Basic Physics 3A. Lecture 15.

Okay, I think we'll go ahead and get started. The first thing I want to do is make a quick comment on one of the homework problems. This is the first time I've received multiple e-mails about one. And the question is drawing the free body diagram for the masses in this situation where we had one hanging over the end and a pulley and another one. And in the homework problem, this surface was frictionless and mass A was bigger than mass B. I forget if they gave you the exact numbers or just a ratio. Many people missed the following fact. They would get the free body diagram for mass A correct. They would draw a tension, they would draw a weight, and they would draw a normal force. They'd get that right and then they'd proceed to draw the free body diagram for mass B. And they would draw a weight and they would draw a tension. They would try and get those two about equal. I failed a little bit there. But then they would get it wrong with a drawing that looked like this. What is wrong with that current drawing besides the fact that T is a little crooked because I can't draw straight? What's wrong with that drawing? Mass B is accelerating. In fact, both masses are accelerating. How do we know that? Now, a number of people wrote back, but mass A is bigger than mass B. Why does that not matter? What free body diagram here makes it absolutely obvious that mass A has to be accelerating? What in this free body diagram makes it completely clear just by looking at the free body diagram that mass A has to be accelerating? Yes? Yeah, there's a force in the X direction and there's nothing anywhere to cancel it. So there is a net force. And if I have a net force, I have an acceleration always. And this is why I say don't use your gut feeling, oh, something's more massive so it's not going to move, right? You have to use F equals MA. It is how the world behaves. Now, if I had added friction over here, oh, so I should say since A accelerates and it's attached to B by a string, B has to accelerate. So the tension has to be less than the weight. And that's the key piece there. Now, suppose I add friction. What is the key feature I'm going to worry about in this problem to determine if the masses accelerate or not? There's a key physical number that's going to be very important that I need to know, to know if these things are going to accelerate or not. What number must I know? Mu s, why? The maximum that the friction can be is equal to, since I'm doing the maximum, mu s times F n. And since mu s is always less than 1, that force of friction will actually be less than the weight of mass A. And it's the weight of mass B that is pulling on everything. So depending how much of a fraction this is and how much bigger this is than that, it still might accelerate. So if m a, for instance, is twice m b, but mu k, say, is something like 0.1, this will still accelerate because the net force will still be positive. Yes? Actually, I should have said mu s here, sorry. But yeah, go on. Basically that has to do with the fact that I can't suddenly generate larger forces than the weight due to friction. When I say it's always less than 1, I have to be a little careful because, of course, there's a gray area from what I would just consider normal friction going over to gluing stuff. And once you glue stuff, you can think of it as friction greater than 1. And so you can imagine some super, super sticky surfaces where you have a bigger mu s or mu k. At that point, you would tend to start to think of it as the things being stuck together. So I was very glad I got those questions because it really points to, I think, one of the big misconceptions you can run into when things are going over corners, on ramps, and connected to each other. It's not just the weights. You have to ask what are the real forces and what direction are they pulling. Any other questions on that? Now for this week, we're going to take all the forces we had, the normal force, friction, weight, tensions, and our pushes and our pulls, and we're going to ask the question, does where they act matter? So I've got two clicker questions to start us off with. Now I forgot that in this lecture hall I can't access a door easily that you can see. So I'm going to use my little demo device here, which we can think of as a door. Here's the hinge of the door, right? And the question is to make it rotate, or to open or close the door, do I push on the end of the door? Do I push kind of perpendicular or far away from the hinge, or do I open the door at the hinge? Right? I mean that's kind of a weird sideways door, right? You've all opened a door. The hinge is where it's attached to the wall. Do I push inward on the door, like the edge of the door, where the little thing is that latches into the door? Do I push perpendicular to the way I want to rotate it, or do I go over to the hinge and I push on the hinge? How do I push to open a door? And I hope you know what I mean by the difference between into the edge of the door versus on the face of the door. Because apparently some of you are having trouble opening doors. About 30% of you right now, which is not good. Okay, if my door was here and opened