# Buoyancy

The forces at work in buoyancy. Note that, because the upward force of buoyancy is equal to the downward force of gravity, the object is floating.

In science, buoyancy is an upward force exerted by a fluid that opposes the weight of an immersed object. In a column of fluid, pressure increases with depth as a result of the weight of the overlying fluid. Thus a column of fluid, or an object submerged in the fluid, experiences greater pressure at the bottom of the column than at the top. This difference in pressure results in a net force that tends to accelerate an object upwards. The magnitude of that force is proportional to the difference in the pressure between the top and the bottom of the column, and (as explained by Archimedes' principle) is also equivalent to the weight of the fluid that would otherwise occupy the column, i.e. the displaced fluid. For this reason, an object whose density is greater than that of the fluid in which it is submerged tends to sink. If the object is either less dense than the liquid or is shaped appropriately (as in a boat), the force can keep the object afloat. This can occur only in a reference frame which either has a gravitational field or is accelerating due to a force other than gravity defining a "downward" direction (that is, a non-inertial reference frame). In a situation of fluid statics, the net upward buoyancy force is equal to the magnitude of the weight of fluid displaced by the body.[1]

The center of buoyancy of an object is the centroid of the displaced volume of fluid.

## Archimedes' principle

A metallic coin (one British pound coin) floats in mercury due to the buoyancy force upon it and appears to float higher because of the surface tension of the mercury.

Archimedes' principle is named after Archimedes of Syracuse, who first discovered this law in 212 B.C.[2] For more objects, floating and sunken, and in gases as well as liquids (i.e. a fluid), Archimedes' principle may be stated thus in terms of forces:

Any object, wholly or partially immersed in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object.

with the clarifications that for a sunken object the volume of displaced fluid is the volume of the object, and for a floating object on a liquid, the weight of the displaced liquid is the weight of the object.

More tersely: Buoyancy = weight of displaced fluid.

Archimedes' principle does not consider the surface tension (capillarity) acting on the body,[3] but this additional force modifies only the amount of fluid displaced, so the principle that Buoyancy = weight of displaced fluid remains valid.

The weight of the displaced fluid is directly proportional to the volume of the displaced fluid (if the surrounding fluid is of uniform density). In simple terms, the principle states that the buoyancy force on an object is going to be equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. Thus, among completely submerged objects with equal masses, objects with greater volume have greater buoyancy. This is also known as upthrust.

Suppose a rock's weight is measured as 10 newtons when suspended by a string in a vacuum with gravity acting upon it. Suppose that when the rock is lowered into water, it displaces water of weight 3 newtons. The force it then exerts on the string from which it hangs would be 10 newtons minus the 3 newtons of buoyancy force: 10 − 3 = 7 newtons. Buoyancy reduces the apparent weight of objects that have sunk completely to the sea floor. It is generally easier to lift an object up through the water than it is to pull it out of the water.

Assuming Archimedes' principle to be reformulated as follows,

${\text{apparent immersed weight}}={\text{weight}}-{\text{weight of displaced fluid}}\,$

then inserted into the quotient of weights, which has been expanded by the mutual volume

${\frac {{\text{density}}}{{\text{density of fluid}}}}={\frac {{\text{weight}}}{{\text{weight of displaced fluid}}}},\,$

yields the formula below. The density of the immersed object relative to the density of the fluid can easily be calculated without measuring any volumes:

${\frac {{\text{density of object}}}{{\text{density of fluid}}}}={\frac {{\text{weight}}}{{\text{weight}}-{\text{apparent immersed weight}}}}\,$

(This formula is used for example in describing the measuring principle of a dasymeter and of hydrostatic weighing.)

Example: If you drop wood into water, buoyancy will keep it afloat.

Example: A helium balloon in a moving car. During a period of increasing speed, the air mass inside the car moves in the direction opposite to the car's acceleration (i.e., towards the rear). The balloon is also pulled this way. However, because the balloon is buoyant relative to the air, it ends up being pushed "out of the way", and will actually drift in the same direction as the car's acceleration (i.e., forward). If the car slows down, the same balloon will begin to drift backward. For the same reason, as the car goes round a curve, the balloon will drift towards the inside of the curve.

