 Uh, quiz. Here we go. Mark your own. And, Cirquey says the book is sitting on a level desk. Which of the following statements are true? There are no forces acting on the book because it's not accelerating. True or false? What forces are acting on it? Give me the obvious one. And, now if I had said the net force is zero because it's not accelerating, that would be true. The net force is the combined combination of all the forces. False. The magnitude of the normal force between the book and the desk equals the magnitude of the gravitational force on the book. I think we can assume it's true. The net force on the book is equal to zero. I think that's true. Friction force is what's holding the book on the desk. False. Could it sit on the desk if the desk was made out of ice? Long as the desk was level, just fine. False. Half mark for each of those to get your two marks. Example two says find the acceleration of the entire system. Okay. What are the forces acting? I have M1g. Normal force one. Tension. M2g. Normal force two. Tension. Applied force. And it says mu equals zero, at least in this case, so no friction. Who's winning? 24. Minus tension plus tension equals M1 plus M2 times A. Oh, tensions cancel. Relaxing. How do I get the A by itself? Divide by M1 plus M2. Okay. A is going to be 24 divided by M1 plus M2, which is going to be 24 divided by 20, which is going to be divided by 4.1.2 meters per second squared. If you got the right answer, two marks. Otherwise, I would probably give one mark if I saw that, a half mark for that, and then a half mark for the answer. Find the tension between the two masses. It doesn't matter which mass I use. I'm going to look at one mass, though. Matt, I could go winner minus loser equals 12A, or I could go winner. Oh, is there a loser on this mass? I'm going to use this one. Winner equals M1A. Or you could go winner minus loser equals M2A. But regardless, here the tension is already by itself. Mass 1 is 8. Acceleration is 1.2. 8 times 1 is 8. 8 times 0.2 is 1.9.6. 9.6. And you'll get the same answer over here with a bit more work. What I do both on a test to check my answer is probably later on. I'd go through the whole test first, though, and then I'd say, hey, can I find some of these forces in different ways? Check my answers. All right. It says, find the acceleration of the entire system in question two if now there is a coefficient of friction, and the pulling force is increased from 24 to 50 N. So I'll add this for the next one. I would have friction force 1 and friction force 2. So now my equation's going to look like this. Winner minus tension minus friction force 2 plus tension minus friction force 1, that equals M1 plus M2 times A. Ah, tensions to cancel. 50 minus friction is what times what? I don't know the normal force. Oh, but look, look, look, look. I know another force the same size as the normal force. Force what? By the way, today is the day that that changes. Today is the day I'm going to say to you, oh, but look, look, look, look. I know another force the same size as the normal force, and the answer will not be Mg, which is why I've never, ever taught you friction equals mu Mg. In fact, I've gone ballistic anytime anybody has ever said that. It's that way as long as you're on level ground. Today is the day we go on slants. Today is really where physics 12 starts in forces. This is going to be mu M2g minus mu M1g, and if I divide by M1 plus M2, that will give me A. A is going to be 50 minus 0.12 times 12 times 9.8 minus 0.12 times 8 times 9.8, all divided by 20. What do you get for your acceleration? 1.32? Anyone else? In terms of part marks, I would probably go one mark if I saw that, one mark if I saw that, half mark for the numbers, and a half mark for the answer. You'll notice you can get 2 out of 3 and get the answer completely wrong. I really want to reward you for knowing how to do it. Find the tension between the two masses. Okay. I'll use this one because tension is a winner. My equation is going to be tension minus friction 1 equals M1A. You could also use 50 minus friction 2 minus tension equals M2A, which will get you there just fine with a bit more work. Here, tension is going to be M1A plus friction 1, which is going to be M1A plus mu M1g. What did you get for the tension? 20 newtons? 20, and I won't be fussy on the quiz, but if you wrote that on a test, I would have to take a half mark off. Why? How many sig figs is that? Yes, I'd look for either this stupid thing or 20.0 times 10 to the 1 right at the scientific notation, then you'll always be fine. Elevator questions. Okay. What does a scale read? I actually told you in the question. I won't on a test. I'll expect you to know that a scale does not measure mass. What did we say a scale measures? Normal force. Person and the scale are inside a elevator. Find the scale reading in each case. Here's my free body diagram. Mg, why did I draw the normal force larger than Mg and A? Because the elevator is accelerating which way? Who's winning? My equation is going to be winner minus loser equals M a. The normal force, which is the scale reading in newtons, is going to be M a plus Mg, which is going to be 60 times 1.2 plus 60 times 9.8. What did you get for 4a? 660. What if the elevator accelerates downwards? Well, my free body diagram would have Mg normal force would be lower because Mg is winning. My equation is going to be Mg minus normal force equals M a. The normal force is going to be plus that to this side, our vendor, minus that to this side. Mg minus M a. 60 times 9.8 minus 60 times 2.25. What did you get for the normal force here, Sean? 723? I think it's got to be less. He's going down. He's got to weigh less than going up. By the way, you probably put a plus sign there. Just be thinking. Pardon me? Why? We said we take care of the negatives in our winner minus loser. We'll never worry about direction. We take care of it here and here. I'm going to give you 1.5, by the way. Sorry, what's the correct answer? 