Figure shows a block of mass m placed on a smooth wedge of mass M. Calculate the minimum value of M' and tension in the string, so that the block of mass m will move vertically downward with acceleration 10 m s^-2 .

$\begin{array}{l}{\mathrm{M}}^{\mathrm{\prime }}\mathrm{g}-\mathrm{T}={\mathrm{M}}^{\mathrm{\prime }}\mathrm{a} ...\left(\mathrm{i}\right)\\ \mathrm{T}=\mathrm{Ma} ...\left(\mathrm{ii}\right)\\ {\mathrm{M}}^{\mathrm{\prime }}\mathrm{g}=\mathrm{a}\left(\mathrm{M}+{\mathrm{M}}^{\mathrm{\prime }}\right)\\ \Rightarrow \mathrm{a}=\frac{{\mathrm{M}}^{\mathrm{\prime }}\mathrm{g}}{\left(\mathrm{M}+{\mathrm{M}}^{\mathrm{\prime }}\right)}\end{array}$

$\mathrm{ma} \sin \mathrm{\theta }=\mathrm{mg} \cos \mathrm{\theta }\to$so that normal force is zero.

$\begin{array}{l}\mathrm{a}=\mathrm{g} \cot \mathrm{\theta }\\ \mathrm{g} \cot \mathrm{\theta }=\frac{{\mathrm{M}}^{\mathrm{\prime }}\mathrm{g}}{\left(\mathrm{M}+{\mathrm{M}}^{\mathrm{\prime }}\right)}\Rightarrow \cot \mathrm{\theta } \mathrm{M}+\cot \mathrm{\theta } {\mathrm{M}}^{\mathrm{\prime }}={\mathrm{M}}^{\mathrm{\prime }}\end{array}$

or      ${\mathrm{M}}^{\mathrm{\prime }}=\frac{\mathrm{M} \cot \mathrm{\theta }}{(1-\cot \mathrm{\theta })}, \mathrm{T}=\mathrm{Ma}=\mathrm{Mg} \cot \mathrm{\theta }=\mathrm{Mg}/\tan \mathrm{\theta }$