Question

Three points A , B and C each of mass 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

Hence the net force along PB is F(net) = (√2 +1 /√2) Gm^2 / d^2

Explanation:

Force due to B will be  

F( B) = Gm^2 / d^2 (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

F(A) = F(C) = Gm^2 / d^2

Thus, resultant force is

F(net) = 2(F1 Cos(45∘) + F2

F(net) =  √ 2  F1 + F2 =  ( √ 2 )Gmm  / ( √ 2d )^2  +  Gmm  / d^2

F(net) = (√2 +1 /√2) Gm^2 / d^2  (Along PB)

Hence the net force along PB is F(net) = (√2 +1 /√2) Gm^2 / d^2

 

Answer 2

Given :

Mass of each three particles :

= m

They are placed in  a line and the distance between each two particle i.e. ;

AB= BC = d

Fourth particle P is placed perpendicular to AC and at a distance :

= d from B

Mass of fourth particle :

m

To Find :

Gravitational force on the fourth particle P = ?

Solution :

Distance of P from A , B and C are :

$\sqrt{2d} , d$ and $\sqrt{2d}$ respectively

∴A and C are at a equal distance from P , so magnitude of force due to A and C will be equal which is ;

$F_{A}=F_{B}=\frac{Gm^{2} }{2d^{2} }$

Since both are perpendicular to each other so their resultant will be :

$F_{AB}=\sqrt{F_{A}^{2}+F_{B}^{2} } \\\\F_{AB}=\frac{Gm^{2} }{\sqrt{2} d}$

Now force due to B at C will be :

$F_{C}=\frac{Gm^{2} }{d^{2} }$

So by superposition theorem net gravitational acting on p will be :

$F_{net}=F_{AC}+F_B}$

$F_{net}=\frac{Gm^{2} }{\sqrt{2} d^{2} } +\frac{Gm^{2} }{d^{2} } \\\\F_{net}=(\frac{1}{\sqrt{2} } +1)\frac{Gm^{2} }{d^{2} }$

So the gravitational force on P is $\frac{(\sqrt{2}+1) }{\sqrt{2} } \times \frac{Gm^{2} }{d^{2} }$ N.