this way, this is the face of the door. This is the edge of the door. And that's a person going through the door. Five, four, three, two, one. Okay, how many of you have ever seen Farside cartoons? They're really old, right? Yeah, where the kid is pushing on the door that says pull. Those of you with B, I'm curious how you get in and out of doors. Not to be insulting here, but if I come up to a door and push on it like this, yeah, that's the edge of the door. Oh, they don't open very well, right? You have to push the door, or use the door handle, right, which is on the face of the door, and you pull and push on the face of the door. We do not push parallel to the thing we're trying to rotate. The reason for this question is to highlight that if I want to make something rotate, I have to worry about where the force is applying and the direction the force is applying. Notice forces like this are not making this rotate. A force like this is. So the direction and where I apply matters. If I come up to this rod and I push right here, not really getting any rotation, okay? So now let me ask you another question. So away from the door, away from the hinge on the face of the door. Now, let me draw an object for you. Think of it as like a stick on ice, so on a frictionless surface. The first thing I'm going to do is I'm going to hit it right on the end there. So answer this question. If I hit it right on the end, and by the way, just so you know, clockwise is like that, if you've forgotten, and counterclockwise is that, because I know you all use just digital clocks on your iPods and phones, so you don't know what clockwise and counterclockwise is. No, I know, that's insulting. Only now, since all of them are analog, right, you can get the app that's analog, and we're back to analog clocks. But what happens? A, it rotates the object clockwise, B, it rotates it counterclockwise, C, it accelerates the object to the right, D would be both A and C, negating the only, and E would be both A and B, whoops, sorry, B and C. E should be both B and C. That was a typo. Sorry about that. Remember, we're on a frictionless surface, it's just sitting there on ice, we're out in outer space, and I knock the end of it. What happens? Five more seconds, four, three, two, one. Most people are having it rotate and move, okay? Now let's answer the question again. Suppose I hit it right in the middle, okay, that doesn't look right in the middle. I'll try, get a little bit closer to the middle, there. I hit it right in the middle. Now what does it do? It should be quicker, 10 seconds, 13, oh wait, that's going the wrong way, 9, 8, 7, 6, 5, 4, 3, 2, 1. Excellent. Perfect. So, pretty much everyone got that right. If I hit it right in the middle, it's not going to rotate. If I hit it at the end, and I meant to bring a bigger ruler, we'll try this here. I don't know if I can balance this on my finger. Okay, I'll balance it on two fingers. If I hit it at the end, it goes up and spins. So whenever I hit it with a force, no matter where I hit it, whoops, it still moves. It still accelerates. And the reason for this is very important. As we go into this next section, we have to realize that all forces do two things. One, they act as if they are applied to the center of mass, and we'll talk about that in a second, and cause an acceleration of the center of mass. This is what we mean by treating it as a point particle. And that is what F total equals ma really does for us. It's really the acceleration of the center of mass. So any object, and that's why we were doing little boxes all last week, any object acts like it's all gathered together at its center of mass, and no matter where I apply the force, that center of mass will move the same way. So all that spinning was extra stuff. And if I hit it in the center, or if I hit it at the end, it'll go to the exact same height. Because it'll have the same initial velocity, because it has the same acceleration. Now, if...so that's what we've been doing up to now. Now in this class, we actually do not worry about calculating this. If we had more time, we would do a whole section on how to calculate the center of mass. For everything we have, it'll be uniform objects that are perfectly symmetric, and the center of mass will be the center of the object. So all of your examples that you'll be asked to do this week, in this section, on the exams, will have that nice property. If you're someone who has not yet taken the MCATs and plan to, that may be something you'll have to look up on your own and study a little bit. I'm sorry, we just don't have time in the ten weeks to cover that. It's fairly straightforward mathematically to do. It's not too hard, but it's not something we do in the class. So you just need to know that this is the center of the object. And everything we did last week was this part of forces. What we're doing now is adding the other feature for an extended object, something with size. Forces can also cause rotation about an axis. And the way we talk about this, we're going to define a new object called the torque, which will have a symbol capital T or the Greek letter tau. And it's really the torque that causes rotational