
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669
!
PROBLEM 5.105
For the machine element shown, locate the x coordinate of
the center of gravity.
SOLUTION
First assume that the machine element is homogeneous so that its center of gravity will coincide with the
centroid of the corresponding volume.
V , mm3
x , mm
y , mm
xV, mm 4
yV, mm 4
I
(100)(18)(90) = 162000
50
9
8100000
1458000
II
(16)(60)(50) = 48000
92
48
4416000
2304000
III
π (12) 2 (10) = 4523.9
105
54
475010
244290
IV
−π (13) 2 (18) = −9556.7
28
9
–267590
–86010
Σ
204967.2
12723420
3920280
We have
X ΣV = Σ xV
X (204967.2 mm3 ) = 12723420 mm 4
X = 62.1 mm
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
670
!
PROBLEM 5.106
Locate the center of gravity of the sheet-metal form shown.
SOLUTION
First, assume that the sheet metal is homogeneous so that the center of gravity of the form will coincide with
the centroid of the corresponding area.
1
yI = − (1.2) = −0.4 m
3
1
zI = (3.6) = 1.2 m
3
4(1.8)
2.4
m
xIII = −
=−
3π
π
A, m 2
x, m
y, m
z, m
xA, m3
yA, m3
zA, m3
I
1
(3.6)(1.2) = 2.16
2
1.5
−0.4
1.2
3.24
−0.864
2.592
II
(3.6)(1.7) = 6.12
0.75
0.4
1.8
4.59
2.448
11.016
0.8
1.8
−3.888
4.0715
9.1609
3.942
5.6555
22.769
III
Σ
We have
π
2
(1.8)2 = 5.0894
−
2.4
π
13.3694
X ΣV = Σ xV : X (13.3694 m 2 ) = 3.942 m3
or X = 0.295 m
Y ΣV = Σ yV : Y (13.3694 m 2 ) = 5.6555 m3
or Y = 0.423 m
Z ΣV = Σ zV : Z (13.3694 m 2 ) = 22.769 m3
or Z = 1.703 m
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
671
PROBLEM 5.107
Locate the center of gravity of the sheet-metal form shown.
SOLUTION
First assume that the sheet metal is homogeneous so that the center of gravity of the form will coincide with
the centroid of the corresponding area. Now note that symmetry implies
X = 125 mm
yII = 150 +
2 × 80
π
= 200.93 mm
2 × 80
zII =
π
yIII
I
II
III
Σ
We have
A, mm 2
y , mm
(250)(170) = 42500
75
π
2
(80)(250) = 31416
π
2
(125)2 = 24544
= 50.930 mm
4 × 125
= 230 +
3π
= 283.05 mm
z , mm
yA, mm3
zA, mm3
40
3187500
1700000
200.93
50930
6312400
1600000
283.05
0
6947200
0
16447100
3300000
98460
Y Σ A = Σ y A: Y (98460 mm 2 ) = 16447100 mm3
or Y = 1670 mm
Z Σ A = Σ z A: Z (98460 mm 2 ) = 3.300 × 106 mm3
or Z = 33.5 mm
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
672
PROBLEM 5.108
A wastebasket, designed to fit in the corner of a room, is 16 in. high
and has a base in the shape of a quarter circle of radius 10 in. Locate
the center of gravity of the wastebasket, knowing that it is made of
sheet metal of uniform thickness.
SOLUTION
By symmetry:
X =Z
For III (Cylindrical surface)
x=
A=
For IV (Quarter-circle bottom)
x=
A=
2r
π
π
2
=
2(10)
rh =
π
π
2
= 6.3662 in.
(10)(16) = 251.33 in.2
4r 4(10)
=
= 4.2441 in.
3π
3π
π
4
r2 =
π
4
(10) 2 = 78.540 in.2
A, in.2
x , in.
x , in.
xA, in.3
yA, in.3
I
(10)(16) = 160
5
8
800
1280
II
(10)(16) = 160
0
8
0
1280
III
251.33
6.3662
8
1600.0
2010.6
IV
78.540
4.2441
0
333.33
0
Σ
649.87
2733.3
4570.6
X Σ A = Σ x A:
X (649.87 in.2 ) = 2733.3 in.3
X = 4.2059 in.
Y Σ A = Σ y A:
X = Z = 4.21 in.
Y (649.87 in.2 ) = 4570.6 in.3
Y = 7.0331 in.
Y = 7.03 in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
673
!
PROBLEM 5.109
A mounting bracket for electronic components is formed from
sheet metal of uniform thickness. Locate the center of gravity
of the bracket.
SOLUTION
First, assume that the sheet metal is homogeneous so that the center of gravity of the bracket coincides with
the centroid of the corresponding area. Then (see diagram)
4(0.625)
3π
= 1.98474 in.
zV = 2.25 −
π
(0.625) 2
2
= −0.61359 in.2
AV = −
A, in.2
x , in.
y , in.
z , in.
xA, in.3
yA, in.3
zA, in.3
I
(2.5)(6) = 15
1.25
0
3
18.75
0
45
II
(1.25)(6) = 7.5
2.5
–0.625
3
18.75
–4.6875
22.5
III
(0.75)(6) = 4.5
2.875
–1.25
3
12.9375
–5.625
13.5
IV
5!
− " # (3) = − 3.75
$4%
1.0
0
3.75
3.75
0
–14.0625
V
− 0.61359
1.0
0
1.98474
0.61359
0
–1.21782
Σ
22.6364
46.0739
10.3125
65.7197
We have
X Σ A = Σx A
X (22.6364 in.2 ) = 46.0739 in.3
or
X = 2.04 in.
Y ΣA = Σy A
Y (22.6364 in.2 ) = −10.3125 in.3
or Y = − 0.456 in.
Z Σ A = Σz A
Z (22.6364 in.2 ) = 65.7197 in.3
or
Z = 2.90 in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
674
PROBLEM 5.110
A thin sheet of plastic of uniform thickness is bent to
form a desk organizer. Locate the center of gravity of the
organizer.
SOLUTION
First assume that the plastic is homogeneous so that the center of gravity of the organizer will coincide with
the centroid of the corresponding area. Now note that symmetry implies
Z = 30.0 mm
x2 = 6 −
2× 6
x4 = 36 +
x8 = 58 −
π
= 2.1803 mm
2×6
π
2× 6
π
x10 = 133 +
= 39.820 mm
= 54.180 mm
2×6
π
= 136.820 mm
y2 = y4 = y8 = y10 = 6 −
y6 = 75 +
2×5
π
2×6
π
= 2.1803 mm
= 78.183 mm
π
× 6 × 60 = 565.49 mm 2
2
A6 = π × 5 × 60 = 942.48 mm 2
A2 = A4 = A8 = A10 =
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
675
PROBLEM 5.110 (Continued)
We have
A, mm 2
x , mm
y , mm
xA, mm3
yA, mm3
1
(74)(60) = 4440
0
43
0
190920
2
565.49
2.1803
2.1803
1233
1233
3
(30)(60) = 1800
21
0
37800
0
4
565.49
39.820
2.1803
22518
1233
5
(69)(60) = 4140
42
40.5
173880
167670
6
942.48
47
78.183
44297
73686
7
(69)(60) = 4140
52
40.5
215280
167670
8
565.49
54.180
2.1803
30638
1233
9
(75)(60) = 4500
95.5
0
429750
0
10
565.49
136.820
2.1803
77370
1233
Σ
22224.44
1032766
604878
X Σ A = Σ x A: X (22224.44 mm 2 ) = 1032766 mm3
or X = 46.5 mm
Y Σ A = Σ y A: Y (22224.44 mm 2 ) = 604878 mm3
or Y = 27.2 mm
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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676
PROBLEM 5.111
A window awning is fabricated from sheet metal of uniform
thickness. Locate the center of gravity of the awning.
SOLUTION
First, assume that the sheet metal is homogeneous so that the center of gravity of the awning coincides with
the centroid of the corresponding area.
(4)(25)
= 14.6103 in.
3π
(4)(25) 100
zII = zVI =
=
in.
3π
3π
(2)(25)
yIV = 4 +
= 19.9155 in.
yII = yVI = 4 +
π
zIV =
(2)(25)
π
AII = AVI =
AIV =
π
2
π
4
=
50
π
in.
(25) 2 = 490.87 in.2
(25)(34) = 1335.18 in.2
A, in.2
y , in.
z , in.
yA, in.3
zA, in.3
I
(4)(25) = 100
2
12.5
200
1250
II
490.87
14.6103
7171.8
5208.3
III
(4)(34) = 136
2
272
3400
IV
1335.18
19.9155
26591
21250
V
(4)(25) = 100
2
200
1250
VI
490.87
14.6103
7171.8
5208.3
Σ
2652.9
41607
37567
100
3π
25
50
π
12.5
100
3π
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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677
PROBLEM 5.111 (Continued)
X = 17.00 in.
