Question

A block of mass $m$ is lying on a wedge having inclination angle $\alpha =ta{n}^{-1}\left(\frac{1}{5}\right)$. Wedge is moving with a constant acceleration $a=2m/s^2$ horizontally. Find the minimum value of coefficient of friction $\mu ,$ so that $m$ remains stationary with respect to wedge.

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

Answer 1

The correct option is B $\frac{5}{12}$

We have the free body diagram as shown in the figure.
The horizontal force balance gives us
$macos\alpha +mgsin\alpha =\mu N$....................(i)
The vertical force balance gives us
$N+masin\alpha =mgcos\alpha$
or
$N=[mgcos\alpha -masin\alpha ]$
Substituting this N in (i) we get
$macos\alpha +mgsin\alpha =\mu [mgcos\alpha -masin\alpha ]$
$\left[\frac{acos\alpha +gsin\alpha }{gcos\alpha -asin\alpha }\right]=\mu$
$\Rightarrow \mu =\left[\frac{a+gtan\alpha }{g-atan\alpha }\right]=\frac{5}{12}$