A homogeneous bar of length L and mass M is a distance h from a point mass m as shown- The force on m is F- ThenF- dfrac-GMm-left - h-L right -2-F- dfrac-GMm-h-2-F- dfrac-GMm-hleft - h-L right -F- dfrac-GMm-L-2-

Question

Toppr Admin · Accepted Answer

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 $d M = \frac{M}{L} d x,$
Distance of this small element from mass $m$ is $x + h$
Force on m due to this small element is $d F = \frac{G m d M}{(x + h)^{2}}$
To get the total force $F$ on the mass $m$ , we integrate $d F$ from $x = 0$ to $x = L$
$F = \int_{x = 0}^{x = L} \frac{G m \frac{M}{L} d x}{(x + h)^{2}} = \frac{- G m M}{L} (\frac{1}{(L + h)} - \frac{1}{h}) = \frac{G M m}{h (L + h)}$
So, the force on m due to homogeneous bas is $= \frac{G M m}{h (L + h)} .$

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 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

$F = \frac{G M m}{{(h + L)}^{2}}$
$F = \frac{G M m}{h^{2}}$
$F = \frac{G M m}{h (h + L)}$
$F = \frac{G M m}{L^{2}}$