Six particles each of mass 'm' are placed at six verties A,B,C,D,Eand F of a regular hexagon of side 'a'. A seventh particle of mass 'M' is kept at center 'O' of the hexagon.
(a)Find net force on 'M'.
(b)Find net force on 'M' if particles at A is removed.
(c) Find net force on 'M' if particles at A and C are removed .

Answer: A::B::DSolution: (a) $F_A=F_B=F_D=F_E=F_F=\frac{GmM}{a^2}=F$ (say)

So, net force will be zero, as three pairs of equal and opposite forces are acting on 'M' at O.
(b) When paricles at A is removed then $F_A$ is removed . So, there is no force to cancel $F_D$.
$\therefore F_{\text{net}}=F_D=\frac{GMm}{a^2}$ (towards D)
(c) When particles at A and C are removed then, $F_A$ and $F_C$ are removed.$F_B$ and $F_B$ and $F_E$ are still cancelled. So, net force is the resultant of two forces $F_D$ and $F_F$ of equal magnitudes acting at ${120}^{\circ }$. So, the resultant will pass through the centre ot towards E. magnitude of this resultant is
$F_{\text{net}}=\sqrt{F^2+F^2+2\left(F\right)\left(F\right){\cos 120}^{\circ }}$
$=F=\frac{GMm}{a^2}$ (towards E)