Question

A block of mass $m$ is kept on an inclined plane inside a lift moving down with acceleration of $2{\text{m/s}}^2$. What should be the coefficient of friction between block and inclined plane to let the block move down with constant velocity relative to lift?
(Take $g=10{\text{m/s}}^2$)

• A. $\frac{1}{\sqrt{3}}$
• B. $\frac{1}{\sqrt{5}}$
• C. $\frac{1}{\sqrt{6}}$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

The correct option is A $\frac{1}{\sqrt{3}}$

$F.B.D$ of block from lift frame.


Pseudo force,
$F_p=ma=2m\text{N}$
For constant velocity in accelerating frame.
Applying equilibrium condition perpendicular to inclined plane.
$N+ma\cos \theta =mg\cos \theta$
$\Rightarrow N=m\cos \theta (g-a)$
Kinetic friction will act on block.
$f=\mu N=\mu m\cos \theta (g-a)$
Applying equilibrium condition along the inclined plane.
$mg\sin \theta =f+ma\sin \theta$
$\Rightarrow mg\sin \theta =\mu m\cos \theta (g-a)+ma\sin \theta$
$\Rightarrow \mu =\frac{g\sin \theta -a\sin \theta }{\cos \theta (g-a)}$
$\Rightarrow \mu =\frac{10\sin {30}^{\circ }-2\sin {30}^{\circ }}{\cos {30}^{\circ }(10-2)}$
$\Rightarrow \mu =\frac{5-1}{8\times \frac{\sqrt{3}}{2}}$
$\Rightarrow \mu =\frac{1}{\sqrt{3}}$