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| Meta Title | Fluid mechanics - Hydrostatics, Pressure, Buoyancy | Britannica |
| Meta Description | Fluid mechanics - Hydrostatics, Pressure, Buoyancy: It is common knowledge that the pressure of the atmosphere (about 105 newtons per square metre) is due to the weight of air above the Earthās surface, that this pressure falls as one climbs upward, and, correspondingly, that pressure increases as one dives deeper into a lake (or comparable body of water). Mathematically, the rate at which the pressure in a stationary fluid varies with height z in a vertical gravitational field of strength g is given by If Ļ and g are both independent of z, as is more or less the case in lakes, then This means that, |
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| Boilerpipe Text | It is common knowledge that the
pressure
of the
atmosphere
(about 10
5
newtons per square metre) is due to the weight of air above the Earthās surface, that this pressure falls as one climbs upward, and, correspondingly, that pressure increases as one dives deeper into a lake (or
comparable
body of water). Mathematically, the rate at which the pressure in a stationary
fluid
varies with height
z
in a vertical gravitational
field
of strength
g
is given by
If Ļ and
g
are both independent of
z
, as is more or less the case in lakes, then
This means that, since Ļ is about 10
3
kilograms per cubic metre for water and
g
is about 10 metres per second squared, the pressure is already twice the atmospheric value at a depth of 10 metres. Applied to the atmosphere, equation (
124
) would imply that the pressure falls to zero at a height of about 10 kilometres. In the atmosphere, however, the variation of Ļ with
z
is far from
negligible
and (
124
) is unreliable as a consequence; a better approximation is given below in the section
Hydrodynamics: Compressible flow in gases
.
Differential
manometers
Instruments for comparing pressures are called differential manometers, and the simplest such instrument is a U-tube containing
liquid
, as shown in
Figure 1A
. The two pressures of interest,
p
1
and
p
2
, are transmitted to the two ends of the liquid column through an inert gasāthe
density
of which is negligible by comparison with the liquid density, Ļāand the difference of height,
h
, of the two menisci is measured. It is a consequence of (
124
) that
Torricelli, Evangelista
Italian physicist and mathematician Evangelista Torricelli, inventor of the mercury barometer.
A
barometer
for measuring the pressure of the atmosphere in absolute terms is simply a manometer in which
p
2
is made zero, or as close to zero as is
feasible
. The barometer invented in the 17th century by the Italian physicist and mathematician
Evangelista Torricelli
, and still in use today, is a U-tube that is sealed at one end (see
Figure 1B
). It may be filled with liquid, with the sealed end downward, and then inverted. On inversion, a negative pressure may momentarily develop at the top of the liquid column if the column is long enough; however, cavitation normally occurs there and the column falls away from the sealed end of the tube, as shown in the figure. Between the two exists what Torricelli thought of as a vacuum, though it may be very far from that condition if the barometer has been filled without
scrupulous
precautions to ensure that all dissolved or adsorbed gases, which would otherwise collect in this space, have first been removed. Even if no contaminating
gas
is present, the Torricellian vacuum always contains the vapour of the liquid, and this exerts a pressure which may be small but is never quite zero. The liquid conventionally used in a Torricelli barometer is of course mercury, which has a low
vapour pressure
and a high density. The high density means that
h
is only about 760 millimetres; if water were used, it would have to be about 10 metres instead.
Figure 1C
illustrates the principle of the
siphon
. The top container is open to the atmosphere, and the pressure in it,
p
2
, is therefore atmospheric. To balance this and the weight of the liquid column in between, the pressure
p
1
in the bottom container ought to be greater by Ļ
g
h
. If the bottom container is also open to the atmosphere, then
equilibrium
is clearly impossible; the weight of the liquid column prevails and causes the liquid to flow downward. The siphon operates only as long as the column is continuous; it fails if a bubble of gas collects in the tube or if cavitation occurs. Cavitation therefore limits the level differences over which siphons can be used, and it also limits (to about 10 metres) the depth of wells from which water can be pumped using suction alone.
Archimedesā principle
Consider now a cube of side
d
totally immersed in liquid with its top and bottom faces horizontal. The pressure on the bottom face will be higher than on the top by Ļ
g
d
, and, since pressure is
force
per unit area and the area of a cube face is
d
2
, the resultant upthrust on the cube is Ļ
g
d
3
. This is a simple example of the so-called Archimedesā principle, which states that the upthrust experienced by a submerged or floating body is always equal to the weight of the liquid that the body displaces. As
Archimedes
must have realized, there is no need to prove this by detailed examination of the pressure difference between top and bottom. It is obviously true, whatever the
bodyās shape
. It is obvious because, if the solid body could somehow be removed and if the cavity thereby created could somehow be filled with more fluid instead, the whole system would still be in equilibrium. The extra fluid would, however, then be experiencing the upthrust previously experienced by the solid body, and it would not be in equilibrium unless this were just
sufficient
to balance its weight.
Archimedesā problem was to discover, by what would nowadays be called a nondestructive test, whether the crown of King
Hieron II
was made of pure gold or of gold diluted with silver. He understood that the pure metal and the alloy would differ in density and that he could determine the density of the crown by weighing it to find its mass and making a separate measurement of its volume. Perhaps the inspiration that struck him (in his bath) was that one can find the volume of any object by submerging it in liquid in something like a measuring cylinder (
i.e.,
in a container with vertical sides that have been suitably graduated) and measuring the
displacement
of the liquid surface. If so, he no doubt realized soon afterward that a more elegant and more accurate method for determining density can be based on the principle that bears his name. This method involves weighing the object twice, first, when it is suspended in a vacuum (suspension in air will normally suffice) and, second, when it is totally submerged in a liquid of density Ļ. If the density of the object is Ļā², the ratio between the two weights must be
If Ļā² is less than Ļ, then
W
2
, according to equation (
126
), is negative. What that means is that the object does not submerge of its own accord; it has to be pushed downward to make it do so. If an object with a mean
density
less than that of water is placed in a lake and not subjected to any downward force other than its own weight, it naturally floats on the surface, and Archimedesā principle shows that in equilibrium the volume of water which it displaces is a fraction Ļā²/Ļ of its own volume. A
hydrometer
is an object graduated in such a way that this fraction may be measured. By floating a hydrometer first in water of density Ļ
0
and then in some other liquid of density Ļ
1
and comparing the readings, one may determine the ratio Ļ
1
/Ļ
0
ā
i.e.,
the
specific gravity
of the other liquid.
