 So, number nine says find the acceleration in the chord tension. Free body diagram, I would say this is going to have M2G and tension. Yes? And this is going to have M1G, normal force number one, tension pulling to the right. And it does mention a coefficient of friction right here. So, I'll assume that there's a friction force pulling to the left. Whose wing? Well, if this thing is accelerating, can it possibly be accelerating to the left? Do masses without an external force suddenly pull things up cliffs? I don't think so. If anything is winning, it's moving to the right and down. I think that's my winner. Which means when I walk along the rope, anything that ends up pointing down, winner or positive, anything that ends up pointing up, loser, negative. Speaking of, so this tension ends up being loser, winner. Oh, this friction when I follow it around ends up being loser. And that's going to equal M1 plus M2 times A. Despite the fact that two students just walked in late, I lose some tension. It's amazing. It's going to be M2G. Friction is what times what? Mu times a normal force. I don't know the normal force. Oh, but look, look, look, look, look, look, look. I don't know the force same size as the normal force. So it's going to be minus mu M1G. And since they want me to find the acceleration, Lea, do you mind if I just divide by M1 plus M2 right now? Okay. Did you get that far or no? Did you ask me this question? Yeah, you had this far or no? Did you get, did you get this far or no? Oh, okay. Do you want me to keep, did you get, when you were doing the homework, did you get to this point? Or, okay. So that'll get you A. I'll let you plug in the numbers and crunch it yourself. Okay. Then to find the tension, I could either go looking at one mass, winner minus loser, or to be honest, I'd probably go winner minus loser. You know what? Actually, I'd go winner minus loser. If tension's the winner, it's an easier equation. So I'd look at mass one, and I'd go tension minus friction equals, now I'm just looking at one mass, M1A, where that A is that A down there. Is that okay? I'll let you try the rest of it on your own or you can snag it later. 13, Brett asked. I'm going to guess you got the 3.6. I'm going to guess what's bugging you is the 0.36. Yeah? Because to get the 0.36, treat this as a 14. Treat it as one mass and that'll give you the friction. So that'll give you the acceleration. Treat that as 8. Treat that as 14. You're good to go. Okay. Let's look at the 2 kilogram mass only. Right there. What are the forces acting on it? Get the obvious ones. Yeah? Now, which way is it accelerating? Which way is... So it stopped. If the entire thing is accelerating to the right, which way is it accelerating? To the right. There has to be a force accelerating it to the right. Which force? Tension. Is the rope touching it? No. You know what? Let's do a thought experiment. Suppose here and here, suppose this and this were actually made of ice so that there was no friction between these two. When you pulled on the bottom, what would happen to this mass here? I think it would stay in place. Yeah? So you know which force is moving it to the right? I think friction. I think it's this. I think that's my free body diagram. Positive. Is that okay so far? Who's winning? Think about it. Who's the only possible winner? Friction. Loser? None. My equation is going to be for mass 2, friction equals mA. Which A? The A that you found, the 3.6. Friction is what times what? No, friction is what times what? Mu times a normal force. I don't know the normal force. Oh, but look, look, look, look, look. I know another force the same size as the normal force. Mu mg is going to be mA. Oh, conveniently in this case as a matter of fact, this time the mass is going to be mA, which is kind of nice. Happens sometimes. And I think friction, the coefficient of friction, sorry, ends up being the overall acceleration divided by g. Because if there was no friction, if there was wheels right here, it would just stay in one spot. This would yank out from underneath it and it would drop straight down. It must be friction that's moving it to the right. I'd consider that fair game as a nasty multiple choice. I'd give you one as a written like this. Certainly this is fair game. That, well, if I made this as a written and it was made out of 7 marks and I made this 5 and I made that 2, so if you didn't get it and kill it, maybe I'd live with it. Anyways, just trying to give you an idea of coming attractions. Okay, pause for a second. So I've given you lesson 5 and I've given you the take home quiz. Do next class. We'll mark it right away at the beginning of class. Lesson 5, the Atwood machine. And again, in theory this should be reviewed, but we're going to work a few other things into here as well. It says this, the Atwood machine is a pulley system as seen below. The Mass M2 makes it easier for a person or a motor to lift Mass M1. In fact, Mass M2 is called the counterweight. Every elevator has something like this. Any elevator that you go into, somewhere off in a side column, side shaft, there is a large counterweight and that weight is the weight of the elevator empty so that all the electric motor has to move is the person, the people. Example 1, equal masses are placed on both sides of the Atwood machine and then one mass is lifted up higher. When I let go of this higher mass, how does the system move? A, it remains at rest. B, the higher mass falls. C, the higher mass goes higher. Excuse me, we'll sneeze there. Using principles of physics, explain your answer. Once again, we're going to vote once again how high you hold your hand up is how sure you are of the answer. So I'd like you to imagine, this is my Atwood machine right here, okay? Except this has a lot of friction. That that pulley at the top is frictionless and massless in our magic physics world. If I raise this mass higher and then I were to let it go when there was no friction, what would happen? Would the mass fall down? Would the mass go up? Or would the mass stay where it is? Those are our options. Ready? Who says A, the system remains at rest? 