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(g) The condition for the chair to slide is
,max
where
0.42.
s
xs
N
s
Ff
F
μμ
>=
=
For
θ
= 0°, we have
,max
100 N
(0.42)(245 N) 103 N
=<
=
=
so the crate remains at rest.
(h) For
= 30.0°, we find
,max
86.6 N
(0.42)(195 N)
81.9 N
=>
=
=
so the crate slides.
(i) For
= 60°, we get
,max
50.0 N
(0.42)(158 N)
66.4 N
=
=
which means the crate must remain at rest.
86. (a) Our +
x
direction is horizontal and is chosen (as we also do with +
y
) so that the
components of the 100 N force
G
F
are nonnegative. Thus,
F
x
=
F
cos
= 100 N, which
the textbook denotes
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 Spring '08
 Any
 Physics, Acceleration, Force

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