E PROBLEM 8.
The 20-lb block A hangs from a cable as shown. Pulley C is connected by a
C
D short link to block E, which rests on a horizontal rail. Knowing that the
θ coefficient of static friction between block E and the rail is ms = 0.35 and
T neglecting the weight of block E and the friction in the pulleys, determine
the maximum allowable value of θ if the system is to remain in equilibrium.
B
A 20 lb
SOLUTION
Free-body diagrams:
ms = 0.35
fs = tan-1 0.35 = 19.29
Force triangle for pulley C:
Law of sines:
sin a sin fs
=
2T T
sin a = 2sin fs = 2 sin19.29o
a = 41.35o
q = fs + a = 19.29 + 41.35
q = 60.6o ◀
Note: Answer is independent of WA .
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1241
PROBLEM 8.11
The 50-lb block A and the 25-lb block B are supported by an incline
that is held in the position shown. Knowing that the coefficient of
A static friction is 0.15 between the two blocks and zero between block
B B and the incline, determine the value of q for which motion is
impending.
θ
SOLUTION
Since motion impends, F = ms N between A + B
Free body: Block A
Impending motion: SFy = 0: N1 - 50 cos q = 0
N1 = 50 cos q
SFx = 0: T - 50 sin q + m1 N1 = 0
T = 50 sin q - m1 (50)cos q (1)
Free body: Block B
Impending motion: SFy = 0: N 2 - N1 - 25 cos q = 0
N 2 = 75cos q
SFx : T - m1N1 - m2 N 2 - 25sin q
T = m1 (50) cos q + m2 (75) cos q + 25sin q (2)
Eq. (1) - Eq. (2): 0 = 25sin q - m1 (100) cos q - m2 (75) cos q = 0
Substituting in for m1 = 0.15, m2 = 0, we have:
15
15cos q = 25sin q : tan q = ; q = 31.0 ◀
25
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1249
500 N
PROBLEM 8.32
45 mm 45 mm
90 mm 90 mm
A 500-N concrete block is to be lifted by the pair of tongs shown.
Determine the smallest allowable value of the coefficient of static
A B 75 mm friction between the block and the tongs at F and G.
C D
105 mm
360 mm
315 mm
F 500 N G
SOLUTION
Free body: Members CA, AB, BD
1
C y = Dy = (500) = 250 N
2
By symmetry:
Since CA is a two-force member,
Cx Cy 250 N
= =
90 mm 75 mm 75 mm
C x = 300 N
SFx = 0: Dx = C x
Dx = 300 N
Free body: Tong DEF
SM E = 0: (300 N)(105 mm) + (250 N)(135 mm)
+ (250 N)(157.5 mm) - Fx (360 mm) = 0
Fx = + 290.625 N
Fy 250 N
Minimum value of ms : ms = =
Fx 290.625 N
ms = 0.860 ◀
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the prior written consent of McGraw-Hill Education.
1275
y PROBLEM 5.8
Locate the centroid of the plane area shown.
r = 38 in.
16 in.
x
20 in.
SOLUTION
A, in 2 x , in. y , in. xA, in3 yA, in3
p
1 (38)2 = 2268.2 0 16.1277 0 36,581
2
2 - 20 ´ 16 = 320 -10 8 3200 -2560
S 1948.23 3200 34,021
S xA 3200 in 3
Then X= = X = 1.643 in. ◀
SA 1948.23 in 2
S yA 34,021 in3
Y= = Y = 17.46 in. ◀
SA 1948.23 in 2
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566
