(9.)

The constituent particles of a liquid are distinguished from those of solids by having little or no coherence; so that unless the mass be confined by the sides of the vessel which contains it, the particles will fall asunder by their gravity. A mass of liquid, therefore, unlike a solid, can never retain any particular form, but will accommodate itself to the form of the vessel in which it is placed. It will press against the bottom of the vessel which contains it with the whole force of its weight, and it will press against the sides with a force proportional to the depth of the particles in contact with the sides measured from the surface of the liquid above. This lateral pressure also distinguishes liquids from solids. Let us take for illustration the case of a square or a cubical vessel, A B C D, fig. 3. If a solid body, such as a piece of lead, be cut to the shape of this vessel, so as to fit in it without pressing with any force against its sides, the mechanical effect which would be produced by it when placed in the vessel, would be merely a pressure upon the bottom, B C, the amount of which would be equal to the weight of the metallic mass. No pressure would be exerted against the sides; for the coherence of the particles of the solid maintaining them in their position, the removal of the sides would not subject the solid body contained in the vessel to any change.

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.