A uniform rod of mass `M` and length `L` rests on two supports. A ball of mass `m=M//3` is placed at distance `L//8` from the center as shown. Find reactions at supports.

Explanation

Since the rod is in equilibrium
`SigmaF_(y)=0implies R_(A)+R_(B)=Mg+mg=(4mg)/(3)` ......(i)
Taking torque about `B`
`Sigma tau_(B)=0implies R_(A)(L)/(2)=mgL/4+mgL/8` ......(ii)
`R_(A)=(Mg)/(2)+(mg)/(4)=(Mg)/(2)+(Mg)/(12)=(7Mg)/(12)`
`R_(B)=(4Mg)/(3)-(7Mg)/(12)=(9Mg)/(12)=(3)/(4)Mg`
The reactions at supports are
`(7Mg)/(12)` and `(3Mg)/(4)`