Question

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

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

Answer 1

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

As shown in fig.

$ma$ is pseudo force towars left.

If block A moves with constant velocity then net force is zero.

Force along the surface of bock B is;

$ma\cos \theta =f+mg\sin \theta$ ..................(1)

where $f-$ frictional force

force opposite to the movement of block A

Let $\mu =$ coefficient of frictional force

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

(1) becomes;

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

where $a=g,\theta ={30}^{\circ }$

$mg\cos {30}^{\circ }=\mu (mg\sin {30}^{\circ }+mg\cos {30}^{\circ })+mg\sin {30}^{\circ }$

$g\left(\frac{\sqrt{3}}{2}\right)=\mu (\frac{g}{2}+\frac{\sqrt{3}g}{2})+\left(\frac{g}{2}\right)$

$\mu \left(\frac{1+\sqrt{3}}{2}\right)=\frac{\sqrt{3}-1}{2}$

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