In what orientation an object
floats
is a matter of grave concern to those who design boats and those who travel in them. A simple example will
suffice
to illustrate the factors that determine orientation.
Figure 2
shows three of the many possible orientations that a uniform square prism might adopt when floating, with half its volume submerged in a liquid for which Ļ = 2Ļā²; they are separated by rotations of 22.5Ā°. In each of these diagrams, C is the centre of mass of the prism, and B, a point known as the
centre of buoyancy
, is the centre of mass of the displaced water. The distributed forces acting on the
prism
are equivalent to its weight acting downward through C and to the equal weight of the displaced water acting upward through B. In general, therefore, the prism experiences a torque. In
Figure 2B
the torque is counterclockwise, and so it turns the prism away from
2A
and toward
2C
. In
2C
the torque vanishes because B is now vertically below C, and this is the orientation that corresponds to stable equilibrium. The torque also vanishes in
2A
, and the prism can in principle remain indefinitely in that orientation as well; the equilibrium in this case, however, is unstable, and the slightest disturbance will cause the prism to topple one way or the other. In fact, the
potential energy
of the system, which increases in a linear fashion with the difference in height between C and B, is at its smallest in orientation
2C
and at its largest in orientation
2A
. To improve the
stability
of a floating object one should, if possible, lower C relative to B. In the case of a boat, this may be done by redistributing the load inside.
Surface tension of liquids
Of the many hydrostatic phenomena in which the
surface tension
of liquids plays a role, the most significant is probably
capillarity
. Consider what happens when a tube of narrow bore, often called a capillary tube, is dipped into a liquid. If the liquid āwetsā the tube (with zero contact angle), the liquid surface inside the tube forms a concave
meniscus, which is a virtually spherical surface having the same radius,
r
, as the inside of the tube. The tube experiences a downward force of magnitude 2Ļ
r
d
Ļ, where Ļ is the surface
tension
of the liquid, and the liquid experiences a reaction of equal magnitude that lifts the meniscus through a height
h
such that
ā
i.e.,
until the upward force for which surface tension is responsible is balanced by the weight of the column of liquid that has been lifted. If the liquid does not wet the tube, the meniscus is convex and depressed through the same distance
h
(see
Figure 3
). A simple method for determining surface tension involves the measurement of
h
in one or the other of these situations and the use of equation (
127
) thereafter.
It follows from equations (
124
) and (
127
) that the pressure at a point P just below the meniscus differs from the pressure at Q by an amount
it is less than the pressure at Q in the case to which
Figure 3A
refers and greater than the pressure at Q in the other case. Since the pressure at Q is just the
atmospheric pressure
, it is equal to the pressure at a point immediately above the meniscus. Hence, in both instances there is a pressure difference of 2Ļ/
r
between the two sides of the curved meniscus, and in both the higher pressure is on the inner side of the curve. Such a pressure difference is a requirement of equilibrium wherever a liquid surface is curved. If the surface is curved but not spherical, the pressure difference is
where
r
1
and
r
2
are the two principal radii of curvature. If it is cylindrical, one of these radii is
infinite
, and, if it is curved in opposite directions, then for the purposes of (
129
) they should be treated as being of opposite sign.
The diagrams in
Figure 3
were drawn to represent cross sections through cylindrical tubes, but they might equally well represent two vertical parallel plates that are partly submerged in the liquid a small distance apart. Consideration of how the pressure varies with height shows that over the range of height
h
the plates experience a greater pressure on their outer surfaces than on their inner surfaces; this is true whether the liquid wets both plates or not. It is a matter of observation that small objects floating near one another on the surface of a liquid tend to move toward one another, and it is the pressure difference just referred to that makes them behave in this way.
One other problem having to do with surface tension will be considered here. The diagrams in
Figure 4
show stages in the growth of a liquid drop on the end of a tube which the liquid is supposed to wet. In passing from stage A to stage B, by which time the drop is roughly hemispheric in shape, the radius of curvature of the drop diminishes; and it follows from (
128
) that, to bring about this growth, one must slowly increase the pressure of the liquid inside the tube. If the pressure could be held steady at the value corresponding to B, the drop would then become unstable, because any further growth (
e.g.,
to the more or less spherical shape indicated in
Figure 4C
) would involve an increase in radius of curvature. The applied pressure would then
exceed
that required to hold the drop in equilibrium, and the drop would necessarily grow bigger still. In practice, however, it is easier to control the rate of flow of water through the tube, and hence the rate of growth of the drop, than it is to control the pressure. If the rate of flow is very small, drops will form the nonspherical shapes suggested by
Figure 4D
before they detach themselves and fall. It is not an easy matter to analyze the shape of a drop on the point of detachment, and there is no simple formula for the volume of the drop after it is detached. |
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# [Hydrostatics](https://www.britannica.com/science/hydrostatics)
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Written by
[Thomas E. Faber Lecturer in Physics, University of Cambridge, 1959ā93; Fellow of Corpus Christi College, Cambridge. Author of *Fluid Dynamics for Physicists* and others.](https://www.britannica.com/contributor/Thomas-E-Faber/3527)
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It is common knowledge that the [pressure](https://www.britannica.com/science/pressure) of the [atmosphere](https://www.britannica.com/science/atmosphere) (about 105 newtons per square metre) is due to the weight of air above the Earthās surface, that this pressure falls as one climbs upward, and, correspondingly, that pressure increases as one dives deeper into a lake (or [comparable](https://www.britannica.com/dictionary/comparable) body of water). Mathematically, the rate at which the pressure in a stationary [fluid](https://www.britannica.com/science/fluid-physics) varies with height *z* in a vertical gravitational [field](https://www.britannica.com/science/field-physics) of strength *g* is given by
If Ļ and *g* are both independent of *z*, as is more or less the case in lakes, then
This means that, since Ļ is about 103 kilograms per cubic metre for water and *g* is about 10 metres per second squared, the pressure is already twice the atmospheric value at a depth of 10 metres. Applied to the atmosphere, equation ([124](https://www.britannica.com/science/fluid-mechanics/Hydrostatics#ref-15113)) would imply that the pressure falls to zero at a height of about 10 kilometres. In the atmosphere, however, the variation of Ļ with *z* is far from [negligible](https://www.britannica.com/dictionary/negligible) and ([124](https://www.britannica.com/science/fluid-mechanics/Hydrostatics#ref-15113)) is unreliable as a consequence; a better approximation is given below in the section [Hydrodynamics: Compressible flow in gases](https://www.britannica.com/science/fluid-mechanics/Waves-on-shallow-water#ref77486).