acceleration. Now at this point, we could then do a whole chapter on constant rotational acceleration with angular motion and angular spinning and all of the kinematics and all of the forces that we did with linear motion. Again, we're not going to do that. There really isn't time in ten weeks to do that. The part we're going to do is the special case where the net torque is zero, so there is no rotational acceleration. So if an object is in rotational rest, not rotating, and there's no net torque, it stays not rotating. That's the special case we're going to focus on as part of the special case of no net force. But to be able to do this special case, we do need to know how to calculate torques, and that's what we're going to focus on today. And then Wednesday and Friday, we'll do a lot of examples of what we call static equilibrium, and that will be the subject of the homework, and then we're done with what will be on the next midterm. So the next midterm will cover all the chapters on forces that you did the last two weeks plus this week. So it'll be three weeks of material on the midterm. Now one thing to watch for, this midterm, because of the way of the timing, throws things off just a little bit. So this homework assignment is due Sunday. The following homework assignment is not due until a Tuesday. And then the quiz for this week, you hopefully noticed, was due on Wednesday. So the exam and the timing of stuff, due dates are shifting slightly from the regular what they've been, which is usually Sunday and Monday. So pay close attention to the due dates on the webpage. You don't want to get off. Luckily, if anything, they're moving later, not earlier. So worst case scenario, you do the homework early usually doesn't get you in trouble. Any questions on where we're going and what we're doing? So again, where is it? It's all of these forces that are going to be generating the torques that we'll be studying. And where does that come from? The basic rule is that torque equals r cross f. So we introduce a new type of multiplication, the vector cross product. So again, that's read as r cross f. How many of you have done the cross product before? Okay, a few of you. Everybody say, again, if you've done the cross product, raise your hand. Everybody else look around for someone with their hand up and become their friend. So you're going to have lots of friends if you raise your hand. I'm going to say it right now. This is critical in 3B. So it's a good idea to learn it now. And in 3B, you use it with magnetic fields, which are very hard to picture. Right now, it turns out it'll be actually kind of intuitive and physical. This is one of the places where you can usually use your intuition to get very far. And there are three pieces to the cross product. Piece one, and to computing torques, is you need to pick an axis of rotation. So in some cases, there's an obvious one for you. In this case, this thing is only going to rotate around here. In this case, this is made to rotate around here. But there will be some objects where you'll see you're kind of free to pick what it rotates around. And you'll have to make that choice. So you'll get to do that. The next thing is the R. You need a distance from the axis to the force. That's R. And it's really a displacement. It's a straight line distance. And then the third thing is you need an angle between R and the force. And this is where the cross product really comes in. The cross product is telling you that the angle matters. And one way we write this is the magnitude of the cross product, we'll worry about the direction in a second, is the magnitude of R times the magnitude of F times the sine of the angle between them. Now what is the sine of zero degrees? Zero. So if this is R, the distance to my axis, and my force is like this, or my force is like this, I'm pulling on it or pushing on it, that's an angle of zero or 180, depending on how you're defining your angles. The sine of zero and the sine of 180 is zero. So pulling or pushing in the same direction of R makes it not rotate, there's no torque, and in fact my equation gives me that. So if you're trying to remember if this is sine or cosine, your intuition can help you here. Because it's designed to get the correct answer when you push or pull along the axis. What is the most effective way to make this thing rotate? What angle of force? If I'm going to push at the end, what angle do you think I want to use my force to get the best rotation? 90. And what is the sine of 90? 