CHAPTER 473
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
+
SM B = 0 : + (9319.5 N )(1000 mm) - ( 4000 N )(1250 mm) - 2 A(1500 mm) = 0
A = 1440 N ≠ 
A = +1439.83 N (b)
Front wheels
+
SM A : - (9319.5 N )(500 mm) - ( 4000 N )(2750 mm) + 2 B(1500 mm) = 0
B = +5219.92 N Check:
B = 5220 N ≠ b
+ SFy = 0 : 2 A + 2 B - 9320 N - 4000 N = 0
2(1440) + 2(5220) - 9320 - 4000 = 0 0 = 0 (Checks) PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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3
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 A = 0 : (2 F )(1 m) - (60 N )(0.15 m) - (250 N )(0.3 m) = 0
F = 42.0 N ≠ 
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4
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:
+
SMA = 0 : 2(75 N )(1 m) - (60 N )(0.15 m) - (250 N )(0.3 m) - (250 N ) x = 0
x = 0.264 m 
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5
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
SFx = 0 : Ax = 0
+
SMB = 0 : (75 N )(700 mm) + (100 N )(550 mm) + (175 N )(350 mm)
+ (100 N )(150 mm
m) - Ay (150 mm) = 0
Ay = +1225 N A = 1225 N ≠ 
(b)
Tension in BC
+
S MA = 0 : (75 N )(550 mm) + (100 N )( 400 mm) + (175 N )(200 mm)
- (75 N )(150 mm)) - FBC (150 mm) = 0
FBC = +700 N Check:
+
FBC = 700 N 
SFy = 0 : - 75 N - 100 N - 175 N - 100 N - 75 N + A - FBC = 0
-525 + 1225 - 700 = 0
0 = 0 (Checks)
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6
PROBLEM 4.5
Two crates, each of mass 350 kg, are placed as shown in the bed
of a 1400-kg pickup truck. Determine the reactions at each of
the two (a) rear wheels A, (b) front wheels B.
SOLUTION
Free-Body Diagram:
W = (350 kg)(9.81 m/s2 ) = 3.434 kN
Wt = (1400 kg)(9.81 m/s2 ) = 13.734 kN
(a)
Rear wheels
+
SM B = 0 : W (1.7 m + 2.05 m) + W (2.05 m) + Wt (1.2 m) - 2 A(3 m) = 0
(3.434 kN )(3.75 m) + (3.434 kN )(2.05 m)
+ (13.734 kN )(1.2 m) - 2 A(3 m) = 0
A = +6.0663 kN (b)
Front wheels
+
A = 6.07 kN ≠ 
SFy = 0 : - W - W - Wt + 2 A + 2 B = 0
-3.434 kN - 3.434 kN - 13.734 kN + 2(6.0663 kN) + 2B = 0 B = + 4.2347 kN B = 4.23 kN ≠ 
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7
PROBLEM 4.6
Solve Problem 4.5, assuming that crate D is removed and that
the position of crate C is unchanged.
PROBLEM 4.5 Two crates, each of mass 350 kg, are placed
as shown in the bed of a 1400-kg pickup truck. Determine the
reactions at each of the two (a) rear wheels A, (b) front wheels B.
SOLUTION
Free-Body Diagram:
W = (350 kg)(9.81 m/s2 ) = 3.434 kN
Wt = (1400 kg)(9.81 m/s2 ) = 13.734 kN
(a)
Rear wheels
+
SM B = 0 : W (1.7 m + 2.05 m) + Wt (1.2 m) - 2 A(3 m) = 0
(3.434 kN )(3.75 m) + (13.734 kN )(1.2 m) - 2 A(3 m) = 0 A = + 4.893 kN (b)
Front wheels
+
A = 4.89 kN ≠ 
SM y = 0 : - W - Wt + 2 A + 2 B = 0
-3.434 kN - 13.734 kN + 2(4.893 kN) + 2B = 0 B = +3.691 kN B = 3.69 kN ≠ 
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8
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:
+
SFx = 0 : Bx = 0
SMB = 0 : (200 N )(150 mm) - (150 N )a - (50 N )( a + 200 mm) + (300 mm) A = 0 A=
+
(175000 + 200 a)
300
Bx = 0 B =
Since
(b)
(1)
SMA = 0 : - (200 N )(150 mm) - (250 N )(300 mm) - (150 N )( a + 300 mm)
-(50 N )( a + 500 mm) + (300 mm)B y = 0
By =
(a)
(200a - 20000)
300
(175000 + 200 a)
300
(2)
For a = 250 mm
Eq. (1):
A=
(200 ¥ 250 - 20000)
= +100 N 300
Eq. (2):
B=
(175000 + 200 ¥ 250)
= +750 N 300
Eq. (1):
A=
(200 ¥ 175 - 20000)
= +50 N 300
Eq. (2):
B=
(175000 + 200 ¥ 175)
= +700 N 300
A = 100.0 N Ø 
B = 750 N Ø 
For a = 175 mm
A = 50.0 N Ø 
B = 700 N ≠ 
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9
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 B = 0 : (200 N )(150 mm) - (150 N )a - (50 N )( a + 200 mm) + (300 mm) A = 0 A=
(200 a - 20000)
300
A = 0 : (200a - 20000) = 0 a = 100.0 mm 
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10
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 Fx = 0 : Bx = 0 B = By +
SM A = 0 : (50 N )d - (100 N )(0.45 m - d ) - (150 N )(0.9 m - d ) + B(0.9 m - d ) = 0 50d - 45 + 100d - 135 + 150 d + 0.9 B - Bd d=
+
180 N ◊ m - (0.9 m) B
300 A - B
(1)
SM B = 0 : (50 N )(0.9 m) - A(0.9 m - d ) + (100 N)(0.45 m) = 0 45 - 0.9 A + Ad + 45 = 0 d=
Since B 180 N, Eq. (1) yields
d
(2)
180 - (0.9)180 18
=
= 0.15 m 300 - 180
120
d 150.0 mm v
(0.9)180 - 90 72
=
= 0.40 m 180
180
d 400 mm v
Since A 180 N, Eq. (2) yields
d
Range:
(0.9 m) A - 90 N ◊ m
A
150.0 mm d 400 mm 
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11
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
SFx = 0 : Bx = 0
B = By +
SM A = 0 : (80 N )d - (100 N )(0.45 m - d ) - (150 N )(0.9 m - d ) + B(0.9 m - d ) = 0 80d - 45 + 100d - 135 + 150 d + 0.9 B - Bd = 0 d=
+
180 N ◊ m - 0.9 B
330 N - B
SM B = 0 : (80 N )(0.9 m) - A(0.9 m - d ) + (100 N)(0.45 m) = 0 d=
Since B 180 N, Eq. (1) yields
d (180 - 0.9 ¥ 180) / (330 - 180) =
Since A 180 N, Eq. (2) yields
d (0.9 ¥ 180 - 112) /180 =
Range:
(1)
0.9 A - 117
A
18
= 0.12 m 150
45
= 0.25 m 180
120.0 mm d 250 mm (2)
d = 120.0 mm v
d = 250 mm v

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12
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
SFx = 0 : Bx = 0 B = By ≠
+
SM A = 0 : - P (0.9 m) + B(2.7 m) - (27 kN )(3.3 m) - (27 kN)(3.9 m) = 0 P = 3B - 216 kN +
(1)
SM B = 0 : - A(2.7 m) + P (1.8 m) - (27 kN )(0.6 m) - (27 kN)(1.2 m) = 0 P = 1.5 A + 27 kN (2)
Since B 135 kN, Eq. (1) yields
P (3)(135) - 216 kN P 189.0 kN v
Since 0 A 135 kN, Eq. (2) yields
0 + 27 kN P (1.5)(135) + 27 27 kN P 229.5 kN v
Range of values of P for which beam will be safe:
27.0 kN P 189.0 kN 
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13
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 C = 0 : P (2 m) - ( 4 kN )(3 m) - (20 kN )(8 m) + D(6 m) = 0 P = 86 kN - 3D +
(1)
SM D = 0 : P (8 m) + ( 4 kN )(3 m) - (20 kN )(2 m) - C (6 m) = 0 P = 3.5 kN + 0.75C (2)
For no motion C 0 and D 0
For C 0 from (2) P 3.50 kN
For D 0 from (1) P 86.0 kN
Range of P for no motion:
3.50 kN P 86.0 kN 
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14
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 C = 0 : P (2 m) - ( 4 kN )(3 m) - (20 kN )(8 m) + D(6 m) = 0
P = 86 kN - 3D +
(1)
SM D = 0 : P (8 m) + ( 4 kN )(3 m) - (20 kN )(2 m) - C (6 m) = 0
P = 3.5 kN + 0.75C (2)
For C 0, from (2):
P 3.50 kN v
For D 0, from (1):
P 86.0 kN v
P 3.5 kN + 0.75(50 kN )
P 41.0 kN
v
P 86 kN - 3(50 kN )
P -64.0 kN
v
For C 50 kN, from (2):
For D 50 kN, from (1):
Comparing the four criteria, we find
3.50 kN P 41.0 kN 
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15
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 D = 0 : (1350 N )(200 mm - a) - (1350 N )( a - 50) - (225 N )(100 mm) + ( 400) B = 0 -2700a + 315000 + 400B = 0 a=
(315000 + 400B)
2700
(1)
For B = 450 N Ø = -450 N, Eq. (1) yields:
a
[315000 + 400( -450)]
2700
a 50.0 mm v
a
[315000 + 400(900)]
2700
a 250 mm v
For B = 900 N ≠ = +900 N, Eq. (1) yields:
Required range: 50.0 mm a 250 mm 
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16
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 C = 0 : FAB (100 mm) - FDE (120 mm) = 0 FDE =
(a)
(1)
FAB = 720 N
For
5
FDE = (720 N ) 6
+
(b)
5
FAB 6
+
FDE = 600 N 
3
SFx = 0 : - (720 N ) + C x = 0
5
C x = +432 N 4
SFy = 0 : - (720 N ) + C y - 600 N = 0
5
C y = +1176 N
C = 1252.84 N a = 69.829∞
C = 1253 N
69.8° 
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17
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)
5
FDE = FAB 6
3
3
+
SFx = 0 : - FAB + C x = 0 C x = FAB 5
5
4
+ SFy = 0 : - FAB + C y - FDE = 0 5
(1)
4
5
- FAB + C y - FAB = 0
5
6
49
Cy =
FAB
30
C = C x2 + C y2
1
( 49)2 + (18)2 FAB 30
C = 1.74005FAB
=
For
C = 1600 N, 1600 N = 1.74005FAB FAB = 920 N 
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18
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 C = 0 : P (375 mm) - (900 N )(151.5544 mm) = 0 +
P = 363.731 N (b)
P = 364 N Ø 
C x = 900 N Æ
+
SFx = 0 : C x - 900 N = 0
+
SFy = 0 : C y - P = 0 C y - 363.731 N = 0
C y = 364 N ≠
a = 22.0∞
C = 970.722 N
C = 971 N
22.0∞ 
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19
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 C = 0 : P (375 mm) - T (151.5544 mm) = 0
P = 0.40415T SFy = 0 : - P + C y = 0 - 0.40415T + C y = 0 C y = 0.40415T +
SFx = 0: -T + C x = 0
Cx = T C = C x2 + C y2 = T 2 + (0.40415T )2
C = 1.0786T
For
C = 1125 N
1125 N = 1.0786T fi T = 1043.02 N
T = 1043 N 
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20
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
Ty
At B:
Tx
=
0.18 m
0.24 m
3
Ty = Tx 4
(a)
(1)
SM C = 0 : Tx (0.18 m) - (240 N )(0.4 m) - (240 N )(0.8 m) = 0 +
Tx = +1600 N Ty =
Eq. (1)
3
(1600 N ) = 1200 N 4
T = Tx2 + Ty2 = 16002 + 12002 = 2000 N (b)
+
SFx = 0: C x - Tx = 0
C x - 1600 N = 0 C x = +1600 N +
T = 2.00 kN 
C x = 1600 N Æ
SFy = 0 : C y - Ty - 240 N - 240 N = 0 C y - 1200 N - 480 N = 0 C y = +1680 N a = 46.4∞
C = 2320 N C y = 1680 N ≠
C = 2.32 kN
46.4° 
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21
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
Ty
At B:
Tx
=
0.32 m
0.24 m
4
Ty = Tx
3
SM C = 0 : Tx (0.32 m) - (240 N )(0.4 m) - (240 N )(0.8 m) = 0 +
Tx = 900 N 4
Ty = (900 N ) = 1200 N 3
Eq. (1)
T = Tx2 + Ty2 = 9002 + 12002 = 1500 N +
SFx = 0: C x - Tx = 0 C x - 900 N = 0 C x = +900 N
+
T = 1.500 kN 
C x = 900 N Æ
SFy = 0 : C y - Ty - 240 N - 240 N = 0
C y - 1200 N - 480 N = 0
C y = +1680 N
a = 61.8∞
C = 1906 N C y = 1680 N ≠
C = 1.906 kN
61.8° 
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22
PROBLEM 4.21
Determine the reactions at A and B when (a) h = 0, (b) h = 200 mm.
SOLUTION
Free-Body Diagram:
+
(a)
SMA = 0 : ( B cos 60∞)(0.5 m) - ( B sin 60∞)h - (150 N )(0.25 m) = 0 37.5
B=
0.25 - 0.866h
When h = 0:
B=
Eq. (1):
+
37.5
= 150 N 0.25
+
B = 150.0 N
30.0° 
SFy = 0 : Ax - B sin 60∞ = 0
Ax = (150)sin 60∞ = 129.9 N A x = 129.9 N Æ
SFy = 0 : Ay - 150 + B cos 60∞ = 0 Ay = 150 - (150) cos 60∞ = 75 N a = 30∞
A = 150.0 N (b)
(1)
A y = 75 N ≠
A = 150.0 N
30.0° 
B = 488 N
30.0° 
When h = 200 mm = 0.2 m
37.5
= 488.3 N 0.25 - 0.866(0.2)
SFx = 0 : Ax - B sin 60∞ = 0 B=
Eq. (1):
+
Ax = ( 488.3)sin 60∞ = 422.88 N +
A x = 422.88 N Æ
SFy = 0 : Ay - 150 + B cos 60∞ = 0 Ay = 150 - ( 488.3) cos 60∞ = -94.15 N a = 12.55∞
A = 433.2 N A y = 94.15 N Ø
A = 433 N
12.6° 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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23
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 A = 0 : ( E sin a )(200 mm) + ( E cos a )(125 mm) - (90 N )(250 mm) - (90 N )(75 mm
m) = 0
E=
(a)
When a = 30°:
E=
29250
200 sin a + 125 cos a
29250
= 140.454 N 200 sin 30∞ + 125 cos 30∞
+
E = 140.5 N
60.0° 
SFx = 0 : Ax - 90 N + (140.454 N) sin 30∞ = 0 Ax = +19.773 N
+
A x = 19.77 N Æ
SFy = 0 : Ay - 90 N + (140.454 N ) cos 30∞ = 0
Ay = -31.637 N
A y = 31.7 N Ø
A = 37.3 N
58.0° 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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24
PROBLEM 4.22 (Continued)
(b)
When a = 45∞ :
E=
+
29250
= 127.279 N 200 sin 45∞ + 125 cos a
45.0° 
SFx = 0 : Ax - 90 N + (127.279 N )sin 45∞ = 0 Ax = 0
+
E = 127.3 N
Ax = 0
SFy = 0 : Ay - 90 N + (127.279) cos 45∞ = 0 Ay = 0
Ay = 0 A=0 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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25
PROBLEM 4.23
For each of the plates and loadings shown, determine the reactions at A and B.
SOLUTION
(a)
Free-Body Diagram:
+
SMA = 0 : B(500 mm) - (250 N )(100 mm) - (200 N )(250 mm) = 0 B = +150 N +
B = 150.0 N ≠
SFx = 0 : Ax + 200 N = 0 Ax = -200 N
+
A x = 200 N ¨
SFy = 0 : Ay + B - 250 N = 0 Ay + 150 N - 250 N = 0 Ay = +100 N
A y = 100.0 N ≠
a = 26.56∞
A = 223.607 N A = 224 N
26.6° 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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26
PROBLEM 4.23 (Continued)
(b)
Free-Body Diagram:
+
SM A = 0 : ( B cos 30∞)(500 mm) - (200 N )(250 mm) - (250 N )(100 mm) = 0 B = 173.205 N +
B = 173.2 N
60.0° 
SFx = 0 : Ax - B sin 30∞ + 200 N Ax - (173.205 N )sin 30∞ + 200 N = 0 Ax = -113.3975 N
+
A x = 113.4 N ¨
SFy = 0 : Ay + B cos 30∞ - 250 N = 0 Ay + (173.205) cos 30∞ - 250 N = 0 Ay = +100 N
A y = 100.0 N ≠
a = 41.4∞
A = 151.192 N A = 151.2 N
41.4° 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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27
PROBLEM 4.24
For each of the plates and loadings shown, determine the reactions at A and B.
SOLUTION
(a)
Free-Body Diagram:
+
SM B = 0 : - A(500 mm) + (250 N )( 400 mm) - (200 N )(250 mm) = 0 A = +100 N +
A = 100.0 N ≠ 
SFx = 0 : 200 N + Bx = 0 Bx = -200 N +
B x = 200 N ¨
SFy = 0 : A + By - 250 N = 0 100 N + By - 250 N = 0 By = +150 N B y = 150.0 N ≠
a = 36.87∞
B = 250 N B = 250 N
36.9° 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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28
PROBLEM 4.24 (Continued)
(b)
+
SM A = 0 : -( A cos 30∞)(500 mm) - (200 N )(250 mm) + (250 N )( 400 mm) = 0 A = 115.47 N +
A = 115.5 N
60.0° 
SFx = 0 : A sin 30∞ + 200 N + Bx = 0 (115.47 N )sin 30∞ + 200 N + Bx = 0 Bx = -257.735 N
+
B x = 258 N ¨
SFy = 0 : A cos 30∞ + By - 250 N = 0 (115.47 N ) cos 30∞ + By - 250 N = 0 By = +150 N
a = 30.2∞
B = 298.207 N B y = 150.0 N ≠
B = 298 N
30.2° 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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29
PROBLEM 4.25
Determine the reactions at A and B when (a) a = 0, (b) a = 90∞,
(c) a = 30∞.
SOLUTION
(a)
a = 0∞
+
SM A = 0 : B(0.5 m) - 100 N ◊ m = 0 B = 200 N +
+
SFx = 0 : Ax = 0 SFy = 0 : Ay + 200 = 0 Ay = -200 N A = 200 N Ø
(b)
B = 200 N ≠ 
a = 90∞
+
SMA = 0 : B(0.3 m) - 100 N ◊ m = 0 B=
+
+
1000
N = 333.33 N 3
SFA = 0 : Ax - 333.33 N = 0,
Ax = 333.33 N SFy = 0 : Ay = 0 A = 333 N Æ
(c)
B = 333 N ¨ 
a = 30∞
+
SM A = 0 : ( B cos 30∞)(0.5 m) + ( B sin 30∞)(0.3 m) - 100 N ◊ m = 0 B = 171.523 N +
SFx = 0 : Ax - (171.523 N )sin 30∞ = 0 Ax = 85.762 N +
SFy = 0 : Ay + (171.523 N) cos 30∞ = 0
Ay = -148.543 N A = 171.523 N a = 60.0∞ A = 171.5 N
60.0°
B = 171.5 N
60.0° 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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30
PROBLEM 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 A = 0 : (T sin 45∞)(0.2 m) + (90 N )(0.1 m) + (90 N )(0.2 m) = 0 T = 190.92 N (b)
+
SFx = 0 : Ax - (190.92 N ) cos 45∞ = 0
Ax = +135 N +
T = 190.9 N 
A x = 135.0 N Æ
SFy = 0 : Ay - 90 N - 90 N + (190.92 N) sin 45∞ = 0 Ay = +45 N A y = 45.0 N ≠
A = 142.3 N
18.43°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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31
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
; a = 33.69∞ 15
Move T along BD until it acts at Point D.
