Three particles A, B, and C each having mass m are kept along a straight line with AB = BC = l. A fourth particle D is kept on the perpendicular bisecter of AC at a distance l from B. Determine the gravitational force on D.

Explanation

`CD = AD = sqrt(AB^2 + BD^2) = sqrt2 l`
Gravitational force on D = Vector sum of gravitational forces due to A, B and C.

Force due to A` = (Gmm)/((AD)^2) = (Gm^2)/(2l^2).` This will be along `bar(DA)`
Force due to `C = (Gmm)/((CD)^2) = (Gm^2)/(2l^2) `. This is along `bar(DC)`
Force due to `B = (Gmm)/((BD)^2) = (Gm^2)/(l^2) `. This is along `bar(DB)`
We can resolve the forces along horizontal and vertical directions.
Let the unit vector along horizontal direction`bar( AC)` be hati and the vertical direction `bar(BD)` be `hatj`
Net horizontal force on D
` = (Gm^2)/(2l^2) cos 45^@ (-hati) + (Gm^2)/(l^2) cos 90^@ (hati) + (Gm^2)/(2l^2) cos 45^@ (hati)`
` = (-Gm^2)/(2sqrt2 l^2 ) +(Gm^2)/(2sqrt2 l^2) = 0`
Net vertical force on D
` = (Gm^2)/(2l^2) cos 45^@ (- hatj) + (Gm^2)/(l^2) (-hatj) + (Gm^2)/(2l^2) cos 45^@ (-hatj)`
`= (-Gm^2)/(l^2) ((1)/(sqrt2) + 1) (hatj)`
` = (Gm^2)/(l^2) ((1)/(sqrt2)+ 1) (-hatj)`
`(-hatj)` shows that the net force is directed along DB.