 Hey, we looked at friction last day, are there any questions from the homework that you would like me to go over that you were going, ah, I'm not so sure I quite wrap my brain around this. I heard somebody murmuring about an elevator question, but maybe they figured it out. Yes. I'd love to do number eight. Anything before number eight? Number six? Okay. Any others? By the way, six? At a higher level, I'm asking you to prove a new formula, but there's some great thinking that goes on in number six. Number eight also a fairly challenging question, anything before those or after those? So number six says prove that the stopping distance, prove that the stopping distance is given by that there, okay? And what do you see on the V right here, Brett? I'm going to write the V squared equation as my starting point, I think, but I'm going to get the D by itself. Okay? I'm going to say that D equals V final squared minus V initial squared all over 2A. Except I know one more thing. If I'm stopping distance, if I'm stopping, I know V final. What's V final? Has to be. So I'm starting with this as my, hey, this is how I'm going to approach this. How did I pick that up? Well, they said that there was a squared on the V. So I said, well, I start with the one that's got a V squared. I'm willing to bet the rest of this is going to be finding an expression for the acceleration. If I can find the expression for the acceleration of an object when friction is the only force, I'll plug that in here. And also they said distance not displacement. So I'm willing to bet they don't care that it's going to be negative or positive. I think they just want the magnitude. Is that okay so far? So I'm starting there and now I'm going to say, okay, here's my car. What are the forces acting on it? Get the obvious ones. Absolutely gravity down. And since we're on level ground, we'd have a normal force up. What else? Well, it says the car is stopping. What force is making it stop? I think friction to the left. Friction to the left. In fact, I would say who's winning friction? No loser. There's no force in the forwards direction when you're trying to stop your car. All that makes you want to keep going is Newton's first law inertia. You feel like there's a force because you feel like you're getting pushed forward. It's actually the seatbelt pushing you backwards. Is that okay so far? So when I go winner minus loser, it's going to be friction equals MA. Friction is what times what? That's forces mass times acceleration. Friction specifically is what? Mu times the normal force. So we're going to try this again. Scene one, act one, take two and action. Brett, oh bright child of mine. Friction is what times what? Oh, you're so smart and clever, Brett. I'm so proud of you, excellent. Mu times the normal force. So I'm going to write this. Mu times the normal force equals MA. I don't know the normal force. Oh, but look, look, look, look, look at my free body diagram. CC, CC, CC, CC, CC. I know another force the same size as the normal force. So if I hear you, I can say this. Is that okay? What happens to the Ms? See, as it turns out, the stopping distance for a semi-truck and the stopping distance for a Volkswagen don't, no difference, the mass is canceled as it turns out. You know what the acceleration here is? That's your acceleration when you're coming to a stop. It's G times whatever your tires are between the tires and the road, but G times the coefficient of friction. So now I can modify this equation. It's going to be V squared on the top. Yeah, VI, but I'll just say V divided by two. Oh, what's A? I've just proved that the stopping distance of a car is that. I believe that is what the RCMP use when they're measuring skid marks. Except what you would do there is you would measure the skid marks. So you know D, you're much more interested in VI. You want to find out if the guy was speeding before they crashed. They have a machine. I saw a video of it somewhere. It's a weight on a pendulum and it hangs so that when it swings, it hits the pavement and how far it slides on the pavement. When it swings down, that's how they measure the coefficient of friction. It's got a calibrated scale and everything so they can measure it. Because in real life, the coefficient of friction changes based on the temperature, time of day, rain or not moisture. There's lots of factors that affect the coefficient of friction of pavement and between pavement and rubber. But there you go. The RCMP map. Is that okay? Am I going to give you one like that? I'd feel okay if that was the nasty multiple choice question and I'd have different formulas for your picker. Which of these is the formula that describes the stopping distance of a car? And I have various ones for you to pick from. But I would also, I would