What is the buoyant force or buoyant force? Archimedes’ principle

The buoyant force, buoyant force or buoyant force, is a force that points in the opposite direction to gravity and that acts on any solid that is partially or totally submerged in a fluid, be it a liquid or a gas. This force was first discovered and characterized by the Greek mathematician, physicist and engineer Archimedes in the 3rd century BC and, as the story goes, was the cause of the famous cry of Eureka ! that so characterizes the aforementioned Hellenic scholar.

Although they do not have the same origin, we can think of the buoyant force as the normal force exerted by liquids and other fluids on the bodies with which they come into contact.

Eureka! and Archimedes’ Principle

According to the account of the Roman architect Vitruvius, the buoyant force was discovered by Archimedes while he was in the bathtub. Archimedes had been commissioned by King Hieron of Syracuse to determine if the crown he had commissioned from his goldsmiths was made of pure gold, or if, on the contrary, he had been deceived by combining gold with silver or some other less valuable metal.

Apparently, Archimedes thought a lot about how to solve this problem without being able to find the solution, until one day, while he was getting into a bathtub, he noticed that, when he immersed himself in the water, his body displaced part of the liquid, causing it to fall down the drain. edge. Then he came up with what we know today as Archimedes’ Principle: when submerging a body in water (or any other liquid), it will feel a push force that will reduce its weight by an amount equivalent to the volume of water displaced.

The difference between the original weight of the body and the weight of the body submerged in water corresponds to the buoyant force or buoyant force. In equation form, Archimedes’ principle can be written as:

Where B represents the buoyant force (in some texts it is represented as F _B ) and W _f corresponds to the weight of the fluid displaced by the submerged body.

Archimedes knew that gold was a heavier (denser) metal than any other metal that goldsmiths could use to make the crown, so if the crown were made of solid pure gold, it would have to displace the same mass of water as any other solid gold object of equal mass, so the apparent weight or weight reduced by buoyant force should be the same for the crown and the control object.

On the other hand, if gold were mixed with silver or another metal, then, being less dense, it should displace a greater volume (and therefore a weight) of water, thus obtaining a lower apparent weight than that of the control object ( since the buoyant force will be greater).

According to Vitruvius’s account, Archimedes was so excited by the solution to the problem that he ran out of his bath through the streets of Syracuse towards the king’s palace shouting Eureka! Eureka! (which translates to “Got it! Got it!”) without even realizing that he was completely naked.

Explanation of Archimedes’ Principle

Archimedes’ Principle can be easily explained in terms of Newton’s laws. The form of the Archimedean Principle equation shown above proves that the buoyant force is independent of the characteristics of the submerged object since it only depends on the mass of the fluid (not the object) displaced. That is, it does not depend on the composition, density or shape of the body.

So, the buoyant force felt by, for example, a cube of wood, must be the same as that felt by a cube made of the same fluid. Now, if we imagine a cube made of the same fluid and that is submerged, like the one shown in the following figure, it is evident that it will be in mechanical equilibrium with the liquid that surrounds it (otherwise we would see streams of water spontaneously form in any glass of water). According to Newton’s first law, the only way for a body to be in mechanical equilibrium (that is, at rest or moving at constant velocity) is if no net force acts on it. This can only happen if there is no force acting on the body or if all the forces acting on it cancel each other (their vector sum is zero).

Since we know that the block of fluid has mass, it must then feel the force of gravity, so the only way for it to be in equilibrium is for some other force that pushes it in the opposite direction to be acting on the block. This force must be the buoyant force that Archimedes proposed.

So, since the only two forces acting on our imaginary block of fluid are its weight and the buoyant force, these must have the same magnitude and be directed in opposite directions, so the buoyant force on the block of fluid is equal to its weight and points up. Now, since this force is independent of the characteristics of the object, if we replace the block of fluid with a block of the same shape and size of some other material, the buoyant force that the new block will feel must be exactly the same as that felt by the new block. block of fluid that we had to remove to make room for the second block to be put in its place, and this force is equal to the weight of this displaced fluid.

Origin of the buoyant force

The buoyant force is generated due to the increase in hydrostatic pressure as we are submerged in a fluid. This is because moving down in a fluid increases the height (and therefore the mass) of the column of fluid above us, so the pressure increases roughly linearly with depth (for at least in the case of non-compressible fluids).

The pressure is the force per unit area and it is applied perpendicular to the contact surface between the body and the fluid. This means that each section of the surface of a submerged body is feeling a pressure that tries to crush it from all directions. As we will see below, this crushing force is greater in the lower part of a submerged body than in the part closest to the surface.

