Suppose, as in this post, we have a fluid of density *ρ*, with some arbitrary three-dimensional object immersed in the fluid, occupying a volume V. This object has surface ∂V. We saw that for any point on the surface, the pressure is , and the force on an area element is:

. Thus, the torque on that element is

,

where **r** is the coordinate vector to the element. Recalling that for vectors **v** and **w** and scalar *a*,

, so since *P*(*x*,*y*,*z*) is a scalar, we see

and so the total torque (about the origin) is

;

and one form of the divergence theorem tells us that for vector field **A**,

.

Thus

Now, the product rule for the curl of the product of a scalar field *ψ* and a vector field **a** is

.

Now, applying that to , and noting that (as it is a radial vector field), and , we see

.

Now, since is a constant vector, it can be “factored out” of the integral:

.

We recall that the total buoyant force on the object is ; plugging this into the above, we see

.

Now, recall that the center of mass of a region with density function *ρ*(*x*,*y*,*z*) is . If the density is a constant, it factors out of the integrals, and one gets . Thus, if we consider the volume as if it were filled with the fluid; that is to say, the volume of fluid displaced, its center of mass would be , and then we see that

,

which is equivalent to the torque if the buoyant force acted entirely on the point **r**_{b}; this point is called the center of buoyancy. Just as the force of gravity on an extended object can be treated as if it acts entirely on the center of mass, the buoyant force can be treated as if it acts entirely on the center of buoyancy.

### Like this:

Like Loading...

*Related*

Tags: Buoyancy, Center of Buoyancy, Center of Mass, Curl, Divergence Theorem, Friday Physics, physics, Torque

This entry was posted on November 6, 2009 at 1:21 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed.
You can leave a response, or trackback from your own site.

November 13, 2009 at 1:23 am |

[…] us that ; combining, we see , and to keep the upper face “dry,” we have . Now, the center of buoyancy is located at the centroid> of the trapezoidal prism. By symmetry, this center will be in the […]