consider that a fairly challenging one. It'd be like maybe two of these as a multiple choice, not 10 of these as multiple choices. Is that okay, Brett? Let's do number eight. Eight. Okay, can I do it all? I think I can do it down here. Says a 50 kilogram block is pushed with a force of 60 Newtons across a 20 meter long floor. The first 10 meters are frictionless but the last 10 meters have some friction. I guess we go from ice to a small, rougher surface. Okay? Says describe the motion of the block. Now that phrase describe the motion is too general. In real life, if I gave this to you, I'd say tell me the total, the velocity for the first, the velocity for the second. I don't know, I'd be more specific. But I glanced at the answers and I said, oh, it looks like they want me to find the velocity in the first chunk. And then it looks like they want me to describe how far it goes after that. I see them before and after. On the frictionless section, what are the forces acting on this block? Get the obvious ones? Absolutely. What else? What else? Says there's a force pushing it forward. How big is the force pushing it forward? 60 Newtons, that's my F applied. Yes? Friction? Not in the first chunk. So nothing. When I go winner minus loser, who's winning? I can't hear you. I heard the force. I think you're right. Yeah, the 60. Who's losing? No one. In fact, in the before section, it's going to be 60 equals ma. Did they tell me the mass? Yeah, what? Oh, so I think the acceleration on that first section is gonna be 60 divided by 50. The acceleration on the first section is gonna be 1.2 meters per second squared. Does it say I'm starting at rest? I think I'm going to have to assume that. So I'm going to assume vi is zero. I know that a is 1.2. The distance, how far do we travel for that first frictionless section? It tells me that. 10. Could I find the final from this? Yeah, this is the squared one. This is gonna be vf squared equals vi squared plus 2ad. Oh, and vi is zero. That's kind of convenient. Vf is gonna be the square root of two times 1.2 times 10. It's gonna be 1.2 times 10 is 12 times two. It's gonna be a square root of 24. Square root of 25 is five. It's gonna be like 4.9 or something like that, which I notice it does actually say. Square root of 24 is that four? I should check. Yeah, 4.8, 4.9. That's how they got the velocity for the first part. And again, I like the physics here. I don't like the wording of the question. When it says describe the motion, I would be much more specific. I would say part A, find the final velocity after the first 10 meters. Part B, how far does the block travel once it hits the friction surface? That would make a clearer question. Speaking of which, after, let's go back to my diagram right here. What are the forces acting on this? Still the 60, oh, friction. Who's winning? Now, because it comes to a stop, I know that friction must be winning, but what if they hadn't told me that? I'm gonna show you, what if I had assumed the 60 was winning? There is something that will tell me that I guessed wrong and it becomes very, very obvious. I'm gonna assume the 60 is winning, you know it's not. I'm gonna go winner minus loser equals MA. Is that okay? Friction is what times, sorry. Friction is what times what? I don't know the normal force. Oh, but look, look, look, look, look, look, look at my free bar diagram. I know the force same size as the normal force. What, what, what, what, what, what? MG. So I can go like this. 60 minus mu MG equals, and I'm gonna go A and I'll divide by M right here because I wanna get the acceleration. I'm pretty sure anyways, I may as well do that on this last line. Do the M's cancel by the way? No, I need an M in everything for it to cancel. Sometimes happen, but not this time. I think the acceleration is gonna be 60 minus, I've scrolled down, what was the coefficient of friction? What was mu? Oh, you have to, you don't have a friend of you either? Fine, 0.35. Mass was 50, G is 9.8. Now remember, I've told you, we've actually guessed wrong divided by 50. Crunch the numbers, what do you get for the acceleration and there's something that's gonna tell you that you got your winner and loser mixed up. Bracket 60 minus 0.35 times 50 times 9.8 divided by 50. You know what tells you that I guessed wrong? Negative acceleration says, you know what? You let the positive be in the wrong direction. Now in this case, actually, it's still the right answer. It's still the right magnitude. If you have multiple friction sources, more than one mass, you might have to quickly redo the question, although it's a fairly easy redo. All I would do is this, look, here's my quick fix, look up. That's a loser, that's a winner. That's, you just put pluses and minuses, make everything that was a winner a