 going very much along the same kind of pattern of the physics one. In fact, if you remember I said this is sort of advanced physics one. What we've just done is particle kinematics, that's the, just the business of the details of where the thing is and where it's, what it's doing while it's there. It's just nothing more really than the time, the acceleration, the velocity and the position, especially as a factor we look at a couple of different coordinate systems that do better for different problems than some others. But it's basically kind of the same stuff we did in physics one. Now we're going to do the next part of it, which is just like what we did in physics one at this time. In fact, it's today that we're doing it there. And that's that we now look at the particle kinetics. This is the simple fact of addressing if we have some acceleration, how do we either maintain that, how do we change it, how do we avoid acceleration? That's certainly all we do in statics. So it's all, it's all basically about the acceleration and the other things come from it. But the problems can come from either direction. If you remember from physics one, we had three different ways that we've addressed these problems. And we'll look at those three again in this class just again in a little bit more advanced way. We have other coordinate systems we can use in this class. We have more sophisticated techniques we can use and problems we can solve. But the one we'll address today for a day or two is the very familiar Newton's law. Newton has three laws. The first two are really just two different forms of this one. We'll refresh them a little bit in a bit here just to make sure we've got them. We'll do a couple straightforward problems. Maybe a little bit more than we would have done in physics one. Then the second one we'll again come back to the work energy equation. This is kind of like it was in the, using the different coordinate systems for us. Each one of these tends to solve a different type of problem a little bit easier. But all of these, especially these last two, the work energy and the impulse momentum equations. They spring right from f equals m a. So these aren't exclusive of each other. These will really just be different ways to look at problems where we short out some of the things ahead of time that may have been a little more cumbersome if we try to get to them through there. So we'll spend a day or two here looking back at f equals m a. The simple Newton's laws as he first stated them. His first law was stated, and you've heard it in various forms. In fact, it's even running on a commercial right now that an object in motion remains in motion. If you go out on the street, talk to the clowns out there and ask them, what are one of Newton's laws? That's as much as they'll come up with for you. For our purposes, it's not, it's not precise enough. When you say an object in motion will remain in motion. What we mean more precisely is that an object in motion will remain in that precise type of motion in the absence of unbalanced forces. But tell us what that means is that the acceleration will be zero. The only way the acceleration is going to be zero is if the sum of the forces is zero. So Newton's first law is really that where those two really mean to us the very same thing. It can't be that one is true and the other doesn't directly follow either direction. So maybe we'll put a magical arrow there that just reminds us that both of those things must be true and for the other one to also be true. Of course what this then means to us if the acceleration is zero then the velocity vector is constant. And so that's a little bit more of what you'll hear from Newton's first law that an object at rest will stay at rest. So that's a particular case of the velocity being zero which is certainly a constant, sometimes one of our favorite constants. Or an object moving in a straight line at constant speed will remain in that kind of motion which again is a constant velocity situation. Remember that velocity of course is a vector. Then Newton's second law and personally I don't see that they're all that different from each other. It's just the case when A is not equal to zero then if the forces are unbalanced then a mass will accelerate. The mass upon which those forces are acting. So this is of course then a situation where A is not equal to zero and V is not equal to a constant. And that's for the most part of what dynamics is about. However we will look at some problems where the linear acceleration is constant and the angular acceleration might not be and that will also fall into our category of dynamics types of equations. Probably what we're looking at here. And then of course the third law is more subtle. Everybody out of the street can say it but they don't necessarily understand what it means or where it comes from. And that's the law of action reaction. That any force exerted on one object causes that object to exert an equal and opposite force back on whatever the first thing was. This is the I guess the wonderful law that makes boxing the important sport it is. And we're going to have women in the Olympics boxing this year in London. So you're going to boxing? He's boxing. Personally women certainly have the right to box if they'd like. I thought they'd have more sense than that. So there's going to be women's boxing. For those women they're no brighter than the men who box. God bless them all. So we've got for our full three dimensional problems we right now have three equations. Which of