- #1

leafy

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- TL;DR Summary
- These videos seems like static force can do work. Can anyone explains it?

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- Thread starter leafy
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- #1

leafy

- 73

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- TL;DR Summary
- These videos seems like static force can do work. Can anyone explains it?

- #2

Dale

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The string does the work.

- #3

leafy

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I can understand spring does work, but string doesn't store energy. Work is force x distance, so one end of the string the displacement is zero. The other end of the string, it displaced with the cylinder through space. But what is the source of energy?The string does the work.

- #4

Dale

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No, the displacement is very clearly not zero on both ends. I am not sure why you would think otherwise. It is very clear that the string moves.one end of the string the displacement is zero.

Obviously the thing pulling the string. There is no mystery here.But what is the source of energy?

The interesting thing about these wheels is how the string winds up under tension. That is a simple matter of choosing the size of the axle, but it is interesting anyway. But there is nothing unusual about the work.

- #5

leafy

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- #6

Dale

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Why do you think that. The string looks like normal string which is fairly inextensible. The displacement will be approximately the same on both ends.the displacement is many times higher

You are getting fooled by the fact that the string is being wound up on the axle, but that is a perpendicular displacement and not the displacement of interest.

This is silly. Of course it does.Obviously, the energy doesn't add up.

- #7

leafy

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Not really, if you think a force moving through space, both ends have the same force moving through different space length. That's why the string wound up, to compensate for the translation movement.The displacement will be approximately the same on both ends.

- #8

sophiecentaur

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Haha. Nothing is obvious in this world. Add up all the energy components involved (go as deep as you like) and the total will not change. Your mental experiment is leaving out some vital (not obvious) factors.Obviously, the energy doesn't add up.

Any string will stretch with any applied force (however small) so that involves doing work. If the stretch is elastic than mechanical energy of that work will be returned in some way. If there are internal losses, some of the work will be 'lost' and turn up as heating up the string.

- #9

Dale

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Nope. The important quantity is not the force moving through space, it is the distance that the material moves at the point of application of the force. This is hard to see if you haven’t worked these types of problems before because the force is applied to different pieces of material as the axle winds up. But with a bit of experience you will see immediately that it is the same distance on both ends.Not really, if you think a force moving through space, both ends have the same force moving through different space length. That's why the string wound up, to compensate for the translation movement.

- #10

leafy

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work done by the weight = mgd

work done on the cylinder = mgD

- #11

Dale

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So this problem is straightforward to work out, though a little tedious. Here is a free-body diagram for the wheel:Not really, if you think a force moving through space, both ends have the same force moving through different space length. That's why the string wound up, to compensate for the translation movement.

We have the radius of the wheel ##R## and the radius of the axle ##r## with ##R=r+\Delta r##. ##a## and ##v## are the acceleration and velocity of the center of mass, and ##\alpha## is the torque about the center of mass. The wheel (including the axle) has a mass ##m## and a moment of inertia ##I=kmR^2## where ##k## is some number that depends on the exact geometry of the wheel and axle assembly.

We are only interested in horizontal forces so there is the tension ##T## provided by the string and the friction force ##F## from the ground. The wheel rolls without slipping so ##a=R\alpha##. From Newton's 2nd law and Newton's 2nd law for rotation we have $$T-F = ma$$ $$FR-Tr=kmR^2 \alpha$$ Solving we get $$a=\frac{T \Delta r}{m R (1+k)}$$ $$\alpha=\frac{a}{R}$$

The rate of work is given by the power which is the force times the velocity of the material at the point of application of the force. The velocity of the material of the wheel goes from ##0## at the ground to ##v## at the center, so at the point of application of ##T## the velocity is $$v_T=\frac{\Delta r}{R} v$$ so the rate of work done on the wheel by the string is ##T\cdot v_T=T \frac{\Delta r}{R} v##.

