In the arrangement shown in figure the masses of the wedge M and the body m are known. The appreciable friction exists only between the wedge and the body m, the friction coefficient being equal to k. The masses of the pulley and the thread are negligible. Find the acceleration of the body m relative to the horizontal surface on which the wedge slides.

Explanation

Let us draw free body diagram of each body and fix the coordinate system, as shown in the figure. After analysing the motion of M and m on the basis of force diagram, let us draw the kinematical diagram for accelerations (figure).
As the length of threads are constant so,
`ds_(mM)=ds_M` and as `vecv_(mM)` and `vecv_M` do not change their directions that why
`|vecw_(mM)|=|vecw_M|=w` (say) and
`vecw_(mM)uarr uarr vecv_M` and `vecw_Muarr uarr vecv_M`
As `vecw_M=vecw_(mM)+vecw_M`
so, from the traingle law of vector addition
`w_m=sqrt2w` (1)
From the Eq. `F_x=mw_x`, for the wedge and block:
`T-N=Mw`, (2)
and `N=mw` (3)
Now, from the Eq. `F_y=mw_y`, for the block
`mg-T-kN=mw` (4)
Simultaneous solution of Eqs. (2), (3) and (4) yields:
`w=(mg)/((km+2m+M))=(g)/((k+2+M//m))`
Hence using Eq. (1)
`w_m=(gsqrt2)/((2+k+M//m))`