A smooth cylinder of mass m and radius R is r | vedclass

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(D) For the cylinder to be in equilibrium,the net horizontal force must be zero.Let $N_A$ be the normal reaction at edge $A$ and $N_B$ be the normal r

Vedclass · Accepted Answer

$N_B = \sqrt{3} N_A$

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(D) For the cylinder to be in equilibrium,the net horizontal force must be zero.Let $N_A$ be the normal reaction at edge $A$ and $N_B$ be the normal reaction at edge $B$.The normal force acts perpendicular to the surface of the cylinder,i.e.,along the radius passing through the contact point.Resolving the normal forces into horizontal and vertical components:The horizontal component of $N_A$ is $N_A \sin 60^{\circ}$ (directed towards the right).The horizontal component of $N_B$ is $N_B \sin 30^{\circ}$ (directed towards the left).For horizontal equilibrium:$N_A \sin 60^{\circ} = N_B \sin 30^{\circ}$Substituting the values of trigonometric functions:$N_A \cdot \frac{\sqrt{3}}{2} = N_B \cdot \frac{1}{2}$Multiplying both sides by $2$:$\sqrt{3} N_A = N_B$

$A$ smooth cylinder of mass $m$ and radius $R$ is resting on two corner edges $A$ and $B$ as shown in the figure. The relation between the normal reactions at the edges $A$ and $B$ is

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