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Consider weight-lifting machines—machines which have the property that they lift one weight by lowering another. Let us also make a hypothesis: that there is no such thing as perpetual motion with these weight-lifting machines. (In fact, that there is no perpetual motion at all is a general statement of the law of conservation of energy.) We must be careful to define perpetual motion. First, let us do it for weight-lifting machines. If, when we have lifted and lowered a lot of weights and restored the machine to the original condition, we find that the net result is to have lifted a weight, then we have a perpetual motion machine because we can use that lifted weight to run something else. That is, provided the machine which lifted the weight is brought back to its exact original condition, and furthermore that it is completely self-contained—that it has not received the energy to lift that weight from some external source. We imagine that there are two classes of machines, those that are not reversible, which includes all real machines, and those that are reversible, which of course are actually not attainable no matter how careful we may be in our design of bearings, levers, etc. We suppose, however, that there is such a thing—a reversible machine—which lowers one unit of weight (a pound or any other unit) by one unit of distance, and at the same time lifts a three-unit weight. Call this reversible machine, Machine A. Suppose this particular reversible machine lifts the three-unit weight a distance X. Then suppose we have another machine, Machine B, which is not necessarily reversible, which also lowers a unit weight a unit distance, but which lifts three units a distance Y. We can now prove that Y is not higher than X; that is, it is impossible to build a machine that will lift a weight any higher than it will be lifted by a reversible machine. Let us see why. Let us suppose that Y were higher than X. We take a one-unit weight and lower it one unit height with Machine B, and that lifts the three-unit weight up a distance V. Then we could lower the weight from Y to X, obtaining free power, and use the reversible Machine A, running backwards, to lower the three-unit weight a distance X and lift the one-unit weight by one unit height. This will put the one-unit weight back where it was before, and leave both machines ready to be used again! We would therefore have perpetual motion if Y were higher than X, which we assumed was impossible. With those assumptions, we thus deduce that Y is not higher than X, so that of all machines that can be designed, the reversible machine is the best.

This is a part of the fist book of "Feynman's Lectures". Its purpose is to derive the formula that calculates the Gravitational potential energy of a body near the surface of the earth through a series of "simple" thoughts-part of which, is the following thought-.

Although beautiful, it baffles me. Although i thought i understood it, after giving it much thought, i found some parts that i don't get.

The first thing is the definition of a perpetual weight lifting machine.

If, when we have lifted and lowered a lot of weights and restored the machine to the original condition, we find that the net result is to have lifted a weight, then we have a perpetual motion machine because we can use that lifted weight to run something else.

My problem is that i cannot imagine such thing when it comes to simple weight lifting machines such as the one in the following picture.

enter image description here

The other is the "obtaining free power"

Let us suppose that Y were higher than X. We take a one-unit weight and lower it one unit height with Machine B, and that lifts the three-unit weight up a distance V. Then we could lower the weight from Y to X, obtaining free power, and use the reversible Machine A, running backwards, to lower the three-unit weight a distance X and lift the one-unit weight by one unit height. This will put the one-unit weight back where it was before, and leave both machines ready to be used again!

  1. How are we lowering the weights? Is it because it's a thought experiment we don't have to ask ourselves such question?

  2. As I can understand it by lowering the three units of weight from height Y to X we are lifting the one unit of weight to a height H, so that energy is conserved. So where do we get this free power from? Although I know that he is not wrong, I would assume that after the main assumption that he makes-Y>X- he makes an aposteriori assumption that energy in not conserved and he ends up in a false conclusion. So we couldn't know which assumption is wrong.

So could you please explain to me the whole thought process stressing my queries?

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    $\begingroup$ Just for reference link to a book for this chapter: Feynman Lectures. $\endgroup$ – m0nhawk Jun 22 '15 at 15:49
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The black machine is a weight lifting machine. It is self contained with no power source.

If it can lift an external weight and return to its original state as shown below, it is a perpetual motion machine. Suppose the blue weight is water. We could add a water wheel and generator on the right. You start at the top and work your way to the bottom. Then you pour the water onto the water wheel and generate electricity, This leaves you back at the top illustration, except that you have generated free power in between.

Feynman is assuming that a perpetual motion machine is impossible.

enter image description here

Feynman is defining a reversible machine as one like the black machine with the blue weight left out. Without a power source, it can lower 3 weights as it raise one weight 3 times as far. Likewise it can run the other way without a power source.

A reversible machine is also impossible, but only just. You can make a machine where it will lower 1 weight with a very tiny blue weight to overcome friction. But now it will only run one way.


He assumes for the sake of argument that we do manage to make a reversible machine like the black one, He uses it to show that no machine without a power source can outperform a reversible machine.

For example, the illustration above shows a proposed machine that does outperform a reversible machine. It lifts more weight. Because it lifts more weight, it is a perpetual motion machine, and perpetual motion machines cannot exist.


Another proposed machine might lift 1 weight farther than a reversible machine. Any machine like this without a power source has the same problem. Because it outperforms a reversible machine, it is a perpetual motion machine.

This proposed black and blue machine has no power source. It has a jack underneath one weight that lifts it higher than the black reversible part can do alone. You can get free power by lowering the weight back where it was. So the black and blue machine is a perpetual motion machine.

enter image description here

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  • $\begingroup$ Ok as i get it, this explains why reversible weight lifting machines cannot exist! I still cannot see why non-reversible machines cannot lift weights higher that a reversible one. My main problem is the "getting free power" part. Maybe you can make a similar illustration involving the non-reversible machine that is referred on the description. Nonetheless i like it! $\endgroup$ – George Smyridis Jun 29 '15 at 0:11
  • $\begingroup$ I actually had labelled things poorly. I added some explanation. I hope this is clearer. $\endgroup$ – mmesser314 Jun 29 '15 at 1:08
  • $\begingroup$ So lifting more weight is equivalent to lifting higher. Lifting more weight concludes that the machine is perpetual, which cannot exist. Although i understand the lifting more weight case, i still have problem understanding the lifting higher case it self. $\endgroup$ – George Smyridis Jun 29 '15 at 16:36
  • $\begingroup$ I added a second diagram that shows the second case. $\endgroup$ – mmesser314 Jun 30 '15 at 14:33
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    $\begingroup$ @mmesser314 I understand the crux of your answer with one exception, How exactly, do reversible machines work? How given the initial configuration in the diagram, the weights rise up and down by themselves without any external force? $\endgroup$ – Omar Nagib Jun 30 '15 at 15:47

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