453? Newtons. C. Elevator moves up at a constant velocity. That really means winner minus loser equals, why can I put a zero there if we're at a constant velocity? My acceleration is zero, so when I go M a, it's going to be M times zero. In fact, what it really means is that the normal force equals Mg. This is what Einstein said. Einstein said, you can't tell if your eyes are closed, whether you're moving at a constant speed or standing still. Everything is relative. In fact, for all we know, all of us, well, we are all moving right now. Is the earth standing still? No. Can we tell? No. You don't know, is it? Even though we know intellectually it's spinning around, well, it's spinning around its axis and it's spinning around the sun, everything around us is moving at the same rate as us, so we can't tell that we're moving. What did you get? Mg is what? 588? What's the answer for D? Prove it. I would have said this. I would have said, okay, gravity. I'm not sure that the normal force is zero. I'll put a small one there. I know it's losing, and when I go winner minus loser equals M a, when I get the normal force by itself, I would go 60 times 9.8 minus 60 times 9.8. Was the mass 60 or 65? I can't even remember. 60. Oh, that's definitely zero. So there's the proof. By the way, zero, if you don't go to two sig figs, I'm not fussy, but I will be fussy if you didn't put zero newtons, I would take a half mark off. If you said 0.0, yeah, good, you know, two sig figs. This is what we call apparent weightlessness. True weightlessness is if you're floating in space away from any planets and there's no gravity. You're truly weightless. Apparent weightlessness is you have no normal force. Give yourself a score, please, out of 19. Hey, can you get the homework from last day? We looked at the Atwood machine. Probably did. Here you go. The Atwood machine. I said try this one for homework, and then I told you that I was going to do it last night. I totally forgot. Who thinks they have an answer, and if we all have the same answer, then we're probably right. I don't know. Zay, what'd you get? Oh, okay. John, would you get 10.2? Anyone else? 8.2? 9.6? Okay, 8.54. Three. Okay, so here's what I'm going to do right now. Right click, new sticky, complete physics scholarship question, and email out the solution. Sleep for, oh, let's sleep for about six hours. I'll do it tonight and I'll send out the solution. I don't have time to do it and get through today's lesson, unfortunately. I wish I did, or I'll do it at lunch today if I think of it. From the rest of the homework, were there any questions that you were going, oh, I can't get these answers? Number two? Here? Okay. Why is number two strange? Okay, so there's several ways to do this. One would be to say, well, look, if the mass is 3M and M, why don't I let M be, and I'm just making this up, two. Then the bigger mass would be six, and if you crunch the numbers, it'll work and give you the same answer, or you could let little M be five and big M be 15, and believe it or not, you'll get the same, you'll get the same acceleration no matter what. Yeah? 6.74? From the previous question, you're still on that? Okay, we've moved on, but thank you for following along. I'll solve it later. Whoever has answers, keep track of them, and then I'll do it next class, and we'll talk about it. So that's one way of doing it. Kate, let it give you an algebraic one. Don't be scared to make up reasonable numbers. The only thing I don't ever want to use is a one, because I'm scared if I plug in a one, something might cancel when it wasn't supposed to. So I'll use two, or five, or ten are my standard fallbacks, because those are nice easy numbers to do math with. Okay? You could do it algebraically, by the way, by doing exactly what you did, factoring out an M and gathering like terms, and you'll find M's cancer, and it actually works, but that's the high tech way. The low tech way, oh, okay, let M be 17.3, which will be dumb, but you could, and then big M would be, 3M would be whatever 17.3 times 3 is, and the numbers would work, and you'll still get the 4.9. Did I answer your question? Okay. I'll let you try the rest of it. Any others? Okay, class number eight. Okay, I'll call this first one mass one, I'll call this one mass two. I'll start with mass two first, because it's easier. What are the forces acting on the 3.2? Get the obvious ones. Absolutely. Is it touching the ground? Okay, is it in free fall? Tension. What are the forces acting on mass one? Get the obvious