## Forces and equilibrium

This is the equation to calculate the pressure inside a fluid in equilibrium. The corresponding equilibrium equation is:

${\mathbf {f}}+\operatorname {div}\,\sigma =0$

where f is the force density exerted by some outer field on the fluid, and σ is the Cauchy stress tensor. In this case the stress tensor is proportional to the identity tensor:

$\sigma _{{ij}}=-p\delta _{{ij}}.\,$

Here δij is the Kronecker delta. Using this the above equation becomes:

${\mathbf {f}}=\nabla p.\,$

Assuming the outer force field is conservative, that is it can be written as the negative gradient of some scalar valued function:

${\mathbf {f}}=-\nabla \Phi .\,$

Then:

$\nabla (p+\Phi )=0\Longrightarrow p+\Phi ={\text{constant}}.\,$

Therefore, the shape of the open surface of a fluid equals the equipotential plane of the applied outer conservative force field. Let the z-axis point downward. In this case the field is gravity, so Φ = −ρfgz where g is the gravitational acceleration, ρf is the mass density of the fluid. Taking the pressure as zero at the surface, where z is zero, the constant will be zero, so the pressure inside the fluid, when it is subject to gravity, is

$p=\rho _{f}gz.\,$

So pressure increases with depth below the surface of a liquid, as z denotes the distance from the surface of the liquid into it. Any object with a non-zero vertical depth will have different pressures on its top and bottom, with the pressure on the bottom being greater. This difference in pressure causes the upward buoyancy forces.

The buoyancy force exerted on a body can now be calculated easily, since the internal pressure of the fluid is known. The force exerted on the body can be calculated by integrating the stress tensor over the surface of the body which is in contact with the fluid:

${\mathbf {B}}=\oint \sigma \,d{\mathbf {A}}.$

The surface integral can be transformed into a volume integral with the help of the Gauss divergence theorem:

${\mathbf {B}}=\int \operatorname {div}\sigma \,dV=-\int {\mathbf {f}}\,dV=-\rho _{f}{\mathbf {g}}\int \,dV=-\rho _{f}{\mathbf {g}}V$

where V is the measure of the volume in contact with the fluid, that is the volume of the submerged part of the body. Since the fluid doesn't exert force on the part of the body which is outside of it.

The magnitude of buoyancy force may be appreciated a bit more from the following argument. Consider any object of arbitrary shape and volume V surrounded by a liquid. The force the liquid exerts on an object within the liquid is equal to the weight of the liquid with a volume equal to that of the object. This force is applied in a direction opposite to gravitational force, that is of magnitude:

$B=\rho _{f}V_{{\text{disp}}}\,g,\,$

where ρf is the density of the fluid, Vdisp is the volume of the displaced body of liquid, and g is the gravitational acceleration at the location in question.

If this volume of liquid is replaced by a solid body of exactly the same shape, the force the liquid exerts on it must be exactly the same as above. In other words the "buoyancy force" on a submerged body is directed in the opposite direction to gravity and is equal in magnitude to

$B=\rho _{f}Vg.\,$

The net force on the object must be zero if it is to be a situation of fluid statics such that Archimedes principle is applicable, and is thus the sum of the buoyancy force and the object's weight

$F_{{\text{net}}}=0=mg-\rho _{f}V_{{\text{disp}}}g\,$

If the buoyancy of an (unrestrained and unpowered) object exceeds its weight, it tends to rise. An object whose weight exceeds its buoyancy tends to sink. Calculation of the upwards force on a submerged object during its accelerating period cannot be done by the Archimedes principle alone; it is necessary to consider dynamics of an object involving buoyancy. Once it fully sinks to the floor of the fluid or rises to the surface and settles, Archimedes principle can be applied alone. For a floating object, only the submerged volume displaces water. For a sunken object, the entire volume displaces water, and there will be an additional force of reaction from the solid floor.

