Question

In the arrangement shown in the figure above 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. ($Friction=k\times NormalReaction$)

• A. $w_m=\frac{g\sqrt{3}}{(1+k+\frac{M}{m})}$
• B. $w_m=\frac{2g\sqrt{2}}{(k+\frac{M}{m})}$
• C. $w_m=\frac{g\sqrt{2}}{(2+k+\frac{M}{m})}$
• D. $w_m=\frac{2g\sqrt{3}}{(1+k+\frac{M}{m})}$

Answer 1

Answer: (C)

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

Let us draw free body diagram of each body and fix the coordinate system, as shown in figure below. After analysing the motion of $M$ and $m$ on the basis of force diagrams, let us draw the kinematical diagram for accelerations (figure shown below).
As the length of threads are constant so,
$d{s}_{mM}=d{s}_M$ and as $\overrightarrow{v_{mM}}$ and $\overrightarrow{v_M}$ do not change their directions that why
$|\overrightarrow{w_{mM}}|=|\overrightarrow{w_M}|=w$ (say) and
$\overrightarrow{w_{mM}}\upuparrows \overrightarrow{v_M}$ and $\overrightarrow{w_M}\upuparrows \overrightarrow{v_M}$
As $\overrightarrow{w_m}=\overrightarrow{w_{mM}}+\overrightarrow{w_M}$
so from the triangle law of vector addition
$w_m=\sqrt{2}w$ (1)
From the equation $F_x=m{w}_x$, for the wedge and block
$T-N=Mw$, (2)
and $N=mw$ (3)
Now, from the equation $F_y=m{w}_y$, for the block
$mg-T-kN=mw$ (4)
Simultaneous solution of equations (2), (3) and (4) yields
$w=\frac{mg}{(km+2m+M)}=\frac{g}{(k+2+\frac{M}{m})}$
Hence using equation (1)
$w_m=\frac{g\sqrt{2}}{(2+k+\frac{M}{m})}$