Question

Four particles having masses m,2m,3m and 4m are placed at the four corners of a square of edge a.

Find gravitational force acting on a particle of mass m place at the center.

Answer 1

Gravitational force on mass 'm' at center is

$F_x\hat{i}+F_x\hat{j}=\overrightarrow{F}=\frac{G\times m\times m}{({\frac{a}{\sqrt{2}})}^2}+\frac{G\times m\times 2m}{({\frac{a}{\sqrt{2}})}^2}+\frac{G\times m\times 3m}{({\frac{a}{\sqrt{2}})}^2}+\frac{G\times m\times 4m}{({\frac{a}{\sqrt{2}})}^2}$

[Note that above is vector sum]

$\left|\overrightarrow{F_1}\right|=\frac{2G{m}^2}{a^2};\left|\overrightarrow{F_2}\right|=\frac{4G{m}^2}{a^2};\left|\overrightarrow{F_3}\right|=\frac{6G{m}^2}{a^2};\left|\overrightarrow{F_4}\right|=\frac{8G{m}^2}{a^2}$

By resolving the force along x-direction,

$F_x=\left(F_2\cos 45°+F_3\cos 45°\right)-\left(F_1\cos 45°+F_4\cos 45\right)F_x=(\frac{10G{m}^2}{a^2}-\frac{10G{m}^2}{a^2})\times \cos 45°=0$

By resolving the force along y-direction,

$F_y=[(F_1+F_2)-(F_4+F_3)]\sin 45°=[\frac{6G{m}^2}{a^2}-\frac{14G{m}^2}{a^2}]\sin 45°=\frac{{-7m}^2}{\sqrt{2}{a}^2}$

Net force on m(at center) is $\left(\frac{{-7m}^2}{\sqrt{2}{a}^2}\right)\hat{j}$ [along negative y-direction]