## Differential [manometers](https://www.britannica.com/technology/manometer)
[](https://cdn.britannica.com/08/2508-004-2696813B/representations-manometer-barometer-Torricellian.jpg)
[differential manometer, Torricellian barometer, and siphon](https://cdn.britannica.com/08/2508-004-2696813B/representations-manometer-barometer-Torricellian.jpg)Figure 1: Schematic representations of (A) a differential manometer, (B) a Torricellian barometer, and (C) a siphon.
(more)
Instruments for comparing pressures are called differential manometers, and the simplest such instrument is a U-tube containing [liquid](https://www.britannica.com/science/liquid-state-of-matter), as shown in Figure 1A. The two pressures of interest, *p*1 and *p*2, are transmitted to the two ends of the liquid column through an inert gasāthe [density](https://www.britannica.com/science/density) of which is negligible by comparison with the liquid density, Ļāand the difference of height, *h*, of the two menisci is measured. It is a consequence of ([124](https://www.britannica.com/science/fluid-mechanics/Hydrostatics#ref-15113)) that
[](https://cdn.britannica.com/21/195421-050-FC98CA10/Evangelista-Torricelli-physicist-Italian-mercury-barometer.jpg)
[Torricelli, Evangelista](https://cdn.britannica.com/21/195421-050-FC98CA10/Evangelista-Torricelli-physicist-Italian-mercury-barometer.jpg)Italian physicist and mathematician Evangelista Torricelli, inventor of the mercury barometer.
(more)
A [barometer](https://www.britannica.com/technology/barometer) for measuring the pressure of the atmosphere in absolute terms is simply a manometer in which *p*2 is made zero, or as close to zero as is [feasible](https://www.merriam-webster.com/dictionary/feasible). The barometer invented in the 17th century by the Italian physicist and mathematician [Evangelista Torricelli](https://www.britannica.com/biography/Evangelista-Torricelli), and still in use today, is a U-tube that is sealed at one end (see Figure 1B). It may be filled with liquid, with the sealed end downward, and then inverted. On inversion, a negative pressure may momentarily develop at the top of the liquid column if the column is long enough; however, cavitation normally occurs there and the column falls away from the sealed end of the tube, as shown in the figure. Between the two exists what Torricelli thought of as a vacuum, though it may be very far from that condition if the barometer has been filled without [scrupulous](https://www.britannica.com/dictionary/scrupulous) precautions to ensure that all dissolved or adsorbed gases, which would otherwise collect in this space, have first been removed. Even if no contaminating [gas](https://www.britannica.com/science/gas-state-of-matter) is present, the Torricellian vacuum always contains the vapour of the liquid, and this exerts a pressure which may be small but is never quite zero. The liquid conventionally used in a Torricelli barometer is of course mercury, which has a low [vapour pressure](https://www.britannica.com/science/vapor-pressure) and a high density. The high density means that *h* is only about 760 millimetres; if water were used, it would have to be about 10 metres instead.
Figure 1C illustrates the principle of the [siphon](https://www.britannica.com/technology/siphon-instrument). The top container is open to the atmosphere, and the pressure in it, *p*2, is therefore atmospheric. To balance this and the weight of the liquid column in between, the pressure *p*1 in the bottom container ought to be greater by Ļ*g**h*. If the bottom container is also open to the atmosphere, then [equilibrium](https://www.merriam-webster.com/dictionary/equilibrium) is clearly impossible; the weight of the liquid column prevails and causes the liquid to flow downward. The siphon operates only as long as the column is continuous; it fails if a bubble of gas collects in the tube or if cavitation occurs. Cavitation therefore limits the level differences over which siphons can be used, and it also limits (to about 10 metres) the depth of wells from which water can be pumped using suction alone.
## [Archimedesā principle](https://www.britannica.com/science/Archimedes-principle)
Consider now a cube of side *d* totally immersed in liquid with its top and bottom faces horizontal. The pressure on the bottom face will be higher than on the top by Ļ*g**d*, and, since pressure is [force](https://www.britannica.com/science/force-physics) per unit area and the area of a cube face is *d*2, the resultant upthrust on the cube is Ļ*g**d*3. This is a simple example of the so-called Archimedesā principle, which states that the upthrust experienced by a submerged or floating body is always equal to the weight of the liquid that the body displaces. As [Archimedes](https://www.britannica.com/biography/Archimedes) must have realized, there is no need to prove this by detailed examination of the pressure difference between top and bottom. It is obviously true, whatever the [bodyās shape](https://www.britannica.com/science/somatotype). It is obvious because, if the solid body could somehow be removed and if the cavity thereby created could somehow be filled with more fluid instead, the whole system would still be in equilibrium. The extra fluid would, however, then be experiencing the upthrust previously experienced by the solid body, and it would not be in equilibrium unless this were just [sufficient](https://www.britannica.com/dictionary/sufficient) to balance its weight.
