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, D

Explanation

(a) `F_(A) =F_(B) =F_(D) =F_(E) =F_(F) =(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)`.
`:.F_("net") = F_(D )= (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^(@)`. So, the resultant will pass through the centre ot towards E. magnitude of this resultant is
`F_("net") = sqrt(F^(2)+F^(2)+2(F)(F)cos120^(@))`
`= F=(GMm)/(a^(2))` (towards E)