loser, make everything that was a loser a winner, but double check to make sure you got friction moving in the right way. This time, I know it's still moving to the right, so I know friction is to the left. On more complicated diagrams, you might have had it moving to the right when actually, no, it's moving to the left. Anyways, I get, I'll write the negative. Negative 2.23, the negative tells me that it's slowing down. Now that you know this, could you solve for how far it travels if you know the acceleration and you know your initial velocity is 4.9 because that's what it was going over here and you do know your final velocity. What's your final velocity going to be? Use vf squared equals vi squared plus 2a to get the d by itself. Is that okay? So again, I like the physics there. I think that part of it's great. I just, the wording of the question, I would have a part a and a part b on a test if I gave that to you. Any others? So let's continue my children with lesson four. Multiple body problems, many body problems. What if you got more than one mass? The nice thing is this tug of war approach can be generalized. All we're gonna do is we're gonna tweak one small thing. So in many body problems, we have several objects connected together by, well strings or even something like a train where train car one, train car two is not a string, it's a connect, but several different masses that are connected. And the easy approach is to realize we can treat this as one great big object if all of the objects are accelerating at the same rate and they are. I'm gonna argue with you that if I have two objects attached by a string, there is no way that I can make these two masses accelerate at different rates. Unless I do this, let's pretend we're not gonna do this. We're gonna have nice smooth acceleration, which means if they're attached by a string, they may be accelerating in different directions, but the rate is going to be the same no matter what. Why does that help? Oh, huge help. Example one says, find the acceleration and the string tensions and we're gonna treat it as one single system. This is going to be our approach. What are the forces acting on this object? Get the obvious ones. Absolutely gravity down. Now, because I have more than one mass, I'm gonna call the first mass M1 and I'll call it M1g. And my notation is I just go from left to right. You don't have to use the same notation as me, but so all of ours look the same. I'm gonna suggest Brett start from left to go to right and first one is mass one, next one is mass two, next one is mass three, et cetera, et cetera, et cetera. Is this mass sinking into the ground? Then there must be another force in the opposite direction. Is it flying into the air? Then these two forces must be the same size. What force is this right here? And since there's more than one mass on a surface, I'm gonna call this normal force number one. There is one more force as well. I heard someone say it, two syllables. I thought I heard it. Tension. Is there a rope right there? Then tension is pulling it to the right. So far so good. What are the forces acting on this two kilogram mass? Look at the obvious ones. Oh, by the way, is there friction? No, but it's an easy, it'd be one extra force that way, new times the normal force, times normal force number one. What are the forces acting here? I heard someone say gravity. I'll call it M2G, because it's a second mass. Is this mass sinking into the ground? Is it flying into the air? Then there must be a balanced force in the opposite direction. I'll call that normal force number two. That's the vertical forces. What are the horizontal forces acting on it? Well, there's one obvious one to the right. What? The applied force, 20. And there is one more to the left. Yeah. You know why? Forces come in pairs. Here's Newton's third law. I can't pull this way without pulling this way at exactly the same amount. Who's winning? Vertically we're in balance, yes? Horizontally along the string, who's winning? The applied force. So my equation's gonna look like this. Anything to the right is winner, positive. Anything pointing to the left is loser, negative. So winner. And I'm gonna walk down the whole string. I run in tension. Which way is tension pointing to the left or to the right? Loser. I run into tension again. Which way is tension pointing to the left or to the right? Winner. Oh, that's all of the forces on the string. Equals. The only difference here is instead of MA, since we're treating this as one great big mass mathematically, it's gonna be the mass of both of them times A. Hey, what do you notice about the tension? Cancels, oh. So relaxing when tension vanishes. I love it when tension vanishes. I'm in my happy place right now. You know why? No tension. Okay, got the point