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Feynman irreversible and reversible weightlifting machine
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[QUOTE="vela, post: 6466871, member: 221963"] Not exactly. Start with the machine as shown in the lecture. With masses ##3m## on one end and ##m## on the other, there would be no net torque on the lever, so the machine wouldn't do any lifting either way. If you reduce the ##3m## mass by a little bit ##\delta m##, there'd be a net torque that would lift the remaining ##3m-\delta m##. To restore the machine to the initial state, you'd have to do some work to lift the ##\delta m## to restore the mass to ##3m##, reduce ##m## by ##\delta m## to lift ##m-\delta m## up, and then do a little more work the lift ##\delta m## to restore the mass on the right to ##m##. So there's some amount of work proportional to ##\delta m## to get the machine to run through one cycle. Now it doesn't matter how small ##\delta m## is. It's just that the smaller it is, the longer it will take the machine to go through one cycle. In a real machine, however, there's always some amount of friction, which is what makes it irreversible, and a finite amount of work is needed to overcome this friction. You can't make the work required arbitrarily small. That's the little extra actual machines need to work. [/QUOTE]
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Feynman irreversible and reversible weightlifting machine
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