PHYS 200: Fundamentals of Physics I
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Fundamentals of Physics I
PHYS 200 - Lecture 4 - Newton's Laws (cont.) and Inclined Planes
Chapter 1. Continuation of Types of External Forces [00:00:00]
Professor Ramamurti Shankar: I was just saying that I am beginning to recognize some of you people. I would ideally like to have a class with maybe just ten students so that we can have a lot of dialogue and I can find out what you’re thinking about, address your questions. That’ll make a lot of fun but that’s not possible in an introductory course. But I think you should feel free to interrupt or ask questions or discuss any reasonably related issue that I didn’t get to. Okay? But that’ll make it interesting for everybody because I told you many times the subject is not new to me. It’s probably–it’s new to you. So what makes it interesting for me is the fact that it’s a different class with a different set of people with different questions. And trust me, your ability to surprise me is limitless because even on a subject which I thought is firmly fixed in my mind, some of you guys come up with some point of view or some question that’s always of interest to me, of people who’ve been practicing this business for a long time. So I welcome that, and that’ll make it more lively.
All right, so I think last time I got off on a rant about all the different pedagogical techniques, some of which I don’t endorse fully. I’m going to go back now 300 years in time, to one of the greatest laws that we have because–look at the power of this law, right? Here is the equation, and all the mechanical phenomena that we see, around the world, can be understood with this law. And I was starting to give you examples on how to put this law to work, because I think I at least made you realize that simply writing down the law does not give you a good feeling for how you actually use it. So, maybe you have understood it, but I’m going to remind you one more time on how you’re supposed to use this law. So, I’m going to take a concrete example. The use of any law of physics is to be able to predict something about the future, given something about the present. So, all problems that we solve can be categorized that way.
So I’m going to take a very simple problem, a problem to which I will return in great detail later on; but let me first start with–let me start with this problem, which I will do very quickly now and we’ll come back and do it more slowly later. But at least it’s a concrete problem. The problem looks like this. So, here is a table and here’s a spring and here’s the mass m. There’s a force constant k. I want to pull it by some amount A, and let it go. So that’s the knowledge of the present. The question is, when I let it go, what’s this guy going to do? That’s the typical physics problem. It can get more and more complex. You can replace the mass by a planet; you can replace the spring by the Sun, which is attracting the planet; you can put many planets, you can make it more and more complicated. But they all boil down to a similar situation. I know some information now and I want to be able to say what’ll happen next. So, here I pull the mass, when I want A and I want to know what’ll it do next.
Remember, when we go back to the laws of Newton, the laws of Newton only tell you this–and we’ve been talking about this is useful information. The first thing you have to know in order to use the laws of Newton, you have to separately know the left-hand side. You have to know what force is going to act on a body. You cannot simply say, “Oh, I know the force on the body, it is m times a; ma is not a force acting on a body; a is the response to a force; you got to have some other means of finding the force. And in this case, the force on the mass is due to the spring. So, I pull the spring by various amounts and I see what force it exerts. Now, I think you know now in practice how I know what force it exerts, right? I pull it by some amount, attach, say, the one kilogram standard mass. I see what acceleration it experiences, and m times that acceleration, or 1 times the acceleration, is the force the spring exerts. So I pull it by various amounts and I study the spring. And I’ve learned, by studying the spring, that the force it exerts is some number k, called a force constant, times the amount by which I pull it. If I start off the mass in a position where the spring is neither expanded nor contracted, that’s what we like to call x = 0. So I pulled it to x = A. Now, what I’m told is when you pull it to any point x, that’s the force the spring exerts.
So, this is part of an independent study. People who work in spring physics will study springs and they will find out from you, find out and tell you that any time you buy a spring from me it’ll exert this force. And they have done–they figured that out by pulling the spring and attaching it to various entities and seeing what acceleration it produces. There, the masses are taken to be known, because you can always borrow the mass from the Bureau of Standards, or we discussed last time how if you have an unknown mass you can then compare it to this mass and find out what its value is. So every object’s mass can be measured. And then the guys making the spring have studied what it does to different masses and figured this out.
Now, you come with this mass and you say, what happens when I connect it? Well, I’m assuming the mass of this guy can also be found by comparisons, the way I described to you last time. So we can always find a mass of any object, as we went into in some length. Then, Newton’s law says this is equal to ma, but I want to write a as the second derivative of x. So, you now go from a physical law, which is really a postulate. There is no way to derive F = ma. You cannot just think about it and get it. So, whenever I do physics I will sometimes tell you this is a law; that means don’t even try to derive it. It just summarizes everything we know in terms of some new terms, but it cannot be deduced. On the other hand, the fate of this mass can now be deduced by applying Newton’s law to this equation.
Now, this is a new equation, you may not have seen this equation before. For example, if I told you – forget the left-hand side – if I told you the right-hand side is 96, I think you guys know how to solve that, right? You have to find a function whose second derivative is 96 divided by m, and you all know how to do that; it’s t2 times a number, and you can fudge the number so it works. This is more complicated.
The time derivative of this unknown function is not a given number but the unknown function itself; in other words, x itself is a function of time. This is called a differential equation. A differential equation is an equation that tells you something about an unknown function in terms of its derivatives. You can have a differential equation involving the second derivative or the first derivative or the fourteenth derivative, whatever it is. You are supposed to find out what x (t) is, given this information. So, one thing is, you can go to the Math Department and say, “Hey look, I got this equation, what’s the solution?” and they will tell you.
Now, sometimes we have to do our own work and we can solve this equation by guessing. In fact, the only way to solve a differential equation is by guessing the answer; there is no other way. You can make a lot of guesses and every time it works you keep a little table; then you publish it, called Table of Integrals. So, I have in my office a huge table, Mark Caprio has got his own integral, we don’t leave home without our Table of Integrals. I got one at home, I got one at work, I may want to keep one in the car because you just don’t know when you will need an integral. Okay? So people have tabulated them over hundreds of years. But how do they find them? They’re going to find them in the way I’m going to describe to you now. You look at the equation and you guess the answer.