across. Yeah, tension cancels and how would I get the acceleration? How would I get the A by itself? Here's what we end up with. A equals M1G divided by the mass of both, which is M1 plus M2. Let's crunch the numbers. M1, sorry, huh? Oh, what happened to the 20, Mr. Dewick? How about Mr. Dewick do this one? 20, I was doing an Atwood machine already, like an idiot, sorry. Thank you. Who pointed out the 20? Was it Nicole first? Was it a tie between Nicole and Brett? Who was first? Someone gets a candy. Who said it out loud first because they had enough integrity and guts to point it out? I think it was Nicole. You get a candy later. Katie, let's fill in the rest of this though. Katie, do I know mass one? What? Do I know mass two? What? Do I need a calculator? Say no, because it's 20 divided by five. I can do this in my head, but to two sig figs, this is going to accelerate at 4.0 meters per second squared. If they give me nothing and I want to find the acceleration, Jeanette treated all as one big mass. If they want specific forces, namely tension, now we're going to look at a single solitary mass. Consider the three kilogram mass. What were the forces acting on it? Get the obvious ones. Normal force and tension horizontally along the string. Who's winning here? Not meant that, what's moving it to the right? Yeah, by the way, you'll notice that 20 is not moving it to the right because that 20 is not touching this mass. The only that's touching this mass is this string right here, which is tension. Winner, tension. Any loser? That that means this just equals ma. Do I know the mass? How big? Do I know the acceleration? What did I just finish figuring it out? That's why I had to go to the system as a whole to find the acceleration first and then focus on an individual mass to find the tension. Turns out the tension is gonna be three times four. Katie, do I need a calculator for that? What's the tension in Newtons? 12 Newtons. Start out looking at the whole system to find the acceleration and then look at an individual mass to find the tension. Oh, here's a nice thing. Emily, it doesn't matter which mass you looked at. You could have also looked at the two kilogram mass. I probably would have picked the three kilogram because it's the easiest free body diagram. The two kilogram mass has more forces on it, but just for consistency, let's see if we get the same answer. Now on the two kilogram mass, we had mg normal force, 20 and tension. Who's winning in the two kilogram mass? Not tension. Who's winning? The 20. In fact, your equation here, that's gonna be 20 minus tension equals ma. Caitlyn, to get the tension by itself, I think I would plus this over this side and minus this over this side. Is that okay in one step? I think tension is gonna be 20 minus ma. It's gonna be 20 minus. Which mass am I looking at this time? The two, do I know? Oh yeah, I figured that out in the earlier question. What was the acceleration? And is 20 minus eight for tension the same answer as I got looking at mass number one? Also 12 newtons. So the nice thing about this is, Brad, it doesn't matter which one you pick. Pick the mass with the least junk on it. That's my strategy over there. Turn the page. What if we have more than two masses? If we have several masses connected by strings, no problem. We need to realize the first string accelerates the first mass, but the second string accelerates the both of the first two masses. What we need to realize is, the tensions will be different. Here we go. Says, find the acceleration and chord tensions. Again, we're gonna treat this like a single solitary mass. What are the forces acting on this guy? Get the obvious ones. M1g, normal force number one. What force is pulling it to the right? And since there's two strings, I'm gonna call it tension one. You know what I'm gonna call the other string? Tension eight. No, tension two. That would make way more sense. Some of you, what? I think Brianna, this would be M2g. Normal force number two. Tension one, slowing it down. Tension two, speeding it up, accelerating it. And you know what? I'm pretty sure this is gonna be mass three times g. Normal force number three. Tension two, slowing it down. 20 newtons pulling it to the right. Kayla, is that okay so far? So far so good. By the way, if I had friction, all I would have is an extra loser this way and extra loser this way, extra loser that way. The equation gets yuckier, but not really more complicated. Hey, who's winning? Who's winning? It's gotta be the 20. So my equation's gonna look like this. Winner, and now that I've decided that to the right is a winner, I'm gonna walk down this whole string. Anything to the left is loser, and anything to the right is winner. And I get this. Minus tension two, plus tension two. Minus tension