Question

A block of mass M is tied to one end of a massless rope. The other end of the rope is in the hands of a man of mass 2M as shown in Fig. The block and the man are resting on a rough wedge of mass M. The whole system is resting on a smooth horizontal surface. The man starts walking towards right while holding the rope in his hands. Pulley is massless and frictionless. Find the displacement (in m) of the wedge when the block meets the pulley. Assume wedge is sufficiently long so that man does not fall down.

Answer 1

Let x be displacment of wedge w.r.t ground, y be displacmenet of man w.r.t ground and z be displacement of block w.r.t. ground. As the length of string will remain same, so $(l_1+z-x)+(l_2+y-x)=l_1+l_2$
$\Rightarrow z+y=2x\left(i\right)$

Also, if the pulley and the block meet, then we can write
$x=z+2\left(ii\right)$
$\Delta x_{cm}=0\Rightarrow \frac{Mx+2My+Mz}{4M}=0$
$\Rightarrow x+2y+z=0\left(iii\right)$
From Eqs.(i),(ii) and (iii),
$x=\frac{-1}{2}m,y=\frac{3}{2}m,z=\frac{-5}{2}m$