Question

A block of mass m lies on the wedge of mass M as shown in the figure. Answer following parts separately

(a) With what minimum acceleration must the wedge be moved towards right horizontally so that block of mass 'm' falls freely
(b) Find the minimum friction coefficient required between wedge M and ground so that it does not move while block of mass 'm' slips down on it.

• A. $gcot\theta$ and ${\mu }_{min}=\frac{msin\theta cos\theta }{mco{s}^2\theta +M}$
• B. $gcot\theta$ and ${\mu }_{min}=\frac{msi{n}^2\theta }{msin\theta cos\theta +M}$
• C. $gtan\theta$ and ${\mu }_{min}=\frac{mco{s}^2\theta }{mcos\theta sin\theta +M}$
• D. $gtan\theta$ and ${\mu }_{min}=\frac{msin\theta cos\theta }{msi{n}^2\theta +M}$

Answer 1

The correct option is A $gcot\theta$ and ${\mu }_{min}=\frac{msin\theta cos\theta }{mco{s}^2\theta +M}$

Given the block and the wedge, we have to find he acceleration such that the block is under free fall,
Under free fall we will have acceleration components along the common normal of both the block and wedge to be same.
$g\times cos\theta =a\times sin\theta$
$a=g\times cot\theta$
from FBD of mass m, writing the force equation in direction parallel to wedge inclination
$mgsin\theta =ma$ ...(i)
the writing the force equation in direction perpendicular to wedge inclination
$mgcos\theta =N_1$...(ii)
from FBD of wedge mass M, writing the force equation in horizontal direction
For wedge to be static, friction force exerted by ground surface should be greater than applied force on wedge
$\Rightarrow f_{min}\ge N_1\sin \theta$
writing the force equation for wedge in vertical direction
$N_{1}cos\theta +Mg=N_2$
$f_{min}={\mu }_{min}{N}_2={\mu }_{min}(N_{1}cos\theta +Mg)$
substituting $N_1$ from equation(ii)
${\mu }_{min}⩾\frac{mgsin\theta cos\theta }{mg{\cos }^2\theta +Mg}$
${\mu }_{min}⩾\frac{msin\theta cos\theta }{m{\cos }^2\theta +M}$