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Question

(A) Let $R$ be the radius of the hemisphere. The contact point on the inclined plane is $O$. For the plane surface to remain horizontal, the net torqu

Vedclass · Accepted Answer

$\mu = \frac{Q}{\sqrt{P(P + 2Q)}}$

Vedclass · Answer

(A) Let $R$ be the radius of the hemisphere. The contact point on the inclined plane is $O$. For the plane surface to remain horizontal, the net torque about the contact point $O$ must be zero.Let $\phi$ be the angle of inclination of the plane. The weight $P$ acts vertically downwards through the center of the hemisphere, and $Q$ acts vertically downwards at the rim.From the geometry of the system, the horizontal distance of the line of action of $P$ from $O$ is $R \sin \phi$, and the horizontal distance of the line of action of $Q$ from $O$ is $R - R \sin \phi$.For rotational equilibrium, the sum of torques about $O$ is zero: $P(R \sin \phi) = Q(R - R \sin \phi)$.Simplifying, $P \sin \phi = Q - Q \sin \phi \implies \sin \phi (P + Q) = Q \implies \sin \phi = \frac{Q}{P+Q}$.For the hemisphere to remain in equilibrium on a rough inclined plane, the coefficient of friction $\mu$ must satisfy $\mu \ge 	an \phi$.Since $\sin \phi = \frac{Q}{P+Q}$, we have $\cos \phi = \sqrt{1 - \sin^2 \phi} = \sqrt{1 - \left(\frac{Q}{P+Q}ight)^2} = \sqrt{\frac{(P+Q)^2 - Q^2}{(P+Q)^2}} = \frac{\sqrt{P^2 + 2PQ}}{P+Q}$.Thus, $	an \phi = \frac{\sin \phi}{\cos \phi} = \frac{Q/(P+Q)}{\sqrt{P^2 + 2PQ}/(P+Q)} = \frac{Q}{\sqrt{P(P+2Q)}}$.Therefore, the minimum coefficient of friction is $\mu = \frac{Q}{\sqrt{P(P+2Q)}}$.

$A$ solid hemisphere of weight $P$ rests with its curved surface in contact with a rough inclined plane. $A$ weight $Q$ is placed at some point on the rim of the hemisphere to keep its plane surface horizontal. Then its minimum coefficient of friction is

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