 put together now our rigid body dynamics and what we've put together so far was the two separate situations of either pure translation which wasn't terribly different than an awful lot of what we did with the particle dynamics in translation other than the fact we had to use the rigid body part and actually set the rotation to not be zero, the guarantee we only had translation. Then we did look at the rotation, pure rotation on Friday and then now we'll put the two of them together and look at what we call general motion studies. So that's the deal that if we've got some object with a center of gravity somewhere and subject to any types of forces, any magnitude, any number, any direction that those are going to cause assuming that there's no net, assuming that there is a net force after all those are added up, there will be some acceleration of the body and we've looked at that in the translation part. But if there's also no, if there is a net torque then there can also be some kind of rotation. We'll look at the possibility of the two of those now together. So if you remember this is a kind of a combination of a free body diagram where we show the object plus any forces on it and a kinetic diagram where we show the result of the resulting motion if those forces are unbalanced. And the little piece that we put together last week still is applicable in that if the points, the points that we're interested in might be different, well if it be consistent with this blue, I don't like combining different vectors, the pink ones are the force vectors. If we're looking at some other point then this business of using the new point, maybe call it point P, if we sum the torques with respect to G, that would be IG, alpha. But if we want to do it with respect to some other point, I guess this class uses M, don't we? Every book's different, every author's different. I don't know why I don't sit down and talk about it one day. If we do that, if we sum the moments about some other point, we need to use the moment of inertia with respect to that point, which we may not have, but we can go ahead with the parallel axis theorem and use the moment of inertia about the object's center of gravity, which we typically do have. So that kind of business we set up the other day still holds. It's just with the translation part, the angular rotation, the angular acceleration was zero in this equation, but not in this one. Now it's non-zero in both, and so we'll be able to put the whole general idea of this motion together. Alright, so let's do a couple problems and step through doing exactly this. So imagine we have a drum of some kind, maybe just sitting on a tabletop, it doesn't matter because what happens is we have a cord wrapped around it such that there's a force applied to that cord and that will lift that drum off the floor and start to turn it. So we want to find both of those, both of those possible, both of those quantities. So we want to find, actually we want to find three things. Find the acceleration, the translational acceleration of the drum, find the angular acceleration, and also find the acceleration of the cord or cable itself. So that's a general setup. Put a couple of points to it, a couple values, let's put 180 Newton force on there, a radius of half a meter, a mass of 15 kilograms. So all this is, we want to find those three things assuming it all starts for rest. So a very simple setup, very simple problem. Start with a free body diagram and then a kinetic diagram. Joe? What do you say the drum looks? Is it just lying on the ground or is it just lying on the ground? Well it's got the cord wrapped around it, that just guarantees the no slip condition if we need it and it's just sitting there and this force is applied to it. So because it's an upward force, you know, when it's just sitting there it's an equilibrium. Now there's an upward force applied to it, it's going to be unbalanced, it'll be in no longer static equilibrium, it'll be in dynamic motion upwards, so we just want to find out what that is. And it becomes pretty obvious why that should be the case once we put the only two forces in the problem on there. There's no normal force with the floor because it's just started to be lifted, just being lifted off of that. We could add more to it I guess if we want friction with the floor and the light but we're not going to bother with that. So because of this there'll be, we suspect, some acceleration in the y direction. There could be some in the x direction, doesn't appear likely, there aren't any forces in the x direction but just to be complete and then there's also now an angular acceleration that we need to find. Once we've got that business we can find the acceleration of the cord just based on the acceleration of a particular point on the rim where the cord last touches, where the cord's coming off from there, whatever the acceleration is of that point will be the acceleration of the cord as well. So very simple setup, let's see what we have. If the forces aren't balanced in the x