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Question

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. Then
6316_c9168799bc77408ca633294a401d1c21.png
  1. F=GMm(h+L)2
  2. F=GMmh2
  3. F=GMmh(h+L)
  4. 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+h
Force on m due to this small element is dF=GmdM(x+h)2
To get the total force F on the mass m, we integrate dF from x=0 to x=L
F=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).

633267_6316_ans_29c72ec9e84a437686414a86acc9a016.png

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