Let’s make our life simple by taking a case where the forced constant is just 1, okay? It takes 1 Newton per meter to stretch that spring. I let the mass of the object be 1 kilogram. This is just to keep the algebra simple. Later on, you can put any m and k, and we’ll do all that. Then what am I saying? I’m saying, find me a function whose second derivative is minus that function. Now, as a word problem maybe it rings a bell, right? Do you know such a function?
Professor Ramamurti Shankar: Exponential is good. Another one? Yes?
Student: Any trig functions.
Professor Ramamurti Shankar: Trigonometric functions are good. So I’m going to temporarily dismiss the exponential. I will tell you why. An exponential function is actually a very good guess. You can say x = et, for example. The trouble is, this is growing exponentially, and we know this spring mass system, the x is not growing exponentially. Or you might say, well, let me fix that by making it minus t; then it quickly comes to rest after a time and doesn’t move. We sort of feel we want this thing to be jiggling around for a long time. So, the exponential function is even better than this function because it says, take one derivative and I reproduce myself. But that’s too much; we only want a function that reproduces itself when you take two derivatives, and that’s where the trigonometric functions come in. So, I make a guess, x (t) is cos t, and you can check if you take two derivatives, it obeys the equation, the second derivative of x is -x itself. But do I like this solution or is there something not fully satisfactory? Yes?
Student: But you can’t help the negative when you [inaudible]
Professor Ramamurti Shankar: Pardon me?
Professor Ramamurti Shankar: Oh it will be negative x. So, let’s do that. So let’s take dx/dt = -sin (t). Then, take one more derivative, d2x over dt2, derivative of sine is cosine. So, if this is -cos (t), but x was cos (t); so it’s really minus of the x I started with. Yes? So you have a question or–?
Student: I was thinking, you wrote it up there [inaudible]
Professor Ramamurti Shankar: Oh, this one?
Professor Ramamurti Shankar: Caught on film. Yes, that’s a mistake. Okay? That should be minus. Very good. Look, it makes me immensely happy when you guys catch me. Of course, it becomes an addiction, you try to catch me; sometimes it doesn’t work, but keep trying. This time you’re absolutely right. That’s why you were looking confused. Anything else? Everyone follow the way this very elementary problem is solved? It’s solved by making a guess. So, you’re right. So, the minus sign was wrong but now you agree everything’s working. This is the answer. But no, I meant something else is not quite okay with this. Yes?
Student: Your function also works if you added a factor of the sine, also?
Professor Ramamurti Shankar: Yes. But first of all, I can add some sin (t) and it’ll still work. But more importantly, if I put t = 0, I get x = 1. Why should it be true that I pulled it by exactly one meter? I could have pulled it by 2 or 3 or 9 meters. I want to be able to tell how much I pulled it by at t = 0. Suppose it was pulled to 5 meters and released. I want x (0) to be 5. The point is, you can easily fix that, by putting a 5 here. Does it screw up everything? It doesn’t because the 5 just comes along for the ride, everywhere. And don’t think it’s 5x, because this 5 is part of x. So you can multiply your answer by any number you like, and you pick that number, that reproduces your initial displacement. So if you pulled it by 5, you put a 5. So, now we’re very happy to describe this problem. It is true, you can put sin (t) but I will, in this problem we don’t need it.
Here’s an answer that does everything you want it to do. At the initial time, it gives you 5 times cos 0, which is 5; and that’s what you told me was the initial displacement. Initial velocity, if you take a dx/dt, it’s proportional to sin (t) and a t = 0, that vanishes, that’s correct; you pulled it and you let it go. So, the instant you released it, it had no velocity. And it satisfies Newton’s laws, and that’s your answer. I want to do this simple example in totality because this is the paradigm. This is the example after which everything else is modeled. Yes?
Student: How do you know that it’s not sin (t) at the end?
Professor Ramamurti Shankar: Ah. You need a t = 0, if the answer was sin (t); well sin (t) vanishes a t = 0. And cos (t), it doesn’t vanish, has a robust value of 1. But I want a value of 5, I put a 5 in front. Look, you guys may have a few more things to say about this problem because this is not the final and most general solution to the problem. It is a solution that certainly works for the one example I had, which is a mass being pulled to 5 meters and released. So, carry this in your mind. This is really to me the most important thing I can tell you right now. The way F = ma is applied in real life is half the world is working on the left-hand side, finding the forces that act on bodies under various conditions. In this example, the force is due to a spring and by playing with a spring you can study the force it exerts the various displacements of various amounts by which you pull it, and in this example if you pull the spring by an amount x, it exerts the force -kx, and you can measure k and you put the label on your product and you sell the spring with a given k. Other people are measuring masses of various objects and they’ve decided the mass of this is whatever, maybe 1 kilogram. You put the two sides together and you find the response of the system for all future times. And what this tells you then, you can now plot this. This is a graph that looks like this; it goes from 5 to -5 and repeats itself and it’s very nice. That’s what we think the spring would do.
Chapter 2. Kinetic and Static Friction [00:14:50]
Now, you think this is a real spring or am I missing something? This is the whole story? Yes?
Student: There’s friction.
Professor Ramamurti Shankar: There’s friction because we know if you pull a real spring it’s not going to oscillate forever. It’s going to slowly stop shaking and pretty much come to rest. That just means–Who do you want to blame for this equation? Do we conclude now there’s something wrong with Newton’s laws? Or what’s the problem?
Student: The k coefficients should not be [inaudible]
Professor Ramamurti Shankar: Pardon me?
Student: The k coefficient is wrong.
Professor Ramamurti Shankar: The fault is not with the k. Yes?
Student: We didn’t take into account all the force that’s friction and the amount of force.