+
SM A = 0 : (T sin 33.69∞)(0.15 m) - (90 N )(0.1 m) - (90 N )(0.2 m) = 0 T = 324.5 N (b)
+
T = 324 N 
SFx = 0 : Ax - (324.99 N ) cos 33.69∞ = 0
Ax = +270 N +
A x = 270 N Æ
SFy = 0 : Ay - 90 N - 90 N + (324.5 N) sin 33.69∞ = 0 Ay = 0
Ay = 0 A = 270 N Æ 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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32
PROBLEM 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
1
Triangle ACD is isosceles with S C = 90∞ + 30∞ = 120∞ S A = S D = (180∞ - 120∞) = 30∞
2
Thus DA forms angle of 60° with horizontal.
(a)
We resolve FAD into components along AB and perpendicular to AB.
+
(b)
+
+
SM C = 0 : ( FAD sin 30∞)(250 mm) - (500 N )(100 mm) = 0 SFx = 0 : - ( 400 N ) cos 60∞ + C x - 500 N = 0
SFy = 0 : - ( 400 N) sin 60∞ + C y = 0
FAD = 400 N 
C x = +300 N
C y = +346.4 N
C = 458 N
49.1° 
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33
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 A = 0 : - (1260 N )(75 mm)
+
T (300 mm) -
(b)
+
SFx = 0 :
7
T (300 mm)
25
24
T (200 mm) = 0
25
24T = 94500
T = 3937.5 N T = 3940 N 
7
(3937.5) + 3937.5 + Ax = 0
25
Ax = -5040 N
+
SFy = 0 : Ay - 1260 N Ay = +5040 N
A x = 5040 N ¨
24
(3937.5 N ) = 0
25
A y = 5040 N ≠
A = 7128 N
45.0° 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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34
PROBLEM 4.30
Neglecting friction, determine the tension in cable ABD and the reaction
at support C.
SOLUTION
Free-Body Diagram:
+
SM C = 0 : T (0.25 m) - T (0.1 m) - (120 N )(0.1 m) = 0 T = 80.0 N 
+
SFx = 0 : C x - 80 N = 0 C x = +80 N
C x = 80.0 N Æ
+
SFy = 0 : C y - 120 N + 80 N = 0 C y = +40 N
C y = 40.0 N ≠
C = 89.4 N
26.6° 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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35
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 D = 0 : C x ( R) - P ( R) = 0
+
Cx = + P
+
SFx = 0 : C x - B sinq = 0 P - B sin q = 0
B = P/sin q P
q
sinq
+ SFy = 0 : C y + B cosq - P = 0 B=
C y + ( P/sin q ) cos q - P = 0
1 ˆ
Ê
C y = P Á1 Ë tan q ˜¯
For q = 30∞ :
(a)
B = P/sin 30∞ = 2 P (b)
Cx = + P Cx = P Æ
C y = P (1 - 1/tan 30∞) = - 0.732/P B = 2P
C y = 0.7321P Ø
C = 1.239 P
60.0° 
36.2° 
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36
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:
Cx = + P
1 ˆ
Ê
C y = P Á1 Ë tan q ˜¯
B=
P
sin q
For q = 60∞ :
(a)
(b)
B = P/sin 60∞ = 1.1547 P B = 1.155 P
30.0° 
Cx = + P Cx = P Æ
C y = P (1 - 1/tan 60∞) = + 0.4226 P C y = 0.4226 P Ø
C = 1.086 P
22.9° 
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37
PROBLEM 4.33
Neglecting friction, determine the tension in cable ABD and the reaction
at C when q = 60°.
SOLUTION
+
SM C = 0 : T (2a + a cos q ) - Ta + Pa = 0 T=
+
+
P
1 + cosq
(1)
SFx = 0 : C x - T sinq = 0 P sin q
C x = T sin q =
1 + cos q
SFy = 0 : C y + T + T cosq - P = 0 C y = P - T (1 + cos q ) = P - P
Cy = 0
C y = 0, C = C x Since
1 + cos q
1 + cos q
C=P
sin q
Æ (2)
1 + cos q
For q = 60∞ :
P
P
=
1 + cos 60∞ 1 + 12
Eq. (1):
T=
Eq. (2):
C=P
sin 60∞
0.866
=P
1 + cos 60∞
1 + 12
T=
2
P 
3
C = 0.577 P Æ 
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38
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 C = 0 : - T ( a) - P ( a) + (T sin 45∞)(2a sin 45∞) + (T cos 45∞)( a + 2a cos 45∞) = 0
or T = 0.586 P 
T = 0.58579 +
SFx = 0 : C x + (0.58579 P )sin 45∞ = 0 C x = 0.41422 P +
SFy = 0 : C y + 0.58579 P - P + (0.58579 P ) cos 45∞ = 0 Cy = 0 or C = 0.414 P Æ 
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39
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:
SFx = 0 : A cos 30∞ - (600 N ) cos 60∞ = 0 A = 346 N Æ 
A = 346.4102 N +
SM B = 0 : C (200 mm) - (600 N )( 400 mm) cos 30∞ + (346.4102 N )(200 mm)sin 30∞ = 0
C = 866.0253 N +
C = 866 N
60.0° 
SM C = 0 : B(200 mm) - (600 N )(200 mm) cos 30∞ + (346.4102 N )( 400 mm)sin 30∞ = 0
B = 173.205 N B = 173.2 N
60.0° 
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40
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:
SFx = 0:
A= D
SFy = 0: TBE = 250 N 
We note that the forces shown form two couples.
+
SM = 0 : A(200 mm) - (250 N )(75 mm) = 0 A = 93.75 N A = 93.75 N Æ
D = 93.75 N ¨ 
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41
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
(a)
+
0.15 m
AE BE
BE
=
;
=
; BE = 0.075 m 0.5 m 0.25 m
AD CD
Ê 0.075 ˆ
SM A = 0 : Tx (0.25 m) - Á
T (0.5 m) - ( 400 N )(0.1 m) - ( 400 N))(0.4 m) = 0 Ë 0.35 x ˜¯
Tx = 1400 N 0.075
(1400 N )
0.35
= 300 N
Ty =
T = 1432 lb 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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42
PROBLEM 4.37 (Continued)
(b)
+
SFy = 0 : A - 300 N - 400 N - 400 N = 0 A = +1100 N (c)
+
A = 1100 N ≠ 
SFx = 0 : -C + 1400 N = 0
C = + 1400 N C = 1400 N ¨ 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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43
PROBLEM 4.38
Determine the tension in each cable and the reaction at D.
SOLUTION
0.08 m
0.2 m
a = 21.80∞
0.08 m
tan b =
0.1 m
b = 38.66∞ tan a =
+
SM B = 0 : (600 N )(0.1 m) - (TCF sin 38.66∞)(0.1 m) = 0 TCF = 960.47 N +
SM C = 0 : (600 N )(0.2 m) - (TBE sin 21.80∞)(0.1 m) = 0
TBE = 3231.1 N +
TCF = 96.0 N 
TBE = 3230 N 
SFy = 0 : TBE cos a + TCF cos b - D = 0 (3231.1 N ) cos 21.80∞ + (960.47 N ) cos 38.66∞ - D = 0 D = 3750.03 N D = 3750 N ¨ 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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44
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:
+
+
SFx = 0 : 75 N - B sin 30∞ = 0 B = 150 N
60.0° 
SM A = 0 : - (150 N )(100 mm) + B sin 30∞(75 mm) + B cos 30∞(275 mm) - F (325 mm) = 0 -15000 + 5625 + 35723.55 - F (325) = 0 F = +81.0725 N +
F = 81.1 N Ø 
SFy = 0 : A - 150 N + B cos 30∞ - F = 0 A - 150 N + (150 N ) cos 30∞ - 81.0725 N = 0 A = +101.169 N A = 101.2 N ≠ 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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45
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:
+
+
SFx = 0 : P - B sin 30∞ = 0 B = 2P
60°
SM A = 0 : - (150 N )(100 mm) + B sin 30∞(75 mm) + B cos 30∞(275 mm) - F (325 mm) = 0
-15000 N ◊ mm+ 2 P sin 30∞(75 mm) + 2 P cos 30∞(275 mm) - F (325 mm) = 0
-15000 + 75P + 476.314 P - 325F = 0
P=
325F + 15000
551.314 For F = 0:
P=
15000
= 27.208 N 551.314
For P = 100 N:
P=
32500 + 15000
= 86.158 N 551.314
For
0 F 100 N (1)
27.2 N P 86.2 N 
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46
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 = 0 : -T cos 30∞ + (80 N ) cos 30∞ = 0 y
T = 80 N +
T = 80.0 N 
SM C = 0 : ( A sin 30∞)(0.4 m) - (80 N )(0.2 m) - (80 N )(0.2 m) = 0 A = +160 N A = 160.0 N
+
30.0° 
SM A = 0 : (80 N )(0.2 m) - (80 N )(0.6 m) + (C sin 30∞)(0.4 m) = 0 C = +160 N C = 160.0 N
30.0° 
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47
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:
+
SFy = 0 : - T + (80 N ) cos 30∞ = 0 T = 69.282 N +
T = 69.3 N 
SM C = 0 : - (69.282 N ) cos 30∞(0.2 m)
-(80 N )(0.2 m) + ( A sin 30∞)(0.4 m) = 0
A = +140.000 N +
A = 140.0 N
30.0° 
C = 180.0 N
30.0° 
SM A = 0 : + (69.282 N ) cos 30∞(0.2 m)
- (80 N )(0.6 m) + (C sin 30∞)(0.4 m) = 0
C = +180.000 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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48
PROBLEM 4.43
An 8-kg mass can be supported in the three different ways shown. Knowing that the pulleys have a 100-mm
radius, determine the reaction at A in each case.
SOLUTION
W = mg = (8 kg)(9.81 m/s2 ) = 78.48 N SFx = 0 : Ax = 0
(a)
+
+
SFy = 0 : Ay - W = 0 A y = 78.48 N ≠
SM A = 0 : M A - W (1.6 m) = 0 M A = + (78.48 N )(1.6 m) M A = 125.56 N ◊ m
A = 78.5 N ≠
(b)
M A = 125.6 N ◊ m
+
SFx = 0 : Ax - W = 0 A x = 78.48 ≠
+
SFy = 0 : Ay - W = 0 A y = 78.48 ¨
A = (78.48 N ) 2 = 110.99 N
+

45°
SM A = 0 : M A - W (1.6 m) = 0 M A = + (78.48 N )(1.6 m)
A = 111.0 N
45°
M A = 125.56 N ◊ m M A = 125.6 N ◊ m

SFx = 0 : Ax = 0
(c)
+
SFy = 0 : Ay - 2W = 0
Ay = 2W = 2(78.48 N ) = 156.96 N ≠
+
SM A = 0 : M A - 2W (1.6 m) = 0 M A = + 2(78.48 N )(1.6 m)
A = 157.0 N ≠
M A = 125.1 N ◊ m
M A = 125 N ◊ m

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49
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
+
SFx = 0 : C x + (22.5 N ) = 0 C x = -22.5 N +
SFy = 0 : C y - (22.5 N ) = 0 C y = 22.5 N C = ( C x ) 2 + (C y )2
Then
= (22. 5) 2 + (22.5)2
= 31.8198 N
Ê +22.5 ˆ
= -45∞ or C = 31.8 N
q = tan -1 Á
Ë -22.5 ˜¯
and
+
45.0∞ 
SM C = 0 : M C + (22.5 N )(160 mm) + (22.5 N )(60 mm) = 0 M C = -4950 N ◊ mm
or
MC = 4950 N ◊ mm

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50
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
+
SFx = 0 : C x + (22.5 N ) = 0 C x = -22.5 N +
SFy = 0 : C y - (22.5 N ) = 0 C y = 22.5 N C = (C x )2 + (C y )2
Then
= (22.5)2 + (22.5)2
= 31.8198 N
Ê 22.5 ˆ
= -45.0∞ or C = 31.8 N
q = tan -1 Á
Ë -22.5 ˜¯
and
+
45.0∞ 
SM C = 0 : M C + (22.5 N )(165 mm) + (22.5 N )(65 mm) = 0 M C = -5175 N ◊ mm
or
MC = 5180 N ◊ mm

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51
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:
D
A
B
C
6m
20kN 20kN 20 kN 20 kN
1.8m 1.8 m 1.8 m 1.8 m
Ex
E
ME
F
4.5 m
My
Equilibrium Equations
150 kN
DF = ( 4.5 m)2 + (6 m)2 = 7.5m,
4.5
+
SFx = 0 : E x +
(150 kN ) = 0 7.5
E x = -90.0 kN +
SFy = 0 : E y - 4(20 kN) -
E x = 90.0 kN ¬ 
6
(150 kN ) = 0 7.5
E y = +200 kN +
E y = 200 kN  
SM E = 0 : (20 kN)(7.2 m) + (20 kN )(5.4 m) + (20 kN )(3.6 m) 6
(150 kN)(4.5 m) + M E = 0
+ (20 kN)(1.8 m) 7.5
M E = +180.0 kN ◊ m M E = 180.0 kN ◊ m 
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52
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
W = 500 N (a)
From f.b.d. of beam AD
+
SFx = 0 : Dx = 0 +
SFy = 0 : D y - 200 N - 200 N + 500 N = 0 D y = -100 N +
or
D = 100.0 N Ø 
SM D = 0 : M D - (500 N )(1.5 m) + (200 N )(2.4 m) + (200 N)(1.2 m) = 0
M D = 30 N ◊ m or M D = 30.0 N ◊ m

W = 450 N
(b)
From f.b.d. of beam AD
+
SFx = 0 : Dx = 0 +
SFy = 0 : D y + 450 N - 200 N - 200 N = 0 D y = -50 N
+
or
D = 50.0 N Ø 
SM D = 0 : M D - ( 450 N )(1.5 m) + (200 N )(2.4 m) + (200 N )(1.2 m) = 0
M D = -45 N ◊ m or M D = -45.0 N ◊ m

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53
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
M D = -60 N ◊ m For Wmin ,
From f.b.d. of beam AD
+
SM D = 0 : (200 N )(2.4 m) - Wmin (1.5 m) + (200 N )(1.2 m) - 60 N ◊ m = 0
Wmin = 440 N M D = 60 N ◊ m For Wmax,
From f.b.d. of beam AD
+
SM D = 0 : (200 N )(2.4 m) - Wmax (1.5 m) + (200 N )(1.2 m) + 60 N ◊ m = 0
Wmax = 520 N or 440 N W 520 N 
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54
PROBLEM 4.49
Knowing that the tension in wire BD is 1300 N, determine the
reaction at the fixed support C of the frame shown.
SOLUTION
T = 1300 N
5
T
13
= 500 N
12
Ty = T
13
= 1200 N Tx =
+
+
SM x = 0 : C x - 450 N + 500 N = 0
C x = 50 N ¨
C x = -50 N SFy = 0 : C y - 750 N - 1200 N = 0 C y = +1950 N C y = 1950 N ≠
C = 1951 N
+
88.5° 
SM C = 0 : M C + (750 N )(0.5 m) + ( 4.50 N )(0.4 m) - (1200 N )(0.4 m) = 0
M C = -75.0 N ◊ m MC = 75.0 N ◊ m

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55
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
+
Ê5 ˆ
SM C = 0 : (750 N)(0.5 m) + ( 450 N )(0.4 m) - Á T ˜ (0.6 m) Ë 13 ¯
Ê 12 ˆ
- Á T ˜ (0.15 m) + M C = 0
Ë 13 ¯
Ê 4.8 ˆ
375 N ◊ m + 180 N ◊ m - Á
m T + MC = 0 Ë 13 ˜¯
13
(555 + M C ) 4.8
13
M C = -100 N ◊ m: T =
(555 - 100) = 1232 N 4.8
13
M C = +100 N ◊ m: T =
(555 + 100) = 1774 N 4.8
T=
For
For
For
| M C | 100 N ◊ m : 1.232 kN T 1.774 kN 
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56
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 = 2W .
SOLUTION
(a)
Triangle ABC is isosceles.
CD = ( BC ) cos
We have
+
Setting
q
q
= l cos
2
2
qˆ
Ê
SM C = 0 : W Á l cos ˜ - P (l sin q ) = 0
Ë
2¯
q
q
q
q
q
sin q = 2 sin cos : Wl cos - 2 Pl sin cos = 0
2
2
2
2
2
W - 2 P sin
For P = 2W :
(b)
sin
q
=0
2
q W
W
=
=
= 0.25
2 2 P 4W
q
= 14.5∞ 2
or
ÊW ˆ
q = 2 sin -1 Á ˜ b
Ë 2P ¯
q = 29.0∞ b
q
= 165.5∞ q = 331∞(discard )
2
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57
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
+
ÈÊ 1 ˆ
˘
SM C = 0 : T (l sin q ) + W ÍÁ ˜ cos q ˙ - T (l cos q ) = 0
ÎË 2 ¯
˚
W cos q
T=
2(cos q - sin q )
Dividing both numerator and denominator by cos q,
T=
(b)
For T = 3W ,
or
3W =
1 ˆ
WÊ
˜
Á
2 Ë 1 - tanq ¯
or T =
(W2 )
(1 - tan q )
b
(W2 )
(1 - tan q )
1
1 - tan q =
6
Ê 5ˆ
q = tan -1 Á ˜ = 39.806∞ Ë 6¯
or q = 39.8∞ b
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58
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 C = 0 : P (l cos q ) + P (l sin q ) - M = 0
M
b
Pl
M = 1500 N ◊ m, P = 200 N, and l = 600 mm
1500 N ◊ m
5
sin q + cos q =
= = 1.25
(200 N)(600 mm) 4
or sinq + cosq =
(b)
For
sin 2 q + cos2 q = 1
Using identity
sin q + (1 - sin 2 q )1/ 2 = 1.25
(1 - sin 2 q )1/ 2 = 1.25 - sin q
1 - sin 2 q = 1.5625 - 2.5 sin q + sin 2 q
2 sin 2 q - 2.5 sin q + 0.5625 = 0
Using quadratic formula
sin q =
=
or
-( -2.5) ± (625) - 4(2)(0.5625)
2(2)
2.5 ± 1.75
4
sin q = 0.95572 and sin q = 0.29428
q = 72.886∞ and q = 17.1144∞
or q = 17.11∞ and q = 72.9∞ b
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59
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:
+
SFy = 0 : C cosq - P - Q = 0
C=
(a)
+
SM A = 0 : C
P+Q
cosq
a
- Pl cos q = 0
cos q
P+Q a
◊
- Pl cos q = 0 cos q cos q
(b)
For
cos3 q =
a( P + Q )
b
Pl
P = 80 N, Q = 60 N, l = 500 mm, and a = 125 mm
(125 mm)(80 N + 60 N )
(80 N )(500 mm)
= 0.4375
cos3 q =
cos q = 0.75915 q = 40.6∞ b
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60
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 = spring constant
s = elongation of spring
l
=
-l
cos q
l
=
(1 - cos q )
cos q
kl
T=
(1 - cos q )
cos q
(a)
(b)
 Fy = 0 : T sinq - W = 0
From f.b.d. of collar B
+
or
kl
(1 - cos q )sin q - W = 0 cos q
For
or tan q - sin q =
W
b
kl
W = 300 N
l = 500 mm
k = 800 N/m
500 mm
= 0.5 m
2400 N/m
300 N
= 0.75
tan q - sin q =
(800 N/m)(0.5 m)
l=
Solving numerically,
q = 57.957∞ or q = 58.0∞ b
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61
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 = 14 kl.
