ARCHIMEDES' PRINCIPLE: FORMULA, PROOF, APPLICATIONS - PHYSICAL - 2025

The Archimedes ' principle states that a body immersed wholly or partly, receives a vertical upward force called thrust, which is equivalent to the weight of the volume of liquid displaced by the body.

Some objects float in water, some sink, and some partially submerge. To sink a beach ball it is necessary to make an effort, because immediately that force is perceived that tries to return it to the surface. Instead a metal sphere sinks rapidly.

Figure 1. Floating balloons: Archimedes' principle in action. Source: Pixabay.

On the other hand, submerged objects seem lighter, therefore there is a force exerted by the fluid that opposes the weight. But it can't always fully compensate for gravity. And, although it is more evident with water, gases are also capable of producing this force on objects immersed in them.

History

Archimedes of Syracuse (287-212 BC) was the one who must have discovered this principle, being one of the greatest scientists in history. They say that King Hiero II of Syracuse ordered a goldsmith to make a new crown for him, for which he gave him a certain amount of gold.

Archimedes

When the king received the new crown, it was the correct weight, but he suspected that the goldsmith had deceived him by adding silver instead of gold. How could he prove it without destroying the crown?

Hiero called Archimedes, whose reputation as a scholar was well known, to help him solve the problem. The legend states that Archimedes was submerged in the bathtub when he found the answer and, such was his emotion, that he ran naked through the streets of Syracuse to search for the king, shouting "eureka", which means "I found him".

What did Archimedes find? Well, when taking a bath, the water level in the bathtub rose when he entered, which means that a submerged body displaces a certain volume of liquid.

And if he immersed the crown in water, this also had to displace a certain volume of water if the crown was made of gold and a different one if it was made of alloy with silver.

Formula

The lifting force referred to by Archimedes' principle is known as hydrostatic thrust or buoyant force and, as we have said, it is equal to the weight of the volume of fluid displaced by the body when submerged.

The displaced volume is equal to the volume of the object that is submerged, either totally or partially. Since the weight of anything is mg, and the mass of the fluid is density x volume, denoting the magnitude of the thrust as B, mathematically we have:

B = m _fluid xg = density of fluid x Submerged volume x gravity

B = ρ _fluid x V _submerged xg

Where the Greek letter ρ ("rho") denotes density.

Apparent weight

The weight of objects is calculated using the familiar mg expression, however things feel lighter when submerged in water.

The apparent weight of an object is what it has when it is immersed in water or another liquid and knowing it, the volume of an irregular object such as the crown of King Hiero can be obtained, as will be seen below.

To do this, it is completely submerged in water and attached to a string attached to a dynamometer - an instrument fitted with a spring used to measure forces. The greater the weight of the object, the greater the elongation of the spring, which is measured on a scale provided in the apparatus.

Figure 2. Apparent weight of a submerged object. Source: prepared by F. Zapata.

Applying Newton's second law knowing that the object is at rest:

ΣF _y = B + T - W = 0

The apparent weight W _a equals the tension in the string T:

Since the thrust compensates for the weight, since the fluid portion is at rest, then:

From this expression it follows that the thrust is due to the pressure difference between the upper face of the cylinder and the lower face. Since W = mg = ρ fluid. V. g, it has to:

Which is precisely the expression for the thrust mentioned in the previous section.

Applications

Archimedes' principle appears in many practical applications, among which we can name:

- The aerostatic balloon. Which, due to its average density less than that of the surrounding air, floats in it due to the thrust force.

- The ships. The hull of ships is heavier than water. But if the whole hull plus the air inside it is considered, the ratio between the total mass and the volume is less than that of the water and that is the reason why ships float.

- Life jackets. Being constructed of lightweight and porous materials, they are able to float because the mass-volume ratio is lower than that of water.

- The float to close the filling tap of a water tank. It is a large volume air-filled sphere that floats on top of the water, which causes the pushing force - multiplied by the lever effect - to close the cap of the filling tap of a water tank when it has reached the level. total.

Examples

Example 1

Legend has it that King Hiero gave the goldsmith a certain amount of gold to make a crown, but the distrustful monarch thought that the goldsmith may have cheated by placing a metal less valuable than gold inside the crown. But how could he know without destroying the crown?

The king entrusted the problem to Archimedes and this, seeking the solution, discovered his famous principle.

Suppose the corona weighs 2.10 kg-f in air and 1.95 kg-f when completely submerged in water. In this case, is there or is there no deception?

Figure 5. Free-body diagram of King Heron's crown. Source: prepared by F. Zapata

The diagram of the forces is shown in the figure above. These forces are: the weight P of the crown, the thrust E and the tension T of the rope hanging from the scale.

It is known P = 2.10 kg-f and T = 1.95 kg-f, it remains to determine the magnitude of the thrust E:

On the other hand, according to Archimedes' principle, the thrust E is equivalent to the weight of the water displaced from the space occupied by the crown, that is, the density of the water times the volume of the crown due to the acceleration of gravity:

From where the volume of the crown can be calculated:

The density of the crown is the quotient between the mass of the crown out of the water and its volume:

The density of pure gold can be determined by a similar procedure and the result is 19300 kg / m ^ 3.

