Mechanics Engineering - Statics
Chapter 3
Problem 3-1 Determine the magnitudes of F1 and F2 so that the particle is in equilibrium. Given: F = 500 N
1 = 45 deg 2 = 30deg
Solution: Initial Guesses F 1 = 1N Given
+ Fx = 0; +
F 2 = 1N
F 1 cos ( 1 ) + F2 cos ( 2 ) - F = 0 F 1 sin ( 1 ) - F 2 sin ( 2 ) = 0
Fy = 0;
F1 = Find ( F1 , F2) F2
F1 259 = N F2 366
Problem 3-2 Determine the magnitude and direction of F so that the particle is in equilibrium. Units Used: kN = 10 N Given: F 1 = 7 kN F 2 = 3 kN c = 4 d = 3
132
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2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 3
Solution: The initial guesses: Given Equations of equilibrium:
+ Fx = 0;
F = 1kN
= 30deg
-d F + F cos ( ) = 0 2 2 1 c +d
c 2 2 F1 - F2 - F sin ( ) = 0 c +d F = 4.94 kN
+
F y = 0;
F = Find ( F , )
= 31.8 deg
Problem 3-3 Determine the magnitude of F and the orientation of the force F 3 so that the particle is in equilibrium. Given: F 1 = 700 N F 2 = 450 N F 3 = 750 N
1 = 15 deg 2 = 30 deg
Solution: Initial Guesses: Given
+ Fx = 0;
F = 1N
= 10deg
F 1 cos ( 1 ) - F2 sin ( 2 ) - F3 cos ( ) = 0
133
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Engineering Mechanics - Statics
Chapter 3
+
Fy = 0;
F + F2 cos ( 2 ) + F 1 sin ( 1 ) - F 3 sin ( ) = 0
F = Find ( F , )
F = 28.25 N
= 53.02 deg
Problem 3-4 Determine the magnitude and angle of F so that the particle is in equilibrium. Units Used: kN = 10 N Given: F 1 = 4.5 kN F 2 = 7.5 kN F 3 = 2.25 kN
3
= 60 deg = 30 deg
Solution: Guesses: F = 1 kN Given Equations of Equilibrium:
+
+
= 1
F x = 0;
F cos ( ) - F2 sin ( ) - F 1 + F 3 cos ( ) = 0 F sin ( ) - F 2 cos ( ) - F 3 sin ( ) = 0 F = 11.05 kN
F y = 0;
F = Find ( F , )
= 49.84 deg
134
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Engineering Mechanics - Statics
Chapter 3
Problem 3-5 The members of a truss are connected to the gusset plate. If the forces are concurrent at point O, determine the magnitudes of F and T for equilibrium. Units Used: kN = 10 N Given: F 1 = 8 kN F 2 = 5 kN
3
1 = 45 deg = 30 deg
Solution:
+ Fx = 0;
-T cos ( ) + F 1 + F 2 sin ( 1 ) = 0 F1 + F2 sin ( 1 ) cos ( )
T =
T = 13.3 kN
+
Fy = 0;
F - T sin ( ) - F 2 cos ( 1 ) = 0 F = T sin ( ) + F2 cos ( 1 ) F = 10.2 kN
Problem 3-6 The gusset plate is subjected to the forces of four members. Determine the force in member B and its proper orientation for equilibrium. The forces are concurrent at point O. Units Used: kN = 10 N
135
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Engineering Mechanics - Statics
Chapter 3
Given: F = 12 kN F 1 = 8 kN F 2 = 5 kN
1 = 45 deg
Solution: Initial Guesses T = 1kN
= 10deg
Given
+ Fx = 0; +
F 1 - T cos ( ) + F 2 sin ( 1 ) = 0 -T sin ( ) - F2 cos ( 1 ) + F = 0
Fy = 0;
T = Find ( T , )
T = 14.31 kN
= 36.27 deg
Problem 3-7 Determine the maximum weight of the engine that can be supported without exceeding a tension of T1 in chain AB and T2 in chain AC. Given:
= 30 deg
T1 = 450 lb T2 = 480 lb Solution: Initial Guesses F AB = T1 F AC = T2 W = 1lb
136
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Engineering Mechanics - Statics
Chapter 3
Given Assuming cable AB reaches the maximum tension F AB = T1.
+ Fx = 0; +
F AC cos ( ) - F AB = 0 F AC sin ( ) - W = 0
Fy = 0;
FAC1 = Find ( FAC , W) W1
Given
W1 = 259.81 lb
Assuming cable AC reaches the maximum tension FAC = T2. F AC cos ( ) - F AB = 0 F AC sin ( ) - W = 0
+ Fx = 0; +
Fy = 0;
FAB2 = Find ( FAB , W) W2
W = min ( W1 , W2 )
W2 = 240.00 lb
W = 240.00 lb
Problem 3-8 The engine of mass M is suspended from a vertical chain at A. A second chain is wrapped around the engine and held in position by the spreader bar BC. Determine the compressive force acting along the axis of the bar and the tension forces in segments BA and CA of the chain. Hint: Analyze equilibrium first at A, then at B. Units Used: kN = 10 N Given: M = 200 kg
3
1 = 55 deg
g = 9.81 m s
2
137
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Engineering Mechanics - Statics
Chapter 3
Solution: Initial guesses: Given Point A F CA cos ( 1 ) - FBA cos ( 1 ) = 0 M( g) - FCA sin ( 1 ) - FBA sin ( 1 ) = 0 F BA = 1 kN F CA = 2 kN
+ Fx = 0; +
Fy = 0;
FBA = Find ( FBA , FCA) FCA
At point B:
+ Fx = 0;
FBA 1.20 = kN FCA 1.20
F BA cos ( 1 ) - F BC = 0 F BC = FBA cos ( 1 ) F BC = 687 N
Problem 3-9 Cords AB and AC can each sustain a maximum tension T. If the drum has weight W, determine the smallest angle at which they can be attached to the drum. Given: T = 800 lb W = 900 lb Solution:
+
Fy = 0;
W - 2T sin ( ) = 0
= asin
W 2T
= 34.2 deg
138
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Engineering Mechanics - Statics
Chapter 3
Problem 3-10 The crate of weight W is hoisted using the ropes AB and AC. Each rope can withstand a maximum tension T before it breaks. If AB always remains horizontal, determine the smallest angle to which the crate can be hoisted. Given: W = 500 lb T = 2500 lb Solution: Case 1: Assume The initial guess Given
+ F x = 0;
TAB = T
= 30 deg
TAC = 2000 lb
TAB - TAC cos ( ) = 0 TAC sin ( ) - W = 0
+
F y = 0;
TAC1 = Find ( TAC , ) 1
Case 1: Assume The initial guess Given
+ F x = 0;
1 = 11.31 deg
TAC = T
TAC1 = 2550 lb
= 30 deg
TAB = 2000 lb
TAB - TAC cos ( ) = 0 TAC sin ( ) - W = 0
+
F y = 0;
TAB2 = Find ( TAB , ) 2
= max ( 1 , 2 )
2 = 11.54 deg = 11.54 deg
TAB2 = 2449 lb
139
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Engineering Mechanics - Statics
Chapter 3
Problem 3-11 Two electrically charged pith balls, each having mass M, are suspended from light threads of equal length. Determine the resultant horizontal force of repulsion, F, acting on each ball if the measured distance between them is r. Given: M = 0.2 gm r = 200 mm l = 150 mm d = 50 mm Solution: The initial guesses : T = 200 N Given
+
F = 200 N
F x = 0;
F - T
r - d = 0 2l
2 2 r - d l - 2 + - Mg = 0 Fy = 0; T l
T = Find ( T , F) F
F = 1.13 10
-3
N
T = 0.00 N
Problem 3-12 The towing pendant AB is subjected to the force F which is developed from a tugboat. Determine the force that is in each of the bridles, BC and BD, if the ship is moving forward with constant velocity. Units Used: kN = 10 N
3
140
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Engineering Mechanics - Statics
Chapter 3
Given: F = 50 kN
1 = 20 deg 2 = 30 deg
Solution: Initial guesses: Given
+ Fx = 0; +
TBC = 1 kN
TBD = 2 kN
TBC sin ( 2 ) - TBD sin ( 1 ) = 0 TBC cos ( 2 ) + TBD cos ( 1 ) - F = 0
Fy = 0;
TBC = Find ( TBC , TBD) TBD
TBC 22.32 = kN TBD 32.64
Problem 3-13
Determine the stretch in each spring for equilibrium of the block of mass M. The springs are shown in the equilibrium position.