direction we'll have x direction acceleration but there aren't any x forces. So there can't be any x acceleration by whatever notation method you prefer to go. We're pretty quickly able to establish that kind of no-brainer. We do expect though there to be y acceleration because there is now a net force upwards and we need to see if we can figure out what that is. So subtracting the two forces will give us the acceleration in the y direction. However, as usual to this point there's two unknowns. Sorry? Oh, t is known. Yeah, I was amazed how easy a problem I gave you. So we came up with a fine ay here at this point because t is given to think Travis, in a way. Appreciate Travis. And so we know all those pieces, w, remember we don't consider as an unknown. We put in all those pieces, we get an acceleration of 2.19. Then we can sum the moments about some point. We know t, we know w, it doesn't really matter. We can solve something about either one. Probably be easier about g because it's just a simple drum and so i, g is just straight out of the book. So summing the moments with respect to g then the only unbalanced torque is that due to the cable. So that's a distance r away of course. And then the moment of inertia of a drum right out of the book. I think it's, yeah, 1 half m r squared alpha. And so it's all for alpha. Alpha equals minus 100. Now we already know what direction it's going to be. And that comes out to be 48 radians per second. That's all, in fact that's physics one stuff for the most part. This is a problem, wouldn't be any trouble doing physics one. What about the acceleration though of the chord? How are we going to do that? Be the same as the acceleration of point a? We're going to do that. That would be fine if what was true. Joey said just make this a equals r alpha. That's the acceleration of point a relative to g but g itself is accelerating. The object as a whole is accelerating. But we can use those very ideas. We can find out the acceleration of point a using our relative acceleration equation. So a g, that we already have. It's a 219 meters per second squared in the plus j direction. Because it was an acceleration in the y direction. What about the relative acceleration that we've got? What about the relative acceleration part? Yeah, you can do it as a cross product. But remember with these two dimensional problems everything's perpendicular. Almost all of our vectors are perpendicular anyway. So there's two parts to it. There's one part in the tangential direction and one part in the normal direction. Which implies of course a normal and tangential coordinates. Where do those go? Well what point determines what these are? Remember it's a point in rotation that determines where those directions are and how it's moving as the moment determines those directions. Remember tangential is generally in the direction in which that point is moving about some center point and the normal is towards that center point. Point a will have a normal direction there. It's going to move up like that. It will be a tangential direction. So just as Chris said, a normal tangential direction in the y. How big? Let's see, a y we've got 2.19 j. I'll put the units on at the end meters per second squared. What's the tangential component of the acceleration of a relative to g? That is our alpha and we have both of those. So r was 25 meters, alpha plus we have is 48 gradients per second squared. So the units are going to match meters per second squared and that's in the j direction. How about the normal direction? r alpha tangential and that r alpha comes from the cross product. How about the normal direction? Any point that has any circular component to its motion is going to have some centripetal component to its acceleration. That's what we're trying to register here. The full cross product if you remember is omega cross quantity omega cross r. If that helps any, from the looks it doesn't. r omega squared and that would be in the i direction which is 0. Why? It's not rotating yet. We're at the instant that this force was applied. There's been no chance for any angular velocity to make itself present. So we only have that last little bit then. And this part would be about 24 meters per second squared. So the whole thing together, 26.2 meters per second squared. About 10 times the acceleration of the drum itself. The drum remember has an acceleration of about 2. This is more than 10 times that. So if you were designing this, maybe you're some kind of winch system, you're going to have to take into account the acceleration of that cable just to get a certain acceleration on the drum itself. A lot bigger, a lot more acceleration there. So the middle part couldn't have any one of my physics 1 students do, well not any one of them, but several of them. This part is a part we need to pay attention to more now that we're sort of here in the advanced physics one. Alright, I have a question. Do another one sort of like this one? Well let's see. Maybe I'll, yeah we'll do this one first and then we'll roll