Question

A block of mass 5 kg is resting on an inclined plane. The inclination of the plane to the horizontal direction is gradually increased. It is found that, when the angle of inclination is ${30}^0$, the block just begins to slide down the plane. The coefficient of sliding friction ${\mu }_s$ between the block and the plane is :-

• A. ${\mu }_s=sin{30}^0$
• B. ${\mu }_s=cos{30}^0$
• C. ${\mu }_s=tan{30}^0$
• D. ${\mu }_s=cot{30}^0$

Answer 1

Answer: (C)

${\mu }_s=tan{30}^0$

When forces are in equilibrium, $f_s=w\sin \theta$ and $N=w\cos \theta$
$\therefore {\mu }_s=\frac{f_s}{N}=\frac{wsin\theta }{wcos\theta }=tan{30}^0$