141
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Engineering Mechanics - Statics
Chapter 3
Given: M = 2 kg a = 3m b = 3m c = 4m kAB = 30 N m N m N m m s Solution: The initial guesses: F AB = 1 N F AC = 1 N Given F AB c b 2 2 - FAC 2 2 = 0 a +c a +b a a 2 2 + FAB 2 2 - M g = 0 a +b a +c
2
kAC = 20
kAD = 40
g = 9.81
+
+
F x = 0;
F y = 0;
F AC
FAC = Find ( FAC , FAB) FAB
xAC = F AC kAC FAB kAB
FAC 15.86 = N FAB 14.01
xAC = 0.79 m
xAB =
xAB = 0.47 m
142
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Engineering Mechanics - Statics
Chapter 3
Problem 3-14 The unstretched length of spring AB is . If the block is held in the equilibrium position shown, determine the mass of the block at D. Given:
= 2m
a = 3m b = 3m c = 4m kAB = 30 N m N m N m m s Solution: F AB = kAB
2
kAC = 20
kAD = 40
g = 9.81
(
a +c -
2
2
)
The initial guesses: mD = 1 kg Given
+ Fx = 0;
F AC = 1 N F AB c b 2 2 - FAC 2 2 = 0 a +c a +b a a 2 2 + FAB 2 2 - mD g = 0 a +b a +c F AC = 101.8 N mD = 12.8 kg
+
F y = 0;
F AC
FAC = Find ( FAC , mD) mD
143
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Engineering Mechanics - Statics
Chapter 3
Problem 3-15 The springs AB and BC have stiffness k and unstretched lengths l/2. Determine the horizontal force F applied to the cord which is attached to the small pulley B so that the displacement of the pulley from the wall is d. Given: l = 6m k = 500 N m
d = 1.5 m
Solution: T = k
l + d2 - 2
2
l
2 d d +
2
T = 177.05 N
+ Fx = 0;
( 2T) - F = 0
l 2
2
F =
d
( 2T)
F = 158.36 N
l + d2 2
2
Problem 3-16 The springs AB and BC have stiffness k and an unstretched length of l. Determine the displacement d of the cord from the wall when a force F is applied to the cord.
144
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Engineering Mechanics - Statics
Chapter 3
Given: l = 6m k = 500 N m
F = 175 N Solution: The initial guesses: d = 1m Given
+ Fx = 0;
T = 1N
-F + ( 2T)
2
d d +
=0
l 2
2
Spring
T = k d +
2
2 l - l 2 2
T = Find ( T , d) d
T = 189.96 N
d = 1.56 m
Problem 3-17 Determine the force in each cable and the force F needed to hold the lamp of mass M in the position shown. Hint: First analyze the equilibrium at B; then, using the result for the force in BC, analyze the equilibrium at C. Given: M = 4 kg
1 = 30 deg 2 = 60 deg 3 = 30 deg
Solution: Initial guesses:
145
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Engineering Mechanics - Statics
Chapter 3
TBC = 1 N Given At B:
+ Fx = 0; +
TBA = 2 N
TBC cos ( 1 ) - TBA cos ( 2 ) = 0 TBA sin ( 2 ) - TBC sin ( 1 ) - M g = 0
Fy = 0;
TBC = Find ( TBC , TBA) TBA
At C: Given
+ Fx = 0; +
TBC 39.24 = N TBA 67.97
F = 2N
TCD = 1 N
-TBC cos ( 1 ) + TCD cos ( 3 ) = 0 TBC sin ( 1 ) + TCD sin ( 3 ) - F = 0
Fy = 0;
TCD = Find ( TCD , F) F
TCD 39.24 = N F 39.24
Problem 3-18 The motor at B winds up the cord attached to the crate of weight W with a constant speed. Determine the force in cord CD supporting the pulley and the angle for equilibrium. Neglect the
146
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Engineering Mechanics - Statics
Chapter 3
size of the pulley at C. Given: W = 65 lb Solution: The initial guesses: = 100 deg Given Equations of Equilibrium:
+ Fx = 0;
c = 12
d = 5
F CD = 200 lb
F CD cos ( ) - W F CD sin ( ) - W
=0 2 2 c +d
d c
+
Fy = 0;
-W=0 2 2 c +d
= Find ( , FCD) FCD
= 78.69 deg
F CD = 127.5 lb
Problem 3-19 The cords BCA and CD can each support a maximum load T. Determine the maximum weight of the crate that can be hoisted at constant velocity, and the angle for equilibrium.
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Engineering Mechanics - Statics
Chapter 3
Given: T = 100 lb c = 12 d = 5 The maximum will occur in CD rather than in BCA. Solution : The initial guesses: = 100 deg Given Equations of Equilibrium:
+ Fx = 0;
W = 200 lb
T cos ( ) - W
=0 c +d
d
2 2
+
F y = 0; T sin ( ) - W
-W=0 c +d
c
2 2
= Find ( , W) W
= 78.69 deg
W = 51.0 lb
148
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Engineering Mechanics - Statics
Chapter 3
Problem 3-20 The sack has weight W and is supported by the six cords tied together as shown. Determine the tension in each cord and the angle for equilibrium. Cord BC is horizontal. Given: W = 15 lb
1 = 30 deg 2 = 45 deg 3 = 60 deg
Solution: Guesses TBE = 1 lb TBC = 1 lb TCD = 1 lb Given At H:
+
TAB = 1 lb TAC = 1 lb TAH = 1 lb
= 20deg
Fy = 0;
TAH - W = 0
At A:
+ Fx = 0; +
-TAB cos ( 2 ) + TAC cos ( 3 ) = 0 TAB sin ( 2 ) + TAC sin ( 3 ) - W = 0
Fy = 0;
At B:
+ Fx = 0; +
TBC - TBE cos ( 1 ) + TAB cos ( 2 ) = 0 TBE sin ( 1 ) - TAB sin ( 2 ) = 0
Fy = 0;
At C:
+ Fx = 0; +
TCD cos ( ) - TBC - TBE cos ( 3 ) = 0 TCD sin ( ) - TAC sin ( 3 ) = 0
Fy = 0;
149
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Engineering Mechanics - Statics
Chapter 3
TBE TAB TBC TAC = Find ( TBE , TAB , TBC , TAC , TCD , TAH , ) T CD TAH
TBE 10.98 TAB 7.76 TBC = 4.02 lb TAC 10.98 13.45 TCD TAH 15.00
= 45.00 deg
Problem 3-21 Each cord can sustain a maximum tension T. Determine the largest weight of the sack that can be supported. Also, determine of cord DC for equilibrium. Given: T = 200 lb
1 = 30 deg 2 = 45 deg 3 = 60 deg
Solution: Solve for W = 1 and then scale the answer at the end. TBE = 1 TBC = 1 TCD = 1 TAB = 1 TAC = 1 TAH = 1
Guesses
150
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Engineering Mechanics - Statics
Chapter 3
= 20 deg
Given At H:
+
Fy = 0;
TAH - W = 0
At A:
+ Fx = 0; +
-TAB cos ( 2 ) + TAC cos ( 3 ) = 0 TAB sin ( 2 ) + TAC sin ( 3 ) - W = 0
Fy = 0;
At B:
+ Fx = 0; +
TBC - TBE cos ( 1 ) + TAB cos ( 2 ) = 0 TBE sin ( 1 ) - TAB sin ( 2 ) = 0
Fy = 0;
At C:
+ Fx = 0; +
TCD cos ( ) - TBC - TBE cos ( 3 ) = 0 TCD sin ( ) - TAC sin ( 3 ) = 0
Fy = 0;
TBE TAB TBC TAC = Find ( TBE , TAB , TBC , TAC , TCD , TAH , ) T CD TAH
W = max ( TBE , TAB , TBC , TAC , TCD , TAH ) T
TBE 0.73 TAB 0.52 TBC = 0.27 TAC 0.73 0.90 TCD TAH 1.00
W = 200.00 lb
= 45.00 deg
151
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Engineering Mechanics - Statics
Chapter 3
Problem 3-22 The block has weight W and is being hoisted at uniform velocity. Determine the angle for equilibrium and the required force in each cord. Given: W = 20 lb
= 30 deg
Solution: The initial guesses:
= 10 deg
Given
TAB = 50 lb
TAB sin ( ) - W sin ( ) = 0 TAB cos ( ) - W - W cos ( ) = 0
= Find ( , TAB) TAB
= 60.00 deg
TAB = 34.6 lb
Problem 3-23 Determine the maximum weight W of the block that can be suspended in the position shown if each cord can support a maximum tension T. Also, what is the angle for equilibrium?