up. We'll do something a little bit different. Alright, same kind of thing. Only this time and this might be exactly the type of design you want to do if you're making a thing that will lower and raise and lower the curtain in the theater. So imagine a spool type thing. There's a larger disc on the outside and a smaller disc on the inside. The larger disc, that might represent the curtain there, is pinned at the top and then wrapped around the outer part of the spool and then the inner part, a hundred new force on it. So some of the other pieces we need. A total mass of this system of 8 kilograms, radius of gyration. Sounds like how much room you have to leave between you and the next person on the dance floor. So it sounds like the radius of gyration is .35. We talked about that on Friday. And the two radii of the two different discs, inner one and the outer one, outer one I called 1 is .5 meters, the inner one I called 2 is .2 meters. And so we want to figure out a couple things. The tension in the line over here, if you were actually trying to design some curtain raising system, you want to make sure that the curtain wouldn't rip and the whole thing would drop. Also find the acceleration of the system as a whole. And it's angular velocity. Angular acceleration, sorry. Alright, so find those two bits. Start with a free body diagram and then a kinetic diagram. And then we can just start setting things up and solve for the unknowns. Alright, so you get started on those two diagrams. What is that? This morning? Recycle. So was that your can? We don't have to get it out. I heard you say it's an LV, but we can get it out of hand. They seem to be trying to dump it there. They're feeling guilty. Alright, well the rest of us will get a lot of work done while we're waiting. So draw the free body diagram, the kinetic diagram they can be simple, can put them on the same one except none of you are using a colored pen so if you're not using a pencil you can get away with it. Be very careful, you can make different types of arrows. In general I wouldn't recommend very similar looking arrows. It stands for completely different things on the diagrams. Looks good, not too difficult. We've got the applied hundred Newton force. We also have the unknown tension over that edge and of course the weight of the system that we're lifting. And the kinetic diagram seems pretty apparent. There won't be any Y acceleration but there should be some, sorry there won't be any X acceleration. There should be some Y acceleration and that's the acceleration of centered gravity. And then because of the setup we certainly expect there to be an angular acceleration. The kinetic diagram is the result of the unbalanced forces and torques that are causing the motion and some of the other forces like the tension. Alright, once you've got that setup then start setting up the equations. Solve them as best you can. The hundred Newton's and the tension up, the weight down and the difference between those, where is the weight? I think we have it in 78.5. Between those is the acceleration of the object. So two unknowns there both A, G and T are unknown. Some of the forces of X direction won't help. We've already done that by simplifying the kinetic diagram anyway. So you're going to have to sum the moments about, make sense to do it about G since that's where we have the radius of the acceleration. The positive in the direction we expect the acceleration, angular acceleration though, not absolutely necessary but just to be consistent. So we have 100 at R2, R2 is 0.2. That will be in our positive direction. The unknown tension will be at R1 in the opposite direction. So have a minus sign on it. And then of course W will cause no, no, sorry W will cause no torque with respect to G. We have the radius generation, not the moment of inertia but the two are directly related. Not even worth considering a separate unknown. Yeah, so we've got all those pieces then. What do we got? Three unknowns. One, two, and three. So we need a third equation. Summing the forces in the Y direction won't work. Could sum the moments about some other point, I guess. It's not always an independent equation. What's the third equation then? Three unknowns, only two equations. Third equation. David? I want some of the forces acting out in the third in that right cable. Well we could but all that's going to tell us is whatever tension is pulling it down has got to be equal to the reaction of the roof holding it to the top. So it's not going to help us. The no slip condition, how do we apply it here? Where? Actually we're going to do it on this diagram here. So this point on that. You, I guess you could, it's just that point itself is accelerating. So it's going to make things a little bit more problematic. But it's the right idea. What? This one. Yeah. That makes more sense because this is, if this is a no slip condition, remember that point at the instant no matter what the motion is, if there's no slip then that point's not moving. It has no velocity at the point