 In analyzing structures, we will often start with some known design forces, which we call loads. These loads might come from things like wind. They will come from things like the weight of the structure. They will come from things like snow or rain. And we'll need to determine unknown forces at places where the structure is supported. These unknown forces are called reactions. In two dimensions, each unknown force has two pieces of unknown information, either the magnitude and direction or components of the magnitude and direction. So for example, if we have a reaction, we might either know what the reaction is, the reaction at point A at some angle, or we might break the reaction down into two components. The reaction in the x direction and the reaction in the y direction, but usually we'll also need to label those as corresponding to whatever reaction A is. Now geometry might supply some of the information. For example, we might happen to know the angle or the ratio between the two slopes or something along those lines to help us make a relationship between the x component and the y component. We also have three equations of static equilibrium to help us relate our loads to our reactions. Again, our static equilibrium equations, the sum of the forces in the x direction need to be 0, the sum of forces in the y direction need to be 0, and the sum of the moments around some point in that plane also need to be 0 if we're working in two dimensions. There are additional ones when we work in three dimensions. If the number of equations we have is equal to the number of unknowns, then we have a system that is known as being statically determinate. So if we're looking at a problem where we know some loads and we know some reactions, we're going to use a strategy to walk through solving problems like this. Let's lay out our strategy here. First thing we might want to do is draw our free body diagram to separate the body of interest from the various supports. Identify the loads. What information do we know? What forces do we know? Identify our reactions. What forces do we not know but are interested in knowing? Then we usually want to establish our components in some sort of global basis. Traditionally, we'll use some sort of vertical and horizontal to establish components for each of our loads and each of our reactions so that we can compare them in the appropriate dimensions. We apply geometry where it's helpful. For example, if we know there's some relationship between RAX and RAY, we include those equations. We apply static equilibrium. And then we solve for the reactions. Let's put this strategy into action, shall we? Consider a simplified little structure of a campfire. In this case, we're supporting our cauldron of whatever we are choosing to cook up that day over in two dimensions over the campfire. And there's a little bit of information here. You'll notice there's some geometry information here about how high it is and where it is. We're actually going to assume that it's equal distances on each side over the campfire. So there's a symmetry here that we can take advantage of. And we're going to need a particular weight. Let's go ahead and establish that the weight of the cauldron here is, let's give it a weight of 50 pounds. And we've labeled our points A, point A, point B, and point C. Well, now that we see what's going on here, let's go ahead and draw for ourselves a free body diagram. In order to do so, we're going to need to figure out what thing makes the most sense. Where does it make the most sense to make our cuts? Well, here's what I'm going to do. I'm actually going to choose to make a free body diagram around point A. So if I recognize that there's point A here, and what I see is I have my extension of one of my supports, my extension of the second support, and then my cable that's holding up my cauldron. And we're cutting through each of those three. So I first draw the free body diagram, and now I need to actually provide some reactions, well, and some loads in the places where I cut things. Well, first of all, we know the load. We have the 50 pounds that is pulling down at that center point. And we can also see that there must be some sort of support holding us up. Let's call this reaction B and call the other one on the other side reaction C. So at the moment, we have two unknowns. Although, well, it seems like we have two unknowns. But let's sort of consider this from the point of view of our components. If we take R, B, and R, C and break them down into separate components, what we end up having is, here's R, B. We have R, B in the x direction and R, B in the y direction. Now notice I've automatically assumed that we had some sort of basis. Let me go ahead and draw that basis. We'll call the vertical y and the horizontal x. Let's similarly look at the components for R, C, where we'll define R, C, x and R, C, y. And we have a sense for which direction they're pointing, so we'll go ahead and define them in that particular direction. Notice now that we've done broken this up into components, we essentially have four unknowns. R, B, x, R, B, y, R, C, x, and R, C, y. And we have one known load here. However, we can apply some geometry. We