Question

A block of mass 10 Kg is placed on an inclined plane. When the angle of inclination is ${30}^o$, the block just beings to slide down the plane. find the static force of friction if $\mu =0.5$.

Answer 1

Given

Angle of inclination of $\theta ={30}^o$

Normal reaction, $N=mg\cos {30}^o=10\times 10\times \frac{\sqrt{3}}{2}=50\sqrt{3}$

Limiting Friction force, $F_r=\mu N=\mu 50\sqrt{3}N$

Hence, during motion limiting friction force is $=0.5\times 50\sqrt{3}=25\sqrt{3}$