at the instant shown. So that's the one where it's clear to make the no slip condition. So that would be then our no slip condition of A G equals R, in this case it would be R one direction as shown. So there's the three equations, the three unknowns and it can now be solved. So that's a matter of algebra. That's not why we're here. So I'm going to give you the solution of that algebra. The physics is all done, 10.3 radians per second and then what called the tension in the, tension in the line is 19. And that's not simply the difference between the tension up and the weight down. If that was the case then we wouldn't have acceleration. All right, do this. Set up the alternate solution by summing the moments with respect to that point out there in the ring that we call it A. Just set that equation up. It gives you the same solution. It happens to do it, you can get alpha and A G with just that and the no slip condition and then T you'd still have to get in some other way. Well, the first equation. So it's not going to save you an equation on this, in this case, but I'd like you to set it up to make sure you have all the pieces to it. The purpose of doing this, if you did, was to eliminate T from the equation. And remember when we do this, when we sum the moments about some point, we need to take the moment of inertia with respect to that point, which we don't have, but you can handle it. So with respect to point A we have both the weight and the 100 Newton force, our exerting moments with respect to point A. That's the positive direction. We have 100 Newton's and its moment arm is R1 plus R2 with respect to point A. So it's 0.5 plus 0.2 meters is the moment arm for the 100 Newton force. The weight is exerting a moment in the opposite direction and it's only R1. So a minus sign and the weight with the 785. So those are the two moments being exerted. Equals the moment of inertia with respect to an axis through point A. We don't have that moment of inertia though. We have the radius of gyration with respect to G, which we use to get the moment of inertia with respect to G. We can use that if we also apply the apparel of inertia to the apparel of axis theorem, which with the no slip condition yields MAGD. So that we can do. We have the radius of gyration. The mass was 8 plus AGMAGD and that will be in the positive direction as shown. So the positive sign here is okay. Everybody comfortable with that positive sign? D is R1. Oops, I have an alpha in there. That would have been nice because we're, that would have given us only one unknown. But we still have two unknowns in there. Oh, I put it in. Yeah, I put it in the R twice. Gotta stop talking while I'm writing. Yeah. So that's I alpha plus MAGD. This one doesn't happen to make the solution really any easier. I don't know if this equation was in any way more difficult than that one probably another way around. But it is very often you don't know that until you've actually set them up, which one is easier. But it is a way to check problems when you do have the time. All right. Time to lay down some bets. Everybody ready? All caught up on this? All right. So everybody get your spirit changed out. We're going to make some bets here. What I've got is a simple spool with string wrap wrapped around the axle. Something like this. First thing I'm going to do is let it sit on the table and I'm going to just simply pull on the string. Not very hard. I don't want the thing to slip. If I pulled it hard enough, it would just yank it right off the table and that's no fun. So I'm just going to pull on it just lightly. We have to lay down our bets now as to what will happen. I'm not going to pull very hard. Remember there's a static friction between it and the fly. I want to pull harder than that because then it will just start sliding off and that's not what we're going to do. So I'm just going to pull lightly on the string and you get a bet what you think will happen. I guess there's three choices. Nothing. I'll pull on the string and nothing will happen. I'll have to keep pulling and pulling and pulling and pulling until the string breaks. I get tired for class ends. Two. Not in any particular order. Rolls. Other possibilities. Leaps straight up. Goes down through the table. That'd be pretty cool. It's possible. Yeah. But I don't know if we actually want it up there. Actually Alan, you probably have four or five more entries before. It depends on if you didn't know. It depends on if you want to. We're going to back you up. Four is Alan. Thank you. Yeah. I didn't know I was possible. But you do not have to vote for that if you don't want to. All right. Phil got a choice. Don't give it yet. Make sure everybody's got a choice. Anthony, you ready? Joey, you ready? David? Thank you. Tommy? Everybody's ready? Samantha? Okay. Here, I'll just put down just a mark for who votes for which one. John? Come on, give me an answer. You're right. One, two, or three? Two. Ken? Two. Samantha. Travis? Two. Anthony. How hard are you pulling? Just lightly. Very