Quote:Original post by arithma
Take this example and you will immediatly know where I want to take you regarding the resultant point of application (used for converting forces in angular uses):

A cylinder is immersed perpendicularly with respect to the surface of water. Since the forces on the (curved) sides of the cylinder cancel each other out, the resultant force is exclusively the pressure applied to the base surface. This in other words leads us to the conclusion that the point of application is also in that plane (containing the button surface). In conclusion, we cannot take the center of mass of displaced water as the point of application.

Not really. All points on the direction of force (on the line that pass "application point" and is directed in direction of force) is equally valid as points of application (if body is rigid (that is, if you are not interested in deformation/etc.)). It seems somewhat counter-intuitive at first, but it is so. There's actually a line of application that pass some point, and we can use this point as point of application. But effect of forces placed anywhere on line of application is equal. In this case, force is direted upward and center of top surface finally gives same result as center of cylinder or center of bottom surface or center of immersed volume, because them all is on same line that contains force vector.

Torque is equal to cross product of point of application and force. If point is P and force is F, torque around zero point is F x P (or P x F, matter of convention) Note that
F x (P+a*F) = F x P + F x a*F = F x P
That is, addition of scaled vector F to P doesn't change torque.

Buoyancy force is not the result of pressure difference it is the result of weight difference.
A body immense in fluid experiences the same upward buoyancy force at 1 meter depth than at 1000 meter depth

I think there is confusion with the buoyancy force and the fluid pressure, Archimedes principle says that the buoyancy force is equal to the weight of the displaced fluid, point in the direction opposite to the gravity, and is applied at the center of gravity of the displaced fluid.

So for bodies fully immerse the center is the same (assuming equal density) and the force is just the weight difference. For bodies partially immerse gets more complicated because the volume of the immerse body have to be integrated, but you can do it by populating the volume with pointe evenly spaced and just counting how many are under the surface and determining the volume and the center for the submerged ones.

Or you can just integrate the face areas using the Green, and divergence theorem by applying the flow traveling along the gravity vector. But this is more complex.

Quote:
Buoyancy force is not the result of pressure difference

it's exactly result of pressure difference. If you have 1x1x1 cube aligned top side to top, you can easily see that buoyancy force is equal to pressure difference between top side and bottom side multiplied by area. Same for any other shape. (And, indeed, pressure difference * area is shown to be equal to weight of water in that volume). In incompressible fluid in constant gravitation field, indeed, pressure difference is same at depth 1 meter as at depth 1000 meters.

Heck, if you have a bottle with water, increased weight of bottle is a result of pressure difference :-) and pressure difference is result of weight of water.

Quote:
But it is just that, I know that the point of application is down there (different from where the center of mass is) but because it is displaced along gravity it has no rotational effect. But if you take another example (ie a tilted bowling ball where the more volumous part is downward) we observe rotational effects. It is like such since the point of application (which is in conclusion not the CM itself) causes that rotation (in conjuntion with the resultant force of course).

i'm unsure what you mean, but you can easily see that equal forces applied along line parallel to force produce exactly same torque. In case of ball staying on surface, for any orientation, line of application of surface force passes ball center. You can use ball center in calculations as point of application of force, instead of point of contact with surface, and will get exactly same torque in result.

(again, assuming objects we talk about is perfectly rigid)

Actually it is not result of a pressure difference. It is a result of weight difference.
If the force is the same a regardless of the depth then it means it is independent of the depth. Furthermore more in any demonstration of the principle the concept of pressure in not required at all what is used is the weigh difference of a column of water at depth h and a column of water of depth h1.

Why making more confusing when there is a principle that says how to do it?
Just calculate the volume by whatever mean you can, determine how much this water volume weights and the different between that weight and the solid weight it the buoyancy force. The center of application for the force is the geometry center of the submerge volume (which means that for fully submerged volumes the torque is zero since the center coincides with the rigid body), and the direction of the force a vector parallel to the gravity. The problem is completely well defined.

