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(A) The forces acting on the insect are the gravitational force $mg$ acting vertically downwards,the normal reaction $N$ acting radially outwards,and

Vedclass · Accepted Answer

$\cot\, \alpha = 3$

Vedclass · Answer

(A) The forces acting on the insect are the gravitational force $mg$ acting vertically downwards,the normal reaction $N$ acting radially outwards,and the frictional force $f$ acting tangentially upwards along the surface.Resolving the gravitational force $mg$ into two components:$1$. $mg \cos \alpha$ acting radially inwards (opposite to $N$).$2$. $mg \sin \alpha$ acting tangentially downwards along the surface.Since the insect crawls very slowly,it is in equilibrium. Thus:$N = mg \cos \alpha$For the insect to not slip,the frictional force $f$ must balance the tangential component of gravity:$f = mg \sin \alpha$We know that the maximum static friction is $f_{max} = \mu N$.At the maximum angle $\alpha$,$f = f_{max} = \mu N$.Substituting the expressions for $f$ and $N$:$mg \sin \alpha = \mu (mg \cos \alpha)$Dividing both sides by $mg \cos \alpha$:$	an \alpha = \mu$Given $\mu = 1/3$,we have:$	an \alpha = 1/3$Therefore,$\cot \alpha = 1/	an \alpha = 3$.

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