Archimedes' principle is a law that explains buoyancy or upthrust. It states that

When a body is completely or partially immersed in a fluid it experiences an upthrust, or an apparent loss in weight, which is equal to the weight of fluid displaced.According to a tale, Archimedes discovered this law while taking a bath. After making this discovery, he is said to have leapt out of his bathtub and ran through the streets of Syracuse naked shouting "Eureka!".

An object experiences upthrust due to the fact that the pressure exerted by a fluid on the lower surface of a body being greater than that on the top surface, since pressure increses with depth. Pressure,

*p*is given by

*p*=

*hρg,*where:

*h*is the height of the fluid column

*ρ*(rho) is the density of the fluid

*g*is the acceleration due to gravity

Let us confirm this principle theoretically. On the figure on the left, a solid block is immersed completely in a fluid with density

*ρ.*The difference in the force exerted,

*d*on the top and bottom surfaces with area

*a*is due to the difference in pressure, given by

*d*=

*h*

_{2}

*aρg*–

*h*

_{1}

*aρg*= (

*h*

_{2}–

*h*

_{1})

*aρg*

But (

*h*

_{2}–

*h*

_{1}) is the height of the wooden block. So, (

*h*

_{2}–

*h*

_{1})

*a*is the volume of the solid block,

*V*.

*d*=

*Vρg*

∴ Upthrust =

*Vρg*

In any situation, the volume of fluid displaced (or the volume of the object submerged) is considered to calculate upthrust, because (

*h*

_{2}–

*h*

_{1}) is the height of the solid block only when it is completely immersed. Furthermore, the pressure difference of the fluid acts only on the immersed part of an object.

Now, moving back to

*Vρg*. Since

*V*is the volume of fluid displaced, then the product of

*V*,

*ρ*and

*g*is the weight of the fluid displaced. So, we can say that

Upthrust = Weight of the fluid displaced

Compare this conclusion with the statement above summarising Archimedes' principle. Are they the same? Well, not totally. The “apparent loss in weight” was not mentioned in my explanation.

In the figure on the left, there are arrows on the top and bottom of the solid block. The downward arrow represent the weight of the block pulling it downwards and the upward arrow represent the upthrust pushing it upwards. If one were to measure the weight of the solid block when it is immersed in the fluid, he will find that the weight of the block is less than that in air. There is a so-called “apparent loss in weight”, because the buoyant force has supported some of the block’s weight.

Weight in air – Upthrust | = | Weight in fluid |

Upthrust | = | Weight in air - Weight in fluid |

∴ Upthrust = Apparent loss in weight

**Objects Floating Freely**

When an object is floating freely (i.e. neither sinking nor moving vertically upwards), then the upthrust must be fully supporting the object’s weight. We can say

Upthrust on body = Weight of floating body

By Archimedes’ principle,

Upthrust on body = Weight of fluid displaced

Therefore,

Weight of floating body = Weight of fluid displaced

This result, sometimes called the “principle of floatation”, is a special case of Archimedes’ principle and can be stated:

A floating body displaces its own weight of fluid.If a body cannot do this, even when completely immersed, it sinks.

**See Also**

- Problem: Materials #1 - a sample problem involving Archimedes' principle.
- Materials - for related articles and problems.

## 8 comments:

Can you just explain how to measure the volume of an irregularly shaped object which floats in water? Can we suse archimedes principle? How? Please send me the answer to my email nairjula@yahoo.com

Anonymous:

I have e-mailed you my answer earlier on regarding this question. Others can view my answer here:

Q&A: Finding Volume of a Floating Object

a body displaces equal volume of water when it is immersed completely in water but by what volume of mercury gets displaced when the same body is immersed

Anonymous:

The same body completely immersed will displace an equal amount of mercury as in water because both are fluids.

I don't understand why the volume of the water displaced = the volume of the body. Please can someone help me

why is it not advisable to allow the object to touch the bottom or sides of the displacement can?

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