CHAPTER 3CHAPTER 7CHAPTER 4 - · PDF fileCHAPTER 3CHAPTER 7CHAPTER 4. 3. ... or distributed...
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CHAPTER 3CHAPTER 7CHAPTER 7CHAPTER 3CHAPTER 4CHAPTER 4
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PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.1
A tractor of mass 950 kg is used to lift gravel weighing 4 kN. Determine the reaction at each of the two (a) rear wheels A, (b) front wheels B.
SOLUTION
(a) Rear wheels + S - -M AB = + =0 9319 5 1000 4000 1250 2 1500 0: ( . )( ) ( )( ) ( )N mm N mm mm
A = +1439 83. N A = 1440 N ≠
(b) Front wheels + SM BA: ( . )( ) ( )( ) ( )- - + =9319 5 500 4000 2750 2 1500 0N mm N mm mm
B = +5219 92. N B = 5220 N ≠
Check: + = + - - =SF A By 0 2 2 9320 4000: N N 0
2 1440 2 5220 9320 4000 0( ) ( )+ - - =
0 0= ( )Checks
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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:
+ SM FA = - - =0 2 1 60 0 15 250 0 3 0: ( )( ) ( )( . ) ( )( . )m N m N m
F = 42 0. N ≠
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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:
+ SM
xA = -
- - =0 2 75 1 60 0 15
250 0 3 250 0
: ( )( ) ( )( . )
( )( . ) ( )
N m N m
N m N
x = 0 264. m
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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 SF Ax x= =0 0:
+ SMB = + ++
0 75 700 100 550 175 350
100 150
: ( )( ) ( )( ) ( )( )
( )(
N mm N mm N mm
N mmm mm) ( )- =Ay 150 0
Ay = +1225 N A = 1225 N ≠
(b) Tension in BC + SMA = + +-
0 75 550 100 400 175 200
75 150
: ( )( ) ( )( ) ( )( )
( )(
N mm N mm N mm
N mm)) ( )- =FBC 150 0mm
FBC = +700 N FBC = 700 N
Check: + SF A Fy BC= - - - - - + - =
- + - =
0 75 100 175 100 75 0
525 1225 700 0
: N N N N N
0 0= ( )Checks
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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
Wt
= =
= =
( )( . ) .
( )( . ) .
350 9 81 3 434
1400 9 81 13 734
kg m/s kN
kg m/s kN
2
2
(a) Rear wheels + SM W W W AB t= + + + - =0 1 7 2 05 1 2 2 3 0: ( . ) ( . ) ( . ) ( )m 2.05 m m m m
( . )( . ) ( . )( . )
( . )( . ) ( )
3 434 3 75 3 434 2 05
13 734 1 2 2 3
kN m kN m
kN m m
++ - =A 00
A = +6 0663. kN A = 6 07. kN ≠
(b) Front wheels + SF W W W A By t= - - - + + =0 2 2 0:
- - - + + =3 434 3 434 13 734 0. . .kN kN kN 2(6.0663 kN) 2B
B = +4 2347. kN B = 4 23. kN ≠
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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
Wt
= =
= =
( )( . ) .
( )( . ) .
350 9 81 3 434
1400 9 81 13 734
kg m/s kN
kg m/s kN
2
2
(a) Rear wheels + SM W W AB t= + + - =0 1 7 1 2 2 3 0: ( . ) ( . ) ( )m 2.05 m m m
( . )( . ) ( . )( . ) ( )3 434 3 75 13 734 1 2 2 3 0kN m kN m m+ - =A
A = +4 893. kN A = 4 89. kN ≠
(b) Front wheels + SM W W A By t= - - + + =0 2 2 0:
- - + + =3 434 13 734 0. .kN kN 2(4.893 kN) 2B
B = +3 691. kN B = 3 69. kN ≠
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PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.7
A T-shaped bracket supports the four loads shown. Determine the reactions at A and B (a) if a = 250 mm, (b) if a = 175 mm. Neglect the weight of the bracket.
SOLUTION
Free-Body Diagram:
+
SF Bx x= =0 0:
+ SM a a AB = - - + + =0 200 150 150 50 200 300 0: ( )( ) ( ) ( )( ) ( )N mm N N mm mm
Aa= -( )200 20000
300 (1)
+ SM a
aA = - - - +
-0 200 150 250 300 150 300
50
: ( )( ) ( )( ) ( )( )
( )(
N mm N mm N mm
N ++ + =500 300 0mm mm) ( )By
Ba
y = +( )175000 200
300
Since B Ba
x = = +0
175000 200
300
( ) (2)
(a) For mma = 250
Eq. (1): A = ¥ - = +( )200 250 20000
300100 N A = 100 0. N Ø
Eq. (2): B = + ¥ = +( )175000 200 250
300750 N B = 750 N Ø
(b) For mma = 175
Eq. (1): A = ¥ - = +( )200 175 20000
30050 N A = 50 0. N Ø
Eq. (2): B = + ¥ = +( )175000 200 175
300700 N B = 700 N ≠
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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 = 250 mm, (b) if a = 175 mm. Neglect the weight of the bracket.
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.
+ SM a a AB = - - + + =0 200 150 150 50 200 300 0: ( )( ) ( ) ( )( ) ( )N mm N N mm mm
Aa= -( )200 20000
300
A a= - =0 200 20000 0: ( ) a = 100 0. mm
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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
S F Bx x= =0 0:
B By=
+ SM d d d B dA = - - - - + - =0 50 100 0 45 150 0 9 0 9 0: ( ) ( )( . ) ( )( . ) ( . )N N m N m m
50 45 100 135 150 0 9d d d B Bd- + - + + -.
dB
A B=
◊ --
180 0 9
300
N m m( . ) (1)
+ SM A dB = - - + =0 50 0 9 0 9 100 0: ( )( . ) ( . ) (N m m N)(0.45 m)
45 0 9 45 0- + + =. A Ad
dA
A=
- ◊( . )0 9 90m N m (2)
Since B �180 N, Eq. (1) yields
d �180 0 9 180
300 180
18
1200 15
--
= =( . ). m d �150 0. mm v
Since A�180 N, Eq. (2) yields
d �( . )
.0 9 180 90
180
72
1800 40
- = = m d � 400 mm v
Range: 150 0. mm 400 mm� �d
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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
SF Bx x= =0 0:
B By=
+ SM d d d B dA = - - - - + - =0 80 100 0 45 150 0 9 0 9 0: ( ) ( )( . ) ( )( . ) ( . )N N m N m m
80 45 100 135 150 0 9 0d d d B Bd- + - + + - =.
dB
B=
◊ --
180 0 9
330
N m
N
. (1)
+ SM A dB = - - + =0 80 0 9 0 9 100 0: ( )( . ) ( . ) (N m m N)(0.45 m)
dA
A= -0 9 117.
(2)
Since B �180 N, Eq. (1) yields
d � ( . ) ( ) .180 0 9 180 330 18018
1500 12- ¥ - = =/ m d = 120 0. mm v
Since A�180 N, Eq. (2) yields
d � ( . ) .0 9 180 11245
1800 25¥ - = =/180 m d = 250 mm v
Range: 120 0. mm 250 mm� �d
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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 135 kN and that the reaction at A must be directed upward.
SOLUTION
SF Bx x= =0 0:
B = By ≠
+ SM P BA = - + - - =0 0 9 2 7 27 3 3 27: ( . ) ( . ) ( )( . ) (m m kN m kN)(3.9 m) 0
P B= -3 216 kN (1)
+ SM A PB = - + - - =0 2 7 1 8 27 0 6 27 0: ( . ) ( . ) ( )( . ) (m m kN m kN)(1.2 m)
P A= +1 5 27. kN (2)
Since B �135 kN, Eq. (1) yields
P � ( )( )3 135 216- kN P �189 0. kN v
Since 0 135� �A kN, Eq. (2) yields
0 27 1 5 135 27+ +kN � �P ( . )( )
27 229 5kN kN� �P . v
Range of values of P for which beam will be safe:
27 0. kN 189.0 kN� �P
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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:
+ SM P DC = - - + =0 2 4 3 20 8 6: ( ) ( )( ) ( )( ) (m kN m kN m m) 0
P D= -86 3kN (1)
+ SM P CD = + - - =0 8 4 3 20 2 6 0: ( ) ( )( ) ( )( ) (m kN m kN m m)
P C= +3 5 0 75. .kN (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 86.0 kN� �P
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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:
+ SM P DC = - - + =0 2 4 3 20 8 6: ( ) ( )( ) ( )( ) (m kN m kN m m) 0
P D= -86 3kN (1)
+ SM P CD = + - - =0 8 4 3 20 2 6 0: ( ) ( )( ) ( )( ) (m kN m kN m m)
P C= +3 5 0 75. .kN (2)
For C � 0, from (2): P � 3 50. kN v
For D � 0, from (1): P � 86 0. kN v
For C � 50 kN, from (2):
P
P
�
�
3 5 0 75 50
41 0
. . ( )
.
kN kN
kN
+ v
For D � 50 kN, from (1):
P
P
�
�
86 3 50
64 0
kN kN
kN
--
( )
. v
Comparing the four criteria, we find
3 50. kN 41.0 kN� �P
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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 450-N downward or 900-N upward.
SOLUTION
Assume B is positive when directed ≠
Sketch showing distance from D to forces.
+ SM a a BD = - - - - + =0 1350 200 1350 50 225 100 400: ( )( ) ( )( ) ( )( ) ( )N mm N N mm 00
- + + =2700 315000 400 0a B
aB
=+( )315000 400
2700 (1)
For B = 450 N Ø = -450 N, Eq. (1) yields:
a �[ ( )]315000 400 450
2700
+ - a � 50 0. mm v
For B = 900 N ≠ = +900 N, Eq. (1) yields:
a �[ ( )]315000 400 900
2700
+ a � 250 mm v
Required range: 50 0 250. mm mm� �a
17
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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:
+ SM F FC AB DE= - =0 100 120 0: ( ) ( )mm mm
F FDE AB= 5
6 (1)
(a) For FAB = 720 N
FDE = 5
6720( )N FDE = 600 N
(b) +
SF Cx x= - + =03
5720 0: ( )N
Cx = +432 N
+SF C
C
y y
y
= - + - =
= +
04
5720 600
1176
: ( )N N 0
N
C == ∞
1252 84
69 829
.
.
N
a
C = 1253 N 69.8°
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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)
F FDE AB= 5
6 (1)
+ SF F C C Fx AB x x AB= - + = =03
50
3
5:
+ SF F C Fy AB y DE= - + - =04
50:
- + - =
=
4
5
5
60
49
30
F C F
C F
AB y AB
y AB
C C C
F
C F
x y
AB
AB
= +
= +
=
2 2
2 21
3049 18
1 74005
( ) ( )
.
For C FAB= =1600 1 74005N, 1600 N . FAB = 920 N
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PROBLEM 4.17
The required tension in cable AB is 900 N. Determine (a) the vertical force P that must be applied to the pedal, (b) the corresponding reaction at C.
SOLUTION
Free-Body Diagram:
BC = 175 mm
(a) + SM PC = - =0 375 900 151 5544 0: ( ) ( )( . )mm N mm
P = 363 731. N P = 364 N Ø
(b) + SF Cx x= - =0 900: N 0 Cx = 900 N Æ
+SF C P Cy y y= - = - =0 0 363 731: . N 0 Cy = 364 N ≠
a = ∞=
22 0
970 722
.
.C N
C = 971 N 22.0∞
20
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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 1125 N.
SOLUTION
Free-Body Diagram:
BC = 175 mm
+ SM P T P TC = - = =0 375 151 5544 0 0 40415: ( ) ( . ) .mm mm
+ SF P C T Cy y y= - + = - + =0 0 0 40415: . 0
C Ty = 0 40415.
+ SF T Cx x= - + =0: 0 C Tx =
C C C T T
C T
x y= + = +
=
2 2 2 20 40415
1 0786
( . )
.
For C = 1125 N
1125 1 0786 1043 02N N= fi =. .T T T = 1043 N
21
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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: T
Ty
x
= 0 18
0 24
.
.
m
m
T Ty x= 3
4 (1)
(a) + SM TC x= - - =0 0 18 240 0 4 240 0 8 0: ( . ) ( )( . ) ( )( . )m N m N m
Tx = +1600 N
Eq. (1) Ty = =3
41600 1200( )N N
T T Tx y= + = + =2 2 2 21600 1200 2000 N T = 2 00. kN
(b) + SF C Tx x x= - =0: 0
C Cx x- = = +1600 1600N 0 N Cx = 1600 N Æ
+SF C Ty y y= - - - =0 240 240 0: N N
Cy - - =1200 480 0N N
Cy = +1680 N Cy = 1680 N ≠
a = ∞=
46 4
2320
.
C N C = 2 32. kN 46.4°
22
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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: T
T
T T
y
x
y x
=
=
0 32
0 24
4
3
.
.
m
m
+ SM TC x= - - =0 0 32 240 0 4 240 0 8 0: ( . ) ( )( . ) ( )( . )m N m N m
Tx = 900 N
Eq. (1) Ty = =4
3900 1200( )N N
T T Tx y= + = + =2 2 2 2900 1200 1500 N T = 1 500. kN
+ SF C Tx x x= - =0: 0
C Cx x- = = +900 900N 0 N Cx = 900 N Æ
+SF C Ty y y= - - - =0 240 240 0: N N
Cy - - =1200 480 0N N
Cy = +1680 N Cy = 1680 N ≠
a = ∞=
61 8
1906
.
C N C = 1 906. kN 61.8°
23
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PROBLEM 4.21
Determine the reactions at A and B when (a) h = 0, (b) h = 200 mm.
SOLUTION
Free-Body Diagram:
+ SM B B hA = ∞ - ∞ - =0 60 0 5 60 150 0 25 0: ( cos )( . ) ( sin ) ( )( . )m N m
Bh
=-37 5
0 25 0 866
.
. . (1)
(a) When h = 0:
Eq. (1): B = =37 5
0 25150
.
.N B = 150 0. N 30.0°
+ SF A By x= - ∞ =0 60 0: sin
Ax = ∞ =( )sin .150 60 129 9 N A x = 129 9. N Æ +SF A By y= - + ∞ =0 150 60 0: cos
Ay = - ∞ =150 150 60 75( )cos N A y = 75 N ≠
a = ∞=
30
150 0A . N A = 150 0. N 30.0°
(b) When h = 200 mm = 0.2 m
Eq. (1): B =-
=37 5
0 25 0 866 0 2488 3
.
. . ( . ). N B = 488 N 30.0°
+ SF A Bx x= - ∞ =0 60 0: sin
Ax = ∞ =( . )sin .488 3 60 422 88 N A x = 422 88. N Æ
+SF A By y= - + ∞ =0 150 60 0: cos
Ay = - ∞ = -150 488 3 60 94 15( . )cos . N A y = 94 15. N Ø
a = ∞=
12 55
433 2
.
.A N A = 433 N 12.6°
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 distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.22
For the frame and loading shown, determine the reactions at A and E when (a) a = ∞30 , (b) a = ∞45 .
SOLUTION
Free-Body Diagram:
+ SM E EA = +- -
0 200 125
90 250 90 75
: ( sin )( ) ( cos )( )
( )( ) ( )(
a amm mm
N mm N mmm) = 0
E =+
29250
200 125sin cosa a
(a) When a = 30°: E =∞ + ∞
=29250
200 30 125 30140 454
sin cos. N E = 140 5. N 60.0°
+ SF Ax x= - + ∞ =0 90 30 0: sinN (140.454 N)
Ax = +19 773. N A x = 19 77. N Æ
+SF Ay y= - + ∞ =0 90 140 454 30 0: ( . )cosN N
Ay = -31 637. N A y = 31 7. N Ø
A = 37 3. N 58.0°
25
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.22 (Continued)
(b) When a = ∞45 :
E =∞ +
=29250
200 45 125127 279
sin cos.
aN E = 127 3. N 45.0°
+ SF Ax x= - + ∞ =0 90 127 279 45 0: ( . )sinN N
Ax = 0 A x = 0
+SF Ay y= - + ∞ =0 90 127 279 45 0: ( . )cosN
Ay = 0 Ay = 0 A = 0
26
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.23
For each of the plates and loadings shown, determine the reactions at A and B.
SOLUTION
(a) Free-Body Diagram:
+ SM BA = - - =0 500 250 100 200 250 0: ( ) ( )( ) ( )( )mm N mm N mm
B = +150 N B = 150 0. N ≠
+ SF Ax x= + =0 200 0: N
Ax = -200 N A x = 200 N ¨
+SF A By y= + - =0 250: N 0
Ay + - =150 250 0N N
Ay = +100 N A y = 100 0. N ≠
a = ∞=
26 56
223 607
.
.A N A = 224 N 26.6°
27
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.23 (Continued)
(b) Free-Body Diagram:
+ SM BA = ∞ - - =0 30 500 200 250 250 100 0: ( cos )( ) ( )( ) ( )( )mm N mm N mm
B = 173 205. N B = 173 2. N 60.0°
+ SF A Bx x= - ∞ +0 30 200: sin N
Ax - ∞ + =( . )sin173 205 30 200 0N N
Ax = -113 3975. N A x = 113 4. N ¨
+ SF A By y= + ∞ - =0 30 250: cos N 0
Ay + ∞ - =( . )cos173 205 30 250 0N
Ay = +100 N A y = 100 0. N ≠
a =
=41 4
151 192
.
.
∞A N A = 151 2. N 41.4°
28
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.24
For each of the plates and loadings shown, determine the reactions at A and B.
SOLUTION
(a) Free-Body Diagram:
+ SM AB = - + - =0 500 250 400 200 250 0: ( ) ( )( ) ( )( )mm N mm N mm
A = +100 N A = 100 0. N ≠
+ SF Bx x= + =0 200 0: N
Bx = -200 N Bx = 200 N ¨
+SF A By y= + - =0 250: N 0
100 250 0N N+ - =By
By = +150 N B y = 150 0. N ≠
a = ∞=
36 87
250
.
B N B = 250 N 36.9°
29
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.24 (Continued)
(b)
+ SM AA = - ∞ - + =0 30 500 200 250 250 400 0: ( cos )( ) ( )( ) ( )( )mm N mm N mm
A = 115 47. N A = 115 5. N 60.0°
+ SF A Bx x= ∞ + + =0 30 200 0: sin N
( . )sin115 47 30 200 0N N∞ + + =Bx
Bx = -257 735. N Bx = 258 N ¨
+SF A By y= ∞ + - =0 30 250: cos N 0
( . )cos115 47 30 250N N 0∞ + - =By
By = +150 N B y = 150 0. N ≠
a = ∞=
30 2
298 207
.
.B N B = 298 N 30.2°
30
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.25
Determine the reactions at A and B when (a) a = 0, (b) a = 90 ,∞ (c) a = 30 .∞
SOLUTION
(a) a = 0∞
+ SM BA = - ◊ =0 0 5 100 0: ( . )m N m
B = 200 N
+ SF Ax x= =0 0:
+SF Ay y= + =0 200 0:
Ay = -200 N
A = 200 N Ø B = 200 N ≠
(b) a = ∞90
+ SM BA = - ◊ =0 0 3 100: ( . )m N m 0
B = =1000
3333 33N N.
+ SF A AA x x= - = =0 333 33 0 333 33: . , .N N
+ SF Ay y= =0 0:
A = 333 N Æ B = 333 N ¨
(c) a = ∞30
+ SM B BA = ∞ + ∞ - ◊ =0 30 0 5 30 0 3 100: ( cos )( . ) ( sin )( . )m m N m 0
B = 171 523. N
+ SF Ax x= - ∞ =0 171 523 30 0: ( . )sinN
Ax = 85 762. N
+ SF A Ay y y= + ∞ = = -0 171 523 0 148 543: ( . cos .N) 30 N
A = = ∞171 523. N 60.0a
A = 171 5. N 60.0° B = 171 5. N 60.0°
31
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
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.
+ SM TA = ∞ + + =0 45 0 2 90 0 1 90 0 2 0: ( sin )( . ) ( )( . ) ( )( . )m N m N m
T = 190 92. N T = 190 9. N
(b) + SF Ax x= - ∞ =0 190 92 45 0: ( . )cosN
Ax = +135 N A x = 135 0. N Æ
+ SF Ay y= - - + ∞ =0 90 90 0: sinN N (190.92 N) 45
Ay = +45 N A y = 45 0. N ≠
A = 142 3. N 18.43°
32
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
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 ;a a= = ∞10
1533.69
(a) Move T along BD until it acts at Point D.