Now, symmetry implies
and
Y Σ A = Σ y A: Y (2652.9 in.2 ) = 41607 in.3
or Y = 15.68 in.
Z Σ A = Σ z A: Z (2652.9 in.2 ) = 37567
or Z = 14.16 in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
678
PROBLEM 5.112
An elbow for the duct of a ventilating system is made of sheet
metal of uniform thickness. Locate the center of gravity of the
elbow.
SOLUTION
First, assume that the sheet metal is homogeneous so that the center of gravity of the duct coincides with the
centroid of the corresponding area. Also, note that the shape of the duct implies
Y = 38.0 mm !
Note that
xI = zI = 400 −
xII = 400 −
zII = 300 −
2
π
2
π
2
π
(400) = 145.352 mm
(200) = 272.68 mm
(200) = 172.676 mm
xIV = zIV = 400 −
4
(400) = 230.23 mm
3π
4
(200) = 315.12 mm
3π
4
zV = 300 −
(200) = 215.12 mm
3π
xV = 400 −
Also note that the corresponding top and bottom areas will contribute equally when determining x and z .
Thus
x , mm
z , mm
xA, mm3
zA, mm3
(400)(76) = 47752
145.352
145.352
6940850
6940850
(200)(76) = 23876
272.68
172.676
6510510
4122810
A, mm 2
I
II
π
2
π
2
III
100(76) = 7600
200
350
1520000
2660000
IV
π!
2 " # (400) 2 = 251327
$4%
230.23
230.23
57863020
57863020
V
π!
−2 " # (200)2 = −62832
$4%
315.12
215.12
–19799620
–13516420
VI
−2(100)(200) = −40000
300
350
–12000000
–14000000
Σ
227723
41034760
44070260
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
679
PROBLEM 5.112 (Continued)
We have
X Σ A = Σ x A: X (227723 mm 2 ) = 41034760 mm3
or X = 180.2 mm
Z Σ A = Σ z A: Z (227723 mm 2 ) = 44070260 mm3
or Z = 193.5 mm
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
680
PROBLEM 5.113
An 8-in.-diameter cylindrical duct and a 4 × 8-in. rectangular
duct are to be joined as indicated. Knowing that the ducts were
fabricated from the same sheet metal, which is of uniform
thickness, locate the center of gravity of the assembly.
SOLUTION
Assume that the body is homogeneous so that its center of gravity coincides with the centroid of the area.
A, in.2
x , in.
y , in.
xA, in.3
yA, in.3
π (8)(12) = 96π
0
6
0
576π
10
−128
−160π
4(4)
16
=−
3π
3π
12
− 42.667
96π
6
12
576
1152
6
8
576
768
4(4) 16
=
3π
3π
8
− 42.667
− 64π
7
(4) 2 = − 8π
2
(4)(12) = 48
6
10
288
480
8
(4)(12) = 48
6
10
288
480
Σ
539.33
1514.6
4287.4
1
2
−
π
2
π
3
Then
(4) 2 = 8π
4
2
(8)(12) = 96
5
(8)(12) = 96
6
−
2(4)
(8)(4) = −16π
π
π
−
=
8
π
X =
Σ x A 1514.67
=
in.
ΣA
539.33
or X = 2.81 in.
Y =
Σ y A 4287.4
=
in.
ΣA
539.33
or Y = 7.95 in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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681
PROBLEM 5.114
A thin steel wire of uniform cross section is bent into the shape shown.
Locate its center of gravity.
SOLUTION
First assume that the wire is homogeneous so that its center of gravity will coincide with the centroid of the
corresponding line.
x1 = 0.3sin 60° = 0.15 3 m
z1 = 0.3cos 60° = 0.15 m
0.6sin 30° !
0.9
m
## sin 30° =
π
π
6
%
0.6sin 30° !
0.9
3m
z2 = "
## cos 30° =
π
"
π
6
$
%
x2 = "
"
$
π!
L2 = " # (0.6) = (0.2π ) m
$3%
L, m
x, m
y, m
z, m
xL, m 2
yL, m 2
zL, m 2
1
1.0
0.15 3
0.4
0.15
0.25981
0.4
0.15
2
0.2π
0.18
0
0.31177
3
0.8
0
0.4
0.6
0
0.32
0.48
4
0.6
0
0.8
0.3
0
0.48
0.18
Σ
3.0283
0.43981
1.20
1.12177
We have
0.9
π
0.9 3
0
π
X Σ L = Σ x L: X (3.0283 m) = 0.43981 m 2
or X = 0.1452 m
Y Σ L = Σ yL:
Y (3.0283 m) = 1.20 m 2
or
Y = 0.396 m
Z Σ L = Σ z L:
Z (3.0283 m) = 1.12177 m 2
or
Z = 0.370 m
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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682
PROBLEM 5.115
Locate the center of gravity of the figure shown, knowing that it is
made of thin brass rods of uniform diameter.
SOLUTION
Uniform rod
AB 2 = (1 m) 2 + (0.6 m) 2 + (1.5 m) 2
AB = 1.9 m
L, m
x, m
y, m
z, m
xL, m 2
yL, m 2
ΣL , m 2
AB
1.9
0.5
0.75
0.3
0.95
1.425
0.57
BD
0.6
1.0
0
0.3
0.60
0
0.18
DO
1.0
0.5
0
0
0.50
0
0
OA
1.5
0
0.75
0
0
1.125
0
Σ
5.0
2.05
2.550
0.75
X Σ L = Σ x L : X (5.0 m) = 2.05 m 2
X = 0.410 m
Y Σ L = Σ y L : Y (5.0 m) = 2.55 m 2
Y = 0.510 m
Z Σ L = Σ z L : Z (5.0 m) = 0.75 m 2
Z = 0.1500 m
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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683
!
PROBLEM 5.116
Locate the center of gravity of the figure shown, knowing that it is
made of thin brass rods of uniform diameter.
SOLUTION
X =0
By symmetry:
L, in.
y , in.
z , in.
yL, in.2
zL, in.2
AB
302 + 162 = 34
15
0
510
0
AD
302 + 162 = 34
15
8
510
272
AE
302 + 162 = 34
15
0
510
0
BDE
π (16) = 50.265
0
0
512
Σ
152.265
1530
784
2(16)
π
= 10.186
Y ΣL = Σ y L : Y (152.265 in.) = 1530 in.2
Y = 10.048 in.
!
Y = 10.05 in.
!
Z = 5.15 in.
!
Z ΣL = Σ z L : Z (152.265 in.) = 784 in.2 !
Z = 5.149 in. !
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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684
PROBLEM 5.117
The frame of a greenhouse is constructed from uniform aluminum
channels. Locate the center of gravity of the portion of the frame
shown.
SOLUTION
First assume that the channels are homogeneous so that the center of gravity of the frame
will coincide with the centroid of the corresponding line.
x8 = x9 =
2×3
π
y8 = y9 = 5 +
6
π
2×3
π
ft
= 6.9099 ft
L, ft
x , ft
y , ft
z , ft
xL, ft 2
yL, ft 2
zL, ft 2
1
2
3
0
1
6
0
2
2
3
1.5
0
2
4.5
0
6
3
5
3
2.5
0
15
12.5
0
4
5
3
2.5
2
15
12.5
10
5
8
0
4
2
0
32
16
6
2
3
5
1
6
10
2
7
3
1.5
5
2
4.5
15
6
6.9099
0
9
32.562
0
6.9099
2
9
32.562
9.4248
8
1
0
16
2
69
163.124
53.4248
8
9
π
2
π
2
× 3 = 4.7124
× 3 = 4.7124
10
2
Σ
39.4248
We have
=
6
π
6
π
0
X Σ L = Σ x L : X (39.4248 ft) = 69 ft 2
or X = 1.750 ft
Y Σ L = Σ y L : Y (39.4248 ft) = 163.124 ft 2
or
Y = 4.14 ft
!
Z Σ L = Σ z L : Z (39.4248 ft) = 53.4248 ft 2
or Z = 1.355 ft
!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
685
PROBLEM 5.118
A scratch awl has a plastic handle and a steel blade and shank. Knowing that the density of plastic is 1030 kg/m3
and of steel is 7860 kg/m3, locate the center of gravity of the awl.
SOLUTION
Y =Z =0
First, note that symmetry implies
5
xI = (12.5 mm) = 7.8125 mm
8
2π !
