# Archimedes' Principle

### Learning Objectives

By the end of this section, you will be able to:- Define buoyant force.
- State Archimedes’ principle.
- Understand why objects float or sink.
- Understand the relationship between density and Archimedes’ principle.

*buoyant force*on any object in any fluid. (See Figure 2.) If the buoyant force is greater than the object’s weight, the object will rise to the surface and float. If the buoyant force is less than the object’s weight, the object will sink. If the buoyant force equals the object’s weight, the object will remain suspended at that depth. The buoyant force is always present whether the object floats, sinks, or is suspended in a fluid.

**Buoyant Force**

The buoyant force is the net upward force on any object in any fluid.

*w*

_{fl}. This weight is supported by the surrounding fluid, and so the buoyant force must equal

*w*

_{fl}, the weight of the fluid displaced by the object. It is a tribute to the genius of the Greek mathematician and inventor Archimedes (ca. 287–212 B.C.) that he stated this principle long before concepts of force were well established. Stated in words,

*Archimedes’ principle*is as follows: The buoyant force on an object equals the weight of the fluid it displaces. In equation form, Archimedes’ principle is

*F*_{B }= *w*_{fl},

*F*

_{B}is the buoyant force and

*w*

_{fl}is the weight of the fluid displaced by the object. Archimedes’ principle is valid in general, for any object in any fluid, whether partially or totally submerged.

**Archimedes’ Principle**

According to this principle the buoyant force on an object equals the weight of the fluid it displaces. In equation form, Archimedes’ principle is

*F*_{B }= *w*_{fl},

*F*_{B}is the buoyant force and*w*_{fl}is the weight of the fluid displaced by the object.*Humm …*High-tech body swimsuits were introduced in 2008 in preparation for the Beijing Olympics. One concern (and international rule) was that these suits should not provide any buoyancy advantage. How do you think that this rule could be verified?

**Making Connections: Take-Home Investigation**

The density of aluminum foil is 2.7 times the density of water. Take a piece of foil, roll it up into a ball and drop it into water. Does it sink? Why or why not? Can you make it sink?

## Floating and Sinking

Substance | $\rho \left({\text{10}}^{3}{\text{kg/m}}^{3}\text{or}\text{g/mL}\right)\\$ |
Substance | $\rho \left({\text{10}}^{3}{\text{kg/m}}^{3}\text{or}\text{g/mL}\right)\\$ |
Substance | $\rho \left({\text{10}}^{3}{\text{kg/m}}^{3}\text{or}\text{g/mL}\right)\\$ |
---|---|---|---|---|---|

Solids |
Liquids |
Gases |
|||

Aluminum | 2.7 | Water (4ºC) | 1.000 | Air | 1.29 × 10^{−3} |

Brass | 8.44 | Blood | 1.05 | Carbon dioxide | 1.98 × 10^{−3} |

Copper (average) | 8.8 | Sea water | 1.025 | Carbon monoxide | 1.25 × 10^{−3} |

Gold | 19.32 | Mercury | 13.6 | Hydrogen | 0.090 × 10^{−3} |

Iron or steel | 7.8 | Ethyl alcohol | 0.79 | Helium | 0.18 × 10^{−3} |

Lead | 11.3 | Petrol | 0.68 | Methane | 0.72 × 10^{−3} |

Polystyrene | 0.10 | Glycerin | 1.26 | Nitrogen | 1.25 × 10^{−3} |

Tungsten | 19.30 | Olive oil | 0.92 | Nitrous oxide | 1.98 × 10^{−3} |

Uranium | 18.70 | Oxygen | 1.43 × 10^{−3} |
||

Concrete | 2.30–3.0 | Steam (100º C) | 0.60 × 10^{−3} |
||

Cork | 0.24 | ||||

Glass, common (average) | 2.6 | ||||

Granite | 2.7 | ||||

Earth’s crust | 3.3 | ||||

Wood | 0.3–0.9 | ||||

Ice (0°C) | 0.917 | ||||

Bone | 1.7–2.0 |

### Example 1. Calculating buoyant force: dependency on shape

(a) Calculate the buoyant force on 10,000 metric tons (1.00 × 10

To find the buoyant force, we must find the weight of water displaced. We can do this by using the densities of water and steel given in Table 1. We note that, since the steel is completely submerged, its volume and the water’s volume are the same. Once we know the volume of water, we can find its mass and weight.

