Question

A thin ring of mass 30 kg and radius 1 m
rests on a rough inclined plane as shown in
the figure.
The minimum coefficient of friction needed
for the system to remain in equilibrium is
3
30°​

Answer 1

i think answer for this question is 1/2√3

Answer 2

Answer:

The minimum coefficient of friction is $\frac{1}{2\sqrt{3} }$.

Explanation:

Let the coefficient of friction be $\mu$.

The force by upper rope be $F$;

Friction force be $F_{f}$.

The normal force on the hoop = $\mathrm{mg} \cos \theta$

Therefore,

Maximum friction force = $\mu \mathrm{mg} \cos \theta$

Hoop at rest. So moment balance on hoop where friction is acting)=

$\sum \overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}=0 2 \mathrm{rF} \hat{\mathrm{j}}-\mathrm{rmg} \sin \theta \hat{\mathrm{j}}=0 \mathrm{F}=\frac{\mathrm{mg} \sin \theta}{2}$

Force balance along the incline:

$\mathrm{F}_{\mathrm{f}}+\mathrm{F}=\mathrm{mg} \sin \theta \therefore \mathrm{F}_{\mathrm{f}}=\frac{\mathrm{mg} \sin \theta}{2}$

$\therefore \mathrm{F}_{\mathrm{f}}=\frac{\mathrm{mg} \sin \theta}{2} For \mu to be minimum. \mathrm{F}_{\mathrm{f}}=\mu \mathrm{mg} \cos \theta=\frac{\mathrm{mg} \sin \theta}{2} \therefore \quad \mu=\frac{\tan \theta}{2} ( \\ \tan \theta=\tan #30 )\\\\\quad \mu=\frac{\tan \ 30}{2}\\= \frac{1}{2\sqrt{3} }$