Question

A block of mass $m$ is lying on a rough inclined plane having an inclination $\alpha=\tan ^{-1}\left(\frac{1}{5}\right)$. The inclined plane is moving horizontally with a constant acceleration of $a=2\, ms ^{-2}$ as shown in the figure. The minimum value of coefficient of friction, so that the block remains stationary with respect to the inclined plane is (Take, $g=10\, ms ^{-2}$ )

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

Answer 1

Answer: (B)

The block-plane system is shown in the figure,

So, from the above free body diagram (FBD), downward acceleration,
$a_{\text {down }}=g\, \sin \,\alpha + a\, \cos \,\alpha \ldots $(i)
and the upward or opposing acceleration,
$a_{\text {up }}=\mu(g \,\cos \,\alpha-a \,\sin\, \alpha)$
For block to be stationary,
$a_{\text {down }}=a_{\text {up }} $ [from Eqs. (i) and (ii)]
$\Rightarrow \mu=\frac{g\, \sin\, \alpha + a \,\cos \,\alpha}{g \,\cos \,\alpha-a\, \sin\, \alpha} \dots$ (iii)
Divided the Eq. (iii) by $\sin \alpha$ and putting the values,
$\mu=\frac{10+2 \frac{1}{\tan \alpha}}{10 \frac{1}{\tan \alpha}-2}$
$=\frac{10+2 \times 5}{10 \times 5-2} $
$\mu=\frac{20}{48}=\frac{5}{12}$