- #1
christian0710
- 409
- 9
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?
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?