 Classmates, questions, comments, concerns. We're sailing through this term. So for some of you that have missing grades, remember everything is on angel. So you can't say how I didn't know I was missing something that's right there if you aren't. So catch up quickly so you can have a relaxed spring break. You need at least the second half of it to sober up from the first half of it anyway. So it's not the way you want to spend your spring break, I don't think. So get caught up, stay caught up, do all right. Any questions before we get going? All right, yesterday we had a problem. We'll take it up again here in a slightly different form. We had one of my personal rocket ships. I had a certain mass, also had a certain thrust to it. I think I called that T, is that right? I think it's kind of 10 to the fifth. Now as we go through this stuff, now the things are changing a little bit. We only have four things, the whole four weeks coming up there. Now we're adding more stuff as we go along. One of the things I'm going to expect you to do as we go along is recognize what we're talking about, what the quantity is in terms of what the units are. I won't always make it so obvious as to say M equals so that you know that's the mass. I might just say a 20,000 kilogram rocket ship. You have to know from that wording that that's the mass you were given, not the weight. What's the difference between mass and weight? Weight is the gravitational force on an object. That object being of mass, M, whatever that is, in a gravitational field, which we designate with G. Now we typically take G to be 9.81 meters per second squared. It does not mean the object's actually moving at that acceleration. You are subject to this very same relationship. You have a weight because you have mass in a gravitational field, but you're not accelerating right now. Some of you in certain ways are decelerating right now. Feeling it, Joey? Yeah, yeah, because we're past noon on Wednesday. So we're heading downhill, decelerating. So don't take it means that that is your acceleration. It is, it's the gravitational field of strength giving us an idea of what the gravitational force is on you in that gravitational field. If you go to some other planet, which some of you seem to do frequently, that G will change because it depends upon, it also depends upon, if you remember what we were looking at in some of the earlier chapters, where you are on the earth and even the altitude above the earth, the higher altitude you go, in general, the less the gravitational field strength. So we use that to calculate weight, but this will have units of meters per second. That will have units of kilograms. And that will have units of newtons, which we, because we're lazy, we're not efficient just right hand. We don't write out the word newtons. Too much right, they hate the right, so we don't do it if they don't have it. So starting up here, yeah, I'll put M equals for you, so you know that means mass. T, I don't know, maybe you have to think about what that might stand for. I could have put F because it is a force. I might have put F sub T, meaning it's a thrust force. Those kind of things you can do, like the origin, it's optional, what you use for these variable designations. I try to do something that's both simple and clear. So instead of putting F sub T, it makes sense to me to just put T, paying attention to that being a force, even so, it's gotta be a force, because it's got units of force. So start gathering this stuff, start becoming more conversant with it. In some ways, part of what you're doing here is learning a little bit of a language you may not have spoken before. You're, I'm trying to teach you some of the language that we use in science, whether it's written, or verbal, or arithmetic, algebraic, and calculus, hey, right, that should take notes. Write down what I say. So we had that problem yesterday. What do we have to, oh, we had to find the acceleration, and we did so. How do we do so? How did we find the, oh, is that the acceleration right there? That would be the acceleration of an object in free fall, an object where the only force is gravity. And we're gonna talk in a little bit about why there's another force on you right now that's preventing you from being in free fall. What do we use? What do we use when there's more than one force, more than the gravitational force acting on something to figure out what its acceleration is? We've gotten mass, we've got something about the force that we wanted to figure out what the acceleration is. What do we use, Phil? Yeah? We use F net. All right, he said F net equals who? Times the acceleration. This is a complete, good start. You get the rest of the day off because you gave us a real good start, but it's incomplete. You should be sitting there going, uh, fix it, quick. If you're gonna be good engineers, with someone you wanna be, you've gotta have this anal retentive thing, this compulsive disorder dwelling up in you that you can't stand to see something on the board that's not right. Billy would've given him more if I'd given you more time. That's why your skin's not crawling right now. Do you have that, that, that oaky feeling? It's just not right. Don't leave it like that. Something you don't look like you're feeling oaky. Oh, Billy, you do. You just kind of an ookiness about you. An aura of oaky. Lem's doing his best to suppress it, but he feels it. What's wrong? This is a vector equation. Force is a vector, and acceleration is a vector. We can't have an equal sign if we don't have equal things on either side of it. So we gotta have both sides being a vector, if it is and it is in this case. If you don't believe force is a