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Conservation of energy: a mass and pulley system

  • Thread starter smashueatu
  • Start date
18
4
Homework Statement
For the apparatus shown in the figure, find the equilibrium angle θ in terms of the two masses.
Homework Equations
(1) ΔE = ΔU + ΔK = 0 (Conservation of energy)

(2) ΔE = 2My₁ + mgy₂ = 0
Problem 18.PNG


The solution is an application of the law of conservation of energy.

Start with equation (1). The masses are in equilibrium and are not accelerating. This implies that ΔK = 0, because the kinetic energy will not change without acceleration. Thus, we are left to find equation (2) in terms of θ.

Finally, we set up a geometry problem to find substitutions for the heights in equation (2).

p18 geometry.jpg


This is where I fail. I can't find the heights in terms of the angle without introducing extra variables. I have tried constructing multiple triangles and applying some basic geometry and trigonometry, but to no success (pythagorean, law of sines, etc...).

Any hints or ideas would be greatly appreciated.
 

phinds

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Question: can you get an answer without considering how far apart the pulleys are? I mean, suppose they are 1 foot apart. Will the answer be the same if they are 100 feet apart?

Forget this post. I was solving the wrong problem. See below
 
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jbriggs444

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Start with equation (1). The masses are in equilibrium and are not accelerating. This implies that ΔK = 0,
For an equilibrium to be attained in the first place, kinetic and potential energy must be dissipated into heat. You'll not be able to solve this with conservation of mechanical energy.

Rather, you need to look at the conditions required for equilibrium. If nothing is accelerating, what force balance equations can you write down?
 
18
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Question: can you get an answer without considering how far apart the pulleys are? I mean, suppose they are 1 foot apart. Will the answer be the same if they are 100 feet apart?
I don't think you can. If you have an enclosed triangle like the picture above, then sinθ = y/r.

r relates to x (the distance between the masses), so the angle changes with x.
 

phinds

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I don't think you can. If you have an enclosed triangle like the picture above, then sinθ = y/r.

r relates to x (the distance between the masses), so the angle changes with x.
Yes. See post #2
 
18
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For an equilibrium to be attained in the first place, kinetic and potential energy must be dissipated into heat. You'll not be able to solve this with conservation of mechanical energy.

Rather, you need to look at the conditions required for equilibrium. If nothing is accelerating, what force balance equations can you write down?
Since this problem is in the conservation of energy chapter, I think it's assuming friction is negligible.

However, if I approach it with forces, then I would say that the Fnet = 0 for all masses and then just analyze the forces on the mass in the middle to get an equation in terms of theta. If I include the friction of the pulley, then I will be left with some unknown constant in my solution, which I don't think fulfills the problem statement.
 

jbriggs444

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I think it's assuming friction is negligible.
If friction is zero, equilibrium will never be attained. Minimizing energy is a possible approach. Conserving it is not.

But yes, a friction that is negligible with respect to force balance but which is non-zero so that the bounces die out and equilibrium is attained will work.
 

Delta2

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I think this problem is very easy to analyze in terms of force balances, I wonder why it is in the conservation of energy chapter.

If m=0 then it is ##\theta=0##
if m=M then it is ##\theta=\pi/6##
The bigger the m while the M masses remain constant the bigger ##\theta will be##
I think if ##m\geq 2M## then the system cant achieve an equilibrium

The above hold even if there is no friction.
 

phinds

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if m=M then it is ##\theta=\pi/6##
Are you saying that that is true regardless of the distance between the pulleys? I don't see how that can be. See post #2

Forget this post. I was solving the wrong problem. See below
 
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Delta2

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Are you saying that that is true regardless of the distance between the pulleys? I don't see how that can be. See post #2
Yes distance doesn't matter as long as mass m hangs from the middle of that distance. if not then distance might matter.
 

phinds

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Yes distance doesn't matter as long as mass m hangs from the middle of that distance. if not then distance might matter.
Ah yeah. I see how I was looking at the problem incorrectly. For some reason, I was interpreting "equilibrium" to mean the middle weight would hang at the same vertical position as the side weights so I was solving the wrong problem.
 

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