Three point mass bodies of masses `m, 2m` and `4m` are placed at `A, B` and `C` as shown in fig. where `AB = BC = l`.

Find the magnitude of the resulatant gravitational pull on body at `A` due to bodies at `B` and `C`.

Explanation

Refer to Fig. `AC = (AB^(2) + BC^(2))^(1//2)`
`= (l^(2) + l^(2))^(1//2) = l sqrt(2)`
Gravitational pull on body at `A` due to body at `B` is
`F_(1) = (Gm xx 2m)/(l^(2)) = (2Gm^(2))/(l^(2)) along AB`.
Gravitational pull on body at `A` due to body at `C` is
`F_(2) = (Gm xx 2m)/((lsqrt(2))^(2)) = (2Gm^(2))/(l^(2)) along AC`.
As, `AB = AC` and `/_ABC = 90^(@)` , so `/_BAC = 45^(@)`.
i.e., angule `theta` between `vec(F_(1))` and `vec(F_(2))` is `45^(@)`
Using parelogram law of forces, the resultant gravitaitonal force on body at `A` is
`F = sqrt(F_(1)^(2) + F_(2)^(2) + 2F_(1)F_(2) cos 45^(@))`
`=[((2Gm^(2))/(l^(2))) + ((2Gm^(2))/(l^(2)))^(2) + 2 xx (2Gm^(2))/(l^(2)) xx (2Gm^(2))/(l^(2)) xx (1)/(sqrt(2))]^(1//2)`
`= - (2Gm^(2))/(l^(2)) (1 + 1+ sqrt(2))^(1//2) = (3.696 Gm^(2))/(l^(2))`