 All right, any questions? Let's see if there's homework due Wednesday, right? That's when this next assignment's due. Any questions on it? Due Wednesday? No? You got that one okay? You just came in and asked about it? Yeah, I think you're right. Okay. Wednesday? I think it's in the due March 2nd? Yeah. All right. All right, somewhere around in there, so. All right, no questions then, off we go. All right, what we're working with is Newton's Laws. And I believe where we left off on last Wednesday is we were talking about just what types of forces we're gonna be coming across in this class. The thing is remember that force is a vector. So any time we're either using a force in a problem or trying to find force in a problem, we need to find both its direction and its magnitude. A lot of these forces that we're gonna have in these problems, their direction's going to be already understood because of the nature of the problem. There's no guesswork involved, but if you don't recognize that, then you're not gonna see that you already have part of the unknowns you're looking for. If we're looking for a force vector, we need to find both the direction and the magnitude. So some of these forces, the direction's already known and you don't have to worry about that, but you gotta pay attention to these. So we had forces like pushes. You can reach in and pull on something, I guess, as well. So those are generally applied forces, forces of a different nature than some other forces we're gonna come across. So those, the direction and the magnitude is set by whatever the problem is. You just have to read it carefully and see if it says a force is applied at an angle of 30 degrees or something like that. You have to pay attention to. Pulls, there can be a different part to those if we're talking about pulls by strings, ropes, cables, all of those kind of things. If that's the case, the direction is always along the length of the string, which means anytime we have a problem that may have a rope running in it, maybe it goes over a pulley and then somebody's pulling on that, on that rope, in terms of what the force on the object is, the force is directed right along the direction of the string itself. So you don't have to worry about the angle of a force exerted by a string or a rope if you know the angle of the string or the rope itself. And so that's a pretty easy force to put in. You may need to just find the magnitude for that particular problem. So watch out for those. We talked about most of those. I think then, then what did we talk about? Oh, we talked about gravitational force, also known as, go ahead, don't wait. We know, we know, also know, why didn't you just say something? Also known as, we know it as the weight of an object. Because of the close relationship between the force due to gravity and the mass, we won't consider really either one of those to be unknown. If you know the mass, just take it automatically to know the weight. We might not have calculated that right in front of us. Excuse me. But we don't need to complicate things. We've got other things to look at without having to worry about the fact that we have mass. Excuse me. If we have mass, just assume that we know what the weight of an object is. Remember, this is the strength of the gravitational field as felt by that mass. It is not necessarily the acceleration the object is undergoing. You have weight yourself right now, but you're not accelerating at all. This is, ideally, understood as the strength of the gravitational field. It's only when all other forces are gone from a problem that you would actually accelerate at that rate, is always straight down. So you already know, for this one, you essentially know both the magnitude and the direction for all problems. If you're just given mass and a problem, no sweat, we already know what the weight is, maybe just haven't calculated it yet, but don't worry that that's something that we don't know. We just may not have it at the moment. We also talked about, I believe, the normal force. One of the more subtle ones that we see in problems. Right, I talked about that, I believe. What does the word normal mean to us in physics and engineering? Orthogonal or perpendicular. This is as the one that's most associated with Newton's third law, that for every action, there's an equal and opposite reaction. Most easily seen if an object is resting quietly on a surface, as you yourselves are right now, the gravitational field you have weight, there must be a force pushing back to counteract that force, otherwise there would be acceleration. And there's not, it's just sitting there doing nothing. That force back is the normal force. It's perpendicular to the surfaces causing it, the two surfaces in contact. That's where the word normal comes from. So it's always a push on the object. It's always away from the surface, an inner surface like the chair of your desk cannot pull on you, it can only push on you. Always away, always a push, and always perpendicular to the two surfaces in contact. If there are no surfaces in contact in the