E2 Equilibrium of drop forming substances
Fluid level in communicating vessels
In vessels, which are interconnected below the fluid level by channels (Fig. 183), the fluid has the same level everywhere, provided, of course, they contain the same fluid. Why? A fluid, which is only acted upon by gravity, demands for its equilibrium that the pressure has the same magnitude on all planes (dyn/cm²) at the same level. The magnitude of the force acting on a single cm² only depends on how far it lies below the free surface. In other words: The equilibrium does not depend on the number of cm², that is, the width and shape of the vessel (disregarding capillary tubes which are discussed below). There must act on every cm² of the same horizontal plane the same force, for example, at AB as at CD as at EF, etc. This pressure is numerically equal to the product of 1 cm², the height in cm and the specific weight (1·h·s). However, the specific weight is in all vessels the same, whence 1·h·s can only have the same value, if also h has the same value, that is, the fluid is equally high over every cm² of the cross-section, that is, it is equally high in all vessels. Thus, communicating vessels effectively represent a single vessel; their free surfaces lie in the same horizontal plane.
The employment of a water level glass tube at steam containers depends on this principle. The tube communicates with the steam container; the height of the water in the tube indicates that in the steam container. In the gauging apparatus of surveyors, the employment of two communicating vessels, filled with the same fluid, also depends on this principle (Fig. 184). You fix with it the point M which must share the free surface with D and E. You can also measure differences in height with this gadget and a ruler.
However, the free surfaces do not lie in the same horizontal plane, but at different heights, if the free surfaces belong to fluids of different specific weight, that is, communicating vessels can then not be viewed to be a single vessel; in fact, the surface in one vessel lies lower with respect to that in another one the larger is the specific weight of the fluid below its free surface.
Pour water into a U-shaped vessel (Fig. 185) with communicating legs S1 and S2 and subsequently into S1, on top of the free surface, a fluid which is lighter than water and does not mix with it, for example, oil. The fluids will touch each other at the cross-section a. Then the free surface in S2 has water below it, that in S1 oil, and the oil level O lies higher than that of the water W. Since there is equilibrium, the pressure (dyn/cm²) at a equals that at b. The pressure of the oil column (its height is h1, its specific weight s1) at a must be h1·s1 = h2·s2, that is, you must have h1/h2 = s2/s1. The heights above the horizontal plane, at which the two fluids meet, are thus inversely related to the specific weights. This mechanism can be used to compare specific weights. Dulong and Alexis Thérèse 1791-1820 have thus compared the specific weight of mercury at different temperatures.
What is the pressure on a not horizontal plane area - the side pressure? Every point of such an area experiences the same pressure which acts at a point at the same horizontal level. Hence there acts on such an area a variety of differing, parallel, equi-directed forces, the magnitudes and points of action of which are known. Our task is then: Compute from the single, known, parallel forces the magnitude and point of action of their resultant.
An example of the task of finding the point of attack will clarify the problem. A hollow cube (Fig. 186) is filled to its rim with water. What is the pressure on the entire vertical wall AB and at which point must you apply a force against this wall, if it is not connected rigidly to the other walls and the bottom and you want to keep it in place against the acting pressure?
The computation yields: The magnitude of the resultant, that is, the pressure on the (not horizontal) area equals the pressure which it would be exposed to if it were lying horizontally at that horizontal level of the fluid, at which is located its centre of gravity.
The point of action of the resultant lies lower than the centre of gravity of the area; its location must be computed specially. It cannot be identical to the centre of gravity, because that is the centre of equally large parallel forces; in this case, the forces are not equally large. The solution of the task in Fig. 186 follows: Since the wall is a square, it is subject to the same pressure as the horizontal cross-section through its centre S - this is equal to the size of the wall, because the vessel is a cube - that is, the cross-section which halves the vessel and on which presses half the fluid. The magnitude of the pressure force lies vertically under the centre of the wall, its distance from the bottom is one third of the cube's edge length. This is where the force must act from outside.
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.
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.
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.
For the sake of simplicity, let the body (Fig. 187) 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.
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.
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.
A proof of Archimedes' Principle is given by the equal-armed lever balance of special shape, the hydrostatic balance (Fig. 188). 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.
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 (Fig. 189) 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.
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).
1 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).
