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# 12.4: Archimedes’ Law

Difficulty Level: At Grade Created by: CK-12

## Objectives

The student will:

• Understand buoyancy and how it applies to Archimedes’ Law.
• Be able to solve problems using Archimedes’ Law.

## Vocabulary

• buoyancy: The force that pushes upward on an object that is partially or wholly submerged in water.

## Introduction

Objects float in water or sink based on their density.  Buoyancy is the force that pushes upward on an object that is partially or wholly submerged in water.  This happens because the water lower down has higher pressure than the water around the top, so the bottom of the object is pushed up more than the top of the object is pushed down.

For an object of irregular shape, calculating the force as pressure times area, F=P×A\begin{align*}F = P \times A\end{align*}, could potentially be difficult.  However, there is a simple principle for how much force comes from buoyancy, attributed to the ancient Greek engineer and physicist Archimedes (287 BC – 212 BC).  Archimedes' principle states the following:

The force of buoyancy on an object immersed in liquid is equal to the weight of liquid displaced by that object.

Expressed mathematically, the buoyant force FB\begin{align*}F_B\end{align*} is equal to the weight of the liquid displaced, and we know that weight is mg\begin{align*}mg\end{align*}.  The mass of the liquid displaced is equal to mass times density, ρV\begin{align*}\rho V\end{align*}.  This gives us:

FB=mg=(ρliqV)g\begin{align*}F_B = mg = (\rho_{\text{liq}} V) g\end{align*}

This is phrased simply, but can be shown to have many immediate consequences.

• If an object is more dense than water, then water with equal volume to it has less mass.  If so, FB\begin{align*}F_B\end{align*} is less than the object's weight, and it will sink.
• If an object is less dense than water, then water with equal volume to it has more mass.  If so, FB\begin{align*}F_B\end{align*} is more than the object's weight, and it will float.

We can derive this by looking at a case of a simple cube floating upright in the water. The pressure of the water is always P=ρgh\begin{align*}P = \rho g h\end{align*}, where h\begin{align*}h\end{align*} is the depth in the water.  The cube has area A\begin{align*}A\end{align*} for all faces.  It is pushed down by the pressure on the top, Ftop=PtopA\begin{align*}F_{\text{top}} = P_{\text{top}} A\end{align*}, while is it pushed up by the higher pressure on the bottom, Fbottom=PbottomA\begin{align*}F_{\text{bottom}} = P_{\text{bottom}} A\end{align*}.  The difference between these is:

FbottomFtop=PbottomAPtopA=(ρghbottom)A(ρghtop)A=ρg (hbottomhtop)A=ρgV

Note that the area of a face times the height of the cube is equal to the volume of the cube V\begin{align*}V\end{align*}

Show that the buoyant force is equal to the weight of the fluid displaced.

Recall that the density is given by ρ=mV\begin{align*}\rho = \frac{m}{V}\end{align*}.

Substitute mV\begin{align*}\frac{m}{V}\end{align*} for ρ\begin{align*}\rho\end{align*} into the equation Fb=ρgV\begin{align*}F_b = \rho gV\end{align*}.

Fb=ρgV=(mV)gV=mgFb=mg\begin{align*}F_b = \rho gV = \left(\frac{m}{V} \right) gV = mg \rightarrow F_b = mg\end{align*}, where m\begin{align*}m\end{align*} is the mass of the fluid displaced and mg\begin{align*}mg\end{align*} is its weight.

#### Illustrative Example 1

a. A cube of iron with side length 25.0 cm is submerged in a tank of water with temperature 4C\begin{align*}4^\circ C\end{align*} such that the bottom of the cube is 1.00 m below the surface of the water Figure below. Calculate the buoyant force on the iron cube.

Answer: The volume of the cube is V=(0.25 m)3=0.0156250.0156 m3\begin{align*}V = (0.25 \ m)^3 = 0.015625 \rightarrow 0.0156 \ m^3\end{align*}. This is the same volume as the displaced water. Recall that water at a temperature of 4C\begin{align*}4^\circ C\end{align*} has a density of 1,000kgm3\begin{align*}1,000 \frac{kg}{m^3}\end{align*}. The buoyant force is therefore Fb=mg=ρVg=(1,000kgm3)(0.01563 m3)(9.81ms2)=153.3153 N\begin{align*}F_b = mg = \rho Vg = \left(1,000 \frac{kg}{m^3} \right)(0.01563 \ m^3) \left(9.81 \frac{m}{s^2} \right) = 153.3 \rightarrow 153 \ N\end{align*}.

b. What is the minimum force required to hold the cube in equilibrium while it remains in water? The density of iron is 7,860kgm3\begin{align*}7,860 \frac{kg}{m^3}\end{align*}.

Answer: We begin with the Free-Body-Diagram of Figure below.

Free-Body-Diagram.

As the Free Body-Diagram in Figure above shows, the normal force equals the difference of the buoyant force and the weight of the cube.

F=0FN+Fbmg=0FN=mgFb\begin{align*}\sum F = 0 \rightarrow F_N+F_b-mg = 0 \rightarrow F_N = mg-F_b\end{align*}.

A force equal to the normal force FN\begin{align*}F_N\end{align*} applied to the cube will keep the cube in equilibrium anywhere beneath the water surface. The weight of the iron block is

mirongFN=ρironVg=(7,860kgm3)(0.01563 m3)(9.81ms2)=1,205.21,205 N=mgFb=1,205153=1,052 N.

The links below demonstrate Archimedes Principle.

### Floating Objects

Archimedes’ principle is no different for objects that are only partially submerged. The weight of the volume of fluid displaced is still equal to the buoyant force. But since the object is in equilibrium while floating, the upward force on the object is the buoyant force.

