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(b) Solving the above equation algebraically for
F
, with
W
denoting the weight, we obtain
(0.42)(180 N)
76 N
.
cos
sin
cos
(0.42) sin
cos
(0.42) sin
s
s
W
F
μ
θ
θθ
==
=
++
+
(c) We minimize the above expression for
F
by working through the condition:
2
(sin
cos )
0
(cos
sin
)
ss
s
W
dF
d
−
+
which leads to the result
= tan
–1
s
= 23°.
(d) Plugging
= 23° into the above result for
F
, with
s
= 0.42 and
W
= 180 N, yields
70 N
F
=
.
32. We use the familiar horizontal and vertical axes for
x
and
y
directions, with rightward
and upward positive, respectively. The rope is assumed massless so that the force exerted
by the child
G
F
is identical to the tension uniformly through the rope. The
x
and
y
components of
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Force, Mass

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