Question

Three particles A,B and C, each of mass 'm' are placed in a line AB = BC = d. Find the gravitational force on a fourth particle P of same mass,placed at a distance d from the particle on the perpendicular bisector of line AC.

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Answer 1

Force due to B will be  

FB=d2Gm2 downwards.

Horizontal components of forces due to A and C will be cancelled by each other and vertical components will add up.

Force due to A and C is

FA=FC=2d2Gm2

Thus, resultant force is

F=FB+2FA+2FC=d2Gm2+2d2Gm2=d2Gm2(1+21)

Answer is d2Gm2(1+21).

Answer 2

Answer:

HELLO MATE.

Given Three particles A, B and C, each of mass m, are placed in a line with AB = BC = d. Find the  gravitational force on a fourth particle P of same mass, placed at a distance d from the particle B on  the perpendicular bisector of the line AC.  

Fa = Gm^2 / (AP)^2

     = Gm^2 / 2d^2 along PA

The force at P due to C is

Fc = Gm^2 / (CP)2

     = Gm^2 / 2d^2 along PC.

The force at P due to B is

Fb = Gm^2 / d^2 along PB

Now resultant of Fa, Fb and Fc will be along PB

So angle APB = angle BPC = 45 degree

So Fa along PB = Fa cos 45 degree = Gm^2 / 2√2d^2

So Fc along PB = Fc cos 45 degree = Gm^2 / 2√2d^2

So Fb along PB = Gm^2 / d^2

The resultant of three forces will be  

Gm^2 / d^2 (1 / 2√2 + 1/2√2 + 1)

Gm^2 / d^2 (1 + 1 / √2) net force along PB.

Hope it helps you mate.

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