+ SM TA = ∞ - - =0 33 69 0 15 90 0 1 90 0 2 0: ( sin . )( . ) ( )( . ) ( )( . )m N m N m
T = 324 5. N T = 324 N
(b) + SF Ax x= - ∞ =0 324 99 33 69 0: ( . )cos .N
Ax = +270 N A x = 270 N Æ
+SF Ay y= - - + ∞ =0 90 90 33 69 0: sin .N N (324.5 N)
Ay = 0 A y = 0 A = 270 N Æ
33
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
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 SC = ∞ + ∞ = ∞90 30 120 S SA D= = ∞ - ∞ = ∞1
2180 120 30( )
Thus DA forms angle of 60° with horizontal.
(a) We resolve FAD into components along AB and perpendicular to AB.
+ SM FC AD= ∞ - =0 30 250 500 100 0: ( sin )( ) ( )( )mm N mm FAD = 400 N
(b) + SF Cx x= - ∞ + - =0 400 60 500 0: ( )cosN N Cx = +300 N
+ SF Cy y= - ∞ + =0 400 0: ( sinN) 60 Cy = +346 4. N
C = 458 N 49.1°
34
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.29
A force P of magnitude 1260 N 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 = 75 mm, determine (a) the tension in the cable, (b) the reaction at A.
SOLUTION
Free-Body Diagram:
(a) + SM
T T
T
A = -
-
- =
0 1260 75
3007
25300
24
25200 0
: ( )( )
( ) ( )
( )
N mm
mm mm
mm
24 94500
3937 5
T
T
== . N T = 3940 N
(b) + SF Ax x= + + =07
253937 5 3937 5 0: ( . ) .
Ax = -5040 N A x = 5040 N ¨
+ SF Ay y= - - =0 126024
253937 5 0: ( . )N N
Ay = +5040 N A y = 5040 N ≠
A = 7128 N 45.0°
35
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.30
Neglecting friction, determine the tension in cable ABD and the reaction at support C.
SOLUTION
Free-Body Diagram:
+ SM T TC = - - =0 0 25 0 1 120 0 1 0: ( . ) ( . ) ( )( . )m m N m T = 80 0. N
+ SF C Cx x x= - = = +0 80 80: N 0 N Cx = 80 0. N Æ
+ SF C Cy y y= - + = = +0 120 40: N 80 N 0 N Cy = 40 0. N ≠
C = 89 4. N 26.6°
36
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.31
Rod ABC is bent in the shape of an arc of circle of radius R. Knowing the q = 30°, determine the reaction (a) at B, (b) at C.
SOLUTION
Free-Body Diagram: + SM C R P RD x= - =0 0: ( ) ( )
C Px = +
+ SF C Bx x= - =0 0: sinq
P B
B P
- ==
sin
/sin
0
B = P
sinq q
+SF C B Py y= + - =0 0: cosq
C P P
C P
y
y
+ - =
= -ÊËÁ
ˆ¯̃
( /sin )cos
tan
q q
q
0
11
For q = ∞30 :
(a) B P P= ∞ =/sin30 2 B = 2P 60.0°
(b) C P C Px x= + = Æ
C P Py = - ∞ = -( /tan ) . /1 1 30 0 732 Cy P= 0 7321. Ø
C = 1 239. P 36.2°
37
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.32
Rod ABC is bent in the shape of an arc of circle of radius R. Knowing the q = 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:
C P
C P
BP
x
y
= +
= -ÊËÁ
ˆ¯̃
=
11
tan
sin
q
q
For q = ∞60 :
(a) B P P= ∞ =/sin .60 1 1547 B = 1 155. P 30.0°
(b) C P C Px x= + = Æ
C P Py = - ∞ = +( /tan ) .1 1 60 0 4226 Cy P= 0 4226. Ø
C = 1 086. P 22.9°
38
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.33
Neglecting friction, determine the tension in cable ABD and the reaction at C when q = 60°.
SOLUTION
+ SM T a a Ta PaC = + - + =0 2 0: ( cos )q
TP=
+1 cosq (1)
+ SF C Tx x= - =0 0: sinq
C TP
x = =+
sinsin
cosq q
q1
+ SF C T T Py y= + + - =0 0: cosq
C P T P P
C
y
y
= - + = - ++
=
( cos )cos
cos1
1
10
q qq
Since C C Cy x= =0, C =+
Psin
cos
qq1
Æ (2)
For q = ∞60 :
Eq. (1): TP P=
+ ∞=
+1 60 1 12cos
T P= 2
3
Eq. (2): C P P= ∞+ ∞
=+
sin
cos
.60
1 60
0 866
1 12
C = 0 577. P Æ
39
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.34
Neglecting friction, determine the tension in cable ABD and the reaction at C when q = 45°.
SOLUTION
Free-Body Diagram:
Equilibrium for bracket:
+ SM T a P a T a
T a aC = - - + ∞ ∞
+ ∞ + ∞0 45 2 45
45 2 45
: ( ) ( ) ( sin )( sin )
( cos )( cos ) == 0
T = 0 58579. or T P= 0 586.
+ SF C Px x= + ∞ =0 0 58579 45 0: ( . )sin
C Px = 0 41422.
+SF C P P Py y= + - + ∞ =0 0 58579 0 58579 45 0: . ( . )cos
Cy = 0 or C = 0 414. P Æ
40
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
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 600 N force is applied at D. Determine the reactions at A, B, and C.
SOLUTION
Free-Body Diagram:
SF Ax = ∞ - ∞ =0 30 600 60 0: cos ( )cosN
A = 346 4102. N A = 346 N Æ
+ SM CB = - ∞+
0 200 600 400 30
346 4102 200 30
: ( ) ( )( )cos
( . )( )sin
mm N mm
N mm ∞∞ = 0
C = 866 0253. N C = 866 N 60.0°
+ SM BC = - ∞+
0 200 600 200 30
346 4102 400 30
: ( ) ( )( )cos
( . )( )sin
mm N mm
N mm ∞∞ = 0
B = 173 205. N B = 173 2. N 60.0°
41
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.36
A light bar AD is suspended from a cable BE and supports a 250-N 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:
SF A Dx = =0:
SFy = 0: TBE = 250 N
We note that the forces shown form two couples.
+ SM A= - =0 200 250 75 0: ( ) ( )( )mm N mm
A = 93 75. N
A = 93 75. N Æ D = 93 75. N ¨
42
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
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
AD
BE
CD
BEBE= = =;
.
. .; .
0 15
0 5 0 250 075
m
m mm
(a) + SM T TA x x= - ÊËÁ
ˆ¯̃
- -0 0 250 075
0 350 5 400 0 1 400: ( . )
.
.( . ) ( )( . ) (m m N m N))( . )0 4 0m =
Tx = 1400 N
Ty =
=
0 075
0 351400
300
.
.( )N
N T = 1432 lb
43
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.37 (Continued)
(b) + SF Ay = - - - =0 300 400 400 0: N N N
A = +1100 N A = 1100 N ≠
(c) + SF Cx = - + =0 1400 0: N
C = + 1400 N C = 1400 N ¨
44
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.38
Determine the tension in each cable and the reaction at D.
SOLUTION
tan.
.
tan.
.
a
a
b
b
=
= ∞
=
= ∞
0 08
21 80
0 08
38 66
m
0.2 m
m
0.1 m
+ SM TB CF= - ∞ =0 600 0 1 38 66 0 1 0: ( )( . ) ( sin . )( . )N m m
TCF = 960 47. N TCF = 96 0. N
+ SM TC BE= - ∞ =0 600 0 2 21 80 0 1 0: ( )( . ) ( sin . )( . )N m m
TBE = 3231 1. N TBE = 3230 N
+ SF T T Dy BE CF= + - =0 0: cos cosa b
( . )cos . ( . )cos .3231 1 21 80 960 47 38 66 0N N∞ + ∞ - =D
D = 3750 03. N D = 3750 N ¨
45
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
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 = 75 N, determine (a) the force each pin exerts on the plate, (b) the reaction at F.
SOLUTION
Free-Body Diagram:
+ SF Bx = - ∞ =0 75 30 0: sinN B = 150 N 60.0°
+ SM B B FA = - + ∞ + ∞ -0 150 100 30 75 30 275 325: ( )( ) sin ( ) cos ( ) ( )N mm mm mm mm == 0
- + + - =15000 5625 35723 55 325 0. ( )F
F = +81 0725. N F = 81 1. N Ø
+ SF A B Fy = - + ∞ - =0 150 30 0: cosN
A - + ∞ - =150 150 30 81 0725 0N N N( )cos .
A = +101 169. N A = 101 2. N ≠
46
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.40
For the plate of Problem 4.39 the reaction at F must be directed downward, and its maximum allowable value is 100 N. 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 = 75 N, determine (a) the force each pin exerts on the plate, (b) the reaction at F.
SOLUTION
Free-Body Diagram:
+ SF P Bx = - ∞ =0 30 0: sin B = 2P 60°
+ SM B B FA = - + ∞ + ∞ -0 150 100 30 75 30 275 325: ( )( ) sin ( ) cos ( ) ( )N mm mm mm mm == 0
- ◊ ∞ + ∞ - =-15000 2 30 75 2 30 275 325 0
15000
N mm+ mm mm mmP P Fsin ( ) cos ( ) ( )
++ + - =75 476 314 325 0P P F.
PF= +325 15000
551 314. (1)
For F = 0: P = =15000
551 31427 208
.. N
For P = 100 N: P = + =32500 15000
551 31486 158
.. N
For 0 100� �F N 27 2. N 86.2 N� �P
47
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
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 (a = 0), determine the tension in the cord and the reactions at A and C.
SOLUTION
Free-Body Diagram:
+ SF Ty = - ∞ + ∞ =0 30 80 30 0: cos ( )cosN
T = 80 N T = 80 0. N
+ SM AC = ∞ - - =0 30 0 4 80 0 2 80 0 2 0: ( sin )( . ) ( )( . ) ( )( . )m N m N m
A = +160 N A = 160 0. N 30.0°
+ SM CA = - + ∞ =0 80 0 2 80 0 6 30 0 4 0: ( )( . ) ( )( . ) ( sin )( . )N m N m m
C = +160 N C = 160 0. N 30.0°
48
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.42
Solve Problem 4.41 if the cord BE is parallel to the rods (a = 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 (a = 0), determine the tension in the cord and the reactions at A and C.
SOLUTION
Free-Body Diagram:
+ SF Ty = - + ∞ =0 80 30 0: ( )cosN
T = 69 282. N T = 69 3. N
+ SM
AC = - ∞
- + ∞ =0 69 282 30 0 2
80 0 2 30 0 4 0
: ( . )cos ( . )
( )( . ) ( sin )( . )
N m
N m m
A = +140 000. N A = 140 0. N 30.0°
+ SM
CA = + ∞
- + ∞ =0 69 282 30 0 2
80 0 6 30 0 4 0
: ( . )cos ( . )
( )( . ) ( sin )( . )
N m
N m m
C = +180 000. N C = 180 0. N 30.0°
49
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
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 9 81 78 48kg m/s N2
(a) SF Ax x= =0 0:
+SF A Wy y= - =0 0: A y = 78 48. N ≠
+ SM M WA A= - =0 1 6 0: ( . )m
M A = + ( . )( . )78 48 1 6N m MA = ◊125 56. N m
A = 78 5. N ≠ MA = ◊125 6. N m
(b) + SF A Wx x= - =0 0: A x = 78 48. ≠
+ SF A Wy y= - =0 0: A y = 78 48. ¨
A = =( . ) .78 48 2 110 99N N 45°
+ SM M WA A= - =0 1 6 0: ( . )m
M A A= + = ◊( . )( . ) .78 48 1 6 125 56N m N mM
A = 111 0. N 45° MA = ◊125 6. N m
(c) SF Ax x= =0 0:
+ SF A Wy y= - =0 2 0:
A Wy = = =2 2 78 48 156 96( . ) .N N ≠
+ SM M WA A= - =0 2 1 6 0: ( . )m
M A = +2 78 48 1 6( . )( . )N m MA = ◊125 1. N m
A = 157 0. N ≠ MA = ◊125 N m
50
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.44
A tension of 22.5 N is maintained in a tape as it passes through the support system shown. Knowing that the radius of each pulley is 10 mm, determine the reaction at C.
SOLUTION
From f.b.d. of system
+ SF Cx x= + =0 22 5 0: ( . )N
Cx = -22 5. N
+ SF Cy y= - =0 22 5 0: ( . )N
Cy = 22 5. N
Then C C Cx y= +
= +=
( ) ( )
( . ) ( . )
.
2 2
2 222 5 22 5
31 8198 N
and q = +-
ÊËÁ
ˆ¯̃
= - ∞-tan.
.1 22 5
22 545 or C = 31 8. N 45.0∞
+ SM MC C= + + =0 22 5 160 22 5 60 0: ( . )( ) ( . )( )N mm N mm
MC = - ◊4950 N mm or MC = ◊4950 N mm
51
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.45
Solve Problem 4.44, assuming that 15 mm radius pulleys are used.
PROBLEM 4.44 A tension of 22.5 N is maintained in a tape as it passes through the support system shown. Knowing that the radius of each pulley is 10 mm, determine the reaction at C.
SOLUTION
From f.b.d. of system
+ SF Cx x= + =0 22 5 0: N( . )
Cx = -22 5. N
+ SF Cy y= - =0 22 5 0: N( . )
Cy = 22 5. N
Then C C Cx y= +
= +=
( ) ( )
( . ) ( . )
.
2 2
2 222 5 22 5
31 8198 N
and q =-
ÊËÁ
ˆ¯̃
= - ∞-tan.
..1 22 5
22 545 0 or C = 31 8. N 45 0. ∞
+ SM MC C= + + =0 22 5 165 22 5 65 0: N mm N mm( . )( ) ( . )( )
MC = - ◊5175 N mm or MC = ◊5180 N mm
52
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.46
The frame shown supports part of the roof of a small building. Knowing that the tension in the cable is 150 kN, determine the reaction at the fixed end E.
SOLUTION
Free-Body Diagram:
4.5m
1.8m
20kN
D
C
6m
150kNME
Ex
My
E F
BA
20kN20kN20kN
1.8m1.8m1.8m
Equilibrium Equations DF = + =( . ( .4 5 6 7 52 2m) m) m,
+ SF Ex x= + =04 5
7 5150 0:
.
.( )kN
Ex = -90 0. kN Ex = 90 0. kN ¬
+ SF Ey y= - - =0 4 206
7 5150 0: (
.( )kN) kN
Ey = +200 kN Ey = 200 kN
+ SME = + +
+
0 20 20 5 4 20 3 6
20
: ( ( )( . ) ( )( . )
(
kN)(7.2 m) kN m kN m
kN)(1.8m) -- 6
7 5150 0
.( kN)(4.5m)+ ME =
ME = + ◊180 0. kN m ME = ◊180 0. kN m
53
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.47
Beam AD carries the two 200-N 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 = 500 N, (b) W = 450 N.
SOLUTION
(a) W = 500 N
From f.b.d. of beam AD
+ SF Dx x= =0 0:
+ SF Dy y= - - + =0 200 200 500: N N N 0
Dy = -100 N or D = 100 0. N Ø
+ SM MD D= - ++ =
0 500 1 5 200 2 4
200
: ( )( . ) ( )( . )
(
N m N m
N)(1.2 m) 0
MD = ◊30 N m or MD = ◊30 0. N m
(b) W = 450 N
From f.b.d. of beam AD
+ SF Dx x= =0 0:
+ SF Dy y= + - - =0 450 200 200: N N N 0
Dy = -50 N or D = 50 0. N Ø
+ SM MD D= - ++ =
0 450 1 5 200 2 4
200 1 2 0
: ( )( . ) ( )( . )
( )( . )
N m N m
N m
MD = - ◊45 N m or MD = - ◊45 0. N m
54
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 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 60 N · m.
SOLUTION
For Wmin , MD = - ◊60 N m
From f.b.d. of beam AD + SM WD = -+ - ◊ =
0 200 2 4 1 5
200 1 2 60 0
: ( )( . ) ( . )
( )( . )minN m m
N m N m
Wmin N= 440
For Wmax, MD = ◊60 N m
From f.b.d. of beam AD + SM WD = -+ + ◊ =
0 200 2 4 1 5
200 1 2 60 0
: ( )( . ) ( . )
( )( . )maxN m m
N m N m
Wmax = 520 N or N 520 N440 � �W
55
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 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
T T
T T
x
y
=
=
=
=
=
1300
5
13500
12
131200
N
N
N
+ SM C Cx x x= - + = = -0 450 500 50: N N 0 N Cx = 50 N ¨
+ SF C Cy y y= - - = = +0 750 120 1950: N 0 N 0 N Cy = 1950 N ≠
C = 1951 N 88.5°
+ SM MC C= + +- =
0 750 0 5 4 50 0 4
1200 0 4 0
: ( )( . ) ( . )( . )
( )( . )
N m N m
N m
MC = - ◊75 0. N m MC = ◊75 0. N m
56
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 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
+ SM T
T
C = + - ÊËÁ
ˆ¯̃
-ÊËÁ
0 750 450 0 45
130 6
12
13
: ( ( )( . ) ( . )N)(0.5 m) N m m
ˆ̂¯̃
+ =( .0 15 0m) MC
375 1804 8
130N m N m m◊ + ◊ - Ê
ËÁˆ¯̃
+ =.T MC
T MC= +13
4 8555
.( )
For M TC = - ◊ = - =10013
4 8555 100 1232N m: N
.( )
For M TC = + ◊ = + =10013
4 8555 100 1774N m: N
.( )
For | | :MC �100 N m◊ 1 232. kN 1.774 kN� �T
57
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.51
A vertical load P is applied at end B of rod BC. (a) Neglecting the weight of the rod, express the angle q corresponding to the equilibrium position in terms of P, l, and the counterweight W. (b) Determine the value of q corresponding to equilibrium if P W= 2 .
SOLUTION
(a) Triangle ABC is isosceles.
We have CD BC l= =( )cos cosq q2 2
+ SM W l P lC = ÊËÁ
ˆ¯̃
- =02
0: cos ( sin )q q
Setting sin sin cos : cos sin cosq q q q q q= - =22 2 2
22 2
0Wl Pl
W P- =22
0sinq
q = ÊËÁ
ˆ¯̃
-22
1sinW
P b
(b) For P W= 2 : sin .q2 2 4
0 25= = =W
P
W
W
q2
14 5= ∞. q = ∞29 0. b
or q q2
165 5 331= ∞ = ∞. ( )discard
58
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 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 q. (b) Determine the value of q for which the tension in the cord is equal to 3W.
SOLUTION
(a) From f.b.d. of rod AB
+ SM T l W T lC = + ÊËÁ
ˆ¯̃
ÈÎÍ
˘˚̇
- =01
20: ( sin ) cos ( cos )q q q
TW=
-cos
(cos sin )
qq q2
Dividing both numerator and denominator by cos q,
TW=
-ÊËÁ
ˆ¯̃2
1
1 tanq or T
W
=( )
-2
1( tan )q b
(b) For T W= 3 , 31
11
6
2WW
=( )
-
- =
( tan )
tan
q
q
or q = ÊËÁ
ˆ¯̃
= ∞-tan .1 5
639 806 or q = ∞39 8. b
59
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.53
Rod AB is acted upon by a couple M and two forces, each of magnitude P. (a) Derive an equation in q, P, M, and l that must be satisfied when the rod is in equilibrium.(b) Determine the value of q corresponding to equilibrium when M = 1500 N m,� P = 200 N, and l = 600 mm.
SOLUTION
Free-Body Diagram: (a) From free-body diagram of rod AB
+ SM P l P l MC = + - =0 0: ( cos ) ( sin )q q
or sin cosq q+ = M
Pl b
(b) For M P l= ◊ = =
+ =◊
=
1500 600
1500
N m, 200 N, and mm
N m
(200 N)(600 mm)sin cosq q 55
41 25= .
Using identity sin cos
sin ( sin ) .
( sin ) . sin
2 2
2 1 2
2 1 2
1
1 1 25
1 1 25
1
q q
q q
q q
+ =
+ - =
- = -
/
/
-- = - +sin . . sin sin2 21 5625 2 5q q q
2 2 5 0 5625 02sin . sin .q q- + =
Using quadratic formula
sin( . ) ( ) ( )( . )
( )
. .
q =- - ± -
= ±
2 5 625 4 2 0 5625
2 2
2 5 1 75
4
or sin . sin .
