Question

The coefficient of friction between wedge A and block B is $\mu$. The wedge A accelerates at rate $am/s^2$ such that block B will remain at rest with respect to the wedge. Then for what value of $\theta$, we get $a=g\left[\frac{1+\mu }{1-\mu }\right]$?

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${53}^{\circ }$

Answer 1

The correct option is B ${45}^{\circ }$

Free Body Diagram of Block B. $ma$ is the pseudoforce.


EoM perpendicular to wedge

$\Sigma f_y=(mg\cos \theta +ma\sin \theta -N)=0$

EoM parallel to wedge

$\Sigma f_x=(mg\sin \theta +\mu N-ma\cos \theta )=0$

(when B is just about to slip, frictional force $f=\mu N)$

$\Rightarrow mg\sin \theta +\mu (mg\cos \theta +ma\sin \theta )-ma\cos \theta =0$

(substituting $N$)

$\Rightarrow a=\frac{(g\sin \theta +\mu g\cos \theta )}{\cos \theta -\mu \sin \theta }$
$=g\left(\frac{1+\mu cot\theta }{1-\mu tan\theta }\right)$

For $\theta ={45}^{\circ }$,

$a=g\left(\frac{1+\mu }{1-\mu }\right)$