In order for Archimedes' principle to be used alone, the object in question must be in equilibrium (the sum of the forces on the object must be zero), therefore;

$mg=\rho _{f}V_{{\text{disp}}}g,\,$

and therefore

$m=\rho _{f}V_{{\text{disp}}}.\,$

showing that the depth to which a floating object will sink, and the volume of fluid it will displace, is independent of the gravitational field regardless of geographic location.

(Note: If the fluid in question is seawater, it will not have the same density (ρ) at every location. For this reason, a ship may display a Plimsoll line.)

It can be the case that forces other than just buoyancy and gravity come into play. This is the case if the object is restrained or if the object sinks to the solid floor. An object which tends to float requires a tension restraint force T in order to remain fully submerged. An object which tends to sink will eventually have a normal force of constraint N exerted upon it by the solid floor. The constraint force can be tension in a spring scale measuring its weight in the fluid, and is how apparent weight is defined.

If the object would otherwise float, the tension to restrain it fully submerged is:

$T=\rho _{f}Vg-mg.\,$

When a sinking object settles on the solid floor, it experiences a normal force of:

$N=mg-\rho _{f}Vg.\,$

It is common to define a buoyancy mass mb that represents the effective mass of the object as can be measured by a gravitational method. If an object which usually sinks is submerged suspended via a cord from a balance pan, the reference object on the other dry-land pan of the balance will have mass:

$m_{b}=m_{{\mathrm {o}}}\cdot \left(1-{\frac {\rho _{{\mathrm {f}}}}{\rho _{{\mathrm {o}}}}}\right)\,$

where mo is the true (vacuum) mass of the object, and ρo and ρf are the average densities of the object and the surrounding fluid, respectively. Thus, if the two densities are equal, ρo = ρf, the object is seemingly weightless, and is said to be neutrally buoyant. If the fluid density is greater than the average density of the object, the object floats; if less, the object sinks.

Another possible formula for calculating buoyancy of an object is by finding the apparent weight of that particular object in the air (calculated in Newtons), and apparent weight of that object in the water (in Newtons). To find the force of buoyancy acting on the object when in air, using this particular information, this formula applies:

'Buoyancy force = weight of object in empty space − weight of object immersed in fluid'

The final result would be measured in Newtons.

Air's density is very small compared to most solids and liquids. For this reason, the weight of an object in air is approximately the same as its true weight in a vacuum. The buoyancy of air is neglected for most objects during a measurement in air because the error is usually insignificant (typically less than 0.1% except for objects of very low average density such as a balloon or light foam).

### Simplified model

Pressure distribution on an immersed cube
Forces on an immersed cube
Approximation of an arbitrary volume as a group of cubes

A simplified explanation for the integration of the pressure over the contact area may be stated as follows:

Consider a cube immersed in a fluid with the upper surface horizontal.

The sides are identical in area, and have the same depth distribution, therefore they also have the same pressure distribution, and consequently the same total force resulting from hydrostatic pressure, exerted perpendicular to the plane of the surface of each side.

There are two pairs of opposing sides, therefore the resultant horizontal forces balance in both orthogonal directions, and the resultant force is zero.

The upward force on the cube is the pressure on the bottom surface integrated over its area. The surface is at constant depth, so the pressure is constant. Therefore, the integral of the pressure over the area of the horizontal bottom surface of the cube is the hydrostatic pressure at that depth multiplied by the area of the bottom surface.

Similarly, the downward force on the cube is the pressure on the top surface integrated over its area. The surface is at constant depth, so the pressure is constant. Therefore, the integral of the pressure over the area of the horizontal top surface of the cube is the hydrostatic pressure at that depth multiplied by the area of the top surface.

As this is a cube, the top and bottom surfaces are identical in shape and area, and the pressure difference between the top and bottom of the cube is directly proportional to the depth difference, and the resultant force difference is exactly equal to the weight of the fluid that would occupy the volume of the cube in its absence.

This means that the resultant upward force on the cube is equal to the weight of the fluid that would fit into the volume of the cube, and the downward force on the cube is its weight, in the absence of external forces.

This analogy is valid for variations in the size of the cube.

