In figure, man A is standing on a movable plank while man B is standing on a stationary platform. Both are pulling the string down such that the plank moves slowly up. As a result of this the string slips through the hands of the men. Find the ratio of length of the string that slips through the hands of A and B. 

Let the plank move up by x, then pulley 2 will move down by x. Let the end of string C moves down by a distance y.

Let the initial length of string passing over pulley 2,

   ${\mathrm{l}}_1+{\mathrm{l}}_2 ...\left(\mathrm{i}\right)$

After displacements x and y mentioned above, the lengths become

  $\left({\mathrm{l}}_1-2\mathrm{x}\right)+\left({\mathrm{l}}_2+\mathrm{y}-\mathrm{x}\right) ...\left(\mathrm{ii}\right)$

Equating Eqs. (i) and (ii), we get y = 3x

Length of string that slips through A is y + x = 4x and through B is y = 3x.

Required ratio$=\frac{4\mathrm{x}}{3\mathrm{x}}=\frac{4}{3}$