Three equal masses m are placed at the three corners of an equilateral triangle of side a. The force exerted by this system on another particle of mass m placed at the mid-point of a side.

Explanation

If m is placed at the mid point of a side .AB.
Then `vec(F_(OA)) = (4Gm^(2))/(a^(2))` in OA direction
and `vec(F_(OB)) = (4Gm^(2))/(a^(2))` in O1B direction

Since both are equal in opposite, cancel each other
`vec(F_(OC)) = (Gm^(2))/(((sqrt(3))/(2)a)^(2)) = (4Gm^(2))/(3a^(2))` in OC direction
`therefore` Net gravitational force on `m = (4Gm^(2))/(3a^(2))`