Professor Ramamurti Shankar: That is correct. So, we will say we have missed something. There’s another force acting on this mass, besides the spring; that’s the force of friction that’ll oppose the motion of the mass. So, you can say one of two things. Either you can say something is wrong with Newton’s laws, or you can say we’ve not applied Newton’s laws properly because we haven’t identified all the forces. So, if you never knew about friction, you do this and you find, hey, it doesn’t work. The mass comes to rest, whereas the theory says it should oscillate forever. So, you are at the fork. Either you can say Newton’s laws are right; I’m missing some forces, I’m going to look around and find out what they are. Then, you will derive friction from that. Or maybe it’s time to give up Newton’s laws. It turns out, even if there is no friction this is not the correct law of physics. The correct law is given by Einstein’s relativistic dynamics. This does not satisfy the relativistic dynamics; by that I mean, if you really pull a real mass by 5 meters or whatever and release it, its motion will not be exactly what I said; or it’ll be off by a very, very, very tiny amount that you probably will not discover in most laboratories. But if the mass begins to move at a velocity that is comparable to that of light, then this equation will be wrong.
So, people say, oh, what kind of business are you in? Every once in a while somebody is wrong. I’ll tell you right now. Everything we know is wrong, you know? And it’s no secret; you can publish it, you can tell the whole world, this practitioner said everything is wrong. But it’s wrong in this limited sense. Is Newton right or wrong? Well, Newton didn’t try to describe things moving at speeds comparable to light. He dealt with what problem he could deal with at that time. So, it’s a law that has a limited domain of validity. You can always push the frontiers of observation until you come to a situation where the law doesn’t work. But the specialty of relativity doesn’t also work all the time. If the mass becomes very tiny, it becomes of atomic dimensions, then you need the laws of quantum mechanics. That’s wrong too. So, things work in a certain domain and sometimes you abandon the formalism; but don’t rush to do that. In this case, the problem is not with formalism but due to friction. So, you have to make sure you got all the forces. Once you got all the forces, it still doesn’t work, then of course the experimentalists are very happy because there’s something wrong with the theory; then you got to find the new theory, till that works.
Okay, so that’s the cycle. So it’s an ongoing process; physics is still a work in progress. This is work as of 300 years ago. And no one’s found anything wrong with Newton’s laws, provided you don’t violate two conditions. You don’t deal with objects moving at speeds comparable to that of light, and you don’t deal with objects which are very, very tiny. The notion of what’s very tiny will be very clear when you learn quantum mechanics next term. Right now, there is something rough, called “tiny,” there’s things like atomic dimensions.
All right. So now we are going to do more concrete problems, but I wanted to write that up for you as a simple, solvable, worked out – an example that displays everything that happens in physics.
All right. So, by the way, people are always looking for new forces, the electrical forces. You find a charge on a table, it’s not connected to a spring; I bring another charge next to it, in this case it starts moving. What do I do? A body accelerating without any force… I can say Newton is wrong or I can say maybe there’s a new force between that body and this body, and that will turn out to be the electrical force.
Electrical force occurs whenever you take something, you take a cat and you take a piece of amber or you rub it, you do various things, then you put them next to other things and they’re attracted or repelled, and you discover you’re charging things, and that’s how electricity was discovered. But you don’t have to have a cat; if you have a buffalo, you can take it with a buffalo, it still works. Then, you got to find out what you really need to find electric forces. But you find there is a new force and you know you’ve got a whole new force to study; then you got to find out what’s the cause of the force. It turns out, it’s not the mass of the object, it’s something else called electric charge. Then, how does the force vary with distance? Well, it varies like 1/r2. You know that. So that’s how you slowly study and identify new forces. So, the forces were found in the order of gravitation first, and then electricity, magnetism and now of course there are new forces called “strong forces.” There are weak forces. They all act on the subnuclear level; so that’s why that business is not done. We are not finished with the left-hand side. We’re still looking for new forces.
Okay, so now I’m going to start doing problems in two dimensions. Everything I did last time was 1D. You’re pulling something, you’re pushing something, you got one block connected to a second block to a third block and you pull it. We’re done all that. So we’re going to elevate this to high dimensions. And as you know, in high dimensions, acceleration is a vector, and so force better be a vector because when you take a vector and multiply it by a number you get something else. And there, by that process, you’ve gone to higher dimensions.
So let’s take again simple problems and try to make them more and more complex. And you cannot tell me eventually you agree the problem is complex enough, or I will crank it up as we go along. If at some point you plead for mercy, then we’ll stop. But there is no limit to how difficult mechanics problems can be. If you go back and read Cambridge University exams, in 1600 and 1700, there were really difficult problems. That’s why they invented quantum mechanics, which is a lot easier than those problems from Newtonian mechanics. You can make it as hard as you like.
So, I’m just going to do a few, then once you got the idea, you stop. But again, we’re going to start with easy one. Easy one is a table on which this mass is sitting, doing absolutely nothing. And we want to study that guy and understand it in terms of Newton’s laws. Because we’re in two dimensions, you got to have two axes. So here is my x and here is my y. So, the first rule is if F = ma, we’re really saying i times fx + j times fy, is miax + mjay. It’s a vector equation. The force has to be a combination of i and j because nobody–By the way, I’m working in the xy plane. I told you I don’t need to go to 3D, for a long time. In 3D, in every vector there’s some amount of i and some amount of j, that’s, what’s true for f, that’s true for a. You write it out. And I told you, if two vectors are equal, then the x components have to match and the y parts also have to match. You can see that. If I draw two arrows on the plane and I tell you this arrow is the same as that arrow, it’s the same. It’s got the same horizontal part and the same vertical part. So, this is really two equations, and I’m going to apply them to this block.
So, when I look at this block, in the x direction, it’s not moving at all. There are no known forces acting in the x direction, and therefore it’s a case of 0 = 0. Did I miss something? You guys look a little worried. Yes? Okay, now I look in the y direction. If you look in the y direction, I take this block and I draw its free body diagram, and I write all the forces on it. So, there is the force of gravity, mg; however, this M is a big M or a small m? Well, let’s change the figure to small m, so that the rest of the equations are right. The little mg is acting down. If that’s all you had, the block would fall through the table; it’s not doing that, so the table we know is exerting a force.
