Question

A particle inside a hollow sphere of radius r, having coefficient of friction $\left(1/\sqrt{3}\right)$, up to what height can particle be at rest?

Answer 1

The friction force equals to

$F_f=\mu mg\cos \theta$

where $\mu =\frac{1}{\sqrt{3}}$ and $\cos \theta =\frac{r-h}{r}$

h is the height.

The friction force prevent particle to slide down under the component of gravitational force

$F_1=mgsin\theta$

$F_f=F_1=\mu mgcos\theta =mgsin\theta$

$tan\theta =\mu =\theta ={30}^{\circ }$

$cos\theta =\frac{\sqrt{3}}{2}=\frac{r-h}{r}=h=\frac{2-\sqrt{3}}{2}r$