 Alright, if you remember, when we started the course, we started with particle motion and broke it into two parts. And we're going to do the very same thing with the rigid body motion that we've gotten now. And the reason I'm bringing that up is because we're going to make the first transition here just like we did in the particle. We looked at particle kinematics first. We looked solely at the acceleration, the velocity, the position, and of course the time that went with it. But treating everything as a particle. Now that we're treating everything as a rigid body, we're going to again start with the kinematics. But that's what we just finished. And if you notice, I think we did that whole business in three days. And I think the particle kinematics was about three weeks because it's all very much similar. Remember, as much as anything, you just swap out the variables and you've got the same concepts that we did with particle kinematics and translation that we do in rotation. We needed to take the greater steps in here in looking at general motion. So we had to put the translation and the rotation together. But that was still all just kinematics. Then we went to the kinetics. Remember the difference? What was the kinetics before us? Well, yeah. What we were doing here is finally figuring out how do we make sure we get the kinematics we want. A lot of what we did in the kinematics was acceleration. The kinetics is how do we get that acceleration? So we started with at equals ma as one way to solve kinetics problems. Then we used the work energy method. And then we looked at impulse momentum. We're going to do very much the same pattern only, again, in a monorapid fashion because so much of what we're doing we've already learned. But we're going to be doing the kinetics starting today. And we'll be doing the Newton's Law as force balance. Remember, these are not all mutually exclusive of each other. They're just different ways of looking at very much a lot of the same stuff. And then these have to lay themselves out as different ways to solve different problems in a little bit more efficient way. And we're going to find the same thing is true as we get over here at the rigid body motion. So we'll look now at rigid body kinetics. So there's a rigid body of ours. We'll break it into the different possible ways we can approach this. So we've got some rigid body subject to all kinds of possible forces. Any direction, any place, any magnitude, however many there are naturally and most of our problems will be down to a little bit fewer than these and or they'll be in more regular directions. But for the most part we want to keep this very general. Might also have a concern with the weight of the object. It's not always a concern in problems, but it often is. And so we tend to use that as a single force at the center of gravity rather than a distributed force throughout the object as it really is because each molecule that makes it up has its own weight. And we know in pure translation a rigid body will act such that the sum of the forces in whichever direction will cause the center of gravity itself to accelerate in that direction. And the sum of the forces in the y direction will cause the center of gravity to possibly have a y component of that acceleration. All we're saying there is we've got some forces on there and it's unbalanced then the center of gravity will accelerate and that will be our description for the subsequent motion of the object due to these unbalanced forces. And there will be x and y components to that and we can find out the x and y components in the summation of the forces in those particular directions. If it is in pure translation though, it's got to be true then that the moment the moment summed on the object will themselves result, will sum this here otherwise it would have some angular acceleration. If you remember from Physics 1, I guess we could be a little more complete. We can put in that the moment of inertia of the object times its angular acceleration but if it's in pure translation that is zero. So if the moments are in balance it's in pure translation. If it's in pure translation then the moments balance. Those two things are essentially one and the same. These are collectively known as the equations of motion. They're the equations that describe what the subsequent motion or lack thereof could have a static situation. These equations determine all that we're going to need to know about the object or our purpose in the next couple days is we've got to finish out the semesters to apply those equations of motion as appropriate. So that's what we need for translational motion, pure rotational motion. Again we've got the possibility of any kinds of forces, any number of forces acting on this object but in this case in the possible case of pure rotation then the forces must sum to zero in both directions so that there's no subsequent acceleration and we'll have only some kind of rotational motion about the center of gravity. If it's not about the center of gravity we'll deal with that possible case in another way in a second. But the equations of motion are such that the x and the y result in no acceleration and