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 due to masses at 1 and 3 are `F_(1) and F_(3)` with
`F_(1)=F_(3)=(GM)/(a^(2))`
Resultant of `F_(1)and F_(3)` is `F_(r) =sqrt(2)(GM)/(a^(2))`
and its direction is along the diagonal, i.e., toward 2.
Force on m due to mass m at 2 is `F_(2)=(GM)/((sqrt(2a))^(2))=(GM)/(2a^(2)) and F_(r)` and `F_(2)` act in the same direction
Resultant of these two is 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 figure.