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Consider a stack of wood chips: each 0.5cm thick and 2x2 cm in length and width. There are 200 of them all stacked on each other. For some reason they all instantly fall. Evwn though their centre of gravity is at the centre of the stack and they have the extra added help of friction the further down the chips you go (because the second to bottom chip is squashed against the first because of the heavy load). So why does it fall?

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  • $\begingroup$ It wouldn't if the blocks were perfectly shaped. But they aren't. $\endgroup$
    – pela
    Oct 21 '18 at 17:56
  • $\begingroup$ ... And if the floor was perfectly level $\endgroup$
    – mmesser314
    Oct 21 '18 at 18:01
  • $\begingroup$ So what makes the building more delicate the taller it gets $\endgroup$
    – yolo
    Oct 21 '18 at 18:14
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First, let's assume the tower is a single solid rod. The tower will fall over when the center of mass extends past the base. For a longer tower, a smaller angular displacement is needed for this. This is because we want the horizontal displacement of the center of mass to be $$x=\frac 12 H\sin\theta<\frac 12 w$$ where $H$ is the length of the tower, $w$ is the width of a block, and $\theta$ is the angle the tower makes with the vertical.

Therefore: $$\sin\theta<\frac HL$$

The larger $H$ is, the smaller $\theta$ needs to be. This means that for a taller tower, smaller disturbances can cause it to fall.


Now our actual tower is not a single solid rod, but the same idea applies. Additionally, if the tower fails this at any section it will fall over, with taller portions falling onto ones below. This is why it seems like everything falls at once, as you have said. I'm not sure the friction between the blocks really matters, since there probably isn't a lot of shearing going on here.

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  • $\begingroup$ Could you explain why those formulae are the way they are. $\endgroup$
    – yolo
    Oct 21 '18 at 21:05
  • $\begingroup$ @yolo The $\frac 12 H\sin\theta$ is just from the triangle formed by the COM, the point directly below the COM on the ground, and the point where the center of the vertical tower would touch the ground. The $\frac 12 w$ is just the distance from the center of the tower to its edge. $\endgroup$ Oct 21 '18 at 23:24
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An additional factor that I haven't seen mentioned yet is the fact that the environment acts upon a structure such as that. Errant air currents would have a deleterious effect on such a stack the higher it went. Even a tiny breeze would topple a perfect stack if it was beyond a certain height.

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  • $\begingroup$ This is mentioned generally in my answer "This means that for a taller tower, smaller disturbances can cause it to fall." $\endgroup$ Oct 23 '18 at 2:13
  • $\begingroup$ Ahhh... I didn't see that. However, I did cover the topic in tiny bit more detail, in case other people don't see your full comment, or ruminate on the specifics of it, such as I did. Thanks though. $\endgroup$
    – Ace Hall
    Oct 24 '18 at 7:36

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