course are those three equations. Obviously that's not going to work as three. Well it's the three coordinate directions of our three dimensional space. And if we're in Cartesian coordinates it's like that. If it's in polar or cylindrical coordinates, cylindrical coordinates being the 3D version of polar, it's our spherical coordinates. We again have three it's just the subscripts would be different. So for now we can handle any problem is actually limited by how many coordinate directions we're using in our problems. If we have three coordinate directions then we have the three kinetics equations to use. We're doing two coordinate directions, two dimensional space. We only have two directions. However, we also have at our disposal any of the kinematics things we looked at in the opening weeks of the term. Which means we can solve problems more unknowns than just however many coordinate directions there are. We can use any of the kinematics questions that would apply. Alright, a little bit of a rehash to make sure we got all these pieces. Since we're looking at forces we need to make sure we know what all the possibilities are. All these forces we're working with are vectors. Which means we need to know to define the vector both its direction and its magnitude. So some of these forces we already know one or the other of those makes the problem a little bit easier. Remember you're trying to get as many equations as you can for each of the unknowns in the problems. So we have different types of forces. Of course there are general types. These are just pushes, pulls, those kind of things. Something reaching in and moving the object around in some way. Exclusive of any of the others that are going to come up on the list. So that involves a lot of different things. Sometimes it's just somebody reaching in and pushing. It can also be the subclass on pulls which are ropes, strings, cables, chains. All of those things that you might find in your bedroom. All of those things are those type of things. The deal is you can't push with any of these. You can only pull with these. If you don't believe me, by golly you're welcome to go try. I guess you could soak a rope and water, freeze it, and then push with it in some measure. But that's not typically what we're doing. These only pull and they only do so in their own direction. So that's great when we're looking for forces. We're trying to decrease the number of unknowns. We generally already know the direction of a pole doing one of these items. Always pull, never push, and in their own direction. So if you know where the string lies in the problem, you already know the direction. Not only the angle but the which end, the arrowhead goes on for any problem involving ropes, strings, and the like. All of these tend to be what we might consider outside or other type of forces as we'll explain in a little bit. Other possibilities. We're certainly going to have some problems with friction in them. Remember that friction is a contact force. It's only due to two different surfaces in contact with each other. It could be that one of the surfaces is a fluid surface as we have when we have air friction because of air passing over a solid body. But it's still two surfaces in contact. It's always parallel to those surfaces. It's at the surfaces in contact and it's parallel to those surfaces in contact. So that's again good because that takes up at least part of the direction concern with the force vector. What we don't necessarily know is which end of the force vector, the direction that the arrowhead goes on. But we can get some help for it because friction always opposes the relative motion of those two surfaces. You have to look right at the two surfaces in contact where the friction exists, figure out what would happen if there was no friction, then friction acts to oppose that kind of motion. And you have to be careful, especially as we'll see several times in this class, that the object we're talking about might be moving but the two surfaces in contact are not. That's exactly what happens with your car tires. The automobile is moving, the car tires are turning, but if you're driving carefully there's no slippage where the two surfaces are in contact. So the two surfaces are not moving with respect to each other, relative to each other, but the object itself is undergoing quite a bit of motion. So we'll look at that in some detail and even come up with some things that are pretty interesting and pretty non-intuitive I think. So going with the friction force is the normal force. This is an action-reaction pair. It's the business of an object exerting a force on some surface and that surface exerts a force back on the object. So this, too, is a contact force. You will not have a normal force unless you have two objects in contact. It's not uncommon for students to throw a normal force into a problem when there's not even another thing in contact with the object we're talking about, but they just feel like a normal force should be in there. So careful sticking those in where they don't belong. These are always perpendicular to the surface, to the surfaces in contact. That's great. Again, we can find out something about the angle at which these forces are acting because we know what the angle of the surface is in contact with. We can do a little better than with friction. Friction is not, we can get an idea of where the arrowhead should go and we can figure out what the motion is going to be. With the normal force, it's always much easier. The normal force