# Math Help - [SOLVED] Uniform Plank - friction?

1. ## [SOLVED] Uniform Plank - friction?

On the top of a fixed rough cylinder, of radius $r$, rests a thin uniform plank, and a man stands on the plank just above the point of contact. Show that he can walk slowly a distance $(n + 1)r\epsilon$ along the plank without its slipping off the cylinder, if the weight of plank is $n$ times that of the man and $\epsilon$ is the angle of friction between the plank and the cylinder.

2. ## Solution

Here's how to do it:

Spoiler:

From the description of the problem we can assume that initially the plank is balanced with its centre of gravity at the top point of the circular profile of the cylinder. As he walks along the plank it will roll on the cylinder to a certain degree and then he will be able to walk a little further past this point such that his weight is balanced by the weight of the plank about the point of contact.

Let the top point (and so initial location of man) be $O$ and let the final point of contact of the plank with the cylinder surface be $P$. Also let the centre of the circular profile of the cylinder be $C$. The angle $\hat{OCP}$ we will then call $\alpha$. Finally, let the mass of the man be $m$ and so that of the plank is $n m$.

Note from this that the plank must be inclined at an angle $\alpha$ to the horizontal since it is tangent to the surface of the cylinder. Also at $P$ there must be a frictional force, $F_{\mu}$, given by

$F_{\mu} = \mu R = R \tan \epsilon$ (limiting equilibrium)

where $R$ is the reaction of the cylinder on the plank and must be normal to the surface of the plank.

Hence resolving forces perpendicularly to the plank, we have

$R = (n+1) mg \cos \alpha$

and resolving parallel to the plank we have

$F_{\mu} = (n+1) mg \sin \alpha$,

$\Leftrightarrow R \tan \epsilon = (n+1) mg \sin \alpha$,

so eliminating $R$ we have

$(n+1) mg \cos \alpha \, \tan \epsilon = (n+1) mg \sin \alpha$,

$\Leftrightarrow \tan \epsilon \, \cos \alpha = \sin \alpha$,

$\Rightarrow \tan \epsilon = \tan \alpha$,

$\Rightarrow \epsilon = \alpha$.

Now we simply need to resolve moments. To do this we need to know the distance from $O$ to $P$ along the plank. Since the plank rolls on the cylinder's surface this distance must be $r \epsilon$ . Let the total distance the man travels from $O$ be $x$ so then (taking moments) we have

$r \epsilon \, n mg \cos \epsilon = (x-r \epsilon) mg \cos \epsilon$,

$\Leftrightarrow n r \epsilon = x - r \epsilon$,

$\color[rgb]{0,0,1} \boxed{\iff x = (n+1) r \epsilon}$.