Sorry, this is going to be lengthy, skip the first and second paragraphs if you haven’t the patients.
Yes, the amount of work required to bring your table to the point of tipping is, or at least I agree that it is, a good measure of how stable your table is. However considering how much force it will take to make the table move, round its fulcrum, is also valid, after all, you need to know what force you’re applying at any given moment in order to find work. For the time being, let’s simplify, and consider torque around the fulcrum rather than force applied to the side, as that’s another matter. This torque is just the torque applied by gravity at any given angle of your table. It can be expressed as cos(t+u)*m*g*l, using variables from before, and u being the angle between the leaver arm made from fulcrum to CM, and the bottom horizontal of the table (I forgot to add this in my equation before, thus, a, is 90-u).This is finding the component of gravity, acting perpendicular to what amounts to a leaver arm, from the fulcrum, to the center of mass, and multiplying it by the length of that leaver arm, l.
Now it’s not really meaningful to consider the torque alone, however it is somewhat meaning full to considered the amount of force required to get the table moving, this is simply the force you need to apply to overcome the torque applied by gravity. First you have to consider how you’re going to apply this force, F. I’m going to make the assumption that we’re considering the force to be purely horizontal, applied to some point on the side opposite to the fulcrum, some point that is, say, d, distance off the bottom of the table. Now because were applying a force to overcome a torque, we work out what torque that force becomes, so we have a second, as it were leaver arm, going from the fulcrum to the point at which were applying our force, so we have an angle that leaver makes with the horizontal, call it t2, and a length of that leaver, call it l2. t2 will be arctan(d/w), where w is the width of the table. L2 will be ((w)^2*(d)^2)^0.5. The force being applied perpendicular to the leaver, should be F*sin(t2). So the torque applied around the fulcrum by our horizontal force, F, acting on the side, w width form fulcrum, at distance, d, from the bottom, is F*sin(arctan(d/w))* ((w)^2*(d)^2)^0.5, which in order to move the table, has to be greater than the torque applied by gravity, cos(t+u)*m*g*l.
This is overly complicated; it’s not practical to evaluate stability based on this. However we can pick up on a few things, first, the force and torque required to overcome gravity does indeed rely on the height of the CM, more generally the position of CM relative to the fulcrum, as this is the only variable (assuming constant mass and g) which changes how much torque gravity applies, the torque we have to overcome. Now we can make the assumption that for any large variance in CM, the dimensions of your table, more importantly, w and d, change somewhat respectively, which is where I think handleman is coming from, that, practically speaking, for whatever height of your table, the CM will be roughly in about the same relative position. But we need to be careful with that kind of assumption, as it won’t always hold true, I’d say it basically never holds true, but that all depends on how much you want to reduce theory to rules of thumb. But the general point I’m trying to make here is that it’s impractical to consider this horizontal force, too much math, and too many assumptions. All we really have to consider is the torque gravity is applying on the CM, cos(t+u)*m*g*l. We can use this torque alone, and get an idea of how much more it takes to get one table going, than another, all relatively speaking. Or we can consider the about of work required to tip it over fully, which is that torque, cos(t+u)*m*g*l, integrated, with respect to t, from 0 to a, where, again, a is the angle t has to be in order for the CM to be above the fulcrum, the point at which the table will tip over. This integration gives us angle*torque, basically work, the amount of work we have to do throughout the range of 0 to a, to completely overcome gravity. The integral becomes sin(a+u)*m*g*l- sin(u)*m*g*l, again, I forgot to include u before. However, personally I would just go with evaluating cos(t+u)*m*g*l for some table of minimum stability, and using that as a minimum for your other tables, this is really easy, if you already know your CM, all you have to do is work out what angle it makes with the horizontal, or u, and how far it is from the fulcrum, l, and the mass, which I assume you know. Personally if I’m sitting at a table I’d rather the table not move at all, rather than just not tip far enough to fall completely over, it’s a bit more practical. So if you have some table that doesn’t move at all, using cos(t+u)*m*g*l for it, as a minimum, means that all tables following that, wont tip either.
Longwindedly yours,
Nick
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