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 side .AB.
Then `bar(F_(OA))=(4"Gm"^2)/a^2` in
OA direction and
`bar(F_(OB))=(4Gm^(2))/a^2` in O1B direction
Since both are equal in opposite , cancel each other
`bar(F_(OC))=(Gm^2)/((sqrt3/2a)^2)=(4Gm^2)/(3a^2)` in OC direction
`:.` Net gravitational force on m `=(4Gm^2)/(3a^2)`