 Hi, I'm Zor. Welcome to Unisor Education. I will continue talking about superposition of the functions, which is basically called the principle of superposition of the forces, not functions, forces, in dynamics. Now, this lecture is part of the course called Physics for Teens, presented on Unisor.com website over there. I do suggest you to watch this lecture from the website rather than from, let's say, YouTube or some other source because every lecture has very detailed notes and also the website has certain educational functionality. Like, for instance, if you are working with a supervisor or a parent, they can actually enroll you into a particular part of the whole course, or the whole course, whatever, and they can take a look at the exams, which you can really take. There are exams in the course. Now, also there is a prerequisite to Physics for Teens. It's called Mass for Teens on the same side. And there is even another course, which is Acidics course, US Law for Teens. Now, the website is free, no advertisement, so I do suggest you to take a look at this. So, now back to principle of superposition of forces. So, we know that if you have more than one force acting on an object, then we can always replace the result of acting these forces, results of the action by vector sum of these two forces. Now, obviously, these forces are collinear, they are acting in the same direction, then the absolute value of the result will be the sum of these two forces. That's the easiest part. In more complicated cases, we can always, as I was saying, replace the action of two forces with one being their vector sum. Now, well, from this obviously we can extrapolate to n forces by again combining them together according to the rules of vector algebra. Now, there is one very important consequence from this. Not only we can add two forces together to get their resultant, that's the name of the resulting force, resultant. So, not only you can combine these two forces into one, you can always, if it's convenient for some reason, you can always represent one force if there is some one force acting on the object as a sum of two or even more forces which, if combined together, give our force. So, instead of these two I can put their sum or instead of this one I can put these two forces and the result will be the same. And under many scenarios in the real life that's very, very convenient to do this. Now, what I'm going to do is I'm going to present a couple of examples which really show how this particular approach of representing one force as the combination of two really helps us to solve certain problems. Alright, so my first problem is related to the movement on the slope. So, there is a slope phi, no friction there is an object. Now, my purpose right now is basically to tell how this object moves on this slope under whatever the conditions I have and what kind of conditions I have. Well, first of all, there is a force which pulls down this particular object. That's its weight. So, the weight goes down. Let's call it W. Now, is that the only force which acts on this particular object? Well, obviously not. Because if that was the only force it would just drop through this slope down. So, there are some other forces which keeps the object on the slope and obviously the object will slide down the slope. Well, we all know from the children's playground, for instance, they have a slope. A child sits on the top and then it goes down the slope. Why? I mean, obviously the force which gravitates down to earth, its weight, obviously it plays the important role. But again, it should be something else. So, what other forces are acting on this object which result in its moving along the slope? Well, obviously there is something which we call a reaction force. Now, whenever somebody is or something sits on the slope, we also have the pressure which this particular object makes on the surface where it's supposed to slide down. So, there is this direction which is always perpendicular to the surface. And along this direction I have the force which this object acts on this particular surface of the slope. Okay, but this is the force which acts on the slope itself. And slope is fixed on the ground so it's not moving, obviously. Now, this force is not the force which acts on the object. So, which force is acting on the object? Well, obviously reaction force. You know the third law, Newton's third law. Whatever the pressure the object exerts on the slope, the slope goes back with the same force. Okay, fine. So, we know this. The problem is, we really don't know the value of this force. But now let's think about it this way. We know that the object will move downhill. So, this is the direction the object will move. Now, we know that there are two forces, this one and this one. The weight and the force which we can call reaction force. Now, the combination of these two forces, obviously is the force which goes down the slope in this particular direction. So, we know that this force goes vertically down and we know its magnitude and direction. We know this force goes perpendicularly to the slope and we know its direction. And we also know the direction of the resulting force. This is direction, the resulting force. So, some of this and this should go down this way. Now, how do we actually make the combination of two forces in this particular case? Well, from this point, I have to put the line. We have to build the parallelogram, right? So, how to build the parallelogram? If you basically know the side, you know this angle. So, you have this direction and you know the diagonal where the result is supposed to be. Well, obviously, parallelogram should be built this way. I draw a line parallel to this one and