To see how this generates the buoyant force, consider the following figure showing a cubic-shaped block submerged in any fluid. To simplify the analysis, we will assume that the top and bottom caps are parallel to the water surface (that is, they are perpendicular to the vertical) and that the four side caps are perpendicular to the first.

Since the pressure exerts a force perpendicular to the surface, there will be six different resultant forces pushing one on each of the six faces of the cube. Since the lateral faces are vertical, the forces resulting from the pressure on them will be parallel to the surface of the liquid and therefore do not contribute to the buoyant force which must be vertical (as we saw above). So we only need to consider the forces on the top and bottom cap. The pressure on the upper face pushes the body down, while the pressure on the lower face pushes up.

Now, when comparing the pressure on the upper face, we can verify that it is at a lower depth than the lower face. Since pressure is proportional to depth, then the pressure on the top face must be less than the pressure felt by the bottom face. Finally, since both faces have the same area, then the relative force exerted by the pressure on both faces will only depend on the pressure and we conclude that the body feels a greater push force from below than from above. The vector sum of these two forces gives a resultant that points upwards and that corresponds to the buoyant force.

Despite the fact that the analysis was done on a body with a very simple shape, this same reasoning can be extrapolated to any body with any shape.

Where does the buoyant force act?

As we have just seen, the buoyant force is actually the result of the pressure exerted on the surface of a submerged body. However, just as weight is the sum of the attractive force felt by each particle that makes up a body and, even so, we can represent weight by means of a single vector that acts on the center of gravity, the same we can do with the buoyant force.

But where do we place this force?

The answer is found again from Newton’s laws. The mechanical equilibrium of a body floating at rest on a liquid implies not only that the net force is zero, but also that there is no torque or twisting force, since the body is not rotating. As a consequence, the buoyant force must not only counteract the weight so that the body does not accelerate up or down, but it must also act on the same line of action of the weight. For this reason, we can assume that the buoyant force also acts on the center of mass.

Buoyant force formulas

Although the basic equation of the buoyant force is the one proposed by Archimedes, it can be manipulated in different ways to obtain other more useful expressions.

First of all, by Newton’s Second Law, we know that the weight of the displaced fluid is equal to its mass times the acceleration due to gravity (W=mg). Furthermore, we also know that mass is related to volume through density. The combination of these formulas with the previous one gives the following results:

Where m _f represents the mass of the displaced fluid, g is the acceleration due to gravity, ρ _f is the density of the fluid, and V _f is the volume of the displaced fluid.

In addition, we can also express the buoyant force as a function of the apparent weight of a body immersed in a fluid:

Where W _real is the real weight of the submerged body that is approximately equal to its weight in air while W _apparent is the reduced weight that we would feel when trying to lift the body when it is submerged.

On the other hand, equation 3 can also be expressed as a function of the volume of the submerged body, since the displaced volume of the fluid must be equal to the volume of the fraction of the body that is submerged. This gives rise to two different cases:

Buoyant force on totally submerged bodies

If a body of volume V _o is totally submerged, then the displaced volume of the liquid will be equal to the volume of the body. Thus, equation 3 remains:

Buoyant force on partially submerged bodies

If, on the contrary, only a fraction of the body is submerged, then the volume of fluid displaced will be equal to the part of the volume of the body that is submerged ( V _s ):

Formula for floating bodies

Finally, we have the special case in which a body floats on the surface of a fluid, supported only by the buoyant force. In this case, we can say that the apparent weight of the body is zero and that therefore the buoyant force is exactly equal to the actual weight of the body (a conclusion that we could also have reached by a simple analysis of forces on a diagram). free body). In this case, only part of the body’s volume is submerged, so Equation 5 also applies.

So, combining this with the body weight formulas, we can arrive at the following equation:

where ρ _c is the density of the body and the other variables are the same as before. This equation makes it possible to easily find the submerged fraction of any floating body from the relationship between its density and that of the fluid in which it floats.

Examples of calculations with the buoyant force

Example 1: Icebergs or ice floes

The expression “just the tip of the iceberg” refers to the fact that the part of an iceberg that we can see above the surface of the water is only a small fraction of the total mass of the iceberg. But how much exactly is this fraction? We can calculate this from equation 6. The additional information we need is that the density of ice at 0 °C is 0.920 g/mL and that of seawater is approximately 1.025 g/mL since it is about salty, cold water that is denser than pure water.