You can even make a material with specific weight larger than that of a fluid swim naturally in it by giving it a suitable shape. For example, a iron plate does not swim on or float in water, even if it is fully submerged, since it weighs more than the water it displaces. But in the shape of a ship, it will swim, because the submerged part of the ship with its curved form displaces a volume of water which is equal to its weight. A swimming elastic plate, which is deformed by a weight at its centre, exhibits a strange phenomenon (Hertz 1884). Computations show that the buoyancy which the water exerts on the plate due to its deformation equals the weight on it.
Hertz says: "How ever large is the weight, it will always be carried by the buoyancy which a plane unloaded plate experiences. If you place a small round disk of stiff paper on water, you can deposit several hundred grams on it while the buoyancy of the paper is only a few grams. Thus, if someone swims on a large plate of ice, it is, exactly speaking, more correct to say, he swims because the plate of ice is deformed by his weight into a very shallow boat than to say that he swimsm, because the ice is light enough to carry him in addition to its own weight. For he would swim anyhow, even if the ice were not lighter than the water; if instead of a person you place arbitrarily large weights on the ice, they might break through the ice and sink, but they would never sink with the ice. The limit of the load depends on the strength of the plate and not on the weight of the ice. It is quite different when persons or weights are distributed uniformly over the area." (Hertz's Collected Works, 1, 292)
Since a living being weighs more than the water it displaces (its empty spaces may be disregarded), it will sink in water1. It compensates for its sinking with swimming motions, by which it exerts a pressure downwards and the resistance of the water against this pressure lifts its body, that is, it swims artificially. Mainly through the developing gases in its inside, a dead person is specifically lighter than water and swims naturally. Birds swim naturally in their cover consisting of feathers and air. The floating of fish is artificial: the pressure of their muscles on their swim bladder is required - for rising and sinking - as is seen from the fact that dead fish (also not decaying ones) swim naturally on the water surface. There are fish which are heavier than water (those without swim bladders such as sharks and rays) and fish the weight of which equals that of the water they displace, because they balance the excess weight of their bodies by means of an air filled swim bladder (most of the bone fish teleostae)2. The former, for example, sharks, sink when they do not move along (like an aeroplane), the latter can stop in place, for example, goldfish and carps (like balloons). A bone fish can swim as slowly as it likes, a shark must have a minimum velocity, in order to generate a resistance of the water against its lower half, the upwards component of which balances its overweight and thus carries it - these are the same differences as arise between aeroplanes and air ships (Hesse).
1. Except in the Dead Sea the salt content of 25 % of which makes the water too heavy. (ordinary sea water has a content of 3.5%). However, one can anyhow drown in it, because ordinary swimming is next to impossible as the legs cannot submerge and a person has little control over its body.
2. Telocasta have bone skeletons
Stability of the swimming body. Meta-centre
Its own weight and buoyancy act all the time on a swimming body, whence it is continuously acted upon by two forces, which are equally large and have opposite directions. Their equality allows only vertical displacement, but admits rotation. For the body to remain at rest, the forces must yet fulfil one condition as shows the following example (Figs. 190/191).
The weight of a swimming body is replaced by a force Gt which acts vertically downwards at its centre of gravity G . The buoyancy equals the weight of the fluid volume displaced by the body. This weight can also be replaced by a force. If the centre of gravity of the fluid prior to being displaced by the body lies at A, the swimming body is acted upon at A by the vertically upwards force AB (= Gt). For it to remain at rest, the two forces AB and Gt must lie on the same straight line; in other words, the centre of gravity of the swimming body and the point of attack of the buoyancy must lie above each other (Figs. 190a/191/a); otherwise they form a couple and will rotate the body.
Imagine that a body has been displaced from its position of rest, say, by sudden wind action to the position of Figs. 190b/191b, so that G and A no longer lie on the same vertical line. The centre of gravity G maintains, of course, its position in the body, but the point of action of the buoyancy does not, for in every new position of the body its submerged portion has another volume, that is, the displaced fluid volume changes and therefore the position of its centre of gravity. It now lies at A'. The forces Gt and A'B' then form a couple and cause the body to rotate.