The standard name for that is N, and N stands for normal. And normal is a mathematical term for perpendicular. You say this vector is normal with that vector, you mean they are perpendicular, and here we mean this force is perpendicular to the table. There are the forces in the y direction. In the y direction, N is a positive force, it’s with a plus sign; mg is a negative force, and that’s equal to m times ay. It’s interesting that in this application of the Newtonian equation, I’m going to say that I know the right-hand side. I know the right hand side is 0 because I know this block is neither sinking into the table nor flying off the table. It’s sitting on the table; it has no velocity, it has no acceleration, vertically. So that has to be 0 because I know the right-hand side. So, in Newton’s laws, quite often either I will know the left or I will know the right; either I will know the force or I will know the acceleration. So, if this is 0, I come to the conclusion N = mg. This is a very simple example, but this is how you figure out how the Newton’s laws work in this particular problem.
Okay, now I’m ready to introduce another force, and that’s the force of friction. As you all seem to be aware, there is generally a force of friction between this mass m, and this table. So, how do we learn there is a force of friction? Suppose someone said, how do you know there is friction? Here is a mass. What experiment will you do that tells you, hey, there is another force called friction? Okay, anybody tell me something you would do? Yes?
Student: Maybe put the mass on an incline.
Professor Ramamurti Shankar: Okay, but suppose it’s on this table. But you are around, you can do what you want. Yes? Go ahead.
Student: [inaudible] constant v; in order to move with a constant v, you’d have to apply force.
Professor Ramamurti Shankar: That’s possible.
Student: But moving forward, over the constant v, there’s no force.
Professor Ramamurti Shankar: That’s a very good answer. One answer is, you find that to keep it moving at constant velocity, you have to apply a force. That means the force you’re applying is cancelled by somebody else, because there is zero acceleration, there’s got to be zero force. Yes?
Student: Just apply force to it and push it on the table and it’ll stop eventually, and that’s how you think there must be a force, that’s out there.
Professor Ramamurti Shankar: Very good. But even before that, even before it starts moving, you find that if you push it, it doesn’t move. Right? I try to push the podium, well, I don’t know my own strength, but I’m going to imagine, I push it, it’s not moving. But if I push hard enough it will move. What’s happening before it moves? I’m applying a force and I’m getting nothing in return for it. So I know there is another force opposing what I do, and that’s called a force of static friction. So, static friction is when the body is not moving. So, I’m applying a force F. Let me give it some other name, little f. No, that’s also bad. Little f was the name for friction. So, I’m going to apply a force called Fme. You can label these things any way you like. Okay? This particular notation has not caught on because people are not as interested in what I think as I am. But right now I’m going to call it F sub-me. I invite you to invent any notation you want because there’s nothing sacred about a notation, at least I want to convey to you that notion. So, I call it Fme, I’m applying to the right, then I know that it’s not moving; therefore, there has to be another force that cancels it. But how much is the other force? Yes?
Student: Equal to Fme.
Professor Ramamurti Shankar: That’s correct. So, it is not a fixed force. Static friction is not a fixed force; it’s whatever it takes to keep it from moving. It will not be less than what I apply, because then it’ll move; it cannot be more than what I apply because then it’ll start moving backwards. So, it will do what it takes to keep it from moving. So, static friction is a force which has a range from 0 to some maximum. So, let’s call it the force of friction is less than or equal to some maximum. And that maximum turns out to be a number called the coefficient of static friction times the normal force. The force of friction seems to depend on how much normal force the table is exerting on the object, which in our example, if you want in our example, not always, this is always true, but in our example you may replace it by μs times mg, because N = mg. So the force of friction seems to depend on how heavy the object is that’s sitting on the table. But what does it not depend on that it could depend on? Can you think of something else it could depend on but doesn’t? Yes?
Student: The amount of surface area.
Professor Ramamurti Shankar: It doesn’t depend on the area of contact. In other words, if this was a rectangular block, it doesn’t seem to matter whether it’s doing that or whether it’s doing that. You might think, hey, there’s going to be more friction because there’s more contact. But it doesn’t. So, it’s interesting not only what it is but what it’s not. Okay, this is called a coefficient of static friction. So you guys should remember that the force that friction applies is equal to the force that I apply, up to a maximum of this. And once I beat the maximum, it’ll give up and it’ll start moving. Once it starts moving, you’re going to have a situation like this. I apply Fme this way, and friction will apply a force which now is called new μ sub-kinetic times N. So, the force of friction is different when it’s moving as compared to when it’s not moving, it’s slightly less, once a body is in motion. So, it’s always true for all situations; μ sub-kinetic is less than μ sub-static.
Okay? So, this is how you characterize friction. You study various objects and you find out friction certainly depends on the nature of the two surfaces. Given that constant, it seems to depend only on the normal force and not on the area. Yes?
Student: So is μ sub-kinetic less than the maximum [inaudible]
Professor Ramamurti Shankar: Remember, μ sub-kinetic is a fixed number.
Professor Ramamurti Shankar: Maybe .2. μ sub-static is a fixed number, .25. If it’s .25 for static, it means if the normal force is some mg, up to a quarter of the weight is the frictional force it can apply to prevent its motion. Suppose I apply a force on a body, which is originally 1/10th of its weight; it won’t move; 2/10th of its weight it won’t move; a quarter of its weight it’s a tie; .26 of its weight it’ll start moving. Once it starts moving, frictional force will be .2 times its weight, not .25, because the kinetic friction is less. This is just the summary of friction. Okay, so far even though I used vectors, most of the action is just in one or the other direction. Yes?
Professor Ramamurti Shankar: No, it again seems to–It can actually depend on many things. Friction is something we don’t understand fully. But it depends only on the normal force, just the way I described it.
Chapter 3. Inclined Planes [00:31:34]
Okay. Now, we are going to do the one problem that has sent more people away from physics than anything else. Do you know what it is? You haven’t heard about the problem that sends people away? Okay, this is called the inclined plane. A lot of people may not remember where they were during the Kennedy assassination, but they say, “I remember the day I saw the inclined plane; that’s the day I said I’m not going into physics.” Because this is very bad publicity for our field. You come into a subject hearing about relativity and quantum mechanics and then–or maybe gravitation and astrophysics, and we hit you with this. So why am I doing this? Well, if you can do this and you find it boring, you can move on. But if you couldn’t do this, you have no right to complain because this is the entry ticket into the business. You’ve got to be able to do this. Only then I can tell you other things. Okay, so we have to. We’ll do this quickly. I’m watching the clock. I won’t linger over this, but we have to do this problem. Okay? So let’s quickly finish this problem.
