 Hi, I'm Zor. Welcome to a new Zor education. Today's lecture is about motion on the inclined plane. Well, this lecture is part of the course, which is called Physics for Teens. It's basically a high school level course of physics, which I am recording right now, one lecture at a time. Right now we are in the middle of Mechanics part and there are others. This course is presented on the Unisor.com website. If you find this lecture on Unitube or anywhere else, I do suggest you actually to switch to Unisor.com. It's a free site without any advertisement and the advantages of listening to this lecture through this website is that, number one, you have a complete picture of whatever the topics will be provided. Number two, every lecture has detailed notes. Number three, there is an educational process if you would like to get involved in it for either self-study or supervised study. And it allows you basically to take exams and if you are working just by yourself or with a supervisor, you can take a look at the results of your exams and do something about it. Anyway, let's get back to our motion on the inclined plane. First, let me just draw a picture which would probably explain everything. This is something which we call inclined plane and this is an angle phi. So it's a plane which is not horizontal but at a certain angle. And there is an object there which obviously has certain weight and it slides down this inclined plane. Well, sometimes I use the word inclined plane, sometimes I'm using the word slide because it really looks like a slide on the children's playground. So on the slide, this object is sliding under its own weight. So my task is to determine how fast it will go down. Now, obviously, if the angle phi is greater, it's steeper, it should go faster, right? And if angle phi is 90 degree, which means basically there is no inclined plane because there is nothing which holds the object from underneath, the object should fall down with acceleration of the free fall, which is on planet Earth called G, which is approximately 9.8 meters per second square. Alright, so let's see. So in our extreme case, this would be the maximum. Now, in our other extreme case, when phi is equal to zero, basically we have a horizontal plane, right? Then the object will just sit by itself, right? You just put something in the horizontal plane and nothing happens. It just does not move. The weight is completely compensated with the reaction of the supporting plane, right? So in this particular case, in the middle between these two extremes, we understand that there should be some kind of reaction force which should force our object to slide exactly on this plane, on an inclined plane. So the combination of this force and this force, the weight and the reaction of the supporting plane are the only two forces which are acting on our object. And their resultant force should be directed along the slope, right? So this is supposed to be parallelogram. This, this, this and this is supposed to be parallelogram, right? Because if it's not, then our object either would fly off the inclined plane, which does not happen. Now, if it's turned this way, it should actually go through the plane, which also doesn't happen. The reaction force is exactly of such a value that this is parallelogram and the diagonal is directed along the slope, right? Okay, now we do know that this angle is straight, right angle, sorry, right, right angle, right? Why? Because the reaction force is always perpendicular to the supporting plane. So no matter how the plane is positioned at angle or whatever, the reaction of this supporting plane always on a perpendicular or we're saying normal to the plane. Okay, so this is normal to this plane. So this angle is right angle. Alright, now if this angle is right angle, then this angle is also right angle, right? This one. Which means if this is w, this is the right triangle and we can actually determine two different forces. One is the resulting force, this one, which actually pushes the object down the slope, right? Well, if this is phi and this is the right angle, then obviously this is phi as well, right? Because this is perpendicular to this, this is perpendicular to that. So if two sides of one angle are corresponding with perpendicular of two sides and other, then the angles are the same. So this is angle phi. This is w. Now when I'm talking about the value w, obviously it's a vector, but if I don't put this vector thing on the top, I actually mean that we all know the direction of this vector and I'm talking only about its absolute value, about its magnitude. And this positive, because if it's directed somewhere else, then I will use not in the positive direction of some axis, but in the negative direction I will use the minus sign basically. So we're assuming that all values are positive, but we do know that their direction is such and such. In this case, this goes down and this goes this way. Alright, fine. So in theory, we can very easily determine the strength of this force, the magnitude of this force, because it's just the right triangle and this obviously w times sine of phi, right? This ketitus is hypotenuse times the sine of this angle. Alright, fine. Now this is the force which moves, let's call it wf. Now what other forces are involved here, which I didn't really mention before. Now obviously if this is a reaction force, it's supposed to be a reaction on something, on what? Obviously on the pressure which this object exhorts on the plane, which must be exactly equal to in magnitude to this r. Well, it's not equal maybe on my drawing, but in any case, it's supposed to be equal. So this is wr. So this force which goes down can always be