Question

Three identical particles each of mass $m$ are placed at the vertices of an equilateral triangle of side $a$. Find the force exerted by this system on a particle $P$ of mass $m$ placed at the
(a) The mid point of $AB$ side
(b) centre of the triangle.

Answer 1

Using the superposition principle, the net gravitational force on $P$ is $\overrightarrow{F}={\overrightarrow{F}}_A+{\overrightarrow{F}}_B+{\overrightarrow{F}}_C$
(a) As shown in the figure, when $P$ is at the mid point of a side, ${\overrightarrow{F}}_A$ and ${\overrightarrow{F}}_B$ will be equal in magnitude but opposite in direction. Hence they will cancel each other. So the net force on the particle $P$ will be the force due to the particle placed at $C$ only.
$\Rightarrow F=F_C=G\frac{m.m}{(CP{)}^2}=G\frac{m^2}{(a\sin {60}^{\circ }{)}^2}=\frac{4G{m}^2}{3a^2}$ along $PC$.
(b) At the centre of the triangle $O$, the force ${\overrightarrow{F}}_A,{\overrightarrow{F}}_B$ and ${\overrightarrow{F}}_C$ will be equal in magnitude and will subtend ${120}^{\circ }$ with each other. Hence the resultant force on $P$ at $O$ is $\overrightarrow{F}={\overrightarrow{F}}_A+{\overrightarrow{F}}_B+{\overrightarrow{F}}_C=0$.