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, $a=2 ms ^{-2}$. 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

Answer: (B)

FBD of $m$ in frame of wedge

$N=m g \cos (\alpha)-m a \sin (\alpha)$
Now, $f=\mu N=\operatorname{macos} \alpha+\operatorname{mg} \sin \alpha$
$\Rightarrow \mu=\frac{a \cos \alpha+g \sin \alpha}{g \cos \alpha-a \sin \alpha}$
$\Rightarrow \mu=\frac{a+g \tan \alpha}{g-a \tan \alpha}=\frac{6}{12} $
$\Rightarrow \mu=\frac{5}{12}$