if we sum the moments about g we can determine what that angular rotation, the rotational acceleration is that we might have in these problems. So that then becomes the equations of motion. Just because any of them happen to be zero doesn't mean they're not of use to us. They most certainly will be as we go through this. And then of course we'll very quickly get to the case of general motion. We'll spend the day looking at translation of rigid bodies then we'll look at rotational rigid bodies and then the general motion of rigid bodies. This has with it other possibilities for, at least for the analysis of these that can make some of the problems somewhat easier, not necessarily significantly easier. But in this case we have a situation such that the forces result in some acceleration of the body in a translational fashion. But there are also the possibility of unbalanced moments that will leave us with some acceleration, angular acceleration. And the equations of motion are the full, at least for our purposes, of three non-zero equations now that will give us general motion. Hopefully you recognize that these are all 2D problems which is why we only have three equations of motion. If they were full 3D equations we could have as many as six equations of motion, three in each of the coordinate directions for the two, the forces and the general rotation causing moments. We do have one other possibility and I'll show you how it applies today. It's just another possible way for us to need to look at this business of general motion like we've got here. We could have the possibility that two or more of the forces happen to intersect at a certain point. Well, two of the forces almost always intersect unless they're parallel. But we could have the case that two forces happen to intersect at a particular point, that could be one of great interest in the problem as we'll see when we apply this in a little bit. There could be other forces in the problem. But we can use that to our advantage in some cases to actually make the problems a little bit easier. Once we sum all the forces and we know that there's some acceleration of the object itself, we can actually sum the moments about a different point by using the parallel axis theorem on the moment of inertia. And so the equations of motion for us then can be somewhat expanded for our purposes. However, you're going to have to be careful as you go to apply this. So we could sum the moments about this point p. That means that the two forces I've drawn there, f1 and f2, don't happen to exert any moment and that's how the problem can become simpler. But when we do that, we need to understand that then we need to know the moment of inertia with respect to that point p, not the moment of inertia with respect to its own center of gravity, which is usually the one we have at our disposal in the first place. However, we can relate the two and you can do exactly this with the parallel axis theorem. And so depending upon which one of the two moments of inertia that we have, it could be that we have Ip, it could be that we have Ig instead, or it could be that things become so much simpler if we do this about point p. And I'll show you a case of exactly that after we look at a couple cases of it. It can be tricky. I don't think it's any coincidence that this second form has MAD on it because if you screw this up and it's fairly easy to do, it'll become quite mad. And you'll try to express that anger in my direction and I'll just callously deflect what I do. Alright, so let's see how this works. We'll do one problem in this general motion form and then we'll do another one in this general motion. Well, actually we're not doing a general motion problem. We'll do a translational motion problem only. But these two different approaches do come in because for translation we want the angular acceleration to be zero so we need to apply that equation. But first let me lay out a little bit of the procedure of how to go through these things. Just to help lay it out a little bit. As you can imagine, one of the most important things to us is going to be an accurate and useful free body diagram. By accurate I mean you've got to have all the pertinent forces. You can't have just some of them. You need to have all of them. You also, though, don't want forces that don't matter in there unless you can make it clear that they don't matter with that drawing. In other words, there's sometimes where the weight hasn't mattered because we weren't doing any movement in that direction. We'll find this time now though that most of this will apply. You'll need to identify the unknowns. Not just the ones asked for but all of the ones in the free body diagram or other parts of the problem that are unknown so that you can get the right number of equations. Those equations to start with are the equations of motion. The very ones I just wrote down, the sum of the forces and the sum of the moments. In translation, the sum of the forces generally in the x-direction won't be zero. In the y-direction they will be. And the sum of the moments about g generally will also be zero because if we're looking at pure translation we don't want an angular