Six point masses of mass $m$ each are at the vertices of a regular hexagon of side $l$. Calculate the net gravitational force on any one of the masses.

(N/A) Let the vertices of the regular hexagon be $A, B, C, D, E, F$ in order. Consider the mass at vertex $A$. The forces acting on it due to the other five masses are:
$1$. Force due to mass at $B$: $F_1 = \frac{Gm^2}{l^2}$ along $\vec{AB}$.
$2$. Force due to mass at $F$: $F_5 = \frac{Gm^2}{l^2}$ along $\vec{AF}$.
$3$. Force due to mass at $C$: $F_2 = \frac{Gm^2}{(AC)^2} = \frac{Gm^2}{3l^2}$ along $\vec{AC}$.
$4$. Force due to mass at $E$: $F_4 = \frac{Gm^2}{(AE)^2} = \frac{Gm^2}{3l^2}$ along $\vec{AE}$.
$5$. Force due to mass at $D$: $F_3 = \frac{Gm^2}{(AD)^2} = \frac{Gm^2}{4l^2}$ along $\vec{AD}$.
By symmetry,the components of $F_1$ and $F_5$ perpendicular to the diagonal $AD$ cancel out. Their components along $AD$ are $F_1 \cos 30^{\circ} + F_5 \cos 30^{\circ} = 2 \cdot \frac{Gm^2}{l^2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}Gm^2}{l^2}$.
Similarly,the components of $F_2$ and $F_4$ perpendicular to $AD$ cancel out. Their components along $AD$ are $F_2 \cos 30^{\circ} + F_4 \cos 30^{\circ} = 2 \cdot \frac{Gm^2}{3l^2} \cdot \frac{\sqrt{3}}{2} = \frac{Gm^2}{\sqrt{3}l^2}$.
The force $F_3$ is already along $AD$.
Total force $F_{net} = \frac{\sqrt{3}Gm^2}{l^2} + \frac{Gm^2}{\sqrt{3}l^2} + \frac{Gm^2}{4l^2} = \frac{Gm^2}{l^2} (\sqrt{3} + \frac{1}{\sqrt{3}} + \frac{1}{4}) = \frac{Gm^2}{l^2} (\frac{3+1}{\sqrt{3}} + \frac{1}{4}) = \frac{Gm^2}{l^2} (\frac{4}{\sqrt{3}} + \frac{1}{4}) = \frac{Gm^2}{l^2} (\frac{16+\sqrt{3}}{4\sqrt{3}})$.