An insect crawls up a hemispherical surface very slowly. The coefficient of friction between the insect and the surface is $1 / 3$. If the line joining the centre of the hemispherical surface to the insect makes an angle $\alpha$ with the vertical, the maximum possible value of $\alpha$ so that the insect does not slip is given by

Question

Aicharya · Accepted Answer

The insect crawls up the bowl upto a certain height $h$ only till the component of its weight along the bowl is balanced by limiting frictional force. For limiting condition at point $\mathrm{A}$ $R=m g \cos \alpha$ $F_1=m g \sin \alpha$ Dividing eq. (ii) by (i) $\tan \alpha=\frac{1}{\cot \alpha}=\frac{F_1}{R}=\mu\left[\right.$ As $\left.F_1=\mu R\right]$ $\Rightarrow \quad \tan \alpha=\mu=\frac{1}{3}\left[\because \mu=\frac{1}{3}(\right.$ Given $\left.]\right)$ $\therefore \quad \cot \alpha=3$

An insect crawls up a hemispherical surface very slowly. The coefficient of friction between the insect and…

Solution