First, we use the definition of density

Because the steel is completely submerged, this is also the volume of water displaced,

By Archimedes’ principle, the weight of water displaced is

Here we are given the maximum volume of water the steel boat can displace. The buoyant force is the weight of this volume of water.

The mass of water displaced is found from its relationship to density and volume, both of which are known. That is,

The maximum buoyant force is ten times the weight of the steel, meaning the ship can carry a load nine times its own weight without sinking.

^{7}kg) of solid steel completely submerged in water, and compare this with the steel’s weight. (b) What is the maximum buoyant force that water could exert on this same steel if it were shaped into a boat that could displace 1.00 × 10^{5}m^{3}of water?**Strategy for (a)**

To find the buoyant force, we must find the weight of water displaced. We can do this by using the densities of water and steel given in Table 1. We note that, since the steel is completely submerged, its volume and the water’s volume are the same. Once we know the volume of water, we can find its mass and weight.**Solution for (a)**

First, we use the definition of density $\rho =\frac{m}{V}\\$

to find the steel’s volume, and then we substitute values for mass and density. This gives${V}_{\text{st}}=\frac{{m}_{\text{st}}}{{\rho }_{\text{st}}}=\frac{1\text{.}\text{00}\times {\text{10}}^{7}\text{ kg}}{7\text{.}8\times {\text{10}}^{3}{\text{ kg/m}}^{3}}=1\text{.}\text{28}\times {\text{10}}^{3}{\text{m}}^{3}\\$

.*V*_{w}. We can now find the mass of water displaced from the relationship between its volume and density, both of which are known. This gives$\begin{array}{lll}{m}_{w}& =& {\rho }_{w}{V}_{w}=\left(\text{1.000}\times {\text{10}}^{3}{\text{kg/m}}^{3}\right)\left(1.28\times {\text{10}}^{3}{m}^{3}\right)\\ & =& \text{1.28}\times {\text{10}}^{6}\text{kg.}\end{array}\\$

*m*_{w}*g*, so the buoyant force is$\begin{array}{lll}{F}_{B}& =& {w}_{w}={m}_{w}g=\left(\text{1.28}\times {\text{10}}^{6}\text{kg}\right)\left(9.80{\text{m/s}}^{2}\right)\\ & =& 1.3\times {\text{10}}^{7}\text{N}\end{array}\\$

.
The steel’s weight is ${m}_{\text{w}}g=9\text{.}\text{80}\times {\text{10}}^{7}\text{N}\\$

, which is much greater than the buoyant force, so the steel will remain submerged. Note that the buoyant force is rounded to two digits because the density of steel is given to only two digits.**Strategy for (b)**

Here we are given the maximum volume of water the steel boat can displace. The buoyant force is the weight of this volume of water.**Solution for (b)**

The mass of water displaced is found from its relationship to density and volume, both of which are known. That is,$\begin{array}{lll}{m}_{w}& =& {\rho }_{w}{V}_{w}=\left(\text{1.000}\times {\text{10}}^{3}{\text{kg/m}}^{3}\right)\left(\text{1.00}\times {\text{10}}^{5}{m}^{3}\right)\\ & =& \text{1.00}\times {\text{10}}^{8}\text{kg}\end{array}\\$

.
The maximum buoyant force is the weight of this much water, or$\begin{array}{lll}{F}_{B}& =& {w}_{w}={m}_{w}g=\left(\text{1.00}\times {\text{10}}^{8}\text{kg}\right)\left(\text{9.80}{\text{m/s}}^{2}\right)\\ & =& \text{9.80}\times {\text{10}}^{8}\text{N}\end{array}\\$

.
**Discussion**

The maximum buoyant force is ten times the weight of the steel, meaning the ship can carry a load nine times its own weight without sinking.**Making Connections: Take-Home Investigation**

A piece of household aluminum foil is 0.016 mm thick. Use a piece of foil that measures 10 cm by 15 cm. (a) What is the mass of this amount of foil? (b) If the foil is folded to give it four sides, and paper clips or washers are added to this “boat,” what shape of the boat would allow it to hold the most “cargo” when placed in water? Test your prediction.

