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Under the chapter entitled Conservation of energy, under the section of Gravitational potential energy, there is this following paragraph :

A very simple weight-lifting machine is shown in Fig. 4–1. This machine lifts weights three units “strong.” We place three units on one balance pan, and one unit on the other. However, in order to get it actually to work, we must lift a little weight off the left pan. On the other hand, we could lift a one-unit weight by lowering the three-unit weight, if we cheat a little by lifting a little weight off the other pan. Of course, we realize that with any actual lifting machine, we must add a little extra to get it to run. This we disregard, temporarily. Ideal machines, although they do not exist, do not require anything extra. A machine that we actually use can be, in a sense, almost reversible: that is, if it will lift the weight of three by lowering a weight of one, then it will also lift nearly the weight of one the same amount by lowering the weight of three.

enter image description here

According to my understanding based on this diagram, this weight-lifting machine works just like a see-saw does. You consider the torque generated by the weights placed on the two pans. If the two torques are equal, then they're in equilibrium, otherwise the rod starts rotating in the direction of the greater torque. And assuming there is a ground, this rotating rod will be stopped once it hits the ground.

Firstly, What does he mean by: "However, in order to get it actually to work, we must lift a little weight off the left pan"?since we want to lift weight three units strong(those three blocks) Why not just change the position of the fulcrum(without removing anything) until system rotates?

"Of course, we realize that with any actual lifting machine, we must add a little extra to get it to run", what he's talking about? and what is this "little extra"?

and then : "A machine that we actually use can be, in a sense, almost reversible: that is, if it will lift the weight of three by lowering a weight of one, then it will also lift nearly the weight of one the same amount by lowering the weight of three.". I don't understand this part at all. How can it be reversible? I mean, if we have a certain configuration like that of the diagram, where we fix the position of the fulcrum and the weights on both sides, the system will rotate in the direction of the greater torque and will stop once the rods hit the ground, How you can reverse the system's motion with its configuration(the fulcrum position and the weights) unchanged? and why he says "almost reversible"? why "almost" not totally?

I can't make out what Feynman is trying to convey in this paragraph.

Omar Nagib
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  • Related: http://physics.stackexchange.com/q/190738/37364 And a link to the books. http://feynmanlectures.caltech.edu/I_04.html#Ch4-S2 – mmesser314 Jun 29 '15 at 02:13
  • I have the impression that one should add a little weight on the left pan to get it started to lift the right pan, as we look at the page. If we take a bit off the left the right will sink , not lift. It is probably a confusion between left and right. – anna v Jun 29 '15 at 07:41

1 Answers1

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This chapter is on the conservation of energy. Ideally the machine he shows will work. If we jump ahead in the book to where we know about conservation of energy, we see that energy gained by one side is lost by the other.

But in practice, some energy is lost to friction. The little extra weight is needed to overcome friction. You could also overcome friction by adjusting the fulcrum.

By reversible, he means an idealized frictionless machine where the weights and distances are just right so that it can run either way without a power source. If there is too much weight on the right, the right side will sink. But it would not rise without adding power. If it is perfectly balanced, it will run either way without a power source.

mmesser314
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    If it's perfectly balanced, then by definition, it should be in equilibrium forever, shouldn't it? So how, if it is balanced, it will run either way without a power sorce? – Omar Nagib Jun 29 '15 at 11:33
  • It's an idealization. With a small weight on the right, the right side will slowly sink. You would have to add a small amount energy to raise the right. Perfectly balanced is the limit as the extra weight goes to 0. The time it takes for the pan to sink all by itself goes to infinity. The amount of energy you would have to add goes to 0. – mmesser314 Jun 29 '15 at 13:01
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    It seems to me that reversible machines are not only impossible in real life, but in an ideal situation as well. What dictates the motion of this system is the net torque acting on it. If it's zero, then the system stays at rest in equilibrium forever. If the net torque is non-zero, then the system rotates in the direction of the greater torque. Let's assume the torque done by the weight on the left is the greater one, therefore the system will rotate anti-clock wise until it hits the ground and stops. Given this configuration, how can you reverse its motion without external supply of energy? – Omar Nagib Jun 29 '15 at 16:14
  • The picture of the pan balance isn't quite right. The fulcrum has to be on the line connecting the centers of mass. As drawn, the balance is unstable. Also, you could replace the fulcrum with a hinge so it won't fall off if it rotates too far. It isn't really correct to say that the pan balance would stay at rest forever. It would rotate with its initial speed forever. A torque causes an angular acceleration, which changes the angular velocity. But we are assuming an initial angular velocity of 0. – mmesser314 Jun 29 '15 at 20:53
  • Given those changes, you are right. If one pan is heavier it would sink. That would not be reversible. Reversible is when the balance is perfect. – mmesser314 Jun 29 '15 at 20:54
  • But if it's perfectly balanced, and you wish to reverse its motion, you have to apply a torque momentarily on it, don't you? and I presume applying a torque for some time interval should not cost energy, is that right? – Omar Nagib Jun 29 '15 at 23:18
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    If you are willing to run it infinitely slowly, the amount of energy you need to add to run it either way is 0. The point about reversibility is not stopping and running the other way. It is running either way at 0 cost. – mmesser314 Jun 30 '15 at 14:00
  • yes, I agree. It can run either way with constant angular velocity with no energy expenditure. This is what do you mean by reversible? – Omar Nagib Jun 30 '15 at 14:06
  • Yes. For some machines you need to add energy to run them one way. You can get it back by running them the other way. For a reversible machine, the amount of energy is 0. – mmesser314 Jun 30 '15 at 14:37
  • So what does Feynman mean, when he says, this is a weight-lifting machine, how it does lift things up? – Omar Nagib Jun 30 '15 at 14:40
  • Hey @mmesser314 not sure if this is the best place to ask this but I would be willing to pay an hour of consultancy with you if you were so kind to take some time off and explain this chapter of Feynman's to me. Can you email me at hello@paulrberg DOT com, please? – Paul Razvan Berg Nov 22 '20 at 21:40
  • @PaulRazvanBerg - I am busy. If you can explain where you are having problems understanding that chapter, the you might try posting another question. – mmesser314 Nov 23 '20 at 00:37
  • I see, thanks anyways! I posted FLCH42NO2, FLCH42NO3 and FLCH42NO4 and I will probably keep posting questions as I dive deeper into the chapter. There are many details that are missed or implied in Feynman's explanation. – Paul Razvan Berg Nov 23 '20 at 23:11