1. So that's the maximum my torque can be, is when I'm at 90 degrees. Any questions on that? Now there's a couple other useful ways to think about it, and we're going to show it in the demos in a moment, but I want to just draw it here. So here's an R, here's some F at some angle. Now if I look at this and I put the two at the same point, that's how I know what the angle theta between them is. I generally draw the force where it's acting so I know how far away it is, but if I really want to know the angle between them, I need to move them like I can move vectors so they're both starting at the same origin. And when you do that, you realize that the only part of the force, the only component of the force making anything rotate is what we call the component perpendicular. It's only this part of the force that is causing the rotation. The parallel component can't cause any rotation. If you look at that diagram, we see that F perpendicular is just F sine theta. And so our torque, the magnitude of it, we can write as RF perpendicular. A little harder to picture, but just as useful, is the idea of the perpendicular R to the line of the force. So we can also equivalently write this as R perpendicular times F. There are times where that is actually easier than picturing F perpendicular, but most problems, F perpendicular is easier to picture. And the reason this can be very useful, as I said, let's make it a little more clear, I said right here is the angle between them. Sometimes you're not really given that angle. For some reason, you might know this angle in the problem. Notice now, F perpendicular is F cosine of that angle. But it's still F perpendicular. So you can either use geometry to find this angle, which is just 90 minus theta, or you can use the fact that you need to use F perpendicular. So you've got multiple choices here. Again, that's one of the confusing things. There's really only one idea, but it's got multiple ways of doing it. So you can pick your favorite, or be aware of all of them. Any questions on this fun geometry before we now do a little exercise? Excellent. Now we need to find the direction of the torque. And first step, does everybody know which hand is their right hand? This may sound silly. How many of you are currently writing with your right hand? This is what you're going to do in the test. You're going to have to use the right hand rule. You're going to be writing. Your right hand will have a pen in it. And you go, oh, what the heck, I'll use this hand. Is this your right hand? No. You will put your pen down. You will use your right hand, and you will do the right hand rule. Or you will get it wrong. Is that clear? So let's go over what the right hand rule is. The idea is the following. Here's my r. Here's my axis that I want to do rotation about. If I push down in that direction with my force, which way is the bar going to rotate? Is it going to rotate clockwise or counterclockwise? Clockwise. If I push up with a force, which way is it rotating? Counterclockwise. Now we have a convention in physics. This is almost always positive. This is negative. Because we have what we call is our standard right hand coordinate system. We pick x and y, and then z is out of the board. The way we get this is we line our fingers of our right hand with x, and now your right hand, unless you're highly, highly double jointed, can only curl one way. It can't curl the other way. So you can either go like this and curl down, or like this and curl up. So there's only two ways your thumb can point, out of the board or into the board. For a right handed coordinate system, your fingers have to curl from x to y. And then your thumb tells you which way is positive, which way is z. Everyone got that? So if I look up here, so first of all, positive is direction your thumb points when your fingers curl from x to y. Now for the cross product, you put your fingers along r, and you curl toward f. So let's go to a clicker question. I'm going to draw a particular force. So we're going to pivot about a. Here's our axis. This is r. And let's make that force f. Is the torque into or out of the board? So is the direction of this torque coming towards you or away from you, meaning is your thumb this way or is it your thumb that way? Yeah, yeah, there's going to be like a lot of this. During the test, the room will be full. Please don't poke the person next to you when you're doing the torque problems. Okay, five more seconds. Four, three, two, one. Most people went out of the board, excellent. Notice what we have to do. One thing to do is it helps if you slide the force and picture it in your mind at the origin because then it's easier to figure out what you're spinning into. My fingers are in r and the only way I can curl into f is this way. You're not allowed to curl more than 180 degrees. So you can't do this and spin all the way around and say, oh, I got f. That's not allowed. The other thing you could do is if this was a bar and that was the force, which way would it rotate? It would rotate up. Your fingers will curl in the direction it's going to rotate from that force. And then your thumb will tell you whether you're into or out of the board. Now let's do another one. Let's do that force. Is that into or out of the board? I'm only going to give you ten seconds on this one. I'll give you fifteen. Do it real quick. Into or out of the board? Doing a little yoga warm up before coming to the exam can be helpful. Five, four, three, two, one. Excellent. Ninety percent of us correctly have, oh, it's covered up now, r and f. Notice if I go like this, I can't curl into f. And so that's the only way to do it. We'll do one more. Suppose f is like that. Into or out of the board? Ten seconds. Five seconds. Four. Three. Two. One. Excellent. Now this is a slightly tricky one because notice I didn't attach the force to the r. But if I redraw