one, plus tension one. Equals M1 plus M2 plus M3. The mass of all of them times eight. Why is Mr. Duick so relaxed right now? What cancels? I love it when I lose tension. So relaxing. Mr. Duick, are you gonna use that joke for the rest of the unit? Maybe. Mitch, how would I get the A by itself? More specific, divide by what? Yep. The acceleration is gonna be 20 divided by M1 plus M2 plus M3. I'm pretty sure it ends up being 20 over 10. We've picked some nice numbers here because Mitchell, I think even you don't need a calculator for this. What is 20 divided by 10? Did you really reach for your calculator, honestly and truly, or are you just joking? I hope you're, let's pretend you were joking, and even if you weren't joking, let's pretend you were joking. By the way, why is two wrong? That's one sig fig. How many sig figs? So if they work out evenly, Mitchell, don't forget to do that, right? This whole thing is gonna accelerate forwards at 2.0 meters per second squared. Now I can find the tensions. Doesn't matter which masses I look at, although I'll try and pick ones that are easier. I cannot start here though, because here I don't know tension one or tension two. I'm gonna find tension one from this guy, and I'll probably find tension two from this guy. So tension one is this guy here. Last time we redrew the diagram, Trevor, I still have it right here, so let's see if I can, without redrawing it, just use this. Who's winning on this mass if we're only looking at this mass? Who's winning? Who's losing on this mass? Nothing. If I had friction, it would be still be losing. No friction. And I'm just looking at one mass, so it's M1A. Do I know mass one? Yes, it's three. Do I know A? Just figured it out. It's two. Mitchell, without a calculator, what is three times two? Six is wrong. Why is six wrong? Thank you. Two sig figs. I have no idea what you said. Oh, the screen froze? Okay, this is my wireless. So I get 6.0 newtons for tension one. Okay, why did I use mass one? Because when I looked at my free body diagram, tension one, when I went winner minus loser was gonna be easy to go by itself. In mass two, it would have been tension winner minus loser since I didn't know both of those, but that wouldn't have helped me because I'd have two unknowns. In fact, to find tension two, I'm gonna go to mass three or I could go to mass two, but then I'd be using tension one to find tension two, if I made a mistake, it might make another mistake. So I'm gonna go over here. Is that okay so far? Core two. I'm actually going to do this one down here first. Turn the page if you haven't already. So I'm gonna do the consistency check first because that's what I would have done first. I would have said, since we've turned the page, it was M3G normal forest three 20 and tension two. I would have said, hey, looking at this mass, who's winning? Sorry, Brandon. Oh, sorry, I thought you said my bad. Who's winning, Mitch? Who's winning? Boys and girls, wake up now. I've turned the page. Who's winning? Mitchell, who's winning? 20. Who's losing? And since I'm only looking at one mass right now, it's gonna be M3A. How would I get tension two by itself? I think the easiest way, Mitch, would be plus that there, minus that there. I think tension two is gonna be 20 minus M3A, which is gonna be 20 minus five times. What was the acceleration? I think two from the previous page. Was it not? Someone needs to check. Two, in my head, no calculator required. Oh, by the way, why is 10 technically wrong? Okay, I'll either do that to two sig figs or 10.0. Sorry, do you have a question? You're good. Now, above here, let's use the two kilogram mass and see if we also get 10 for tension two. So the two kilogram mass, it would have been mass two G down, normal force two up, tension two to the right, tension one to the left. That's what we had from our free body diagram from the previous page. Jacob, here, who's winning? Can't hear you. Why? You're right, why? Which way is this whole thing accelerating? To the, so the force to the right must be the winner. So it would be tension two minus tension one equals M two, sorry, A. Is that okay, Connor? How do I get tension two by itself? So if I hear you correctly, tension two is gonna be M two A plus tension one. So this is our consistency check to see if we get the same answer. We should get 10. Let's see, mass two was two, A was also two, plus what was tension one from the previous page? How many Newtons? Next Newtons? I also get 10 Newtons. So now we have several different approaches. Caitlin, if they give me no information, but all the masses start treated as one and work your way down to individual forces. If they give you tension, start with one mass, work your way out overall to find acceleration or missing masses or whatever. Hanging mass, if