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. $\frac{2}{5}$

Answer 1

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

In the FBD the pseudo force will be acting on the block towards left as the wedge is accelerating towards right,


Balancing the forces acting perpendicular to the plane:
$N+ma\sin \alpha =mg\cos \alpha$
$\Rightarrow N=mg\cos \alpha -ma\sin \alpha$
Along the plane
$mg\sin \alpha +ma\cos \alpha =f=\mu N$
$mg\sin \alpha +ma\cos \alpha =\mu (mg\cos \alpha -ma\sin \alpha )$
$\mu =\frac{g\tan \alpha +a}{g-a\tan \alpha }$,
$\mu =\frac{g\left(\frac{1}{5}\right)+a}{g-a\left(\frac{1}{5}\right)}$
putting the values we get,
$\mu =\frac{5}{12}$