Pushed of Archimedes

The pushed of Archimedes is the particular force which undergoes a body plunged in all or partly in a Fluide (Liquide or Gaz) subjected to a field of Gravité.

This force comes from the increase in the pressure of the fluid with the depth (effect of gravity on the fluid, to see the article Hydrostatique): the pressure being stronger on the lower part of an object immersed than on its upper part, it results an overall vertical push directed from it upwards.

This push defines the Flottabilité of a body.

History and legend

Archimedes

See also: Archimedes

Archimedes is a Greek scientist who lived with Syracuse (Sicily) of 287 av. J. - C. with 212 av. J. - C.. It is known for its multiple scientific, theoretical or practical work, whether it is in mathematical or Physique. Among the latter, its Treaty of the floating bodies provides the foundations of what will be later the named science Hydrostatique. It is in particular in this work that he studies with rigor the immersion of a body, solid or fluid, in a fluid of Densité lower, equal or higher. The theorem which will bear the name of the scientist later there is thus stated (this theorem was then shown at the 16th century).

The crown of the king Hiéron II

Vitruve reports that the king Hiéron II of Syracuse (306-214) would have asked his young friend and scientific adviser Archimedes (old then 22 years only) to check if a crown of Or, that it had been made make like offering with Jupiter, were completely out of gold or if the craftsman had put money at it. The checking had of course as a constraint not to deteriorate the crown. The form of this one was moreover too complex to carry out a calculation of the volume of the ornament. Archimedes would have found the means of checking if the crown were really out of gold, whereas it was with the public baths, by observing how objects floated there. It would then have left in the street while exclaiming celebrates it “Eureka” (I found).

What notes Archimedes with the public baths is that, for the same given volume, the bodies do not have same the apparent Poids, i.e. a mass per different unit of volume. One nowadays speaks about Density. The money (density: 10500 kg·m^-3) being less dense than gold (density: 19300 kg·m^-3), it thus has a lower density. From there, Archimedes deduces that if the craftsman hid money in the crown of the king, then it has a lower density. Thus the trickery of the jeweller was discovered.

The solution with the problem

To answer the question of king Hiéron, Archimedes thus could compare the volumes of water moved by the crown and an identical mass of gold. If both move the same volume of water, their density is then equal and one can conclude from it that both are made up of the same metal. To carry out the experiment, one can imagine to plunge in a container filled at short-nap cloth-edge the gold mass. A certain quantity of water will overflow then of the container. Then, one withdraws gold and one replaces it by the crown to be studied. If the crown is well completely out of gold, then water will not overflow. On the other hand, if its density is lower, of additional water will overflow.

This method presents two disadvantages. First is that it utilizes here of nothing the Archimedes' principle. The second problem is that with realistic conditions, because of the form of the crown and the density of gold, the height of moved water is very low (lower than the millimetre). It is thus not very probable that Archimedes could draw the significant conclusions starting from such an experiment.

A more realistic method is the following one. By laying out on each arm of a balance the crown on a side and its equal gold weight, balance is initially obtained. Then, one can immerse the two arms in water. If the crown and gold have the same density, then the push of Archimedes will be equal on the two arms of the balance and balance will be respected. If the crown does not contain only gold, then it will undergo a push of more important Archimedes and an imbalance will be then visible.

Other proposals of the treaty of the floating bodies

The treaty of the floating bodies contains other proposals relating to the theorem of Archimedes:

• Proposal III: a of the same solid volume and of the same weight (makes voluminal of the same mass of it) as the liquid in which it is there abandoned will insert in order to not emerge by no means above surface, but not to go down low.
• Proposal IV: Any body lighter than the liquid where it is abandoned will not be completely immersed, but will remain partly above the surface of the liquid.
• Proposal V: a solid lighter than the liquid in which one gives up it inserts there in such way that a volume of liquid equal to the immersed part has the same weight as the whole solid.
• Proposal VI: When a body is lighter than the liquid where one inserts it and goes back to surface, the force which pushes in top this body has as a measurement the quantity whose weight of an equal volume of liquid exceeds the weight even body.
• Proposal VII: a body heavier than the liquid where it is given up will go down at the bottom and its weight, in the liquid, will decrease by a measured quantity, by what a volume of liquid equal to that of the body weighs.