So here is the problem. There is a mass, m, sitting on its inclined plane, it’ll make some angle here. You want to know what it’s going to do. All right, we know it’s going to slide down the hill, but we want to be more precise, and the whole purpose of Newton’s Laws is to quantify things for which you already have an intuition. So the only novel thing about the inclined plane is that for the first time we are going to pick our x and y axes not along the usual directions, but along and perpendicular to the incline. That’s going to be my x and that’s going to be my y. And I ask, what are the forces on this mass? Well, I’ve told you, first deal with contact forces. But the only thing in contact with the mass is the plane. The inclined plane can exert a force, in general, along its own surface and perpendicular to its own surface. But I’m going to take a case where there is no friction. If there is no friction, by definition it cannot exert a force along its own length. So it can only exert a force in this direction and we’re going to call that N. Then, there’s only one other force; that’s the force of gravity which we agreed we have to remember. Even though the Earth is not touching this block, the Earth is somewhere down here, it’s able to reach out and pull this block down. So, let me show you the forces. So here is N, and here is mg, these are your forces. And with no friction, that’s it. That’s the key to doing any problem I give you, in any exam. No matter how stressful, I guarantee you’ll get it from Newton’s laws, provided you write every force that is there and don’t write any force that is not there. So, this is it. These are the two forces, and the mass will do what these forces tell it to do.
So, what we want to do then, since my axes are in this direction, I have to take this mg and write it as the sum of two vectors, one which is pointing in this direction and one which is pointing in that direction. So, if I add these two arrows, I get mg. That’s called resolving the force into some other direction. Now, the key to all of this is to know that this angle, here, is the same as this angle here. In other words, if mg is acting down, that angle θ is the same as this angle θ here. This is going to be used all the time, so I’ll tell you a little trick that tells you how I figured that out. θ is the angle between the horizontal and the incline. Now, you’ve got to agree that if I draw perpendicular to the horizontal and I draw perpendicular to the incline, the angle between those two perpendiculars will be the same as the angle between the two lines I began with, because perpendicular means rotate by 90. If you rotate both lines by 90, the angle between the rotated lines is the same. The vertical is perpendicular to this horizontal, and this is the perpendicular to the plane. Once you got that, the rest is very easy.
So, in the x direction, I’m going to write the following equation: mg times sin θ. So, you’ve got to understand that this part here will be mg times sin θ, and this part here will be mg times cos θ. You’ve got–This is something you better get really used to. If a vector is like this and you want the force in that direction, the angle between is θ, then you put the cosine. The component that’s adjacent to the vector, is the cosine, this one here is a sin θ. So, I got the horizontal part in the x direction, I got mg sin θ, and that’s it. There’s nothing else. That’s got to be equal to m times ax. Remember ax is not horizontal, ax is along the incline, downhill.
In the y direction I write N - mg cos θ = m times ay. That’s simply writing down Newton’s laws as two laws, one along x and one along y, and putting in whatever we know. What happens next is up to what else we know. Well, we know that this block is sliding down the hill, it’s not going into the block, nor is it flying out of the block; it’s moving along the block. That’s the reason we chose a coordinate to be perpendicular to the block, because that coordinate, the y coordinate, it’s not changing. Even though the block is sliding down the hill, when it’s sliding down the hill the usual x and y coordinates, if you define them this way, they’re all changing. But by this clever choice, it’s always at y = 0. So, it has no y coordinate, there’s no y velocity, there’s no y acceleration. So, we know this ay = 0. So, this is an example where if we know the acceleration, by this kind of argument, from that you deduce that N has to be mg cos θ. Okay?
Then, you come to this fellow here. There we find out that ax, you cancel the m, and you find it’s equal to g sin θ. So, that’s the big result. It’ll slide down the hill with an acceleration g sin θ. This is a good way to measure g, by the way, because if you drop something, it falls too fast for you to time it. But if you let it go down an incline, by making θ very small so sin θ is very small and the grading is very low, the body can accelerate with a much slower acceleration downhill, because it’s reduced by this factor, sin θ.
Okay, here’s another thing I should tell you right now. Most of us, when we become professionals, don’t put in numbers till the very end. So, if you are told the incline is an angle of 37 degrees, g is 9.8, don’t start putting numbers into the first equation. We don’t do that and I know for some of you it’s pretty traumatic to work with symbols, and sometimes on exams they say, oh, you had all these problems with symbols.
But I want to sell you another idea. It’s much better to work with symbols than with numbers till the end. Why is that? First of all, if you put numbers in, and I suddenly tell you, “Hey, I was wrong about the slope, it’s really 39 degrees and not 37,” you’re going to do the whole calculation again. But this way, you come to the formula and you say, “What’s your θ?” You change your mind, okay, that’s the new θ. You change the value of g, you made a better measurement, that’s the answer. Then, you can see if your problem had some mistakes in it. Suppose you got g2 of sin θ. You will know it’s wrong because this is an acceleration and that’s an acceleration. The units have to match and sin θ, of course, has no units. But maybe you got the trigonometry wrong. Maybe it’s really g cos θ that should be the right answer.
How do I know it’s sin θ? You can do a test. We know, for example, as the incline becomes less and less inclined, you got to get less and less acceleration, because it’s going downhill less and less, and when θ goes to small values, ax also diminishes. Or, if you take an incline and you really make it almost vertical, and in the end you make it completely vertical, the block is just falling under gravity. So, when θ is 90, a becomes g. So, you can test your results. So, be prepared for that part, if you are in this course. I will give you some homework problems in which numbers are not put in till the very end, or I may not even ask you to put the numbers in, because the thing is to get this formula because anyone can take the numbers and put them in the calculator. So, that’s going to be part of our agreement that we will work with symbols. And if you look at any calculation we do, if you caught me in my office, I’m not putting numbers in very early. I mean, I cannot wait to find out what the answer is but I keep the symbols till the end, then I put the numbers. Then, you can vary different numbers. You can say, “Well, what if gravity was weaker? I go up to Denver and do the experiment. Will I get a different answer?” You will. So, how things depend on the input parameters is interesting. Here’s another interesting result. What is missing in this formula that could have been there?
