Question

A sphere is rotating between two tough inclined walls as shown in figure. Coefficient of friction between each wall and the sphere is 1/3. If $f_1$ and $f_2$ be the friction forces at P and Q. Then $\frac{f_{!}}{f_2}$ is:

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

Answer 1

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

$f_1=\mu N_1{f}_2=\mu N_2\Rightarrow \frac{f_1}{f_2}=\frac{N_1}{N_2}$

now after breaking component of f1 f2 N1 N2

we get the equationby balancing horizontal forces

$F_1\cos 30+N_2\sin 60=N_2\sin 30+F_2\cos 60\therefore \frac{\mu N_1\sqrt{3}}{2}+\frac{N_2\sqrt{3}}{2}=\frac{N_1}{2}+\frac{\mu N_2}{2}\Rightarrow \mu \sqrt{3}{N}_1+\sqrt{3}{N}_2=N_1+\mu N_2(\mu \sqrt{3}-1)N_1=(\mu -\sqrt{3})N_2\Rightarrow \frac{N_1}{N_2}={\left(\frac{(\mu \sqrt{3}-1)}{(\mu -\sqrt{3})}\right)}^{-1}={\left(\frac{\sqrt{3}-\frac{1}{\mu }}{1-\frac{\sqrt{3}}{\mu }}\right)}^{-1}={\left(\frac{\sqrt{3}-3}{1-\sqrt[3]{33}}\right)}^{-1}={\left(\frac{3-\sqrt{3}}{\sqrt[3]{33}-1}\right)}^{-1}=\frac{4}{\sqrt{3}}+1$