Question

Find the maximum value of $M/m$ in the situation shown in figure (6-W10) so that the system remains at rest. Friction coefficient at both the contacts is $\mu$. Discuss the situation when $\tan \theta <\mu$.

Answer 1

From FBD of block on inclined , for maximum ratio block should at the condition of just starting the motion

therefore : $Mg\sin \theta =T+f_1$

here $f_1=\mu Mg\cos \theta$

also block on horizontal plane is in the condition of just start

therefore : $T=f_2$ and $f_2=\mu mg$

using these we get: $Mg\sin \theta =\mu mg+\mu Mg\cos \theta$

$Mg(\sin \theta -\mu \cos \theta )=\mu mg$

$\frac{M}{m}=\frac{\mu }{\sin \theta -\mu \cos \theta }$

for maximum value of $M/m$ ,$\sin \theta -\mu \cos \theta$ should be minimum whih is equal to $-\sqrt{1+{\mu }^2}$

so minimum value of $M/m=\frac{\mu }{\sqrt{1+{\mu }^2}}$

If $\tan \theta <\mu$

$\Rightarrow Mg\sin \theta <\mu Mg\cos \theta$

This implies that block on inclined plane will not move as limiting frictional force is greater than component of gravitational force along inclination , therefore $T=0$ , for any ratio of $M/m$ block will not move and frictional force acting on body on inclined will be $Mg\sin \theta$ and block on horizontal will be $0$.