SOLUTION
First note
T = tension in spring = ks
where
s = elongation of spring
= ( AB)q - ( AB)q = 90∞
Êqˆ
Ê 90∞ ˆ
= 2l sin Á ˜ - 2l sin Á
Ë 2¯
Ë 2 ˜¯
È Êqˆ Ê 1 ˆ˘
= 2l Ísin Á ˜ - Á
˜˙
Î Ë 2¯ Ë 2 ¯ ˚
È Êqˆ Ê 1 ˆ˘
T = 2kl Ísin Á ˜ - Á
˜˙
Î Ë 2¯ Ë 2 ¯ ˚
(a)
(1)
È
Êqˆ˘
SM C = 0 : T Íl cos Á ˜ ˙ - P (l sin q ) = 0
Ë 2¯ ˚
Î
Substituting T from Equation (1)
From f.b.d. of rod BC
+
È Êqˆ Ê 1 ˆ˘È
Êqˆ˘
2kl Ísin Á ˜ - Á
l cos Á ˜ ˙ - P (l sin q ) = 0
˙
˜
Í
Ë 2¯ ˚
Î Ë 2¯ Ë 2 ¯ ˚ Î
È Êqˆ Ê 1 ˆ˘
È
Êqˆ
Êqˆ
Êqˆ˘
2kl 2 Ísin Á ˜ - Á
˜¯ ˙ cos ÁË 2 ˜¯ - Pl Í2 sin ÁË 2 ˜¯ cos ÁË 2 ˜¯ ˙ = 0
¯
Ë
Ë
2
2 ˚
Î
˚
Î
Factoring out
Êqˆ
2l cos Á ˜ , leaves
Ë 2¯
È Êqˆ Ê 1 ˆ˘
Êqˆ
- P sin Á ˜ = 0
kl Ísin Á ˜ - Á
˙
˜
Ë 2¯
Î Ë 2¯ Ë 2 ¯ ˚
or
È
˘
kl
q = 2 sin -1 Í
˙ b
Î 2 ( kl - P ) ˚
1 Ê kl ˆ
Êqˆ
sin Á ˜ =
˜
Á
Ë 2¯
2 Ë kl - P ¯
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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62
PROBLEM 4.56 (Continued)
(b)
P=
kl
4
kl
È
˘
q = 2 sin -1 Í
kl ˙
Î 2 kl - 4 ˚
È kl Ê 4 ˆ ˘
= 2 sin -1 Í
Á ˜˙
Î 2 Ë 3kl ¯ ˚
Ê 4 ˆ
= 2 sin -1 Á
Ë 3 2 ˜¯
(
)
= 2 sin -1 (0.94281)
= 141.058∞
or q = 141.1∞ b
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63
PROBLEM 4.57
Solve Sample Problem 4.5, assuming that the spring is unstretched when q = 90∞.
SOLUTION
First note
T = tension in spring = ks
where
s = deformation of spring
= rb
F = kr b
From f.b.d. of assembly
+
SM 0 = 0 : W (l cos b ) - F ( r ) = 0
Wl cos b - kr 2 b = 0
or
cos b =
For
k = 45 N/mm
r = 75 mm
l = 200 mm
W = 1800 N
cos b =
or
kr 2
b
Wl
( 45 N/mm)(75 mm)2
b
(1800 N )(200 mm)
cos b = 0.703125b
Solving numerically,
b = 0.89245 rad
or
b = 51.134∞
Then
q = 90∞ + 51.134∞ = 141.134∞ or q = 141.1∞ b
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64
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:
Fs = ks = k (l - l cos q ) = kl (1 - cos q )
Spring force:
(a)
(b)
ˆ
Êl
SM D = 0 : Fs (l sin q ) - W Á cos q ˜ = 0
¯
Ë2
W
kl (1 - cos q )l sin q - l cos q = 0
2
W
kl (1 - cos q ) tan q - = 0 2
W = 375 N
l = 750 mm = 0.75m
k = 600 N/m
(1 - cos q ) tan q = tan q - sin q
375 N
=
2(600 N/m)(0.75 m)
= 0.41667
+
For given values of
Solving numerically:
q = 49.710∞
or (1 - cos q ) tan q =
or
W
b
2kl
q = 49.7∞ b
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65
PROBLEM 4.59
Eight identical 500 ¥ 750-mm rectangular plates, each of mass m = 40 kg, are held in a vertical plane as
shown. All connections consist of frictionless pins, rollers, or short links. In each case, determine whether
(a) the plate is completely, partially, or improperly constrained, (b) the reactions are statically determinate or
indeterminate, (c) the equilibrium of the plate is maintained in the position shown. Also, wherever possible,
compute the reactions.
SOLUTION
1.
Three non-concurrent, non-parallel reactions
(a)
Plate: completely constrained
(b)
Reactions: determinate
(c)
Equilibrium maintained
A = C = 196.2 N ≠
2.
Three non-concurrent, non-parallel reactions
(a)
Plate: completely constrained
(b)
Reactions: determinate
(c)
Equilibrium maintained
B = 0, C = D = 196.2 N ≠
3.
Four non-concurrent, non-parallel reactions
(a)
Plate: completely constrained
(b)
Reactions: indeterminate
(c)
Equilibrium maintained
A x = 294 N Æ, D x = 294 N ¨
(A y + D y = 392 N ≠ )
4.
Three concurrent reactions (through D)
(a)
Plate: improperly constrained
(b)
Reactions: indeterminate
(c)
No equilibrium
( SM D π 0)
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66
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
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
53.1°
B = D y = 196.2 N ≠
(C + D x = 0)
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67
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
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
56.3°, B = 333 N ¨
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 = 333 N Æ
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68
PROBLEM 4.60 (Continued)
5.
Four non-concurrent, non-parallel reactions
(a)
Bracket: complete constraint
(b)
Reactions: indeterminate
(c)
Equilibrium maintained
( SM C = 0) A y = 250 N ≠
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 y + B 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)
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69
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.
DBCD:
tan b =
Free-Body Diagram:
(Three-force member)
240
180
b = 53.13∞
Force triangle
A = (300 N ) tan 53.13∞
= 400 N
300 N
cos 53.13∞
= 500 N
A = 400 N ≠ b
B=
B = 500 N
53.1∞ b
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70
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 B = 600 N)
300 N
= 0.5
600 N
b = 60.0∞
cos b =
Eq. (1)
240 mm
tan 60.0∞
= 138.56 mm
a=
For B 600 N a ≥ 138.6 mm b
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71
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 force
intersect.
tan b =
151.554 m
375 mm
b = 22.006∞
Force triangle
(a)
P = (900 N )tan 22.006∞
P = 363.7 (b)
C=
P = 364 N Ø b
900 N
= 970.7 N cos 22.006∞
C = 971 N
22.0∞ b
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72
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 =
151.554 mm
375 mm
b = 22.006∞
Force triangle
T = (1125 N ) cos 22.006∞
T = 1043.04 N
T = 1043 N b
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73
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.
75 sin 50∞
75 cos 50∞ + 375
= 0.135756
a = 7.7310∞
tan a =
Force triangle
300 N
sin 7.7310∞
= 2230.11 N
300 N
B=
tan 7.7310∞
= 2209.842 N
A=
A = 2230 N
7.73∞ b
B = 2210 N Æ b
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74
PROBLEM 4.66
Determine the reactions at B and D when b = 60 mm.
SOLUTION
Since CD is a two-force member, the line of action of reaction at D must pass through Points C and D.
Free-Body Diagram:
(Three-Force body)
Reaction at B must pass through E, where the reaction at D and 80 N force intersect.
220 mm
250 mm
b = 41.348∞
tan b =
Force triangle
Law of sines
B
D
80 N
=
=
sin 3.652∞ sin 45∞ sin 131.348∞
B = 888.0 N
B = 888 N
D = 942.8 N
41.3∞
D = 943 N
45.0∞ b
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75
PROBLEM 4.67
Determine the reactions at B and D when b = 120 mm.
SOLUTION
Since CD is a two-force member, line of action of reaction at D must pass through C and D.
Free-Body Diagram:
(Three-Force body)
Reaction at B must pass through E, where the reaction at D and 80-N force intersect.
280 mm
250 mm
b = 48.24∞
tan b =
Force triangle
Law of sines
B
D
80 N
=
=
sin 3.24∞ sin 135∞ sin 41.76∞
B = 1000.9 N
D = 942.8 N B = 1001 N
48.2∞ D = 943 N
45.0∞ b
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76
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:
Triangle EAD:
Triangle EGB:
Force triangle
75
= 0.6
125
a = 30.964∞
tan a =
AE = 250 tan a = 150 mm
GE = AE - AG = 150 - 37.5 = 112.5 mm
GB
75
=
GE 112.5
b = 33.690∞
tan b =
B
D
250 N
=
=
sin 120.964∞ sin 33.690∞ sin 25.346∞
B = 500.773 N
D = 323.943 N B = 501 N
56.3∞ b
C = D = 324 N
31.0∞ b
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77
PROBLEM 4.69
A 50-kg crate is attached to the trolley-beam system shown.
Knowing that a = 1.5 m, determine (a) the tension in cable CD,
(b) the reaction at B.
SOLUTION
Free-Body Diagram:
Three-Force body: W and TCD intersect at E.
0.7497 m
1.5 m
b = 26.56∞
tan b =
Force triangle 3 forces intersect at E.
W = (50 kg) 9.81 m/s2
= 490.5 N
Law of sines
TCD
490.5 N
B
=
=
sin 61.56∞ sin 63.44∞ sin 55∞
TCD = 498.9 N
B = 456.9 N
TCD = 499 N b
(a)
B = 457 N
(b)
26.6∞ b
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78
PROBLEM 4.70
Solve Problem 4.69, assuming that a = 3 m.
PROBLEM 4.69 A 50 kg crate is attached to the trolley-beam
system shown. Knowing that a = 1.5 m, determine (a) the tension
in cable CD, (b) the reaction at B.
SOLUTION
W and TCD intersect at E
Free-Body Diagram:
Three-Force body:
AE 0.301 m
=
AB
3m
b = 5.722∞
tan b =
Force Triangle (Three forces intersect at E.)
W = (50 kg) 9.81 m/s2
= 490.5 N
Law of sines
TCD
490.5 N
B
=
=
sin 29.278∞ sin 95.722∞ sin 55∞
TCD = 997.99 N
B = 821.59 N
TCD = 998 N b
(a)
B = 822 N
(b)
5.72∞ b
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79
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 a =
EF 575
=
AF 300
a = 62.447∞
tan b =
EH 125
=
DH 300
b = 22.620∞
Force triangle
A
T
200 N
=
=
sin 67.380∞ sin 27.553∞ sin 85.067∞
A = 185.3 N
62.4∞ b
T = 92.9 N b
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80
PROBLEM 4.72
Determine the reactions at A and D when b = 30∞.
SOLUTION
From f.b.d. of frame ABCD
SM D = 0 : - A(0.18 m) + [(150 N) sin 30∞](0.10 m)
+ [(150 N) cos 30∞](0.28 m)) = 0
+
A = 243.74 N
or A = 244 N Æ b
+
SFx = 0 : (243.74 N) + (150 N) sin 30∞ + Dx = 0
Dx = -318.74 N
+
SFy = 0 : D y - (150 N) cos 30∞ = 0
D y = 129.904 N
Then
D = ( Dx )2 + Dx2
= (318.74)2 + (129.904)2
= 344.19 N
and
Ê Dy ˆ
q = tan -1 Á
Ë Dx ˜¯
Ê 129.904 ˆ
= tan -1 Á
Ë -318.74 ˜¯
= -22.174∞ or D = 344 N
22.2∞ b
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81
PROBLEM 4.73
Determine the reactions at A and D when b = 60∞.
SOLUTION
From f.b.d. of frame ABCD
SM D = 0 : - A(0.18 m) + [(150 N) sin 60∞](0.10 m)
+ [(150 N) cos 60∞](0.28 m)) = 0
+
A = 188.835 N
or A = 188.8 N Æ b
+
SFx = 0 : (188.835 N ) + (150 N )sin 60∞ + Dx = 0
Dx = -318.74 N
+
SFy = 0 : D y - (150 N ) cos 60∞ = 0
D y = 75.0 N
Then
D = ( Dx ) 2 + ( D y ) 2
= (318.74)2 + (75.0)2
= 327.44 N
and
Ê Dy ˆ
q = tan -1 Á
Ë Dx ˜¯
Ê 75.0 ˆ
= tan -1 Á
Ë -318.74 ˜¯
= -13.2409∞
or D = 327 N
13.24∞ b
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82
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 kg)(9.81 m/s2 ) = 196.2 N
Free-Body Diagram:
P
100 mm
92 mm
D
30°
Ê 92 mm ˆ
a = cos -1 Á
= 23.074∞
Ë 100 mm ˜¯
W
f.b.d
N
P
From the geometry of the three forces acting on the roller
Applying the law of sines to the force triangle,
T
120°
N
W
q = 90∞ - 30∞ - a = 60∞ - 23.074∞ = 36.926∞
and
or
D
W
P
=
sin q sin a
196.2 N
P
=
sin 36.926∞ sin 23.074∞
\ P = 127.991 N
or P = 128.0 N
Free-Body Diagram:
100 mm P
30°
G
92 mm
(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 kg)(9.81 m/s2 ) = 196.2 N
From the geometry of the three forces acting on the roller
Ê 92 mm ˆ
a = cos -1 Á
= 23.074∞
Ë 100 mm ˜¯
α
W
N
α
N
W
30∞ b
q = 90∞ + 30∞ + a = 120∞ - 23.074∞ = 96.926∞
and
Applying the law of sines to the force triangle,
W
P
=
sin q sin a
or
θ
60°
196.2 N
P
=
sin 96.926∞ sin 23.074
\ P = 77.460 N
P
or P = 77.5 N
30∞ b
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83
PROBLEM 4.75
Member ABC is supported by a pin and bracket at B and by an
inextensible cord attached at A and C and passing over a frictionless
pulley at D. The tension may be assumed to be the same in portions
AD and CD of the cord. For the loading shown and neglecting the size
of the pulley, determine the tension in the cord and the reaction at B.
SOLUTION
Free-Body Diagram:
Reaction at B must pass through D.
tan a =
175 mm
300 mm
a = 30.256∞
tan b =
175 mm
600 mm
b = 16.26∞
Force triangle
Law of sines
T
T - 360 N
B
=
=
sin 59.744∞ sin 13.996∞ sin 106.26∞
T (sin 13.996∞) = (T - 360 N)(sin 59.744∞)
T (0.24185) = (T - 360)(0.86378)
T = 500 N B = (500 N)
T = 500 N b
sin 106.26∞
sin 59.744
= 555.695 N
B = 556 N
30.3∞ b
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84
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 =
120
; a = 36.9∞
160
T T - 75 lb B
=
=
4
3
5
3T = 4T - 300; T = 300 lb
5
5
B = T = (300 lb) = 375 lb 4
4
B = 375 lb
36.9°
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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85
PROBLEM 4.77
Rod AB is supported by a pin and bracket at A and rests against a frictionless peg
at C. Determine the reactions at A and C when a 170-N vertical force is applied
at B.
SOLUTION
Free-Body Diagram:
(Three-Force body)
The reaction at A must pass through D where C and 170-N force intersect.
160 mm
300 mm
a = 28.07∞
tan a =
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∞ - a ) - 2a = 90∞ - a
Force triangle is isosceles and we have
A = 170 N
C = 2(170 N )sin a
= 160.0 N
A = 170.0 N
33.9°
C = 160.0 N
28.1° b
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86
PROBLEM 4.78
Solve Problem 4.77, assuming that the 170-N force applied at B is horizontal
and directed to the left.
PROBLEM 4.77 Rod AB is supported by a pin and bracket at A and rests
against a frictionless peg at C. Determine the reactions at A and C when a 170-N
vertical force is applied at B.
SOLUTION
Free-Body Diagram: (Three-Force body)
The reaction at A must pass through D, where C and the 170-N force intersect.
160 mm
300 mm
a = 28.07∞
tan a =
We note that triangle ADB is isosceles (since AC = BC). Therefore S A = S 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 = 170.0 N
C = 2(170 N ) cos a = 300 N
A = 170.0 N
56.1°
C = 300 N
28.1° b
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87
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
55.662
250
b = 12.552∞
tan b =
Force triangle
Law of sines
150 N
A
B
=
=
sin17.448∞ sin 60∞ sin 102.552∞
A = 433.247 N
B = 488.31 N
A = 433 N
B = 488 N
12.55° b
30.0° b
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88
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∞
1
S A = S D = (180∞ - 120∞) = 30∞
2
On the other hand, from triangle BCF:
and
Dimensions in mm
CF = ( BC )sin 30∞ = 200 sin 30∞ = 100 mm
FD = CD - CF = 250 - 100 = 150 mm
From triangle EFD, and since S D = 30∞ :
EF = ( FD ) tan 30∞ = 150 tan 30∞ = 86.60 mm
From triangle EFC:
Force triangle
Law of sines
100 mm
CF
=
EF 86.60 mm
a = 49.11∞
tan a =
FAD
500 N
C
=
=
sin 49.11∞ sin 60∞ sin 70.89∞
FAD = 400 N, C = 458 N
(a)
FAD = 400 N b
C = 458 N
(b)
49.1° b
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89
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:
( 3 - 1) R
R
= 3 -1
b = 36.2∞
tan b =
Force triangle
Law of sines
P
B
C
=
=
sin 23.8∞ sin 126.2∞ sin 30∞
B = 2.00 P
C = 1.239 P
(a)
B = 2P
60.0° b
(b)
C = 1.239 P
36.2° b
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90
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 =
R-
R
3
Free-Body Diagram:
(Three-Force body)
R
1
=13
b = 22.9∞
Force triangle
Law of sines
P
B
C
=
=
sin 52.9∞ sin 67.1∞ sin 60∞
B = 1.155P
C = 1.086 P
(a)
B = 1.155 P
30.0° b
(b)
C = 1.086 P
22.9° b
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91
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
yED = xED = a,
Since
Slope of ED is
slope of HC is
Free-Body Diagram:
45∞
45∞
Also
DE = 2a
and
a
Ê 1ˆ
DH = HE = Á ˜ DE =
Ë 2¯
2
For triangles DHC and EHC
Now
a
2
a
R
2R
c = R sin( 45∞ - b )
sin b =
=
For
a = 20 mm and R = 100 mm
20 mm
sin b =
2 (100 mm)
= 0.141421
b = 8.1301∞
and
c = (100 mm)sin( 45∞ - 8.1301∞)
= 60.00 mm or c = 60.0 mm b
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92
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 force
geometry
Free-Body Diagram:
tan b =
xGB
y AB
where
y AB = L cosq
and
xGB =
tan b =
=
1
L sinq
2
1
2
L sin q
L cos q
1
tan q 2
or tan q = 2 tan b b
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93
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 =
xCB
y BC
where
xCB =
and
1
L sinq
2
y BC = L cos q
or
1
tan q
2
tan q = 2 tan b
For
b = 30∞
tan b =
tan q = 2 tan 30∞
= 1.15470
q = 49.107∞ or q = 49.1∞ b
W = mg = (8 kg)(9.81 m/s2 ) = 78.480 N
(b)
From force triangle
A = W tan b
= (78.480 N ) tan 30∞
= 45.310 N and
B=
W
78.480 N
=
= 90.621 N cos b
cos 30∞
or A = 45.3 N ¬ b
or B = 90.6 N
60.0∞ b
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94
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 BE = h
For triangle ACE:
S 2 = ( AE )2 + (2h)2 (1)
For triangle ABE:
L2 = ( AE )2 + ( h)2 (2)
Subtracting Equation (2) from Equation (1)
S 2 - L2 = 3h2 (3)
S 2 - L2
b
3
As length S increases relative to length L, angle q increases until
h=
or
rod AB is vertical. At this vertical position:
h+ L = S
or h = S - L
Therefore, for all positions of AB
hS - L or
S 2 - L2
S - L
3
or
S 2 - L2 3( S - L)2
(4)
= 3( S 2 - 2SL + L2 )
= 3S 2 - 6SL + 3L2
or
0 2S 2 - 6SL + 4 L2
and
0 S 2 - 3SL + 2 L2 = ( S - L)( S - 2 L)
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95
PROBLEM 4.86 (Continued)
For
S-L=0 S=L
Minimum value of S is L
For
S - 2L = 0 S = 2L
Maximum value of S is 2L
Therefore, equilibrium does not exist if S 2 L b
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96
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 BE = h
For triangle ACE:
S 2 = ( AE )2 + (2h)2 (1)
For triangle ABE:
L2 = ( AE )2 + ( h)2 (2)
Subtracting Equation (2) from Equation (1)
S 2 - L2 = 3h2
(a)
S 2 - L2
3
L = 500 mm and S = 750 mm
h=
or
For
(750)2 - (500)2
3
= 322.7486 mm
h=
or h = 322 mm b
W = 4.5 ¥ 9.81 = 44.145 N
(b) We have
Ê 2h ˆ
q = sin -1 Á ˜
Ë s¯
and
È 2(322.7486) ˘
= sin -1 Í
˙˚
750
Î
q = 59.391∞
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97
PROBLEM 4.87 (Continued)
From the force triangle
W
sin q
44.145 N
=
sin 59.391∞
= 51.2919 N T=
(c)
or T = 51.3 N b
W
tan q
44.145 N
=
tan 59.391∞
B=
= 26.1166 N or B = 26.1 N ¬ b
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98
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 angle q of
triangle DCA.
a = 2q
The horizontal projections of AE , ( x AE ), and AG, ( x AG ), are equal.
x AE = x AG = x A
or
( AE ) cos 2q = ( AG ) cos q
and
(2 R) cos 2q = R cos q
Now
then
or
cos 2q = 2 cos2 q - 1
4 cos2 q - 2 = cos q
4 cos2 q - cos q - 2 = 0
Applying the quadratic equation
cos q = 0.84307 and cos q = - 0.59307
q = 32.534∞ and q = 126.375∞( Discard ) or q = 32.5∞ b
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99
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:
(CE )2 + ( BE )2 = ( BC )2
(2 L cos q )2 + ( L sin q )2 = R2
2
Ê Rˆ
2
2
ÁË ˜¯ = 4 cos q + sin q
L
2
Ê Rˆ
2
2
ÁË ˜¯ = 4 cos q + 1 - cos q
L
2
Ê Rˆ
2
ÁË ˜¯ = 3 cos q + 1
L
2
˘
1È
cos2 q = ÍÊÁ R ˆ˜ - 1˙ b
3 ÎË L ¯
˚
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100
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
2
˘
1È
cos2 q = ÍÊÁ R ˆ˜ - 1˙
3 ÎË L ¯
˚
For L = 375 mm, R = 500 mm, and W = 45 N
2
˘
1 ÈÊ 500 mm ˆ
˙ ; q = 59.39∞
cos q = ÍÁ
1
3 ÍË 375 mm ˜¯
˙˚
Î
2
(a)
In DABC:
q = 59.4∞ b
BE
L sin q
1
=
= tan q
CE 2 L cos q 2
1
tan a = tan 59.39∞ = 0.8452
2
a = 40.2∞
tan a =
Force triangle
A = W tan a = ( 45 N ) tan 40.2∞ = 38.028 N
( 45 N )
W
=
= 58.916 N
B=
cos a cos 40.2∞
(b)
A = 38.0 N ® b
B = 58.9 N
(c)
49.8° b
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101
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
1.125 0.41758
i+
j + 0.3 k
2
2
We have 5 unknowns and six equations of equilibrium.
rG /B =
Plywood sheet is free to move in z direction, but equilibrium is maintained ( SFz = 0).
