4 particles having masses m,2m,3m and 4m are placed at the 4 corners of a square of edge a.Find the gravitational acting on a particle of mass m placed at the center?

 We have a square ABCD. Suppose the masses m,2m,3m and 4m are placed at the corners of square ABCD respectively.
At the centre 'O' a mass m is placed.

Force due to the mass at the centre is given by,

$$\begin{gathered} {\mathrm{F}}_{\mathrm{A}}=\frac{\mathrm{G}\times \mathrm{m}\times \mathrm{m}}{({\displaystyle \frac{\mathrm{a}}{\sqrt{2}}}{)}^2}=\frac{2{\mathrm{Gm}}^2}{{\mathrm{a}}^2} \mathrm{along} \mathrm{OA} \\ \mathrm{Force} \mathrm{due} \mathrm{to} \mathrm{mass} \mathrm{at} 3\mathrm{m} \mathrm{on} \mathrm{the} \mathrm{mass} \mathrm{at} \mathrm{the} \mathrm{centre} \mathrm{is} \mathrm{given} \mathrm{by}, \\ {\mathrm{F}}_{\mathrm{C}}=\frac{\mathrm{G}\times \mathrm{m}\times 3\mathrm{m}}{({\displaystyle \frac{\mathrm{a}}{\sqrt{2}}}{)}^2}=\frac{6{\mathrm{Gm}}^2}{{\mathrm{a}}^2} \mathrm{along} \mathrm{OC} \\ \mathrm{Resultant} \mathrm{force}, {\mathrm{F}}_{\mathrm{AC}}={\mathrm{F}}_{\mathrm{C}}-{\mathrm{F}}_{\mathrm{A}} \\ =\frac{6{\mathrm{Gm}}^2}{{\mathrm{a}}^2}-\frac{2{\mathrm{Gm}}^2}{{\mathrm{a}}^2}=\frac{4{\mathrm{Gm}}^2}{{\mathrm{a}}^2} \mathrm{along} \mathrm{OC} \\ \mathrm{Similarly}, \\ \mathrm{Force} \mathrm{due} \mathrm{to} \mathrm{mass} 2\mathrm{m} \\ {\mathrm{F}}_{\mathrm{B}}=\frac{\mathrm{G}\times \mathrm{m}\times 2\mathrm{m}}{({\displaystyle \frac{\mathrm{a}}{\sqrt{2}}}{)}^2}=\frac{4{\mathrm{Gm}}^2}{{\mathrm{a}}^2} \mathrm{along} \mathrm{OB} \\ \mathrm{Force} \mathrm{due} \mathrm{to} \mathrm{mass} 4\mathrm{m} \\ {\mathrm{F}}_{\mathrm{D}}=\frac{\mathrm{G}\times \mathrm{m}\times 4\mathrm{m}}{({\displaystyle \frac{\mathrm{a}}{\sqrt{2}}}{)}^2}=\frac{8{\mathrm{Gm}}^2}{{\mathrm{a}}^2} \mathrm{along} \mathrm{OD} \\ \mathrm{Resultant} \mathrm{force},{\mathrm{F}}_{\mathrm{DB}}={\mathrm{F}}_{\mathrm{D}}-{\mathrm{F}}_{\mathrm{B}}=\frac{4{\mathrm{Gm}}^2}{{\mathrm{a}}^2} \mathrm{along} \mathrm{OD} \\ \mathrm{Now}, \\ \mathrm{The} \mathrm{magnitude} \mathrm{of} \mathrm{the} \mathrm{resultant} \mathrm{of} {\mathrm{F}}_{\mathrm{AC}} \mathrm{and} {\mathrm{F}}_{\mathrm{DB}} \\ \mathrm{F}=\sqrt{({\mathrm{F}}_{\mathrm{AC}}{)}^2+({\mathrm{F}}_{\mathrm{DB}}{)}^2} (\mathrm{Theses} \mathrm{forces} \mathrm{are} \mathrm{at} \mathrm{right} \mathrm{angles}) \\ \mathbf{F}\mathbf{=}\mathbf{4}\frac{\sqrt{\mathbf{2}}{\mathbf{Gm}}^{\mathbf{2}}}{{\mathbf{a}}^{\mathbf{2}}} \end{gathered}$$

$$\begin{gathered} \mathrm{tan\alpha }=\frac{{\mathrm{F}}_{\mathrm{AC}}}{{\mathrm{F}}_{\mathrm{DB}}} \\ \mathrm{\alpha }={45}^0 \\ \mathrm{So}, \frac{{\mathrm{F}}_{\mathrm{AC}}}{{\mathrm{F}}_{\mathrm{DB}}}=\tan {45}^0=1 \\ \mathrm{i}.\mathrm{e}.\mathrm{parallel} \mathrm{to} \mathrm{AD}. \end{gathered}$$