Question

A thin hoop of weight $500N$ and radius $1m$ rests on a rough inclined plane as shown in the figure. The minimum coefficient of friction needed for this configuration is :

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

Answer 1

The correct option is D $\frac{1}{2\sqrt{3}}$

Let the coefficient of friction be $\mu$. The force by upper rope be $F$; Fricition force be $F_f$
The normal force on the hoop is $mgcos\theta$
Therefore maximum friction force can be $\mu mgcos\theta$
As the hoop is in rest. Therefore moment balance on hoop about its bottom point(where friction is acting) is:
$\sum \overrightarrow{r}\times \overrightarrow{F}=02rF\hat{j}-rmgsin\theta \hat{j}=0$
$F=\frac{mgsin\theta }{2}$
Also force balance along the incline:
$F_f+F=mgsin\theta \therefore F_f=\frac{mgsin\theta }{2}$
For $\mu$ to be minimum.
$F_f=\mu mgcos\theta =\frac{mgsin\theta }{2}\therefore \mu =\frac{tan\theta }{2}=\frac{1}{2\sqrt{3}}$