Law of Mechanical Equilibrium for Bodies

Problem 3: Law of Mechanical Equilibrium for Bodies

Two bodies are attached to a lever. At what point will they remain in equilibrium?

Thus, I seek the point where, if the lever moves slightly, the quantity of action is as small as possible.

Let L be the length of the lever (which I suppose to be massless), and let two masses and be placed at either end.

z is the distance from the m1 to the equilibrium point being sought.

L - z is the distance to m2.

If the lever rotates slightly around a point, the 2 masses describe geometrically similar arcs.

Their size is proportional to their respective distances from the point of rotation.

Thus, these arcs are the distances traveled by the bodies and also represent their speeds per unit time.

Hence, the quantity of action is proportional to the product of the mass of each body multiplied by the square of its arc length.

Or, equivalently (since the two arcs are geometrically similar), the quantity of action is proportional to the product of the mass of each body multiplied by the square of its distance to the point of rotation, i.e. m1z^2 and m2(L - z)^2.

The sum of these two terms should be minimized, giving:

…

or, rather

…

from which the equilibrium position may be derived:

…

This is the basic law of mechanical equilibrium.