The two cases of Figs. 190b/191b differ totally. In the second case,. the couple tends to return the body to its position of rest, that is, to straighten it out, in the second case, it tries to turn it over. Thus, in the first case, the equilibrium is unstable, in the second case, stable. If a ship were to swim in an unstable state, as in Fig. 190a, the smallest gust would turn it over, whence we demand that it swims in a stable state as in Fig. 191a.
You can formulate the condition for a body to swim in a stable state as follows: Place in the body deflected from its position of rest (Figs.190b/191b) through G and through the earlier point of action A of the buoyancy a straight line. It intersects (in the deflected body) the line of action of the buoyancy at M, the meta-centre of Pierre Bouguer 1698-1758 1746. If the equilibrium is stable, the centre of gravity of the body lies below, if it is unstable above the centre of gravity of the body. In order to make a swimming ship stable, one must have the centre of gravity as low as possible (for example, by means of ballast), so that it will also lie below the meta-centre when it leans over very much.
Hydro-static equilibrium of Earth's crust (Isostasy).
Two equally heavy massive cylinders which swim on a fluid displace two equally heavy cylinders of fluid. If the cylinders of the displaced fluid have the same diameter, they also have the same height, that is, they sink equally deep into the fluid. If such cylinders have different specific weights, they stick out of the fluid inversely proportional to the ratio of their specific weights. Fig. 192 shows several equally heavy .cylinders with the same cross-section, but different specific weights swimming in a fluid which supports. They demonstrate the fundamental geophysical problem of isostasy:
Earth's interior is most probably to a certain degree plastic, so that the firm strata of its surface effectively swim on the lower layers, which must therefore have a larger specific weight. (The geologists speak of two layers in Earth's crust; the upper, Sal, consisting predominantly of rocks with silicon and aluminium, the lower of rocks with silicon and magnesium, Sima,. the former swimming on the latter.) Submergence of the lighter matter in the heavier one generates mass defect below the projecting part. Accordingly, the visible elevations of mass above Earth's surface correspond to subterranean mass defects. The visible elevations, according to Archimedes' Principle, are equal to what is missing below or, in other words, the mass defects are equal to the visible masses; they compensate each other. All of the subterranean masses, the defects of which compensate all the elevations, are bounded by a surface which corresponds to a common depth of submergence like in Fig. 192. It is called the compensation surface and is defined by the fact that on each unit area lies the same mass. Following C.E. Dutton 1841-1912, the state of equilibrium is referred to as Isostasy.
The compensation surface lies most probably 118 km below Earth's surface. The surfaces of equal density coincide then with the level surfaces - only not in the top layers of Earth with their mixture of masses of different forms and densities. One level surface will therefore be the last (counting from inside outwards) which corresponds to hydro-static equilibrium; it must have the property that the same pressure is on each of its units of area - the characteristic of the compensation layer. Geological events (formation of mountains, vulcanism, , fracture formation) take place above it, in Earth's crust.
Measurement of density according to Archimedes' Principle
You call the density of a body the ratio of its mass to its volume (dimensional formula: m·l-³), that is, in the cm-g-sec system, the ratio g/cm³. In order to obtain its density, you must therefore find 1. its mass in grams, 2. its volume in cm³, 3. divide the number of grams by the number of cm³. You determine its mass by weighing, its volume, if it cannot be found by measurement, indirectly: You then find out how much weight it loses, if it is fully submerged in a fluid (Fig. 188). For example:
If a piece of copper weighs in air 11.378 g, in distilled water at 4º C 10.100 g, then its loss of weight in water is 1.278 g, that is, it has displaced 1.278 g of water at 4ºC, that is, 1.278 cm³, but has itself a volume of 1.278 cm³. 1.278 cm³ copper contain 11.387 g, whence the weight of 1 cm³ copper is given by 11.378/1.278 = 8.903 g; the density of copper is 8.903 g/cm³.
Methods of determining the density of solids differ essentially by the means employed for the determination of their loss of weight in a fluid, that is, on the method of determination of their volume. You use for this purpose a hydrostatic spring-balance or a weight-hydrometer.
1. You use the hydrostatic balance (Fig. 188) to determine a body's loss of weight by weighing it normally and then again when it is submerged in a fluid.