If two cubes are placed alongside each other with a face of each in contact, the pressures and resultant forces on the sides or parts thereof in contact are balanced and may be disregarded, as the contact surfaces are equal in shape, size and pressure distribution, therefore the buoyancy of two cubes in contact is the sum of the buoyancies of each cube. This analogy can be extended to an arbitrary number of cubes.

An object of any shape can be approximated as a group of cubes in contact with each other, and as the size of the cube is decreased, the precision of the approximation increases. The limiting case for infinitely small cubes is the exact equivalence.

Angled surfaces do not nullify the analogy as the resultant force can be split into orthogonal components and each dealt with in the same way.

### Stability

A floating object is stable if it tends to restore itself to an equilibrium position after a small displacement. For example, floating objects will generally have vertical stability, as if the object is pushed down slightly, this will create a greater buoyancy force, which, unbalanced by the weight force, will push the object back up.

Rotational stability is of great importance to floating vessels. Given a small angular displacement, the vessel may return to its original position (stable), move away from its original position (unstable), or remain where it is (neutral).

Rotational stability depends on the relative lines of action of forces on an object. The upward buoyancy force on an object acts through the center of buoyancy, being the centroid of the displaced volume of fluid. The weight force on the object acts through its center of gravity. A buoyant object will be stable if the center of gravity is beneath the center of buoyancy because any angular displacement will then produce a 'righting moment'.

The stability of a buoyant object at the surface is more complex, and it may remain stable even if the centre of gravity is above the centre of buoyancy, provided that when disturbed from the equilibrium position, the centre of buoyancy moves further to the same side that the centre of gravity moves, thus providing a positive righting moment. If this occurs, the floating object is said to have a positive metacentric height. This situation is typically valid for a range of heel angles, beyond which the centre of buoyancy does not move enough to provide a positive righting moment, and the object becomes unstable. It is possible to shift from positive to negative or vice versa more than once during a heeling disturbance, and many shapes are stable in more than one position.

## Compressible fluids and objects

The atmosphere's density depends upon altitude. As an airship rises in the atmosphere, its buoyancy decreases as the density of the surrounding air decreases. In contrast, as a submarine expels water from its buoyancy tanks, it rises because its volume is constant (the volume of water it displaces if it is fully submerged) while its mass is decreased.

### Compressible objects

As a floating object rises or falls, the forces external to it change and, as all objects are compressible to some extent or another, so does the object's volume. Buoyancy depends on volume and so an object's buoyancy reduces if it is compressed and increases if it expands.

If an object at equilibrium has a compressibility less than that of the surrounding fluid, the object's equilibrium is stable and it remains at rest. If, however, its compressibility is greater, its equilibrium is then unstable, and it rises and expands on the slightest upward perturbation, or falls and compresses on the slightest downward perturbation.

#### Submarines

Submarines rise and dive by filling large tanks with seawater. To dive, the tanks are opened to allow air to exhaust out the top of the tanks, while the water flows in from the bottom. Once the weight has been balanced so the overall density of the submarine is equal to the water around it, it has neutral buoyancy and will remain at that depth.

#### Balloons

The height to which a balloon rises tends to be stable. As a balloon rises it tends to increase in volume with reducing atmospheric pressure, but the balloon itself does not expand as much as the air on which it rides. The average density of the balloon decreases less than that of the surrounding air. The weight of the displaced air is reduced. A rising balloon stops rising when it and the displaced air are equal in weight. Similarly, a sinking balloon tends to stop sinking.

#### Divers

Underwater divers are a common example of the problem of unstable buoyancy due to compressibility. The diver typically wears an exposure suit which relies on gas filled spaces for insulation, and may also wear a buoyancy compensator, which is a variable volume buoyancy bag which is inflated to increase buoyancy and deflated to decrease buoyancy. The desired condition is usually neutral buoyancy when the diver is swimming in mid-water, and this condition is unstable, so the diver is constantly making fine adjustments by control of lung volume, and has to adjust the contents of the buoyancy compensator if the depth varies.