. .
q qq q
= == ∞ = ∞
0 95572 0 29428
72 886 17 1144
and
and
or andq q= ∞ = ∞17 11 72 9. . b
60
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 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 q that must be satisfied when the rod is in equilibrium. (b) Determine the value of q corresponding to equilibrium when P = 80 N, Q = 60 N, l = 500 mm, and a = 125 mm.
SOLUTION
Free-Body Diagram:
+SF C P Qy = - - =0 0: cosq
CP Q= +cosq
(a) + SM Ca
PlA = - =0 0:cos
cosq
q
P Q a
Pl+ ◊ - =
cos coscos
q qq 0 cos
( )3 q = +a P Q
Pl b
(b) For P Q l a= = = =80 60 500 125N, N, mm and mm,
cos
( )( )
( )( )
.
3 125 80 60
80 500
0 4375
q =+
=
mm N N
N mm
cos .q = 0 75915 q = ∞40 6. b
61
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 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 q = 0. (a) Derive an equation in q, 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 q corresponding to equilibrium.
SOLUTION
First note T ks=
where k
s
ll
l
==
= -
= -
spring constant
elongation of spring
cos
cos( co
q
q1 ss )
cos( cos )
q
qq T
kl= -1
(a) From f.b.d. of collar B + Â = - =F T Wy 0 0: sinq
or kl
Wcos
( cos )sinq
q q1 0- - = or tan sinq q- = W
kl b
(b) For W
l
k
l
===
= =
- =
300
500
800
5000 5
300
N
mm
N m
mm
2400 N/mm
N
800
/
.
tan sin(
q qNN/m m)( . )
.0 5
0 75=
Solving numerically, q = ∞57 957. or q = ∞58.0 b
62
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 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 q = ∞90 . (a) Neglecting the weight of the rod, express the angle q corresponding to equilibrium in terms of P, k, and l. (b) Determine the value of q corresponding to equilibrium when P kl= 1
4 .
SOLUTION
First note T ks= =tension in spring
where s
AB AB
l l
=
= -
= ÊËÁ
ˆ¯̃
-
= ∞
elongation of spring
( ) ( )
sin sin
q q
q90
22
2990
2
22
1
2
∞ÊËÁ
ˆ¯̃
= ÊËÁ
ˆ¯̃
- ÊËÁ
ˆ¯̃
È
ÎÍ
˘
˚˙l sin
q
T kl= ÊËÁ
ˆ¯̃
- ÊËÁ
ˆ¯̃
È
ÎÍ
˘
˚˙2
2
1
2sin
q
(1)
(a) From f.b.d. of rod BC + SM T l P lC = ÊËÁ
ˆ¯̃
ÈÎÍ
˘˚̇
- =02
0: cos ( sin )q q
Substituting T from Equation (1)
22
1
2 20kl l P lsin cos sin
q q qÊËÁ
ˆ¯̃
- ÊËÁ
ˆ¯̃
È
ÎÍ
˘
˚˙
ÊËÁ
ˆ¯̃
ÈÎÍ
˘˚̇
- ( ) =
222
1
2 22
22kl Plsin cos sin c
q q qÊËÁ
ˆ¯̃
- ÊËÁ
ˆ¯̃
È
ÎÍ
˘
˚˙
ÊËÁ
ˆ¯̃
- ÊËÁ
ˆ¯̃ oos
q2
0ÊËÁ
ˆ¯̃
ÈÎÍ
˘˚̇
=
Factoring out 22
l cos ,qÊ
ËÁˆ¯̃
leaves
kl Psin sinq q2
1
2 20Ê
ËÁˆ¯̃
- ÊËÁ
ˆ¯̃
È
ÎÍ
˘
˚˙ - Ê
ËÁˆ¯̃
=
or sinq2
1
2ÊËÁ
ˆ¯̃
=-
ÊËÁ
ˆ¯̃
kl
kl P q =
-È
ÎÍ
˘
˚˙-2
21sin
( )
kl
kl P b
63
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.56 (Continued)
(b) Pkl
kl
kl
kl
kl
kl
=
= -( )È
ÎÍ
˘
˚˙
= ÊËÁ
ˆ¯̃
ÈÎÍ
˘˚̇
=
-
-
4
22
22
4
3
2
1
4
1
q sin
sin
ssin- ÊËÁ
ˆ¯̃
1 4
3 2
== ∞
-2 0 94281
141 058
1sin ( . )
. or 141.1q = ∞ b
64
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.57
Solve Sample Problem 4.5, assuming that the spring is unstretched when q = ∞90 .
SOLUTION
First note T ks= =tension in spring
where s
r
F kr
===
deformation of spring
bb
From f.b.d. of assembly + SM W l F r0 0 0= - =: ( cos ) ( )b
or Wl kr
kr
Wl
cos
cos
b b
b b
- =
=
2
2
0
For k
r
l
W
====
=
45
75
200
1800
45 75
1800 20
2
N mm
mm
mm
N
N/mm mm
N
/
cos( )( )
( )(b
00 mm)b
or cos .b b= 0 703125
Solving numerically, b = 0 89245. rad
or b = ∞51 134.
Then q = ∞ + ∞ = ∞90 51 134 141 134. . or 141.1q = ∞ b
65
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 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 q = 0. (a) Neglecting the weight of the blocks, derive an equation in W, k, l, and q that must be satisfied when the rod is in equilibrium. (b) Determine the value of q when W = 375 N, l = 750 mm, and k = 600 N/m.
SOLUTION
Free-Body Diagram:
Spring force: F ks k l l kls = = - = -( cos ) ( cos )q q1
(a) + SM F l Wl
D s= - ÊËÁ
ˆ¯̃
=02
0: ( sin ) cosq q
kl lW
l( cos ) sin cos12
0- - =q q q
klW
( cos ) tan12
0- - =q q or ( cos ) tan12
- =q q W
kl b
(b) For given values of W
l
k
== ==
- = -
=
375
750 0 75
600
1
375
2 60
N
mm m
N/m
N
.
( cos ) tan tan sin
(
q q q q
00 0 75
0 41667
N/m m)( . )
.=
Solving numerically: q = ∞49 710. or q = ∞49 7. b
66
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 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 C D= = =0 196 2, . N ≠
3. Four non-concurrent, non-parallel reactions
(a) Plate: completely constrained
(b) Reactions: indeterminate
(c) Equilibrium maintained
A x = 294 N Æ, Dx = 294 N ¨
(A Dy y+ = 392 N ≠ )
4. Three concurrent reactions (through D)
(a) Plate: improperly constrained
(b) Reactions: indeterminate
(c) No equilibrium ( )SMD π 0
67
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 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 Æ, D = 491 N 53.1°
7. Two reactions
(a) Plate: improperly constrained
(b) Reactions determined by dynamics
(c) No equilibrium ( )SFy π 0
8. 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
68
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 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 500 N.
SOLUTION
1. Three non-concurrent, non-parallel reactions
(a) Bracket: complete constraint
(b) Reactions: determinate
(c) Equilibrium maintained
A = 601N 56.3°, B = 333 N ¨
2. Four concurrent, reactions (through A)
(a) Bracket: improper constraint
(b) Reactions: indeterminate
(c) No equilibrium ( )SM A π 0
3. Two reactions
(a) Bracket: partial constraint
(b) Reactions: determinate
(c) Equilibrium maintained
A = 250 N ≠, C = 250 N ≠
4. Three non-concurrent, non-parallel reactions
(a) Bracket: complete constraint
(b) Reactions: determinate
(c) Equilibrium maintained
A = 250 N ≠, B = 417 N 36.9°, C = 333N Æ
69
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.60 (Continued)
5. Four non-concurrent, non-parallel reactions
(a) Bracket: complete constraint
(b) Reactions: indeterminate
(c) Equilibrium maintained ( )SMC y= =0 250 NA ≠
6. Four non-concurrent, non-parallel reactions
(a) Bracket: complete constraint
(b) Reactions: indeterminate
(c) Equilibrium maintained
A x = 333 N ®, B = 333 N ¨
(A By y+ = 500 N ≠ )
7. Three non-concurrent, non-parallel reactions
(a) Bracket: complete constraint
(b) Reactions: determinate
(c) Equilibrium maintained
A C= = 250 N ≠
8. Three concurrent, reactions (through A)
(a) Bracket: improper constraint
(b) Reactions: indeterminate
(c) No equilibrium ( )SM A π 0
70
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 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. Free-Body Diagram: (Three-force member)
DBCD : tan b
b
=
= ∞
240
180
53 13.
Force triangle
A = ∞=
( ) tan .300 53 13
400
N
N A = 400 N ≠ b
B =
=
300
500
N
cos 53.13 N
∞ B = 500 N 53 1. ∞ b
71
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 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 b (1)
Force Triangle
(with N)B = 600
cos .
.
b
b
= =
= ∞
3000 5
60 0
N
600 N
Eq. (1) a =
=
240
138 56
mm
tan 60.0 mm
∞.
For B � 600 138 6 N mma ≥ . b
72
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.63
Using the method of Section 4.7, solve Problem 4.17.
PROBLEM 4.17 The required tension in cable AB is 900 N. 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 P and 900 N forceintersect.
tan.
.
b
b
=
=
151 554
375
22 006
m
mm
∞
Force triangle
(a) P = ( )900 N tan 22.006∞
P = 363 7. P = 364 N Ø b
(b) C =∞
=900
22 006970 7
N N
cos .. C = 971 N 22 0. ∞ b
73
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 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 1125 N.
SOLUTION
Free-Body Diagram: (Three-Force body)
Reaction at C must pass through E, where D and the force T intersect.
tan
.
.
b
b
=
= ∞
151 554
375
22 006
mm
mm
Force triangle
T
T
= ∞=
( )cos .
.
1125 22 006
1043 04
N
N T = 1043 N b
74
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 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 300-N 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.
tansin
cos.
.
a
a
= ∞∞ +
== ∞
75 50
75 50 3750 135756
7 7310
Force triangle A
B
=
=
=
=
300
2230 11
300
2209 842
N
sin 7.7310 N
N
tan 7.7310 N
∞
∞
.
.
A = 2230 N 7 73. ∞ b
B = 2210 N Æ b
75
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
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.
tan
.
b
b
=
= ∞
220
41 348
mm
250 mm
Force triangle
Law of sines
80
45 131 348888 0
942 8
N
sin 3.652 N
N
∞ ∞ ∞= =
==
B D
B
D
sin sin ..
. B = 888 N 41 3. ∞ D = 943 N 45 0. ∞ b
76
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
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.
tan
.
b
b
=
= ∞
280
48 24
mm
250 mm
Force triangle
Law of sines 80
135 41 76
N
sin 3.24∞=
∞=
∞B D
sin sin .
B
D
==
1000 9
942 8
.
.
N
N B = 1001 N 48 2 943. ∞ =D N 45 0. ∞ b
77
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.68
Determine the reactions at B and C when a = 37 5. mm.
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 250 N load intersect.
Triangle CFD: tan .
.
a
a
= =
= ∞
75
1250 6
30 964
Triangle EAD: AE
GE AE AG
= == - = - =
250 150
150 37 5 112 5
tan
. .
a mm
mm
Triangle EGB: tan.
.
b
b
= =
= ∞
GB
GE
75
112 533 690
Force triangle B D
sin . sin .120 964 33 690
250
∞=
∞=
∞ N
sin 25.346
B
D
==
500 773
323 943
.
.
N
N B = 501 N 56 3. ∞ b
C D= = 324 N 31 0. ∞ b
78
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
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.
tan.
..
b
b
=
= ∞
0 7497
1 526 56
m
m
Force triangle 3 forces intersect at E.
W ==
( )
.
50
490 5
kg 9.81 m/s
N
2
Law of sines 490 5
63 44 55
498 9
456 9
.
sin . sin
.
.
N
sin 61.56
N
N
∞=
∞=
∞
=
=
T B
T
B
CD
CD
(a) TCD = 499 N b
(b) B = 457 N 26 6. ∞ b
79
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
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:
tan.
.
b
b
= =
= ∞
AE
AB
0 301
5 722
m
3 m
Force Triangle (Three forces intersect at E.)
W ==
( )50
490
kg 9.81 m/s
.5 N
2
Law of sines 490 5
95 722 55997 99
821
.
sin . sin.
.
N
sin 29.278 N
∞=
∞=
∞==
T B
T
B
CD
CD
559 N
(a) TCD = 998 N b
(b) B = 822 N 5 72. ∞ b
80
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
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 200-N 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 200 N load intersect.
tan
.
tan
.
a
a
b
b
= =
= ∞
= =
= ∞
EF
AF
EH
DH
575
300
62 447
125
300
22 620
Force triangle A T
sin . sin .67 380 27 553
200
∞=
∞=
∞ N
sin 85.067
A = 185 3. N 62 4. ∞ b
T = 92 9. N b
81
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.72
Determine the reactions at A and D when b = ∞30 .
SOLUTION
From f.b.d. of frame ABCD
+ SM AD = - + ∞+ ∞
0 0 18 150 30 0 10
150 30 0 28
: ( . [( sin ]( .
[( cos ]( .
m) N) m)
N) m)) = 0
A = 243 74. N
or A = 244 N Æ b
+ SF Dx x= + ∞ + =0 243 74 150 30 0: ( . ( sinN) N)
Dx = -318 74. N
+ SF Dy y= - ∞ =0 150 30 0: ( cosN)
Dy = 129 904. N
Then D D Dx x= +
= +=
( )
( . ) ( . )
.
2 2
2 2318 74 129 904
344 19 N
and q =ÊËÁ
ˆ¯̃
=-
ÊËÁ
ˆ¯̃
-
-
tan
tan.
.
1
1 129 904
318 74
D
Dy
x
= - ∞22 174. or D = 344 N 22 2. ∞ b
82
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.73
Determine the reactions at A and D when b = ∞60 .
SOLUTION
From f.b.d. of frame ABCD
+ SM AD = - + ∞+ ∞
0 0 18 150 60 0 10
150 60 0 28
: ( . [( sin ]( .
[( cos ]( .
m) N) m)
N) m)) = 0
A = 188 835. N
or A = 188 8. N Æ b
+ SF Dx x= + ∞ + =0 188 835 150 60 0: ( . ) ( )sinN N
Dx = -318 74. N
+ SF Dy y= - ∞ =0 150 60 0: ( )cosN
Dy = 75 0. N
Then D D Dx y= +
= +=
( ) ( )
( . ) ( . )
.
2 2
2 2318 74 75 0
327 44 N
and q =ÊËÁ
ˆ¯̃
-tan 1 D
Dy
x
=-
ÊËÁ
ˆ¯̃
= - ∞
-tan.
.
.
1 75 0
318 74
13 2409 or D = 327 N 13 24. ∞ b
83
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.74
A 20-kg roller, of diameter 200 mm, 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 8 mm, 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
(a) Based on the roller having impending motion to the left, the only contact between the roller and floor will be at the edge of the tile.
First note W mg= = =( )( . ) .20 9 81 196 2kg m/s N2
From the geometry of the three forces acting on the roller
a =ÊËÁ
ˆ¯̃
=-cos .1 92
10023 074
mm
mm∞
and q a= - - = - =90 30 60 23 074 36 926∞ ∞ ∞ ∞ ∞. .
Applying the law of sines to the force triangle,
W P
sin sinq a=
or 196 2
36 926 23 074
.
sin . sin .
N
∞=
∞P
\ =P 127 991. N
or NP = 128 0. 30∞ b
(b) Based on the roller having impending motion to the right, the only contact between the roller and floor will be at the edge of the tile.
First note W mg= = ( )( . )20 9 81kg m/s2 = 196 2. N
From the geometry of the three forces acting on the roller
a =ÊËÁ
ˆ¯̃
=-cos .1 92
10023 074
mm
mm∞
and q a= + + = - =90 30 120 23 074 96 926∞ ∞ ∞ ∞ ∞. .
Applying the law of sines to the force triangle,
W P
sin sinq a=
or 196 2
96 926 23 074
.
sin . sin .
N
∞= P
\ =P 77 460. N
or NP = 77 5. 30∞ b
Free-Body Diagram:
Free-Body Diagram:
100 mm
f.b.d
30°
92 mm
N
P
W
120°
WN
P
100 mm P
W
N
αG
30°
92 mm
NW
P
60°θ
α
84
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
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
Reaction at B must pass through D. Free-Body Diagram:
tan
.
tan
.
a
a
b
b
=
= ∞
=
= ∞
175
30 256
175
16 26
mm
300 mm
mm
600 mm
Force triangle
Law of sines T T B
T T
sin . sin . sin .
(sin . ) (
59 744
360
13 996 106 26
13 996
∞= -
∞=
∞ = -
N
∞
3360
0 24185 360 0 86378
N)(sin 59.744 )∞
T T( . ) ( )( . )= -
T = 500 N T = 500 N b
B =
=
(sin .
.
50059 744
555 695
N)sin 106.26
N
∞
B = 556 N 30 3. ∞ b
85
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
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 ; .a a= = ∞120
16036 9
T T B
4
75
3 5=
-=
lb
3 4 300 300
5
4
5
4300 375
T T T
B T
= - =
= = =
;
( )
lb
lb lb B = 375 lb 36.9°
86
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
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.
tan
.
a
a
=
= ∞
160
28 07
mm
300 mm
We note that triangle ABD is isosceles (since AC = BC ) and, therefore
SCAD = = ∞a 28 07.
Also, since CD CB^ , reaction C forms angle a = ∞28 07. with horizontal.
Force triangle
We note that A forms angle 2a with vertical. Thus A and C form angle
180 90 2 90∞ - ∞ - - = ∞ -( )a a a
Force triangle is isosceles and we have
A
C
===
170
2 170
160 0
N
N
N
( )sin
.
a
A = 170 0. N 33.9° C = 160 0. N 28.1° b
87
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
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.
tan
.
a
a
=
= ∞
160
28 07
mm
300 mm
We note that triangle ADB is isosceles (since AC = BC). Therefore S SA B= = ∞ -90 a.
Also S ADB = 2a
Force triangle
The angle between A and C must be 2a a a- =
Thus, force triangle is isosceles and
a = ∞28 07.
A
C
== =
170 0
2 170 300
.
( )cos
N
N Na A = 170 0. N 56.1° C = 300 N 28.1° b
88
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
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
A = 150 0. N 30.0° b
B = 150 0. N 30.0° b
(b) h = 200 mm
tan.
.
b
b
=
= ∞
55 662
25012 552
Force triangle
Law of sines
150
60 102 552433 247
488 31
N
sin17.448N
N
∞=
∞=
∞==
A B
A
B
sin sin ..
.
A = 433 N 12.55° b
B = 488 N 30.0° b
89
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
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 SC = ∞ + ∞ = ∞90 30 120
and S SA D= = ∞ - ∞ = ∞1
2180 120 30( ) Dimensions in mm
On the other hand, from triangle BCF:
CF BC
FD CD CF
= ∞ = ∞ == - = - =
( )sin30 200 30 100
250 100 150
sin mm
mmFrom triangle EFD, and sinceSD = ∞30 :
EF FD= ∞ = ∞ =( ) tan tan .30 150 30 86 60 mm
From triangle EFC: tan
.
a
a
= =
= ∞
CF
EF
100
49 11
mm
86.60 mm
Force triangle
Law of sines F C
F C
AD
AD
sin . sin49 11 60
500
400 458
∞=
∞=
= =
N
sin 70.89
N, N
∞
(a) FAD = 400 N b
(b) C = 458 N 49.1° b
90
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.81
Knowing that q = 30°, determine the reaction (a) at B, (b) at C.
SOLUTION Free-Body Diagram: (Three-Force body)Reaction at C must pass through D where force P and reaction at B intersect.
In DCDE: tan( )
.
b
b
= -
= -= ∞
3 1
3 1
36 2
R
R
Force triangle
Law of sines P B C
B P
C P
sin . sin . sin.
.
23 8 126 2 302 00
1 239
∞=
∞=
∞==
(a) B = 2P 60.0° b
(b) C = 1 239. P 36.2° b
91
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.82
Knowing that q = 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 DCDE: tan
.
b
b
=-
= -
= ∞
R
R
R3
11
322 9
Force triangle
Law of sines P B C
B P
C P
sin . sin . sin.
