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 $a=2m{s}^{-2}$ 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, $g=10m{s}^{-2}$ )

• 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,
$a_{\text{down }}=g\sin \alpha +a\cos \alpha \dots$(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,
$\mu =\frac{10+2\frac{1}{\tan \alpha }}{10\frac{1}{\tan \alpha }-2}$
$=\frac{10+2\times 5}{10\times 5-2}$
$\mu =\frac{20}{48}=\frac{5}{12}$