A plank of mass m rests symmetrically on two wedges B and C of mass M. What is the acceleration of the plank? Neglect friction between all the contact surfaces.

Explanation

When the plank is released, let it fall through a distance and let both the wedges move through a distance x.
(i)`y= x tan theta`...(i)

On differnetiating this expression twice, we obtain `a=A tan theta` FBD of rod and wedge

Equation of wedge
`Sigma F_(x)=N sin theta=MA` ...(ii)
`Sigma F_(y)=N'-Ncos theta - Mg=0` ...(iii)
Equations of plank:
`Sigma F_(x)=N sin theta-N sin theta=0`....(iv)
`Sigma F_(y)=mg-2N cos theta =ma` ...(v)
From (ii), `N=(MA)/(sin theta)`
On substituting expression for N and a in (v), we obtain
`mg-(2Ma cos theta)/(sin theta)=(mA sin theta)/(cos theta)`
`implies A=(mg sin theta cos theta)/(m sin^(2)theta+2M cos^(2)theta)`.