In the arrangement shown in figure the masses `m` of the bar and M of the wedge, as well as the wedge angle `alpha`, are known. The masses of the pulley and the thread are negligible. The friction is absent. Find the acceleration of the wedge M.

Explanation

To analyse the kinematic relations between the bodies, sketch the force diagram of each body as shown in the figure.
On the basis of force diagram, it is obvious that the wedge M will move towards right and the block will move down along the wedge. As the length of the thread is constant, the distance travelled by the block on the wedge must be equal to the distance travelled by the wedge on the floor. Hence `ds_(mM)=ds_M`. As `vecv_(mM)` and `vecv_M` do not change their directions and acceleration that's why `vecw_(mM)uarruarr vecv_(mM)` and `vecw_M uarr uarr vecv_M` and `w_mM=w_M=w` (say) and according the diagram of kinematical dependence is shown in figure.


As `vecw_m=vecw_(mM)+vecw_M`, so from traingle law of vector addition.
`w_m=sqrt(w_M^2+w_(mM)^2-2w_(mM)w_Mcos alpha)=wsqrt(2(1-cos alpha))` (1)
From `F_x=mw_x`, (for the wedge),
`T=Tcos alpha+N sin alpha=Mw` (2)
For the bar m let us fix (x-y) coordinate system in the frame of ground Newton's law in projection form along x and y axes (figure) gives
`mgsin alpha-T=mw_(m(x))=m[w_(mM(x))+w_(M(x))]`
`=m[w_(mM)+w_Mcos (pi-alpha)]=mw(1-cos alpha]` (3)
`mg cos alpha-N=mw_(m(y))=m[w_(mM(y))+w_(M(y))]=m[0+w sin alpha]` (4)
Solving the above Eqs. simultaneously, we get
`w=(mg sin alpha)/(M+2m(1-cos alpha))`
Note: We can study the motion of the block m in the frame of wedge also, alternately we may solve this problem using conservation of mechanical energy.