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Two smooth uniform right circular cylinders, each of mass $m$ and radius $a$, are placed symmetrically in contact with each other and with 2 planes, each inclined at an angle $\alpha$ to the horizontal. The axes of the cylinders lie in the same horizontal plane, and are parallel to the line of intersection of the 2 inclined planes. A third smooth uniform circular cylinder, of mass $2m$ and radius $a$, is placed symmetrically on top of the other 2. If the 2 lower cylinders are forced apart, prove that $\tan (\alpha) \lt \frac{1}{2\sqrt{3}}$

Rio Alvarado
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J132
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1 Answers1

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Consider the limiting case when the two lower spheres have only just separated, so that the normal reaction between them is zero.

The diagram is symmetrical, and there is no friction.

Let $R$ be the reaction between each of the lower spheres and the upper sphere.

Considering the forces on the upper sphere alone, we have $$2R\cos30=2mg\Rightarrow R=\frac{2mg}{\sqrt{3}}$$

Now consider the forces acting on one of the lower spheres alone, and take moments about the point of contact of the sphere and the plane. We now have $$mga\sin \alpha=Ra\sin(30-\alpha)$$

Now you can substitute for $R$ and get $\tan \alpha=\frac {1}{2\sqrt{3}}$

The inequality follows from the fact that the lower spheres are actually forced apart.

David Quinn
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  • Thanks. I would put 2Rcos30 = 2mga for the upper cylinder as it's radius is also a. This allows you to cancel the mga terms later on. Also for sin(30-α) use sin(30-α) = sin30cosα - cos30sinα – J132 Aug 02 '15 at 22:18
  • Okay but your first statement is not correct. Feel free to accept my answer if it is satisfactory rather than leave your question unanswered :) – David Quinn Aug 03 '15 at 06:35
  • Use $mgsinα = Rsin(30-α)$ for one of the lower spheres or set a = 1. – J132 Aug 12 '15 at 21:09
  • Yes I missed out a factor of $a$ in the moments equation and I have edited my answer. – David Quinn Aug 12 '15 at 22:22