vector, well, of course it is. You push left on something. It's very different than if you pushed down on something. Even with the same amount of force, you push in different directions, different things are going to happen. Phil gave me this F net. It's a little bit different than I personally choose to write. This is perfectly fine. It's just not what I choose to write. So what's another option for writing instead of F net? The summation. I'd like to do that just as a reminder that we've got Anna all the forces in the problem. So we've got this thrust driving this rocket ship up. That's a force. If we divide it by the mass which we've got, we should get the acceleration, didn't we? Is that what we did yesterday? Or on Monday? I don't think that's what we did on Monday. Don't make me go back and check the tapes in front of a court of law. We did, there it is. What forces do we have to put in this summation? Every single possible little force there is. Well there's lots of forces acting on the odds. It's made up of a bunch of atoms. And all those atoms are working on each other and all the nuclei are working on all the electrons and the nuclei, the neutrons are working on the protons and there's all kinds of happy stuff going on in there. Do we need all those forces? Well when we drive ourselves crazy, if we had to take into account the forces, all the atomic and nuclear forces, we can't. Those in fact are all forces inside the thing. None of those forces are coming from outside acting on the object. None of those forces are going out in any way acting on, those are all internal forces. And in fact, those all cancel each other. Remember we talked about Newton's third law? A little bit on Monday, the action-reaction pair. This is an object, of course, from one object on another gets an equal and opposite force back on itself. Remember that? We talked about it a little bit. We'll hit it some more in a second. Well that's what's going on with all the electrons and the neutrons and all that stuff. As a nucleus pulls on an electron, the electron pulls on the nucleus with an equal and opposite force and those two forces cancel out each other. So that's why we don't have to worry about those. They're internal, they have always come in an equal and opposite pair, they always cancel each other out, so let's not mess with them. But this thrust isn't that way. This is a force on the rocket that isn't canceled by anything directly. So we have to account for it. But, do the math real quick. Take this force, divide it by m, and we don't get the acceleration we got on Monday. It was 5.1, wasn't it? 5.2. You divide this force by that mass, just like we could do here, we don't get this acceleration. Did we goof up on Monday? It happens, so it's pretty easy to goof up at the board. Are you checking it? Check it for us, Joey. Three times 10 to the fifth, divide it by 20,000? You don't get that. You can just do it in your head. 300,000 divided by 20,000 is about what? So how can we don't get the same numbers now that we got on Monday? Yeah, check your notes. I don't have all the forces on this picture that were in this problem. And we can't do this till we have all the forces in the problem. All the, what did I call? What did I say? I said something about what forces to use. Huh? I said something, I think. Check your notes, did you write it down? Who's my favorite student who takes good notes? No? In this case, we could call that the why I called. But I said something else. Oh, there's no forces. Huh? Of course, they gotta be real. You guys have so many imaginary forces going on in your head, we've never finished a problem. The forces, like you think, oh, I gotta get a new cell phone. That's a force on your life. That doesn't count here. That's not the kind of stuff we're at. Yeah, it'd accelerate you in a certain way, but that's not what we're talking about. We're talking about real forces. All the forces on the object. Yeah, all the forces on the object because we don't want forces on something else. To help us figure out how the rocket's gonna accelerate, that wouldn't make any sense. It's gonna be with forces on the rocket, but I did use a word. Excuse me. All the forces. Did I say external? I think I said something else. Yeah, they gotta be external because I just talked about the fact that all the internal forces cancel each other out in pairs, all the molecular, all the nuclear forces we don't care about. The force is a vector anyway, so that's absolutely vital, but I wouldn't have said it a step. I'm looking for, I did say something else here. Holy. No. Maybe I didn't. I'm pretty sure. I thought I said it. In dynamics, we're doing kind of the same thing in parallel that we do in here as we go along. It's just in dynamics, we do it in much greater depth. So I may have said it there and I didn't say it here. All the pertinent forces. Did I not use that term for it? Okay, there you go. All the pertinent forces. All the ones that matter. There might be other forces in a problem. They just don't matter. Getting, judging whether a force is pertinent or not may take a little practice as we go along. A good example, if we have some object sitting on a frictionless surface, sitting on ice that is then greased, sprayed with silicone, and then you rubbed it with your hair. That makes that slippery. If we push on something like that, its weight is not necessarily a pertinent force. Because it's that way, the object can only move that way. So the weight may not be a pertinent force. We might need the weight to figure out what the mass is and then figure out how it's accelerating, but the weight does nothing in terms of the acceleration directly itself. So there are