Tipping The Toy

My brother phoned to remind me how much more generally nervous I should be about things, as well as to ask my opinion in an utterly pointless dispute he was having with his significant other. The dispute was over no stakes whatsoever and had no consequences of any practical value so I can see why it’d call for an outside expert. It’s more one of physics, but I did major in physics long ago, and it’s easier to treat mathematically anyway, and it was interesting enough that I spent the rest of the night working it out and I’m still not positive I’m unambiguously right. I could probably find out for certain with some simple experiments, but that would be precariously near trying, and so is right out. Let me set up the problem, though, since it’s interesting and should offer room for people to argue I’m completely wrong.

Their child has this toy. It’s basically a box, a cube, really. Naturally the child spends considerable time trying to tip the toy over. Sometimes the child manages it by pushing hard at the center of the edge and sending it rolling. Sometimes the child manages by pushing at one corner until the whole thing tips over. Neither seems to put the child to more strain than the other method, but the question is: which is harder to do, tip the box over by pushing from the center of an edge, or by pushing from the corner?

I thought of three ways to tackle this problem, and might come up with more before I’m done, particularly if readers have other ideas about ways to model the box and the tipping-over. I want to lay out first the assumptions I’m making, the ways I simplify the box so that I can treat it without doing hard work.

First of all, I’m going to pretend the toy really is a box, six sides, each of them a square, each square with a side of length $a$ . I picked the letter $a$ to have some useful label; $l$ would probably be about as good except that it looks too much like a “1” in some typefaces or in my handwriting. $s$ is similarly good for a side length but can be blurred into “5”. And since I don’t really care how big the toy is I could set the size to 1 or 5 or 10 or whatever just as easily, but I have an instinct that says not to tie myself down to any one length if I don’t have to.

The next thing I’m assuming is that the box’s contents are homogeneous. This is just the same idea as in homogenized milk; there’s no particularly dense or hollow parts of the box. It’s all the same contents throughout the whole innards. This makes all its physical properties a lot easier to work out.

For example, since the box is a cube, and it’s homogenous inside, then we know just where its center of mass is. If we rest the box on the floor — which we’ll assume is horizontal, since I haven’t lived in the place with noticeably non-horizontal floors in years (it was a fine apartment even if the doors couldn’t necessarily be closed) — then that center is a distance $\frac{1}{2}a$ off the floor, and for that matter the center is a distance $\frac{1}{2}a$ away from the center of each face of the box. The center of mass is worth knowing because for a lot of physics problems we can pretend a solid object is just a dot of mass at the center of mass, and come out with the right answers anyway.

While it’s a distance of $\frac{1}{2}a$ from the center of mass to the center of any face, it’s a distance of $\frac{1}{\sqrt{2}}a$ from the center of mass to the center of any edge, that is, the straight lines that marks the ends of the squares of each face. And finally it’s a distance of $\frac{\sqrt{3}}{2}a$ from the center of mass to any of the vertices, the corners, of the box. We’ll need the distances later on. Also, since it’ll be needed, the diagonal distance from, say, the upper-left to the lower-right edge is $\sqrt{2}a$ , while the long diagonal from one corner to the farthest other corner is $\sqrt{3}a$ . That’s too dull to bother proving, but it’s the Pythagorean Theorem ruthlessly applied. Or it’s the L² norm, if you want to make it sound more intimidating than it is.

And, what the heck. Let’s suppose that the box isn’t so enormously large that we have to pay attention to the curvature of the Earth. That is, the acceleration due to gravity is some constant number, which convention names $g$ . Also, the box has to have some mass; physics’s convention is to call that m unless there’s good reason not to, such as there’s a second object which also has a mass that we care about. There isn’t, so we don’t have any reason to go with anything but $m$ here.