Question

In the situation shown in figure, a wedge of mass m is placed on a rough surface, on which a block of equal mass is placed on the inclined plane of wedge. The friction coefficient between the wedge and the block and also between the ground and the wedge is $\mu$. An external force F is applied horizontally on the wedge towards the right, assuming block does not slide on the wedge. Find the minimum value of Force $F$ at which the block will start slipping.

Answer 1

$N_1=mg+mg=2mg$

$N_2=mg\cos \theta +ma\sin \theta$

Here, $f_1$ and $f_2$ are the frictional force and force $ma\sin \theta$ is pseudo force and acceleration of the system is $a$

$F-f_1=(m+m)a$

$a=\frac{F-2\mu mg}{2m}$

If block start slipping then

$ma\cos \theta =mg\sin \theta +f_2$

$ma\cos \theta =mg\sin \theta +\mu (mg\cos \theta +ma\sin \theta )$

$a(\cos \theta -\mu \sin \theta )=g(\sin \theta +\mu \cos \theta )$

$a=\frac{g(\sin \theta +\mu \cos \theta )}{(\cos \theta -\mu \sin \theta )}$

$\frac{F-2\mu mg}{2m}=\frac{g(\sin \theta +\mu \cos \theta )}{(\cos \theta -\mu \sin \theta )}$

$F=2m[\mu g+\frac{g(\sin \theta +\mu \cos \theta )}{(\cos \theta -\mu \sin \theta )}]$