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11. We denote the magnitude of 110 N force exerted by the worker on the crate as
F
. The
magnitude of the static frictional force can vary between zero and
,max
s
sN
f
F
μ
=
.
(a) In this case, application of Newton’s second law in the vertical direction yields
N
Fm
g
=
. Thus,
()
(
)
22
0.37 35kg (9.8m /s ) 1.3 10 N
ss
N
s
fF
m
g
μμ
===
=
×
which is greater than
F
.
(b) The block, which is initially at rest, stays at rest since
F
<
f
s
, max
. Thus, it does not
move.
(c) By applying Newton’s second law to the horizontal direction, that the magnitude of
the frictional force exerted on the crate is
2
1.1 10 N
s
f
=×
.
(d) Denoting the upward force exerted by the second worker as
F
2
, then application of
Newton’s second law in the vertical direction yields
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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, Friction, Work

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