3
WI = (1030 kg/m3 ) "
# (0.0125 m)
$ 3 %
= 4.2133 × 10−3 kg
xII = 52.5 mm
π!
WII = (1030 kg/m3 ) " # (0.025 m) 2 (0.08 m)
$4%
−3
= 40.448 × 10 kg
xIII = 92.5 mm − 25 mm = 67.5 mm
π!
WIII = −(1030 kg/m3 ) " # (0.0035 m)2 (.05 m)
$4%
−3
= −0.49549 × 10 kg
xIV = 182.5 mm − 70 mm = 112.5 mm
π!
WIV = (7860 kg/m3 ) " # (0.0035 m) 2 (0.14 m) 2 = 10.5871 × 10−3 kg
$4%
1
xV = 182.5 mm + (10 mm) = 185 mm
4
π!
WV = (7860 kg/m3 ) " # (0.00175 m) 2 (0.01 m) = 0.25207 × 10−3 kg
$3%
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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686
!
PROBLEM 5.118 (Continued)
We have
W, kg
x, mm
xW, kg ⋅ mm
I
4.123 × 10−3
7.8125
32.916 × 10−3
II
40.948 × 10−3
52.5
2123.5 × 10−3
III
−0.49549 × 10−3
67.5
−33.447 × 10−3
IV
10.5871 × 10−3
112.5
1191.05 × 10−3
V
0.25207 × 10−3
185
46.633 × 10−3
Σ
55.005 × 10−3
3360.7 × 10−3
X ΣW = Σ xW : X (55.005 × 10−3 kg) = 3360.7 × 10−3 kg ⋅ mm
X = 61.1 mm
or
(From the end of the handle)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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687
!
PROBLEM 5.119
A bronze bushing is mounted inside a steel sleeve. Knowing that the
specific weight of bronze is 0.318 lb/in.3 and of steel is 0.284 lb/in.3,
determine the location of the center of gravity of the assembly.
SOLUTION
X =Z =0
First, note that symmetry implies
Now
W = ( pg )V
& π!
'
yI = 0.20 in. WI = (0.284 lb/in.3 ) *" # ( 1.82 − 0.752 in.2 ) ( 0.4 in.) + = 0.23889 lb
.$ 4 % ,
/
(
)
& π!
'
yII = 0.90 in. WII = (0.284 lb/in.3 ) *" # ( 1.1252 − 0.752 in.2 ) (1 in.) + = 0.156834 lb
.$ 4 % ,
/
(
)
& π!
'
yIII = 0.70 in. WIII = (0.318 lb/in.3 ) *" # ( 0.752 − 0.52 in.2 ) (1.4 in.) + = 0.109269 lb
,
4
.$ %
/
(
We have
)
Y ΣW = Σ yW
Y =
(0.20 in.)(0.23889 lb) + (0.90 in.)(0.156834 lb) + (0.70 in.)(0.109269 lb)
0.23889 lb + 0.156834 lb + 0.109269 lb
Y = 0.526 in.
or
(above base)!
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reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
688
PROBLEM 5.120
A brass collar, of length 2.5 in., is mounted on an aluminum rod
of length 4 in. Locate the center of gravity of the composite body.
(Specific weights: brass = 0.306 lb/in.3, aluminum = 0.101 lb/in.3)
SOLUTION
Aluminum rod:
W = γV
(π
)
= (0.101 lb/in.3 ) 0 (1.6 in.) 2 (4 in.) 1
,4
= 0.81229 lb
Brass collar:
W = γV
π
= (0.306 lb/in.3 ) [(3 in.)2 − (1.6 in.)2 ](2.5 in.)
4
= 3.8693 lb
Component
W(lb)
y (in.)
yW (lb ⋅ in.)
Rod
0.81229
2
1.62458
Collar
3.8693
1.25
4.8366
Σ
4.6816
6.4612
Y ΣW = Σ y W : Y (4.6816 lb) = 6.4612 lb ⋅ in.
Y = 1.38013 in.
Y = 1.380 in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
689
!
PROBLEM 5.121
The three legs of a small glass-topped table are equally spaced and are
made of steel tubing, which has an outside diameter of 24 mm and a
cross-sectional area of 150 mm 2 . The diameter and the thickness of
the table top are 600 mm and 10 mm, respectively. Knowing that the
density of steel is 7860 kg/m3 and of glass is 2190 kg/m3, locate the
center of gravity of the table.
SOLUTION
X =Z =0
First note that symmetry implies
Also, to account for the three legs, the masses of components I and II will each bex
multiplied by three
yI = 12 + 180 −
2 × 180
mI = ρ ST VI = 7860 kg/m3 × (150 × 10−6 m 2 ) ×
π
2 × 280
mII = ρ ST VII = 7860 kg/m3 × (150 × 10−6 m 2 ) ×
π
mIII = ρGLVIII = 2190 kg/m3 ×
yIII = 24 + 180 + 280 + 5
= 489 mm
or
(0.180 m)
π
2
(0.280 m)
= 0.51855 kg
= 370.25 mm
We have
2
= 0.33335 kg
= 77.408 mm
yII = 12 + 180 +
π
π
4
(0.6 m) 2 × (0.010 m)
= 6.1921 kg
m, kg
y , mm
ym, kg ⋅ mm
I
3(0.33335)
77.408
77.412
II
3(0.51855)
370.25
515.98
III
6.1921
489
3027.9
Σ
8.7478
3681.3
Y Σm = Σ ym : Y (8.7478 kg) = 3681.3 kg ⋅ mm
Y = 420.8 mm
The center of gravity is 421 mm
(above the floor)!
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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690
PROBLEM 5.122
Determine by direct integration the values of x for the two volumes obtained by passing a vertical cutting
plane through the given shape of Figure 5.21. The cutting plane is parallel to the base of the given shape and
divides the shape into two volumes of equal height.
A hemisphere
SOLUTION
Choose as the element of volume a disk of radius r and thickness dx. Then
dV = π r 2 dx, xEL = x
The equation of the generating curve is x 2 + y 2 = a 2 so that r 2 = a 2 − x 2 and then
dV = π ( a 2 − x 2 ) dx
V1 =
Component 1
and
2x
1
EL dV
2
a/2
0
a/2
(
x3 )
π (a − x )dx = π 0 a 2 x − 1
3 -0
,
2
=
11 3
πa
24
=
2
a/2
0
2
x (,π (a 2 − x 2 ) dx )a/2
( x2 x4 )
= π 0a2
− 1
2
4 -0
,
7
= π a4
64
Now
x1V1 =
2x
1
EL dV :
11
! 7
x1 " π a 3 # = π a 4
$ 24
% 64
or
x1 =
21
a
88
Component 2
V2 =
2
a
a /2
(
π (a 2 − x 2 )dx = π 0 a 2 x −
,
a
x3 )
1
3 - a/2
&
(
a 3 )'
a3 ) 0 2 a ! ( 2 ) 1 3
3( 2
= π * 0 a (a ) − 1 − a " # −
+
3 - 0 $2%
3 13
3,
,
-/
.
5
= π a3
24
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
691
PROBLEM 5.122 (Continued)
and
2
a
( x 2 x4 )
− 1
xEL dV =
x (,π ( a 2 − x 2 )dx )- = π 0 a 2
a/2
2
2
4 - a/2
,
2
a
2
4 )'
&
(
a
a
2
(
(a ) 4 ) 0 2 ( 2 )
3 ( 2 (a)
2 ) 13
= π *0a
−
−
+
1− a
2
4 - 0
2
4 13
3,
,
-/
.
9
= π a4
64
Now
x2V2 =
2
5
! 9
xEL dV : x2 " π a3 # = π a 4
2
$ 24
% 64
or
x2 =
27
a
40
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
692
!
PROBLEM 5.123
Determine by direct integration the values of x for the two volumes obtained by passing a vertical cutting
plane through the given shape of Figure 5.21. The cutting plane is parallel to the base of the given shape and
divides the shape into two volumes of equal height.
