 That's one where there's two pulleys connected by a belt. One pulleys starts up and starts to turn the other one because of the belt around them. Now, and for a little bit, there's slipping going on and then after a little bit, they don't slip. It's exactly the same kind of thing that goes on in your car when you start it up. The engine starts turning the fan belt. There's always a little bit of slippage before the water pump and whatever, the radiator fan, before they get up to non-slip speed that the engine starts pulling on the belt, there's a little bit of slipping. In fact, sometimes you'll even hear that sort of squealing. So, you know, say that one's turning this way, that will cause, of course, this one to start turning that way. The way the problem is written up, one of them is already operating at full speed, they engage the belt and it starts to bring the other one up to full speed. In doing so, one will slow down a little bit, the other will speed up a little bit and tell then there's no slipping and then they both run at whatever speeds, is there steady state speed. So, if you look at those two, let's say this one's running that way first, but it'll start to slow down a little bit as it tries to accelerate that one. So, if you look at that bigger one, it's already running this way, but it's going to accelerate that way as it starts to slow down, that's because as it's pulling on the belt for the lower one, the smaller one, there's a difference in the tensions of the belt on either side of that. So, I just labeled them A and B to call them something. So, as the big wheel is pulling on the little one, that'll cause more tension on the upper line than on the lower line, that serves as an imbalance in the torques, which will tend to cause the bigger one to decelerate. So, if its speed is that way, its acceleration will tend to be that way as it tries to drag the little wheel up to some speed. Comfortable with that, that makes some sense. It's not terribly different than when we're looking at particle velocities and accelerations and object moving in one direction, maybe trying to tow something behind it that something behind it will cause it to decelerate even though its velocity is forward. Look at the little one and it's got those same forces acting on it because it's the same belt but in opposite directions. So, I tried to draw them bigger if they're bigger forces and smaller if they're smaller forces. So, that will cause this one to have then a net acceleration in that direction because there's an imbalance in the torques. Comfortable with that. Well, you can write that business for this one. Remember the sum of the torques is going to equal the moment of inertia times the acceleration. That's the rotational equivalent of that equals MA. So, you can write that for this wheel and you can write that for that wheel too. The forces will be the same but the torques are different because both of these have different radii which means the torques are different and each has a different moment of inertia because they're different size and they'll have different acceleration. This one, of course, is decelerating. This one is accelerating but even not only are the signs opposite but it's very unlikely the magnitudes would be the same. So, you can set up these two equations using the forces, the moment of inertia and the like. You can combine these two. Let's just label for useful purposes that one and two. So, that's one, two, two. And this will be the sum of the torques on one. Some of the torques on two just to keep those kind of separated. You can find a ratio, something like alpha one over alpha two when you have those two from those equations but that ratio also will equal, well, let's see. Alpha is delta omega over delta t. Delta t's the same for both of those so that'll cancel. So, this will be delta omega for the first one over delta omega for the second one. Are you giving those omegas? I forget. So, you just wanna find out when they're finally running at the same when there's no more slipping, right? You're looking for that point. So, you'll have this equation with a couple unknown omegas in it but you can eliminate the unknown omegas by using the no slip condition that it'll finally reach when that's true, the pointer right there will have the same velocity as a point right there because they're connected by the same belt when there's no slipping. So, the no slip condition occurs when finally d one equals v two. That's not true when there's slipping going on. It is true when they're not slipping or r one omega one equals r two. Omega two. And I think if I remember you have both the r's, you have I think the initial angular velocity of not even that so maybe the answer's. What do you do about it turns great. Oh yeah, it'll be in terms of that. So, when you set all those up it should be, it should be algebra left over. Let me make sure I call it one and two the same one just so the pictures all match and I'll give it to you. The final angular velocity of the big one, the big wheel will be the original angular speed. So, so maybe here, because this one's a still, it's going to still over one plus, isn't it right? Velocity of the little one, the second one, will be m one r one over m two r two. Times that speed. It's just the ratio of the moments of inertia. I think that's everything