times when certain forces are in a problem, certain values may be part of the problem, but they might be in pertinent to the acceleration to the summation that we're doing here. So what's missing here, Alan? I think you said it a long time ago. The weight. We also have weight in this problem, acting down. We know the mass, we know G, we can figure out what the weight is. Once we have the weight, it kind of makes sense, I think, that the thrust is working against the weight, and so we have to use the two of them together to find out the acceleration. All right, so that's what we did on Monday, right? All right, we're gonna add to this problem a little bit, a little bit later in the flight. Same thrust, same engine, still squirting along at the same amount, but now the acceleration is six meters per second. What was 5.2? Now it's six, neglecting air resistance, which we will do in almost all of our problems. So I'll say that right now, neglecting air resistance. Why would the acceleration change, huh? What do you mean a decrease in weight? Did they open the door, kick my brother-in-law out? I would, tried to do it in the car once, grabbed on, held on, and they took me in the eye. That ain't fair. Why, Tyler? Yeah, it's burning up fuel. The ship is getting lighter, so if they're still the same thrust, but it's losing mass, because it's burning fuel and squirting it out the tail, its mass is going to be less. How much less? In fact, you figure out for me, what's the change in mass? You know, there's how much fuel has it burned by the time it reaches an acceleration of six. NASA does this on launches. They check the measured acceleration against the expected rate at which fuel is burning, make sure they've got the all the right values as things are going along, and find the change in mass. How much fuel was burned? Did you say yes or no? Huh? Yes. You think so? Did you stop working? Yeah. What was the thing you got? First acceleration that we had on money, right at takeoff, full fuel tanks was 5.2. Do you need that number? I don't think you need that number. Plus we're doing the same problem, so if you get an answer, find somebody else who has an answer, see if you got the same thing. Good engineering expectation. For a second opinion when you can't. You can't on tests. Anybody's name? Anybody? What's his name? You're wrong, he's not. What's his name? Okay, good. You can ask John. Or you can introduce yourself. No, he doesn't look at me. He didn't want to play, does he? Yeah. So you're trying to be friendly, but Chris won't let you. Thing to do, we've got acceleration. We've got thrust. You can find out the weight now, find out the mass now, find out what the change in mass was. So you don't need the original acceleration. You do need the original mass because you want to find out the mass changed. Did you get something? Is your name Chris? Somebody's name is Chris? Joey? I don't know. Is it he? You don't even like to be Chris? All your folks get your name changed? Now you got something left? You just know he's back there or somewhere. Cause I keep looking over your head when I talk to Len. Maybe there's no one there and I'm just faking you out. Do you know who Len is? Check with Len. No, that's better. Oh, that's good. I know. I made a wrong turn. I think you're in your car. Find delta M. You see delta means two minus one. We have mass one. That was mass take off 20,000 kilograms. Need to find mass two now. Anybody agree yet? Len, did he agree with you? Was he wrong? Is that something that we're talking about? Then check with Len. Everybody, does anybody not know Joey's name? Joey. Joey, everybody knows your name. And class with me. That's a bit difficult. Actually, it's third, cause you're in IED as well. We did know that. Check with John now. There's a one name you know. Anybody who checked, did you and Alan agree? Evidently, it didn't agree, cause otherwise we'd just start over. Tyler's not playing with anybody. He's not being anything fun. Come on, Tyler. Calc's our name is the boot of this Calculator. It's a bunch of a fan. Here's a big idea. We're doing the same problem. Should get the same answer. You guys agree? Joey's going, Cal, all these names, people. There's an Andrew, and a Mark, and Tyler, and I don't even know how these people are. Nobody agreed yet? All right, let's see, let's do it here together, see if we can agree. We need to find M2. M1 we were given, that's the way to take off, or the massive take off, that we already had the 20,000 kilograms. If we can find M2 now, the mass now, that goes with this acceleration, under the same thrust as before, we could find delta M, so we need to find M2 to find delta M. Let's see, we need to, we have some of the forces to get the acceleration. We have a new acceleration because we're accelerating a new mass. Is the sum of the forces the same as it was on Monday? And that's the only thing that's changed. I said that thrust is the same, is the sum of the forces the same. Why not? The weights change because the mass has changed. It could even be that this rocket's high enough, now G has even changed, but I'm not pushing my luck here in terms of what we can do on hand. So let's see, I think we picked up as positive because we got the forces in different directions. So we've got T, which remains the same, minus W2, because they're in opposite directions. My recommendation to you is, do these directions algebraically as you go. So I put this minus sign in here, indicating that the weight is down. That way, every time I put in weight, I have a positive weight, which makes more sense. Have you ever gotten on the scale and had negative weight? You may have lost some weight, but your reading was