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

Explanation

Force on `m=1` due to masses at corners 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 i.e. towards corner 2
Force on m due to mass M at 2 is `F_(2)=(GM)/((sqrt(2)a)^(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.