# Homework Help: Tension of a rope on a cone, fallacious solution

1. Nov 26, 2014

### ZetaOfThree

1. The problem statement, all variables and given/known data
A rope of mass $m$ forming a circle is placed over a smooth round cone with half angle $\theta$. Find the tension in the rope.
2. Relevant equations
$\sum{F}=0$

3. The attempt at a solution
I know how to solve the problem, but I have another way that I think should work but it doesn't. I don't know why.
The way that this problem is done is to consider the tension on a small element spanning an angle $\alpha$ of the rope as shown in the following image: . From there, we balance the forces on that small element as shown in the following image: . We find that $$T={m \over 2 \pi}g \cot{\theta}$$
My question is why do we have to consider a small element of the rope? Why can't we consider half of the rope (that is, an element of the rope spanning $\alpha = \pi$)? Therefore, the tension acts on the rope as in the following figure: , leading to a free-body diagram as in the following image: .
This leads to the result $$T={m \over 4}g \cot{\theta}$$ What's going wrong?

2. Nov 26, 2014

### Staff: Mentor

There is no tension at the end of the rope, and tension is not constant over the length.

3. Nov 26, 2014

### Bystander

In your alternate approach, I'll give you "2T" or "mg/2," but not both; "pi" ain't obvious to me yet.

4. Nov 26, 2014

### Orodruin

Staff Emeritus
How is the normal force acting on different parts of your string?
The rope is a full circle and has no end ;)

5. Nov 26, 2014

### Staff: Mentor

Oh, I misread how the rope is placed.
In that case: the force is not acting in the same direction everywhere.

6. Nov 27, 2014

### Orodruin

Staff Emeritus
Just as a comment regarding alternative ways of approaching the problem: I would actually prefer to solve this one by using energy methods. The answer drops out quite naturally without having to consider individual parts of the rope.

7. Nov 27, 2014

### ZetaOfThree

I know what's wrong with my alternative solution. The component of the normal force that opposes $2T$ is a sum of small normal forces acting in the direction opposite to $2T$, which act on each element of the rope perpendicular to the surface of the cone. The force that opposes $2T$ can be found using integration and you get the correct answer.

Yeah, virtual work does give the result in like two lines. I was trying to find as many solutions to this problem as a could. I should have been more careful with my alternative solution above. I assumed the the force from the cone would be in a direction at angle $\alpha$ from the horizontal. For half the rope, this is not the case.

Thanks for the responses everyone!