 Today, let's discuss static equilibrium from another perspective. Let's talk about the static equilibrium, where we include moments as part of our static equilibrium. So we've already talked about having a condition where we have two force vectors that are in line with each other, or collinear, that balance each other. But there are conditions where we might have two force vectors that are balanced in one dimension, so they're equal and opposite, but they might be displaced in a second dimension. In other words, they're not perfectly lined up. They're not along the same line. They are not collinear. They are parallel. And in that particular case, that will change things a little bit. Let's draw a picture. Here I have the edge of a table and a couple of water sitting on the edge of the table. Notice that's a little bit dangerous. The weight of the water might be considered as being at the center of the cup, whereas the normal force holding up the cup would be applied to the bottom of the cup, but not in the center of the cup if the cup's moved too far over. Notice there's a little bit of a displacement of very small distance between the line of the center of mass of the cup and the line of the normal force. This status or this situation is going to cause a change in the motion, but this change in the motion is going to be a rotation, some sort of spin. Not only will the cup drop, but this cup is likely to turn over, spin around, and pour its contents onto the floor. However, we already know that static equilibrium implies no change in motion. So let's talk a little bit about how moments affect this. This impetus to rotate or this tendency to rotate depends on the amount of the force, how much is pushing, and it also depends on how much the line of force is displaced with respect to some axis in space, some point in two-dimensional space. So if I apply a force here, actually, let's consider a force applied at an angle here. If I have some force and I want to apply it about some point, here's the point, the amount, the tendency for it to rotate is going to depend both on the force itself, the magnitude of the force, how hard you're pushing, and the distance of the point you're thinking about, rotating about, we'll call it point P, from the line of action of the force. And this distance we always take to be the minimum distance from the line of action, which is also the perpendicular distance, the distance along a perpendicular from the line of action. And our relationship there, moment, is equal to force times this distance. And this distance has a name. It's called the moment arm. Let's consider the example of a wrench. Notice many wrenches have two different sizes on either side. If I take the wrench and I apply it to a bolt point A, I could push at two different locations. I could push very close to point A, or I could push further away from point A. If I want to get the most moment around point A, I would push at this further point. Notice the direction that we tend to spin, we can define as being either clockwise, moving in the same direction as standard clocks, or counterclockwise, moving in the opposite direction of standard clocks. ccw for counterclockwise and cw for clockwise. So if I push at the point on the right, I'm going to get a greater moment, and it's going to tend to make a spin in a clockwise direction. If I instead consider point B, if I instead apply the wrench to a nut on another location, point B, well, then that other point, the point further along the handle away from point B, the one here on the left, this is where we're going to push for a greater moment around B. But notice if we push in this direction, it's going to be a counterclockwise moment. It's going to be a tendency to make it spin the other direction around B. Notice in both of these cases, the reason why the moment is greater is because of the length of the moment arm, because of how far the force is from the point we're choosing to rotate around. So a spin, some form of rotation, a spin will begin unless clockwise moment is equivalent to counterclockwise moment. So if you apply a moment to something, it will begin to spin unless there's a moment in another direction. In other words, if I want something to spin one way, or if I don't want something to spin one way, I have to make sure that there's something trying to make it spin the other way. The moment in my clockwise direction has to be equal to the moment in my counterclockwise direction if I don't want it to spin. And again, the idea of static equilibrium is that we don't have motion. If I do a little algebra here, we could also think about this as being the moment in one direction minus the moment in the other direction has to add up to some sort of zero moment. Now, we notice that this is automatically true if you have two balanced, collinear forces like we had before. If I have one force that's pushing down on an object and another equal force, equal and opposite force pushing up on an object, it doesn't matter what point we choose to pick. That point will be a distance from the line of action of both forces. But we can note that the moment applied in the clockwise direction is applied by the red force here is equal to the force times the distance. Well, the moment applied in the counterclockwise direction by the green force is also equal to the force times the distance, but in the other direction. So the other way we can look at it is we can say the force times the distance in the clockwise direction and the force times the distance in the counter clockwise direction, which we represent with a negative in this case, we see that that's equal to zero. In other words, the sum of the moments taking into account the directions they're going is equal to zero. The sum of the moments around some point, let's go ahead and label this as point P. We usually represent the point by putting a little subscript there on the sum of the moments about that point. However, that can be any points. So again, the idea is sum of moments is equal to zero. That's, again, automatically true about two forces that are collinear. However, it's