Archimedesā problem was to discover, by what would nowadays be called a nondestructive test, whether the crown of King [Hieron II](https://www.britannica.com/biography/Hieron-II) was made of pure gold or of gold diluted with silver. He understood that the pure metal and the alloy would differ in density and that he could determine the density of the crown by weighing it to find its mass and making a separate measurement of its volume. Perhaps the inspiration that struck him (in his bath) was that one can find the volume of any object by submerging it in liquid in something like a measuring cylinder (*i.e.,* in a container with vertical sides that have been suitably graduated) and measuring the [displacement](https://www.britannica.com/dictionary/displacement) of the liquid surface. If so, he no doubt realized soon afterward that a more elegant and more accurate method for determining density can be based on the principle that bears his name. This method involves weighing the object twice, first, when it is suspended in a vacuum (suspension in air will normally suffice) and, second, when it is totally submerged in a liquid of density Ļ. If the density of the object is Ļā², the ratio between the two weights must be
If Ļā² is less than Ļ, then *W*2, according to equation ([126](https://www.britannica.com/science/fluid-mechanics/Hydrostatics#ref-15111)), is negative. What that means is that the object does not submerge of its own accord; it has to be pushed downward to make it do so. If an object with a mean [density](https://www.britannica.com/dictionary/density) less than that of water is placed in a lake and not subjected to any downward force other than its own weight, it naturally floats on the surface, and Archimedesā principle shows that in equilibrium the volume of water which it displaces is a fraction Ļā²/Ļ of its own volume. A [hydrometer](https://www.britannica.com/technology/hydrometer) is an object graduated in such a way that this fraction may be measured. By floating a hydrometer first in water of density Ļ0 and then in some other liquid of density Ļ1 and comparing the readings, one may determine the ratio Ļ1/Ļ0ā*i.e.,* the [specific gravity](https://www.britannica.com/science/specific-gravity) of the other liquid.
[](https://cdn.britannica.com/09/2509-050-305BFADB/orientations-prism-liquid-orientation-density.jpg)
[possible orientations of a square prism in liquid](https://cdn.britannica.com/09/2509-050-305BFADB/orientations-prism-liquid-orientation-density.jpg)Figure 2: Three possible orientations of a uniform square prism floating in liquid of twice its density. The stable orientation is (C) (see text).
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In what orientation an object [floats](https://www.britannica.com/science/buoyancy) is a matter of grave concern to those who design boats and those who travel in them. A simple example will [suffice](https://www.merriam-webster.com/dictionary/suffice) to illustrate the factors that determine orientation. Figure 2 shows three of the many possible orientations that a uniform square prism might adopt when floating, with half its volume submerged in a liquid for which Ļ = 2Ļā²; they are separated by rotations of 22.5Ā°. In each of these diagrams, C is the centre of mass of the prism, and B, a point known as the [centre of buoyancy](https://www.britannica.com/science/centre-of-buoyancy), is the centre of mass of the displaced water. The distributed forces acting on the [prism](https://www.britannica.com/dictionary/prism) are equivalent to its weight acting downward through C and to the equal weight of the displaced water acting upward through B. In general, therefore, the prism experiences a torque. In Figure 2B the torque is counterclockwise, and so it turns the prism away from 2A and toward 2C. In 2C the torque vanishes because B is now vertically below C, and this is the orientation that corresponds to stable equilibrium. The torque also vanishes in 2A, and the prism can in principle remain indefinitely in that orientation as well; the equilibrium in this case, however, is unstable, and the slightest disturbance will cause the prism to topple one way or the other. In fact, the [potential energy](https://www.britannica.com/science/potential-energy) of the system, which increases in a linear fashion with the difference in height between C and B, is at its smallest in orientation 2C and at its largest in orientation 2A. To improve the [stability](https://www.britannica.com/dictionary/stability) of a floating object one should, if possible, lower C relative to B. In the case of a boat, this may be done by redistributing the load inside.
## Surface tension of liquids
[1 of 2](https://www.britannica.com/video/Explanation-surface-tension/-159381)
The science of surface tension explainedLearn about surface tension and compare the surface tensions of different liquids, including water, alcohol, mercury, and soap bubbles.
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[See all videos for this article](https://www.britannica.com/science/fluid-mechanics/images-videos)
[2 of 2](https://cdn.britannica.com/10/2510-004-EF2E6BED/Capillarity.jpg)
[capillarity](https://cdn.britannica.com/10/2510-004-EF2E6BED/Capillarity.jpg)Figure 3: Capillarity.
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Of the many hydrostatic phenomena in which the [surface tension](https://www.britannica.com/science/surface-tension) of liquids plays a role, the most significant is probably [capillarity](https://www.britannica.com/science/capillarity). Consider what happens when a tube of narrow bore, often called a capillary tube, is dipped into a liquid. If the liquid āwetsā the tube (with zero contact angle), the liquid surface inside the tube forms a concave meniscus, which is a virtually spherical surface having the same radius, *r*, as the inside of the tube. The tube experiences a downward force of magnitude 2Ļ*r**d*Ļ, where Ļ is the surface [tension](https://www.britannica.com/dictionary/tension) of the liquid, and the liquid experiences a reaction of equal magnitude that lifts the meniscus through a height *h* such that ā*i.e.,* until the upward force for which surface tension is responsible is balanced by the weight of the column of liquid that has been lifted. If the liquid does not wet the tube, the meniscus is convex and depressed through the same distance *h* (see Figure 3). A simple method for determining surface tension involves the measurement of *h* in one or the other of these situations and the use of equation ([127](https://www.britannica.com/science/fluid-mechanics/Hydrostatics#ref-15110)) thereafter.
It follows from equations ([124](https://www.britannica.com/science/fluid-mechanics/Hydrostatics#ref-15113)) and ([127](https://www.britannica.com/science/fluid-mechanics/Hydrostatics#ref-15110)) that the pressure at a point P just below the meniscus differs from the pressure at Q by an amountit is less than the pressure at Q in the case to which Figure 3A refers and greater than the pressure at Q in the other case. Since the pressure at Q is just the [atmospheric pressure](https://www.britannica.com/science/atmospheric-pressure), it is equal to the pressure at a point immediately above the meniscus. Hence, in both instances there is a pressure difference of 2Ļ/*r* between the two sides of the curved meniscus, and in both the higher pressure is on the inner side of the curve. Such a pressure difference is a requirement of equilibrium wherever a liquid surface is curved. If the surface is curved but not spherical, the pressure difference iswhere *r*1 and *r*2 are the two principal radii of curvature. If it is cylindrical, one of these radii is [infinite](https://www.merriam-webster.com/dictionary/infinite), and, if it is curved in opposite directions, then for the purposes of ([129](https://www.britannica.com/science/fluid-mechanics/Hydrostatics#ref-15108)) they should be treated as being of opposite sign.
The diagrams in Figure 3 were drawn to represent cross sections through cylindrical tubes, but they might equally well represent two vertical parallel plates that are partly submerged in the liquid a small distance apart. Consideration of how the pressure varies with height shows that over the range of height *h* the plates experience a greater pressure on their outer surfaces than on their inner surfaces; this is true whether the liquid wets both plates or not. It is a matter of observation that small objects floating near one another on the surface of a liquid tend to move toward one another, and it is the pressure difference just referred to that makes them behave in this way.