## Density and Archimedes’ Principle

$\text{fraction submerged =}\frac{{V}_{\text{sub}}}{{V}_{\text{obj}}}=\frac{{V}_{\text{fl}}}{{V}_{\text{obj}}}\\$

.
The volume submerged equals the volume of fluid displaced, which we call *V*

_{fl}. Now we can obtain the relationship between the densities by substituting

$\rho =\frac{m}{V}\\$

into the expression. This gives$\frac{{V}_{\text{fl}}}{{V}_{\text{obj}}}=\frac{{m}_{\text{fl}}/{\rho }_{\text{fl}}}{{m}_{\text{obj}}/{\overline{\rho }}_{\text{obj}}}$

,${\overline{\rho }}_{\text{obj}}\\$

is the average density of the object and *ρ*

_{fl}is the density of the fluid. Since the object floats, its mass and that of the displaced fluid are equal, and so they cancel from the equation, leaving

$\text{fraction submerged}=\frac{{\overline{\rho }}_{\text{obj}}}{{\rho }_{\text{fl}}}\\$

.*specific gravity*:

$\text{specific gravity}=\frac{\overline{\rho }}{{\rho }_{\text{w}}}\\$

,$\overline{\rho }\\$

is the average density of the object or substance and *ρ*

_{w}is the density of water at 4.00°C. Specific gravity is dimensionless, independent of whatever units are used for

*ρ*. If an object floats, its specific gravity is less than one. If it sinks, its specific gravity is greater than one. Moreover, the fraction of a floating object that is submerged equals its specific gravity. If an object’s specific gravity is exactly 1, then it will remain suspended in the fluid, neither sinking nor floating. Scuba divers try to obtain this state so that they can hover in the water. We measure the specific gravity of fluids, such as battery acid, radiator fluid, and urine, as an indicator of their condition. One device for measuring specific gravity is shown in Figure 5.

**Specific Gravity**

Specific gravity is the ratio of the density of an object to a fluid (usually water).

### Example 2. Calculating Average Density: Floating Woman

Suppose a 60.0-kg woman floats in freshwater with 97.0% of her volume submerged when her lungs are full of air. What is her average density?

We can find the woman’s density by solving the equation

Entering the known values into the expression for her density, we obtain

Her density is less than the fluid density. We expect this because she floats. Body density is one indicator of a person’s percent body fat, of interest in medical diagnostics and athletic training. (See Figure 6.)

**Strategy**

We can find the woman’s density by solving the equation$\text{fraction submerged}=\frac{{\overline{\rho }}_{\text{obj}}}{{\rho }_{\text{fl}}}\\$

for the density of the object. This yields${\overline{\rho }}_{\text{obj}}={\overline{\rho }}_{\text{person}}=\left(\text{fraction submerged}\right)\cdot {\rho }_{\text{fl}}\\$

.
We know both the fraction submerged and the density of water, and so we can calculate the woman’s density.**Solution**

Entering the known values into the expression for her density, we obtain${\overline{rho }}_{\text{person}}=0\text{.}\text{970}\cdot \left({\text{10}}^{3}\frac{\text{kg}}{{\text{m}}^{3}}\right)=\text{970}\frac{\text{kg}}{{\text{m}}^{3}}\\$

.
**Discussion**

Her density is less than the fluid density. We expect this because she floats. Body density is one indicator of a person’s percent body fat, of interest in medical diagnostics and athletic training. (See Figure 6.)## More Density Measurements

*appears*to weigh less when submerged; we call this measurement the object’s

*apparent weight*. The object suffers an

*apparent weight loss*equal to the weight of the fluid displaced. Alternatively, on balances that measure mass, the object suffers an

*apparent mass loss*equal to the mass of fluid displaced. That is

apparent weight loss = weight of fluid displaced

orapparent mass loss = mass of fluid displaced.

The next example illustrates the use of this technique.### Example 3. Calculating Density: Is the Coin Authentic?

The mass of an ancient Greek coin is determined in air to be 8.630 g. When the coin is submerged in water as shown in Figure 7, its apparent mass is 7.800 g. Calculate its density, given that water has a density of 1.000 g/cm

To calculate the coin’s density, we need its mass (which is given) and its volume. The volume of the coin equals the volume of water displaced. The volume of water displaced

The volume of water is

You can see from Table 1 that this density is very close to that of pure silver, appropriate for this type of ancient coin. Most modern counterfeits are not pure silver.

