 Hi, I'm Zor. Welcome to New Zor Education. I would like to solve a little bit more complex problem related to friction. And this problem actually related to another problem without the friction, which I have presented in the lecture about inclined plane in the topic called Superposition of Forces in this course. Now, this course is Physics for Teens. It's presented on Unizor.com website. I suggest you to pay attention to the website because it contains mass for teens, which is mathematics for teenagers, kind of a prerequisite course for this one. And also, this course is presented and every course has a base menu and all the lectures are interconnected. So if you, by chance, found this lecture somewhere else, as a lecture only, as a video only on YouTube, for instance, just make sure that you understand that the whole course is presented on Unizor.com in a consecutive set of lectures. The site is free and there are no advertisements. Now, the physics course is still in the making, actually. So right now I'm in the mechanics and it goes for a few years probably until I finish everything. Okay, so today's problem is as follows. You have an inclined plane which is on some kind of a table. This is a table and it can slide back and forth. Now there is a mass of this inclined plane and on it there is some kind of an object which also can slide on it, on this side of the... Well, you can actually consider this as a triangular prism on the side. Okay, so there is a mass of the object on the top. There is an angle and, again, so far this is exactly the same problem as it was presented in the lecture called Inclined Plane in the superposition of forces topic of this course. Now we will complicate the picture. We assume that there is a friction here and a friction here. So this object slides along this inclined with certain coefficient of friction K1 and the whole slide, this inclined plane is sliding along this table, horizontal table with the coefficient of friction K2. So what we have to find out is, well, what happens with this inclined plane? Well, obviously as the object slides down, it pushes the plane towards that direction and our task is to determine the acceleration of this slide, of this inclined plane. All right, now, what's very important in any problem of this type is to determine what kind of forces are acting on what kind of object. Now, the easiest thing is, obviously, the weight. This has the weight. If m is the mass, m times g, where g is acceleration of the free fall, is the gravity of the earth, right? And for this object also we have m times g as the weight. Now, let's consider first this object, the small object, what kind of forces are acting on this object as it slides down? Well, obviously there is a weight. So there is a force down, which is equal to m times g. That's fine. Now, obviously it pushes on the surface of the inclined plane and the pressure is always perpendicular to the surface. So we have certain pressure, which this object, I will put it slightly differently. I will put it this way. Okay, this object pushes the supporting inclined plane with some kind of a pressure force, n1, towards this direction. Now, I assume that all my letters signify positive numbers. So if I'm doing this type of things, this is n1 and this is n1, it's not two vectors equal to each other. It's two vectors opposite to each other in direction and equal in magnitude, right? So obviously this object pushes the inclined plane with the force n1, but then the plane pushes the object in an opposite direction according to the Newton's third law with the same force. So what is the trajectory of this particular object? Well, if this inclined plane is fixed, then it would be parallel to the inclined plane, basically. I mean along the inclined plane. But since as it moves here, the plane itself is pushed towards that direction, the resulting force, which will move this particular object, will not be in this direction, it will be tilted down, right? All right. Now, what can we say about this particular object? I would like to consider the reference frame, which is convenient for the movement of this particular small object. And my reference frame would be as follows. Let's say this is the beginning, the original position of my inclined plane, but eventually it will move that direction, but that's of no importance for us at this particular time. This is the system of coordinates, which I'm going to consider right now. This is x, and this is y. Now, that's more convenient to investigate the movement of this particular object. And let me explain you why. Because if I will consider only the forces which are working along the y-axis, I can very easily construct the second Newton's law for this particular object. Here is why. In this particular direction of the y-axis, what kind of forces are acting? Well, obviously, towards positive y-direction I have a reaction of the inclined plane, which is n1. Now, towards the negative side relative to the y-axis, the component of the mg is working since it's a negative direction, I will put a minus sign, mg times cosine, right? Cosine of phi, because this angle is obviously phi. So if I'm talking about projection of this vector, vector of weight of this object onto this axis, it will be something like this, which is hypotenuse times cosine of the phi will give me that particular force. So, I shouldn't really put it here. Now, n1 and mg times cosine have two different values, obviously. So these are components of my forces which I have projected on my y-axis. Now, what is the movement? Well, the