Homework Help: Tension in Rope on Cone

1. Sep 13, 2014

KleZMeR

1. The problem statement, all variables and given/known data

I am given the weight (force) of the rope as W. It sits on a cone about halfway down, with the cone's top angle ø. Radius at a given placement is r, and h is our height at a given placement.
I need to find the tension, T, in the rope.

2. Relevant equations

W=mg
Integral (F * dr) = 0
I am taking r to be along the x axis.
L = sqrt(r^2 + h^2)
X = L*cos(ø)
Y = L*sin(ø)
dX = dø*L*-sin(ø)
dY = dø*L*cos(ø)

3. The attempt at a solution

Expressing my equilibrium as:

T*L*dø*cos(ø)-m*g**dø*L*-sin(ø) = 0

I get: T = W*tan(ø)

This seems over simplified? Or am I over-thinking it? It's around a circle radius r and each element of T summed over the circle would be 2*pi*T but the gravitational force is also summed over 2*pi. Perhaps I skipped over the line integral of this? I am very interested in the correct integral setup of this problem because it looks like a future test question, and I also want to know how my 2*pi factor disappears (if it was ever present?) Any help is appreciated.

Last edited: Sep 13, 2014
2. Sep 13, 2014

Simon Bridge

You need a diagram.

You are given the "top angle" of the cone - which wording suggests that the cone is oriented with the apex upwards, and the rope is draped over the outside. If the rope is draped anywhere below the apex, then it will be draped over a hyperbola - but nothing dangles.

The mention of circles suggests that the cone is tilted so the central axis is horizontal.
In which case, what is stopping you from using a previous result for tension from being draped over a circle?

3. Sep 13, 2014

Staff: Mentor

I suspect that the rope is in fact a circular ring of rope and the OP is meant to find the tension in the rope that results from it being placed on a cone (apex up). The weight of the rope results in an outward radial force all around the circumference which must be countered by a tension in the rope.

4. Sep 14, 2014

KleZMeR

Image

Thanks guys, here's the diagram.

I get that the weight contributes to the tension, but it is not the full weight that is equal to the tension, but rather the radial component.

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5. Sep 14, 2014

Simon Bridge

Cool - so walk us through the reasoning that leads to your result.

6. Sep 17, 2014

KleZMeR

Ok, well I consider the constraint, being the solid cone. Any contribution from gravity must then be in the radial direction, normal to the cone. I am trying to find the tension as a function of the angle, and the tangential component of the angle is definitely in the normal direction.