In figure if $\mathrm{\mu }=0.1$ is the coefficient of friction between the wedge and the horizontal surface then for what maximum value of m in kg, the wedge remains at rest [take M =18 kg and $\mathrm{\theta }={45}^{\circ }$]

F.B.D of m
 

$\mathrm{N}=\mathrm{mgcos}\mathrm{\theta }$

$\mathrm{mgsin}\mathrm{\theta }=\mathrm{ma}\Rightarrow \mathrm{a}=\mathrm{gsin}\mathrm{\theta }$
F.B.D. of M

$\begin{array}{l}\mathrm{R}=\mathrm{Ncos}\mathrm{\theta }+\mathrm{Mg}\mathrm{\mu R}=\mathrm{Nsin}\mathrm{\theta }\\ \Rightarrow \mathrm{\mu }\cdot (\mathrm{Ncos}\mathrm{\theta }+\mathrm{Mg})=\mathrm{Nsin}\mathrm{\theta }\\ \Rightarrow \mathrm{\mu Mg}=\mathrm{N}[\sin \mathrm{\theta }-\mathrm{\mu cos}\theta ]\\ \Rightarrow \mathrm{\mu Mg}=\mathrm{mgcos}\mathrm{\theta }[\sin \mathrm{\theta }-\mathrm{\mu cos}\mathrm{\theta }]\\ \Rightarrow \mathrm{m}=\frac{\mathrm{\mu M}}{\cos \mathrm{\theta }[\sin \mathrm{\theta }-\mathrm{\mu cos}\mathrm{\theta }]}\end{array}$
After substituting the values we get m = 4kg .