High School Balancing on Pulley system

[curiousEngineer]

TL;DR

Pulley balancing question

This is a basic conceptual question. Consider having a pulley mounted with identical mass of block A and B. If i pull the block A slightly by say 10mm, and then release it. Will these weights try to balance itself due to potential energy difference or will be in the same state?

 

[anuttarasammyak]

Say the rope has zero mass, total potential energy is constant during the process. If you pull 10 mm or with any other length, keep them at rest with your hand and release your hold, they balance there. If we cannot have zero mass rope, the downer side would go down more.

 

[kuruman]

Pull the block how? Sideways like a pendulum and release from rest or pull it down? If pull it down, release while the block is moving or stop the blocks and then release?

 

[curiousEngineer]

Pull the block how? Sideways like a pendulum and release from rest or pull it down? If pull it down, release while the block is moving or stop the blocks and then release?

Im pulling it down by 10mm, after small rest im releasing it again.

 

[curiousEngineer]

Say the rope has zero mass, total potential energy is constant during the process. If you pull 10 mm or with any other length and keep them at rest, they balance there. If we cannot have zero mass rope, the downer side would go down more.

Thank you for the explanation! Say there is friction also in the pulley. That will also affect this imbalance right?

 

[anuttarasammyak]

You refer friction. Both friction between rope and pulley and friction of pulley to rotate would matter.

 

[A.T.]

... the downer side would go down more.

True in many ways, but I think "lower side" fits better here.

 

[Lnewqban]

Will these weights try to balance itself due to potential energy difference or will be in the same state?

I believe that self-balancing and potential energy difference are not related, at least in this case.

Balance, or lack of it, is the result of the net force (mass x acceleration) acting on a system of masses.

Amount of accumulated potential energy is the result of the altitude (weight x height above ground).

 

[A.T.]

I believe that self-balancing and potential energy difference are not related, at least in this case.

Self-balancing (stable equilibrium) happens around a local minimum of the total potential energy, for the whole system within which potential energy can be transferred (here both weights & rope).

In this case you don't have such a local minimum:
- For a mass-less rope it is a neutral equilibrium.
- For a massive rope it is an unstable equilibrium.

From: https://theshawclassroom.weebly.com/energy-and-equilibiria-13.html

 

[Lnewqban]

Welcome, @curiousEngineer

What makes you believe that balance of the blocks and their potential energies (20 mm height difference) are related in the case of the represented pulley?

Why is our pulley and equal masses (and massless string) remaining in any placed position, while a weighting device (simple balance) loaded with equal masses would always go back to a horizontal position?

https://cdn.dribbble.com/users/1167...50/media/32ed8c6a3144ee35c46ff19b70842f3a.gif

 

[curiousEngineer]

Welcome, @curiousEngineer

What makes you believe that balance of the blocks and their potential energies (20 mm height difference) are related in the case of the represented pulley?

Why is our pulley and equal masses (and massless string) remaining in any placed position, while a weighting device (simple balance) loaded with equal masses would always go back to a horizontal position?

https://cdn.dribbble.com/users/1167...50/media/32ed8c6a3144ee35c46ff19b70842f3a.gif

Thank you for the reply. In case of weighing balance, i guess its because of horizontal bar which is connected to the weights on either side. And when one side is pulled due to the net torque difference it balances itself. Am i correct?

But in the case of pulley, there is no net torque regardless position, so the weights stay put if the net force and net torque are zero. But technically the rope mass on downer side will be high. So due to net force value the weight should come down but its compensated by the rope friction with pulley? Correct me if im wrong. Thank you again!

 

[Lnewqban]

Thank you for the reply. In case of weighing balance, i guess its because of horizontal bar which is connected to the weights on either side. And when one side is pulled due to the net torque difference it balances itself. Am i correct?

Not exactly.
That case is like case (a) shown in Figure 1 of post #9: stable equilibrium.

But in the case of pulley, there is no net torque regardless position, so the weights stay put if the net force and net torque are zero.

But we could simplify the pulley-masses system to be seen as a "horizontal bar which is connected to the weights on either side" (the bar being the diameter of the pulley).
However, our pulley behaves like case (c) above: neutral equilibrium.

The net torque in each case is what makes the difference.
We have one torque in case (a), but none in case (c).

Case (a): The weighting device has the pivot point purposely located a little above the imaginary straight line that joints the centers of masses of both blocks.

Case (b): The pulley has the pivot point located exactly in line with the imaginary straight line that joints the points at which the strings supporting the blocks are tangent to it.

 

[A.T.]

pulley and equal masses (and massless string) remaining in any placed position, while a weighting device (simple balance) loaded with equal masses would always go back to a horizontal position?

https://cdn.dribbble.com/users/1167...50/media/32ed8c6a3144ee35c46ff19b70842f3a.gif

View attachment 355026

The balance in the animation above would NOT go back to a horizontal position, because it's not how these balances actually look like. See the video below:

 

[curiousEngineer]

In this same problem, consider if I'm applying a pulling force F1 to make it displace down for x distance. Now alternatively i'm applying pushing force F2 to displace it up for same x distance. Now the force which i need to apply i.e F1 and F2 will be the same or something different. Its because, i recently done a small experiment with pulley where i felt the force F1 which i need to give was slightly higher than force F2. Can someone please explain the math? and explain if im correct or wrong?
Cheers!

 

[anuttarasammyak]

F_1 keeps the wire tightened but F_2 looses the wire for a while. I think it could make difference.

 

[curiousEngineer]

F_1 keeps the wire tightened but F_2 looses the wire for a while. I think it could make difference.

Yes. But any technical explanation for this?

 

[A.T.]

Now the force which i need to apply ...

Need to overcome static bearing friction at the pulley axle, I assume.

Yes. But any technical explanation for this?

Explanation for why pulling on a rope increases its tension? Seems obvious, but try a free body diagram with all the forces balanced (pulling at constant speed against friction).

Or explanation for why more tension means more axle friction? Check out what friction depends on.