Question

A sphere is rotating between two rough inclined walls as shown in figure. Coefficient of friciton between each wall and the sphere is $\frac{1}{3}$. If $f_1$ amd $f_2$ are the friction forces at $P$ and $Q$, then $\frac{f_1}{f_2}$ is

• A. $\frac{4}{\sqrt{3}}+1$
• B. $\frac{1}{\sqrt{3}}+2$
• C. $\frac{1}{2}+\sqrt{3}$
• D. $1+2\sqrt{3}$

Answer 1

The correct option is A $\frac{4}{\sqrt{3}}+1$

Let $\mu$ be the friction coefficient between the sphere and each wall. FBD of the sphere is given below. $f_1$ and $f_2$ will act in the directions opposite to tendency of slipping.


Net force on the sphere in horizontal direction is zero.
$\therefore N_1\cos {60}^{\circ }+\mu N_2\cos {60}^{\circ }=N_2\cos {30}^{\circ }+\mu N_1\cos {30}^{\circ }$

$\Rightarrow$ $N_1+\mu N_2=\sqrt{3}(N_2+\mu N_1)$

$\Rightarrow$ $N_1(1-\sqrt{3}\mu )=N_2(\sqrt{3}-\mu )$

$\therefore$ $\frac{N_1}{N_2}=\frac{\sqrt{3}-\mu }{1-\sqrt{3}\mu }$

Substituting $\mu =\frac{1}{3}$, we get

$\frac{N_1}{N_2}=\frac{\sqrt{3}-\frac{1}{3}}{1-\frac{\sqrt{3}}{3}}=\frac{3\sqrt{3}-1}{3-\sqrt{3}}=1+\frac{4}{\sqrt{3}}$

Hence, $\frac{f_1}{f_2}=\frac{\mu N_1}{\mu N_2}=1+\frac{4}{\sqrt{3}}$

Why this question? This question will help students to grasp the concept of wedge constraint and friction on a rotating wheel.