CHAPTER 4
PROBLEM 4.1
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/s 2 ) = 3.4335 kN
Wt = (1400 kg)(9.81 m/s 2 ) = 13.7340 kN
(a)
Rear wheels:
ΣM B = 0: W (1.7 m + 2.05 m) + W (2.05 m) + Wt (1.2 m) − 2 A(3 m) = 0
(3.4335 kN)(3.75 m) + (3.4335 kN)(2.05 m)
+ (13.7340 kN)(1.2 m) − 2 A(3 m) = 0
A = +6.0659 kN
(b)
Front wheels:
A = 6.07 kN W
ΣFy = 0: − W − W − Wt + 2 A + 2 B = 0
−3.4335 kN − 3.4335 kN − 13.7340 kN + 2(6.0659 kN) + 2B = 0
B = +4.2346 kN
B = 4.23 kN W
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345
PROBLEM 4.2
Solve Problem 4.1, assuming that crate D is removed and that
the position of crate C is unchanged.
PROBLEM 4.1 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/s 2 ) = 3.4335 kN
Wt = (1400 kg)(9.81 m/s 2 ) = 13.7340 kN
(a)
Rear wheels:
ΣM B = 0: W (1.7 m + 2.05 m) + Wt (1.2 m) − 2 A(3 m) = 0
(3.4335 kN)(3.75 m) + (13.7340 kN)(1.2 m) − 2 A(3 m) = 0
A = + 4.8927 kN
(b)
Front wheels:
A = 4.89 kN W
ΣM y = 0: − W − Wt + 2 A + 2 B = 0
−3.4335 kN − 13.7340 kN + 2(4.8927 kN) + 2B = 0
B = +3.6911 kN
B = 3.69 kN W
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346
PROBLEM 4.3
A T-shaped bracket supports the four loads shown. Determine the
reactions at A and B (a) if a = 10 in., (b) if a = 7 in.
SOLUTION
Free-Body Diagram:
ΣFx = 0: Bx = 0
ΣM B = 0: (40 lb)(6 in.) − (30 lb)a − (10 lb)(a + 8 in.) + (12 in.) A = 0
(40a − 160)
12
A=
(1)
ΣM A = 0: − (40 lb)(6 in.) − (50 lb)(12 in.) − (30 lb)(a + 12 in.)
− (10 lb)(a + 20 in.) + (12 in.) B y = 0
By =
Since
(a)
(b)
(1400 + 40a)
12
Bx = 0, B =
(1400 + 40a )
12
(2)
For a = 10 in.,
Eq. (1):
A=
(40 × 10 − 160)
= +20.0 lb
12
Eq. (2):
B=
(1400 + 40 × 10)
= +150.0 lb
12
Eq. (1):
A=
(40 × 7 − 160)
= +10.00 lb
12
Eq. (2):
B=
(1400 + 40 × 7)
= +140.0 lb
12
A = 20.0 lb W
B = 150.0 lb W
For a = 7 in.,
A = 10.00 lb W
B = 140.0 lb W
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347
PROBLEM 4.4
For the bracket and loading of Problem 4.3, determine the smallest
distance a if the bracket is not to move.
PROBLEM 4.3 A T-shaped bracket supports the four loads shown.
Determine the reactions at A and B (a) if a = 10 in., (b) if a = 7 in.
SOLUTION
Free-Body Diagram:
For no motion, reaction at A must be downward or zero; smallest distance a for no motion
corresponds to A = 0.
ΣM B = 0: (40 lb)(6 in.) − (30 lb)a − (10 lb)(a + 8 in.) + (12 in.) A = 0
A=
(40a − 160)
12
A = 0: (40a − 160) = 0
a = 4.00 in. W
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348
PROBLEM 4.5
A hand truck is used to move two kegs, each of mass 40 kg.
Neglecting the mass of the hand truck, determine (a) the vertical
force P that should be applied to the handle to maintain
equilibrium when α = 35°, (b) the corresponding reaction at each
of the two wheels.
SOLUTION
Free-Body Diagram:
W = mg = (40 kg)(9.81 m/s 2 ) = 392.40 N
a1 = (300 mm)sinα − (80 mm)cosα
a2 = (430 mm)cosα − (300 mm)sinα
b = (930 mm)cosα
From free-body diagram of hand truck,
Dimensions in mm
ΣM B = 0: P(b) − W ( a2 ) + W (a1 ) = 0
(1)
ΣFy = 0: P − 2W + 2 B = 0
(2)
α = 35°
For
a1 = 300sin 35° − 80 cos 35° = 106.541 mm
a2 = 430 cos 35° − 300sin 35° = 180.162 mm
b = 930cos 35° = 761.81 mm
(a)
From Equation (1):
P (761.81 mm) − 392.40 N(180.162 mm) + 392.40 N(106.54 mm) = 0
P = 37.921 N
(b)
From Equation (2):
37.921 N − 2(392.40 N) + 2 B = 0
or P = 37.9 N W
or B = 373 N W
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349
PROBLEM 4.6
Solve Problem 4.5 when α = 40°.
PROBLEM 4.5 A hand truck is used to move two kegs, each of
mass 40 kg. Neglecting the mass of the hand truck, determine
(a) the vertical force P that should be applied to the handle to
maintain equilibrium when α = 35°, (b) the corresponding reaction
at each of the two wheels.
SOLUTION
Free-Body Diagram:
W = mg = (40 kg)(9.81 m/s 2 )
W = 392.40 N
a1 = (300 mm)sinα − (80 mm)cosα
a2 = (430 mm)cosα − (300 mm)sinα
b = (930 mm)cosα
From F.B.D.:
ΣM B = 0: P(b) − W ( a2 ) + W (a1 ) = 0
P = W ( a2 − a1 )/b
(1)
ΣFy = 0: − W − W + P + 2 B = 0
B =W −
For
1
P
2
(2)
α = 40°:
a1 = 300sin 40° − 80 cos 40° = 131.553 mm
a2 = 430 cos 40° − 300sin 40° = 136.563 mm
b = 930cos 40° = 712.42 mm
(a)
From Equation (1):
P=
392.40 N (0.136563 m − 0.131553 m)
0.71242 m
P = 2.7595 N
(b)
From Equation (2):
B = 392.40 N −
1
(2.7595 N)
2
P = 2.76 N W
B = 391 N W
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350
PROBLEM 4.7
A 3200-lb forklift truck is used to lift a 1700-lb crate. Determine
the reaction at each of the two (a) front wheels A, (b) rear wheels B.
SOLUTION
Free-Body Diagram:
(a)
Front wheels:
ΣM B = 0: (1700 lb)(52 in.) + (3200 lb)(12 in.) − 2 A(36 in.) = 0
A = +1761.11 lb
(b)
Rear wheels:
ΣFy = 0: − 1700 lb − 3200 lb + 2(1761.11 lb) + 2 B = 0
B = +688.89 lb
A = 1761 lb W
B = 689 lb W
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351
PROBLEM 4.8
For the beam and loading shown, determine (a) the reaction at A,
(b) the tension in cable BC.
SOLUTION
Free-Body Diagram:
(a)
Reaction at A:
ΣFx = 0: Ax = 0
ΣMB = 0: (15 lb)(28 in.) + (20 lb)(22 in.) + (35 lb)(14 in.)
+ (20 lb)(6 in.) − Ay (6 in.) = 0
Ay = +245 lb
(b)
Tension in BC:
ΣM A = 0: (15 lb)(22 in.) + (20 lb)(16 in.) + (35 lb)(8 in.)
− (15 lb)(6 in.) − FBC (6 in.) = 0
FBC = +140.0 lb
Check:
A = 245 lb W
FBC = 140.0 lb W
ΣFy = 0: − 15 lb − 20 lb = 35 lb − 20 lb + A − FBC = 0
−105 lb + 245 lb − 140.0 = 0
0 = 0 (Checks)
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352
PROBLEM 4.9
For the beam and loading shown, determine the range of the
distance a for which the reaction at B does not exceed 100 lb
downward or 200 lb upward.
SOLUTION
Assume B is positive when directed .
Sketch showing distance from D to forces.
ΣM D = 0: (300 lb)(8 in. − a ) − (300 lb)(a − 2 in.) − (50 lb)(4 in.) + 16 B = 0
−600a + 2800 + 16B = 0
a=
(2800 + 16B)
600
(1)
For B = 100 lb = −100 lb, Eq. (1) yields:
a≥
[2800 + 16( −100)] 1200
=
= 2 in.
600
600
a ≥ 2.00 in. Y
For B = 200 = +200 lb, Eq. (1) yields:
a≤
Required range:
[2800 + 16(200)] 6000
=
= 10 in.
600
600
2.00 in. ≤ a ≤ 10.00 in.
a ≤ 10.00 in. Y
W
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353
PROBLEM 4.10
The maximum allowable value of each of the reactions is
180 N. Neglecting the weight of the beam, determine the range
of the distance d for which the beam is safe.
SOLUTION
ΣFx = 0: Bx = 0
B = By
ΣM A = 0: (50 N) d − (100 N)(0.45 m − d ) − (150 N)(0.9 m − d ) + B(0.9 m − d ) = 0
50d − 45 + 100d − 135 + 150d + 0.9 B − Bd = 0
d=
180 N ⋅ m − (0.9 m) B
300 A − B
(1)
ΣM B = 0: (50 N)(0.9 m) − A(0.9 m − d ) + (100 N)(0.45 m) = 0
45 − 0.9 A + Ad + 45 = 0
d=
(0.9 m) A − 90 N ⋅ m
A
(2)
Since B ≤ 180 N, Eq. (1) yields
d≥
180 − (0.9)180 18
=
= 0.15 m
300 − 180
120
d≤
(0.9)180 − 90 72
=
= 0.40 m
180
180
d ≥ 150.0 mm Y
Since A ≤ 180 N, Eq. (2) yields
Range:
150.0 mm ≤ d ≤ 400 mm
d ≤ 400 mm Y
W
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354
PROBLEM 4.11
Three loads are applied as shown to a light beam supported by
cables attached at B and D. Neglecting the weight of the beam,
determine the range of values of Q for which neither cable
becomes slack when P = 0.
SOLUTION
ΣM B = 0: (3.00 kN)(0.500 m) + TD (2.25 m) − Q (3.00 m) = 0
(1)
Q = 0.500 kN + (0.750) TD
ΣM D = 0: (3.00 kN)(2.75 m) − TB (2.25 m) − Q(0.750 m) = 0
Q = 11.00 kN − (3.00) TB
(2)
For cable B not to be slack, TB ≥ 0, and from Eq. (2),
Q ≤ 11.00 kN
For cable D not to be slack, TD ≥ 0, and from Eq. (1),
Q ≥ 0.500 kN
For neither cable to be slack,
0.500 kN ≤ Q ≤ 11.00 kN W
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355
PROBLEM 4.12
Three loads are applied as shown to a light beam supported by
cables attached at B and D. Knowing that the maximum allowable
tension in each cable is 4 kN and neglecting the weight of the beam,
determine the range of values of Q for which the loading is safe
when P = 0.
SOLUTION
ΣM B = 0: (3.00 kN)(0.500 m) + TD (2.25 m) − Q (3.00 m) = 0
(1)
Q = 0.500 kN + (0.750) TD
ΣM D = 0: (3.00 kN)(2.75 m) − TB (2.25 m) − Q(0.750 m) = 0
Q = 11.00 kN − (3.00) TB
(2)
For TB ≤ 4.00 kN, Eq. (2) yields
Q ≥ 11.00 kN − 3.00(4.00 kN)
Q ≥ −1.000 kN
For TD ≤ 4.00 kN, Eq. (1) yields
Q ≤ 0.500 kN + 0.750(4.00 kN)
Q ≤ 3.50 kN
For loading to be safe, cables must also not be slack. Combining with the conditions obtained in Problem 4.11,
0.500 kN ≤ Q ≤ 3.50 kN W
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356
PROBLEM 4.13
For the beam of Problem 4.12, determine the range of values of Q
for which the loading is safe when P = 1 kN.
PROBLEM 4.12 Three loads are applied as shown to a light beam
supported by cables attached at B and D. Knowing that the maximum
allowable tension in each cable is 4 kN and neglecting the weight of
the beam, determine the range of values of Q for which the loading
is safe when P = 0.
SOLUTION
ΣM B = 0: (3.00 kN)(0.500 m) − (1.000 kN)(0.750 m) + TD (2.25 m) − Q(3.00 m) = 0
Q = 0.250 kN + 0.75 TD
(1)
ΣM D = 0: (3.00 kN)(2.75 m) + (1.000 kN)(1.50 m)
− TB (2.25 m) − Q (0.750 m) = 0
Q = 13.00 kN − 3.00 TB
(2)
For the loading to be safe, cables must not be slack and tension must not exceed 4.00 kN.
Making 0 ≤ TB ≤ 4.00 kN in Eq. (2), we have
13.00 kN − 3.00(4.00 kN) ≤ Q ≤ 13.00 kN − 3.00(0)
1.000 kN ≤ Q ≤ 13.00 kN
(3)
Making 0 ≤ TD ≤ 4.00 kN in Eq. (1), we have
0.250 kN + 0.750(0) ≤ Q ≤ 0.250 kN + 0.750(4.00 kN)
0.250 kN ≤ Q ≤ 3.25 kN
Combining Eqs. (3) and (4),
(4)
1.000 kN ≤ Q ≤ 3.25 kN W
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357
PROBLEM 4.14
For the beam of Sample Problem 4.2, determine the range of values
of P for which the beam will be safe, knowing that the maximum
allowable value of each of the reactions is 30 kips and that the
reaction at A must be directed upward.
SOLUTION
ΣFx = 0: Bx = 0
B = By
ΣM A = 0: − P(3 ft) + B(9 ft) − (6 kips)(11 ft) − (6 kips)(13 ft) = 0
(1)
P = 3B − 48 kips
ΣM B = 0: − A(9 ft) + P (6 ft) − (6 kips)(2 ft) − (6 kips)(4 ft) = 0
P = 1.5 A + 6 kips
Since B ≤ 30 kips, Eq. (1) yields
P ≤ (3)(30 kips) − 48 kips
(2)
P ≤ 42.0 kips Y
Since 0 ≤ A ≤ 30 kips, Eq. (2) yields
0 + 6 kips ≤ P ≤ (1.5)(30 kips)1.6 kips
6.00 kips ≤ P ≤ 51.0 kips
Range of values of P for which beam will be safe:
6.00 kips ≤ P ≤ 42.0 kips
Y
W
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358
PROBLEM 4.15
The bracket BCD is hinged at C and attached to a control
cable at B. For the loading shown, determine (a) the tension
in the cable, (b) the reaction at C.
SOLUTION
At B:
Ty
Tx
=
0.18 m
0.24 m
3
Ty = Tx
4
(a)
(1)
ΣM C = 0: Tx (0.18 m) − (240 N)(0.4 m) − (240 N)(0.8 m) = 0
Tx = +1600 N
From Eq. (1):
Ty =
3
(1600 N) = 1200 N
4
T = Tx2 + Ty2 = 16002 + 12002 = 2000 N
(b)
T = 2.00 kN W
ΣFx = 0: Cx − Tx = 0
Cx − 1600 N = 0 C x = +1600 N
C x = 1600 N
ΣFy = 0: C y − Ty − 240 N − 240 N = 0
C y − 1200 N − 480 N = 0
C y = +1680 N
C y = 1680 N
α = 46.4°
C = 2320 N
C = 2.32 kN
46.4° W
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359
PROBLEM 4.16
Solve Problem 4.15, assuming that a = 0.32 m.
PROBLEM 4.15 The bracket BCD is hinged at C and
attached to a control cable at B. For the loading shown,
determine (a) the tension in the cable, (b) the reaction at C.
SOLUTION
At B:
Ty
Tx
=
0.32 m
0.24 m
4
Ty = Tx
3
ΣM C = 0: Tx (0.32 m) − (240 N)(0.4 m) − (240 N)(0.8 m) = 0
Tx = 900 N
From Eq. (1):
Ty =
4
(900 N) = 1200 N
3
T = Tx2 + Ty2 = 9002 + 12002 = 1500 N
T = 1.500 kN W
ΣFx = 0: C x − Tx = 0
C x − 900 N = 0 C x = +900 N
C x = 900 N
ΣFy = 0: C y − Ty − 240 N − 240 N = 0
C y − 1200 N − 480 N = 0
C y = +1680 N
C y = 1680 N
α = 61.8°
C = 1906 N
C = 1.906 kN
61.8° W
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360
PROBLEM 4.17
The lever BCD is hinged at C and attached to a control rod at B. If
P = 100 lb, determine (a) the tension in rod AB, (b) the reaction at C.
SOLUTION
Free-Body Diagram:
(a)
ΣM C = 0: T (5 in.) − (100 lb)(7.5 in.) = 0
(b)
3
ΣFx = 0: C x + 100 lb + (150.0 lb) = 0
5
T = 150.0 lb W
C x = 190 lb
C x = −190 lb
4
ΣFy = 0: C y + (150.0 lb) = 0
5
C y = 120 lb
C y = −120 lb
α = 32.3°
C = 225 lb
C = 225 lb
32.3° W
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361
PROBLEM 4.18
The lever BCD is hinged at C and attached to a control rod at B.
Determine the maximum force P that can be safely applied at D if the
maximum allowable value of the reaction at C is 250 lb.
SOLUTION
Free-Body Diagram:
ΣM C = 0: T (5 in.) − P (7.5 in.) = 0
T = 1.5P
3
ΣFx = 0: P + C x + (1.5P) = 0
5
C x = −1.9 P
ΣFy = 0: C y +
C x = 1.9 P
4
(1.5P) = 0
5
C y = −1.2 P
C y = 1.2 P
C = C x2 + C y2
= (1.9 P) 2 + (1.2 P) 2
C = 2.2472 P
For C = 250 lb,
250 lb = 2.2472P
P = 111.2 lb
P = 111.2 lb
W
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362
PROBLEM 4.19
Two links AB and DE are connected by a bell crank as
shown. Knowing that the tension in link AB is 720 N,
determine (a) the tension in link DE, (b) the reaction at C.
SOLUTION
Free-Body Diagram:
ΣM C = 0: FAB (100 mm) − FDE (120 mm) = 0
FDE =
(a)
For
(1)
FAB = 720 N
FDE =
(b)
5
FAB
6
FDE = 600 N W
5
(720 N)
6
3
ΣFx = 0: − (720 N) + C x = 0
5
C x = +432 N
4
ΣFy = 0: − (720 N) + C y − 600 N = 0
5
C y = +1176 N
C = 1252.84 N
α = 69.829°
C = 1253 N
69.8° W
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363
PROBLEM 4.20
Two links AB and DE are connected by a bell crank as
shown. Determine the maximum force that may be safely
exerted by link AB on the bell crank if the maximum
allowable value for the reaction at C is 1600 N.
SOLUTION
See solution to Problem 4.15 for F.B.D. and derivation of Eq. (1).
FDE =
5
FAB
6
(1)
3
ΣFx = 0: − FAB + C x = 0
5
ΣFy = 0: −
Cx =
3
FAB
5
4
FAB + C y − FDE = 0
5
4
5
− FAB + C y − FAB = 0
5
6
49
Cy =
FAB
30
C = C x2 + C y2
1
(49) 2 + (18) 2 FAB
30
C = 1.74005FAB
=
For C = 1600 N, 1600 N = 1.74005FAB
FAB = 920 N W
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364
PROBLEM 4.21
Determine the reactions at A and C when (a) α = 0, (b) α = 30°.
SOLUTION
(a)
α =0
From F.B.D. of member ABC:
ΣM C = 0: (300 N)(0.2 m) + (300 N)(0.4 m) − A(0.8 m) = 0
or
A = 225 N
A = 225 N W
ΣFy = 0: C y + 225 N = 0
C y = −225 N or C y = 225 N
ΣFx = 0: 300 N + 300 N + C x = 0
C x = −600 N or C x = 600 N
Then
C = C x2 + C y2 = (600) 2 + (225) 2 = 640.80 N
and
θ = tan −1 ¨
§ Cy ·
−1 § −225 ·
¸ = tan ¨
¸ = 20.556°
© −600 ¹
© Cx ¹
or
(b)
C = 641 N
20.6° W
α = 30°
From F.B.D. of member ABC:
ΣM C = 0: (300 N)(0.2 m) + (300 N)(0.4 m) − ( A cos 30°)(0.8 m)
+ ( A sin 30°)(20 in.) = 0
A = 365.24 N
or
A = 365 N
60.0° W
ΣFx = 0: 300 N + 300 N + (365.24 N) sin 30° + C x = 0
C x = −782.62
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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365
PROBLEM 4.21 (Continued)
ΣFy = 0: C y + (365.24 N) cos 30° = 0
C y = −316.31 N or C y = 316 N
Then
C = C x2 + C y2 = (782.62) 2 + (316.31) 2 = 884.12 N
and
θ = tan −1 ¨
§ Cy ·
−1 § −316.31 ·
¸ = tan ¨
¸ = 22.007°
© −782.62 ¹
© Cx ¹
or
C = 884 N
22.0° W
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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366
PROBLEM 4.22
Determine the reactions at A and B when (a) α = 0, (b) α = 90°,
(c) α = 30°.
SOLUTION
(a)
α =0
ΣM A = 0: B(20 in.) − 75 lb(10 in.) = 0
B = 37.5 lb
ΣFx = 0: Ax = 0
+ ΣFy = 0: Ay − 75 lb + 37.5 lb = 0
Ay = 37.5 lb
(b)
A = B = 37.5 lb W
α = 90°
ΣM A = 0: B(12 in.) − 75 lb(10 in.) = 0
B = 62.5 lb
ΣFx = 0: Ax − B = 0
Ax = 62.5 lb
ΣFy = 0: Ay − 75 lb = 0
Ay = 75 lb
A = Ax2 + Ay2
= (62.5 lb) 2 + (75 lb) 2
= 97.6 lb
75
62.5
θ = 50.2°
tan θ =
A = 97.6 lb
50.2°; B = 62.51 lb
W
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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367
PROBLEM 4.22 (Continued)
(c)
α = 30°
ΣM A = 0: ( B cos 30°)(20 in.) + ( B sin 30°)(12 in.)
− (75 lb)(10 in.) = 0
B = 32.161 lb
ΣFx = 0: Ax − (32.161) sin 30° = 0
Ax = 16.0805 lb
ΣFy = 0: Ay + (32.161) cos 30° − 75 = 0
Ay = 47.148 lb
A = Ax2 + Ay2
= (16.0805) 2 + (47.148) 2
= 49.8 lb
47.148
16.0805
θ = 71.2°
tan θ =
A = 49.8 lb
71.2°; B = 32.2 lb
60.0° W
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368
PROBLEM 4.23
Determine the reactions at A and B when (a) h = 0,
(b) h = 200 mm.