Professor Ramamurti Shankar: Pardon me? The mass. It doesn’t seem to matter what the object is. That’s an interesting property of the result. Right? And that’s because m cancelled out of the two sides of that equation. So, these are all interesting features you would not know if you kept numbers everywhere and you got a = .62 meters per second squared. To me, that number doesn’t tell me as much. At the end, I want to know the number but the formula is very interesting.
All right. So, now we are going to make it a little more complicated. We are going to now add friction to this. So, here is the block but now there’s a coefficient, the friction, and let’s say the block is–Let’s ask the following simple question. I think it was related to what one of you guys said. Let the block be at rest and θ is an angle I can vary. Maybe there is a little thingy here, you can turn it back and forth. Let me crank up θ and see where it’ll start slipping. In other words, there is a coefficient of static friction between the block and the plane, and I want to know, like parking a car in San Francisco, what angle can it take before it starts sliding down. So, let’s write the forces and let’s write F = ma. So, the forces are again here. Let me cut to the chase and write mg cos θ here, N here, mg sin θ here. What shall I write for friction?
Student: The first one.
Professor Ramamurti Shankar: Do not write μs times N, because the frictional force is anything you need, up to μs times N. In other words, if θ was very small and mg sin θ is very small, the frictional force will be in fact mg sin θ. So, it’ll equal whatever it takes to keep it from slipping. But if I’ve cranked up the angle to that maximum angle, beyond which it cannot stay there, let’s give a name to that angle that is sin θ*. Well, at the maximum angle friction is doing the maximum thing it can do; then, we can certainly say mg sin θ is then equal to the static friction times N, which is now μs times mg cos θ. Again, the mass cancels. So, it doesn’t matter what kind of car you parked on the slope, it only depends on the following. But g cancels too. So, whether you park your car on the Earth or park it on another planet, this is going to be the same restriction, which is that tan θ has to be less than or equal to μs; equal to μs is the critical value, less than that is acceptable. Okay, that’s a simple problem. If you tilt it to more than that it’ll topple over.
So, now let’s take another problem where the block has started moving; it’s going down the hill. Now, what formula do I write? You notice that I’m not writing every step, every time. Each time I deal with it I’m going to do some things faster. For example, I already resolved gravity to two parts, without going through the whole song and dance, because you know how to do that. Now, I put in the friction. So now, I’ve got μ sub kinetic not equal to 0, mass is moving. You can ask, “What’s the acceleration now?” So, the horizontal equation along x is going to be mg sin θ minus μ kinetic times a normal force, but allow me to write for the normal force mg cos θ, and that’s got to be equal to m times ax.
Now, can you do this? That’s the test. It is not enough to watch me do this and say, okay, he seems to know what he’s doing. The question is, can you do this? Would you have thought about this? Do you agree that I’m not doing anything beyond Newton’s laws when I do this? Yes, each one of you should ask. If you cannot do that, you have to deal with that now because it’s only going to get compounded. On the other hand, if you can do this, you know everything I know about Newton’s laws. This is all there is to Newton’s laws and how to apply them. So, you cancel the m. So, notice here, this is the normal force, but I’ve taken the y equation that says normal force is mg cos θ. Put that here. Then, I find acceleration is g times sin θ - μ sub-kinetic cos θ.
Okay, here is the result: ax is what I’m calling a now. I want to study this in various limits. The first limit I take is no coefficient of kinetic friction, I get back to my sin θ. Well, let me try the following interesting limits. I crank up the coefficient of kinetic friction, more and more and more. Look what happens here: sin θ and cos θ are fixed numbers. If this becomes larger than some amount, than say 10, what happens to the expression? a is negative. And what does that mean?
Student: It’s going uphill.
Professor Ramamurti Shankar: The block is moving uphill. And do you buy that? So, what’s wrong? I mean, I took the equation, right? I told you all the time how you can study various limits. Something is wrong. It says, if kinetic friction is very large, the block will move uphill.
Professor Ramamurti Shankar: That’s not the reason. Yes?
Student: It can’t be greater than the static one.
Professor Ramamurti Shankar: No, that’s not the reason. I mean, you can also vary the angles, maybe keeping k at some value, vary the angles so that this number times that can beat this. Then what happens? Yes?
Student: Where do you find the direction of sin θ downhill?
Professor Ramamurti Shankar: That is the point. Okay? Let me repeat his answer; that’s the correct answer. You’re right about kinetic being less than static and so on. But you can always imagine keeping that number constant but changing the angle to the cosine gets bigger and sine gets smaller. At some point, you agree, cosine will beat the sine, because sine is going towards 0 and cosine’s going towards 1. But the correct answer is, in writing this forced law, I’m assuming the force of friction points to the left, uphill. That is correct only if the body is moving downhill. But if you got an answer where a was negative, therefore it’s going uphill, the starting premise is wrong. So, one very useful lesson to learn here is when you apply formula, don’t forget the conditions under which you derived it, and don’t apply the result to a case where the answer does not belong to the domain that you assumed you’re applying it to.
So, the trouble with friction is the following. Friction is not a definite force with a definite magnitude or direction. The magnitude is definite, but direction is not definite. It is uphill if you’re going downhill; it’s downhill if you’re going uphill. It’s like the constant opposition to motion. There is force of gravity, the normal force, everything else is a fixed direction; you can draw it once and for all. So in a formula, where the forces are what they are in any context, you can apply the formula in any limit. But if the frictional force assumed a downward motion, don’t apply it when the block is going up.
Chapter 4. Pulleys [00:49:04]
Okay, so I’m going to do one more of these inclined planes and then pretty much done with the inclined plane. Okay, this is a problem of two masses. There’s a rope that goes over the pulley, let’s call this m, let’s call that M, and this angle again, it’s θ, and no friction. You can mix and match various things but let’s keep friction out because I’m trying to do something different here. So, you want to know what these guys will do now. So first guess is, well, M looks so impressive compared to small m, so maybe it’ll go downhill. But I realize that’s of course fallacy because M is a symbol; the value associated with it maybe 1/10th that of m. But if M really is bigger than m, then am I assured it’ll go downhill?