## Buoyancy of air

Similar to objects at the bottom of its ocean of water looking upward at objects floating above it, humans live at the bottom of an "ocean" of air and look upward at balloons drifting above us. A balloon is suspended in air, and a jellyfish is suspended in water for the same reason: each is buoyed upward by a force equal to the weight of fluid that would occupy its volume; when that buoyant force equals its own weight, it neither rises nor falls. In one case, the displaced fluid is air; and in the other case, the fluid is water. Objects in water are buoyed up because the pressure acting up against the bottom of the object exceeds the pressure acting down against the top. Likewise, air pressure acting up against an object in air is greater than the pressure above pushing down. The buoyancy, in both cases, is equal to the weight of fluid displaced - Archimedes' principle holds for air just as it does for water.

A cubic meter of air at ordinary atmospheric pressure and room temperature has a mass of about 1.2 kilograms, so its weight is about 12 newtons. Therefore, any 1-cubic-meter object in air is buoyed up with a force of 12 newtons. If the mass of the 1-cubic-meter object is greater than 1.2 kilograms (so that its weight is greater than 12 newtons), it falls to the ground when released. If an object of this size has a mass less than 1.2 kilograms, it rises in the air. Any object that has a mass that is less than the mass of an equal volume of air will rise in air - in other words, any object less dense than air will rise. Gas-filled balloons that rise in air, thus, are less dense than air.

Greatest buoyancy would be achieved if the balloon were evacuated. The weight of a structure needed to keep an evacuated balloon from collapsing would more than offset the advantage of the extra buoyancy. Thus, hot air balloons are filled with gas less dense than ordinary air, which keeps the balloon from collapsing while keeping it light. In sport balloons, the gas is simply heated air. In balloons intended to reach high altitudes or stay up for extended periods of time, helium is generally used. The density of helium is small enough so that the combined weight of helium, balloon, and whatever the cargo happens to be is less than the weight of air it displaces. Hydrogen is the least dense gas, but it is highly flammable and thus seldom used. Low-density gas is used in a balloon for the same reason that cork or closed cell foam buoyancy material is used in a swimmer's life preserver. The cork or foam possesses no strange tendency to rise. Both are buoyed upward like anything else. They are simply light enough for the buoyancy to be significant.

Unlike water, the atmosphere has no discernible surface (there is no "top"). Furthermore, unlike water, the atmosphere becomes less dense with altitude. Whereas a cork will float to the surface of water, a released helium-filled balloon does not rise to any atmosphere surface. With regards to how high a balloon will rise, a balloon will rise only so long as it displaces a weight of air greater than its own weight. Air becomes less dense with altitude, so, when the weight of displaced air equals the total weight of the balloon, upward acceleration ends. We can also say that, when the buoyant force on the balloon equals its weight, the balloon will cease to rise. Equivalently, when the average density of the balloon (including its load) equals the density of the surrounding air, the balloon will cease rising. Helium-filled toy balloons usually break when released in the air because, as the balloon rises to regions of less pressure, the helium in the balloon expands, increasing the volume and stretching the foil until it breaks.

## Density

A density column containing some common liquids and solids. From top: baby oil, rubbing alcohol (with red food coloring), vegetable oil, wax, water (with blue food coloring), and aluminum.

If the weight of an object is less than the weight of the displaced fluid when fully submerged, then the object has an average density that is less than the fluid and when fully submerged will experience a buoyancy force greater than its own weight. If the fluid has a surface, such as water in a lake or the sea, the object will float and settle at a level where it displaces the same weight of fluid as the weight of the object. If the object is immersed in the fluid, such as a submerged submarine or air in a balloon, it will tend to rise. If the object has exactly the same density as the fluid, then its buoyancy equals its weight. It will remain submerged in the fluid, but it will neither sink nor float, although a disturbance in either direction will cause it to drift away from its position. An object with a higher average density than the fluid will never experience more buoyancy than weight and it will sink. A ship will float even though it may be made of steel (which is much denser than water), because it encloses a volume of air (which is much less dense than water), and the resulting shape has an average density less than that of the water.