ones. Absolutely M1g. Is it in free fall? Then there's tension, and is it on a surface? So I'm going to put tension, and there's also a normal force. Must be. And in this case, by the way, normal force is not Mg. I'm going to have to figure it out. You know why I know it's not Mg? Because there's two upwards forces and only one downwards force, and I'm assuming tension isn't zero. You know how I know tension isn't zero? Because this thing is hanging, and the tensions I know have to be the same. Who's winning? Now it's a tie, but let's let this be the winner for now. So I'm going to go winner minus loser. Now when I follow this tension around, though, it becomes a winner. And when I follow normal force around, it also ends up being a winner. When it gets the other side, it's pointing down. And I have minus M1g, and that equals, why zero? It's an equilibrium, if you want the fancy physics. Yeah, it's at rest. Oh, what happens to the tension, Caitlyn? Excellent. Hey, do I know M2? Check. Do I know g? Check. Do I know M1? Check. Do I know g? Check. I can solve for normal force, plus this over, minus that over. I think the normal force is going to be M1g minus M2g. I think the normal force is going to be, heck, I'll even crunch the numbers really, really quickly. M1, 5 times 9.8 minus 3.2 times 9.8. 17.6, 17.6, 18 Newtons if I go to two sig figs. Is that all right? Actually, looks ugly, but turns out in some ways to be easier than a lot of the friction ones, certainly, because everything is vertical still. Is that all right? I would love to do more, but I got to get today's lesson done. Lesson 6. Lesson 6, what we call the inclined plane. What we're really saying is, okay, what if we're not on level ground anymore? Suppose a person is standing on a ramp and they're standing on a scale, on a Newton scale that's attached to a rope. As the angle of the ramp increases, what happens to the scale reading? A, no change. B, increases. C, decreases. So if you were standing on a scale on a ramp and the ramp was getting steeper and steeper and steeper and you were looking down at the scale, what would you notice? Would the scale not change at all? Would the scale reading increase or would the scale reading decrease? By the way, this has nothing to do with the scale. I'm really asking this, what is a scale measure? You see, what I'm really asking you is, what happens to the normal force in this situation? That's what you want to be thinking. Don't let the scale trip you up. Instead, be thinking, what will be happening to the normal force? Once again, I'm going to ask you to vote. Who thinks the answer is, A, no change. Who thinks the answer is, B, increases. 7, 8, 9, 10, 11. Who thinks the answer is C, decreases. Okay, I'm going to give you about 30 seconds to try and convince each other right now before we go over this. So, who has the convincing argument that they think will sway the entire class? Yeah. First, what do you think the answer is? I don't think you have a convincing argument that will sway the entire class. Who picked C and has a convincing argument that will sway the entire class? I think he, I was watching his hands and trying to hear, I think you approached it the way I would approach it. What was your argument? Okay, I like, what he's doing is he's saying, look, instead of dealing with the ramp, why don't I start with the one that I'm familiar with level? And are you going to compare it to vertical? And then are you going to ask what happened? And it must go in between, nice. Here's a great argument. Go ahead. So, on level ground, okay, so on level ground, we agree normal force is MG. Uh-huh. If he goes vertical, what's his normal force? He's not even touching this, in fact, he's in free fall, so his normal force would be, so we would go from full normal force to zero. How could that happen? Do you think it suddenly jumps to zero? I think it gradually decreases, decreases, decreases, decreases, decreases until you're in free fall. Is that a fair argument? By the way, the reason I wanted to do that is I'm trying to show you strategies that I use when they throw these at me. And one of the big ones is go to the two simplest cases that you know to try and figure out what happens between them. I can kind of imagine level ground. I'm familiar with that. We've talked a lot about free fall. Okay, what we're really talking about is what happens as you go from level ground to free fall. Decreases. Explain your answer using principles of physics. The black to be theorem. There, we've named it after you. Level ground. Did you really have to ask? Don't I always say don't write this down if I don't want you to write something down? I'm pretty consistent on that. Level ground. Normal force is MG. Vertical wall. Normal force is zero. How does it get from MG to zero? It must slowly decrease. And how would I write that? I think if I was on a provincial exam or on a test, I might say therefore it must decrease from MG to zero. I'm going to try and write it shorthand here because this is in our notes. Therefore, F, what does a downward arrow mean? I've used this before. Decreases. What does an upward arrow mean? From MG to zero. Have we convinced you, Emily? Okay. Okay. By the way, math's approach works very well, I