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4-1.Newton's Laws of Motion
normal
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
A
${N_A} = \sqrt 2 {N_B}$
B
${N_B} = \frac{{2\sqrt 3 {N_A}}}{5}$
C
${N_A} = \frac{{{N_B}}}{2}$
D
${N_B} = \sqrt 3 {N_A}$
Solution
$\mathrm{N}_{\mathrm{A}} \sin 60^{\circ}=\mathrm{N}_{\mathrm{B}} \sin 30^{\circ}$
$\mathrm{N}_{\mathrm{A}} \cdot \frac{\sqrt{3}}{2}=\mathrm{N}_{\mathrm{B}} \frac{1}{2}$
$\sqrt{3} \mathrm{N}_{\mathrm{A}}=\mathrm{N}_{\mathrm{B}}$
$\sqrt{3} \mathrm{N}_{\mathrm{A}}=\mathrm{N}_{\mathrm{B}}$
Standard 11
Physics