Now on the other end of the string, assuming that the string is ideal (flexible, massless, inextensible), the external force is applying tension ##T##. If the string is a length ##L## then the position of the end of the string ##x_S## is equal to the position of the center of the wheel ##x## plus ##L## and minus the amount of thread that has been wound up onto the spool ##r\theta##. So that is ##x_S=x+L-r\theta##. Now, since we know ##a## and ##\alpha## from above we can solve $$x=\frac{a}{2}t^2$$ $$v=at$$ $$\theta = \frac{\alpha}{2}t^2$$ $$\omega = \alpha t$$ and with those we can obtain $$x_S=x-r \theta + L$$ $$x_S=\frac{a}{2}t^2-r\frac{\alpha}{2}t^2 + L$$ $$v_S=at-r\alpha t =\frac{R a}{R}t-\frac{ra}{R} t $$ $$v_S=\frac{\Delta r}{R}at=\frac{\Delta r}{R}v = v_T$$

So the rate of work done on the string by the external force supplying the tension is also ##T\cdot v_T=T \frac{\Delta r}{R} v## and hence energy is conserved at all points in time.

No, it is not, see above. As I explained in post 9 it is the motion of the material at the point of application of the force that is the relevant quantity. Not the motion of the point of application of the force.work done on the cylinder = mgD

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- #12

leafy

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You're only half correct. Consider the case of a block mass moving through a flat friction surface.No, it is not, see above. As I explained in post 9 it is the motion of the material at the point of application of the force that is the relevant quantity. Not the motion of the point of application of the force.

Yes, there is the motion of the block at the point of application of the friction force, a relevant quantity.

But, there is also the motion of the point of application of the friction force moving through the flat surface.

- #13

Dale

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The statement (mechanical power, the rate of mechanical work, is the product of the force and the velocity of the material at the point of application of the force) is completely correct and applies completely for all mechanical forces including friction. However, the sliding block is not relevant to the current scenario. If you would like it please open a new thread on that topic.You're only half correct. Consider the case of a block mass moving through a flat friction surface.

Yes, there is the motion of the block at the point of application of the friction force, a relevant quantity.

But, there is also the motion of the point of application of the friction force moving through the flat surface.

For this scenario it is clear that energy is conserved and the work balances out. There is no mystery, it behaves exactly as I described above.

- #14

leafy

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- #15

Dale

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By all means, be my guest. But if you are right then energy is not conserved and the wheel will have substantially more energy than was given to the string. Specifically, the input power is ##T \frac{\Delta r}{R}v## and if you are right then you will have an output power of ##Tv## for a factor ##R/\Delta r## gain in energy. This is not a small violation of energy conservation.

In any case, go ahead and do the experiment. It will be a good test of whether or not you are capable of performing good experimental measurements. I predict that if you do it correctly not only will you see nowhere near a gain of ##R/\Delta r## you will actually get a gain slightly less than 1 due to friction and other losses.

- #16

sophiecentaur

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Experiments, along these lines, have been carried out for centuries and, within the limits of experimental error (improving all the time) they have affirmed the principle of Energy Conservation. I am not sure why you don't just accept that your suggested experiment would give the 'accepted' answer.

My standard reply to this sort of question is to say that, if you really think you have spotted a mistake in established Physics, then you would need to start at square one and provide actual evidence (plus, of course, fully understanding all the basic Physics that has developed over centuries).

I'm not just saying you should accept what you are told. What I am saying is that you would have to prove that what you've been told is actually wrong.

- #17

leafy

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My thought is instead of a tension force pulling on the wheel, you can imagine yourself running and pushing on the same point on the axle. Now compare that to just sit and pull on the wheel. Your intuition would say something is off about input/output. Yes, the gain factor would be huge. It's crazy but I have to go with my intuition.

Another situation is imagine you're on a moving train pushing on a mass with force F. If you stick with mechanical power, it would be F x speed of the mass. But if you calculate the energy gain from a absolute frame, it would be F x speed of train + mass. The force acting on the mass has to move through much larger space in absolute frame. That's where my logic about force moving through space.

- #18

Dale

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I told you. The important thing is not the motion of the point of application of the force, but the motion of the material at the point of application of the force. That is where you misunderstood.I just don't get where I misunderstood things.

Going with your intuition only works if you have good intuition. Good intuition is something that is built over time. If you do something right, then do it right again, and again and again, then you build good intuition. If you have bad intuition (which you do here) then you need to focus on correct practice to tear down your incorrect intuition and then begin building good intuition in its place.It's crazy but I have to go with my intuition.

There is no such thing as absolute space. However, I would indeed recommend that you open up new threads to discuss how these different scenarios work. All of them follow the rule I posted earlier.Another situation is imagine you're on a moving train pushing on a mass with force F. If you stick with mechanical power, it would be F x speed of the mass. But if you calculate the energy gain from a absolute frame, it would be F x speed of train + mass. The force acting on the mass has to move through much larger space in absolute frame. That's where my logic about force moving through space.