think. So the black to be theorem, the black to be approach, the black to be strategy works very well, I think. For example, two. So example two says same situation, but instead of the normal force, the scale. What happens to the tension in the rope? No change increases or decreases. Can you see how I hope using his strategy? I think the answer becomes reasonably obvious. What's the tension if you're on the ground in the rope? Basically, probably hanging. What's the tension if you're vertical? In fact, I think it'd be MG, wouldn't it? You'd be hanging from the rope. So, oh, it goes from zero to maximum. Increases. Use the black to be theorem. Yes, this is going online. You can show your parents that now you've got a theorem named after you. We're so proud. We're just disappointed it took so long. Here's what I want you to realize. What we're trying to get you to believe, Jeanette, is that when you're on a ramp, gravity is doing two things. It's pulling you against the ramp, squeezing you. And it's also pulling you down the ramp. In fact, here's a ramp. And here's a mass. Right now, whoop, got to find the equilibrium point because it's fairly slippery. It's not level right now. Can you see this at a slight angle? Gravity is pulling it down, but gravity is also pulling it against the ramp and holding it there. But as the angle increases, there's a normal force. As the angle increases, the normal force decreases. Friction decreases because friction is new tons of normal force and gravity's downward pull increases until eventually it'll start to slide. And if I did this, the ramp wouldn't come into play. This thing would just tumble on its own, wouldn't be touching anymore. And if I do this, normal force equals energy. Nothing wants to coincide with it. In terms of the other forces that act on the ramp, there will be a normal force. But the normal force, by definition, is always perpendicular to the surface. It's always 90 degrees to the surface. In fact, the word normal in math, if we say two objects are normal to each other, we say they're 90 degrees to each other. Oh, I thought that just meant that they were like normal, normal, like not weird. No, actually it has a mathematical meaning. If the surface is rough, there will be friction parallel to the surface of the ramp. So here it is. It says, draw a force diagram for an object on a rough ramp that is sliding down the ramp and then B will do up the ramp. So here's our first angled free body diagram. I'm going to say, what are the forces acting on this? And I'm going to start out with my usual, get the obvious ones. There should be no hesitation and no questions in your voice when I say, get the obvious ones. And which way does gravity always act? So MG and draw it nice and long. We're going to exaggerate it for a reason because we're going to use nice big diagrams here. That's supposed to be straight down, if it looks a bit slanted, sorry. What else? Is the ramp sitting on a surface? Then there is a normal force. Now the normal force acts 90 degrees to the surface. So the normal force is going to go out at a right angle like that. Which way does it say this block is sliding down the ramp? So which way is friction acting? That's the free body diagram for an object sliding down the ramp or an object at rest where friction is exactly canceling out gravity. What about sliding up the ramp? Well, what are the forces acting on this? Get the obvious ones. Absolutely. What else? A normal force at a 90 degree angle. Which way would friction be acting this time? Downwards. By the way, this is why it's way tougher to push something, to slide something up a hill than down a hill. Up a hill, you've got gravity and friction tugging against you. Down a hill, gravity is helping. It's really only friction that you're trying to overcome. Here's what we're going to do. Here is our strategy. We're going to pretend this right here is level ground. We're going to tilt our heads because that makes the normal force vertical like it used to be, which I like. And it makes friction parallel to the ground which I like. The problem is gravity is neither vertical or nor horizontal. You know what we're going to do with gravity? We're going to break it into something. You know what we're going to break it into? Components. Components. But we're going to always do it the same way so that our diagrams all look the same. There are many ways to do it. We're just going to memorize the easiest, simplest approach. And then our tug-of-war will work again. Turn the page. I just fibbed. Go back a page, actually. I'll show you what I mean. I'll change colors. You guys can use a dotted line. What I'm going to do, I like the normal force. I'm happy with that. I like friction. I'm happy with that. I'm going to break gravity up. I'm going to break gravity up into that and that with the lovely right angle right there. I'm going to break gravity up. I'm going to break gravity up into this force and this force. By the way, do you see how I made them add to each other tip to give me the hypotenuse? And I'm going to give these names. I'm going to call this MG perpendicular. The math symbol for perpendicular is an upside down capital T because it's at 90 degrees to the ramp. And I'm going to call this MG