acceleration in the problem. Obviously, you've got to get the same number of those. You need the same number of equations that you have unknowns. We only have, for our 2D problems, three equations of motion. There might be more than that unknown in the problem and in that case you need to bring in the kinematics. That will supply you with more equations. For rotational motion problems it could be those arc length type equations. v equals r omega, alpha equals r, now a equals r alpha. That might be all you need to apply. It could be the constant acceleration equations, any of the ones that you might need to know, might need to apply to have enough equations to satisfy the number of unknowns that you've got. This is your job as undergraduates. You've got to get the numbers here to match. The number of unknowns, the number of equations have to match. Okay, so we'll do a problem here with a darn fine looking sports car and a problem. Look at that, that's a beauty. Nothing more than a hot car. Jetsons? No, no. Alright, so there's a nice sports car. We want to find out what the acceleration will be because guys, if you don't know it now, you'll confirm it. If you want to track a lady, you've got to up the acceleration in your car. It's the only way it works. If you can do that while peeling out, all the better. Because she might be on the other side of the parking lot, if she hears your tires squealing, it's like the mating call of the American male. Am I right? Look at her, she's back to her feet. The acceleration while peeling out, you guys know that phrase, don't you? It's not just a phrase from my 60s. The tires are slipping against the ground. Yeah, the tires will be slipping. So we need to look at kinetic friction in this problem. The co-efficient of kinetic friction will take to be 0.25. And some of the other things we need, the center of gravity. Now this is not something we've had to worry about before. We need to know where it is in relation to some of the other parts and pieces. It's a certain distance back from the front wheels and in front of the back wheels. These are in meters. 1.25 meters and 0.75 meters. So the center of gravity is back a little bit. We're going to assume that this is a rear wheel drive car. Typically they do put the center of gravity back farther towards the drive wheels just so the normal force goes up and that's the friction goes up. And we also need to know that that center of gravity is 0.3 meters above the road surface itself. On this, man, that's silly. We got rid of those and put in a multi-track CPU player with iPod. What's the kind of stuff that you need to see in a guy's car? I'm on airbags. Gun rack. Gun rack in the back wing earlier. That's not a Corvette. It looks like a mid-action Corvette. Anyway, back to the problem class. All right, so we need a free body diagram to start with. So very simple sketch of the car. Starts looking a little perchoonish in the free body diagrams. All right, obviously it's got weight. Oh, by the way, 2,000 kilograms. So it's got some weight and we'll take it to act at that point, G. Other forces, well, there must be normal forces due to contact with the ground. So we'll call those N-A at the front, N-B at the back. And if it's a rear-wheel drive car, then the motive force is actually the friction at the back wheel. We'll assume that the free wheels are light enough that there's no appreciable friction there. It's actually the friction with the roadway that applies the torque that makes the front wheels start spinning. But we'll assume that those wheels are so light that their contribution to the problem is negligible. Also, if we have the mass, we're not going to consider the weight as an unknown. Let's just take that as a given. Now, in some of these problems, it can be very helpful to use what's called a kinetic diagram. If you're careful, it can just be a part of the free body diagram. If you're not careful, it should be a separate diagram. But all that is is a drawing of the object with the expected resultant motion sketched in. So for our amazing car, it would be nothing more in this case than the acceleration drawn in. And when I say you need to be careful when you do this, you can put this on the free body diagram. It would put nothing more than like that. What happens, though, if you're not careful, some of you, you can take this to be a vector that needs to be added in with all the others if you're not careful. Most of you might see that you need to do MA for that to actually be a force. But that's not a force that we add into the force balance. That in itself is just simply the result of the forces being out of balance. So you can do either one of those. It works very nicely for me because I'm smart enough to do all my work in pink and blue and yellow chalk. But most of you don't do your work in chalk. So be careful with the kinetic diagram. All right, so we'll do this in two ways. We'll do this using the general motion equation in this way. And then we'll do it using the general motion equation in that way. And see whether one or the other would have been simpler. So some of the forces in the x direction, that's where we're going to get this acceleration we're