Formulation of the theorem of Archimedes

In a uniform field of gravity, the push of Archimedes P _A is always given by the following formula:

$\backslash \; vec\; \{P\}\; \_\; \{\backslash \; rm\; has\; \backslash ,\}\; =\; -\; \backslash ,\; M\_\; \{\backslash \; rm\; F\; \backslash ,\}\; \backslash \; vec\; \{G\}$,

where M _f is the Masse fluid contained in moved volume V , and G the value of the Pesanteur.

If the Density ρ of the fluid is it also uniform, one will have:

$\backslash \; vec\; \{P\}\; \_\; \{\backslash \; rm\; has\; \backslash ,\}\; =\; -\; \backslash ,\; \backslash \; rho\; \backslash ,\; V\; \backslash \; vec\; \{G\}$

or, if one considers only the sizes of the forces:

P _A = ρ V   G .

The push of Archimedes P _A will be expressed in newton (NR) if the density ρ is in kg/m ³, the volume of fluid moved V in m ³ and the value of gravity G in N/kg (or m/s ²).

Demonstration

Experiment of thought

Let us consider a fluid at rest. Let us delimit, by an experiment of thought, a certain volume of an unspecified form within this fluid. This volume is him also at rest: in spite of its weight, this volume does not fall. That thus means that its weight is rigorously balanced by a force, equal and opposed, which maintains it on the spot, and which comes only from outside. Let us replace now, always in our experiment of thought, this volume by an unspecified body: the force which maintained the fluid is always there, it does not have any reason to have changed: she always equal and is opposed to the weight of moved fluid. It is the force of Archimedes.

Idea of calculation

Let us suppose a Cube edge has entirely immersed in a liquid, its face top horizontal and being located at a depth Z ₁ > 0 (the positive direction is downwards).

In the case of an incompressible liquid at rest subjected to a field of uniform Gravity, the absolute Pression p is worth

p = p _o + p _h,

where p _o is the Atmospheric pressure and p _h the hydrostatic Pression.

To a depth Z , the hydrostatic pressure corresponds to the Poids P of a column of liquid (which one can imagine cylindrical) height Z and of Base has , divided by the base. However

P = m G = ('' Z has '') G ,

where m is the Masse of the column, zA its volume, ρ the Density (presumedly uniform) of the liquid and G the Accélération of the Gravité, which gives

p _h = P/has = ρ G Z .

The absolute pressure is worth thus

p = p _o + ρ G Z .

By Symmetry, the compressive forces exerted on the four vertical faces of the cube cancel two to two.

The force F ₁ exerted downwards on the face top, of surface has = has ², is worth

F ₁ = p ₁ has = ( p _o + ρ   G   Z ₁) has ².

The force F ₂ exerted upwards on the face of bottom, located at the depth Z ₂ = Z ₁ + has , is worth

F ₂ = p ₂ has = ( p _o + ρ   G   Z ₂)   has ² = + '' ρ   G ''   ('' Z '' ₁ + '' has '')   has ².

The resultant F of all the compressive forces is worth thus

F = F ₁ - F ₂ = - ( ρ   G   has )   has ² = - ρ   G   has ³ = - ρ   G   V = - ρV   G = - M _F G ,

where V = has ³ is the volume of the cube, i.e. in fact immersed volume, and M _f the mass of the fluid contained in a volume V . The size of the resulting force is thus quite equal to that of the weight M _F G of the volume of moved fluid; this force being negative, it is well directed vertically upwards.

It is possible to generalize the preceding demonstration with a volume of an unspecified form. It is enough to break up surface bordering volume into an infinity of infinitesimal elements dS presumedly plane, then to make the sum, using the integral calculus, of all the infinitesimal forces df exerted on each element of surface.

