(9.)
Now let us suppose this solid mass of lead to be rendered liquid by being melted. The constituent particles will then be deprived of that cohesion by which they were held together; they will accordingly have a tendency to separate, and fall asunder by their gravity, and will only be prevented from actually doing so by the support afforded to them by the sides, [Pg026] A B, D C, of the vessel. They will therefore produce a pressure against the sides, which was not produced by the lead in its solid state. This pressure will vary at different depths: thus a part of the side of the vessel at P will receive a pressure proportional to the depth of the point P below the surface of the lead. If, for example, we take a square inch of the inner surface of the side of the vessel at P, it will sustain an outward pressure equal to the weight of a column of lead having a square inch for its base, and a height equal to P A. And, in like manner, every square inch of the sides of the vessel will sustain an outward pressure equal to the weight of a column of lead having a square inch for its base, and a height equal to the depth of the point below the surface of the lead.