Student: The mass has to be greater. [inaudible]
Professor Ramamurti Shankar: No, no, I’m saying, if the big mass M is bigger than the small mass m, is that enough to say it’ll go downhill? Yes?
Student: No, it depends on the angle.
Professor Ramamurti Shankar: It depends on the angle because part of the mass is not helpful in going downhill. See, if I had a pulley like this, then you’re absolutely right, the big guy wins. Now, it’s not clear because this has got all of gravity pulling it down; this has got only part of gravity pulling it down. So, let’s skip this direction, it’s not very interesting. But let’s look at this direction, for which the equation is the following. If I just took that block, it’s got mg sin θ acting down, and this rope will exert a certain tension T, here, which I don’t know. Under the combined effect of these two, I can say mg sin - T = ma, where a is assumed to be positive going downhill. That’s one equation. Then, I come to this fellow here, this block; well, it’s got mg acting down and tension acting up. I’m sorry guys, I made one mistake here. I think you know what it is. These should be M, that should be M, sorry for that, and M (Mg sin θ - T = Ma). For m, the equation is T - mg. Now, let’s be careful with what I do next. I’m going to say m times a, which is the same a.
You should all understand why the acceleration in magnitude is the same. We’re not saying the direction is the same. This is accelerating downhill, at some angle; that’s accelerating uphill. These are not equations which are vectors; these stand for equations along certain directions. This is the up/down direction here, and that is up the slope, down the slope direction, for that one. But I think you realize if the rope is inelastic, if this mass moves down one inch, that’s got to go up one inch. Therefore, if it goes one inch in one second, so will the other guy, they’ll have the same velocity and they’ll have the same acceleration. So, that’s why there’s only one unknown acceleration, a. Once you realize that you’re done because now we know what to do. Whenever you see a -T and a + T, you add them and you get g times M sin θ - m. On this side, it’s going to be M + m times a. Can you see that? So, let me short circuit one step, and divide by M + m. So, that’s the formula here. And again, you got to ask, does it make sense? Now you notice that for it to be positive, and to go downhill, it’s not enough if M is bigger than m. M sin θ should be bigger than m; that’s because M sin θ is the part that’s really pulling downhill. M cos θ is trying to ram it into the inclined plane, and that’s being countered by the normal force. The second thing to notice is: can I use this formula when m is bigger than M sin θ? In other words, can I use this formula when a becomes a negative? After all, I did all my thinking saying it’s going down, but can I use it when it’s going up?
Professor Ramamurti Shankar: Yes, and the reason? Anybody want to say why that’s okay here? Yes?
Student: No friction.
Professor Ramamurti Shankar: Right.
Student: And friction is a force, depending upon which [inaudible]
Professor Ramamurti Shankar: That’s correct. All the forces I drew here, they’re not going to change their mind when you change the body’s motion. Gravity is always going to pull down. So, all these forces are fixed in magnitude and direction, and therefore once you got a formula for positive a, you can apply it to negative a, but you cannot do it when you have friction because in friction, you assume the direction of motion.
All right, so now I’m going to leave the plane. This is it. We are finished with the plane. You are not finished because, oh don’t clap here, I got a homework problem, by popular demand. Two masses, one pulley and friction. That’s to really say we’re finished with the plane. Once you can do that, you’ve graduated from this boot camp; you can handle anything else. But that’s not something I want to do in class. That’s something you do in the privacy of your room, put all these things in and crank it up. And you also have enough time. But that’s a very–that’s got some interesting angles you should think about.
Okay, so now I’m going to do physics that basically involves rotational motion. But let me do one Mickey-Mouse problem, and let it go, because I assigned it to you. The typical problem is: there’s a tree here, there’s a tree here, and that is–there are two ropes and there’s a backpack hanging here with some weight. You are told what that angle is; you’re told what that angle is. You are supposed to find the tension on these two ropes. I think it’s fairly simple so I won’t do the details but you know what you’re supposed to do. You take these two tensions, break them into the horizontal parts, with sines and cosines, and into vertical parts; horizontal parts that cancel because of this mass. This backpack is not going anywhere horizontally; the vertical parts of these two, which will be in fact additive, should cancel the weight of the object. That gives you two simultaneous equations for T1 and T2, and you can fiddle with them, and solve them.
Well, by the way I should tell you, if you got some equation like T1 sin θ 1 = T2, sin θ 2, or that - T2 sin θ is 0, and another equation involving T1 with some numbers and T2. Some of you seemed to have trouble the other day when there were sin θs here; whereas, if these had been numbers like 3 and 4, you knew how to solve these simultaneous equations. But I want you to realize that the solving of the simultaneous equations for T1 and T2 involve forming combinations of these coefficients to eliminate either T1 or T2. So, when you add them, one of them drops out. And that doesn’t depend on whether these are trigonometric functions or any other numbers, they’re eliminated in the same way. Likewise on the right hand side, instead of some numbers like 3 and 4, you have the weight, W, of this mass; well, if you multiply this equation by 3, multiply everything by 3 and juggle them. So, you should get used to eliminating coefficients which are not simply numbers, which are symbols, sines, cosines, but you know how to–If you had a sine here and a cosine there, for example, how will you combine them? Maybe you’ll multiply this by a cosine and that by a sine, and if you subtract it’ll drop out. You’ve got to do stuff like that. Okay? But it’s important that in a Physics 200 Level course you are able to solve those things.