in there. I want you to put all that together. But I'll have these posted overnight anyway so you can see them. Or another one on that says that proton zips around the two kilometer fairy lab particle accelerator. That's a large circular track that can keep charged particles like electrons. No, it says protons, isn't it? Not that it matters terribly. It says protons. It can keep charged particles moving in a circle by the presence of a magnetic field. You'll cover this in physics three. But the deal is as a charged particle is moving with certain velocity through a magnetic field, there's a force exerted on that particle perpendicular to the velocity. And we should recognize, I hope, that anytime there's a force in a particular direction, there's acceleration in that direction if it's unbalanced. And if we have acceleration perpendicular to velocity, we have circular motion. So that's what's going on with the particle accelerator. The angular momentum, what's our symbol for angular momentum? L, angular momentum is the momentum, the object in question, in this case the proton, has with respect to the center of the motion. So the momentum is mass times velocity and it's at a distance r from the center. So the angular momentum is r and v. That works, it's the product of these when they're perpendicular, as they always are with circular motion. No, force at an equal on v, units wouldn't work. Velocity is not equal mass times velocity. Velocity is not equal mass times velocity. They're in completely different colors. This one's separated over here a little bit. This is a particle of some mass that has a velocity, therefore it has a momentum, mv. This is the momentum. We use p in this class for momentum, right? So this is the momentum. Oh, I got a p in there for proton. This is a momentum, p, and that's a momentum a certain distance from the center. So that's the angular momentum. It's the linear momentum with the moment arm, just like we do a force with a moment arm gives us a torque. Others, this last chapter that we started on last week on Wednesday, it has two parts to it. The second part we're kind of running out of time, so I'd rather not introduce it and have it just cloud the issue. Plus we'll get to it in great detail in a class that the engineers will take a year from now. The elasticity of solips, the fact that as we exert forces on objects, real objects, it tends to compress them a little bit. If it's a compressive force, or it even stretches them a little bit, if it's a tensile force. And then when that force is relieved, the things return to their original size, usually. Unless you've done it so far that they actually change the nature of the object chambers, either fractures or shatters or deforms permanently, or even the crystalline nature of them can change, and then it's a different material. So if you choose to do the optional homework, that's an OPT stands for, if you choose to do this week, this chapter's homework, chapter 14 homework as optional homework, then you also have to do that section for it to count as part of the optional homework. But I'm not gonna put it on the test for tomorrow. Just all we're gonna look at on, the only part that will be on the test tomorrow is this part that we started last Wednesday about equilibrium. And so that's what we'll look at then today or something. There's only two conditions that need to be met for an object to be in equilibrium. Well, what do we mean by equilibrium in the first place? It'd be nice to know what we're even shooting for when we're shooting for equilibrium. The object's static, probably. Or accelerating. Well, it's what? It's not accelerating. All those things go together. It's balanced in that whatever forces are on it aren't enough to make it do anything than what it's just doing anyway, which is usually just sitting there. So the thing we want to be true is that acceleration is zero and angular acceleration is zero. Usually that extends even a little bit farther because generally the velocity is zero to start with. So if the acceleration is zero, then the velocity stays zero. And that's what we want, buildings and structures and other big objects like that for the most part to do, which is stay where they are. Don't move, don't go anywhere, especially when I'm driving my car across it. So we usually have the other situation in the problem that the velocity is zero and the angular velocity is zero. That's not always the case. You have to look at the problem. But if we're talking about bridges and buildings and the like, yeah, they have no velocity to start with and we want them to have no velocity to finish with. So to achieve this very important condition, what do we need to have happen? Well, take care of part of it. Forces, which of course we know when we add up the forces, that tells us what the acceleration is. But we want the acceleration to be zero. So we want the sum of the forces then to be zero. For every left pushing force, there's got to be an equal and opposite amount of force pushing right. For every up pushing force, I mean equal and opposite amount of down pushing force. If not, if either of those conditions or both aren't met, if there's some imbalance there, then there's going to be acceleration because that's what happens when forces are unbalanced. So we're looking for the condition where all of the