 So the term static equilibrium relates a physical observation that something stays at rest or in motion to a theoretical observation that the forces which cause motion change are equal and balanced. So if something is not in motion, it must mean that the forces acting on it are equal and balanced. So notice a single non-zero force, something sitting alone by itself, cannot be balanced. It's going to create some sort of motion if acting by itself. Two forces, however, can be balanced if they are co-linear. In other words, if they're acting on the same line and opposite in sense. Consider, for example, our Statue of Liberty. We know that the Statue of Liberty has a certain amount of weight. We also notice that she doesn't move very much. We consider her to be in static equilibrium. Well, if she is in static equilibrium, it must mean that there is more than one force. And we've observed that there is a force that she's putting a force on the pedestal, which is acting in a direction that's in line with her weight and in the opposite direction. We know in this particular case that the weight has to be equal to the normal force, where n normal is the force perpendicular to the pedestal surface, and w is the weight of the Statue of Liberty itself. If we do a little algebra here, we can see the relation that the normal force minus the weight is equal to 0. If we consider the normal force to be positive, up is positive, and down being negative. Concurrent vectors are vectors that have lines of action that meet at a single point. For example, if I have some sort of ring with cables tied to it, and all of those cables are aligned in such a way that their point of action meets in the center of the ring, these would be an example of concurrent vectors. And usually, we represent the pole of each of those cables with a letter t representing the word tension. Tension is the word for something that is being pulled on both ends. In two dimensions, these vectors each can be described with two components, one associated with each basis. So in order to view that, I need to create a basis. Let me create a Cartesian basis. We'll call an x direction as positive to the right, and a y direction as positive up and down. And we'll make that at a right angle. And then we can look at each of these vectors as having two components. Let me move them off to the side. This first vector, t1, can be considered to have two components. An x component, we can call t1x. And a y component, we'll call t1y. Similarly, for vector t2, we can have a component t2x and t2y. And finally, for our vector t3, we can have a component t3y and also a component t3x. But notice if t3 is aligned with our y-axis, it may not actually have a component in the horizontal direction. Once we've broken these down into components, we can think of each component as acting along the line of action that is parallel to each basis. For example, we can consider only the components that line up along the x-axis. If we do that, we realize that just this t2x is pointing in a positive direction. And the t1x is pointing in a negative direction. We can say that in the x direction, positive forces must balance with the negative forces. In other words, that t2x, which is in our positive direction, is equal to t1x, which is in our negative direction. Or the other way to express that is to do a little bit of algebra and demonstrate that t2x minus t1x is equal to 0. In other words, that the sum of all the vectors in the x direction is equal to 0. If we are careful and make sure that we keep track of the signs to represent which direction the force is acting in. Similarly, if we do that in the y direction, we take all the positive and set them equal to the negative. We recognize that t1 in the y direction is pointing up, t2 in the y direction is pointing up. And that's equal to t3, which is in the y direction, pointing down. We could say t3y, but that's equal to t3. Or, again, we can do the algebra and express them as a sum of all of them, noticing that ty is positive, t2y is positive, and t3 in a downward direction is negative. So that's equal to 0. In short, what we recognize here is that the sum of forces in the x direction is equal to 0, and the sum of forces in the y direction is equal to 0. Using a notation called sigma notation, using the Greek letter sigma, which means we're adding up all the forces in the x direction and adding all the forces in the y direction. If we're in 3D, we can also recognize the sum of forces in the z direction would also be equal to 0, but most of our examples are going to be in 2D. So these are the conditions for static equilibrium if all of our forces are concurrent, if they're all acting through a common point. Let's consider a practical example, shall we? Here's a stop light hanging from a couple of cables. And each of those cables are attached to rings labeled A, B, C, and D. Actually, A, B, and C, D is actually the connection to the stop light itself. Well, we might be interested in knowing the amount of tension in the cables, knowing how much each cable is pulled, because certain cables can only hold a certain amount of weight, and it might be important to know how much weight that is. Let me go ahead and provide some information about this particular system. First of all, we're probably going to need to know the information about the geometry of the system. So we're going to say in this particular case that this length from A to directly above C is going to be 10 feet, and the remaining length is going to be 20 feet, and that a distance at C is below that straight line is going to be 3 feet. That geometry is going to be important to figure out what's happened. If we decide to change the length of the chain or the position where the stop light is, it's going to change the forces that are inside. We also want to identify how heavy the light is itself, and we're going to say that it has a weight of 30 pounds. So if we want to solve a problem like this, the first thing we're going to do is go ahead and create a free body diagram. So where's the best place for a free body diagram? Well, if we're interested in the tension or how much forces each of the cables, we're going to draw a free body diagram around the ring where all three cables come together. That will separate the ring out, and it will replace each of the cables with a force. Let's take a look at a close-up of that free body diagram. Here we have the tension in each of the three cables. The tension from A to C, the tension from C to B, and the tension from C to D. We happen to know what the tension from C to D is. That's going to be equal to our 30 pounds. The amount that the cable has to pull