Equilibrium of a particle

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Question

Three identical rigid circular cylinders A, B and C are placed on smooth inclined plane shown in the figure. The minimum value of tan $θ$ that prevent the arrangement from collapse is $\frac{1}{3 \sqrt{n}}$ . The value of n is

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Solution

From free body diagram of cylinder A, $2 N \cos 30^{\circ} = m g$
$∴ N = \frac{m g}{\sqrt{3}}$
From free body diagram of cylinder B,
$\begin{array}{l} N_{0} \cos θ = mg + Ncos 30^{\circ} \\ N_{0} \sin θ = Nsin 30^{\circ} \\ \tan θ = \frac{N \sin 30^{\circ}}{m g + N \cos 30^{\circ}} \\ \Rightarrow \tan θ = \frac{\frac{m g}{\sqrt{3}} \sin 30^{\circ}}{m g + \frac{m g}{\sqrt{3}} \cos 30^{\circ}} = \frac{1}{3 \sqrt{3}} \end{array}$

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Equilibrium of a particle

Remember concepts with our Masterclasses.

Three identical rigid circular cylinders A, B and C are placed on smooth inclined plane shown in the figure. The minimum value of tanθ that prevent the arrangement from collapse is 13n. The value of n is

From free body diagram of cylinder A, 2Ncos⁡30∘=mg∴ N=mg3From free body diagram of cylinder B,N0cos⁡θ=mg+Ncos⁡30∘N0sin⁡θ=Nsin⁡30∘tan⁡θ=Nsin⁡30∘mg+Ncos⁡30∘⇒tan⁡θ=mg3sin⁡30∘mg+mg3cos⁡30∘=133

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Three identical rigid circular cylinders A, B and C are placed on smooth inclined plane shown in the figure. The minimum value of tan $θ$ that prevent the arrangement from collapse is $\frac{1}{3 \sqrt{n}}$ . The value of n is