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 dmof thickness drat a distance rfrom 1m. The gravitational force between dmand 1mis1122(/)Gm dmGm M L drdFrr==,where we have substituted )dmM L dr=since mass is uniformly distributed. The direction of dFGis to the right (see figure). The total force
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