is only a push on the object upon which we're doing this force balance. So it's only a push by the surface on the object we're talking about. If I have something sitting on the table, the table can only push back on it. You cannot pull back on the book. So for accelerating a book, we need to know the normal force. We know its direction and of course the normal force and the friction force go together. We don't have friction without the normal force. And by the way, normal is the physics and engineering term for perpendicular. Now, here's my advice to you. After having taught this at an RPI in Hudson Valley and up here probably 10 or 12 times, here's my advice for finding the normal force. Use a free body diagram to find it. If you use a free body diagram that's been drawn correctly, you'll be able to find out what the normal force is. It's magnitude and its direction. The direction's not as difficult because that's pretty easy to find just from the picture itself. But the magnitude, a lot of times, is harder. Use a free body diagram. Too many students just assume the normal force and the weight are always equal and you're not going to get it right if you do that. So the best way I've ever found of finding the normal force with some consistency and the more consistent we are, the better you do is to use a free body diagram. So from now on we'll be doing free body diagrams with some joy, I can tell. All right, other forces we might see. Forces due to springs or other elastic media. However, we'll mostly use just springs. These can push or pull depending on what happens to the spring. It comes at some rest length. If you compress it, then it pushes back. If you stretch it, it pulls back. Either way, it will take all the springs to be linear springs unless otherwise mentioned. And this del is the difference of the spring from its rest length. So we did a few problems with those in Physics 1. We'll do more problems in here with those and more complicated problems. Now, the thing about springs as we take them, the amount of stretch or compression we keep in the linear region, which if we relate that to what happens in strength of materials, we don't stretch the spring to the point where it no longer recovers back to its original length when the force is removed. So this is then one of the things we call a conservative force. Any force that can return back to its original state once everything's been run backwards. The force stretching it is removed. It will return to its original length. If you let the problem run backwards again to its original spot, the spring will return to exactly what it was in the original situation. Springs, as we take them in this class, do that. That's not the case with other forces like friction force. If you take a problem, run it one way, turn it around, and run it back to its original position, you've doubled the amount of friction, not returned to zero where there was no friction. So that's an example of a non-conservative force. Our other best example of conservative forces are those due to gravity. These forces, of course, if we return something to its original place, it will have the same force of gravity on it, nothing will have changed. And we don't need to do it any more with it than W and Cosm G, which gives us its magnitude and, of course, its direction, which is always straight down. That's where Mother Earth is, and that's what causes gravity. Now, I say that because I know some of you have had well-meaning high school physics professors before, and when you did an inclined plane-type problem, they told you to take the plane, leave it level, and move the weight force over at the angle. Anybody have a high school physics professor or teacher that did that for you? Okay, not in here, because to me, that's as if you have moved the Earth over to the side somewhere. As far as I'm concerned, I like the Earth where I left, where I found it this morning, where I'd like to find it this evening and tomorrow morning when I wake up. If you do this kind of thing, and I were your boss, if you did something like this and I were your boss in an engineering firm, I would think, yeah, I'm not sure that they can handle very sophisticated problems if they can't even handle this in its real situation. They have to make it something artificial, something that doesn't look right, something that could cause other problems and other mistakes. So failure, if you don't want to shook your head, we'll be watching you very carefully. Leave the Earth where it is, put the weight straight down, leave the plane where it is. We're trying to get more sophisticated as we do these problems, so let's address things as realistically as possible so that we can do better with it all. All right, I think that's our catalog of forces for this class. We'll see some, there'll be some problems where we even have them all, I think. It's not too difficult to come up with fairly useful problems that have them all. All right, one last thing. Our book addresses it a little bit in the early part. I'll simplify it now, so it's much less problem. General equation F equals MA. From that, we define our units of force. In SI system, of course, it's 1 Newton, and that's 1 kilogram accelerated at 1 meter per second squared or half a kilogram at 2 meters per second squared or any other multiple thereof. That's the magnitude of a force of a Newton. And hopefully you all remember how poetic it is when we have a weight of 1 Newton is about the same weight as an apple. That's just the poetry of the universe, I