^{3}and that effects caused by the wire suspending the coin are negligible.**Strategy**

To calculate the coin’s density, we need its mass (which is given) and its volume. The volume of the coin equals the volume of water displaced. The volume of water displaced *V*_{w}can be found by solving the equation for density$\rho =\frac{m}{V}\\$

for *V*.**Solution**

The volume of water is ${V}_{\text{w}}=\frac{{m}_{\text{w}}}{{\rho }_{\text{w}}}\\$

where *m*_{w}is the mass of water displaced. As noted, the mass of the water displaced equals the apparent mass loss, which is*m*_{w }= 8.630 g − 7.800 g = 0.830 g. Thus the volume of water is${V}_{\text{w}}=\frac{0\text{.}\text{830 g}}{1\text{.}\text{000 g}{\text{/cm}}^{3}}=0\text{.}\text{830}{\text{cm}}^{3}\\$

. This is also the volume of the coin, since it is completely submerged. We can now find the density of the coin using the definition of density:${\rho }_{\text{c}}=\frac{{m}_{\text{c}}}{{V}_{c}}=\frac{8\text{.}\text{630 g}}{0\text{.830 c}{\text{m}}^{3}}=10.4\text{ g/cm}^{3}\\$

.
**Discussion**

You can see from Table 1 that this density is very close to that of pure silver, appropriate for this type of ancient coin. Most modern counterfeits are not pure silver.**PhET Explorations: Buoyancy**

When will objects float and when will they sink? Learn how buoyancy works with blocks. Arrows show the applied forces, and you can modify the properties of the blocks and the fluid.

## Section Summary

- Buoyant force is the net upward force on any object in any fluid. If the buoyant force is greater than the object’s weight, the object will rise to the surface and float. If the buoyant force is less than the object’s weight, the object will sink. If the buoyant force equals the object’s weight, the object will remain suspended at that depth. The buoyant force is always present whether the object floats, sinks, or is suspended in a fluid.
- Archimedes’ principle states that the buoyant force on an object equals the weight of the fluid it displaces.
- Specific gravity is the ratio of the density of an object to a fluid (usually water).

### Conceptual Questions

1. More force is required to pull the plug in a full bathtub than when it is empty. Does this contradict Archimedes’ principle? Explain your answer.

2. Do fluids exert buoyant forces in a "weightless" environment, such as in the space shuttle? Explain your answer.

3. Will the same ship float higher in salt water than in freshwater? Explain your answer.

4. Marbles dropped into a partially filled bathtub sink to the bottom. Part of their weight is supported by buoyant force, yet the downward force on the bottom of the tub increases by exactly the weight of the marbles. Explain why.

### Problem & Exercises

1. What fraction of ice is submerged when it floats in freshwater, given the density of water at 0°C is very close to 1000 kg/m

^{3}?
2. Logs sometimes float vertically in a lake because one end has become water-logged and denser than the other. What is the average density of a uniform-diameter log that floats with 20.0% of its length above water?

3. Find the density of a fluid in which a hydrometer having a density of 0.750 g/mL floats with 92.0% of its volume submerged.

4. If your body has a density of 995 kg/m

^{3}, what fraction of you will be submerged when floating gently in: (a) Freshwater? (b) Salt water, which has a density of 1027 kg/m^{3}?
5. Bird bones have air pockets in them to reduce their weight—this also gives them an average density significantly less than that of the bones of other animals. Suppose an ornithologist weighs a bird bone in air and in water and finds its mass is 45.0 g and its apparent mass when submerged is 3.60 g (the bone is watertight). (a) What mass of water is displaced? (b) What is the volume of the bone? (c) What is its average density?

6. A rock with a mass of 540 g in air is found to have an apparent mass of 342 g when submerged in water. (a) What mass of water is displaced? (b) What is the volume of the rock? (c) What is its average density? Is this consistent with the value for granite?

7. Archimedes’ principle can be used to calculate the density of a fluid as well as that of a solid. Suppose a chunk of iron with a mass of 390.0 g in air is found to have an apparent mass of 350.5 g when completely submerged in an unknown liquid. (a) What mass of fluid does the iron displace? (b) What is the volume of iron, using its density as given in Table 1. (c) Calculate the fluid’s density and identify it.

8. In an immersion measurement of a woman’s density, she is found to have a mass of 62.0 kg in air and an apparent mass of 0.0850 kg when completely submerged with lungs empty. (a) What mass of water does she displace? (b) What is her volume? (c) Calculate her density. (d) If her lung capacity is 1.75 L, is she able to float without treading water with her lungs filled with air?

