Three equal masses m are placed at the three corners of an equilateral tringle 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

(a) As shows in Fig. `D` is the mid point of side `BC`. Gravitational force on a particle at `D` due to particle at `B` is equal and opposite to the graviational force on particle at `D` due to particle at `C`, i.e., `F_(B) = F_(C)`. So their resultant is zero.

The gravitational force on particle at `D` due to particle at `A` is
`F_(A) = (Gm m)/((AD)^(2)) = (G m^(2))/((r sin 60^(@))^(2))`
`= (4 G m^(2))/(3 r^(2))` along `DA`
`:.` resultant force on particle at `D`
`= (4 G m^(2))/(3 r^(2))` along `DA`
(b) The gravitational force of three particles at
`A, B` and `C` on particle at `O` are `vec(F_(A)), vec(F_(B))` and `vec(F_(C))`. They willl be equal in magnitude but inclined at an angle of `120^(@)` from each other, so the resultant force on mass `m` at `O` is
`vec(F) = vec(F_(A)) + vec(F_(B)) + vec(F_(C)) = 0`