looking for. Because we already anticipate that there will be an x direction acceleration. Well, it's just what we knew anyway. That's the general motion of a car. But we confirmed that with our kinetic diagram. So taking the expected direction of acceleration as positive just to make things simpler. So what have we got? We've got FB. And that's it. Equals M. And I can leave off the x there because we're anticipating no y direction acceleration. The whole acceleration is in the x direction here. And we know that to equal UK NB. For this particular situation. That's where the friction is acting as that back wheel. So we know that it would be the normal force times the coefficient of friction. So two unknowns so far. AG and NB. So we don't want to increase that by too much if we can help it. So we'll sum the forces in the y direction. But we anticipate no y direction acceleration. Now if the point of this acceleration was not only to peel out and make a lot of noise. But also to pop the front end in the air like a wheelie. Then that would not be zero. Because we actually want the center of gravity to rise as the car went into a wheelie. A sort of a desperate situation for a guy to do that. If the original wheeling out didn't attract enough women he might have to do the wheelie business. Motorcycle guys do it more. We'll get to them in a minute. Alright. We'll take up as positive. So we have NA plus NB. So there's another unknown. NA is now on the equation. Minus W equals W. We know that up forces equal to down forces since they sum to zero. They didn't sum to zero. It's not worth doing that. But since they sum to zero like we did in statics just makes things easier. So we have three unknowns now. AG, NB, and now NA is in it. So we need at least one more equation. We need more than that if this equation brings into it a new unknown. So we'll sum the moments about G. Don't want any angular acceleration with respect to the center of gravity. So we know that's zero. And we'll take the, well since they sum to zero we'll do clockwise moments equal counterclockwise moments. We won't have the minus signs. So about G we have NA times 1.25. That's the moment arm for the normal force NA. That's a clockwise torque about G. So is the friction force which we're making mu K NB. The friction we don't need to take as an unknown since it's kind of like a mass in the weight. And its moment arm is, I want you to pick it out. This is with respect to G. It's what? Bill says 0.3. Anybody agree with them or disagree with them? Remember what we're looking for. About a point G, the moment arm for FB. And that's the minimum distance between the two which is the perpendicular distance. And that's the 0.3 in the diagram. The center of gravity is 0.3 above the roadway. It's very easy to mess these up. These have not been of terrible importance to us in previous problems because we were taking things as particles. And then NB is a moment in the opposite direction and its moment arm is the 0.75. So we didn't introduce any new unknowns. We don't need another equation. These three should suffice. They are independent. There's three of them. We have three unknowns. So that's sufficient to solve the problem. And since it's not an algebra class, I'll just give you those. So solving that system of equations, we get an acceleration of 1.59 meters per second squared. Fellows, you've got to be over two to attract the girls. So this isn't going to be adequate. The car is a nice looking car. I'll tell you in a minute. It's just not going to do it. Of course, that's pretty low coefficient of friction. So maybe we shouldn't do this on gravel. NA is 6.88. 6.88. That would kill him. It was a pretty light car. And this is just the algebraic solution of that system of equations. It was 12.7. That's part of the reason for having the center of gravity more near the back of the car, more near the drive wheels. This makes NB bigger, which increases the force of friction, which increases the acceleration. Okay. Chris, those numbers work out? Frown. Slightly different or grossly different? 0.75, 0.3. Those numbers are okay. 2,000 kilograms, that's okay. All right. Well, if anything, what's left is algebra. Because I think all the physics is right. Check the algebra, but I'm pretty sure that's okay. Then I want to redo the problem from the beginning, not summing the moments about G, but summing them about some other point. So we'll start over again with the car. Same basic setup. All the dimensions are the same. We've got W and this friction force FB. Now, we're only asked to find the acceleration in this, so if we can get rid of some of these forces, maybe the solution will be a little bit easier. Notice that FB and NA both intersect right there. So if we sum them forces, sum them up, sorry, sum them moments about that point, then the problem might be a little bit easier. But I warn you, you have to do it properly. Here's what happens, and it's a very easy thing to do. And once I tell you not to do it, it's still going to be a little difficult to see why you can't do it. We do not want this to have any angular acceleration. The temptation is to say then that this is zero, but that's not the