not necessarily true of two forces that are balanced but parallel. And it's pretty simple to see why. If I again pick some point, now you can see that the distance from the line of action from that point to each of the forces is different. In the first case, we have a red distance. Let's call that distance 1. And we have a green distance. Let's call that distance 2. Well, in that case, the moment trying to make us go in the clockwise direction is equal to the force times d1, whereas the moment in the counterclockwise direction is equal to the force times distance 2. And you can see that those are not equal to each other because the distances are different. So it's impossible to have static equilibrium of moments if you have two forces that are not collinear. However, if we have more than two forces, for example, three forces that are all co-planar, in other words, they're all lying in the same plane like our piece of paper, if we take three or more forces, we could potentially determine what their moment is about a point. Each of them will have moments about a point determined by their distances from the line of action of each of those forces. And we can determine whether or not those might be in static equilibrium. Let's consider an example. Here we have a picture of a friendly parakeet sitting on a wire. Let's go ahead and see if that friendly parakeet might be in static equilibrium. Well, he doesn't seem to be falling. He's not flapping his wings. So let's see what conditions might exist so that he can sit there comfortably without having to, well, spin. Notice he seems to be a little displaced, so he's not sitting in the middle of his little swing there. He's sitting off to one side a little bit. So let's draw a free body diagram representing, well, we won't represent the parakeet itself. Let's represent his little swing. Here's his little swing. And what things are acting on his little swing? Well, first of all, there's a wire on one side holding up the swing. And there's a wire on the other side holding up the swing. And we believe that those wires are being pulled or stretched so they're in something we call tension. So we're going to replace them with a force tension 1 on the left side and tension 2 on the right side. So two forces that are holding up the swing. And then we also recognize that the parakeet has some weight. And we're going to assume that this weight is substantially more than the weight of the swing. So for the moment, we're going to neglect the weight of the swing, although we could include that if we wanted to. But let's assume that the weight of the parakeet is applied at some point that's not the middle of the swing. And there is our weight of the parakeet at that point. So let's go ahead and label these particular points here. Let's see, this is going to be the weight of the parakeet. All right. So if we want this system to be in equilibrium, we have to apply some rules. Well, our first rule of static equilibrium is that the sum of the forces in the x direction is equal to 0. Well, we haven't exactly established what x and y are. So let's go ahead and create, before we go any further, create our basis. Let's call the x direction the horizontal direction. Let's call the y direction the vertical direction. And I'm going to add one more definition here. We're going to stick with this idea that moments that are clockwise, we're going to call those positive moments. So I'm going to draw a little arrow here in a little circle and demonstrate that clockwise is going to be what we consider to be positive. So when I say the sum of the forces in the x direction is 0, we notice in this case there are no forces acting in the x direction. So that doesn't really help us solve our equation at all. So we can use that. It's already satisfied. Everything is 0 in the x direction. Let's then consider the next equilibrium equation. Sum of forces in the y direction has to be equal to 0. Well, let's add up the forces in the y direction. The first force in the y direction is our tension number 1, which is pointing up and we'll label it as positive. We also have tension number 2, which is pointing up and will keep us positive. And then we have the weight of the parakeet, which is pointing down. And we know that those have to add up to give us 0. So now we've established that. And if we know the weight of the parakeet, we can find, well, we have one equation, but we have two unknowns. We don't know T1 and we don't know T2. If the parakeet was sitting at the middle, we could make an argument about the symmetry and sort of say that they must be equal. But in this case, because the parrot is sitting a little bit closer to one side, it's pretty evident that the tension on that side is going to be a little more than the tension on the other. So let's go ahead and define a couple distances here. Let's go ahead and define this as being distance 1 and define this as being distance 2. The distance is from the weight of the parakeet to each side. And now we have a choice. We can actually find the moments around any number of possible points. We're going to actually do this. We're going to choose three different points to try to do this around and show that it doesn't really matter which point you choose. Let me go ahead and choose three points. Here's the first point. We're going to call this point A. We'll call this point B. And we'll call this point C. Let's assume that we decided to choose the moments around point A, the sum of moments around point A. Well, if we look at point A, we notice the point A is actually on the line of action for T1. So in other words, force T1 is multiplied by a distance of 0. There is no moment arm. So in this case, force T1 doesn't really come into play if we're doing the moment around point A. If we do the moment around, if we continue with the moment around point A, let's look at how the weight affects it. Well, in this case, if we're looking at point A, our weight is creating a moment that's going to make this spin in our clockwise direction. So that's going to be a positive contribution. So we're going to add the weight times