think, because we know Newton's story with him probably hung over underneath an apple tree and the apple hit him on the head. I don't know if that's his story, something like that. In SI system, things get a little bit stinky in the English system, as usual. We have two sets of units really in the English system. One is the force unit is a pound force, and that's enough force to cause one pound mass to accelerate at 32.2 feet per second squared. The trouble with this is that the pound as a force and the pound as a mass were defined independently of each other, so that's why we've got that part in there. So I believe they tried to fix it once. I'm not of the opinion that they did, so I'm trying to fix it one more time. And that's 32 pounds of force is a mass of one slug accelerated at one foot per second squared, which I think is quite possibly the god-awfulest, ugliest unit ever contrived this slug. So I believe there was some mention of that early in the book in the first chapter. Is that right? A couple of those problems had slugs on there. I can't remember if there was pound mass or not. So that's the English system. It came about because the mass and the force were defined independently of each other, then had to be brought back together. But here's how we're going to be able to handle it in this course. It all comes because of f equals ma, so we're going to work it from that direction. Anytime in this course when the term a pound is used, whether it's in a problem or whether I use it, pound will always mean a force in this class, which obviously that is that one. This class we don't use the little subscript f. Some situations you'll see that those aren't actually subscripts. They'll write it as a full-size letter and write this item. However, in this class, anytime you see the unit of a pound, it means a force. So if we're talking about the weight of something, you know that if it says it's 85 pounds, you take that to be the force of gravity on that object, always be a pound. So as we do it then, well, for example, handle it in this way. And if you do it this way, it'll always work out in all our problems and we won't have to deal with any of that stuff. Because if you notice as I was writing these down, I had to keep looking at them because I can't keep these straight for the life of me. They're terrible units. I don't like them. So for example, imagine we have a problem where they say we have a 40-pound object. We will know that to mean the weight because pound will always be a force in this class and so that will refer to its weight. So the mass will be then 40 pounds, W equals mg, so over g in the English system is that. So if you ever get to a problem in English units where you need the mass, then I recommend you do just this. Of course, 40 divided by 32.2 is 1.24. And then the units are... Oh, I put meters there, I meant feet. Then the units are pounds, seconds squared per foot. If I were you, I wouldn't do anything else with it but that. What normally happens in these problems is we have to find the mass and then use that in some other part of the problem. If you find the mass, take it over with those units and put it in the other part of the problem and the units just automatically return just like they would have to the original form. That way you never have to worry what's a slug, what's a pound force, what's a pound mass. It will always work out in this class in that way. So just leave them like that, carry it through the problem and the problem will clean up itself if you're watching your units, which you always do anyway. That's my recommendation to you. If you need to call that something, call it a man or something. Name it after a dead white male German physicist. I won't leave it like that. But that's my recommendation, that's what I'll do in problems. I'll just leave the English mass units like that and I'll straighten up later in the problem and then I'll come back together. Very rarely on these problems for some reason do we actually have to find the mass as the whole point of the problem. Okay, so let's do a couple. Most of these are warm-up and flexing the old physics-worn muscles that have long since atrophied, so we'll do a couple of problems. Lots of pushing crates across the floor if you remember in physics one. We have a 60-kilogram crate there with an angled force on it, 300 newtons, 30 degrees, and a coefficient of friction between the box and the surface at rest on of 0.3. So the coefficient of friction, if you remember, it's generally around that. I hope you remember what the k implies in this problem then. What's this little k imply? Oh, you were being cute, coming off guard. He's always cute. Well, Chris, you aren't. Always. What's the k stand for? What's the k meaning is happening in this problem that might not happen in other problems of the same type of friction? No, I got it. It's a tangible meaning. Well, be more specific like I told you with friction. What if the surface is resting on is just some great big wheel itself and now we're pushing on it? It can move. No, these surfaces are moving relative to each other. Meaning it's sliding over this surface. Whether the surface itself is moving, we have no idea. We don't particularly care yet in this problem. We need more information. But it's the fact that the surfaces themselves are slipping over each other because we have this k here for kinetic. Meaning that two surfaces are in motion relative to each other. All right, so from that we want to find the accelerations. If I only gave you static friction, you