represented as the sum of this, which is wf, and this, which is wr. Which is r, I'm talking about again magnitude, that's w times cosine of phi, right? Obviously. Because this is w, so this is w, this is also angle phi, so r is equal to w times the cosine of this angle. Well, basically that's quite sufficient to determine how our object will move, because if I have the force which moves it down the slope and I have the mass, well, if I have the mass, I don't have the mass, but if I have the mass, then this force must be equal to the mass times acceleration, right? That's the second law of Newton. Now on the other hand, since I know what wf is, it's w times sine f, fine, and I know that the weight is related to mass and free acceleration, because weight is basically the force which forces the object of mass m to go down, vertically down with acceleration g, right? So it's mg times sine f, fine, sine phi. So from this and this, we deduce that a is equal to g times sine of phi. So it doesn't depend actually on mass, which we don't know. The acceleration depends only on some constant and the slope. Now let's just analyze. If phi is equal to zero, which means my slope is not really slow, but the horizontal plane, sine is equal to, zero is equal to zero, so I have no acceleration, right? Which is right. I mean, the object will just lie on the flat surface. If my sine is equal to one for angle equal to 90 degree, right? Then my acceleration will be just free falling, g. And that's exactly what corresponds. Now let me just make a little logical deviation. I did not talk about the reference frame. I'm kind of assuming that the reference frame is my y-axis goes down and my x-axis goes to the right, right? I mean, not down, but vertically. So I assume basically this is my directions. This is x and this is y, which is fine. In this particular case, it's fine. However, I would like to mention one might actually be a simplification, if you wish. You see, there are usually many forces which are acting on the object. Now in this case we have two forces, right? Now the reference frame might actually result in easier approach if the reference frame is catering to the movement of the object, not to the forces. Now a major force is the gravity. So my system of coordinates is basically catering to the fact that my gravity looks simpler. It has only the y-direction and no x, right? So it's vertically, so there is a y-coordinate. So if I'm talking about the vectors, it's basically zero comma w, right? So no projection on the x-axis and there is only projection on the y-axis, which is equal to an absolute value of w. And if I'm using w as my absolute value, the magnitude, then I really should put minus here because the force of the gravitation is directed downwards, right? And my y goes up. So that's kind of a natural thing. However, it's not the simplest thing. You might consider slightly different system of coordinates. And here is the one which I prefer in this particular case. And when we will talk about the next problem, you will see how much more important this is. I prefer to simplify the movement of our object. For this, knowing that the object is moving parallel to the slope, I will use this slope as my x-coordinate. And perpendicular to this would be y. It seems a little unnatural from the first viewpoint. However, it greatly simplifies the movement of our body. Now, I was calculating the acceleration along the slope. What if I would like to measure acceleration as a vector which has certain coordinates? Well, it would have two coordinates in the old system of coordinates, right? Because it goes under angle, so it will have some acceleration along the x-axis and some acceleration along the y-axis. Now, in this particular case, with this particular coordinate system, I do have only movement along the x-axis. And there is no movement along the y-axis. It's always equal to zero, which is a little bit more convenient. Now, what's less convenient is our weight. Now, the weight becomes a little bit more complicated. But reaction force is not. So, if I'm talking about real vector algebra, I'm simplifying a little bit more. I'm simplifying movement and the reaction force. And I'm still a little bit more complicated with the weight. And it does make sense, because in this particular case, now I can say that my acceleration has only x-component, and it's equal to whatever I was just calculating, which is g times sine of phi. And my y-component is zero. So that's my vector of acceleration of this object. And it's a little bit easier to represent it this way. And now, well, basically, that's it about this particular problem. And I would like to switch to a little bit more complicated, where these coordinates will be really important. Now, consider we have exactly the same setting. However, there is one important consideration. This is not an inclined plane or a slide, which is, well, fixed on the surface of the earth anymore. Let's consider that it sits on some kind of a table. Table is fixed, but this thing can slide without friction along this table. Now, why is it more complicated? Well, just think about it. If the object forces the whole thing down, well, it means that this slide might slide from under the object to the right. So the object will not move exactly parallel to the x-axis, and I'm still retaining this coordinate system because it will be easier for me. So the object is not moving this way. It's moving basically under some kind of an angle. Why? Because on one hand, it slides the slope, and on the other hand, the slope goes from under it. So the movement would be like this, basically, right? So that complicates the picture, significantly complicates. But we