Question

A block of mass $m$ is lying on the edge having inclination angle $\alpha ={\tan }^{-1}\left(\frac{1}{5}\right)$. Wedge is moving with a constant acceleration, $a=2m{s}^{-2}$. The minimum value of coefficient of friction $\mu$, so that $m$ remains stationary with respect to wedge is :

• A. $\frac{2}{a}$
• B. $\frac{5}{12}$
• C. $\frac{1}{5}$
• D. $\frac{2}{5}$

Answer 1

The correct option is B $\frac{5}{12}$

$FBD$ of $m$ in frame of wedge

$N=mg\cos \left(\alpha \right)-ma\sin \left(\alpha \right)$

Now, $f=\mu N=ma\cos \alpha +mg\sin \alpha$

$\Rightarrow \mu =\frac{a\cos \alpha +g\sin \alpha }{g\cos \alpha -a\sin \alpha }$

$\Rightarrow \mu =\frac{a+g\tan \alpha }{g-a\tan \alpha }=\frac{5}{12}\Rightarrow \mu =\frac{5}{12}$