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. What are the odds of him being late? 10. Who says B, the higher mass falls down? 1, 2, 3, 4, 5, 6. Who says C, the higher mass goes up? 1, 2, 3. Once again, defend your answer or attack someone else's. Defender attack. What do you got? Convince me. That you're right or that someone else has to be wrong. How would I go about presenting a convincing argument here? You know what the first thing that I would do would be? I'd do a free body diagram. I would label the forces on these masses. What's the force on this one? Get the obvious one. Is it in free fall? No, something must be holding it up. Oh, what? Tension. What's the force on this one? Get the obvious ones. Since they're the same mass, I don't need to call it M1 and M2. It says the masses are identical, so MGMG. Katie, is this one in free fall? Tension. Who's winning? Those of you that picked B, which was the biggest answer, you said that this one was winning. Let's set up an equation, Brett. It would look like this. Winner. Tension is a loser. If I follow this tension all the way around, what happens? That tension is up. Winner. If I follow this MG all the way around, I know it's pointing down over here, but if I follow it over this side, which way will it end up pointing? Loser. That equals M1 plus M2 times A. Well, let's carry on with the physics. Zero, as it turns, everything cancels on the left side. Yejul equals, how would I get the A by itself? So here's my question. What's the acceleration going to be an Atwood machine with equal masses? It's going to be zero. So what's the correct answer then if it's not going to accelerate? Therefore, A equals zero. By the way, that's how I would answer it. I might write down, I wrote it next to here so that you could see. And if you did write it next to the diagram, I'd give you full marks on a test. Sometimes it's easier too. But for a force question, 90% of the time I'm falling back on a free-body diagram. And I'm just going to see if I can write a force equation and spot what's going on. Here, when everything canceled, I went, oh, there's no net force. There's no unbalanced force. Now we're Newton's first law. If you have no unbalanced force, you're either moving at a constant speed or you're standing still. In this case, since we were standing still at the beginning, I think we're still standing still. You can see, still standing, still standing. Is that okay? So those of you that picked A, well done. Those of you that picked something else, not so well done. Speaking of not so well done, we're back, sorry, somebody was using their phone so I had to go yell at them and love. Does that make sense? Okay. The key is going to be when we have unequal masses. Hey, let's be obvious. Then the heavier one is going to fall down. The lighter one is going to move up. And this in particular is where not letting down be negative all the time and not letting up be positive all the time, the winter minuses approach much better here. You can solve these using systems of equations like you learned in Math 11 where you had two equations with two unknowns and actually that's how I was taught. That's how I taught at my first couple of years this tug-of-war approach from another teacher in district and I went, ooh, much better. So example two, a classic Atwood machine. I like this question, I like this question, I like this question. It says, assume mass two equals 5.8 and mass one equals 2.3 kilograms. Solve for the acceleration and then find the tension just like the last day when we were looking at many body problems, our approach is going to be the same. Treat everything as one mass to find the acceleration and then if they want to find specific smaller forces I'll look at individual masses. Which is the heavier one? What are the forces acting on mass two? Get the obvious ones. And I'll call that M2G since the masses are different. What about here? Is there a normal force? Are we touching a surface? So no, no. What are the forces acting on mass one? Also M1G. Intention. Who's winning? Read the question, which is the heavier mass? Anything that ends up pointing downwards on the right side is winner. Which way is this pointing on the right side, Trevor? Loser, I mean the force. Are you a Bruins fan? Okay, maybe, you know, I still mean the force. What about this tension? If you follow it all the way around to this, I know it looks like it's pointing up on this side, but if you follow it all the way around, which way does it end up pointing? No, winner. Sorry, it's hard for me to say that word with you. Yeah, let's make this a little better. By the way, you may notice when I do a lowercase t, I always put a curl on it because at least once I've done something where a t became a plus sign later on, when I looked at it later on and forgot what I was doing and spent hours staring at it and finally figured out, no, it's not