152
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Engineering Mechanics - Statics
Chapter 3
Given: T = 80 lb
= 30 deg
The maximum load will occur in cord AB. Solution: TAB = T The initial guesses:
= 100deg W = 200lb
Given
+
Fy = 0;
TAB cos ( ) - W - W cos ( ) = 0 TAB sin ( ) - W sin ( ) = 0
+ Fx = 0;
= Find ( , W) W
W = 46.19 lb
= 60.00 deg
Problem 3-24 Two spheres A and B have an equal mass M and are electrostatically charged such that the repulsive force acting between them has magnitude F and is directed along line AB. Determine the angle , the tension in cords AC and BC, and the mass M of each sphere.
153
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Engineering Mechanics - Statics
Chapter 3
Unit used: mN = 10 Given: F = 20 mN g = 9.81 m s
2 -3
N
1 = 30 deg 2 = 30 deg
Solution: Guesses TB = 1 mN TA = 1 mN Given For B:
+ Fx = 0; +
M = 1 gm
= 30 deg
F cos ( 2 ) - TB sin ( 1 ) = 0 F sin ( 2 ) + TB cos ( 1 ) - M g = 0
Fy = 0;
For A:
+ Fx = 0; +
TA sin ( ) - F cos ( 2 ) = 0 TA cos ( ) - F sin ( 2 ) - M g = 0
Fy = 0;
TA TB = Find ( T , T , , M) A B M
TA 52.92 = mN TB 34.64
= 19.11 deg
M = 4.08 gm
Problem 3-25 Blocks D and F weigh W1 each and block E weighs W2. Determine the sag s for equilibrium. Neglect the size of the pulleys.
154
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Engineering Mechanics - Statics
Chapter 3
Given: W1 = 5 lb W2 = 8 lb a = 4 ft Solution: Sum forces in the y direction Guess Given s = 1 ft 2
W - W = 0 1 2 s +a
s
2 2
s = Find ( s)
s = 5.33 ft
Problem 3-26
If blocks D and F each have weight W1, determine the weight of block E if the sag is s. Neglect the size of the pulleys. Given: W1 = 5 lb s = 3 ft a = 4 ft Solution: Sum forces in the y direction
2
W - W = 0 1 s +a
s
2 2
155
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Engineering Mechanics - Statics
Chapter 3
W = 2
W1 2 2 s +a
s
W = 6.00 lb
Problem 3-27 The block of mass M is supported by two springs having the stiffness shown. Determine the unstretched length of each spring. Units Used: kN = 10 N Given: M = 30 kg l1 = 0.6 m l2 = 0.4 m l3 = 0.5 m kAC = 1.5 kAB = 1.2 g = 9.81 Solution: Initial guesses: Given
+ Fx = 0;
3
kN m kN m
m s
2
F AC = 20 N
F AB = 30 N
l2 FAB l2 + l3 l3 FAB l2 + l3
2 2 2 2
-
l1 F AC l1 + l3 l3 F AC l1 + l3
2 2 2 2
=0
+
F y = 0;
+
-Mg=0
156
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Engineering Mechanics - Statics
Chapter 3
FAC = Find ( FAC , FAB) FAB
Then guess LAB = 0.1 m
FAC 183.88 = N FAB 226.13
LAC = 0.1 m
Given
F AC = kAC l1 + l3 - LAC
2 2
F AB = kAB l2 + l3 - LAB
2 2
LAB = Find ( LAB , LAC) LAC
LAB 0.452 = m LAC 0.658
Problem 3-28 Three blocks are supported using the cords and two pulleys. If they have weights of WA = WC = W, WB = kW, determine the angle for equilibrium. Given: k = 0.25
Solution:
+ Fx = 0; +
W cos ( ) - k W cos ( ) = 0 W sin ( ) + k W sin ( ) - W = 0 cos ( ) = k cos ( ) sin ( ) = 1 - k sin ( )
Fy = 0;
157
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Engineering Mechanics - Statics
Chapter 3
1 = k cos ( ) + ( 1 - k sin ( ) ) = 1 + k - 2k sin ( )
2 2 2 2
k = asin
2
= 7.18 deg
Problem 3-29 A continuous cable of total length l is wrapped around the small pulleys at A, B, C, and D. If each spring is stretched a distance b, determine the mass M of each block. Neglect the weight of the pulleys and cords. The springs are unstretched when d = l/2. Given: l = 4m k = 500 N m
b = 300 mm g = 9.81 m s
2
Solution: Fs = k b Guesses T = 1N Given 2T sin ( ) - F s = 0 -2T cos ( ) + M g = 0 b+ l 4 sin ( ) = l 4 F s = 150.00 N
= 10 deg
M = 1 kg
T = Find ( T , , M) M
158
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Engineering Mechanics - Statics
Chapter 3
T = 107.14 N
= 44.43 deg
M = 15.60 kg
Problem 3-30 Prove Lami's theorem, which states that if three concurrent forces are in equilibrium, each is proportional to the sine of the angle of the other two; that is, P/sin = Q/sin = R/sin .
Solution: Sine law: sin ( 180deg - ) R = sin ( 180deg - ) Q = sin ( 180deg - ) P
However, in general sin ( ) R = sin ( ) Q =
sin ( 180deg - ) = sin ( ) , hence sin ( ) P Q.E.D.
Problem 3-31 A vertical force P is applied to the ends of cord AB of length a and spring AC. If the spring has an unstretched length , determine the angle for equilibrium. Given: P = 10 lb
= 2 ft
k = 15 lb ft
a = 2 ft
159
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Engineering Mechanics - Statics
Chapter 3
b = 2 ft Guesses
= 10 deg
T = 1 lb x = 1ft Given
= 10 deg
F = 1 lb
-T cos ( ) + F cos ( ) = 0 T sin ( ) + F sin ( ) - P = 0 F = k( x - ) a sin ( ) = x sin ( ) a cos ( ) + x cos ( ) = a + b
T = Find ( , , T , F , x) F x
T 10.30 = lb F 9.38
= 35 deg
Problem 3-32 Determine the unstretched length of spring AC if a force P causes the angle for equilibrium. Cord AB has length a. Given: P = 80 lb
= 60 deg
k = 50 lb ft
a = 2 ft
160
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Engineering Mechanics - Statics
Chapter 3
b = 2 ft Guesses
= 1ft
T = 1 lb x = 1ft Given
= 10 deg
F = 1 lb
-T cos ( ) + F cos ( ) = 0 T sin ( ) + F sin ( ) - P = 0 F = k( x - ) a sin ( ) = x sin ( ) a cos ( ) + x cos ( ) = a + b
T = Find ( , , T , F , x) F x
T 69.28 = lb F 40.00
= 2.66 ft
Problem 3-33 The flowerpot of mass M is suspended from three wires and supported by the hooks at B and C. Determine the tension in AB and AC for equilibrium. Given: M = 20 kg l1 = 3.5 m l2 = 2 m l3 = 4 m
161
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Engineering Mechanics - Statics
Chapter 3
l4 = 0.5 m g = 9.81 Solution: Initial guesses: TAB = 1 N TAC = 1 N m s
2
= 10 deg
Given
= 10 deg
-TAC cos ( ) + TAB cos ( ) = 0 TAC sin ( ) + TAB sin ( ) - M g = 0 l1 cos ( ) + l2 cos ( ) = l3 l1 sin ( ) = l2 sin ( ) + l4
TAB TAC = Find ( T , T , , ) AB AC
53.13 = deg 36.87
TAB 156.96 = N TAC 117.72
Problem 3-34 A car is to be towed using the rope arrangement shown. The towing force required is P. Determine the minimum length l of rope AB so that the tension in either rope AB or AC does not exceed T. Hint: Use the equilibrium condition at point A to determine the required angle for attachment, then determine l using trigonometry applied to triangle ABC. Given: P = 600 lb T = 750 lb
= 30 deg
162
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Engineering Mechanics - Statics
Chapter 3
d = 4 ft Solution: The initial guesses TAB = T TAC = T
= 30 deg
l = 2 ft Case 1: Given
+ Fx = 0; +
Assume TAC = T
TAC cos ( ) - TAB cos ( ) = 0 P - TAC sin ( ) - TAB sin ( ) = 0 l d
F y = 0;
sin ( )
=
sin ( 180deg - - )
TAB = Find ( TAB , , l) l 1
Case 2: Given
+ Fx = 0; +
TAB = 687.39 lb
= 19.11 deg
l1 = 2.65 ft
Assume TAB = T TAC cos ( ) - TAB cos ( ) = 0 P - TAC sin ( ) - TAB sin ( ) = 0 sin ( ) l = sin ( 180deg - - ) TAC = 840.83 lb d
F y = 0;
TAC = Find ( TAC , , l) l 2
l = min ( l1 , l2 )
= 13.85 deg
l2 = 2.89 ft
l = 2.65 ft
163
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Engineering Mechanics - Statics
Chapter 3
Problem 3-35 Determine the mass of each of the two cylinders if they cause a sag of distance d when suspended from the rings at A and B. Note that s = 0 when the cylinders are removed. Given: d = 0.5 m l1 = 1.5 m l2 = 2 m l3 = 1 m k = 100 g = 9.81 N m m s Solution: TAC = k
2
( l 1 + d) 2 + l 2 2 -
l1 + l2
2
2
TAC = 32.84 N
= atan
l1 + d l2
= 45 deg
TAC sin ( ) g
M =
M = 2.37 kg
164
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Engineering Mechanics - Statics
Chapter 3
Problem 3-36 The sling BAC is used to lift the load W with constant velocity. Determine the force in the sling and plot its value T (ordinate) as a function of its orientation , where 0 90 .
