Question

find force of gravity on M rod.​

Answer 1

Answer:

In this type of questions , we need to consider a small segment of rod exerting a small force on mass m

Let that small force be dF.

If we go on integrating such small forces due to various small units , we get the total force on mass m

Let the small segment of rod have mass

dM and length dx , located at x distance from mass m.

$\sf{ \red{ \therefore \: dF = \dfrac{Gm \: (dM)}{ {x}^{2} }}}$

$\sf{ \red{ \implies \: dF = \dfrac{Gm \: ( \dfrac{M}{L} dx)}{ {x}^{2} }}}$

$\sf{ \red{ \implies \: dF = \dfrac{GmM}{ L } \times \dfrac{dx}{ {x}^{2} } }}$

Integrating both sides with limits from :

• Upper limit : a+L
• Lower limit : a

$\sf{ \red{ \implies \: \int_{0}^{F} \: (dF) = \dfrac{GmM}{ L } \times \int_{a}^{a+L}(\dfrac{dx}{ {x}^{2} }) }}$

$\sf{ \red{ \implies \: F= \dfrac{GmM}{ L } \times \{ \dfrac{ - 1}{x} \}_{a}^{a+L} }}$

$\sf{ \red{ \implies \: F= \dfrac{GmM}{ L } \times \{ \dfrac{1}{a} - \dfrac{1}{a+L} \} }}$

$\sf{\red{\implies \: F= \dfrac{GMm}{a(a+L)}}}$

So final answer is as follows :

$\boxed{\sf{\red{\bold{ \: F= \dfrac{GMm}{a(a+L)}}}}}$