SOLUTION
Free-Body Diagram:
ΣM A = 0: ( B cos 60°)(0.5 m) − ( B sin 60°)h − (150 N)(0.25 m) = 0
37.5
B=
0.25 − 0.866h
(a)
When h = 0,
From Eq. (1):
B=
37.5
= 150 N
0.25
B = 150.0 N
(1)
30.0° W
ΣFy = 0: Ax − B sin 60° = 0
Ax = (150)sin 60° = 129.9 N
A x = 129.9 N
ΣFy = 0: Ay − 150 + B cos 60° = 0
Ay = 150 − (150) cos 60° = 75 N
A y = 75 N
α = 30°
A = 150.0 N
(b)
A = 150.0 N
30.0° W
When h = 200 mm = 0.2 m,
From Eq. (1):
B=
37.5
= 488.3 N
0.25 − 0.866(0.2)
B = 488 N
30.0° W
ΣFx = 0: Ax − B sin 60° = 0
Ax = (488.3) sin 60° = 422.88 N
A x = 422.88 N
ΣFy = 0: Ay − 150 + B cos 60° = 0
Ay = 150 − (488.3) cos 60° = −94.15 N
A y = 94.15 N
α = 12.55°
A = 433.2 N
A = 433 N
12.55° W
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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369
PROBLEM 4.24
A lever AB is hinged at C and attached to a control cable at A. If the
lever is subjected to a 75-lb vertical force at B, determine (a) the
tension in the cable, (b) the reaction at C.
SOLUTION
Free-Body Diagram:
Geometry:
x AC = (10 in.) cos 20° = 9.3969 in.
y AC = (10 in.)sin 20° = 3.4202 in.
Ÿ yDA = 12 in. − 3.4202 in. = 8.5798 in.
§ yDA ·
−1 § 8.5798 ·
¸ = tan ¨
¸ = 42.397°
© 9.3969 ¹
© x AC ¹
α = tan −1 ¨
β = 90° − 20° − 42.397° = 27.603°
Equilibrium for lever:
ΣM C = 0: TAD cos 27.603°(10 in.) − (75 lb)[(15 in.)cos 20°] = 0
(a)
TAD = 119.293 lb
TAD = 119.3 lb W
ΣFx = 0: C x + (119.293 lb) cos 42.397° = 0
(b)
C x = −88.097 lb
ΣFy = 0: C y − 75 lb − (119.293 lb) sin 42.397° = 0
C y = 155.435
Thus,
C = C x2 + C y2 = (−88.097) 2 + (155.435) 2 = 178.665 lb
and
θ = tan −1
Cy
Cx
= tan −1
155.435
= 60.456°
88.097
C = 178.7 lb
60.5° W
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370
PROBLEM 4.25
For each of the plates and loadings shown, determine the reactions at A and B.
SOLUTION
(a)
Free-Body Diagram:
ΣM A = 0: B(20 in.) − (50 lb)(4 in.) − (40 lb)(10 in.) = 0
B = +30 lb
B = 30.0 lb W
ΣFx = 0: Ax + 40 lb = 0
Ax = −40 lb
A x = 40.0 lb
ΣFy = 0: Ay + B − 50 lb = 0
Ay + 30 lb − 50 lb = 0
Ay = +20 lb
A y = 20.0 lb
α = 26.56°
A = 44.72 lb
A = 44.7 lb
26.6° W
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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371
PROBLEM 4.25 (Continued)
(b)
Free-Body Diagram:
ΣM A = 0: ( B cos 30°)(20 in.) − (40 lb)(10 in.) − (50 lb)(4 in.) = 0
B = 34.64 lb
B = 34.6 lb
60.0° W
ΣFx = 0: Ax − B sin 30° + 40 lb
Ax − (34.64 lb) sin 30° + 40 lb = 0
Ax = −22.68 lb
A x = 22.68 lb
ΣFy = 0: Ay + B cos 30° − 50 lb = 0
Ay + (34.64 lb) cos 30° − 50 lb = 0
Ay = +20 lb
A y = 20.0 lb
α = 41.4°
A = 30.2 lb
A = 30.24 lb
41.4° W
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372
PROBLEM 4.26
For each of the plates and loadings shown, determine the reactions at A and B.
SOLUTION
(a)
Free-Body Diagram:
ΣM B = 0: A(20 in.) + (50 lb)(16 in.) − (40 lb)(10 in.) = 0
A = +20 lb
A = 20.0 lb W
ΣFx = 0: 40 lb + Bx = 0
Bx = −40 lb
B x = 40 lb
ΣFy = 0: A + By − 50 lb = 0
20 lb + By − 50 lb = 0
By = +30 lb
α = 36.87°
B = 50 lb
B y = 30 lb
B = 50.0 lb
36.9° W
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373
PROBLEM 4.26 (Continued)
(b)
ΣM A = 0: − ( A cos 30°)(20 in.) − (40 lb)(10 in.) + (50 lb)(16 in.) = 0
A = 23.09 lb
A = 23.1 lb
60.0° W
ΣFx = 0: A sin 30° + 40 lb + Bx = 0
(23.09 lb) sin 30° + 40 lb + 8 x = 0
Bx = −51.55 lb
B x = 51.55 lb
ΣFy = 0: A cos 30° + By − 50 lb = 0
(23.09 lb) cos 30° + By − 50 lb = 0
By = +30 lb
B y = 30 lb
α = 30.2°
B = 59.64 lb
B = 59.6 lb
30.2° W
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374
PROBLEM 4.27
A rod AB hinged at A and attached at B to cable BD supports
the loads shown. Knowing that d = 200 mm, determine (a) the
tension in cable BD, (b) the reaction at A.
SOLUTION
Free-Body Diagram:
(a)
Move T along BD until it acts at Point D.
ΣM A = 0: (T sin 45°)(0.2 m) + (90 N)(0.1 m) + (90 N)(0.2 m) = 0
T = 190.919 N
(b)
T = 190.9 N W
ΣFx = 0: Ax − (190.919 N) cos 45° = 0
Ax = +135.0 N
A x = 135.0 N
ΣFy = 0: Ay − 90 N − 90 N + (190.919 N) sin 45° = 0
Ay = +45.0 N
A y = 45.0 N
A = 142.3 N
18.43°W
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375
PROBLEM 4.28
A rod AB, hinged at A and attached at B to cable BD, supports
the loads shown. Knowing that d = 150 mm, determine (a) the
tension in cable BD, (b) the reaction at A.
SOLUTION
Free-Body Diagram:
tan α =
(a)
10
; α = 33.690°
15
Move T along BD until it acts at Point D.
ΣM A = 0: (T sin 33.690°)(0.15 m) − (90 N)(0.1 m) − (90 N)(0.2 m) = 0
T = 324.50 N
(b)
T = 324 N W
ΣFx = 0: Ax − (324.50 N) cos 33.690° = 0
A x = 270 N
Ax = +270 N
ΣFy = 0: Ay − 90 N − 90 N + (324.50 N) sin 33.690° = 0
A = 270 N
Ay = 0
W
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376
PROBLEM 4.29
A force P of magnitude 90 lb is applied to member ACDE, which is
supported by a frictionless pin at D and by the cable ABE. Since the
cable passes over a small pulley at B, the tension may be assumed to be
the same in portions AB and BE of the cable. For the case when a = 3 in.,
determine (a) the tension in the cable, (b) the reaction at D.
SOLUTION
Free-Body Diagram:
(a)
ΣM D = 0: (90 lb)(9 in.) −
5
12
T (9 in.) − T (7 in.) + T (3 in.) = 0
13
13
T = 117 lb
(b)
ΣFx = 0: Dx − 117 lb −
T = 117.0 lb W
5
(117 lb) + 90 = 0
13
Dx = +72 lb
ΣFy = 0: D y +
12
(117 lb) = 0
13
Dy = −108 lb
D = 129.8 lb
56.3° W
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377
PROBLEM 4.30
Solve Problem 4.29 for a = 6 in.
PROBLEM 4.29 A force P of magnitude 90 lb is applied to
member ACDE, which is supported by a frictionless pin at D and
by the cable ABE. Since the cable passes over a small pulley at B,
the tension may be assumed to be the same in portions AB and BE
of the cable. For the case when a = 3 in., determine (a) the tension
in the cable, (b) the reaction at D.
SOLUTION
Free-Body Diagram:
(a)
ΣM D = 0: (90 lb)(6 in.) −
5
12
T (6 in.) − T (7 in.) + T (6 in.) = 0
13
13
T = 195 lb
(b)
ΣFx = 0: Dx − 195 lb −
T = 195.0 lb W
5
(195 lb) + 90 = 0
13
Dx = +180 lb
ΣFy = 0: D y +
12
(195 lb) = 0
13
Dy = −180 lb
D = 255 lb
45.0° W
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378
PROBLEM 4.31
Neglecting friction, determine the tension in cable ABD and the
reaction at support C.
SOLUTION
Free-Body Diagram:
ΣM C = 0: T (0.25 m) − T (0.1 m) − (120 N)(0.1 m) = 0
ΣFx = 0: C x − 80 N = 0
C x = +80 N
ΣFy = 0: C y − 120 N + 80 N = 0
C y = +40 N
T = 80.0 N W
C x = 80.0 N
C y = 40.0 N
C = 89.4 N
26.6° W
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379
PROBLEM 4.32
Neglecting friction and the radius of the pulley, determine
(a) the tension in cable ADB, (b) the reaction at C.
SOLUTION
Free-Body Diagram:
Dimensions in mm
Geometry:
Distance:
AD = (0.36) 2 + (0.150) 2 = 0.39 m
Distance:
BD = (0.2)2 + (0.15)2 = 0.25 m
Equilibrium for beam:
ΣM C = 0:
(a)
§ 0.15 ·
§ 0.15 ·
(120 N)(0.28 m) − ¨
T ¸ (0.36 m) − ¨
T ¸ (0.2 m) = 0
© 0.39 ¹
© 0.25 ¹
T = 130.000 N
§ 0.36 ·
§ 0.2 ·
ΣFx = 0: C x + ¨
¸ (130.000 N) + ¨ 0.25 ¸ (130.000 N) = 0
0.39
©
¹
©
¹
(b)
or
T = 130.0 N W
C x = − 224.00 N
§ 0.15 ·
§ 0.15 ·
(130.00 N) + ¨
ΣFy = 0: C y + ¨
¸
¸ (130.00 N) − 120 N = 0
© 0.39 ¹
© 0.25 ¹
C y = − 8.0000 N
Thus,
C = C x2 + C y2 = (−224) 2 + (− 8) 2 = 224.14 N
and
θ = tan −1
Cy
Cx
= tan −1
8
= 2.0454°
224
C = 224 N
2.05° W
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380
PROBLEM 4.33
Rod ABC is bent in the shape of an arc of circle of radius R. Knowing the
θ = 30°, determine the reaction (a) at B, (b) at C.
SOLUTION
Free-Body Diagram:
ΣM D = 0: C x ( R) − P( R) = 0
Cx = + P
ΣFx = 0: C x − B sin θ = 0
P − B sin θ = 0
B = P/sin θ
B=
P
sin θ
θ
ΣFy = 0: C y + B cos θ − P = 0
C y + ( P/sin θ ) cos θ − P = 0
1 ·
§
C y = P ¨1 −
¸
tan θ ¹
©
For θ = 30°,
(a)
(b)
B = P/sin 30° = 2 P
Cx = + P
B = 2P
60.0° W
Cx = P
C y = P(1 − 1/tan 30°) = − 0.732/P
C y = 0.7321P
C = 1.239P
36.2° W
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381
PROBLEM 4.34
Rod ABC is bent in the shape of an arc of circle of radius R. Knowing that θ = 60°,
determine the reaction (a) at B, (b) at C.
SOLUTION
See the solution to Problem 4.33 for the free-body diagram and analysis leading to the following expressions:
Cx = + P
1 ·
§
C y = P ¨1 −
tan θ ¸¹
©
P
B=
sin θ
For θ = 60°,
(a)
(b)
B = P/sin 60° = 1.1547 P
Cx = + P
B = 1.155P
30.0° W
Cx = P
C y = P(1 − 1/tan 60°) = + 0.4226 P
C y = 0.4226 P
C = 1.086P
22.9° W
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382
PROBLEM 4.35
A movable bracket is held at rest by a cable attached at C and by
frictionless rollers at A and B. For the loading shown, determine (a) the
tension in the cable, (b) the reactions at A and B.
SOLUTION
Free-Body Diagram:
(a)
ΣFy = 0: T − 600 N = 0
(b)
ΣFx = 0: B − A = 0
T = 600 N W
∴ B=A
Note that the forces shown form two couples.
ΣM = 0: (600 N)(600 mm) − A(90 mm) = 0
A = 4000 N
∴ B = 4000 N
A = 4.00 kN
; B = 4.00 kN
W
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383
PROBLEM 4.36
A light bar AB supports a 15-kg block at its midpoint C. Rollers at A
and B rest against frictionless surfaces, and a horizontal cable AD is
attached at A. Determine (a) the tension in cable AD, (b) the reactions
at A and B.
SOLUTION
Free-Body Diagram:
W = (15 kg)(9.81 m/s 2 )
= 147.150 N
(a)
ΣFx = 0: TAD − 105.107 N = 0
(b)
ΣFy = 0: A − W = 0
TAD = 105.1 N W
A − 147.150 N = 0
A = 147.2 N W
ΣM A = 0: B(350 mm) − (147.150 N) (250 mm) = 0
B = 105.107 N
B = 105.1 N
W
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384
PROBLEM 4.37
A light bar AD is suspended from a cable BE and supports a
50-lb block at C. The ends A and D of the bar are in contact
with frictionless vertical walls. Determine the tension in cable
BE and the reactions at A and D.
SOLUTION
Free-Body Diagram:
ΣFx = 0:
A= D
TBE = 50.0 lb W
ΣFy = 0:
We note that the forces shown form two couples.
ΣM = 0: A(8 in.) − (50 lb)(3 in.) = 0
A = 18.75 lb
A = 18.75 lb
D = 18.75 lb
W
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385
PROBLEM 4.38
A light rod AD is supported by frictionless pegs at B and C and
rests against a frictionless wall at A. A vertical 120-lb force is
applied at D. Determine the reactions at A, B, and C.
SOLUTION
Free-Body Diagram:
ΣFx = 0: A cos 30° − (120 lb) cos 60° = 0
A = 69.3 lb
A = 69.28 lb
W
ΣM B = 0: C (8 in.) − (120 lb)(16 in.) cos 30°
+ (69.28 lb)(8 in.)sin 30° = 0
C = 173.2 lb
C = 173.2 lb
60.0° W
B = 34.6 lb
60.0° W
ΣM C = 0: B(8 in.) − (120 lb)(8 in.) cos 30°
+ (69.28 lb)(16 in.) sin 30° = 0
B = 34.6 lb
Check:
ΣFy = 0: 173.2 − 34.6 − (69.28)sin 30° − (120)sin 60° = 0
0 = 0 (check) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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386
PROBLEM 4.39
Bar AD is attached at A and C to collars that can move freely
on the rods shown. If the cord BE is vertical (α = 0), determine
the tension in the cord and the reactions at A and C.
SOLUTION
Free-Body Diagram:
ΣFy = 0: − T cos 30° + (80 N) cos 30° = 0
T = 80.0 N W
T = 80 N
ΣM C = 0: ( A sin 30°)(0.4 m) − (80 N)(0.2 m) − (80 N)(0.2 m) = 0
A = + 160 N
A = 160.0 N
30.0° W
C = 160.0 N
30.0° W
ΣM A = 0: (80 N)(0.2 m) − (80 N)(0.6 m) + (C sin 30°)(0.4 m) = 0
C = + 160 N
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387
PROBLEM 4.40
Solve Problem 4.39 if the cord BE is parallel to the rods (α = 30°).
PROBLEM 4.39 Bar AD is attached at A and C to collars that
can move freely on the rods shown. If the cord BE is vertical
(α = 0), determine the tension in the cord and the reactions at A
and C.
SOLUTION
Free-Body Diagram:
ΣFy = 0: − T + (80 N) cos 30° = 0
T = 69.3 N W
T = 69.282 N
ΣM C = 0: − (69.282 N) cos 30°(0.2 m)
− (80 N)(0.2 m) + ( A sin 30°)(0.4 m) = 0
A = + 140.000 N
A = 140.0 N
30.0° W
C = 180.0 N
30.0° W
ΣM A = 0: + (69.282 N) cos 30°(0.2 m)
− (80 N)(0.6 m) + (C sin 30°)(0.4 m) = 0
C = + 180.000 N
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388
PROBLEM 4.41
The T-shaped bracket shown is supported by a small wheel at E and pegs at C
and D. Neglecting the effect of friction, determine the reactions at C, D, and E
when θ = 30°.
SOLUTION
Free-Body Diagram:
ΣFy = 0: E cos 30° − 20 − 40 = 0
E=
60 lb
= 69.282 lb
cos 30°
E = 69.3 lb
60.0° W
ΣM D = 0: (20 lb)(4 in.) − (40 lb)(4 in.)
− C (3 in.) + E sin 30°(3 in.) = 0
−80 − 3C + 69.282(0.5)(3) = 0
C = 7.9743 lb
C = 7.97 lb
W
D = 42.6 lb
W
ΣFx = 0: E sin 30° + C − D = 0
(69.282 lb)(0.5) + 7.9743 lb − D = 0
D = 42.615 lb
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389
PROBLEM 4.42
The T-shaped bracket shown is supported by a small wheel at E and pegs at C
and D. Neglecting the effect of friction, determine (a) the smallest value of θ for
which the equilibrium of the bracket is maintained, (b) the corresponding reactions
at C, D, and E.
SOLUTION
Free-Body Diagram:
ΣFy = 0: E cos θ − 20 − 40 = 0
E=
60
cos θ
(1)
ΣM D = 0: (20 lb)(4 in.) − (40 lb)(4 in.) − C (3 in.)
§ 60
·
+¨
sin θ ¸ 3 in. = 0
© cos θ
¹
1
C = (180 tan θ − 80)
3
(a)
For C = 0,
180 tan θ = 80
tan θ =
From Eq. (1):
E=
θ = 24.0° W
4
θ = 23.962°
9
60
= 65.659
cos 23.962°
ΣFx = 0: −D + C + E sin θ = 0
D = (65.659) sin 23.962 = 26.666 lb
(b)
C = 0 D = 26.7 lb
E = 65.7 lb
66.0° W
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390
PROBLEM 4.43
Beam AD carries the two 40-lb loads shown. The beam is held by a
fixed support at D and by the cable BE that is attached to the
counterweight W. Determine the reaction at D when (a) W = 100 lb,
(b) W = 90 lb.
SOLUTION
(a)
W = 100 lb
From F.B.D. of beam AD:
ΣFx = 0: Dx = 0
ΣFy = 0: D y − 40 lb − 40 lb + 100 lb = 0
Dy = −20.0 lb
or
D = 20.0 lb W
ΣM D = 0: M D − (100 lb)(5 ft) + (40 lb)(8 ft)
+ (40 lb)(4 ft) = 0
M D = 20.0 lb ⋅ ft
(b)
or M D = 20.0 lb ⋅ ft
W
W = 90 lb
From F.B.D. of beam AD:
ΣFx = 0: Dx = 0
ΣFy = 0: D y + 90 lb − 40 lb − 40 lb = 0
Dy = −10.00 lb
or
D = 10.00 lb W
ΣM D = 0: M D − (90 lb)(5 ft) + (40 lb)(8 ft)
+ (40 lb)(4 ft) = 0
M D = −30.0 lb ⋅ ft
or M D = 30.0 lb ⋅ ft
W
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391
PROBLEM 4.44
For the beam and loading shown, determine the range of values of W for
which the magnitude of the couple at D does not exceed 40 lb ⋅ ft.
SOLUTION
For Wmin ,
From F.B.D. of beam AD:
M D = − 40 lb ⋅ ft
ΣM D = 0: (40 lb)(8 ft) − Wmin (5 ft)
+ (40 lb)(4 ft) − 40 lb ⋅ ft = 0
Wmin = 88.0 lb
For Wmax ,
From F.B.D. of beam AD:
M D = 40 lb ⋅ ft
ΣM D = 0: (40 lb)(8 ft) − Wmax (5 ft)
+ (40 lb)(4 ft) + 40 lb ⋅ ft = 0
Wmax = 104.0 lb
or 88.0 lb ≤ W ≤ 104.0 lb W
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392
PROBLEM 4.45
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 ) = 78.480 N
(a)
ΣFy = 0: Ay − W = 0
A y = 78.480 N
ΣM A = 0: M A − W (1.6 m) = 0
M A = 125.568 N ⋅ m
M A = + (78.480 N)(1.6 m)
M A = 125.6 N ⋅ m
A = 78.5 N
ΣFx = 0: Ax = 0
(b)
ΣFx = 0: Ax − W = 0
A x = 78.480
ΣFy = 0: Ay − W = 0
A y = 78.480
45°
A = (78.480 N) 2 = 110.987 N
ΣM A = 0: M A − W (1.6 m) = 0
M A = 125.568 N ⋅ m
M A = + (78.480 N)(1.6 m)
A = 111.0 N
W
(c)
M A = 125.6 N ⋅ m
45°
W
ΣFx = 0: Ax = 0
ΣFy = 0: Ay − 2W = 0
Ay = 2W = 2(78.480 N) = 156.960 N
ΣM A = 0: M A − 2W (1.6 m) = 0
M A = + 2(78.480 N)(1.6 m)
A = 157.0 N
M A = 251.14 N ⋅ m
M A = 251 N ⋅ m
W
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393
PROBLEM 4.46
A tension of 20 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
Free-Body Diagram:
ΣFx = 0: C x + (20 N) = 0
C x = −20 N
ΣFy = 0: C y − (20 N) = 0
C y = +20 N
C = 28.3 N
45.0° W
ΣM C = 0: M C + (20 N)(0.160 m) + (20 N) (0.055 m) = 0
M C = −4.30 N ⋅ m
M C = 4.30 N ⋅ m
W
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394
PROBLEM 4.47
Solve Problem 4.46, assuming that 15-mm-radius pulleys are used.
PROBLEM 4.46 A tension of 20 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
Free-Body Diagram:
ΣFx = 0: C x + (20 N) = 0
C x = −20 N
ΣFy = 0: C y − (20 N) = 0
C y = +20 N
C = 28.3 N
45.0° W
ΣM C = 0: M C + (20 N) (0.165 m) + (20 N) (0.060 m) = 0
M C = −4.50 N ⋅ m
M C = 4.50 N ⋅ m
W
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395
PROBLEM 4.48
The rig shown consists of a 1200-lb horizontal member ABC and
a vertical member DBE welded together at B. The rig is being used
to raise a 3600-lb crate at a distance x = 12 ft from the vertical
member DBE. If the tension in the cable is 4 kips, determine the
reaction at E, assuming that the cable is (a) anchored at F as
shown in the figure, (b) attached to the vertical member at a point
located 1 ft above E.
