An insect crawls up a hemispherical surface v | vedclass

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

Vedclass · Accepted Answer

$cot\,\alpha = 3$

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(A) The forces acting on the insect are the gravitational force $mg$ acting downwards,the normal reaction $R$ acting radially outwards,and the frictional force $f$ acting tangentially along the surface.Since the insect moves very slowly,it is in equilibrium at any point. Resolving the forces along the tangent and normal to the surface:$1$. Along the normal: $R = mg \cos \alpha$$2$. Along the tangent: $f = mg \sin \alpha$For the insect to not slip,the frictional force must satisfy $f \leq \mu R$. The maximum angle $\alpha$ occurs when the limiting friction is reached,i.e.,$f = \mu R$.Substituting the expressions:$mg \sin \alpha = \mu (mg \cos \alpha)$Dividing both sides by $mg \cos \alpha$:$	an \alpha = \mu$Given $\mu = 1/3$,we have $	an \alpha = 1/3$,which implies $\cot \alpha = 1/\mu = 3$.

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