Question

Three particles of identical masses $m$ are kept at the vertices of an equilateral triangle of each side length '$a$'. Find the gravitational force of attraction on any one of the particles.

Answer 1

Force on $C$ due to $A$

$F_A=\frac{G{m}^2}{a^2}$ along $CA$

Force of $C$ due to $B$

$F_B=\frac{G{m}^2}{a^2}$ along $CB$

$\therefore F=\sqrt{{F_A}^2+{F_B}^2+2F_A{F}_B\cos \theta }$

$\Rightarrow F=\sqrt{(\frac{G{m}^2}{a^2}{)}^2+(\frac{G{m}^2}{a^2}{)}^2+2(\frac{G{m}^2}{a^2}\left)(\frac{G{m}^2}{a^2}\right)\cos 60\%}$

$\Rightarrow F=\sqrt{3}\frac{G{m}^2}{a^2}$