problem, there is no normal force. It's tempting for students a lot since it's kind of a subtle force to just kind of throw it into a problem thinking it's going to be there. You just feel like there's got to be another force, you put it in and call it the normal force, when in fact there aren't any other surfaces in contact that would cause the normal force. So either there is no other force in the problem than your intuition is tricking you, or you've got to come up with something else as the normal force. We will also have friction as a force in a lot of our problems. Not all of them, a lot of our problems will neglect friction, either consider it negligible or simply take it out of the problem to make the problem a little more straightforward for the time being. We'll talk about that a lot more on Wednesday. Maybe get to it today, depends on how other things go. This is also a contact force. So if you don't have two surfaces in contact, you will not have friction. We'll deal with this more specifically on Wednesday, but it always opposes motion. If you're pushing something to the right across the tabletop, friction's pushing to the left. That's just the way it goes, that's the nature of friction. I think, oh no, we did have one other type of force, forces exerted by springs. If we have a spring attached to an object, object to the right some distance and hold it there, the spring is going to pull back, of course, with an opposite force. And in fact, it depends upon how far the spring has been stretched from its rest position, there's a minus there because if we pull the spring, the object with the spring attached out to the right, the spring's gonna pull back to the left. If we push in on the spring to the left, it's gonna be compressed and it's gonna push back to the right, it's always doing the opposite of the motion, the displacement we put into it. And in fact, it happens to be proportional to that amount of displacement. And so that proportionality of constant is typically given the letter K, you might remember this from high school physics. As the spring constant, it's easy to design springs where that constant isn't, constant, in fact, it can change with how much stretch is put in the spring. Our problem, for all our problems, we'll keep it as a constant, also known as the spring modulus or spring strength. Any one of those terms will do and will all mean the same thing. But we will take it to be a constant in this class. What are the units on that spring constant? And by the way, this is determined by the manufacturer how big the coils are, how big the wire is used for the spring, what material it's made out of, how long it is, all kinds of things. It's always determined by the manufacturer himself. If you go to the ACE hardware, ask for Earl back in the spring section, tell him I sent you, he'll send you the spring and right there on the box will be marked the spring constant. So it's not something that changes with the problem and it's usually part of the original design itself, wherever it's used. But the units on it, if we have newtons over here and meters here, the spring constant has to have units of newtons per meter or pounds per foot or pounds per inch or something like that. The idea being if I want to stretch a spring out a meter, I've got to exert that many newtons to do it. So if it's a 200 newton per meter spring strength, to stretch it a meter, I have to exert a 200 newton force to do so. You probably know from your experience to stretch a spring a little bit doesn't take much. The more you stretch it, the more force it takes. Some of you probably have those springy Charles Atlas machines, have I got one? No, sold it on eBay. All right, we'll deal with each of these a little bit more as we get to each one of them as we go along. All right, I think that's the only types of forces will come. I don't think I skipped anything on the list. All right, dealing with each one of them is a little bit of a different problem. We have to get to it as we go. And I think we just finished a problem on Wednesday, hadn't we? We just had a box on, I think on an incline and we were looking at some of the forces still. We'll do another real quick problem that at first glance might look like it's kind of tricky, but then if we dissect it carefully, look at it carefully, it comes out to be a pretty straightforward problem. So imagine on a surface we have a cart, a very well-made cart, which means the wheels let it move frictionlessly. And on the top there, we have a pulley and we have two ropes running to that cart. One goes around the pulley and comes back. The other runs off the back over another pulley from which is hanging some kind of large mass. All right, so that's our setup, that's our fixture. Mike, isn't it easier to do these now that you're halfway through technical free-hand sketching? Put some details to it just for help here. Let's attach that cable at one end and we'll pull on the other end with some force T. Actually, we'll call that T one. We have two ropes in the problem. That's a single rope wrapping around there. Then