SOLUTION
Free-Body Diagram:
M E = 0: M E + (3600 lb) x + (1200 lb) (6.5 ft) − T (3.75 ft) = 0
M E = 3.75T − 3600 x − 7800
(a)
(1)
For x = 12 ft and T = 4000 lbs,
M E = 3.75(4000) − 3600(12) − 7800
= 36, 000 lb ⋅ ft
ΣFx = 0 ∴ Ex = 0
ΣFy = 0:
E y − 3600 lb − 1200 lb − 4000 = 0
E y = 8800 lb
E = 8.80 kips ; M E = 36.0 kip ⋅ ft
W
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396
PROBLEM 4.48 (Continued)
(b)
ΣM E = 0: M E + (3600 lb)(12 ft) + (1200 lb)(6.5 ft) = 0
M E = −51, 000 lb ⋅ ft
ΣFx = 0 ∴ Ex = 0
ΣFy = 0:
E y − 3600 lb − 1200 lb = 0
E y = 4800 lb
E = 4.80 kips ; M E = 51.0 kip ⋅ ft
W
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397
PROBLEM 4.49
For the rig and crate of Prob. 4.48, and assuming that cable is
anchored at F as shown, determine (a) the required tension in cable
ADCF if the maximum value of the couple at E as x varies from 1.5
to 17.5 ft is to be as small as possible, (b) the corresponding
maximum value of the couple.
SOLUTION
Free-Body Diagram:
M E = 0: M E + (3600 lb) x + (1200 lb)(6.5 ft) − T (3.75 ft) = 0
M E = 3.75T − 3600 x − 7800
(1)
For x = 1.5 ft, Eq. (1) becomes
( M E )1 = 3.75T − 3600(1.5) − 7800
(2)
For x = 17.5 ft, Eq. (1) becomes
( M E ) 2 = 3.75T − 3600(17.5) − 7800
(a)
For smallest max value of |M E |, we set
( M E )1 = − ( M E )2
T = 11.20 kips W
3.75T − 13, 200 = −3.75T + 70,800
(b)
From Equation (2), then
M E = 3.75(11.20) − 13.20
|M E | = 28.8 kip ⋅ ft W
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398
PROBLEM 4.50
A 6-m telephone pole weighing 1600 N is used to support the ends of two
wires. The wires form the angles shown with the horizontal axis and
the tensions in the wires are, respectively, T1 = 600 N and T2 = 375 N.
Determine the reaction at the fixed end A.
SOLUTION
Free-Body Diagram:
ΣFx = 0: Ax + (375 N) cos 20° − (600 N) cos10° = 0
Ax = +238.50 N
ΣFy = 0: Ay − 1600 N − (600 N)sin10° − (375 N) sin 20° = 0
Ay = +1832.45 N
A = 238.502 + 1832.452
1832.45
θ = tan −1
238.50
A = 1848 N
82.6° W
ΣM A = 0: M A + (600 N) cos10°(6 m) − (375 N) cos 20°(6 m) = 0
M A = −1431.00 N ⋅ m
M A = 1431 N ⋅ m
W
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399
PROBLEM 4.51
A vertical load P is applied at end B of rod BC. (a) Neglecting
the weight of the rod, express the angle θ corresponding to the
equilibrium position in terms of P, l, and the counterweight W.
(b) Determine the value of θ corresponding to equilibrium if
P = 2W.
SOLUTION
Free-Body Diagram:
(a)
Triangle ABC is isosceles. We have
§θ ·
§θ ·
CD = ( BC ) cos ¨ ¸ = l cos ¨ ¸
©2¹
©2¹
θ·
§
ΣM C = 0: P(l cos θ ) − W ¨ l cos ¸ = 0
2¹
©
Setting cos θ = 2 cos 2
θ
2
− 1:
θ
θ
§
·
Pl ¨ 2 cos 2 − 1¸ − Wl cos = 0
2 ¹
2
©
cos 2
θ 1
§W ·
−¨
¸ cos − = 0
2 © 2P ¹
2 2
θ
cos
θ
2
=
·
1 §W
W2
¨ ±
¸
8
+
¸
4¨ P
P2
©
¹
ª1 §W
θ = 2cos −1 « ¨
«4¨ P
¬ ©
±
·º
W2
+ 8 ¸ » W
2
¸»
P
¹¼
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400
PROBLEM 4.51 (Continued)
(b)
For P = 2W ,
cos
cos
θ
2
θ
2
θ
2
=
· 1
1§1
1
+ 8 ¸ = 1 ± 33
¨¨ ±
¸ 8
4©2
4
¹
(
= 0.84307 and cos
= 32.534°
θ = 65.1°
θ
2
θ
2
)
= −0.59307
= 126.375°
θ = 252.75° (discard)
θ = 65.1° W
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401
PROBLEM 4.52
A vertical load P is applied at end B of rod BC. (a) Neglecting the weight of
the rod, express the angle θ corresponding to the equilibrium position in
terms of P, l, and the counterweight W. (b) Determine the value of θ
corresponding to equilibrium if P = 2W.
SOLUTION
(a)
Triangle ABC is isosceles. We have
CD = ( BC ) cos
θ
2
= l cos
θ
2
θ·
§
ΣM C = 0: W ¨ l cos ¸ − P(l sin θ ) = 0
2¹
©
Setting sin θ = 2sin
θ
θ
θ
θ
W − 2 P sin
(b)
For P = 2W ,
sin
θ
2
θ
2
or
θ
cos : Wl cos − 2 Pl sin cos = 0
2
2
2
2
2
θ
2
=
θ
2
=0
θ = 2sin −1 §¨
W ·
¸ W
© 2P ¹
W
W
=
= 0.25
2 P 4W
θ = 29.0° W
= 14.5°
= 165.5° θ = 331° (discard)
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402
PROBLEM 4.53
A slender rod AB, of weight W, is attached to blocks A and B,
which move freely in the guides shown. The blocks are connected
by an elastic cord that passes over a pulley at C. (a) Express the
tension in the cord in terms of W and θ. (b) Determine the value of
θ for which the tension in the cord is equal to 3W.
SOLUTION
(a)
From F.B.D. of rod AB:
ª§ 1 ·
º
ΣM C = 0: T (l sin θ ) + W «¨ ¸ cos θ » − T (l cos θ ) = 0
¬© 2 ¹
¼
T=
W cos θ
2(cosθ − sin θ )
Dividing both numerator and denominator by cos θ,
T=
(b)
For T = 3W ,
or
3W =
W
2
1
§
·
¨ 1 − tan θ ¸
©
¹
or T =
( W2 )
(1 − tan θ )
W
( W2 )
(1 − tan θ )
1
1 − tan θ =
6
§5·
© ¹
θ = tan −1 ¨ ¸ = 39.806°
6
or
θ = 39.8° W
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403
PROBLEM 4.54
Rod AB is acted upon by a couple M and two forces, each of magnitude P.
(a) Derive an equation in θ, P, M, and l that must be satisfied when the rod
is in equilibrium. (b) Determine the value of θ corresponding to equilibrium
when M = 150 N · m, P = 200 N, and l = 600 mm.
SOLUTION
Free-Body Diagram:
(a)
From free-body diagram of rod AB:
ΣM C = 0: P(l cos θ ) + P(l sin θ ) − M = 0
or sinθ + cosθ =
(b)
M
W
Pl
For M = 150 lb ⋅ in., P = 20 lb, and l = 6 in.,
sin θ + cos θ =
150 lb ⋅ in.
5
= = 1.25
(20 lb)(6 in.) 4
sin 2 θ + cos 2 θ = 1
Using identity
sin θ + (1 − sin 2 θ )1/2 = 1.25
(1 − sin 2 θ )1/2 = 1.25 − sin θ
1 − sin 2 θ = 1.5625 − 2.5sin θ + sin 2 θ
2sin 2 θ − 2.5sin θ + 0.5625 = 0
Using quadratic formula
sin θ =
=
or
−( −2.5) ± (625) − 4(2)(0.5625)
2(2)
2.5 ± 1.75
4
sin θ = 0.95572 and sin θ = 0.29428
θ = 72.886° and θ = 17.1144°
or θ = 17.11° and θ = 72.9° W
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404
PROBLEM 4.55
Solve Sample Problem 4.5, assuming that the spring is unstretched
when θ = 90°.
SOLUTION
First note:
T = tension in spring = ks
where
s = deformation of spring
= rβ
F = kr β
From F.B.D. of assembly:
or
ΣM 0 = 0: W (l cos β ) − F (r ) = 0
Wl cos β − kr 2 β = 0
cos β =
For
kr 2
β
Wl
k = 250 lb/in.
r = 3 in.
l = 8 in.
W = 400 lb
cos β =
or
(250 lb/in.)(3 in.)2
β
(400 lb)(8 in.)
cos β = 0.703125β
Solving numerically,
β = 0.89245 rad
or
β = 51.134°
Then
θ = 90° + 51.134° = 141.134°
or θ = 141.1° W
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405
PROBLEM 4.56
A slender rod AB, of weight W, is attached to blocks A and B that
move freely in the guides shown. The constant of the spring is k,
and the spring is unstretched when θ = 0. (a) Neglecting the weight
of the blocks, derive an equation in W, k, l, and θ that must be
satisfied when the rod is in equilibrium. (b) Determine the value
of θ when W = 75 lb, l = 30 in., and k = 3 lb/in.
SOLUTION
Free-Body Diagram:
Spring force:
(a)
Fs = ks = k (l − l cos θ ) = kl (1 − cos θ )
§l
·
ΣM D = 0: Fs (l sin θ ) − W ¨ cos θ ¸ = 0
©2
¹
kl (1 − cos θ )l sin θ −
kl (1 − cos θ ) tan θ −
(b)
For given values of
W
l cos θ = 0
2
W
=0
2
or (1 − cos θ ) tan θ =
W
W
2kl
W = 75 lb
l = 30 in.
k = 3 lb/in.
(1 − cos θ ) tan θ = tan θ − sin θ
75 lb
=
2(3 lb/in.)(30 in.)
= 0.41667
Solving numerically,
θ = 49.710°
or
θ = 49.7° W
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406
PROBLEM 4.57
A vertical load P is applied at end B of rod BC. The constant of the
spring is k, and the spring is unstretched when θ = 60°. (a) Neglecting
the weight of the rod, express the angle θ corresponding to the
equilibrium position terms of P, k, and l. (b) Determine the value of θ
1
corresponding to equilibrium if P = 4 kl.
SOLUTION
Free-Body Diagram:
(a)
Triangle ABC is isosceles. We have
§θ ·
§θ ·
AB = 2( AD ) = 2l sin ¨ ¸ ; CD = l cos ¨ ¸
©2¹
©2¹
Elongation of spring:
x = ( AB)θ − ( AB )θ = 60°
§θ ·
= 2l sin ¨ ¸ − 2l sin 30°
©2¹
§ θ 1·
T = k x = 2kl ¨ sin − ¸
2 2¹
©
θ·
§
ΣM C = 0: T ¨ l cos ¸ − P(l sin θ ) = 0
2¹
©
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407
PROBLEM 4.57 (Continued)
θ
θ
θ·
§ θ 1·
§
2kl ¨ sin − ¸ l cos − Pl ¨ 2sin cos ¸ = 0
2 2¹
2
2
2¹
©
©
cos
θ
2
=0
2(kl − P ) sin
or
θ = 180° (trivial)
sin
θ
2
θ
2
− kl = 0
=
1
2
kl
kl − P
ª1
¬2
º
¼
θ = 2sin −1 « kl /( kl − P) » W
(b)
For P =
1
kl ,
4
sin
θ
2
θ
2
=
1
2
3
4
kl
kl
=
2
3
θ = 83.6° W
= 41.8°
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408
PROBLEM 4.58
A collar B of weight W can move freely along the vertical rod shown.
The constant of the spring is k, and the spring is unstretched when
θ = 0. (a) Derive an equation in θ, W, k, and l that must be satisfied
when the collar is in equilibrium. (b) Knowing that W = 300 N,
l = 500 mm, and k = 800 N/m, determine the value of θ corresponding
to equilibrium.
SOLUTION
First note:
T = ks
where
k = spring constant
s = elongation of spring
l
=
−l
cos θ
l
(1 − cos θ )
=
cos θ
kl
T=
(1 − cos θ )
cos θ
(a)
From F.B.D. of collar B:
or
(b)
For
ΣFy = 0: T sin θ − W = 0
kl
(1 − cos θ )sin θ − W = 0
cos θ
or tan θ − sin θ =
W
W
kl
W = 3 lb
l = 6 in.
k = 8 lb/ft
6 in.
l=
= 0.5 ft
12 in./ft
tan θ − sin θ =
Solving numerically,
3 lb
= 0.75
(8 lb/ft)(0.5 ft)
θ = 57.957°
or
θ = 58.0° W
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409
PROBLEM 4.59
Eight identical 500 × 750-mm rectangular plates, each of mass m = 40 kg, are held in a vertical plane as
shown. All connections consist of frictionless pins, rollers, or short links. In each case, determine whether
(a) the plate is completely, partially, or improperly constrained, (b) the reactions are statically determinate or
indeterminate, (c) the equilibrium of the plate is maintained in the position shown. Also, wherever possible,
compute the reactions.
SOLUTION
1.
Three non-concurrent, non-parallel reactions:
(a)
Plate: completely constrained
(b)
Reactions: determinate
(c)
Equilibrium maintained
A = C = 196.2 N
2.
Three non-concurrent, non-parallel reactions:
(a)
Plate: completely constrained
(b)
Reactions: determinate
(c)
Equilibrium maintained
B = 0, C = D = 196.2 N
3.
Four non-concurrent, non-parallel reactions:
(a)
Plate: completely constrained
(b)
Reactions: indeterminate
(c)
Equilibrium maintained
,
A x = 294 N
D x = 294 N
( A y + D y = 392 N )
4.
Three concurrent reactions (through D):
(a)
Plate: improperly constrained
(b)
Reactions: indeterminate
(c)
No equilibrium
(ΣM D ≠ 0)
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410
PROBLEM 4.59 (Continued)
5.
Two reactions:
(a)
Plate: partial constraint
(b)
Reactions: determinate
(c)
Equilibrium maintained
C = D = 196.2 N
6.
Three non-concurrent, non-parallel reactions:
(a)
Plate: completely constrained
(b)
Reactions: determinate
(c)
Equilibrium maintained
B = 294 N
7.
8.
, D = 491 N
53.1°
Two reactions:
(a)
Plate: improperly constrained
(b)
Reactions determined by dynamics
(c)
No equilibrium
(ΣFy ≠ 0)
Four non-concurrent, non-parallel reactions:
(a)
Plate: completely constrained
(b)
Reactions: indeterminate
(c)
Equilibrium maintained
B = D y = 196.2 N
(C + D x = 0)
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411
PROBLEM 4.60
The bracket ABC can be supported in the eight different ways shown. All connections consist of smooth pins,
rollers, or short links. For each case, answer the questions listed in Problem 4.59, and, wherever possible,
compute the reactions, assuming that the magnitude of the force P is 100 lb.
SOLUTION
1.
Three non-concurrent, non-parallel reactions:
(a)
Bracket: complete constraint
(b)
Reactions: determinate
(c)
Equilibrium maintained
56.3°, B = 66.7 lb
A = 120.2 lb
2.
3.
Four concurrent, reactions (through A):
(a)
Bracket: improper constraint
(b)
Reactions: indeterminate
(c)
No equilibrium
(ΣM A ≠ 0)
Two reactions:
(a)
Bracket: partial constraint
(b)
Reactions: indeterminate
(c)
Equilibrium maintained
A = 50 lb , C = 50 lb
4.
Three non-concurrent, non-parallel reactions:
(a)
Bracket: complete constraint
(b)
Reactions: determinate
(c)
Equilibrium maintained
A = 50 lb , B = 83.3 lb
36.9°, C = 66.7 lb
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412
PROBLEM 4.60 (Continued)
5.
6.
Four non-concurrent, non-parallel reactions:
(a)
Bracket: complete constraint
(b)
Reactions: indeterminate
(c)
Equilibrium maintained
(ΣM C = 0) A y = 50 lb
Four non-concurrent, non-parallel reactions:
(a)
Bracket: complete constraint
(b)
Reactions: indeterminate
(c)
Equilibrium maintained
A x = 66.7 lb
B x = 66.7 lb
( A y + B y = 100 lb )
7.
Three non-concurrent, non-parallel reactions:
(a)
Bracket: complete constraint
(b)
Reactions: determinate
(c)
Equilibrium maintained
A = C = 50 lb
8.
Three concurrent, reactions (through A)
(a)
Bracket: improper constraint
(b)
Reactions: indeterminate
(c)
No equilibrium
(ΣM A ≠ 0)
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413
PROBLEM 4.61
Determine the reactions at A and B when a = 150 mm.
SOLUTION
Free-Body Diagram:
Force triangle
80 mm 80 mm
=
a
150 mm
β = 28.072°
tan β =
A=
320 N
sin 28.072°
B=
320 N
tan 28.072°
A = 680 N
28.1° W
B = 600 N
W
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414
PROBLEM 4.62
Determine the value of a for which the magnitude of the reaction at B
is equal to 800 N.
SOLUTION
Free-Body Diagram:
Force triangle
tan β =
80 mm
a
a=
80 mm
tan β
(1)
From force triangle:
tan β =
From Eq. (1):
a=
320 N
= 0.4
800 N
80 mm
0.4
a = 200 mm W
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415
PROBLEM 4.63
Using the method of Sec. 4.7, solve Problem 4.22b.
PROBLEM 4.22 Determine the reactions at A and B when (a) α = 0,
(b) α = 90°, (c) α = 30°.
SOLUTION
Free-Body Diagram:
(Three-force body)
The line of action at A must pass through C, where B and the 75-lb load intersect.
In triangle ACE:
Force triangle
tan θ =
10 in.
12 in.
θ = 39.806°
B = (75 lb) tan 39.806°
= 62.5 lb
75 lb
A=
= 97.6°
cos 39.806°
A = 97.6 lb
50.2°; B = 62.5 lb
W
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416
PROBLEM 4.64
A 500-lb cylindrical tank, 8 ft in diameter, is to be raised over a 2-ft
obstruction. A cable is wrapped around the tank and pulled horizontally as
shown. Knowing that the corner of the obstruction at A is rough, find the
required tension in the cable and the reaction at A.
SOLUTION
Free-Body Diagram:
Force triangle
cos α =
GD 2 ft
=
= 0.5
AG 4 ft
α = 60°
1
( β = 60°)
2
T = (500 lb) tan 30°
T = 289 lb
θ = α = 30°
A=
500 lb
cos 30°
A = 577 lb
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417
PROBLEM 4.65
For the frame and loading shown, determine the reactions at A and C.
SOLUTION
Since member AB is acted upon by two forces, A and B, they must be colinear, have the same magnitude, and
be opposite in direction for AB to be in equilibrium. The force B acting at B of member BCD will be equal in
magnitude but opposite in direction to force B acting on member AB. Member BCD is a three-force body with
member forces intersecting at E. The F.B.D.’s of members AB and BCD illustrate the above conditions. The
force triangle for member BCD is also shown. The angle β is found from the member dimensions:
§ 6 in. ·
¸ = 30.964°
© 10 in. ¹
β = tan −1 ¨
Applying the law of sines to the force triangle for member BCD,
30 lb
B
C
=
=
sin(45° − β ) sin β sin135°
or
30 lb
B
C
=
=
sin14.036° sin 30.964° sin135°
A= B=
(30 lb)sin 30.964°
= 63.641 lb
sin14.036°
or
and
C=
A = 63.6 lb
45.0° W
C = 87.5 lb
59.0° W
(30 lb) sin135°
= 87.466 lb
sin14.036°
or
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418
PROBLEM 4.66
For the frame and loading shown, determine the reactions at C and D.
SOLUTION
Since BD is a two-force member, the reaction at D must pass through Points B and D.
Free-Body Diagram:
(Three-force body)
Reaction at C must pass through E, where the reaction at D and the 150-lb load intersect.
Triangle CEF:
tan β =
4.5 ft
3 ft
β = 56.310°
Triangle ABE:
tan γ =
1
2
γ = 26.565°
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419
PROBLEM 4.66 (Continued)
Force Triangle
Law of sines:
150 lb
C
D
=
=
sin 29.745° sin116.565° sin 33.690°
C = 270.42 lb,
D = 167.704 lb
C = 270 lb
56.3°; D = 167.7 lb
26.6° W
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420
PROBLEM 4.67
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 the 80-N force intersect.
220 mm
250 mm
β = 41.348°
tan β =
Force triangle
Law of sines:
80 N
B
D
=
=
sin 3.652° sin 45° sin131.348°
B = 888.0 N
D = 942.8 N
B = 888 N
41.3°
D = 943 N
45.0° W
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421
PROBLEM 4.68
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 the 80-N force intersect.
280 mm
250 mm
β = 48.24°
tan β =
Force triangle
Law of sines:
80 N
B
D
=
=
sin 3.24° sin135° sin 41.76°
B = 1000.9 N
D = 942.8 N B = 1001 N
48.2° D = 943 N
45.0° W
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422
PROBLEM 4.69
A T-shaped bracket supports a 300-N load as shown. Determine
the reactions at A and C when α = 45°.
SOLUTION
Free-Body Diagram:
(Three-force body)
The line of action of C must pass through E, where A and the 300-N force intersect.
Triangle ABE is isosceles:
EA = AB = 400 mm
In triangle CEF:
tan θ =
150 mm
CF
CF
=
=
EF EA + AF 700 mm
θ = 12.0948°
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423
PROBLEM 4.69 (Continued)
Force Triangle
Law of sines:
A
C
300 N
=
=
sin 32.905° sin135° sin12.0948°
A = 778 N ;
C = 1012 N
77.9° W
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424
PROBLEM 4.70
A T-shaped bracket supports a 300-N load as shown. Determine
the reactions at A and C when α = 60°.
SOLUTION
Free-Body Diagram:
EA = (400 mm) tan 30°
= 230.94 mm
In triangle CEF:
tan θ =
CF
CF
=
EF EA + AF
150
230.94 + 300
θ = 15.7759°
tan θ =
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425
PROBLEM 4.70 (Continued)
Force Triangle
Law of sines:
300 N
A
C
=
=
sin 44.224° sin120° sin15.7759°
A = 770 N
C = 956 N
A = 770 N ;
C = 956 N
74.2° W
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426
PROBLEM 4.71
A 40-lb roller, of diameter 8 in., which is to be used on a tile floor, is resting
directly on the subflooring as shown. Knowing that the thickness of each
tile is 0.3 in., determine the force P required to move the roller onto the tiles
if the roller is (a) pushed to the left, (b) pulled to the right.
SOLUTION
Geometry: For each case as roller comes into contact with tile,
3.7 in.
4 in.
α = 22.332°
α = cos −1
(a)
Roller pushed to left (three-force body):
Forces must pass through O.
Law of sines:
Force Triangle
40 lb
P
=
; P = 24.87 lb
sin 37.668° sin 22.332°
P = 24.9 lb
(b)
Roller pulled to right (three-force body):
Forces must pass through O.
