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Archimedes Eureka!

The story of the discovery of "Archimedes' Principle" may be apochryphal
The Fields metal for outstanding achievment in mathematics carries a bust of Archimedes[1]
[2]. Archimedes was purportedly given the task of determining whether King Hiero's goldsmith was embezzling gold by replacing the valuable metal with other metals in an alloy usd to create a laurel wreath dedicated to the gods[3].

Archimedes knew that alloys could be distinguished by their densities, but he could not determine the density of the wreath because its irregular shape did not allow an easy calculation of the volume.

As the story goes, the frustrated Archimedes took a relaxing bath, and noticed the rise of water in the tub as he entered it. He realized that the increase in water height would provide a means of measuring the volume of his body. Allegedly, upon this discovery, he went running naked though the streets shouting, εὑρηκα! εὑρηκα! ("Eureka! Eureka!" --Greek "I found it"). As a result, the term "eureka" entered common parlance and is used today to indicate a moment of surprised discovery.

The drawing is based on the correlations of ideal human proportions with geometry described by the ancient Roman architect Vitruvius in Book III of his treatise De Architectura. Vitruvius described the human figure as being the principal source of proportion among the Classical orders of architecture. Other artists had attempted to depict the concept, with less success. The drawing is traditionally named in honour of the architect.
. This story is recorded in de Architectura, Book IX, written by Marcus Vitruvius Pollio (born c. 80–70 BC, died after c. 15 BC)two centuries after Archimedes experiment allegedly took place. The original text and modern translation are available on the web, and the relevant section is quoted below.
11. After this, he is said to have taken two masses, each of a weight equal to that of the crown, one of them of gold and the other of silver. Having prepared them, he filled a large vase with water up to the brim, wherein he placed the mass of silver, which caused as much water to run out as was equal to the bulk thereof. The mass being then taken out, he poured in by measure as much water as was required to fill the vase once more to the brim. By these means he found out what quantity of water was equal to a certain weight of silver.

12. He then placed the mass of gold in the vessel, and, on taking it out, found that the water which ran over was lessened, because, as the magnitude of the gold mass was smaller than that containing the same weight of silver. After again filling the vase by measure, he put the crown itself in, and discovered that more water ran over then than with the mass of gold that was equal to it in weight; and thus, from the superfluous quantity of water carried over the brim by the immersion of the crown, more than that displaced by the mass, he found, by calculation, the quantity of silver mixed with the gold, and made manifest the fraud of the manufacturer.

Some scholars have doubted the accuracy of this tale, saying among other things that the method would have required precise measurements that would have been difficult to make at the time. [3][4]If the report is accurate, we can replicate the experiment to see if it is likely that Archimedes could have done it.

Earlier we showed how to use unity factors can be used to express quantities in different units of the same parameter. For example, a density can be expressed in g/cm3 or lb/ft3. Now we will also how conversion factors representing mathematical functions, like D = m/v, can be used to transform quantities into different parameters. For example, what is the volume of a given mass of gold? Unity factors and conversion factors are conceptually different, and we'll see that the "dimensional analysis" we develop for unit conversion problems must be used with care in the case of functions.

When we are referring to the same object or sample of material, it is often useful to be able to convert one parameter into another. Conversion of one kind of quantity into another is usually done with what can be called a conversion factor, but the conversion factor is based on a mathematical function (D = m / V) or mathematical equation that relates parameters.

Suppose Archimedes submerged the laurel wreath and found that it took 428 cm3 to replace the water displaced from the vase. Knowing the density, we can calculate the mass of gold, then find the volume of 14K gold (which contains other metals and has a density of ) that would have the same mass. Is this volume difference large enough to be determined by Archimedes? It's likely that he could compare masses accurately, so the volume measurement will be the critical one.

To determine the mass of gold equivalent to a volume of 428 cm3. This can be done by manipulating Eq. (1.1) which defines density. If we multiply both sides by V, we obtain

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

m = V × ρ or mass = volume × density       (1.2)

Density and Composition of Gold and alloys
Metal ‰ Gold content Density /g/cm3
24K Gold 99.0% Minimum allowed 19.3
22K Gold 91.6 15.6
18K Gold 75% Au, 16% Ag and 9% Cu [1]
14K Gold 58.5% Au,4.0% Ag, 31.2% Cu, and 6.3% Zn12.9 - 14.6
Silver 10.49

Taking the density of gold from the Table, we can now calculate the mass:

$\text{Mass}=\text{m}=\text{V}\rho =\text{428 cm}^{3}\times \frac{\text{10}\text{.32 g}}{\text{1 cm}^{3}}=8.27\times \text{10}^{3}\text{g}=\text{8}\text{.27 kg}$

This is more than 18 lb of gold. At todays prices, it would be worth over 40 000 dollars!

