Question

21. A homogeneous bar of length L and mass M is at
a distance h from a point mass m a shown. The
force on m is F, then find F.
M
т
h
L
-
>
GMM
GMm
(1) (h+L)?
(2) om
GMm
(3) h(1 + L)
GMm
12​

Answer 1

Answer:

The answer is GMmh/L(L +h).

Explanation:

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 = M/Ldx

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=L and x = 0 GmM/L dx/ (x + h)^2 = - GMm/L (1/L + h - 1/L)

                         = GmMh/L(L +h)

Thus the force on m due to homogeneous bar is GMmh/L(L +h).

Answer 2

Answer:

see the attached image

Explanation:

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