Let us consider the following diagram in which size point masses are placed at six sides.

A, B, C, D, E, F.

AC = AG + GC + = 2AG

= 2I cos 30º

= 2I\(\sqrt{\frac{3}{2}}\) = \(\sqrt3\) I = AE

AD = AH + HJ + JD

= I sin 30º + I + I sin 30º

= 2 I

Force on A due to B, F₁ = \(\frac{GMm}{I^2}\) along A to B

= \(\frac{Gm^2}{I^2}\)

Force on A due to C,F₂ = \(\frac{Gm.m}{(\sqrt3I)^2}\)

= \(\frac{Gm^2}{3I^2}\)  along A to C [\(\because\) AC = \(\sqrt3I\)]

Force on A due to D,F₃ = \(\frac{Gm.m}{(\sqrt2I)^3}\)

= \(\frac{Gm^2}{4I^2}\) along A to D [\(\because\) AD = 2I]

Force on A due to E,, F₄ = \(\frac{Gm.m}{(\sqrt3I)^2}\)

= \(\frac{Gm^2}{3I^2}\) along A to E

Force on A due to F, F₅ = \(\frac{Gm.m}{(I)^2}\) along A to F

= \(\frac{Gm^2}{I^2}\) along A to F

Resultant force due to F₁ and F₅,

F₁ = \(\sqrt{F_1^2+F^2_5+2F_2F_5\cos120^o}\)

= \(\frac{Gm^2}{I^2}\) along A to D [Angle b/w F₁ and F₅= 120º]

Resultant force due to F₂ and F₄,

F₁ = \(\sqrt{F_1^2+F^2_4+2F_2F_4\cos60^o}\)

= \(\frac{\sqrt3Gm^2}{3I^2}\) = \(\frac{Gm^2}{\sqrt3I^2}\) along A to D

\(\because\) Net force along A to D = F₁ + F₂ + F₃

= \(\frac{Gm^2}{I^2}+\frac{Gm^2}{\sqrt3I^2}+\frac{Gm^2}{4I^2}\)

= \(\frac{Gm^2}{I^2}\)\(\big(1+\frac{1}{\sqrt3}+\frac{1}{4}\big)\)