that, this looks like that. F3 is going in that direction, r is in that direction. So I curl that way, r cross f, and I come out of the board. Again, if you think of hitting something from underneath, if I hit it from underneath, it rotates that way. And that's what that force is. Any questions on that? So that's the direction, oh yeah, sorry, yes, exactly. So the torque is always perpendicular to the plane the object is rotating in. That is how it's defined. And the reason being, and this is very useful, we don't do, like I said, we only do the static so it's not as obvious why you need it to be perpendicular, but imagine an object spinning. Which direction is that spinning? Well down here it's spinning this way. Up here it's spinning that way. Down here it's spinning this way. So over here it's spinning that way. So in the plane, it doesn't have a well-defined direction. But the right-hand rule gives us one. If we all agree on clockwise and counterclockwise, in and out of the board is well-defined. Now we have to agree on using our right hand to determine in and out of the board. So you do have to make one decision, but once you've agreed on that, then it's well-defined. So the best and absolute best way to give the direction is to draw your z-axis and give it as plus or minus. So draw your z-axis and then your plus if in your positive z, negative in the negative z. So in and out of the board is pretty standard. Into and out of the paper is pretty standard. Clockwise and counterclockwise is reasonably standard because we're all reading the piece of paper we're looking at. In the lecture hall, of course, it's not because if I stand here and spin it this way, which way is it going? No, it's going counterclockwise for you guys, I believe. Yeah, it's going counterclockwise for you, but for me it's going clockwise because we're on two different sides. Luckily, it's very rare that I'll pick up your piece of paper and turn it around and grade it from the back. So if you say clockwise or counterclockwise, it's pretty unambiguous when we're both looking at the same piece of paper. I just wanted to give you that warning for when we're in real 3D space. Now the final piece in torque that gets confusing is selecting the axis because imagine that I just had a rod out in space and I'm going to apply a force like that. I can look at three distinct points, A, B, and C, and suppose this rod is five meters long and B is right in the middle. I get three completely different answers for the torque whether I look at the torque about point A, because remember I'm doing RF perpendicular, so there'll be some angle theta here, and F perpendicular will be sine theta, and R now is my five meters, so I get to do five meters times F sine theta. If I do it about B, now I'm only two and a half meters away, so my torque is two and a half meters times F sine theta. And finally if I do it about C, R is what? Zero. And so I would get zero. And the units, by the way, are newton meters. It's a force times a meter. And torque doesn't get an abbreviation, it just stays newton meters. This is important because we're going to do energy next, and energy is also a force times a distance, which is a newton times a meter, but we call it a joule. And torque and energy are very different things, and so we don't actually want to mix those units. And I think that's one of the reasons torque didn't get its own unit. We call it a newton meter so it's clear that you're dealing with a torque. And with energy we'll call it a joule so it's clear that you're dealing with an energy. But notice we now see two ways to get zero torque. We can either have R equals zero, or theta equals zero, or 180. There's of course the trivial way to just have no force. So that's a third way. If I have no force I can't have any torque. Notice I can have zero net force and still have a net torque, however. That is possible because my forces might be at different R's. So the magnitude of the force might cancel, but the torque might not cancel. And where do you see that? Well, we would see it right here. When I apply a force at the end to make this spin, how do I know there's zero net force on my rod when I do that? The center of mass doesn't accelerate anywhere, right? The center of mass, I told you for something uniform like this is right at the center, it never goes anywhere. Remember what happened when I hit my pen, right? When I hit my pen at the end, it went flying up in the air. When I hit this at the end, it doesn't go anywhere, it just spins. Zero net force, but not zero net torque. And that's because one force is acting here at R equals zero. In fact, there's two forces here. There's the weight down and the force from this object, the part that it's attached to, and then there's this force here. So these three forces can cancel to be zero net force up or down, but this one, R cross F, is the only one that provides a torque. This is R equals zero. So that's what you got for torques. Now let's look at some of these in more detail. So here, how many of you are familiar with a ratchet wrench? Like three of you, excellent. So this is holding up right now because it's a nice ratchet wrench. You saw me earlier. I can