never negative, so it just makes sense to leave the values positive and do the minus signs algebraically. It's gonna cause whatever mass it is we're looking for to go the acceleration that I gave you. That's what this is here, this A2, we can call it now. M2 appears in two places. It's already here, because that's the mass we're accelerating, but it's in here because the thrust has to push against the weight. So we've got all of those numbers in, except M2, should have come up with an M2 of what? You're gonna have it? Is that a hand up? Samantha, you have it? Okay. Oops, that's not what I had. Phil, what'd you have? A $19,000. Yeah, that's what I had. Check your numbers, maybe a minus sign or something. M2, remember in here? Well, let's see. W2 is M2G, so we can make this T equals M2 times A2 plus G. Is that the right algebra? The thrust we're given, taking the thrust to be the same. The same engine, running the full power divided by A2 plus G. Is that what worked? That's what you did? Samantha, did you catch? Similar, would you have maybe a minus sign? A lot of students think the accelerations in the opposite direction of G, how come there's a plus sign here? There's a plus sign, because this worked out, these are not vectors here that we're adding. This is just quantities that we're adding right there, algebraically. So don't put in a minus sign when one isn't there. You should get just under 19 kilograms. So the change in mass, and burn about 1,000 kilograms of fuel so far. About as simple as they're gonna get, because they're gonna get more complicated from here. Because you're smarter now. You're smarter now than you were Monday. So let's keep going. All right, I can erase this, start clean. Are you okay? And I hope you believe me, whether you may have put it to use yet, that one of the most useful tools in doing these problems, and it will be so for the rest of the term. For those of you continuing on in engineering, who will have statics in the fall, and dynamics and mechanics materials in the spring, this will continue to be true. So let's get a handle on this now while we can. That's this business of drawing free body diagrams. There are very few problems you should do for me for the next year and a half that don't have a free body diagram with them. That's part of why I even started doing that course. We have the technical pre-hand sketching course, is to help with this. It's that important, free body diagram. Simple and large. If they're too small, they're not gonna be helpful after this. Once you've drawn it, and then have to do something with it. If you've done it too small, it's not gonna be helpful. So why do it? You've gotta make them big enough because there's more stuff we gotta get in there once we've drawn the basic free body diagram. We need to use it to help us solve the problems. And if you make it too small, it's not gonna help. Make it simple because you don't wanna waste your time on the drawing, don't put in windows and bumpers and decals and paint jobs and the like. Just make it nice and simple so you can get to the physics stuff, finish that early, catch an earlier flight to Daytona Beach. So make them large and simple. Just the object undergoing the forces. So if we're talking about a simple problem like a crate on some greasy ice, which we are pushing across the ice, very simple problem. When I go to draw the free body diagram, I draw the object we're pushing with the forces free of anything else, free even of the ground. It's just, I'm not pushing, I'm not accelerating the ground. I'm trying to accelerate the object itself. So just the object, that's the free part. Just the object, free of anything else. There's a rope attached to it and somebody's pulling on the rope. I don't draw the rope, the object. Free of everything else. To put down on it as accurately as possible. The more accurate you do this, the more helpful this drawing's gonna be. If you're sloppy about it, that's the kind of help you're gonna get from it. Sloppy help. All who might not know yet if a force is pertinent or not, but that's fine. As you go through the calculation, it just won't be used and then you'll know, okay, that one was an impertinent force. But there's no harm if it's on the diagram and you just don't have them to use it later. What you can't do is be missing any forces. If you're missing any forces, you're not doing the right problem. You're doing a different problem. You gotta have all the pertinent forces. So let's talk about the types of forces we're gonna see in here. So you'll recognize them in the problem and know how to apply them to the free body diagram so they're useful to you. You mess these up at all and it's just gonna mess up the problem for you. You're gonna get frustrated, obligated by law to take points away, get the red pen out and starts flashing it and slashing it all in place. Red because it looks like blood, like I'm drawing blood. So we gotta get all the pertinent forces out. Don't forget, when we're looking for forces, whether we're solving for them or we're applying them in a problem, these are vectors. So you've got to have all of the parts of a vector to do these problems right. What are the parts of a vector? The characteristics of a vector. There's three of them. Direction, the free body diagram, the direction and the magnitude are the most crucial. It's in the calculation itself where the units really come into play but pay attention to all of them. So as we go through these forces, remember at any time we're looking for as much as we possibly can about the direction and the magnitude of the forces. The more we can figure out ahead of time, the less we have to figure out later in the problem. The