Law of sines:
Force Triangle
40 lb
P
=
; P = 15.3361 lb
sin 97.668° sin 22.332°
P = 15.34 lb
30.0° W
30.0° W
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427
PROBLEM 4.72
One end of rod AB rests in the corner A and the other end is attached to
cord BD. If the rod supports a 40-lb load at its midpoint C, find the reaction
at A and the tension in the cord.
SOLUTION
Free-Body Diagram: (Three-force body)
The line of action of reaction at A must pass through E, where T and the 40-lb load intersect.
Force triangle
EF 23
=
AF 12
α = 62.447°
5
EH
tan β =
=
DH 12
β = 22.620°
tan α =
A
T
40 lb
=
=
sin 67.380° sin 27.553° sin 85.067°
A = 37.1 lb
62.4° W
T = 18.57 lb W
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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428
PROBLEM 4.73
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
Three-force body: W and TCD intersect at E.
0.7497 m
1.5 m
β = 26.556°
tan β =
Three forces intersect at E.
W = (50 kg) 9.81 m/s 2
= 490.50 N
Law of sines:
Force triangle
TCD
490.50 N
B
=
=
sin 61.556° sin 63.444° sin 55°
TCD = 498.99 N
B = 456.96 N
TCD = 499 N W
(a)
(b)
B = 457 N
26.6° W
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429
PROBLEM 4.74
Solve Problem 4.73, assuming that a = 3 m.
PROBLEM 4.73 A 50-kg crate is attached to the trolley-beam
system shown. Knowing that a = 1.5 m, determine (a) the tension in
cable CD, (b) the reaction at B.
SOLUTION
W and TCD intersect at E.
Free-Body Diagram:
Three-Force Body
AE 0.301 m
=
AB
3m
β = 5.7295°
tan β =
Three forces intersect at E.
Force Triangle
W = (50 kg) 9.81 m/s 2
= 490.50 N
Law of sines:
TCD
490.50 N
B
=
=
sin 29.271° sin 95.730° sin 55°
TCD = 998.18 N
B = 821.76 N
(a)
(b)
TCD = 998 N W
B = 822 N
5.73° W
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430
PROBLEM 4.75
Determine the reactions at A and B when β = 50°.
SOLUTION
Free-Body Diagram: (Three-force body)
Reaction A must pass through Point D where the 100-N force
and B intersect.
In right ∆ BCD:
α = 90° − 75° = 15°
BD = 250 tan 75° = 933.01 mm
Dimensions in mm
In right ∆ ABD:
AB
150 mm
=
BD 933.01 mm
γ = 9.1333°
tan γ =
Force Triangle
Law of sines:
100 N
A
B
=
=
sin 9.1333° sin15° sin155.867°
A = 163.1 N; B = 257.6 N
A = 163.1 N
74.1° B = 258 N
65.0° W
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431
PROBLEM 4.76
Determine the reactions at A and B when β = 80°.
SOLUTION
Free-Body Diagram:
(Three-force body)
Reaction A must pass through D where the 100-N force and B intersect.
In right triangle BCD:
α = 90° − 75° = 15°
BD = BC tan 75° = 250 tan75°
BD = 933.01 mm
In right triangle ABD:
tan γ =
AB
150 mm
=
BD 933.01 mm
γ = 9.1333°
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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432
PROBLEM 4.76 (Continued)
Force Triangle
Law of sines:
100 N
A
B
=
=
sin 9.1333° sin15° sin155.867°
A = 163.1 N
B = 258 N
55.9° W
65.0° W
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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433
PROBLEM 4.77
Knowing that θ = 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 ∆ CDE:
( 3 − 1) R
R
= 3 −1
β = 36.2°
tan β =
Force Triangle
Law of sines:
P
B
C
=
=
sin 23.8° sin126.2° sin 30°
B = 2.00 P
C = 1.239 P
(a)
B = 2P
(b)
C = 1.239P
60.0° W
36.2° W
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434
PROBLEM 4.78
Knowing that θ = 60°, determine the reaction (a) at B, (b) at C.
SOLUTION
Reaction at C must pass through D where force P and reaction at B intersect.
In ∆CDE:
tan β =
R−
=1−
R
3
Free-Body Diagram:
(Three-force body)
R
1
3
β = 22.9°
Force Triangle
Law of sines:
P
B
C
=
=
sin 52.9° sin 67.1° sin 60°
B = 1.155P
C = 1.086 P
(a)
B = 1.155P
(b)
C = 1.086P
30.0° W
22.9° W
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435
PROBLEM 4.79
Using the method of Section 4.7, solve Problem 4.23.
PROBLEM 4.23 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 the 150-N weight
and B interect.
Force triangle is equilateral.
A = 150.0 N
B = 150.0 N
(b)
30.0° W
30.0° W
h = 200 mm
55.662
250
β = 12.5521°
tan β =
Law of sines:
A
B
150 N
=
=
sin17.4480° sin 60° sin102.552°
A = 433.24 N
B = 488.31 N
Force Triangle
A = 433 N
B = 488 N
12.55° W
30.0° W
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436
PROBLEM 4.80
Using the method of Section 4.7, solve Problem 4.24.
PROBLEM 4.24 A lever AB is hinged at C and attached to a control
cable at A. If the lever is subjected to a 75-lb vertical force at B,
determine (a) the tension in the cable, (b) the reaction at C.
SOLUTION
Reaction at C must pass through E, where the 75-lb force
and T intersect.
9.3969 in.
8.5798 in.
α = 47.602°
tan α =
14.0954 in.
24.870 in.
β = 29.543°
tan β =
Force Triangle
Law of sines:
Free-Body Diagram:
Dimensions in in.
75 lb
T
C
=
=
sin18.0590° sin 29.543° sin132.398°
(a)
(b)
T = 119.3 lb W
C = 178.7 lb
60.5° W
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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437
PROBLEM 4.81
Member ABC is supported by a pin and bracket at B and by an
inextensible cord attached at A and C and passing over a frictionless
pulley at D. The tension may be assumed to be the same in portions AD
and CD of the cord. For the loading shown and neglecting the size of
the pulley, determine the tension in the cord and the reaction at B.
SOLUTION
Free-Body Diagram:
Reaction at B must pass through D.
7 in.
12 in.
α = 30.256°
7 in.
tan β =
24 in.
β = 16.26°
tan α =
Force Triangle
Law of sines:
T
T − 72 lb
B
=
=
sin 59.744° sin13.996° sin106.26
T (sin13.996°) = (T − 72 lb)(sin 59.744°)
T (0.24185) = (T − 72)(0.86378)
T = 100.00 lb
sin 106.26°
sin 59.744°
= 111.14 lb
T = 100.0 lb W
B = (100 lb)
B = 111.1 lb
30.3° W
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438
PROBLEM 4.82
Member ABC is supported by a pin and bracket at B and by an
inextensible cord attached at A and C and passing over a frictionless
pulley at D. The tension may be assumed to be the same in portions
AD and CD of the cord. For the loading shown and neglecting the size
of the pulley, determine the tension in the cord and the reaction at B.
SOLUTION
Free-Body Diagram:
Force Triangle
Reaction at B must pass through D.
tan α =
120
; α = 36.9°
160
T T − 75 N B
=
=
4
3
5
3T = 4T − 300; T = 300 N
5
5
B = T = (300 N) = 375 N
4
4
B = 375 N
36.9°
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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439
PROBLEM 4.83
A thin ring of mass 2 kg and radius r = 140 mm is held against a frictionless
wall by a 125-mm string AB. Determine (a) the distance d, (b) the tension in
the string, (c) the reaction at C.
SOLUTION
Free-Body Diagram:
(Three-force body)
The force T exerted at B must pass through the center G of the ring, since C and W intersect at that point.
Thus, points A, B, and G are in a straight line.
(a) From triangle ACG:
d = ( AG ) 2 − (CG )2
= (265 mm)2 − (140 mm) 2
= 225.00 mm
d = 225 mm W
Force Triangle
W = (2 kg)(9.81 m/s 2 ) = 19.6200 N
Law of sines:
T
C
19.6200 N
=
=
265 mm 140 mm 225.00 mm
(b)
(c)
T = 23.1 N W
C = 12.21 N
W
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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440
PROBLEM 4.84
A uniform rod AB of length 2R rests inside a hemispherical bowl of radius
R as shown. Neglecting friction, determine the angle θ corresponding to
equilibrium.
SOLUTION
Based on the F.B.D., the uniform rod AB is a three-force body. Point E is the point of intersection of the three
forces. Since force A passes through O, the center of the circle, and since force C is perpendicular to the rod,
triangle ACE is a right triangle inscribed in the circle. Thus, E is a point on the circle.
Note that the angle α of triangle DOA is the central angle corresponding to the inscribed angle θ of
triangle DCA.
α = 2θ
The horizontal projections of AE , ( x AE ), and AG , ( x AG ), are equal.
x AE = x AG = x A
or
( AE ) cos 2θ = ( AG ) cos θ
and
(2 R) cos 2θ = R cos θ
Now
then
or
cos 2θ = 2 cos 2 θ − 1
4 cos 2 θ − 2 = cos θ
4 cos 2 θ − cos θ − 2 = 0
Applying the quadratic equation,
cos θ = 0.84307 and cos θ = − 0.59307
θ = 32.534° and θ = 126.375° (Discard)
or θ = 32.5° W
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441
PROBLEM 4.85
A slender rod BC of length L and weight W is held by two cables as
shown. Knowing that cable AB is horizontal and that the rod forms an
angle of 40° with the horizontal, determine (a) the angle θ that cable
CD forms with the horizontal, (b) the tension in each cable.
SOLUTION
Free-Body Diagram:
(Three-force body)
(a)
The line of action of TCD must pass through E, where TAB and W intersect.
CF
EF
L sin 40°
= 1
L cos 40°
2
tan θ =
= 2 tan 40°
= 59.210°
(b)
Force Triangle
θ = 59.2° W
TAB = W tan 30.790°
= 0.59588W
TAB = 0.596W W
W
cos 30.790° = 1.16408W
TCD =
TCD = 1.164W W
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442
PROBLEM 4.86
A slender rod of length L and weight W is attached to a collar at A and is
fitted with a small wheel at B. Knowing that the wheel rolls freely along
a cylindrical surface of radius R, and neglecting friction, derive an
equation in θ, L, and R that must be satisfied when the rod is in
equilibrium.
SOLUTION
Free-Body Diagram (Three-force body)
Reaction B must pass through D where B and W intersect.
Note that ∆ABC and ∆BGD are similar.
AC = AE = L cos θ
In ∆ ABC:
(CE ) 2 + ( BE )2 = ( BC )2
(2 L cos θ ) 2 + ( L sin θ )2 = R 2
§R·
2
2
¨ ¸ = 4cos θ + sin θ
©L¹
2
§R·
2
2
¨ ¸ = 4cos θ + 1 − cos θ
©L¹
2
§R·
2
¨ ¸ = 3cos θ + 1
©L¹
2
2
º
1ª
cos 2 θ = «§¨ R ·¸ − 1» W
3 ¬© L ¹
¼
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443
PROBLEM 4.87
Knowing that for the rod of Problem 4.86, L = 15 in., R = 20 in., and
W = 10 lb, determine (a) the angle θ corresponding to equilibrium,
(b) the reactions at A and B.
SOLUTION
See the solution to Problem 4.86 for the free-body diagram and analysis leading to the following equation:
2
º
1ª
cos 2 θ = «¨§ R ¸· − 1»
3 ¬© L ¹
¼
For L = 15 in., R = 20 in., and W = 10 lb,
2
º
1 ª§ 20 in. ·
cos θ = «¨
¸ − 1» ; θ = 59.39°
3 «© 15 in. ¹
»
¬
¼
2
(a)
In ∆ ABC:
θ = 59.4° W
1
BE
L sin θ
=
= tan θ
CE 2 L cos θ 2
1
tan α = tan 59.39° = 0.8452
2
α = 40.2°
tan α =
Force Triangle
A = W tan α = (10 lb) tan 40.2° = 8.45 lb
(10 lb)
W
=
= 13.09 lb
B=
cos α cos 40.2°
(b)
A = 8.45 lb
B = 13.09 lb
W
49.8° W
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444
PROBLEM 4.88
Rod AB is bent into the shape of an arc of circle and is lodged between two
pegs D and E. It supports a load P at end B. Neglecting friction and the
weight of the rod, determine the distance c corresponding to equilibrium
when a = 20 mm and R = 100 mm.
SOLUTION
Free-Body Diagram:
Since yED = xED = a,
slope of ED is
45°;
slope of HC is
45°.
Also
and
For triangles DHC and EHC,
DE = 2 a
a
§1·
DH = HE = ¨ ¸ DE =
2
2
© ¹
sin β =
a
2
R
=
a
2R
Now
c = R sin(45° − β )
For
a = 20 mm and
sin β =
R = 100 mm
20 mm
2(100 mm)
= 0.141421
β = 8.1301°
and
c = (100 mm) sin(45° − 8.1301°)
= 60.00 mm
or c = 60.0 mm W
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445
PROBLEM 4.89
A slender rod of length L is attached to collars that can slide freely along
the guides shown. Knowing that the rod is in equilibrium, derive an
expression for the angle θ in terms of the angle β.
SOLUTION
As shown in the free-body diagram of the slender rod AB, the three forces intersect at C. From the force
geometry:
Free-Body Diagram:
tan β =
xGB
y AB
where
y AB = L cos θ
and
xGB =
tan β =
1
L sin θ
2
1
2
L sin θ
L cos θ
1
= tan θ
2
or tan θ = 2 tan β W
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446
PROBLEM 4.90
An 8-kg slender rod of length L is attached to collars that can slide freely
along the guides shown. Knowing that the rod is in equilibrium and that
β = 30°, determine (a) the angle θ that the rod forms with the vertical,
(b) the reactions at A and B.
SOLUTION
(a)
As shown in the free-body diagram of the slender rod AB, the three forces intersect at C. From the
geometry of the forces:
Free-Body Diagram:
tan β =
xCB
yBC
where
xCB =
1
L sin θ
2
and
yBC = L cos θ
tan β =
1
tan θ
2
or
tan θ = 2 tan β
For
β = 30°
tan θ = 2 tan 30°
= 1.15470
θ = 49.107°
or
W = mg = (8 kg)(9.81 m/s2 ) = 78.480 N
(b)
From force triangle:
A = W tan β
= (78.480 N) tan 30°
or
= 45.310 N
and
θ = 49.1° W
B=
78.480 N
W
=
= 90.621 N
cos β
cos 30°
A = 45.3 N
or B = 90.6 N
W
60.0° W
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447
PROBLEM 4.91
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
ΣM C = 0: (−120k ) × ( Dx i + Dy j) + (120 j − 160k ) × T i + (80k − 200i ) × (−720 j) = 0
−120 Dx j + 120 D y i − 120T k − 160Tj + 57.6 × 103 i + 144 × 103 k = 0
Equating to zero the coefficients of the unit vectors:
k:
−120T + 144 × 103 = 0
i:
120 Dy + 57.6 × 103 = 0
j: − 120 Dx − 160(1200 N) = 0
(b)
ΣFx = 0:
C x + Dx + T = 0
ΣFy = 0:
C y + Dy − 720 = 0
ΣFz = 0:
Cz = 0
(a) T = 1200 N W
Dy = −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 W
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448
PROBLEM 4.92
Solve Problem 4.91, assuming that the axle has been
rotated clockwise in its bearings by 30° and that the
720-N load remains vertical.
PROBLEM 4.91 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.
ΣM C = 0: (−120k ) × ( Dx i + Dy j) + (120 j − 160k ) × T i + (80k − 173.21i ) × (−720 j) = 0
−120 Dx j + 120 D y i −120T k −160T j + 57.6 × 103 i + 124.71 × 103 k = 0
Equating to zero the coefficients of the unit vectors,
k : − 120T + 124.71 × 103 = 0
i:
T = 1039 N W
120 Dy + 57.6 × 103 = 0 Dy = −480 N
j: − 120 Dx − 160(1039.2)
(b)
T = 1039.2 N
ΣFx = 0:
C x + Dx + T = 0
ΣFy = 0:
C y + Dy − 720 = 0
ΣFz = 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 W
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449
PROBLEM 4.93
A 4 × 8-ft sheet of plywood weighing 40 lb has been temporarily
propped against column CD. It rests at A and B on small wooden
blocks and against protruding nails. Neglecting friction at all
surfaces of contact, determine the reactions at A, B, and C.
SOLUTION
Free-Body Diagram:
We have five unknowns and six equations of equilibrium. Plywood sheet is free to move in x direction, but
equilibrium is maintained (ΣFx = 0).
ΣM A = 0: rB /A × ( B y j + Bz k ) + rC /A × C k + rG /A × ( −40 lb) j = 0
i j
5 0
0 By
k
i
j
k
i
j
k
0 + 4 4sin 60° −4cos 60° + 2 2sin 60° −2 cos 60° = 0
Bz 0
C
−40
0
0
0
(4C sin 60° − 80 cos 60°) i + (−5Bz − 4C ) j + (5B y − 80)k = 0
Equating the coefficients of the unit vectors to zero,
i:
4C sin 60° − 80 cos 60° = 0
j:
−5Bz − 4C = 0
k:
5B y − 80 = 0
ΣFy = 0:
Ay + B y − 40 = 0
ΣFz = 0:
Az + Bz + C = 0
C = 11.5470 lb
Bz = 9.2376 lb
B y = 16.0000 lb
Ay = 40 − 16.0000 = 24.000 lb
Az = 9.2376 − 11.5470 = −2.3094 lb
A = (24.0 lb)j − (2.31 lb)k ; B = (16.00 lb) j − (9.24 lb)k ; C = (11.55 lb)k W
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450
PROBLEM 4.94
Two tape spools are attached to an axle supported by bearings
at A and D. The radius of spool B is 1.5 in. and the radius of
spool C is 2 in. Knowing that TB = 20 lb 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
Free-Body Diagram:
We have six unknowns and six equations of equilibrium.
ΣM A = 0: (4.5i + 1.5k ) × (−20 j) + (10.5i + 2 j) × (−TC k ) + (15i) × ( Dx i + Dy j + Dz k ) = 0
−90k + 30i + 10.5TC j − 2TC i + 15D y k − 15Dz j = 0
Equate coefficients of unit vectors to zero:
K
i:
30 − 2TC = 0
K
10.5TC − 15Dz = 0 10.5(15) − 15Dz = 0
j:
K
k:
−90 + 15 Dy = 0
ΣFx = 0:
Dx = 0
ΣFy = 0:
Ay + D y − 20 lb = 0
Ay = 20 − 6 = 14 lb
ΣFz = 0:
Az + Dz − 15 lb = 0
Az = 15 − 10.5 = 4.5 lb
TC = 15 lb
Dz = 10.5 lb
Dy = 6 lb
A = (14.00 lb) j + (4.50 lb)k ; D = (6.00 lb) j + (10.50 lb)k W
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451
PROBLEM 4.95
Two transmission belts pass over a double-sheaved
pulley that is attached to an axle supported by bearings
at A and D. The radius of the inner sheave is 125 mm
and the radius of the outer sheave is 250 mm. Knowing
that when the system is at rest, the tension is 90 N in
both portions of belt B and 150 N in both portions of
belt C, determine the reactions at A and D. Assume that
the bearing at D does not exert any axial thrust.
SOLUTION
We replace TB and TB′ by their resultant (−180 N)j and TC and TC′ by their resultant (−300 N)k.
Dimensions in mm
We have five unknowns and six equations of equilibrium. Axle AD is free to rotate about the x-axis, but
equilibrium is maintained (ΣMx = 0).
ΣM A = 0: (150i ) × (−180 j) + (250i) × ( −300k ) + (450i ) × ( Dy j + Dz k ) = 0
−27 × 103 k + 75 × 103 j + 450 Dy k − 450 Dz j = 0
Equating coefficients of j and k to zero,
j:
75 × 103 − 450 Dz = 0
Dz = 166.7 N
k:
− 27 × 103 + 450 Dy = 0
Dy = 60.0 N
ΣFx = 0:
Ax = 0
ΣFy = 0:
Ay + Dy − 180 N = 0
ΣFz = 0: Az + Dz − 300 N = 0
Ay = 180 − 60 = 120.0 N
Az = 300 − 166.7 = 133.3 N
A = (120.0 N) j + (133.3 N)k ; D = (60.0 N) j + (166.7 N)k W
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452
PROBLEM 4.96
Solve Problem 4.95, assuming that the pulley rotates at a
constant rate and that TB = 104 N, T B = 84 N, TC = 175 N.
PROBLEM 4.95 Two transmission belts pass over a
double-sheaved pulley that is attached to an axle supported
by bearings at A and D. The radius of the inner sheave is
125 mm and the radius of the outer sheave is 250 mm.
Knowing that when the system is at rest, the tension is 90 N
in both portions of belt B and 150 N in both portions of
belt C, determine the reactions at A 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
ΣM A = 0: (150i + 250k ) × (−104 j) + (150i − 250k ) × ( −84 j)
+ (250i + 125 j) × (−175k ) + (250i − 125 j) × (−TC )
+ 450i × ( D y j + Dz k ) = 0
−150(104 + 84)k + 250(104 − 84)i + 250(175 + TC′ ) j − 125(175 − TC′ )
+ 450 D y k − 450 Dz j = 0
Equating the coefficients of the unit vectors to zero,
i : 250(104 − 84) − 125(175 − TC′ ) = 0
175 = TC′ = 40
j: 250(175 + 135) − 450 Dz = 0
Dz = 172.2 N
k : − 150(104 + 84) + 450 D y = 0
Dy = 62.7 N
TC′ = 135;
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453
PROBLEM 4.96 (Continued)
ΣFx = 0:
Ax = 0
ΣFy = 0:
Ay − 104 − 84 + 62.7 = 0
Ay = 125.3 N
ΣFz = 0:
Az − 175 − 135 + 172.2 = 0
Az = 137.8 N
A = (125.3 N) j + (137.8 N)k; D = (62.7 N) j + (172.2 N)k W
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454
PROBLEM 4.97
Two steel pipes AB and BC, each having a mass per unit
length of 8 kg/m, are welded together at B and supported by
three wires. Knowing that a = 0.4 m, determine the tension
in each wire.
SOLUTION
W1 = 0.6m′g
W2 = 1.2m′g
ΣM D = 0: rA/D × TA j + rE/D × (−W1 j) + rF/D × (−W2 j) + rC/D × TC j = 0
(−0.4i + 0.6k ) × TA j + (−0.4i + 0.3k ) × (−W1 j) + 0.2i × (−W2 j) + 0.8i × TC j = 0
−0.4TAk − 0.6TA i + 0.4W1k + 0.3W1i − 0.2W2 k + 0.8TC k = 0
Equate coefficients of unit vectors to zero:
1
1
i : − 0.6TA + 0.3W1 = 0; TA = W1 = 0.6m′g = 0.3m′g
2
2
k : − 0.4TA + 0.4W1 − 0.2W2 + 0.8TC = 0
−0.4(0.3m′g ) + 0.4(0.6m′g ) − 0.2(1.2m′g ) + 0.8TC = 0
TC =
(0.12 − 0.24 − 0.24)m′g
= 0.15m′g
0.8
ΣFy = 0: TA + TC + TD − W1 − W2 = 0
0.3m′g + 0.15m′g + TD − 0.6m′g − 1.2m′g = 0
TD = 1.35m′g
m′g = (8 kg/m)(9.81m/s 2 ) = 78.48 N/m
TA = 0.3m′g = 0.3 × 78.45
TB = 0.15m′g = 0.15 × 78.45
TC = 1.35m′g = 1.35 × 78.45
TA = 23.5 N W
TB = 11.77 N W
TC = 105.9 N W
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455
PROBLEM 4.98
For the pipe assembly of Problem 4.97, determine (a) the
largest permissible value of a if the assembly is not to tip,
(b) the corresponding tension in each wire.