I do not now what the incompressibility, gravity field, objects are perfectly rigid, etc has to do with it since it was not stated as part of the problems.
I can easily see engineers never making anything if all these things were in the way

Your are right it is teh rsult of presure difference?
But presure diffrence is not part of the calculation in any way.

seems you don't understand what is pressure and what is difference.

Put 1mx1mx1m cube into water so top is at depth 1 m. You have pressure on top = 9.8 kN/M2 and pressure on bottom = 18.6 kN/M2 . Pressure difference is 9.8 kN/M2 and buoyancy is 9.8 kN
Put that cube at some other depth where pressure is A (say, at depth 1000 m where A=9.8*106 N/M2). On top you'll have A and on bottom you'll have A+9.8 kN/M2, and difference is again 9.8 kN/M2 and buoyancy is again 9.8 kN. Understand now? If pressure would be same on top and bottom, there would be no force.

Basically, buoyancy force (and in general, any force caused by liquid) is equal to integral of pressure*dA , (where "dA" stands for vector that is normal to surface and have length equal to area), and it could be nonzero only if on different parts of object you have different pressures. Also, that integral of pressure*dA is equal to [minus] weight of same water in volume of object, it can be proven in numerous ways(e.g. classically by observing that if there would be water, forces would be compensated, or using calculus)

(Speaking of gravity, and compressibility, in real water, there will be bit bigger buoyancy (for same volume of object) at 10 000 m depth than on surface, because water is densier and gravity is stronger. But it of course doesn't matter. Still, it is better to say that explicitly)

[Edited by - Dmytry on March 30, 2005 3:41:56 PM]

Oh I do, but it seems you do not have common sense. I said yes to keep you happy and get over with.

MotorHerp answer completely covered the question, if it any clarification was needed; it was perhaps how to go about implementation details for calculating the buoyancy center.

A simple explanation of how to implement the Archimedes principle is more than sufficient. However you brought gratuitous confusion by adding the variables pressure, gravity, compressibility, deformable bodies, turbulence, and how knows what else (why not gravity variation due to elevation, oh wait you did bring it up in the end, what next quamtum variations).

The problem with you is that more often than not you go completely off the tangent answering the question that was not asked. What you like is to grandstand displaying how much you have know about math and physics and you have the tendency to think you invented these things.

Now since you brought it up saying I do not underneath what pressure is, I am going to say I think you are right. In my time pressure was not a quantity it was a convenient scalar function of force that just expresses force per unit area. To have pressure you need weight and in fact you can completely disregard the concept of pressure when calculating the buoyancy force. All it is needed to know is the weight difference between the fluid and the body.

It is very simple if at depth H1 the force, and buoyancy center is the same than at depth H2, However the pressure at H1 is very different than at H2 then I do not know how the pressure influence the calculation. In the old days when we run experiments with such outcome we use to conclude something like “Buoyancy force is independent of the surface pressure”

Even if the force per unit area is different at the body top than it is a the box bottom, the fact is that the bouyance force is same regarless of the fuild depth.

If we did the same experiment using two fluids with different density and we found that at depth H1 the buoyancy force in fluid A is different than the Buoyancy force at same depth H1 in fluid B, then we would conclude “buoyancy force is a function of the fluid density, (which is relate to the fluid specific weight)”

But that was in the old days maybe the science is different now.
Or perhaps you can show the variable pressure is applied in the Archimedes principle.

I even gave example HOW at depth of 1000M you have same buoyancy as on depth 1M, as you have same pressure difference. As buoyancy is a direct result of *difference* of pressure (to stay simple, difference between top and bottom of object, that's is!), as i have said really many times. And no, addition of constant pressure does not change difference (you don't understand that a-b = (a+c) - (b+c) ?) And, as I also said many times, buoyancy can be computed as weight of liquid in volume of object. My point is that you can not assume that pressure on surface is equal when computing buoyancy by integrating pressure over surface to find "application point". If you don't understand what does word "difference" mean.... well.

and, read that , BTW.

(note that it's the case when you can't assume that point of application of force is in center of submerged volme, as net force is not vertical.)