Tower of building blocks

Question

Why does a tower of building blocks (cubes) fall?

Theoretically, as long as the center-of-mass is above the blocks' bottom faces, and as long as one does not shake the tower too much, it should not fall. This means that, with sufficient care, one can build a tower infinitely high. @Bernhard has suggested the link confirming this.

Nevertheless, any real tower of building blocks collapses, if it is too high. I am interested in the reasons for this collapse, the mathematical model for this instability, and the experimental evidence confirming this or that model of collapse.

Possible explanations
Scenario 1
The collapse of the tower could be due to the fact that the supporting surface is not perfectly horizontal. In this case there exists a limiting height where the center of mass is not supported anymore and the tower flips over.

The condition of stability in this case: $$ \tan\alpha < \frac{a}{h} = \frac{1}{N}, $$ where $s$ is the side of a cube, $h=Na$ is the height of the tower of $N$ cubes. For small angles ($\tan\alpha\approx\alpha$) we get $$ N<\frac{1}{\alpha}=\frac{180^\circ}{\pi \alpha^\circ}\approx \frac{60^\circ}{\alpha^\circ}, $$ which means that for incline of $1^\circ\sim 10^\circ$ the critical height of tower ranges from 60 to 6 cubes. In my experiments it is about 20-30 cubes, which corresponds to an angle of $2^\circ-3^\circ$ - not too big to exclude on the basis of visual observations, but also not too small to be a routine thing in home conditions.

Scenario 2
Alternative scenario is imperfections of the blocks themselves: if their faces are not perfectly parallel. It seems that in that case the tower does not necessarily turn over, but a part of it may simply slip. I.e., one could distinguish different scenarios experimentally.

Scenario 3
@BillWatts proposed the self-buckling instability as a possible explanation. In this case the critical height its (in length units): $$ l_{max} = \left(7.8373\frac{EI}{\rho g A}\right)^{\frac{1}{3}} $$ The cubes that I am experimenting with are wooden cubes with side $d=4.5$cm and mass ranging from 30 to 50 gramms. Taking the Young's modulus of the wood as $10$GPa, the equation above predicts the tower height of 17.5 meters, i.e. nearly 400 cubes, whereas in practice it collapses at 20-25 cubes. Still, this hypothesis cannot be ruled out, since the equation above was derived for a solid wooden tower, rather than disjoint cubes - i.e., the effective Young's modulus can be much smaller.

Scenario 4
@RogerWood proposed an estimate of instability arising from the fact that the blocks are not be perfectly aligned. While I do not believe this to be a correct explanation, it is the most likely cause of collapse of the towers built by a child.

Scenario 5
@GTos suggests that one necessarily provides a small torque when putting the last cube, which might be crucial for higher towers.

Some experimental evidence
My testing cubes have four smooth faces and two faces with a carved pattern. When stacking the cubes by smooth faces, the towers typically collapse at 20 cubes. However, when the cubes are stacked by the rugged faces, I can exhaust all the 28 cubes available. Ruggedness of the cube faces may affect friction between the cubes, but also the effective Young modulus. I have repeated this test in different locations, and putting cubes in random order, so I believe that it allows us to exclude the first two scenarios, and speaks in favor of scenario 3.

Update
This article addresses a very similar system, but the experimental methods are somewhat different from what one could do with already made wooden cubes that I have.

Note for moderators: this article was originally published in the American Journal of Physics, but the version cited here is from the open access HAL archive (French equivalent of arXiv).

Update from 14/12/2020

• If we go on with the buckling theory, then here is yet another very interesting article, which discusses the buckling of a column as a second order phase transition.
• @AndersSandbreg has brought to my attention a book-stacking experiment, which demonstrates how different scenarios could be in play.
• As could be seen from the above, the major difficulty in confirming/rejecting the buckling hypothesis is not knowing the Young modulus of a composite object, such as a tower of cubes.
• • One solution could be to measure Young modulus from the frequency of oscillations of towers of different height (which requires some technical prowess, but should be doable).
• • Another solution is to associate the tower collapse with the critical slowing of the oscillations (see the paper cited above).

Answer 1

Let's call the $n$th block from the bottom $n$. If the center of mass of blocks greater than $n$ is not over $n$, the tower collapses. If one thinks of a tower that goes left to right to left again, etc. it will collapse even if the total center of mass of $n>1$ is over $n=1$.

Answer 2

It could be due to the torque which is directly proportional to the height of the tower. In fact when you lay the $n_{th}$ on the top it is impossible for your fingers to give it initial velocities equal to zero (due to vibrations caused by blood pressure) along the x-y directions. So we would have an impulse $I = mv=F dt$. So higher the tower higher the torque until everything falls down.

Answer 3

It's interesting to pose the problem as everything being level but each block having a small random Gaussian displacement from the one below. The displacements accumulate as the tower rises, so the position of the top block does a random walk away from the position of the bottom block. It seems sufficient to consider this in just one dimension.
Let the blocks be 2x2x2 and let $\sigma^2$ be the variance of the relative position of two adjacent blocks. For only two blocks, if the relative position is smaller than 1, the top block will not fall off. This occurs with probability erf(1/($\sigma\sqrt2$). For blocks n apart, the variance is n$\sigma^2$. The variance of the center of gravity for a stack of N blocks is
(1/N)$\Sigma$_n=1,N n$\sigma^2$ = ((N+1)/2)$\sigma^2$ relative to the block they are stacked on. The probability of that entire stack not falling off the block below is erf(1/($\sigma\sqrt(N+1)$).
For a stack of height N+1 (including the bottom block), the probability that not any part of the stack topples is $\Pi$_n=1,N erf(1/($\sigma\sqrt(n+1)$). You can square this if you want the probability that nothing topples in either direction.
As I'm writing this, I'm realizing that the probabilities of different portions of the stack not falling are not independent since they contain some of the same blocks with the same displacements. So I'm very uncertain whether that product of probabilities is really valid ...