lightly. Chris? Two. Three. Bill? Two. David? Two, but I'm not betting anything. Yes, you are. Alan? Two. Two. Nobody took four. We got one each on the other two. All right. So, let's set up just like we got there. With a string coming over the top, just sitting there and I pull this way. And it rolls that way. Not a big deal if we think about it. So, what? Got to it? No. Well, hang on, we're not done. This is part A. If we look at it, if we look at the kinetic diagram, we have the force in the string that way. There's static friction back that way. It's actually the torque those are applying that cause it to roll. There's other forces. Of course, it's weight and it's normal force. Those are not dynamic forces in this problem. Those are static forces in this problem. The normal force is important because it determines the size of the friction force. But in terms of the kinetics of what's happening, the dynamics of what's happening, there's no concern with those forces. And so we have then, if we sum the forces about that point or some of the moments about that point, we very clearly have a moment in that direction. All right. That's part A. Part two is the same thing. Only now I flip it over and the string comes out the bottom. So, very same thing. I'm just going to flip it over so the string comes out the bottom like that. Pull lightly. I'm not, I don't want it to skid, so I'm not going to pull very hard. Just pull very lightly. Same choices. Should we start with that one this time? Same choices. Nothing, rolls to the right, rolls to the left, rolls to the right in reference to just how it's pictured there. All right. Allen. I'm just going to pull on it lightly, right? Yeah, just pull lightly. All right. There's one vote for two. Phil. David. Two. Joey. Anthony. Two. Two. Oh, this is much more exciting. John. Ken. Three. Nobody votes one. I guess you're all assuming I'm actually strong enough to pull hard enough. All right. So, Joey, you're close as you verify strings on the bottom. There it's just sitting there, so now it's just this setup. And I'm going to pull it lightly to the right and it moves to the right. So whoever said two is correct. Why is it though? Because an awful lot of people certainly thought it was going to go to the left. Let's look at the free body diagram. We now have a force like that. We still have friction like that. Huh? There's a proof right there. That picture is not significantly different than this picture, especially if we look at the torques about the bottom point. That's the point that doesn't move. So that was a good place to set things up. So there's less torque around that point. And I don't know how apparent it is. It's kind of dependent on me, my ability to pull with the same force. I don't know if it's apparent which one's faster. And I really don't think. There is less torque here about that point. But it's still in the same direction as it was for this situation. So both answers were at rolls to the right. So those seven owe those four lunch. I figure. Well David wasn't better. What? David wasn't better. I'm going to take one out. Okay. Which would you bet three or four? Two. I mean three or two. Oh, so you don't get lunch, David. That's fine. Because you didn't bet. You know, nothing ventured, nothing gained. I know. So those three get an even better lunch. Because David has to sit out. You have to go though and watch. No, I'll just do my own lunch. No, you have to. That's what you've got. No hall passes for this, David. I think lunch is worth having to buy something. Okay. So let the physics explain to you what's going to happen. Don't always trust your intuition, especially when your intuition is a bit immature, no, malformed, young. Accurate? Just not ready yet. You're just not ready yet. You know, it takes a lot of years to get a real accurate intuition. In fact, only a few select geniuses actually do. And then they become teachers. All right, so here's another problem. What? A truck, my truck, so it's an Austin truck, with a unsecured barrel on the truck. All right, some of the details for you here. Big barrel? Has a radius of gyration of two feet, a weight of 200 pounds. Is what an R. That? No, that's a K. I've only gone through this before. It's a car owner. Radius of gyration, that's a K. Don't make me man just before I write the final exams for the term. Radius of three feet. The acceleration of the truck is five feet per second squared. And coefficient of friction between the tank and the floor. Static coefficient, 0.15, magnetic coefficient, 0.1. I think that's all the pieces. So find the angular acceleration of the tank on the truck bed. Remember, it's not tied down. The truck starts to pull away. It's going to sort of pull out from underneath that tank and cause it to start to turn, as usual. A free body diagram and a kinetic diagram of the tank. I think I gave you all the pieces you need. I'm going to go through the equation K, W, R, and acceleration and then the coefficients of friction. All right, so set the equations up on how to replace