up is equal to the weight that's being pulled down by the stop light. And that same tension is going to be 30 pounds pulling down on our ring. So now that we've created our free body diagram, let's go ahead and think about each of these vectors in terms of components. We're going to recognize that they all meet at a common point. We'll need a basis to create a basis. Let's define the X direction as positive to the right and the Y direction as positive vertically. We'll make it perpendicular. And now we can break each of these up into the component parts. We're going to recognize that CD only has a component in the Y direction. Let's do the components in the X direction for CA. So we have TCA in the X direction. And we have a component here, TCB also in the X direction. And then we have components TCA in the Y direction and TCB in the Y direction. Notice we're also going to need a little bit more information here. Let's go ahead and give the angles the directions of these each a name. This will be theta or angle CA and let's call this theta angle CB. Now that we've identified component vectors, let's go ahead and apply static equilibrium. In other words, let's make the assumption that all the sum of forces in the X direction is equal to zero. Now that we have that sum, we notice that we have a force in the X direction, TCB in the X direction. And we have a force also in the X direction, TCA in the X direction, but in the opposite direction, so we'll give it a negative value. And we know those two together must sum up to be equal to zero. Now that's not particularly useful because we don't know what those values are. However, we can relate those values to things that we do know what they are that we might want to know what they are. We can figure out the angle. So let's go ahead and use a little bit of our trigonometry to recognize that each of these X components is related for the tension itself through the cosine of the angle. So TCBX is equal to TCB times the cosine of angle CB. Similarly, TCAX is equal to TCA times the cosine of the angle CA. We can also then apply the sum of forces in the Y direction. Noticing that TCAY is in a positive direction. We add TCBY, which is also in a positive direction. We also have to include in this case the 30 pounds for TCD, which we recognize as being downward in a negative direction. And when we sum those together, we recognize that they are equal to zero. The sum of forces in the Y direction is equal to zero, so those are equal to Y. We also recognize that their relationship in the Y dimension uses the sine of each angle. So TCAY is TCA times the sine of the angle CA plus TCBY, the yogurt place, is equal to TCB times the sine of the angle CB. We subtract our 30 pounds and that entire sum is equal to zero. So what do we do with this information? Well, now we have to refer to the information we know about the geometry of our system. We have to see if we can figure out what these angles actually are. For example, here is my angle theta theta minus theta. CB, and here is my angle theta CA. See if we can figure out what those angles are by the relationships in the geometry. We'll notice those are both parts of triangles that we know the sides of. I'm gonna do a little calculation here. Let's first of all find the hypotenuse of each triangle. That length AC is going to be a squared plus b squared equals c squared is gonna be the square root of 10 feet, which is that horizontal length right there plus three feet, which is the vertical distance. But in both of these cases, we have to square each of those, a squared plus b squared equals c squared. And we find out that the length of segment AC is about 10.44 feet. If we do a similar calculation for length BC, we find that length BC is the square root of 20 feet squared plus the three feet squared. And we find the length of BC is 20.22 feet. Well, how is that useful? Well, now we can use our trigonometric relationship, socotoa, over here on the left, I'm gonna write some angle relationships that would be useful. For example, the sine of angle theta CA, which we'll use below, is gonna be equal to the opposite. Here's the opposite side, which is the three feet, over the hypotenuse, which is the 10.44 feet. The cosine of that angle is gonna be equal to the adjacent of 10 feet over the 10.4 feet. So that's the adjacent side, so that's the adjacent side over the hypotenuse. We can do a similar thing to determine the sine of angle CB, which is now our three feet over 20.22 feet, and the cosine of that angle, which is equal to 20 feet over 20.22 feet. Notice all of the feet units are going to cancel out here, feet divided by feet, and we just get ratios. But we don't necessarily need to calculate these ratios in a calculator right this moment. We can just simply use those ratios in our formula down below. Or you could type them into the calculator if you would like. But notice we found these not by actually calculating the angle. We don't need the angle. All we really need are the trigonometric relationships. We need to know the sines and the cosines for our formula down below. So let's finish this problem. Here's our two formulas. We have TCB times the cosine of CB. That value is 20 over 20.22 minus TCA times the cosine of angle CA. Well, that's the 10 over 10.44. And we know that's equal to zero. For our forces in the y direction, TCA, the sine of CA, let's see here, this is the CA. So this is going to be the three over 10.44 minus TCB times three over 20.22. That's a plus minus 30 pounds is equal to zero. Now that we've added the geometry into play, we now have an equation where we have two unknowns, TCA and TCB, but we have two equations. Now it's simply a matter of algebra. Two equations in two unknowns. If we solve both of those equations and I'll leave that solution up to you, we come up with the tension in CA is equal to 69.7 pounds and the tension in CB is 67.5 pounds. Notice this tension is significantly more than the 30 pounds, it's more than twice as much as the 30 pounds of the stop light itself, which demonstrates why it's important for us to do this calculation. Because in order to hold this stop light up, the two cables have to both pull up, but at the same time they have to add additional force where they pull against each other. They have to balance the vertical force together, but they have to oppose each other and that additional pulling leads to these additional values. So just to recall, static equilibrium, sum of forces in the x direction is zero, sum of forces in the y direction is equal to zero and that should suffice as long as you have concurrent force vectors, all meeting their lines of action at the same point.