Chapter 5. Friction and Circular Motion: Roundabouts, Loop-the-Loop [00:57:30]
Okay, now I’m coming to the interesting problems in which motion involves going in a circle. So here’s one problem. This is a string on which there is a mass and the mass is going around in a circle. This is like an amusement park, so that they have these dangling little, tiny, baby rockets and you sit there and start spinning, and instead of being vertical, it starts lifting up, and you want to know why it’s lifting up. So, if this angle here is θ, and this is some mass m, and it’s going around in a circle with velocity v, and the radius of the circle is say r; we want to find some relation between this angle θ and these other parameters in the problem. So what do I do? I apply F = ma, that’s it; and I’m going to get everything by applying F = ma, to this guy. So, here is that mass. The tension on the rope can only be along the rope. The rope is not a rigid object. It cannot exert a force perpendicular to itself. It can only pull it along each direction. That’s the tension on the rope. There is mg, and that’s it. Under these things it’s going to obey F = ma. So, what I do with this tension is I trade this guy for two forces. This is T. If that angle is θ; let me see, that angle is θ; so this is T sin θ, and this is T cos θ. So, you trade that oblique T for two Ts that are in the actual simple directions. The fact that it’s not going up or down vertically tells me mg = T sin θ.
Professor Ramamurti Shankar: Oh, I’m sorry, yes. Okay, then in the other direction T sin θ is the horizontal force, that’s got to be mass times acceleration. At the instant–but I’ve shown the object, because it’s going in a circle, it has an acceleration, in this direction, with size v2/R. Therefore, if you divide this equation by that equation you will get v2/Rg = tan θ, and that’s the angle that the string will assume. So, if you know the velocity you can do this problem.
Okay, then here’s another interesting problem. When you go on these roads, if you’re going in a circle–so for this is your car, I’m looking at you from the top, you’re going in a circle. Anything wrong? Going in a circle, you’re accelerating towards the center; someone’s got to provide the force, that’s what Newton says. That force is the friction between your tire and the road. In fact, it’s a case of static friction. You might think it’s kinetic because the car is moving, but it’s not moving in this direction. So, it’s a static friction that keeps you from slipping. So, you need a certain static friction; so that N times μ static – that N is equal to mg – must be equal to mv2/R. If you don’t have the static friction, your car will not be able to make the curve, it’ll fly off.
So, what people have done is to find a clever way in which you don’t have to have any friction and you can still make the turn; that is, to bank your road. So imagine, now you’re going into the blackboard and you’re turning left, and the road looks like this, going away from me. Now, I don’t have to draw this thing. This road is a track, a racetrack going around and round like this. And I gave it some angle θ. This road has no friction so it can only exert a force that way. But let me resolve that force into two parts, one like this and one like this. Again, if these two have an angle θ, that is the same angle θ there, and there is the mass of the car this way. So, what do I find? I find N cos θ = mg and N sin θ = mv2/R. Therefore, if you again take the ratios you will find tan θ = v2/Rg. I think this may be a little fast now but you guys can do the algebra.
What that means is the following. If you want the car to go around the bend, at a certain speed, 40 miles per hour or so many kilometers per hour, and the road itself is part of a circle of radius r; it doesn’t have to be a full circle. At that instant it has to be a part of a circle. Then, if you bank your road at that angle, you don’t need any friction to make the turn. You got to understand how we beat the system. If your road instead of being flat is tilted, then even though the road can only exert a frictionless normal force, now part of the normal force is pointing to the center. In fact, this picture is not drawn to scale. If I drew it that way, this part of the vector is horizontal and directed towards the center of the circle, on which you orbit. So, that’s how you do the banking. In any good system of roads you will find there is banking.
Okay, I’m going to do the last problem because it’s in the homework, and maybe I’m a little late today, I apologize for that. The last problem is a very famous, very important, and that’s the loop-the-loop problem. As you know, the loop-the-loop is when you come down on a track, you go on a circle, and for awhile you’re upside down. And the famous question is, “Why don’t you fall down?” What’s the trick? Does it violate Newton’s laws, and so on? We’ll find we can understand it fully with Newton’s laws. But we’ll find out that in order for this thing to pull it off, its speed at that instant has to have a minimum value. And the application of Newton’s law to this object is very simple. Here is gravity. What about the track on which it’s going? What’s the direction of the force the track exerts?
Professor Ramamurti Shankar: Pardon me? Down. Anybody said up? A track cannot exert a force upwards, down. So everybody, track, mg and they’re all acting down. So you see, that’s when you get a little worried. Okay, this guy’s up there, all the forces are down. So, we got to ask, so what does that mean? Well, I have to agree they’re all down. I say T + mg is the downward force, that’s got to be mv2/r. Therefore, T = m times (v2/R - g). So, what does that mean? If this T comes out negative, you’re dead, because if it’s negative, you want the track to exert an upward force and it cannot. In a real amusement park they have other T-brackets and so on to support you. But if you really believe in the laws of physics, you don’t need any of that, you just got to make sure this number is positive. For it to be positive, you need v2 bigger than or equal to Rg. So, you got to make sure when you’re on the top, you have this minimum velocity. If you go faster than that, that is just fine; if you go faster than that, T will be some positive number. So, here’s the interesting thing. Have we escaped the pull of gravity? How come we are not falling down? And the track is not helping us, it’s also pushing us down. Yes?
Student: The velocity vector is moving towards the left and the force; well, the full force is changing the direction of the velocity towards the center, but it’s still going along a circular path.
Professor Ramamurti Shankar: Okay.
Student: You can make that for forced radials today.
Professor Ramamurti Shankar: Yes. So, the point is, if you have an apple, with two forces, both pushing down, that apple is accelerating down, and accelerating down means really falling towards the ground, towards the center of the circle here. In this example, it is definitely accelerating, but the acceleration is not added to zero velocity, in which case it’ll pick up more and more speed; it’s added to a huge horizontal velocity. So, in a tiny time, you give it a little velocity like this, your new velocity points at a new angle. That just means your body has come here and has stopped going like that and started going like this. Therefore, downward acceleration does not mean coming closer to the ground, it only means your velocity vector is changing its direction. On the other hand, if you had no velocity vector, downward acceleration means what you think it is. That is why, if you drop an apple from here, it has an acceleration due to mg; if the track was pushing it down even more, it’ll fall towards the ground and hit the ground. Here it actually has an acceleration but it doesn’t mean that you are any closer to the center. Going around in a circle is an example of constantly accelerating towards the center but not getting any closer. That’s the way vectors work and you should think about that. Ok, so that’s it for now.
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