Solution: W - 2T cos ( ) = 0 T=
W 2 cos ( )
1
Problem 3-37 The lamp fixture has weight W and is suspended from two springs, each having unstretched length L and stiffness k. Determine the angle for equilibrium. Units Used: kN = 10 N Given: W = 10 lb L = 4 ft k = 5 lb ft
3
a = 4 ft
165
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Engineering Mechanics - Statics
Chapter 3
Solution: The initial guesses: T = 200 lb Given Spring
+
= 10 deg
T = k
a - L cos ( )
Fy = 0;
2T sin ( ) - W = 0
T = Find ( T , )
T = 7.34 lb
= 42.97 deg
Problem 3-38 The uniform tank of weight W is suspended by means of a cable, of length l, which is attached to the sides of the tank and passes over the small pulley located at O. If the cable can be attached at either points A and B, or C and D, determine which attachment produces the least amount of tension in the cable.What is this tension? Given: W = 200 lb l = 6 ft a = 1 ft b = 2 ft c = b d = 2a Solution: Free Body Diagram: By observation, the force F has to support the entire weight of the tank. Thus, F = W. The tension in
166
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Engineering Mechanics - Statics
Chapter 3
cable is the same throughout the cable.
Equations of Equilibrium F y = 0; Attached to CD W - 2T sin ( ) = 0
1 = acos 2 = acos
l l
2 a
1 = 70.53 deg 2 = 48.19 deg
Attached to AB
2 b
We choose the largest angle (which will produce the smallest force)
= max ( 1 , 2 )
T = 1 W 2 sin ( )
= 70.53 deg
T = 106 lb
Problem 3-39 A sphere of mass ms rests on the smooth parabolic surface. Determine the normal force it exerts on the surface and the mass mB of block B needed to hold it in the equilibrium position shown. Given: ms = 4 kg a = 0.4 m b = 0.4 m
= 60 deg
g = 9.81 Solution: k = a b
2
m s
2
Geometry: The angle 1 which the surface make with the horizontal is to be determined first.
167
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Engineering Mechanics - Statics
Chapter 3
tan ( 1 ) =
dy dx
=2kx
evaluated at x = a
1 = atan ( 2 k a)
1 = 63.43 deg
Free Body Diagram : The tension in the cord is the same throughout the cord and is equal to the weight of block B, mBg. The initial guesses: mB = 200 kg Given
+ Fx = 0; +
F N = 200 N
mB g cos ( ) - FN sin ( 1 ) = 0 mB g sin ( ) + F N cos ( 1 ) - ms g = 0
Fy = 0;
mB = Find ( mB , FN) FN
F N = 19.66 N mB = 3.58 kg
Problem 3-40 The pipe of mass M is supported at A by a system of five cords. Determine the force in each cord for equilibrium. Given: M = 30 kg g = 9.81 m s
2
c = 3 d = 4
= 60 deg
Solution: Initial guesses: TAB = 1 N TBC = 1 N Given TAB sin ( ) - M g = 0 TAE - TAB cos ( ) = 0
168
TAE = 1 N TBD = 1 N
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Engineering Mechanics - Statics
Chapter 3
TBD TBD
c 2 2 - TAB sin ( ) = 0 c +d d 2 2 + TAB cos ( ) - TBC = 0 c +d
TAB TAE = Find ( T , T , T , T ) AB AE BC BD TBC TBD TAB 339.8 TAE = 169.9 N TBC 562.3 TBD 490.5
Problem 3-41 The joint of a space frame is subjected to four forces. Strut OA lies in the x-y plane and strut OB lies in the y-z plane. Determine the forces acting in each of the three struts required for equilibrium. Units Used: kN = 10 N Given: F = 2 kN
3
1 = 45 deg 2 = 40 deg
Solution: F x = 0; -R sin ( 1 ) = 0 R = 0
169
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Engineering Mechanics - Statics
Chapter 3
F z = 0;
P sin ( 2 ) - F = 0 P = sin ( 2 ) F
P = 3.11 kN F y = 0; Q - P cos ( 2 ) = 0 Q = P cos ( 2 ) Q = 2.38 kN
Problem 3-42 Determine the magnitudes of F1, F 2, and F3 for equilibrium of the particle. Units Used: kN = 10 N Given: F 4 = 800 N
3
= 60 deg = 30 deg = 30 deg
c = 3 d = 4
Solution: The initial guesses: F 1 = 100 N Given F 2 = 100 N F 3 = 100 N
cos ( ) 0 + F1 sin ( )
-cos ( ) 0 c -d + F -sin ( ) + F sin ( ) = 0 3 4 2 2 c +d 0 0 -cos ( )
F2
170
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Engineering Mechanics - Statics
Chapter 3
F1 F2 = Find ( F1 , F2 , F3) F 3
F1 800 F2 = 147 N F 564 3
Problem 3-43 Determine the magnitudes of F1, F 2, and F3 for equilibrium of the particle. Units Used: kN = 1000 N Given: F 4 = 8.5 kN F 5 = 2.8 kN
= 15 deg = 30 deg
c = 7 d = 24
Solution: Initial Guesses: Given F 1 = 1 kN F 2 = 1 kN F 3 = 1 kN
-cos ( ) 0 + F1 sin ( )
-sin ( ) -c 1 0 -d + F 0 + F cos ( ) + F 0 = 0 3 4 5 2 2 c +d 0 0 -1 0
F2
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Engineering Mechanics - Statics
Chapter 3
F1 F2 = Find ( F1 , F2 , F3) F 3
F1 5.60 F2 = 8.55 kN F 9.44 3
Problem 3-44 Determine the magnitudes of F1, F 2 and F3 for equilibrium of particle the F = {- 9i - 8j - 5k}. Units Used: kN = 10 N
3
Given:
-9 F = -8 kN -5
a = 4m b = 2m c = 4m
1 = 30 deg 2 = 60 deg 3 = 135 deg 4 = 60 deg 5 = 60 deg
Solution: Initial guesses: F 1 = 8 kN F 2 = 3 kN F 3 = 12 kN
172
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Engineering Mechanics - Statics
Chapter 3
Given
cos ( 2 ) cos ( 1) cos ( 3 ) F 1 -cos ( 2 ) sin ( 1 ) + F 2 cos ( 5 ) + cos sin ( 2 ) ( 4) F1 F2 = Find ( F1 , F2 , F3) F 3
a c +F=0 2 2 2 a + b + c -b
F3
F1 8.26 F2 = 3.84 kN F 12.21 3
Problem 3-45 The three cables are used to support the lamp of weight W. Determine the force developed in each cable for equilibrium. Units Used: kN = 10 N Given: W = 800 N a = 4m b = 4m c = 2m
3
Solution: Initial Guesses: F AB = 1 N F AC = 1 N F AD = 1 N
173
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Engineering Mechanics - Statics
Chapter 3
Given
0 1 1 + F 0 + F AB AC 0 0
-c 0 -b + W 0 = 0 2 2 2 a +b +c a -1
F AD
FAB FAC = Find ( F AB , FAC , FAD) F AD
FAB 800 FAC = 400 N F 1200 AD
Problem 3-46 Determine the force in each cable needed to support the load W. Given: a = 8 ft b = 6 ft c = 2 ft d = 2 ft e = 6 ft W = 500 lb
Solution: Initial guesses: F CD = 600 lb F CA = 195 lb F CB = 195 lb
174
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Engineering Mechanics - Statics
Chapter 3
Given
c -e + 2 2 c +e 0
FCA
-d -e + 2 2 d +e 0
F CB
0 0 b + W 0 = 0 2 2 a + b a -1
F CD
FCD FCA = Find ( FCD , F CA , F CB) F CB
Problem 3-47 Determine the stretch in each of the two springs required to hold the crate of mass mc in the equilibrium position shown. Each spring has an unstretched length and a stiffness k. Given: mc = 20 kg
FCD 625 FCA = 198 lb F 198 CB
= 2m
k = 300 a = 4m b = 6m c = 12 m Solution: Initial Guesses F OA = 1 N F OB = 1 N F OC = 1 N Given N m
