Find the Tension in a Flexible Chain Resting on a Cone

AI Thread Summary
The discussion revolves around finding the tension in a flexible chain resting on a frictionless cone. The user derived the tension using Newton's laws, resulting in the formula T = (W/2π) * (h/r). There is a debate about the applicability of the virtual work principle for this static problem, with some users suggesting it may not be necessary. One participant confirms that the static solution is valid and matches their own answer. The conversation highlights the tension between using traditional methods versus virtual work for solving static equilibrium problems.
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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