In fig, no relative motion takes place between the wedge and the block placed on it. The rod slides down wards over the wedge and pushes the wedge to move in horizontal direction, The mass of wedge is same as that of the block and is equal to M. If `tan theta=(1//sqrt(3))`, find the mass of rod. (Neglect rotation of the rod).

Explanation

Acc.of wedge should be `a_(1)=g tan theta`
`N sin theta=2 Ma_(1)=2Mg tan theta`
`implies N=(2 mg)/(cos theta)`
`mg-N cos theta=ma_(2)`
where `a_(2)=a_(1)tan theta`
solving, we get `m=(2M)/(1-tan^(2)theta)`