Question

Block $\text{A}$ is placed on a rough wedge $\text{B}$ which is placed on a smooth surface. The wedge has angle of inclination of ${30}^{\circ }$ and is imparted a horizontal acceleration $g$ towards right. Block $\text{A}$ is given an initial velocity $v_0$ from rest. Find the coefficient of friction for which block $\text{A}$ moves with constant velocity $v_0$ with respect to wedge $(g=10{\text{m/s}}^2)$

• A. $\frac{(3+\sqrt{3})}{(3-\sqrt{3})}$
  
• B. $\frac{(2-\sqrt{3})}{(2+\sqrt{2})}$
• C. $\frac{(\sqrt{2}-1)}{(\sqrt{2}+1)}$
  
• D. $\frac{(\sqrt{3}-1)}{(\sqrt{3}+1)}$

Answer 1

The correct option is D $\frac{(\sqrt{3}-1)}{(\sqrt{3}+1)}$

Free body diagram:


From free body diagram normal reaction,
$N=ma\sin \theta +mg\cos \theta$

$\Rightarrow N=mg\sin \theta +mg\cos \theta [\because a=g]$

$N=mg[\sin \theta +\cos \theta ]$

So, the frictional force acting on the contact surfaces,
$f=\mu N=\mu mg[\sin \theta +\cos \theta ]...\left(1\right)$

For block $\text{A}$ to move with constant velocity,
$ma\cos \theta =f+mg\sin \theta$

Substituting the values, we get
$mg\cos \theta =\mu mg[\sin \theta +\cos \theta ]+mg\sin \theta$

$\Rightarrow mg[\cos \theta -\sin \theta ]=\mu mg[\sin \theta +\cos \theta ]$

$\Rightarrow [\cos 30-\sin 30]=\mu [\sin 30+\cos 30]$

$\Rightarrow (\frac{\sqrt{3}}{2}-\frac{1}{2})=\mu [\frac{1}{2}+\frac{\sqrt{3}}{2}]$

$\therefore \mu =\frac{(\sqrt{3}-1)}{(\sqrt{3}+1)}$

Hence, option (d) is correct answer.