we have another rope over here. It's gonna have some tension in it too. Wanna find both of those tensions for this problem. Give it a little bit more detail. That mass hanging off the backside there is 200 kilograms. This pulley has an outside diameter OD of 1.5 meters and I think that's all the pieces there except that I'll tell you that move with constant velocity which could be zero, could be to the right, could be to the left. I'll just tell you that it's moving with constant velocity. T one and T two. Find the tension in those lines because if you don't buy stout enough lines they're gonna break. If you buy too strong you're just wasting money so you wanna buy just the right kind of rope. Wanna find the tension in the two lines and do what we'll use to do that. How's that gonna give us the tension? Well, what are we gonna do with the forces? Why find the forces if we don't need them? And why find the weight if we don't need it? What are we gonna do? How is that gonna help us find either of those two tensions? You're right, we will need that but I wanna see if we can understand first why we're gonna need that. Move forces. There we go. Well, now hang on. Some of the forces equals what? Mass times acceleration. What's the acceleration of either the cart? Well, in fact, won't the cart's acceleration and the mass's acceleration be perfectly linked because of the connection between the two? If one accelerates the other's gonna accelerate and that right, it's gotta be because they're tied together. So if the cart takes off that way, the weight's gonna take off that way. And vice versa. Do we know anything about the acceleration? Now I didn't say the acceleration was zero. Is it? Why is the acceleration zero? He said that. Maybe over there, over there. Because the velocity's constant. If the velocity's constant, the acceleration is zero. Automatically. Now, I didn't put a vector on this but it doesn't really matter if it's constant because that one's moving one direction, this is gonna move in another direction. They can only move in those directions. So we know that the acceleration is zero. We know that some of the forces must be zero. Some of the forces on what? On the cart? On the cart and the mass? Or on the mass? Or heck, even on the cliff side there, nothing is supposed to accelerate in this problem. The sum of the forces on anything in this problem should be zero. You can use that to your advantage in problems. Sometimes it's easy to put things together. Sometimes it's easier to take them apart and separate them. So let's go with what Samantha said in the cart. We'll look at a free body diagram of the cart. If it's not quite enough to solve the problem, we'll take it a little farther. But we know the cart shouldn't accelerate. We know it's got some forces on it and it's got forces on it that we're interested in. So here's the cart, here's the pulley. Any forces on it, we need to sum the forces together. We might as well figure out what they are. So we're gonna draw a free body diagram with all of the forces on it. Any forces on it? Joey, you're nodding. T2 or T1 are on it. Remember, those are ropes. Ropes do what? They only pull and in which direction? Whatever direction the rope is in itself. It's pretty easy here since that one's parallel to the ground there. So there's T2. The way to imagine this is I cut through the rope because remember I'm trying to draw a free body diagram, which means I take the object and it's free of everything else. It's free of the cliff side here and it's free of the rope. So I cut through the rope and I replace it with the force it exerts. That's what ropes do. They pull. What about T1? Do the same thing. Cut through the rope. Put in the force it exerts. I gotta cut through this rope too. Remember I've got, it's an imaginary cut. I've got to make this diagram a free body diagram free of everything else, which means I have to get rid of the wall and the rope and replace it with the force. So how do I replace this rope with a force? What do ropes do? They pull on their own length. Here's a rope. It's level to the ground. They can only pull. It's got to be doing that as well. It's the same rope that goes around the pulley either side. It must be the same tension on either side. What would there to be in the rope to change the tension? So in fact, that's why we use pulleys. You take a pulley like that, wrap it around, and it will double the force you exert. Everybody's used pulleys, or at least seen them before, block and tackle, pulleys. You know some of you guys, when you go into your backyard to pull that engine out of one of the AMC Eagles you have in your backyard, don't you do it with a block and tackle? Now you do it. Or you just go into Glens Falls and you can find a car sitting there and take the engine from it. That's how you do it. You pull out with a block and tackle. That's why block and tackles, that's why pulleys work is they allow us to double the amount of force exerted. But don't let it fool you in any other way than if there's a rope there, ropes only pull and they only pull in their