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.
Side
pressure
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.
Archimedes'
Principle
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.
Hydrostatic
Balance
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).
Swimming
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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