forces have some counterpart equal and opposite pushing back by the same amount. It may not always be terrifically obvious in a problem just because the way some forces line up. We might have a force like that and a force like that and a force like that. That can be in equilibrium as long as the horizontal component of that force is equal to that one. And the vertical component of that force is equal to that one. If that's the case, then even though there's three forces, if we look at the components, then we see that all the up forces are canceled. All the sideways forces are also canceled. We get a net acceleration of zero. Whatever the object is. So these problems tend to be assisted by a free body diagram. So you can see what the forces are. You can even look at the diagram and see I've got a vertical force there and no way there's another vertical force counteracting that I must be missing something in the problem. So you know you're not done checking up on making sure you've got all the forces you're supposed to and you might as well not sum the forces until you've got them all. This does not alone guarantee equilibrium or doesn't. Gotta sum the torques as well. This just guarantees there's not gonna be any linear acceleration. In other words, the linear velocity will remain constant. But if we've got something just setting on a wall and it's turning, we don't want it to have angular acceleration either. Just the fact that it's pinned to the wall will guarantee it's not going anywhere. But we also don't want it to have an angular acceleration. So we also have to sum the torques. We know that if the torques sum to zero, then the angular acceleration is zero. And those are our two equilibrium conditions. How many equations there for your use? Three, are you blind? One, two. He said three, sits up here as teacher's pet. He's right? There are only three equations there which is sufficient for the problems we do. The problems we've been working on where these three equations are, does everybody see why it's three equations? It's two-dimensional. We do two-dimensional problems. Everything that we have that does any turning turns only in a single plane. Either the plane on the board if I draw or the plane in your paper if you write it down. We don't have things that can turn in all three dimensions, like an object would if it could wobble. Not only rotate, but the axis of rotation could change shape. So this is okay. This is sufficient for our two-dimensional problems. Three equations in two-dimensional problems. If we had three-dimensional problems, we'd need the full three-dimensional version of those two equations. We started a problem on Wednesday. Did we not finish it? I don't think we did. We got it set up. I'll leave those there, because we need those. We had an eight-meter long platform hinged at one end with somebody standing at the quarter point. Oh, there's half, there's a quarter. Somebody was right there. 600 newtons and that form itself weighed something, right? I think 200 newtons. The force for something nice in uniform density, we always put the force representing its weight at its center of gravity, which for something as uniform density and nice symmetric shape is just right down the center. Then we had, I believe we had a cable to actually get to the roof with an angle of 53 degrees. The problem as we set it up, but we did nothing lower with it than that. We had that much, but didn't actually quite get to solve it. Get to solving it, is that true, John? We started, okay. This is not the free body diagram. Why not? It's the object we want to keep in equilibrium, which is the platform is not free of some of the other things in the problem. It's still got the cables drawn there. It's still got the wall support there. So we free it up of everything else. We've already got a couple of forces on there, so we'll go ahead and put those in. I think we knew the weight to be 200 newtons. You just look at that and you know it's not gonna be in equilibrium. So you know you're not done. That's not always the case. There's sometimes a picture can kind of look like it could be in equilibrium, so you gotta think about those ones. This one's so obviously not in equilibrium. There's got to be something else going on. There's two forces going down, no forces going up. So what did we have that was going up? We have a wire there and wires only pull and they only pull along their own length. So we know that to be 53 degrees, but we don't know what the tension is. In fact, I think we were supposed to find that. You've got to know what that tension is gonna be so you can get down to the store and buy a good enough cable that'll hold it. Still, it doesn't look like this is in equilibrium because we could have some torques pulling that way. In fact, all the torques could cause it to go counterclockwise about a particular point. If we pick a point right there, every one of those forces is trying to turn that thing counterclockwise around that point. We don't want any sum of the torques. We don't want any angular acceleration anywhere, much less that one point. So you know you're still not done. There's got to be some other forces. And what else did we come up with? What we call the reaction force, the force exerted by the support over here. We call that reaction forces. Did I draw it in actually? Yeah, yeah. We