Force to lift a chain: Conservative or not?

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SUMMARY

The discussion focuses on calculating the force required to lift a chain with linear mass density λ at a constant velocity v_0. The potential energy U is derived as U = λgy^2/2, indicating that the force F required to lift the chain is F = -λgy, confirming that this force is conservative when the chain remains on the ground. The conversation also clarifies that the work done on the chain is zero over a closed path, reinforcing the conservative nature of gravitational force. The final equation for lifting the chain when it leaves the surface is F = -λLg, which remains constant.

PREREQUISITES
  • Understanding of linear mass density (λ)
  • Familiarity with potential energy concepts in physics
  • Knowledge of conservative forces and their properties
  • Basic calculus for differentiation (∂U/∂y)
NEXT STEPS
  • Study the implications of non-constant mass in conservative force scenarios
  • Explore the concept of potential energy in multi-dimensional systems
  • Learn about the mechanics of lifting objects with varying mass distributions
  • Investigate the mathematical modeling of finite-sized links in chains
USEFUL FOR

Physics students, mechanical engineers, and anyone interested in the dynamics of lifting systems and conservative forces.

AJKing
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Homework Statement



You are lifting a chain straight up at a constant velocity v_0. The chain has a linear mass density λ. What is the force required to lift the chain as a function of height?

The Attempt at a Solution



U = mgh = λygh

The height in the potential energy is the same as the potential energy at the center of mass

h = y/2

U = λgy^2/2

This is a conservative potential energy in one dimension

F = -∂U/∂y= -λgy

Is this correct?

Can conservative forces consider objects of non-constant mass as I've done?
 
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Is the chain initially lying on the ground then.

Per your question:
By definition of a conservative force, the work should be zero on a closed path right?
So check.
 
Simon Bridge said:
Is the chain initially lying on the ground then.

Per your question:
By definition of a conservative force, the work should be zero on a closed path right?
So check.

Yes, it's on the ground.

Right, the work done on the object should be zero about a closed path. Regardless of the fact that my work is not zero.
The work done on this chain will certainly be zero about any closed path.
The change in mass does not matter.

Then I suppose my force is conservative and that my equation is correct in the case that the chain doesn't leave the surface. As soon as it does, I've got to rewrite to:

F = -λLg

which is constant.

Is that all correct?
 
The applied force here is not really from a field though is it?

Technically it is gravity that is the conservative force - since it can be described as the gradient of a potential function. Your applied force only exists at a point - it (or rather, whatever is applying the force) is the thing doing the work.

But your math looks fine from here.

To check this sort of thing all you need is to check the reasoning - to lift at constant speed, the applied force has to be equal to gravity. Presumably it was briefly larger than gravity at some earlier stage in order to accelerate to the constant speed.

This is one would modify the model to allow for finite sized links in the chain.
 

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