.
52 9 67 1 601 155
1 086
∞=
∞=
∞==
(a) B = 1 155. P 30.0° b
(b) C = 1 086. P 22.9° b
Free-Body Diagram:(Three-Force body)
92
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
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
Since y x aED ED= = , Free-Body Diagram:
Slope of ED is 45∞
slope of isHC 45∞
Also DE a= 2
and DH HE DEa= = Ê
ËÁˆ¯̃
=1
2 2
For triangles DHC and EHC sin b = =a
R
a
R2
2Now c R= ∞ -sin( )45 b
For a R= =
=
== ∞
20 100
2 100
0 141421
8 1301
mm and mm
20 mm
mmsin
( )
.
.
b
b
and c = ∞ - ∞( )sin( . )100 mm 45 8 1301
= 60 00. mm or c = 60 0. mm b
93
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
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 q in terms of the angle b.
SOLUTION
As shown in the free-body diagram of the slender rod AB, the three forces intersect at C. From the forcegeometry
Free-Body Diagram:
tan b =x
yGB
AB
where
y LAB = cosq
and
x LGB = 1
2sinq
tansin
cos
tan
bqq
q
=
=
12
1
2
L
L
or tan tanq b= 2 b
94
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
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 b = 30°, determine (a) the angle q 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 b =x
yCB
BC
where
x LCB = 1
2sinq
and y LBC =
=
cos
tan tan
q
b q1
2
or tan tanq b= 2
For b = ∞30
tan tan
.
.
q
q
= ∞== ∞
2 30
1 15470
49 107 or q = ∞49 1. b
(b) W mg= = =( )( ) .8 78 480kg 9.81 m/s N2
From force triangle A W== ∞
tan
( . ) tan
b78 480 30N
= 45 310. N or A = 45 3. N ¬ b
and BW= =
∞=
cos
.
cos.
b78 480
3090 621
NN or B = 90 6. N 60 0. ∞ b
95
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
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,
y hBE =
For triangle ACE: S AE h2 2 22= +( ) ( ) (1)
For triangle ABE: L AE h2 2 2= +( ) ( ) (2)
Subtracting Equation (2) from Equation (1)
S L h2 2 23- = (3)
or hS L= -2 2
3 b
As length S increases relative to length L, angle q increases until
rod AB is vertical. At this vertical position:
h L S h S L+ = = -or
Therefore, for all positions of AB
h S L� - (4)
or S L
S L2 2
3
- -�
or S L S L
S SL L
S SL L
2 2 2
2 2
2 2
3
3 2
3 6 3
- -
= - +
= - +
� ( )
( )
or 0 2 6 42 2� S SL L- +
and 0 3 2 22 2� S SL L S L S L- + = - -( )( )
96
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.86 (Continued)
For S L S L- = =0
Minimum value of S is L
For S L S L- = =2 0 2
Maximum value of S is 2L
Therefore, equilibrium does not exist if S L� 2 b
97
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.87
A slender uniform rod of length L = 500 mm is held in equilibrium as shown, with one end against a frictionless wall and the other end attached to a cord of length S = 750 mm. Knowing that the mass of the rod is 4.5 kg, 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,
y hBE =
For triangle ACE: S AE h2 2 22= +( ) ( ) (1)
For triangle ABE: L AE h2 2 2= +( ) ( ) (2)
Subtracting Equation (2) from Equation (1)
S L h2 2 23- =
or hS L= -2 2
3(a) For L S= =500 750mm and mm
h = -
=
( ) ( )
.
750 500
3322 7486
2 2
mm or h = 322 mm b
(b) We have W = ¥ =4 5 9 81 44 145. . . N
and q
q
= ÊËÁ
ˆ¯̃
= ÈÎÍ
˘˚̇
= ∞
-
-
sin
sin( . )
.
1
1
2
2 322 7486
750
59 391
h
s
98
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PROBLEM 4.87 (Continued)
From the force triangle
TW=
=∞
=
sin.
sin ..
q44 145
59 39151 2919
N
N or T = 51 3. N b
(c) BW=
=∞
tan.
tan .
q44 145
59 391
N
= 26 1166. N or B = 26 1. N ¬ b
99
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.88
A uniform rod AB of length 2R rests inside a hemispherical bowl of radius R as shown. Neglecting friction, determine the angle q 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 a of triangle DOA is the central angle corresponding to the inscribed angleq of triangle DCA.
a q= 2
The horizontal projections of AE xAE, ( ), and AG xAG, ( ), are equal.
x x xAE AG A= =
or ( )cos ( )cosAE AG2q q=
and ( )cos cos2 2R Rq q=
Now cos cos2 2 12q q= -
then 4 22cos cosq q- =
or 4 2 02cos cosq q- - =
Applying the quadratic equation
cos . cos .q q= = -0 84307 0 59307and
q q= ∞ = ∞32 534 126 375. . ( )and Discard or q = ∞32 5. b
100
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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 q, 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 DABC and DBGD are similar.
AC AE L= = cosq
In DABC: ( ) ( ) ( )
( cos ) ( sin )
cos sin
CE BE BC
L L R
R
L
2 2 2
2 2 2
22
2
4
+ =
+ =
ÊËÁ
ˆ¯̃ = +
q q
q 22
22 2
22
4 1
3 1
q
q q
q
R
L
R
L
ÊËÁ
ˆ¯̃ = + -
ÊËÁ
ˆ¯̃ = +
cos cos
cos
cos221
31q = Ê
ËÁˆ¯̃ -
È
ÎÍ
˘
˚˙
R
L b
101
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.90
Knowing that for the rod of Problem 4.89, L = 375 mm, R = 500 mm, and W = 45 N, determine (a) the angle q 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
cos221
31q = Ê
ËÁˆ¯̃ -
È
ÎÍ
˘
˚˙
R
LFor L R W= = =375 500 45mm, mm, and N
(a) cos ; .22
1
3
5001 59 39q q=
ÊËÁ
ˆ¯̃
-È
ÎÍÍ
˘
˚˙˙
= ∞mm
375 mm q = ∞59 4. b
In DABC: tansin
costan
tan tan . .
.
a qq
q
a
a
= = =
= ∞ =
= ∞
BE
CE
L
L2
1
21
259 39 0 8452
40 2
Force triangle
A W
BW
= = ∞ =
= =∞
=
tan ( ) tan . .
cos
( )
cos ..
a
a
45 40 2 38 028
45
40 258 916
N N
NN
(b) A = 38 0. N ® b
(c) B = 58 9. N 49.8° b
102
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PROBLEM 4.91
A 1 2. m 2.4 m¥ sheet of plywood of mass 17 kg 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
W = 17 × 9.81 = 166.77 N
r i j kG B/ = + +1 125
2
0 41758
20 3
. ..
We have 5 unknowns and six equations of equilibrium.
Plywood sheet is free to move in z direction, but equilibrium is maintained ( ).SFz = 0
SM r A A r C r wB A B x y C B G B= ¥ + + ¥ - + ¥ - =0 0: ( ) ( ) ( )/ / /i j i j
i j k i j k i j k
0 0 1 2
0
1 125 0 41758 0 6
0 0
0 5625 0 20879 0 3
0
. . . . . . .
A A Cx y
+-
+-1166 77 0
0
.
=
- + - + + - =1 2 1 2 0 6 0 41758 50 031 93 808125 0. . . . . .A A C Cy xi j j k i k
Equating coefficients of unit vectors to zero: i: - + =1 2 50 031 0. .Ay Ay = 41 6925. N
j: - + =0 6 1 2 0. .C Az A Cx = = =1
2
1
2224 64 112 32( . ) . N
k: 0 41758 93 808125 0. .C - = C = 224 64. N C = 225 N b
SFx = 0: A B Cx x+ - = 0: Bx = - =224 64 112 32 112 32. . . N
SFy = 0: A B Wy y+ - = 0: By = - =166 77 41 6925 125 08. . . N
A i j B i j C i= + = + = -( . ( . ) ( . ) ( . ) ( )112 3 41 7 112 3 125 1 225N) N N N N b
103
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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 equations of equilibrium.
SM T D D DA C x y= + ¥ - + + ¥ - + ¥ + +0 90 30 80 210 40 300: ( ) ( ) ( ) ( ) ( ) (i k j i j k i i j zzk) = 0
- + + - + - =7200 2400 210 40 300 300 0k i j i k jT T D DC C y z
Equate coefficients of unit vectors to zero:
i : 2400 40 0- =TC TC = 60 N
j: ( )( )210 300 0 210 60 300 0T D DC z z- = - = Dz = 42 N
k : - + =7200 300 0Dy Dy = 24 N
SFx = 0: Dx = 0
SF A Dy y y= + - =0 80: N 0 Ay = - =80 24 56 N
SF A Dz z z= + - =0 60: N 0 Az = - =60 42 18 N
A j k D j k= + = +( . ) ( . ) ( . ) ( . )56 0 18 00 24 0 42 0N N N N b
104
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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 equations of equilibrium.
SM T D D DA C x y= + ¥ - + + ¥ - + ¥ + +0 90 30 80 210 50 300: ( ) ( ) ( ) ( ) ( ) (i k j i j k i i j zzk) = 0
- + + - + - =7200 2400 210 50 300 300 0k i j i k jT T D DC C y z
Equate coefficients of unit vectors to zero:
i : 2400 50 0- =TC TC = 48 N
j: ( )( )210 300 0 210 48 300 0T D DC z z- = - = Dz = 33 6. N
k : - + =7200 300 0Dy Dy = 24 N
SFx = 0: Dx = 0
SF A Dy y y= + - =0 80: N 0 Ay = - =80 24 56 N
SFz = 0: A Dz z+ - =48 0 Az = - =48 33 6 14 4. . N
A j k D j k= + = +( . ) ( . ) ( . ) ( . )56 0 14 40 24 0 33 6N N N N b
105
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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 50 mm, and the sheave at C has a radius of 40 mm. 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) SM Tx-axis : N N mm N mm= - + - =0 120 90 100 150 80 0( )( ) ( )( ) T = 187 5. N b
(b) SF Bx x= =0 0:
SM BD z y( ) ( )( ) ( )-axis : N 7.5 N mm mm= + - =0 150 18 150 300 0
By = 168 75. N
SM BD y z( ) ( )( ) ( )-axis : N N mm mm= + + =0 120 90 500 300 0
Bz = -350 N
or N NB j k= -( . ) ( )168 8 350 b
SM DB z y( ) ( )( ) ( )-axis : N 7.5 N mm mm= - + + =0 150 18 150 300 0
NDy = 168 75.
SM DB y z( ) ( )( ) ( )-axis : N N mm mm= + + =0 120 90 200 300 0
Dz = -140 N
or N ND j k= -( . ) ( . )168 8 140 0 b
106
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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 equations of equilibrium—OK
SM k i j j k i k i jC x yD D T= - ¥ + + - ¥ + - ¥ - =0 120 120 160 80 200 720: ( ) ( ) ( ) ( ) ( ) 00
- + - - + ¥ + ¥ =120 120 120 160 57 6 10 144 10 03 3D D T Tx yj i k j i k.
Equating to zero the coefficients of the unit vectors:
k : - + ¥ =120 144 10 03T (a) T = 1200 N b
i : 120 57 6 10 0 4803D Dy y+ ¥ = = -. N
j: ( )- - =120 160 1200 0Dx N Dx = -1600 N
SFx = 0: C D Tx x+ + = 0 Cx = - =1600 1200 400 N
SFy = 0: C Dy y+ - =720 0 Cy = + =480 720 1200 N
SFz = 0: Cz = 0
(b) C i j D i j= + = - -( ) ( ) ( ) ( )400 1200 1600 480N N N N b
107
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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-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 equations of equilibrium.
SM k i j j k i k iC x yD D T= - ¥ + + - ¥ + - ¥ -0 120 120 160 80 173 21 720: ( ) ( ) ( ) ( . ) ( jj) = 0
- + - - + ¥ + ¥ =120 120 120 160 57 6 10 124 71 10 03 3D D T Tx yj i k j i k. .
Equating to zero the coefficients of the unit vectors:
k : . .- + ¥ = =120 124 71 10 0 1039 23T T N T = 1039 N b
i : .120 57 6 10 0 4803D Dy y+ ¥ = = - N
j: ( . )- -120 160 1039 2Dx Dx = -1385 6. N
SFx = 0: C D Tx x+ + = 0 Cx = - =1385 6 1039 2 346 4. . .
SFy = 0: C Dy y+ - =720 0 Cy = + =480 720 1200 N
SFz = 0: Cz = 0
(b) C i j D i j= + = - -( ) ( ) ( ) ( )346 1200 1386 480N N N N b
108
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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
r i
r i k
r i k
B A
C A
G A
/
/
/
== += +
0 6
0 8 1 05
0 3 0 6
.
. .
. .
W mg
W
= ==
( ) .
.
18 9 81
176 58
kg
N
SM B C WA B A C A G A= ¥ + ¥ + ¥ - =0 0: ( )r j r j r j/ / /
( . ) ( . . ) ( . . ) ( )0 6 0 8 1 05 0 3 0 6 0i j i k j i k j¥ + + ¥ + + ¥ - =B C W
0 6 0 8 1 05 0 3 0 6 0. . . . .B C C W Wk k i k i+ - - + =
Equate coefficients of unit vectors of zero:
i : . ..
..1 05 0 6 0
0 6
1 05176 58C W C+ = = Ê
ËÁˆ¯̃
=N 100.90 N
k : . . .0 6 0 8 0 3 0B C W+ - =
0 6 0 8 100 90 0 3 176 58 0 46 24. . ( . . ( . ) .B B+ - = = -N) N N
SF A B C Wy = + + - =0 0:
A A- + + = =46 24 121 92. .N 100.90 N 176.58 N 0 N
( ) . ( ) . ( ) .a A b B c C= = - =121 9 46 2 100 9N N N b
109
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 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
r i
r i k
r i k
B A
C A
G A
/
/
/
== += +
0 6
0 65 1 2
0 3 0 6
.
. .
. .
W mg
W
= ==
( ) .
.
18 9 81
176 58
kg m/s
N
2
SM B C WA B A C A G A= ¥ + ¥ + ¥ - =0 0: ( )r j r j r j/ / /
0 6 0 65 1 2 0 3 0 6 0. ( . . ) ( . . ) ( )i j i k j i k j¥ + + ¥ + + ¥ - =B C W
0 6 0 65 1 2 0 3 0 6 0. . . . .B C C W Wk k i k i+ - - + =
Equate coefficients of unit vectors to zero:
i : . ..
..- + = = Ê
ËÁˆ¯̃
=1 2 0 6 00 6
1 2176 58C W C N 88.29 N
k : . . .0 6 0 65 0 3 0B C W+ - =
0 6 0 65 88 29 0 3 176 58 0 7 36. . ( . . ( . ) .B B+ - = = -N) N N
SF A B C Wy = + + - =0 0:
A A- + - = =7 36 95 648. .N 88.29 N 176.58 N 0 N
( ) . ( ) . ( ) .a A b c= -95 6 7 36 88 3N N N b
110
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PROBLEM 4.99
The rectangular plate of mass 40 kg and is supported by three vertical wires. Determine the tension in each wire.
SOLUTION
Free-Body Diagram: W = 40 × 9.81 N = 392.4 N
SM r j r j jB A B A C B C G BT T r= ¥ + ¥ + ¥ - =0 392 4 0: ( . )/ / / N
( ) [( ( ) ] [( ) ( ) ]1500 1500 375 750 750mm mm) mm mm mmk j i k j i k¥ + + ¥ + +T TA C ¥¥ - =( . )392 4 0N j
- + - - + =1500 1500 375 294300 294300 0T T TA C Ci k i k i
Equating to zero the coefficients of the unit vectors:
k : 1500 294300 0 196 2T TC C- = fi = . N TC = 196 2. N b
i : - - + = fi =1500 375 196 2 294300 0 147 15T TA A( . ) . N TA = 147 2. N b
SFy = 0: T T TA B C+ + - =392 4 0. N
147 15 196 2 392 4 0 49 05. . . .N N+ + - = fi =T TB B TB = 49 1. N b
111
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.100
The rectangular plate of mass 40 kg 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: W = 40 × 9.81 = 392.4 N
Let -Wb j be the weight of the block and x and z the block’s coordinates.
Since tensions in wires are equal, let
T T T TA B C= = =
SM T T T W x z WA B C G b0 0 0= ¥ + ¥ + ¥ + ¥ - + + ¥ - =: ( ) ( ) ( ) ( ) ( ) ( )r j r j r j r j i k j
or, ( ) ( ) ( ) ( ) ( ) (1875 375 1500 750 750 1125k j k j i k j i k j¥ + ¥ + + ¥ + + ¥ - +T T T W xii k j+ ¥ - =z Wb) ( ) 0
or, - - + - - + - + =1875 375 1500 750 750 1125 0T T T T W W xW zWb bi i k i k i k i
Equate coefficients of unit vectors to zero:
i : - + + =3000 1125 0T W zWB (1)
k : 1500 750 0T W xWB- - = (2)
Also, SFy = 0: 3 0T W Wb- - = (3)
Eq. (1) + 1000 Eq. (3): 125 1000 0W z Wb+ - =( ) (4)
Eq. (2) – 500 Eq. (3): - - - =250 500 0W x Wb( ) (5)
112
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.100 (Continued)
Solving (4) and (5) for W Wb / and recalling of 0 1500� �x mm, 0 2250� �z mm,
Eq.(4): W
W zb =
- -=125
1000
125
1000 00 125� .
Eq.(5): W
W xb =
- -=250
500
250
500 00 5� .
Thus, ( ) . . ( . ) .minW Wb = = =0 5 0 5 392 4 196 2N N ( ) .minWb = 196 2 N b
Making W Wb = 0 5. in (4) and (5):
125 1000 0 5 0W z W+ - =( )( . ) z = 750 mm b
- - - =250 500 0 5 0W x W( )( . ) x = 0 mm b
113
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 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
W m g
W m g1
2
0 6
1 2
= ¢= ¢
.
.
SM T W W TD A D A E D F D C D C= ¥ + ¥ - + ¥ - + ¥ =0 01 2: ( ) ( )r j r j r j r j/ / / /
( . . ) ( . . ) ( ) . ( ) .- + ¥ + - + ¥ - + ¥ - + ¥ =0 4 0 6 0 4 0 3 0 2 0 81 2i k j i k j i j i jT W W TA C 00
- - + + - + =0 4 0 6 0 4 0 3 0 2 0 8 01 1 2. . . . . .T T W W W TA A Ck i k i k k
Equate coefficients of unit vectors to zero:
i : . . ; . .- + = = = ¢ = ¢0 6 0 3 01
2
1
20 6 0 31 1T W T W m g m gA A
k : . . . .- + - + =0 4 0 4 0 2 0 8 01 2T W W TA C
- ¢ + ¢ - ¢ + =0 4 0 3 0 4 0 6 0 2 1 2 0. ( . ) . ( . ) . ( .m g m g m g TC) 0.8
Tm g
m gC = - - ¢ = ¢( . . . )
.0 12 0 24 0 24
0 150.8
SF T T T W Wy A C D= + + - - =0 01 2:
0 3 0 15 0 6 1 2 0
1 35
. . . .
.
¢ + ¢ + - ¢ - ¢ == ¢
m g m g T m g m g
T m gD
D ¢ = =m g (8 kg/m)(9.81m/s ) N/m2 78 48.
T m gA = ¢ = ¥0 3 0 3 78 45. . . TA = 23 5. N b
T m gB = ¢ = ¥0 15 0 15 78 45. . . TB = 11 77. N b
T m gC = ¢ = ¥1 35 1 35 78 45. . . TC = 105 9. N b
114
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 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
W m g
W m g1
2
0 6
1 2
= ¢= ¢
.
.
