Question

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.​

Answer 1

Explanation:

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.

Answer 2

Answer:

Explanation:

Answer :

A::B::D

Solution :

Fnet=2(F1cos45∘)+F2

=2–√F1+F2=(2–√)Gmm(2d−−√)2+Gmmd2

=(2–√+12–√)Gm2d2 (Along PB)