case. Because this remember equals IG alpha plus MAGD. This remember comes from the parallel axis theorem, applied to the moment of inertia. Just like before IG alpha is zero, but notice this part is a remainder here, which shows that this alpha is not zero. At that instant they do have a different angular acceleration about those different points. So don't say back here that this is zero. If you're going to use just the, some of the moments equals I alpha, make sure that these subscripts match. AA here, I had GG here. Make those subscripts don't match. A here and a G here, then you need to apply the parallel axis theorem. And this entire quantity is not zero. So it's very, very easy to say alpha is zero back here, but you won't get the right results. That's what you did over there. Thank you. Thank you for falling on your sword like that. We appreciate that. I almost have a bad car, so I had to make up for it in some other brave way. That's not pink. That's not pink. That's red. That's candy apple red. When I draw a car, my pink mark and my pink chalk becomes candy apple red. There's a little bit of metallic gold flake in. All right, so here's what we're going to do. One other thing that I think helps with this, but it's not absolutely required, that's to use the direction of positive as the same direction of the acceleration we're looking for with respect to the point we've selected. In this case we've selected point A. Two forces intersect there. We're not concerned with just what the two forces are. So it makes sense to pick point A. Notice that with respect to point A, the acceleration is counterclockwise with respect to point A, so that's the easiest way to pick the positive direction. But you don't need to. You can pick either way as the positive direction. It's just if you do it this way there's fewer minus signs and that generally makes solutions go a little bit better. All right, so sum the moments with respect to point A and A doesn't apply. W is counterclockwise, so that's minus 1.25 meters W. And remember, W is not an unknown. We have the mass so we have the weight. The minus, because it's counterclockwise with respect to our point A when counterclockwise is defined as the positive direction, just as that's the way H is going. So let's W, F. The friction force goes right through A, which is why we didn't pick point A. So we need then to add on the moment due to B. Now, that's with respect to point A. So it's the full two meters back of the front wheel. Point A is our front wheel. So this is then two meters. And that equals IA alpha, but we don't have IA. So we need to go to this form where the moment of inertia doesn't matter. And that equals M, which we've got A, G, which we're looking for. And D, I'll just go ahead and write it in. D is the distance, the minimum distance between A, G and our point of interest, which is the point three. So one, two unknowns. So we will need another equation, but if we have another equation that only has those two in it, then we haven't introduced any unknowns. We have a little bit simpler situation. And notice our sum of the forces in the x-direction has only those two unknowns. A and F don't come into it. And so that's our best bet for the second equation. And it's in simplest form. MAG equals mu K and B. That way we don't need to solve for NA. Not that it was a terribly good deal, but this is easier to solve. This is a very simple two-equation system. That was a little bit more complicated. Three are always more complicated. Well, I don't know if it's always, but my little brain is always. And there's only two unknowns, and that very simply goes into that one. We get the same answer we would have gotten before. Watch this. If nothing else, everybody should realize that it's crisp in it, and I can do it. David, okay? I understand why you don't want that. Isn't that odd? But it's because of the angular momentum of a point G with respect to point A. If you want more detail on where that comes from, it's in the book in a brief discussion. I don't think it's nearly as clear a discussion where it comes from as it is to use it if I give you the appropriate caveat, which I believe I did. Okay? You can see where that came from in the book if you want. I don't think it's worth it for us to develop that in any extensive way to derive that. Okay, so let's do another problem. I'll leave more of this one to you. We have a motorcycle with rider. So we'll just sketch it in very simply. Remember, we don't need a big deal for the drawing here. So there's the motorcycle. It's center of gravity. We'll put right there G1, and the rider's center of gravity will put up about there. Now, we've got two choices in the problem. We can combine those two and find a combined center of gravity. As long as we know where those two are, we can combine them and get a single center of gravity for the entire vehicle. But it's just as easy to leave them separately and apply the moments that they cause and any forces they cause separately, just comes out to be the same in the wash. So a couple of the dimensions that we need, again with respect to the contact of the back tire and the front tire. So we have these three horizontal distances we're concerned with, just like we were in the other problem. This one's 0.4. This one is also 0.4. But that one's 0.7. And these are all meters. So nice big drawing. Gosh, you need a bigger motorcycle there, Joe. But who doesn't need a bigger motorcycle? That's what I always say. And then some of the vertical dimensions we need, we need to know the height of those two centers of gravity above the ground. As we saw, we did in the other problem, just because this is a motorcycle instead of a car doesn't mean that the same ideas don't apply. So the rider center of gravity is 0.3 above the motorcycles and the motorcycle center of gravity is 0.6 above the ground. So what are the two pieces? We need the mass. M1 and M2, where one is the bike, two is the rider. 