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leafy
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- TL;DR Summary
- These videos seems like static force can do work. Can anyone explains it?
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?Dale said:The string does the work.
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.leafy said:one end of the string the displacement is zero.
Obviously the thing pulling the string. There is no mystery here.leafy said:But what is the source of energy?
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.leafy said:the displacement is many times higher
This is silly. Of course it does.leafy said:Obviously, the energy doesn't add up.
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.Dale said:The displacement will be approximately the same on both ends.
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.leafy said:Obviously, the energy doesn't add up.
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.leafy said: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.
So this problem is straightforward to work out, though a little tedious. Here is a free-body diagram for the wheel:leafy said: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.
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.leafy said:work done on the cylinder = mgD
You're only half correct. Consider the case of a block mass moving through a flat friction surface.Dale said: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.
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.leafy said: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.
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.leafy said:The only solution to our argument is to let the block drop a height h, after height h, measure the rotional energy of the wheel, the translational energy of the wheel, and the kinetic energy of the weight. If it add up to the drop height potential, then I submit.
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.leafy said:The only solution to our argument is to let the block drop a height h, after height h, measure the rotional energy of the wheel, the translational energy of the wheel, and the kinetic energy of the weight. If it add up to the drop height potential, then I submit.
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.leafy said: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.leafy said: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.leafy said: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.
And I said the motion of the point of application of the force is also important and gave you example which you said irrelevant.Dale said: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.
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.leafy said:And I said the motion of the point of application of the force is also important and gave you example which you said irrelevant.
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.leafy said:And I said the motion of the point of application of the force is also important and gave you example which you said irrelevant.
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.Dale said: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.
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.leafy said: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.
Not in principle. It would only take more work and energy if the chasing system is less efficient somehow.leafy said:one takes more work and energy
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).Dale said:Not in principle. It would only take more work and energy if the chasing system is less efficient somehow.
You are just randomly throwing numbers here. Physics isn’t a plate of pasta that you throw on the wall and see what sticks.leafy said: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)
You have to understand that imparting a force while moving cost a lot of energy.Dale said:You are just randomly throwing numbers here.
Not in the scenario you described. (Unless there are some inefficiencies involved)leafy said:You have to understand that imparting a force while moving cost a lot of energy.
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.leafy said:If the wheel is moving at 1m/s, imparting a force while moving is equivalent of throwing at rock at greater than 1m/s and transfer energy. Matter of fact, let make the wheel move at 100m/s. You have to throw rock faster than that. Let's say the station pusher throw rock too to move the wheel, they only need to throw rock at approximately 0 m/s to impart the same force.
I have regard for math and carefully worked physics.leafy said:You have no regard for translation speed of the wheel.
leafy said: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
leafy said:Let's say I was pulling the string. I felt a tiny displacement. So my work done is the force I pull times the displacement my hand made. On the cylinder end, the same force carry by the string, but the displacement is many times higher. The work done is then force time big displacement. Obviously, the energy doesn't add up.