He **Archimedes’ principle** says that a totally or partially submerged body receives an upward vertical force called *push*which is equal to the weight of the volume of fluid displaced by the body.

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

On the other hand, submerged objects appear 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.

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**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 him a new crown, for which he gave him a certain amount of gold.

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 check without destroying the crown?

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

https://giphy.com/gifs/StiTO3EChtlNBvLIZ3

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

And if he immersed the crown in water, it 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 silver alloy.

## Archimedes principle formula

The buoyant force referred to by Archimedes’ principle is known as push hydrostatic either buoyant force and, as we have said, it is equivalent to the weight of the volume of fluid displaced by the body when it is 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 mgand the mass of the fluid is density x volumedenoting as B the magnitude of the push, mathematically it is had that:

B=mfluent xg = fluid density x Submerged volume x gravity

B = ρfluid x Vsubmerged xg

Where the Greek letter ρ (“rho”) denotes density.

### apparent weight

The weight of objects is calculated using the well-known expression *mg*however things feel lighter when submerged in water.

He apparent weight of an object is the one it has when it is immersed in water or another liquid and knowing it, it is possible to obtain the volume of an irregular object like King Hiero’s crown, as will be seen below.

To do this, it is completely submerged in water and attached to a rope attached to a *dynamometer* -an instrument provided with a spring that serves 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 device.

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

ΣFand =B+T–W=0

The apparent weight Wa is equal to the tension in the rope T:

*T = Wa*

Wa = mg – ρfluid . V.g

If the immersed volume V is required, it is solved as:

V = (W – Wa ) / ρfluid . g

## Demonstration

https://giphy.com/gifs/mCPHppgtNPbHl4Cgaq

When a body is submerged, the push is the resultant force of all the forces that are exerted on the body through the pressure caused by the surrounding fluid:

### pressure and depth

Since pressure increases with depth, the resultant of these forces is always directed vertically upward. Therefore, Archimedes’ principle is a consequence of the fundamental theorem of hydrostatics, which relates the pressure P exerted by a fluid with the depth z as:

P = ρ.gz

### Forces on a fluid in static equilibrium

To demonstrate Archimedes’ principle, a small cylindrical portion of fluid is taken at rest to analyze the forces exerted on it, as shown in the following figure. The forces on the curved surface of the cylinder cancel each other.

The magnitudes of the vertical forces are *F**1* = *P1*.A and *F2* = *P2*.A, there is also the weight W. Since the fluid is in equilibrium, the sum of the forces must vanish:

*∑Fy = P2.A- P1.A- W = 0*

*P2.A- P1.A= W*

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

*B = P2.A- P1.A = W*

From this expression it follows that the thrust is due to the difference in pressure between the upper face of the cylinder and the lower face. As W = mg = ρfluent. V.gyou have to:

*B = ρ**fluent. Vsubmerged. g*

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

## Applications of Archimedes’ principle

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

– The aerostatic balloon. Which, because it has a lower average density than that of the surrounding air, floats on it due to the buoyant force.

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

– Life jackets. Being made of light and porous materials, they are able to float because the mass-volume ratio is less 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 water, which causes the pushing force – multiplied by the lever effect – to close the filler tap cap 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 might 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 he, looking for the solution, discovered his famous principle.

Suppose the crown 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 not deceit?

The force diagram is shown in the figure above. These forces are: the weight* P* of the crown, the thrust

*and the tension*

**AND***the rope hanging from the scale.*

**you**P = 2.10 kg-f and T = 1.95 kg-f are known, it remains to determine the magnitude of the thrust * AND*:

*T + E = P ⇒ E = P – T = (2.10 – 1.95) kg-f = 0.15 kg-f*

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 times the acceleration of gravity:

*E = ρwater⋅V⋅g = 1000 kg/m^3 ⋅ V ⋅ 9.8m/s^2 = 0.15 kg ⋅ 9.8 m/s^2*

From where the volume of the crown can be calculated:

*V = 0.15 kg / 1000 kg/m^3 = 0.00015 m^3*

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

*Crown density = 2.10 kg / 0.00015 m^3 = 14000 kg/m^3*

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 It’s 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 10500 kg/m^3.

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

The total mass of the crown is:

M = ρc⋅V = ρo⋅Vo + ρp⋅Vp

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

V = Vo + Vp ⇒ Vp = V – Vo

Substituting into the mass equation is:

ρc⋅V = ρo⋅Vo + ρp⋅(V – Vo) ⇒ (ρo – ρp)Vo = (ρc – ρp)V

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

Vo = V⋅(ρc – ρp)/(ρo – ρp) = …

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

To know the weight in gold that the crown contains, we multiply Vo by the density of 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 capable of holding 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 needed for this to happen? 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, a volume of helium capable of providing a thrust of 70 kg + the weight of the helium will be required.

Thrust is the product of the volume of helium times the density of the helium times the acceleration of gravity. That thrust must offset the weight of the helium plus the weight of everything else.

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 at atmospheric pressure are required 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, a 5 m diameter balloon filled with helium will be required.

**Exercise 2**

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

Find what fraction of the total volume remains outside the water and what height it protrudes from the surface of the water, taking the density of the latter to be 1000 kilograms per cubic meter.

**Solution **

Buoyancy occurs when the weight of the body is equal to the push due to the water:

E = M⋅g

The weight is the density of the body Dc multiplied by its volume V and by the acceleration due to gravity g.

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

That is:

D⋅V’⋅g = Dc⋅V⋅g

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

*(V’/V) = (Dc/D) *

That is, the protruding volume fraction (V»/V) is

*(V»/V) = 1 –…*