we have a hanging mass and no external force, then we need to understand that it's the weight of the hanging mass that accelerates the system. Example three says, identify the net force and then find the acceleration and the core tensions. Okay, what are the forces acting on this hanging mass? Get the obvious ones. And I'm gonna call that M two G because it's the second mass as I read right to left. Is this in free fall? Oh, then what's slowing it down? Tension. Is there a normal force? Is this mass touching a surface when it's hanging there? So no normal force. What are the forces acting on this guy? Get the obvious ones. I'll call that M one G. What else? Normal force? What else? Friction? Nope, otherwise I'd have friction going that way. No problem. Trevor, who's winning? Which way do you think this whole system is moving? So who has to be winning then? M two G. I know it's definitely not moving to the left. There's no way you hang something from a cliff and magically suddenly it starts going back up the cliff. Right now without a motor or something like that and there's nothing like that here. So I would argue that's winner, which means Trevor, everything when I follow it along the rope everything ends up pointing down as winner. Everything ends up pointing up as loser. So loser. Oh, but what about this guy? When I follow it around the pulley it ends up pointing down winner and there's no more forces along the rope. That equals M one plus M two times A. Relaxing, getting rid of tension. They said they want me to find the net force. There it is. Now they want me to find the acceleration. How do I get the A by itself killer? Nice. The acceleration is gonna be M two G divided by M one plus M two. It's gonna be, what's M two? Oh, the big mass four times 9.8 divided by six. I can have those two together in my head. Oh, 19.6 divided by three. It's gonna be six points. I don't know, six point what? Good to see you finally reaching for, oh no we haven't needed a calculator so far. What happened? Four times 9.8 divided by six. Six point five, three, three, three, three, three, three, three. So six point five, three. Units are under. Darn right, don't over complicate it. There's a reason I told you in physics 11 to memorize units, okay? What's the third thing they wanted me to find? Tension, I think. Yeah, I'll do it right here. Well, we'll do it over here. Look at my diagram here. Which one is gonna be easier to find the tension with? Mass one or mass two? And you said mass two, I actually think mass one because look at your string horizontal forces. How many forces are acting on mass one? Who's winning? In fact, the equation is gonna be tension equals M1A or Brett, if we'd used mass two, it would be M2G minus tension equals M2A. Both of those will give you the same answer for tension. Here I went winner minus loser. Which equation's gonna be easier to solve though? The first one. Do I know mass one? Do I know A? Six point five, three repeating. What do you get for the tension? Sorry, I've been doing one. Oh, sorry, 13.1, is that right? Condition three sig figs? Newtons. By the way, over on this side, you would have got tension equals M2G minus M2A. Tension would have been four times 9.8 minus four times 6.5, 3, 2, 3, 3, 3, 3. Do you also get 13 point, oh yeah. Same answer both ways. Definitely one of the equations will almost always be cleaner. For problems that involve friction, we can still use this system approach but the tension calculations are a bit trickier. How would I find the acceleration of each system? Well, you know what I would do first? I would label the diagrams. What are the forces acting on this? Get the obvious ones. M1G, what else? Normal force, number one. What else? Tension? What else? There is one more now. Now they're mentioning friction. Which way is friction acting? Well, which way are we moving? So which way is friction acting? And I'll call that friction force, number one. I'm also gonna have M2G, normal force, number two. Applied force, tension and friction force. Force, number two. Does that make sense, right? There's two things slowing this thing down. The tension from the previous math, math and friction. Who's winning? Force F, the applied force. Winner. Which means now I'm committed. Katie, anything to the right is gonna be winner positive and anything to the left is gonna be loser negative. So winner, then I'm gonna have minus loser, minus loser. Does that make sense, Sam? Keep going along the string. Then I have plus winner, minus loser. And that's gonna equal M1 plus M2 times A. Oh, why is Mr. Dewick relaxed? Tension cancels. I'm gonna have F dropping down. Minus, Joel, can you read that out to me please? Friction is what times what, Brett? I don't know the normal force. Oh, but look, look, look, look, look, look, look, look. I