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(A) Let the mass of the insect be $m$. The forces acting on the insect are its weight $mg$ (downwards),the normal reaction $R$ (radially outwards),and

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 $mg$ (downwards),the normal reaction $R$ (radially outwards),and the frictional force $f$ (tangentially upwards along the surface).At any angle $\alpha$ with the vertical,the components of the weight are $mg \cos \alpha$ (radially inwards) and $mg \sin \alpha$ (tangentially downwards).For the insect to be in equilibrium without slipping,the forces must be balanced:$R = mg \cos \alpha$ $(i)$$f = mg \sin \alpha$ (ii)For the limiting condition of friction,$f = \mu R$,where $\mu = 1/3$.Substituting equations $(i)$ and (ii) into the limiting friction condition:$mg \sin \alpha = \mu (mg \cos \alpha)$$	an \alpha = \mu = 1/3$Since $	an \alpha = 1/3$,we have $\cot \alpha = 3$.

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