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

## Important Formula

What do the formula symbols mean?

## Buoyant Force

Unit
Force experienced by a body in the opposite direction to gravity when this body is immersed in a liquid (e.g. water). The body experiences a buoyancy.

## Density

Unit
Mass density of the liquid. It tells you how heavy one cubic meter of the liquid is.

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

## Volume

Unit
Volume of the liquid displaced by the immersed body. This is exactly the volume of the body immersed in the liquid.

## Gravitational acceleration

Unit
The gravitational acceleration is experienced by a body in the gravitational field of a planet. On the earth the gravitational acceleration is: $$g ~=~ 9.8 \, \frac{\text m}{\text{s}^2}$$.
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.
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### 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 dependence of air density.
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 $r ~=~ \sqrt[3]{ \frac{ 3 ~\cdot~ 400 \, \text{kg} }{4\pi \left( 1.2466 \, \frac{\text{kg}}{\text{m}^3} ~-~ 1.644 \, \frac{\text{kg}}{\text{m}^3} \right) }} ~=~ 10.51 \, \text{m}$