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Feynman's reversible lifting machine analogy- help

  1. Oct 21, 2012 #1
    Hi I'm having a very hard time trying to picture an argument in Feynmans physics volume one ). I can't picture this example he is talking about, so I wrote it down from the book and really hope someone can help me answer the question in the bottom.

    From the book (everything needed to know is included)

    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.


    I don't get this part:
    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.


    Why can't Y (the distance lifted by machine A) be highter than X (distance lifted by machine B)?? I assume both machines have same proportions (are able to reach the same maximum height) and both machines lift 3 units of weight a distance by lowering one unit a distance. So what physical law says that Machine B lift 3 units as high as machine A?
     
  2. jcsd
  3. Oct 21, 2012 #2
    And what kind of a picture am I to Imagine in this sentence
    "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."
    Is it that we lower the one unit weight of machine A with machine B, or do we just lower the weight of machine B one unit?
     
  4. Oct 21, 2012 #3
    I have tried to draw and illustrate my interpretation of A and then
    drawn machine B according to this sentence

    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.


    [PLAIN]http://i50.photobucket.com/albums/f317/christian0710/Computer/distanceup_zpsf7b187cc.jpg[/PLAIN]

    But what happens after we have lifted the 3 units of machine B the height of v?
    This is the sentence I cant visualize

    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.
     
  5. Oct 21, 2012 #4
    Basicaly he is saying a balance of the weights is the most efficient that can be acheived.Or the weights can at best arrive at a position where they are balanced and no more.Otherwise it would be possible to build a perpetual motion machine if you could lift a weight higher in the reverse or other direction when the same force is applied.
     
  6. May 1, 2014 #5
    The clue to understand

    If you read carefully the statement , you see the term "lower" is written in slight italics , which has deeper meaning in it, lets see what.

    1. Reversible machine : can be restored without any external work done on it.

    2. Irreversible machine : need some work to lift it to restore it . That is someone has to do work on the system to make it work again.

    3. Perpetual machine : This machine lifts more than the reversible machine that means , to restore it
    you have to lower it ,implies there is some power left in the machine , because when you lower it from Y to X the other side get lifted too to some height which is less than a foot (say this height k from ground). and therefore the machine can lift something else, for instance to restore the reversible machine.


    Another way to think this Perpetual machine (Machine B) , and why they can't exist .
    suppose i have a perpetual machine which contrary to reversible machine(machine A) lifts it to a greater height ,alright lets bring it down we now have some free work which can be used , but we will use it to the machine itself , Now lets say Y-X=C which is the difference in height it lifts ,which implies when we restore and again drop it down to the ground. Now clearly this device is going to lift the same higher than c (say c>>) , then in next step we make it down and lift again , this time the lifting goes c>>> , and so on it goes higher and higher.

    The conclusion is that each time the machine lifts C units higher , now this can only happen that each time some external agency is working , and iff the machine is self sustaining then it has to be a perpetual machine.

    Next post , I shall post the mathematical translation of it .

    thanks
     
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