SM T W W TD A D A E D F D C D C= ¥ + ¥ - + ¥ - + ¥ =0 01 2: ( ) ( )r j r j r j r j/ / / /
( . ) ( . ) ( ) ( . ) ( ) ( . )- + ¥ + - + ¥ - + - ¥ - + - ¥a T a W a W aAi k j i k j i j i0 6 0 3 0 6 1 21 2 TTC j = 0
- - + + - - + - =T a T W a W W a T aA A Ck i k i k k0 6 0 3 0 6 1 2 01 1 2. . ( . ) ( . )
Equate coefficients of unit vectors to zero:
i : . . ; . .- + = = = ¢ = ¢0 6 0 3 01
2
1
20 6 0 31 1T W T W m g m gA A
k : ( . ) ( . )- + - - + - =T a W a W a T aA C1 2 0 6 1 2 0
- ¢ + ¢ - ¢ - + - =0 3 0 6 1 2 1 2 0. . . ( . )m ga m ga m g a T aC(0.6 )
Ta a a
aC = - + --
0 3 0 6 1 2 0 6
1 2
. . . ( . )
.For Max a and no tipping, TC = 0
(a) - + - =- + - =0 3 1 2 0 6 0
0 3 0 72 1 2 0
. . ( . )
. . .
a a
a a
1 5 0 72. .a = a = 0 480. m b
115
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.102 (Continued)
(b) Reactions: ¢ = =m g ( ) . .8 9 81 78 48kg/m m/s N/m2
T m gA = ¢ = ¥ =0 3 0 3 78 48 23 544. . . . N TA = 23 5. N b
SF T T T W Wy A C D= + + - - =0 01 2:
T T m g m gA D+ + - ¢ - ¢ =0 0 6 1 2 0. .
T m g TD A= ¢ - = ¥ - =1 8 1 8 78 48 23 544 117 72. . . . . TD = 117 7. N b
116
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.
SOLUTION
r i k
r i k
r i k
B A
C A
G A
a
a/
/
/
= += += +
750
750
375 375
By symmetry: B C=
SM B C WA B A C G A= ¥ + ¥ + ¥ - =0 0: ( )r j r j r j/ /
( ) ( ) ( ) ( )a B a B Wi k j i k j i k j+ ¥ + + ¥ + + ¥ - =750 750 375 375 0
Ba B B Ba W Wk i k i k i- + - - + =750 750 375 375 0
Equate coefficient of unit vector i to zero:
i : - - + =750 375 0B Ba W
BW
aC B
W
a=
+= =
+375
750
375
750 (1)
SF A B C Wy = + + - =0 0:
AW
aW A
aW
a+
+ÈÎÍ
˘˚̇
- = =+
2375
7500
750; (2)
PROBLEM 4.103
The 120 N square plate shown is supported by three vertical wires. Determine (a) the tension in each wire when a = 250 mm, (b) the value of a for which the tension in each wire is 40 N.
117
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.103 (Continued)
(a) For a = 250 mm, W = 120 N
Eq. (1) C B= = =375 120
100045
¥N
Eq. (2) A = =250 120
100030
( )N A B C= = =30 0 45 0. .N N b
(b) For tension in each wire = 40 N
Eq. (1) 40375 120
750N =
+¥
( )a
750 1125mm mm+ =a a = 375 mm b
118
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.104
The table shown weighs 150 N and has a diameter of 1.2 m. It is supported by three legs equally spaced around the edge. A vertical load P of magnitude 500 N 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 b r= = ∞ =0 6 30 0 3. sin .m m
We shall sum moments about AB.
( ) ( )b r C a b P bW+ + - - = 0
( . . ) ( . ) ( . )0 3 0 6 0 3 500 0 3 150 0+ + - - =C a
C a= - -1
0 945 0 3 500
.[ ( . ) ]
If table is not to tip, C � 0
[ ( . ) ]
( . )
45 0 3 500 0
45 0 3 500
- --
a
a
�
�
a a a- =0 3 0 09 0 39 0 39. . . .� � m m
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
119
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.105
A 3-m boom is acted upon by the 3.4-kN 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 equations of equilibrium but equilibrium is maintained ( ).SMx = 0
Free-Body Diagram:
BD BD
BE
u ruu
= - + + == - + +
( . ) ( . ) ( . ) .
( . ) ( . )
1 8 2 1 1 8 3 3
1 8 2 1
m m m m
m m
i j k
i j (( . ) .
.( . . . )
1 8 3 3
3 31 8 2 1 1 8
m mk
i j k
BE
T TBD
BD
T TBD BD
BD BD
=
= = - + + =u ruu
1116 7 6
3 31 8 2 1 1 8
1
( )
.( . . . )
- + +
= = - + - =
i j k
i j kT TBE
BE
T TBE BE
BE BE
u ruu
116 7 6( )- + -i j k
SM T TA B BD B BE C= ¥ + ¥ + ¥ - =0 3 4 0: ( . )r r r j
1 811
6 7 6 1 811
6 7 6 3 3 4 0. ( ) . ( ) ( . )i i j k i i j k i j¥ - + + + ¥ - + - + ¥ - =T TBD BE
12 6
11
10 8
11
12 6
11
10 8
1110 2 0
. . . ..T T T TBD BD BE BEk j k j k- + + - =
Equate coefficients of unit vectors to zero.
j:. .- + = =10 8
11
10 8
110T T T TBD BE BE BD
k :. .
.12 6
11
12 6
1110 2 0T TBD BE+ - =
212 6
1110 2
..TBD
ÊËÁ
ˆ¯̃
= TBD = 4 45. kN b
TBD
= 4.4524 kN TBE = 4 45. kN b
120
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PROBLEM 4.105 (Continued)
SF Ax x= - - =06
114 4524
6
114 4524 0: ( . ) ( . )kN kN
Ax = 4 8572. kN
SF Ay y= + + - =07
114 4524
7
114 4524 3 4 0: ( . ) ( . ) .kN kN kN
Ay = -2 2667. kN
SF Az z= + - =06
114 4524
6
114 4524 0: ( . ) ( . )kN kN
Az = 0 A i j= -( . ) ( . )4 86 2 27kN kN b
121
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 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 equations of equilibrium. Equilibrium is maintained ( ).SMAC = 0
r k
r kB
A
==
1 2
2 4
.
.
AD AD
AE AE
u ruu
u ruu
= - + - =
= + - =
0 8 0 6 2 4 2 6
0 8 1 2 2 4 2 8
. . . .
. . . .
i j k
i j k
m
m
T
AD
AD
T
TAE
AE
T
ADAD
AEAE
= = - + -
= =
u ruu
u ruu
2 60 8 0 6 2 4
2 80
.( . . . )
.( .
i j k
88 1 2 2 4i j k+ -. . )
SMC A AD A AE B= ¥ + ¥ + ¥ - =0 3 0: ( )r T r T r jkN
i j k i j k
k0 0 2 4
0 8 0 6 2 42 6
0 0 2 4
0 8 1 2 2 42 8
1 2.
. . ..
.
. . ..
. (
- -+
-+ ¥ -
T TAD AE 33 6 0. )kN j =
Equate coefficients of unit vectors to zero.
i
j
: . . .
: . .
- - + =- + =
0 55385 1 02857 4 32 0
0 73846 0 68671 0
T T
T TAD AE
AD AE
(1)
T TAD AE= 0 92857. (2)
122
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.106 (Continued)
Eq. (1): - - + =0 55385 0 92857 1 02857 4 32 0. ( . ) . .T TAE AE
1 54286 4 32
2 800
. .
.
T
TAE
AE
== kN TAE = 2 80. kN b
Eq. (2): TAD = =0 92857 2 80 2 600. ( . ) . kN TAD = 2 60. kN b
S
S
F C C
F C
x x x
y y
= - + = =
= +
00 8
2 62 6
0 8
2 82 8 0 0
00 6
2 6
:.
.( . )
.
.( . )
:.
.(
kN kN
22 61 2
2 82 8 3 6 0 1 800
02 4
2 62
. ).
.( . ) ( . ) .
:.
.(
kN kN kN kN+ - = =
= -
C
F C
y
z zS .. ).
.( . ) .6
2 4
2 82 8 0 4 80kN kN kN- = =Cz
C j k= +( . ) ( . )1 800 4 80kN kN b
123
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 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 equations of equilibrium. Equilibrium is maintained ( ).S MAC = 0
AD AD
AE AE
u ruu
u ruu
= - + - =
= + - =
0 8 0 6 2 4 2 6
0 8 1 2 2 4 2 8
. . . .
. . . .
i j k
i j k
m
m
T
AD
AD
T
TAE
AE
T
ADAD
AEAE
= = - + -
= =
u ruu
u ruu
2 60 8 0 6 2 4
2 80
.( . . . )
.( .
i j k
88 1 2 2 4i j k+ -. . )
SMC A AD A AE A= ¥ + ¥ + ¥ -0 3 6: ( . )r T r T r jkN
Factor rA: r T T jA AD AE¥ + -( ( . ) )3 6 kN
or: T T jAD AE+ - =( )3 0kN (Forces concurrent at A)
Coefficient of i: - + =T TAD AE
2 60 8
2 80 8 0
.( . )
.( . )
T TAD AE= 2 6
2 8
.
. (1)
Coefficient of j: T T
T
AD AE
AE
2 60 6
2 81 2 3 6 0
2 6
2 8
0 6
2 6
1 2
2
.( . )
.( . ) .
.
.
.
.
.
.
+ - =
ÊËÁ
ˆ¯̃
+
kN
883 6 0
0 6 1 2
2 83 6
T
T
AE
AE
- =
+ÊËÁ
ˆ¯̃
=
.
. .
..
kN
kN
TAE = 5 600. kN TAE = 5 60. kN b
124
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.107 (Continued)
Eq. (1): TAD = =2 6
2 85 6 5 200
.
.( . ) . kN TAD = 5 20. kN b
S
S
F C C
F C
x x x
y y
= - + = =
= +
00 8
2 65 2
0 8
2 85 6 0 0
00 6
2 6
:.
.( . )
.
.( . ) ;
:.
.
kN kN
(( . ).
.( . ) .
:.
.( . )
5 21 2
2 85 6 3 6 0 0
02 4
2 65 2
2
kN kN kN
kN
+ - = =
= - -
C
F C
y
z zS ..
.( . ) .
4
2 85 6 0 9 60kN kN= =Cz
C k= ( . )9 60 kN b
Note: Since forces and reaction are concurrent at A, we could have used the methods of Chapter 2.
125
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PROBLEM 4.108
A 300-kg crate hangs from a cable that passes over a pulley B and is attached to a support at H. The 100-kg 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:
W
WC
G
= ¥ == ¥ =
300 9 81 2943
100 9 81 981
.
.
N
N
We have five unknowns ( , , , , )T T A A ADE DF x y z and five equilibrium equations. The boom is free to spin about the AB axis, but equilibrium is maintained, since SM AB = 0.
We have BH BH
DE
u vuu
u vuu
= - =
= -
( . ) ( ) .
( . ).
.(
9 15 7 11 5205
4 152 65
3 657
m m m
m
i j
i mm m
m m m m
) ( )
( . ) ( . ) ( ) .
( .
j k
i j k
+
= - + =
=
2
4 15 5 0822 2 6 8594
4 15
DE
DFu vuu
mm m m m) ( . ) ( ) .i j k- - =5 0822 2 6 8594DF
126
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.108 (Continued)
Thus: T Ti j
iBH BHBH
BH= = - = -
u vuu
( ).
.( . ) ( .2943
9 15 7
11 52052337 438 1788N N 2204
6 85944 15 5 0822 2
N)
.( . . )
j
T T i j k
T T
DE DEDE
DF DF
DE
DE
T
D
= = - +
=
u vuu
FF
DF
TDF
u vuu
= - -6 8594
4 15 5 0822 2.
( . . )i j k
(a) SM r W r W r T r T r T
iA J C K G H BH E DE F DF= ¥ + ¥ + ¥ + ¥ + ¥ =
-0 0
3 65
: ( ) ( ) ( ) ( ) ( )
( . ) ¥¥ - - ¥ - + ¥ -( ) ( . ) ( ) ( . ) ( . . )2943 1 825 981 5 5 2337 438 1788 204j i j i i j
+-
+ --
T TDE DF
6 85941 5 0 2
4 15 5 0822 26 8594
1 5 0 2
4 15 5 08.
.
. ..
.
. .
i j k i j k
222 2
0
-=
or, 10741 95 1790 325 9835 1222 5 0822
6 8594. . . ( )
.
.k k k i+ - + -
¥( )T TDE DF
+ - - + =5 3
6 8594
7 6233
6 85940
.
.( )
.
.( )T T T TDE DF DE DFj k
Equating to zero the coefficients of the unit vectors:
i or j: T T T TDE DF DE DF- = =0 *
k : . . ..
.( )10741 95 1790 325 9835 122
7 6233
6 85942 0+ - - =TDE TDE = 1213 441. N
T TDE DF= = 1213 N b
(b) SF A Ax x x= + + ÊËÁ
ˆ¯̃
= = -0 2337 438 24 15
6 85941213 441 0 3805 72: .
.
.( . ) . NN
SF Ay y= - - - - ÊËÁ
ˆ¯̃
0 2943 981 1788 204 25 0822
6 85941213 441: .
.
.( . ) == =0 7510 31Ay . N
SF Az z= =0 0: A i j= - +( ) ( )3810 7510N N b
*Remark: The fact that T TDE DF= could have been noted at the outset from the symmetry of structure with respect to xy plane.
127
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.
SOLUTION
Free-Body Diagram:
By symmetry with xy plane
T T T
CD
CD
CD CE= =
= - + +=
u ruu
3 1 5 1 2
3 562
i j k. .
. m
T T
CD
CDT
T T
CD
CE
= = - + +
= - + -
u vuu
3 1 5 1 2
3 5623 1 5 1 2
3 562
i j k
i j k
. .
.. .
.
r i r iB A C A/ /= =2 3
SMA C A CD C A CE B A= ¥ + ¥ + ¥ - =0 5 0: ( )r T r T r j/ / / kN
i j k i j k i j k
3 0 0
3 1 5 1 23 562
3 0 0
3 1 5 1 23 562
2 0 0
0 5 0
0
-+
- -+
-=
. ..
. ..
T T
Coefficient of k: 2 3 1 53 562
10 0 3 958¥ ¥ÈÎÍ
˘˚̇
- = =..
.T
T kN
SF A T TCD CE= + + - =0 5 0: j
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 ( ),f = 0 determine (a) the tension in cables CD and CE, (b) the reaction at A.
128
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.109 (Continued)
Coefficient of k: Az = 0
Coefficient of i: A Ax x- ¥ = =2 3 958 3 3 562 0 6 67[ . . ] ./ kN
Coefficient of j: A Ay y+ ¥ - = =2 3 958 1 5 3 562 5 0 1 667[ . . . ] ./ kN
(a) T TCD CE= = 3 96. kN
(b) A i j= +( . ) ( . )6 67 1 667kN kN b
129
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 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 f = ∞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 ismaintained ( )SM AC = 0
r i
r iB A
C A
/
/
==
2
3Load at B. = - +
= - +( cos ) ( sin )
. .
5 30 5 30
4 33 2 5
j k
j k
CD CD
TCD
CD
TCD CD
u ruu
u ruu
= - + + =
= = - +
3 1 5 1 2 3 562
3 5623 1
i j k
T i
. . .
.( .
m
55 1 2j k+ . )
Similarly, T i j kCET= - + -
3 5623 1 5 1 2
.( . . )
SMA C A CD C A CE B A= ¥ + ¥ + ¥ - + =0 4 33 2 5 0: ( . . )r T r T r j k/ / /
i j k i j k i j k
3 0 0
3 1 5 1 23 562
3 0 0
3 1 5 1 23 562
2 0 0
0 4 3-+
- -+
-. ..
. ..
.
T TCD CE
33 2 5
0
.
=
Equate coefficients of unit vectors to zero.
j: ..
..
- + - =3 63 562
3 63 562
5 0T TCD CE
- + - =3 6 3 6 17 810 0. . .T TCD CE (1)
130
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.110 (Continued)
k : ..
..
.4 53 562
4 53 562
8 66 0T TCD CE+ - =
4 5 4 5 30 846. . .T TCD CE+ = (2)
( ) . ( ):2 1 25 1+ 9 53 11 0 5 901T TCE CE- = =. . kN
Eq. (1): - + - =3 6 3 6 5 901 17 810 0. . ( . ) .TCD
TCD = 0 954. kN
SF CD CE= + + - + =0 4 33 2 5 0: . .A T T j k
i :.
.( )
.
.( )Ax + - + - =0 954
3 5623
5 901
3 5623 0
Ax = 5 77. kN
j:.
.( . )
.
.( . ) .Ay + + - =0 954
3 5621 5
5 901
3 5621 5 4 33 0
Ay = 1 443. kN
k :.
.( . )
.
.( . ) .Az + + - + =0 954
3 5621 2
5 901
3 5621 2 2 5 0
Az = -0 833. kN
Answers:
(a) TCD = 0 954. kN TCE = 5 90. kN b
(b) A i j k= + -( . ) ( . ) ( . )5 77 1 443 0 833kN kN kN b
131
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.111
A 1200 mm 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 equations of equilibrium but equilibrium is maintained ( ).S MAC = 0
T = Tension in both parts of cable DAE.
r k
r kB
A
==
750
1200
mm
mm
AD AD
AE AE
BF
u ruu
u ruu
u ru
= - - =
= - =
500 1200 1300
500 1200 1300
i k
j k
mm
mmuu
= - =400 750 850i k mmBF
T i k i k
T
AD
AE
TAD
AD
T T
TAE
AE
= = - - = - -
=
u ruu
u ruu
1300500 1200
135 12( ) ( )
== - = -
= =
T T
TBF
BF
TBF BF
BF
1300500 1200
135 12
850400
( ) ( )
(
j k j k
T i
u ruu
-- = -75017
8 15k i k) ( )TBF
SM r T r T r T r jC A AD A AE B BF B= ¥ + ¥ + ¥ + ¥ - =0 1600 0: ( )N
i j k i j k i j k
0 0 1200
5 0 1213
0 0 1200
0 5 1213
0 0 750
8 0 1517
7
- -+
-+
-+T T TBF ( 550 1600 0k j) ( )¥ - =
Coefficient of i: - + = =6000
131 200 000 0 2600T T, , N
132
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.111 (Continued)
Coefficient of j: - + =6000
13
6000
170T TBD
T T TBF BF= = =17
13
17
132600 3400( ) N
SF T T T j= + + - + =0 1600 0: AD AE BF C
Coefficient of i: - + + =5
132600
8
173400 0( ) ( ) Cx
Cx = -600 N
Coefficient of j: 5
132600 1600 0( ) - + =Cy
Cy = +600 N
Coefficient of k: - - - + =12
132600
12
132600
15
173400 0( ) ( ) ( ) Cz
Cz = 7800 N
Answers: T TDAE = TDAE = 2600 N b
TBD = 3400 N b
C i j k= - + +( ) ( ) ( )600 600 7800N N N b
133
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.112
Solve Problem 4.111, assuming that the 1600 N load is applied at A.
PROBLEM 4.111 A 1200 mm 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 equations of equilibrium but equilibrium ismaintained ( ).SMAC = 0
T = tension in both parts of cable DAE.
r k
r kB
A
==
750
1200
mm
mm
AD AD
AE AE
BF
u ruu
u ruu
u ru
= - - =
= - =
500 1200 1300
500 1200 1300
i k
j k
mm
mmuu
= - =400 750 850i k BF mm
T i k i k
T
AD
AE
TAD
AD
T T
TAE
AE
= = - - = - -
=
u ruu
u ruu
1300500 1200
135 12( ) ( )
== - = -
= =
T T
TBF
BF
TBF BF
BF
1300500 1200
135 12
850400
( ) ( )
(
j k j k
T i
u ruu
-- = -75017
8 15k i k) ( )TBF
SMC A AD A AE B BF A= ¥ + ¥ + ¥ + ¥ - =0 1600 0: ( )r T r T r T r jN
i j k i j k i j k
0 0 1200
5 0 1213
0 0 1200
0 5 1213
0 0 750
8 0 1517
12
- -+
-+
-+T T TBF 000 1600 0k j¥ - =( )
Coefficient of i: - + = =6000
131 920 000 0 4160T T, , N
134
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.112 (Continued)
Coefficient of j: - + =6000
13
6000
170T TBD
T T TBD BD= = =17
13
17
134160 5440( ) N
SF T T T j C= + + - + =0 1600 0: AD AE BF
Coefficient of i: - + + =5
134160
8
175440 0( ) ( ) Cx
Cx = -960 N
Coefficient of j: 5
134160 1600 0( ) - + =Cy
Cy = 0
Coefficient of k: - - - + =12
134160
12
134160
15
175440 0( ) ( ) ( ) Cz
Cz = 12480 N
Answers: T TDAE = TDAE = 4160 N b
TBD = 5440 N b
C i k= - +( ) ( )960 12480N N b
135
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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 i j
F i j
= ∞ + ∞= +
= =
F F
F
W mg
(sin ) (cos )
( . . )
( .
