A block of mass 2 kg rests on a rough inclined plane making an angle of 30⁰ with the horizontal. The coefficient of static friction between the block and the plane is 0.7. The frictional force on the block is

$\begin{array}{l}{\mathrm{F}}_{\mathrm{l}}=\mathrm{\mu mgcos}\mathrm{\theta }\\ {\mathrm{F}}_{\mathrm{l}}=0.7\times 2\times 10\times \cos {30}^{\circ }=12\mathrm{N}\end{array}$

(approximately) but when the block is lying on the inclined plane then component of weight down the plane = $\mathrm{mgsin\theta }$

$=2\times 9.8\times \sin {30}^{\circ }=9.8\mathrm{N}$, It means the body is stationary so static friction will work on it, Static friction = applied force = 9.8N