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

Explanation

The force ast P due to A is
`F_A=(Gm^2)/((AP)^2)=(Gm^2)/(2d^2)`
along PA. the force at P due to C is
`F_C=(Gm^2)/((CP)^2)=(Gm^2)/(2d^2)`
along PC. The force at P due to B is
`F_B=(Gm^2)/d^2` along PB
The resultant of `F_A,F_B and F_c` will be along PB. Clearly `/_APB=/_BPC=45^@`
Component of `F_A` and `PB=F_Acos5^@ =(Gm^2)/(2sqrt2d^2)`
component of `F_C` along `PB=F_Ccos45^@=(Gm^2)/(2sqrt2d^2)`
component of `F_B along PB=(Gm^2)/d^2`
Hence the resultant of three forces is
`(Gm^2)/d^2)1/(2sqrt2)+1/2sqrt2)+1)=(Gm^2)/d^2(1+1/sqrt2)` along PB.