75 75
0 2588 0 9659
20 9 81kg m//s N2
/
/
/
) .
.
. .
. .
( .
=== += +=
196 2
0 6
0 9 0 6
0 45 0 3
0 25
r k
r i k
r i k
F
A B
C B
G B
F 888 0 9659i j+ . )
SM r j r F r AB G B C B A B= ¥ - + ¥ + ¥ =0 196 2 0: ( . )/ / /
i j k i j k i j k
0 45 0 0 3
0 196 2 0
0 9 0 0 6
0 2588 0 9659 0
0 0 0 6. .
.
. .
. .
.
-+
++F
A Ax y 00
0=
Coefficient of i: + - - =58 86 0 5796 0 6 0. . .F Ay (1)
Coefficient of j: + + =0 1553 0 6 0. .F Ax (2)
Coefficient of k: - + = =88 29 0 8693 0 101 56. . : .F F N
Eq. (2): + - - = =58 86 0 5796 101 56 0 6 0 0. . ( . ) . A Ay y
Eq. (3): + + = = -0 1553 101 56 0 6 0 26 29. ( . ) . .A Ax x N
F = 101 6. N A i= -( . )26 3 N b
SF A B F j: + + - =W 0
Coefficient of i: 26 29 0 2588 101 56 0 0. . ( . )+ + = =B Bx x
Coefficient of j: B By y+ - = =0 9659 101 56 196 2 0 98 1. ( . ) . . N
Coefficient of k: Bz = 0 B j= ( . )98 1 N b
136
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.
SOLUTION
Free-Body Diagram:
W = −mgj = −(30 kg)(9.81 m/s2)j = −(294 N)j
DC DC
TDC
DCT
u ruu
u ruu
= - + - =
= = -
( ( (480 240 160 560
67
mm) mm) mm) mmi j k
T ii j k+ -37
27T T
Equilibrium Equations:
SF i j k i j T j
i
= + + + + + - =
+ - + + +
0 294 0
67
37
: ( )
( ) (
A A A B B
A B T A B
x y z x y
x x y y
N
TT A Tz- + - =294 027N) ( )j k (1)
S SM r FB = ¥ =( :) 0
2 2 29467
37
27r A A A r r T T T r rx y zk i j k i k i j k i k¥ + + + + ¥ - + - + + ¥ -( ) ( ) ( ) ( ) ( NN
N) N
)
( ( ) ( )
j
i j k
=
- - + + - + - =
0
2 294 2 294 037
27
67A T r A T r T ry x (2)
Setting the coefficients of the unit vectors equal to zero in Eq. (2)
Ax = +49.0 N A
y = +73.5 N T
= 343 N b
Setting the coefficients of the unit vectors equal to zero in Eq. (1), we obtain three more scalar equations. After substituting the values of T, A
x and A
y into these equations we obtain
Az = +98.0 N B
x = +245 N B
y = +73.5 N
The reactions at A and B therefore
A = +(49.0 N)i + (73.5 N)j + (98.0 N)k b
B = +(245 N)i + (73.5 N)j b
PROBLEM 4.114
A uniform pipe cover of radius r = 240 mm and mass 30 kg is held in a horizontal position by the cable CD. Assuming that the bearing at B does not exert any axial thrust, determine the tension in the cable and the reaction at A and B.
r = 240 mm
r = 240 mm160 mm
80 mm
240 mm
r = 240 mm
W = – (294 N) j
T
D
A
C
Axi
Azkz
Bxi
Ayj
Byj
x
B
y
137
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 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 Dimensions in mm
r i i
r i k
i k
r
B A
G A
C A
/
/
/
( )960 180 780
960
290
450
2
390 225
- =
= -ÊËÁ
ˆ¯̃
+
= +== +600 450i k
T = Tension in cable DCE
CD CD
CE CE
u ruu
u ruu
= - + - =
= + - =
690 675 450 1065
270 675 450 855
i j k
i j k
mm
mmm
T i j k
T i j k
W i
CD
CE
T
T
mg
= - + -
= + -
= - = -
1065690 675 450
855270 675 450
( )
( )
(( )( . ) ( )100 9 81 981kg m/s N2 j j= -
SM r T r T r j r BA C A CD C A CE G A B AW= ¥ + ¥ + ¥ - + ¥ =0 0: ( )/ / / /
i j k i j k
i j k
600 0 450
690 675 4501065
600 0 450
270 675 450855
39
- -+
-+
+
T T
00 0 225
0 981 0
780 0 0
0
0
-+ =
i j k
B By z
138
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.115 (Continued)
Coefficient of i: - - + ¥ =( )( ) ( )( ) .450 6751065
450 675855
220 725 10 03T T
T = 344 6. N T = 345 N b
Coefficient of j: ( ).
( ).- ¥ + ¥ + ¥ + ¥ -690 450 600 450
344 6
1065270 450 600 450
344 6
855780Bzz = 0
Bz = 185 49. N
Coefficient of k: ( )( ).
( )( ).
.600 675344 6
1065600 675
344 6
855382 59 10 7803+ - ¥ + =B By y 1113 2. N
B j k= +( . ) ( . )113 2 185 5N N b
SF A B T T W= + + + + =0 0: CD CE
Coefficient of i: Ax - + =690
1065344 6
270
855344 6 0( . ) ( . ) Ax = 114 4. N
Coefficient of j: A Ay y+ + + - = =113 2675
1065344 6
675
855344 6 981 0 377. ( . ) ( . ) N
Coefficient of k: Az + - - =185 5450
1065344 6
450
855344 6 0. ( . ) ( . ) Az = 141 5. N
A i j k= + +( . ) ( ) ( . )114 4 377 144 5N N N b
139
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 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
CE
==
1065
855
mm
mm
Now: T i j k
T i j k
W i
CD
CE
T
T
mg
= - + -
= + -
= - = -
1065690 675 450
855270 675 450
( )
( )
(( ) (100 981kg)(9.81 m/s N)2 j j= -
SM r r j rA C A CE G A B AT W B= ¥ + ¥ - + ¥ =0 0: / / /( )
i j k i j k i j k
600 0 450
270 675 450855
390 0 225
0 981 0
780 0 0
0
0
-+
-+ =T
B By z
Coefficient of i: - + ¥ =( )( ) .450 675855
220 725 10 03T
T = 621 3. N T = 621 N b
Coefficient of j: ( ).
.270 450 600 450621 3
855980 0 364 7¥ + ¥ - = =B Bz z N
Coefficient of k: ( )( ).
. .600 675621 3
855382 59 10 780 0 113 23- ¥ + = =B By y N
B j k= +( . (113 2 365 N) N) b
140
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.116 (Continued)
SF A B T W= + + + =0 0: CE
Coefficient of i: A Ax x+ = = -270
855621 3 0 196 2( . ) . N
Coefficient of j: A Ay y+ + - = =113 2675
855621 3 981 0 377 3. ( . ) . N
Coefficient of k: Az + - =364 7450
855621 3 0. ( . ) Az = -37 7. N
A i j k= - + -( . ( ( .196 2 377 37 7 N) N) N) b
141
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PROBLEM 4.117
The rectangular plate shown weighs 40 kg 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
Weight of plate = 392.4 N
r i i
r i k
i k
r
B A
E A
G A
/
/
/
= - == - += +
=
( )
( )
1000 200 800
750 100 500
650 500
10000
2250
500 250
250 625 500
838 525
i k
i k
i j k
T
+
= +
= + -=
=
EF
EF
TA
u ruu
. mm
FF
AFT
TA E A G A
u ruu
( . . . )
(
0 29814 0 74536 0 59629
0 3
i j k
M r r
+ -
= ¥ + ¥ -S : / / 992 4 0. )j r+ ¥ =B A B/
i j k i j k i j k
650 0 500
0 29814 0 74536 0 59629
500 0 250
0 392 4 0
80
. . . .-+
-+T 00 0 0
0
0
B By z
=
Coefficient of i: - + = fi =372 68 98100 0 263 23. .T T N T = 263 N b
Coefficient of j: ( . . )( . ) :387 5885 149 07 263 23 800+ - =Bz 176.581 N
Coefficient of k: ( . )( . ) :484 484 263 23 196200 800 0- + =By By = 85 837. kN
B j k= +( . ( .85 8 176 6 N) N) b
142
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PROBLEM 4.117 (Continued)
SF A B T j= + + - =0 392 4: N) 0( .
Coefficient of i: A Ax x+ = = -0 29814 263 23 0 78 479. ( . ) . N
Coefficient of j: A Ay y+ + - = =85 837 0 74536 263 23 392 4 0 110 362. . ( . ) . . N
Coefficient of k: A Az z+ - = = -176 581 0 59629 263 23 0 19 6196. . ( . ) . N
A i j k= - + -( . ( . ( .78 5 110 4 19 62 N) N) N) b
143
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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 40 kg 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
r i i
r i kB A
E A
/
/
= - == +
( )1000 200 800
650 500
r i kG A/ = +500 250
EH
EH
u ruu
= - + -=
750 300 500
950
i j k
mm
T i j k= = - + -TEH
EH
Tu ruu
950750 300 500( )
SM r T r r BA E A G A B A= ¥ + ¥ - + ¥ =0 392 4 0: / / /( . )
i j k i j k i j k
650 0 500
750 300 500950
500 0 250
0 392 4 0
800 0 0
0- -+
-+T
B By z.
== 0
Coefficient of i: - + =( )( ) ( . )( )300 500950
392 4 250 0T
T = 621.3 N T = 621N b
Coefficient of j: ( ).
325000 375000621 3
950800 0-
( )- =Bz Bz = - 40 875. N
Coefficient of k: ( )( )( . )
( )( . )650 300621 3
950500 392 4 800 0- + =By By = 85 8375. N
B j k= -( . ( .85 8 40 9N) N) b
144
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PROBLEM 4.118 (Continued)
SF = 0: A B T j+ + - =( .392 4 N) 0
Coefficient of i: A Ax x- = =750
950621 3 0 490 5( . ) . N
Coefficient of j: A Ay y+ + - = =85 8375300
950621 3 392 4 0 110 3625. ( . ) . . N
Coefficient of k: A Az z- - = =40 875500
950621 3 0 367 875. ( . ) . N
A i j k= + +( ( . (491 110 4 368 N) N) N) b
145
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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: DABH is Equilateral
Dimensions in mm
r i j
r i k
T j k j
H C
F C
T T T
/
/
= - += += ∞ - ∞ =
50 250
350 250
30 30 0 5(sin ) (cos ) ( . -- 0 866. )k
SM r j r j kC F C H C C y C zT M M= ¥ - + ¥ + + =0 400 0: / /( ) ( ) ( )
i j k i j k
j k350 0 250
0 400 0
50 250 0
0 0 5 0 866
0
-+ -
-+ + =
. .
( ) ( )T M MC y C z
Coefficient of i: + ¥ - = =100 10 216 5 0 461 93 . .T T N T = 462 N b
Coefficient of j: - + =43 3 461 9 0. ( . ) ( )MC y
( )
) .
M
M
C y
C y
= ¥ ◊
= ◊
20 10
20 0
3 N mm
( N m
146
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.119 (Continued)
Coefficient of k: - ¥ - + =140 10 25 461 9 03 ( . ) ( )MC z
( ) .
) .
M
MC z
C z
= ¥ ◊= ◊
151 54 10
151 5
3 N mm
( N m
SF C T= + - =0 400 0: j
M j kC = ◊ + ◊( . ( .20 0 151 5 N m) N m) b
Coefficient of i: Cx = 0
Coefficient of j: C Cy y+ - = =0 5 461 9 400 0 169 1. ( . ) . N
Coefficient of k: C Cz z- = =0 866 461 9 0 400. ( . ) N
C j k= +( . (169 1 400 N) N) b
147
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SOLUTION
Weight of plate = 40 × 9.81 = 392.4 N
r i k
r i k
i j k
E A
G A
EF
EF
/
/
= += +
= + -=
650 500
500 250
250 625 500
838 52
u ruu
. 55
0 29814 0 75436 0 59629
mm
T TEF
EFT= = + -
u ruu
( . . . )i j k
SM r T r j j kA E A G A A y A zM M= ¥ + ¥ - + + =0 75 0: / / ( ) ( ) ( )
i j k i j k
650 0 500
0 29814 0 75430 0 59629
500 0 250
0 392 4 0. . . .
( )
-+
-+T M A y jj k+ =( )M A z 0
Coefficient of i: - + = =372 68 98100 0 263 23. .T T N T = 263N b
Coefficient of j: ( . . )( . ) ( ) ( ) .387 5885 149 07 263 23 0 141264 6+ + = = ◊M MA y A y N mm
Coefficient of k: ( . )( . ) ( ) ( ) .484 484 263 23 196200 0 68669 3- + = = ◊M MA z A z N mm
SF A T= + - =0 392 44 0: . j
M j kA = - ◊ + ◊( (141300 68700 N mm) N mm) b
Coefficient of i: A Ax x+ = = -0 29814 263 23 0 78 479. ( . ) . N
Coefficient of j: A Ay y+ - = =( . )( . ) . .0 74536 263 23 392 4 0 196 199 N
Coefficient of k: Az – . ( . )0 59629 263 23 Az = 156 961. N
A i j k= - + +( . ( . ( .78 5 196 2 157 0N) N) N) b
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 40 kg 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.
148
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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:
r j k
r i jA C
E C
/
/
= += -
105 50
40 60
SM T M MC A C E C C y C z= ¥ - + ¥ + + + =0 3 0: ( ) ( ) ( ) ( )r j r i k j k/ /
( ) ( ) ( ( ) ( )105 50 3 40 60 0 ) ( )j k j i j i k j k+ ¥ - + - ¥ + + + =T M MC y C z
Coefficient of i: 150 60 0- =T T = 2 5. N b
Coefficient of j: - + = = ◊40 2 5 0 100( . ) ( ) N) ( N mmM MC y C y
Coefficient of k: 60 2 5 0 150( . ( ) ( ) N) N mm+ = = - ◊M MC z C z
M j kC = ◊ - ◊( . ( .100 0 150 0 N mm) N mm) b
SF C C Cx y z= + + - + + =0 3 2 5: N) N) (2.5 N) 0i j k j i k( ( .
Equate coefficients of unit vectors to zero.
C C Cx y z= - = = -2 5 3 2 5. . N N N b
C i j k= - + -( . ( . ( .2 50 3 00 2 50 N) N) N) b
149
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SOLUTION
Free-Body Diagram:
First note Ti j
CF CF CF CF
CF
T T
T
= = - +
+= -
l ( . ( . )
( . ) ( . )
(
0 08 0 06
0 08 0 06
0
2 2
m) m
m
.. . )
( . ( . )
( . ) ( . )
8 0 6
0 12 0 09
0 12 0 092 2
i j
Tj k
+
= = -
+DE DE DET Tl m) m
mDDE
DET= -( . . )0 8 0 6j k
(a) From f.b.d. of assembly
SF T Ty CF DE= + - =0 0 8 480: 0.6 N 0.
or 0 6 0 8 480. .T TCF DE+ = N (1)
SM T Ty CF DE= - + =0 0 8 0 135 0 6 0 08: m) m) 0( . )( . ( . )( .
or T TDE CF= 2 25. (2)
Substituting Equation (2) into Equation (1)
0 6 0 8 2 25 480. . [( . ) ]T TCF CF+ = N
TCF = 200 00. N
or TCF = 200 N b
and from Equation (2) TDE = =2 25 200 00. ( . N) 450.00
or TDE = 450 N b
PROBLEM 4.122
The assembly 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.
150
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PROBLEM 4.122 (Continued)
(b) From f.b.d. of assembly
SF A Az z z= - = =0 0 6 450 00 0 270 00: N) N( . )( . .
SF A Ax x x= - = =0 0 8 200 00 160 000: N) 0 N( . )( . .
or A i k= +( . (160 0 270 N) N) b
SM Mx Ax= + -
-
0 480 200 00
45
: N)(0.135 m) N)(0.6)](0.135 m)( [( .
[( 00 N)(0.8)](0.09 m) 0=
M Ax= - ◊16 2000. N m
SM Mz Az
= - +
+
0 480 200 00: N)(0.08 m) N)(0.6)](0.08 m
[(450
( [( . )
NN 0.8)](0.08 m 0)( ) = M Az
= 0
or M iA = - ◊( .16 20 N m) b
151
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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 2 kN load is applied at F, determine the tension in each cable.
SOLUTION
Free-Body Diagram:
In this problem: a = 500 mm
We have CD CD
BD
u ruu
u ruu
= - =
= - +
( (
( (
600 800 1000
1000 60
mm) mm) mm
mm)
j k
i 00 800 1414 21
1000 800
mm) mm) mm
mm) mm)
j k
i k
- =
= -
( .
( (
BD
BE Bu ruu
EE = 1280 63. mm
Thus T TCD
CDT
T TBD
BDT
CD CD CD
BD BD BD
= = -
= = -
u ruu
u ruu
( . . )
( .
0 6 0 8
0 7071
j k
i ++ -
= = -
0 4243 0 5657
0 7809 0 6247
. . )
( . . )
j k
i kT TBE
BETBE BE BE
u ruu
SM r T r T r T r WA C CD B BD B BE F= ¥ + ¥ + ¥ + ¥ =0 0: ( ) ( ) ( ) ( )
152
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PROBLEM 4.123 (Continued)
Noting that r i k
r k
r i k
C
B
F a
= - +== - +
( (
(
(
1000 800
800
800
mm) mm)
mm)
mm)
and using determinants, we write
i j k i j k
i j
--
+- -
+
1000 0 800
0 0 6 0 8
0 0 800
0 7071 0 4243 0 5657. . . . .
T TCD BD
kk i j k
0 0 800
0 7809 0 0 6247
0 800
0 2 0
0
. .-+ -
-=T aBE
Equating to zero the coefficients of the unit vectors:
i: - - + =480T TCD BD339 44 1600 0. (1)
j: - - + =800 565 68 624 72 0T T TCD BD BE. . (2)
k: - + =600 2 0T aCD (3)
Recalling that a = 500 mm, Eq. (3) yields
TCD = =2 500
600
5
3
( )kN TCD = 1 667. kN b
From (1): - ÊËÁ
ˆ¯̃
- + =4805
3339 44 1600 0. TBD
TBD = 2 35682. kN TBD = 2 36. kN b
From (2): - ÊËÁ
ˆ¯̃
- + =8005
3565 68 2 35682 624 72 0. ( . ) . TBE
TBE = 4 2684. kN TBE = 4 27. kN b
153
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PROBLEM 4.124
Solve Problem 4.123, assuming that the 2 kN 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 2 kN 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):
- - + =480 339 44 1600 0T TCD BD. (1)
- - + =800 565 68 624 72 0T T TCD BD BE. . (2)
- + =600 2 0T aCD (3)
In this problem, the 2kN load is applied at C and we have a = 1000 mm. Carrying into (3) and solving for TCD ,
TCD = 10
3kN TCD = 3 33. kN b
From (1): - ÊËÁ
ˆ¯̃
- + =48010
3339 44 1600 0. TBD TBD = 0 b
From (2): - ÊËÁ
ˆ¯̃
- + = =80010
30 624 72 0. T TBE BEfi 4.2685 kN TBE = 4 27. kN b
154
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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 Ti j
DG DG DG DGT T= =- +
+l
( . ( .
( . ) ( . )
0 48 0 14
0 48 0 142 2
m) m)
m
= - +
= +
0 48 0 14
0 50
2524 7
. .
.
( )
i j
i j
T
T
DG
DG
Ti k
BE BE BE BET T= =- +
+l
( . ( .
( . ) ( . )
0 48 0 2
0 48 0 22 2
m) m)
m
= - +
= - +
0 48 0 2
0 52
1312 5
. .
.