a plus sign, it's a letter. So a little trick of the trade for you. Oh, and if you ever write a lowercase l, don't you do your lowercase l like that? Because what does that look like? A one, how do you know? I always do my lowercase ls as written ls. Oh, what about gravity? Now, gravity looks like a winner because it's pointing down, but if I follow it all the way around to this side, Kayla, which way does it end up pointing actually? It's a loser. And we're still looking at all the masses at once because they're attached by a string. We know they have to be accelerating at the same rate. Even though three students walked in late today, I can lose some tension. But not much. Conor, how would I get the A by itself? I think I would. I think the acceleration is going to be M2G minus M1G all divided by M1 plus M2. Now it's plug and chug. Mass 2, 5.8. Mass 1, 9.8. Minus mass 1, 2.3, 9.8. All divided by 5.8 plus 2.3. Get your calculators out. What's the acceleration? What'd you get? See that little motivational tool there? Okay, I did a little doldrum. Yeah, I know, but you need to practice this. I think I've said something like, if you're not careful, you won't know how to do these on your calculator. I think I also said something in Math 12, like I'll laugh at people that don't know how to put their calculator into its case properly, not mentioning any names. See if it's like, isn't that nice how it fits like that? No, better. That's interesting. All I know is we're laughing at somebody. My Math 12 students are going, holy smokes, he actually knew that was going to happen during the year sometime. Yeah. And he spotted that from a distance. What'd you get, M? Okay, by the way, there's a reason I haven't bothered doing it up here because first time many of you will do it wrong, which is why I'm trying to push some of you to do this on your calculator. Don't clear it or you did, that's okay. What'd you get? Anybody else, 4.23? Okay, we got 4. Now, how many numbers on the top? By the way, did I say something about not using a pen or a pencil to you? Is that mine? Or is that your calculator? Okay, next time I'll lovingly break the pen. I won't, but it damages the buttons. How many numbers on top? Meant to be easy. Yeah, how many numbers more than one on top? Brackets, so bracket. And on my graphing calculator, even though I write brackets between numbers, that would be going bracket, bracket, too much typing. I just go 5.8 times 9.8 minus 2.3 times 9.8. That's the top. Divided by, is there more than one thing on the bottom? Bracket, 5.8 plus 2.3. That's how you type it in. And you get 4.23, 4.23. Although that's kind of a neat decimal. 2, 3, 4, 5, 6, 7, 9. 10, 11. No, I'm the only nerd that noticed that. Okay, 4.23. Units, yeah. Whoever said Newton's, come on, ask what you're solving. Now that I know the acceleration, I can find the tension. Doesn't matter which mass we use, although if I was clever Jacob, I would use the one where tension is a winner. Look at the two masses. In which one is tension the winner? Well, which way is this guy accelerating, up or down? Meant to be really easy, folks. Which way is this mass accelerating, up or down? Which way is tension pulling? Oh, so tension is losing this battle. Which way is this guy accelerating, up or down? Up, which way is tension pulling? Ah, tension's the winner on this side. I'll use this mass. Winner, tension's the winner. Minus loser, M1G is the loser. And I'm just looking at a single solitary mass now. So it's going to be M1A. By the way, do you see why I said it's easier to let tension be the winner because to get the T by itself, Brett, all I need to do is plus that over. If I'd done the other equation on this side, still works, I'll get the same answer. I've got to do a little bit more math. Tension is going to be M1A plus M1G. What was M1? 2.3. 2.3 times 4.23 plus 2.3 times 9.8. 2.3 times, and I still have that answer on my calculator, so I'll use that, plus 2.3 times 9.8. Do you get 32.3? And that's a tension, Newton's. Now, how can I tweak this? How can I tweak this? Supposing I gave you tension and only mass 1 but not mass 2, could you use tension and mass 1 to find the acceleration? Yeah? Once you knew the acceleration, could you write a tension equation over here to solve for mass 2? It would be M2G minus tension equals M2A. Yeah, so I can mix and match. I can. Example 3, turn the page. Example 3 says, consider the following system. Which of the following is the best force diagram for mass M2? Okay. Is there friction between these masses in the ground, and how do you know? They put a little mu equals 0 there. Try doing your own little free body diagram on mass 2 over here. I'm not going to do one. Then look at A, B, or C. And once again, we're going to vote. What are the forces acting on mass 2, do you think? Hmm. Hmm. All right. Who says A? Who says B? Who says C? Okay. Once again, attack or defend. Why can't C possibly be correct? Okay. And by the way, your argument is okay but you're missing a step. Which way are both of these masses going to be accelerating? Can mass C possibly accelerate to the right, or are the horizontal forces in balance in that diagram? See, it looks like to me in mass, in diagram C, I think those two forces cancel each other out. I'm not so sure mass C is accelerating to the right. I'm leaning towards A or B. Let's see. I'm going to label my forces. What are the forces acting here? Get the obvious ones. I have M1G normal force number one. And I have a force pushing it this