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Let's say you have a mildly elliptical bowl balanced on top of a stick. This bowl is filled with water, not necessarily to the top but fairly full (I'm not 100% sure this makes a difference). You then place a floating weight on one end of the bowl, what happens to the bowl?

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As long as the weight is placed slowly, the bowl will remain balanced.

The forces on the bowl are

  • Gravity
  • The pole
  • The pressure from the water

Note the lack of anything about the added weight. That's because it can only effect the bowl by raising the water level.

All three of these will remain centered on the pole as the water level rises, as long as the incremental center of area remains in the center.

If any of the axes of the ellipsoid is aligned with the pole this is trivially true.

The specific criteria that must hold true is:

$$\oint \! \vec r \times d \vec t \times \vec r=0$$

Where $\vec t$ is tangent vector to the path along the water-air-bowl interface, and $\vec r$ is the shortest vector from the axis of the pole to the position along the path.

Of course if the weight is placed quickly into the bowl, then the momentum of the weight would transfer to the water and the bowl. With no way to counteract an off-center impulse the bowl would then tip off the pole.

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  • $\begingroup$ I think there needs to be some constraint on the shape of the bowl as well (such as that it is symmetrical about the position of the pole). I'm not sure that "mildly elliptical" is sufficient. $\endgroup$ – BowlOfRed Sep 24 '15 at 20:50
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    $\begingroup$ @Rick what about the floating weight placed not inline with the axis of the supporting pole? $\endgroup$ – Armadillo Sep 24 '15 at 20:59
  • $\begingroup$ @jakemcgregor It doesn't matter where the weight is placed as the water/weight system only effects the bowl through the pressure of the water which is only determined by the height of the water. $\endgroup$ – Rick Sep 24 '15 at 21:01
  • $\begingroup$ @Rick Your constraint is just an expression. Shouldn't it be an equation? ($=0$) $\endgroup$ – Bernhard Sep 26 '15 at 9:19
  • $\begingroup$ @jakemcgregor the weight displaces an equal weight of water, which causes the water level to rise uniformly in the bowl. The center of mass of the bowl will not be affected. $\endgroup$ – Ernie Sep 30 '15 at 0:02

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