( )
i k
j k
T
T
BE
BE )
From f.b.d. of frame ABCD
SM Tx DG= ÊËÁ
ˆ¯̃
- =07
250 3 350 0 15 0: m) N)( m)( . ( .
or TDG = 625 N b
SM Ty BE= ¥ÊËÁ
ˆ¯̃ - Ê
ËÁˆ¯̃
=024
25625 0 3
5
130 48 0: N m) m)( . ( .
or TBE = 975 N b
SM Tz CF= + ¥ÊËÁ
ˆ¯̃
- =0 0 147
25625 0 48 350 0 48 0: m) N m) N)( m)( . ( . ( .
or TCF = 600 N b
155
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PROBLEM 4.125 (Continued)
SF A T T Tx x CF BE x DG x= + + + =0 0: ( ) ( )
Ax - - ¥ÊËÁ
ˆ¯̃
- ¥ÊËÁ
ˆ¯̃
=600 97524
25625 0N
12
13N N
Ax = 2100 N
SF A Ty y DG y= + - =0 350 0: N( )
Ay + ¥ÊËÁ
ˆ¯̃
- =7
25625 350 0N N
Ay = 175 0. N
SF A Tz z BE z= + =0 0: ( )
Az + ¥ÊËÁ
ˆ¯̃
=5
13975 0N
Az = -375 N
Therefore A i j k= + -( ) ( . ) ( )2100 175 0 375 N N N b
156
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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 Ti j
DG DG DG DGT T= =- +
+l
( . ( .
( . ) ( . )
0 48 0 14
0 48 0 142 2
m) m)
m
= - +
= +
0 48 0 14
0 50
2524 7
. .
.
( )
i j
i j
T
T
DG
DG
Ti k
BE BE BE BET T= =- +
+l
( . ( .
( . ) ( . )
0 48 0 2
0 48 0 22 2
m) m)
m
= - +
= - +
0 48 0 2
0 52
1312 5
. .
.
( )
i k
i k
T
T
BE
BE
From f.b.d. of frame ABCD
SM Tx DG= ÊËÁ
ˆ¯̃
- =07
250 3 350 0 3 0: m) N)( m)( . ( .
or TDG = 1250 N b
SM Ty BE= ¥ÊËÁ
ˆ¯̃ - Ê
ËÁˆ¯̃
=024
251250 0 3
5
130 48 0: N m m)( . ) ( .
or TBE = 1950 N b
SM Tz CF= + ¥ÊËÁ
ˆ¯̃
- =0 0 147
251250 0 48 350 0 48 0: m) N m) N)( m)( . ( . ( .
or TCF = 0 b
157
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PROBLEM 4.126 (Continued)
SF A T T Tx x CF BE x DG x= + + + =0 0: ( ) ( )
Ax + - ¥ÊËÁ
ˆ¯̃
- ¥ÊËÁ
ˆ¯̃
=0 195024
251250 0
12
13N N
Ax = 3000 N
SF A Ty y DG y= + - =0 350 0: N( )
Ay + ¥ÊËÁ
ˆ¯̃
- =7
251250 350 0N N
Ay = 0
SF A Tz z BE z= + =0 0: ( )
Az + ¥ÊËÁ
ˆ¯̃
=5
131950 0N
Az = -750 N
Therefore A i k= -( (3000 750N) N) b
158
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SOLUTION
From f.b.d. of weldment
SM r A r B r CO A O B O C O= ¥ + ¥ + ¥ =0 0: / / /
i j k i j k i j k
300 0 0
0
0 200 0
0
0 0 250
0
0
A A B B C Cy z x z x y
+ + =
( ) ( ) ( )- + + - + - + =300 300 200 200 250 250 0A A B B C Cz y z x y xj k i k i j
From i-coefficient 200 250 0B Cz y- =
or B Cz y= 1 25. (1)
j-coefficient - + =300 250 0A Cz x
or C Ax z= 1 2. (2)
k-coefficient 300 200 0A By x- =
or B Ax y= 1 5. (3)
SF A B C= + + - =0 0: P
or ( ) ( . ) ( )B C A C A Bx x y y z z+ + + - + + =i j k1 2 0kN
From i-coefficient B Cx x+ = 0
or C Bx x= - (4)
j-coefficient A Cy y+ - =1 2 0. kN
or A Cy y+ = 1 2. kN (5)
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 = 1 2. kN , a b= =300 200mm, mm and c = 250 mm.
159
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PROBLEM 4.127 (Continued)
k-coefficient A Bz z+ = 0
or A Bz z= - (6)
Substituting Cx from Equation (4) into Equation (2)
- =B Az z1 2. (7)
Using Equations (1), (6), and (7)
CB A B B
yz z x x= =
-= Ê
ËÁˆ¯̃
=1 25 1 25
1
1 25 1 2 1 5. . . . . (8)
From Equations (3) and (8)
CA
C Ayy
y y= =1 5
1 5
.
.or
and substituting into Equation (5)
2 1 2Ay = . kN
A Cy y= = 600 N (9)
Using Equation (1) and Equation (9)
Bz = =1 25 600 750. ( )N N
Using Equation (3) and Equation (9)
Bx = =1 5 600 900. ( )N N
From Equation (4) Cx = -900 N
From Equation (6) Az = -750 N
Therefore A j k= -( ) ( )600 750N N b
B i k= +( ) ( )900 750N N b
C i j= - +( ) ( )900 600N N b
160
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PROBLEM 4.128
Solve Problem 4.127, assuming that the force P is removed and is replaced by a couple M j= + ◊(75 N mm) 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 = 1 2. kN, a b= =300 200mm, mm, and c = 250 mm.
SOLUTION
From f.b.d. of weldment
SM r A r B r C MO A O B O C O= ¥ + ¥ + ¥ + =0 0: / / /
i j k i j k i j k
j300 0 0
0
0 200 0
0
0 0 250
0
75000 0
A A B B C Cy z x z x y
+ + + ◊ =( )N mm
( ) ( ) ( ) (- + + - + - + + ◊300 300 200 200 250 250 75000A A B B C Cz y z x y xj k j k i j N mmm)j = 0
From i-coefficient 200 250 0B Cz y- =
or C By z= 0 8. (1)
j-coefficient - + + =300 250 75 000 0A Cz x ,
or C Ax z= -1 2 300. (2)
k-coefficient 300 200 0A By x- =
or B Ax y= 1 5. (3)
SF A B C= + + =0 0:
( ) ( ) ( )B C A C A Bx x y y z z+ + + + + =i j k 0
From i-coefficient C Bx x= - (4)
j-coefficient C Ay y= - (5)
k-coefficient A Bz z= - (6)
161
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PROBLEM 4.128 (Continued)
Substituting Cx from Equation (4) into Equation (2)
AB
zx= - Ê
ËÁˆ¯̃
2501 2.
(7)
Using Equations (1), (6), and (7)
C B A By z z x= = - = ÊËÁ
ˆ¯̃
-0 8 0 82
3200. . (8)
From Equations (3) and (8)
C Ay y= - 200
Substituting into Equation (5) 2 200Ay =
Ay = 100 0. N
From Equation (5) Cy = -100 0. N
Equation (1) Bz = -125 0. N
Equation (3) Bx = 150 0. N
Equation (4) Cx = -150 0. N
Equation (6) Az = 125 0. N
Therefore A j k= +( . ) ( . )100 0 125 0N N b
B i k= -( . ) ( . )150 0 125 0N N b
C i j= - -( . ) ( . )150 0 100 0N N b
162
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SOLUTION
From f.b.d. of pipe assembly ABCD
SF Bx x= =0 0:
SM BD x z( ) : ( )( . ) ( )-axis N m m= - =0 48 2 5 2 0
Bz = 60 0. N
and B k= ( . )60 0 N b
SM CD z y( ) : ( )-axis m N m= - ◊ =0 3 90 0
Cy = 30 0. N
SM CD y axis z( ) : ( ) ( . )( ) ( )( )- = - - + =0 3 60 0 4 48 4 0m N m N m
Cz = -16 00. N
and C j k= -( . ) ( . )30 0 16 00N N b
SF Dy y= + =0 30 0 0: .
Dy = -30 0. N
SF Dz z= - + - =0 16 00 60 0 48 0: . .N N N
Dz = 4 00. N
and D j k= - +( . ) ( . )30 0 4 00N N b
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 k M k= - = - ◊( , ( .48 90N) N m) 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.
163
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PROBLEM 4.130
Solve Problem 4.129, assuming that the plumber exerts a force F k= -(48 N) 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 k M= - = - ◊( , ( ) .48 90N) 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
SF Bx x= =0 0:
SM BD x z( ) : ( )( . ) ( )-axis N m m= - =0 48 2 5 2 0
Bz = 60 0. N and B k= ( . )60 0 N b
SM C BD z y x( ) : ( ) ( )-axis m 2 m= - =0 3 0
Cy = 0
SM CD y z( ) : ( ) ( . )( ) ( )( )-axis m N m N m= - + =0 3 60 0 4 48 4 0
Cz = -16 00. N and C k= - ( . )16 00 N b
SF D Cy y y= + =0 0:
Dy = 0
SF D B C Fz z z z= + + - =0 0:
Dz + - - =60 0 16 00 48. .N N N 0
Dz = 4 00. N and D k= ( . )4 00 N b
164
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.
SOLUTION
r i j
T i j r j k
T j k r
E F
B B B F
C C C F
T
T
/
/
/
( ) /
( ) /
= - +
= - = -
= - + =
200 80
2 80 200
2 2000 80
2 80 200
i j
T i j r j k
+
= - + = +D D D ET ( ) / /
SM T PF B F B C F C D F D E F= ¥ + ¥ + ¥ + ¥ - =0 0: ( )/ / / /r T r T r r j
i j k i j k i j k i j k
0 80 200
1 1 02
200 80 0
0 1 12
0 80 200
1 1 02
2--
+-
+- -
+ -T T TB c D 000 80 0
0 0
0
-=
P
Equate coefficients of unit vectors to zero and multiply each equation by 2.
i : - + + =200 80 200 0T T TB C D (1)
j: - - - =200 200 200 0T T TB C D (2)
k : - - + + =80 200 80 200 2 0T T T PB C D (3)
80
2002 80 80 80 0( ): - - - =T T TB C D (4)
Eqs. ( ) ( ):3 4+ - - + =160 280 200 2 0T T PB C (5)
Eqs. ( ) ( ):1 2+ - - =400 120 0T TB C
T T TB C C= - -120
4000 3. (6)
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.
165
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PROBLEM 4.131 (Continued)
Eqs. ( )6 Æ ( ):5 - - - + =160 0 3 280 200 2 0( . )T T PC C
- + =232 200 2 0T PC
T PC = 1 2191. T PC = 1 219. b
From Eq. (6): T P PB = - = - =0 3 1 2191 0 36574. ( . ) . T PB = -0 366. b
From Eq. (2): - - - - =200 0 3657 200 1 2191 200 0( . ) ( . )P P T Dq
T PD = -0 8534. T PD = -0 853. b
SF = 0: F T T T j+ + + - =B C D P 0
i :( . ) ( . )
FP P
x +-
--
=0 36574
2
0 8534
20
F P F Px x= - = -0 3448 0 345. .
j:( . ) ( . ) ( . )
FP P P
y --
- --
- =0 36574
2
1 2191
2
0 8534
2200 0
F P F Py y= =
k :( . )
FP
z + =1 2191
20
F P F Pz z= - = -0 8620 0 862. . F i j k= - + -0 345 0 862. .P P P b
166
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SOLUTION
Free-Body Diagram:
Five unknowns and six equations of equilibrium. But equilibrium is maintained ( )SMAB = 0
W mg
W
=
==
( ) .
.
10 9 81
98 1
kg m/s
N
2
GC GC
TGC
GC
T
u ruu
u ruu
= - + - =
= = - + -
300 200 225 425
425300 200
i j k
T i j
mm
( 2225
600 400 150
300 200 75
k
r i j
r i j
)
/
/
B A
G A
= - + +
= - + +
mm
mm
SM WA B A G A G A= ¥ + ¥ + ¥ - =0 0: ( )/ / /r B r T r j
i j k i j k i j k
- + -- -
+ -600 400 150
0 0
300 200 75
300 200 225425
300 200 75
0B
T
--98 1 0.
Coefficient of i : ( . . ) .- - + =105 88 35 29 7357 5 0T
T = 52 12. N T = 52 1. N b
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.
167
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PROBLEM 4.132 (Continued)
Coefficient of j : ( ).
150 300 75 300 22552 12
4250B - ¥ + ¥ =
B = 73 58. N B i= ( . )73 6 N b
SF A B T j= + + - =0 0: W
Coefficient of i : . .Ax + - =73 58 52 15300
4250 Ax = 36 8. N b
Coefficient of j: . .Ay + - =52 15200
42598 1 0 Ay = 73 6. N b
Coefficient of k : .Az - =52 15225
4250 Az = 27 6. N b
168
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SOLUTION
Free-Body Diagram:
DF DF
TDE
DF
T
u ruu
u ruu
= - + - =
= = - + -
400 275 200 525
525400 275
i j k
T i j
mm
( 2200
400
400 350
175 600
625
k
r i
r i k
i k
)
/
/
D E
C E
EAEA
EA
=
= -
= = -lu ruu
S l lMEA EA D E EA C E= ◊ ¥ + ◊ ◊ - =0 300 0: ( ) ( ( ))/ /r T r j
175 0 600
400 0 0
400 275 200525 625
175 0 600
400 0 350
0 300 0
1
62
-
- -¥
+--
-
T
550
600 400 275
525 625
175 300 350 600 400 300
6250
880
=
- ¥ ¥¥
+ - ¥ ¥ + ¥ ¥ =
-
T
0000
753625000 0
426 5625
T
T
+ =
= . N T = 427 N b
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 300-N load is applied at C as shown, determine the tension in the cable.
169
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PROBLEM 4.134
Solve Problem 4.133, assuming that cable DF is replaced by a cable connecting B and F.
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 300-N load is applied at C as shown, determine the tension in the cable.
SOLUTION
Free-Body Diagram:
r i
r i kB A
C A
/
/
=
= +
225
225 250
BF BF
TBF
BF
T
u ruu
u ruu
= - + + =
= = = -
400 275 400 628 987
25 1616
i j k
T i
.
.(
mm
++ +
= = -
11 16
7 24
j k
i k25
)
l AEAE
AE
u ruu
S lMAE AF B A AE C A= ◊ ¥ + ◊ ◊ - =0 300 0: ( ) ( ( ))/ /r T r jl
7 0 24
225 0 0
16 11 1625 25 16
7 0 24
225 0 250
0 300 0
1
250
-
-¥
+-
-=T
.
- ¥ ¥¥
+ ¥ ¥ + ¥ ¥24 225 11
25 25 16
24 225 300 7 250 300
25.T
2360 89 2 145 000 0. , ,T - = T = 909 N b
170
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SOLUTION
Free-Body Diagram:
W mg
W
CE
CE
= ==
= - + -=
( )( . )
.
50 9 81
490 50
240 600 400
760
kg m/s
N
2
u ruu
i j k
mmm
T i j k
i j
= = - + -
= = -
TCE
CE
T
AB
ABAB
u ruu
u ruu
760240 600 400
480 200
( )
l5520
1
1312 5= -( )i j
S l lM r r jAB AB E A AB G AT W= ◊ ¥ + ◊ ¥ - =0 0: ( ) ( )/ /
r i j r i j kE A G A/ /;= + = - +240 400 240 100 200
12 5 0
240 400 0
240 600 40013 20
12 5 0
240 100 200
0 0
1
130
12
-
- -¥
+-
--
=
-
T
W
( ¥¥ ¥ - ¥ ¥ + ¥ =400 400 5 240 400760
12 200 0)T
W
T W= =0 76 0 76 490 50. . ( . )N T = 373 N b
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.
171
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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
W
DE
DE
= ==
= - + -=
( )( . )
.
50 9 81
490 50
240 400 400
614
kg m/s
N
2
u ruu
i j k
..
.( )
5
614 5240 400 400
480 2
mm
T i j k
i
= = + -
= = -
TDE
DE
T
AB
ABAB
u ruu
u ruu
l 000
520
1
1312 5
ji j= -( )
r i j r i j kE A G A/ /;= + = - +240 400 240 100 200
12 5 0
240 400 0
240 400 40013 614 5
12 5 0
240 100 200
0 0
1
130
1
-
-¥
+ --
=
-
T
W.
( 22 400 400 5 240 400614 5
12 200 0¥ ¥ - ¥ ¥ + ¥ ¥ =).
TW
T W= =0 6145 0 6145 490 50. . ( . )N
T = 301 N b
172
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SOLUTION
First note lBD =- - +
+ +
=
( ) ( ) ( )
( ) ( ) ( )
(
150 225 300
150 225 300
1
29
2 2 2
mm mm mm
mm
i j k
-- - +
= -
==
=
2 3 4
150
400
200
i j k
r i
P k
r i
C j
)
( )
( )
( )
( )
/
/
A B
C D
C
mm
N
mm
From the f.b.d. of the plates
S l lMBD BD A B BD C D= ◊ ¥ + ◊ ¥ =0 0: ( ) ( )/ /r rP C
- -- È
Î͢˚̇
+- -
ÈÎÍ
˘˚̇
=2 3 4
1 0 0
0 0 1
150 400
29
2 3 4
1 0 0
0 1 0
200
290
( ) ( )
(
C
-- + =3 150 400 4 200 0)( )( ) ( )( )C
C = 225 N or NC j= ( )225 b
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.
173
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SOLUTION
Free-Body Diagram:
AF AF
AF
G A
u ruu
= - - =
= - -
= -
1 2 0 6 1 2 1 8
1
32 2
0 6 0 31
. . . .
( )
. .
i j k
i j k
r i j
m
/
l
rr i j k
r iG A
B A
21 2 0 3 0 6
1 2
/
/
= - -
=
. . .
.
S l l lM TAF AF G A AF G A AF B A= ◊ ¥ - + ◊ ¥ - + ◊ ¥ =0 60 60 02: / / /( ( ) ( ( )) ( )r j r j r
2 1 2
0 6 0 3 0
0 60 0
1
3
2 1 2
1 2 0 3 0 6
0 60 0
1
3
- ---
+- -
- --
+ ◊ ¥ =. . . . . ( )l AF B Ar T/ 00
( . ) ( . . ) ( )2 0 6 601
32 0 6 60 2 1 2 60
1
30¥ ¥ + - ¥ ¥ + ¥ ¥ + ◊ ¥ =l AF B Ar T/
l lAF B A A F B A◊ ¥ = - ◊ ¥ = -( ) ( )r T T r/ / /or48 48 (1)
PROBLEM 4.138
Two 0 6 1 2. .¥ m plywood panels, each of weight 60 N, 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.
174
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PROBLEM 4.138 (Continued)
Projection of T on ( )l AF B A¥ r / is constant. Thus, Tmin is parallel to
l AF B A¥ = - - ¥ = - +r i j k i j k/1
32 2 1 2
1
32 4 1 2( ) . ( . . )
Corresponding unit vector is 15
2( )- +j k
T Tmin ( )= - +21
5j k (2)
Eq. (1): T
T5
21
32 2 1 2 48
52
1
32 4 1 2
( ) ( ) .
( ) ( . .
- + ◊ - - ¥ÈÎÍ
˘˚̇
= -
- + ◊ - +
j k i j k i
j k j kk) = -48
T
T3 5
4 8 1 2 483 5 48
624 5( . . )
( )+ = - = - =
T = 53 6656. N
Eq. (2) T Tmin
min
( )
( )
( (
= - +
= - +
= - +
21
5
24 5 21
548 24
j k
j k
T j k N) N )
Since Tmin has no i component, wire BH is parallel to the yz plane, and x = 1 2. m
(a) x y= =1 200 2 40. . m; m b
(b) Tmin .= 53 7 N b
175
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.
SOLUTION
Free-Body Diagram:
AF AF
AF
G A
u ruu
= - - =
= - -
= -
1 2 0 6 1 2 1 8
1
32 2
0 6 0 31
. . . .
( )
. .
i j k
i j k
r i j
m
/
l
rr i j k
r iG A
B A
21 2 0 3 0 6
1 2
/
/
= - -
=
. . .
.
S l l lM TAF AF G A AF G A AF B A= ◊ ¥ - + ◊ ¥ - + ◊ ¥ =0 60 60 02: / / /( ( ) ( ( )) ( )r j r j r
2 1 2
0 6 0 3 0
0 60 0
1
3
2 1 2
1 2 0 3 0 6
0 60 0
1
3
- ---
+- -
- --
+ ◊ ¥ =. . . . . ( )l AF B Ar T/ 00
( . ) ( . . ) ( )2 0 6 601
32 0 6 60 2 1 2 60
1
30¥ ¥ + - ¥ ¥ + ¥ ¥ + ◊ ¥ =l AF B Ar T/
l AF B A◊ ¥ = -( )r T/ 48 (1)
BH y BH yu ruu
= - + - = +1 2 1 2 2 88 2 1 2. . ( . )i j k /
Ti j k= = - + -
+T
BH
BHT
y
y
u ruu
1 2 1 2
32 2 1 2
. .
