Once upon a time, a man named Archimedes was tasked with figuring out how to tell if a crown was made of gold or if it was a fake - without ruining it. Because of the crown's odd shape, he didn't know its volume to tell how dense it was. One day, Archimedes took a bath and noticed that the bathwater rose according to how much of his body was in the water. The volume of the submerged part of his body was the same volume as the water that rose out of the way - or, in other words, was displaced. He realized that he could determine the crown's volume this way and compare its weight to the same volume of pure gold to see if they weighed the same. This idea struck him so much that he ran naked through the town shouting, "Eureka!" This lightbulb moment led to Archimedes' Principle.
Archimedes' Principle Definition
Whether the above story happened like that or not is controversial. No matter how the idea came to be, we need to precisely define what we mean by Archimedes' principle before we can dig into it.
Archimedes' principle states that the upward buoyant force on a fully or partially submerged object equals the weight of the fluid the object displaces.
We'll discuss why this is true both intuitively and mathematically.
Intuitive Explanation
If we submerge a cube made of a weightless plastic filled with water, such as the one on the left in the image below, it will float in equilibrium with the surrounding water because all the water has the same density. The forces acting on the cube are the downward force of gravity and the upward buoyant force. Because the cube isn't accelerating, due to Newton's Second Law, \(\sum F = ma\), these forces added together equal zero. This means the buoyant force equals the weight of the water in the cube.
What if we switched the cube to a metal cube of the same size, like the rightmost image above? The water around the cube wouldn't know it was any different than the water-filled cube, so the buoyant force acting on it would equal the weight of water the cube could contain. But now the cube's weight is greater, so it would fall to the bottom of the glass. If you picked the cube up off the bottom, it would feel lighter than it really is because of the buoyant force pushing upwards on it.
Mathematical Explanation
When we immerse an object in a fluid, the fluid exerts pressure on all sides of the object. In the image below, we submerge a cube in water.
We simplified the forces due to the pressure of the water into a single downward force and a single upward force. The horizontal pressures acting on the cube are equal and opposite, so they sum to equal zero, and we exclude them from the image.
The pressure is equal to force per unit area, or
$$P=\frac{F}{A}.$$
We also know that the pressure at a point in a fluid is equal to the density of the fluid multiplied by the gravitational acceleration and the height of the fluid above the point:
$$P=\rho_\text{f} gh.$$
This is why the force acting on the bottom of the cube is greater than the force on the top - because pressure increases as the depth increases. Combining these two equations, the equation for the force acting on the top of the cube would be as follows:
$$F_1=\rho_\text{f} gh_1 A,$$
and the force acting on the bottom of the cube would be:
$$F_2=\rho_\text{f} g h_2 A.$$
To find the buoyant force, we want to find the difference between the force acting on the top and the force acting on the bottom:
$$F_2 - F_1 = \rho_\text{f} g (h_2 - h_1)A.$$
Notice that \(h_2-h_1\) is just the height of the cube, and by multiplying it by the face of the cube, \(A\), we get the volume of the cube, or rather, the volume of water that the cube displaced. Now we get the following equation for buoyant force:
$$F_\text{b} = \rho_\text{f} V_\text{f} g$$
Mass is equal to density times volume,
$$m=\rho V,$$ so we can substitute the mass of the liquid to replace the density and volume of the liquid:
$$F_b=m_\text{f} g$$
Since weight is equal to mass times gravity, this result means that the buoyant force is equal to the weight of the displaced fluid, just as Archimedes said.
Pressure increases as depth in the liquid increases, but that doesn't mean that the buoyant force increases. The object's height stays the same, so the pressure difference between the top and bottom of the object stays constant no matter how deep the object is in the fluid. The buoyant force depends only on the weight of the liquid displaced and the gravitational acceleration, not on the object's depth.
Archimedes' Principle - Formula/Equation
As was just proven above, Archimedes' principle results in the following formula for buoyancy:
$$F_\text{b}=m_\text{f} g.$$
You can also use the following equation, substituting the mass for density times volume, as we described above:
$$F_b=\rho V g.$$
Both of these equations mean the same thing; the one you use depends on what information you have. One crucial point is that you use the mass, density, or volume of the fluid, not of the object.
This is the most important thing to remember about buoyancy and where most mistakes occur. The fluid volume is not always the same as the object's volume.
Archimedes' Principle and the Law of Flotation
What if our cube floats? If we know the object is fully submerged, then we know the volume of the fluid that the object displaces is the same as the object's volume. But if it floats, this isn't the case. This is why it's important to remember that the volume you use is that of the fluid displaced by the liquid, not the object's volume.
