Question

If the coefficient of friction between the wedge $A$ and block $B$ shown in the figure is $\mu $ , then the maximum possible horizontal acceleration of $A$ for which $B$ doesn't slip is [angle of inclination of wedge = $45^\circ $ ]

• A. $\mu g$
• B. $g\left(\frac{1 + \mu }{1 - \mu }\right)$
• C. $\frac{g}{\mu }$
• D. $g\left(\frac{1 - \mu }{1 + \mu }\right)$

Answer 1

Answer: (B)

$FBD$ of block $B$ w.r.t. wedge $A$ , for maximum $'a'$ :

Perpendicular to wedge:
$\Sigma f_{y}=\left(\right.mgcos\theta +masin\theta -N\left.\right)=0$
and $\Sigma f_{x}=mgsin\theta +\mu N-macos\theta =0$ (for maximum a)
$\Rightarrow mgsin\theta +\mu \left(\right.mgcos\theta +masin\theta \left.\right)-macos\theta =0$
$\Rightarrow a=\frac{\left(\right. g sin\theta + \mu g cos\theta \left.\right)}{cos\theta - \mu sin\theta }$
for $\theta =45^\circ $
$a=g\left(\frac{tan 45 ^\circ + \mu }{cot 45 ^\circ - \mu }\right);a=g\left(\frac{1 + \mu }{1 - \mu }\right)$