don't know the force the same size as the normal force here, what? In fact, it's gonna be mu M2g. And then I have another minus. Joel, read this one to me. Friction is what times what? I don't know the normal force. I don't know, no, it's not, never, no, never, ever. I've never, ever, ever said mu times Mg, ever. Friction is what times what? Mu times the normal force. You okay? Keep looking out there, it's not that important. This is important, you hear? Friction is what times what? I don't know the normal force. Oh, but look, look, look, look, in this case, I don't know the force the same size as the normal force. What? That's not a force, that's a mass. No, that's not the, okay. M1g, not mass, right? I wanted the force, okay? Oh, look it up, ladies. You better. And if I wanted to get the A by itself, I would divide by mass one plus M2. There's my diagram for this one. Two masses on a rough surface. What about here? What are the forces acting? Let's do M2 first. What are the forces acting on M2? Get the obvious ones. M2g, what else? Tension, what are the forces? Is there a normal force? It's not touching a surface. What are the forces acting on mass one? Get the obvious ones. Gravity, normal force. What's pulling it to the right? Anything slowing it down? In fact, I'm gonna argue these two are identical. The only difference is, this is not experiencing any friction. It's gonna be exactly the same equation, but no friction too. Watch, I'll show you. Now, winner this time is M2g. That's minus free force. Minus tension plus tension. Minus friction equals M1 plus M2 times A. Once again, my tensions cancel. Zay, how would I get the A by itself? How would I get this here A by itself? Yeah, I won't bother doing that because I'm not sure what they want me to find. There is my setup. Okay? That's how I would find the acceleration of each of these. So let's try one with friction here. Example five. Find the overall acceleration of the system and then find the tension on each core. Are we doing two more? Good. M1g, normal force number one. Emily, what are the other forces acting on this one? I absolutely agree with you. Tension, what else? Friction, which way? Thank you. How do I know this friction, by the way, because mu is not equal to what? Zero, zero, no friction. Emily, what are the forces on the three kilogram? Get the obvious one. M2g, what else? Number two, what else? 20 pulling to the right. What else? And one more. It's also on the rough surface. Friction force, number two. Who's winning? Who's winning? Which is what? How big? So, winner. Now I'm committed, our vendor. Anything to the right? Winner, plus. Anything to the left? Loser, negative. Walk down the string and I run into minus tension, minus friction, two, because they're losers. Keep walking down the string. I run into plus tension, minus friction, one. That equals M1 plus M2 times A. So far, so good? Yay. Jewel of friction is what times what? Okay, I don't know the normal force in here, but look, look, look, look, look, look. I don't know the force the same size as the normal force. What? And M2g. So in this case, our final result is gonna be friction equals mu Mg, but that's never where we start. We're gonna get 20 minus mu M2g minus mu M1g. And if I divide all of that by M1 plus M2, that's my acceleration. What does A equal? 20 minus 0.15 times three, Mr. Dewick, times 9.8 minus 0.15 times five times 9.8 all over eight. What's my acceleration? 14.7, anybody else? 14.7 seems really high to me, because that's almost two Gs. I heard a 1.03, anybody else? 1.03, 1.03, yeah. Okay, part B says find the tension. How can I find the tension? Look at one mass. Which one? Doesn't matter, but I picked the one that has the least number of horizontal forces on it. I think I would pick the first mass, the five kilo. What are the forces acting on this guy? Who's winning? Brett, I think I heard it. No? Who's winning? Who's winning? Tension, who's losing? So to find the tension, my equation's gonna be tension minus friction one equals M1a. Tension equals M1a. Plus friction. Dual friction is what times what? I don't know the normal force. Oh, but look, look, look, look, look. I don't know the force, same side, so the normal force, what? This is gonna be M1a plus M1g. What's the tension in the chord? Five times 1.03 plus five times 9.8. What's the tension in the chord? 12.5. To the page! Example six, I like, I'd love to do it, but you guys are zoning out and I wanna give you a chance to work on the homework. What's your homework? Number one, number three. So I skipped two, four, five, six, nine B, 10 B, 11 B, I'm sorry, 12 B, 13, and I'm done there. Okay. You got about 10 minutes left to work on the homework. Sorry, I went a bit longer because the projector thingy conked out.