When an object floats in a fluid, the only forces acting on it are the buoyant and gravitational forces. We can see the two forces acting on the floating cube in the image above. Since a floating object isn't accelerating, according to Newton's Second Law, the sum of the two forces equals zero. This means that for floating objects, the buoyant force (weight of the displaced fluid) equals the gravitational force (or weight) of the object. We refer to this as the law of flotation.
Archimedes' Principle Examples
Below are some examples to demonstrate the principles discussed above.
Let's look at our submerged cube from above. It sinks to the bottom of the water. If each side is \(0.25\,\mathrm{m}\) long, it weighs \(16\,\mathrm{kg}\), and the density of water is \(1000\,\mathrm{kg/m^3}\), what is the buoyant force acting on the cube?
Using the second equation for buoyant force, we can plug in the density of the water, the volume of the water displaced by the cube (which in this case is the same as the volume of the cube since we know the cube is fully submerged), and the gravitational acceleration:
\begin{align}F_\text{b} &= \rho V g \\&= (1000\,\mathrm{kg/m^3})(0.25\,\mathrm{m})^3(9.81\,\mathrm{m/s^2}) \\&= 153\,\mathrm{N}\end{align}
We can compare this number to the gravitational force, or weight, of the cube to make sure it's fully submerged:\begin{align}F_g &= (16\,\mathrm{kg})(9.81\,\mathrm{m/s^2}) \\&= 157\,\mathrm{N}\end{align}
Since the gravitational force is greater than the buoyant force, the cube is fully submerged, which suggests we used the correct volume.
Next, let's consider a floating cube.
Let's say our same cube from the example above weighs \(13\,\mathrm{kg}\) instead of \(16\,\mathrm{kg}\). This causes the cube to float, but we don't know how much of it sticks out of the water. What percentage of the cube is below the water?
We can write the buoyant force equation we used above, but this time we can't use the same cube volume since we don't know how deep the cube is submerged. We will split the volume into the area of the bottom of the cube, \(A\), which we know, multiplied by our unknown height, \(h\):
$$F_\text{b} = \rho (Ah) g$$
We can also set the buoyant force equal to the weight of the object (the mass of the object, \(m_\mathrm{o}\), times the gravitational acceleration):
$$F_\text{b} = m_\mathrm{o} g.$$
We'll substitute the second equation into the first, so we can solve for our unknown height:
\begin{align}m_\mathrm{o}g &= \rho (Ah) g \\h &= \frac{m_\mathrm{o}}{\rho A} \\h &= \frac{13\,\mathrm{kg}}{(1000\,\mathrm{kg/m^3})(0.25\,\mathrm{m})^2}\end{align}
Now we have our height of the cube that is submerged:
$$h=0.208\,\mathrm{m}$$
To know how much of the cube is submerged, we can create a ratio between the volume under the water and the total volume. We'll use a \(\mathrm{w}\) subscript for the variable in the water, and we'll use a \(\mathrm{t}\) subscript for the total cube variables:
$$\frac{V_\mathrm{w}}{V_\mathrm{t}}=\frac{Ah_\mathrm{w}}{Ah_\mathrm{t}}$$
The areas cancel out since they are the same, so we can plug in the values for the heights:
\begin{align}\frac{V_\mathrm{w}}{V_\mathrm{t}} &= \frac{0.208\,\mathrm{m}}{0.25\,\mathrm{m}} =0.83.\end{align}
The cube is 83% submerged in the water.
Archimedes' Principle Applications
Archimedes' principle is important in many engineering designs. Some applications of Archimedes' principle are the following.
- Understanding Archimedes' principle allows engineers to design ships that float even though they are made of heavy materials.
- Hydrometers use Archimedes' principle to determine the density of fluids.
- Submarines use Archimedes' principle to control how they rise and dive within the water.
- Archimedes' principle allows engineers to design life jackets that can keep a human body afloat.
Archimedes' Principle Conclusion
Archimedes' principle is an intuitive and useful tool when dealing with physics problems involving buoyancy. In addition, it allows us to find the volume that would otherwise be very difficult to analyze. Knowing the volume and mass, we can then get the density of objects and analyze their material properties as well.
Archimedes Principle - Key takeaways
- Archimedes' principle states that the upward buoyant force on an object is equal to the weight of the fluid that the object displaced (\(F_\text{b} = m_\text{f} g\)).
- When finding the buoyant force, always use the mass, or density and volume, of the fluid, rather than of the object.
- When an object floats in a fluid with no other external forces, the buoyant force equals the object's weight.
- Archimedes' principle is relevant to many engineering designs related to density and floating.
References
- Fig. 1 - Submerged cubes, StudySmarter Originals.
- Fig. 2 - Pressure around a submerged cube, StudySmarter Originals.
- Fig. 3 - Floating cube, StudySmarter Originals.
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