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- #19

leafy

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And I said the motion of the point of application of the force is also important and gave you example which you said irrelevant.I told you. The important thing is not the motion of the point of application of the force, but the motion of the material at the point of application of the force. That is where you misunderstood.

- #20

Dale

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Yes, you said that and that is your misunderstanding. It is always the motion of the material at the point of application that is important for mechanical energy, not the motion of the point of application. Always.And I said the motion of the point of application of the force is also important and gave you example which you said irrelevant.

The example is irrelevant to the scenario in the OP, but it is relevant to your misconception. I would recommend opening new threads on those topics.

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- #21

sophiecentaur

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Have you done a course on Newtonian mechanics? If your intuition was not gained from that sort of background then you need to consider it's possibly wrong. Why are you arguing with @Dale ? He does know what he's talking about and it's no one's job here to nail exactly where you are going wrong. Try to bend to the subject and not expect the subject to bend to your intuition.And I said the motion of the point of application of the force is also important and gave you example which you said irrelevant.

- #22

leafy

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Let's say you're right that the motion of the material is important. How can you be sure that that at time t+delta(t) it will be the same as the time t? The point of material also move upward and at time t+delta(t), or half delta(t), the force already jump to the next point in the arc.Yes, you said that and that is your misunderstanding. It is always the motion of the material at the point of application that is important for mechanical energy, not the motion of the point of application. Always.

- #23

Dale

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That is why we use the velocity. That gives us the displacement of the material over an infinitesimal time. This is one of the reasons that we use calculus in Newtonian physics.Let's say you're right that the motion of the material is important. How can you be sure that that at time t+delta(t) it will be the same as the time t? The point of material also move upward and at time t+delta(t), or half delta(t), the force already jump to the next point in the arc.

- #24

leafy

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One is you have to station pushers along the path the wheel travel and each time the wheel pass by, each pusher going to push the wheel.

Second is you only need one pusher, but he must chase the wheel to impart a force.

Both yields the same result but one takes more work and energy. Hm...

- #25

Dale

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Not in principle. It would only take more work and energy if the chasing system is less efficient somehow.one takes more work and energy

- #26

leafy

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oh? Let's say we have 10 pushers station along the path. The wheel translate at 1m/s. The pusher push on material moving at small v(almost zero).Not in principle. It would only take more work and energy if the chasing system is less efficient somehow.

P=Fv x 10

Or we have one pusher moving at 1m/s also push on the material moving at small v.

P=Fv x 10 + F(1m/s)

It is a big difference.

- #27

Dale

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You are just randomly throwing numbers here. Physics isn’t a plate of pasta that you throw on the wall and see what sticks.Or we have one pusher moving at 1m/s also push on the material moving at small v.

P=Fv x 10 + F(1m/s)

- #28

leafy

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You have to understand that imparting a force while moving cost a lot of energy.You are just randomly throwing numbers here.

- #29

Dale

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Not in the scenario you described. (Unless there are some inefficiencies involved)You have to understand that imparting a force while moving cost a lot of energy.

- #30

leafy

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- #31

Dale

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None of this matters. If something is pushing the wheel in a given spot then the ##\vec v## is fixed to match that velocity regardless of whether it is one thing or many.

If you believe otherwise you will need to rigorously work the math. This pasta on the wall physics is pointless.

- #32

leafy

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You have no regard for translation speed of the wheel.

- #33

Dale

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I have regard for math and carefully worked physics.You have no regard for translation speed of the wheel.

- #34

leafy

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It's ok.

- #35

Lnewqban

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Here I drew the the system. The cylinder moved a distance D at a later time. The force (in red) moves the object for a distance D. The weight on the right also displaced a smaller distance d.

work done by the weight = mgd

work done on the cylinder = mgDView attachment 294549

Using your well drawn diagram, it seems to me that combining D and mg is not correct.

The force doing effective work should be the one applied to the center of mass of the cylinder, which is the effective distance represented by D.

The magnitude of that force is smaller than mg, as many times as the distance of each force to the contact point with the table is different.

If you replace the cylinder with a lever pivoted at the table, you could visualize that you are losing force intensity at D, as you have less and less mechanical advantage as you apply resistance to mg higher and higher.

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