don't know what direction it's going to be. So what we typically do is just make the simple move of already breaking it into X and Y components. Sometimes you just have to guess which direction it's going to be. Seems pretty obvious that that thing's got to hold that end from falling down so it better have a force up. So I might want to call that Ry. Is that what I called it last Wednesday just so I don't switch things up? Yeah, it's the reaction in the Y direction. That's all we're saying. And then you can look at this. You see the tension's trying to pull it to the left. So the reaction has got to work against that. We've got to have some X reaction against the tension. Otherwise it's going to go left. Now it's starting to look like we might have the possibility of getting this thing into equilibrium. It's starting to think maybe we've got all the forces. Sooner or later you do, you can't just add them forever but do we have all the forces? Remember the rule. Any force you put up on a free body diagram what's the rule about it? Any force, whether it's in this part of the class where these things always sum to zero or the earlier part of the class where we let things accelerate what had to be true about any force that went on a free body diagram has to be caused by something real. Something that I can go point to and put my hand on. I can hold, give it a hug. It's got to be something real. So you can't say that that force is caused by the motion. I can't, so you can't touch motion. You can see it but it's not real enough for you to go touch. That's the rule with the forces. So any others? No. If there is, there's got to be something in the problem that causes it and we pretty much have everything in here. So then we sum the forces. How many unknowns? I heard three. Three. Doesn't hurt to count them up. Make sure you know what you're looking for. In a problem we may not be asked to find all the unknowns but that doesn't mean they aren't counting as unknowns. Because they still affect all the equations. So all three of those things are unknowns. Remember that knowing the components of any force is the same as knowing its magnitude and direction. Either way, that's two unknowns. Either the two components are unknown or the direction of the magnitude is unknown but still two things unknown. All right, so sum the forces in the x direction. We know those forces should sum to zero. Just to make the algebra easier, I think it's wise to look at that fact as also saying all left facing forces equal all right facing forces. When you write it up that way, there's no minus signs. You don't get confused. Just look at your picture and you write it up that way. All the left facing things equal all the right facing things. So we've got rx pointing in one direction. What's pointing in the opposite direction to balance it? t in the x direction or to keep it simple, t cosine 53. You want to try to set up these equations so no more unknowns come into it. Every time a new unknown comes in, besides one of these three you need another equation for it. The forces in the y direction, that must also sum to zero. So every up force must have a down force to beat it. So you write it out. You write out the sum of the forces in the y direction. See if we get the same thing. You can use the symbols if you want or you can use the numbers if we have them. Watch your units. Too big a deal on this one, but it could be on other ones. Don't show me. Show Andrew, he's working with you. We're doing the same problem. Your algebra might be a little bit different. You may have just laid them out in a slightly different order, but just look at the diagram. All the up ones have got to equal all the down ones. They're going to soon, yes. I think they'll be able to work it out though. We're in the last week of the term and then start working on when these again. Love that sum of the job. You guys have come over and you're going to try to pull in and throw something. It's not before it's just one. I'm trying to have them. I haven't tried to do it. I don't know what you guys do on a basis of the land drives in place called the police on them. I'd say fine, I'll serve you. You got the money, I'll serve you. I don't care. I didn't see what I did. I didn't see what I did. You agree? Yeah. Yeah. Yeah. Alan, would you agree with anybody? Boycotting? All right, let's see. What do we have pointing up? RY. Of course, RY is pointing up. Anything else pointing up? T. Yeah, of course. We got the little piece of T there which would be T sine theta is pointing up. T sine 53 is also an up force. You don't have to do it this way and I won't grade it wrong if you don't do it this way but it's the way I find over the years just to make things the most simple. All the up forces equal all the down ones. Don't mind the sines. It's more difficult to miss any of these when you do it that way. That's all the up ones, all the down ones. We got the 600 and the 200. So that's all three unknowns and only two equations. So we can't solve it yet. I can, you can, and Superman can. I'd like to see that. So we use our one other equation where there's some of the torques. This one's a little bit trickier. Well, not trickier. I guess it is. This is where students will screw up a little bit more than others because this stuff, we started this months ago when we were