Question

The minimum force required to move a body up an inclined plane of inclination ${30}^0$ is found to be thrice the minimum force required to prevent it from sliding down the plane. The coefficient of friction between the body and the plane is

• A. $\frac{1}{\sqrt{3}}$
• B. $\frac{1}{2\sqrt{3}}$
• C. $\frac{1}{3\sqrt{3}}$
• D. $\frac{1}{4\sqrt{3}}$

Answer 1

Answer: (B)

$\frac{1}{2\sqrt{3}}$

Force required to move body upward, $F_1=mgsin\theta +\mu mgcos\theta$

Force required to prevent sliding, $F_2=mgsin\theta -\mu mg\cos \theta$
$mgsin\theta +\mu mgcos\theta =3(mgsin\theta -\mu mg\cos \theta )$

As, $F_1=3F_2\Rightarrow \mu =\frac{1}{2\sqrt{3}}$