parallel because that section is parallel to the ramp. And the symbol for parallel is two parallel lines. I'm going to call this MG perpendicular and MG parallel. Friction is what times what? New times a normal force. I don't know the normal force. Oh, but look, look, look, look, look, look. I know another force the same size as the normal force, not MG. What? That component. So we're going to be doing some trig here. We're going to end up with lots of signs and cosines in our equations for the remainder of this unit. But it's actually right angle trig, so no sine law, cosine law. And you'll find it gets pretty standard. You'll start to memorize shortcuts. Unfortunately, this time, y and sine don't go together. So now turn the page. Which of the following is the best force diagram for a mass that is at rest on a ramp? Which of the following is the best force diagram for that mass there? Well, what are the forces acting on this mass? Get the obvious ones. Gravity. Does that help at all? No, they both have gravity. Normal force. Oh, they both have normal force. A or B? Why A? Okay, because it's at rest, if B was correct, if I break this into components, part of gravity is pulling down the hill and there's nothing up the hill to resist it. So B can't be right. If I break this into components, oh, you know how big friction is? It's exactly the same size, Sean, as that component there, because it's at rest. I'll call this mg perpendicular. I'll call this mg parallel. Do you see why I said I want to draw a bigger diagonal? This is why Mitchell, I drew a nice big diagram here, because if you're doing them small, it gets really awkward to label things. But in my answer, I would say friction has to equal the parallel to the ramp component of gravity to B at rest. B would have to be sliding. In fact, B is a frictionless surface. B is ice. In order to determine the size of the squeeze perpendicular and slide parallel components, we need to find the angle in our sketched triangle. Now, the angle they will always give you, Connor, is the angle of the ramp, that one. They'll give that to you in 99% of the diagrams. Where does this angle appear in my little sketch? So remember, first of all, we call this mg perpendicular and we call this mg parallel. We're gonna do this once and then we're gonna say, look, as long as we always draw our diagram the same, just memorize the angle goes there and you don't have to walk through this derivation every time. So walk through it once. Here's our angle. Look at the diagram that I've drawn. I extended gravity nice and long. How big is angle one? Look at it. 90. Why? Well, gravity is perpendicular to the ground and I extended it so I could notice that. How big is angle two? You know how big angle two is? 90 minus theta. Why? What is every triangle add to? 180. If that's 90, those two have to add to 90. 90 minus that would give you that. Okay with that? How big is angle three? Same size as angle two. Why? See the Z? Can I get rid of the Z now? Because it really comes with a diagram. Alt, int, or Z angles, alternate interior is the fancy schmancy name. So you're ready? Instead of angle three, I'm just gonna say angle two because they're the same size. Angle two plus angle four add to 90. You know how I know? Because that's 90 there. Angle two plus theta add to 90. If angle two plus angle four add to 90, and angle two plus angle theta add to 90, what can you tell me about angle four and theta? Cara, they're the same. Angle four is theta. So what I'm gonna say to you Cara is this, if you draw gravity nice and long and exaggerated and then always go perpendicular to parallel, this angle will always be that angle. That's honestly what I just memorized. I don't walk through this step-by-step process anymore. That means always draw perpendicular and then parallel. And that way also our diagrams look the same. There's nothing wrong Emily with drawing parallel first and then perpendicular. It just means your theta's gonna be somewhere else. Let's try one. Example six says find the acceleration of a two kilogram mass sliding down a frictionless 25 degree angle. Okay. What are the forces acting on this mass? Get the obvious one. Gonna draw that nice and big and exaggerated. MG. What else? Normal force which is always at right angles. That's it for your free body diagram. If I ever say draw the free body diagram, that's it. We're gonna put the components on but I'm gonna put them on as dotted lines. So I'm gonna break it and I'll change colors so it stands out. I'm gonna break this up into a perpendicular and a parallel with the 90 degree angle right there. So MG is the hypotenuse. This is gonna be MG perpendicular. This is gonna be MG parallel. Pause for a moment Kayla and just let that sink in. Is that okay Kara? Oh and by the way how big are these two forces compared to each other? Did you see I was actually able to estimate how long to draw this by just kind of visualizing the perpendicular component. I knew to stop my little arrow when I drew it right there and you'll get good at that too eventually Brett. But you can kind of almost visualize how long to draw the normal force. Just imagine how long the parallel is gonna be when you sketch it in and or the perpendicular is gonna be when