[](https://cdn.britannica.com/11/2511-004-97E4DF70/Stages-formation-drop.jpg)
[formation of a liquid drop](https://cdn.britannica.com/11/2511-004-97E4DF70/Stages-formation-drop.jpg)Figure 4: Stages in the formation of a liquid drop (see text).
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One other problem having to do with surface tension will be considered here. The diagrams in Figure 4 show stages in the growth of a liquid drop on the end of a tube which the liquid is supposed to wet. In passing from stage A to stage B, by which time the drop is roughly hemispheric in shape, the radius of curvature of the drop diminishes; and it follows from ([128](https://www.britannica.com/science/fluid-mechanics/Hydrostatics#ref-15109)) that, to bring about this growth, one must slowly increase the pressure of the liquid inside the tube. If the pressure could be held steady at the value corresponding to B, the drop would then become unstable, because any further growth (*e.g.,* to the more or less spherical shape indicated in Figure 4C) would involve an increase in radius of curvature. The applied pressure would then [exceed](https://www.britannica.com/dictionary/exceed) that required to hold the drop in equilibrium, and the drop would necessarily grow bigger still. In practice, however, it is easier to control the rate of flow of water through the tube, and hence the rate of growth of the drop, than it is to control the pressure. If the rate of flow is very small, drops will form the nonspherical shapes suggested by Figure 4D before they detach themselves and fall. It is not an easy matter to analyze the shape of a drop on the point of detachment, and there is no simple formula for the volume of the drop after it is detached.
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# [Hydrodynamics](https://www.britannica.com/science/hydrodynamics-physics)
## [Bernoulliās law](https://www.britannica.com/science/Bernoullis-theorem)
Up to now the focus has been fluids at rest. This section deals with fluids that are in [motion](https://www.britannica.com/science/motion-mechanics) in a steady fashion such that the [fluid](https://www.britannica.com/science/fluid-physics) [velocity](https://www.britannica.com/science/velocity) at each given point in space is not changing with time. Any flow pattern that is steady in this sense may be seen in terms of a set of [streamlines](https://www.britannica.com/science/streamline), the trajectories of imaginary particles suspended in the fluid and carried along with it. In steady flow, the fluid is in motion but the streamlines are fixed. Where the streamlines crowd together, the fluid velocity is relatively high; where they open out, the fluid becomes relatively stagnant.
When [Euler](https://www.britannica.com/biography/Leonhard-Euler) and [Bernoulli](https://www.britannica.com/biography/Daniel-Bernoulli) were laying the foundations of hydrodynamics, they treated the fluid as an idealized inviscid substance in which, as in a fluid at rest in [equilibrium](https://www.merriam-webster.com/dictionary/equilibrium), the shear stresses associated with viscosity are zero and the [pressure](https://www.britannica.com/science/pressure) *p* is isotropic. They arrived at a simple law relating the variation of *p* along a streamline to the variation of *v* (the principle is credited to Bernoulli, but Euler seems to have arrived at it first), which serves to explain many of the phenomena that real fluids in steady motion display. To the inevitable question of when and why it is justifiable to neglect viscosity, there is no single answer. Some answers will be provided later in this article, but other matters will be taken up first.
Consider a small element of fluid of mass *m*, whichāapart from the [force](https://www.britannica.com/science/force-physics) on it due to gravityāis acted on only by a pressure *p*. The latter is isotropic and does not vary with time but may vary from point to point in space. It is a well-known [consequence](https://www.britannica.com/dictionary/consequence) of [Newtonās laws of motion](https://www.britannica.com/science/Newtons-laws-of-motion) that, when a particle of mass *m* moves under the influence of its weight *m**g* and an additional force *F* from a point P where its speed is *v*P and its height is *z*P to a point Q where its speed is *v*Q and its height is *z*Q, the [work](https://www.britannica.com/science/work-physics) done by the additional force is equal to the increase in kinetic and [potential energy](https://www.britannica.com/science/potential-energy) of the particleā*i.e.,* that
In the case of the fluid element under consideration, *F* may be related in a simple fashion to the [gradient](https://www.britannica.com/science/gradient-mathematics) of the pressure, and one finds
If the variations of fluid [density](https://www.britannica.com/science/density) along the streamline from P to Q are negligibly small, the factor Ļā1 may be taken outside the [integral](https://www.merriam-webster.com/dictionary/integral) on the right-hand side of ([131](https://www.britannica.com/science/fluid-mechanics/Hydrostatics#ref-15106)), which thereupon reduces to Ļā1(*p*Q - *p*P). Then ([130](https://www.britannica.com/science/fluid-mechanics/Hydrostatics#ref-15107)) and ([131](https://www.britannica.com/science/fluid-mechanics/Hydrostatics#ref-15106)) can be combined to obtain
[ Britannica Quiz Physics and Natural Law](https://www.britannica.com/quiz/physics-and-natural-law)
Since this applies for any two points that can be visited by a single element of fluid, one can immediately deduce Bernoulliās (or Eulerās) important result that along each streamline in the steady flow of an inviscid fluid the quantityis constant.
Under what circumstances are variations in the density negligibly small? When they are very small compared with the density itselfā*i.e.,* whenwhere the symbol Ī is used to represent the extent of the change along a [streamline](https://www.britannica.com/dictionary/streamline) of the quantity that follows it, and where *V*s is the [speed of sound](https://www.britannica.com/science/speed-of-sound-physics) (see below [Compressible flow in gases](https://www.britannica.com/science/fluid-mechanics/Waves-on-shallow-water#ref77486)). This condition is satisfied for all the flow problems having to do with water that are discussed below. If the fluid is air, it is adequately satisfied provided that the largest excursion in *z* is on the order of metres rather than kilometres and provided that the fluid velocity is everywhere less than about 100 metres per second.
Bernoulliās law indicates that, if an inviscid fluid is flowing along a pipe of varying [cross section](https://www.britannica.com/science/cross-section-physics), then the pressure is relatively low at constrictions where the velocity is high and relatively high where the pipe opens out and the fluid stagnates. Many people find this situation paradoxical when they first encounter it. Surely, they say, a constriction should increase the local pressure rather than diminish it? The [paradox](https://www.merriam-webster.com/dictionary/paradox) evaporates as one learns to think of the pressure changes along the pipe as cause and the velocity changes as effect, instead of the other way around; it is only because the pressure falls at a constriction that the pressure gradient upstream of the constriction has the right sign to make the fluid accelerate.
