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:
$$ rSsin30 = rμScos30 + rμS $$
Best Answer
Taking moments at the point of contact between one of the lower cylinders and the plane gives $$Sr\sin 30=\mu S(r+r\cos 30)$$ $$\Rightarrow \frac 12 S=\mu S(1+\frac{\sqrt{3}}{2})$$ $$\Rightarrow \mu =2-\sqrt{3}$$