Question

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 placed at the fourth corner.

Answer 1

$F_A=F_C=\frac{GM}{a^2}$

And $F_B=\frac{GM}{(\sqrt{2}a{)}^2}=\frac{GM}{2a^2}$

Now, the net force is resultant of the three forces.

The resultant of $F_A$ and $F_C$ is along DB that is along $F_B$ and as angle between them is ${90}^o$, hence their resultant is

$\sqrt{2}\frac{GM}{a^2}$

Hence, net force $F=\frac{\sqrt{2}GM}{a^2}+\frac{GM}{2a^2}$

$⟹F=\frac{GM}{2a^2}(2\sqrt{2}+1)$ towards B