Question

If the coefficient of friction between A and B is $\mu$, the maximum acceleration of the wedge A for which B will remain at rest with respect to wedge is

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

Answer 1

The correct option is B $g\left[\frac{1+\mu }{1-\mu }\right]$

By placing the frame of reference on block A, FBD of B is

Let, $N$ be the normal force acting. Now taking components of all the forces perpendicular to the plane
$N-mg\cos \theta -ma\sin \theta =0$
$N=\frac{mg+ma}{\sqrt{2}}$
As, $f=\mu N$
$f=\mu \frac{m(g+a)}{\sqrt{2}}$
Taking $mg$ and $ma$ along $f$
$mg\sin \theta +f=ma\cos \theta$
$\frac{mg}{\sqrt{2}}+\mu \frac{m(g+a)}{\sqrt{2}}=\frac{ma}{\sqrt{2}}$
$a=g\left[\frac{1+\mu }{1-\mu }\right]$