ch13-p016 - 16. Since the rod is an extended object, we...

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16. Since the rod is an extended object, we cannot apply Equation 13-1 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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