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(A) Let the mass of the insect be $m$. The forces acting on the insect are its weight $mg$ (acting vertically downwards),the normal reaction $N$ (acti

Vedclass · Accepted Answer

$\cot \alpha = 3$

Vedclass · Answer

(A) Let the mass of the insect be $m$. The forces acting on the insect are its weight $mg$ (acting vertically downwards),the normal reaction $N$ (acting radially outwards),and the frictional force $f$ (acting tangentially upwards along the surface).Resolving the weight $mg$ into two components: $mg \cos \alpha$ along the radial direction and $mg \sin \alpha$ along the tangential direction.Since the insect crawls very slowly,it is in equilibrium at any point.For the radial direction: $N = mg \cos \alpha$.For the tangential direction: $f = mg \sin \alpha$.For the insect to not slip,the frictional force must satisfy $f \le \mu N$.At the maximum angle $\alpha$,the limiting friction condition is reached: $f = \mu N$.Substituting the expressions for $f$ and $N$: $mg \sin \alpha = \mu (mg \cos \alpha)$.This simplifies to $	an \alpha = \mu$.Given $\mu = 1/3$,we have $	an \alpha = 1/3$.Therefore,$\cot \alpha = 1/	an \alpha = 3$.

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