Question

If the acceleration is towards right the frictional force exerted by wedge on the block will be: (coefficient of friction between wedge and block $=\frac{\sqrt{3}}{2}$)

• A. mg
• B. $\frac{mg}{2}$
• C. $\frac{3mg}{2}$
• D. $2mg$

Answer 1

The correct option is A mg

Draw $FBD$ of block with respect to wedge

Given: $a=\sqrt{3}g,\mu =\frac{\sqrt{3}}{2}$
Since $FBD$ is with respect to non-inertial frame of reference. Hence a pseudo force will act on the block towards left as wedge is accelerating towards right
Component of weight perpendicular to plane:
$mg\cos \theta =mg\cos {30}^{\circ }=\frac{\sqrt{3}mg}{2}$
Component of weight along the plane:
$mg\sin \theta =mg\sin {30}^{\circ }=\frac{mg}{2}$
Component of pseudo force along the plane will be
$ma\cos {30}^{\circ }=\frac{\sqrt{3}ma}{2}=\frac{mg}{2}$
and perpendicular to plane will be
$ma\sin {30}^{\circ }=\frac{ma}{2}=\frac{\sqrt{3}mg}{2}$

Find friction acting on the block
Force equation in perpendicular to plane direction
$N-\frac{\sqrt{3}mg}{2}-\frac{\sqrt{3}mg}{2}=0$
$N=\sqrt{3}mg$
$f_L=\mu N=\frac{3mg}{2}$
Force equation along the plane of wedge
$f+\frac{mg}{2}=\frac{3mg}{2}$
$\therefore f=mg$
$\because$ required friction is less than limiting friction
$\therefore$ block will not slide

Final answer: Option(a)