Question

A block of mass $m$ is lying on a wedge having inclination angle $\alpha ={\tan }^{-1}\left(\frac{1}{5}\right)$. The floor is smooth and friction coefficient is $\mu$ between block and inclined plane. The wedge is moving with a constant acceleration $a$ $=2m/s^2$ horizontally. The minimum value of coefficient of friction $\mu ,$ so that $m$ remains stationary with respect to wedge is

• A. $\frac{5}{6}$
• B. $\frac{5}{12}$
• C. $\frac{1}{5}$
• D. $\frac{2}{5}$

Answer 1

Answer: (B)

$\frac{5}{12}$

$macos\alpha +mgsin\alpha =\mu N$
$N+masin\alpha =mgcos\alpha$
$N=[mgcos\alpha -masin\alpha ]$
$macos\alpha +mgsin\alpha =\mu [mgcos\alpha -masin\alpha ]$
$\left[\frac{acos\alpha +gsin\alpha }{gcos\alpha -asin\alpha }\right]=\mu \Rightarrow \mu =\left[\frac{a+gtan\alpha }{g-atan\alpha }\right]=\frac{5}{12}.$