SOLUTION
W1 = 0.6m′g
W2 = 1.2m′g
ΣM D = 0: rA/D × TA j + rE/D × (−W1 j) + rF/D × (−W2 j) + rC/D × TC j = 0
(− ai + 0.6k ) × TA j + (− ai + 0.3k ) × (−W1 j) + (0.6 − a)i × (−W2 j) + (1.2 − a)i × TC j = 0
−TA ak − 0.6TA i + W1ak + 0.3W1i − W2 (0.6 − a)k + TC (1.2 − a )k = 0
Equate coefficients of unit vectors to zero:
1
1
i : − 0.6TA + 0.3W1 = 0; TA = W1 = 0.6m′g = 0.3m′g
2
2
k : − TA a + W1a − W2 (0.6 − a ) + TC (1.2 − a) = 0
−0.3m′ga + 0.6m′ga − 1.2m′g (0.6 − a ) + TC (1.2 − a) = 0
TC =
(a)
0.3a − 0.6a + 1.2(0.6 − a)
1.2 − a
−0.3a + 1.2(0.6 − a) = 0
−0.3a + 0.72 − 1.2a = 0
1.5a = 0.72
For maximum a and no tipping, TC = 0.
a = 0.480 m W
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456
PROBLEM 4.98 (Continued)
(b)
Reactions:
m′g = (8 kg/m) 9.81 m/s 2 = 78.48 N/m
TA = 0.3m′g = 0.3 × 78.48 = 23.544 N
TA = 23.5 N W
ΣFy = 0: TA + TC + TD − W1 − W2 = 0
TA + 0 + TD − 0.6m′g − 1.2m′g = 0
TD = 1.8m′g − TA = 1.8 × 78.48 − 23.544 = 117.72
TD = 117.7 N W
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457
PROBLEM 4.99
The 45-lb square plate shown is supported by three vertical wires.
Determine the tension in each wire.
SOLUTION
Free-Body Diagram:
ΣM B = 0: rC /B × TC j + rA/B × TA j + rG /B × (−45 lb) j = 0
K
[−(20 in.)i + (15 in.)k ] × TC j + (20 in.)k × TA j
+ [−(10 in.)i + (10 in.)k ] × [−(45 lb)j] = 0
−20TC k − 15TC i − 20TAi + 450k + 450i = 0
Equating to zero the coefficients of the unit vectors,
k:
−20TC + 450 = 0
i:
−15(22.5) − 20TA + 450 = 0
ΣFy = 0:
TA + TB + TC − 45 lb = 0
5.625 lb + TB + 22.5 lb − 45 lb = 0
TC = 22.5 lb W
TA = 5.625 lb W
TB = 16.875 lb W
TA = 5.63 lb; TB = 16.88 lb; TC = 22.5 lb W
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458
PROBLEM 4.100
The table shown weighs 30 lb and has a diameter of 4 ft. It is supported by
three legs equally spaced around the edge. A vertical load P of magnitude
100 lb is applied to the top of the table at D. Determine the maximum
value of a if the table is not to tip over. Show, on a sketch, the area of the
table over which P can act without tipping the table.
SOLUTION
r = 2 ft b = r sin 30° = 1 ft
We shall sum moments about AB.
(b + r )C + (a − b) P − bW = 0
(1 + 2)C + (a − 1)100 − (1)30 = 0
1
C = [30 − (a − 1)100]
3
If table is not to tip, C ≥ 0.
[30 − ( a − 1)100] ≥ 0
30 ≥ (a − 1)100
a − 1 ≤ 0.3 a ≤ 1.3 ft a = 1.300 ft
Only ⊥ distance from P to AB matters. Same condition must be satisfied for each leg. P must be located
in shaded area for no tipping.
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459
PROBLEM 4.101
An opening in a floor is covered by a 1 × 1.2-m sheet of
plywood of mass 18 kg. The sheet is hinged at A and B and
is maintained in a position slightly above the floor by a small
block C. Determine the vertical component of the reaction
(a) at A, (b) at B, (c) at C.
SOLUTION
rB/A = 0.6i
rC/A = 0.8i + 1.05k
rG/A = 0.3i + 0.6k
W = mg = (18 kg)9.81
W = 176.58 N
ΣM A = 0: rB/A × Bj + rC/A × Cj + rG/A × (−Wj) = 0
(0.6i ) × Bj + (0.8i + 1.05k ) × Cj + (0.3i + 0.6k ) × ( −Wj) = 0
0.6 Bk + 0.8Ck − 1.05Ci − 0.3Wk + 0.6Wi = 0
Equate coefficients of unit vectors to zero:
§ 0.6 ·
i : 1.05C + 0.6W = 0 C = ¨
¸176.58 N = 100.90 N
© 1.05 ¹
k : 0.6 B + 0.8C − 0.3W = 0
0.6 B + 0.8(100.90 N) − 0.3(176.58 N) = 0 B = −46.24 N
ΣFy = 0: A + B + C − W = 0
A − 46.24 N + 100.90 N + 176.58 N = 0
A = 121.92 N
(a ) A = 121.9 N (b) B = −46.2 N (c) C = 100.9 N W
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460
PROBLEM 4.102
Solve Problem 4.101, assuming that the small block C is
moved and placed under edge DE at a point 0.15 m from
corner E.
PROBLEM 4.101 An opening in a floor is covered by a
1 × 1.2-m sheet of plywood of mass 18 kg. The sheet is
hinged at A and B and is maintained in a position slightly
above the floor by a small block C. Determine the vertical
component of the reaction (a) at A, (b) at B, (c) at C.
SOLUTION
rB/A = 0.6i
rC/A = 0.65i + 1.2k
rG/A = 0.3i + 0.6k
W = mg = (18 kg) 9.81 m/s 2
W = 176.58 N
ΣM A = 0: rB/A × Bj + rC/A × Cj + rG/A × (−Wj) = 0
0.6i × Bj + (0.65i + 1.2k ) × Cj + (0.3i + 0.6k ) × (−Wj) = 0
0.6 Bk + 0.65Ck − 1.2Ci − 0.3Wk + 0.6Wi = 0
Equate coefficients of unit vectors to zero:
§ 0.6 ·
C =¨
¸176.58 N = 88.29 N
© 1.2 ¹
i : −1.2C + 0.6W = 0
k : 0.6 B + 0.65C − 0.3W = 0
0.6 B + 0.65(88.29 N) − 0.3(176.58 N) = 0 B = −7.36 N
ΣFy = 0: A + B + C − W = 0
A − 7.36 N + 88.29 N − 176.58 N = 0
A = 95.648 N
(a ) A = 95.6 N (b) B = − 7.36 N (c) C = 88.3 N W
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461
PROBLEM 4.103
The rectangular plate shown weighs 80 lb and is supported by three
vertical wires. Determine the tension in each wire.
SOLUTION
Free-Body Diagram:
ΣM B = 0: rA/B × TA j + rC/B × TC j + rG/B × (−80 lb) j = 0
(60 in.)k × TA j + [(60 in.)i + (15 in.)k ] × TC j + [(30 in.)i + (30 in.)k ] × ( −80 lb) j = 0
−60TAi + 60TC k − 15TC i − 2400k + 2400i = 0
Equating to zero the coefficients of the unit vectors,
i:
60TA − 15(40) + 2400 = 0
k:
60TC − 2400 = 0
ΣFy = 0:
TA + TB + TC − 80 lb = 0
30 lb + TB + 40 lb − 80 lb = 0
TA = 30.0 lb W
TC = 40.0 lb W
TB = 10.00 lb W
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462
PROBLEM 4.104
The rectangular plate shown weighs 80 lb and is supported by three
vertical wires. Determine the weight and location of the lightest block
that should be placed on the plate if the tensions in the three wires are
to be equal.
SOLUTION
Free-Body Diagram:
Let −Wb j be the weight of the block and x and z the block’s coordinates.
Since tensions in wires are equal, let
TA = TB = TC = T
ΣM 0 = 0: (rA × Tj) + (rB × Tj) + (rC × Tj) + rG × (−Wj) + ( xi + zk ) × ( −Wb j) = 0
or
(75 k ) × Tj + (15 k ) × Tj + (60i + 30k ) × Tj + (30i + 45k ) × (−Wj) + ( xi + zk ) × (−Wb j) = 0
or
−75T i − 15T i + 60T k − 30T i − 30W k + 45W i − Wb × k + Wb z i = 0
Equate coefficients of unit vectors to zero:
i:
−120T + 45W + Wb z = 0
(1)
k:
60T − 30W − Wb x = 0
(2)
ΣFy = 0:
3T − W − Wb = 0
(3)
Eq. (1) + 40 Eq. (3):
5W + ( z − 40)Wb = 0
(4)
Eq. (2) – 20 Eq. (3):
−10W − ( x − 20)Wb = 0
(5)
Also,
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463
PROBLEM 4.104 (Continued)
Solving Eqs. (4) and (5) for Wb /W and recalling that 0 ≤ x ≤ 60 in., 0 ≤ z ≤ 90 in.,
Eq. (4):
Wb
5
5
=
≥
= 0.125
W 40 − z 40 − 0
Eq. (5):
Wb
10
10
=
≥
= 0.5
W 20 − x 20 − 0
Thus, (Wb ) min = 0.5W = 0.5(80) = 40 lb
Making Wb = 0.5W in Eqs. (4) and (5):
5W + ( z − 40)(0.5W ) = 0
−10W − ( x − 20)(0.5W ) = 0
(Wb ) min = 40.0 lb W
z = 30.0 in. W
x = 0 in. W
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464
PROBLEM 4.105
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:
(ΣMAC = 0).
Five unknowns and six equations of equilibrium, but equilibrium is maintained
rB = 1.2k
TAD
TAE
rA = 2.4k
JJJG
AD = −0.8i + 0.6 j − 2.4k
AD = 2.6 m
JJJG
AE = 0.8i + 1.2 j − 2.4k
AE = 2.8 m
JJJG
AD TAD
=
=
(−0.8i + 0.6 j − 2.4k )
AD 2.6
JJJG
AE TAE
=
=
(0.8i + 1.2 j − 2.4k )
AE 2.8
ΣM C = 0: rA × TAD + rA × TAE + rB × (−3 kN) j = 0
i
j
k
i
j
k
TAD
T
+ 0
0
0
2.4
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 : − 0.55385TAD − 1.02857TAE + 4.32 = 0
(1)
j : − 0.73846TAD + 0.68671TAE = 0
TAD = 0.92857TAE
From Eq. (1):
(2)
−0.55385(0.92857)TAE − 1.02857TAE + 4.32 = 0
1.54286TAE = 4.32
TAE = 2.800 kN
TAE = 2.80 kN W
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465
PROBLEM 4.105 (Continued)
From Eq. (2):
TAD = 2.60 kN W
TAD = 0.92857(2.80) = 2.600 kN
0.8
0.8
(2.6 kN) +
(2.8 kN) = 0
2.6
2.8
0.6
1.2
(2.6 kN) +
(2.8 kN) − (3.6 kN) = 0
ΣFy = 0: C y +
2.6
2.8
2.4
2.4
(2.6 kN) −
(2.8 kN) = 0
ΣFz = 0: C z −
2.6
2.8
ΣFx = 0: C x −
Cx = 0
C y = 1.800 kN
C z = 4.80 kN
C = (1.800 kN) j + (4.80 kN)k W
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466
PROBLEM 4.106
Solve Problem 4.105, assuming that the 3.6-kN load is applied at
Point A.
PROBLEM 4.105 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:
(ΣMAC = 0).
Five unknowns and six equations of equilibrium, but equilibrium is maintained
JJJG
AD = −0.8i + 0.6 j − 2.4k
AD = 2.6 m
JJJG
AE = 0.8i + 1.2 j − 2.4k
AE = 2.8 m
JJJG
AD TAD
(−0.8i + 0.6 j − 2.4k )
TAD =
=
AD 2.6
JJJG
AE TAE
(0.8i + 1.2 j − 2.4k )
TAE =
=
AE 2.8
ΣM C = 0: rA × TAD + rA × TAE + rA × (−3.6 kN) j
Factor rA :
or
Coefficient of i:
rA × (TAD + TAE − (3.6 kN) j)
TAD + TAE − (3 kN) j = 0
−
(Forces concurrent at A)
TAD
T
(0.8) + AE (0.8) = 0
2.6
2.8
TAD =
Coefficient of j:
2.6
TAE
2.8
(1)
TAD
T
(0.6) + AE (1.2) − 3.6 kN = 0
2.6
2.8
2.6
§ 0.6 · 1.2
TAE ¨
TAE − 3.6 kN = 0
¸+
2.8
© 2.6 ¹ 2.8
§ 0.6 + 1.2 ·
TAE ¨
¸ = 3.6 kN
© 2.8 ¹
TAE = 5.600 kN
TAE = 5.60 kN W
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467
PROBLEM 4.106 (Continued)
From Eq. (1):
TAD =
TAD = 5.20 kN W
2.6
(5.6) = 5.200 kN
2.8
0.8
0.8
(5.2 kN) +
(5.6 kN) = 0
2.6
2.8
0.6
1.2
ΣFy = 0: C y +
(5.2 kN) +
(5.6 kN) − 3.6 kN = 0
2.6
2.8
2.4
2.4
ΣFz = 0: Cz −
(5.2 kN) −
(5.6 kN) = 0
2.6
2.8
ΣFx = 0: C x −
Cx = 0
Cy = 0
Cz = 9.60 kN
C = (9.60 kN)k W
Note: Since the forces and reaction are concurrent at A, we could have used the methods of Chapter 2.
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468
PROBLEM 4.107
A 10-ft boom is acted upon by the 840-lb force shown. Determine the
tension in each cable and the reaction at the ball-and-socket joint at A.
SOLUTION
We have five unknowns and six equations of equilibrium, but equilibrium is maintained (ΣM x = 0).
Free-Body Diagram:
JJJG
BD = (−6 ft)i + (7 ft) j + (6 ft)k BD = 11 ft
JJJG
BE = (−6 ft)i + (7 ft) j − (6 ft)k BE = 11 ft
JJJG
BD TBD
TBD = TBD
=
(−6i + 7 j + 6k )
BD 11
JJJG
BE TBE
TBE = TBE
(−6i + 7 j − 6k )
=
BE 11
ΣM A = 0: rB × TBD + rB × TBE + rC × ( −840 j) = 0
6i ×
TBD
T
(−6i + 7 j + 6k ) + 6i × BE (−6i + 7 j − 6k ) + 10i × (−840 j) = 0
11
11
42
36
42
36
TBD k − TBD j + TBE k + TBE j − 8400k = 0
11
11
11
11
Equate coefficients of unit vectors to zero:
i: −
k:
36
36
TBD + TBE = 0 TBE = TBD
11
11
42
42
TBD + TBE − 8400 = 0
11
11
§ 42
·
2 ¨ TBD ¸ = 8400
© 11
¹
TBD = 1100 lb W
TBE = 1100 lb W
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469
PROBLEM 4.107 (Continued)
ΣFx = 0: Ax −
6
6
(1100 lb) − (1100 lb) = 0
11
11
Ax = 1200 lb
ΣFy = 0: Ay +
7
7
(1100 lb) + (1100 lb) − 840 lb = 0
11
11
Ay = −560 lb
ΣFz = 0: Az +
6
6
(1100 lb) − (1100 lb) = 0
11
11
Az = 0
A = (1200 lb)i − (560 lb) j W
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470
PROBLEM 4.108
A 12-m pole supports a horizontal cable CD and is held by a ball
and socket at A and two cables BE and BF. Knowing that the
tension in cable CD is 14 kN and assuming that CD is parallel to
the x-axis (φ = 0), determine the tension in cables BE and BF and
the reaction at A.
SOLUTION
Free-Body Diagram:
There are five unknowns and six equations of equilibrium. The pole is free to rotate about the y-axis, but
equilibrium is maintained under the given loading (ΣM y = 0).
JJJG
JJJG
Resolve BE and BF into components:
JJJG
BE = (7.5 m)i − (8 m) j + (6 m)k
JJJG
BF = (7.5 m)i − (8 m) j − (6 m)k
Express TBE and TBF in terms of components:
TBE = TBE
TBF = TBF
JJJG
BE
= TBE (0.60i − 0.64 j + 0.48k )
BE
JJJG
BF
= TBF (0.60i − 0.64 j − 0.48k )
BF
BE = 12.5 m
BF = 12.5 m
(1)
(2)
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471
PROBLEM 4.108 (Continued)
ΣM A = 0: rB/A × TBE + rB/A × TBF + rC/A × (−14 kN)i = 0
8 j × TBE (0.60i − 0.64 j + 0.48k ) + 8 j × TBF (0.60i − 0.64 j − 0.48k ) + 12 j × (−14i ) = 0
−4.8 TBE k + 3.84 TBE i − 4.8TBF k − 3.84TBF i + 168k = 0
Equating the coefficients of the unit vectors to zero,
i:
3.84TBE − 3.84TBF = 0
k:
−4.8TBE − 4.8TBF + 168 = 0
ΣFx = 0:
Ax + 2(0.60)(17.50 kN) − 14 kN = 0
Ax = 7.00 kN
ΣFy = 0:
Ay − z (0.64) (17.50 kN) = 0
Ay = 22.4 kN
ΣFz = 0:
Az + 0 = 0
TBE = TBF
TBE = TBF = 17.50 kN W
Az = 0
A = −(7.00 kN)i + (22.4 kN) j W
Because of the symmetry, we could have noted at the outset that TBF = TBE and eliminated one unknown.
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472
PROBLEM 4.109
Solve Problem 4.108, assuming that cable CD forms an angle φ = 25°
with the vertical xy plane.
PROBLEM 4.108 A 12-m pole supports a horizontal cable CD and
is held by a ball and socket at A and two cables BE and BF.
Knowing that the tension in cable CD is 14 kN and assuming that
CD is parallel to the x-axis (φ = 0), determine the tension in cables
BE and BF and the reaction at A.
SOLUTION
Free-Body Diagram:
JJJG
BE = (7.5 m)i − (8 m) j + (6 m)k
BE = 12.5 m
JJJG
BF = (7.5 m)i − (8 m)j − (6 m)k
BF = 12.5 m
JJJG
BE
= TBE (0.60i − 0.64 j + 0.48k )
TBE = TBE
BE
JJJG
BF
= TBF (0.60i − 0.64 j − 0.48k )
TBF = TBF
BF
ΣM A = 0: rB/A × TBE + rB/A × TBF + rC/A × TCD = 0
8 j × TBE (0.60i − 0.64 j + 0.48k ) + 8 j × TBF (0.60i − 0.64 j − 0.48k )
+ 12 j × (19 kN)(− cos 25° i + sin 25°k ) = 0
−4.8TBE k + 3.84TBE i − 4.8TBF k − 3.84TBF i + 152.6 k − 71.00 i = 0
Equating the coefficients of the unit vectors to zero,
Solving simultaneously,
i: 3.84TBE − 3.84TBF + 71.00 = 0;
TBF − TBE = 18.4896
k: − 4.8TBE − 4.8 TBF + 152.26 = 0;
TBF + TBE = 31.721
TBE = 6.6157 kN;
TBF = 25.105 kN
TBE = 6.62 kN; TBF = 25.1 kN W
ΣFx = 0: Ax + (0.60) (TBF + TBE ) − 14 cos 25° = 0
Ax = 12.6883 − 0.60(31.7207)
Ax = −6.34 kN
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473
PROBLEM 4.109 (Continued)
ΣFy = 0: Ay − (0.64) (TBF + TBE ) = 0
Ay = 0.64(31.721)
Ay = 20.3 kN
ΣFz = 0: Az − 0.48(TBF − TBE ) + 14sin 25° = 0
Az = 0.48(18.4893) − 5.9167
Az = 2.96 kN
A = −(6.34 kN)i + (20.3 kN) j + (2.96 kN) k W
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474
PROBLEM 4.110
A 48-in. boom is held by a ball-and-socket joint at C and
by two cables BF and DAE; cable DAE passes around a
frictionless pulley at A. For the loading shown, determine
the tension in each cable and the reaction at C.
SOLUTION
Free-Body Diagram:
Five unknowns and six equations of equilibrium, but equilibrium is
maintained (ΣMAC = 0).
T = Tension in both parts of cable DAE.
rB = 30k
rA = 48k
JJJG
AD = −20i − 48k
AD = 52 in.
JJJG
AE = 20 j − 48k
AE = 52 in.
JJJG
BF = 16i − 30k
BF = 34 in.
JJJG
AD T
T
TAD = T
= (−20i − 48k ) = ( −5i − 12k )
AD 52
13
JJJG
AE T
T
TAE = T
= (20 j − 48k ) = (5 j − 12k )
AE 52
13
JJJG
T
BF TBF
TBF = TBF
=
(16i − 30k ) = BF (8i − 15k )
BF
34
17
ΣM C = 0: rA × TAD + rA × TAE + rB × TBF + rB × (−320 lb) j = 0
i j k
i j k
i j k
T
T
T
+ 0 0 48
+ 0 0 30 BF + (30k ) × (−320 j) = 0
0 0 48
13
13
17
−5 0 −12
0 5 −12
8 0 −15
Coefficient of i:
−
240
T + 9600 = 0
13
T = 520 lb
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475
PROBLEM 4.110 (Continued)
Coefficient of j:
−
240
240
T+
TBD = 0
13
17
TBD =
17
17
T = (520) TBD = 680 lb
13
13
ΣF = 0: TAD + TAE + TBF − 320 j + C = 0
Coefficient of i:
−
20
8
(520) + (680) + Cx = 0
52
17
−200 + 320 + Cx = 0
Coefficient of j:
20
(520) − 320 + C y = 0
52
200 − 320 + C y = 0
Coefficient of k:
Cx = −120 lb
−
C y = 120 lb
48
48
30
(520) − (520) − (680) + Cz = 0
52
52
34
−480 − 480 − 600 + Cz = 0
Cz = 1560 lb
Answers: TDAE = T
TDAE = 520 lb W
TBD = 680 lb W
C = −(120.0 lb)i + (120.0 lb) j + (1560 lb)k W
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476
PROBLEM 4.111
Solve Problem 4.110, assuming that the 320-lb load is
applied at A.