The formula which defines density can also be used to convert the mass of a sample to the corresponding volume. If both sides of Eq. (1.2) are multiplied by 1/ρ, we have

\begin{align} & \frac{\text{1}}{\rho }\times \text{m}=\text{V}\rho \times \frac{\text{1}}{\rho }=\text{V} \\ & \text{ V}=\text{m}\times \frac{\text{1}}{\rho } \\ \end{align}       (1.3)

So we can calculate the volume of 14K gold (assuming a density of ρ = 14.0) with the same mass as 428 cm3 of pure gold (8.27 kg):

$\text{ V}=\text{m}\times \frac{\text{1}}{\rho }$ = $\frac {8270 g}{14.0 \frac {g}{cm^3}} = 591 cm^3$.

The difference (591-428= 163 g or 38% greater than pure gold) seems easily measurable. Obviously, if the laurel wreath was much smaller, or if the goldsmith had been less greedy and alloyed the gold with a smaller mass of silver, it might have gone undetected.

A simpler method is shown in the figure.

The Archimedes principle may have been used to determine whether the golden crown was less dense than gold. If the laurel wreath (left) is less dense than the reference weight (right), its larger volume will displace more water and thus experience a larger upward buoyant force, causing it to weigh less in the water [4]

Notice that we used the mathematical function D = m/V to convert parameters from mass to volume or vice versa in these examples. How does this differ from the use of unity factors to change units of one parameter?

An Important Caveat

A mistake sometimes made by beginning students is to confuse density with concentration, which also may have units of g/cm3. By dimensional analysis, this looks perfectly fine. To see the error, we must understand the meaning of the function

C = $\frac{m}{V}$.

In this case, V refers to the volume of a solution, which contains both a solute and solvent.

Given a concentration of an alloy is 10g gold in 100 cm3 of alloy, we see that it is wrong (although dimensionally correct as far as conversion factors go) to incorrectly calculate the volume of gold in 20 g of the alloy as follows:

20 g     x     $\frac{100 cm^3}{10 g}$     = 200 cm3

It is only possible to calculate the volume of gold if the density of the alloy is known, so that the volume of alloy represented by the 20 g could be calculated. This volume multiplied by the concentration gives the mass of gold, which then can be converted to a volume with the density function.

The bottom line is that using a simple unit cancellation method does not always lead to the expected results, unless the mathematical function on which the conversion factor is based is fully understood.

EXAMPLE

Gold can be extracted from low grade gold ore by the "cyanide process". A gold ore with a concentration of 0.060 g / cm3 has a density of 8.25 g / cm3. What is the volume of gold (D = 19.32 g / cm3)in 100 cm3 of the ore?

The volume of 100 g of ore is

V = m / D = 100 g /8.25 g cm-3 = 12.12 cm3.

The mass of gold in this volume is

m = V x C = 12.12 cm3 x 0.060 g / cm3 = 0.727 g.

The volume of gold = m / D = 0.727 g / 19.32 g / cm3 = 0.0376 cm3.

Note that we cannot calculate the volume of gold by

$\frac {\frac{8.25 g}{cm^3} x 100 cm^3}{\frac {19.32 g}{cm^3}}$ = 42.7 cm3

even though this is dimensionally correct.

Note that this result required when to use the function C = m/V, and when to use the function D=m/V as conversion factors. Pure dimensional analysis could not reliably give the answer, since both functions have the same dimensions.

EXAMPLE 1.10 Find the volume occupied by a 4.73-g sample of benzene.

According to Table 1.4, the density of benzene is 0.880 g cm–3. Using Eq. (1.3),

$\text{Volume = }V\text{ = }m\text{ }\times \text{ }\frac{\text{1}}{\rho }\text{ = 4}\text{.73 g }\times \text{ }\frac{\text{1 cm}^{\text{3}}}{\text{0}\text{.880 g}}\text{ = 5}\text{.38 cm}^{\text{3}}$

(Note that taking the reciprocal of $\tfrac{\text{0}\text{.880 g}}{\text{1 cm}^{3}}$ simply inverts the fraction ― 1 cm3 goes on top, and 0.880 g goes on the bottom.)

The two calculations just done show that density is a conversion factor which changes volume to mass, and the reciprocal of density is a conversion factor changing mass into volume. This can be done because of the mathematical formula, Eq. (1.1), which relates density, mass, and volume. Algebraic manipulation of this formula gave us expressions for mass and for volume [Eq. (1.2) and (l.3)], and we used them to solve our problems. If we understand the function D = m/V and heed the caveat above, we can devise appropriate converstion factors by unit cancellation, as the following example shows:

EXAMPLE 1.11 A student weights 98.0 g of mercury. If the density of mercury is 13.6 g/cm3, what volume does the sample occupy?

