Archimedes' Principle
Ø Buoyancy is a fundamental aspect of the Physics of Fluids. For example, it allows to explain natural swimming of bodies, when they are supported by a fluid at rest.
Ø Natural swimming - swimming by swimming motions is artificial. Like your own swimming, it is a lasting fight with sinking. Rudders, sails and propellers are means of propagation of naturally swimming bodies. At rest! A body can also be supported by an upwards jet. However, it does not swim then, but dances.
Ø A body which is freely movable in a fluid at rest is pulled vertically downwards by gravity and pushed upwards by its buoyancy. Its behaviour depends on the relative magnitudes of these two forces. If its weight is larger than its buoyancy, it sinks below - it drops; if its buoyancy is equal to its weight, it can neither rise nor fall - it swims.
Ø For the sake of simplicity, let the body be a rectangular prism and its base lie horizontally, parallel to the free level of the fluid. (The treatment of arbitrarily shaped and located bodies demands Infinitesimal Calculus!) Every point of the surface of the prism is subject to a pressure which is determined by its depth below the free surface of the fluid. However, the pressure against its sides causes nothing, because at the same level the pressures on opposite sides are the same, but opposite, and therefore balance. Only the pressures on the horizontal faces need be considered.
Ø The pressure on the top face is q·k·s, that on the bottom face q·(k + h)·s, the weight of the body q·h·S, where q is the prism's cross-section, h its height, k the depth of the upper face below the free surface of the fluid, s the specific weight of the fluid and S that of the prism. Hence the vertical forces are q·k·s + q·h·S downwards and q·(k + h)·s upwards. The result depends on whether q·k·s + q·h·S is larger than, equal to or smaller than q·(k + h)·s. The prism has the weight q·h·S , a body with the volume q·h of the prism, but with the density of the fluid, has the weight q·h·s. However, in order that the prism can occupy the place in the fluid, it must displace an equal volume of fluid: q·h·s is therefore the weight of fluid displaced by it. Hence:
Weight of the submerged body > = < weight of the fluid it displaces.
Ø The weight of a body drops on submersion in a fluid as much as the weight of the fluid which it displaces (Archimedes' Principle) We have chosen a rectilinear prism, because the demonstration of the principle is then simpler. However, it can be shown theoretically and experimentally that it is valid for bodies of any shape, so that q·h = V can be interpreted as the volume of any body.
Hydrostatic Balance
v A proof of Archimedes' Principle is given by the equal-armed lever balance of special shape, the hydrostatic balance . The body to be weighed hangs below one scale and immerses completely in the fluid, in which its loss of weight is to be found. C is a hollow cylinder, the internal volume of which equals that of of the filled cylinder D. You first establish equilibrium of the balance, while D is surrounded by air and C is empty. If you now place the container with the fluid below D, so that it is completely immersed, the balance deflects to the right, that is, D has lost weight. If you now fill C completely with the same fluid as is already in D, equilibrium is restored. The loss of weight is thus compensated by the weight of a volume of fluid, which is equal to the volume in the cylinder D. However, that is the volume of the fluid, which D has displaced by taking its place.
Referring now to the work of the preceding section, you have the results:
1. If qhS > qhs, that is, the body is heavier than the fluid it displaced, a downwards force acts: The body becomes submerged.
2. If qhS = qhs, that is, the body has the same weight as the fluid it displaces; the two forces are equal: The body floats in the fluid.
3. If qhS < qhs, that is, the body is lighter than the fluid it displaces, an upwards force acts: The body begins to rise and sticks out of the fluid and displaces less fluid than when it is fully immersed. The volume of the part of the body, sticking out of the fluid, reduces the buoyancy, qhs. In the end, as much of the body sticks out of the fluid so that the already submerged part of fluid weighs as much as the entire body; it does not rise further, but swims on the surface of the fluid.
v In order to show that the fluid displaced by the submerged part of a body weighs as much as the entire swimming body, you fill the vessel V up to the opening o with fluid and then place into it a body A which will swim on the fluid. By weighing of the displaced fluid, you convince yourself that the fluid which flowed out of the vessel at o and the body have the same weight.
v The densities of a body and a fluid decide how much of the volume of a body gets immersed and how much sticks out as, for example, in the case of an iceberg1. At 0º, fresh water has the density 0.9167, sea water with 3.4 % salt the density 1.0273. Let V be the volume of the iceberg; its weight is then V·0.9167·g. Let V·x denote the submerged part of the iceberg, where x is a real fraction; then the weight of the displaced sea water is V·x·1.0273·g. From the equality of the two weights follows that x = 0.9167/1.0273 = ~ 9/10. Hence only 1/10 of the volume of an iceberg sticks out of the ocean. However, the slimmer, less massive part of the volume will sticks out, the broader, more massive part will be submerged, because the iceberg (by lowering its centre of gravity) will always have the maximum possible stability. As a rule, the visible part of an iceberg is estimated at 1/7 - 1/8 of the total height (frequently 40 - 60 m).
Icebergs are the ends of glaciers (fresh water) which have broken off and been carried away by currents and wind; these glaciers lie on polar main land and islands (Inland ice).
