Three equal particles each of mass m are placed at the three comers of as equilateral triangle of side
A Find the force exerted by this system on another particle of mass m placed at (a) the mid-point of a side, (b) at the
centre of the triangle.

Explanation

As gravitaional force is a two body interaction, the principle of superposition is valid, i.e., resultant force
on particle of mass m at P,
`overset rarr F=overset rarr F_(A)+overset rarr F_(B)+ overset rarr F_(C)`
(a) As shown in figure, when P is at the mid-point of a side, `F_(A)" and"F_(B)` will be equal in magnitude but opposite in
direction so will cancel each other . So Point mass m at P will experience a force due to C only, i.e.,

`F=F_(C)=G("mm")/(("CP")^(2))=("Gm"^(2))/(3a^(2))`
(b) When P is at the centre of the triangle, O, the forces of three particles `overset rarr F_(A),overset rarr F_(B)"and overset rarr F_(C)` Will be equal in
magnitude and will subtend equal angles with each other `(120^(@))` , so that the resultant force on m at O
`overset rarr F=overset rarr F_(A)+overset rarr F_(B)+ overset rarr F_(C)=0`