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

 

$\frac{9}{a}=\tan {37}^0=\frac{3}{4}$

$\therefore a=12m/s^2$

Let N = normal reaction between rod and wedge. Then N $\sin {37}^0$ will provide the  necessary ma force to the wedge

$\therefore N\sin {37}^0=ma=\left(10\right)\left(12\right)=120$

$\therefore N=\frac{120}{\sin {37}^0}=\frac{120}{0.6}=200N$