# Buoyancy

The forces at work in buoyancy

In physics, buoyancy (pronounced /ˈbɔɪ.ənsi/) is an upward acting force, caused by fluid pressure, that reduces an object's weight. 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] This is the force that enables the object to float.

## Archimedes' principle

Archimedes' principle is named after Archimedes of Syracuse, who first discovered this law.[2] His treatise, On floating bodies, proposition 5 states, that

Any floating object displaces its own weight of fluid.

For more general 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.

Archimedes of Syracuse

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.[4]

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 buoyant 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. Thus, among completely submerged objects with equal masses, objects with greater volume have greater buoyancy.

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 buoyant 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,

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

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

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

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:

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

(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. In increasing speed or driving a curve, the air moves in the opposite direction of the car's acceleration. The balloon however, is pushed due to buoyancy "out of the way" by the air, and will actually drift in the same direction as the car's acceleration.

## 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 stress tensor. In this case the stress tensor is proportional to the identity tensor:

$\sigma_{ij}=-p\delta_{ij}.\,$

Here $\delta_{ij}\,$ 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:

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

Hence 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

$p=\rho_f g z.\,$

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 buoyant 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–Ostrogradsky 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 buoyant 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 the exact same shape, the force the liquid exerts on it must be exactly the same as above. In other words the "buoyant force" on a submerged body is directed in the opposite direction to gravity and is equal in magnitude to

$B = \rho_f V g. \,$

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 buoyant force and the object's weight

$F_\text{net} = 0 = m g - \rho_f V_\text{disp} g \,$

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:

$T = \rho_f V g - m g . \,$

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

$N = m g - \rho_f V g . \,$

It is common to define a buoyant 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 $m_{\mathrm{o}}\,$ 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).

### 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 buoyant 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 buoyant 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'.

## 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. As a submarine expels water from its buoyancy tanks (by pumping them full of air) it rises because its volume is constant (the volume of water it displaces if it is fully submerged) as its weight 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 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. Normally, precautions are taken to ensure that no air has been left in the tanks. If air were left in the tanks and the submarine were to descend even slightly, the increased pressure of the water would compress the remaining air in the tanks, reducing its volume. Since buoyancy is a function of volume, this would cause a decrease in buoyancy, and the submarine would continue to descend.

The height of a balloon tends to be stable. As a balloon rises it tends to increase in volume with reducing atmospheric pressure, but the balloon's cargo does not expand. The average density of the balloon decreases less, therefore, than that of the surrounding air. The balloon's buoyancy decreases because the weight of the displaced air is reduced. A rising balloon tends to stop rising. Similarly, a sinking balloon tends to stop sinking.

## Density

A pound coin floats in mercury due to the buoyant force upon it.

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 force buoyancy 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.

## Beyond Archimedes Principle

Archimedes' principle is a fluid statics concept. In its simple form, it applies when the object is not accelerating relative to the fluid. To examine the case when the object is accelerated by buoyancy and gravity, the fact that the displaced fluid itself has inertia as well must be considered.[5]

This means that both the buoyant object and a parcel of fluid (equal in volume to the object) will experience the same magnitude of buoyant force because of Newton's third law, and will experience the same acceleration, but in opposite directions, since the total volume of the system is unchanged. In each case, the difference between magnitudes of the buoyant force and the force of gravity is the net force, and when divided by the relevant mass, it will yield the respective acceleration through Newton's second law. All acceleration measures are relative to the reference frame of the undisturbed background fluid.

### Atwood's machine analogy

Atwood's Machine Analogy for dynamics of buoyant objects in vertical motion. The displaced parcel of fluid is indicated as the dark blue rectangle, and the buoyant solid object is indicated as the gray object. The acceleration vectors (a) in this visual depict a positively buoyant object which naturally accelerates upward, and upward acceleration of the object is our sign convention.

The system can be understood by analogy with a suitable modification of Atwood's machine, to represent the mechanical coupling of the displaced fluid and the buoyant object, as shown in the diagram right.

• The solid object is represented by the gray object
• The fluid being displaced is represented by dark blue object
• Undisturbed background fluid is analogous to the inextensible massless cord
• The force of buoyancy is analogous to the tension in the cord
• The solid floor of the body of fluid is analogous to the pulley, and reverses the direction of the buoyancy force, such that both the solid object and the displaced fluid experience their buoyancy force upward.

### Results

It is important to note that this simplification of the situation completely ignores drag and viscosity, both of which come in to play to a greater extent as speed increases, when considering the dynamics of buoyant objects. The following simple formulation makes the assumption of slow speeds such that drag and viscosity are not significant. It is difficult to carry out such an experiment in practice with speeds close to zero, but if measurements of acceleration are made as quickly as possible after release from rest, the equations below give a good approximation to the acceleration and the buoyancy force.

A system consists of a well-sealed object of mass m and volume V which is fully submerged in a uniform fluid body of density ρf and in an environment of a uniform gravitational field g. Under the forces of buoyancy and gravity alone, the "dynamic buoyant force" B acting on the object and its upward acceleration a are given by:

Buoyant force
$B = \frac{2 g m \rho_f V}{m + \rho_f V}$
Upward acceleration
$a = \frac{g (\rho_f V - m)}{m + \rho_f V}$

Derivations of both of these equations originates from constructing a system of equations by means of Newton's second law for both the solid object and the displaced parcel of fluid. An equation for upward acceleration of the object is constructed by dividing the net force on the object (Bmg) by its mass m. Due to the mechanical coupling, the object's upward acceleration is equal in magnitude to the downward acceleration of the displaced fluid, an equation constructed by dividing the net force on the displaced fluid (BρfVg) by its mass ρfV.

Should other forces come in to play in a different situation (such as spring forces, human forces, thrust, drag, or lift), it is necessary for the solver of problem to re-consider the construction of Newton's second law for both bodies, now involving these other forces.

In the case of neutral buoyancy, m is equal to ρfV. Thus B reduces to mg and the acceleration is zero. If the object is much denser than the fluid, then B approaches zero and the object's upward acceleration is approximately −g, i.e. it is accelerated downward due to gravity as if the fluid were not present. Similarly, if the fluid is much denser than the object, then B approaches 2mg and the upward acceleration is approximately g.