We know that volume is related to mass through density.

Therefore

V = m × conversion factor

Since the mass is in grams, we need to get rid of these units and replace them with volume units. This can be done if the reciprocal of the density is used as a conversion factor. This puts grams in the denominator so that these units cancel:

$V=m\times \frac{\text{1}}{\rho }=\text{98}\text{.0 g}\times \frac{\text{1 cm}^{3}}{\text{13}\text{.6 g}}=\text{7}\text{.21 cm}^{3}$

If we had multiplied by the density instead of its reciprocal, the units of the result would immediately show our error:

$V=\text{98}\text{.0 g}\times \frac{\text{13,6 }g}{\text{1 cm}^{3}}=\text{1}\text{.333}{\text{g}^{2}}/{\text{cm}^{3}}\;$ (no cancellation!)

It is clear that square grams per cubic centimeter are not the units we want.

Using a conversion factor is very similar to using a unity factor—we know the factor is correct when units cancel appropriately. A conversion factor is not unity, however. Rather it is a physical quantity (or the reciprocal of a physical quantity) which is related to the two other quantities we are interconverting. The conversion factor works because of that relationship [Eqs. (1.1), (1.2), and (1.3) in the case of density, mass, and volume], not because it is equal to one. Once we have established that a relationship exists, it is no longer necessary to memorize a mathematical formula. The units tell us whether to use the conversion factor or its reciprocal. Without such a relationship, however, mere cancellation of units does not guarantee that we are doing the right thing.

A simple way to remember relationships among quantities and conversion factors is a “road map“of the type shown below:

$\text{Mass }\overset{density}{\longleftrightarrow}\text{ volume or }m\overset{\rho }{\longleftrightarrow}V\text{ }$

This indicates that the mass of a particular sample of matter is related to its volume (and the volume to its mass) through the conversion factor, density. The double arrow indicates that a conversion may be made in either direction, provided the units of the conversion factor cancel those of the quantity which was known initially. In general the road map can be written

$\text{First quantity }\overset{\text{conversion factor}}{\longleftrightarrow}\text{ second quantity}$

As we come to more complicated problems, where several steps are required to obtain a final result, such road maps will become more useful in charting a path to the solution.

EXAMPLE 1.12 Black ironwood has a density of 67.24 lb/ft3. If you had a sample whose volume was 47.3 ml, how many grams would it weigh? (1 lb = 454 g; 1 ft = 30.5 cm).

$V\xrightarrow{\rho }m\text{ }$

tells us that the mass of the sample may be obtained from its volume using the conversion factor, density. Since milliliters and cubic centimeters are the same, we use the SI units for our calculation:

Mass = m = 47.3 cm3 × $\frac{\text{67}\text{.24 lb}}{\text{1 ft}^{3}}$

Since the volume units are different, we need a unity factor to get them to cancel:

$m\text{ = 47}\text{.3 cm}^{\text{3}}\text{ }\times \text{ }\left( \frac{\text{1 ft}}{\text{30}\text{.5 cm}} \right)^{\text{3}}\text{ }\times \text{ }\frac{\text{67}\text{.24 lb}}{\text{1 ft}^{\text{3}}}\text{ = 47}\text{.3 cm}^{\text{3}}\text{ }\times \text{ }\frac{\text{1 ft}^{\text{3}}}{\text{30}\text{.5}^{\text{3}}\text{ cm}^{\text{3}}}\text{ }\times \text{ }\frac{\text{67}\text{.24 lb}}{\text{1 ft}^{\text{3}}}$

We now have the mass in pounds, but we want it in grams, so another unity factor is needed:

$m\text{ = 47}\text{.3 cm}^{\text{3}}\text{ }\times \text{ }\frac{\text{1 ft}^{\text{3}}}{\text{30}\text{.5}^{\text{3}}\text{ cm}^{\text{3}}}\text{ }\times \text{ }\frac{\text{67}\text{.24 lb}}{\text{1 ft}^{\text{3}}}\text{ }\times \text{ }\frac{\text{454 g}}{\text{ 1 lb}}\text{ = 50}\text{.9 g}$

In subsequent chapters we will establish a number of relationships among physical quantities. Formulas will be given which define these relationships, but we do not advocate slavish memorization and manipulation of those formulas. Instead we recommend that you remember that a relationship exists, perhaps in terms of a road map, and then adjust the quantities involved so that the units cancel appropriately. Such an approach has the advantage that you can solve a wide variety of problems by using the same technique.