479 Horizontal Motion: The horizontal component of velocity ... - Wuala

91962_05_R1_p0479-0512 6/5/09 3:54 PM Page 498
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R1–27. The 150-lb man lies against the cushion for which
the coefficient of static friction is m s = 0.5. Determine the
resultant normal and frictional forces the cushion exerts on
him if, due to rotation about the z axis, he has a constant
speed v = 20 ft>s. Neglect the size of the man. Take
u = 60°.
z
8 ft
G
u
+a ©F y = m(a n ) y ; N - 150 cos 60° = 150
32.2 a 202 b sin 60°
8
N = 277 lb
Ans.
+b©F x = m(a n ) x ; -F + 150 sin 60° = 150
32.2 a 202 b cos 60°
8
F = 13.4 lb
Ans.
Note: No slipping occurs
Since m s N = 138.4 lb 7 13.4 lb
*R1–28. The 150-lb man lies against the cushion for which
the coefficient of static friction is m s = 0.5. If he rotates
about the z axis with a constant speed v = 30 ft>s, determine
the smallest angle u of the cushion at which he will begin to
slip up the cushion.
z
8 ft
G
u
; + ©F n = ma n ; 0.5N cos u + N sin u = 150
32.2 a (30)2
8 b
+ c ©F b = 0; -150 + N cos u - 0.5 N sin u = 0
N =
150
cos u - 0.5 sin u
(0.5 cos u + sin u)150
(cos u - 0.5 sin u)
= 150
32.2 a (30)2
8 b
0.5 cos u + sin u = 3.493 79 cos u - 1.746 89 sin u
u = 47.5°
Ans.
498