125 and 75. The bike is 125 kilograms. The rider is 75 kilograms. If you gave technical free-hand sketching, you can do an awesome motorcycle event. Really? Oh, can you? You took it. You're all excited about this. All right, so there's a picture. I'll give you a couple seconds to put that together. We're going to need another one, another sketch of it because we are going to do a free-body diagram. So just sketch it in. It's not any great shape to what that looks like. We're just going to apply the forces as we need to. And this one will address a different method for young men to try to attract young women in case the car didn't work, so I pulled the car earlier. I don't see many red sports cars because teenagers wrap them around trees. All right, what we want to do is find the minimum kinetic, no, static. We're not going to spin the wheels on this one. I guess that wouldn't work. We want the minimum static coefficient of friction so that the rider pops a wheelie. And you just got to love that phrase for all of the Americana that lives in that one phrase, pops a wheelie. I'm reliving my childhood right now as we speak. We used to do it on a stingray bicycle. I had a can of bars with a banana, a secret leopard skin on my front. It wasn't retro then. That was the cutting edge. Let's also then find the acceleration of the bike just to make sure it's going to register on the babe meter that all men have in their heads. So you know how useful that is. All right, everybody up to speed here. Billy all set. All right, free body diagram. We've got, of course, the two weights, the weight of the rider and the weight of the bike. Don't consider those unknown. We've got the masses. We also know that there's a normal force at each and there's a friction force on the rear wheel. There's a few two-wheel drive motorcycles around, but I don't know that there's any front-wheel drive motorcycles around. So it's just a standard bike with rear-wheel drive. We want to find the minimum coefficient of static friction. If the static friction we have is too small when the wheels will start spinning and the friction force is lower, the acceleration is less and the bike won't pop a wheeler. If the static friction is more than that, then it's sufficient for finding the wheel. It's just an extra static friction in the bank if we've got more than this static friction. To figure out just where that limit is, where the minimum static friction is, we need to take the normal force at the front-wheel to be zero. A little bit more acceleration, a little bit more force at the back wheel with the back wheel sticking and that will start to pop up when we've got our wheeler. So we're going to assume that we're just at the point where anything more will cause the wheel to pop up off the ground. We don't want to go there yet because then we have to look at the angular acceleration of the bike as it does its wheel. All right, so there's our setup. As for a kinetic diagram, what we expect at these two masses will not only accelerate, but they'll have the same acceleration. It's been known for bikes to accelerate a little more than the bike does, which is you can find some YouTube videos of that, I'm sure, where the bike goes down the road and the rider doesn't. That's not going to work for picking up babes. You've got to stay on the bike, fellas. All right, so we need to find that acceleration as part of it and not considering those as separate unknowns. The bike and the motorcycle have the same acceleration. Okay, so there's our free body diagram. Depending on how you yourself did it, if you did the kinetic diagram, it's pretty obvious in this case what we want to happen. If we want to actually have the wheelie, then there would be a vertical component to the two of these, but we're just at the limit of doing that wheelie. So we're going to take those to be, just to be a translational problem. So set up our equations of motion. We'll take them in the direction of acceleration to be positive, just fewer minus signs that way. So we have FB, the friction force, which is mu s times NB. Mu s we're looking for. NB we don't happen to know yet. That will cause the bike to accelerate, which is the two masses, both at the acceleration that we're looking for. That looked like it in the x direction. In the y direction, we don't want any acceleration, so we should sum to zero. So we'll just take all the up forces, recall the down forces, and