less work for you to do. The less work for you to screw up. Which means the less points you're gonna lose. All the better. All the better. All I'm trying to do here is keep you on your dot given points. All right, so some of the forces, not in any particular order, but some of the forces that we're gonna come across in our class here. Things like pushes. Well, that's what the thrust was doing. It was pushing on the rocket. Lots of problems we're gonna see in here where somebody you can pretend it's you if you want is pushing on an object to move it. Whether it's a crate or a girlfriend or anything else. There's pushes. Generally, we don't know anything about the direction of magnitude other than what's given. And it may be nothing is given, but it may say somebody's pushing on a crate with a 30 degree angle. Great, now you know the direction. Now you know the angle, you just may not know the magnitude depends on how the problem's written. A lot of times though, we will know the direction. If you have a push on a problem and it says it's at 30 degrees, it usually means something like that. That's good. Half of the part of that force vector is taken care of. You know the direction. It's acting on directly towards the object because that's how pushes work. And if they give you the angle, put it in. Look at the drawing if one comes with it. Look at the wording carefully. We'll have some forces like this. We'll have some that are levels, some that are vertically. There's lots of possibilities. This is just one example. Cables, anything you might find in your bedroom. Anything that you wanna throw on that list? You look like, it's not on my list. No, okay, you're all right. The thing that's good about anything in a problem that pulls is you do know the direction of it because, well, rope, strings, cables, chains can only pull. You can't push with a cable. I guess you could get a rope, wet it, freeze it and then push with it. But in general, any of these things attached to an object, it can only be a pull force. It can only be directed away from the object. So if we have some object and somebody's pulling on it with a rope, it's got to be a force away from the object. That's how pulls work. The force is directed right along where the rope itself is pointing. You can't push with a rope sideways. You can't put a rope on something, be pulling on it and direct the object sideways. If you move sideways, will the rope move sideways with you too and you're still pulling only in the direction the rope points, direction. So if you've got a rope, a string, a cable, a chain in a problem, you know that the force is away from the object in exactly that direction of the rope itself. It can't do anything else. That's great. That's all of the direction is already known. It's one less thing for you to have to figure out. If they tell you to find what's the force in the rope required to give it a certain acceleration, you already know the direction. You just have to work on the magnitude in that problem. Half of it's done. If you recognize ropes only pull and they only pull in their own direction. Those kind of, these are sort of what we might call maybe a better word is applied forces. Because these are forces we have to come in and do something. We have to come in and push it. We have to come in and pull it. There's other forces we're gonna have that just kind of happen because they're there. See some other forces. Let's see. All right. Oh, here, here's something you're doing right now. Let's figure out why. Here's you to a simple free body diagram. This is how I see all of you. When I look at it because this is what I see. Bunch of blocks. Oops, dead of that, I don't know. Mike's already quick dialing his lawyer. Any forces on you right now? Well, of course, you're sitting here, you have some mass and you're in a gravitational field so you have weight. An object, forces on a mass will cause that mass to accelerate. Why aren't you accelerating right now? That must be some force pushing back up against the weight that cancels it. That's the only way the sum of forces acting on a mass, in this case you, to have no acceleration. If that's zero, then that must be zero. There's no other two ways about it. This, F equals MA, this is not a simplification, an estimation. This is the way it works. This equation has never been proven wrong. It's the way of the universe. This side is zero, which it is. Not one of you is accelerating right now. The forces must sum to zero. So something's gotta be missing here. There must be something pushing back up on you with exactly the same amount as your weight. There's just gotta be. It's the chair. Actually what it is is the molecular forces of the chair. You're on the chair being pulled down by the earth. The earth is trying to pull you through the chair. The chair is fighting back with its own molecular forces. Materials in general don't like being torn apart. All of those little forces in the chair are pushing back on you, hold themselves together. If you sit on a really weak chair where those molecular forces are too weak, you will accelerate. It's pretty funny when it happens. And I'm thinking sometimes it might be fun to bring in some trick chairs just to watch y'all accelerate. Look how he accelerated. But the administration asked me to stop doing that. So these chairs are stout enough to push back on you with enough force that they maintain their integrity and they counteract the weight. This is also an example of Newton's third law. You're pushing down on the chair. The chair's pushing back up on you with an equal and opposite force. That's an action reaction chair. Sorry, force. Action reactions parent. Your weight is