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Is there an accepted proof of the Law of the Lever?

I know of several attempted proofs, such as Archimedes's, but these either are incomplete or do not transfer to the setting of Newton's Laws. (It should be noted that there are two claimed proofs that I have come across that I do not understand enough to deny: Mach's in his _Science of Mechanics__ and Newton's in his Principia.)

In the book Newtonian Mechanics by French, the author claims that there is not an accepted proof.

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  • $\begingroup$ the wikipedia quotes archimedes proof as proof. $\endgroup$ – anna v Oct 15 '18 at 19:29
  • $\begingroup$ @annav I do not know quite what you are saying, but Mach's book Science of Mechanics states several reasons for Archimede's proof not being accepted. $\endgroup$ – user109871 Oct 15 '18 at 19:34
  • $\begingroup$ en.wikipedia.org/wiki/Lever#Law_of_the_lever , a paragraph in your link $\endgroup$ – anna v Oct 16 '18 at 4:09
  • $\begingroup$ link to Mach's book math.harvard.edu/archive/hist_206r_2009/Mach.pdf . lever and archimedes first at page 34 of pdf, page 8 of book $\endgroup$ – anna v Oct 16 '18 at 4:25
  • $\begingroup$ Based on sciencedirect.com/science/article/abs/pii/0039368172900027 Mach critiqued his own understanding of Archimedes' proof, not the actual proof. After reading the Heath's translation of the proof by Archimedes I also think that's the case. Mach confused finding the centres of mass with attaching masses to the lever. Archimedes' proof is based on finding the centres of mass first. The lever law is simply a consequence. The connection between the centre of mass and the lever is not explained in the proof though. Maybe it was proven elsewhere and it didn't survive to our times. $\endgroup$ – Mateusz Grotek Jan 31 at 23:36
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Of course there's an accepted proof. If the lever's at equilibrium, there must be zero net torque on it. Write down the equation for the torque of masses pushing on a lever and it's algebraically equivalent to the "law of the lever". So that's a proof.

What Archimedes was trying to do was to prove the law of the lever without any appeal to physical theories, the same way we might, for example, prove the Pythagorean theorem without any reference to real triangles, but just by using some axioms.

This is a non-starter. Math proves things about formal mathematical systems, not about real-world objects like levers. Trying to prove the law of the lever from pure math is a category error. For example, Archimedes used symmetry arguments, but there's no reason the universe has to obey the implicit axioms behind his symmetry arguments. A perfectly-symmetric lever could, for example, decide at random which way to fall. No purely-mathematical proof could ever exist because the universe isn't obliged to follow any particular axioms. Likewise, the Pythagorean theorem doesn't prove anything about real triangles. Indeed, it is false in general relativity.

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  • $\begingroup$ The equation/definition of torque is from the Law of the Lever. This is not a proof. Also, classical mechanics is still a defined theory with axioms, and statements can be proved. $\endgroup$ – user109871 Oct 15 '18 at 19:48
  • $\begingroup$ No, torque is defined by $F\times r$. $\endgroup$ – Mark Eichenlaub Oct 15 '18 at 19:50
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    $\begingroup$ No. Write down the usual definition of angular momentum $L$. If something's at equilibrium, $L=0$ from the definition. But from $F=ma$ you can prove that $\dot{L} = \tau$ using $\tau = F \times r$, so in equilibrium $\tau = 0$. It's got nothing to do with levers. It's just from Newtonian mechanics. $\endgroup$ – Mark Eichenlaub Oct 15 '18 at 20:04
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    $\begingroup$ Physics must have postulates to pick up the mathematical solutions ( which are axiomatic) that fit the data. They are physics axioms and can have no proof other than their validation by data.The same as in mathematical axioms a theorem can become an axiom and an axiom a theorem, physics laws can change, but have to be consistent, @user109871 , that is what you are arguing about, which statement is a postulate/law of physics and which can be proven from it. One of them can be set up as a postulate and the other proven. $\endgroup$ – anna v Oct 16 '18 at 4:31
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    $\begingroup$ the pythagorean theorem fails in spherical surfaces , no need for general relativity for failure. $\endgroup$ – anna v Oct 16 '18 at 4:33
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I found explanations of this unsatisfactory as well, including in answers to related questions like this one. Here is a direct explanation in terms of equilibrium of forces.

The green rectangle is the lever, with thickness $h$. (It's upside-down from a typical picture, but that doesn't matter.)

Diagram of lever internals

We are here reducing the law of the lever to just three simple forces: two stretching (along the top sides of the triangle), and one compressing (along the bottom). See below for an experiment showing this at work in real life.