A semiellipsoid of revolution
SOLUTION
Choose as the element of volume a disk of radius r and thickness dx. Then
dV = π r 2 dx, xEL = x
x2 y2
+
= 1 so that
h2 a 2
The equation of the generating curve is
r2 =
a2 2
(h − x 2 )dx
h2
dV = π
and then
V1 =
Component 1
=
and
a2 2
(h − x 2 )
h2
2
1
xEL dV =
2
h/2
0
h/2
a2
a2 (
x3 )
π 2 (h 2 − x 2 )dx = π 2 0 h 2 x − 1
3 -0
h
h ,
11 2
πa h
24
2
h/2
0
( a2
)
x 0π 2 (h 2 − x 2 ) dx 1
, h
h/2
a2 ( x2 x4 )
= π 2 0 h2 − 1
4 -0
h , 2
7
= π a 2 h2
64
Now
Component 2
x1V1 =
2x
1
EL dV :
11
! 7
x1 " π a 2 h # = π a 2 h 2
$ 24
% 64
or
x1 =
21
h
88
h
a2
a2 (
x3 )
π 2 (h 2 − x 2 )dx = π 2 0 h 2 x − 1
V2 =
h/2 h
3 - h/2
h ,
2
h
a2
=π 2
h
=
&
(
h 3 )'
( h )3 ) 0 2 h ! ( 2 ) 1 3
3( 2
*0h ( h ) −
+
1− h " #−
3 - 0 $2%
3 13
3,
,
-/
.
5
π a2h
24
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693
PROBLEM 5.123 (Continued)
and
2
2
xEL dV =
h
a2
=π 2
h
( 2 x2 x4 )
− 1
0h
4 - h/2
, 2
a2
=π 2
h
&
(
h 2
h 4 )'
2
(
( h) 4 ) 0 2 ( 2 )
3 ( 2 ( h)
2 ) 13
−
−
*0h
+
1− h
2
4 - 0
2
4 13
3,
,
-/
.
=
Now
2
( a2
)
x 0π 2 (h 2 − x 2 ) dx 1
h/2
, h
h
x2V2 =
9
π a 2 h2
64
2
5
! 9
xEL dV : x2 " π a 2 h # = π a 2 h 2
2
$ 24
% 64
or
x2 =
27
h
40
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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694
!
PROBLEM 5.124
Determine by direct integration the values of x for the two volumes obtained by passing a vertical cutting
plane through the given shape of Figure 5.21. The cutting plane is parallel to the base of the given shape and
divides the shape into two volumes of equal height.
A paraboloid of revolution
SOLUTION
Choose as the element of volume a disk of radius r and thickness dx. Then
dV = π r 2 dx, xEL = x
The equation of the generating curve is x = h −
h 2
a2
2
y
so
that
(h − x)
r
=
h
a2
a2
(h − x)dx
h
dV = π
and then
V1 =
Component 1
2
h/2
0
π
a2
( h − x) dx
h
h/2
a2 (
x2 )
=π
0 hx − 1
h ,
2 -0
3
= π a2 h
8
2
and
1
xEL dV =
2
h/2
0
( a2
)
x 0π
(h − x)dx 1
, h
h/2
a 2 ( x 2 x3 )
1
2 2
=π
0h − 1 = π a h
h , 2
3 -0
12
x1V1 =
Now
2x
1
EL dV :
3
! 1
x1 " π a 2 h # = π a 2 h 2
8
$
% 12
or
x1 =
2
h
9
Component 2
V2 =
2
h
h/2
h
π
a2
a2 (
x2 )
(h − x)dx = π
0 hx − 1
h
h ,
2 - h/2
&
(
h 2 )'
( h) 2 ) 0 h ! ( 2 ) 1 3
a2 3(
=π
* 0 h( h ) −
+
1 − h" # −
h 3,
2 - 0 $2%
2 13
,
-/
.
1 2
= πa h
8
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695
PROBLEM 5.124 (Continued)
and
2
2
a2
=π
h
=
Now
h
( a2
)
a 2 ( x 2 x3 )
xEL dV =
x 0π
( h − x)dx 1 = π
0h − 1
h/2
2
h , 2
3 - h/2
, h
h
x2V2 =
&
( h 2 ( h )3 ) '
2
( h )3 ) 0 ( 2 )
3 ( ( h)
3
−
− 2 1+
*0h
1− h
0
1
2
3
2
3
- ,
3,
- 3/
.
1
π a 2 h2
12
2
2
1
! 1
xEL dV : x2 " π a 2 h # = π a 2 h 2
$8
% 12
or
x2 =
2
h
3
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696
!
PROBLEM 5.125
Locate the centroid of the volume obtained by rotating the shaded area
about the x axis.
SOLUTION
First note that symmetry implies
y =0
and
z =0
We have
y = k ( X − h) 2
at
x = 0,
or
k=
y = a : a = k ( − h) 2
a
h2
Choose as the element of volume a disk of radius r and thickness dx. Then
dV = π r 2 dx, X EL = x
r=
Now
a
( x − h) 2
h2
so that
dV = π
Then
V=
2
a2
( x − h) 4 dx
4
h
h
0
π
a2
π a2 (
5 h
4
(
)
x
h
dx
x − h) )
−
=
4
4 ,(
-0
5h
h
1
= π a2h
5
and
2x
EL dV
( a2
)
x 0π 4 ( x − h) 4 dx 1
0
, h
2
h
a
=π 4
( x5 − 4hx 4 + 6h 2 x3 − 4h3 x 2 + h4 x)dx
0
h
=
2
h
2
2
=π
=
Now
a
h4
h
(1 6 4 5 3 2 4 4 3 3 1 4 2 )
0 6 x − 5 hx + 2 h x − 3 h x + 2 h x 1
,
-0
1
π a 2 h2
30
π
! π 2 2
xV = xEL dV : x " a 2 h # =
a h
5
$
% 30
2
or
x=
1
h
6
!
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697
PROBLEM 5.126
Locate the centroid of the volume obtained by rotating the shaded area
about the x axis.
SOLUTION
y =0
First note that symmetry implies
z =0
Choose as the element of volume a disk of radius r and thickness dx. Then
dV = π r 2 dx, xEL = x
r = kx1/3
Now
dV = π k 2 x 2/3 dx
so that
at x = h, y = a :
a = kh1/3
or
k=
a
h1/3
a 2 2/3
x dx
h 2/3
dV = π
Then
V=
and
2
h
0
π
a 2 2/3
x dx
h 2/3
a2
h 2/3
=π
h
( 3 5/3 )
05 x 1
,
-0
3
= π a2 h
5
Also
2
xEL dV =
2
h
0
!
a2
a2
x "" π 2/3 x 2/3 dx ## = π 2/3
h
$ h
%
h
( 3 8/3 )
08 x 1
,
-0
3
= π a 2 h2
8
Now
2
xV = xdV :
3
! 3
x " π a 2 h # = π a 2 h2
5
$
% 8
5
or x = h
8
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698
PROBLEM 5.127
Locate the centroid of the volume obtained by rotating the shaded
area about the line x = h.
SOLUTION
x =h
First, note that symmetry implies
z =0
Choose as the element of volume a disk of radius r and thickness dx. Then
dV = π r 2 dy, yEL = y
Now x 2 =
Then
and
Let
Then
h 2
h2 2
a − y2
(a − y 2 ) so that r = h −
2
a
a
dV = π
V=
2
h2
a − a2 − y 2
a2
(
a
0
π
2
) dy
h2
a − a2 − y2
2
a
(
2
) dy
y = a sin θ 4 dy = a cos θ dθ
h2
a2
h2
=π 2
a
V =π
= π ah 2
2
2
π /2
0
π /2
0
2
π /2
0
(
a − a 2 − a 2 sin 2 θ
2
) a cosθ dθ
( a 2 − 2a (a cos θ ) + (a 2 − a 2 sin 2 θ ) ) a cos θ dθ
,
-
(2cos θ − 2 cos 2 θ − sin 2 θ cos θ ) dθ
(
θ sin 2θ
= π ah 2 0 2sin θ − 2 " +
4
$2
,
π /2
! 1 3 )
# − 3 sin θ 1
%
-0
π !
(
1)
= π ah 2 0 2 − 2 "" 2 ## − 1
0,
$ 2 % 3 1-
= 0.095870π ah 2
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699
PROBLEM 5.127 (Continued)
and
2
yEL dV =
2
=π
a
0
( h2
y 0π 2 a − a 2 − y 2
, a
h2
a2
(
2
a
0
( 2a y − 2ay
2
) dy )12
)
a 2 − y 2 − y 3 dy
a
2
1 )
h2 (
= π 2 0 a 2 y 2 + a (a 2 − y 2 )3/2 − y 4 1
3
4 -0
a ,
=π
=
Now
h2
a2
&( 2 2 1 4 ) ( 2
2 3/2 ) '
* 0 a (a ) − a 1 − 0 a( a ) 1 +
4 - ,3
-/
.,
1
π a 2 h2
12
2
y (0.095870π ah2 ) =
yV = yEL dV :
1
π a 2 h2
12
or y = 0.869a
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700
!
PROBLEM 5.128*
Locate the centroid of the volume generated by revolving the portion
of the sine curve shown about the x axis.