pulling down on the chair. The chair's pushing back with the same amount. Not more, otherwise you'd take off. Not less, otherwise you'd go down. Just the right amount to leave you where you are. We call this the normal force. Typically give it the letter big N. Be careful because we also have big N for Newton. This is the normal force. Physics and engineering, this word normal has a particular meaning. It's not the meaning, it's not the normal meaning of what you're thinking where ordinary, it doesn't mean that at all to us in physics and engineering. It means one thing and one thing only. It means perpendicular. But I need to explain it more carefully. Perpendicular to what is the question? The normal force, it only occurs when one thing is in contact with another. Your butt is in contact with the chair. If you stand up and you're off of the chair, there's no longer a normal force between you and the chair. It's gone. Contact force between two objects, between two surfaces. If those two surfaces aren't in contact, there's no normal force. Now, careful with that. I have students in problems, a lot of times we'll try to put a normal force in a problem just because they think there should be one there. But then they can't tell me what's the object that's causing it. Where's that normal force coming from? The object we're talking about is not in contact with anything. If it's not in contact with anything, there's no normal force. That rocket problem we did, there was no normal force. That rocket wasn't in contact with anything. Not the way it laid out the problem. It's a contact force. Object causing the normal force can only push. So that's what I drew here. Here's you. Here's the normal force coming from the chair because it acts on you, it's only pushing. So it's always directed towards the object. Only perpendicular to the two surfaces in contact. So if you know where those two surfaces are, you know the full direction of the normal force. Only perpendicular to surfaces in contact. Just look at a problem, see what surfaces are in contact. If any, you know the normal force is acting towards the object because normal forces can only push. And you know it's perpendicular to the two surfaces in contact. Completely important because we gotta get this normal force right in these just like we got any other force right. We're gonna get the problem wrong. Here's an object on an incline. Two surfaces in contact? Of course, the box, whatever it might be. Maybe it's still you sitting on a slope. So we draw a free body diagram. I make it big. I try to give it the same amount of slant so the drawing helps me. It's got some weight. I figure, I didn't say where it was, but let's figure it's on the earth or somewhere or some planet. Normal force, let's see, it's a contact force. We do have two forces, surfaces in contact. It's only a push so it's only gonna act towards the object and it's only perpendicular to the surfaces that were in contact. There's the surfaces, there's the perpendicular, and I can draw the normal force. I don't know how big it is, but I know it's direction. That's great, that's half the problem. You know it's direction. It's not the magnitude of the normal force. There's only one thing you can do and I don't know any other way to come up with the normal force than to do this. I'm actually on a free body diagram and then sum the forces on the object, but normal force will be one of those forces and that's how you can figure out its magnitude. There's no other way. Occasionally, occasionally the force will be equal in magnitude to the weight of the object, but only occasionally. Do not assume that right into a problem or you'll screw up. Occasionally it happens, but not always. Let's see, let's do a little bit more with this problem. There's gotta be something keeping the box from actually sliding down the plane. You put a block on a slippery, greasy, icy, Adirondack hillside, it's gonna slide down. So there's gotta be something keeping it from sliding down. So let's say that is some force parallel to the, let's say that's u, so that's f u. Do that because it would hurt your wrist. Let's figure out then. Let's say that's, let's put some numbers on here. Let's say the weight is, make it something nice and round. We'll say 200 newts. Let's say this slope here, this incline, just to give it a nice fat round number is 30 degrees. You need to exert because we need to know is Mike strong enough to do this? Or does he have to call in reinforcements? So let's figure out how big a force must you exert to hold the box there? Too much force, you'll push it up the hill. Not enough force, it'll go ahead and come down the whole hill. Remember this is a greasy, icy, frictionless surface. Who frictions a force, we just haven't talked about it yet. It's coming up. Is that a problem we could solve? How many unknowns are in this problem? How many things do we not know? Well, we don't know the mass, but we're gonna say if you know the mass, you know the weight, if you know the weight, you know the mass. We're not gonna consider that an unknown. It's always w equals mg, so if you know one, you know the other. We'll take it as that. Make things a little simpler for our discussion. What do we not know in this problem? Because we need as many equations as we have unknowns that we couldn't solve it anyway. Mike, one thing we don't know. We don't know the magnitude of this force. So there's one thing. We don't know the magnitude of that force. Do we know the direction of that force? Yeah, I said it's parallel to the slope, so we already know its direction. Remember, all forces are vectors. You've got no direction of magnitude, or you don't know the force. So