Buoyancy Force

[grikmok]

Hi,

EK says that buoyancy force do not change with depth. I don't understand how come it doesn't change. For example, air. As we go higher and higher, the density goes down. Wouldn't the force change accordingly too, since Fb = Density (fluid) * Volume * Height?

thanks,

grik

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[VCU07]

Hi,

EK says that buoyancy force do not change with depth. I don't understand how come it doesn't change. For example, air. As we go higher and higher, the density goes down. Wouldn't the force change accordingly too, since Fb = Density (fluid) * Volume * Height?

thanks,

grik

The water displaced will be equal in magnitude to the upward boyant force.

 

[WashMe]

The pressure of a fluid increases linearly with depth according to rho(density)*gravity*depth(from surface) (part of bernoulli equation). As a fully submerged object sinks, since the increase is linear, the difference in pressure above and below the object is always the same regardless of depth. Thus, if the object wasn't supported at the surface, it won't "float" an any depth.

That is why a fully submerged object will sink to the bottom of a container rather than reach a certain depth and sort of hover there.

*MCAT MCAT MCAT* This is the key to remember:
1)An object that is at least partially floating displaces fluid equal to its mass.
2)A fully submerged object displaces fluid equal to its volume.

Consider taking a floating styrofoam block the size of your car and putting a bunch of gold bricks on it until it is half submerged. Mass of water displaced = mass of styrofoam+gold bricks , which is a lot. Now, remove the styrofoam and just put the gold bricks in the water, adding a little extra to account for the mass of the styrofoam from the previous experiment; they won't displace much water in comparison. Volume of water displaced = volume of gold bricks. This example works not only for gold bricks, of course, but for any object that could be placed on the styrofoam that has a density greater than the density of water (or other fluid in question).

Also, you'll want to know the floating equation, rho(object) / rho(fluid). The resulting number, from ~0 to ~1 (not >1 because the object would sink), tells you what percentage of the object is submerged. Think, an object with 0.4g/cm^3 would be 40% submerged in water, which is 1.0g/cm^3. The other 60%, of course, would be above the surface.

 

[grikmok]

Thanks very much, everyone!

 

[grikmok]

The pressure of a fluid increases linearly with depth according to rho(density)*gravity*depth(from surface) (part of bernoulli equation). As a fully submerged object sinks, since the increase is linear, the difference in pressure above and below the object is always the same regardless of depth. Thus, if the object wasn't supported at the surface, it won't "float" an any depth.

That is why a fully submerged object will sink to the bottom of a container rather than reach a certain depth and sort of hover there.

Is this the same principle as a rising balloon?

 

[loveoforganic]

Not really. A similar concept to a rising balloon would be if you dragged some styrofoam to the bottom of the ocean then released it.

 

[WashMe]

Not really. A similar concept to a rising balloon would be if you dragged some styrofoam to the bottom of the ocean then released it.

Except that the density of air (and the pressure) decreases with increasing altitude, whereas the MCAT considers the density of liquids to be the same throughout (this is why hydraulic lifts work). The density of helium in the rising balloon, of course, will decrease as well (because outside pressure is dropping) as the balloon expands to equalize pressure--- until the latex or mylar ruptures.

 

[grikmok]

Except that the density of air (and the pressure) decreases with increasing altitude, ...

So, in air, buoyancy force is also decreases with increasing altitude, right?

 

[WashMe]

So, in air, buoyancy force is also decreases with increasing altitude, right?

Well, if you're talking about a helium balloon, then:

F(buoyant) = [rho(air) - rho(helium)] * V * g

As the balloon rises...
the rho(air) decreases
the rho(helium) decreases (the balloon expands)
the V increases (the balloon expands)
g technically decreases eventually... (G*mEarth)/r^2

I am not a physics major and I took the MCAT over a year ago, but here is what I think:

The balloon would not allow the helium gas to expand forever, so at the point where the balloon physically constrained the helium gas to keep it at a higher pressure/density (thus matching the pressure/density of the outside atmosphere), it would "float" at a certain altitude. This is analogous to having three fluid layers in cylinder: non-polar high density, polar medium density, and non-polar low density; a medium density object placed in the cylinder would "float" at the interface between the high and medium density fluids. I suspect that the helium would escape the balloon before this could be at some equilibrium though.

I would have to say that the buoyant force decreases. It would be easier to figure out if I knew exactly how to take into account the object's changing volume. If the volume were constant, the buoyant force would certainly decrease according to rho*V*g.

Physics majors please chime in.

I don't think you'll need to answer a question like that, but maybe you will...