Here we're going to talk about how if you take any object and you hang it from a single point, it ends up hanging so that it's center of mass is directly below the point. For instance, if I had this weird forceps here, and I wanted to know where's the center of mass. I'd start by hanging it here, and then I would hang it here, and then I could even hang it here. I'm going to show you in detail how you calculate this and how you can use it to find a center of mass. But let's not do those forceps. Let's find something simpler. Let's start with four masses connected by light rods in a square. Because that's an ideal object to start with. We know what the center of mass of this is by symmetry and intuition, I guess we note that the center of mass is there. If these all have the same mass, and it's a square, that's where it would be. Now, let's imagine that we put a pivot point or we put it on something that allows it to rotate around that axis, and just let it go. What you can see it's going to fall this way. It's actually going to be an oscillator. It's going to go back and forth forever. Let's say there's some damping force where eventually it ends up in its most stable position. That's what we think, that stable position should have this directly below that. That's what we want to find out. That's generally true in statics. To approach it mathematically, we're going to say it could be at some angle. Let's draw it at some angle. Here we go, like that. It could be like this. Where these are solid mass m, a side length L. Now we say, oh, it's rotated to an angle Theta. The question is stable at what Theta or it's static at what Theta? Here we go. Let's say statics, you say sum of the forces equals 0 and the sum of the torques equals 0. Often you're building up a bunch of equations because you have a bunch of unknowns, but here we only have one unknown, our only unknown is the angle Theta. The force is actually aren't going to tell us anything in this case. We really just want to do a torque condition around one axis to find one angle, this is really all we're doing. Let's do that. Let's do torque around that axis, and we need to do it for really just three of the masses, because this one is on the pivot point, the gravitational force on that mass will create no torque. We're going to number them, I believe, 1, 2, and 3, and figure out what angle gives us no torque. Here we go. Let's see, we could start and let's run it like this, 2 and 3 and start summing them up. For one, let's really do it in terms r cross Fs. This is a good chance to practice r cross F. For one this is the axis. There's r and then F is the mg straight down. I'll draw them over here to separate them out for you, so that r is actually a length L. This is the vector r, but its magnitude is L. Mg is that way. We going to think, let's see if we were to add these tail to tail, then the angle we want for r cross mg, it looks like it's 90 minus Theta, because there's Theta, because there's Theta, and this is 90. That angle that's left over that's r is 90 minus Theta. That one must be L because that's the vector mg sine of 90 minus Theta. But then it's going to tend to turn it clockwise so it's negative. They did the cross-product manually. We just put in the sign of the angle who made it negative because it's actually a negative angle from r to mg. Now let's see. We also want the second one, I messed up my position on the board here, but that's okay. The second one is here. Now here is r for the second one. It's down like that and then mg is straight down. Let's draw the second one? It's like this, r and then mg. Let's think what would that be? I'm going to go ahead and add it here. This one is also going to tend to make it go clockwise, it would also be negative. The length here is square root of 2L It's across the diagonal minus square root of 2L. The force is mg, so mg. Let's see how I'm doing here. I'm really getting it right. Now we just need that angle. That was a weird angle. The sine of what is it? It's like this one 90 minus Theta, but it's 45 more because it's across the square. I called it 90 minus Theta minus 45. It's minus 45 because that 45 was making this angle smaller. You have 90 minus Theta take off 45 more and that's that term. Then finally there's this one. Again let's see this one. R is L again and this is going to drive it this way. It's actually positive. We would say plus and I can't quite fit it. I'll put it down here. Plus let's see. It's L mg and then the sine of what horrible angle have we built up this time? Let's see was zero when you turned it Theta. This one is just Theta times the sine of Theta. The sum of those three terms we're trying to make zero. What Theta is the sum of those terms zero. Let's do some easy stuff. Let's see, let's cancel the L mg. I'll do that. You do the rest. L mg. What are we left with here? Let's see the sine of 90 minus Theta is equal sine of Theta. We can do that. We'll bring this call this cosine Theta. That was a good one. Now we have the square root of two left and we have the sine of 90 minus 45 minus Theta. Let's do 90 minus 45. I'll call that pi over four. This one we can't turn from a sine to a cosine or something because it's got a 45 in it. It's not just pi and pi over two. Then this we're just going to leave this plus sine Theta equals zero. What a mess. Let's see. What we're left with is minus cosine Theta minus the square root of two. Now we actually have to turn this from sign A minus B and use the identity where I would see if its sign A minus B, it's sine A cosine B. Sine pi over four, cosine Theta is minus cosine pi over four sine Theta. Because that's just an identity to break up that sum and the argument of those two. Then plus sine Theta. It's really complicated. But it's about to get simple because let's see what is this? This is square root of two over two. What is this? This is square root of two over two. We can pull out a square root of two over two because those two are the same thing. What is square root of two times square root of two over two it's one. That just went away. We're left with cosine minus cosine plus sine plus sine is equal to zero. Let's see sine plus sine. Let's bring it up over here. Two sine Theta minus cosine minus cosine minus two cosine Theta equals zero. Let's cancel the two and say sine Theta equals cosine Theta. Sine Theta over cosine Theta equals one. When does that happen? 45 degrees. You could go further and take the tangent. The main place that happens that we care about for how this falls is 45 degrees and when it falls 45 degrees, it looks like this. It's hanging as you would expect. Where if you have your pivot point here and you draw a line straight down with gravity, the center of mass will be under the pivot point. That's all we needed. We needed to find an angle of 45 degrees there to confirm this sentence. Now to use that for a complicated object like the forceps or let's draw a more complicated object. Let's draw a nice two-dimensional complicated object like something like, oh, I should have thought of this ahead of time. A house. You have a little thin house. You want to know where the center of mass is. You could hang it like this and symmetrically you can see it's going to hang straight down and you could draw a line on it like that. Then you could hang it again from say this part. Hang it here, what's it going to do? It's going to fall a little bit. It's going to end up like that. It's going to do something like that. You can draw a line there because you know the center of mass is somewhere along that line. But from your other one you knew the center of mass is somewhere along this line. You can do more and more and eventually you can find where they cross. You really just need two. Where those two cross, that is the center of mass of the object. Works well for two-dimensional objects, three-dimension, you have to draw a line through an object. A little more complicated. But we definitely saw, we know it's true that the only way you can have the torque equal to zero, is if the center of mass falls directly under the pivot point.