there's one thing we don't know. We need the magnitude. Let's do it. Question mark, just to remind us we don't know it. So we need at least one equation. What else don't we know? You could say, well, I don't know what we don't know, and that way, I guess you answered the question. No, no, no, it's perpendicular to the surface. So don't we know it? Yeah, we don't know the magnitude. Again, on this one, we know the direction because we know the surface, the angle of the surface. We don't know the magnitude here either. Things we don't know. We need at least two equations to solve this problem. Anything else we don't know? Anything else we don't know? We know the direction, not the magnitude. Direction, not the magnitude. We know the... There isn't anything else we don't know. There's only two things we don't know. We only need two equations. I'll give you one, just to help. So here's one equation. What's the acceleration? Yeah, zero. We don't want it to accelerate the problem. I said exert enough force so it stays put. So we know the acceleration here. We must know the sum of the forces, but to do that we need some of the magnitude of the business, too, so there's one equation. That the sum of the forces equals zero. Substitute into what? We need a second equation. In terms of what other one? There's only one equation there. It's the same equation here. What do you have, weight? Your job as undergraduates is to get as many equations as you have unknowns. That's your job. We have two unknowns. Maybe it helped. I numbered them. There's two unknowns. We only have one equation. I gave you that. What's the other equation? That's a force, sum's right in there. We don't need the mass because it's not accelerating. So w equals mg is not an extra equation we need. We already got it and we don't need it anyway. We already got the weight. We don't need the mass, so w equals mg won't help us. Let's see other equation. This is your job as undergraduates. You're failing at the one thing I need you to do is just come up with as many equations as you have unknown. That's all the homework's ever been. Joe, what do you mean an equation for magnitude? This is the only force equation we've got. We don't have an equation for magnitude. Do we? We don't be nice if we did. Equation of magnitude. It's kind of like pulling out a great sword. Well, okay, so I can do trigonometry on this 30 degree angle. So what? I still don't have two equations. That's just mechanics. That's just algebra. That's why we let the math teachers teach basketball. Where's the second equation? Give up. Do you surrender? You cry uncle? Do I have you in a headlock? Simply speaking, that's true. They're perpendicular, but again, that's just mechanics. All right, you give up. You want the second equation? Patrick does. Patrick did about 20 minutes ago. Here's the second equation. We have two things, magnitude and direction. So that equation can handle two things at once. Here's how we do it. We break it into the cosine and the trigon, the dot product if you want. Those are all the mechanics to actually do this. Remember when I told you in projectile motion that the two directions, the x, y directions are kept separate. Same thing here. We keep the x and the y directions separate. There's things going on in both. There's intertwining between the two. Then you know it's the same mass in each of these. But this is actually two equations so we can always handle two unknowns in our problem. We have two equations. What is the x direction? What is the y direction? Whatever you say. Remember the origin and the coordinate system, that's arbitrary. You can set any one. That's not gonna change to physics. That's you. That's a human invention. So put x and y wherever you want. Here's my advice. Take it or leave it. My advice is if possible, put the x and y directions in the direction of the unknowns you're looking for, the unknown parts. We don't know the magnitude of either of those. If we could put the x and y in those directions, that's just less complication. The answer's gonna pop right out because each force then is entirely in one of the directions. It's just easier. But you don't have to. You can put x and y anywhere you want. You don't even have to make them perpendicular if you don't want it. I think it'd be a fool if you didn't. But so my advice, if possible, look at the problem. Put the component directions in the direction of the unknowns. Well, we know the directions. We don't know the magnitude. So let's go ahead and line them up like that. That would work. That's okay. Another thing that does for us, only one of the forces is off angle. The other two are right on angle. They don't even have any angle to go into them. Cosine of 90 is one. The cosine of 90 is zero and everything's easier. Do not take this problem and retilt it so that there is no incline. Don't you dare do this. It's hot to do that in high school. Do simple. Some of, I guess nobody here, I'm glad. Some well-meaning high school teachers will teach students to do this. It hasn't changed the problem any. Except it's made it physically unrealistic. Because now where's the earth? Not where I last thought, which is right down there. It's now over there somewhere. And you now have a physically unrealistic problem. So I don't want to see that. And it doesn't sound like I'm going to if nobody learned that from their physics teachers. There's a reason that some teachers only teach high school. All right, so let's do this problem here. Some of the forces in the x direction will determine x acceleration except we don't expect there to be any. Right, so I like to write this down and then actually do the summing of the forces below. So I put