you sketch it in. Okay. Which way is this ramp accelerating? Sorry. Which way is this mass accelerating down the ramp? Look at your forces. What's the only force that's acting in the same direction as down the ramp or let me ask us another way. Who's winning? Winner. Who's losing? It's a trick question. No one. If I had friction that might be losing but we'll get there eventually. This equals ma. How big is this angle? How big is this angle? The proof that we just did is that one is 25. You can write the 25 or even if you just go boom and kind of color the occluant that you've clued in that these two are the same. Opposite adjacent or hypotenuse? Adjacent or hypotenuse? Because mg is the one that I know. Adjacent or hypotenuse? Which trig function? Sine. Did you say sin? The only sin is saying sin instead of sine. Anyways, sine 25 equals opposite over hypotenuse. Can you get the mg parallel by itself? What's the mathematical expression for mg parallel? What would I do with this down here? Yeah, turns out mg sine 25 equals mg parallel. And I usually do this off in the margin. Actually, that's a fib. Eventually you'll memorize some of these or reach the point where you can do these in your head. But until I get good, I usually do the little trig soketoa off in the margin. And Andrew, here's the key. Instead of mg parallel, you know what I'm going to write? Instead of mg parallel, you know what I'm going to write? Instead of mg parallel, you know what I'm going to write? Is that okay? Oh, I noticed something else kind of nice. What cancels? Turns out they didn't need to tell me the mass. The mass cancels. The acceleration is g sine 25. Make sure you're in degrees, especially if you have a graphing calculator and more of you have graphing calculators now. Make sure you're in degrees. What is g sine 25? Sine 30, I know, is 0.5. So it's going to be a little less than 4.9. I'm going to take a guess of 4.5, 6. I don't know. What? 4. what? I'm that far off? Okay. 4.14. Trig angles are tough to guess in your head. Units, meters per second squared. This is going to be our strategy. Anytime they give us a ramp, we're going to say, uh-uh, I'm going to go components, and that way I'll go parallel and perpendicular to the ramp. Did we use the normal force or mg perpendicular at all? There was no friction, so we didn't need to. I do know that they're the same size, though. Otherwise, it would be sinking into the ramp or flying off the ramp because I'd have unbalanced forces in those directions. Is 4.14 right? Yeah, okay. Example seven. Consider a 5 kilogram mass sliding down a rough with a coefficient of friction of 0.2. 35 degree angle ramp. A, find the force of friction. B, find the acceleration. Okay. What are the forces acting on this block? Get the obvious ones. Jacob, get the obvious ones. They aren't right. 90 degrees, by the way, about that long, so about this long. See how I did that? About that long right there, that's where I'm going to start going 90 degrees. What else? Which way? How do you know? You're right. How do you know? Sliding down a friction is the opposite direction of the motion. Friction. There's my free-body diagram. Now I'm going to add components. Mg perpendicular and Mg parallel. Oh, and this angle is the same size as that angle. That's the 35 right here. Look very carefully. Who's winning? Oh, wait a minute. A, said find the force of friction. Okay, let's do that. A, friction is what times what? Let's write that down just to jog our memory. I don't know the normal force. Oh, but look, look, look, look. I know another force the same size as the normal force. What? What's the same size as this guy? Mg perpendicular. Friction is going to be new Mg perpendicular. Let's see. With this angle right here, Mg perpendicular, is it opposite adjacent or hypotenuse? Hypotenuse. Which trig function am I going to end up using for perpendicular? For what it's worth, parallel is going to be sine almost all the time, and perpendicular is going to be cos almost all the time. I can only say almost all the time, not every single time. So if I hear you correctly, you're saying this. Cosine of what's the angle? 35 equals adjacent over hypotenuse. Trevor, how would I get Mg perpendicular by itself in this little equation right here? Can you see it? Move the Mg up. Okay, can I do that right here without rewriting it? Let's see if we can. I think friction ends up being mu Mg cos 35. Friction ends up being 0.2 times 5 times 9.8 times the cos sine of 35. How big is the force of friction? I'm going to make sure I'm in degrees. Hey, you get 8.03? 