A block of mass m = 4 kg is placed over a rough inclined plane as shown in figure. The coefficient of friction between the block and the plane is $\mathrm{\mu } =$0.5. A force F= I 0 N is applied on the block at an angle ${30}^0$. The friction force between the block and wedge is

Drawing free body diagram of block, $\sum _{ }\mathrm{Fy} = 0$

$\Rightarrow \mathrm{N}+\mathrm{F} \sin {30}^0 = \mathrm{mg} \cos {37}^0$

or, $\mathrm{N} = \mathrm{mg} \cos {37}^0-\mathrm{F} \sin {30}^0$

    = $\left(4\right)\left(10\right)\left(\frac{4}{5}\right)-\left(10\right)\left(\frac{1}{2}\right)$

or, $\mathrm{N}\hspace{0.17em}= 27 \mathrm{N}--------\left(\mathrm{i}\right)$

    ${\mathrm{f}}_{\max } = \mathrm{\mu N} = 0.5\times 27 = 13.5 \mathrm{N}$

$\mathrm{mg} \sin {37}^0 = \left(4\right)\left(10\right)\left(\frac{4}{5}\right) = 32 \mathrm{N}$

and $\mathrm{F} \cos {30}^0 = \left(10\right)\left(\frac{\sqrt{3}}{2}\right) = 8.66 \mathrm{N}$

Now since $\mathrm{mg} \sin {37}^0 > {\mathrm{f}}_{\max } + \mathrm{F} \cos {30}^0$
Therefore block will slide down and friction will be kinetic.