PROBLEM 4.110 A 48-in. boom is held by a ball-andsocket joint at C and by two cables BF and DAE; cable
DAE passes around a frictionless pulley at A. For the
loading shown, determine the tension in each cable and
the reaction at C.
SOLUTION
Free-Body Diagram:
Five unknowns and six equations of equilibrium, but equilibrium is
maintained (ΣMAC = 0).
T = tension in both parts of cable DAE.
rB = 30k
rA = 48k
JJJG
AD = −20i − 48k
AD = 52 in.
JJJG
AE = 20 j − 48k
AE = 52 in.
JJJG
BF = 16i − 30k
BF = 34 in.
JJJG
AD T
T
TAD = T
= (−20i − 48k ) = ( −5i − 12k )
AD 52
13
JJJG
AE T
T
TAE = T
= (20 j − 48k ) = (5 j − 12k )
AE 52
13
JJJG
T
BF TBF
TBF = TBF
=
(16i − 30k ) = BF (8i − 15k )
BF
34
17
ΣM C = 0: rA × TAD + rA × TAE + rB × TBF + rA × ( −320 lb) j = 0
i j k
i j k
i j k
T
T
T
+ 0 0 48
+ 0 0 30 BF + 48k × (−320 j) = 0
0 0 48
13
13
17
−5 0 −12
0 5 −12
8 0 −15
Coefficient of i:
−
240
T + 15,360 = 0
13
T = 832 lb
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477
PROBLEM 4.111 (Continued)
Coefficient of j:
−
240
240
T+
TBD = 0
13
17
TBD =
17
17
T = (832)
13
13
TBD = 1088 lb
ΣF = 0: TAD + TAE + TBF − 320 j + C = 0
−
Coefficient of i:
20
8
(832) + (1088) + C x = 0
52
17
−320 + 512 + Cx = 0
20
(832) − 320 + C y = 0
52
Coefficient of j:
320 − 320 + C y = 0
Coefficient of k:
−
Cy = 0
48
48
30
(832) − (852) − (1088) + Cz = 0
52
52
34
−768 − 768 − 960 + Cz = 0
Answers:
Cx = −192 lb
TDAE = T
Cz = 2496 lb
TDAE = 832 lb W
TBD = 1088 lb W
C = −(192.0 lb)i + (2496 lb)k W
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478
PROBLEM 4.112
A 600-lb crate hangs from a cable that passes over a
pulley B and is attached to a support at H. The 200-lb
boom AB is supported by a ball-and-socket joint at A
and by two cables DE and DF. The center of gravity
of the boom is located at G. Determine (a) the tension
in cables DE and DF, (b) the reaction at A.
SOLUTION
Free-Body Diagram:
WC = 600 lb
WG = 200 lb
We have five unknowns (TDE , TDF , Ax , Ay , Az ) and five equilibrium equations. The boom is free to spin about
the AB axis, but equilibrium is maintained, since ΣM AB = 0.
JJJJK
BH = (30 ft)i − (22.5 ft) j
BH = 37.5 ft
We have
JJJK
8.8
DE = (13.8 ft)i −
(22.5 ft) j + (6.6 ft)k
12
= (13.8 ft)i − (16.5 ft) j + (6.6 ft)k
DE = 22.5 ft
JJJK
DF = (13.8 ft)i − (16.5 ft) j − (6.6 ft)k
DF = 22.5 ft
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479
PROBLEM 4.112 (Continued)
Thus:
TBH = TBH
TDE = TDE
TDF = TDF
(a)
JJJJK
BH
30i − 22.5 j
= (600 lb)
= (480 lb)i − (360 lb) j
BH
37.5
JJJK
DE TDE
=
(13.8i − 16.5 j + 6.6k )
DE 22.5
JJJK
DF TDE
=
(13.8i − 16.5 j − 6.6k )
DF 22.5
ΣM A = 0: (rJ × WC ) + (rK × WG ) + (rH × TBH ) + (rE × TDE ) + (rF × TDF ) = 0
− (12i ) × (−600 j) − (6i ) × (−200 j) + (18i ) × (480i − 360 j)
i
j
k
i
j
k
TDE
TDF
5
0
6.6 +
5
0
+
−6.6 = 0
22.5
22.5
13.8 −16.5 6.6
13.8 −16.5 −6.6
7200k + 1200k − 6480k + 4.84(TDE − TDF )i
or
+
58.08
82.5
(TDE − TDF ) j −
(TDE + TDF )k = 0
22.5
22.5
Equating to zero the coefficients of the unit vectors,
i or j:
TDE − TDF = 0
k : 7200 + 1200 − 6480 −
(b)
TDE = TDF *
82.5
(2TDE ) = 0
22.5
TDE = 261.82 lb
§ 13.8 ·
ΣFx = 0: Ax + 480 + 2 ¨
¸ (261.82) = 0
© 22.5 ¹
§ 16.5 ·
ΣFy = 0: Ay − 600 − 200 − 360 − 2 ¨
¸ (261.82) = 0
© 22.5 ¹
ΣFz = 0: Az = 0
TDE = TDF = 262 lb W
Ax = −801.17 lb
Ay = 1544.00 lb
A = −(801 lb)i + (1544 lb) j W
*Remark: The fact is that TDE = TDF could have been noted at the outset from the symmetry of structure with
respect to xy plane.
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480
PROBLEM 4.113
A 100-kg uniform rectangular plate is supported in the position
shown by hinges A and B and by cable DCE that passes over a
frictionless hook at C. Assuming that the tension is the same in
both parts of the cable, determine (a) the tension in the cable,
(b) the reactions at A and B. Assume that the hinge at B does
not exert any axial thrust.
SOLUTION
rB/A (960 − 180)i = 780i
Dimensions in mm
§ 960
· 450
rG/A = ¨
k
− 90 ¸ i +
2
© 2
¹
= 390i + 225k
rC/A = 600i + 450k
T = Tension in cable DCE
JJJG
CD = −690i + 675 j − 450k
JJJG
CE = 270i + 675 j − 450k
CD = 1065 mm
CE = 855 mm
T
(−690i + 675 j − 450k )
1065
T
(270i + 675 j − 450k )
TCE =
855
W = −mgi = −(100 kg)(9.81 m/s 2 ) j = −(981 N) j
TCD =
ΣM A = 0: rC/A × TCD + rC/A × TCE + rG/A × (−Wj) + rB/A × B = 0
i
j
k
i
j
k
T
T
600
0
450
450
+ 600 0
1065
855
270 675 −450
−690 675 −450
i
+ 390
0
j
0
−981
k
i
225 + 780
0
0
j
0
k
0 =0
By
Bz
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481
PROBLEM 4.113 (Continued)
Coefficient of i:
−(450)(675)
T
T
− (450)(675)
+ 220.73 × 103 = 0
1065
855
T = 345 N W
T = 344.64 N
Coefficient of j:
(−690 × 450 + 600 × 450)
344.64
344.64
+ (270 × 450 + 600 × 450)
− 780 Bz = 0
1065
855
Bz = 185.516 N
Coefficient of k: (600)(675)
344.64
344.64
+ (600)(675)
− 382.59 × 103 + 780 By = 0 By = 113.178 N
1065
855
B = (113.2 N) j + (185.5 N)k W
ΣF = 0: A + B + TCD + TCE + W = 0
690
270
(344.64) +
(344.64) = 0
1065
855
Coefficient of i:
Ax −
Coefficient of j:
Ay + 113.178 +
675
675
(344.64) +
(344.64) − 981 = 0
1065
855
Ay = 377 N
Coefficient of k:
Az + 185.516 −
450
450
(344.64) −
(344.64) = 0
1065
855
Az = 141.5 N
Ax = 114.5 N
A = (114.5 N)i + (377 N) j + (144.5 N)k W
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482
PROBLEM 4.114
Solve Problem 4.113, assuming that cable DCE is replaced
by a cable attached to Point E and hook C.
PROBLEM 4.113 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 4.113 for free-body diagram and analysis leading to the following:
CD = 1065 mm
CE = 855 mm
T
(−690i + 675 j − 450k )
1065
T
(270i + 675 j − 450k )
TCE =
855
W = −mgi = −(100 kg)(9.81 m/s 2 ) j = −(981 N)j
Now,
TCD =
ΣM A = 0: rC/A × TCE + rG/A × (−W j) + rB/A × B = 0
i
j
k
i
j
k
i
j
T
600 0
450
0
225 + 780 0
+ 390
855
270 675 −450
0 −981 0
0 By
Coefficient of i:
−(450)(675)
k
0 =0
Bz
T
+ 220.73 × 103 = 0
855
T = 621 N W
T = 621.31 N
Coefficient of j:
Coefficient of k:
(270 × 450 + 600 × 450)
(600)(675)
621.31
− 780 Bz = 0 Bz = 364.74 N
855
621.31
− 382.59 × 103 + 780 By = 0 By = 113.186 N
855
B = (113.2 N)j + (365 N)k W
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483
PROBLEM 4.114 (Continued)
ΣF = 0: A + B + TCE + W = 0
270
(621.31) = 0
855
Coefficient of i:
Ax +
Coefficient of j:
Ay + 113.186 +
Coefficient of k:
Az + 364.74 −
Ax = −196.2 N
675
(621.31) − 981 = 0
855
450
(621.31) = 0
855
Ay = 377.3 N
Az = −37.7 N A = −(196.2 N)i + (377 N)j − (37.7 N)k W
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484
PROBLEM 4.115
The rectangular plate shown weighs 75 lb and is held in the
position shown by hinges at A and B and by cable EF. Assuming
that the hinge at B does not exert any axial thrust, determine
(a) the tension in the cable, (b) the reactions at A and B.
SOLUTION
rB/A = (38 − 8)i = 30i
rE/A = (30 − 4)i + 20k
rG/A
= 26i + 20k
38
= i + 10k
2
= 19i + 10k
JJJG
EF = 8i + 25 j − 20k
EF = 33 in.
JJJG
AE T
T =T
= (8i + 25 j − 20k )
AE 33
ΣM A = 0: rE/A × T + rG/A × (−75 j) + rB/A × B = 0
i
j
k
i
j
k
i
j
T
+ 19 0 10 + 30 0
26 0 20
33
8 25 −20
0 −75 0
0 By
−(25)(20)
Coefficient of i:
Coefficient of j:
Coefficient of k:
(160 + 520)
(26)(25)
T
+ 750 = 0:
33
k
0 =0
Bz
T = 49.5 lb W
49.5
− 30 Bz = 0: Bz = 34 lb
33
49.5
− 1425 + 30 By = 0: By = 15 lb
33
B = (15.00 lb)j + (34.0 lb)k W
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485
PROBLEM 4.115 (Continued)
ΣF = 0: A + B + T − (75 lb)j = 0 Coefficient of i:
Coefficient of j:
Coefficient of k:
Ax +
Ay + 15 +
8
(49.5) = 0
33
25
(49.5) − 75 = 0
33
Az + 34 −
20
(49.5) = 0
33
Ax = −12.00 lb
Ay = 22.5 lb
Az = −4.00 lb A = −(12.00 lb)i + (22.5 lb)j − (4.00 lb)k W
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486
PROBLEM 4.116
Solve Problem 4.115, assuming that cable EF is replaced by a
cable attached at points E and H.
PROBLEM 4.115 The rectangular plate shown weighs 75 lb and
is held in the position shown by hinges at A and B and by cable
EF. Assuming that the hinge at B does not exert any axial thrust,
determine (a) the tension in the cable, (b) the reactions at A and B.
SOLUTION
rB/A = (38 − 8)i = 30i
rE/A = (30 − 4)i + 20k
= 26i + 20k
38
i + 10k
2
= 19i + 10k
rG/A =
JJJJG
EH = −30i + 12 j − 20k
EH = 38 in.
JJJJG
EH T
T=T
= (−30i + 12 j − 20k )
EH 38
ΣM A = 0: rE/A × T + rG/A × (−75 j) + rB/A × B = 0
i
j
k
i
j
k
i
j
T
+ 19 0 10 + 30 0
26 0 20
38
−30 12 −20
0 −75 0
0 By
−(12)(20)
Coefficient of i:
Coefficient of j:
Coefficient of k:
(−600 + 520)
(26)(12)
T
+ 750 = 0
38
T = 118.75
k
0 =0
Bz
T = 118.8lb W
118.75
− 30 Bz = 0 Bz = −8.33lb
38
118.75
− 1425 + 30 By = 0 B y = 15.00 lb
38
B = (15.00 lb)j − (8.33 lb)k W
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487
PROBLEM 4.116 (Continued)
ΣF = 0:
Coefficient of j:
Coefficient of k:
30
(118.75) = 0
38
Ax = 93.75 lb
12
(118.75) − 75 = 0
38
Ay = 22.5 lb
Ax −
Coefficient of i:
Ay + 15 +
A + B + T − (75 lb)j = 0
Az − 8.33 −
20
(118.75) = 0
38
Az = 70.83 lb
A = (93.8 lb)i + (22.5 lb)j + (70.8 lb)k W
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488
PROBLEM 4.117
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 CE:
F = F (cos 75°)i + F (sin 75°) j
F = F (0.25882i + 0.96593j)
W = mg = 20 kg(9.81 m/s 2 ) = 196.2 N
rA /B = 0.6k
rC /B = 0.9i + 0.6k
rG /B = 0.45i + 0.3k
F = F (0.25882i + 0.96593j)
ΣM B = 0: rG/B × (−196.2 j) + rC/B × F + rA/B × A = 0
(a)
i
j
k
i
j
k
i
0.45
0
0.3 + 0.9
0
0.6 F + 0
0
−196.2 0
0.25882 +0.96593 0
Ax
j
0
Ay
k
0.6 = 0
0
Coefficient of i :
+58.86 − 0.57956 F − 0.6 Ay = 0
(1)
Coefficient of j:
+0.155292 F + 0.6 Ax = 0
(2)
Coefficient of k:
−88.29 + 0.86934 F = 0:
From Eq. (2):
+58.86 − 0.57956(101.56) − 0.6 Ay = 0
From Eq. (3):
+0.155292(101.56) + 0.6 Ax = 0
(b)
Coefficient of i:
Coefficient of j:
F = 101.56 N
Ay = 0
Ax = −26.286 N
F = (101.6 N) W
ΣF : A + B + F − Wj = 0
26.286 + Bx + 0.25882(101.56) = 0 Bx = 0
B y + 0.96593(101.56) − 196.2 = 0 B y = 98.1 N
Coefficient of k:
Bz = 0
A = −(26.3 N)i; B = (98.1 N) j W
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489
PROBLEM 4.118
The bent rod ABEF is supported by bearings at C and D and by wire
AH. Knowing that portion AB of the rod is 250 mm long, determine
(a) the tension in wire AH, (b) the reactions at C and D. Assume that
the bearing at D does not exert any axial thrust.
SOLUTION
∆ABH is equilateral.
Free-Body Diagram:
Dimensions in mm
rH/C = −50i + 250 j
rD/C = 300i
rF/C = 350i + 250k
T = T (sin 30°) j − T (cos 30°)k = T (0.5 j − 0.866k )
ΣM C = 0: rH/C × T + rD × D + rF/C × (−400 j) = 0
i
j
k
i
j
−50 250
0 T + 300 0
0
0.5 −0.866
0 Dy
Coefficient i:
k
i
j
k
0 + 350
0
250 = 0
Dz
0 −400 0
−216.5T + 100 × 103 = 0
T = 461.9 N
Coefficient of j:
T = 462 N W
−43.3T − 300 Dz = 0
−43.3(461.9) − 300 Dz = 0
Dz = −66.67 N
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490
PROBLEM 4.118 (Continued)
Coefficient of k:
−25T + 300 D y − 140 × 103 = 0
−25(461.9) + 300 Dy − 140 × 103 = 0
D y = 505.1 N
D = (505 N) j − (66.7 N)k W
ΣF = 0: C + D + T − 400 j = 0
Coefficient i:
Cx = 0
Coefficient j:
C y + (461.9)0.5 + 505.1 − 400 = 0 C y = −336 N
Coefficient k:
Cz − (461.9)0.866 − 66.67 = 0
Cx = 0
Cz = 467 N
C = −(336 N) j + (467 N)k W
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491
PROBLEM 4.119
Solve Problem 4.115, 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 4.115 The rectangular plate shown weighs 75 lb and
is held in the position shown by hinges at A and B and by cable
EF. Assuming that the hinge at B does not exert any axial thrust,
determine (a) the tension in the cable, (b) the reactions at A and B.
SOLUTION
rE/A = (30 − 4)i + 20k = 26i + 20k
rG/A = (0.5 × 38)i + 10k = 19i + 10k
JJJG
AE = 8i + 25 j − 20k
AE = 33 in.
JJJG
AE T
T =T
= (8i + 25 j − 20k )
AE 33
ΣM A = 0: rE/A × T + rG/A × (−75 j) + ( M A ) y j + ( M A ) z k = 0
i
j
k
i
j
k
T
+ 19 0 10 + ( M A ) y j + ( M A ) z k = 0
26 0 20
33
8 25 −20
0 −75 0
−(20)(25)
Coefficient of i:
Coefficient of j:
Coefficient of k:
(160 + 520)
(26)(25)
T = 49.5 lb W
T
+ 750 = 0
33
49.5
+ ( M A ) y = 0 ( M A ) y = −1020 lb ⋅ in.
33
49.5
− 1425 + ( M A ) z = 0
33
( M A ) z = 450 lb ⋅ in.
ΣF = 0: A + T − 75 j = 0
Coefficient of i:
Ax +
8
(49.5) = 0
33
M A = −(1020 lb ⋅ in.)j + (450 lb ⋅ in.)k W
Ax = 12.00 lb
Coefficient of j:
Ay +
25
(49.5) − 75 = 0
33
Ay = 37.5 lb
Coefficient of k:
Az −
20
(49.5) = 0
33
Az = 30.0 lb
A = −(12.00 lb)i + (37.5 lb)j + (30.0 lb)k W
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492
PROBLEM 4.120
Solve Problem 4.118, 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 4.118 The bent rod ABEF is supported by bearings
at C and D and by wire AH. Knowing that portion AB of the rod
is 250 mm long, determine (a) the tension in wire AH, (b) the
reactions at C and D. Assume that the bearing at D does not exert
any axial thrust.
SOLUTION
Free-Body Diagram:
∆ABH is equilateral.
Dimensions in mm
rH/C = −50i + 250 j
rF/C = 350i + 250k
T = T (sin 30°) j − T (cos 30°)k = T (0.5 j − 0.866k )
ΣM C = 0: rF/C × (−400 j) + rH/C × T + ( M C ) y j + ( M C ) z k = 0
i
j
k
i
j
k
350
0
250 + −50 250
0 T + (M C ) y j + (M C ) z k = 0
0 −400 0
0
0.5 −0.866
Coefficient of i:
Coefficient of j:
+100 × 103 − 216.5T = 0 T = 461.9 N
T = 462 N W
−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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493
PROBLEM 4.120 (Continued)
Coefficient of k:
−140 × 103 − 25(461.9) + ( M C ) z = 0
( M C ) z = 151.54 × 103 N ⋅ mm
(M C ) z = 151.5 N ⋅ m
ΣF = 0: C + T − 400 j = 0
M C = (20.0 N ⋅ m)j + (151.5 N ⋅ m)k W
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 W
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494
PROBLEM 4.121
The assembly shown is welded to collar A that fits on the vertical
pin shown. The pin can exert couples about the x and z axes but
does not prevent motion about or along the y-axis. For the loading
shown, determine the tension in each cable and the reaction at A.
SOLUTION
Free-Body Diagram:
First note:
−(0.08 m)i + (0.06 m) j
TCF = λ CF TCF =
(0.08) 2 + (0.06)2 m
TCF
= TCF (−0.8i + 0.6 j)
TDE = λ DE TDE =
(0.12 m)j − (0.09 m)k
(0.12) 2 + (0.09)2 m
TDE
= TDE (0.8 j − 0.6k )
(a)
From F.B.D. of assembly:
ΣFy = 0: 0.6TCF + 0.8TDE − 480 N = 0
or
(1)
0.6TCF + 0.8TDE = 480 N
ΣM y = 0: − (0.8TCF )(0.135 m) + (0.6TDE )(0.08 m) = 0
or
TDE = 2.25TCF
(2)
Substituting Equation (2) into Equation (1),
0.6TCF + 0.8[(2.25)TCF ] = 480 N
TCF = 200.00 N
or
and from Equation (2):
TDE = 2.25(200.00 N) = 450.00
or
TCF = 200 N W
TDE = 450 N W
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495
PROBLEM 4.121 (Continued)
(b)
From F.B.D. of assembly:
ΣFz = 0: Az − (0.6)(450.00 N) = 0
Az = 270.00 N
ΣFx = 0: Ax − (0.8)(200.00 N) = 0
Ax = 160.000 N
ΣM x = 0:
or A = (160.0 N)i + (270 N)k W
MAx + (480 N)(0.135 m) − [(200.00 N)(0.6)](0.135 m)
− [(450 N)(0.8)](0.09 m) = 0
M Ax = −16.2000 N ⋅ m
ΣM z = 0:
MAz − (480 N)(0.08 m) + [(200.00 N)(0.6)](0.08 m)
+ [(450 N)(0.8)](0.08 m) = 0
M Az = 0
or M A = −(16.20 N ⋅ m)i W
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496
PROBLEM 4.122
The assembly shown is used to control the tension T in a tape that
passes around a frictionless spool at E. Collar C is welded to rods ABC
and CDE. It can rotate about shaft FG but its motion along the shaft is
prevented by a washer S. For the loading shown, determine (a) the
tension T in the tape, (b) the reaction at C.
SOLUTION
Free-Body Diagram:
rA/C = 4.2 j + 2k
rE/C = 1.6i − 2.4 j
ΣM C = 0: rA/C × (−6 j) + rE/C × T (i + k ) + ( M C ) y j + ( M C ) z k = 0
(4.2 j + 2k ) × (−6 j) + (1.6i − 2.4 j) × T (i + k ) + ( M C ) y j + ( M C ) z k = 0
Coefficient of i:
Coefficient of j:
Coefficient of k:
T = 5.00 lb W
12 − 2.4T = 0
−1.6(5 lb) + (M C ) y = 0 ( M C ) y = 8 lb ⋅ in.
2.4(5 lb) + ( M C ) z = 0 ( M C ) z = −12 lb ⋅ in.
M C = (8.00 lb ⋅ in.)j − (12.00 lb ⋅ in.)k W
ΣF = 0: Cx i + C y j + Cz k − (6 lb)j + (5 lb)i + (5 lb)k = 0
Equate coefficients of unit vectors to zero.
Cx = −5 lb C y = 6 lb Cz = −5 lb W
C = −(5.00 lb)i + (6.00 lb)j − (5.00 lb)k W
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497
PROBLEM 4.123
The rigid L-shaped member ABF is supported by a ball-and-socket
joint at A and by three cables. For the loading shown, determine the
tension in each cable and the reaction at A.
SOLUTION
Free-Body Diagram:
rB/A = 12i
rF/A = 12 j − 8k
rD/A = 12i − 16k
rE/A = 12i − 24k
rF/A = 12i − 32k
JJJG
BG = −12i + 9k
BG = 15 in.
