# The coefficient of friction between the block A of mass m and block B of mass 2 m is $\mu$. There is no friction between block B and the inclined plane. If the system of blocks A and B is released from rest and there is no slipping between A and B then

a)$2 \theta \leq \cos^{-1} (2 \mu)$ b)$2 \theta \leq \sin ^{-1}(2 \mu)$ c)$2 \theta \leq \tan ^{-1} (\mu/2)$ d)$\theta \leq \tan^{-1}\mu$

The bodies slide down with out slipping
So the acceleration of bodies A and B is along the incline=a
$a=g \sin \theta$
Vertical component of a ie $a_v=a \sin \theta$
and horizontal component of a $a_H=a \cos \theta$
$a_H=g \sin \theta \cos \theta$
$a_v=g \sin ^2 \theta$
Normal reaction between A and B =V
$mg-N=ma_v$
$N=mg-ma_v$
$\quad=mg-mg \sin^2\theta$
$N=mg \cos ^2 \theta$
Now for B not slipping from A
frictional force $f \geq ma_H$
$\mu N \geq ma_H$
$\mu mg \cos^2 \theta \geq mg \cos \theta \sin \theta$
$\mu \geq \tan \theta$
$\theta \leq \tan^{-1} \mu$
Hence d is the correct answer

edited Jan 26, 2014 by meena.p