Find the Tension in a Flexible Chain Resting on a Cone

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SUMMARY

The discussion centers on calculating the tension in a flexible chain resting on a smooth, frictionless right circular cone. The total weight of the chain is denoted as W, with the solution derived using static equilibrium principles. The tension T is conclusively expressed as T = (W / 2π) * (h / r), where h is the height and r is the base radius of the cone. While the principle of virtual work was suggested, participants agreed that a static approach is more appropriate for this problem.

PREREQUISITES
  • Understanding of static equilibrium in physics
  • Familiarity with the concepts of tension and normal force
  • Basic knowledge of circular motion and angles
  • Ability to apply Newton's laws, specifically F = ma
NEXT STEPS
  • Study the principles of static equilibrium in greater detail
  • Explore the application of virtual work in static systems
  • Learn about the mechanics of chains and their tension in various configurations
  • Investigate the relationship between potential energy and static configurations
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Students and educators in physics, particularly those focusing on mechanics and static equilibrium problems, as well as anyone interested in the application of tension in physical systems.

tobix10
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Homework Statement


A loop of flexible chain, of total weight W, rests on a smooth, frictionless right circular cone of base radius r and height h. The chain rests in a horizontal circle on the cone, whose axis is vertical. Find the tension in the chain.

Homework Equations


Virtual work, but I've done it with Newton's ## F = ma ##.

The Attempt at a Solution


I considered a part of a chain that spread on an arch of angle ## \Delta \theta ## (angle is very small) The forces ## T ## on each end of arch exert horizontal force ## 2T \sin(\frac{\Delta \theta}{2}) ## which has to be equalized by horizontal component ## N_{x} ## of normal force.
Equations are:
## N_{x} = T \Delta \theta ##
## N_{y} = W \frac{\Delta \theta}{2 \pi} ##
## \frac{N_{x}}{N_{y}} = \frac{h}{r} ##
Solution is:
## T = \frac{W}{2 \pi} \cdot \frac{h}{r} ##

Is this answer correct? Any tips how to handle this problem using principle of virtual work?
 
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Hello tobix and welcome to physicsforums.

My answer matches yours.

Why do you want to use virtual work? The problem is a static one, and virtual work typically involves movement, or at least potential movement. It might be useful if they had asked you to prove that the described configuration has the lowest potential energy. But they haven't asked you to do that.
 
I was told to use virtual work principle, but I don't see any starting point. Nevertheless I am going to stick with static solution. Thank you for your answer.
 

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