8.03. What do you think of this? Nice friction. B, find the acceleration. Now we're going to go winner minus loser. B. Which way is this thing accelerating according to the question? Down. So look closely. Who's winning? And don't say Mg because it's not Mg. Who's winning? Mg parallel. See it. So anything that ends up pointing down the ramp, winner positive. Anything that ends up pointing up the ramp, loser negative. I think my equation is going to look like this. Mg parallel minus friction equals Ma. Now I know that friction is 8.03, but actually I'm going to plug in the algebraic expression to show you something kind of nerdily cool. Mu Mg cos 35 equals Ma. What about Mg parallel? What's that? Well, here's my angle, Breanne. Opposite adjacent to hypotenuse. Which trig function? In fact, without showing work, are you okay if I go, I'm pretty sure this is Mg sin 25. Is that okay? 35. Sorry, I'm thinking the angle from last time. 35. Here's what I want you to notice. What do you notice here? Is there an M in the first term? Is there an M in the second term? Is there an M in the third term? Is there an M in every single thing? Then you know what happens to those M's? They cancel and I already have A by itself. The acceleration is going to be g sin 35 minus mu g cos 35. It is 9.8 sin 35 minus 0.29.8 cos 35. What's this thing going to accelerate at? 4.02? Anyone else? Connor, what'd you get? Oh, okay. 4.02 seems a bit high to me. Maybe not. 4.02? Okay. By the way, can it possibly be bigger than 9.8? Because 9.8 would mean the ramp was house deep. Free fall and no friction. So there's a built-in error check here. If you get something bigger than 9.8, I don't think so. Right? Sorry, what was it? Brett? 4.03? 0.02? Direction in the winning direction. Which way is that? Down the ramp. Which you could argue is positive or negative. You know what? I got everything in my setup. Example 7. Example. No number on it apparently. A 7.5 kilogram block remains motionless on a surface that is inclined at an angle of 21 degrees. What's the minimum coefficient of friction on the surface? Okay. I like this question. What am I going to do first here, Dolp? Absolutely. I'm going to draw a little picture. And I'm going to cheat because I can. Don't they have a different, don't they have a ramp shape? I guess I don't. Never mind. Off to freehand. Well, I'll do this. Here's my ramp. There. That looks vaguely straight. The angle is 21 degrees. There's my block. The mass is 7.5. I don't think I'm going to draw that on there because I want to write a bunch of forces on there. What are the forces acting on this block? Get the obvious ones. Mg. What else? Normal force, perpendicular. What else? Got to be friction acting up the hill to keep it from sliding down the hill. And that's it. If I say draw the free body diagram, that's all I want to see. Ooh, but I'm actually solving this. So now I'm going to break gravity into its components. Mg perpendicular and Mg parallel. Who's winning? It's a trick question. Do you know who's really winning? It's a tie because it says it's in equilibrium. It says it's not moving. We want the minimum to hold it in place. So you know what? Let's let down be negative. Gravity is usually the winner. So let's go. Here's our winner. Winner minus loser equals M... Oh, wait a minute. Why can't I put a zero over here? Acceleration is zero. Mg parallel. Oh, this is 21. So this is 21 degrees right there. Opposite adjacent to hypotenuse. This is the hypotenuse. Which trig function am I going to put in for Mg parallel? Because I know the hypotenuse. It's going to be just the hypotenuse sine 21 minus. Friction is what times what? Mu fn. Oh, instead of equals zero, how about I plus friction over to this side and just do that? Is that okay, Mitchell, or did it lose you? No. 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. What? And opposite adjacent to hypotenuse, which trig function relates Mg to Mg perpendicular relates h and a. Cos. In fact, my equation is going to look like this. Mg sine 21 equals mu Mg cos 21. What happens to the masses? Turns out they cancel. It doesn't matter how heavy this mass is. It'll stay in equilibrium under a 21 degree angle with this coefficient of friction, whether it's a semi-truck or a feather. Yeah. Not only does the mass not matter, if you somehow beamed this ramp to Mars where the G is different, this would still stay in equilibrium. Or if you beamed it to Jupiter, this would still stay in equilibrium. As it turns out, this one here doesn't depend on any of those things. That's kind of nerdly cool. In fact, it seems to me the coefficient of friction is sine 21 divided by cos 21. And Matt, you did math 12 last year. Do you remember what sine over cosine was? Turns out it's just a tangent. They don't know that yet. So we won't scare them. But for what it's worth, if you ever wondered where the heck does tangent appear in physics? Because at least it's almost always sine and cosine. Turns out it's a minimum coefficient of friction of a ramp. Bigger would be better because then you'd have room to spare. But right now, this coefficient of friction, which is what? Anyone? 0.38? Units? Caitlyn? Oh, that's right. Coefficient of friction. A lot of people, because it's coefficient of friction, they want to write newtons. No, no, no, no, it's not. Last one. I'm going real slow here. If it seems like it's a long lesson, I want this. This is the, by the way, almost all your questions on your test are going to be on the written section are going to be at angles because that's physics 12. I like number eight. I like number eight. I like number eight. Number eight is a nice question. I like number eight. I really, really, really, really, really, really, really, really, really like number eight. This one here. I like this question. I like this question. It says, consider a six kilogram mass sliding up a rough surface. How can it be sliding up a rough surface? I