can say now that the reaction force is no longer, the reaction force is no longer the WR with a minus sign, where WR is basically W times sine of this angle. Fine. And because of this imbalance, you see, if you are pressing something on the plane and it's not moving, then the weight is actually exactly equal to the reaction force. But if you press something on the plane and the plane by itself is moving down, the pressure would be less on the plane, and that's why the reaction force would be less. Now, if this is an incline plane, again, I'm not only slowing down by sliding down, my plane goes this way, and that basically weakens the pressure, because the plane goes from under the object. So, R now, by its absolute value, is less than W sine pi. So if this is WR, like here, you know what, let me just draw it a little bit better. So, again, we have this thing on the table. I'm still using this as an x-axis, and perpendicularly to this would be y. Now, this is my object. So this is the W, and this would be my WR. This is the real R, and this is minus R. It's smaller by magnitude than this, because the plane slides out. Well, that complicates the picture, because right now I don't have the reaction force. I know that reaction force and the weight are two components, and now this thing goes this way. The resultant of this force and this goes this way, and that's the real trajectory in my system of coordinates of the object, because, again, object goes down, but slope goes to the right, and that would be the real trajectory. Again, that complicates the picture. It's still parallelogram, but it's no longer right triangles here. So, what do we do? Okay, let's just think about it. And here my system of coordinates plays a very important role. Let's just think about the movement of this object separately as a projection to the x-axis and as a projection to the y-axis. Actually, my first is how it's projected to the y-axis. Well, let's just think about the forces which are acting along the y-coordinates. Now, what are these forces? Well, actually I have this force, reaction force, right? And I have a projection of the weight on this line which is parallel to the y-axis, which is WR. So WR is acting along the y-axis and WR is acting along the y-axis, right? So along the y-axis, my forces are to the negative side, it goes WR and this is negative, right? Because, again, I'm assuming that all these numbers, numerical values are just magnitude and that's why I put minus because it's to the negative direction of the y-axis. Y-axis goes this way, my force goes this way. Now, R, on the other hand, is a positive and my acceleration is negative actually, right? My AY is negative, so if all these numbers M, AY, WR and R are positive then I have to put these numbers, these minus signs to compensate. Okay. Now, what is WR? Well, WR I know what it is because this is really projection on the y-axis so this angle is right angle. So I know that WR is equal to WR times cosine phi, right? This is phi, this is phi and WR is WR times cosine of this angle. Well, R is R, I can't say anything about R right now but I do can say something about AY. Well, let's just think about it. By how much in the direction of the y-axis my object moves if my entire plane slides this way? So let's consider my entire plane slides this way. So from this it goes to this. So this is the movement of the plane. Now, what is the vertical, what is the component of the movement of this object in this particular direction? So if this moves by D, how much this is moved? So this is D, this is the same as this, this is phi so this is D times sine of phi. So for every movement of the plane to the right by distance D my object, if it does not have this component along the x if it has only y-component, I'm talking only about its y-component the y-component moves by this. Now, if the distances are related this way so the speeds are related because they are derivatives from the distance and the acceleration are related this way. So this is equal to a plane times sine of phi but this should actually be very clear because it's very important for everything else. If a is acceleration of the plane and we are talking about only in this direction because this is a table on which it slides and the table is horizontal, it doesn't move anywhere so there is a reaction force which prevents it to jump up or down it's only to the right. So as this slide moves from this position to this position vertical component of the movement of this object only vertical, I'm not talking about because any movement like this one has two components along the x-component which is this and along the y-component which is this perpendicular to the plane. That's why it's important to have this system of co-ordinate one is along the plane and another is perpendicular because this is really very important we are projecting everything on the y-axis which is perpendicular to this plane. So this would be my acceleration of the object the y-component of acceleration of the object in terms of acceleration of the plane and again its angle. So let's just leave the letter p here that would be easier and now I can write this particular equation in the following form minus m A p sin phi equals 2 r minus r minus w r and w r is w times cos i. Okay, this is a very important equation and we will call it equation number one. Now what's wrong about this equation but there is nothing wrong about this but what's insufficient because we have two different unknowns r and acceleration of the plane. Well let's assume we have to determine acceleration of the plane and this is the whole problem actually. What is my acceleration of the plane as the object pulls down? Well