could then very easily say, no sweat, man, it's not accelerating. Because it's static friction. All right, simple as this is. Remember, we've got to have all forces acting on these objects before we can sum the forces to find the acceleration. So I want everybody to draw a free body diagram. You don't draw a free body diagram and get these wrong. You can't come crying at me. Remember my other students, previous students, what to keep in mind for free body diagrams? No? How do they line up? Large. Think. Large. Yeah, large and? Large. No, large and? Yeah, don't leave any forces out and don't put any extra in and then you don't need. But, large and what? Large and simple. Think football player. Large and simple. So the way it's acting straight down, of course we have the applied force there. We don't consider that two forces. We might break it into forces later, but we don't consider it two. It just has two different components to it. It's direction and it's magnitude. The components are just saying the same thing. So get all the forces in there. Oh, I'm glad we're going over this. I don't see one of them correct yet. Good. Chris is too busy jazzing up the picture. You've got to have all of the forces in the picture before you bother going to here. If you're going to here without all the forces, you're not doing the right problem. You're doing some other problem. So we've got the friction force. You can either draw it right like that. The surface, if you wish, or you can just draw it in the y direction some other way. Do we have all the forces? A couple of no's. As sooner or later you're done putting up the forces and it's time to then solve the problem. Do we have all the forces? Yeah? No? Phil? And don't say no if you don't know which one's missing. Say yes if you don't think any are missing. Because sooner or later we are done. We do have all the forces and it's time to solve the problem. Phil says no, Joey? David? No. Travis already said no. Samantha? No looking at me? No you're not looking at me. No you won't talk to me. No. John? Tom? Phil what's missing? The normal force. You can almost always look at a problem and tell if something's going to happen with the forces you've got there that shouldn't happen, you need to fix it with more forces. We have two forces down. Both of those are correct. We know we haven't gotten those wrong. There couldn't be any other way. There's nothing to keep it from going down. Except we know that there is. There's a table there. So we're missing the normal force perpendicular to the surface which means straight up. You can draw it down at the surface if you want but it kind of makes the picture a little goofy I think. Two just as well. Now a question for my static students which is, I think everybody right? No, Alan you're the only one who didn't have statics yet. So we'll let Alan answer this one. Actually I'll let everybody answer because I know at least three of you are going to get caught on this question. As drawn, what's to keep the box with the friction going this way underneath the force going that way over what's to keep the box from just spinning? Tipping over? Yeah. You pushed on something and it tipped rather than slid, haven't you? Yeah. But the angle's 30 degrees so it's more. So what? So what if it wasn't 30 degrees it was just you pushing on the top corner the friction's down there. What's to keep this from going like that? Would tip over if I did it that way. Nope. It wouldn't? Nope. Not here. It wouldn't. Not right now. Oh is the moment of the box? I mean the words that it's... What causes the moment? I'm not exactly sure. I think it would tip over. It won't. Why not? This is complete. Who has this as their free body diagram or something awful close to it? That's complete. There are no other forces. But we have a force down here and a component of the force up there and they're separated by some distance. Doesn't that cause a moment? Why isn't there a moment in this class, in this problem? Why is this sufficient for what we're doing now in this section? Just the f equals mA. David? It's because we're assuming it to be a particle. Exactly right. We're assuming this to be a particle which means its own dimensions do not matter. So if you can draw it like that or you can draw it like that if you'd wish but remember we're only looking right now at particle kinetics. We just finished particle kinematics so we don't worry about the sizes on this. We don't consider that there's any moment there. Later we will because we'll start looking at rigid body motion and then the size very, very much matters. Now there could be problems where somehow the dimensions come into it in some other way but in this case we're only looking at those. So this is a complete free body diagram. Remember if you didn't have me for Physics 1 I don't know if you heard this warning explicitly. Do not put any forces up on a picture that what do you have to say about each force? My former students? Pertinent. Which means we don't include all the forces between atoms and all the internal forces that have action reaction pairs and all cancel anyway. But what else? What forces, what must you be able to say about any force you put on these diagrams? Man, you guys should sue your Physics 1 professor. It was terrible. What? It has to be termed as... The forces on our diagrams are all caused by something real. Something that you can point to, I can put my hand on, that you can describe to me directly that we can all see and agree on. There are no forces that are just there