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

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Question

$\mathrm{N}_{\mathrm{A}} \sin 60^{\circ}=\mathrm{N}_{\mathrm{B}} \sin 30^{\circ}$

$\mathrm{N}_{\mathrm{A}} \cdot \frac{\sqrt{3}}{2}=\mathrm{N}_{\ma

Vedclass · Accepted Answer

${N_B} = \sqrt 3 {N_A}$

Vedclass · Answer

$\mathrm{N}_{\mathrm{A}} \sin 60^{\circ}=\mathrm{N}_{\mathrm{B}} \sin 30^{\circ}$

$\mathrm{N}_{\mathrm{A}} \cdot \frac{\sqrt{3}}{2}=\mathrm{N}_{\mathrm{B}} \frac{1}{2}$

$\sqrt{3} \mathrm{N}_{\mathrm{A}}=\mathrm{N}_{\mathrm{B}}$

$\sqrt{3} \mathrm{N}_{\mathrm{A}}=\mathrm{N}_{\mathrm{B}}$

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

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