SOLUTION
y =0
First, note that symmetry implies
z =0
Choose as the element of volume a disk of radius r and thickness dx.
dV = π r 2 dx, xEL = x
Then
r = b sin
Now
πx
2a
dV = π b 2 sin 2
so that
V=
Then
&
2a
a
πx
2a
dx
π b 2 sin 2
πx
2a
dx
2a
x sin π x !
= π b 2 " − π2 a #
2 a %#
$" 2
a
= π b 2 $( 22a ) − ( a2 ) !%
1
= π ab 2
2
and
&x
EL dV
=
&
2a
a
πx (
'
x ) π b 2 sin 2
dx
2a *,
+
Use integration by parts with
dV = sin 2
u=x
du = dx
V=
x
−
2
πx
2a
sin πax
2π
a
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701
PROBLEM 5.128* (Continued)
Then
&
2a
- '
.
2a ' x
sin π x ( /
x sin πax ( !
/
xEL dV = π b 0 " x ) − 2π * # −
− 2π a * dx 1
)
*
*
a )2
" )2
a
a
, %# a
+
, 3/
2/ $ +
2a
1 2 a2
π x ! ./
' 2a (
' a (!
2/
= π b 0 " 2a ) * − a ) * # − " x + 2 cos # 1
a %a /
2π
+ 2 ,
+ 2 ,% $ 4
2/ $
3
&
2
-/' 3 ( 1
a2
a 2 ! ./
1
= π b 2 0) a 2 * − " (2a) 2 + 2 − (a) 2 + 2 # 1
4
2π
2π % /3
/2+ 2 , $ 4
'3 1 (
= π a 2b 2 ) − 2 *
+4 π ,
= 0.64868π a 2b 2
Now
'1
(
xV = xEL dV : x ) π ab 2 * = 0.64868π a 2b 2
+2
,
&
or x = 1.297a
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702
!
PROBLEM 5.129*
Locate the centroid of the volume generated by revolving the
portion of the sine curve shown about the y axis. (Hint: Use a
thin cylindrical shell of radius r and thickness dr as the element
of volume.)
SOLUTION
x =0
First note that symmetry implies
z =0
Choose as the element of volume a cylindrical shell of radius r and thickness dr.
Then
Now
so that
Then
dV = (2π r )( y )(dr ),
y = b sin
&
2a
a
1
y
2
πr
2a
dV = 2π br sin
V=
yEL =
πr
2a
dr
2π br sin
πr
2a
dr
Use integration by parts with
u = rd
dv = sin
dr
2a
2a
πr
v = − cos
2a
π
du = dr
Then
πr
2a
-/
π r (!
' 2a
V = 2π b 0 "( r ) ) − cos * # −
2a , % a
/2 $ + π
&
2a
π r ( ./
) π cos 2a * dr 1
+
, 3/
2a '
a
2a
- 2a
π r ! ./
4a 2
/
= 2π b 0 − [ (2a)(−1) ] + " 2 sin # 1
2a % a /
π
$π
2/
3
' 4a 2 4 a 2
V = 2π b ))
− 2
π
+ π
1(
'
= 8a 2 b ) 1 − *
+ π,
(
**
,
= 5.4535a 2b
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703
PROBLEM 5.129* (Continued)
Also
&y
EL dV
π r ('
πr (
1
dr *
) 2 b sin
*) 2π br sin
2
a
2a ,
+
,+
2a
πr
r sin 2
dr
= π b2
a
2a
=
&
2a '
a
&
Use integration by parts with
dv = sin 2
u=r
du = dr
Then
&
v=
r
−
2
πr
dr
2a
sin πar
2π
a
2a
.
2a ' r
' r sin πar ( !
sin π r ( /
/
yEL dV = π b 0 "( r ) ) − 2π * # −
− 2π a * dr 1
)
*
*
a )2
" )2
a
a
, %# a
+
, 3/
2/ $ +
2a
π r ! ./
r2
a2
' 2a (
' a (!
2/
= π b 0 "(2a) ) * − ( a) ) * # − " + 2 cos # 1
a %a /
+ 2 ,
+ 2 , % $ 4 2π
2/ $
3
&
2
(2a) 2
(a) 2
a2
a 2 ! /.
/- 3
= π b2 0 a 2 − "
+ 2−
+ 2 #1
4
2π
2π % 3/
$ 4
2/ 2
'3 1 (
= π a 2b 2 ) − 2 *
+4 π ,
= 2.0379a 2 b 2
Now
&
yV = yEL dV : y (5.4535a 2b) = 2.0379a 2b 2
or y = 0.374b
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704
!
PROBLEM 5.130*
Show that for a regular pyramid of height h and n sides (n = 3, 4,
is located at a distance h/4 above the base.
) the centroid of the volume of the pyramid
SOLUTION
Choose as the element of a horizontal slice of thickness dy. For any number N of sides, the area of the base of
the pyramid is given by
Abase = kb 2
where k = k ( N ); see note below. Using similar triangles, have
s h− y
=
b
h
s=
or
b
(h − y )
h
dV = Aslice dy = ks 2 dy = k
Then
V=
and
h
&
0
b2
(h − y ) 2 dy
h2
h
b2
b2
1
!
k 2 ( h − y ) 2 dy = k 2 " − (h − y )3 #
h
h $ 3
%0
1
= kb 2 h
3
yEL = y
Also
So then
&
yEL dV =
&
h
0
=k
Now
Note
!
b2
b2
y " k 2 ( h − y ) 2 dy # = k 2
h
$ h
%
b2
h2
&
h
0
( h 2 y − 2hy 2 + y 3 ) dy
h
1 2 2 2 3 1 4!
1 2 2
" 2 h y − 3 hy + 4 y # = 12 kb h
$
%0
'1
( 1
yV = yEL dV : y ) kb 2 h * = kb 2 h 2
3
+
, 12
&
'1
Abase = N )) × b ×
+2
N
=
b2
4 tan πN
b
2
tan πN
or y =
1
h Q.E.D.
4
(
**
,
= k ( N )b 2
!
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705
PROBLEM 5.131
Determine by direct integration the location of the centroid of one-half of a
thin, uniform hemispherical shell of radius R.
SOLUTION
x =0
First note that symmetry implies
The element of area dA of the shell shown is obtained by cutting the shell with two planes parallel to the xy
plane. Now
dA = (π r )( Rdθ )
2r
yEL = −
π
r = R sin θ
Where
dA = π R 2 sin θ dθ
2R
sin θ
yEL = −
so that
π
A=
Then
π /2
&
0
π R 2 sin θ dθ = π R 2 [− cos θ ]π0 /2
= π R2
and
&y
EL dA
=
&
π /2'
0
2R
(
sin θ * (π R 2 sin θ dθ )
)−
π
+
,
π /2
sin 2θ !
= −2 R " −
4 #% 0
$2
3
=−
Now
Symmetry implies
π
2
θ
R3
&
yA = yEL dA: y (π R 2 ) = −
π
2
R3
1
or y = − R
2
1
z =− R
2
z=y
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
706
!
PROBLEM 5.132
The sides and the base of a punch bowl are of
uniform thickness t. If t ,, R and R = 250 mm,
determine the location of the center of gravity of
(a) the bowl, (b) the punch.
SOLUTION
(a)
Bowl
x =0
First note that symmetry implies
z =0
for the coordinate axes shown below. Now assume that the bowl may be treated as a shell; the center of
gravity of the bowl will coincide with the centroid of the shell. For the walls of the bowl, an element of
area is obtained by rotating the arc ds about the y axis. Then
dAwall = (2π R sin θ )( Rdθ )
and
Then
( yEL ) wall = − R cos θ
Awall =
π /2
&π
/6
2π R 2 sin θ dθ
π /2
= 2π R 2 [− cos θ ]π /6
= π 3R 2
and
&
π
=&
π
ywall Awall = ( yEL ) wall dA
/2
/6
(− R cos θ )(2π R 2 sin θ dθ )
= π R3 [cos 2 θ ]ππ /2
/6
3
= − π R3
4
π
R2 ,
By observation
Abase =
Now
y ΣA = Σ yA
or
or
4
ybase = −
3
R
2
'
3
3 (
π
π
'
(
y ) π 3R 2 + R 2 * = − π R3 + R 2 ) −
R*
)
*
4
4
4
+
,
+ 2 ,
y = − 0.48763R R = 250 mm
y = −121.9 mm
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707
PROBLEM 5.132 (Continued)
(b)
Punch
x =0
First note that symmetry implies
z =0
and that because the punch is homogeneous, its center of gravity will coincide with the centroid of
the corresponding volume. Choose as the element of volume a disk of radius x and thickness dy. Then
dV = π x 2 dy,
Now
yEL = y
x2 + y 2 = R2
dV = π ( R 2 − y 2 )dy
so that
V=
Then
&
0
− 3/2 R
π ( R 2 − y 2 ) dy
0
1 !