doing particle motion. This is new in the last couple of weeks. Torques are always forces operating some distance from a central point. What central point? The hinge? What? The middle where the weight is? Wishy-washy answer? What? Even wishy-washier? It's our chair. You guys are right. We don't want this thing to have angular acceleration in any way. We don't want it to turn about any point. It doesn't matter what point. We can look at any point we want. We don't want it to have any angular acceleration about that point. So calculate your torques about any point you want. And you'll do fine. You'll get the right answer as long as you set it upright. However, there are easier points to pick than others that just make the mathematics, the algebra. Because we've got to solve these three. You're going to have a system of three equations. The easier you can make it, the better, the less likely you are to screech it up. How do you pick a good point? We have a couple possibilities. Let's label that one A. That was the hinge. You said that. Joey or Bill said the weight. Do it about the center. So we'll call that point B. Here's a point over here. Point C. Any other point you want to pick is fair game. Because we don't want an angular acceleration of any part of this thing anywhere. How do we choose one of those points or some other point that might make things a little bit easier? What do you look for? Huh? You've already got a hinge on it. Well, yeah, I'm not going to put another hinge on it. That doesn't answer my question. Because you've already got a hinge on it. It's not necessarily obvious to me. That's not the thing you look for. You don't look for the place you're going to go and get a drill out and put some screws in the wall. What you look at is this drawing and you pick a place that gives you a simpler solution. Simpler solution means less chance of error. What do you look for? Remember, torque is force applied at some distance from a central point. Some distance that's perpendicular to that force. So any point that has forces going through it, that force doesn't exert any torque there because it's going right through that point. Remember, that's how we set up torque in the very first place. I said we have some object here and some point there. If the force goes right through a point, it doesn't exert any torque. So pick the point if there is one where more forces go through that point than any other and they exert no torque, they're not going to be in the equation. It's just simpler. We'll set up a couple just to illustrate it. So we can all see what I mean. Let's see. Let's do a couple of examples. We'll pick the easiest one and then we'll write it down there. So summing the torques with respect to point A. Normally too, I advise you start at one end of the object and go to the other end so you don't miss any forces. You've got to look at every single one of them. How much torque does Rx exert with respect to point A? It goes right through point A. So Rx can't exert any torque. It's not going to cause the object to twist about point A because it goes right through point A. So Rx is in this equation. Step across the thing, go to the next one. Ry. How much torque does it exert about point A? Y0. Goes right through that point. 600. Does it exert any torque about point A? It does because it's trying to twist A around it. So how much torque does that force 600 newtons cause? What's its moment arm? The moment arm, remember, is the perpendicular distance from the force to the point we're interacting with. Not just any distance, the perpendicular distance. And if you need to, you can imagine that the force has a line of action. The line of action is the shortest distance from that line of action to the point, which in this case would be two meters. In which direction is the torque being caused by that 600 newtons? Not down. Down's not the right answer for torque. Clockwise. Clockwise. If you're pinned here at point A, that force is going to try to make things go that way. We're going to stop it somehow, but that's what it would tend to do. So this will be all the clockwise torques on this side. Remember, we want to make all the clockwise torques equal all the counterclockwise torques. The 200. Does it cause torque? Yes or no? Yeah, counterclockwise. Yes? What's the size of the torque? 200 causes it. Force times moment arm. The moment arm here, again, is the perpendicular distance out to the line of the force. That's four meters, so this is 210s four meters. And in what direction? It's pulling down point A's over here. It's also going to try to turn the object that way. We're going to stop it, but it is also a clockwise torque. 