Paradoxical or not, predictions based on Bernoulliās law are well-verified by experiment. Try holding two sheets of paper so that they hang vertically two centimetres or so apart and blow downward so that there is a current of air between them. The sheets will be drawn together by the reduction in pressure associated with this current. Ships are drawn together for much the same reason if they are moving through the water in the same direction at the same speed with a small distance between them. In this case, the current results from the [displacement](https://www.britannica.com/dictionary/displacement) of water by each shipās bow, which has to flow backward to fill the space created as the stern moves forward, and the current between the ships, to which they both contribute, is stronger than the current moving past their outer sides. As another simple experiment, listen to the hissing sound made by a tap that is almost, but not quite, turned off. What happens in this case is that the flow is so constricted and the velocity within the constriction so high that the pressure in the constriction is actually negative. Assisted by the dissolved gases that are normally present, the water cavitates as it passes through, and the noise that is heard is the sound of tiny bubbles collapsing as the water slows down and the pressure rises again on the other side.
[](https://cdn.britannica.com/12/2512-004-1BB4C5D1/representation-venturi-tube-pitot.jpg)
[venturi tube and pitot tube](https://cdn.britannica.com/12/2512-004-1BB4C5D1/representation-venturi-tube-pitot.jpg)Figure 5: Schematic representation of (A) a venturi tube and of (B) a pitot tube.
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Two practical devices that are used by hydraulic engineers to monitor the flow of liquids though pipes are based on Bernoulliās law. One is the [venturi tube](https://www.britannica.com/technology/venturi-tube), a short length with a constriction in it of standard shape (see Figure 5A), which may be inserted into the pipe proper. If the velocity at point P, where the tube has a cross-sectional area *A*P, is *v*P and the velocity in the constriction, where the area is *A*Q, is *v*Q, the [continuity](https://www.britannica.com/science/continuity-principle) conditionāthe condition that the mass flowing through the pipe per unit time has to be the same at all points along its lengthāsuggests that ĻP*A*P*v*P = ĻQ*A*Q*v*Q, or that *A*P*v*P = *A*Q*v*Q if the difference between ĻP and ĻQ is [negligible](https://www.britannica.com/dictionary/negligible). Then Bernoulliās law indicates 
Thus one should be able to find *v*P, and hence the quantity *Q* (= *A*P*v*P) that engineers refer to as the rate of discharge, by measuring the difference of level *h* of the fluid in the two side tubes shown in the diagram. At low velocities the pressure difference (*p*P - *p*Q) is greatly affected by viscosity (see below [Viscosity](https://www.britannica.com/science/fluid-mechanics/Viscosity#ref77487)), and equation ([135](https://www.britannica.com/science/fluid-mechanics/Hydrostatics#ref-15102)) is unreliable in consequence. The venturi tube is normally used, however, when the velocity is large enough for the flow to be turbulent (see below [Turbulence](https://www.britannica.com/science/fluid-mechanics/Turbulence#ref77498)). In such a circumstance, equation (135) predicts [values](https://www.britannica.com/dictionary/values) for *Q* that agree with values measured by more direct means to within a few parts percent, even though the flow pattern is not really steady at all.
The other device is the [pitot tube](https://www.britannica.com/technology/pitot-tube), which is illustrated in Figure 5B. The fluid streamlines divide as they approach the blunt end of this tube, and at the point marked Q in the diagram there is complete stagnation, since the fluid at this point is moving neither up nor down nor to the right. It follows immediately from Bernoulliās law that
As with the venturi tube, one should therefore be able to find *v*P from the level difference *h*.
One other simple result deserves mention here. It concerns a jet of fluid [emerging](https://www.britannica.com/dictionary/emerging) through a hole in the wall of a vessel filled with [liquid](https://www.britannica.com/science/liquid-state-of-matter) under pressure. Observation of jets shows that after emerging they narrow slightly before settling down to a more or less uniform cross section known as the vena contracta. They do so because the streamlines are converging on the hole inside the vessel and are obliged to continue converging for a short while outside. It was [Torricelli](https://www.britannica.com/science/Torricellis-law) who first suggested that, if the pressure excess inside the vessel is generated by a head of liquid *h*, then the velocity *v* at the vena contracta is the velocity that a free particle would reach on falling through a height *h*ā*i.e.,* that
This result is an immediate consequence, for an inviscid fluid, of the principle of [energy](https://www.britannica.com/science/energy) conservation that Bernoulliās law enshrines.
In the following section, Bernoulliās law is used in an indirect way to establish a formula for the speed at which disturbances travel over the surface of shallow water. The explanation of several interesting phenomena having to do with water waves is buried in this formula. [Analogous](https://www.merriam-webster.com/dictionary/Analogous) phenomena dealing with sound waves in gases are discussed below in [Compressible flow in gases](https://www.britannica.com/science/fluid-mechanics/Waves-on-shallow-water#ref77486), where an [alternative](https://www.merriam-webster.com/dictionary/alternative) form of Bernoulliās law is introduced. This form of the law is restricted to gases in steady flow but is not restricted to flow velocities that are much less than the speed of sound. The complication that viscosity represents is again ignored throughout these two sections.
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| Readable Markdown | It is common knowledge that the [pressure](https://www.britannica.com/science/pressure) of the [atmosphere](https://www.britannica.com/science/atmosphere) (about 105 newtons per square metre) is due to the weight of air above the Earthās surface, that this pressure falls as one climbs upward, and, correspondingly, that pressure increases as one dives deeper into a lake (or [comparable](https://www.britannica.com/dictionary/comparable) body of water). Mathematically, the rate at which the pressure in a stationary [fluid](https://www.britannica.com/science/fluid-physics) varies with height *z* in a vertical gravitational [field](https://www.britannica.com/science/field-physics) of strength *g* is given by
If Ļ and *g* are both independent of *z*, as is more or less the case in lakes, then
This means that, since Ļ is about 103 kilograms per cubic metre for water and *g* is about 10 metres per second squared, the pressure is already twice the atmospheric value at a depth of 10 metres. Applied to the atmosphere, equation ([124](https://www.britannica.com/science/fluid-mechanics/Hydrostatics#ref-15113)) would imply that the pressure falls to zero at a height of about 10 kilometres. In the atmosphere, however, the variation of Ļ with *z* is far from [negligible](https://www.britannica.com/dictionary/negligible) and ([124](https://www.britannica.com/science/fluid-mechanics/Hydrostatics#ref-15113)) is unreliable as a consequence; a better approximation is given below in the section [Hydrodynamics: Compressible flow in gases](https://www.britannica.com/science/fluid-mechanics/Waves-on-shallow-water#ref77486).
