# Archimedes' Principle: The Buoyancy of a Body in Water

## Important Formula

## What do the formula symbols mean?

## Buoyant Force

`$$ F_{\text A} $$`Unit

`$$ \mathrm{N} $$`

## Density

`$$ \class{blue}{\rho} $$`Unit

`$$ \frac{ \mathrm{kg} }{ \mathrm{m}^3 } $$`

For example, the density of water is: \( \rho ~=~ 10^{-3} \, \frac{ \text{kg} }{ \text{m}^3} \).

## Volume

`$$ \class{red}{V} $$`Unit

`$$ \mathrm{m}^3 $$`

## Gravitational acceleration

`$$ g $$`Unit

`$$ \frac{\mathrm{m}}{\mathrm{s}^2} $$`

## Table of contents

The Archimedes principle states that a body submerged in a fluid (or gas) experiences an **buoyant force** \(F_{\text A}\) equal to the gravitational force \(F_{\text g}\) of the displaced fluid (or gas). This principle was named after the ancient Greek mathematician and physicist Archimedes, who discovered it.

Here, \( \class{blue}{\rho} \) is the **mass density** of the fluid. It indicates how heavy one cubic meter of the fluid is. For example, the density of water is approximately \( \class{blue}{\rho} ~=~ 10^3 \, \frac{ \text{kg} }{ \text{m}^3} \). And \( \class{red}{V} \) is the **volume** of the fluid displaced by the submerged body, that is the volume of the body submerged in the fluid. \( g ~=~ 9.8 \, \frac{\mathrm m}{\mathrm{s}^2} \) represents the gravitational acceleration.

## What is the hydrostatic paradox?

Consider, for example, communicating vessels. These have *different shapes*. They are *open at the top* and *connected at the bottom*.

If you fill them with water, the **filling height** \( h \) is the *same* in all tubes, even though one might think that due to the different shapes of the tubes, there is a different amount of water in each tube, and the amount of water in each tube would push downwards with different strengths, resulting in a greater filling height in thinner tubes. This phenomenon is referred to as the **hydrostatic paradox**.

In reality, the filling height is independent of the shape of the tube, as revealed by the formula for hydrostatic pressure: $$ h ~=~ \frac{\Pi}{\rho \, g} $$

## Exercises with Solutions

Use this formula eBook if you have problems with physics problems.### Exercise #1: Size of the Hot Air Balloon for Ascending

A hot air balloon has a **total weight** of \( m = 400 \, \mathrm{kg}\). The interior of the balloon is filled with air. The air has a **temperature** of \(T = 30^{\circ}\mathrm{C}\).

What **radius** \(r\) must a hot air balloon have to ascend at \(10^{\circ} \mathrm{C}\)?

The air density \(\rho_{\text T}\) for the corresponding temperature \(T\) (at normal pressure) is given in the following table.

Temperature \(T\) in \(^{\circ}\text{C}\) | Density \( \rho_{\text T} \) in \(\frac{\text{kg}}{\text{m}^3}\) |
---|---|

30 | 1.1644 |

20 | 1.2041 |

10 | 1.2466 |

5 | 1.2690 |

0 | 1.2920 |

#### Solution to Exercise #1

In order for the hot air balloon to ascend, the gravitational force acting on it \(m\,g\) must be compensated by the buoyant force \(\rho \, V \, g\). In addition to the mass \(m\) of the hot air balloon, the mass of the air \(m_{\text L}\) inside the balloon must also be considered. So it must hold:

`1 \[ m \, g ~+~ m_{\text L} \, g ~=~ \rho_{10} \, V \, g \]`

The gravitational acceleration \(g\) cancels out in 1

:

`2 \[ m ~+~ m_{\text L} ~=~ \rho_{10}\, V \]`

Here, \( \rho_{10} \) is the density of the air at \(10^{\circ} \, \text{C}\), which was taken from the table in the solution hint. And \( V \) is the volume of the air displaced by the hot air balloon, precisely the volume of the balloon. The volume of the gondola can be neglected here.

Since the mass \(m_{\text L}\) of the air inside the balloon is not known, it is expressed using the known density \(\rho_{30}\) (at 30 degrees Celsius): \( m_{\text L} ~=~ \rho_{30}\, V \):

`3 \[ m ~+~ \rho_{30} \, V ~=~ \rho_{10} \, V \]`

The volume \(V\) must be expressed with the sought radius \(r\) of the hot air balloon. It is assumed that the balloon is approximately *spherical*. Then the volume is: \( V ~=~ \frac{4}{3} \, \pi \, r^3 \). Substituting into 3

:

`4 \[ m ~+~ \rho_{30} \, \frac{4}{3} \, \pi \, r^3 ~=~ \rho_{10} \, \frac{4}{3} \, \pi \, r^3 \]`

Form equation 4

for the radius:

`5 \[ r ~=~ \sqrt[3]{ \frac{3m}{4\pi (\rho_{10} - \rho_{30} )} } \]`

The density of the air at 10 degrees Celsius is according to the table in the solution hint \( \rho_{10} ~=~ 1.2466 \, \frac{\text{kg}}{\text{m}^3} \). And at 30 degrees: \( \rho_{30} ~=~ 1.644 \, \frac{\text{kg}}{\text{m}^3} \). Substituting the given values into 5

gives the radius:
`6
\[ r ~=~ \sqrt[3]{ \frac{ 3 ~\cdot~ 400 \, \text{kg} }{4\pi \left( 1.2466 \, \frac{\text{kg}}{\text{m}^3} ~-~ 1.644 \, \frac{\text{kg}}{\text{m}^3} \right) }} ~=~ 10.51 \, \text{m} \]
`