200 newtons. A moment arm of perpendicular distance of four meters. Keep on down. We don't want to miss any of these. We get down here, we have this force. Is it causing a torque? Yeah, part of it. Part of it. What do you mean part of it? Part of it acts like that, not TY. TX. That's what we had right here. We've already used it. And part of it acts like TY. That's what we have right here. We've got to look at both of those to see if they cause any torque. You know by now, we're at the last available force in this problem. We need some counterclockwise torque because we've only had clockwise torque so far, so you've got to be expecting to have some counterclockwise torque coming up here. How much torque is caused by TX? What do you mean none? It's way down here. It's all the way at the end of the beam. It's still from directly through point A. So it's moment arm, it's perpendicular distance between the line of that force and the point is zero. TX causes no torque. What about TY? It's got that little perpendicular distance to all the way down to A. So we know that TY, which is T sine 30, sorry 53, what's its moment arm? It's the full eight meters. It's perpendicular distance back to point A is a full eight meters and it's trying to turn things counterclockwise. Clockwise torques got to equal all the counterclockwise torques. If not, there's going to be not over and it's going to start spinning in that direction. Which we don't want it to do. We want this to be an equilibrium. How many unknowns in that equation? Just one. That's as simple as these can be. If we do the moment about any other point, we're going to have more unknowns. It's just a little bit more difficult to solve. If the algebra is a little more difficult. If we sum the torques about point B, notice W causes no torque, but now Ry does and Ty does. So you have two unknowns in the one equation. It's just a little bit more difficult to set up. If we did it about point C, some of the torques about point C, how many unknowns would there be in the equation? If we sum about, let's see, let's put that in here. If we do it about B, there'll be two unknowns in the equation. Ry and Ty. Even if we use Ty in that form, T is still unknown. Sum the torques about C. How many unknowns would there be if we did that? And you're one. What is it? If we sum the torques about C, Ty goes through it, Tx goes through it, Rx goes through it, or Y doesn't. So we have Ry as the only unknown here. None of those three will be wrong. It's just some are easier than others. The A and the C equation just a little bit easier, that's all. Especially if the problem asks you for the tension but doesn't ask you for the hinge forces. This equation will give you the tension and you forget about the hinge forces if it doesn't ask for them. Because this one will give you the hinge forces, but then you can use it to solve for the tension. Just more steps. More steps, more chance to screw it up. So that will be three equations and three unknowns, whichever one you choose, whether it's A or C, it doesn't really matter. After that, once you get all three of those set up, it's just algebra. It's left to you to get the algebra right. Which of course you would, wouldn't you? Would anybody do this problem? Get some numbers? Right, the tension's 313. So you can do these and double check this. Actually, there's two components to it. X and Y, 180, I plus 250. Really? I have 550 written down. But yeah, we can, that's the Y component. I may have just written that down. I'm sorry, I forgot to subtract it from the 800. You can check those numbers. Any questions about that? Be careful with the algebra. It's easy to screw up. You don't want to lose points there in a physics class. Let's look at maybe a homework problem or two that were assigned. Now, you know what, let's do some others because the homework ones I'll post, you can just go look at those. Let's do another problem. A traffic signal. Let's see. It doesn't matter. You guys just think everything's green light anyway. Yellow, that means greener. Hit the gas. Red means only suckers stop. You figure that at least half the population is older than you are and more mature. That at least half the population will stop and move light so you don't need to. It's a pretty good chance, I guess. 30 kilograms, 37 degrees here. Presumably, you want that light to remain in equilibrium. You don't want it to accelerate. Now on here under it, if we don't want it to accelerate, we know then the forces must sum to zero and any torques will only do a 2D problem so we can skip these. The last thing I recommend is a free body diagram. Of what? The diagram of an object. The street light. Let's try that. So here's the street light. You draw any forces on it. Look at your picture and tell whether or not that it's good and fake. You can look at your picture and tell whether or not you've got a complete free body diagram. Otherwise, accelerates are quick. You've got to get all the forces on the page. There's a gravity arc acting on it. Street down. Do you have that number or is that an unknown? Yeah, we don't have it, but we've got the mass so don't consider that an unknown. Just do w equals mg and don't fill or make too many at my nose. There's got to be some upward force or that thing's going to accelerate downward. We've got to hurry. The highway department wants it left where it is. Joe, another force? 