own direction. And the tension in a rope is the same anywhere along that same rope. So that's why there's two T1s there now. So that simple rule, ropes pull and they pull in their own direction. It's easy to apply as that. Any other forces on this cart, let's put them in anyway while we're working here until we know what's pertinent and what isn't. Joey, what were you gonna say? Gravity. The force of gravity, the weight of the cart. Forces are pink, by the way. What is the weight of the cart? Is there anything opposing this weight? If not, you can tell by this picture it's gonna accelerate down. The normal force that caused by what? The cart's pushing down on the ground, the ground's pushing back on the cart. Remember, you gotta make these drawings kind of big. We're getting lots of stuff in on them. So don't be skimpy with the size of your drawings. What is the weight? Well, why would you say that? It's not, only it's not as if you're not moving up. Now what's more important is it's not accelerating up or down. There's no acceleration up or down. Therefore, well in fact what we've done very quickly is sum the forces in the y direction. There is no y direction acceleration. So all the y direction forces must sum to zero. In this case, n equals w. If you remember from Wednesday, n did not equal w. The best way to tell is with a free body diagram. Do not assume n equals w for every problem. Three of you will on the next exam. Who are those three? There's one, two, and you will? Okay, so there's our three volunteers. The rest of you can relax, you'll get the problem right. Thank you to you three for throwing yourselves under the bus for the rest of the class. Appreciate that. Notice though, just like it was for projectile motion, we separate the two directions. We sum the y direction forces together and we can sum the x direction forces together. Again, in this case, the acceleration happens to be zero. It's moving at constant velocity. So you just look at the problem well obviously then. T two equals two, T one. We don't know what either of those are yet though. In fact, that's what we were asked to find. Find T one and T two. We're a little bit closer, at least we know how they're related. If we find one, we'll know the other automatically. How do we find the, if I can find T two, I'll know T one. If I can find T one, I'll know T two. What else can I do? A free body with a... Of a pulley? That piece of it there. Well, we gotta be specific. You know what we put in the drawing determines what forces will be on it. The tension, the weight, the weight is creating the tension on that side. Sort of. That rope runs to that mass. In fact, we have to know it's 200 kilograms. Any forces on it. It's got a mass, it's in a gravitational field. It must have a weight. Now we gotta be careful. We've got two W's here. So let's be creative, maybe call that W one. We'll call this W two. Is that weight in free fall? Couple heads shaking, couple heads doing nothing. Is this weight in free fall? Well, at the moment it is, but for the problem, this problem, it isn't in free fall. Because it's also attached to this core T two. Now we can sum the forces in the y direction on that. What's its acceleration? Y zero. If the carts move in a constant velocity, so is the mass, because they're connected. So it's moving a constant velocity as well. It's acceleration is zero as well. You do that easy sum. We just know that T two equals W two. Do we have W two? Now we haven't calculated it, but we do have the mass. Remember this is the mass, 200 kilograms. Remember W is MG, what do I put in there for G? What do you say Phil? Because you're not, which means you agree with him. What did you say? He said it's positive. You said it's negative. Why is it positive? G, G as a value we insert into equations is always positive. Where's the negative come from? Because you were putting it in. However, we've already taken account the direction. T two's up, it must equal W two down. These are already taken care of. This was, the two are the same thing. So the negative is already taken care of. You do not put a negative on G itself. You know that anyway. When you go to the doctor's office and he weighs you, he doesn't say, oh, you're negative 812 pounds. So what's that come out to be? 1,800 or something? What? Newtons, unit of force in the SI system. And for all purposes, we're done. Because now we know what T one is. T one is half of that, or what, eight, and then something, oh, 981. Careful, draw yourself a decent three-body diagram. Well, they are connected. The problem's connected. I wouldn't want to keep this rope together, because if I keep this rope together, then I have to account for the force being exerted by the pulley. It's making the rope bend like that. And it's a lot easier just to cut the rope into two separate pieces, remembering the tension is the same anywhere along the rope. But notice that rope only pulls, and it only pulls. Those two ropes pull. That's all ropes do. They only pull. We're gonna see lots of problems like this. The homework has several different