= π " R y − y3 #
3 %−
$
2
3/2 R
3
'
( 1'
( ! 3
3
3
R* − )−
R * # = π 3R3
= −π " R ) −
" )+ 2 *, 3 )+ 2 *, # 8
$
%
2
and
&
yEL dV =
&
0
− 3/2 R
( y ) π R 2 − y 2 dy !
$
%
(
)
0
1
1 !
= π " R2 y 2 − y4 #
2
4 %−
$
3/2 R
4!
1 2'
3 ( 1'
3 ( #
15
"
R )−
R* − )−
R * = − π R4
= −π
)
*
)
*
"2 + 2 ,
4+ 2 , #
64
$
%
2
Now
15
'3
(
yV = yEL dV : y ) π 3 R3 * = − π R 4
8
64
+
,
&
y =−
or
5
8 3
R R = 250 mm
y = − 90.2 mm
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708
!
PROBLEM 5.133
After grading a lot, a builder places four stakes to designate the
corners of the slab for a house. To provide a firm, level base for
the slab, the builder places a minimum of 3 in. of gravel
beneath the slab. Determine the volume of gravel needed and
the x coordinate of the centroid of the volume of the gravel.
(Hint: The bottom surface of the gravel is an oblique plane,
which can be represented by the equation y = a + bx + cz.)
SOLUTION
The centroid can be found by integration. The equation for the bottom of the gravel is:
y = a + bx + cz , where the constants a, b, and c can be determined as follows:
For x = 0, and z = 0:
y = − 3 in., and therefore
−
3
1
ft = a, or a = − ft
12
4
For x = 30 ft, and z = 0: y = − 5 in., and therefore
−
5
1
1
ft = − ft + b(30 ft), or b = −
12
4
180
For x = 0, and z = 50 ft: y = − 6 in., and therefore
−
6
1
1
ft = − ft + c(50 ft), or c = −
12
4
200
Therefore:
1
1
1
y = − ft −
x−
z
4
180
200
Now
x =
&x
EL dV
V
A volume element can be chosen as:
dV = | y| dxdz
1'
1
1 (
1+
x+
z dx dz
)
4+
45
50 *,
or
dV =
and
xEL = x
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709
PROBLEM 5.133 (Continued)
Then
&
xEL dV =
50
30
0
0
& &
x'
1
1 (
1+
x+
z dx dz
4 )+
45
50 *,
30
x2
z 2!
1 3
x +
x # dz
" +
100 % 0
$ 2 135
1
=
4
&
=
1
4
&
=
1
9 !
650 z + z 2 #
"
4$
2 %0
50
0
50
0
(650 + 9 z )dz
50
= 10937.5 ft 4
The volume is:
&
V dV =
50
30
0
0
& &
1'
1
1 (
1+
x+
z dx dz
)
4+
45
50 *,
30
1 2 z !
" x + 90 x + 50 x # dz
$
%0
=
1
4
=
1
4
=
1
3 !
40 z + z 2 #
"
4$
10 % 0
&
&
50
0
50 '
0
3 (
) 40 + z * dz
5 ,
+
50
= 687.50 ft 3
Then
x =
&x
EL dV
V
=
10937.5ft 4
= 15.9091 ft
687.5 ft 3
V = 688 ft 3
Therefore:
x = 15.91 ft
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710
!
PROBLEM 5.134
Determine by direct integration the location of the centroid of
the volume between the xz plane and the portion shown of the
surface y = 16h(ax − x2)(bz − z2)/a2b2.
SOLUTION
First note that symmetry implies
x=
a
2
z=
b
2
Choose as the element of volume a filament of base dx × dz and height y. Then
dV = ydxdz ,
or
Then
dV =
yEL =
1
y
2
16h
(ax − x 2 )(bz − z 2 ) dx dz
2 2
a b
b
a
0
0
V=
&&
V=
16h
a 2b2
16h
(ax − x 2 )(bz − z 2 ) dx dz
2 2
a b
&
a
a
1 !
(bz − z 2 ) " x 2 − x3 # dz
0
z
3 %0
$
b
b
=
16h a 2 1 3 ! b 2 1 3 !
( a ) − (a) # " z − z #
3
3 %0
a 2 b 2 "$ 2
% $2
8ah b 2 1 3 !
(b) − (b) #
3
3b 2 "$ 2
%
4
= abh
9
=
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711
PROBLEM 5.134 (Continued)
and
&y
EL dV
b
a
0
0
=
&&
=
128h 2
a 4b 4
=
128h 2
a 2b 4
1 16h
! 16h
!
( ax − x 2 )(bz − z 2 ) # " 2 2 (ax − x 2 )(bz − z 2 ) dx dz #
"
2 2
2 $a b
%$a b
%
b
a
0
0
&&
&
(a 2 x 2 − 2ax3 + x 4 )(b 2 z 2 − 2bz 3 + z 4 )dx dz
a
a2
a
1 !
(b 2 z 2 − 2bz 3 + z 4 ) " x3 − x 4 + x5 # dz
0
2
5 %0
$3
b
b
! b2
128h 2 a 2
a
1
b
1 !
= 4 4 " ( a )3 − ( a ) 4 + ( a )5 # " Z 3 − Z 4 + Z 5 #
2
5
Z
5 %0
a b $3
%$ 3
=
Now
64ah 2 b3 3 b 4 1 5 ! 32
abh 2
(b) − (b) + (b) # =
4 "
2
5
15b $ 3
% 225
'4
( 32
yV = yEL dV : y ) abh * =
abh 2
+9
, 225
&
8
or y = 25 h
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712
!
PROBLEM 5.135
Locate the centroid of the section shown, which was cut from a thin circular
pipe by two oblique planes.
SOLUTION
x =0
First note that symmetry implies
Assume that the pipe has a uniform wall thickness t and choose as the element of volume A vertical strip of
width adθ and height ( y2 − y1 ). Then
dV = ( y2 − y1 )tadθ ,
Now
y1 =
=
and
h
3
2a
z+
h
6
yEL =
1
( y1 + y2 ) z EL = z
2
y2 = −
h
( z + a)
6a
=
2h
3
2a
z+
2
h
3
h
( − z + 2a )
3a
z = a cos θ
h
h
(−a cos θ + 2a) −
(a cos θ + a )
3a
6a
h
= (1 − cos θ )
2
Then
( y2 − y1 ) =
and
( y1 + y2 ) =
h
h
(a cos θ + a) + (− a cos θ + 2a)
6a
3a
h
= (5 − cos θ )
6
aht
h
dV =
(1 − cos θ )dθ yEL = (5 − cos θ ), z EL = a cos θ
2
12
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713
PROBLEM 5.135 (Continued)
and
π
aht
(1 − cos θ )dθ = aht[θ − sin θ ]π0
2
= π aht
Then
&y
V =2
&
=2
&
EL dV
0
h
aht
!
(5 − cos θ ) "
(1 − cos θ ) dθ #
12
$ 2
%
π
0
2
π
=
ah t
12
=
ah 2 t
θ sin 2θ !
5θ − 6sin θ + +
"
12 $
2
4 #% 0
=
11
π ah2 t
24
&
0
(5 − 6 cos θ + cos 2 θ ) dθ
π
&z
EL dV
=2
&
π
0
aht
!
a cos θ "
(1 − cos θ ) dθ #
$ 2
%
θ
= a 2 ht "sin θ − −
2
$
1
= − π a 2 ht
2
&
π
sin 2θ !
4 #% 0
11
π ah 2t
24
Now
yV = yEL dV : y (π aht ) =
and
1
zV = zEL dV : z (π aht ) = − π a 2 ht
2
&
or y =
11
h
24
1
or z = − a
2
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714
!
PROBLEM 5.136*
Locate the centroid of the section shown, which was cut from an elliptical
cylinder by an oblique plane.
SOLUTION
x=0
First note that symmetry implies
Choose as the element of volume a vertical slice of width zx, thickness dz, and height y. Then
dV = 2 xydz ,
x=
and
y=−
Then
V=
Then
&
h/2
h h
z+ =
(b − z )
b
2 2b
b
' a 2
!
2 ( h
) 2 b b − z * " 2b (b − z ) # dz
,$
%
−b +
z = b sin θ
V=
1
, z EL = z
24
a 2
b − z2
b
Now
Let
yEL =
ah
b2
π /2
&π
= abh
/2
dz = b cos θ dθ
(b cos θ )[b(1 − sin θ )] b cos θ dθ
π /2
&π
− /2
θ
(cos 2 θ − sin θ cos 2 θ ) dθ
= abh " +
$2
π /2
sin 2θ 1
!