## Differential [manometers](https://www.britannica.com/technology/manometer)
Instruments for comparing pressures are called differential manometers, and the simplest such instrument is a U-tube containing [liquid](https://www.britannica.com/science/liquid-state-of-matter), as shown in Figure 1A. The two pressures of interest, *p*1 and *p*2, are transmitted to the two ends of the liquid column through an inert gasāthe [density](https://www.britannica.com/science/density) of which is negligible by comparison with the liquid density, Ļāand the difference of height, *h*, of the two menisci is measured. It is a consequence of ([124](https://www.britannica.com/science/fluid-mechanics/Hydrostatics#ref-15113)) that
[Torricelli, Evangelista](https://cdn.britannica.com/21/195421-050-FC98CA10/Evangelista-Torricelli-physicist-Italian-mercury-barometer.jpg)Italian physicist and mathematician Evangelista Torricelli, inventor of the mercury barometer.
A [barometer](https://www.britannica.com/technology/barometer) for measuring the pressure of the atmosphere in absolute terms is simply a manometer in which *p*2 is made zero, or as close to zero as is [feasible](https://www.merriam-webster.com/dictionary/feasible). The barometer invented in the 17th century by the Italian physicist and mathematician [Evangelista Torricelli](https://www.britannica.com/biography/Evangelista-Torricelli), and still in use today, is a U-tube that is sealed at one end (see Figure 1B). It may be filled with liquid, with the sealed end downward, and then inverted. On inversion, a negative pressure may momentarily develop at the top of the liquid column if the column is long enough; however, cavitation normally occurs there and the column falls away from the sealed end of the tube, as shown in the figure. Between the two exists what Torricelli thought of as a vacuum, though it may be very far from that condition if the barometer has been filled without [scrupulous](https://www.britannica.com/dictionary/scrupulous) precautions to ensure that all dissolved or adsorbed gases, which would otherwise collect in this space, have first been removed. Even if no contaminating [gas](https://www.britannica.com/science/gas-state-of-matter) is present, the Torricellian vacuum always contains the vapour of the liquid, and this exerts a pressure which may be small but is never quite zero. The liquid conventionally used in a Torricelli barometer is of course mercury, which has a low [vapour pressure](https://www.britannica.com/science/vapor-pressure) and a high density. The high density means that *h* is only about 760 millimetres; if water were used, it would have to be about 10 metres instead.
Figure 1C illustrates the principle of the [siphon](https://www.britannica.com/technology/siphon-instrument). The top container is open to the atmosphere, and the pressure in it, *p*2, is therefore atmospheric. To balance this and the weight of the liquid column in between, the pressure *p*1 in the bottom container ought to be greater by Ļ*g**h*. If the bottom container is also open to the atmosphere, then [equilibrium](https://www.merriam-webster.com/dictionary/equilibrium) is clearly impossible; the weight of the liquid column prevails and causes the liquid to flow downward. The siphon operates only as long as the column is continuous; it fails if a bubble of gas collects in the tube or if cavitation occurs. Cavitation therefore limits the level differences over which siphons can be used, and it also limits (to about 10 metres) the depth of wells from which water can be pumped using suction alone.
## [Archimedesā principle](https://www.britannica.com/science/Archimedes-principle)
Consider now a cube of side *d* totally immersed in liquid with its top and bottom faces horizontal. The pressure on the bottom face will be higher than on the top by Ļ*g**d*, and, since pressure is [force](https://www.britannica.com/science/force-physics) per unit area and the area of a cube face is *d*2, the resultant upthrust on the cube is Ļ*g**d*3. This is a simple example of the so-called Archimedesā principle, which states that the upthrust experienced by a submerged or floating body is always equal to the weight of the liquid that the body displaces. As [Archimedes](https://www.britannica.com/biography/Archimedes) must have realized, there is no need to prove this by detailed examination of the pressure difference between top and bottom. It is obviously true, whatever the [bodyās shape](https://www.britannica.com/science/somatotype). It is obvious because, if the solid body could somehow be removed and if the cavity thereby created could somehow be filled with more fluid instead, the whole system would still be in equilibrium. The extra fluid would, however, then be experiencing the upthrust previously experienced by the solid body, and it would not be in equilibrium unless this were just [sufficient](https://www.britannica.com/dictionary/sufficient) to balance its weight.
Archimedesā problem was to discover, by what would nowadays be called a nondestructive test, whether the crown of King [Hieron II](https://www.britannica.com/biography/Hieron-II) was made of pure gold or of gold diluted with silver. He understood that the pure metal and the alloy would differ in density and that he could determine the density of the crown by weighing it to find its mass and making a separate measurement of its volume. Perhaps the inspiration that struck him (in his bath) was that one can find the volume of any object by submerging it in liquid in something like a measuring cylinder (*i.e.,* in a container with vertical sides that have been suitably graduated) and measuring the [displacement](https://www.britannica.com/dictionary/displacement) of the liquid surface. If so, he no doubt realized soon afterward that a more elegant and more accurate method for determining density can be based on the principle that bears his name. This method involves weighing the object twice, first, when it is suspended in a vacuum (suspension in air will normally suffice) and, second, when it is totally submerged in a liquid of density Ļ. If the density of the object is Ļā², the ratio between the two weights must be
If Ļā² is less than Ļ, then *W*2, according to equation ([126](https://www.britannica.com/science/fluid-mechanics/Hydrostatics#ref-15111)), is negative. What that means is that the object does not submerge of its own accord; it has to be pushed downward to make it do so. If an object with a mean [density](https://www.britannica.com/dictionary/density) less than that of water is placed in a lake and not subjected to any downward force other than its own weight, it naturally floats on the surface, and Archimedesā principle shows that in equilibrium the volume of water which it displaces is a fraction Ļā²/Ļ of its own volume. A [hydrometer](https://www.britannica.com/technology/hydrometer) is an object graduated in such a way that this fraction may be measured. By floating a hydrometer first in water of density Ļ0 and then in some other liquid of density Ļ1 and comparing the readings, one may determine the ratio Ļ1/Ļ0ā*i.e.,* the [specific gravity](https://www.britannica.com/science/specific-gravity) of the other liquid.