0 -1 -1 + F 0 + F OA OB 0 0
b 0 a + m g 0 = 0 c 2 2 2 a + b + c c -1
F OC
175
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Engineering Mechanics - Statics
Chapter 3
FOA FOB = Find ( FOA , FOB , FOC) F OC
OA =
F OA k F OB k
FOA 65.40 FOB = 98.10 N F 228.90 OC
OA = 218 mm
OB =
OB = 327 mm
Problem 3-48
If the bucket and its contents have total weight W, determine the force in the supporting cables DA, DB, and DC. Given: W = 20 lb a = 3 ft b = 4.5 ft c = 2.5 ft d = 3 ft e = 1.5 ft f = 1.5 ft
176
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Engineering Mechanics - Statics
Chapter 3
Solution: The initial guesses: F DA = 40 lb Given F x = 0; b-e e F - F = 0 DC DA 2 2 2 2 2 2 ( b - e) + f + a e + d + ( c - f) -f c- f F + F - F = 0 DC DB DA 2 2 2 2 2 2 ( b - e) + f + a e + d + ( c - f) a d F + F - W = 0 DC DA 2 2 2 2 2 2 ( b - e) + f + a e + d + ( c - f) F DB = 20 lb F DC = 30 lb
F y = 0;
F z = 0;
FDA FDB = Find ( FDA , FDB , FDC) F DC
FDA 10.00 FDB = 1.11 lb F 15.56 DC
Problem 3-49
The crate which of weight F is to be hoisted with constant velocity from the hold of a ship using the cable arrangement shown. Determine the tension in each of the three cables for equilibrium.
177
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Engineering Mechanics - Statics
Chapter 3
Units Used: kN = 10 N Given: F = 2.5 kN a = 3m b = 1m c = 0.75 m d = 1m e = 1.5 m f = 3m Solution: The initial guesses F AD = 3 kN Given
2 3
F AC = 3 kN -c c +b +a
2 2
F AB = 3 kN d d +e +a
2 2 2
F AD +
F AC +
d d + f +a -f d + f +a
2 2 2 2 2 2
F AB = 0
b c +b +a -a c +b +a
2 2 2 2 2
F AD +
e d +e +a
2 2 2
F AC +
F AB = 0
2
F AD +
-a d +e +a
2 2 2
F AC +
-a d + f +a
2 2 2
F AB + F = 0
FAD FAC = Find ( F AD , F AC , F AB) F AB
FAD 1.55 FAC = 0.46 kN F 0.98 AB
178
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Engineering Mechanics - Statics
Chapter 3
Problem 3-50
The lamp has mass ml and is supported by pole AO and cables AB and AC. If the force in the pole acts along its axis, determine the forces in AO, AB, and AC for equilibrium. Given: ml = 15 kg a = 6m b = 1.5 m c = 2m Solution: The initial guesses: F AO = 100 N F AB = 200 N F AC = 300 N Given Equilibrium equations: c c +b +a
2 2 2
d = 1.5 m e = 4m f = 1.5 m g = 9.81 m s
2
F AO -
c+e ( c + e) + ( b + f) + a
2 2 2
F AB -
c c + ( b + d) + a
2 2 2
F AC = 0
-
2
b c +b +a a c +b +a
2 2 2 2 2
F AO +
b+ f ( c + e) + ( b + f) + a a ( c + e) + ( b + f) + a
2 2 2 2 2 2
F AB +
b+d c + ( b + d) + a a c + ( b + d) + a
2 2 2 2 2 2
F AC = 0
F AO -
F AB -
F AC - ml g = 0
FAO FAB = Find ( FAO , FAB , FAC) F AC
FAO 318.82 FAB = 110.36 N F 85.84 AC
179
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Engineering Mechanics - Statics
Chapter 3
Problem 3-51 Cables AB and AC can sustain a maximum tension Tmax, and the pole can support a maximum compression Pmax. Determine the maximum weight of the lamp that can be supported in the position shown. The force in the pole acts along the axis of the pole. Given: Tmax = 500 N Pmax = 300 N a = 6m b = 1.5 m Solution: Lengths AO = AB = AC = a +b +c
2 2 2 2 2 2
c = 2m d = 1.5 m e = 4m f = 1.5 m
a + ( c + e) + ( b + d) a + c + ( b + d)
2 2 2
The initial guesses: F AO = Pmax F AB = Tmax F AC = Tmax W = 300N
Case 1 Assume the pole reaches maximum compression Given
F AO
c F -c - e F -c 0 + AB b + f + AC b + d + W 0 = 0 -b AC AO AB a -a -a -1 W1 138.46 FAB1 = 103.85 N F AC1 80.77
W1 FAB1 = Find ( W , FAB , FAC) F AC1
180
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Engineering Mechanics - Statics
Chapter 3
Case 2 Assume that cable AB reaches maximum tension Given
F AO
c F -c - e F -c 0 + AB b + f + AC b + d + W 0 = 0 -b AC AO AB a -a -a -1 W2 666.67 FAO2 = 1444.44 N F AC2 388.89
W2 FAO2 = Find ( W , FAO , FAC) F AC2
Case 3 Assume that cable AC reaches maximum tension Given
F AO
c F -c - e F -c 0 + AB b + f + AC b + d + W 0 = 0 -b AC AO AB a -a -a -1 W3 857.14 FAO3 = 1857.14 N F AB3 642.86
W = 138.46 N
W3 FAO3 = Find ( W , FAO , FAB) F AB3
Final Answer
W = min ( W1 , W2 , W3 )
Problem 3-52 Determine the tension in cables AB, AC, and AD, required to hold the crate of weight W in equilibrium. Given: W = 60 lb a = 6 ft b = 12 ft c = 8 ft
181
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Engineering Mechanics - Statics
Chapter 3
d = 9 ft e = 4 ft f = 6 ft
Solution: The initial guesses: TB = 100 lb TC = 100 lb TD = 100 lb Given F x = 0; TB - b b +c +d F y = 0; d b +c +d F z = 0; Solving -W +
2 2 2 2 2 2 2
TC -
b b +e + f e
2 2 2
TD = 0
TC -
b +e + f TC +
2
2
2
TD = 0
c b +c +d
2 2
f b +e + f
2 2 2
TD = 0
TB TC = Find ( TB , TC , TD) T D
TB 108.84 TC = 47.44 lb T 87.91 D
Problem 3-53 The bucket has weight W. Determine the tension developed in each cord for equilibrium. Given: W = 20 lb
182
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Engineering Mechanics - Statics
Chapter 3
a = 2 ft b = 2 ft c = 8 ft d = 7 ft e = 3 ft f = a Solution: Initial Guesses: Given F x = 0; c-e ( c - e) + b + a -b ( c - e) + b + a F y = 0; a ( c - e) + b + a
2 2 2 2 2 2 2 2 2
F DA = 20 lb
F DB = 10 lb
F DC = 15 lb -e -e e +b + f -b e +b + f F DC + f e +b + f
2 2 2 2 2 2 2 2 2
F DA +
e + ( d - b) + f d-b e + ( d - b) + f F DA + f e + ( d - b) + f
2 2 2 2
2
2
2
F DC +
F DB = 0
F x = 0;
F DA +
2
F DC +
F DB = 0
2
F DB - W = 0
FDA FDB = Find ( FDA , FDB , FDC) F DC FDA 21.54 FDB = 13.99 lb F 17.61 DC
Problem 3-54 The mast OA is supported by three cables. If cable AB is subjected to tension T, determine the tension in cables AC and AD and the vertical force F which the mast exerts along its axis on the collar at A. Given: T = 500 N
183
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Engineering Mechanics - Statics
Chapter 3
a = 6m b = 3m c = 6m d = 3m e = 2m f = 1.5 m g = 2m Solution: Initial Guesses: Given F x = 0; e e +d +a F y = 0; d e +d +a F z = 0; -a e +d +a
2 2 2 2 2 2 2 2 2
F AC = 90 N
F AD = 350 N
F = 750 N
T-
2
f f +g +a
2 2
F AC -
b b +c +a
2 2 2
F AD = 0
T+
2
g f +g +a
2 2
F AC -
c b +c +a
2 2 2
F AD = 0
T-
2
a f +g +a
2 2
F AC -
a b +c +a
2 2 2
F AD + F = 0
FAC FAD = Find ( F AC , F AD , F ) F
FAC 92.9 FAD = 364.3 N F 757.1
Problem 3-55 The ends of the three cables are attached to a ring at A and to the edge of the uniform plate of mass M. Determine the tension in each of the cables for equilibrium. Given: M = 150 kg e = 4m