problems of these types of things. Biggest warning is watch your units. Watch your minus signs. Don't put in forces that don't exist. Sometimes very tempting to do that. We're gonna see that again in a second. With the next thing we're gonna do. Draw a three-body diagram and take your time with it, because if you rush through the three-body diagram, get it wrong, you'll get the problem wrong. If you hadn't paid attention to these two forces there, you would have gotten the problem wrong. Do you remember what John said, though? About just before we put on the Y forces, what did you call them? Yeah, these were impertinent forces. In this problem, which is why I never gave you the mass of the cart. If we needed the weight, we couldn't have found it because I didn't give you the mass. The reason we didn't need the mass is because it was an impertinent force, in this case, because all of the motion was in the horizontal direction and we didn't have any friction in the problem. All right, any questions about that before I clear the board? Alan? What's the diameter of the other point? Why do I give you the diameter of the upper pulley? Red airing, just screw it down. Just a... You've got to know when the forces don't matter, you've also got to know when other values don't matter. I don't do that often, just when I'm in a bad mood. Mike? Why are you in a bad mood? And my same question is, why is it all? Why is it all worth it? And why does a T1, why does it come for two? Just because there's two ropes, actually? That's the last one I need. There's two ropes attached to this. All right. They have to be attached by a pulley, but they're still attached to this. Ropes can only pull. If there's two ropes, there's two ropes pulling. All right. You can't ignore one of those ropes. We can't just choose to ignore this one, because if we could ignore it, then let's just take it out of the problem. But if we do that, we have an entirely different problem. If we take off this rope, and the rope only wraps partway around the pulley, now what happens? Nothing in the pulley spins, and the rope comes off, and the problem's over. So we needed that to make sense out of the rest of the problem. It's got to be there. All right, so the next little type of problem we'll look at, and we'll need this one for tomorrow's lap. We're gonna look at some things moving in circles tomorrow, so let's just remind ourselves what the deal with that is. All of our problems will be UCM problems. Remember what that means? UCM, universal code of military. It all fits, might as well. Now UCM means, well, take a wild guess at what the C means. Circular. So we're halfway done. What else? Uniform. Uniform. So it was military. Uniform, circular, motion. So we're gonna look at objects going in uniform, circular, motion. So there's an object going in some uniform, circular, motion. What's the uniform mean? Circular, obviously, is r equals a constant. Constant radius, motion is circular motion. What's the uniform mean? Speed is constant. No, is speed or is velocity constant? V equals constant, so V at some little point in time later there's V like that. And so on all the way around for as long as we let it go. Is this accelerated motion? A couple yeses, couple nods, a couple eyebrows go up. What's the question looks? Now come on, V is a constant. Is this accelerated motion? Yes or no? It's a simple question, simple answer. Zero directed inward. Well which is it? I heard some noes. V is a constant. Says so right there. If V is a constant, then A is, it's not zero. If V is a constant, then why isn't the acceleration zero? Notice there's no vector sign here. There is a vector sign here. The magnitude of the velocity never changes. The length of these two vectors is the same. At least with my ability to draw in a cartoon fashion. But the direction's changing. So this is V squared over R. What was the rest? Towards the center. And in fact I can't remember if I gave you this word or not, see that makes it a vector. We've got magnitude, we've got direction. We call this the centripetal acceleration. The circle means center seeking. Always pointed towards the center. The acceleration is in that direction. And in fact those are always perpendicular aren't they? Because the velocity vector is always tangential to the circle, it's the velocity at that instant is only moving in that direction. And the center is always along the radius and the tangent and the radius are always perpendicular. So no matter where we are, we've got that kind of motion. It's not even uncommon to put a little C subscript on that A. Just to remind us it's different than some of the other accelerations we might have. It's magnitude is constant. Because V is a constant and R is a constant. That's our definition of uniform circular motion. But its direction is always changing. All right, let's put that together is what we just had in the last two days. Which is simply this, that is still true. That is always true. I believe on last Wednesday I talked to you about the direction of the sum of the forces and the direction of the