archimedes law

HWGXX7

Why isn't the influence of air pressure assimilated in Archimedes law?

If an obect is more dense than the fluid, it wil sink. That is clear, but suppose I could lower down de air pressue, so less force that tends to push the object tot the surface. Is there any situation possible of adjusting the air pressure to get an object just floating? Or is the influence of the air pressure in incompressible fluids not great, so change in pressure wouldn't be noticed very much?

But , think of it as an theoretical question.

thank in advance

mathman

The key factors are the density of the fluid and the object, with gravity the controlling force. Air pressure plays no role.

HWGXX7

Air pressure plays no role.

The reason is that it can be neglected, is that correct?

mathman

Quote by HWGXX7

The reason is that it can be neglected, is that correct?

Your question is confusing. The air pressure is the same on both the object and the water.

jim hardy

An object immersed in air experiences bouyancy.

I think it IS accounted for in Archimedes' law.

You answered your question in OP.

Or is the influence of the air pressure in incompressible fluids not great, so change in pressure wouldn't be noticed very much?

Have more faith in yourself !

In slide rule days the effect was too small to warrant consideration, but with today's umpteen-digit calculators i'm sure it could be included. Try it and see how many digits out it shows up.

From wiki,
http://en.wikipedia.org/wiki/Properties_of_water

Compressibility

The compressibility of water is a function of pressure and temperature. At 0 °C, at the limit of zero pressure, the compressibility is 5.1×10−10 Pa−1.[28] At the zero-pressure limit, the compressibility reaches a minimum of 4.4×10−10 Pa−1 around 45 °C before increasing again with increasing temperature. As the pressure is increased, the compressibility decreases, being 3.9×10−10 Pa−1 at 0 °C and 100 MPa.

The bulk modulus of water is 2.2 GPa.[29] The low compressibility of non-gases, and of water in particular, leads to their often being assumed as incompressible. The low compressibility of water means that even in the deep oceans at 4 km depth, where pressures are 40 MPa, there is only a 1.8% decrease in volume.[29]

*Bob S*

Quote by mathman

The key factors are the density of the fluid and the object, with gravity the controlling force. Air pressure plays no role.

Air pressure plays a role in the buoyancy of Cartesian divers. See

http://www.google.com/webhp?rlz=1C1R....,cf.osb&cad=b

HWGXX7

I think it IS accounted for in Archimedes' law.

I re-analysed it and atmospherical pressure cannot influence the dyncamic behaviour of a particle that is submerged complete. So there Archimedes law cannot account for it at all.

Air pressure plays a role in the buoyancy of Cartesian divers.

Correct, in general air pressure will influence the behaviour of any particle in contact with the air pressure. Once complete submerged this isn't the case anymore.
http://en.wikipedia.org/wiki/Buoyancy

grtz

nasu

Quote by HWGXX7

Correct, in general air pressure will influence the behaviour of any particle in contact with the air pressure. Once complete submerged this isn't the case anymore.
http://en.wikipedia.org/wiki/Buoyancy
grtz

It does not have to be "in contact". The atmospheric pressure is transmitted through water. The pressure acting on the submerged body is the sum of atmospheric pressure and hydrostatic pressure. If the submerged body is flexible, the increase in atmospheric pressure may produce a change of its volume.

The results that atmospheric pressure does not directly influence buoyancy is due to the buoyancy depending on the pressure difference (or gradient) along the vertical direction.
Atmospheric pressure just adds a constant factor to the hydrostatic pressure and this has no contribution to the gradient.

HWGXX7

It does not have to be "in contact".

Correct, made a mistake there. Difference in air pressure wil result in greater hydrostatic pressure, and therefore a greater upward force.
The object remains in static equilibrium because of difference in de weight of the air pushing downwards.

thank for all the help

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