move it up, but it didn't come back down unless I twisted the knob here because it moves in one direction. It's also got a really nice Teflon bolt, so there's a uniform torque applied by this, which I'm going to have to overcome to make it rotate. And so I can use masses to generate forces down due to their weight. So for instance, I can hang the mass here and it doesn't move, right? And if I was to compute this, so what we have is this bar here. There is a force up from the center. There's actually, because this center has some finite size and there's a force over here at the edge, there is a torque here as well that's keeping it from rotating. Now I apply a mass down in the center here over distance half the rod, and notice the torque is still canceling. It's not starting to accelerate. This is what we would call static equilibrium. Now if I hang the mass here, it accelerates it. It starts to rotate. And what I did, notice I didn't change the force any, right? The force is still the same, same mass. I just made R bigger, right? So if I come over here and I have now a force over here, just this force, I'm now using L. So I'm comparing the torque equal to L over 2 times Fw versus the torque being L times Fw. And of course, T2 is bigger than T1 and it starts to bend. Now this is a case where the idea of R perpendicular makes a lot more sense than force perpendicular because the force is always straight down because it's going to be from weight, and now if I put it up high enough, you can see it doesn't slide down, right? Now the torque is not great enough because I'm at an angle, and you can either think of it as this component of the force is the only component that's making it rotate or this force is now acting over a shorter distance, right? It's acting over the perpendicular distance, not the full arm. We call this the lever arm. It's the leverage that you're getting to make this work, okay? So let's go ahead and draw that in our picture, right? So that was hanging the mass at different places, but I can also be up at an angle, hang my force down, and now this is the effect of R perpendicular and my torque about that point A is R perpendicular times F, and this is less than the magnitude of R. Now, of course, if I make the force bigger, it will go down. So it wasn't that I locked it in place. It was literally just that I had it at a bigger angle, and if I make the angle even bigger, the two masses will support it. Oh, I guess not. Phew! Good thing I moved. The Teflon isn't always as strong as it could be. Let's make it really definite. So now I have a really, really short lever arm, okay? It's just this long now, and that'll stay up there. Yes. Okay. Now, but of course, the other thing I can do is I can go in close and ask, okay, how many masses can it support, because again, here's my really short lever arm. I'm going a shorter distance. It's holding two, it's holding three, and it's supposed to, if all goes well, turn with four. Apparently not. It will turn with four. There we go. Okay. So again, it's the combination of all three things. The total force that's acting, where the force is acting, and the angle between the force and R. You need all three of them. So this demo is really a great visual for that. Any questions on that aspect of it? Yes. So we don't actually, notice the rule is the following. If this is R, the angle between F and R is always the angle you get from going less than 180 is the easiest way to do it. Because if I'm here, that's rotating it this way. If I'm here, which is a negative theta, that's rotating it the other way. And notice that minus sign then gives you the correct answer in your RF sine theta, because the sine of negative theta is negative the sine of positive theta. So you will get correctly that it's negative. Now, for most calculators, this is the safest way to do it, but most calculators these days, they never used to be that smart. But if I put it in that angle, I'll also get the negative answer that I should. So you do want to be careful when plugging in your angle values into your calculator that you're correctly getting the positive and negative. And that's why I always recommend checking the sine, SIGN, by using your right-hand rule and not just relying on your calculator for that, because that RF sine theta is really the magnitude of the torque. That's the safest way to do it. Now before you decide class is over, since I still have two minutes, where are we going next? Something we're going to be doing now for the next two days is about a special case of static equilibrium. Static means not moving, so velocity is zero. Equilibrium means not changing, so a equals zero. Notice we can have v not equal to zero and a equal to zero. That would just be equilibrium. We're going to be doing static equilibrium and, at the same time, no rotation or rotational acceleration. So everything's going to be about setting F equal to zero, the net, and the net torque equal to zero. And that's why we needed to be able to calculate torques. That's the context in which we're going to use it. So we're just going to do two days of examples, and they'll let us review friction, tension, and all of that. See you Wednesday.