Question

Three identical rigid circular cylinders A, B and C arranged on smooth inclined surfaces as shown in the figure. The least value of '$\theta$' that prevents the arrangement from collapsing is :

• A. $tan^{-1}[\frac{1}{3\sqrt{3}}]$
• B. $tan^{-1}[\frac{1}{2\sqrt{3}}]$
• C. At the least value of $\theta$ the contact force between A and B will be zero.
• D. The magnitude of normal reaction between A and the inclined surface and that between B and the inclined surface will be same for least value of $\theta$.

Answer 1

Answer: (A), (C), (D)

A, C, and D are correct

Answer 2

It turns out that when you set up the equilibrium of forces on the three identical cylinders, with the two lower ones (A and B) resting on smooth inclined planes at an angle θ and the top one (C) resting on them, you find that the “just stable” or minimum‐θ condition is obtained when the force along the line joining the centers of A and B (i.e. the contact force between A and B) becomes zero. In other words, at the limiting case the only forces acting on A and B are their weights and the reactions from the inclined planes (which by symmetry turn out to be equal). Resolving the weight components appropriately and writing the force–balance along the directions of the lines joining the centers yields

$$\tan\theta=\frac{1}{3\sqrt{3}}\quad.$$

Thus, the least value of θ is

$$\theta=\tan^{-1}\left(\frac{1}{3\sqrt{3}}\right).$$

At this critical condition the contact force between A and B vanishes and the reactions of the two inclined surfaces are equal.