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

Adjacent pair (each at $60°$ to the inward line): $2\times\dfrac{Gm^2}{l^2}\cos60°=\dfrac{Gm^2}{l^2}$. Next pair (each at $30°$): $2\times\dfrac{Gm^2}{3l^2}\cos30°=\dfrac{Gm^2}{\sqrt3\,l^2}$. Opposite mass (along the line): $\dfrac{Gm^2}{4l^2}$. …

Distances from the chosen mass

Place the six masses at the vertices of a regular hexagon of side $l$; for a regular hexagon, the centre-to-vertex distance also equals $l$. Pick one mass, call it $A$. The other five sit at three distinct distances from $A$:

• the two adjacent vertices are at distance $l$;
• the two vertices one further around are at distance $l\sqrt3$;
• the single diametrically opposite vertex is at distance $2l$.

Force from each group

Using $F=Gm^2/r^2$:

Taking components toward the centre

By the hexagon's symmetry, the two adjacent forces each make $60°$ with the inward line to the centre, so their sideways parts cancel and their radial parts add:

The two one-further forces each make $30°$ with the inward line:

The opposite mass pulls straight along the inward line:

Adding the contributions …

Step 1: regular hexagon side l = circumradius; from one vertex, neighbours at l, next-nearest at lsqrt3, opposite at 2l. Step 2: use mirror symmetry -- perpendicular components cancel, only centre-directed components survive. Step 3: adjacent pair (dist l, 60 deg each): combined = 2(Gm^2/l^2)cos60 = Gm^2/l^2. Step 4: next-nearest pair (dist lsqrt3, 30 deg each): combined = 2*(Gm^2/ …