SM B = 0 : rA/B ¥ ( Ax i + Ay j) + rC /B ¥ ( -Ci ) + rG /B ¥ ( - wj) = 0
i
0
Ax
j
k
i
j
k
i
j
k
0 1.2 + 1.125 0.41758 0.6 + 0.5625 0.20879 0.3 = 0
Ay 0
-C
0
0
0
-1166.77 0
-1.2 Ay i + 1.2 Ax j - 0.6C j + 0.41758C k + 50.031i - 93.808125 k = 0
Equating coefficients of unit vectors to zero:
i:
-1.2 Ay + 50.031 = 0
j:
Ay = 41.6925 N
1
1
Ax = C = (224.64) = 112.32 N
2
2
- 0.6C + 1.2 Az = 0
k: 0.41758C - 93.808125 = 0
C = 224.64 N SFx = 0: Ax + Bx - C = 0:
Bx = 224.64 - 112.32 = 112.32 N
SFy = 0: Ay + By - W = 0:
By = 166.77 - 41.6925 = 125.08 N
C = 225 N b
A = (112.3 N)i + ( 41.7 N ) j B = (112.3 N )i + (125.1 N ) j C = -(225 N )i b
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102
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 A = 0 : (90i + 30k ) ¥ ( -80 j) + (210i + 40 j) ¥ ( -TC k ) + (300i ) ¥ ( Dx i + D y j + Dz k) = 0
-7200k + 2400i + 210TC j - 40TC i + 300 D y k - 300 Dz j = 0
Equate coefficients of unit vectors to zero:
i:
j:
2400 - 40TC = 0
TC = 60 N
210TC - 300 Dz = 0 (210)(60) - 300 Dz = 0
k : -7200 + 300 D y = 0
Dz = 42 N
D y = 24 N
SFx = 0:
Dx = 0
SFy = 0 :
Ay + D y - 80 N = 0
Ay = 80 - 24 = 56 N
SFz = 0 : Az + Dz - 60 N = 0
Az = 60 - 42 = 18 N
A = (56.0 N ) j + (18.00 N )k D = (24.0 N ) j + ( 42.0 N )k b
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103
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 A = 0 : (90i + 30k ) ¥ ( -80 j) + (210i + 50 j) ¥ ( -TC k ) + (300i ) ¥ ( Dx i + D y j + Dz k) = 0
-7200k + 2400i + 210TC j - 50TC i + 300 D y k - 300 Dz j = 0
Equate coefficients of unit vectors to zero:
i:
j:
2400 - 50TC = 0
TC = 48 N
210TC - 300 Dz = 0 (210)( 48) - 300 Dz = 0
k : - 7200 + 300 D y = 0
SFx = 0:
Dz = 33.6 N
D y = 24 N
Dx = 0
SFy = 0 : Ay + D y - 80 N = 0
Ay = 80 - 24 = 56 N
SFz = 0:
Az = 48 - 33.6 = 14.4 N
Az + Dz - 48 = 0
A = (56.0 N ) j + (14.40 N )k D = (24.0 N ) j + (33.6 N )k b
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104
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)
(b)
SM x-axis = 0 : (120 N - 90 N )(100 mm) + (150 N - T )(80 mm) = 0 T = 187.5 N b
SFx = 0 : Bx = 0
SM D ( z -axis ) = 0 : (150 N + 187.5 N )(150 mm) - By (300 mm) = 0
By = 168.75 N
SM D ( y -axis ) = 0 : (120 N + 90 N )(500 mm) + Bz (300 mm) = 0
Bz = -350 N
or B = (168.8 N ) j - (350 N )k b
SM B ( z -axis ) = 0 : - (150 N + 187.5 N )(150 mm) + D y (300 mm) = 0
D y = 168.75 N
SM B ( y -axis ) = 0 : (120 N + 90 N )(200 mm) + Dz (300 mm) = 0
Dz = -140 N
or D = (168.8 N ) j - (140.0 N )k b
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105
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
SMC = 0 : ( -120k ) ¥ ( Dx i + D y j) + (120 j - 160k ) ¥ T i + (80k - 200 i ) ¥ ( -720 j) = 0
-120 Dx j + 120 D y i - 120T k - 160T j + 57.6 ¥ 103 i + 144 ¥ 103 k = 0
Equating to zero the coefficients of the unit vectors:
k:
i:
-120T + 144 ¥ 103 = 0 120 D y + 57.6 ¥ 103 = 0
j: - 120 Dx - 160(1200 N ) = 0
(b)
SFx = 0:
C x + Dx + T = 0
SFy = 0:
C y + D y - 720 = 0
SFz = 0:
Cz = 0
(a) T = 1200 N b
D y = -480 N
Dx = -1600 N
C x = 1600 - 1200 = 400 N
C y = 480 + 720 = 1200 N
C = ( 400 N )i + (1200 N ) j D = -(1600 N )i - ( 480 N ) j b
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106
PROBLEM 4.96
Solve Problem 4.95, assuming that the axle has been
rotated clockwise in its bearings by 30° and that the
720-N load remains vertical.
PROBLEM 4.95 A 200-mm lever and a 240-mmdiameter pulley are welded to the axle BE that is
supported by bearings at C and D. If a 720-N vertical
load is applied at A when the lever is horizontal,
determine (a) the tension in the cord, (b) the reactions
at C and D. Assume that the bearing at D does not exert
any axial thrust.
SOLUTION
Dimensions in mm
We have six unknowns and six equations of equilibrium.
SMC = 0 : ( -120k ) ¥ ( Dx i + D y j) + (120 j - 160k ) ¥ T i + (80k - 173.21i ) ¥ ( -720 j) = 0
-120 Dx j + 120 D y i -120T k - 160T j + 57.6 ¥ 103 i + 124.71 ¥ 103 k = 0
Equating to zero the coefficients of the unit vectors:
k : - 120T + 124.71 ¥ 103 = 0
i:
T = 1039 N b
120 D y + 57.6 ¥ 103 = 0 D y = -480 N
j: - 120 Dx - 160(1039.2)
(b)
T = 1039.2 N SFx = 0:
C x + Dx + T = 0
SFy = 0:
C y + D y - 720 = 0
SFz = 0:
Cz = 0
Dx = -1385.6 N
C x = 1385.6 - 1039.2 = 346.4
C y = 480 + 720 = 1200 N
C = (346 N )i + (1200 N ) j D = -(1386 N )i - ( 480 N ) j b
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107
PROBLEM 4.97
An opening in a floor is covered by a 1 ¥ 1.2-m sheet of
plywood of mass 18 kg. The sheet is hinged at A and B and
is maintained in a position slightly above the floor by a small
block C. Determine the vertical component of the reaction
(a) at A, (b) at B, (c) at C.
SOLUTION
rB /A = 0.6i
rC /A = 0.8i + 1.05k
rG /A = 0.3i + 0.6k
W = mg = (18 kg)9.81
W = 176.58 N
SM A = 0 : rB /A ¥ Bj + rC /A ¥ Cj + rG /A ¥ ( -Wj) = 0
(0.6i ) ¥ Bj + (0.8i + 1.05k ) ¥ Cj + (0.3i + 0.6k ) ¥ ( -Wj) = 0
0.6 Bk + 0.8Ck - 1.05Ci - 0.3Wk + 0.6Wi = 0
Equate coefficients of unit vectors of zero:
Ê 0.6 ˆ
i: 1.05C + 0.6W = 0 C = Á
176.58 N = 100.90 N
Ë 1.05 ˜¯
k : 0.6 B + 0.8C - 0.3W = 0
0.6 B + 0.8(100.90 N) - 0.3(176.58 N ) = 0 B = -46.24 N
SFy = 0 : A + B + C - W = 0
A - 46.24 N + 100.90 N + 176.58 N = 0
( a) A = 121.9 N (b) B = -46.2 N (c) C = 100.9 N A = 121.92 N
b
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108
PROBLEM 4.98
Solve Problem 4.97, assuming that the small block C is moved
and placed under edge DE at a point 0.15 m from corner E.
PROBLEM 4.97 An opening in a floor is covered by a
1 ¥ 1.2-m sheet of plywood of mass 18 kg. The sheet is hinged
at A and B and is maintained in a position slightly above the
floor by a small block C. Determine the vertical component of
the reaction (a) at A, (b) at B, (c) at C.
SOLUTION
rB /A = 0.6i
rC /A = 0.65i + 1.2k
rG /A = 0.3i + 0.6k
W = mg = (18 kg) 9.81 m/s2
W = 176.58 N
S MA = 0 : rB /A ¥ Bj + rC /A ¥ Cj + rG /A ¥ ( -Wj) = 0
0.6i ¥ Bj + (0.65i + 1.2k ) ¥ Cj + (0.3i + 0.6k ) ¥ ( -Wj) = 0
0.6 Bk + 0.65Ck - 1.2Ci - 0.3Wk + 0.6Wi = 0
Equate coefficients of unit vectors to zero:
Ê 0.6 ˆ
176.58 N = 88.29 N
i: - 1.2C + 0.6W = 0 C = Á
Ë 1.2 ˜¯
k : 0.6 B + 0.65C - 0.3W = 0
0.6 B + 0.65(88.29 N) - 0.3(176.58 N ) = 0 B = -7.36 N
SFy = 0 : A + B + C - W = 0
A - 7.36 N + 88.29 N - 176.58 N = 0
( a) A = 95.6 N (b) - 7.36 N (c) 88.3 N A = 95.648 N
b
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109
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 B = 0 : rA/B ¥ TA j + rC /B ¥ TC j + rG /B ¥ ( -392.4 N ) j = 0
(1500 mm)k ¥ TA j + [(1500 mm)i + (375 mm)k ] ¥ TC j + [(750 mm)i + (750 mm)k ] ¥ ( -392.4 N ) j = 0
-1500TAi + 1500TC k - 375TC i - 294300k + 294300i = 0
Equating to zero the coefficients of the unit vectors:
k:
1500TC - 294300 = 0 fi TC = 196.2 N TC = 196.2 N b
i:
-1500TA - 375 (196.2) + 294300 = 0 fi TA = 147.15 N
TA = 147.2 N b
SFy = 0:
TA + TB + TC - 392.4 N = 0
147.15 N + TB + 196.2 - 392.4 = 0 fi TB = 49.05 N TB = 49.1 N b
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110
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
TA = TB = TC = T
SM 0 = 0 : (rA ¥ Tj) + (rB ¥ Tj) + (rC ¥ Tj) + rG ¥ ( -Wj) + ( xi + zk ) ¥ ( -Wb j) = 0
or,
(1875 k ) ¥ Tj + (375 k ) ¥ Tj + (1500i + 750k ) ¥ Tj + (750i + 1125k ) ¥ ( -Wj) + ( xii + zk ) ¥ ( -Wb j) = 0
or,
-1875Ti - 375T i + 1500T k - 750T i - 750W k + 1125W i - xWb k + zWb i = 0
Equate coefficients of unit vectors to zero:
i:
- 3000T + 1125W + zWB = 0 (1)
k:
1500T - 750W - xWB = 0 (2)
SFy = 0:
3T - W - Wb = 0 (3)
Eq. (1) + 1000 Eq. (3):
125W + ( z - 1000)Wb = 0 (4)
Eq. (2) – 500 Eq. (3):
-250W - ( x - 500)Wb = 0 (5)
Also,
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111
PROBLEM 4.100 (Continued)
Solving (4) and (5) for Wb /W and recalling of 0 x 1500 mm, 0 z 2250 mm,
Eq.(4):
Eq.(5):
Wb
125
125
=
= 0.125
W 1000 - z 1000 - 0
Wb
250
250
=
= 0.5
W 500 - x 500 - 0
(Wb ) min = 196.2 N b
Thus, (Wb ) min = 0.5W = 0.5(392.4 N ) = 196.2 N Making Wb = 0.5W in (4) and (5):
125W + ( z - 1000)(0.5W ) = 0 z = 750 mm b
-250W - ( x - 500)(0.5W ) = 0 x = 0 mm b
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112
PROBLEM 4.101
Two steel pipes AB and BC, each having a mass per unit length
of 8 kg/m, are welded together at B and supported by three
wires. Knowing that a = 0.4 m, determine the tension in each
wire.
SOLUTION
W1 = 0.6m¢ g
W2 = 1.2m¢ g
SM D = 0 : rA/D ¥ TA j + rE /D ¥ ( -W1 j) + rF /D ¥ ( -W2 j) + rC /D ¥ TC j = 0
( -0.4i + 0.6k ) ¥ TA j + ( -0.4i + 0.3k ) ¥ ( -W1 j) + 0.2i ¥ ( -W2 j) + 0.8i ¥ TC j = 0
-0.4TAk - 0.6TAi + 0.4W1k + 0.3W1i - 0.2W2 k + 0.8TC k = 0
Equate coefficients of unit vectors to zero:
1
1
i: - 0.6TA + 0.3W1 = 0; TA = W1 = 0.6m¢ g = 0.3m¢ g
2
2
k : - 0.4TA + 0.4W1 - 0.2W2 + 0.8TC = 0
- 0.4(0.3m¢ g ) + 0.4(0.6m¢ g ) - 0.2(1.2m¢ g ) + 0.8TC = 0
(0.12 - 0.24 - 0.24)m¢ g
= 0.15m¢ g
0.8
SFy = 0 : TA + TC + TD - W1 - W2 = 0
0.3m¢ g + 0.15m¢ g + TD - 0.6m¢ g - 1.2m¢ g = 0
TD = 1.35m¢ g
m¢ g = (8 kg/m)(9.81m/s2 ) = 78.48 N/m
TC =
TA = 0.3m¢ g = 0.3 ¥ 78.45 TA = 23.5 N b
TB = 0.15m¢ g = 0.15 ¥ 78.45 TB = 11.77 N b
TC = 1.35m¢ g = 1.35 ¥ 78.45 TC = 105.9 N b
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113
PROBLEM 4.102
For the pipe assembly of Problem 4.101, determine (a) the
largest permissible value of a if the assembly is not to tip,
(b) the corresponding tension in each wire.
SOLUTION
W1 = 0.6m¢ g
W2 = 1.2m¢ g
SM D = 0 : rA/D ¥ TA j + rE /D ¥ ( -W1 j) + rF /D ¥ ( -W2 j) + rC /D ¥ TC j = 0
( - ai + 0.6k ) ¥ TA j + ( - ai + 0.3k ) ¥ ( -W1 j) + (0.6 - a)i ¥ ( -W2 j) + (1.2 - a)i ¥ TC j = 0
-TA ak - 0.6TAi + W1ak + 0.3W1i - W2 (0.6 - a)k + TC (1.2 - a)k = 0
Equate coefficients of unit vectors to zero:
1
1
i: - 0.6TA + 0.3W1 = 0; TA = W1 = 0.6m¢ g = 0.3m¢ g
2
2
k : - TA a + W1a - W2 (0.6 - a) + TC (1.2 - a) = 0
- 0.3m¢ ga + 0.6m¢ ga - 1.2m¢ g (0.6 - a) + TC (1.2 - a) = 0
0.3a - 0.6a + 1.2(0.6 - a)
1.2 - a
-0.3a + 1.2(0.6 - a) = 0
-0.3a + 0.72 - 1.2a = 0
1.5a = 0.72 TC =
(a)
For Max a and no tipping, TC = 0
a = 0.480 m b
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114
PROBLEM 4.102 (Continued)
(b)
Reactions:
m¢ g = (8 kg/m) 9.81 m/s2 = 78.48 N/m
TA = 0.3m¢ g = 0.3 ¥ 78.48 = 23.544 N TA = 23.5 N b
SFy = 0 : TA + TC + TD - W1 - W2 = 0
TA + 0 + TD - 0.6m¢ g - 1.2m¢ g = 0
TD = 1.8m¢ g - TA = 1.8 ¥ 78.48 - 23.544 = 117.72 TD = 117.7 N b
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115
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.
SOLUTION
rB /A = ai + 750k
rC /A = 750i + ak
rG /A = 375i + 375k
By symmetry: B = C
SM A = 0 : rB /A ¥ Bj + rC ¥ Cj + rG /A ¥ ( -Wj) = 0
( ai + 750k ) ¥ Bj + (750i + ak ) ¥ Bj + (375i + 375k ) ¥ ( -Wj) = 0
Bak - 750 Bi + 750 Bk - Bai - 375Wk + 375Wi = 0
Equate coefficient of unit vector i to zero:
i: - 750 B - Ba + 375W = 0
B=
375W
375W
C=B=
750 + a
750 + a
(1)
SFy = 0 : A + B + C - W = 0
È 375W ˘
A+ 2Í
- W = 0;
Î 750 + a ˙˚
A=
aW
750 + a
(2)
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116
PROBLEM 4.103 (Continued)
(a)
For
Eq. (1)
a = 250 mm,
375 ¥ 120
= 45 N
1000
250(120)
A=
= 30 N 1000
C=B=
Eq. (2)
(b)
W = 120 N
A = 30.0 N
B = C = 45.0 N b
For tension in each wire = 40 N
Eq. (1)
40 N =
375 ¥ 120
(750 + a)
750 mm + a = 1125 mm a = 375 mm b
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117
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 = 0.6 m b = r sin 30∞ = 0.3 m
We shall sum moments about AB.
(b + r )C + ( a - b) P - bW = 0
(0.3 + 0.6)C + ( a - 0.3)500 - (0.3)150 = 0
1
C=
[45 - ( a - 0.3)500]
0.9
If table is not to tip, C 0
[45 - ( a - 0.3)500] 0
45 ( a - 0.3)500
a - 0.3 0.09 a 0.39 m a = 0.39 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
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118
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 ( SM x = 0).