acceleration those forces cause. What did I tell you about those two? Each of those is a vector. They're not the same vector. There's a force vector, there's an acceleration vector. But what did I tell you was true about those two vectors? They're in the same direction. They must be because M is always positive. So if we just multiply a positive number times a vector, we get another vector in exactly the same direction. It's just a different vector, but the direction doesn't change. Multiplying by a positive constant will not change the direction of a vector. Therefore, what's the direction of the centripetal force? It's got to be towards the center as well. And that's how big is it? It's M times a directed towards the center. Directed towards the center. And the acceleration must be in the same direction. The acceleration is towards the center, therefore the force is towards the center. So I'll try to draw it as a different vector, because I only have so many colors. There must be something pulling it towards the center. Couple different ways to look at this. I guess you could look at, well, if the velocity vector tips over, there must be something pushing it that way. And that's the force. That's got to always be pushing towards the center because it's always pushing the velocity vector over. I think everybody's done this kind of thing before. Tied something on a rock and swirled it over your head. Then try to let go of it so it hits your little brother. No, it didn't do that part. You got little brothers in that thing. Yeah, go and hit it, homework time. That's a string. If we take that string off and draw a free body diagram, what kind of forces do strings exert? They only pull and they only pull along their own length. So it's got to be a force towards the center. In fact, any time you have a velocity and a force, and they're perpendicular to each other, you will have circular motion. If those are constants, then you'll have circular motion. If they're not quite constant, you'll have elliptical or parabolic motion. You guys believe this? Haven't you ever gone in a car and you go around the corner? Do you feel like you're getting forced to the center of the circle? So how can this be true? Am I making this up? If you're going around a circle, you don't feel like you're being forced towards the center of that circle. You feel like you're being thrown to the outside of that circle. So am I just making this up? I like to make stuff up once in a while. It's kind of fun to confuse you desperately, make you come back for another term, pay double tuition for the same class. I'm telling you, we haven't figured that trick out. Why do you feel like you're getting thrown to the outside of the circle? When in truth, to go in a circle, you need a force to the inside. Try this diagram here. See if this doesn't help show what's really going on. Imagine here, we have a long glass tube that's spinning around like a centrifuge would do. There's an object right there in that glass tube. It's something that can freely slide up and down the tube wherever it needs to go. A little bit later, a second later, that glass tube is now moved to there. This little object is free to slide. So the only force on it is the force pushing it with the side of the tube there. So all it's gonna do is then go to here. And then when the tube spins a little bit farther, it's gonna go to here. There's no force on it directing it towards the center. So it goes farther and farther away from the center. This is what you feel when you're sitting in your car. Your tendency is to go in a straight line while the car goes in a circle underneath you. Only once you either hit the outside door, and now there is enough force in the center to make you go in a circle, or the seatbelt holds onto you, or there's enough friction between your butt and the chair, only when there's enough force towards the center do you go in a circle. If this force is too small, you won't go in this circle, you'll go in a bigger circle. You can see that from the equation. If this is too small, this will get bigger. And it's only when these are all equal that you actually go in a circle. So if you were this little object, you would think, oh my gosh, I'm getting thrown into the outside. When you're not, all you're doing is traveling in a straight line. So it's this force, it's that force we're gonna look at in tomorrow's lap. You're gonna actually calculate it, measure it in two ways and compare them. But in the meantime, let's do a couple problems with this stuff. Alan, you okay? You're frowning at this. Actually go in a straight line, all right? No, not perfectly. Because now that it's over here and the wall of the thing is pushing on it, now it is pushing it to the side a little bit. But there's not enough force towards the center to make it go in a circle. Not until it reaches the end of here, and then there will be a force towards the center. There's still some acceleration on it, right? But it's just not as elliptical or something like that. Yeah, it's trying to find some spot where these two are finally