More general demonstration

Let us suppose a volume unspecified $\backslash \; mathcal\; \{V\}\; \backslash ,$, delimited by a closed surface $\backslash \; Sigma\; \backslash ,$, entirely plunged in a fluid of density $\backslash \; rho\; \backslash ,$ subjected to a field of uniform Pesanteur $\backslash \; vec\; \{G\}\; \backslash ,$.

One seeks to determine the resultant of the forces of Pression exerted on volume:

$\backslash \; vec\; \{F\}\; =\; \backslash \; int\_\; \{\backslash \; Sigma\}\; D\; \backslash \; vec\; \{F\}$.

By definition of the pressure $p\; \backslash ,$, there is

$d\; \backslash \; vec\; \{F\}\; =\; -\; \backslash ,\; p\; \backslash ,\; D\; \backslash \; vec\; \{S\}$

where $d\; \backslash \; vec\; \{S\}\; \backslash ,$ is an element Infinitésimal surface considered, directed by convention towards the outside of this surface, and $d\; \backslash \; vec\; \{F\}\; \backslash ,$ the infinitesimal element of force which is exerted there. One thus seeks to determine

$\backslash \; vec\; \{F\}\; =\; -\; \backslash \; int\_\; \{\backslash \; Sigma\}\; p\; \backslash ,\; D\; \backslash \; vec\; \{S\}$

For the needs for the demonstration, let us consider now the integral $I\; \backslash ,$ following, where it will be supposed that $\backslash \; vec\; \{U\}\; \backslash ,$ represents a vector Field uniform and not no one:

$I\; =\; \backslash \; left\; (\backslash \; int\_\; \{\backslash \; Sigma\}\; p\; \backslash ,\; D\; \backslash \; vec\; \{S\}\; \backslash \; right)\; \backslash \; cdot\; \backslash \; vec\; \{U\}$

$\backslash \; vec\; \{U\}\; \backslash ,$ being uniform, one can as well write

$I\; =\; \backslash \; int\_\; \{\backslash \; Sigma\}\; p\; \backslash ,\; \backslash \; vec\; \{U\}\; \backslash \; cdot\; D\; \backslash \; vec\; \{S\}$

According to the Theorem of flow-divergence,

$\backslash \; int\_\; \{\backslash \; Sigma\}\; p\; \backslash ,\; \backslash \; vec\; \{U\}\; \backslash \; cdot\; D\; \backslash \; vec\; \{S\}\; =\; \backslash \; int\_\; \{\backslash \; mathcal\; \{V\}\}\; \backslash \; operatorname\; \{div\}\; (p\; \backslash ,\; \backslash \; vec\; \{U\})\; \backslash ,\; FD$

However, according to one of the formulas of Leibniz of the vectorial Analysis,

$\backslash \; operatorname\; \{div\}\; (p\; \backslash ,\; \backslash \; vec\; \{U\})\; =\; \backslash \; vec\; \{\backslash \; operatorname\; \{grad\}\}\; (p)\; \backslash \; cdot\; \backslash \; vec\; \{U\}\; +\; p\; \backslash ,\; \backslash \; operatorname\; \{div\}\; (\backslash \; vec\; \{U\})$

And since the divergence of a uniform vector field is null, there is

$\backslash \; operatorname\; \{div\}\; (p\; \backslash ,\; \backslash \; vec\; \{U\})\; =\; \backslash \; vec\; \{\backslash \; operatorname\; \{grad\}\}\; (p)\; \backslash \; cdot\; \backslash \; vec\; \{U\}$

Consequently,

$I\; =\; \backslash \; int\_\; \{\backslash \; Sigma\}\; p\; \backslash ,\; \backslash \; vec\; \{U\}\; \backslash \; cdot\; D\; \backslash \; vec\; \{S\}\; =\; \backslash \; int\_\; \{\backslash \; mathcal\; \{V\}\}\; \backslash \; vec\; \{\backslash \; operatorname\; \{grad\}\}\; (p)\; \backslash \; cdot\; \backslash \; vec\; \{U\}\; \backslash \; FD$

