For Example 2.23, if Archimedes placed the crown in a container of cross section 0.6 m by...

For Example 2.23, if Archimedes placed the crown in a container of cross section 0.6 m by 0.6 m part-way filled with water, how much would the water rise? If he removed the crown (without losing any water) and put in a brick of gold of the same mass as the crown, how much would the water level rise? Would he have been able to detect this difference with the measurement tools available to him?

EXAMPLE 2.23

The story (perhaps apocryphal) is told that the king of Syracuse paid a craftsman to make a gold crown. When the king received the crown, he was suspicious that the craftsman had not made it out of pure gold, but had substituted some other

FIGURE 2.26 Schematic for Archimedes’ solution

metal and merely gold-plated the crown and kept the extra gold for himself. But how could the king check without destroying the crown? If he cut it open and it turned out to be pure gold, he would be humiliated and would have destroyed his expensive crown. The crown was of intricate design, so the total volume of the crown could not be easily calculated by geometrical means.

SOLUTION Archimedes was called in to solve the problem. He set up a balance scale, shown in Figure 2.26, whereby he could measure the weight of the crown both in plain air and while the crown was submerged in water. He measured the weight of the crown in air to be 8.5 N. (It is doubtful he actually used units of Newtons, since Newton had not been born yet. Perhaps he used units like “stones.” But for the purposes of this problem we will use modern units.) He then submerged the crown in water and measured its weight to be 8.0 N. The weight in water is less due to the buoyant force pushing upward on the crown. Can you guess how he calculated the density of the material in the crown to figure out if it was pure gold or not?

First, the buoyant force will be equal to the difference between the regular weight of the crown and its weight when measured submerged in water. So FB = 8.5 N − 8.0 N = 0.5 N. Since we know from Equation 2.16 that the buoyant force is equal to the volume of the submerged object times the specific weight of water,

(2.16)

the volume of the crown can be calculated by ∀crown = FB/γH2O = 0.5 N/9800 N/m3 = 0.000052 m3. The specific weight of the crown can then be calculated by γcrown = W/∀ = 8.5 N/0.000052 m3 = 163,000 N/m3. Finally, the specific gravity of the crown is SG = 163,000 N/m3/9800 N/m3 = 16.7. Fortunately for Archimedes, gold is a very dense metal, so any filler metal the craftsman may have used would have a lower density than gold. The specific gravity of gold is 19.3 (which Archimedes could have obtained using the same apparatus and procedure described above if he did not already know the value). So the crown obviously was not pure gold, and the craftsman most likely came to an unhappy end.

In case you were wondering, there was a buoyant force on the crown when Archimedes weighed it in plain air, but the specific weight of air at sea level is only 11.5 N/m3, and so the buoyant force of air on the crown would only be FB = γair∀ = (11.5 N/m3)(0.000052 m3) = 0.001 N, which can be safely neglected. (Note: In one version of this story, Archimedes found a piece of gold that weighed the same as the crown and put it in a container of water and measured the water height, and then removed the gold and put the crown in the container and measured the water height again to see if they displaced the same amount. This type of measurement would not be very accurate—see Problem 39 at the end of the chapter.)

Problem 39

To isolate an experiment from the surroundings, a slab of concrete 1 m by 2 m by 2 m is placed in a large tub of mercury of diameter 3.5 m. How much will the level of mercury rise when the block is placed in it?