equal. If the force is too small, the radius gets bigger. All right, so let's do a problem. The thing is, if you're being accelerated as you would be going in a circle, fictitious forces arise. You feel like there's other forces that aren't really there. Because you're being accelerated. It's the type of thing, when you're in an elevator and it starts accelerating upwards, what is it you feel is happening? You feel like you're getting pushed down into the floor. But you're not. You're getting accelerated up at the same rate by the floor pushing on you. So the real force is the force of the floor of the accelerator pushing on you, but you feel like you're getting pushed down into the floor. It's a fictitious force you feel because you're in an accelerated reference frame. And in fact, this was Einstein's first glimpse into the possibility of the relativistic motion. He said that if you're in accelerated reference frame and feel fictitious forces, or if you really were put in some frame where you were being pushed down, there's no way to tell which is which. You're in an elevator and you're being pushed down by some force. You can't tell the difference between that and being in an elevator that's accelerating. There's no test you could do that say, I'm in an elevator or I'm in some kind of greater gravitational pull of some kind. But we're not here to study relativity, so we'll leave it be. So here's a problem. Here's a plane comes in, does a loop to loop. And that's a vertical loop. You're way over at the side looking at it. This was what you'd see. Up here, the plane's upside down. Down here, the plane looks like that as it goes in this loop. Radius of 2.7 kilometer. So it's a big circle. That's because he's going pretty fast. 225 meters per second. What force does the pilot feel at those two pictured positions? Pilot feel at those two pictured positions. Well, which? What do you want? The mass of the pilot, I'll show you why we'll leave it like that in a second. Is this accelerated motion? He's going the same speed all the way around the circle. What's the one tool we have to find unknown forces? One tool to find unknown forces, what is it? What would we do with the free body diagram if we had it? You're right, that's what we'll need. You're always one step ahead of me, Bill. You'd be Allen's job, now you've taken it over. What do you do with the free body diagram? I mean, just drawing one will not solve the problem. We need to sum the forces. To find out a force on the pilot, we need to sum those forces. The best way to do that, then, is a free body diagram. Would the free body diagram be the same for that pilot at both the top and the bottom? We only need to do it once. Or should we do it twice for each place? Well, maybe we don't know, so let's just pick one. We'll pick the top. So there's the plane at the top. Anytime you tell me a force, you have to tell me what causes it. What real object causes that force, or I'm not going to put it up. So any forces on that pilot, actually, we don't want to draw the plane. We want to draw the pilot. He's upside down here, but he's upside down. Any forces on that pilot, wait, Mike is smart. He goes first. It's straight down because that's where gravity pulls. Any other forces on that pilot. No, well, he's got constant velocity. So what would his thrust be? Be zero. Otherwise, he'd be accelerating. His velocity be changing, but his velocity is always constant. The thrust always pushes in the same direction as the velocity. So if the velocity is not changing, then there can't be any thrust at this moment. Phil, caused by what? Pilot's butt is in the seat. Force, normal force is a contact force. How do I drop then? Here's the pilot's seat, right like that. What's the rule with the normal force? Hang on. Normal means perpendicular, perpendicular to what? The two surfaces in contact. There they are, right there. The seat and his butt. Perpendicular to that, then, is along that line that either is pushing out or it's pushing in. Which way is it? Where's the seat cushion? At that instant. Fair enough, there's a pilot's wearing pink, pink jumpsuit. There's the pilot's butt. Two surfaces in contact, that's right there. Perpendicular to that, is like that. So your only choice is the forces down or the forces up. Only two choices. Which is it? Not both. All we're looking for is the normal force. How do we put it in? We do one force at a time. I gotta vote for down. I gotta vote for up. I gotta thumbs up. Was that a thumbs up for Allen's down? Was that a thumbs up for Mike's up? Tyler. Ah, Samantha. Two downs and from the same corner of the room. You've got three downs, a down, right? No, you said up. You said down. Are you waffling? Yeah, I'm waffling. Joey, so left is okay as a vote for you? No, you wouldn't even go that far. Bill, Samantha, you're saying it down. Allen, you're gonna jump ship. I'm just thinking double force is actually the pilot and the seat is supposed to accelerate for a second. No, no, no. How did I tell you flip back to the normal force? I told you a couple things about it. One, as