Question

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 1

(a) Net force on M = 0

(b) Net force on M if particle from A is removed = $\frac{GMm}{a^{2} }$

(c) Net force on M if particles from A and C are removed = $\frac{GMm}{a^{2} }$

Refer the diagram attached

(a) To calculate the net force on M :

1. The forces Fa, Fb, Fc, Fd, Fe and Ff are equal in magnitude because the mass m and distance a is equal.
2. The forces (Fa and Fd), (Fb and Fe), (Fc and Ff) are opposite in direction.
3. Hence we have 3 pairs of equal and opposite forces acting on M.
4. Therefore, the net force on M will be zero.

(b) To calculate net force on M if particle at A is removed :

1. When the particle at A is removed we have two pairs of equal and opposite forces (Fb and Fe), (Fc and Ff).
2. The only unbalanced force is Fd.
3. Net force = Fd

                         = $\frac{GMm}{a^{2} }$ which will act towards D

(c) To calculate net force on M if particles at A and C are removed :

1. We will have one pair of equal and opposite forces ( Fb and Fe).
2. The two unbalanced forces acting are Fd and Ff. The net force will be the resultant of these two forces.
3. Fd and Ff are equal in magnitude and at an angle of 120°. The resultant will pass through the center and towards the point E.
4. $Net force = \sqrt{Ff^{2}+Fd^{2} + 2(Fd)(Ff)cos120}$

                        $= \sqrt{Ff^{2} + Ff^{2} + 2(Ff)(Ff)(\frac{-1}{2}) }$

                        $= \sqrt{2Ff^{2}-Ff^{2} }$

                        $= Ff\\\\= \frac{GMm}{a^{2} }$