Question

A block of mass $m$ is lying on a wedge having inclination of $\alpha ={\tan }^{-1}\left(\frac{1}{5}\right)$. Wedge is moving with a constant acceleration of $2{\text{m/s}}^2$. Find the minimum value of coefficient of friction so that $m$ remains stationary with respect to wedge. Take $g=10{\text{m/s}}^2$.

• A. $\frac{2}{9}$
• B. $\frac{5}{12}$
• C. $\frac{1}{5}$
• D. $\frac{2}{3}$

Answer 1

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

In the frame wedge frame, we have to assume a pseudo force $ma$ as shown in the FBD.

Perpendicular to the plane:
$N+ma\sin \alpha =mg\cos \alpha$
$\Rightarrow N=mg\cos \alpha -ma\sin \alpha$
Along the plane:
$mg\sin \alpha +ma\cos \alpha =f=\mu N$
$mg\sin \alpha +ma\cos \alpha =\mu (mg\cos \alpha -ma\sin \alpha )$
$\therefore \mu =\frac{mg\sin \alpha +ma\cos \alpha }{mg\cos \alpha -ma\sin \alpha }$
(dividing by $m\cos \alpha$)
$=\frac{g\tan \alpha +a}{g-a\tan \alpha }$
(substituting $\tan \alpha =\frac{1}{5}$ and $a=2m/s^2$)
$=\frac{5}{12}$