Geometric proof of why this works

Because the system is in equilibrium, we have $F_x = G_x$ (bottom of the lever resisting the inward push from both sides). Forces $\vec{F}$ and $\vec{G}$ are along the sides of the triangle (pulling between points of contact), and equal to the sum of their horizontal and vertical components:

\begin{equation} \vec{F} = \vec{F_x} + \vec{F_y} \quad \text{and} \quad \vec{G} = \vec{G_x} + \vec{G_y} \end{equation}

The rest is by similarity of triangles, looking at the lengths of corresponding vectors:

\begin{equation} \frac{F_y}{F_x} = \frac{h}{m} \quad \implies \quad m F_y = h F_x \\ \frac{G_y}{G_x} = \frac{h}{n} \quad \implies \quad n G_y = h G_x \end{equation}

But $F_x = G_x$, since there is no net horizontal force, so in fact

\begin{equation} m F_y = n G_y \end{equation}

which is indeed the law of the lever.

It is interesting that thickness $h$ does not affect the result, but is integral to the explanation. I believe this matches the intuition that a rigid rod must have non-zero thickness.

Experimental Setup

Here's an experiment to show that this is indeed sufficient for the law of the lever to work.

Experimental setup

The setup is two dumbbells, with weights 10lb and 4lb, hanging at the ends of a metal rod. There is nothing pulling on the middle of the rod; instead two strings are holding its ends. The lengths of the strings are such that the fulcrum is above a point on the rod that divides it in a ratio 4:10. The system is in equilibrium.

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  • $\begingroup$ I would argue that this is also a "geometrical proof" using forces, and in my opinion Archimedes proof is cleaner and more direct. This answer still fails to provide insight into the microscopic explanation on why this principle works: similar to how the cited answer failed. Good effort nonetheless. $\endgroup$ – bruno Jan 2 '20 at 14:55
  • $\begingroup$ I guess my attempt was to reduce it to simpler physical concepts -- addition of forces as vectors, and equality of opposing forces. The latter (inward forces along the bottom of the beam being equal) is intuitive at microscopic level. Decomposing a force into components along perpendicular vectors is less intuitive, but I think much lower-level than Archimedes's argument based on center of mass. (I'd be interested in a better yet explanation also.) $\endgroup$ – DS. Jan 3 '20 at 15:31
  • $\begingroup$ Here it is assumed that force transmission occurs in a triangle fashion. In reality what occurs is the bending of the lever with a complex stress (and force) distribution. As you can see in my recent answer the force is amplified because when it acts on the longer portion of the beam/ lever it generates the same "pressure" as a larger force applied on the shorter side of the beam. $\endgroup$ – bruno Jan 3 '20 at 16:27
  • $\begingroup$ If a smaller force was applied in the shorter side of the lever, that bending moment would not "pull" as hard as the bending moment from the longer part of the lever. You can compare this to pressure in fluids. What is transferred at the atomic level are not forces, but rather pressures (in our cases bending stresses) that amplify the force of one piston (with a small area) to another one (with a great area) $\endgroup$ – bruno Jan 3 '20 at 16:29
  • $\begingroup$ A rigid rod acting as a lever (with bending moments and such) is a more complicated system. I tried to reduce the principle to its most basic, and just edited my answer to include a photo of a simple experiment. It shows that the three simple forces are enough to produce the effect of the Law of the Lever. $\endgroup$ – DS. Jan 5 '20 at 7:39
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To derive such an expression, consider the following three axioms, which can be shown to be true.

Axiom 1: A body in equilibrium will do no work through time. Axiom 2: Work done by a force is equal to $Force*Distance$ Axiom 3: Arc length, distance along a circle, is equal to $radius*angle$

Consider a uniform rod of length $L$ which has its origin fixed at a point $P$. A force $A$ is applied at a distance $x$ from its origin which would cause the rod to accelerate anticlockwise. A second force $B$ is applied at a distance $r$ from its origin, which will cause it to accelerate clockwise. We can consider $B$ to be negative, as it is acting the opposite direction to $A$. Hence, the other force is $-B$. If we were to consider how much work the two forces would do alone, had the rod accelerated through an angle $q$, we would use the fact that arc length $M$ is equal to the product of the radius $R$ and the angle displaced $Q$. Hence, the work done by the force $A$, had it acted solely on the rod, would be $Force * Distance$, which would also be: $$A.Radius.Angle=A.x.q$$ Doing the same for the force $B$, we would obtain that the work done: $$-B.Radius.Angle=-B.r.q$$. However, it is here that we say that the rod is in equilibrium, it is perfectly balanced. Hence, the resultant work done, or energy gained by the rod is zero. Therefore, by adding the two formed equations for work done, we get: $$\sum Work = 0 = A.x.q-B.r.q=0$$ Hence, rearranging: $$A.x.q=B.r.q$$ Dividing through by $q$: $$A.x=B.r$$ For a body in equilibrium. This is a simple version of the law of the lever, and can be interpreted as for a body in equilibrium, the sum of the product of the magnitudes of the forces acting on it and the distance from the pivot they are acting upon. I hope this helps, at least this is how I understand it. Also, sorry if the formatting is off, this is the first answer I have given.

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