Two equal rough cylinders are lying in contact with each other, with their axes parallel and horizontal, on a rough horizontal plane. A third equal cylinder is placed symmetrically on top of the other two. If equilibrium is about to be broken by the upper cylinder slipping between the other two, prove that the coefficient of friction between any two cylinders is $$2 - √3$$.

Working so far:

Let $P$ be the reaction force between the two bottom cylinders. Also let $S$ be the reaction force between all three cylinders and the plane and $μS$ be the frictional force between one of the bottom and top cylinders.

Looking at the forces on the upper cylinder, with friction:

$$2μScos30 = W$$ so $$S=\frac{W}{μ√3}$$

Now, looking at the forces on one of the 2 cylinders and taking moments between one of the cylinders and the plane:

$$ rSsin30 = rμScos30 + rμS + P $$ where $r$ is the radius of the cylinders.

The bottom cylinders will be forced apart so $P=0$ so the moments equation becomes:

$$ rSsin30 = rμScos30 + rμS $$

  • $\begingroup$ I've added some workings out above. $\endgroup$
    – J132
    Sep 25 '15 at 0:35

Taking moments at the point of contact between one of the lower cylinders and the plane gives $$Sr\sin 30=\mu S(r+r\cos 30)$$ $$\Rightarrow \frac 12 S=\mu S(1+\frac{\sqrt{3}}{2})$$ $$\Rightarrow \mu =2-\sqrt{3}$$

  • $\begingroup$ Thanks for the response. I see that the $μSr$ term in your moments equation comes from the contact between the LHS cylinder, for example, and the rough horizontal plane. As the cylinders are equal we can say that the reaction force $S$ between them is the same as well. Is this situation independent of the weight of the cylinders as we can assume that the weight of each cylinder is the same? $\endgroup$
    – J132
    Sep 25 '15 at 17:16
  • $\begingroup$ I have edited some of the workings out above. Looking at the forces in the upper cylinder was not necessary to get to the result (as the cylinders are equal) but I have left it in. $\endgroup$
    – J132
    Sep 25 '15 at 17:38
  • $\begingroup$ I'm not sure what you're asking. $\endgroup$ Sep 25 '15 at 18:33
  • $\begingroup$ The weight of the cylinders is not included in the equations to get to the answer for this particular problem. In some problems the weight is included in the moments equation. In this case, is the weight of each cylinder immaterial? $\endgroup$
    – J132
    Sep 26 '15 at 9:32
  • $\begingroup$ We are assuming that equilibrium is broken by the upper cylinder slipping down against both lower cylinders at the same time. And this would mean that the reactions between the upper and lower cylinders are the same on both sides. If the weights of the lower cylinders were not equal, then this would not necessarily be the case, as one side could be in limiting equilibrium and the other side not. We are not assuming that the lower cylinders are slipping against the horizontal in order to arrive at the answer. $\endgroup$ Sep 26 '15 at 20:06

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