because you feel like it. No forces that are there because the motion. Motion does not cause forces. Forces cause motion. So all of these forces on here are caused by something very, very real and tangible in the problem. Whoever is pushing the table, the scratching across the table, the earth, all of those are things that you can directly name. And then we can solve the problem in its component directions. Unless stated otherwise, we'll always assume that the x-y coordinates are the usual. But we're not bound to that. It's arbitrary. It's not going to change the physics. So we know it's going to accelerate in the x direction. In fact, we're looking for that. We know it's not going to accelerate in the y direction because that's just not practical for this problem. So the x forces must sum and have some residual leftover that will determine the acceleration for us. So 300 Newtons, cosine 30 in the negative x direction. Oh no, that's positive. It's a friction that's negative. And we know that those must sum to the acceleration that's going to be experienced. So unlike in statics where all of the forces sum to zero, only some of them do now in this part of the class. Then of course the friction comes from the normal force. What is the normal force? Well, the best way to find it is from the free body diagram. It's got to be equal to the weight in some part, but the weight also has the force down acting with it. And those sum to zero. So a very, very simple case where the weight, where the normal force is not equal to the weight. So be careful. Don't get caught just assuming that the normal force always equals the weight. It doesn't. It's very easy to come up with a problem but it's not true. In fact, in terms of friction, you can take the normal force to be the force holding the two surfaces together. Now you can solve the problem. You've got everything down here is given at W. W equals mg. We do not consider it to be an extra equation. We don't consider the weight to be unknown. It's just W equals mg. It's very easy to solve. And so you can finish up that problem. Anybody got it all? It was 635. Yeah. Acceleration. Just solving this now. It's not an algebra problem. I mean, it's now just an algebra problem. You should get 0.635 meters per second squared. All right. Just to warm up from physics one. Maybe you couldn't do it then, but you can do it now. All right. Let's look at another one. All right. A cart running on nicely frictionless wheels. And in the cart, we have a nice steel bar of some kind going diagonally across it. Upon which is a freely sliding collar. That collar can slide up or down depending upon what's happening with the cart. In fact, this would serve as an accelerometer. The pieces here. That's at 30 degrees. That's 30 degrees. And I want you to find the acceleration. There's one particular acceleration that will allow that collar to stay right there. If it accelerates it greater than that right, those collar will slide up. If it accelerates it less than that rate, the collar will slide down. So find the acceleration that allows that collar to stay right there. Remember, this is frictionless. A collar is free to slide on that bar. So what's the acceleration that keeps it right where it is? And we've got all the pieces. Acceleration, obviously, is going to come from a force balance on it. So we need to figure out what forces are on the collar that allow it to accelerate like that. So we'll leave it like that for a second now. Obviously, the cart and the collar are accelerating at the same rate because they're not moving relative to each other. So let's try a free body diagram of the collar. Find out the acceleration of that. It's the same as the acceleration of the cart because they're one and the same. So you draw a free body diagram of the collar with all the pertinent forces, forces exerted by something you can name 0.2. We can see. I'll give you the easy one. It weighs something. Otherwise, it wouldn't slide down if the acceleration was too little. But clearly, that can't be it because if there's no force in the x-direction, there won't be any acceleration in the x-direction. So there must be other forces in this problem. And there must be at least some component thereof unbalanced in the x-direction. If there's any component in the x-direction but it gets balanced, then there wouldn't be an acceleration in the x-direction. What's causing that? Name it, remember? Something I can see, something I can touch, something I can feel. Don't put any forces up there if you can't tell me exactly what's causing it. There must be some force in the x-direction or it's not going to accelerate in the x-direction. Simple as that. What's causing that, Phil? Is that a force? What's causing it? Good thing you put that in pencil. A couple of you have it, a couple of you aren't playing along. We expected to accelerate in that direction. So what's causing those forces? Or is that a component of this one? Okay, what's causing that force? Because you've got to have some force in the direction, the x-direction, there must be something. What's causing that force? No, I said movement doesn't cause forces. Real things cause forces. You need to tell me something concrete, something tangible, something very real. Movement's kind of ephemeral. What, you're going to have to Google that? Ephemeral. It means fleeting, wispy, misty, mysterious. Is that too poetic for you? Beautiful. What other forces could there be? Can't be any springs, strings, ropes, cables. Can't be any friction, because I said