this kind of as a header. It just organizes things for me better. And generally we'll do so for you too. So let's see. Any x forces, let's see. And isn't, it's a y force. It's perpendicular to the x direction so it doesn't have any influence on the x direction. F u's there, but it's in the minus x direction. So I'll put minus F u, Charlie. Is that it? If that was the only force in the x direction, we'd have x direction acceleration. Something must be opposing it, huh? No, the normal vector is perpendicular to the x direction. Remember, these two directions don't mix. So if I have a force entirely in the y direction, it's not even gonna show up here. So there's got to be some force opposing you. What is it? Notice that the weight is pushing a little bit in the x direction. Maybe I'll call that wx there. And the y direction, maybe I'll call that wy. So I've taken the weight out and I've replaced it with its two component directions. That's just trigonometry. Now bring in your Sokotoa, Krakatoa, East of Java or whatever that thing is, I don't even remember. So wx is opposing your force. If you weren't there, that's the force that's gonna pull it down the plane. It's directed down the plane. We know those forces will sum to zero because there is no x acceleration. We know that. You can look at the problem and tell there's no, I told you, I don't want the box to go anywhere. So the magnitude of f u, let's see. So f u equals minus wx. That's just algebra. You can look and see, yeah, that follows out of the picture as well. Nice big free body diagram because we need to do this kind of stuff. We need to have room in there to break forces and do their component directions. Isn't that the opposite direction? I already got the negative in here. So if I do that algebra, yeah, now we're okay. See, remember, that was the directional stuff. What we're looking for now is the magnitudes. And so that'll take care of itself now if we keep our algebra just right. Do we know how big this is? Well, yeah, we would if we, we know the magnitude of w, if we knew these angles, one of these angles, we'd be okay. Here's a little helper for you. I invented this myself, this beautiful thing. This helps us, this represents any incline I want. Notice the weight always points straight down. See the magic of that? Fies are getting heavy, good feeling on a straight line. That's why I didn't make that shine because you would have just conked right out. We could have made you bark like a dog and stuff would be very funny. Notice that any time I put an incline in, the normal force is always perpendicular to that incline. Remember, this represents an incline. So let's see here, here's horizontal. That's about 15 degrees, isn't it? Notice that's the very same angle between the weight and the normal direction. So if you remember that, the incline angle is the same as the angle between the weight and the normal direction, you won't have to worry about it anymore. That's all you have to remember. And if you can't remember, just visualize it or make a little sketch. So that 30 degrees is the angle between the weight and the normal direction. There's the normal, there's the normal direction, there's the 30 degrees. The weight is always the same angle as the slope from the normal direction. The normal is always perpendicular to the slope. So WX is the hypotenuse W times, well you only have two choices, sine or cosine, 30 degrees. We're looking for this side here, but that's the same as this side here, which is the side opposite to 30 degrees. Again, this is why you need a big picture. Otherwise you're not gonna be able to see that well. You make a tiny little picture as if you're working for the post office designing stamps. There, I'm working for the post office, you're working for me. Nice big picture, because there's a lot of stuff to get in here. This problem only has three forces and then what if we had four or five? So be careful, so it's W, sine, 30. You need help seeing that right triangle that we do our trigonometry on. Feel free then to turn your paper a little bit. You don't actually move the earth when you do that. You move the earth when you do this, that's what happened in New Zealand yesterday. Somebody did this in a physics class somewhere. New Zealand wasn't ready. We've got W, it's 200 Newton, sine, 30. Look it up, this is sine of 30. I went numbers, five, didn't even do it right. Why did I think of the 300? Oh, 30, I made it 300 in my head. Now I'm doing it again, three, I'm stealing all of it. 100. Trouble is I'm in three different time zones up here. I just said something, I'm right in something and I'm thinking of what I'm gonna say next. So I'm in three time periods at once. That's how you screw up at the border. There we go. The magnitude of the force you need to apply. We need somebody, maybe it's Mike, maybe it's not, it can push with 100 Newton force. That's about the weight of 100 apples. Could you hold 100 apples? What? There's quite a few apples. Road apples? You have to have had a horse to know what road apples are. Now you got it. He said seven. All right, and then, sum the forces in the y direction, which we can do because we know the y acceleration is also zero. It's not going anywhere, y or x. And we can look at it, we're all done. N, w, ask for that, we're asked for this. But that wasn't unknown. We got other forces to talk about. We'll do it real quick on Monday. Good free body diagrams. If I were you, oh, don't you know when I say that? I'd bring in a little ruler or use a, your student ID or something, a straight edge can help with these pictures a lot. Allow you to make bigger pictures accurate. Well done, my friend. You leaving any dirt this time?