the moment arm, which in this case is that D1, the distance of W from point A. And finally, let's do the same thing with T2. We have the tension T2, the moment arm of T2, the distance it is away from point A is a distance of D1 plus D2. It's the entire distance of the entire length of the swing. And we also notice that that is going to be a rotation opposite our positive direction in the counterclockwise direction. So we're going to put a minus sign out in front of that. And the sum of all those forces or the sum of all those moments is going to be equal to 0. So that's one way that we can sort of write that. That would be one moment equation we could use. I'm going to build a similar moment equation this time around point B. What if we did the sum of moments around point B instead? Let's try that. Well, if we look around point B, now we see from point B, we have T1 tending to make us spin in our positive direction. But now it's a distance of D1 away from that middle point B plus the weight of the bird. Well, the weight of the bird is applied directly at point B, which means it has no distance from point B, which means it has a moment arm of 0. And that will not contribute at all to the moment. And then we look at T2, which is, again, going in a counterclockwise direction. So we use a minus sign, but it is now only a distance of D2 away from point B. Now, we won't normally need both of these equations. I'm going to do a third one as well and see how it works out in point C. Usually, we only need one. And I will show you why momentarily. Let's go ahead and look at the third one. The sum of moments around point C. Can you see what they are going to be in this particular case? Let's start with T1. Again, that's going to be a positive direction. It's clockwise. But now it's going to have a further distance away from point C. In this case, it's the sum of both of those. If we look at the weight of the bird, the weight of the bird is making it move in a counterclockwise direction. And that's at a distance D2. And we'll have a minus sign because it's counterclockwise. And then in the case of T2 itself, since it goes through point C, it actually has no moment arm, so it doesn't contribute at all. But once again, the sum of those is going to be equal to 0. So we have three possible equations to use. Which one do we use? Now we have four equations, but we only had two unknowns. Well, it turns out these three equations down here all effectively say the same thing from an algebraic point of view. Let's check that out and see how that works. Let's see if we can go ahead and solve these equations. We'll start by making a relationship using the sum of forces in the y direction and recognize that the weight is equal to the sum of the two tension forces. And then everywhere we see a weight, we can actually substitute in this sum. So if we now recognize the sum of forces around, sum of moments around point A is equal to, well, the first part goes away because T1 times 0 is 0. Is equal to T1 plus T2 times D1 minus T2 D1 plus D2 equals 0. Well, if we do a little algebra, we'll notice that there is a multiplier here of D1 times T2, D1 times T2, and D1 times T2. So we can cancel out that. They're going to subtract away D1 times T2. And that gives us T1 D1 minus T2 D2 is equal to 0. Or that T1 D1 is equal to T2 D2. Or one other way to, well, yeah. Let's go ahead and leave that that way. So we can see a relationship between T1 and T2. That substitution probably didn't help us a lot. If we were trying to solve for T1 or T2, maybe we could take that back in and plug it into the W equation to figure out how it relates to W. However, I'm going to move on and show that this result here that we see is going to be the same result we get no matter what moment we take. Let's instead look at the sum of moments around B. Sum of moments around B is equal to T1 D1. In this case, the weight goes away minus T2 D2, which is the exact same equation algebraically as we just solved for above. And let's do this one more time. Sum of moments around point C, we recognize that we have T1 D1 plus D2 minus W, which replaced with the T1 plus T2 times D2. And then that last segment up there goes away. So if we've taken the sum of moments around point C, you'll notice we get a third equation. And if we continue with that math and do a little bit of algebra there, you'll notice once again that we can cancel out. We have T1 times D2 and negative T1 times D2. So in those case, those will subtract out and we'll get a relationship again. Once more T1 D1 minus T2 D2, the sum of those moments has to be equal to 0. Or again, we can relate that back to this equation here. So it doesn't matter which point we chose the moment about. They are all algebraically the same. If we were trying to solve for T1 and or T2, we can then plug this relationship back into our weight relationship. We know that our weight is equal to T1 plus T2. We can recognize that T1 is equal to T2 D2 divided by D1, if we divide both sides of that equation. We can substitute that in and we would get that T2 D2 over D1 plus T2, the weight would be equal to both of those, or equal to D2 plus D1 over D1 T2. In other words, I take this T2 and add a fraction D2 over D2 and then add the two portions together, a little bit of algebra. And that would give us a relationship between T2 and W. That T2 is equal to D1 over the sum D1 plus D2 times W. Similarly, T1 is going to be equal to D2 over the sum times W. What that basically means is that this value, which is our T2 times that side is equal to this value times that side, that you have an inverse relationship between the two tensions. But the key point here is that the sum of moments is equal to 0, no matter what choice you make for the point that you rotate about. So to recap, static equilibrium. Sum of forces in the x direction is equal to 0. Sum of forces in the y direction is equal to 0. Sum of the moments, and we're going to put a z here, is equal to 0, where these are moments around some z-axis. But notice a z-axis running through the page is going to simply be a point when we are in two dimensions.