# Understanding Feynman's argument about Stevinus

In the Feynman lectures vol 1, chapter 4, he mentioned a "Brilliant, clever" way of solving a problem named:

A simple example is a smooth inclined plane which is, happily, a three-four-five triangle. We hang a one-pound weight on the inclined plane with a pulley, and on the other side of the pulley, a weight $W$. We want to know how heavy $W$ must be to balance the one pound on the plane.

After solving the problem with a simple insight on conservation of energy, he describes:

It can be deduced in a way which is even more brilliant, discovered by Stevinus and inscribed on his tombstone. Figure 4–4 explains that it has to be $\frac{3}{5}$ of a pound, because the chain does not go around. It is evident that the lower part of the chain is balanced by itself, so that the pull of the five weights on one side must balance the pull of three weights on the other, or whatever the ratio of the legs. You see, by looking at this diagram, that WW must be $\frac{3}{5}$ of a pound. (If you get an epitaph like that on your gravestone, you are doing fine.)

After trying for around three days, I can't understand how you solve the problem with the Stevinus' epitaph (but I understand partially why the three weights on the vertical side should balance the five weights on the slanted, but they are same weight)

• It may be noted that I don't have any such epitaph on my gravestone and this video : youtube.com/watch?v=nDKGHGdXLEg helps a lot to understand partially. – user77648 Apr 1 '16 at 0:46

The ball-chain is intended to represent uniform weight distribution along to two upper sides of the triangle; if they were 10 times smaller you would have 30 on the vertical face, and 50 on the ramp. The hanging loop completes the chain.

Does the chain begin to rotate on its own? If it does, we have found a mechanical perpetual motion machine! So no, it doesn't move.

What about the forces at the two ends of the hanging loop? The tension must be the same, or the hanging loop would be pulled to one side -- and the chain would start to move.

Since the tension is the same on both sides, we can cut off the hanging loop, and the rest of the chain will remain stationary, for it was already balanced.

Also see History of Mechanics of the Inclined Plane, and the many footnotes.

• So how does this argument solves the one pound - pulley problem ? – user77648 Apr 1 '16 at 4:11
• It's the same ratio, 3:5, as the lengths (number of balls); hence W=3/5. – Peter Diehr Apr 1 '16 at 10:15
• Can't resist: But... the masses at the top see a smaller gravitational force because they're farther from the Earth's center! :-) (yes, I know that's incorrect) – Carl Witthoft Apr 1 '16 at 12:16
• @CarlWitthoft: Simon Stevin didn't know that; but it also makes no difference if the weights are uniformly distributed. :) – Peter Diehr Apr 1 '16 at 12:28

Looking at the diagram, you can see there are five "units of weight" along the diagonal, and three units along the vertical. We need to prove that these two sets of weight are in equilibrium; because, if they are, then "five units along the diagonal" is in equilibrium with "three units along the vertical", and thus if I have 1 unit along the diagonal (1/5th of the original), then I need 3/5 of a unit along the vertical.

So - why is this in equilibrium? Let's look at the chain as drawn. Should it move? If it would freely move clockwise, then I can put a very small generator on it to extract power - because as it moves, weights travel up one side as they come down the other, and this means that after it's moved one "click" (one unit of weight spacing) we have the same situation as before. Ditto if it moved anti-clockwise by itself. We conclude that the whole system is in equilibrium, or we have a perpetuum mobile. And we don't...

Would it still be in equilibrium without the bottom part? Well, the chain below the triangle is symmetrical. That means that the tension applied on the left must equal the tension on the right. And if these are the same, then removing the chain at the bottom shouldn't change the equilibrium.

And there you have it. This means that without the chain at the bottom, the thing is still in equilibrium: five along the diagonal equals three along the vertical.

Clever Stevinus.

• I am very sorry, what you wrote I had understood relatively better (please see the video in the comment in the answer) than any other thing in the chapter. What I don't understand is the connection of the problem of the weights with this (and the first paragraph of your answer). – user77648 Apr 1 '16 at 2:13
• I wonder at what point in my answer I lose you? – Floris Apr 1 '16 at 2:20
• Most part is relatively better decipherable except this portion: "if they are, then "five units along the diagonal" is in equilibrium with "three units along the vertical", and thus if I have 1 unit along the diagonal (1/5th of the original), then I need 3/5 of a unit along the vertical.". I am repeating, What I don't understand is the connection of the problem of the weights with this (and this portion of your answer). – user77648 Apr 1 '16 at 3:11
• The original problem asked about the ratio between two weights that would be in equilibrium. If there is a weight of one unit along the diagonal, what weight is needed along the vertical to balance it exactly. The weight "chain" gives us a ratio of 3:5 between vertical and diagonal. – Floris Apr 1 '16 at 10:33

because it is a 345 triangle. If the triangle were rotated and the beads were on the 4 and 5 sides it would be the same, 4/5 of a pound, answer. If it were a flat surface the answer would be 0/5, meaning the beads would go nowhere. Make sense?