as you might have expected then, NB equals W1 plus W2. Remember, we're taking NA to be zero. It's just as the front wheel starts to lift off of the, lift off of the pavement. Sum the moments about which point. We can do it about any point we want, as long as we've got the geometry for it. Now we could, remember, combine these two centers of gravity to be just one for this rider and bike system, but that's extra complication. It's going to be somewhere in between the two, and we'd have to figure out where, and we need to know the geometry of all that, is there something that might be easier to do? Yeah, some of the forces about B, we've got two forces that go through that, and so they're not going to contribute to the moment equations. They become a little bit simpler. So sum the moments about B, that then is IV alpha. Do we know anything about that? Say that that's equal to zero, because we need to apply our parallel axis theorem, and it's the IG alpha we take to be zero. Then that MAG, AG is the same for the both of them, so the M just becomes the, we can do those separately and just add them in. In this case my suggestion, though it's not crucial, you just have to be consistent, my suggestion is for this problem, because of the direction of the acceleration with respect to the point of interest, we'll take clockwise as the positive moment direction. It just makes for a few less minus signs. All right, sum the moments. We have W2, which is four meters in front of point B. So we have all point four meters, W2, that's the rider, and that's a clockwise so it's positive. W1 is 0.8 meters in front of the back wheel of the point B. 0.8 meters, W1, remember W1 to a known, and we're assuming there, we're just at the point where there's no normal force at the front wheel, it's just ready to lift off. So that's all our moments, and now we can do MAGD for the two masses separately. It'd be M1 AG, and what is D for M1? There's 0.1, it's got that kind of acceleration. What is the D we put in there? 0.6 meters. It's the 0.6 meters. It's the minimum distance between the point, minimum distance between the point and the line of action of the acceleration, which in this case is the 0.6 meters. And then we'll add on just do M2 separately. We've got the same acceleration, what is D2? 0.9, it's 0.9. It's the line of action of the acceleration in the equation, and it's at a distance of 0.9 above the point B, 0.3 meters, and the 0.6. Be careful, don't go to the weight vectors, the force vectors represent weight themselves, that's very easy to do as well. What we're looking at is the acceleration vector, that's another reason that it's sometimes wise to do a kinematic diagram. The diagram actually has the accelerations in it. So the masses are known. We've got two unknowns. I guess we'll see if we get NB, AG. I guess we even need this third equation. I don't think we did, but it's a good practice to do it. We didn't need it, didn't give us anything. It would give us confirmation, I guess. Oh, well, it's got AG in it all alone. So we either needed these two, and didn't need the third, or we could have used the third all by itself. But they should confirm it. If those two don't agree, it's not too difficult to solve both of them. If those two don't agree, then you did something wrong. It was likely a minus sign or something. So that should give you AG 895. Is that right? NB, which wasn't asked for, but we do need it to find the coefficient of friction, the minimum coefficient, and it looks like that's 0.192, which is pretty high coefficient of friction. So if you're going to do this boys, you might want to change where your center of gravity is. You think about whether you should lean forward or sit back more. Would it be easier if the tank was almost empty on the... You gotta think about all these things. Nobody said that attracting girls was a hard job. That's why it's meant to do it. I don't think she listens to me anymore. I don't blame her. All right. I think you're ready for one. This one's got something you're going to have to pull all the way out of the statics. So 50 kilogram crate being pushed on the ground. Did I hear? No. Well, it never goes away. It never gets all the way out. No. 50 kilogram crate and you want to find the acceleration without any tipping. The force is 0.8 above the floor and parallel to it and the crate is a meter on the side and zero thickness. Let's say 50 kilogram sheet of paper here is supposed to push across the floor. Classic two-dimensional problem. All right. So I want you to draw the free body diagram and let's see who remembers a very obscure point from statics which means the first time I'll throw it I didn't remember a polite way of saying that. So free body diagram of the crate the weight acts right at the center will assume it's a uniformly packed crate whatever it might be. So those two forces are the obvious ones and we're looking to an accelerate without tipping so we want the acceleration to be horizontal. It's going to tip. There's going to be some vertical component to that. Let's see who remembers the very obscure point of this. Not tipping. Did you always find it on it? Did you always kind of over-remembered