λ BG = −0.8i + 0.6k
JJJJG
DH = −12i + 16 j; DH = 20 in.; λDH = −0.6i + 0.8 j
JJJG
FJ = −12i + 16 j; FJ = 20 in.; λFJ = −0.6i + 0.8 j
ΣM A = 0: rB/A × TBG λBG + rDH × TDH λDH + rF/A × TFJ λFJ
+rF/A × (−24 j) + rE/A × ( −24 j) = 0
i
j k
i
j
k
i
j
k
12 0 0 TBG + 12
0 −16 TDH + 12
0 −32 TFJ
−0.8 0 0.6
−0.6 0.8 0
−0.6 0.8 0
i
j
k
i
j
k
+ 12 0 −8 + 12 0 −24 = 0
0 −24 0
0 −24 0
Coefficient of i:
+12.8TDH + 25.6TFJ − 192 − 576 = 0
(1)
Coefficient of k:
+9.6TDH + 9.6TFJ − 288 − 288 = 0
(2)
3
4
Eq. (1) − Eq. (2):
9.6TFJ = 0
TFJ = 0 W
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498
PROBLEM 4.123 (Continued)
TDH = 60 lb W
12.8TDH − 268 = 0
From Eq. (1):
TBG = 80.0 lb W
−7.2TBG + (16 × 0.6)(60.0 lb) = 0
Coefficient of j:
ΣF = 0:
A + TBG λ BG + TDH λ DH + TFJ − 24 j − 24 j = 0
Coefficient of i:
Ax + (80)( −0.8) + (60.0)(−0.6) = 0
Coefficient of j:
Ay + (60.0)(0.8) − 24 − 24 = 0
Coefficient of k:
Az + (80.0)(+0.6) = 0
Ax = 100.0 lb
Ay = 0
Az = −48.0 lb
A = (100.0 lb)i − (48.0 lb) j W
Note: The value Ay = 0 can be confirmed by considering ΣM BF = 0.
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499
PROBLEM 4.124
Solve Problem 4.123, assuming that the load at C has been removed.
PROBLEM 4.123 The rigid L-shaped member ABF is supported
by a ball-and-socket joint at A and by three cables. For the loading
shown, determine the tension in each cable and the reaction at A.
SOLUTION
Free-Body Diagram:
rB/A = 12i
rB/A = 12i − 16k
rE/A = 12i − 24k
rF /A = 12i − 32k
JJJG
BG = −12i + 9k ; BG = 15 in.; λ BG = −0.8i + 0.6k
JJJJG
DH = −12i + 16 j; DH = 20 in.; λ DH = −0.6i + 0.8 j
JJJG
FJ = −12i + 16 j; FJ = 20 in.; λ FJ = −0.6i + 0.8 j
ΣM A = 0: rB/A × TBG λ BG + rD/A × TDH λ DH + rF /A × TFJ λ FJ + rE/A × (−24 j) = 0
i
j k
i
j
k
i
j
k
i
j
k
12 0 0 TBG + 12
0 −16 TDH + 12
0 −32 TFJ + 12 0 −24 = 0
0 −24 0
−0.8 0 0.6
−0.6 0.8 0
−0.6 0.8 0
i:
+ 12.8TDH + 25.6TFJ − 576 = 0
(1)
k:
+9.6TDH + 9.6TFJ − 288 = 0
(2)
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500
PROBLEM 4.124 (Continued)
Multiply Eq. (1) by
3
4
and subtract Eq. (2):
TFJ = 15.00 lb W
9.6TFJ − 144 = 0
TDH = 15.00 lb W
12.8TDH + 25.6(15.00) − 576 = 0
From Eq. (1):
j:
−7.2TBG + (16)(0.6)(15) + (32)(0.6)(15) = 0
TBG = 60.0 lb W
− 7.2TBG + 432 = 0
ΣF = 0: A + TBG λ BG + TDAλ DH + TFJ λ FJ − 24 j = 0
i : Ax + (60)(−0.8) + (15)( −0.6) + (15)(−0.6) = 0
j:
Ay + (15)(0.8) + (15)(0.8) − 24 = 0
k:
Az + (60)(0.6) = 0
Ax = 66.0 lb
Ay = 0
Az = −36.0 lb
A = (66.0 lb)i − (36.0 lb)k W
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501
PROBLEM 4.125
The rigid L-shaped member ABC is supported by
a ball-and-socket joint at A and by three cables. If
a 1.8-kN load is applied at F, determine the tension in
each cable.
SOLUTION
Free-Body Diagram:
In this problem:
We have
Thus,
Dimensions in mm
a = 210 mm
JJJG
CD = (240 mm)j − (320 mm)k CD = 400 mm
JJJG
BD = −(420 mm)i + (240 mm)j − (320 mm)k BD = 580 mm
JJJG
BE = (420 mm)i − (320 mm)k BE = 528.02 mm
TCD
TBD
TBE
JJJG
CD
= TCD
= TCD (0.6 j − 0.8k )
CD
JJJG
BD
= TBD
= TBD (−0.72414i + 0.41379 j − 0.55172k )
BD
JJJG
BE
= TBE
= TBE (0.79542i − 0.60604k )
BE
ΣM A = 0: (rC × TCD ) + (rB × TBD ) + (rB × TBE ) + (rW × W) = 0
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502
PROBLEM 4.125 (Continued)
rC = −(420 mm)i + (320 mm)k
Noting that
rB = (320 mm)k
rW = − ai + (320 mm)k
and using determinants, we write
i
j
k
i
j
k
−420 0 320 TCD +
0
0
320 TBD
0
0.6 −0.8
−0.72414 0.41379 −0.55172
i
j
k
i
j
k
+
0
0
320 TBE + −a
0
320 = 0
0.79542 0 −0.60604
0 −1.8 0
Equating to zero the coefficients of the unit vectors,
i:
−192TCD − 132.413TBD + 576 = 0
(1)
j:
−336TCD − 231.72TBD + 254.53TBE = 0
(2)
k:
−252TCD + 1.8a = 0
(3)
Recalling that a = 210 mm, Eq. (3) yields
TCD =
From Eq. (1):
1.8(210)
= 1.500 kN
252
−192(1.5) − 132.413TBD + 576 = 0
TBD = 2.1751 kN
From Eq. (2):
−336(1.5) − 231.72(2.1751) + 254.53TBE = 0
TBE = 3.9603 kN TCD = 1.500 kN W
TBD = 2.18 kN W
TBE = 3.96 kN W
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503
PROBLEM 4.126
Solve Problem 4.125, assuming that the 1.8-kN load is
applied at C.
PROBLEM 4.125 The rigid L-shaped member ABC is
supported by a ball-and-socket joint at A and by three
cables. If a 1.8-kN load is applied at F, determine the
tension in each cable.
SOLUTION
See solution of Problem 4.125 for free-body diagram and derivation of Eqs. (1), (2), and (3):
−192TCD − 132.413TBD + 576 = 0
(1)
−336TCD − 231.72TBD + 254.53TBE = 0
(2)
−252TCD + 1.8a = 0
(3)
In this problem, the 1.8-kN load is applied at C and we have a = 420 mm. Carrying into Eq. (3) and
solving for TCD ,
TCD = 3.00
From Eq. (1):
From Eq. (2):
−(192)(3) − 132.413TBD + 576 = 0
−336(3) − 0 + 254.53TBE = 0 TCD = 3.00 kN W
TBD = 0 W
TBE = 3.96 kN W
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504
PROBLEM 4.127
The assembly shown consists of an 80-mm rod
AF that is welded to a cross consisting of four
200-mm arms. The assembly is supported by a
ball-and-socket joint at F and by three short
links, each of which forms an angle of 45° with
the vertical. For the loading shown, determine
(a) the tension in each link, (b) the reaction at F.
SOLUTION
rE/F = −200 i + 80 j
TB = TB (i − j) / 2
rB/F = 80 j − 200k
TC = TC (− j + k ) / 2 rC/F = 200i + 80 j
TD = TD (− i + j) / 2
rD/E = 80 j + 200k
ΣM F = 0: rB/F × TB + rC/F × TC + rD/F × TD + rE/F × (− Pj) = 0
i j
k
i
j k
i
j
k
i
j k
TC
TB
TD
0 80 −200
+ 200 80 0
+ 0 80 200
+ −200 80 0 = 0
2
2
2
1 −1
0
0 −1 1
0
−1 −1 0
−P 0
Equate coefficients of unit vectors to zero and multiply each equation by 2.
i:
−200 TB + 80 TC + 200 TD = 0
(1)
j:
−200 TB − 200 TC − 200 TD = 0
(2)
k:
−80 TB − 200 TC + 80 TD + 200 2 P = 0
(3)
−80 TB − 80 TC − 80 TD = 0
(4)
80
(2):
200
Eqs. (3) + (4):
−160TB − 280TC + 200 2 P = 0
Eqs. (1) + (2):
−400TB − 120TC = 0
TB = −
(5)
120
TC − 0.3TC
400
(6)
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505
PROBLEM 4.127 (Continued)
Eqs. (6)
−160( −0.3TC ) − 280TC + 200 2 P = 0
(5):
−232TC + 200 2 P = 0
TC = 1.219 P W
TC = 1.2191P
From Eq. (6):
TB = −0.3(1.2191P) = − 0.36574 = P
From Eq. (2):
− 200(− 0.3657 P) − 200(1.2191P) − 200Tθ D = 0
TD = − 0.8534 P
ΣF = 0:
TB = −0.366 P W
TD = − 0.853P W
F + TB + TC + TD − Pj = 0
i : Fx +
(− 0.36574 P)
−
2
k : Fz +
Fx = − 0.345P
(− 0.36574 P)
−
2
Fy = P
=0
2
Fx = − 0.3448P
j: Fy −
( − 0.8534 P)
(1.2191P)
−
(− 0.8534 P)
2
2
− 200 = 0
Fy = P
(1.2191P)
2
=0
F = − 0.345P i + Pj − 0.862Pk W
Fz = − 0.8620 P Fz = − 0.862 P
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506
PROBLEM 4.128
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 (ΣMAB = 0).
W = mg
= (10 kg) 9.81m/s 2
W = 98.1 N
JJJG
GC = − 300i + 200 j − 225k GC = 425 mm
JJJG
GC
T
T =T
=
(− 300i + 200 j − 225k )
GC 425
rB/ A = − 600i + 400 j + 150 mm
rG/ A = − 300i + 200 j + 75 mm
ΣMA = 0: rB/ A × B + rG/ A × T + rG/ A × (− W j) = 0
i
j
k
i
j
k
i
j
k
T
75
− 600 400 150 + − 300 200
+ − 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 W
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507
PROBLEM 4.128 (Continued)
Coefficient of j : 150 B − (300 × 75 + 300 × 225)
52.12
=0
425
B = 73.58 N
B = (73.6 N)i W
ΣF = 0: A + B + T − W j = 0
Coefficient of i :
Coefficient of j:
Coefficient of k :
Ax + 73.58 − 52.15
Ay + 52.15
Ax = −36.8 N W
300
=0
425
Ay = 73.6 N W
200
− 98.1 = 0
425
Az − 52.15
Az = 27.6 N W
225
=0
425
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508
PROBLEM 4.129
Three rods are welded together to form a “corner” that is supported
by three eyebolts. Neglecting friction, determine the reactions at
A, B, and C when P = 240 lb, a = 12 in., b = 8 in., and c = 10 in.
SOLUTION
From F.B.D. of weldment:
ΣM O = 0: rA/O × A + rB/O × B + rC/O × C = 0
i
12
0
j
0
Ay
k
i
0 + 0
Az Bx
j k
i
8 0 + 0
0 Bz
Cx
j k
0 10 = 0
Cy 0
(−12 Az j + 12 Ay k ) + (8 Bz i − 8 Bx k ) + (−10 C y i + 10 Cx j) = 0
From i-coefficient:
or
Bz = 1.25C y
j-coefficient:
or
or
(2)
12 Ay − 8 Bx = 0
Bx = 1.5 Ay
ΣF = 0: A + B + C − P = 0 (3)
( Bx + Cx )i + ( Ay + C y − 240 lb) j + ( Az + Bz )k = 0 From i-coefficient:
Bx + Cx = 0
C x = − Bx
or
j-coefficient:
or
(1)
−12 Az + 10 C x = 0
Cx = 1.2 Az
k-coefficient:
or
8 Bz − 10 C y = 0
(4)
Ay + C y − 240 lb = 0
Ay + C y = 240 lb
(5)
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509
PROBLEM 4.129 (Continued)
k-coefficient:
Az + Bz = 0
Az = − Bz
or
(6)
Substituting Cx from Equation (4) into Equation (2),
− Bz = 1.2 Az
Using Equations (1), (6), and (7),
Cy =
Bz
− Az
1 § Bx · Bx
=
=
=
1.25 1.25 1.25 ¨© 1.2 ¸¹ 1.5
(7)
(8)
From Equations (3) and (8):
Cy =
1.5 Ay
1.5
or C y = Ay
and substituting into Equation (5),
2 Ay = 240 lb
(9)
Ay = C y = 120 lb
Using Equation (1) and Equation (9),
Bz = 1.25(120 lb) = 150.0 lb
Using Equation (3) and Equation (9),
Bx = 1.5(120 lb) = 180.0 lb
From Equation (4):
Cx = −180.0 lb
From Equation (6):
Az = −150.0 lb
Therefore,
A = (120.0 lb) j − (150.0 lb)k W
C = −(180.0 lb)i + (120.0 lb) j W
B = (180.0 lb)i + (150.0 lb)k W
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510
PROBLEM 4.130
Solve Problem 4.129, assuming that the force P is removed and is
replaced by a couple M = +(600 lb ⋅ in.)j acting at B.
PROBLEM 4.129 Three rods are welded together to form a “corner”
that is supported by three eyebolts. Neglecting friction, determine the
reactions at A, B, and C when P = 240 lb, a = 12 in., b = 8 in., and
c = 10 in.
SOLUTION
From F.B.D. of weldment:
ΣM O = 0: rA/O × A + rB/O × B + rC/O × C + M = 0
i
12
0
j
0
Ay
k
i
0 + 0
Az Bx
j k
i
8 0 + 0
0 Bz
Cx
j k
0 10 + (600 lb ⋅ in.) j = 0
Cy 0
(−12 Az j + 12 Ay k ) + (8 Bz j − 8 Bx k ) + ( −10C y i + 10C x j) + (600 lb ⋅ in.) j = 0
From i-coefficient:
or
8 Bz − 10 C y = 0
(1)
C y = 0.8Bz
j-coefficient:
−12 Az + 10 Cx + 600 = 0
Cx = 1.2 Az − 60
or
k-coefficient:
or
(2)
12 Ay − 8 Bx = 0
(3)
Bx = 1.5 Ay
ΣF = 0: A + B + C = 0 ( Bx + C x )i + ( Ay + C y ) j + ( Az + Bz )k = 0 From i-coefficient:
C x = − Bx
(4)
j-coefficient:
C y = − Ay
(5)
k-coefficient:
Az = − Bz
(6)
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511
PROBLEM 4.130 (Continued)
Substituting Cx from Equation (4) into Equation (2),
Using Equations (1), (6), and (7),
§B ·
Az = 50 − ¨ x ¸
© 1.2 ¹
(7)
§2·
C y = 0.8 Bz = − 0.8 Az = ¨ ¸ Bx − 40
©3¹
(8)
From Equations (3) and (8):
C y = Ay − 40
Substituting into Equation (5),
2 Ay = 40
Ay = 20.0 lb
From Equation (5):
C y = −20.0 lb
Equation (1):
Bz = −25.0 lb
Equation (3):
Bx = 30.0 lb
Equation (4):
Cx = −30.0 lb
Equation (6):
Az = 25.0 lb
A = (20.0 lb) j + (25.0 lb)k W
Therefore,
B = (30.0 lb)i − (25.0 lb)k W
C = − (30.0 lb)i − (20.0 lb) j W
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512
PROBLEM 4.131
In order to clean the clogged drainpipe AE, a plumber has
disconnected both ends of the pipe and inserted a power
snake through the opening at A. The cutting head of the
snake is connected by a heavy cable to an electric motor that
rotates at a constant speed as the plumber forces the cable
into the pipe. The forces exerted by the plumber and
the motor on the end of the cable can be represented by
the wrench F = −(48 N)k , M = −(90 N ⋅ m)k. Determine the
additional reactions at B, C, and D caused by the cleaning
operation. Assume that the reaction at each support consists
of two force components perpendicular to the pipe.
SOLUTION
From F.B.D. of pipe assembly ABCD:
ΣFx = 0: Bx = 0
ΣM D ( x -axis) = 0: (48 N)(2.5 m) − Bz (2 m) = 0
Bz = 60.0 N
and B = (60.0 N)k W
ΣM D ( z -axis) = 0: C y (3 m) − 90 N ⋅ m = 0
C y = 30.0 N
ΣM D ( y -axis) = 0: − C z (3 m) − (60.0 N)(4 m) + (48 N)(4 m) = 0
Cz = −16.00 N
and C = (30.0 N) j − (16.00 N)k W
ΣFy = 0: Dy + 30.0 = 0
Dy = −30.0 N
ΣFz = 0: Dz − 16.00 N + 60.0 N − 48 N = 0
Dz = 4.00 N
and D = − (30.0 N) j + (4.00 N)k W
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513
PROBLEM 4.132
Solve Problem 4.131, assuming that the plumber exerts a force
F = −(48 N)k and that the motor is turned off (M = 0).
PROBLEM 4.131 In order to clean the clogged drainpipe AE, a
plumber has disconnected both ends of the pipe and inserted a
power snake through the opening at A. The cutting head of the snake
is connected by a heavy cable to an electric motor that rotates at a
constant speed as the plumber forces the cable into the pipe. The
forces exerted by the plumber and the motor on the end of the cable
can be represented by the wrench F = −(48 N)k , M = −(90 N ⋅ m)k.
Determine the additional reactions at B, C, and D caused by the
cleaning operation. Assume that the reaction at each support consists
of two force components perpendicular to the pipe.
SOLUTION
From F.B.D. of pipe assembly ABCD:
ΣFx = 0: Bx = 0
ΣM D ( x -axis) = 0: (48 N)(2.5 m) − Bz (2 m) = 0
Bz = 60.0 N
and B = (60.0 N)k W
ΣM D ( z -axis) = 0: C y (3 m) − Bx (2 m) = 0
Cy = 0
ΣM D ( y -axis) = 0: Cz (3 m) − (60.0 N)(4 m) + (48 N)(4 m) = 0
and C = − (16.00 N)k W
Cz = −16.00 N
ΣFy = 0: Dy + C y = 0
Dy = 0
ΣFz = 0: Dz + Bz + C z − F = 0
Dz + 60.0 N − 16.00 N − 48 N = 0
Dz = 4.00 N
and D = (4.00 N)k W
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514
PROBLEM 4.133
The 50-kg plate ABCD is supported by hinges along edge AB
and by wire CE. Knowing that the plate is uniform, determine
the tension in the wire.
SOLUTION
Free-Body Diagram:
W = mg = (50 kg)(9.81 m/s 2 )
W = 490.50 N
JJJG
CE = − 240i + 600 j − 400k
CE = 760 mm
JJJG
CE
T
=
(− 240i + 600 j − 400k )
T =T
CE 760
JJJG
AB 480i − 200 j 1
λ AB =
=
= (12i − 5 j)
520
13
AB
ΣMAB = 0: λ AB ⋅ (rE/ A × T ) + λ AB ⋅ (rG/ A × − W j) = 0
rE/ A = 240i + 400 j; rG/ A = 240i − 100 j + 200k
12
0
12
0
−5
−5
1
T
240 400
0
+ 240 −100 200
=0
13 × 20
13
0
0
− 240 600 − 400
−W
(−12 × 400 × 400 − 5 × 240 × 400)
T
+ 12 × 200W = 0
760
T = 0.76W = 0.76(490.50 N)
T = 373 N W
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515
PROBLEM 4.134
Solve Problem 4.133, assuming that wire CE is replaced by a
wire connecting E and D.
PROBLEM 4.133 The 50-kg plate ABCD is supported by
hinges along edge AB and by wire CE. Knowing that the plate
is uniform, determine the tension in the wire.
SOLUTION
Free-Body Diagram:
Dimensions in mm
W = mg = (50 kg)(9.81 m/s 2 )
W = 490.50 N
JJJG
DE = − 240i + 400 j − 400k
DE = 614.5 mm
JJJG
DE
T
=
(240i + 400 j − 400k )
T =T
DE 614.5
JJJG
AB 480i − 200 j 1
λ AB =
=
= (12i − 5 j)
520
13
AB
rE/ A = 240i + 400 j; rG/ A = 240i − 100 j + 200k
12 −5
0
12
5
0
1
T
240 400
0
+ 240 −100 200
=0
13 × 614.5
13
240 400 − 400
0
0
−W
(−12 × 400 × 400 − 5 × 240 × 400)
T
+ 12 × 200 × W = 0
614.5
T = 0.6145W = 0.6145(490.50 N)
T = 301 N W
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516
PROBLEM 4.135
Two rectangular plates are welded together to form the assembly
shown. The assembly is supported by ball-and-socket joints at B
and D and by a ball on a horizontal surface at C. For the loading
shown, determine the reaction at C.
SOLUTION
λ BD =
First note:
−(6 in.)i − (9 in.) j + (12 in.)k
(6) 2 + (9) 2 + (12)2 in.
1
(− 6i − 9 j + 12k )
16.1555
= − (6 in.)i
=
rA/B
P = (80 lb)k
rC/D = (8 in.)i
C = (C ) j
From the F.B.D. of the plates:
ΣM BD = 0: λ BD ⋅ (rA/B × P ) + λ BD ⋅ ( rC/D × C ) = 0
−6 −9 12
−6 −9 12
ª 6(80) º
ª C (8) º
−1 0 0 «
+ 1 0 0 «
=0
»
¬16.1555 ¼
¬16.1555 »¼
0 0 1
0 1 0
( −9)(6)(80) + (12)(8)C = 0
C = 45.0 lb
or C = (45.0 lb) j W
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517
PROBLEM 4.136
Two 2 × 4-ft plywood panels, each of weight 12 lb, are nailed
together as shown. The panels are supported by ball-and-socket
joints at A and F and by the wire BH. Determine (a) the location
of H in the xy plane if the tension in the wire is to be minimum,
(b) the corresponding minimum tension.
