Three cylinders on two inclined planes, each inclined at an angle alpha.

Question

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 $\tan (\alpha) \lt \frac{1}{2\sqrt{3}}$

Answer 1

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.