think someone's given it a big push and let go, but its momentum is carrying it up the surface for a little bit. Will it keep going? No, it's going to come to a stop. I know that much. I'd like to know how far up the ramp it will travel before stopping. That's part D. To do that, we're going to have to find part A, the force of friction, part B, the net acceleration, C, how long it takes to come to a stop, and then D, solve for D. This is a combination that, combining in one question of everything we've done all year except for jacked out. First, I'm an adult. Here's my ramp. Here's my block. Oh, if I can give you any advice, by the way, if you're doing adult, don't do a little tiny piddly one. You won't be able to label the triangle and the angles properly. Do a fairly big one. I tried to do mine about, what, quarter of the page wide-ish? Andrew, what are the forces acting on this block? Get the obvious one? Absolutely. What else? Kayla, normal force at a right angle to the surface, about like that. What else? Friction. Acting which way? Why downwards? You're right. Ah, friction. And that's it. Now, I'm going to break gravity into its components. Here is the parallel component. Here is the perpendicular component. Oh, and, Brett, this angle here is 32 degrees. This angle here is 32 degrees. Who's winning? Who's winning? Both are winning. No loser, both winners. So, my equation, Caitlin, you're correct, is going to look like this. Parallel plus friction equals ma. There's my angle parallel, opposite adjacent or hypotenuse. Opposite hypotenuse. Which trig function in this question is parallel going to be when I replace it with mg? It's going to be mg what? Sign. Do you see how you can start to do some of the trig in your head? Pushing you guys a little bit. This is going to be mg sign. What's the angle? 32 plus friction is mu times the normal force. Oh, I don't know the normal force. But look, look, look, look, look, look. I know another force the same size as the normal force. What? What's the same size as this guy? Really? mg perpendicular. Oh, our vendor opposite adjacent to hypotenuse compared to this guy. It's going to be adjacent and hypotenuse. Which trig function do you think I'm going to replace mg perpendicular with? Here's my equation. mg sign 32 plus mu mg cos 32 equals ma. Oh, cool. You know what happens here? This is canceled. By the way, this is why, how many of you have been on a roller coaster? Or any amusement park ride? Been on a roller coaster? Which of you? Do you have to step on a scale before you got on the roller coaster? You know why? Because it turns out the masses cancel. Otherwise, we'd have to have all of, we'd have to balance every single car out. You have to step on scales. We have to make sure that they're all, if the masses didn't cancel, we'd have real issues. Katie, question? Divide by m, divide by m, divide by m. Okay? I'm not canceling them by subtraction. I'm saying, if you would divide everything by, if there was a five here, a five here, and a five here, timesing everything by five, you can divide everything by five. You're timesing everything, good question. Maybe I didn't explain it well. You're timesing everything by an m. Divide everything by an m and they're gone. It's kind of nice. Oh, so it looks like the acceleration is going to be g at 9.8, sine 32 plus, I've scrolled down. What was mu? Point, I'll scroll up, 0.1 times 9.8 times cos 32. What's the acceleration? What'd you get? I don't know, anybody. 6.02, anybody else? Oh, that was part B actually said, find the acceleration. What did part A want us to find? Okay, I'm going to, I got room right here. A, friction is what times what? I don't know the normal force. Oh, but look, look, look, look, look, look. In fact, you know what? Isn't that right there, friction without the mass canceling? Right? Mu mg cos 32. Okay, 0.1 times the mass times g times cos 32. What's the force of friction? 0.83, that seems small. Is that right? How big is the mass? 0.85, 0.1, well, 0.1 is a small collision to friction. Times 6 times 9.8 times the cos of 32. I'm getting something totally different, guys. 4.98, right? The mass doesn't cancel up here. It's this whole thing, including the mass for friction. Right? 4.99, nuisance. C, find time to stop. Oh, T equals question mark. Now we're going back to last unit. What else do I know? Oh, to stop. What does that tell me? Let me pause. For that question, it was 4.99, wasn't it? 4.986. Is that not right? That's right, isn't it? 0.1 times 6 times 9.8 mu mg cos. Right? Are we there? I got to move on. I got 90 seconds. Okay? Time. Oh, what's v final? 0. What's A? Got to be careful. Am I slowing down or speeding up? Then A is not 6.02. Negative 6.02. What's VI? 20? Could you find T? T equals VF minus VI all over A. I'm not going to get a chance to do that, but now that you have that, could you find D the distance? Yes. What's your homework? Sorry, guys. Didn't quite work. Number 1, 3, 8. Patience. How many have I given you? I need to give you more than that. No, I don't. I think you need practice with this, but I don't want you to find angle theta. That's terrifying. Hold on. Sorry, guys. 11. 15 is cool, but I won't assign it right now. I'll pause there. Ramps. Have a great weekend, folks.