Feynman irreversible and reversible weightlifting machine

[Est120]

Homework Statement

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.

Relevant Equations

understand what feynman says

i have tried to understand Feynman's words i think i finally understands what he means by "we must add a little extra to get it to run"
he refers to the "inversion" of the process, that's when we need to add extra work (lifting up a little mass)
please correct me if I am wrong, this is something i want to have clear in my head

 

[vela]

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.