184
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Engineering Mechanics - Statics
Chapter 3
a = 2m b = 10 m c = 12 m d = 2m gravity = 9.81 Solution: The initial guesses: F B = 15 N F C = 16 N F D = 16 N Given F x = 0; m s
2
f = 6m g = 6m h = 6m i = 2m
F B( f - i ) ( f - i) + h + c F B ( - h) ( f - i) + h + c
2 2 2 2 2 2
+
F C( -d - e) ( d + e) + ( h - a) + c F C [ - ( h - a) ] ( d + e) + ( h - a) + c
2 2 2 2 2 2
+
FD( -e) e +g +c FD( g) e +g +c
2 2 2 2 2 2
=0
F y = 0;
+
+
=0
F z = 0;
F B( -c) ( f - i) + h + c
2 2 2
+
FC( -c) ( d + e) + ( h - a) + c
2 2 2
+
FD( -c) e +g +c
2 2 2
+ M gravity = 0
FB FC = Find ( FB , FC , FD) F D
FB 858 FC = 0 N F 858 D
Problem 3-56 The ends of the three cables are attached to a ring at A and to the edge of the uniform plate. Determine the largest mass the plate can have if each cable can support a maximum tension of T.
185
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Engineering Mechanics - Statics
Chapter 3
kN = 10 N Given: T = 15 kN a = 2m b = 10 m c = 12 m d = 2m gravity = 9.81 m s
2
3
e = 4m f = 6m g = 6m h = 6m i = 2m
Solution: The initial guesses: FB = T FC = T FD = T M = 1 kg
Case 1: Assume that cable B reaches maximum tension Given F x = 0; F B( f - i ) ( f - i) + h + c F B ( - h) ( f - i) + h + c
2 2 2 2 2 2
+
F C( -d - e) ( d + e) + ( h - a) + c F C [ - ( h - a) ] ( d + e) + ( h - a) + c
2 2 2 2 2 2
+
FD( -e) e +g +c FD( g) e +g +c
2 2 2 2 2 2
=0
F y = 0;
+
+
=0
F z = 0;
F B( -c) ( f - i) + h + c
2 2 2
+
FC( -c) ( d + e) + ( h - a) + c
2 2 2
+
FD( -c) e +g +c
2 2 2
+ M gravity = 0
M1 FC1 = Find ( M , FC , FD) F D1
FC1 -0.00 = kN FD1 15.00
M1 = 2621.23 kg
186
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Engineering Mechanics - Statics
Chapter 3
Case 2: Assume that cable D reaches maximum tension Given F x = 0; F B( f - i ) ( f - i) + h + c F B ( - h) ( f - i) + h + c
2 2 2 2 2 2
+
F C( -d - e) ( d + e) + ( h - a) + c F C [ - ( h - a) ] ( d + e) + ( h - a) + c
2 2 2 2 2 2
+
FD( -e) e +g +c FD( g) e +g +c
2 2 2 2 2 2
=0
F y = 0;
+
+
=0
F z = 0;
F B( -c) ( f - i) + h + c
2 2 2
+
FC( -c) ( d + e) + ( h - a) + c
2 2 2
+
FD( -c) e +g +c
2 2 2
+ M gravity = 0
M2 FB2 = Find ( M , FB , FC) F C2
FB2 15.00 = kN FC2 0.00
M2 = 2621.23 kg
For this set of number FC = 0 for any mass that is applied. For a different set of numbers it would be necessary to also check case 3: Assume that the cable C reaches a maximum. M = min ( M1 , M2 ) M = 2621.23 kg
Problem 3-57 The crate of weight W is suspended from the cable system shown. Determine the force in each segment of the cable, i.e., AB, AC, CD, CE, and CF. Hint: First analyze the equilibrium of point A, then using the result for AC, analyze the equilibrium of point C. Units Used: kip = 1000 lb Given: W = 500 lb a = 10 ft b = 24 ft c = 24 ft
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Engineering Mechanics - Statics
Chapter 3
d = 7 ft e = 7 ft
1 = 20 deg 2 = 35 deg
Solution:
At A: F AC = 570 lb F AB = 500 lb
Initial guesses: Given
+ Fx = 0; +
F AB cos ( 1 ) - F AC cos ( 2 ) = 0 F AB sin ( 1 ) + FAC sin ( 2 ) - W = 0
Fy = 0;
FAC = Find ( FAC , FAB) FAB
At C: Initial Guesses
FAC 574 = lb FAB 500
F CD = 1 lb Given
F CE = 1 lb
F CF = 1 lb
cos ( 2 ) F AC 0 + -sin ( ) 2
-a 0+ 2 2 a +b b
FCD
0 d+ 2 2 c + d -c
F CE
188
0 -e = 0 2 2 c + e -c
F CF
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Engineering Mechanics - Statics
Chapter 3
FCD FCE = Find ( FCD , F CE , F CF ) F CF
F CD = 1.22 kip
FCE 416 = lb FCF 416
Problem 3-58 The chandelier of weight W is supported by three wires as shown. Determine the force in each wire for equilibrium. Given: W = 80 lb r = 1 ft h = 2.4 ft
Solution: The initial guesses: F AB = 40 lb F AC = 30 lb F AD = 30 lb
Given F x = 0; r r +h F y = 0; -r r +h
2 2 2 2
FAC -
r cos ( 45 deg) r +h
2 2
FAB = 0
FAD +
r cos ( 45 deg) r +h
2 2
FAB = 0
189
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Engineering Mechanics - Statics
Chapter 3
F z = 0;
h r +h
2 2
FAC +
h r +h
2 2
F AD +
h r +h
2 2
FAB - W = 0
FAB FAC = Find ( FAB , FAC , FAD) F AD
FAB 35.9 FAC = 25.4 lb F 25.4 AD
Problem 3-59 If each wire can sustain a maximum tension Tmax before it fails, determine the greatest weight of the chandelier the wires will support in the position shown. Given: Tmax = 120 lb r = 1 ft h = 2.4 ft
Solution: The initial guesses: F AB = Tmax F AC = Tmax F AD = Tmax W = Tmax Case 1 Assume that cable AB has maximum tension Given F x = 0; r r +h F y = 0; -r r +h
2 2 2 2
FAC -
r cos ( 45deg) r +h
2 2
F AB = 0
FAD +
r cos ( 45deg) r +h
2 2
F AB = 0
190
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Engineering Mechanics - Statics
Chapter 3
Fz = 0;
h r +h
2 2
FAC +
h r +h
2 2
F AD +
h r +h
2 2
FAB - W = 0
W1 FAC1 = Find ( W , FAC , FAD) F AD1
W1 267.4 FAC1 = 84.9 lb F AD1 84.9
Case 2 Assume that cable AC has maximum tension Given F x = 0; r r +h F y = 0; -r r +h F z = 0; h r +h
2 2 2 2 2 2
FAC -
r cos ( 45 deg) r +h
2 2
FAB = 0
FAD +
r cos ( 45 deg) r +h
2 2
FAB = 0
FAC +
h r +h
2 2
F AD +
h r +h
2 2
FAB - W = 0
W2 FAB2 = Find ( W , FAB , FAD) F AD2
W2 378.2 FAB2 = 169.7 lb F AD2 120
Case 3 Assume that cable AD has maximum tension Given F x = 0; r r +h F y = 0; -r r +h F z = 0; h r +h
2 2 2 2 2 2
FAC -
r cos ( 45deg) r +h
2 2
F AB = 0
FAD +
r cos ( 45deg) r +h
2 2
F AB = 0
FAC +
h r +h
2 2
F AD +
h r +h
2 2
FAB - W = 0
191
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Engineering Mechanics - Statics
Chapter 3
W3 FAB3 = Find ( W , FAB , FAC) F AC3
W = min ( W1 , W2 , W3 )
W3 378.2 FAB3 = 169.7 lb F AC3 120
W = 267.42 lb
Problem 3-60 Determine the force in each cable used to lift the surge arrester of mass M at constant velocity. Units Used: kN = 10 N Mg = 10 kg Given: M = 9.50 Mg a = 2m b = 0.5 m
3 3
= 45 deg
Solution: Initial guesses: Given F x = 0; FB b b +a FC
2 2
F B = 50 kN
F C = 30 kN b cos ( ) b +a
2 2
F D = 10 kN
- FC
=0
F y = 0;
-b sin ( ) b +a
2 2
+ FD
b b +a
2 2
=0
F z = 0;
M g - FB
a b +a
2 2
- FC
a b +a
2 2
- FD
a b +a
2 2
=0
192
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Engineering Mechanics - Statics
Chapter 3
FB FC = Find ( FB , FC , FD) F D
FB 28.13 FC = 39.78 kN F 28.13 D
Problem 3-61 The cylinder of weight W is supported by three chains as shown. Determine the force in each chain for equilibrium.
