Question

The minimum force required to move a body up an inclined plane is three times the minimum force required to prevent it from 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. ${60}^0$
• B. ${45}^0$
• C. ${30}^0$
• D. ${15}^0$

Answer 1

Answer: (C)

${30}^0$

To hold body stationary, force required: $F_1=mgsin\theta -\mu mgcos\theta$

The force required to body move up is given by, $F_2=mg(sin\theta +\mu cos\theta )$

$F_2=3\times F_1\Rightarrow \theta ={30}^o$
$\Rightarrow mg(\sin \theta +\mu \cos \theta )=3mg(\sin \theta -\mu \cos \theta )$
$\Rightarrow 2mg\sin \theta =4mg\mu \cos \theta$
$\Rightarrow \tan \theta =2\mu =\frac{1}{\sqrt{3}}$
$\Rightarrow \theta ={30}^0$