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 $P$ ofsame mass, placed at a distance $d$ from the particle $B$ on the perpendicular bisector of the line $AC$.

Answer 1

Force due to B will be

$F_B=\frac{G{m}^2}{d^2}$ 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

$F_A=F_C=\frac{G{m}^2}{2d^2}$

Thus, resultant force is

$F=F_B+\frac{F_A}{\sqrt{2}}+\frac{F_C}{\sqrt{2}}=\frac{G{m}^2}{d^2}+\frac{G{m}^2}{\sqrt{2}{d}^2}=\frac{G{m}^2}{d^2}(1+\frac{1}{\sqrt{2}})$

Answer is $\frac{G{m}^2}{d^2}(1+\frac{1}{\sqrt{2}})$.