#### Illustrative Example 2

a. The density of yellow pine is 420kgm3\begin{align*}420 \frac{kg}{m^3}\end{align*}. What volume of water at a temperature of 4C\begin{align*}4^\circ C\end{align*} does a cylindrical log of pine with dimensions r=15.0 cm\begin{align*}r = 15.0 \ cm\end{align*} and length L=7.60 m\begin{align*}L = 7.60 \ m\end{align*} displace in order to float? See Figure below.

Answer: The weight of the log mlogg\begin{align*}m_{log} g\end{align*} must equal the buoyant force (the weight of displaced water) mwaterg\begin{align*}m_{water} g\end{align*}.

mloggVwater=mwatergmlog=mwaterρlogVlog=ρwaterVwater=ρlogρwaterVlog

The volume of the log is Vlog=(πr2)L(3.14)(0.150 m)2(7.60 m)=0.5369 m30.537 m3\begin{align*}V_{log} = (\pi r^2)L \rightarrow (3.14)(0.150 \ m)^2(7.60 \ m) = 0.5369 \ m^3 \rightarrow 0.537 \ m^3\end{align*}.

Vwater=ρlogρwaterVlog=420kgm31000kgm3(0.537 m3)=0.22550.226 m3\begin{align*}V_{water} = \frac{\rho_{log}}{\rho_{water}}V_{log} = \frac{420 \frac{kg}{m^3}}{1000 \frac{kg}{m^3}}(0.537 \ m^3) = 0.2255 \rightarrow 0.226 \ m^3\end{align*}

The atmosphere is also able to exert a buoyant force, otherwise hot air balloons would not rise.

True or False: Steel cannot float because it is denser than water.

Answer: False. Large ships are usually made of steel and other comparatively dense materials. It is true that a solid block of material with a density greater than water will sink. But if the block is molded into the shape of a boat (hollowed out), it will be able to displace a greater volume of water than when it was a solid block. The result is that the hollowed block will be able to displace a volume of water equal to its weight before sinking. It will float.

#### Illustrative Example 3

Let’s return to King Hiero’s problem now that we have an understanding of Archimedes' principle. Recall that the king commissioned the goldsmith to craft a crown of gold from a specific amount of gold given to him by the king. The king then became suspicious that the goldsmith had replaced some of the gold with a cheaper metal, perhaps silver. The problem was that the finished crown weighed the same as the gold given to the goldsmith. How could Archimedes determine if the crown was pure gold without damaging it?

There is no way of knowing the exact amount of gold given to the goldsmith, nor is this particularly relevant. We seek only a method whereby Archimedes could have made a conclusive statement regarding the honesty of the goldsmith. There are several methods he could have used, none of which require any calculation. The most accurate was given by Galileo, which we will use.

To make the problem interesting, we’ll choose a 500 g mass of gold. Furthermore, we’ll assume that the goldsmith kept 20% (100 g) of the gold for himself and replaced it with an equal mass of silver. Note that this method assumes that the volumes of gold and silver are additive.

a. Calculate the volume of 100 grams of gold. Gold has a density of 19.3gcm3\begin{align*}19.3 \frac{g}{cm^3}\end{align*}.

ρ=mVV=mρ=100 g19.3gcm3=5.1815.18 cm3\begin{align*}\rho = \frac{m}{V} \rightarrow V = \frac{m}{\rho} = \frac{100 \ g}{19.3 \frac{g}{cm^3}} = 5.181 \rightarrow 5.18 \ cm^3\end{align*}.

b. Calculate the volume of 100 grams of silver. Silver has a density of 10.5gcm3\begin{align*}10.5 \frac{g}{cm^3}\end{align*}.

ρ=mVV=mρ=100 g10.5gcm3=9.5249.52 cm3\begin{align*}\rho = \frac{m}{V} \rightarrow V = \frac{m}{\rho} = \frac{100 \ g}{10.5 \frac{g}{cm^3}} = 9.524 \rightarrow 9.52 \ cm^3\end{align*}

c. What is the volume of 500 g of gold and 400 grams of gold alloyed with 100 grams of silver?

The method below is one of many.

Volume of 500 grams of gold=5.181 cm3100 g(500 g)=25.90525.9 cm3\begin{align*}\text{Volume of 500 grams of gold} = \frac{5.181 \ cm^3}{100 \ g}(500 \ g) = 25.905 \rightarrow 25.9 \ cm^3\end{align*}

Volume of 400 grams of gold + 100 grams of silver=5.181 cm3100 g(400 g)+9.52 cm3100 g(100 g)=30.24430.2 cm3

It seems clear the crown has a larger volume than the original volume of the gold given to the goldsmith. If a scale balance in air had the crown on one side and the original amount of gold on the other, it would balance since the weights are equal. But if the balance and its load were submerged in water, the buoyant force on the crown would be greater since it displaces a greater volume of water. And since the buoyant force is greater on the crown, the scale would sense a smaller weight on the side with the crown, and so this side of the scale would rise.

It has often been said that Archimedes filled a tank of water to its brim and placed the crown in it. He then either measured the overflow or the amount the water level dropped in the tank compared to when it was full. He repeated the experiment for an identical quantity of gold given to the goldsmith. The problem with this method is that unless the original quantity of gold was substantial, the difference in volume would be difficult to measure. The difference for the problem above is only 4.3 cm3\begin{align*}4.3 \ cm^3\end{align*}. Galileo reasoned that it would be more accurate to simply see if the scale became unbalanced. See if you can show that the buoyant force on the crown is 1.1661.17\begin{align*}1.166 \rightarrow 1.17\end{align*} times greater than the buoyant force on the original volume of gold given the goldsmith.

## Date Created:

Mar 11, 2013

Feb 06, 2015
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