movement is equal to, obviously, mass of this particular object times acceleration in the y-direction. Now, how fast my object is moving along the y-axis? Obviously, there is a component towards x-axis, the real trajectory of this object would be something like this, I guess, right? Down on one hand and to the right on another hand. If the plane was stable, then the trajectory would be along the plane. But since the plane moves, it will tilt downwards. Alright, so what is my acceleration in the vertical, this is trajectory, so acceleration towards within the y-axis would be a projection of acceleration on the y-axis. Now, let's just think about it. It's really very simple. If my plane moves with acceleration a, which we have to determine, that's our task, then the vertical movement perpendicularly to this, I shouldn't say vertical, perpendicularly to the incline, in the direction perpendicular to the incline, which is the y-axis, what will be the movement of this object? Well, let's just think about it. Forget about how it moves along the x-axis, which means within this plane. Down it will be this way. Okay, this is my incline plane and this is how it moves, right? Now, what's the vertical component of this? Now, if my plane moves here, my vertical component, as it moves down, should be a times what, sine of phi, right? The component would be this. Sorry. I'm talking about only perpendicularly to... So, if this is d, this is how my incline has moved to the right. Now, if I will completely forget about movement along the incline line and I will consider only the perpendicular, this would be my next point, right? So, if this is d, then what is this? Yes, then this is d and this is phi. So, this is hypotenuse times sine would be... Yes, I'm right, sine phi, sine phi. So, what I'm saying is that as my incline moves to the right with acceleration a, the point on the incline would move vertically by a times sine phi and this is basically the explanation. This is displacement of the plane and this is displacement of the point which is moving vertically. So, the vertical, not vertical again, it's perpendicular to the plane. Sorry about this. So, the perpendicular to the plane movement is always contributed only to the movement of the entire plane. So, we have an equation in this case which has two, unfortunately, unknowns. But that's okay, that's just one equation. Let's remember it and let's move forward. Now, I would like to switch to my incline and the laws related to its movement. Well, that is a little bit more complicated. Now, what kind of... Now, my reference frame, since my incline moves horizontally, this would be my x-axis and this would be my y-axis. That's more convenient in this particular case. Alright, so, what kind of forces are working along the x-axis and along the y-axis? So, whatever the forces are working along the y-axis must nullify each other because our slide is not moving along the y-axis. It's moving only along x-axis. So, all the forces which are acting on this particular object in a vertical direction, now I can say vertical, means y-direction, okay, they must nullify each other, right? Because the object does not move vertically. So, let's just check what kind of forces are involved here. Now, obviously, there is some kind of a weight which is Mg. Also, we have certain pressure which this object exhorts on the table and the table should push backwards with the same force. So, I will put it here. M2 is a pressure and obviously, again, I should put it on a table. So, I start here. But the table pushes back with the same force M2, okay? Now, is this M2 by absolute value, by magnitude, equal to the weight? The answer is no. M2 is actually greater than weight, which means I should really put my thing should be shorter so my picture would be more realistic. This is Mg. Now, why this M2, why the pressure should be greater than the weight? Well, obviously, because this particular object is pushing down as well, right? And it adds its own weight. It adds to the weight of this inclined plane, right? But again, we should be very, very careful about how this weight actually is applying because it goes down but then there is an incline which means that the pressure goes perpendicular to incline and then from that component I should get only the vertical side, right? So, it's a little bit more complicated here. So, this M1, the pressure which is exhorted on the incline, it must be represented again as vertical and horizontal component. This is very important. So, this M1 must participate in both vertical movement and horizontal movement. This is the vertical component and this is horizontal component. So, we are talking about vertical. So, this thing is supposed to be added to the movement down. So, let's just summarize all together. So, we have M2 up. We have Mg down. I will put it as a minus sign. Now, what else? Well, now we have, on one hand, we have this particular force, right? This is the weight, basically. Well, not the weight, the component of the normal pressure and it goes down as well. So, it's minus M1 times, if this is phi, so it's a cosine. And then there is one more component which we really did not talk yet. But it's very, very important to participate in the whole thing. Look at this this way. This object is sliding this way, right? Well, obviously there is a friction. Now, what does it mean that there is a friction? Well, if both are moving in both directions, it means that the incline is pulling against the movement of this guy, right? And the object is pulling back in its direction, in its horizontal