Free-Body Diagram:
uuur
BD = ( -1.8 m)i + (2.1 m) j + (1.8 m)k BD = 3.3 m
BE = ( -1.8 m)i + (2.1 m) j + (1.8 m)k BE = 3.3 m
uuur
T
BD TBD
TBD = TBD
=
( -1.8i + 2.1j + 1.8k ) = BD ( -6i + 7 j + 6k )
BD 3.3
11
uuur
T
BE TBE
TBE = TBE
=
( -1.8i + 2.1j - 1.8k ) = BE ( -6i + 7 j - 6k )
BE 3.3
11
SM A = 0 : rB ¥ TBD + rB ¥ TBE + rC ¥ ( -3.4 j) = 0
T
T
1.8i ¥ BD ( -6i + 7 j + 6k ) + 1.8i ¥ BE ( -6i + 7 j - 6k ) + 3i ¥ ( -3.4 j) = 0
11
11
12.6
10.8
12.6
10.8
TBD k TBD j +
TBE k +
TBE j - 10.2k = 0
11
11
11
11
Equate coefficients of unit vectors to zero.
10.8
10.8
TBD +
TBE = 0 TBE = TBD
11
11
12.6
12.6
k:
TBD +
TBE - 10.2 = 0
11
11
ˆ
Ê 12.6
= 10.2 2Á
T
Ë 11 BD ˜¯
j: -
TBD = 4.4524 kN
TBD = 4.45 kN b
TBE = 4.45 kN b
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119
PROBLEM 4.105 (Continued)
SFx = 0 : Ax -
6
6
( 4.4524 kN ) - ( 4.4524 kN ) = 0
11
11
Ax = 4.8572 kN
SFy = 0 : Ay +
7
7
( 4.4524 kN ) + ( 4.4524 kN ) - 3.4 kN = 0
11
11
Ay = -2.2667 kN
SFz = 0 : Az +
6
6
( 4.4524 kN ) - ( 4.4524 kN ) = 0
11
11
Az = 0 A = ( 4.86 kN )i - (2.27 kN ) j b
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120
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).
TAD
TAE
rB = 1.2k
rA = 2.4k
uuur
AD = -0.8i + 0.6 j - 2.4k
AD = 2.6 m
uuur
AE = 0.8i + 1.2 j - 2.4k
AE = 2.8 m
uuur
AD TAD
=
=
( -0.8i + 0.6 j - 2.4k )
AD 2.6
uuur
AE TAE
=
=
(0.8i + 1.2 j - 2.4k )
AE 2.8
SM C = 0 : rA ¥ TAD + rA ¥ TAE + rB ¥ ( -3 kN ) j = 0
i
j
k
i
j
k
TAD
T
0
0
2.4
+ 0
0
2.4 AE + 1.2k ¥ ( -3.6 kN ) j = 0
2.6
2.8
-0.8 0.6 -2.4
0.8 1.2 -2.4
Equate coefficients of unit vectors to zero.
i : - 0.55385 TAD - 1.02857 TAE + 4.32 = 0 j : - 0.73846 TAD + 0.68671 TAE = 0
TAD = 0.92857 TAE (1)
(2)
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121
PROBLEM 4.106 (Continued)
Eq. (1):
Eq. (2):
-0.55385(0.92857) TAE - 1.02857 TAE + 4.32 = 0
1.54286 TAE = 4.32
TAE = 2.800 kN TAE = 2.80 kN b
TAD = 0.92857(2.80) = 2.600 kN TAD = 2.60 kN b
SFx = 0 : C x -
0.8
0.8
(2.6 kN ) +
(2.8 kN ) = 0
2.6
2.8
Cx = 0
SFy = 0 : C y +
1.2
0.6
(2.8 kN ) - (3.6 kN ) = 0
(2.6 kN ) +
2.8
2.6
C y = 1.800 kN
SFz = 0 : C z -
2.4
2.4
(2.8 kN ) = 0
(2.6 kN ) 2.8
2.6
C z = 4.80 kN
C = (1.800 kN ) j + ( 4.80 kN )k b
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122
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).
uuur
AD = -0.8i + 0.6 j - 2.4k
AD = 2.6 m
uuur
AE = 0.8i + 1.2 j - 2.4k
AE = 2.8 m
uuur
AD TAD
( -0.8i + 0.6 j - 2.4k )
TAD =
=
AD 2.6
uuur
AE TAE
(0.8i + 1.2 j - 2.4k )
TAE =
=
AE 2.8
SM C = 0 : rA ¥ TAD + rA ¥ TAE + rA ¥ ( -3.6 kN ) j
Factor rA :
rA ¥ (TAD + TAE - (3.6 kN ) j)
or:
Coefficient of i:
Coefficient of j:
TAD + TAE - (3 kN ) j = 0
(Forces concurrent at A)
TAD
T
(0.8) + AE (0.8) = 0
2.6
2.8
2.6
TAD =
TAE 2.8
TAD
T
(0.6) + AE (1.2) - 3.6 kN = 0
2.6
2.8
2.6
Ê 0.6 ˆ 1.2
+
T - 3.6 kN = 0
TAE Á
Ë 2.6 ˜¯ 2.8 AE
2.8
-
(1)
Ê 0.6 + 1.2 ˆ
= 3.6 kN
TAE Á
Ë 2.8 ˜¯
TAE = 5.600 kN TAE = 5.60 kN b
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123
PROBLEM 4.107 (Continued)
Eq. (1):
2.6
(5.6) = 5.200 kN 2.8
0.8
0.8
SFx = 0 : C x (5.2 kN ) +
(5.6 kN ) = 0;
2.6
2.8
1.2
0.6
(5.2 kN ) +
(5.6 kN ) - 3.6 kN = 0
SFy = 0 : C y +
2.8
2.6
2.4
2.4
SFz = 0 : C z (5.6 kN ) = 0
(5.2 kN ) 2.8
2.6
TAD =
TAD = 5.20 kN b
Cx = 0
Cy = 0
C z = 9.60 kN
C = (9.60 kN )k b
Note: Since forces and reaction are concurrent at A, we could have used the methods of Chapter 2.
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124
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:
WC = 300 ¥ 9.81 = 2943 N
WG = 100 ¥ 9.81 = 981 N
We have five unknowns (TDE , TDF , Ax , Ay , Az ) and five equilibrium equations. The boom is free to spin about
the AB axis, but equilibrium is maintained, since SM AB = 0.
uuuv
We have
BH = (9.15 m)i - (7 m) j
BH = 11.5205 m
uuuv
2.65
( 7 m ) j + ( 2 m )k
DE = ( 4.15 m)i 3.65
= ( 4.15 m)i - (5.0822 m) j + (2 m)k DE = 6.8594 m
uuuv
DF = ( 4.15 m)i - (5.0822 m) j - (2 m)k DF = 6.8594 m
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125
PROBLEM 4.108 (Continued)
uuuv
9.15i - 7 j
BH
TBH = TBH
= (2943 N )
= (2337.438 N )i - (1788.204 N) j
11.5205
BH
uuuv
T
DE
TDE = TDE
= DE ( 4.15i - 5.0822 j + 2k )
DE 6.8594
uuuv
T
DF
TDF = TDF
= DF ( 4.15i - 5.0822 j - 2k )
DF 6.8594
Thus:
(a)
SM A = 0 : (rJ ¥ WC ) + (rK ¥ WG ) + (rH ¥ TBH ) + (rE ¥ TDE ) + (rF ¥ TDF ) = 0
- (3.65i ) ¥ ( -2943j) - (1.825i ) ¥ ( -981j) + (5.5i ) ¥ (2337.438i - 1788.204 j)
i
j
k
i
j
k
TDE
TDF
1.5
0
2 +
1.5
0
+
-2 = 0
6.8594
6.8594
4.15 -5.0822 2
4.15 -5.0822 -2
10741.95k + 1790.325k - 9835.122k + (TDE - TDF )
5.3
7.6233
+
(TDE - TDF ) j (TDE + TDF )k = 0
6.8594
6.8594
Equating to zero the coefficients of the unit vectors:
or,
i or j:
TDE - TDF = 0
6.8594
i
TDE = TDF *
k : 10741.95 + 1790.325 - 9835.122 -
7.6233
(2TDE ) = 0
6.8594
(b)
(2 ¥ 5.0822)
Ê 4.15 ˆ
(1213.441) = 0
SFx = 0 : Ax + 2337.438 + 2 Á
Ë 6.8594 ˜¯
TDE = 1213.441 N
TDE = TDF = 1213 N b
Ax = -3805.72 N
Ê 5.0822 ˆ
(1213.441) = 0 Ay = 7510.31 N
SFy = 0 : Ay - 2943 - 981 - 1788.204 - 2 Á
Ë 6.8594 ˜¯
SFz = 0 : Az = 0 A = -(3810 N )i + (7510 N ) j b
*Remark: The fact that TDE = TDF could have been noted at the outset from the symmetry of structure with
respect to xy plane.
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126
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.
SOLUTION
Free-Body Diagram:
By symmetry with xy plane
TCD = TCE = T
uuur
CD = -3i + 1.5 j + 1.2k
CD = 3.562 m
uuuv
CD
-3i + 1.5 j + 1.2k
TCD = T
=T
CD
3.562
-3i + 1.5 j - 1.2k
TCE = T
3.562
rB /A = 2i
rC /A = 3i
SMA = 0 : rC /A ¥ TCD + rC /A ¥ TCE + rB /A ¥ ( -5 kN ) j = 0
i
j
k
i
j
k
i j k
T
T
3 0
0
+ 3 0
0
+ 2 0 0 =0
3.562
3.562
-3 1.5 1.2
-3 1.5 -1.2
0 -5 0
Coefficient of k:
T ˘
È
2 Í3 ¥ 1.5 ¥
- 10 = 0 T = 3.958 kN
3.562 ˙˚
Î
SF = 0 : A + TCD + TCE - 5 j = 0
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127
PROBLEM 4.109 (Continued)
Coefficient of k:
Az = 0
Coefficient of i:
Ax - 2[3.958 ¥ 3/ 3.562] = 0
Ax = 6.67 kN
Coefficient of j:
Ay + 2[3.958 ¥ 1.5/ 3.562] - 5 = 0
Ay = 1.667 kN
(a)
TCD = TCE = 3.96 kN
A = (6.67 kN )i + (1.667 kN ) j b
(b)
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128
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 is
maintained ( SM AC = 0)
rB /A = 2i
rC /A = 3i
Load at B.
= -(5 cos 30) j + (5 sin 30)k
= -4.33j + 2.5k
uuur
CD = -3i + 1.5 j + 1.2k
CD = 3.562 m
uuur
CD
T
TCD = TCD
=
( -3i + 1.5 j + 1.2k )
CD 3.562
Similarly,
TCE =
T
( -3i + 1.5 j - 1.2k )
3.562
SMA = 0 : rC /A ¥ TCD + rC /A ¥ TCE + rB /A ¥ ( -4.33j + 2.5k ) = 0
i
j
k
i
j
k
i
j
k
TCD
TCE
3 0
0
+ 3 0
0
+2
0
0 =0
3.562
3.562
-3 1.5 1.2
-3 1.5 -1.2
0 -4.33 2.5
Equate coefficients of unit vectors to zero.
j: - 3.6
TCD
T
+ 3.6 CE - 5 = 0
3.562
3.562
-3.6TCD + 3.6TCE - 17.810 = 0 (1)
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129
PROBLEM 4.110 (Continued)
k : 4.5
TCD
T
+ 4.5 CE - 8.66 = 0
3.562
3.562
4.5TCD + 4.5TCE = 30.846 (2) + 1.25(1):
Eq. (1):
(2)
9TCE - 53.11 = 0 TCE = 5.901 kN
-3.6TCD + 3.6(5.901) - 17.810 = 0
TCD = 0.954 kN
SF = 0 : A + TCD + TCE - 4.33j + 2.5k = 0
0.954
5.901
( -3) +
( -3) = 0
3.562
3.562
Ax = 5.77 kN
0.954
5.901
j: Ay +
(1.5) +
(1.5) - 4.33 = 0
3.562
3.562
Ay = 1.443 kN
0.954
5.901
k : Az +
(1.2) +
( -1.2) + 2.5 = 0
3.562
3.562
Az = -0.833 kN
i: Ax +
Answers:
(a)
TCD = 0.954 kN TCE = 5.90 kN b
A = (5.77 kN )i + (1.443 kN ) j - (0.833 kN )k b
(b)
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130
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.
rB = 750 mm k
Coefficient of i:
r = 1200 mm k
uuuAr
AD = -500i - 1200k
AD = 1300 mm
uuur
AE = 500 j - 1200k
AE = 1300 mm
uuur
BF = 850 mm
BF = 400i - 750k
uuur
AD
T
T
TAD = T
=
( -500i - 1200k ) = ( -5i - 12k )
AD 1300
13
uuur
T
T
AE
(500 j - 1200k ) = (5 j - 12k )
=
TAE = T
13
AE 1300
uuur
T
BF TBF
TBF = TBF
( 400i - 750k ) = BF (8i - 15k )
=
17
BF 850
SMC = 0 : rA ¥ TAD + rA ¥ TAE + rB ¥ TBF + rB ¥ ( -1600 N ) j = 0
i j
k
i j
k
i j k
T
T
T
0 0 1200
+ 0 0 1200
+ 0 0 750 BF + (750k ) ¥ ( -1600 j) = 0
13
13
17
-5 0 -12
0 5 -12
8 0 -15
-
6000
T + 1, 200, 000 = 0
13
T = 2600 N
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131
PROBLEM 4.111 (Continued)
-
Coefficient of j:
6000
6000
T+
TBD = 0
13
17
17
17
TBF = T = (2600) TBF = 3400 N
13
13
SF = 0 : TAD + TAE + TBF - 1600 j + C = 0
Coefficient of i:
Coefficient of j:
Coefficient of k:
5
8
(2600) + (3400) + C x = 0
13
17
C x = - 600 N
5
(2600) - 1600 + C y = 0
13
C y = + 600 N
12
12
15
- (2600) - (2600) - (3400) + C z = 0
13
13
17
C z = 7800 N
-
Answers: TDAE = T TDAE = 2600 N b
TBD = 3400 N b
C = -(600 N )i + (600 N ) j + (7800 N )k b
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132
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 is
maintained ( SMAC = 0).
T = tension in both parts of cable DAE.
rB = 750 mm k
rA = 1200 mm k
uuur
AD = -500i - 1200k
AD = 1300 mm
uuur
AE = 500 j - 1200k
AE = 1300 mm
uuur
BF = 850 mm
BF = 400i - 750k
uuur
AD
T
T
TAD = T
=
( -500i - 1200k ) = ( -5i - 12k )
AD 1300
13
uuur
T
T
AE
(500 j - 1200k ) = (5 j - 12k )
=
TAE = T
13
AE 1300
uuur
T
BF TBF
TBF = TBF
( 400i - 750k ) = BF (8i - 15k )
=
17
BF 850
SM C = 0 : rA ¥ TAD + rA ¥ TAE + rB ¥ TBF + rA ¥ ( -1600 N ) j = 0
i j
k
i j
k
i j k
T
T
T
0 0 1200
+ 0 0 1200
+ 0 0 750 BF + 1200k ¥ ( -1600 j) = 0
13
13
17
-5 0 -12
0 5 -12
8 0 -15
Coefficient of i:
-
6000
T + 1, 920, 000 = 0
13
T = 4160 N
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133
PROBLEM 4.112 (Continued)
Coefficient of j:
Coefficient of i:
Coefficient of j:
Coefficient of k:
Answers:
6000
6000
T+
TBD = 0
13
17
17
17
TBD = T = ( 4160)
13
13
-
TBD = 5440 N
SF = 0 : TAD + TAE + TBF - 1600 j + C = 0
5
8
- ( 4160) + (5440) + C x = 0
13
17
C x = -960 N
5
( 4160) - 1600 + C y = 0
13
Cy = 0
12
12
15
- ( 4160) - ( 4160) - (5440) + C z = 0
13
13
17
C z = 12480 N
TDAE = T TDAE = 4160 N b
TBD = 5440 N b
C = -(960 N )i + (12480 N )k b
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134
PROBLEM 4.113
A 20-kg cover for a roof opening is hinged at corners A and B.
The roof forms an angle of 30° with the horizontal, and the cover
is maintained in a horizontal position by the brace CE. Determine
(a) the magnitude of the force exerted by the brace, (b) the
reactions at the hinges. Assume that the hinge at A does not exert
any axial thrust.
SOLUTION
Force exerted by CD
F = F (sin 75∞)i + F (cos 75∞) j
F = F (0.2588i + 0.9659 j)
W = mg = 20 kg(9.81 m//s2 ) = 196.2 N
rA/B = 0.6k
rC /B = 0.9i + 0.6k
rG /B = 0.45i + 0.3k
F = F (0.2588i + 0.9659 j)
SM B = 0 : rG /B ¥ ( -196.2 j) + rC /B ¥ F + rA/B ¥ A = 0
i
j
k
i
j
k
i
0.45
0
0.3 + 0.9
0
0.6 F + 0
0
-196.2 0
0.2588 +0.9659 0
Ax
j
0
Ay
k
0.6 = 0
0
Coefficient of i:
+58.86 - 0.5796 F - 0.6 Ay = 0 (1)
Coefficient of j:
+0.1553F + 0.6 Ax = 0 (2)
Coefficient of k:
-88.29 + 0.8693F = 0 : F = 101.56 N
Eq. (2):
+58.86 - 0.5796(101.56) - 0.6 Ay = 0
Eq. (3):
+0.1553(101.56) + 0.6 Ax = 0
Ax = -26.29 N
F = 101.6 N Ay = 0
A = -(26.3 N )i b
SF : A + B + F - Wj = 0
Coefficient of i:
26.29 + Bx + 0.2588(101.56) = 0
Coefficient of j:
By + 0.9659(101.56) - 196.2 = 0 By = 98.1 N
Bz = 0 Coefficient of k:
Bx = 0
B = (98.1 N ) j b
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135
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.
SOLUTION
y
Free-Body Diagram:
80 mm
160 mm
r = 240 mm
C
Bxi
W = −mgj = −(30 kg)(9.81 m/s2)j = −(294 N)j
uuur
DC = -( 480 mm)i + (240 mm)j - (160 mm)k
uuur
DC
T=T
= - 67 Ti + 73 Tj - 27 Tk
DC
240 mm
Byj
B
Ayj
Axi
r = 240 mm
T
A
z
DC = 560 mm
x
Azk
D
r = 240 mm
W = – (294 N) j
Equilibrium Equations:
SF = 0 : Ax i + Ay j + Az k + Bx i + By j + T - (294 N ) j = 0
( Ax + Bx - 67 T )i + ( Ay + By + 73 T - 294 N) j + ( Az - 27 T )k = 0 (1)
SM B = S (r ¥ F) = 0 :
2rk ¥ ( Ax i + Ay j + Az k ) + (2ri + rk ) ¥ ( - 67 Ti + 73 Tj - 27 Tk ) + ( ri + rk ) ¥ ( -294 N ) j = 0
( -2 Ay - 73 T + 294 N)ri + (2 Ax - 27 T )rj + ( 67 T - 294 N )rk = 0 (2)
Setting the coefficients of the unit vectors equal to zero in Eq. (2)
Ax = +49.0 N
Ay = +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, Ax and Ay into these equations we obtain
Az = +98.0 N
Bx = +245 N
By = +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
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136
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
rB /A (960 - 180)i = 780i
450
ˆ
Ê 960
rG /A = Á
k
- 90˜ i +
¯
Ë 2
2
rC /A
T = Tension in cable DCE
= 390i + 225k
= 600i + 450k
uuur
CD = -690i + 675 j - 450k
CD = 1065 mm
uuur
CE = 270i + 675 j - 450k
CE = 855 mm
T
TCD =
( -690i + 675 j - 450k )
1065
T
TCE =
(270i + 675 j - 450k )
855
W = - mgi = -(100 kg)(9.81 m/s2 ) j = -(981 N ) j
SM A = 0 : rC /A ¥ TCD + rC /A ¥ TCE + rG /A ¥ ( -Wj) + rB /A ¥ B = 0
i
j
k
i
j
k
T
T
600
0
450
+ 600 0
450
+
1065
855
-690 675 -450
270 675 -450
i
+ 390
0
j
k
i
j
0
225 + 780 0
-981 0
0 By
k
0 =0
Bz
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137
PROBLEM 4.115 (Continued)
Coefficient of i:
Coefficient of j:
Coefficient of k:
Coefficient of i:
Coefficient of j:
Coefficient of k:
T
T
- ( 450)(675)
+ 220.725 ¥ 103 = 0
1065
855
T = 344.6 N T = 345 N b
344.6
344.6
( -690 ¥ 450 + 600 ¥ 450)
+ (270 ¥ 450 + 600 ¥ 450)
- 780Bz = 0
1065
855
Bz = 185.49 N
344.6
344.6
(600)(675)
+ (600)(675)
- 382.59 ¥ 103 + 780 By B y = 113.2 N
1065
855
B = (113.2 N ) j + (185.5 N )k b
-( 450)(675)
SF = 0 : A + B + TCD + TCE + W = 0
690
270
(344.6) +
(344.6) = 0
1065
855
675
675
Ay + 113.2 +
(344.6) +
(344.6) - 981 = 0
1065
855
Ax -
Az + 185.5 -
Ax = 114.4 N
Ay = 377 N
450
450
(344.6) (344.6) = 0
Az = 141.5 N
1065
855
A = (114.4 N )i + (377 N ) j + (144.5 N )k b
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138
PROBLEM 4.116
Solve Problem 4.115, assuming that cable DCE is replaced by a
cable attached to Point E and hook C.