the point? Obviously there's going to be friction along with the floor. Let me make sure I gave you all the pieces. Point two, coefficient of friction at the base here and we're looking for it to slide so the kinetic coefficient of friction is point two force of friction will be right along that bottom surface of course parallel to at the surface. We wouldn't have that if there was no normal force we'll call it NC so it'll confuse it with the N in Newton's and C for the contact. Where is that? Remember all of these problems we're looking at now that we're doing rigid body problems we need to have not only the right magnitude and the right direction but the right location of all forces. So if we don't put this normal force NC the normal force between the crate and the floor if we don't put it in the right place we've got the wrong problem. David? How about at the front end? Front end here? Yes. It would be there if it's at the point of tipping. If it was just ready to tip then in exaggeration it would be on that corner and that's where the normal force would be. Until that point and we may be at that point we might not be until that point it's somewhere between there and where it would have been when the crate was just simply resting there. Once the force comes on the top here P that force is tending to tip it clockwise there must be some force that prevents it from tipping clockwise if we want to do this without tipping. Trouble is we don't know what that is. I was wondering where did it be and I don't know. It's not a big deal though because that force has got to counterbalance the other direction tipping tendency of the 600 Newton force. So it's fairly easy to find out what it is and as long as it's between zero and point five then we know that we're not going to be tipping. In fact by doing this we're assuming that we're not going to be tipping. Okay, so we can we can sell the forces. What do we have for unknowns? A g x well this is m c by the way A g is unknown n c is unknown and x so we have three unknowns. We have three unknowns. A g which we're looking for we want to find the acceleration without tipping we don't know x and we don't know n c so we're going to need three equations. So we'll start with the x one and that should be a great PD channel the x equation. Alright that's the acceleration we're looking for positive in the direction of the expected acceleration p minus f but f is u k and c so we'll make that substitution not consider f as a separate unknown just keep things simple equals m h and the mass is now 50 kilograms so that's no big deal. Alright you set up the other two equations got a couple minutes here you can get those equations then it's just a matter of the algebra of solving for them so you set up the other two necessary equations we have two needed for three unknowns we've only brought in two unknowns and if we can do without the third unknown we're going to be okay we've got done then what's in the crate that's the next question probably body parts from a crashed motorcycle maybe wheelies get a million hits some guy doing a wheelie wiping out the school part a lot x I think so yeah the other equations the forces in the z start in the y direction that makes sense we don't know what n c is maybe we don't need it we weren't asked for it but we might need to actually find it not actually popping the wheelie want to be right at the limit of that so we don't want any acceleration in the y direction of the center of mass so what's that w equals n c so that would be z it's not always the case that w equals the normal force the normal force equals the weight but it often is and some of the moments about which point we can always do it about g sometimes you don't know which one is easier until you've already gone through them and it's certainly not easy to do the problem over again in this case we can say alpha equals zero f big sum direction is positive doesn't matter which actually since the sum is zero we can just do counter clockwise equals clockwise moments so p times what as the moment arm for that force p because it's this distance here so p times 0.3 p we know 600 newtons times 0.3 meters that's a clockwise moment about the center of gravity g w contributes nothing it's through g f is all the friction force is also a clockwise moment so that's 0.2 times nc 491 times its arm right there the 0.5 meters show on your frown so those are the two clockwise contributions of moment about g and then nc is the only one left it's a counter clockwise moment x we need to find that because we need to check to make sure that it's between 0 and 0.5 which actually confirms that we're not tipping so equals nc which is 491 and so you can get x right from that as 467 meters which confirms that it's not tipping because that's less than 0.5 somewhere we can't go beyond the edge of the box so all those parts are left then you can find the acceleration to be 10 meters per second which if you notice is more than either the car or the motorcycle so if you want to attract the girl I guess push a box and don't let it tip it's my advice to you and I'm your advisor question watch out for that this business of i, g, e, alpha and i, a alpha