Question

In the arrangement show in fig., 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 ground.

Answer 1

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 diagrams, let us draw the kinematical diagram for acceleration (Fig).

As the length of threads are constant so,

$d{s}_{mM}=d{s}_M$ and as $\overrightarrow{v_{mM}}$ and $v_M$ so not change their directions that why

$|\overrightarrow{w_{mM}}|=|\overrightarrow{w_M}|=w$ (say) and

$\overrightarrow{w_{mM}}\uparrow \uparrow \overrightarrow{v_M}$ and $\overrightarrow{w_M}\uparrow \uparrow \overrightarrow{v_M}$

As $\overrightarrow{w_m}=\overrightarrow{w_{mM}}+\overrightarrow{w_M}$

so, from the triangle law of vector addition

From the Eq. $F_x=m{w}_x$, for the wedge and block:

$T-N=Mw$

and $N=mw$

Now, from the Eq. $F_y=m{w}_y$, for the block

$mg-T-kN=mw$

Simultaneous solution of Eqs $\left(2\right),\left(3\right)$ and $\left(4\right)$ yields:

$w=\frac{mg}{(km+2m+M)}=\frac{g}{(k+2+M/m)}$

Hence using Eq. $\left(1\right)$

$w_m=\frac{g\sqrt{2}}{(2+k+M/m)}$