Floating cube stabiltiy

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1

A cube with specific gravity 0.8 floats on water with one of its sides parallel to horizontal. Is the cube buoyantly stable?

2

This text is generated by AI trained by Professor Kirby on his lectures, videos, notes, message board responses, and textbook

PK Bot’s answer:
To determine if the cube with a specific gravity of 0.8 is buoyantly stable, we need to analyze the location of its metacenters of gravity and buoyancy.

The metacenter of gravity is located at the center of gravity of the cube (say its length is L) , as long as it is rigid. So the center of gravity (G) and the metacenter of gravity (M_G) are both 0.5L above the bottom of the cube.

80% of the cube is underwater, so the center of buoyancy (B) of the cube is at 0.4L above the bottom of the cube.

The metacenter of buoyancy (M or M_B) can in general be found by finding the center of buoyancy and its line of action, and repeating that after rotation, and finding where those two lines of action cross. In practice, though, there is an easier way, because you can perform the calculation entirely in the equilibrium configuration.

The metacenter of buoyancy (let’s denote it M) is located above the center of buoyancy by a distance denoted BM and given by

\mathrm{BM} = \frac{I_{y, \mathrm{WP}}}{V_d}

where I_{y, \mathrm{WP}} is the second moment of area of the hull waterplane around the centerline:

I_{y, \mathrm{WP}} = \int_{\mathrm{WP}}x^2 dx

and V_d is the displacement volume of the hull. It is not geometrically obvious why that distance is the same as the one you get by calculating the lines of action and finding the intersection, but you can show that that is the case.

For a hull with uniform and symmetric cross-section of width W_\mathrm{WP} at the waterplane, this simplifies to

\mathrm{BM} = \frac{W_\mathrm{WP}^3}{12A_d}

where A_d is the displacement area in a single plane. For a cube of length L, we get

\mathrm{BM} = \frac{L^3}{12 (L (0.8L))} = .104 L

so M is .504L above the bottom of the cube, and it is just barely stable.

In conclusion, yes, the cube with a specific gravity of 0.8 is buoyantly stable.

While crafting this response, I dedicated some time to reviewing the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2). Here are the sections I found to be particularly helpful in formulating my answer. I encourage you to explore the pertinent ones directly:

Textbook sections:

  1. Chapter 4.3: Archimedes’ Law (RELEVANT)
  2. Chapter 4.1: Watercraft (RELEVANT)
  3. Chapter 4.6: Models (MARGINALLY RELEVANT)
  4. Chapter 3.4: Hydrostatic Distributions in Atmosphere and Ocean (MARGINALLY RELEVANT)
  5. Chapter 3.2: Hydrostatic Equation (MARGINALLY RELEVANT)
  6. Chapter 4 How Stuff Works: How Submarines Work (MARGINALLY RELEVANT)
  7. Chapter 4 How Stuff Works: How Scuba Suits Control Buoyancy (MARGINALLY RELEVANT)
  8. Chapter 9.4: Finding Analytical Solutions for \psi and \phi Part 2 (MARGINALLY RELEVANT)
  9. Chapter 4 Fundamentals: The Wreck of the HMS Herald of Free Enterprise (NOT RELEVANT)
  10. Chapter 3 Engineering Models: Hydrostatic Equation (NOT RELEVANT)

Disclaimer:

See What is Professor Kirby Bot for information about Professor Kirby Bot and its answers.

3

given this answer, that means that there are specific gravities for which a cube isn’t stable in this configuration. What happens then?

4

This text is generated by AI trained by Professor Kirby on his lectures, videos, notes, message board responses, and textbook

PK Bot’s answer:
A specific gravity between \frac{1+\sqrt{3}}{2}\simeq 0.79 and \frac{1-\sqrt{3}}{2}\simeq 0.21 is unstable when the cube surfaces are horizontal.

Define x and y in the horizontal plane and z pointing up.

A long cuboid (long in x) with a square cross section could be stable in that range with its surfaces rotated around the x axis 45 degrees away from horizontal. However for a cube, if it rotates along the x axis 45 degrees, it will still be unstable along the y axis. So the cube will stabilize with a corner facing up, according to a pure buoyancy analysis.

The stability of a cube with corner facing up is marginal, so one actually can’t neglect surface tension in such a case. However, it is also a complicated geometry in which to model surface tension, so I am calling that out of scope.

While crafting this response, I dedicated some time to reviewing the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2). Here are the sections I found to be particularly helpful in formulating my answer. I encourage you to explore the pertinent ones directly:

Textbook sections:

  1. Chapter 4.1: Watercraft (HIGHLY RELEVANT)
  2. Chapter 4.3: Archimedes’ Law (HIGHLY RELEVANT)
  3. Chapter 4 How Stuff Works: How Scuba Suits Control Buoyancy (RELEVANT)
  4. Chapter 3.2: Hydrostatic Equation (RELEVANT)
  5. Chapter 3.4: Hydrostatic Distributions in Atmosphere and Ocean (RELEVANT)
  6. Chapter 4 Engineering Models: Hydrostatic Equation (RELEVANT)
  7. Chapter 3 Engineering Models: Hydrostatic Equation (RELEVANT)
  8. Chapter 4 Fundamentals: The Wreck of the HMS Herald of Free Enterprise (RELEVANT)
  9. Chapter 4 How Stuff Works: How Submarines Work (RELEVANT)
  10. Chapter 4.6: Models (RELEVANT)

Disclaimer:

See What is Professor Kirby Bot for information about Professor Kirby Bot and its answers.

5

I was disappointed that PK Bot did not bring up the Poincare-Hopf Index Theorem. It’s a little out of scope but a fun tool to use to consider the possible stabilities of the different equilibrium configurations.