SOLUTION
Free-Body Diagram:
JJJG
AF = 4i − 2 j − 4k
AF = 6 ft
1
λ AF = (2i − j − 2k )
3
rG1/A = 2i − j
rG2 /A = 4i − j − 2k
rB/A = 4i
ΣM AF = 0: λ AF ⋅ (rG1 /A × ( −12 j) + λ AF ⋅ (rG2 /A × (−12 j)) + λ AF ⋅ (rB /A × T ) = 0
2 −1 −2
2 −1 −2
1
1
+ 4 −1 −2 + λ AF ⋅ (rB/A × T) = 0
2 −1 0
3
3
0 −12 0
0 −12 0
1
1
(2 × 2 × 12) + (−2 × 2 × 12 + 2 × 4 × 12) + λ AF ⋅ (rB/A × T) = 0
3
3
λ AF ⋅ (rB/A × T) = −32 or T ⋅ (λ A/F × rB/A ) = −32
(1)
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518
PROBLEM 4.136 (Continued)
Projection of T on (λ AF × rB/A ) is constant. Thus, Tmin is parallel to
1
1
λ AF × rB/A = (2i − j − 2k ) × 4i = (−8 j + 4k )
3
3
Corresponding unit vector is
1
5
(−2 j + k ).
Tmin = T ( −2 j + k )
From Eq. (1):
T
ª1
º
(−2 j + k ) ⋅ « (2i − j − 2k ) × 4i » = −32
5
¬3
¼
T
1
(−2 j + k ) ⋅ ( −8 j + 4k ) = −32
3
5
T
(16 + 4) = −32
3 5
T =−
1
(2)
5
3 5(32)
= 4.8 5
20
T = 10.7331 lb
From Eq. (2):
Tmin = T ( −2 j + k )
1
5
= 4.8 5( −2 j + k )
Tmin
1
5
= −(9.6 lb)j + (4.8 lb k )
Since Tmin has no i component, wire BH is parallel to the yz plane, and x = 4 ft.
(a)
(b)
x = 4.00 ft;
y = 8.00 ft W
Tmin = 10.73 lb W
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519
PROBLEM 4.137
Solve Problem 4.136, subject to the restriction that H must lie on
the y-axis.
PROBLEM 4.136 Two 2 × 4-ft plywood panels, each of weight 12 lb,
are nailed together as shown. The panels are supported by ball-andsocket joints at A and F and by the wire BH. Determine (a) the
location of H in the xy plane if the tension in the wire is to be
minimum, (b) the corresponding minimum tension.
SOLUTION
JJJG
AF = 4i − 2 j − 4k
Free-Body Diagram:
1
λ AF = (2i − j − 2k )
3
rG1/A = 2i − j
rG2 /A = 4i − j − 2k
rB/A = 4i
ΣMAF = 0: λ AF ⋅ (rG/A × (−12 j) + λ AF ⋅ (rG2 /A × (−12 j)) + λ AF ⋅ (rB/A × T ) = 0
2 −1 2
2 −1 −2
1
1
2 −1 0 + 4 −1 −2 + λ AF ⋅ (rB/A × T) = 0
3
3
0 −12 0
0 −12 0
1
1
(2 × 2 × 12) + (−2 × 2 × 12 + 2 × 4 × 12) + λ AF ⋅ (rB/A × T) = 0
3
3
λ AF ⋅ (rB/A × T) = −32
JJJJG
BH = −4i + yj − 4k
BH = (32 + y 2 )1/2
JJJJG
BH
−4i + yj − 4k
=T
T=T
BH
(32 + y 2 )1/ 2
(1)
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520
PROBLEM 4.137 (Continued)
From Eq. (1):
λ AF
2 −1 −2
T
⋅ (rB/A × T ) = 4 0 0
= −32
3(32 + y 2 )1/2
−4 y −4
(−16 − 8 y )T = −3 × 32(32 + y 2 )1/2
T = 96
(32 + y 2 )1/2
8 y + 16
(2)
(8y +16) 12 (32 + y 2 ) −1/ 2 (2 y ) + (32 + y 2 )1/2 (8)
dT
= 0: 96
dy
(8 y + 16) 2
Numerator = 0:
(8 y + 16) y = (32 + y 2 )8
8 y 2 + 16 y = 32 × 8 + 8 y 2
From Eq. (2):
T = 96
(32 + 162 )1/ 2
= 11.3137 lb
8 × 16 + 16
x = 0 ft; y = 16.00 ft W
Tmin = 11.31 lb W
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521
PROBLEM 4.138
The frame ACD is supported by ball-and-socket joints at A
and D and by a cable that passes through a ring at B and is
attached to hooks at G and H. Knowing that the frame
supports at Point C a load of magnitude P = 268 N, determine
the tension in the cable.
SOLUTION
Free-Body Diagram:
λ AD
λ AD
TBG
TBH
JJJG
AD (1 m)i − (0.75 m)k
=
=
AD
1.25 m
= 0.8i − 0.6k
JJJG
BG
= TBG
BG
−0.5i + 0.925 j − 0.4k
= TBG
1.125
JJJG
BH
= TBH
BH
0.375i + 0.75 j − 0.75k
= TBH
1.125
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522
PROBLEM 4.138 (Continued)
rB /A = (0.5 m)i; rC/A = (1 m)i; P = −(268 N)j
To eliminate the reactions at A and D, we shall write
ΣM AD = 0:
λ AD ⋅ (rB /A × TBG ) + λ AD ⋅ (rB /A × TBH ) + λ AD ⋅ (rC /A × P) = 0
(1)
Substituting for terms in Eq. (1) and using determinants,
−0.6
−0.6
−0.6
0.8
0
0.8
0
0.8
0
TBG
TBH
+ 0.5
+ 1
0.5
0
0
0
0
0
0 =0
1.125
1.125
−0.5 0.925 −0.4
0.375 0.75 −0.75
0 −268
0
Multiplying all terms by (–1.125),
0.27750TBG + 0.22500TBH = 180.900
For this problem,
TBG = TBH = T
(0.27750 + 0.22500)T = 180.900
(2)
T = 360 N W
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523
PROBLEM 4.139
Solve Prob. 4.138, assuming that cable GBH is replaced by
a cable GB attached at G and B.
PROBLEM 4.138 The frame ACD is supported by ball-andsocket joints at A and D and by a cable that passes through a
ring at B and is attached to hooks at G and H. Knowing that the
frame supports at Point C a load of magnitude P = 268 N,
determine the tension in the cable.
SOLUTION
Free-Body Diagram:
λ AD
λ AD
TBG
TBH
JJJG
AD (1 m)i − (0.75 m)k
=
=
1.25 m
AD
= 0.8i − 0.6k
JJJG
BG
= TBG
BG
−0.5i + 0.925 j − 0.4k
= TBG
1.125
JJJG
BH
= TBH
BH
0.375i + 0.75 j − 0.75k
= TBH
1.125
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524
PROBLEM 4.139 (Continued)
rB /A = (0.5 m)i; rC/A = (1 m)i; P = −(268 N)j
To eliminate the reactions at A and D, we shall write
ΣM AD = 0: λ AD ⋅ (rB /A × TBG ) + λ AD ⋅ (rB /A × TBH ) + λ AD ⋅ (rC /A × P) = 0
(1)
Substituting for terms in Eq. (1) and using determinants,
−0.6
−0.6
−0.6
0.8
0
0.8
0
0.8
0
TBG
TBH
+ 0.5
+ 1
0.5
0
0
0
0
0
0 =0
1.125
1.125
−0.5 0.925 −0.4
0.375 0.75 −0.75
0 −268
0
Multiplying all terms by (–1.125),
0.27750TBG + 0.22500TBH = 180.900
(2)
For this problem, TBH = 0.
Thus, Eq. (2) reduces to
0.27750TBG = 180.900
TBG = 652 N W
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525
PROBLEM 4.140
The bent rod ABDE is supported by ball-and-socket joints at A
and E and by the cable DF. If a 60-lb load is applied at C as
shown, determine the tension in the cable.
SOLUTION
JJJG
DF = −16i + 11j − 8k
DF = 21 in.
JJJG
DE T
T=T
= (−16i + 11j − 8k )
DF 21
rD/E = 16i
Free-Body Diagram:
rC/E = 16i − 14k
JJJG
EA 7i − 24k
=
λ EA =
EA
25
ΣM EA = 0: λ EA ⋅ (rB/E × T) + λ EA ⋅ (rC/E ⋅ (− 60 j)) = 0
7
0 −24
7
0 −24
1
T
16 0
0
+ 16 0 −14
=0
21 × 25
25
0 −60 0
−16 11 −8
−
24 × 16 × 11
−7 × 14 × 60 + 24 × 16 × 60
T+
=0
21 × 25
25
201.14 T + 17,160 = 0
T = 85.314 lb
T = 85.3 lb W
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526
PROBLEM 4.141
Solve Problem 4.140, assuming that cable DF is replaced by a
cable connecting B and F.
SOLUTION
Free-Body Diagram:
rB/ A = 9i
rC/ A = 9i + 10k
JJJG
BF = −16i + 11j + 16k
BF = 25.16 in.
JJJG
BF
T
=
(−16i + 11j + 16k )
T =T
BF 25.16
JJJG
AE 7i − 24k
λ AE =
=
AE
25
ΣMAE = 0: λ AF ⋅ (rB/ A × T) + λ AE ⋅ (rC/ A ⋅ (− 60 j)) = 0
7
0 −24
7 0 −24
1
T
+9 0
=0
9
0
0
10
25 × 25.16
25
−16 11 16
0 −60 0
−
24 × 9 × 11
24 × 9 × 60 + 7 × 10 × 60
T+
=0
25 × 25.16
25
94.436 T − 17,160 = 0
T = 181.7 lb W
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527
PROBLEM 4.142
A gardener uses a 60-N wheelbarrow to transport a 250-N bag of
fertilizer. What force must she exert on each handle?
SOLUTION
Free-Body Diagram:
ΣM A = 0: (2 F )(1 m) − (60 N)(0.15 m) − (250 N)(0.3 m) = 0
F = 42.0 N W
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528
PROBLEM 4.143
The required tension in cable AB is 200 lb. Determine (a) the vertical
force P that must be applied to the pedal, (b) the corresponding
reaction at C.
SOLUTION
Free-Body Diagram:
BC = 7 in.
(a)
ΣM C = 0: P(15 in.) − (200 lb)(6.062 in.) = 0
P = 80.83 lb
(b)
P = 80.8 lb W
ΣFy = 0: C x − 200 lb = 0
C x = 200 lb
ΣFy = 0: C y − P = 0 C y − 80.83 lb = 0
C y = 80.83 lb
α = 22.0°
C = 215.7 lb
C = 216 lb
22.0° W
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529
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 500-N horizontal force at B, determine
(a) the tension in the cable, (b) the reaction at C.
SOLUTION
Triangle ACD is isosceles with ) C = 90° + 30° = 120° ) A = ) D =
1
(180° − 120°) = 30°.
2
Thus, DA forms angle of 60° with the horizontal axis.
(a)
We resolve FAD into components along AB and perpendicular to AB.
FAD = 400 N W
ΣM C = 0: ( FAD sin 30°)(250 mm) − (500 N)(100 mm) = 0
(b)
ΣFx = 0: − (400 N) cos 60° + C x − 500 N = 0
C x = +300 N
ΣFy = 0: − (400 N) sin 60° + C y = 0
C y = +346.4 N
C = 458 N
49.1° W
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530
PROBLEM 4.145
A force P of magnitude 280 lb is applied to member ABCD, which is
supported by a frictionless pin at A and by the cable CED. Since the cable
passes over a small pulley at E, the tension may be assumed to be the
same in portions CE and ED of the cable. For the case when a = 3 in.,
determine (a) the tension in the cable, (b) the reaction at A.
SOLUTION
Free-Body Diagram:
(a)
ΣM A = 0: − (280 lb)(8 in.)
7
T (12 in.)
25
24
− T (8 in.) = 0
25
T (12 in.) −
T = 875 lb W
(12 − 11.04)T = 840
(b)
ΣFx = 0:
7
(875 lb) + 875 lb + Ax = 0
25
Ax = −1120
ΣFy = 0: Ay − 280 lb −
Ay = +1120
A x = 1120 lb
24
(875 lb) = 0
25
A y = 1120 lb
A = 1584 lb
45.0° W
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531
PROBLEM 4.146
Two slots have been cut in plate DEF, and the plate has
been placed so that the slots fit two fixed, frictionless pins A
and B. Knowing that P = 15 lb, determine (a) the force each
pin exerts on the plate, (b) the reaction at F.
SOLUTION
Free-Body Diagram:
(a)
(b)
ΣFx = 0: 15 lb − B sin 30° = 0
B = 30.0 lb
60.0° W
ΣM A = 0: − (30 lb)(4 in.) + B sin 30°(3 in.) + B cos 30°(11 in.) − F (13 in.) = 0
−120 lb ⋅ in. + (30 lb) sin 30°(3 in.) + (30 lb) cos 30°(11 in.) − F (13 in.) = 0
F = + 16.2145 lb
(a)
F = 16.21 lb W
ΣFy = 0: A − 30 lb + B cos 30° − F = 0
A − 30 lb + (30 lb) cos 30° − 16.2145 lb = 0
A = + 20.23 lb
A = 20.2 lb W
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532
PROBLEM 4.147
Knowing that the tension in wire BD is 1300 N, determine the
reaction at the fixed support C of the frame shown.
SOLUTION
T = 1300 N
5
Tx = T
13
= 500 N
12
Ty = T
13
= 1200 N
ΣM x = 0: C x − 450 N + 500 N = 0
C x = −50 N
ΣFy = 0: C y − 750 N − 1200 N = 0 C y = +1950 N
C x = 50 N
C y = 1950 N
C = 1951 N
88.5° W
ΣM C = 0: M C + (750 N)(0.5 m) + (4.50 N)(0.4 m)
− (1200 N)(0.4 m) = 0
M C = −75.0 N ⋅ m
M C = 75.0 N ⋅ m
W
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533
PROBLEM 4.148
The spanner shown is used to rotate a shaft. A pin fits in a
hole at A, while a flat, frictionless surface rests against the
shaft at B. If a 60-lb force P is exerted on the spanner at D,
find the reactions at A and B.
SOLUTION
Free-Body Diagram:
(Three-force body)
The line of action of A must pass through D, where B and P intersect.
3sin 50°
3cos 50° + 15
= 0.135756
α = 7.7310°
tan α =
60 lb
sin 7.7310°
= 446.02 lb
60 lb
B=
tan 7.7310°
= 441.97 lb
A=
Force triangle
A = 446 lb
7.73° W
B = 442 lb
W
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534
PROBLEM 4.149
Rod AB is supported by a pin and bracket at A and rests against a frictionless
peg at C. Determine the reactions at A and C when a 170-N vertical force is
applied at B.
SOLUTION
Free-Body Diagram:
(Three-force body)
The reaction at A must pass through D where C and the 170-N force intersect.
160 mm
300 mm
α = 28.07°
tan α =
We note that triangle ABD is isosceles (since AC = BC) and, therefore,
) CAD = α = 28.07°
Also, since CD ⊥ CB, reaction C forms angle α = 28.07° with the horizontal axis.
Force triangle
We note that A forms angle 2α with the vertical axis. Thus, A and C form angle
180° − (90° − α ) − 2α = 90° − α
Force triangle is isosceles, and we have
A = 170 N
C = 2(170 N)sin α
= 160.0 N
A = 170.0 N
33.9°;
C = 160.0 N
28.1° W
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535
PROBLEM 4.150
The 24-lb square plate shown is supported by three vertical
wires. Determine (a) the tension in each wire when a = 10 in.,
(b) the value of a for which the tension in each wire is 8 lb.
SOLUTION
rB/A = ai + 30k
rC/A = 30i + ak
rG/A = 15i + 15k
By symmetry, B = C.
ΣM A = 0: rB/A × Bj + rC × Cj + rG/A × (−W j) = 0
(ai + 30k ) × Bj + (30i + ak ) × Bj + (15i + 15k ) × (−W j) = 0
Bak − 30 Bi + 30 Bk − Bai − 15Wk + 15W i = 0
Equate coefficient of unit vector i to zero:
i : − 30 B − Ba + 15W = 0
B=
15W
30 + a
C=B=
15W
30 + a
(1)
ΣFy = 0: A + B + C − W = 0
ª 15W º
A+ 2«
» − W = 0;
¬ 30 + a ¼
(a)
For
a = 10 in.
From Eq. (1):
C=B=
From Eq. (2):
A=
A=
aW
30 + a
(2)
15(24 lb)
= 9.00 lb
30 + 10
A = 6.00 lb; B = C = 9.00 lb W
10(24 lb)
= 6.00 lb
30 + 10
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536
PROBLEM 4.150 (Continued)
(b)
For tension in each wire = 8 lb,
From Eq. (1):
8 lb =
15(24 lb)
30 + a
30 in. + a = 45
a = 15.00 in. W
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537
PROBLEM 4.151
Frame ABCD is supported by a ball-and-socket joint at A and
by three cables. For a = 150 mm, determine the tension in
each cable and the reaction at A.
SOLUTION
First note:
TDG = λ DG TDG =
−(0.48 m)i + (0.14 m)j
(0.48) 2 + (0.14) 2 m
TDG
−0.48i + 0.14 j
TDG
0.50
T
= DG (24i + 7 j)
25
=
TBE = λ BE TBE =
−(0.48 m)i + (0.2 m)k
(0.48)2 + (0.2)2 m
TBE
−0.48i + 0.2k
TBE
0.52
T
= BE (−12 j + 5k )
13
=
From F.B.D. of frame ABCD:
§ 7
·
ΣM x = 0: ¨ TDG ¸ (0.3 m) − (350 N)(0.15 m) = 0
© 25
¹
or
or
§ 24
·
§ 5
·
ΣM y = 0: ¨ × 625 N ¸ (0.3 m) − ¨ TBE ¸ (0.48 m) = 0
13
© 25
¹
©
¹
§ 7
·
ΣM z = 0: TCF (0.14 m) + ¨ × 625 N ¸ (0.48 m) − (350 N)(0.48 m) = 0
© 25
¹
or
TDG = 625 N W
TBE = 975 N W
TCF = 600 N W
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538
PROBLEM 4.151 (Continued)
ΣFx = 0: Ax + TCF + (TBE ) x + (TDG ) x = 0
§ 12
· § 24
·
Ax − 600 N − ¨ × 975 N ¸ − ¨ × 625 N ¸ = 0
© 13
¹ © 25
¹
Ax = 2100 N
ΣFy = 0: Ay + (TDG ) y − 350 N = 0
§ 7
·
Ay + ¨ × 625 N ¸ − 350 N = 0
25
©
¹
Ay = 175.0 N
ΣFz = 0: Az + (TBE ) z = 0
§ 5
·
Az + ¨ × 975 N ¸ = 0
13
©
¹
Az = −375 N
A = (2100 N)i + (175.0 N) j − (375 N)k W
Therefore,
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539
PROBLEM 4.152
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
JJJG
AE = 480i + 160 j − 240k
AE = 560 mm
JJJG
AE 480i + 160 j − 240k
=
λ AE =
AE
560
6i + 2 j − 3k
λ AE =
7
rB/A = 200i
rD/A = 480i + 160 j
JJJG
DF = −480i + 330 j − 240k ;
DF = 630 mm
JJJG
DF
−480i + 330 j − 240k
−16i + 11j − 8k
= TDF
= TDF
TDF = TDF
630
21
DF
ΣM AE = λ AE ⋅ (rD/A × TDF ) + λ AE ⋅ (rB/A × (−600 j)) = 0
6
2 −3
6
2
−3
TDF
1
480 160 0
0
0 =0
+ 200
21 × 7
7
0 −640 0
−16 11 −8
3 × 200 × 640
−6 × 160 × 8 + 2 × 480 × 8 − 3 × 480 × 11 − 3 × 160 × 16
TDF +
=0
21 × 7
7
−1120TDF + 384 × 103 = 0
TDF = 342.86 N
TDF = 343 N W
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540
PROBLEM 4.153
A force P is applied to a bent rod ABC, which may be supported in four different ways as shown. In each case,
if possible, determine the reactions at the supports.
SOLUTION
(a)
ΣM A = 0: − Pa + (C sin 45°)2a + (cos 45°)a = 0
3
C
2
=P
C=
§ 2 · 1
ΣFx = 0: Ax − ¨
P
¨ 3 ¸¸ 2
©
¹
§ 2 · 1
P
ΣFy = 0: Ay − P + ¨
¨ 3 ¸¸ 2
©
¹
2
P
3
Ax =
P
3
Ay =
2P
3
C = 0.471P
A = 0.745P
(b)
45° W
63.4° W
ΣM C = 0: +Pa − ( A cos 30°)2a + ( A sin 30°)a = 0
A(1.732 − 0.5) = P
A = 0.812 P
A = 0.812 P
ΣFx = 0: (0.812 P )sin 30° + C x = 0
60.0° W
Cx = −0.406 P
ΣFy = 0: (0.812 P) cos 30° − P + C y = 0 C y = −0.297 P
C = 0.503P
36.2° W
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541
PROBLEM 4.153 (Continued)
(c)
ΣM C = 0: + Pa − ( A cos 30°)2a + ( A sin 30°)a = 0
A(1.732 + 0.5) = P
A = 0.448P
A = 0.448P
ΣFx = 0: − (0.448P) sin 30° + Cx = 0
Cx = 0.224 P
ΣFy = 0: (0.448 P) cos 30° − P + C y = 0 C y = 0.612 P
C = 0.652 P
(d)
60.0° W
69.9° W
Force T exerted by wire and reactions A and C all intersect at Point D.
ΣM D = 0: Pa = 0
Equilibrium is not maintained.
Rod is improperly constrained. W
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542
PROBLEM 4.F1
For the frame and loading shown, draw the free-body diagram
needed to determine the reactions at A and E when α = 30°.
SOLUTION
Free-Body Diagram of Frame:
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543
PROBLEM 4.F2
Neglecting friction, draw the free-body diagram needed to determine the
tension in cable ABD and the reaction at C when θ = 60°.
SOLUTION
Free-Body Diagram of Member ACD:
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544
PROBLEM 4.F3
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. Draw the free-body diagram needed to
determine the tension in cable BD and the reactions at
A and C.
SOLUTION
Free-Body Diagram of Bar AC:
Note: By similar triangles
yB = 0.075 m yB
0.15 m
=
0.25 m 0.5 m
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545
PROBLEM 4.F4
Draw the free-body diagram needed to determine the tension
in each cable and the reaction at D.
SOLUTION
Free-Body Diagram of Member ABCD:
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546
PROBLEM 4.F5
A 4 × 8-ft sheet of plywood weighing 34 lb has been temporarily
placed among three pipe supports. The lower edge of the sheet
rests on small collars at A and B and its upper edge leans against
pipe C. Neglecting friction on all surfaces, draw the free-body
diagram needed to determine the reactions at A, B, and C.
SOLUTION
Free-Body Diagram of Plywood sheet:
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547
PROBLEM 4.F6
Two transmission belts pass over sheaves welded to an
axle supported by bearings at B and D. The sheave at A
has a radius of 2.5 in. and the sheave at C has a radius
of 2 in. Knowing that the system rotates at a constant
rate, draw the free-body diagram needed to determine
the tension T and 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
Free-Body Diagram of axle-sheave system:
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548
PROBLEM 4.F7
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. Draw the free-body diagram needed to
determine the tension in each cable and the reaction at A.
SOLUTION
Free-Body Diagram of Pole:
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549