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 ${}^{\prime }{a}^{\prime }$ :

Perpendicular to wedge:
$\Sigma f_y=(mgcos\theta +masin\theta -N)=0$
and $\Sigma f_x=mgsin\theta +\mu N-macos\theta =0$ (for maximum a)
$\Rightarrow mgsin\theta +\mu (mgcos\theta +masin\theta )-macos\theta =0$
$\Rightarrow a=\frac{(gsin\theta +\mu gcos\theta )}{cos\theta -\mu sin\theta }$
for $\theta =45^{\circ }$
$a=g\left(\frac{tan45^{\circ }+\mu }{cot45^{\circ }-\mu }\right);a=g\left(\frac{1+\mu }{1-\mu }\right)$