PROBLEM 4.115 A 100-kg uniform rectangular plate is supported
in the position shown by hinges A and B and by cable DCE that
passes over a frictionless hook at C. Assuming that the tension is
the same in both parts of the cable, determine (a) the tension in the
cable, (b) the reactions at A and B. Assume that the hinge at B does
not exert any axial thrust.
SOLUTION
See solution to Problem 4.115 for free-body diagram and analysis leading to the following:
CD = 1065 mm
CE = 855 mm
T
Now:
TCD =
( -690i + 675 j - 450k )
1065
T
TCE =
(270i + 675 j - 450k )
855
W = - mgi = -(100 kg)(9.81 m/s2 ) j = -(981 N)j
SM A = 0: rC /A ¥ TCE + rG /A ¥ ( -Wj) + rB /A ¥ B = 0
i
j
k
i
j
k
i
j
T
600 0
450
+ 390
0
225 + 780 0
855
270 675 -450
0 -981 0
0 By
Coefficient of i:
Coefficient of j:
Coefficient of k:
k
0 =0
Bz
T
+ 220.725 ¥ 103 = 0
855
T = 621.3 N T = 621 N b
621.3
(270 ¥ 450 + 600 ¥ 450)
- 980 Bz = 0 Bz = 364.7 N
855
621.3
(600)(675)
- 382.59 ¥ 103 + 780 By = 0 By = 113.2 N
855
B = (113.2 N)j + (365 N)k b
-( 450)(675)
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139
PROBLEM 4.116 (Continued)
Coefficient of i:
Coefficient of j:
Coefficient of k:
SF = 0: A + B + TCE + W = 0
270
Ax +
(621.3) = 0 Ax = -196.2 N
855
675
Ay + 113.2 +
(621.3) - 981 = 0 Ay = 377.3 N
855
450
Az + 364.7 (621.3) = 0
Az = -37.7 N
855
A = -(196.2 N)i + (377 N)j - (37.7 N)k b
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140
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
rB /A = (1000 - 200)i = 800i
rE /A = (750 - 100)i + 500k
rG /A
= 650i + 500k
10000
i + 250k
=
2
= 500i + 250k
uuur
EF = 250i + 625 j - 500k
EF = 838.525 mm
uuur
AF
T=T
T (0.29814i + 0.74536 j - 0.59629k )
AF
SM A = 0 : rE /A ¥ T + rG /A ¥ ( -392.4 j) + rB /A ¥ B = 0
i
j
k
i
j
k
i
j
650
0
500 T + 500
0
250 + 800 0
0.29814 0.74536 -0.59629
0 -392.4 0
0 By
Coefficient of i:
-372.68T + 98100 = 0 fi T = 263.23 N
Coefficient of j:
(387.5885 + 149.07)(263.23) - 800 : Bz = 176.581 N
Coefficient of k:
( 484.484)(263.23) - 196200 + 800 By = 0 :
k
0 =0
Bz
T = 263 N b
By = 85.837 kN
B = (85.8 N)j + (176.6 N)k b
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141
PROBLEM 4.117 (Continued)
SF = 0 : A + B + T - (392.4 N)j = 0
Coefficient of i:
Coefficient of j:
Coefficient of k:
Ax + 0.29814(263.23) = 0
Ax = -78.479 N
Ay + 85.837 + 0.74536(263.23) - 392.4 = 0
Az + 176.581 - 0.59629(263.23) = 0
Ay = 110.362 N
Az = -19.6196 N
A = -(78.5 N)i + (110.4 N)j - (19.62 N)k b
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142
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
rB /A = (1000 - 200)i = 800i
rE /A = 650i + 500k
rG / A = 500i + 250k
uuur
EH = -750i + 300 j - 500k
EH = 950 mm
uuur
EH
T
( -750i + 300 j - 500k )
T=T
=
EH 950
SM A = 0 : rE /A ¥ T + rG /A ¥ ( -392.4) + rB /A ¥ B = 0
i
j
k
i
j
k
i
j
T
650
0
500
+ 500
0
250 + 800 0
950
-750 300 -500
0 -392.4 0
0 By
Coefficient of i:
Coefficient of j:
Coefficient of k:
T
+ (392.4)(250) = 0 T = 621.3 N
950
(621.3)
(325000 - 375000)
- 800 Bz = 0 Bz = - 40.875 N
950
-(300)(500)
(650)(300)
k
0 =0
Bz
T = 621N b
(621.3)
- (500)(392.4) + 800 By = 0 By = 85.8375 N
950
B = (85.8 N)j - ( 40.9 N)k b
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143
PROBLEM 4.118 (Continued)
SF = 0:
Coefficient of i:
Coefficient of j:
Coefficient of k:
A + B + T - (392.4 N)j = 0
750
Ax (621.3) = 0 Ax = 490.5 N
950
300
Ay + 85.8375 +
(621.3) - 392.4 = 0 Ay = 110.3625 N
950
500
Az - 40.875 (621.3) = 0 Az = 367.875 N
950
A = ( 491 N)i + (110.4 N)j + (368 N)k b
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144
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
rH /C = -50i + 250 j
rF /C = 350i + 250k
T = T (sin 30∞) j - T (cos 30∞)k = T (0.5 j - 0.866k )
SMC = 0 : rF /C ¥ ( -400 j) + rH /C ¥ T + ( M C ) y j + ( M C ) z k = 0
i
j
k
i
j
k
350
0
250 + -50 250
0 T + ( MC ) y j + ( MC ) z k = 0
0 -400 0
0
0.5 -0.866
Coefficient of i:
Coefficient of j:
+100 ¥ 103 - 216.5T = 0 T = 461.9 N T = 462 N b
-43.3( 461.9) + ( M C ) y = 0
( M C ) y = 20 ¥ 103 N ◊ mm
(M C ) y = 20.0 N ◊ m
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145
PROBLEM 4.119 (Continued)
Coefficient of k: -140 ¥ 103 - 25( 461.9) + ( M C ) z = 0
( M C ) z = 151.54 ¥ 103 N ◊ mm
(M C ) z = 151.5 N ◊ m
SF = 0 : C + T - 400 j = 0
MC = (20.0 N ◊ m)j + (151.5 N ◊ m)k b
Coefficient of i:
Coefficient of j:
Coefficient of k:
Cx = 0
C y + 0.5( 461.9) - 400 = 0 C y = 169.1 N
C z - 0.866( 461.9) = 0 C z = 400 N
C = (169.1 N)j + ( 400 N)k b
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146
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.
SOLUTION
Weight of plate = 40 × 9.81 = 392.4 N
rE /A = 650i + 500k
rG /A = 500i + 250k
uuur
EF = 250i + 625 j - 500k
EF = 838.525 mm
uuur
EF
T =T
= T (0.29814i + 0.75436 j - 0.59629k )
EF
SM A = 0 : rE /A ¥ T + rG /A ¥ ( -75 j) + ( M A ) y j + ( M A ) z k = 0
i
j
k
i
j
k
650
0
500 T + 500
0
250 + ( M A ) y j + ( M A ) z k = 0
0.29814 0.75430 -0.59629
0 -392.4 0
Coefficient of i:
-372.68T + 98100 = 0 T = 263.23 N
T = 263 N b
Coefficient of j:
(387.5885 + 149.07)(263.23) + ( M A ) y = 0 ( M A ) y = 141264.6 N ◊ mm
Coefficient of k:
( 484.484)(263.23) - 196200 + ( M A ) z = 0
( M A ) z = 68669.3 N ◊ mm
SF = 0 : A + T - 392.44 j = 0
M A = -(141300 N ◊ mm)j + (68700 N ◊ mm)k b
Coefficient of i:
Ax + 0.29814(263.23) = 0
Ax = -78.479 N
Coefficient of j:
Ay + (0.74536)(263.23) - 392.4 = 0
Ay = 196.199 N
Coefficient of k:
Az – 0.59629(263.23)
Az = 156.961 N
A = ( -78.5 N) i + (196.2 N)j + (157.0 N)k b
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147
PROBLEM 4.121
The assembly shown is used to control the tension T in a tape that passes
around a frictionless spool at E. Collar C is welded to rods ABC and CDE.
It can rotate about shaft FG but its motion along the shaft is prevented by
a washer S. For the loading shown, determine (a) the tension T in the tape,
(b) the reaction at C.
SOLUTION
Free-Body Diagram:
rA/C = 105 j + 50k
rE /C = 40i - 60 j
SM C = 0 : rA/C ¥ ( -3j) + rE /C ¥ T (i + k ) + ( M C ) y j + ( M C ) z k = 0
(105 j + 50k ) ¥ ( -3j) + ( 40i - 60 j) ¥ T (i + k ) + ( M C ) y j + ( M C ) z k = 0
Coefficient of i:
Coefficient of j:
Coefficient of k:
150 - 60T = 0 T = 2.5 N b
-40(2.5 N) + (M C ) y = 0 ( M C ) y = 100 N ◊ mm
60(2.5 N) + ( M C ) z = 0 ( M C ) z = -150 N ◊ mm
MC = (100.0 N ◊ mm)j - (150.0 N ◊ mm)k b
SF = 0 : C x i + C y j + C z k - (3 N)j + (2.5 N)i + (2.5 N)k = 0
Equate coefficients of unit vectors to zero.
C x = -2.5 N C y = 3 N C z = -2.5 N b
C = -(2.50 N)i + (3.00 N)j - (2.50 N)k b
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148
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.
SOLUTION
Free-Body Diagram:
First note
TCF = lCF TCF =
-(0.08 m)i + (0.06 m) j
(0.08)2 + (0.06)2 m
TCF
= TCF ( -0.8i + 0.6 j)
TDE = l DE TDE =
(0.12 m)j - (0.09 m)k
(0.12)2 + (0.09)2 m
TDE
= TDE (0.8 j - 0.6k )
(a)
From f.b.d. of assembly
SFy = 0 : 0.6TCF + 0.8TDE - 480 N = 0
0.6TCF + 0.8TDE = 480 N or
(1)
SM y = 0 : - (0.8TCF )(0.135 m) + (0.6TDE )(0.08 m) = 0
TDE = 2.25TCF or
(2)
Substituting Equation (2) into Equation (1)
0.6TCF + 0.8[(2.25)TCF ] = 480 N
TCF = 200.00 N
TCF = 200 N b
or
and from Equation (2)
TDE = 2.25(200.00 N) = 450.00
TDE = 450 N b
or
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149
PROBLEM 4.122 (Continued)
(b)
From f.b.d. of assembly
SFz = 0 : Az - (0.6)( 450.00 N) = 0
Az = 270.00 N
SFx = 0 : Ax - (0.8)(200.00 N) = 0
Ax = 160.000 N
or A = (160.0 N)i + (270 N)k b
SM x = 0 :
MAx + ( 480 N)(0.135 m) - [(200.00 N)(0.6)](0.135 m)
- [( 450 N)(0.8)](0.09 m) = 0
M Ax = -16.2000 N ◊ m
SM z = 0 :
M Az = 0
MAz - ( 480 N)(0.08 m) + [(200.00 N)(0.6)](0.08 m)
+ [(450 N )(0.8)](0.08 m) = 0
or M A = -(16.20 N ◊ m)i b
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150
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:
We have
Thus
a = 500 mm
uuur
CD = (600 mm)j - (800 mm)k CD = 1000 mm
uuur
BD = -(1000 mm)i + (600 mm)j - (800 mm)k BD = 1414.21 mm
uuur
BE = (1000 mm)i - (800 mm)k BE = 1280.63 mm
uuur
CD
TCD = TCD
= TCD (0.6 j - 0.8k )
CD
uuur
BD
TBD = TBD
= TBD ( -0.7071i + 0.4243j - 0.5657k )
BD
uuur
BE
TBE = TBE
= TBE (0.7809i - 0.6247k )
BE
SM A = 0 : (rC ¥ TCD ) + (rB ¥ TBD ) + (rB ¥ TBE ) + (rF ¥ W ) = 0
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151
PROBLEM 4.123 (Continued)
rC = -(1000 mm)i + (800 mm)k
Noting that
rB = (800 mm)k
rF = - ai + (800 mm)k
and using determinants, we write
i
j
k
i
j
k
-1000 0 800 TCD +
0
0
800 TBD
0
0.6 -0.8
-0.7071 0.4243 -0.5657
k
i
j
k
i
j
0
800 TBE + - a 0 800 = 0
+ 0
0.7809 0 -0.6247
0 -2 0
Equating to zero the coefficients of the unit vectors:
i: -480TCD - 339.44 TBD + 1600 = 0
(1)
j: - 800TCD - 565.68TBD + 624.72TBE = 0 (2)
k: -600TCD + 2a = 0 (3)
Recalling that a = 500 mm, Eq. (3) yields
TCD =
From (1):
From (2):
2(500) 5
= kN 600
3
Ê 5ˆ
-480 Á ˜ - 339.44TBD + 1600 = 0
Ë 3¯
TBD = 2.35682 kN Ê 5ˆ
-800 Á ˜ - 565.68(2.35682) + 624.72TBE = 0
Ë 3¯
TBE = 4.2684 kN TCD = 1.667 kN b
TBD = 2.36 kN b
TBE = 4.27 kN b
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152
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):
-480TCD - 339.44TBD + 1600 = 0 (1)
-800TCD - 565.68TBD + 624.72TBE = 0 (2)
-600TCD + 2a = 0 (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
kN 3
From (1):
Ê 10 ˆ
-480 Á ˜ - 339.44TBD + 1600 = 0 Ë 3¯
From (2):
Ê 10 ˆ
-800 Á ˜ - 0 + 624.72 TBE = 0 fi TBE = 4.2685 kN Ë 3¯
TCD = 3.33 kN b
TBD = 0 b
TBE = 4.27 kN b
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153
PROBLEM 4.125
Frame ABCD is supported by a ball-and-socket joint at A and by three
cables. For a = 150 mm, determine the tension in each cable and the
reaction at A.
SOLUTION
First note
TDG = l DGTDG =
-(0.48 m)i + (0.14 m)j
(0.48)2 + (0.14)2 m
-0.48i + 0.14 j
=
TDG
0.50
T
= DG (24i + 7 j)
25
TBE = l BE TBE =
-(0.48 m)i + (0.2 m)k
(0.48)2 + (0.2)2 m
TDG
TBE
-0.48i + 0.2k
TBE
0.52
T
= BE ( -12 j + 5k ) )
13
=
From f.b.d. of frame ABCD
ˆ
Ê 7
SM x = 0 : Á TDG ˜ (0.3 m) - (350 N)(0.15 m) = 0
¯
Ë 25
TDG = 625 N b
or
ˆ
Ê5
Ê 24
ˆ
SM y = 0 : Á ¥ 625 N ˜ (0.3 m) - Á TBE ˜ (0.48 m) = 0
Ë 25
¯
¯
Ë 13
TBE = 975 N b
or
ˆ
Ê 7
SM z = 0 : TCF (0.14 m) + Á ¥ 625 N ˜ (0.48 m) - (350 N)(0.48 m) = 0
¯
Ë 25
TCF = 600 N b
or
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154
PROBLEM 4.125 (Continued)
SFx = 0 : Ax + TCF + (TBE ) x + (TDG ) x = 0
ˆ
ˆ Ê 24
Ê 12
Ax - 600 N - Á ¥ 975 N ˜ - Á ¥ 625 N ˜ = 0
¯
¯ Ë 25
Ë 13
Ax = 2100 N
SFy = 0 : Ay + (TDG ) y - 350 N = 0
ˆ
Ê 7
Ay + Á ¥ 625 N ˜ - 350 N = 0
¯
Ë 25
Ay = 175.0 N
SFz = 0 : Az + (TBE ) z = 0
ˆ
Ê5
Az + Á ¥ 975 N ˜ = 0
¯
Ë 13
Az = -375 N
A = (2100 N )i + (175.0 N ) j - (375 N )k b
Therefore
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155
PROBLEM 4.126
Frame ABCD is supported by a ball-and-socket joint
at A and by three cables. Knowing that the 350-N
load is applied at D (a = 300 mm), determine the
tension in each cable and the reaction at A.
SOLUTION
First note
TDG = l DGTDG =
-(0.48 m)i + (0.14 m)j
(0.48)2 + (0.14)2 m
TDG
-0.48i + 0.14 j
TDG
0.50
T
= DG (24i + 7 j)
25
=
TBE = l BE TBE =
-(0.48 m)i + (0.2 m)k
(0.48)2 + (0.2)2 m
TBE
-0.48i + 0.2k
TBE
0.52
T
= BE ( -12i + 5k )
13
=
From f.b.d. of frame ABCD
ˆ
Ê 7
SM x = 0 : Á TDG ˜ (0.3 m) - (350 N)(0.3 m) = 0
¯
Ë 25
TDG = 1250 N b
or
ˆ
Ê5
Ê 24
ˆ
SM y = 0 : Á ¥ 1250 N ˜ (0.3 m) - Á TBE ˜ (0.48 m) = 0
Ë 25
¯
¯
Ë 13
TBE = 1950 N b
or
ˆ
Ê 7
SM z = 0 : TCF (0.14 m) + Á ¥ 1250 N˜ (0.48 m) - (350 N)(0.48 m) = 0
¯
Ë 25
TCF = 0 b
or
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156
PROBLEM 4.126 (Continued)
SFx = 0 : Ax + TCF + (TBE ) x + (TDG ) x = 0
ˆ
ˆ Ê 24
Ê 12
Ax + 0 - Á ¥ 1950 N ˜ - Á ¥ 1250 N ˜ = 0
¯
¯ Ë 25
Ë 13
Ax = 3000 N
SFy = 0 : Ay + (TDG ) y - 350 N = 0
ˆ
Ê 7
Ay + Á ¥ 1250 N ˜ - 350 N = 0
¯
Ë 25
Ay = 0
SFz = 0 : Az + (TBE ) z = 0
ˆ
Ê5
Az + Á ¥ 1950 N ˜ = 0
¯
Ë 13
Az = -750 N
A = (3000 N)i - (750 N)k b
Therefore
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157
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 = 300 mm, b = 200 mm and c = 250 mm.