Given: W = 800 lb r = 1 ft d = 1 ft Solution: The initial guesses: F AB = 1 lb F AC = 1 lb F AD = 1 lb Given F x = 0; r r +d F y = 0; -r r +d d F z = 0; r +d
2 2 2 2 2 2
FAC -
r cos ( 45 deg) r +d
2 2
FAB = 0
FAD +
r sin ( 45 deg) r +d
2 2
F AB = 0
FAD +
d r +d
2 2
F AC +
d r +d
2 2
FAB - W = 0
FAB FAC = Find ( FAB , FAC , FAD) F AD
FAB FAC F = AD
469 331 331 lb
193
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Engineering Mechanics - Statics
Chapter 3
Problem 3-62 The triangular frame ABC can be adjusted vertically between the three equal-length cords. If it remains in a horizontal plane, determine the required distance s so that the tension in each of the cords, OA, OB, and OC, equals F . The lamp has a mass M. Given: F = 20 N M = 5 kg g = 9.81 m s d = 0.5 m Solution:
2
F z = 0;
3F cos ( ) = M g
= acos
3F
M g
= 35.16 deg
= s tan ( )
Geometry
2d cos ( 30 deg) 3 s = 3 tan ( )
2d cos ( 30 deg)
s = 410 mm
Problem 3-63 Determine the force in each cable needed to support the platform of weight W. Units Used: kip = 10 lb
3
194
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Engineering Mechanics - Statics
Chapter 3
Given: W = 3500 lb a = 2 ft b = 4 ft c = 4 ft d = 4 ft e = 3 ft f = 3 ft g = 10 ft
Solution: The initial guesses: F AB = 1 lb Given -b g + ( e - a) + b e-a g + ( e - a) + b -g g + ( e - a) + b
2 2 2 2 2 2 2 2 2
F AC = 1 lb
F AD = 1 lb
FAD +
c-d ( c - d) + e + g
2 2 2
FAC +
c c + f +g
2 2 2
F AB = 0
FAD +
e e + ( c - d) + g
2 2 2
FAC -
f c + f +g
2 2 2
F AB = 0
FAD -
g e + ( c - d) + g
2 2 2
FAC -
g g + f +c
2 2 2
F AB + W = 0
FAB FAC = Find ( FAB , FAC , FAD) F AD
FAB 1.467 FAC = 0.914 kip F 1.42 AD
195
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Engineering Mechanics - Statics
Chapter 3
Problem 3-64 A flowerpot of mass M is supported at A by the three cords. Determine the force acting in each cord for equilibrium. Given: M = 25 kg g = 9.81 m s
2
1 = 30 deg 2 = 30 deg 3 = 60 deg 4 = 45 deg
Solution: Initial guesses: F AB = 1 N F AD = 1 N F AC = 1 N Given F x = 0; F y = 0; F z = 0; F AD sin ( 1 ) - F AC sin ( 2 ) = 0 -F AD cos ( 1 ) sin ( 3 ) - FAC cos ( 2 ) sin ( 3 ) + F AB sin ( 4 ) = 0 F AD cos ( 1 ) cos ( 3 ) + F AC cos ( 2 ) cos ( 3 ) + F AB cos ( 4 ) - M g = 0
FAB FAC = Find ( FAB , FAC , FAD) F AD
Problem 3-65
FAB 219.89 FAC = 103.65 N F 103.65 AD
If each cord can sustain a maximum tension of T before it fails, determine the greatest weight of the flowerpot the cords can support.
196
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Engineering Mechanics - Statics
Chapter 3
Given: T = 50 N
1 = 30 deg 2 = 30 deg 3 = 60 deg 4 = 45 deg
Solution: Initial guesses: F AB = T F AD = T F AC = T W = T Case 1 Given F x = 0; F AD sin ( 1 ) - F AC sin ( 2 ) = 0 F y = 0; -F AD cos ( 1 ) sin ( 3 ) - FAC cos ( 2 ) sin ( 3 ) + F AB sin ( 4 ) = 0 F z = 0; F AD cos ( 1 ) cos ( 3 ) + F AC cos ( 2 ) cos ( 3 ) + F AB cos ( 4 ) - W = 0 Assume that AB reaches maximum tension
W1 FAC1 = Find ( W , FAC , FAD) F AD1
Case 2
W1 55.77 FAC1 = 23.57 N F AD1 23.57
Assume that AC reaches maximum tension
197
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Engineering Mechanics - Statics
Chapter 3
Given F x = 0; F AD sin ( 1 ) - F AC sin ( 2 ) = 0 F y = 0; -F AD cos ( 1 ) sin ( 3 ) - FAC cos ( 2 ) sin ( 3 ) + F AB sin ( 4 ) = 0 F z = 0; F AD cos ( 1 ) cos ( 3 ) + F AC cos ( 2 ) cos ( 3 ) + F AB cos ( 4 ) - W = 0
W2 FAB2 = Find ( W , FAB , FAD) F AD2
Case 3 Given F x = 0; F AD sin ( 1 ) - F AC sin ( 2 ) = 0
W2 118.30 FAB2 = 106.07 N F AD2 50.00
Assume that AD reaches maximum tension
F y = 0; -F AD cos ( 1 ) sin ( 3 ) - FAC cos ( 2 ) sin ( 3 ) + F AB sin ( 4 ) = 0 F z = 0; F AD cos ( 1 ) cos ( 3 ) + F AC cos ( 2 ) cos ( 3 ) + F AB cos ( 4 ) - W = 0
W3 FAB3 = Find ( W , FAB , FAC) F AC3
W = min ( W1 , W2 , W3 )
W3 118.30 FAB3 = 106.07 N F AC3 50.00
W = 55.77 N
198
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Engineering Mechanics - Statics
Chapter 3
Problem 3-66 The pipe is held in place by the vice. If the bolt exerts force P on the pipe in the direction shown, determine the forces F A and F B that the smooth contacts at A and B exerton the pipe. Given: P = 50 lb
= 30 deg
c = 3 d = 4
Solution: Initial Guesses Given
+ Fx = 0; +
F A = 1 lb
F B = 1 lb
F B - FA sin ( ) - P -F A cos ( ) + P
=0 2 2 c +d
d c
Fy = 0;
=0 2 2 c +d
FA = Find ( FA , FB) FB
FA 34.6 = lb FB 57.3
Problem 3-67 When y is zero, the springs sustain force F0. Determine the magnitude of the applied vertical forces F and -F required to pull point A away from point B a distance y1. The ends of cords CAD and CBD are attached to rings at C and D.