37 degree wire. Is that 37 degrees, that angle? What's that force's name? D1 and D2. Keep it simple, but keep it helpful. Look at that, you know that at the moment it's going to take off to the right so there must be something to the left and of course that's the tension on the other line. Two unknowns. D1 and D2, how many equations do you need then? Of course there's two. Which two equations? Just those two? Two unknowns, three equations. You save the extra equations for the next problem. Carry it over. Sometimes you can write it on a little piece of paper and put it in your wallet and pull it out to the final. It's like a monopoly free equation card. Grant the bearer one of the equations of his choice. You do when you have fewer unknowns than available equations. What it usually means is you don't need all the equations. That's all you can solve the problem with two equations. You only need as many equations as there are unknowns even if there are more equations. This is going to be a useful equation. Sure, we got, well you write it out. Is it going to be useful to solve some of the forces in the y direction? It's useful as long as those two equations are independent which they always are. The x equation is always an independent of the y equation for the forces. And as long as it doesn't introduce any more unknowns. Write out that equation. See if we get it laid out in the same form with the same unknowns, same minus signs, plus signs and equal times. You need yellow. Is that yellow? Finish your drawing, yeah. Come on here, Joe. Show some artistic flair. That looks good. Oh, yeah. That makes me want to hit the brakes. Right there. I'm ready to stop. Put it on the brakes. I don't speak in terms of anybody. They don't even recognize you're not here at the time. By the way, everybody, introduce yourself and make them feel welcome here in the last week. x equation. If you do, go on to the y equation. We named both two equations. Hope it's these two. They're our favorites. x direction equation look like. Read it to me. Some of the forces in the x direction is attention to sine of 35. By the way, it's one out. T2 sine 45. Cosine and sine 45 are the same anyway. Numerically, you would have come out, but if you ever went back and changed the angles, as you're going to do as engineers, redesigned things, but didn't change the equation, you'd get caught. Keep it the same, even though cosine and sine are the same. And then what? Because that's a left-going force, so the right-going force is T1 cos 37. Anybody have anything other than that for the x direction equation? If you want to, feel free to say this minus that equals zero, but algebraically, it's the same thing. This way, there's no minus signs. Generally, I've found students have less trouble skipping any. Some of the forces in the y direction. Samantha, you have that? If you write T1y and T2y, it looks like you've introduced more unknowns. If we write it that way, then we're already okay. So that would just be the sine of those two. You can make that step independently if you want. It's not incorrect. Decisions solve that problem. The system is sufficient. Why don't we need the tors equation? Yeah, we don't need it, but is there a physical reason we don't need it? This is a physics class, not an algebra class. Thank God. All of these forces pass through a single point. It doesn't matter where it is. Anywhere, if there's a single point, all forces pass through it. There's no torque exerted at all. If you did the torque equation, all you're going to get is a very helpful solution that's zero equals zero, because all the counterclockwise tors are zero, all the clockwise ones are zero. So zero equals zero. It's comforting to know that that's still true. That hasn't changed over the weekend. Lots of things did, especially last night. Have we got that news? Yeah. Not many, I slept through it. So we don't always need all the equations, but that doesn't mean this doesn't apply. It doesn't just, it has a trivial solution. What would be the standard, the tensile circumference of the corners of the, of the light? Right? Oh. Like this. The cables were actually over there. You tell me. Yeah. You can't necessarily tell from your drawing, because for the drawing to actually tell you, you'd need to know, you need to measure those angles and draw everything just right. That's a little difficult to do it, but in general, yeah, it's not going to make any difference. What if that was hanging down on its own line? What should you do the free body diagram of now? Because doing it up the light is not going to show those two lines, because they don't touch the light. They don't exert any force on the light through the three lines. Nothing can accelerate in this problem. So why not do a free body diagram of the knot? I guess it's not a knot with cable, but the place where those all three come together will still have the same general, same deal. Maybe get out of class question. X on your nice blue shirt, what do you say? You do? What do you say, Mike? Make it a good one. Why wouldn't I make it a good one? All right. Imagine the human arm holding something there. We'll make it my arm. Of course, yeah. I don't come to class in sleeveless shirts. That's why they said, stop showing your intimidating tattoos. It would be intimidating, wouldn't it? So there you go. So we'll model this as a simple set of forces. We have the elbow there about which everything pivots. There's this muscle and tendon that pull up in that direction. We'll also say that whatever this weight is here, we'll call that six kilograms. And let's say the forearm itself weighs one kilogram and just to put some approximate dimensions to this business, we'll call the distance that your bicep comes down and attaches to the front of the forearm there. That's what pulls up this forearm is when this force pulls up this thing because of unbalanced torques. That's what God was thinking when he did it. We've got 3.4 centimeters. The