them. We won't spend the time to actually solve them. But we've got to have enough equations, enough unknowns. We're not going to be able to solve it. And there's a little bit more subtly here because we've got both coefficients of friction, kinetic, and you have to think about what's going to happen with one or the other. Oh, by the way, those seven, I'm included in getting a lunch, too, so I'll take David's place. No weight for the truck? Do you like chicken tenders? The truck weighs a gazillion pounds. Perfect. Got another one. I'm taking my grandmother. My mother not for a drive. Just kidding, Nana. I'm not going to do that anymore. I'm going to do the set of manners. All right. Just by the way, what he does when he's watching. Think about how the static and kinetic friction come into play. Because without paying attention to those, you won't have enough equations. But make an assumption to choose the right one of those equations. So free body diagram. See the truck's pulling forward so it's going to exert a friction force like that. Of course, it also has weight, which we know, and a normal force, which we don't know. But since that's the complete free body diagram, it should be obvious that n equals w in this case. Not always remember, but it certainly does in this case. We'll take n as a separate unknown. And then the kinetic diagram. It would seem obvious there's going to be some angular acceleration, but it's going to move at least somewhat with the truck itself. So if you were standing on the side here, you would see it start to move to the right and start to rotate counterclockwise. What? Friction. Right there? What did you think was missing? It just seemed like there should be one in each direction. Left and right? Well, this is caught, and remember, friction is caught by contact with some other surface. So there's only one other surface that's in contact with which is the truck. The truck's going that way, so it's going to try to drag the drone that way. But we can set up the equations, see how we're going in terms of unknowns and equations. You don't need to do it in a y direction. You can do that in our head, and that just tells us n equals w. So we have nothing more than f equals m, a, g. So that's two unknowns, because f is unknown and a, g is unknown. And we don't have the mass, but we don't consider it an unknown. So we can sum the moments about where. Probably make most sense to do it about g. One, that's where we know the radius of gyration. And two, if we do it by some other point, things get a little funny because those points are moving too. Those points are accelerating too. So I think for a starter it would be easier doing this and pick a direction, let's see. So f, let's put those in the same direction, f times r is the torque with respect to g, and the moment of inertia is from the radius of gyration and then we're looking for alpha. So we have two equations, three unknowns. We don't know f, we don't know a, we don't know alpha. So what do we do? No slip. Well, no slip then is, I can't use the connect, you're making a determination by saying that. And it may or may not be true. There could be an acceleration of the truck that's so great that the truck bed and the tank actually slip over each other. And we don't have an equation for that because that's a matter of degrees depends on how much slipping is happening. So what do we do? Tangential acceleration here? Well, what about it? We'll probably have to consider if both slip at your hand. Yeah, we're going to have to, you have to do something first, you have to make an assumption of one situation or the other and then see what happens. See if there's a self-contradiction. Yeah, if we get down to the bottom and that assumption can't be true because we get something else that isn't true either, then we know that assumption won't hold. The only assumption we can make is, well, we can either assume it slips or assume it doesn't, but only one is quantifiable. In other words, only one will lead to an equation for us. We've got the limit of this, where it starts to slip, using the normal force ring. Well, that's an assumption. That's an assumption of no slip. So we can make that assumption. It leads to a third equation and we need a third equation. We can then find out if that leads to any weirdness. Any what? Weirdness? Yeah, any weirdness, other than what's just normally resident in the classroom in one day. But no slip is the only one that we can use to quantify things in any way. So if we assume no slip, then the acceleration of this point will be the acceleration of the truck. That's given. And then we can use that. You can use that to... Remember, we've got to apply it this way. The acceleration of the point G is equal to the acceleration of the truck plus the acceleration of G relative to the truck. And the R alpha is only part of that. Maybe I shouldn't list it as T, because that's actually G relative to T. We need R alpha plus 8G. Yeah. So to be more complete, I need to actually say that's G relative to T. Because