Buoyancy Pressure

• Jun 6th 2008, 11:06 AM
Abelian
Buoyancy Pressure
Hi,
I am looking to work out the buoyancy pressure of an object of mass 2kg when held in a tank of tap water at a depth of 300mm (final answer in mmHg). Using Archimedes, I have that the buoyancy force is equal to:

P*A*y*g

Where p is the density of the water (I worked this out to be 998.23 kg/m^3 for tap water at 20 degrees, namely room temperature), A is the cross sectional area of the object (the object has dimensions 0.23m x 0.38 m x 0.15 m, the face in the water corresponds to an area of 0.23 x 0.38 = 0.0874m^2), y is the depth at which the object is submerged (0.3 metres) and g is gravity, taken as 9.81.

Then:

998.23*0.0874*0.3*9.81 = 256.76 Newtons.

However, I am then unsure what to do with this. I have buoyancy force, how to I convert to buoyancy pressure?!? Do I use Pressure = Force/Area?!? If I can, which area do I consider, the cross sectional area of the object?!? If I do this, I then have:

Pressure = 2937.757 Pa (Pascals)
so that:

2937.757 Pa = 22.035 Torr.

Then, since 1 mmHg = 1 torr, we have 22.035 mmHg. Is this correct for this problem?!?

Any help would be appreciated.
• Jun 6th 2008, 02:57 PM
topsquark
Quote:

Originally Posted by Abelian
Hi,
I am looking to work out the buoyancy pressure of an object of mass 2kg when held in a tank of tap water at a depth of 300mm (final answer in mmHg). Using Archimedes, I have that the buoyancy force is equal to:

P*A*y*g

Where p is the density of the water (I worked this out to be 998.23 kg/m^3 for tap water at 20 degrees, namely room temperature), A is the cross sectional area of the object (the object has dimensions 0.23m x 0.38 m x 0.15 m, the face in the water corresponds to an area of 0.23 x 0.38 = 0.0874m^2), y is the depth at which the object is submerged (0.3 metres) and g is gravity, taken as 9.81.

Then:

998.23*0.0874*0.3*9.81 = 256.76 Newtons.

However, I am then unsure what to do with this. I have buoyancy force, how to I convert to buoyancy pressure?!? Do I use Pressure = Force/Area?!? If I can, which area do I consider, the cross sectional area of the object?!? If I do this, I then have:

Pressure = 2937.757 Pa (Pascals)
so that:

2937.757 Pa = 22.035 Torr.

Then, since 1 mmHg = 1 torr, we have 22.035 mmHg. Is this correct for this problem?!?

Any help would be appreciated.

I am too tired to make a full response right now but I have something to point out and then a question.

Point: The buoyant force is equal to the weight of the water the object displaces. You cannot calculate it using the pressure equation as you have written. If the entire volume of the object has been submerged then the buoyant force on the object is $\displaystyle B = \rho gV$ where V is the volume of the object and $\displaystyle \rho$ is the density of water.

Question: What are you defining buoyant pressure as? If I am not mistaken the buoyant pressure is the difference in pressure between the top and bottom of the object (if both surfaces are parallel and horizontal.) Please correct me if that isn't the definition you are using.

-Dan
• Jun 6th 2008, 03:06 PM
Abelian
Hi,
Thanks for your reply. I was using my notes from mechanics last year:

"Ignoring friction, the only forces acting on the object are its weight due to gravity and the Archimedes buoyancy force. Archimedes principle says that the buoyancy force acting on a body is equal to the weight of the displaced fluid. The objects weight is W = mg and the buoyancy force is:

F(buoy) = pAyg..."

(with p, A, y and g defined as above, the 'object' was from an example given in the notes of a bottle with sand in it).

I was using what I had above, and adapting it to this case. I couldnt understand why there should be a difference?!? Although, I have read the problem back through and have spotted the word submerged now, so once I work out the buoyancy pressure using:

B=p*V

with p the pressure of the water, V the volume of the object, how would I calculate the pressure at the 'top', assuming that the value of B that I have found above is for the bottom?!? Would I then get an answer in Newtons?
• Jun 7th 2008, 06:56 AM
topsquark
Quote:

Originally Posted by Abelian
Hi,
Thanks for your reply. I was using my notes from mechanics last year:

"Ignoring friction, the only forces acting on the object are its weight due to gravity and the Archimedes buoyancy force. Archimedes principle says that the buoyancy force acting on a body is equal to the weight of the displaced fluid. The objects weight is W = mg and the buoyancy force is:

F(buoy) = pAyg..."

(with p, A, y and g defined as above, the 'object' was from an example given in the notes of a bottle with sand in it).

I was using what I had above, and adapting it to this case. I couldnt understand why there should be a difference?!? Although, I have read the problem back through and have spotted the word submerged now, so once I work out the buoyancy pressure using:

B=p*V

with p the pressure of the water, V the volume of the object, how would I calculate the pressure at the 'top', assuming that the value of B that I have found above is for the bottom?!? Would I then get an answer in Newtons?

The buoyant force is the net force on the object due to the fluid it is immersed in. If we have a simple object like a cube we can calculate it by finding the pressure on the bottom face of the object and find the upward force on it from that and subtract from that the force on the pressure from the top face. If the height of the cube is y then we get the force equation that you provided:
$\displaystyle F_B = \rho gy A = \rho gV$
(Sorry for the typo in my first reply!) To find the pressure at the top of the cube you use
$\displaystyle P = mgd$
where d is the depth of the top face. To find the pressure at the bottom of the cube, etc.

If you are trying to find the force on the bottom of the cube with just the buoyant force and dimensions of the cube then you can't do it: all we can get is the net difference in the pressure.

-Dan
• Jun 7th 2008, 08:32 AM
Abelian
Hi,
Right, so do I then have that if the box is held underwater at a depth of 0.3 metres:

Pressure on top face = 2 * 9.8 * 0.3 = 5.88 Pa

Pressure on bottom face = 2 * 9.8 * 0.45 = 8.82 Pa

Do I then substitute both of these into the force equation for p? Or do I use

Pressure = Force / Area

and since I know the pressure for the top and bottom faces, and the cross sectional area, I can work out the Force's for both?!?