Data:

ρ _c = 0.920 g/mL

ρ _f = 1.025 g/mL

Fraction of ice that protrudes = ?

Solution:

From equation 7 we have that:

Remember that this is the fraction of the volume of a floating body that is submerged, so this result indicates that 89.76% of the volume of the iceberg is under water. At the same time, it implies that only 10.24% is what we see on the surface.

Example 2: The crown of Hieron

Suppose Archimedes takes King Hieron’s crown and weighs it in the air, thus obtaining a weight of 7.45 N. He then ties the crown to a thin thread and immerses it in water (density 1.00 g/ mL) while recording the weight with a scale that now reads 6.86 N. Knowing that the density of gold is 19.30 g/mL and that of silver is 10.49 g/mL, will the goldsmith have tricked King Hieron?

Data:

Wactual = 7.45 N

Wapparent = 6.86 N

ρ _f = 1.00 g/mL

ρ _gold = 19.30 g/mL

ρ _silver = 10.49 g/mL

ρ _crown = ?

Solution:

Density is an intensive and characteristic property of a substance, so to answer the question at hand, what we need to do is determine the density of the corona. If the crown is made of solid gold, it should have the same density of gold. Otherwise, and if the material is mixed with silver, the crown will have a much lower density.

On the other hand, we have the real weight and apparent weight. Furthermore, we know that the crown is completely submerged in the water when the apparent weight is determined, so we can use equations 4 and 5. These can also be combined with the equations for the actual weight as a function of the volume of the body and its density. .

Let’s start by determining the buoyant force:

Then, since the crown is completely submerged, we have that the buoyant force is equal to:

This equation can be combined with the crown density equation and the weight equation obtained from Newton’s second law:

In order to obtain the following equation:

Then, solving the equation to find the density of the crown, we have:

Considering that the density of gold is 19.30 g/mL, it is evident that the King has been fooled. Either the crown is hollow, or it is not made of pure gold.

Example 3: A partially submerged cube

A cube with a volume of 2.0 cm ³ is submerged halfway in water. What is the buoyant force experienced by the cube?

Data

V ₀ = 2.0 cm ³

V _s = ½ V ₀

ρ _f = 1.00 g/mL

B = ?

Solution:

We have the density of the fluid because we know that it is water and that the density of water is 1.00 g/cm ³ . In addition, they give us the volume of the cube, as well as the fraction of it that is submerged, so we can apply equation 5 directly. However, we must consider that, since we are calculating a force, if we want the result of in N, we must carry out some unit conversions:

Therefore, the buoyant force will be 0.0098 N.

Example 4: An unknown cube

A cube with a volume of 2.0 cm3 ^floats on water, leaving a quarter of its volume above the surface. What is the density of the cube?

Data:

V ₀ = 2.0 cm ³

V _{above the surface} = ¼ V ₀

ρ _f = 1.00 g/mL

ρ _cube = ?

Solution:

Again, we have the density of the fluid because we know it is water. In this case they provide us with the fraction of the volume that protrudes, but the one we need is the one that is submerged, which is, therefore, ¾ of V ₀ . Finally, they tell us that the cube floats freely, so we can directly apply equation 6:

Thus, we then know that the cube has a density of 0.750 g/cm ³ .

References

Franco Garcia, A. (sf). Archimedes’ principle . Physics with computer. http://www.sc.ehu.es/sbweb/fisica/fluidos/estatica/arquimedes/arquimedes.htm

González Sánchez, JA (sf). Buoyant Force and Archimedes’ Principle . PhysicsPR. https://physicspr.com/buyont.html

Jewett, JW, & Serway, RA (2006). Physics for Sciences and Engineering – Volume I. Thomson International.

Khan Academy. (nd). What is the buoyant force? https://en.khanacademy.org/science/physics/fluids/buoyant-force-and-archimedes-principle/a/buoyant-force-and-archimedes-principle-article

Organs of Palencia. (2021, December 23). How to determine the buoyant force? https://organosdepalencia.com/biblioteca/articulo/read/16377-como-determinar-la-fuerza-boyante

Ross, R. (2017, April 26). Eureka! The Archimedean Principle . Livescience. Com. https://www.livescience.com/58839-archimedes-principle.html

Zaragoza Palacios, BG (nd). GENERAL PHYSICS . University of Sonora. http://paginas.fisica.uson.mx/beatriz.zaragoza/archivos/05a-fisicageneral.pdf