( ) /
PROBLEM 4.139
Solve Problem 4.138, subject to the restriction that H must lie on the y axis.
PROBLEM 4.138 Two 0 6 1 2. .¥ m plywood panels, each of weight 60 N, 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.
176
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.139 (Continued)
Eq. (1): l AF B A T
y
T
y◊ ¥ =
- -
- - += -( ) .
. .( . )
r /
2 1 2
1 2 0 0
1 2 1 2
1
3 2 8848
2
( . . ) ( . )( . )
( . .- - = - ¥ + = +
+1 44 2 4 3 48 2 88 144
2 88
1 44 2 42 1 2
2 1 2
y T y Ty
y/
/
))
( . )= +
+ÊËÁ
ˆ¯̃
1002 88
153
2 12y
y (2)
dT
dy
yy y y
=
ÊËÁ
ˆ¯̃ + + + Ê
ËÁ-
0
53
12
2 88 2 2 8853
2 1 2 2 1 2
: 100
1+ / /( . ) ( ) ( . ) ˆ̂¯̃
ÏÌÓ
¸˝˛
+ÊËÁ
ˆ¯̃1
53
2y
Numerator = 0: 15
32 88
5
32+Ê
ËÁˆ¯̃
= +yy y( . )
5
3
24
5
5
32 2y y y+ = + y = 4 80. m b
Eq. (2): T = +
+ ¥ÊËÁ
ˆ¯̃
=100 2 88 4 8
15 4 8
3
56 5692 1 2( . . )
..
/
N Tmin .= 56 6 N b
177
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 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
AE
AE
AEAE
u ruu
u ruu
= + -=
= = + -
480 160 240
560
480 160 240
5
i j k
i j k
mm
l660
6 2 3
7200
480 160
l AE
B A
D A
= + -
== +
i j k
r i
r i j/
/
DF DF
TDF
DFTDF DF DF
u ruu
u ruu
= - + - =
= = - +
480 330 240 630
480 3
i j k
Ti
; mm
330 240
630
16 11 8
21
j k i j k- = - + -TDF
S lM AE AE D A DF AE B A= - ¥ + - ¥ - =( ) ( ( ))r T r j/ /l 600 0
6 2 3
480 160 0
16 11 821 7
6 2 3
200 0 0
0 640 0
1
70
-
- -¥
+-
-=
TDE
- ¥ ¥ + ¥ ¥ - ¥ ¥ - ¥ ¥
¥+ ¥ ¥ =6 160 8 2 480 8 3 480 11 3 160 16
21 7
3 200 640
70TDF
- + ¥ =1120 384 10 03TDF
TDF = 342 86. N TDF = 343 N b
178
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 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
AE
AE
AEAE
u ruu
u ruu
= + -=
= = + -
480 160 240
560
480 160 240
5
i j k
i j k
mm
l660
6 2 3
7200
480
480 490 240
l AE
B A
C A
CF C
= + -
==
= - + -
i j k
r i
r i
i j k
/
/u ruu
; FF
TCE
CFCF CF
=
= = - + -
726 70
480 490 240
726 70
.
.
mm
Ti j k
u ruu
S l lMAE AE C A CF AE B A= ◊ ¥ + ◊ ¥ - =0 600 0: ( ) ( ( ))r T r j/ /
6 2 3
480 0 0
480 490 240726 7 7
6 2 3
200 0 0
0 640 0
1
70
-
- + -¥
+-
-=
TCF
.
2 480 240 3 480 490
726 7 7
3 200 640
70
¥ ¥ - ¥ ¥¥
+ ¥ ¥ =.
TCF
- + ¥ =653 91 384 10 03. TCF TCF = 587 N b
179
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.
SOLUTION
Free-Body Diagram:
W mg
a
= = == -
( ) .
( (
40 392 40
300 801
kg)(9.81 m/s N
mm)sin mm)cos
2
a aaa aa
a
b2 430 300
930
= -=
( (
(
mm)cos mm)sin
mm)cos
From free-body diagram of hand truck
Dimensions in mm
+ SM P b W a W aB = - + =0 02 1: ( ) ( ) ( ) (1)
+SF P W By = - + =0 2 2 0: (2)
For a = ∞35
a
a1 300 35 80 35 106 541
430 35 300 35 180
= ∞ - ∞ == ∞ - ∞ =
sin cos .
cos sin
mm
2 ..
cos .
162
930 35 761 81
mm
mmb = ∞ =(a) From Equation (1)
P( . . .761 81 392 40 392 40 mm) N(180.162 mm) N(106.54 mm) 0- + =
P = 37 921. N or P = 37 9. N ≠ b
(b) From Equation (2)
37 921 2 0. N 2(392.40 N)- + =B or B = 373 N ≠ b
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 a = ∞35 , (b) the corresponding reaction at each of the two wheels.
180
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SOLUTION
(a) a = ∞0
From f.b.d. of member ABC
+ SM AC = + - =0 300 300 0 8 0: ( ( ( . N)(0.2 m) N)(0.4 m) m)
A = 225 N or A = 225 N ≠ b
+SF Cy y= + =0 225 0: N
Cy y= - =225 225N or NC Ø
+ SF Cx x= + + =0 300 300 0: N N
Cx x= - =600 600 N or NC ¨
Then C C Cx y= + = + =2 2 2 2600 225 640 80( ) ( ) . N
and q =ÊËÁ
ˆ¯̃
= --
ÊËÁ
ˆ¯̃
= ∞- -tan tan .1 1 225
60020 556
C
Cy
x
or C = 641 N 20 6. ∞ b
(b) a = ∞30
From f.b.d. of member ABC
+ SM A
AC = + - ∞
+0 300 300 30 0 8: ( ( ( cos )( .
( sin
N)(0.2 m) N)(0.4 m) m)
330 20 0∞ =)( )in.
A = 365 24. N or A = 365 N 60 0. ∞ b
+ SF Cx x= + + ∞ + =0 300 300 365 24 30 0: ( . sin N N N)
Cx = -782 62.
PROBLEM 4.143
Determine the reactions at A and C when (a) a = 0, (b) a = ∞30 .
181
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PROBLEM 4.143 (Continued)
+SF Cy y= + ∞ =0 365 24 30 0: ( . cos N)
Cy y= - =316 31 316. N or NC Ø
Then C C Cx y= + = + =2 2 2 2782 62 316 31 884 12( . ) ( . ) . N
and q =ÊËÁ
ˆ¯̃
= --
ÊËÁ
ˆ¯̃
= ∞- -tan tan.
..1 1 316 31
782 6222 007
C
Cy
x
or C = 884 N 22 0. ∞ b
182
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.
SOLUTION
Free-Body Diagram:
Geometry:
xAC = ∞ =
= ∞ =( cos .
( sin .
250 20 234 923
250 20 85 505
mm) mm
mm) mmyAC
fi = - =yDA 300 85 505 214 495 mm mm mm. .
a =ÊËÁ
ˆ¯̃
= ÊËÁ
ˆ¯̃
= ∞- -tan tan.
..1 1 214 495
234 92342 397
y
xDA
AC
b = ∞ - ∞ - ∞ = ∞90 20 42 397 27 603. .
Equilibrium for lever:
(a) + SM TC AD= ∞ - ∞ =0 27 603 250 350 0: cos . ( (mm) N)[(375 mm)cos 20 ]
TAD = 556 703. N TAD = 557 N b
(b) + SF Cx x= + ∞ =0 556 703 42 397 0: ( . cos . N)
Cx = -411 12. N
+SF Cy y= - - ∞ =0 350 42 397 0: sin .N N)(556.703
Cy = 725 364. N
Thus: C C Cx y= + = - + =2 2 2 2411 12 725 364 833 77( . ) ( . ) . N
and q = = = ∞- -tan tan.
..1 1 725 364
411 1260 456
C
Cy
x
C = 834 N 60 5. ∞ b
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 350 N vertical force at B, determine (a) the tension in the cable, (b) the reaction at C.
183
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SOLUTION
Free-Body Diagram:
Dimensions in mm
Geometry:
Distance AD = + =( . ) ( . ) .0 36 0 150 0 392 2 m
Distance BD = + =( . ) ( . ) .0 2 0 15 0 252 2 m
Equilibrium for beam:
(a) + SM T TC = - ÊËÁ
ˆ¯̃
- ÊËÁ
0 1200 15
0 390 36
0 15
0 25: (
.
.( .
.
. N)(0.28 m) m) ˆ̂
¯̃=( .0 2 0m)
T = 130 000. N
or T = 130 0. N b
(b) + SF Cx x= + ÊËÁ
ˆ¯̃
+ ÊËÁ
ˆ¯̃
00 36
0 39130 000
0 2
0 25130 000:
.
.( . )
.
.( . N N)) = 0
Cx = -224 00. N
+SF Cy y= + ÊËÁ
ˆ¯̃
+ ÊËÁ
ˆ¯̃
00 15
0 39130 00
0 15
0 25130 00:
.
.( .
.
.( . N) N) -- =120 0N
Cy = -8 0000. N
Thus: C C Cx y= + = - + -( ) =2 2 2 2224 8 224 14( ) . N
and q = = = ∞- -tan tan .1 1 8
2242 0454
C
Cy
x
C = 224 N 2 05. ∞ b
PROBLEM 4.145
Neglecting friction and the radius of the pulley, determine (a) the tension in cable ADB, (b) the reaction at C.
184
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SOLUTION
Free-Body Diagram:
+SF Ey = ∞ - - =0 30 100 200 0: cos
E =∞
=300346 41
N
cos 30 N. E = 346 N 60 0. ∞ b
+ SM
C ED = -
- + ∞0 200
75 30 75
: (100 N)(100 mm) N)(100 mm)
mm) m
(
( sin ( mm) 0=
C = 39 8717. N C = 39 9. N Æ b
SF E C Dx = ∞ + - =0 30 0: sin
( . )( . ) .346 41 0 5 39 8717 0 N N+ - =D
D = 213 077. N D = 213 N ¨ b
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 q = ∞30 .
185
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SOLUTION
Free-Body Diagram:
+SF Ey = - - =0 100 200 0: cosq
E = 300
cosq (1)
+ SM CD = - -0 200 75: (100 N)(100 mm) N)(100 mm) mm)
+300
cos
( (
sinq
qÊÊËÁ
ˆ¯̃
=75 0mm
C = - +400
3300 tanq
(a) For C = 0, 300400
3tanq =
tan .q q= = ∞4
923 962 q = ∞24 0. b
Eq. (1) E =∞
=300
23 962328 294
cos ..
+ SF D C Ex = - + + =0 0: sinq
D = =( . )sin . .328 294 23 962 133 33 N
(b) C D= =0 133 33. N ¨ E = 328 N 66 0. ∞ b
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 u for which the equilibrium of the bracket is maintained, (b) the corresponding reactions at C, D, and E.
186
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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 b is found from the member dimensions:
b = ÊËÁ
ˆ¯̃
= ∞-tan .1 15030 964
mm
250 mm
Applying of the law of sines to the force triangle for member BCD,
150
45 135
N
sin( ) sin sin∞ -= =
∞b bB C
or 150
14 036 30 964 135
N
sin . sin . sin∞=
∞=
∞B C
A B= =∞
∞=
( )sin .
sin ..
150 30 964
14 036318 205
N N
or A = 318 2. N 45 0. ∞ b
and C = ∞∞
=( )sin
sin ..
150 135
14 036437 33
N N
or C = 437 3. N 59 0. ∞ b
187
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.
SOLUTION
Free-Body Diagram: (Three-force body)
Reaction A must pass through Point D where 100-N forceand B intersect
In right D BCD
a = ∞ - ∞ = ∞
= ∞ =90 75 15
250 75 933 01BD tan . mm
Dimensions in mm
In right D ABD
tan
.
.
g
g
= =
=
AB
BD
150
933 01
9 13
mm
mm
Force Triangle
Law of sines
100
15 155 87163 1 257 6
N
9.13N; N
sin sin sin .. .
∞=
∞=
∞= =
A B
A B
A = 163 1. N 74.1° B = 258 N 65.0° b
PROBLEM 4.149
Determine the reactions at A and B when b = ∞50 .
188
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SOLUTION
Free-Body Diagram:
Five unknowns and six equations of equilibrium, but equilibrium is maintained
( )SM AC = 0
r j
r jB
C
==
3
6
CF CF
BD BD
BE
u ruu
u ruu
u ruu
= - - + =
= - - =
= -
3 6 2 7
1 5 3 3 4 5
1 5 3
i j k
i j k
i
m
m. .
. jj k+ =3 4 5BE . m
P i j k
T i j k
= = - - +
= = - -
PCF
CE
P
TBD
BD
TBD BD
BD
u ruu
u ruu
73 6 2
4 51 5 3 3
( )
.( . )) ( )
( )
= - -
= = = - +
T
TBE
BE
T
BD
BE BEBD
32 2
32 2
i j k
T i j k
u ruu
SM
T T
A B BD B BE C
BD BE
= ¥ + ¥ + ¥ =
- -+
-+
0 0
0 3 0
1 2 23
0 3 0
1 2 23
: r T r T r P
i j k i j k i jj k
0 6 0
3 6 27
0
- -=P
Coefficient of i: - + + =2 212
70T T PBD BE (1)
Coefficient of k: - - + =T T PBD BF18
70 (2)
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.
189
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PROBLEM 4.150 (Continued)
Eq. (1) + 2 Eq. (2): - + = =448
70
12
7T P T PBD BD
Eq. (2): - - + = =12
7
18
70
6
7P T P T PBE BE
Since P TBD= =44512
7455N ( ) TBD = 780 N b
TBE = 6
7455( ) TBE = 390 N b
SF T T P A= + + + =0 0: BD BE
Coefficient of i: 780
3
390
3
455
73 0+ - + =( ) Ax
260 130 195 0 195 0+ - + = =A Ax x . N
Coefficient of j: - - - + =780
32
390
32
455
76 0( ) ( ) ( ) Ay
- - - + = =520 260 390 0 1170A Ay y N
Coefficient of k: - + + + =780
32
390
32
455
72 0( ) ( ) ( ) Az
- + + + = = +520 260 130 0 130 0A Az z . N
A i j k= - + +( . ) ( ) ( . )195 0 1170 130 0N N N b
190
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 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 m,= 3 determine the tension in each cable and the reaction at A.
SOLUTION
Free-Body Diagram:
Five unknowns and six equations of equilibrium but equilibrium is maintained
( )SM AC = 0
r j
r jB
C
==
3
6
CF CF
BD BD
BE
u ruu
u ruu
u ruu
= - - + =
= - - =
=
1 5 6 2 6 5
1 5 3 3 4 5
1
. .
. .
.
i j k
i j k
m
m
55 3 3 4 5i j k- + =BE . m
P i j k i j k
T
= = - - + = - - +
=
PCF
CE
P P
TBD
BD BD
u ruu
u ruu
6 51 5 6 2
133 12 4
.( . ) ( )
BBD
T T
TBE
BE
T
BD BD
BE BEBD
= - - = - -
= = =
4 51 5 3 3
32 2
3
.( . ) ( )
(
i j k i j k
T
u ruu
ii j k- +2 2 )
SM
T T
A B BD B BE C
BD BE
= ¥ + ¥ + ¥ =
- -+
-+
0 0
0 3 0
1 2 23
0 3 0
1 2 23
: r T r T r P
i j k i j k i jj k
0 6 0
3 12 413
0
- - +=P
Coefficient of i: - + + =2 224
130T T PBD BE (1)
Coefficient of k: - - + =T T PBD BE18
130 (2)
191
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.151 (Continued)
Eq. (1) + 2 Eq. (2): - + = =460
130
15
13T P T PBD BD
Eq (2): - - + = =15
13
18
130
3
13P T P T PBE BE
Since P TBD= =44515
13455N ( ) TBD = 525 N b
TBE = 3
13455( ) TBE = 105 0. N b
SF T T P A= + + + =0 0: BD BE
Coefficient of i: 525
3
105
3
455
133 0+ - + =( ) Ax
175 35 105 0 105 0+ - + = =A Ax x . N
Coefficient of j: - - - + =525
32
105
32
455
1312 0( ) ( ) ( ) Ay
- - - + = =350 70 420 0 840A Ay y N
Coefficient of k: - + + + =525
32
105
32
455
134 0( ) ( ) ( ) Az
- + + + = =350 70 140 0 140 0A Az z . N
A i j k= - + +( . ) ( ) ( . )105 0 840 140 0N N N b
192
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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:
r i
r i k
r i k
r i k
r
B A
F A
D A
E A
F A
/
/
/
/
/
== -= -= -=
300
300 200
300 400
300 600
3000 800i k-
BG
BG
BG
u ruu
= - +== - +
300 225
375
0 8 0 6
i k
i k
mm
l . .
DH DH
FJ
DH
u ruuu
u ruu
= - + = = - +
= - +
300 400 500 0 6 0 8
300 400
i k i j
i
; ; . .mm l
kk i j; . .FJ FJ= = - +500 0 6 0 8mm; l
SM r T r T r T
r j rA B A BG BG DA DH DH F A FJ FJ
F A E A
= ¥ + ¥ + ¥+ ¥ - +
0
120
:
( )/ /
/ /
l l l¥¥ - =
-+ -
-+
( )
. . . .
120 0
300 0 0
0 8 0 0 6
300 0 400
0 6 0 8 0
3
j
i j k i j k i j k
T TBG DH 000 0 800
0 6 0 8 0
300 0 200
0 120 0
300 0 600
0 120 0
0
--
+ --
+ --
=
. .
TFJ
i j k i j k
Coefficient of i: + + - - =320 640 24000 72000 0T TDH FJ (1)
Coefficient of k: + + - - =240 240 36000 36000 0T TDH FJ (2)
34 Eq. (1) − Eq. (2):
( )
( )
1 2 300
2 3000
fi + =fi + =
¸˝˛
fi =T T
T TTDH FJ
DH FJFJ TFJ = 0 b
193
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.152 (Continued)
Eq. (1): TDH = 300 N TDH = 300 N b
Coefficient of j: - + ¥ =180 240 300 0TBG ( ) TBG = 400 N b
SF A j j= + + + - - =0 24 24 0: T T TBG BG DH DH FJl l
Coefficient of i: A Ax x+ - + - = =( )( . ) ( )( . )400 0 8 300 0 6 0 500 N
Coefficient of j: Ay + - - =( )( . )300 0 8 120 120 0 Ay = 0
Coefficient of k: Az + + =( )( . )400 0 6 0 Az = -240 N
A i j= -( ) ( )500 240N N b
Note: The value Ay = 0
Can be confirmed by considering SMBF = 0
194
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 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) + SM P C a aA a= - + ∞ + ∞ =0 45 2 45 0: ( sin ) (cos )
32
2
3
CP C P= = C = 0 471. P 45° b
+ SF A P AP
x x x= -Ê
ËÁˆ
¯̃=0
2
3
1
2 3: Æ
+SF A P P AP
y y y= - +Ê
ËÁˆ
¯̃=0
2
3
1
2
2
3: ≠
A = 0 745. P 63.4° b
(b) + SM Pa A a A aC = + - ∞ + ∞ =0 30 2 30 0: ( cos ) ( sin )
A P A P( . . ) .1 732 0 5 0 812- = =
A = 0 812. P 60.0° b
+ SF P C C Px x x= ∞ + = = -0 0 812 30 0 0 406: ( . )sin . +SF P P C C Py y y= ∞ - + = = -0 0 812 30 0 0 297: ( . )cos .
C = 0 503. P 36.2° b
195
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
PROBLEM 4.153 (Continued)
(c) + SM Pa A a A aC = + - ∞ + ∞ =0 30 2 30 0: ( cos ) ( sin )
A P A P( . . ) .1 732 0 5 0 448+ = =
A P= 0 448. 60.0° b
+ SF P C C Px x x= - ∞ + = =0 0 448 30 0 0 224: ( . )sin . Æ
+SF P P C C Py y y= ∞ - + = =0 0 448 30 0 0 612: ( . )cos . ≠
C = 0 652. P 69.9° b
(d) Force T exerted by wire and reactions A and C all intersect at Point D.
+ SM PD a= =0 0:
Equilibrium not maintained
Rod is improperly constrained b