Is Archimedes theory only valid for incompressible fluids?
- 7 months ago
A buoyant force is equal to the weight of the fluid displaced because; if the fluid were not displaced, that same force would have to support the weight of the fluid. This would apply to any kind of fluid.
- StuartLv 67 months ago
Balancing the books
If the universe really is expanding faster than thought, then some kind of new physics would have to provide the extra oomph. Is dark energy more exotic and turbo-charged than we thought? Is dark matter more complex than we imagined? Is there some other kind of unseen particle in the cosmos, such as a “sterile neutrino” that interacts with other types of matter only via gravity?
- busterwasmycatLv 77 months ago
it requires a correction for the compressibility of the fluid, but it is not "invalid". Just not perfect. The basic idea is still true: volume displacement has to be accommodated somehow.
- az_lenderLv 77 months ago
In the problem of a helium balloon floating in air, its volume is equal to the weight of the air it displaces. Although air is compressible, the weight of displaced air can be well approximated by using the density of the air at the balloon's center altitude. This works because the density of air does not decrease SIGNIFICANTLY over the short vertical diameter of the balloon.
- How do you think about the answers? You can sign in to vote the answer.
- Steve4PhysicsLv 77 months ago
EDIT in reply to query in comment.
One way of stating Archimedes’ law (principle) is that the upthrust equals the weight of fluid displaced.
As a simple example, if a sphere of volume 0.1m³ is immersed in water:
volume of displaced water is V = 0.1m³;
mass of displaced water is m= ρV = 1000x0.1 = 100kg (using density ρ =1000kg/m³);
weight of displaced water = mg = 100x9.8 =9800N(using g = 9.8m/s²).
So there is an upthrust of U = 9800N (upwards) and the resultant force on the sphere is the vector-sum of this and the sphere’s weight.
The above calculation assumes the density of water is constant. (It also assumes g is constant, but that’s not usually an issue.)
If the density is not constant, in general it will be a function of position and we must find the mass of fuid displaced (m) in a different way. We do this using calculus. If the density of the fluid as a function of position is ρ(x,y,z) in the absence of the object, then we must perform a volume integral:
m = ∫∫∫ρ.dx.dy.dz where the integral is within the region bounded by the immersed object’s surface.
That might not make sense unless you know some calculus, but I hope it helps.
Valid for all fluids. E.g. air in compressible and a helium balloon rises when released because the upthrust = weight of displaced air (and upthrust is greater than the balloon's weight).
However, in situations where there density varies (because of variations in pressure and/or temperature), the calculation of the weight of displaced fluuid is complicated.
- skeptikLv 77 months ago
It's not that it's only valid for incompressible fluids, it's that you would have to account for the compression otherwise. Which would make the calculation excessively complex.
But since the vast majority of its uses deal with water, it works out ok.