Question

The minimum force required to moves body up an inclined plane is three times for minimum force required to prevent it the sliding down the plane. If the coefficient of friction between the body and the inclined plane is $\frac{1}{2\sqrt{3}}$, the angle of the inclined plane is:

• A. ${60}^o$
• B. ${45}^o$
• C. ${30}^o$
• D. ${15}^o$

Answer 1

Answer: (C)

${30}^o$

Force acting along the inclined plane due to gravity on the body = $mgsin\theta$ (downwards always)

When body is tried to move upwards, the frictional force acts downwards and is equal to $\mu mgcos\theta$.

When body is let to slide down, the frictional force acts upwards and is equal to $\mu mgcos\theta$.

Force required to just start it moving upwards = $F_1=mgsin\theta +\mu mgcos\theta$

Force required to prevent the body from sliding down $=F_2=mgsin\theta -\mu mgcos\theta$

It is given that $F_1=3F_2$
$⟹(mgsin\theta +\mu mgcos\theta )=3(mgsin\theta -\mu mgcos\theta )$

$⟹tan\theta =2\mu =\frac{1}{\sqrt{3}}$

$⟹\theta =ta{n}^{-1}\frac{1}{\sqrt{3}}$

$={30}^{\circ }$