In the system shown in figure all surfaces are smooth. Rod is moved by external agent with acceleration $9{ \mathrm{ms}}^{-2}$ vertically downwards. Force exerted on the rod by the wedge will be 

$\frac{9}{a}=\tan 37°=\frac{3}{4}\Rightarrow a=12 \mathrm{m}/{\mathrm{s}}^2$

Let N= normal reaction between rod and wedge. Then $N \sin 37°$ will provide the necessary ma force to the wedge

$\therefore N\sin 37°=ma=\left(10\right)\left(12\right)=120$

$$\begin{gathered} \therefore N=\frac{120}{\sin 37°}=\frac{120}{0.6} \\ =200 \mathrm{N} \end{gathered}$$