I said, it's a contact force. Two, I said it's perpendicular to surfaces in contact. We've got that already. We've got the contact between two surfaces. Remember, we're looking for the forces on the pilot. So we've got the two surfaces in contact. We've got the perpendicular to those surfaces. What else did I then say? Huh? The normal force can only push. That chair can only push on the pilot. It can't pull on him. It can only push on him. If it pulls away, then they won't be in contact anymore. It's got to push on him. That means the force that the chair exerts has got to be down because that's where the pilot is. That chair can only push. The pilot's below the chair. The normal force has got to be down. The normal force pushing down. It's the force on the pilot. Not the force of the plane on the chair or the chair on the, it's the forces on the pilot. They only push. That's where the pilot is. It can only push down. In fact, that's part of the force towards the center that's allowing him to go in a circle. Any other forces on the pilot? Because sooner or later we're done and then it's time to sum the forces. Any other forces on the pilot? No, sooner or later you're done. So we'll sum the forces on the pilot. We only have two forces and they're both in the up-down direction, so we only need to sum in one direction. So we'll see. Might as well take that direction as positive then we don't have any negative signs on it because they're both in that direction. N plus W equals, equals what? The normal force on the pilot in the same direction as the weight of the pilot. That's why they're added together. Equals what? V squared. Equals MA. A is V squared over R. So let's see, little bit of work here, N equals M times V squared over R. That's MG, so it'll be, and I bring it over, it's minus G. We have those pieces. Well wait, we can do a little bit more with it just to show you how they work these calculations. We'll pull out MG, the reason being because then whatever's left is a multiple of his weight and that's what you know as G forces, that whatever that multiple is. Then we're left with V squared over GR minus one. That's just the algebra, is that right? Just the algebra, the simple algebra? That way, this is known as Gs, whatever that multiple is. We have that number, we have all those numbers. Notice that what I have pulled out now is his weight. So I have whatever's left over. What was V? 225. V squared, don't forget that. Students very often forget to square numbers when we've got them. 9.81 meters per second squared over R, which is 2,700 meters. Getting the units the same as we go. What are the units on that? This is 225, that's what I got. V squared, V, oh, that's not, that's what you meant, yeah, you're right. I shouldn't talk while I write, that's why I don't want you to talk during tests just to mess up. So what are the units on that? What are the units on that? Meters squared over seconds squared on the top. Meters squared over seconds squared on the bottom. What are the units? There are no units. That's why this is just a multiple of G's. This is the number of G forces. The single G force is over there and that's where the units are to give the whole thing the units of force. So this comes out to be a single number of times W and that's the G forces the pilot will feel. Anybody have it? 225 squared over 9.81, over 2,700. You got what? Point? Point 91, that's what I had to do. So that pilot's feeling 0.91 G, so is it feeling like there's less than the usual gravity or more to that pilot? Less, and I hope you know that at the top of a big curve like that, the pilot feels lighter. In fact, you might have heard of the vomit comet. The parabolic path, they'll take astronauts in the plane so that they feel about 40 seconds of weightlessness as they go over this. That's because the curvature of that path is just such that they feel no feel, no gravity, but it's still there. All right, well, we got three minutes of plane. Let's look at at the bottom, here's our pilot at the bottom of the loop now. Imagine you were in that plane, you're going down the bottom, you're curving and coming back up. Are you going to feel real heavy? So we expect this number to come out to be greater than one because it's effectively increasing your weight. Any forces on the pilot at that point? Is he weighing anything? It's not so big a loop, he's left the earth, so his weight will still be down. In fact, that hasn't changed one bit from here except for the fact he might have thrown up in the meantime, reducing his mass. Any other forces on the pilot there? The normal force, how do I draw it? Finish that problem for admission to lab tomorrow. Only slightly different than what we've already done. It's gotta be different, it's a different situation. You know that going over the top, he feels light, going down across the bottom, he feels heavy, so somehow this must be a different problem. So calculate that for tomorrow. We'll do a new lab tomorrow, and we'll have a medium sized report associated with it, and we'll work on another writing skill for tomorrow.