it's frictionless. Got the weight, what else is left? There's a possibility. There's only one other thing I think I had on that list. Normal force, which is a contact force. What is that thing in contact with? The bar. The bar, it's in contact with this bar. It's a free body diagram, free of everything we're not concerned with. We want the acceleration of this, because that'll be the same as the acceleration of the cart. So if we get the acceleration of that, free of the bar, at least in the drawing, but that bar is exerting a force on that collar. How did I say normal forces act? Perpendicular to what? Perpendicular to the surfaces in contact. If this cart's moving forward by whatever means, I didn't say anything about that. We're not looking for that. If it's moving forward, the bar is going to push on the collar to bring it along. So it's actually acting along this bottom surface, just inside where the bar goes through the collar, but it's perpendicular to that surface. So it's got to be like that. The surfaces in contact are inside the collar, right along there. What? We have that? No. Well, you didn't have that. You had to point the other way. Do we have to roll-take back? No, a couple of you had it, though. Ken and you had it. Who else had that? Travis, you had it horizontal. Remember, it's perpendicular to surfaces in contact. However, we don't go to the sum of the forces until we have all of the forces. And I don't mean any components in different degrees. That's just the same force broken into pieces. Any new forces, other forces on this picture. We've got to get them all, but there's just no point going to this early. So you had those two forces down, but what else? Was that a question mark? Yeah, there's nothing else on this list that could have added to the forces that we have any possibility of. So, what about the weight, though? Do we know it? I don't ever leave things off the drawings. But don't make up a number. It's a particle, right? That doesn't mean it weighs down. It's just that one. All right, where's the x direction? It's arbitrary, but let's pick something. Now, on these problems where there's some angle in the problem, you have a choice. You can either angle it in the direction of most of the forces. That just makes the fewer signs and cosines you'll need in there. Well, there's only two forces, and that's not going to help any. So my suggestion is angle the coordinates in the direction of the solution you're looking for. We're looking for the x acceleration. So let's put x in that direction. It'll make the problem a little bit simpler, a little bit more straightforward. All right, so any forces in the x direction are going to cause the x acceleration we're looking for. So we'll sum all the forces and find that acceleration. So let's see. N in the x direction, I think, is N sine theta, sine 30. Is that right? We don't know what N is, but we know what 30 is. Sine 60? No, it would be cosine 60. No, it's sine 30, right? Yeah. I recommend you stay with the angles given. Yeah, you can go to 90, but anytime you jump like that, there's a possibility of making a mistake or confusing someone. So that's all the forces in the x direction. It's going to cause this collar to accelerate. We're looking for that acceleration because it's got to be the same as the cart. Even if we don't know the weight of the cart, it's got to be that. Sum the forces in the y direction, but we know the y direction acceleration is zero. See now, if we inclined the axes, we'd have accelerations on both of those. So again, it's another reason to make it a little bit simpler if we put the axes in the direction of the one thing we're looking for, especially if it's acceleration. So now we have the up forces must equal the down forces because they sum to zero. So N cosine 30 equals W equals mg. We don't know what M is, though. Eb, buy some mass? Find some L and cancel. It'll cancel because these are all linear equations in each other. So when we solve for N, put it up here. The M here will cancel the M there. And you're all done. And interestingly enough, then, for any setup with a cart, with a collar on a frictionless shaft at 30 degrees, we'll always have the same solution, no matter how big or small it is. As long as that's frictionless and that's 30 degrees, you'll get the same solution. Chris, you have it? Route 3G. What's that mean? Square root of 3 times 2. I'm an engineer. I need numbers. I'm going to get you away from the math. It was a real world. David? How about 5.66 meters per second squared? Good enough. 5.7 or so. I don't know. Is that a route 3G? I don't know. If you're going to say route something, make it route 66. Nobody knows that. What a great show. David, your age is showing. All right. Here's one for you. Don't always need to find the mass. It doesn't mean it's not worth asking. I remember once many, many years ago, I accidentally left something off on a drawing and I was expecting the students to do it. Okay, so here's my pickup. Here's my flatbed truck. It was just like my car. What's the maximum deceleration this truck can do such that the cart does not slide on the back of the truck? It decelerates too hard. It slams on the brake too hard. The crate's going to slide forward on the bed. Well, you've got to find the maximum because anything less than that won't slide. So we don't have to go on the other side of this. Coming in at 70 kilometers per hour and the coefficient of friction between the bed of the truck and the crate is 0.3 again. Static or kinetic? I heard a couple say