Three masses, each equal to M, are placed at the three corners of a square of side a. the force of attraction on unit mass at the fourth corner will be

Explanation

Force on m = 1 due to masses at corner 1 and 3 are `vec(F_(1))` and `vec(F_(3))` with `F_(1)=F_(3)=("GM")/(a^(2))` resultant of `vec(F_(1))` and `vec(F_(3))` is `F_(r)=sqrt(2)("GM")/(a^(2))` and its direction is along the diagonal
Force on m due to mass M at 2 is
`F_(2)=("GM")/((sqrt(2a))^(2))=("GM")/(2a^(2))`, `F_(r)` and `F_(2)` act in the same direction. Resultant of these two is the net force:
`F_("net")=(sqrt(2)"GM")/(a^(2))+("GM")/(2a^(2))=("GM")/(a^(2))[sqrt(2)+(1)/(2)]`, it is directed along the diagonal as shown in the figure.