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75. (a) We note that
F
N
=
mg
in this situation, so
f
s
,max
=
μ
s
mg
= (0.52)(11 kg)(9.8 m/s
2
) = 56 N.
Consequently, the horizontal force
G
F
needed to initiate motion must be (at minimum)
slightly more than 56 N.
(b) Analyzing vertical forces when
G
F
is at nonzero
θ
yields
,max
sin
(
sin ).
Ns
s
FF
m
g
f
m
g
F
+=
¡
=−
Now, the horizontal component of
G
F
needed to initiate motion must be (at minimum)
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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, Light

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