SOLUTION
From f.b.d. of weldment
SMO = 0 : rA/O ¥ A + rB/O ¥ B + rC/O ¥ C = 0
i
300
0
j
0
Ay
k
i
0 + 0
Az Bx
j
k
i
200 0 + 0
0 Bz
Cx
j
k
0 250 = 0
Cy 0
( -300 Az j + 300 Ay k ) + (200 Bz i - 200 Bx k ) + ( -250 C y i + 250 C x j) = 0
From i-coefficient
200 Bz - 250 C y = 0
Bz = 1.25C y or
j-coefficient
-300 Az + 250 C x = 0
C x = 1.2 Az or
k-coefficient
(1)
(2)
300 Ay - 200 Bx = 0
Bx = 1.5 Ay or
(3)
SF = 0 : A + B + C - P = 0
or
( Bx + C x )i + ( Ay + C y - 1.2 kN ) j + ( Az + Bz )k = 0
From i-coefficient
Bx + C x = 0
C x = - Bx or
j-coefficient
or
(4)
Ay + C y - 1.2 kN = 0
Ay + C y = 1.2 kN (5)
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158
PROBLEM 4.127 (Continued)
k-coefficient
Az + Bz = 0
Az = - Bz or
(6)
Substituting C x from Equation (4) into Equation (2)
- Bz = 1.2 Az (7)
Using Equations (1), (6), and (7)
Cy =
Bz
- Az
1 Ê Bx ˆ Bx
=
=
Á ˜=
1.25 1.25 1.25 Ë 1.2 ¯ 1.5
(8)
From Equations (3) and (8)
Cy =
1.5 Ay
1.5
or C y = Ay
and substituting into Equation (5)
2 Ay = 1.2 kN
Ay = C y = 600 N (9)
Using Equation (1) and Equation (9)
Bz = 1.25(600 N ) = 750 N
Using Equation (3) and Equation (9)
Bx = 1.5(600 N ) = 900 N
From Equation (4)
C x = -900 N
From Equation (6)
Az = -750 N
Therefore
A = (600 N ) j - (750 N )k b
B = (900 N )i + (750 N )k b
C = -(900 N )i + (600 N ) j b
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159
PROBLEM 4.128
Solve Problem 4.127, assuming that the force P is removed and is
replaced by a couple M = +(75 N ◊ mm)j acting at B.
PROBLEM 4.127 Three rods are welded together to form a “corner”
that is supported by three eyebolts. Neglecting friction, determine the
reactions at A, B, and C when P = 1.2 kN, a = 300 mm, b = 200 mm,
and c = 250 mm.
SOLUTION
From f.b.d. of weldment
SMO = 0 : rA/O ¥ A + rB/O ¥ B + rC/O ¥ C + M = 0
i
300
0
j
0
Ay
k
i
0 + 0
Az Bx
j
k
i
200 0 + 0
0 Bz
Cx
j
k
0 250 + (75000 N ◊ mm) j = 0
Cy 0
( -300 Az j + 300 Ay k ) + (200 Bz j - 200 Bx k ) + ( -250C y i + 250C x j) + (75000 N ◊ mm)j = 0
From i-coefficient
200 Bz - 250 C y = 0
C y = 0.8 Bz or
j-coefficient
or
k-coefficient
or
(1)
- 300 Az + 250 C x + 75, 000 = 0
C x = 1.2 Az - 300 (2)
300 Ay - 200 Bx = 0
Bx = 1.5 Ay (3)
SF = 0 : A + B + C = 0
( Bx + C x )i + ( Ay + C y ) j + ( Az + Bz )k = 0
C x = - Bx (4)
j-coefficient
C y = - Ay (5)
k-coefficient
Az = - Bz (6)
From i-coefficient
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160
PROBLEM 4.128 (Continued)
Substituting C x from Equation (4) into Equation (2)
ÊB ˆ
Az = 250 - Á x ˜ Ë 1.2 ¯
Using Equations (1), (6), and (7)
Ê 2ˆ
C y = 0.8 Bz = - 0.8 Az = Á ˜ Bx - 200 Ë 3¯
(7)
(8)
From Equations (3) and (8)
C y = Ay - 200
Substituting into Equation (5)
2 Ay = 200
Ay = 100.0 N
From Equation (5)
C y = -100.0 N
Equation (1)
Bz = -125.0 N
Equation (3)
Bx = 150.0 N
Equation (4)
C x = -150.0 N
Equation (6)
Az = 125.0 N
Therefore
A = (100.0 N ) j + (125.0 N )k b
B = (150.0 N )i - (125.0 N )k b
C = - (150.0 N )i - (100.0 N ) j b
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161
PROBLEM 4.129
In order to clean the clogged drainpipe AE, a plumber has
disconnected both ends of the pipe and inserted a power
snake through the opening at A. The cutting head of the
snake is connected by a heavy cable to an electric motor
that rotates at a constant speed as the plumber forces the
cable into the pipe. The forces exerted by the plumber and
the motor on the end of the cable can be represented by the
wrench F = -( 48 N)k , M = -(90 N ◊ m)k. Determine the
additional reactions at B, C, and D caused by the cleaning
operation. Assume that the reaction at each support consists
of two force components perpendicular to the pipe.
SOLUTION
From f.b.d. of pipe assembly ABCD
SFx = 0 : Bx = 0
SM D ( x -axis ) = 0 : ( 48 N )(2.5 m) - Bz (2 m) = 0
Bz = 60.0 N
and B = (60.0 N )k b
SM D ( z -axis ) = 0 : C y (3 m) - 90 N ◊ m = 0
C y = 30.0 N
SM D ( y - axis) = 0 :
- C z (3 m) - (60.0 N )( 4 m) + ( 48 N )( 4 m) = 0
C z = -16.00 N
and C = (30.0 N ) j - (16.00 N )k b
SFy = 0 : D y + 30.0 = 0
D y = -30.0 N
SFz = 0 : Dz - 16.00 N + 60.0 N - 48 N = 0
Dz = 4.00 N
and D = - (30.0 N ) j + ( 4.00 N )k b
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162
PROBLEM 4.130
Solve Problem 4.129, assuming that the plumber exerts a force
F = -(48 N)k and that the motor is turned off ( M = 0).
PROBLEM 4.129 In order to clean the clogged drainpipe AE,
a plumber has disconnected both ends of the pipe and inserted a
power snake through the opening at A. The cutting head of the snake
is connected by a heavy cable to an electric motor that rotates at a
constant speed as the plumber forces the cable into the pipe. The
forces exerted by the plumber and the motor on the end of the cable
can be represented by the wrench F = -( 48 N)k , M = -(90 N ◊ m)k.
Determine the additional reactions at B, C, and D caused by the
cleaning operation. Assume that the reaction at each support
consists of two force components perpendicular to the pipe.
SOLUTION
From f.b.d. of pipe assembly ABCD
SFx = 0 : Bx = 0
SM D ( x -axis ) = 0 : ( 48 N )(2.5 m) - Bz (2 m) = 0
Bz = 60.0 N and B = (60.0 N )k b
SM D ( z -axis ) = 0 : C y (3 m) - Bx (2 m) = 0
Cy = 0
SM D ( y -axis ) = 0 : C z (3 m) - (60.0 N )( 4 m) + ( 48 N )( 4 m) = 0
C z = -16.00 N and C = - (16.00 N )k b
SFy = 0 : D y + C y = 0
Dy = 0
SFz = 0 : Dz + Bz + C z - F = 0
Dz + 60.0 N - 16.00 N - 48 N = 0
Dz = 4.00 N and D = ( 4.00 N )k b
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163
PROBLEM 4.131
The assembly shown consists of an 80-mm rod
AF that is welded to a cross consisting of four
200 mm arms. The assembly is supported by
a ball-and-socket joint at F and by three short
links, each of which forms an angle of 45∞ with
the vertical. For the loading shown, determine
(a) the tension in each link, (b) the reaction at F.
SOLUTION
rE/F = -200 i + 80 j
TB = TB (i - j) / 2
rB/F = 80 j - 200k
TC = TC ( - j + k ) / 2
rC/F = 2000i + 80 j
TD = TD ( - i + j) / 2
rD/E = 80 j + 200k
SM F = 0 : rB/F ¥ TB + rC/F ¥ TC + rD/F ¥ TD + rE/F ¥ ( - Pj) = 0
i j
k
i
j k
i
j
k
i
j k
Tc
TB
TD
0 80 -200
+ 200 80 0
+ 0 80 200
+ -200 80 0 = 0
2
2
2
1 -1
0
0 -1 1
-1 -1 0
0
-P 0
Equate coefficients of unit vectors to zero and multiply each equation by 2.
Eqs. (3) + ( 4):
Eqs. (1) + (2):
i: - 200 TB + 80 TC + 200 TD = 0 (1)
j: - 200 TB - 200 TC - 200 TD = 0 (2)
k : - 80 TB - 200 TC + 80 TD + 200 2 P = 0 80
(2): - 80 TB - 80 TC - 80 TD = 0 200
-160TB - 280TC + 200 2 P = 0 (3)
(4)
(5)
-400TB - 120TC = 0
TB = -
120
TC - 0.3TC 400
(6)
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164
PROBLEM 4.131 (Continued)
Eqs. (6) Æ (5):
-160( - 0.3TC ) - 280TC + 200 2 P = 0
-232TC + 200 2 P = 0
TC = 1.2191P TB = -0.3(1.2191P ) = - 0.36574 = P From Eq. (6):
From Eq. (2):
TC = 1.219 P b
TB = -0.366 P b
- 200( - 0.3657 P ) - 200(1.2191P ) - 200Tq D = 0
TD = - 0.8534 P TD = -0.853P b
SF = 0: F + TB + TC + TD - Pj = 0
i: Fx +
( - 0.36574 P )
2
-
Fx = - 0.3448P
j: Fy -
k : Fz +
2
=0
Fx = - 0.345P
( - 0.36574 P )
2
Fy = P
( - 0.8534 P )
-
(1.2191P ) ( - 0.8534 P )
- 200 = 0
2
2
Fy = P
(1.2191P )
=0
2
Fz = - 0.8620 P
Fz = - 0.862 P F = - 0.345 Pi + Pj - 0.862 Pk b
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165
PROBLEM 4.132
The uniform 10-kg rod AB is supported by a ball-and-socket joint
at A and by the cord CG that is attached to the midpoint G of the rod.
Knowing that the rod leans against a frictionless vertical wall at B,
determine (a) the tension in the cord, (b) the reactions at A and B.
SOLUTION
Free-Body Diagram:
Five unknowns and six equations of equilibrium. But equilibrium is
maintained ( SMAB = 0)
W = mg
= (10 kg) 9.81 m/s2
W = 98.1 N
uuur
GC = - 300i + 200 j - 225k GC = 425 mm
uuur
GC
T
T=T
=
( - 300i + 200 j - 225k )
GC 425
rB/ A = - 600i + 400 j + 150 mm
rG/ A = - 300i + 200 j + 75 mm
SMA = 0 : rB/ A ¥ B + rG/ A ¥ T + rG/ A ¥ ( - Wj) = 0
i
j
k
i
j
k
i
j
k
T
- 600 400 150 + - 300 200
75
+ - 300 200 75
425
B
0
0
- 300 200 - 225
0
- 98.1 0
Coefficient of i: ( - 105.88 - 35.29)T + 7357.5 = 0
T = 52.12 N T = 52.1 N b
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166
PROBLEM 4.132 (Continued)
Coefficient of j : 150 B - (300 ¥ 75 + 300 ¥ 225)
52.12
=0
425
B = 73.58 N B = (73.6 N )i b
SF = 0 : A + B + T - Wj = 0
Coefficient of i: Ax + 73.58 - 52.15
300
=0
425
Ax = 36.8 N b
Coefficient of j: Ay + 52.15
200
- 98.1 = 0 425
Ay = 73.6 N b
Coefficient of k : Az - 52.15
225
=0
425
Az = 27.6 N b
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167
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:
uuur
DF = -400i + 275 j - 200k
DF = 525 mm
uuur
DE
T
T=T
=
( -400i + 275 j - 200k )
DF 525
rD/E = 400 i
rC/E = 400i - 350k
uuur
EA 175i - 600k
l EA =
=
EA
625
SM EA = 0 : l EA ◊ (rD/E ¥ T) + l EA ◊ (rC/E ◊ ( - 300 j)) = 0
175
0 -600
175
0
-600
T
1
400
0
0
+ 400
0
-350
=0
525 ¥ 625
625
-400 275 -200
0 -300
0
-
600 ¥ 400 ¥ 275
-175 ¥ 300 ¥ 350 + 600 ¥ 400 ¥ 300
T+
=0
525 ¥ 625
625
-880000
T + 53625000 = 0
7
T = 426.5625 N
T = 427 N b
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168
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 balland-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:
rB/ A = 225i
rC/ A = 225i + 250k
uuur
BF = -400i + 275 j + 400k
BF = 628.987 mm
uuur
BF
T
T=T
=
= ( - 16i + 11j + 16k )
BF 25.16
uuur
AE 7i - 24k
l AE =
=
AE
25
S MAE = 0 : l AF ◊ (rB/ A ¥ T) + l AE ◊ (rC/ A ◊ ( - 300 j)) = 0
7
0 -24
7
0
-24
T
1
225 0
0
+ 225
0
250
=0
25 ¥ 25.16
25
-16 11 16
0 -300 0
-
24 ¥ 225 ¥ 11
24 ¥ 225 ¥ 300 + 7 ¥ 250 ¥ 300
T+
25 ¥ 25.16
25
2360.89 T - 2,145, 000 = 0 T = 909 N b
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169
PROBLEM 4.135
The 50-kg plate ABCD is supported by hinges along edge AB
and by wire CE. Knowing that the plate is uniform, determine
the tension in the wire.
SOLUTION
Free-Body Diagram:
W = mg = (50 kg)(9.81 m/s2 )
W = 490.50 N
uuur
CE = - 240i + 600 j - 400k
CE = 760 mm
uuur
CE
T
T=T
=
( - 240i + 600 j - 400k )
CE 760
uuur
AB 480i - 200 j 1
= (12i - 5 j)
=
l AB =
520
13
AB
SMAB = 0 : l AB ◊ (rE/ A ¥ T ) + l AB ◊ (rG/ A ¥ - Wj) = 0
rE/ A = 240i + 400 j; rG/ A = 240i - 100 j + 200k
12
-5
0
12
-5
0
T
1
240 400
0
+ 240 -100 200
=0
13 ¥ 20
13
- 240 600 - 400
0
-W
0
( -12 ¥ 400 ¥ 400 - 5 ¥ 240 ¥ 400)
T
+ 12 ¥ 200W = 0
760
T = 0.76W = 0.76( 490.50 N ) T = 373 N b
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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170
PROBLEM 4.136
Solve Problem 4.135, assuming that wire CE is replaced by a
wire connecting E and D.
PROBLEM 4.135 The 50-kg plate ABCD is supported by
hinges along edge AB and by wire CE. Knowing that the plate is
uniform, determine the tension in the wire.
SOLUTION
Free-Body Diagram:
Dimensions in mm
W = mg = (50 kg)(9.81 m/s2 )
W = 490.50 N
uuur
DE = - 240i + 400 j - 400k
DE = 614.5 mm
uuur
DE
T
T=T
=
(240i + 400 j - 400k )
DE 614.5
uuur
AB 480i - 200 j 1
=
= (12i - 5 j)
l AB =
AB
520
13
rE/ A = 240i + 400 j; rG/ A = 240i - 100 j + 200k
12 -5
0
12
5
0
T
1
240 400
0
+ 240 -100 200
=0
13 ¥ 614.5
13
240 400 - 400
0
-W
0
( -12 ¥ 400 ¥ 400 - 5 ¥ 240 ¥ 400)
T
+ 12 ¥ 200 ¥ W = 0
614.5
T = 0.6145W = 0.6145( 490.50 N )
T = 301 N b
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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171
PROBLEM 4.137
Two rectangular plates are welded together to form the assembly
shown. The assembly is supported by ball-and-socket joints at B
and D and by a ball on a horizontal surface at C. For the loading
shown, determine the reaction at C.
SOLUTION
l BD =
First note
-(150 mm)i - (225 mm) j + (300 mm)k
(150)2 + (225)2 + (300)2 mm
1
( -2i - 3j + 4k )
29
= - (150 mm)i
=
rA/B
P = ( 400 N )k
rC/D = (200 mm)i
C = (C ) j
From the f.b.d. of the plates
SM BD = 0 : l BD ◊ (rA/B ¥ P) + l BD ◊ (rC/D ¥ C) = 0
-2 -3 4
-2 -3 4
È C (200) ˘
È150( 400) ˘
+ 1 0 0Í
-1 0 0 Í
˙=0
˙
29 ˚
Î 29 ˚
Î
0 0 1
0 1 0
( -3)(150)( 400) + ( 4)(200)C = 0
C = 225 N or C = (225 N ) j b
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172
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.
SOLUTION
Free-Body Diagram:
uuur
AF = 1.2i - 0.6 j - 1.2k
1
l AF = (2i - j - 2k )
3
rG1 /A = 0.6i - 0.3j
AF = 1.8 m
rG2 /A = 1.2i - 0.3j - 0.6k
rB /A = 1.2i
SM AF = 0 : l AF ◊ (rG /A ¥ ( -60 j) + l AF ◊ (rG 2 /A ¥ ( -60 j)) + l AF ◊ (rB /A ¥ T ) = 0
2
-1 -2
2
-1
-2
1
1
0.6 -0.3 0
+ 1.2 -0.3 -0.6 + l AF ◊ (rB /A ¥ T) = 0
3
3
0 -60 0
0 -60
0
1
1
(2 ¥ 0.6 ¥ 60) + ( -2 ¥ 0.6 ¥ 60 + 2 ¥ 1.2 ¥ 60) + l AF ◊ (rB /A ¥ T) = 0
3
3
l AF ◊ (rB /A ¥ T) = -48 or T ◊ (l A/F ¥ rB /A ) = -48 (1)
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173
PROBLEM 4.138 (Continued)
Projection of T on (l AF ¥ rB /A ) is constant. Thus, Tmin is parallel to
1
1
l AF ¥ rB /A = (2i - j - 2k ) ¥ 1.2i = ( -2.4 j + 1.2k )
3
3
Corresponding unit vector is
1
5
( -2 j + k )
Tmin = T ( -2 j + k )
Eq. (1):
1
5
(2)
T
È1
˘
( -2 j + k ) ◊ Í (2i - j - 2k ) ¥ 1.2i ˙ = -48
5
Î3
˚
T
1
( -2 j + k ) ◊ ( -2.4 j + 1.2k ) = -48
3
5
T
3 5 ( 48)
( 4.8 + 1.2) = -48
T == 24 5
6
3 5
T = 53.6656 N
Eq. (2)
Tmin = T ( -2 j + k )
1
5
1
5
= -( 48 N)j + (24 N k )
= 24 5 ( -2 j + k )
Tmin
Since Tmin has no i component, wire BH is parallel to the yz plane, and x = 1.2 m
(a)
(b)
x = 1.200 m;
y = 2.40 m b
Tmin = 53.7 N b
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174
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.
