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!