9. Some fish have a density slightly less than that of water and must exert a force (swim) to stay submerged. What force must an 85.0-kg grouper exert to stay submerged in salt water if its body density is 1015 kg/m

^{3}?
10. (a) Calculate the buoyant force on a 2.00-L helium balloon. (b) Given the mass of the rubber in the balloon is 1.50 g, what is the net vertical force on the balloon if it is let go? You can neglect the volume of the rubber.

11. (a) What is the density of a woman who floats in freshwater with 4.00% of her volume above the surface? This could be measured by placing her in a tank with marks on the side to measure how much water she displaces when floating and when held under water (briefly). (b) What percent of her volume is above the surface when she floats in seawater?

12. A certain man has a mass of 80 kg and a density of 955 kg/m

^{3 }(excluding the air in his lungs). (a) Calculate his volume. (b) Find the buoyant force air exerts on him. (c) What is the ratio of the buoyant force to his weight?
13. A simple compass can be made by placing a small bar magnet on a cork floating in water. (a) What fraction of a plain cork will be submerged when floating in water? (b) If the cork has a mass of 10.0 g and a 20.0-g magnet is placed on it, what fraction of the cork will be submerged? (c) Will the bar magnet and cork float in ethyl alcohol?

14. What fraction of an iron anchor’s weight will be supported by buoyant force when submerged in saltwater?

15. Scurrilous con artists have been known to represent gold-plated tungsten ingots as pure gold and sell them to the greedy at prices much below gold value but deservedly far above the cost of tungsten. With what accuracy must you be able to measure the mass of such an ingot in and out of water to tell that it is almost pure tungsten rather than pure gold?

16. A twin-sized air mattress used for camping has dimensions of 100 cm by 200 cm by 15 cm when blown up. The weight of the mattress is 2 kg. How heavy a person could the air mattress hold if it is placed in freshwater?

17. Referring to Figure 3, prove that the buoyant force on the cylinder is equal to the weight of the fluid displaced (Archimedes’ principle). You may assume that the buoyant force is

*F*_{1}-*F*_{2}and that the ends of the cylinder have equal areas*A*. Note that the volume of the cylinder (and that of the fluid it displaces) )A equals (*h*_{2}-*h*_{1})*A*.
18. (a) A 75.0-kg man floats in freshwater with 3.00% of his volume above water when his lungs are empty, and 5.00% of his volume above water when his lungs are full. Calculate the volume of air he inhales—called his lung capacity—in liters. (b) Does this lung volume seem reasonable?

## Glossary

- Archimedes’ principle:
- the buoyant force on an object equals the weight of the fluid it displaces

- buoyant force:
- the net upward force on any object in any fluid

- specific gravity:
- the ratio of the density of an object to a fluid (usually water)

### Selected Solutions to Problems & Exercises

1. 91.7%3. 815 kg/m

^{3}5. (a) 41.4 g (b) 41.4cm

^{3 }(c) 1.09 g/cm

^{3}7. (a) 39.5 g (b) 50cm

^{3 }(c) 0.79g/cm

^{3}

It is ethyl alcohol.

9. 8.21 N11. (a) 960kg/m

^{3 }(b) 6.34%

She indeed floats more in seawater.

13. (a) 0.24 (b) 0.68 (c) Yes, the cork will float because${\rho }_{\text{obj}}<{\rho }_{\text{ethyl alcohol}}\left(0\text{.}\text{678}{\text{g/cm}}^{3}<0\text{.}\text{79}{\text{g/cm}}^{3}\right)\\$

15. The difference is 0.006%.

17.

${F}_{\text{net}}={F}_{2}-{F}_{1}={P}_{2}A-{P}_{1}A=\left({P}_{2}-{P}_{1}\right)A\\$

$=\left({h}_{2}{\rho }_{\text{fl}}g-{h}_{1}{\rho }_{\text{fl}}g\right)A\\$

$=\left({h}_{2}-{h}_{1}\right){\rho }_{\text{fl}}\text{gA}\\$

where

${\rho }_{\text{fl}}\\$

= density of fluid. Therefore,${F}_{\text{net}}=\left({h}_{2}-{h}_{1}\right){\mathrm{A\rho }}_{\text{fl}}g={V}_{\text{fl}}{\rho }_{\text{fl}}g={m}_{\text{fl}}g={w}_{\text{fl}}\\$

where is

${w}_{\text{fl}}\\$

the weight of the fluid displaced.### Licenses and Attributions

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