whole distance out to there, 30 centimeters. I want you to find the force exerted by the muscle to hold steady in equilibrium. Forearm, essentially horizontal force there. Be vertical, a little bit simpler. And also find the forces in the elbow. Let me make sure I gave you everything. The muscle force vertical, just keep it simple. It's a little bit of an angle there, but don't mess with that. And then get a quick approximation of the force in the muscle and the force in the elbow. You get that right? Go early. Don't study early. Study longer. Study harder. So instead of getting to bed, it's pretty easy to get to bed at 2.20 this morning. You guys that only have 2 hours, if you want to, you can come earlier and get started today. I don't think you'll need it. I hope you won't need it. I've got to try not to make a test to go anywhere near the 3 hours, but just to be fair. So who else has to leave early? Mike does. Well, you're an assistant. Do you need to use your mouth in the room? No, I'm a sleeper. The sleeper won't be the last thing. Okay. So, Mike, John, if you want to come earlier, you have to find me. My class in here will be taking an exam at the same time. I'll just be around. Just call me and let me know you're ready. Like I said, it shouldn't take you that long. It just in case. But it would give you sleep times. That's terrible. Reduce the pressure for both tests. Do better. Put on the forearm. Just as a simple thing like that. Some forces on it. Well, the forces in the elbow. If possible, lifting enough weight, you can bust your elbow, blow your elbow apart. Nothing cooler than watching those weight guys kill themselves. I'm going to press the grove across the gym. Len, we ought to pull up that one picture that was at USC football player that dropped that barbell on his neck. Oh, yeah. Yeah, that was fun. That one. That was the difficult Olympics. It was the one that got me pressed for everything because I was the one that had the best results last time. Oh, no, that other one, too. The guy who was not just talking about it, the guy who was bench pressing. Yeah. And it slipped and dropped on his neck. It's not that guy. We're always looking for Darwin to take out one more football player for us, but not this one. Yeah. So even if he did live, he should say, well, God didn't want me to live, so I better not have children. So make that free body diagram. How is this different from the problem we had with the platform? Free. It's not terribly different. What forces do you have on it? It's weight, which is the fact that it has a kilogram mass. So the arm weighs a little bit of something. All that warm. The W arm. Weight of the ball. There's six kilograms there, but that's just mg. Quick. What else? There's got to be some forces. Yeah, the muscle's pulling up, but we're going to model that as vertical 3.4 centimeters out there. We're looking for that one. Right now, that doesn't look, well, it could be an equilibrium because those are pulling down, that one's pulling up. If we did it about here, those pulled clockwise, that pulled counterclockwise, it might be okay. What other forces? What? Well, we could use some x-direction, this we'll call it, that's felb. There might be an x-component. Who wants to get the size of the x-component here? What is it? Why is it zero? Nope. There aren't any further forces available to counteract this one. It's got to be zero. But you have to make sure there aren't any other forces. That one's got to be zero. Even some of the forces in the y-direction, that's certainly possible. We already did it in the x-direction. That's how we got that to be zero. In the y-direction, we don't even need to call it elbow-y, that's the only part there is. Those two up. Any point's okay. If you make it the elbow, then that force isn't in the equation because it goes right through the point. If you make it here, that force isn't in the equation because it goes through the point. So just for the last little thing to illustrate, anybody see problem? Yeah. But you can find elbow here and just go back and find FM there. There's some problem here. No, we'll take the weight of the arm in the center. That's fine. Just put it in the middle of that 30 centimeters. Good enough. Just an approximation here. Doesn't matter if we have two unknowns. Why are there two unknowns? If we do it through point A, FM's not in the equation. That's the only unknown that'll be in the equation. That's the only one exerting torque. If we do it about point A, does anybody see a problem? About point A, that's doing a clockwise torque. This is doing a clockwise torque. What's this one doing? A clockwise torque. What's wrong? You drew that in the wrong direction. It was just a guess. I'm not sure which direction. Now I know it should be down. You know a little bit more. It's got to be down to counteract the torque of the other two, and you can then finish the problem. Just so you have them. This force is 494. Also, force is 560. You can check those, make sure you get them. By the way, this is actually down. Not up. If you ever pick the wrong direction and you solve the problem, you get a negative number. Just tell us you did it in the wrong direction. It's nothing more than that. Yeah. Or you can just leave it like that. You get a negative number. It's up in the opposite direction. Then I'll try to forget.