For the situation shown in figure, the block is stationary w.r.t. incline fixed in an elevator. The elevator is having an acceleration of $\sqrt{5}{\mathrm{a}}_0$ whose components are shown in the figure. The surface is rough and coefficient of static friction between the incline and block is ${\mathrm{\mu }}_{\mathrm{s}}$. Determine the magnitude of force exerted by incline on the block. (Take ${\mathrm{a}}_0=\mathrm{g}/2\text{ and }\mathrm{\theta }={37}^{\circ }, {\mathrm{\mu }}_{\mathrm{s}}=0.2.$) 

The FBD of block from lift frame is as shown in figure. From given data, as $\mathrm{m}\left(\mathrm{g}+{\mathrm{a}}_0\right)\sin \mathrm{\theta }>2{\mathrm{ma}}_0\cos \mathrm{\theta }$

So friction force acts upwards.

          $\begin{array}{l}\mathrm{f}=\mathrm{m}\left(\mathrm{g}+{\mathrm{a}}_0\right)\sin \mathrm{\theta }-2{\mathrm{ma}}_0\cos \mathrm{\theta }\\ =\frac{9\mathrm{g}}{10}-\frac{4\mathrm{mg}}{5}=\frac{\mathrm{mg}}{10}\\ \mathrm{N}=\mathrm{m}\left(\mathrm{g}+{\mathrm{a}}_0\right)\cos \mathrm{\theta }+2{\mathrm{ma}}_0\sin \mathrm{\theta }=\frac{9\mathrm{mg}}{5}\end{array}$

As       ${\mathrm{f}}_{\mathrm{L}}={\mathrm{\mu }}_{\mathrm{S}}\mathrm{N}=\frac{18\mathrm{mg}}{50}=\frac{9\mathrm{mg}}{25}>\mathrm{f}$ so static friction. 

Reaction force,

          $\mathrm{R}=\sqrt{{\mathrm{f}}^2+{\mathrm{N}}^2}=\frac{\mathrm{mg}}{5}\sqrt{\frac{1}{4}+9^2}=\frac{\mathrm{mg}\sqrt{13}}{2}$

Alternative solution:

Net force, $\overrightarrow{\mathrm{F}}-\mathrm{mg}\hat{\mathrm{i}}=\mathrm{m}\left({\mathrm{a}}_0\hat{\mathrm{j}}-2{\mathrm{a}}_0\hat{\mathrm{i}}\right)$

$\begin{array}{l}\Rightarrow \overrightarrow{\mathrm{F}}=\mathrm{m}\left(-2{\mathrm{a}}_0\widehat{\mathrm{i}}+\left({\mathrm{a}}_0+\mathrm{g}\right)\widehat{\mathrm{j}}\right)=\mathrm{m}\left(-\mathrm{g}\widehat{\mathrm{i}}+\frac{3\mathrm{g}}{2}\widehat{\mathrm{j}}\right)\\ \Rightarrow \mathrm{F}=\mathrm{m}\sqrt{{\mathrm{g}}^2+{\left(\frac{3\mathrm{g}}{2}\right)}^2}=\frac{\sqrt{13}\mathrm{mg}}{2}\end{array}$