way. You know what? Let's redraw these because this pushing to, that doesn't look like, I'd rather have the F right here. So here's the force M1G normal force number one. M2G normal force number two. Those ones I think most of you were okay with. The real question is what's happening here? Well, remember at the beginning of class, I pushed a book against a wall with a meter stick? See, I think this is very similar. This is two surfaces in contact. They are each exerting another normal force against each other. I think that there is a normal force, and I'm going to call this normal force number three. That's this guy pushing back against this guy. And because this is where I agree with part of your argument, forces come in pairs, I think normal force number three is also pushing in this direction. The two surfaces are pushing against each other. Does that make sense? Okay, Brett. Did we explain that? I see that. I think A is the best free body diagram for mass two. And the problem is a lot of us, first of all, a lot of people want to pick B because they somehow want this F to go through this mass and press. No, F is pressing against mass one. F is what's creating the normal force over here, but it's not reaching it through the mass and somehow pressing against mass two on its own. That's a completely different force. Explain your answer. I would do a free body diagram. I think that's a convincing answer. Oh, and you'll notice I tried to draw normal force three a little bit shorter than big force because I'm pretty sure we still have a winner to the right because what I didn't really like about this diagram, I should exaggerate this because I can erase really easy. I'll make sure there's our winner to the right. There's our net four. There's why we're accelerating to the right. How would I solve this question? Are they touching? Say yes. Can they possibly accelerate at different rates if they're touching? I would treat them as one mass just like we did with our many body problems. Example four. Find the acceleration of the system and find the contact force. Oh, that's a great name. I called it a normal force. Number three in the previous question. How about instead we'll call it contact force? That sounds like a good name because the two surfaces are in contact. That works for me. So, what are the forces acting? You know what? Let's do a free body diagram for the six kilogram and a free body diagram for the 8.25 kilogram. What are the forces acting on this six kilogram? Get the obvious ones. By the way, we've left physics 11 behind. This is physics 12. M1g. Mitchell is the six kilogram mass sinking into the ground. Is it flying into the air? Then there must be an equal force in the opposite direction. I'll call that normal force number one. Mitchell, is there friction here? How do you know? Mu zero, so no friction. I could add it in if they needed me to. Oh, I have a 42.5 Newton force pushing to the right. What else? There is one force pushing to the left on mass one. I heard someone say it. Yeah, that contact force. We called it a normal force in the previous question, but the two surfaces are in contact. They're pushing against each other. So, there must be a contact force. I think that's everything on the first mass. What are the forces acting on the second mass? Get the obvious ones, Connor. Yeah, I got you. What are the forces acting here? Get the obvious ones. M2g. I agree, normal force. Oh, I guess I have a 6.75 going this way. Are these two surfaces pushing against each other right there? Say yes. Then that contact force must be the same as this contact force, because that's the direction because forces come up here. Now, step back from this diagram. Look at the forces. Who is most likely winning? Who's winning? I'm pretty sure the 42.5. So, I'm going to walk down all of the... Now, I'm not walking along a string. I'm walking along the horizontal forces. I'm going to walk down all of the horizontal forces and I'm going to let to the right be winner and I'm going to let to the left be loser. So, if I start here, that's a winner, that's a loser. That's a winner, that's a loser. And that equals the mass of both of them times A, because I'm treating them both as a single mass. Oh, just like with the tensions, what happens to the contact forces that I don't know anyways? Oh, that's a good thing. And, Trevor, how would I get the A by itself? So, I'm pretty sure the acceleration is going to be negative 6.75 plus 42.5 divided by 6 plus 8.25. Worth practicing typing this on your calculator, see if you can get the acceleration. Preferably not using a pencil to hit the buttons. Not mentioning any names. I can't believe he also knew that someone would do that. Yeah, I know my audience. What do you get? Brett, what do you get? You're squinting? Sorry? 2.51, anybody else? 2.51? Looks about right. If I got something like 200 meters per second squared, I'd be a little bit nervous. If I got a negative, I'd be nervous, because that would have meant I'd guess the wrong winner. Although, I'm pretty sure we're right. 