On Natural Motions Depending on Gravity

Chapter 2. On the moments The measure of a weight's tendency to produce motion around an axis. of heavy bodies floating in a fluid.

of the cylinder IG. Let us imagine the water first elevated at position T and depressed in the right channel at G. From this state, while the water is raised to I, let it descend from T to X. Let two straight lines AG and BH be joined, intersecting at M. Point M will be located on the horizontal line EL. This occurs because the two cylinders of water AB and HG are equal to each other, since they are halves of the equal cylinders TX and IG. Therefore, the height AB is to HG as the basis of the same cylinder H is to basis A The "basis" refers to the cross-sectional area of the channel.. For the same reason, AE to LG will be as basis H to basis A. Wherefore, the height AE to LG will be as AB to HG. Since the two straight lines AE and GL are perpendicular to the horizontal FG, or EL, they are therefore parallel to each other. Thus, due to the similarity of the triangles, as AM is to MG, so will BM be to MH, and also EM to ML. Therefore, the straight lines AG, BH, and EL will intersect each other at the same point M.

Afterwards, let the distance HB be to BQ as the mass of water XBF together with GHI is to the mass of water IHG. By dividing, as the mass of water XBF is to GHI, so the distance HQ will be to QB. Therefore, from mechanical principles, point Q will be the center of gravity of the water XBF together with GHI. However, when the water was at the top T and the channel GLV was completely empty, then the center of gravity of the whole water TAF, remaining at point A at the middle of the same channel, would act just as if the cylinder were suspended from point A. After the water was depressed as far as Y and elevated as far as L in the opposite channel, the center of gravity of the aforementioned water is found again at point R. Finally, the water being depressed as far as A in the first case and as far as