Comparing the two densities it is evident that the crown is not pure gold!

Example 2

Based on the data and the result of example 1, it is possible to determine how much gold was stolen by the goldsmith in the case that part of the gold has been replaced by silver, which has a density of 10,500 kg / m ^ 3.

We will call the density of the crown ρc, ρo the density of gold and ρ _p the density of silver.

The total mass of the crown is:

M = ρc⋅V = ρo⋅Vo + ρ _p ⋅Vp

The total volume of the crown is the volume of silver plus the volume of gold:

V = Vo + Vp ⇒ Vp = V - Vo

Substituting in the equation for the mass is:

ρc⋅V = ρo⋅Vo + ρ _p ⋅ (V - Vo) ⇒ (ρo - ρ _p) Vo = (ρc - ρ _p) V

That is, the volume of gold Vo that contains the crown of total volume V is:

Vo = V⋅ (ρc - ρ _p) / (ρo - ρ _p) =…

… = 0.00015 m ^ 3 (14000 - 10500) / (19300 - 10500) = 0.00005966 m ^ 3

To find the weight in gold that the crown contains, we multiply Vo by the density of the gold:

Mo = 19300 * 0.00005966 = 1.1514 kg

Since the mass of the crown is 2.10 kg, we know that 0.94858 kg of gold was stolen by the goldsmith and replaced by silver.

Solved exercises

Exercise 1

A huge helium balloon is able to hold a person in balance (without going up or down).

Assume that the weight of the person, plus the basket, ropes, and balloon is 70 kg. What is the volume of helium required for this to occur? How big should the balloon be?

Solution

We will assume that the thrust is produced mainly by the volume of helium and that the thrust of the rest of the components is very small compared to that of helium which occupies much more volume.

In this case, it will require a volume of helium capable of providing a thrust of 70 kg + the weight of helium.

Figure 6. Free-body diagram of the helium-filled balloon. Source: prepared by F. Zapata.

Thrust is the product of the volume of helium times the density of helium and the acceleration of gravity. That push must offset the weight of the helium plus the weight of all the rest.

Da⋅V⋅g = Da⋅V⋅g + M⋅g

from which it is concluded that V = M / (Da - Dh)

V = 70 kg / (1.25 - 0.18) kg / m ^ 3 = 65.4 m ^ 3

That is, 65.4 m ^ 3 of helium is required at atmospheric pressure for there to be lift.

If we assume a spherical globe, we can find its radius from the relationship between the volume and the radius of a sphere:

V = (4/3) ⋅π⋅R ^ 3

From where R = 2.49 m. In other words, it will require a 5 m diameter balloon filled with helium.

Exercise 2

Materials with a lower density than water float in it. Suppose you have polystyrene (white cork), wood, and ice cubes. Their densities in kg per cubic meter are respectively: 20, 450 and 915.

Find what fraction of the total volume is outside the water and how high it stands above the surface of the water, taking 1000 kilograms per cubic meter as the density of the latter.

Solution

Buoyancy occurs when the weight of the body equals the thrust due to the water:

E = M⋅g

Figure 7. Free-body diagram of a partially submerged object. Source: prepared by F. Zapata.

Weight is the density of the body Dc multiplied by its volume V and by the acceleration of gravity g.

The thrust is the weight of the fluid displaced according to Archimedes 'principle and is calculated by multiplying the density D of the water by the submerged volume V' and by the acceleration of gravity.

That is:

D⋅V'⋅g = Dc⋅V⋅g

Which means that the submerged volume fraction is equal to the quotient between the density of the body and the density of the water.

That is, the outstanding volume fraction (V '' / V) is

If h is the overhang height and L the side of the cube the volume fraction can be written as

So the results for the ordered materials are:

Polystyrene (white cork):

(h / L) = (V '' / V) = 1 - (Dc / D) = 1- (20/1000) = 98% out of the water

Wood:

(h / L) = (V '' / V) = 1 - (Dc / D) = 1- (450/1000) = 55% out of the water

Ice:

(h / L) = (V '' / V) = 1 - (Dc / D) = 1- (915/1000) = 8.5% out of the water

References

1. Bauer, W. 2011. Physics for Engineering and Sciences. Volume 1. Mc Graw Hill. 417-455.
2. Cengel Y, Cimbala J. 2011. Fluid Mechanics. Fundamentals and applications. First edition. McGraw Hill.
3. Figueroa, D. (2005). Series: Physics for Science and Engineering. Volume 4. Fluids and Thermodynamics. Edited by Douglas Figueroa (USB). 1 - 42.
4. Giles, R. 2010. Mechanics of fluids and hydraulics. McGraw Hill.
5. Rex, A. 2011. Fundamentals of Physics. Pearson. 239-263.
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