+ cos3 θ #
4
3
% −π /2
1
V = π abh
2
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715
PROBLEM 5.136* (Continued)
and
&y
EL dV
Now
So that
Then
&
=
1 ah 2
4 b3
h
! -' a 2
( h
! .
b − z 2 * " (b − z ) # dz 1
(b − z ) # 0) 2
b
b
2b
2
% 2+
,$
% 3
×
−b "
$2
&
b
−b
z = b sin θ
Let
Then
1
b
=
&
dz = b cos θ dθ
1 ah 2
4 b3
1
= abh 2
4
sin 2 θ cos2 θ =
&y
EL dV
=
π /2
&π
yEL dV =
sin 2 θ =
(b − z ) 2 b 2 − z 2 dz
− /2
[b(1 − sin θ )]2 (b cos θ ) × (b cos θ dθ )
π /2
&π
− /2
(cos 2 θ − 2sin θ cos 2 θ + sin 2 θ cos 2 θ ) dθ
1
1
(1 − cos 2θ ) cos 2 θ = (1 + cos 2θ )
2
2
1
(1 − cos 2 2θ )
4
1
abh 2
4
1
!
cos 2 θ − 2sin θ cos 2 θ + (1 − cos 2 2θ ) # dθ
"
/2 $
4
%
π /2
&π
−
π /2
1
1
1 ' θ sin 4θ ( !
' θ sin 2θ ( 1
= abh 2 ") +
+ cos3 θ + θ − ) +
#
*
4
4 , 3
4
4+ 2
8 *, % −π /2
$+ 2
=
Also
&z
EL dV
5
π abh2
32
- a 2
h
! .
z 02
a − z 2 " (b − z ) # dz 1
−b 2 b
$ 2b
% 3
ah b
z (b − z ) b 2 − z 2 dz
= 2
b −b
=
&
b
&
z = b sin θ
Let
Then
&z
EL dV
=
ah
b2
π /2
&π
− /2
= ab 2 h
Using
sin 2 θ cos2 θ =
&z
EL dV
dz = b cos θ dθ
(b sin θ )[b(1 − sin θ )](b cos θ ) × (b cos θ dθ )
π /2
&π
− /2
(sin θ cos 2 θ − sin 2 θ cos 2 θ ) dθ
1
(1 − cos 2 2θ ) from above
4
= ab 2 h
1
!
sin θ cos 2 θ − (1 − cos 2 2θ ) # dθ
"
/2 $
4
%
π /2
&π
−
π /2
1
1
1 ' θ sin 4θ ( !
1
= ab h " − cos3 θ − θ + ) +
= − π ab 2 h
#
*
4
4+ 2
8 , % −π /2
8
$ 3
2
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716
PROBLEM 5.136* (Continued)
Now
'1
( 5
yV = yEL dV : y ) π abh * = π abh 2
2
+
, 32
or y =
and
1
'1
(
z V = z EL dV : z ) π abh * = − π ab 2 h
2
8
+
,
1
or z = − b
4
&
&
5
h
16
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717
!
PROBLEM 5.137
Locate the centroid of the plane area shown.
SOLUTION
Then
A, mm 2
x , mm
y , mm
xA, mm3
yA, mm3
1
126 × 54 = 6804
9
27
61236
183708
2
1
× 126 × 30 = 1890
2
30
64
56700
120960
3
1
× 72 × 48 = 1728
2
48
−16
82944
−27648
Σ
10422
200880
277020
X ΣA = Σ xA
X (10422 m 2 ) = 200880 mm 2
and
or X = 19.27 mm
Y Σ A = Σ yA
Y (10422 m 2 ) = 270020 mm3
or
Y = 26.6 mm
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718
!
PROBLEM 5.138
Locate the centroid of the plane area shown.
SOLUTION
Dimensions in in.
Then
A, in.2
x , in.
y , in.
xA, in.3
yA, in.3
1
2
(4)(8) = 21.333
3
4.8
1.5
102.398
32.000
2
1
− (4)(8) = −16.0000
2
5.3333
1.33333
85.333
−21.333
Σ
5.3333
17.0650
10.6670
X ΣA = Σ xA
X (5.3333 in.2 ) = 17.0650 in.3
and
or X = 3.20 in.
Y Σ A = Σ yA
Y (5.3333 in.2 ) = 10.6670 in.3
or
Y = 2.00 in.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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719
!
PROBLEM 5.139
The frame for a sign is fabricated from thin, flat steel bar stock of mass
per unit length 4.73 kg/m. The frame is supported by a pin at C and by a
cable AB. Determine (a) the tension in the cable, (b) the reaction at C.
SOLUTION
First note that because the frame is fabricates from uniform bar stock, its center of gravity will coincide with
the centroid of the corresponding line.
L, m
x, m
xL, m 2
1
1.35
0.675
0.91125
2
0.6
0.3
0.18
3
0.75
0
0
4
0.75
0.2
0.15
1.07746
1.26936
5
Σ
Then
π
2
(0.75) = 1.17810
4.62810
2.5106
X ΣL = Σx L
X (4.62810) = 2.5106
or
X = 0.54247 m
The free-body diagram of the frame is then
Where
W = (m′Σ L) g
= 4.73 kg/m × 4.62810 m × 9.81 m/s 2
= 214.75 N
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720
PROBLEM 5.139 (Continued)
Equilibrium then requires
'3
(
ΣM C = 0: (1.55 m) ) TBA * − (0.54247 m)(214.75 N) = 0
+5
,
(a)
TBA = 125.264 N
or
or TBA = 125.3 N
3
ΣFx = 0: C x − (125.264 N) = 0
5
(b)
C x = 75.158 N
or
ΣFy = 0: C y +
4
(125.264 N) − (214.75 N) = 0
5
C y = 114.539 N
or
C = 137.0 N
Then
56.7°
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721
!
PROBLEM 5.140
Determine by direct integration the centroid of the area shown. Express
your answer in terms of a and h.
SOLUTION
y1 = −
By observation
h
x(
'
x + h = h )1 − *
a
+ a,
For y2 : at
x = 0,
y = h : h = k (1 − 0 )
at
x = a,
y = 0: 0 = h(1 − ca 2 ) or C =
or
Then
'
x2
y2 = h ))1 − 2
+ a
Now
dA = ( y2 − y1 )dx
k=h
1
a2
(
**
,
'
x2
= h "))1 − 2
$"+ a
( '
x (!
** − )1 − * # dx
, + a , %#
' x x2 (
= h )) − 2 ** dx
+a a ,
xEL = x
yEL =
Then
1
( y1 − y2 )
2
=
h '
x( '
x2
")1 − * + ))1 − 2
2 "$+ a , + a
=
h'
x x2 (
)) 2 − − 2 **
2+
a a ,
&
A = dA =
=
&
a
0
' x x2
h )) − 2
+a a
(!
** #
, %#
a
(
x2
x3 !
** dx = h " − 2 #
,
$ 2a 3a % 0
1
ah
6
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you are using it without permission.
722
PROBLEM 5.140 (Continued)
and
&x
&
EL dA
a
' x x2 ( !
' x3
x4
− 2
x " h )) − 2 ** dx # = h "))
$" + a a , %#
$"+ 3a 4a
=
&
=
1 2
a h
12
0
( ' x x2
** " h )) − 2
0
, "$ + a a
h2 a ' x
x2 x4 (
=
)) 2 − 3 2 + 4 ** dx
2 0+ a
a
a ,
yEL dA =
&
a
h'
x x2
)) 2 − − 2
2+
a a
a
(!
** #
, %# 0
( !
** dx #
, #%
&
a
1 2
h 2 x 2 x3
x5 !
=
" − 2 + 4 # = ah
2 $a a
5a % 0 10
'1 ( 1
xA = xEL dA: x ) ah * = a 2 h
+ 6 , 12
x=
'1 ( 1
yA = yEL dA: y ) ah * = a 2 h
+ 6 , 10
3
y= h
5
&
&
1
a
2
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reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
723
!
PROBLEM 5.141
Determine by direct integration the centroid of the area shown. Express your
answer in terms of a and b.
SOLUTION
y =b
First note that symmetry implies
x = a,
at
y=b
y1 : b = ka 2
y1 =
Then
k=
or
b
a2
b 2
x
a2
y2 : b = 2b − ca 2
c=
or
'
x2
y2 = b )) 2 − 2
a
+
Then
(
**
,
'
x2
dA = ( y2 − y1 )dx2 = "b )) 2 − 2
a
$" +
'
x2 (
= 2b ))1 − 2 ** dx
+ a ,
Now
( b 2!
