Question

A block mass m is kept on an inclined plane of a lift moving down with acceleration of 2 $m{s}^{-2}$. What should be minimum coefficient of friction to let the block move down with constant velocity?

• A. $\mu =\frac{1}{\sqrt{3}}$
• B. $\mu =0.4$
• C. $\mu =0.8$
• D. $\mu$ =not defined

Answer 1

The correct option is C $\mu =0.8$

Given: a block mass m is kept on an inclined plane of a lift moving down with acceleration of $2m/s^2$.

To find the minimum coefficient of friction to let the block move down with constant velocity

Solution:

From the above figure,

$mg\cos \theta -ma\cos \theta =N⟹N=m(g-a)\cos \theta$

$F_{sin}=\mu N=\mu \left[m\right(g-a\left)\cos \theta \right]⟹mg\sin \theta -ma\sin \theta =\mu \left[m\right(g-a\left)\cos \theta \right]⟹\sin \theta =\mu \cos \theta ⟹\mu =\frac{\sin \theta }{\cos \theta }=\tan \theta ⟹\mu =\tan {37}^{\circ }⟹\mu \approx 0.8$

is the minimum coefficient of friction to let the block move down with constant velocity