static. That it? What do you look at? The two surfaces in contact. Yes, the crate's moving, but we don't want it to move relative to the flatbed of the truck. So your static friction want the maximum deceleration. Find max X double dot so that the crate doesn't slide. And in addition, find how far what the braking distance is. You already got it? I was saying I was good at it. Oh, it's Route 3G. Chris, that's the closest I've come to throwing you out of class. Not that far away. That's about as bad as anything anybody's ever said in class. I thought that was yours. You were better than that in technical freedom sketching. Well, that was mine. Maybe we need to get out of class question. This is the answer. Well, I gave you two parts on this, so maybe we are. No delaying. Let me just double check. Make sure nothing's missing. We'll need a massive box or anything. It's the biggest box in the world. Say my mother-in-law's in sight. Help her move. My mother-in-law, which is... Sorry, Nana, just kidding. What's working? All right, got the initial velocity. What the truck brought to a stop without the crate sliding without tying it down? You got something out? Protesting? Three-body diagram of what? Yeah, the box will work. The box and the truck are going to have the same deceleration, so mine will keep it smaller. Plus, we don't know what any of the normal forces... tricky things happen when four-wheel vehicles are putting on the brakes that make them rigid-body problems as often as not. So it's pretty easy just to do the crate. So three-body diagram of the crate, including all forces. What's that? What are you putting in? No, motion does not cause forces on the way around. It's got to be something, remember, you can touch. You can show me. You can point to. And I can't touch the acceleration. I can see it, I guess. It'll make these two small, generous, big, simple football player type pictures. Yeah, for three-body diagram then, Tom. Is that it? I think so. Let's see. The box has weight. If you want, you can just put energy in there straight. Just skip the W part. Obviously, there's a normal force perpendicular to the bed of the truck. What else? That crate is moving 70 kilometers per hour as well. We also have to bring it to a stop. That means we need some force in the direction opposite the velocity to bring it to a stop. And that's the friction. If you want, you can put it right there. If you'd rather, you can put it at the surface. I like to put it at the surface. Just remind me where it is and what it's parallel to. But again, we're treating the boxes as a particle. Any other forces? No sense summing the forces to find the acceleration if we don't have all the forces. What's causing that friction force? That friction force between the bed and the truck. I don't care which way you draw these. If you draw it like that. Or not, it's the same problem since we're working with particles at this point in the class. The object has no real dimension of its own. What's that? That's the acceleration? D is that a minus? Let's see. And then don't forget to find the distance. Okay, that looks right, Travis. Anybody else? Got something? Yeah, that looks right. So now figure the distance. Shouldn't be. This is another one where the mass of it should cancel. Any crater. As long as it has that friction coefficient with the truck bed, any crate will see the same maximum acceleration. Anybody have anything other than that for the free body diagram? Again, on this one the mass should cancel out just like it did before. If you want, you can also put on the diagram maybe in a different color. The dynamics of the situation that you expect to happen. Some books actually teach that as a separate drawing. The kinematic drawing is the results of that friction. Now, if we draw it like that, then obviously in the acceleration we find that it would be negative. So we certainly don't expect acceleration like that. It's only one horizontal force and it's in the opposite direction of that. Is that it? Okay, looks good. This is Travis. So you're willing to go stand a half a meter farther while you get run over. Actually, you're willing to see the truck much less than that. So double check something. Something went wrong. Something forces in the x direction. That will give us the acceleration we're looking for. So we have minus f, which is mu s n. Remember the static friction? That's the maximum static friction. But we're looking for the maximum acceleration. So we'd be at the maximum static friction. And that will give us the acceleration. So the minus sign makes sense there since it's opposed to what I drew for the positive direction. But we don't know the normal force. So we have to go in the y direction. This time n equals w. But as we've already seen, it doesn't always. So the n's cancel. We get the acceleration as minus mg minus mu s g. Is that right? What? 1.3g. Yeah, but not root 3g. Minus 2.94 meters per second square. Then how about the distance? Assume that acceleration to be constant and then use constant acceleration equations. We've got initial velocity. Now we have acceleration. What's the third thing we know? Final velocity. We're trying to bring it to a stop. So with those three things known, you should do okay with that equation if I'm delta s. Travis, you can come up with a different number. Yeah. I have 64.4 meters. 64.4 meters. So you go ahead and say that 3.3 meters. Well, I'm putting in a little bit so you don't actually get hit. All right. That's it then. If there are any questions, Joe, you okay?