Two wedges each of mass 600 g are placed next to each other on a rough horizontal surface. The coefficient of static friction between the wedges and the surface is 0.4. A cube of mass 'M' is balanced on the wedges as shown in the figue. If there is no friction between the cube and wedges, the largest mass 'M' of the cube that can be balanced without motion of the wedges is ______ kg.

The normal reactions from the wedges must balance the weight of the block in order to keep the system at equilibrium.

$2\mathrm{N} \cos {45}^{\mathrm{o}}=\mathrm{Mg} \Rightarrow \sqrt{2} \mathrm{N}=\mathrm{Mg} (\mathrm{since}, \mathrm{\theta }={45}^{\mathrm{o}})$

the horizontal component of the normal reaction is given by $\mathrm{N} \sin {45}^{\mathrm{o}}=\frac{\mathrm{Mg}}{\sqrt{2}}.\frac{1}{\sqrt{2}}=\frac{\mathrm{Mg}}{2}$...(i)

And the horizontal component of the normal reaction is balanced by the frictional force offered by the wedge.

$\mathrm{f}=\mathrm{\mu }(\mathrm{Ncos\theta }+\mathrm{mg})$

$\mathrm{f}=\mathrm{\mu }\left(\frac{\mathrm{Mg}}{\sqrt{2}}\times \frac{1}{\sqrt{2}}+\mathrm{mg}\right)=\mathrm{\mu g}\left(\frac{\mathrm{M}}{2}+\mathrm{m}\right)$....(ii)

from (i) and (ii)

$\frac{\mathrm{\mu Mg}}{2}+\mathrm{\mu mg}=\frac{\mathrm{Mg}}{2}$

$\mathrm{M}=\frac{2\mathrm{\mu m}}{(1-\mathrm{\mu })}$

On putting the values,

$\mathrm{M}=\frac{2\times 0.4\times 600}{(1-0.4)}=800 \mathrm{g} = 0.8 \mathrm{kg}$