[2013-Dec-02] Physics Problems

[InfiniteHobos]

According to my calculations, the answer is 1/3. That is, the axis should be 1/3 of the way up the level of the liquid (20 m), i.e. 20/3 m.
It doesn't depend on the density of the wall or liquid. The only density that matters is the density difference between the liquid and air, which cancels out, because it acts both above and below the axis. The density of the wall never enters in the first place.

You just need to integrate the torque exerted on the wall (due to pressure) from the ground to the axis, and from the axis to the surface, and set those two quantities equal and opposite to each other, so the net torque is zero. (From the surface to the top of the wall, there is air on both sides so there is no pressure differential and hence no contribution to torque.)
Pressure is proportional to depth, so you basically just integrate (x dx) and get x^2, since all the proportionality constants cancel out.
If 'h' is the height of the liquid and 'A' is the height of the axis, then you evaluate it like this:
x^2 |{from 0 to h-A} = x^2 |{from h-A to h}
This leads to the polynomial
3A^2 - 4hA + h^2 = 0
which is easily solved for
A = h/3

Curiously, the quadratic formula also gives A = h as a second solution, but I can't see any physical validity to this. Looking back at how I formulated the integrals, the only significance of this solution is that it makes them turn into 0=0, so it is safe to discard this solution as spurious.

In short, the problem was fun as it was and I joined just to post that.

[InfiniteHobos]

Oohhhhh oops, the 2nd solution (A = h) doesn't end at just 0=0 based on the limits. It actually leads to the implication that h = 0 and A = 0, i.e. if there is no liquid, there is no axis of rotation.
So it does lead to a valid physical solution after all! I'm so glad. Always trust the math.

The existence of antiparticles was implied like that, by the math, before they were discovered.