SOLUTION
uuur
AF = 1.2i - 0.6 j - 1.2k
1
l AF = (2i - j - 2k )
3
rG1 /A = 0.6i - 0.3j
Free-Body Diagram:
AF = 1.8 m
rG2 /A = 1.2i - 0.3j - 0.6k
rB /A = 1.2i
SM AF = 0 : l AF ◊ (rG /A ¥ ( -60 j) + l AF ◊ (rG 2 /A ¥ ( -60 j)) + l AF ◊ (rB /A ¥ T ) = 0
2
-1 -2
2
-1
-2
1
1
0.6 -0.3 0
+ 1.2 -0.3 -0.6 + l AF ◊ (rB /A ¥ T) = 0
3
3
0 -60 0
0 -60
0
1
1
(2 ¥ 0.6 ¥ 60) + ( -2 ¥ 0.6 ¥ 60 + 2 ¥ 1.2 ¥ 60) + l AF ◊ (rB /A ¥ T) = 0
3
3
l AF ◊ (rB /A ¥ T) = -48 uuur
BH = -1.2i + yj - 1.2k
BH = (2.88 + y 2 )1/ 2
(1)
uuur
BH
-1.2i + yj - 1.2k
T=T
=T
BH
(32 + y 2 )1/ 2
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175
PROBLEM 4.139 (Continued)
l AF
Eq. (1):
2
-1 -2
1
T
0
◊ (rB /A ¥ T ) = 1.2 0
= -48
3 (2.88 + y 2 )
-1.2 y -1.2
(2.88 + y 2 ) 2
(2.88 + y 2 )1/ 2
= 100
T = 144
(1.44 + 2.4 y )
Ê 5yˆ
1
+
ÁË
˜
3¯
1
2 1/ 2
( -1.44 - 2.4 y )T = -3 ¥ 48(2.88 + y )
dT
= 0: 100
dy
Numerator = 0:
ÏÊ 5 y ˆ 1
2 -1/ 2
2 1/ 2 Ê 5 ˆ ¸
ÌÁË1+ ˜¯ (2.88 + y ) (2 y ) + (2.88 + y ) ÁË ˜¯ ˝
3
2
3 ˛
Ó
Ê 5yˆ
ÁË1 + ˜¯
3
2
Ê 5yˆ
2 5
ÁË1 + ˜¯ y = (2.88 + y )
3
3
5 2
24 5 2
y +y=
+ y 3
5 3
Eq. (2):
(2)
T=
100(2.88 + 4.82 )1/ 2
= 56.569 N Ê 5 ¥ 4.8 ˆ
ÁË1 +
˜
3 ¯
y = 4.80 m b
Tmin = 56.6 N b
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176
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
uuur
AE = 480i + 160 j - 240k
AE = 560 mm
uuur
AE 480i + 160 j - 240k
l AE =
=
560
AE
6i + 2 j - 3k
l AE =
7
rB /A = 200i
rD /A = 480i + 160 j
uuur
DF = -480i + 330 j - 240k;
DF = 630 mm
uuur
DF
-16i + 11j - 8k
-480i + 330 j - 240k
= TDF
TDF = TDF
= TDF
DF
630
21
SM AE = l AE - (rD /A ¥ TDF ) + l AE - (rB /A ¥ ( -600 j)) = 0
6
2 -3
6
2
-3
TDE
1
480 160 0
+ 200
0
0 =0
21 ¥ 7
7
-16 11 -8
0 -640 0
3 ¥ 200 ¥ 640
-6 ¥ 160 ¥ 8 + 2 ¥ 480 ¥ 8 - 3 ¥ 480 ¥ 11 - 3 ¥ 160 ¥ 16
TDF +
=0
21 ¥ 7
7
-1120TDF + 384 ¥ 103 = 0
TDF = 342.86 N TDF = 343 N b
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177
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
uuur
AE = 480i + 160 j - 240k
AE = 560 mm
uuur
AE 480i + 160 j - 240k
l AE =
=
560
AE
6i + 2 j - 3k
l AE =
7
rB /A = 200i
rC /A = 480i
uuur
CF = -480i + 490 j - 240k; CF = 726.70 mm
uuur
CE -480i + 490 j - 240k
TCF = TCF
=
CF
726.70
SMAE = 0 : l AE ◊ (rC /A ¥ TCF ) + l AE ◊ (rB /A ¥ ( -600 j)) = 0
6
2
-3
6
2
-3
TCF
1
480
0
0
+ 200
0
0
=0
726.7 ¥ 7
7
-480 +490 -240
0 -640 0
2 ¥ 480 ¥ 240 - 3 ¥ 480 ¥ 490
3 ¥ 200 ¥ 640
TCF +
=0
726.7 ¥ 7
7
-653.91TCF + 384 ¥ 103 = 0 TCF = 587 N b
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178
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.
SOLUTION
Free-Body Diagram:
2
W = mg = ( 40 kg)(9.81 m/s ) = 392.40 N
a1 = (300 mm)sina - (80 mm)cosa
a2 = ( 430 mm)cosa - (300 mm)sina
b = (930 mm)cosa
From free-body diagram of hand truck
Dimensions in mm
+
+
For
(a)
From Equation (1)
SM B = 0 : P (b) - W ( a2 ) + W ( a1 ) = 0 (1)
SFy = 0 : P - 2W + 2 B = 0 (2)
a = 35∞
a1 = 300 sin 35∞ - 80 cos 35∞ = 106.541 mm
a2 = 430 cos 35∞ - 300 sin 35∞ = 180.162 mm
b = 930 cos 35∞ = 761.81 mm
P(761.81 mm) - 392.40 N(180.162 mm) + 392.40 N(106.54 mm) = 0
P = 37.921 N (b)
or P = 37.9 N ≠ b
From Equation (2)
37.921 N - 2(392.40 N) + 2 B = 0 or B = 373 N ≠ b
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179
PROBLEM 4.143
Determine the reactions at A and C when (a) a = 0, (b) a = 30∞.
SOLUTION
(a)
a = 0∞
From f.b.d. of member ABC
+
SM C = 0 : (300 N)(0.2 m) + (300 N)(0.4 m) - A(0.8 m) = 0
A = 225 N +
or
A = 225 N ≠ b
SFy = 0 : C y + 225 N = 0
C y = -225 N or C y = 225 N Ø
+
SFx = 0 : 300 N + 300 N + C x = 0
C x = -600 N or C x = 600 N ¨
Then
C = C x2 + C y2 = (600)2 + (225)2 = 640.80 N
and
Ê Cy ˆ
Ê -225 ˆ
= 20.556∞
q = tan -1 Á ˜ = tan -1 Á
Ë -600 ˜¯
Ë Cx ¯
(b)
or
C = 641 N
20.6∞ b
a = 30∞
From f.b.d. of member ABC
+
SM C = 0 : (300 N)(0.2 m) + (300 N)(0.4 m) - ( A cos 30∞)(0.8 m)
+ ( A sin 30∞)(20 in.) = 0
A = 365.24 N +
or
A = 365 N
60.0∞ b
SFx = 0 : 300 N + 300 N + (365.24 N) sin 30∞ + C x = 0
C x = -782.62
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180
PROBLEM 4.143 (Continued)
+
SFy = 0 : C y + (365.24 N) cos 30∞ = 0
C y = -316.31 N or C y = 316 N Ø
Then
C = C x2 + C y2 = (782.62)2 + (316.31)2 = 884.12 N
and
Ê Cy ˆ
Ê -316.31ˆ
= 22.007∞
q = tan -1 Á ˜ = tan -1 Á
Ë -782.62 ˜¯
Ë Cx ¯
C = 884 N
or
22.0∞ b
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181
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.
SOLUTION
Free-Body Diagram:
Geometry:
x AC = (250 mm) cos 20∞ = 234.923 mm
y AC = (250 mm) sin 20∞ = 85.505 mm
fi yDA = 300 mm - 85.505 mm = 214.495 mm
Êy ˆ
Ê 214.495 ˆ
= 42.397∞
a = tan -1 Á DA ˜ = tan -1 Á
Ë 234.923 ˜¯
Ë x AC ¯
b = 90∞ - 20∞ - 42.397∞ = 27.603∞
Equilibrium for lever:
(a)
+
SM C = 0 : TAD cos 27.603∞(250 mm) - (350 N)[(375 mm)cos 20∞] = 0
TAD = 556.703 N (b)
+
TAD = 557 N b
SFx = 0 : C x + (556.703 N) cos 42.397∞ = 0
C x = -411.12 N
+
SFy = 0 : C y - 350 N - (556.703 N) sin 42.397∞ = 0
C y = 725.364 N
Thus:
C = C x2 + C y2 = ( -411.12)2 + (725.364)2 = 833.77 N
and
q = tan -1
Cy
Cx
= tan -1
725.364
= 60.456∞ 411.12
C = 834 N
60.5∞ b
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182
PROBLEM 4.145
Neglecting friction and the radius of the pulley, determine
(a) the tension in cable ADB, (b) the reaction at C.
SOLUTION
Free-Body Diagram:
Dimensions in mm
Geometry:
Distance
AD = (0.36)2 + (0.150)2 = 0.39 m
Distance
BD = (0.2)2 + (0.15)2 = 0.25 m
Equilibrium for beam:
(a)
+
SM C = 0 :
Ê 0.15 ˆ
Ê 0.15 ˆ
(120 N)(0.28 m) - Á
T (0.36 m) - Á
T (0.2 m) = 0
Ë 0.25 ˜¯
Ë 0.39 ˜¯
T = 130.000 N
or
(b)
+
T = 130.0 N b
Ê 0.2 ˆ
Ê 0.36 ˆ
(130.000 N ) + Á
(130.000 N)) = 0
SFx = 0 : C x + Á
Ë 0.25 ˜¯
Ë 0.39 ˜¯
C x = - 224.00 N
+
Ê 0.15 ˆ
Ê 0.15 ˆ
(130.00 N) + Á
(130.00 N) - 120 N = 0
SFy = 0 : C y + Á
Ë 0.25 ˜¯
Ë 0.39 ˜¯
C y = -8.0000 N
Thus:
C = C x2 + C y2 = ( -224)2 + (- 8) = 224.14 N
and
q = tan -1
2
Cy
Cx
= tan -1
8
= 2.0454∞ 224
C = 224 N
2.05∞ b
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183
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∞.
SOLUTION
Free-Body Diagram:
+
SFy = 0 : E cos 30∞ - 100 - 200 = 0
E=
+
300 N
= 346.41 N cos 30∞
E = 346 N
60.0∞ b
SM D = 0 : (100 N)(100 mm) - (200 N)(100 mm)
- C (75 mm) + E sin 30∞(75 mm) = 0
C = 39.8717 N C = 39.9 N Æ b
SFx = 0 : E sin 30∞ + C - D = 0
(346.41 N )(0.5) + 39.8717 N - D = 0
D = 213.077 N D = 213 N ¨ b
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184
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.
SOLUTION
Free-Body Diagram:
+
SFy = 0 : E cosq - 100 - 200 = 0
E=
+
300
cosq
(1)
SM D = 0 : (100 N)(100 mm) - (200 N)(100 mm) - C (75 mm)
Ê 300
ˆ
+Á
sin q ˜ 75 mm = 0
Ë cos q
¯
C=
(a)
For C = 0,
-400
+ 300 tanq
3
400
3
4
tan q =
q = 23.962∞ 9
300
E=
= 328.294
cos 23.962∞
300 tanq =
Eq. (1)
+
q = 24.0∞ b
SFx = 0: -D + C + E sinq = 0
D = (328.294)sin 23.962 = 133.33 N
(b)
C = 0 D = 133.33 N ¨
E = 328 N
66.0∞ b
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185
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:
Ê 150 mm ˆ
b = tan -1 Á
= 30.964∞
Ë 250 mm ˜¯
Applying of the law of sines to the force triangle for member BCD,
150 N
B
C
=
=
sin( 45∞ - b ) sin b sin 135∞
or
150 N
B
C
=
=
sin 14.036∞ sin 30.964∞ sin 135∞
A= B =
(150 N )sin 30.964∞
= 318.205 N
sin 14.036∞
or
and
C=
A = 318.2 N
45.0∞ b
C = 437.3 N
59.0∞ b
(150 N )sin 135∞
= 437.33 N
sin 14.036∞
or
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186
PROBLEM 4.149
Determine the reactions at A and B when b = 50∞.
SOLUTION
Free-Body Diagram: (Three-force body)
Reaction A must pass through Point D where 100-N force
and B intersect
In right D BCD
a = 90∞ - 75∞ = 15∞
BD = 250 tan 75∞ = 933.01 mm
Dimensions in mm
In right D ABD
AB
150 mm
=
BD 933.01 mm
g = 9.13
tan g =
Force Triangle
Law of sines
100 N
A
B
=
=
sin 9.13∞ sin 15∞ sin 155.87∞
A = 163.1 N; B = 257.6 N
A = 163.1 N
74.1° B = 258 N
65.0° b
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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187
PROBLEM 4.150
The 6-m pole ABC is acted upon by a 455-N force as shown. The
pole is held by a ball-and-socket joint at A and by two cables BD
and BE. For a = 3 m, determine the tension in each cable and the
reaction at A.
SOLUTION
Free-Body Diagram:
Five unknowns and six equations of equilibrium, but equilibrium is maintained
( SM AC = 0)
rB = 3j
rC = 6 j
uuur
CF = -3i - 6 j + 2k CF = 7 m
uuur
BD = 1.5i - 3j - 3k BD = 4.5 m
uuur
BE = 1.5i - 3j + 3k BE = 4.5 m
uuur
CF P
P=P
= ( -3i - 6 j + 2k )
CE 7
uuur
T
BD TBD
TBD = TBD
=
(1.5i - 3j - 3k ) = BD (i - 2 j - 2k )
3
BD 4.5
uuur
BE TBD
TBE = TBE =
=
(i - 2 j + 2k )
3
BE
SM A = 0 : rB ¥ TBD + rB ¥ TBE + rC ¥ P = 0
i j k
i j k
i
j k
TBD
TBE
P
0 3 0
+ 0 3 0
+ 0 6 0
=0
3
3
7
1 -2 -2
1 -2 2
-3 -6 2
Coefficient of i:
-2TBD + 2TBE +
12
P =0
7
(1)
Coefficient of k:
-TBD - TBF +
18
P =0
7
(2)
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188
PROBLEM 4.150 (Continued)
- 4TBD +
Eq. (1) + 2 Eq. (2):
Eq. (2):
-
48
12
P = 0 TBD = P
7
7
12
18
6
P - TBE + P = 0 TBE = P
7
7
7
P = 445 N TBD =
Since
12
( 455) 7
6
TBE = ( 455) 7
TBD = 780 N b
TBE = 390 N b
SF = 0 : TBD + TBE + P + A = 0
780 390 455
+
(3) + Ax = 0
3
3
7
Coefficient of i:
260 + 130 - 195 + Ax = 0
Coefficient of j:
-
780
390
455
( 2) ( 2) (6) + Ay = 0
3
3
7
-520 - 260 - 390 + Ay = 0
Coefficient of k:
-
Ax = 195.0 N
Ay = 1170 N
780
390
455
(2) +
( 2) +
(2) + Az = 0
3
3
7
-520 + 260 + 130 + Az = 0
Az = +130.0 N
A = -(195.0 N )i + (1170 N ) j + (130.0 N )k b
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189
PROBLEM 4.151
Solve Problem 4.150 for a = 1.5 m.
PROBLEM 4.150 The 6-m pole ABC is acted upon by a 455-N
force as shown. The pole is held by a ball-and-socket joint at A and
by two cables BD and BE. For a = 3 m, determine the tension in
each cable and the reaction at A.
SOLUTION
Free-Body Diagram:
Five unknowns and six equations of equilibrium but equilibrium is maintained
( SM AC = 0) rB = 3j
rC = 6 j
uuur
CF = -1.5i - 6 j + 2k CF = 6.5 m
uuur
BD = 1.5i - 3j - 3k BD = 4.5 m uuur
BE = 1.5i - 3j + 3k BE = 4.5 m
uuur
CF
P
P
P=P
=
( -1.5i - 6 j + 2k ) = ( -3i - 12 j + 4k )
CE 6.5
13
uuur
T
BD TBD
=
(1.5i - 3j - 3k ) = BD (i - 2 j - 2k ) TBD = TBD
BD 4.5
3
uuur
BE TBD
=
TBE = TBE =
(i - 2 j + 2k )
BE
3
SM A = 0 : rB ¥ TBD + rB ¥ TBE + rC ¥ P = 0
j
k
i j k
i j k
i
TBD
TBE
P
6
0
=0
0 3 0
+ 0 3 0
+ 0
13
3
3
-3 -12 + 4
1 -2 -2
1 -2 2
Coefficient of i:
-2TBD + 2TBE +
24
P =0
13
(1)
Coefficient of k:
-TBD - TBE +
18
P=0
13
(2)
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190
Eq. (1) + 2 Eq. (2):
Eq (2):
Since
PROBLEM 4.151 (Continued)
60
15
-4TBD +
P = 0 TBD = P 13
13
15
18
3
- P - TBE + P = 0 TBE = P 13
13
13
15
P = 445 N TBD = ( 455) 13
3
TBE = ( 455) 13
TBD = 525 N b
TBE = 105.0 N b
SF = 0 : TBD + TBE + P + A = 0 525 105 455
+
(3) + Ax = 0 3
3
13
Coefficient of i:
175 + 35 - 105 + Ax = 0
Coefficient of j:
-
525
105
455
( 2) ( 2) (12) + Ay = 0 3
3
13
-350 - 70 - 420 + Ay = 0
Coefficient of k:
Ax = 105.0 N -
Ay = 840 N
525
105
455
(2) +
(2) +
( 4) + Az = 0
3
3
13
-350 + 70 + 140 + Az = 0
Az = 140.0 N
A = -(105.0 N )i + (840 N ) j + (140.0 N )k b
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191
PROBLEM 4.152
The rigid L-shaped member ABF is supported by a ball-and-socket
joint at A and by three cables. For the loading shown, determine the
tension in each cable and the reaction at A.
SOLUTION
Free-Body Diagram:
rB /A = 300i
rF /A = 300i - 200k
rD /A = 300i - 400k
rE /A = 300i - 600k
rF /A = 3000i - 800k
uuur
BG = -300i + 225k
BG = 375 mm
l BG = -0.8i + 0.6k
uuuur
DH = -300i + 400k; DH = 500 mm; l DH = -0.6i + 0.8 j
uuur
FJ = -300i + 400k; FJ = 500 mm; l FJ = -0.6i + 0.8 j
SM A = 0 : rB /A ¥ TBG l BG + rDA ¥ TDH l DH + rF /A ¥ TFJ l FJ
+rF /A ¥ ( -120 j) + rE /A ¥ ( -120 j) = 0
i
j k
i
j
k
i
j
k
300 0 0 TBG + 300 0 -400 TDH + 300 0 -800 TFJ
-0.6 0.8
0
-0.8 0 0.6
-0.6 0.8
0
i
j
k
i
j
k
+ 300
0
-200 + 300
0
-600 = 0
0 -120
0
0 -120
0
Coefficient of i:
+320TDH + 640TFJ - 24000 - 72000 = 0 (1)
Coefficient of k:
+240TDH + 240TFJ - 36000 - 36000 = 0 (2)
(1) fi TDH + 2TFJ = 300 ¸
˝ fi TFJ = 0 (2) fi TDH + TFJ = 300 ˛
TFJ = 0 b
3
4
Eq. (1) − Eq. (2):
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192
PROBLEM 4.152 (Continued)
TDH = 300 N Eq. (1):
Coefficient of j:
TDH = 300 N b
-180TBG + 240 ¥ (300) = 0 SF = 0 :
TBG = 400 N b
A + TBG l BG + TDH l DH + TFJ - 24 j - 24 j = 0 Coefficient of i:
Ax + ( 400)( -0.8) + (300)( -0.6) = 0
Ax = 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 = (500 N )i - (240 N ) j b
Note:
The value Ay = 0
Can be confirmed by considering SM BF = 0
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193
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 A = 0 : - Pa + (C sin 45∞)2a + (cos 45∞)a = 0 3
C
2
=P C=
P
3
2
+
Ê 2 ˆ 1
SFx = 0 : Ax - Á
P˜
Ë 3 ¯ 2
+
Ê 2 ˆ 1
SFy = 0 : Ay - P + Á
P˜
Ë 3 ¯ 2
C = 0.471P
Ax =
P
Æ
3
Ay =
2P
≠
3
(b)
+
45° b
A = 0.745P
63.4° b
SM C = 0 : +Pa - ( A cos 30∞)2a + ( A sin 30∞)a = 0 A(1.732 - 0.5) = P
A = 0.812 P A = 0.812 P
+
SFx = 0 : (0.812 P )sin 30∞ + C x = 0 C x = -0.406 P +
SFy = 0 : (0.812 P ) cos 30∞ - P + C y = 0 C y = -0.297 P C = 0.503P
60.0° b
36.2° b
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194
PROBLEM 4.153 (Continued)
(c)
+
SM C = 0 : + Pa - ( A cos 30∞)2a + ( A sin 30∞)a = 0 A(1.732 + 0.5) = P
A = 0.448P
+
SFx = 0 : - (0.448P )sin 30∞ + C x = 0
+
SFy = 0 : (0.448 P ) cos 30∞ - P + C y = 0 C y = 0.612 P ≠
60.0° b
C x = 0.224 P Æ
C = 0.652 P
(d)
A = 0.448P
69.9° b
Force T exerted by wire and reactions A and C all intersect at Point D.
+
SM D = 0 : Pa = 0 Equilibrium not maintained
Rod is improperly constrained b
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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195