Also the spring balance of Philip von Jolly 1809-1889 (Fig. 193) is a hydrostatic balance underneath the scale of which hangs a second scale submerged in the fluid in which the body is to be submerged. The lower end of the spring has a marker which is displaced along a scale during weighing. When you place the body in the upper scale, the marker moves to a definite position on the scale. You then determine a) how many grams you have to place in the upper scale instead of the body, in order to move the marker to the same place on the scale - that is, you find the weight of the body in air - b) how many more grams you have to place when the body lies in the lower scale, that is, how much weight it loses by buoyancy.
2. The weight hydrometer (Fig. 194) is a float B consisting of two rigidly interconnected scales A and C lying above each other, the lower (as in Jolly's spring balance) in the fluid, the upper in air. The marker O, which you move by a load on the float just to the level of the fluid, lies between A and B. The two weightings proceed as for Jolly's instrument.
3. The container hydrometer is a small bottle which you fill to its rim with a fluid. If you then introduce a small body, it displaces the fluid outside. Hence, if you
a) weigh it
filled to the rim with the fluid and with the body next to it on
the scale and
b) when the body is in the bottle,
you find from the difference of the two results the weight of the fluid displaced by the body..
Measurement of the density of fluids (Mohr's scale hydrometer)
In order to measure the density of a fluid, you determine the loss of weight of a solid body first in water and then in the fluid. Its loss of weight in water yields its volume. Its loss of weight in the fluid which is equal to the weight of the displaced fluid thus yields the weight (through the first measurement of a known volume of this fluid. You can employ always the same body (in Figs. 195/196 a small glass container with mercury), whence you need not find its loss of weight in water, that is, its volume is only determined once, in order to find out the volume of the fluid displaced for the second measurement. Thus, the measurement of the density of a fluid is reduced to that of the loss of weight of the small glass bottle in it. Here too you employ a hydrostatic balance or Jolly's spring balance or a weight hydrometer or a scale hydrometer.
For this purpose, the hydrostatic balance of Friedrich Mohr 1806-1879 (Fig. 195) is used. The weighing of the glass bottle for the determination of its loss of weight is done by movement of sliding weights on the lever, subdivided into 10 equal sections.
The hydrometer of Gabriel Daniel Fahrenheit 1686-1736 (Fig. 196) is a hollow glass float, which carries (instead of the lower scale with a load which is the same for all weightings) a mass of mercury, mostly the sphere of a thermometer, since one must take into account the temperature of the fluid. The gadget is loaded for each measurement in such a way that it dips into the fluid to a definite mark. If it weighs P·g in air and has to be loaded, when it swims in water, additionally with p·g, in order that it will submerge to the marker, it receives in water, since the displaced water volume is equally heavy as the swimming body, a buoyancy of (P + p)g, that is, it displaces (P + p) cm³ water, that is, it dips in with a volume of (P + p) cm³. If it must be loaded in the fluid to be investigated with p'g in order to dip in as far as the marker - that is, again to dip in with the volume (P + p) cm³ - it displaces (P + p')g of the fluid. Thus (P + p')g cm³ contain (P + p')g of the fluid, whence 1 cm³ contains (P + p')/(P + p)g.
Scale hydrometers differ from weight hydrometers in the same way as automatic balances differ from non-automatic ones: They only require reading of a scale. A scale hydrometer (Fig. 197) - always a thermometer like float - is a hydrometer with an empirically subdivided and numbered scale. You let it swim in the fluid and read off the number on its scale to which it dips in. (The same weight of fluid is displaced.)
The number read off the scale does not always represent the density. Its significance depends on the purpose for which the hydrometer has been calibrated. For example, scale hydrometers are calibrated as alcohol meter in order to determine the percentage of the weight of pure alcohol in a mixture of alcohol and water (spirits), as alcohol meter for mixtures of absolute alcohol (Gay-Lussac hydrometer), as alkali meter for the determination of the alkali content in lyes, as lactometer of the water content in milk, etc.
There exist also scale hydrometers with arbitrary subdivisions like that of Antoine Baumé 1728-1804. Concentrated sulphuric acid should have 66º B., that is, its density should be such that the Baumé hydrometer dips in up to the subdivision 66; the density of nitric acid in the trade should correspond to 36 B. In order to transform Grade B. into density, you use a table. If n is the number of degrees and d the density, then, depending on whether the fluid (at 12.5º C) is heavier or lighter than water, d=/(146-n) and d=/(146+n).
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