** − 2 x # dx
#%
, a
xEL = x
and
& &
A = dA
Then
and
b
a2
&
a
0
'
x2
2b ))1 − 2
+ a
a
(
4
x3 !
2
dx
b
x
=
−
= ab
"
2#
**
3
a
,
$
%0 3
a
'
x2 ( !
x2
x4 !
1
xEL dA = x " 2b ))1 − 2 ** dx # = 2b " − 2 # = a 2b
0
"$ + a , #%
$ 2 4a % 0 2
'4 ( 1
xA = xEL dA: x ) ab * = a 2b
+3 , 2
&
a
&
3
x= a
8
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
724
PROBLEM 5.142
Knowing that two equal caps have been removed from a 10-in.-diameter wooden
sphere, determine the total surface area of the remaining portion.
SOLUTION
The surface area can be generated by rotating the line shown about the y axis. Applying the first theorem of
Pappus-Guldinus, we have
A = 2π X L = 2π Σ x L
= 2π (2 x1 L1 + x2 L2 )
Now
tan α =
4
3
or
α = 53.130°
Then
x2 =
5 in. × sin 53.130°
π
53.130° × 180
°
= 4.3136 in.
and
π (
'
L2 = 2 ) 53.130° ×
* (5 in.)
180
°,
+
= 9.2729 in.
!
'3 (
A = 2π " 2 ) in. * (3 in.) + (4.3136 in.)(9.2729 in.) #
$ +2 ,
%
A = 308 in.2
or
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
725
PROBLEM 5.143
Determine the reactions at the beam supports for the given loading.
SOLUTION
RI = (3 m)(900 N/m)
= 2700 N
1
RII = (1 m)(900 N/m)
2
= 450 N
Now
ΣFx = 0: Ax = 0
'1 (
ΣM B = 0: − (3 m) Ay + (1.5 m)(2700 N) − ) m * (450 N) = 0
+3 ,
Ay = 1300 N
or
A = 1300 N
ΣFy = 0: 1300 N − 2700 N + By − 450 N = 0
or
By = 1850 N
B = 1850 N
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
726
PROBLEM 5.144
A beam is subjected to a linearly distributed downward
load and rests on two wide supports BC and DE, which
exert uniformly distributed upward loads as shown.
Determine the values of wBC and wDE corresponding to
equilibrium when wA = 600 N/m.
SOLUTION
We have
1
(6 m)(600 N/m) = 1800 N
2
1
RII = (6 m)(1200 N/m) = 3600 N
2
RBC = (0.8 m) (WBC N/m) = (0.8 WBC )N
RI =
RDE = (1.0 m) (WDE N/m) = (WDE ) N
Then
ΣM G = 0: −(1 m)(1800 N) − (3 m)(3600 N) + (4 m)(WDE N) = 0
WDE = 3150 N/m
or
and
or
ΣFy = 0: (0.8WBC ) N − 1800 N − 3600 N + 3150 N = 0
WBC = 2812.5 N/m
WBC = 2810 N/m
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
727
PROBLEM 5.145
The square gate AB is held in the position shown by hinges along its top edge A
and by a shear pin at B. For a depth of water d = 3.5 ft, determine the force
exerted on the gate by the shear pin.
SOLUTION
First consider the force of the water on the gate. We have
1
Ap
2
1
= A(γ h)
2
P=
Then
1
(1.8 ft) 2 (62.4 lb/ft 3 )(1.7 ft)
2
= 171.850 lb
PI =
1
(1.8 ft) 2 (62.4 lb/ft 3 ) × (1.7 + 1.8cos 30°) ft
2
= 329.43 lb
PII =
Now
or
'1
(
'2
(
ΣM A = 0: ) LAB * PI + ) LAB * PII − LAB FB = 0
3
3
+
,
+
,
1
2
(171.850 lb) + (329.43 lb) − FB = 0
3
3
FB = 276.90 lb
or
FB = 277 lb
30.0°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
728
PROBLEM 5.146
Consider the composite body shown. Determine (a) the value
of x when h = L/2, (b) the ratio h/L for which x = L.
SOLUTION
V
x
xV
Rectangular prism
Lab
1
L
2
1 2
L ab
2
Pyramid
1 'b(
a
h
3 )+ 2 *,
1
1 (
'
abh ) L + h *
6
4 ,
+
1
h
4
1 (
'
ΣV = ab ) L + h *
6 ,
+
Then
Σ xV =
1
1 (!
'
ab "3L2 + h ) L + h * #
6 $
4 ,%
+
X ΣV = Σ xV
Now
so that
1 (! 1 '
1 (
'
X " ab ) L + h * # = ab ) 3L2 + hL + h 2 *
6
6
4 ,
,%
+
$ +
h 1 h2 (
' 1 h( 1 '
=
+
+
X )1 +
L
3
)
*
*
L 4 L2 *,
+ 6 L , 6 )+
or
(a)
L+
X = ? when h =
Substituting
1
L
2
h 1
= into Eq. (1)
L 2
1 ' 1 (! 1
'1( 1'1(
X "1 + ) * # = L "3 + ) * + ) *
$ 6 + 2 , % 6 "$ + 2 , 4 + 2 ,
or
(1)
X=
57
L
104
2!
#
#%
X = 0.548L
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
729
PROBLEM 5.146 (Continued)
(b)
h
= ? when X = L
L
Substituting into Eq. (1)
or
or
h 1 h2 (
' 1 h( 1 '
=
+
+
L )1 +
L
3
)
*
*
L 4 L2 *,
+ 6 L , 6 )+
1+
1 h 1 1 h 1 h2
= +
+
6 L 2 6 L 24 L2
h2
= 12
L2
h
=2 3
L
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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730
PROBLEM 5.147
Locate the center of gravity of the sheet-metal form shown.
SOLUTION
First assume that the sheet metal is homogeneous so that the center of gravity of the form will coincide with
the centroid of the corresponding area.
1
yI = 0.18 + (0.12) = 0.22 m
3
1
zI = (0.2 m)
3
2 × 0.18 0.36
m
xII = yII =
=
π
xIV
I
II
IV
Σ
A, m 2
x, m
y, m
z, m
x A, m3
yA, m3
z A, m3
1
(0.2)(0.12) = 0.012
2
0
0.22
0.2
3
0
0.00264
0.0008
0.36
0.36
π
π
0.1
0.00648
0.00648
0.005655
0.26
0
0.1
0.00832
0
0.0032
0.31878
0
0.1
–0.001258
0
–0.000393
0.013542
0.00912
0.009262
π
(0.18)(0.2) = 0.018π
2
(0.16)(0.2) = 0.032
III
−
π
4 × 0.05
= 0.34 −
3π
= 0.31878 m
π
2
(0.05) 2 = −0.00125π
0.096622
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
731
PROBLEM 5.147 (Continued)
We have
X ΣV = Σ xV : X (0.096622 m 2 ) = 0.013542 m3
or X = 0.1402 m
Y ΣV = Σ yV : Y (0.096622 m 2 ) = 0.00912 m3
or Y = 0.0944 m
Z ΣV = Σ zV : Z (0.096622 m 2 ) = 0.009262 m3
or Z = 0.0959 m
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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732
PROBLEM 5.148
Locate the centroid of the volume obtained by rotating the shaded area about
the x axis.
SOLUTION
y =0
First, note that symmetry implies
z =0
Choose as the element of volume a disk of radius r and thickness dx.
dV = π r 2 dx, xEL = x
Then
Now r = 1 −
2
' 1(
dV = π )1 − * dx
x,
+
' 2 1 (
= π )1 − + 2 * dx
x x ,
+
1
so that
x
V=
Then
3
&
1
'
+
π )1 −
3
2 1 (
1!
+ 2 * dx = π " x − 2 ln x − #
x x ,
x %1
$
1( '
1 (!
'
= π ") 3 − 2ln 3 − * − ) 1 − 2 ln1 − * #
3, +
1 ,%
$+
= (0.46944π ) m3
and
&x
EL dV
=
&
3
1
3
!
x2
' 2 1 ( !
x "π )1 − + 2 * dx # = π " − 2 x + ln x #
x x , %
$ +
$2
%1
-/ 32
! 13
! ./
= π 0 " − 2(3) + ln 3# − " − 2(1) + ln1# 1
/2 $ 2
% $2
% /3
= (1.09861π ) m
Now
&
xV = xEL dV : x (0.46944π m3 ) = 1.09861π m 4
or x = 2.34 m
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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733