$\backslash \; vec\; \{U\}\; \backslash ,$ being uniform, one can as well write:

$I\; =\; \backslash \; left\; (\backslash \; int\_\; \{\backslash \; Sigma\}\; p\; \backslash ,\; D\; \backslash \; vec\; \{S\}\; \backslash \; right)\; \backslash \; cdot\; \backslash \; vec\; \{U\}\; =\; \backslash \; left\; (\backslash \; int\_\; \{\backslash \; mathcal\; \{V\}\}\; \backslash \; vec\; \{\backslash \; operatorname\; \{grad\}\}\; (p)\; \backslash ,\; FD\; \backslash \; right)\; \backslash \; cdot\; \backslash \; vec\; \{U\}$

One from of thus deduced that

$\backslash \; vec\; \{F\}\; =\; -\; \backslash \; int\_\; \{\backslash \; Sigma\}\; p\; \backslash ,\; D\; \backslash \; vec\; \{S\}\; =\; -\; \backslash \; int\_\; \{\backslash \; mathcal\; \{V\}\}\; \backslash \; vec\; \{\backslash \; operatorname\; \{grad\}\}\; (p)\; \backslash ,\; FD$

However, according to the fundamental law of the Hydrostatic ,

$\backslash \; vec\; \{\backslash \; operatorname\; \{grad\}\}\; (p)\; =\; \backslash \; rho\; \backslash ,\; \backslash \; vec\; \{G\}$

From where

$\backslash \; vec\; \{F\}\; =\; -\; \backslash \; int\_\; \{\backslash \; mathcal\; \{V\}\}\; \backslash \; rho\; \backslash ,\; \backslash \; vec\; \{G\}\; \backslash ,\; FD\; =\; -\; \backslash \; left\; (\backslash \; int\_\; \{\backslash \; mathcal\; \{V\}\}\; \backslash \; rho\; \backslash ,\; FD\; \backslash \; right)\; \backslash \; vec\; \{G\}\; =\; -\; \backslash ,\; M\_\; \{\backslash \; rm\; F\; \backslash ,\}\; \backslash \; vec\; \{G\}$

The resultant of the compressive forces is thus equal in size to the weight of the volume of fluid moved, but directed in the contrary direction of the weight, i.e. upwards.

Applications

Example of an entirely immersed solid

Entirely let us immerse a solid of volume V , mass m and Density ρ in a uniform fluid of density ρ_f, then slacken it starting from the rest. At the beginning, speed being null, two forces only act on the solid: its weight F _p (downwards) and push of Archimedes F _a (upwards).

F _p = ρ V   G

F _a = ρ_f V   G

F _p/ F _a = ρ/ρ_f

The report/ratio of the densities is in fact equivalent to that of the Densité S.

• If the density of the solid is higher than that of the fluid, then F _p > F _a and the solid runs.
• If the density of the solid is equal to that of the fluid, then F _p = F _a and the solid remains motionless; it is in neutral or indifferent balance.
• If the density of the solid is lower than that of the fluid, then F _p < F _a and the solid goes up towards surface.

In both cases where the solid is not balances some, its later movement is determined by three forces: its weight, push of Archimedes (opposed to the weight) and a force of viscous friction F _f (opposed at the speed).

According to the second law of the movement of Newton, one has then:

F _p - F _a ± F _f = m   has (the positive direction is downwards)

where has is the acceleration of the solid.

As the force of viscous friction is not constant, but that it increases with speed, acceleration decreases gradually, so that the solid reaches a speed limit more or less quickly, when the resultant of the forces is null.