obviously I don't know mass that's important and I don't know the mass of the plane this slide because obviously it's important consider it this way if the slide itself is very massive well most likely the small object which is on its surface it will slide down but it will move only slightly the slide acceleration will be very small because the force is basically the same right so acceleration will be smaller so it actually depends on mass it depends on the mass of the slide itself well we'll see what we will need for complete solution but actually these are kind of complicating factors now let's move on now since I'm actually talking about acceleration of the entire plane I have to somehow relate it to again the second law of Newton I need the force which is actually pushing it in this direction well let's just think about which force is this well obviously it's related to the force exorted by this object and related to its weight but the force actually is only the pressure which is equal to minus r as far as the vector is concerned so the pressure from the object is basically the same as the reaction force of the plane itself into that direction so all I can say is that we are pushing with the force r but in this direction that's why it's minus okay but plane goes this way now how is it supposed to be working well let's just forget about the object and let's just think about only the plane so this is my inclined plane this is my object which is pushing this with the force equal in magnitude of r now obviously this is support why would my slide go this way well obviously I have to represent this as perpendicular and horizontal and the horizontal component pushes the slide to the right and the vertical component is balanced by the reaction force of the table now this is still an angle phi and this is angle phi so my horizontal component is my force of pushing the pressure force which is equal to r in magnitude and the horizontal component would be times what? times sine and phi this is just right triangles so the horizontal component this is phi, this is phi so it's hypotenuse times sine would be the catheters and this is equal to what? well this is equal to mass of this inclined plane the slide itself times its acceleration in this direction the same thing as this one we are talking about this horizontal acceleration now it's much easier now we should really assume that we must know mass and mass so remaining unknown are acceleration of the plane and the reaction force so we have two equations with two unknowns and all we have to do is to solve these two equations to find out what is my r what is my acceleration of the plane and r as well so how can we do that? well it's simple, it's linear anyway so r is equal to from here m a p divided by sine of phi putting it into this we will have minus m a p sine phi equals to m a p divided by sine phi minus now w is again m g because w is a weight this is a mass this is a free fall acceleration times cosine phi well from this we obviously have to multiply it by sine that would be square that would be sine here now what is my a p? a p is here and here so a p is equal to this goes to the right this goes to the left so it would be my m g sine phi cos sine phi divided by m plus this minus goes this way m sine squared phi alright so this is my a r I mean a p from this we can obviously find the r if you want r is equal to m a p divided by sine phi so it would be m m g we will divide by sine phi so it's only cos sine phi and m plus m sine phi squared phi so we have found the reaction force and that's very important now this is the reaction force if you remember from in the y direction right this is reaction force and this is my system of coordinates this is x and this is y this is r so we have found the magnitude of the reaction force what's the x component x component is 0 because it's perpendicular to this particular line right the reaction force is perpendicular to a slope so that's why it's very important actually to use this particular system of coordinates it would be probably possible to do a little bit more I would say habitual system of coordinates like this one vertical y and horizontal x but it's easier in this case that's all I'm saying so we have found these guys and since we have completely expression of r in this coordinates and we obviously have the weight as well it's time sine and sine times cosine we have two real forces so we can find not only the horizontal but also the vertical horizontal in this system along the slope and perpendicular to the slope so we can find the real movement which I'm not going to do it it's kind of a simple thing anyway but the most important is basically this to find the acceleration of the slide because that's what's new about this problem relative to the previous one where the slide was fixed alright so this was a presentation of a little bit more complicated mechanical problem and you know what are probably complicated even further in one of the future lectures when I will introduce the friction we will have some friction friction of this object on the surface of the slope and then the friction of the slide on the surface of the table so with these complications that would be a really big formula as a result in any case that's it for today I suggest you to look at the website dot com go to the physics 14 and to mechanics and to dynamics and this is super position of forces that's one of the lectures in that particular part of this course read the explanation I think it's like textbook basically so you have the advantage of having a textbook and the video lecture together so I do suggest to go again through all this material that's it, thanks a lot and good luck