What is the Correct Integral Setup for Finding Tension in a Rope on a Cone?

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SUMMARY

The correct integral setup for finding the tension in a rope draped over a cone involves understanding the relationship between the weight of the rope (W = mg) and the geometry of the cone. The tension (T) in the rope can be expressed as T = W * tan(ø), where ø is the cone's top angle. The discussion emphasizes the importance of considering the radial components of forces acting on the rope, particularly when the cone is oriented with the apex upwards. A diagram is essential for visualizing the forces and their contributions to tension.

PREREQUISITES
  • Understanding of basic physics concepts such as force and tension.
  • Familiarity with calculus, particularly integral calculus.
  • Knowledge of geometric properties of cones and circles.
  • Ability to interpret and create diagrams for physical problems.
NEXT STEPS
  • Study the derivation of tension in circular motion scenarios.
  • Learn about the application of line integrals in physics problems.
  • Explore the relationship between radial and tangential components of forces.
  • Review the principles of equilibrium in static systems.
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Students studying physics, particularly those focusing on mechanics and statics, as well as educators preparing for teaching related concepts in tension and forces on inclined surfaces.

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Homework Statement



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.

Homework 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)

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:
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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?
 
Simon Bridge said:
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?

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

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