friction of rope wrapped around a cylinder - the Capstan Equation

Question

I have the following problem:

A rope is wound round a fixed cylinder of radius $r$ so as to make n complete turns. The coefficient of friction between the rope and cylinder is $\mu$. Show that if one end of the rope is held by a force $F$, a force $Fe^{2\pi n\mu}$ must be applied to the other end to produce slipping.

I have seen this previous question: Rope wrapped around a cylinder

and followed up the Wikipedia article on the Capstan equation. However there are still several things I am stuck on.

My initial attempt

Before I tried to find help on this question, I thought that I could just say that the total 'downwards' force, which is $F+F'$, $F'$ being the applied force, can be thought of as being distributed over the tops of the rope on the cylinder. This would give me that the maximum normal force of any point is

$(F+F')/n$

So that

$Maximum frictional force, F_r=\mu \frac{F+F'}{n}$

Following this through so that for slipping I have $F'-F>F_r$

I get that $F'>\frac{1+\mu}{1-\mu}F$

I think my approach here is wrong because:

-I got a too large value for the maximum frictional force as I overestimated the maximum normal force by thinking that the normal force acts only at the bits of rope passing over the top of the cylinder.

After looking at the Capstan equation

It had not occured to me to use calculus, and I am still confused as to where infintesimals are needed for use in a question and where you can go without... Any insight into this would be very helpful

More specifically to the problem, there are two things that I can't wrap my head around.

1. The normal force at any point on the cylinder is equal and opposite to the component of the force/tension in the rope that acts radially towards the centre of the circle. The tension at each point in the string is $F'-F$, however the tension at each infintesimal point of string is also acting ALONG A TANGENT to the cylinder, therefore it has no component radially towards the centre of the cylinder. Therefore I get that the normal force, and thus also frictional force on the string is 0 at each point around the cylinder? This must be wrong.

2. When I am resolving vertically so that the total downwards tension in the string (and so upwards force from the cylinder) is equal to the total downwards force ($F'+F$), I also find that since the tension is acting tangentially to the cylinder everywhere, at the top of the cylinder there is no component of the tension acting vertically downwards. So where I thought all of the downwards force of the cylinder could be concentrated there is actually no downwards force?

Answer 1

You've correctly spotted the error in your initial attempt: you're considering that friction only takes place on the top of the cylinder, but actually it takes place all around the cylinder.

Calculus is needed because friction force evolve non-lineraly all around the cylinder, so you have to zoom in to understand how it works.

You say that tension at each point is $F' - F$, but this is false: because of friction, at one end tension is $F$, and at the other end of the rope it will be $F'$, and it evolve exponentially in-between. To find this result, you have to consider a small piece of rope, say between $\theta + \delta \theta$ and $\theta - \delta \theta$. Then, the forces acting on it are tensions in $\theta + \delta \theta$ and $\theta - \delta \theta$, the normal force of the cylinder and, of course, friction. But the cylinder is... cylindric, so tensions in $\theta + \delta \theta$ and $\theta - \delta \theta$ don't have exactly the same direction: since $\delta \theta$ can be arbitrarly small, you can make $\frac{\delta T}{\delta \theta}$ and a normal component appear. Then, you get a first order linear differential equation which gives you after being solved the exponential solution.

Here, you can see what mistake you made in your two last questions: you didn't consider a system, you just said that the sum of the forces somewhere should be null since nothing moved... But your dynamic tools are only valid when applied to a real system, in this case a small piece of rope.

Answer 2

I also spotted this

In the capstan derivation. Whilst sin(0) = 0, the limit of sin(Angle) as Angle tends to 0 is the Angle (in radians). [This is the small angle approximation.]

For an infinitesimal angle there IS an infinitesimal radial force component that contributes to F'-F. Choosing an absolute zero instead is discarding the differentiation information - all the infinitesimal parts have to be put back together to render an answer;- (this is an issue in the area of application of calculus.)

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It's easy to imagine the rope squeezes the cylinder when it is wrapped around at least once. This squeezing force is a linearly related to tangential force from c=2 pi r geometry. The capstan equation, however, gives a ration in which two opposing tangential forces are divided, so the ratio has cancelled out. (The ratio of squeezing force is that of equal proportions of the tangential two forces is the same as the ration of the forces).

Hope that helps.