199
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Engineering Mechanics - Statics
Chapter 3
Given: F 0 = 60 lb k = 40 lb ft
d = 2 ft y1 = 2 ft Solution: Initial spring stretch: s1 = F0 k s1 = 1.50 ft
Initial guesses: F s = 1 lb T = 1 lb F = 1 lb
Given
F-
y1 2d
T=0
y1 d - 2 T-F =0 s
2
2
d
2 y d - d2 - 1 + s1 Fs = k 2
Fs T = Find ( Fs , T , F) F
Fs 70.72 = lb T 81.66
F = 40.83 lb
Problem 3-68 When y is zero, the springs are each stretched a distance . Determine the distance y if a force F is applied to points A and B as shown. The ends of cords CAD and CBD are attached to
200
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Engineering Mechanics - Statics
Chapter 3
pp p rings at C and D. Given:
= 1.5 ft
k = 40 lb ft
d = 2 ft F = 60 lb Solution: Initial guesses: F s = 1 lb Given F- y T=0 2d
2
T = 1 lb
y = 1 ft
y d - 2 T - F = 0 s
2
d
F s = kd -
d -
2
2 y + 2
Fs T = Find ( Fs , T , y) y
Fs 76.92 = lb T 97.55
y = 2.46 ft
Problem 3-69 Cord AB of length a is attached to the end B of a spring having an unstretched length b. The other end of the spring is attached to a roller C so that the spring remains horizontal as it stretches. If a weight W is suspended from B, determine the angle of cord AB for equilibrium.
201
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Engineering Mechanics - Statics
Chapter 3
Given: a = 5 ft b = 5 ft k = 10 lb ft
W = 10 lb Solution: Initial Guesses F BA = 1 lb F sp = 1 lb
= 30 deg
Given F sp - FBA cos ( ) = 0 F BA sin ( ) - W = 0 F sp = k( a - a cos ( ) )
Fsp FBA = Find ( Fsp , FBA , )
Fsp 11.82 = lb FBA 15.49
= 40.22 deg
Problem 3-70 The uniform crate of mass M is suspended by using a cord of length l that is attached to the sides of the crate and passes over the small pulley at O. If the cord can be attached at either points A and B, or C and D, determine which attachment produces the least amount of tension in the cord and specify the cord tension in this case.
202
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Engineering Mechanics - Statics
Chapter 3
Given: M = 50 kg g = 9.81 m s a = 0.6 m b = 1.5 m l = 2m c = a 2 b 2
2
d = Solution:
Case 1 Attached at A and B
Guess
T = 1N
Given
l 2 2 -d 2 Mg - 2T = 0 l 2
Guess
T1 = Find ( T)
T1 = 370.78 N
Case 2 Attached at C and D
T = 1N
Given
l 2 2 -c 2 Mg - 2T = 0 l 2
T2 = Find ( T)
T2 = 257.09 N
Choose the arrangement that gives the smallest tension.
T = min ( T1 , T2 )
T = 257.09 N
Problem 3-71 The man attempts to pull the log at C by using the three ropes. Determines the direction in which he should pull on his rope with a force P, so that he exerts a maximum
203
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 3
p p force on the log. What is the force on the log for this case ? Also, determine the direction in which he should pull in order to maximize the force in the rope attached to B.What is this maximum force? Given: P = 80 lb
= 150 deg
Solution:
+ Fx = 0; +
F AB + P cos ( ) - F AC sin ( - 90 deg) = 0 P sin ( ) - F AC cos ( - 90 deg) = 0 sin ( ) = 1.
Fy = 0;
P sin ( )
F AC =
cos ( - 90deg)
In order to maximize FAC we choose P sin ( )
Thus
= 90 deg
F AC =
cos ( - 90deg)
F AC = 160.00 lb
Now let's find the force in the rope AB. F AB = -P cos ( ) + F AC sin ( - 90 deg) F AB = -P cos ( ) + P sin ( ) sin ( - 90 deg) cos ( - 90 deg) cos ( + - 90deg) cos ( - 90 deg)
F AB = P
sin ( ) sin ( - 90 deg) - cos ( ) cos ( - 90 deg) cos ( - 90 deg)
= -P
204
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Engineering Mechanics - Statics
Chapter 3
In order to maximize the force we set
cos ( + - 90 deg) = -1
+ - 90 deg = 180 deg = 270 deg - = 120.00 deg
F AB = -P
cos ( + - 90 deg) cos ( - 90 deg)
F AB = 160.00 lb
Problem 3-72 The "scale" consists of a known weight W which is suspended at A from a cord of total length L. Determine the weight w at B if A is at a distance y for equilibrium. Neglect the sizes and weights of the pulleys.
Solution:
+
Fy = 0;
2W sin ( ) - w = 0
Geometry
h=
L - 2
y
2
-
d = 1 ( L - y) 2 - d2 2 2
2
205
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Engineering Mechanics - Statics
Chapter 3
1 ( L - y) 2 + d2 2 w = 2W L- y 2
w=
2W L-y
( L - y) - d
2
2
Problem 3-73 Determine the maximum weight W that can be supported in the position shown if each cable AC and AB can support a maximum tension of F before it fails. Given:
= 30 deg
F = 600 lb c = 12 d = 5
Solution: Initial Guesses F AB = F F AC = F W = F
Case 1 Assume that cable AC reaches maximum tension F AC sin ( ) - d c +d F AC cos ( ) + c c +d
2 2 2 2
Given
FAB = 0
FAB - W = 0
W1 = Find ( W , FAB) FAB1
W1 1239.62 = lb FAB1 780.00
Case 2 Assume that cable AB reaches maximum tension
206
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Engineering Mechanics - Statics
Chapter 3
Given
F AC sin ( ) -
d c +d
2 2
FAB = 0
F AC cos ( ) +
c c +d
2 2
FAB - W = 0
W2 = Find ( W , FAC) FAC2
W = min ( W1 , W2 )
W2 953.55 = lb FAC2 461.54
W = 953.6 lb
Problem 3-74 If the spring on rope OB has been stretched a distance . and fixed in place as shown, determine the tension developed in each of the other three ropes in order to hold the weight W in equilibrium. Rope OD lies in the x-y plane. Given: a = 2 ft b = 4 ft c = 3 ft d = 4 ft e = 4 ft f = 4 ft xB = -2 ft yB = -3 ft zB = 3 ft
= 30 deg
k = 20 lb in
= 2 in
W = 225 lb
207
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Engineering Mechanics - Statics
Chapter 3
Solution: Initial Guesses Given c a +b +c -b a +b +c a a +b +c
2 2 2 2 2 2 2 2 2
F OA = 10 lb
F OC = 10 lb
F OD = 10 lb
F OA + k
xB xB + yB + zB
2 2 2
+
2
-d d +e + f
2 2
FOC + F OD sin ( ) = 0
F OA + k
yB xB + yB + zB
2 2 2
+
2
f d +e + f
2 2
FOC + F OD cos ( ) = 0
F OA + k
zB xB + yB + zB
2 2 2
+
2
e d +e + f
2 2
FOC - W = 0
FOA FOC = Find ( FOA , FOC , FOD) F OD
FOA 201.6 FOC = 215.7 lb F 58.6 OD
Problem 3-75 The joint of a space frame is subjected to four member forces. Member OA lies in the x - y plane and member OB lies in the y - z plane. Determine the forces acting in each of the members required for equilibrium of the joint. Given: F 4 = 200 lb
= 40 deg = 45 deg
Solution: The initial guesses : F 1 = 200 lb F 2 = 200 lb F 3 = 200 lb
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2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 3
Given F y = 0; F x = 0; F z = 0; F 3 + F 1 cos ( ) - F 2 cos ( ) = 0 -F 1 sin ( ) = 0 F 2 sin ( ) - F4 = 0
F1 F2 = Find ( F1 , F2 , F3) F 3 F1 0 F2 = 311.1 lb F 238.4 3
209
2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.