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# A homogeneous bar of length L and mass M is at a distance h from a point mass m as shown. The force on m is F. ThenF=GMm(h+L)2F=GMmh2F=GMmh(h+L)F=GMmL2

A
F=GMmh2
B
F=GMm(h+L)2
C
F=GMmh(h+L)
D
F=GMmL2
Solution
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#### Consider a small element dx of the rod at a distance x from the left end of the rod.Mass of this small element is dM=MLdx,Distance of this small element from mass m is x+hForce on m due to this small element is dF=GmdM(x+h)2To get the total force F on the mass m, we integrate dF from x=0 to x=LF=∫x=Lx=0GmMLdx(x+h)2=−GmML(1(L+h)−1h)=GMmh(L+h)So, the force on m due to homogeneous bas is =GMmh(L+h).

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