Finding Buoyant Force of Air: Calculating Volume Displaced

[eddieb340]

How do you find the exact buoyant force of air?


B = pgV

where p = density of air (1.29kg/m³)
g = 9.81 m/s²
V = volume displaced

The problem is, you can't measure Volume directly. You have to do it experimentally by weighing a mass in air and then in water.

I can find the volume of the object in the air, only in water.

 

[olgranpappy]

How do you find the exact buoyant force of air?


B = pgV

where p = density of air (1.29kg/m³)
g = 9.81 m/s²
V = volume displaced

The problem is, you can't measure Volume directly. You have to do it experimentally by weighing a mass in air and then in water.

I can find the volume of the object in the air, only in water.

Why can't you do it directly--oh... is this a homework problem?

 

[eddieb340]

No, its a lab problem and he won't let us calculate it directly. We have to use buoyant forces.
So we weighed the mass in the air and then in the water, found the different and solved for the volume from this following equation:

T + B = mg where B = pgV

but you can't weigh the mass in vacuum so get the volume in a vacuum. This is my dilemma

 

[olgranpappy]

No, its a lab problem and he won't let us calculate it directly. We have to use buoyant forces.
So we weighed the mass in the air and then in the water, found the different and solved for the volume from this following equation:

T + B = mg where B = pgV

but you can't weigh the mass in vacuum so get the volume in a vacuum. This is my dilemma

okay. you weighed the thing twice. so you have *two* T's and *two* B's in your equation.
<br /> T_1+B_{1}=mg<br />
and
<br /> T_2+B_2=mg<br />

The B's are not the same because the densities of the fluid in the two cases are different--indeed, in one of the cases B is quite small. The T's are your measurements. You know 'g', you know the densities. You can solve this set of equations (using algebra) for the volume.

 

[eddieb340]

With all do respect, the volumes are actually different. Think about it. Water is 1000 times more dense than air. So the outside force of the water is crushing the block, therefore making the volume smaller.

What happens when you put a balloon 10,000 feet under water? It will be smaller (compressed because of the pressure) than a balloon in air.

So since the volumes are different we have 3 unknowns, needing 3 equations.

Anyone know where to go from there?

 

[Doc Al]

With all do respect, the volumes are actually different. Think about it. Water is 1000 times more dense than air. So the outside force of the water is crushing the block, therefore making the volume smaller.

What's the block made out of? I seriously doubt that you are expected to account for the change in volume due to the difference in pressure. In any case, you can estimate the change in volume if you know the bulk modulus of the object.

What's the average pressure on the object when it's under water? How deep are you submerging it when you weigh it? Considering that atmospheric pressure is about 10^5 N/m^2 and the density of water is about 10^3 kg/m^3, estimate the additional pressure if you submerged the object 10 cm, say.

You'll find that the change in volume will be far too small to measure or worry about.

 

[olgranpappy]

With all do respect

For future reference, the phrase is:

"With all due respect"

I.e., your respect for me is owed ("due") not performed ("do"). But, thanks being so polite. And cheers.

 

[eddieb340]

Thanks Doc Al and everyone for their help, I actually didnt get the measurements of how deep we submerged the Aluminum block and your right that it would be some a minute change it really doesn't matter. I wish i had taken the measurement it would be nice to see how little the volume changes.

Thanks again