recognize that the triangle that R, B is part of has a geometry associated with it, that the height of that entire triangle is 5 feet, and the width, the base of that entire triangle is 3 feet. Well, if I use Pythagorean theorem, A squared plus B squared equals C squared, I can see that that hypotenuse is equal to the square root of 3 feet squared plus 5 feet squared, which is going to give me 25 plus 9, or the square root of 34 feet. And that means my relationships here with this angle, I can use my Soccatoa relationships to recognize that the sign of that angle is going to be equal to opposite over hypotenuse, which is 5 feet over the square root of 34 feet. Need to make sure I have the units there. And that the cosine of the angle is going to be equal to 3 feet over the square root of 34 feet. And in both of these cases with these ratios, the feet units are going to cancel out. We also happen to know that the two angles, the angles at B and the angles at C are the same because of our symmetry there. So this sign and cosine work for both RB and RC. So that gives us some geometric relationships that basically say RBX is equal to RB cosine of theta B, which is the same as, if I find the cosine here, 3 over root 34 RB. And RBY is equal to RB sine of theta B, which is going to be 5 over square root of 34. This is also going to be true for RC of X, 3 over root 34 RC and RC of Y, which is 5 over root 34 RC. Well, now with these equations, we basically now have two unknowns. Instead of having our components BX, BY, we now have just the values RB and RC are the only things we don't know. Once we know them, we can find the other four. So effectively, we're now down to two unknowns because we use the relationships of geometry. Now that we've applied geometry, let's apply static equilibrium. Our first equation for static equilibrium says the sum of forces in the x direction has to be equal to 0. Well, let's look at our sum of forces in the x direction. If we consider our point A again, we know that we have pushing on it. We have in the x direction our BX coming from the B point. And on our right side, we have our CX. But that our CX is defined to push in the other direction. And we don't have any other things acting in the x direction. So we have our BX in a positive direction, our CX in a negative direction. And we know that those are equal to 0. Well, if I use my relationships, I know that RBX is equal to RB cosine of theta minus RC cosine of theta. In this case, the angles are the same. It doesn't matter what the values are. And if we know that that's equal to 0, we know that RB is equal to RC that they're the same, which we could have argued by symmetry if we had wanted to. So now we've eliminated one of our unknowns. We don't know which either of them are, but we know they're equal to each other. So let's use another equation of static equilibrium that the sum of the forces in the y direction is equal to 0. Well, what are those forces in the y direction? We know that we have RBY pointing up. We have RCY pointing up. And the last piece we have is the weight of our 50-pound cauldron pointing down. Let's put those together. RBY pointing up plus RCY pointing up minus 50 pounds pointing down. Those must add up to equal 0. Well, let's solve that. RBY can be replaced by RB sine of our angle. RCY can be replaced by RC sine of our angle because they're the same angles. And we can set that equal to our 50 pounds. Well, if the angles are the same, RB plus RC must be equal to 50 pounds divided by the sine of our angle. And we've also learned that RB and RC are equal to each other. So now I can replace one with the other. Let's do it this way. RB plus RB is equal to 2RB, which is equal to 50 pounds divided by the sine of our angle. Well, we're very close. RB is equal to 25 pounds divided by the sine of our angle. And if we remember, the sine of our angle was 5 over root 34. So 5 over root 34. Notice we've inverted it because it's in the denominator, times 25 pounds. And now we've solved for RB. That ends up being 29.15. And because RB equals RC, we know that RC is also 29.15 pounds. Let's see if everything balances. Let's just make sure everything balances. We know RBX was equal to RB cosine of theta, which means that it's equal to that 29.15 pounds. And I'm going to use this little number sign for pounds because it's sometimes used there. 29.15 pounds times, in this case, it's the cosine. So that's going to be 3 out of root 34. When I do that math, that gives me 15 pounds. And we've already established that our RBY is equal to RB sine of theta, which using the same arguments, 29.15 pounds times 5 over root 34 is going to be equal to 25 pounds, which makes sense if BY and CY are equal 25 pounds plus 25 pounds is equal to 50 pounds. So let's just check those. Sum of forces in the x direction is RBX minus RCX, which is 15 minus 15 is 0. Sum of forces in the y direction is RBY plus RCY minus 50, 25 plus 25 minus 50 equals 0. Notice we don't need the sum of moments because all of our forces were concurrent. They all ran through the same point. So the moments actually don't have a meaning in this particular case because there is no all of our forces end up being concurrent.