In what orientation an object [floats](https://www.britannica.com/science/buoyancy) is a matter of grave concern to those who design boats and those who travel in them. A simple example will [suffice](https://www.merriam-webster.com/dictionary/suffice) to illustrate the factors that determine orientation. Figure 2 shows three of the many possible orientations that a uniform square prism might adopt when floating, with half its volume submerged in a liquid for which Ļ = 2Ļā²; they are separated by rotations of 22.5Ā°. In each of these diagrams, C is the centre of mass of the prism, and B, a point known as the [centre of buoyancy](https://www.britannica.com/science/centre-of-buoyancy), is the centre of mass of the displaced water. The distributed forces acting on the [prism](https://www.britannica.com/dictionary/prism) are equivalent to its weight acting downward through C and to the equal weight of the displaced water acting upward through B. In general, therefore, the prism experiences a torque. In Figure 2B the torque is counterclockwise, and so it turns the prism away from 2A and toward 2C. In 2C the torque vanishes because B is now vertically below C, and this is the orientation that corresponds to stable equilibrium. The torque also vanishes in 2A, and the prism can in principle remain indefinitely in that orientation as well; the equilibrium in this case, however, is unstable, and the slightest disturbance will cause the prism to topple one way or the other. In fact, the [potential energy](https://www.britannica.com/science/potential-energy) of the system, which increases in a linear fashion with the difference in height between C and B, is at its smallest in orientation 2C and at its largest in orientation 2A. To improve the [stability](https://www.britannica.com/dictionary/stability) of a floating object one should, if possible, lower C relative to B. In the case of a boat, this may be done by redistributing the load inside.
## Surface tension of liquids
Of the many hydrostatic phenomena in which the [surface tension](https://www.britannica.com/science/surface-tension) of liquids plays a role, the most significant is probably [capillarity](https://www.britannica.com/science/capillarity). Consider what happens when a tube of narrow bore, often called a capillary tube, is dipped into a liquid. If the liquid āwetsā the tube (with zero contact angle), the liquid surface inside the tube forms a concave meniscus, which is a virtually spherical surface having the same radius, *r*, as the inside of the tube. The tube experiences a downward force of magnitude 2Ļ*r**d*Ļ, where Ļ is the surface [tension](https://www.britannica.com/dictionary/tension) of the liquid, and the liquid experiences a reaction of equal magnitude that lifts the meniscus through a height *h* such that ā*i.e.,* until the upward force for which surface tension is responsible is balanced by the weight of the column of liquid that has been lifted. If the liquid does not wet the tube, the meniscus is convex and depressed through the same distance *h* (see Figure 3). A simple method for determining surface tension involves the measurement of *h* in one or the other of these situations and the use of equation ([127](https://www.britannica.com/science/fluid-mechanics/Hydrostatics#ref-15110)) thereafter.
It follows from equations ([124](https://www.britannica.com/science/fluid-mechanics/Hydrostatics#ref-15113)) and ([127](https://www.britannica.com/science/fluid-mechanics/Hydrostatics#ref-15110)) that the pressure at a point P just below the meniscus differs from the pressure at Q by an amountit is less than the pressure at Q in the case to which Figure 3A refers and greater than the pressure at Q in the other case. Since the pressure at Q is just the [atmospheric pressure](https://www.britannica.com/science/atmospheric-pressure), it is equal to the pressure at a point immediately above the meniscus. Hence, in both instances there is a pressure difference of 2Ļ/*r* between the two sides of the curved meniscus, and in both the higher pressure is on the inner side of the curve. Such a pressure difference is a requirement of equilibrium wherever a liquid surface is curved. If the surface is curved but not spherical, the pressure difference iswhere *r*1 and *r*2 are the two principal radii of curvature. If it is cylindrical, one of these radii is [infinite](https://www.merriam-webster.com/dictionary/infinite), and, if it is curved in opposite directions, then for the purposes of ([129](https://www.britannica.com/science/fluid-mechanics/Hydrostatics#ref-15108)) they should be treated as being of opposite sign.
The diagrams in Figure 3 were drawn to represent cross sections through cylindrical tubes, but they might equally well represent two vertical parallel plates that are partly submerged in the liquid a small distance apart. Consideration of how the pressure varies with height shows that over the range of height *h* the plates experience a greater pressure on their outer surfaces than on their inner surfaces; this is true whether the liquid wets both plates or not. It is a matter of observation that small objects floating near one another on the surface of a liquid tend to move toward one another, and it is the pressure difference just referred to that makes them behave in this way.
One other problem having to do with surface tension will be considered here. The diagrams in Figure 4 show stages in the growth of a liquid drop on the end of a tube which the liquid is supposed to wet. In passing from stage A to stage B, by which time the drop is roughly hemispheric in shape, the radius of curvature of the drop diminishes; and it follows from ([128](https://www.britannica.com/science/fluid-mechanics/Hydrostatics#ref-15109)) that, to bring about this growth, one must slowly increase the pressure of the liquid inside the tube. If the pressure could be held steady at the value corresponding to B, the drop would then become unstable, because any further growth (*e.g.,* to the more or less spherical shape indicated in Figure 4C) would involve an increase in radius of curvature. The applied pressure would then [exceed](https://www.britannica.com/dictionary/exceed) that required to hold the drop in equilibrium, and the drop would necessarily grow bigger still. In practice, however, it is easier to control the rate of flow of water through the tube, and hence the rate of growth of the drop, than it is to control the pressure. If the rate of flow is very small, drops will form the nonspherical shapes suggested by Figure 4D before they detach themselves and fall. It is not an easy matter to analyze the shape of a drop on the point of detachment, and there is no simple formula for the volume of the drop after it is detached. |
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