Proof of the Law of the Lever 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.
 A: 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.
A: 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.)

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.

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.
A: 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.
A: Not a proof,but intuitively, as stated by the answer above. If I apply a force to a lever closer to the pivot, and move it such that Theta Is some angle. and in the second scenario I apply a force at the end such that theta is the same angle. The total amount of work done by the twoforces must be the same single the Lever starts and ends at the same  position in each case. Since the situation where I apply the force closer to the pivot moves that point a shorter distance, for the same work I would have to apply a greater force