2.51 meters per second squared. Okay. Ah, then it says, so we found the acceleration, we did part A. The second part says, find what? The contact force. How am I going to find that, Nicole? Instead of looking at both masses, you know what I'm going to do? Look at mass, a nice recovery. I'm not sure if you were here or not. You're here? Okay. So, which mass do you want to look at? Doesn't matter. Although, try and pick the mass where the contact force is the winner, because that'll probably be easier to get the contact force by itself. Nicole, and which one is the contact force in the winning direction? Or is it the winner? Really? Oh, yes, it is. Good. Excellent. So, my equation is going to look like this. Winner minus loser equals 8.25 times A. Which A? The one that we just found, right? Contact force is going to be 8.25 times 2.51 plus 6.25. What's the contact force between these two surfaces? And folks, I would now consider something like this. Fair game. Think about it. Did we really do anything new there? I showed you a new diagram. We thought about it. We fell back on a good free body diagram and Newton's Laws to figure out any missing forces. What do you get? What's the contact force between these two guys? 27.4. Anybody else? Yep. 27.4, what? Newton's is the force of it. Example 5. A curling rock is traveling east across a sheet of frictionless ice. So this ice is perfect. It bumps another rock and it experiences a force that acts briefly to the north. They collide. What will be the path of rock number one after the bump? So if we're looking down at it from the top and these two guys collide so it experiences a very brief force up, will it follow path A, path B, path C, or my arrow seems to have vanished here? Path D. Think about it. I'm assuming most of you have watched 30 seconds of curling at one point in your life even when flipping through the channels or during the Olympics. Who says A? Who says B? Who says C? Who says D? So convince me that D is wrong or that C is wrong. Okay? You know what I would do? I would say to myself, after the collision, after the collision, what are the forces acting on this curling rock and normal force? Those are vertical. And since we're looking straight down at it, those are coming out of the page towards us or away from us. What about any horizontal forces? Are there any horizontal forces after the collision? Emily, you're right. No. You know what that means? Newton's first law kicks in. Objects want to continue going at a steady path in a straight line. The only way C could be right is if there was still some kind of an outside force acting on it. But after the collision, there is no longer an outside force acting on it. Example six. A water bottle is drilled with many small holes so that it leaks when it's filled. If I fill this water bottle with water and I drop it while it's in free fall, what happens? A, the water stops leaking out. B, the water leaks out at a slower rate. C, the water leaks out at the same rate. Or D, the water leaks out faster. This time we're not going to vote but what I would like you to do, all of you, is write down what you think the correct answer is on your paper. A, B, C, or D. And then I think we'll test this hypothesis by actually performing the experiment. Write down what you think the correct answer is. Correct answer? A. Why? Oh, come on. What are the forces acting on the bottle when it's in free fall? What are the forces acting on the bottle? I guess a little bit of air resistance but we've ignored that. When it's in free fall, what are the forces acting on the water? Matt, gravity, you're saying they both have the same force acting on them? So what could you say about their acceleration if they both have exactly the same force acting on them? So is it possible for the water to somehow get ahead of the bottle to leak out? No. I would love to do this in real free fall. I wish I had, but you can't really take big objects and drop them from 10,000 feet. They're not too happy with that. Although I think misplusters looked at something like this recently. I got to try and track it down. I've just jogged my memory now. They were doing something in free fall. I can't remember what it was. I'll have to go check it out. So using principles of physics, explain your answer. Same force on both bottle. I think the question says bucket. I used a bottle because it was less messy. Although a bucket actually might have been more of that maybe next time. On both bottle and water. So same acceleration which means the water can't get ahead or behind from the bucket. Example seven. I think this is the last one. Yes it is. This is an old scholarship question. And I'm actually not going to do this. This is going to be part of your homework. And I'll probably talk about it next class. It says this. Mass two is 12.5 kilograms. Mass one is 3.6 kilograms. And both masses are initially at rest 3.2 meters above the ground like the diagram there. Assume the cord and the pulley are both massless and frictionless. Its mass two is released to fall to the ground. What's the maximum height that mass one will reach? Now you might think that the answer to this is well if they're both 3.2 meters and sorry let's do it right here so you guys can see if they're both 3.2 meters above the ground won't mass one hit 6.4? Actually what's going to happen if you do the arithmetic in this it's going to keep going up on its own momentum for a little bit. So what you want to do is your hint for your homework. Find the acceleration first of all. Find the final of mass number one when it's traveled 3.2 meters. Figure out how much further it will travel before it hits a velocity of zero. At least if you can handle that. This would be harder than you'll get asked on the test. Normally I go through this one and I can tell you guys so what's your homework? Example 7 otherwise we already answered number 1 we threw it upwards. Try number 2 3 6 8 and you got to take home quiz. Pause right there.