direction. It pulls the incline back. So, both are, the friction is slowing down both of them. It slows down this guy in its direction this way and it slows down this way in its direction in its trajectory. So, what is this force of friction which we have to get into our equation? Well, obviously this force of friction is directed on the object this way. It slows down and on the incline plane, it slows down in its direction this. But it's supposed to be the same, it's parallel. So, it's parallel to this incline plane. And it's equal to N1 times K1, right? Because K1 is the coefficient of friction. So, since N1 is the magnitude of the pressure, then N1 times K1 is the friction force which slows down this guy on its way on its trajectory and slows down this incline plane on its trajectory. So, this is exactly the same as N1K1. And what's very important, this force N1 times K1 has both vertical and horizontal components, right? And vertical is, this is angle phi. So, vertical is N1K1 times sine, right? And it's supposed to be with a minus sign. So, minus N1 sine phi times K1. And what's very important, it's supposed to be equal to zero because all these forces in summary, all these vertically directed components of our forces are supposed to be equal to zero because the whole pyramid, not pyramid, incline plane is moving only horizontally, there is no vertical movement. All the vertical forces must be equal to zero. So, what did we do actually? We have added another equation which is good. But unfortunately, we added another unknown which is N2. But don't despair, we still have horizontal movement which we can analyze now. The horizontal movement eventually should be equal to mass of this incline plane times its unknown yet acceleration. And it's supposed to be equal to sum of these four of all the forces which are acting horizontally. So, what kind of forces are acting horizontally? Okay, well, first of all, which is very easy, if you have the pressure N2, then backwards is the force which is equal to N2 times K2. Where K2 is the coefficient of friction. So, if you have a pressure N2, then there is a force going this way. And it's supposed to be with a minus sign because it's acting against the movement, right? All right. Now, what kind of force is also acting on this particular object? What force is actually to move this way? Well, obviously this pressure, this is N1, and its component, which is this one, moves the object this way, right? Which is what? It's N1, this is phi times sine. So, this is N1 times sine phi. Okay, what else? Again, don't forget that we have this friction which has both vertical and horizontal component. We have already encountered the vertical component here, right? And horizontal component acting against the motion. So, this guy, as it slides down, it slows down the motion of the inclined. So, the hypotenuse is N1 K1, and we have to multiply by, this is phi, right? By cosine, I guess. So, it's minus. Minus because it moves backwards. N1 cosine phi and K1, right? Now, look at this. Three equations with three unknowns, and what kind of three unknowns? A, N1 and N2. Pressures, right? Again, all numbers are positive, and I'm using these minuses wherever it's necessary to signify the direction. So, it's really three positive unknowns, and the system of equations is linear, you see? N1, N2, N2, A, they're all linear. So, everything else is known. So, sine and cosine are known. Mg, capital Mg, everything is known. K1, K2. So, it's a linear system of three equations with three unknowns, and you can actually solve it using a trivial methodology, something like, for instance, you can find N2 from this and substitute into this, and that immediately goes to only two equations, which is N1 and A. And then you can find N1 as a function of A, substitute into this, and you will have only one equation, linear equation, with A. Granted, the result will be huge, but it's not really important. I mean, I'm not going to do all this little algebra solving linear equation. It's not the purpose of this lecture. The purpose of this lecture is to explain how many different forces are playing together in movement of that thing. And especially important, I would like you to pay attention to this friction force, which is actually acting like slowing down force in this particular direction on this inclined plane, and it has vertical and horizontal component. This one is adding into our vertical movement, and this one is adding in the horizontal movement. So, well, that's it, basically. Obviously, it would be a nice exercise for you if you will just wipe out this thing and put it on a piece of paper, your own drawing, and derive all these equations using relatively the same logic as I'm just suggesting to you, and solve the equation. I think I will probably put on the notes for this lecture. I will put the answer, all this huge formula. Maybe not, I don't know. But in any case, the most importantly is to have this system of three equations with three unknowns, and you have to understand really how to derive these equations. That's a very important exercise. I mean, if you understand that, basically the whole mechanics of all these first, second, third law of Newton, friction, etc., will be absolutely no problem for you. That's probably one particular problem if you completely understand it, all the forces and how they act, at what angles, etc., and how to derive these equations. If you understand that, it means there is absolutely no problem in mechanical movements for you. Okay, that's it. Thank you very much and good luck.