Three particles x, y, z are placed in a line as shown in the figure. A fourth particle at O is also placed at a perpendicular bisector of line xz. Calculate the total gravitational force at O. All the particle are of some mass.

Explanation

`F_(x)` = Force at O due to `x = (Gm^(2))/(2s^(2))` along Ox
`F_(y)` = Force at O due to `y = (Gm^(2))/(2s^(2))` along Oy
`F_(z)` = Force at O due to `z = (Gm^(2))/(2s^(2))` along Oz
The resultant of all forces `F_(x), F_(y), F_(z)` will be along Oy.
Component of `F_(x)` along `Oy = F_(x) cos 45^(@)`
`= (Gm^(2))/(2sqrt(2)s^(2))`
Component of `F^(y)` along Oy
`= (Gm^(2))/(s^(2))`
Component of `F_(y)` along `Oy = F_(y) cos 45^(@) = (Gm^(2))/(2sqrt(2)s^(2))`
Hence, the resultant of all of the forces is
`F = (Gm^(2))/(s^(2)2sqrt(2)) + (Gm^(2))/(s^(2)) + (Gm^(2))/(2sqrt(2)s^(2))`
`= (Gm^(2))/(s^(2))((1)/(2sqrt(2))+(1)/(2sqrt(2))+1)` along Oy