Unbalancing weights on an inclined plane with pulleys

Richard Feynman uses the following figure in Chapter 4.2 of his lectures:

Then he goes on to explain why $$W$$ must weigh 3/5 of a pound for the system to be in balance. There are two explanations and I understand them both. The first applies the principle of conversation of energy, so if between states (a) and (b) the top weight moved 3 distance units up and the bottom weight moved 5 distance units down, we know that weight $$W$$ has to equal $$3/5 lbs$$. The second explanation involves the epitaph of Stevinus, which is even easier to visualise.

Fine, but I'm a noob and there's something I don't understand. Why does the system have to be in balance? Wouldn't the top weight be in the exact same position if we increased the weight of $$W$$? In my mind, I picture this as the top weight colliding with the pulley and stopping there.

I reckon that there should be a threshold above which the rope breaks, but let's suppose that we increase the weight of $$W$$ just a little bit.

• The system does not need to be in balance, which means that nothing moves (strictly, nothing accelerates). If $W>\frac35lb$ then $W$ will accelerate down and the $1lb$ weight accelerates up the slope. Opposite happens if $W<\frac35lb$. However if $W=\frac35lb$ then you can deduce that nothing moves = the system is in balance. Conversely if nothing moves then you can deduce that $W=\frac35lb$. ... The rope breaking has got nothing to do with this. Feynman is assuming the $1lb$ block does not touch the pulley. – sammy gerbil Nov 27 '20 at 22:16
• Thanks, this was useful. How would you square your explanation that if the system is in balance then nothing moves with this statement of Feynman's: "If we say it is just balanced, it is reversible and so can move up and down". – Paul Razvan Berg Nov 27 '20 at 22:22
• If $W=\frac35lb$ then there is no acceleration up or down but the system could still move with (any!) constant speed up or down or stay still. All 3 options are consistent with zero acceleration. – sammy gerbil Nov 27 '20 at 22:30
• @sammygerbil I think your first comment is actually the answer already – Bernhard Nov 27 '20 at 22:32
• Yeah @sammygerbil if you could post both the first comment and the second to account for the "no acceleration" bit I will mark the answer as accepted. – Paul Razvan Berg Nov 27 '20 at 22:33