Example of a solid floating on the surface of a liquid

Let us consider a solid of volume V and density ρ_S floating on the surface of a liquid of density ρ_L. If the solid floats, it is that its weight is balanced by the push of Archimedes:

F _A = F _P.

The push of Archimedes being equal (in size) to the weight of the volume of moved liquid (equivalent with immersed volume V _i), one can write:

ρ_L V _i   G = ρ_S V   G .

Immersed volume is worth thus

V _i = (  ρ_S  /  ρ_L  )   V .

Since V > V _i, it follows that ρ_S < ρ_L  .

Application to the case of a Iceberg

Let us consider a piece of pure Glace to 0   floating °C in Sea water. Either ρ_S = 0,917 kg/dm³ and ρ_L = 1,025 kg/dm³ (one would have ρ_L = 1,000 kg/dm³ for Pure water with 3,98   °C). The ρ_S ratio/  ρ_L (i.e. the relative Density) is of 0,895, so that volume immersed V _i represents nearly 90% of total volume V of the iceberg.

An ice floe which melts in glass

It is easy to check that the cast iron of a piece of pure ice floating on pure water occurs without change of level of water. The volume of ice immersed corresponds indeed to the volume of liquid water necessary to equalize the weight of the ice floe. While melting, the ice floe produced (by conservation of the mass) exactly this volume of water, which “stops the hole left by the disappearance of the solid ice”. The water level remains the same one. On the figure opposite, the volume delimited in dotted line is, in glass of left, volume of immersed ice, and in glass of right-hand side, the liquid volume of water produced by the cast iron of the ice floe.

One can also make following calculation: if one considers, for example, an ice floe of 1 Cm3 and of density 0.917 G cm -3 (which thus contains 0,917 G of water), immersed volume is of 0.917 Cm3 (as for an iceberg, the major part is under water). When the ice floe melts, these 0,917 G of water which will have désomais a density of 1 G cm -3 will occupy exactly the volume which the immersed part of the ice floe occupied.

Other examples of application of Pushed of Archimedes

• the Archimedes' principle applies to Fluide S, i.e. as well with liquids as with gases. It is thus thanks to the push of Archimedes that a Montgolfière or a Dirigeable can rise in the airs (in both cases, a gas of density weaker than the air is used, than it is heated air or Hélium).

• a plunger is put “to run” around -12 m in the Atlantique or the the Mediterranean because its density increases with the depth (because of increasing compression, particularly of the bubbles contained in neoprene of its combination: its mass does not change but its volume decreases) until reaching and exceeding that of the ambient conditions.
• the fresh water having a density lower than salt water, the push of Archimedes is stronger in the Dead Sea (the most salted sea world) that in a lake. It is thus easier to float there.
• the Spationaute S are involved with the exercises in space in swimming pools where, thanks to the push of Archimedes which balances their weight, they can know a state which are connected up to a certain point with the Impesanteur.
• the weight of the ships (and thus their density) variable according to whether they are in load or on Lest, the push of Archimedes also will vary. To maintain a level of floating (Draft) constant and to assume a better stability, the ships are equipped with ballasts which they can fill or empty according to their Cargaison or the salinity of water in which they sail. (See also careens).
• the submarines control their density by also using ballasts.

Not application

All occurs as if the push of Archimedes applied to the Center of hull , i.e. with the Center of gravity of the volume of moved fluid.

This characteristic is important for the calculation of the Stabilité of a Sous-marin in diving or of a Aérostat at low altitude: under penalty of seeing the machine being turned over, it is necessary that the center of hull is located above center of gravity.

As regards a ship or an airship in high-altitude, on the other hand, the center of hull is often located below the center of gravity (for example for a Planche with veil). However, when the penetration of the object in the fluid evolves/moves, the center of hull moves, creating a couple which comes to be opposed to the movement. The stability is then ensured by the position of the metacenter , which is the point of application of the variations of the push. This metacenter must be above the center of gravity.

In an anecdotic way, one can notice that the originators of airships and submarines must make sure simultaneously of two types of balances for their machines.

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