Suppose we have a cube, of edge length *L*, with mass *M* symmetrically distributed so that the geometric center of the cube is the center of mass. Suppose we have a fluid with density *ρ*, such that (the average density of the cube is less than that of the fluid), so that the cube will float in the fluid. If the cube is floating such that the upper face is entirely outside the fluid, and is tipped toward one edge by an angle *θ* from the horizontal, then what is the torque on the cube about its center?

Let us define the ratio of the densities of the cube and the fluid as *xi;*, so that . Then 0<*xi;*<1. Then via Archimedes’ Principle, the weight of fluid displaced equals the weight of the cube, and thus the volume *V*_{d} of fluid displaced can be found:

,

or, using *M*=*ξρL*^{3}, *V*_{d}=*ξL*^{3}, and *xi;* is also the fraction of the cube submerged.

Now, if it is tilted at an angle *θ*, the submerged portion is a trapezoidal prism; if the trapezoid has bases *b*_{1} and *b*_{2}, then it has area , and thus the prism has volume , which, equating to the above, tells us

.

Our geometry tells 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 plane halfway between the trapezoidal faces of the prism. Using *xy* coordinates on this plane with the origin at the center of the cube and with the axes along those of the cube, we find the center of buoyancy to be the centroid of the cross-section trapezoid.

Now, using these coordinates, we see that the trapezoid is the region , ; the area of the trapezoid is *A*=*ξL*^{2}, so performing the double integrals for the centroid, we have

and

.

Now, to convert these to coordinates with horizontal and vertical axes, we rotate coordinates by *θ*. If x’ is our horizontal axis and y’ our vertical axis, then

, so our center of buoyancy in these coordinates (with origin still at the center of our cube) is

And so the torque is, with positive torque being in the direction to decrease *θ*

.

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Tags: Archimedes, Buoyancy, Center of Buoyancy, Friday Physics, physics, Torque

This entry was posted on November 13, 2009 at 1:17 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed.
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