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16. Since the rod is an extended object, we cannot apply Equation 131 directly to find
the force. Instead, we consider a small differential element of the rod, of mass
dm
of
thickness
dr
at a distance
r
from
1
m
. The gravitational force between
dm
and
1
m
is
11
22
(/
)
Gm dm
Gm M L dr
dF
rr
==
,
where we have substituted
)
dm
M L dr
=
since
mass is uniformly distributed. The direction of
dF
G
is to the right (see figure). The total force
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 Spring '08
 Any
 Physics, Force, Mass

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