The system shown in fig, is released from rest. Calculate the tension in the string and the force exerted by the string on the pulleys, assuming pulleys and strings are massless.

Answer

`(mg sin theta)/(M+2 m(1-cos theta))`.

Explanation

Let acceleration of the wedge be a leftwards then acceleration of block, relative to the wedge will also be equal to a but down the incline. Hence, net acceleration of the block will be equal to vector sum of these two as shown in fig.
Now considering FBD's


For the wedge,
`N_(2)=T sin theta+N_(1) cos theta`..(i)
`T-T cos theta+N_(1) sin theta=Ma` ..(ii)
For the block,
`mg-N_(1) cos theta=ma sin theta` ..(iii)
`T cos theta-N_(1) sin theta=m(a-a cos theta)`
Solving above equation, `a=(mg sin theta)/(M+2 m(1-cos theta))`.