that point does have the acceleration of T, but don't forget the tank is rolling as well. So it's not directly the acceleration of G which we need for R alpha. So the acceleration of T we have, the acceleration of G, this is all in the I direction, so it makes it kind of easy. So for here, and then alpha, we'll be able to find from the whole piece. So it turns out all of this is in the I direction. So we don't need to do it strictly as a vector equation. We can do it as a magnitude. And then this is... That's the R alpha part. Is the direction appropriate? No, there's a minus sign in there. Because of the opposite direction in which that's turning. So there, that's the only piece we need. So then I think we can find all the pieces. Yeah, we've got all the pieces now. So why don't you finish that up, and that'll be it for the day. Well, sort of. We need to get this to a certain point and then determine if this assumption was okay with no slip assumption. The way to do that is to find out what the friction force is and see if that's less than the maximum static friction available. If it is less, then it's a good assumption. If it's more than that, our assumption's invalid. So check that f is greater than or less or equal to the maximum static friction. Which we can get because we know the normal force is 200 pounds and the coefficient is 0.15, so that's just 30. So if you solve for the friction force and it's less than the 30, then our assumption is okay. If it's not, if it does slip, then we can't solve it. It's not a problem that can be solved empirically. So finish that up. Give you a little algebra to do and then double check the assumption. Got it, Travis? That's, oh, that's alpha. It seems a little more than I had. I'm sure I've read the right one. And that's more than I had. But it should be positive. We already took care of the fact that it's in the opposite direction. Let's see if somebody else gets your numbers or my numbers. Did you check out the friction force? It worked okay. Good assumption? What did you get from the friction force? Yeah, that's way above 100. I didn't pick it. That's the limit for the static friction. Remember static friction is variable and goes up to a maximum and then things start to slip after that. Remember that? The maximum point right here is the normal force times the static coefficient and that's the 30. We need our force of friction to be less than that down here somewhere because the likelihood it's right at the very limit of static friction is not common. Right at that point. What for an alpha? It's not a one. Travis had just done what I had. Anybody else yet? Chris, you got something? Where's an alpha? Do you have that? Do you have F? F is about almost 21. You had 20. For 90.2. 20.2 is 20.7. I had 9. I had 9.6. I don't know. You tried that on us last week. Tell me, has something else entirely different? David, what do you have? I had Ag equal to 1.54 feet per second squared. Ag. I have 1.54 feet per second squared, too. When I plugged that into the friction formula and checked out the coefficient ratio between friction at 1.4, so this was 1.04. So, repeating. Oh, between these two? Yes. I don't have it that way. I just have this. What about that? Do 30 times the ratio? How did you solve it? Well, we have Ag as an unknown here. We have alpha as an unknown there. And F as an unknown there. So, what's the easiest substitution to make then? You need all three of them in the end because you didn't ask for L, F, but you have to find it to check your assumption. So, whatever's the easiest way to solve those three equations for the three unknowns? Anybody else get it? Phil, what'd you get? That's what I got. Thanks, Phil. Where were you on the lunch boat? Do you get lunch? I don't get lunch. Okay, now you do. We're agreeing with me. David, you're still out. I'm not a gambling person. I guess not. Even when you're right, maybe he has access to food somewhere else. That's just algebra. So, double check the numbers and then we're done. You get five thirds for alpha? Huh? Alpha is 1.154. Ag is 1. Can you show me how you got alpha? I got 1.6. No, that's not right. Five thirds basically 1.6 would be right. That's not right, is it? I don't know. I can show you, but it's just algebra. Physics is done. And f is 9.6, which is less than a third. I think Phil got what I got. David got what I got, so the minus signs must be okay. Oh, yes, okay. That's just not the code. I guess what I'm asking is, did you use your third equation and no slope equation to find alpha? You have to. To find alpha? You have three unknowns, f, ag, and alpha. You have to have this equation, but you can't, you don't, this is not ag equals r alpha. It's agt equals r alpha. Okay, so I guess we needed a minus sign there, but I had it up there. That's because it's accelerating in one direction but turning in the other. So those are, those are the cross-piles would give you that minus. All right, well check your algebra because several of us got the same answer there. We'll start again on Friday.