from this point, I draw a line parallel to this one, right? So, that's my force WR which is the reaction force and this is my force which I can call WF force which pushes my object forward. That's how I built these two forces this one and this one in such a way that it makes the parallelogram because the parallelogram is the way how we add forces together. Now, if this force is shorter I will not be able to build the parallelogram in such a way that diagonal goes along this direction, right? So, if I for instance do it from here, if this is where my reaction force ends and I will build the parallelogram this would be my direction of the result, right? So, that's wrong. Now, this is the only way. Okay, fine. So, let me wipe it out, draw a nice picture my sir picture. I will never be able to draw a nice picture. But anyway, here is the real thing. So, this is my slope, this is my object this is my real force this is down, this is perpendicular and this is my, for instance, this is my weight. So, this is my forward and this is my reaction. So, that's the real picture. Alright, what do we know about this? Well, we know this angle, right? And which means this is the same angle, right? Now, this is the right angle, this is the right angle, right? Because this is perpendicular to the surface of the slope. So, basically I have the right triangle. I know diagonal, that's the weight of the object it's given how can I find out this force, which basically is the resultant force which makes the object moving. So, obviously WF and I'm talking right now not as a vector, but as an absolute where it's a magnitude of this vector, right? So, if I don't put this it's considered to be the magnitude of this particular vector. And obviously it's written as what W times sin phi. Now, this is obviously W cos sin phi. Now, let's be a little bit negligent about frame of reference. I did not really mention what exactly the frame of references. So, I'm talking about magnitude and I'm talking about this particular picture. And this is basically sufficient to write the equation of the motion. Because if I know the force, then my equation of the motion is M times A equals to where M is the mass of the object. A is the acceleration down the slope. Now, this is fixed, right? Weight is fixed and angle phi is fixed. So, my force is constant, which means my acceleration is also constant. Mass is constant. So, this is equal to WF equals to W sin phi. So, this is the most important part. Well, A obviously from this is equal to... Now, what's very important is, and I will definitely go through this when I will talk about gravitation, the connection between the weight and the mass on the surface of the earth. This ratio weight divided by mass is equal to 9.8 meter second square. This is the acceleration of the free fall. So, any object will fall with this acceleration because the gravity actually pulls down within certain, obviously, precision. Because obviously, if you go very far away to the space, that would not be exactly the same. Even on earth, if you go to the top of the mountain, the acceleration will be different. This is on the ground. It's some kind of an average. So, and it's called generally G constant, G, the gravitation constant. So, this is equals to G times sin phi, where G is a known constant. Okay? So, it depends on the angle phi. What's interesting is let's just do some kind of analysis. If the phi is equal to zero, which means slope is actually flat. Well, if the slope is flat, my weight goes down, my reaction force goes up, and there is nothing which moves down the slope. And obviously, this gives us, if phi is equal to zero, then this is equal to zero. There is no acceleration along this slope. On the other hand, if my phi is steeper and steeper, if it's 90 degree, then basically this particular object doesn't really press. You see, if the slope is 90 degree, the object on it doesn't really press. Everything goes down. So, the whole force of weight and the whole acceleration will go to, because sin of 90 degrees is equal to one, so it will go to complete G. So, acceleration will be down as if it's a free falling, because the slope doesn't really resist the movement. It's vertical, right? Well, that's it. That's it about this particular problem. And we will continue with the next problem. Now, my next problem is related to the following situation. Many of you probably saw it in Circus or somewhere else. It's kind of a trick. You have a motorcycle which goes into the loop and then it goes very fast around the loop. So, I would like to examine how this particular thing is working depending on the initial speed. Well, considering it's a constant speed, so v is the speed, constant speed, as it goes along the circle. So, it's basically a linear speed of the motorcycle as it goes. Now, obviously if the radius is equal to r, this is equal to r omega, where omega is angular speed. How many regions per second it makes, right? The angle. Okay, so this is given, but based on this number we have to really make all these calculations. But not everywhere. I'm talking only about two points at the very top and the very button. But I'm mostly interested in kind of a pressure this particular motorcycle exhorts on the loop at this point and on this point. Now, intuitively seems obvious, but as it goes down this way it's kind of pressing down and the pressure should be greater. When it goes up, the pressure should be smaller. It's like levitating a little bit, right? So, let's just examine this using the principle of superposition of forces and our representation of certain forces as a combination of two or more other forces. So, let's talk about, for instance, the top of this. First of all, I would like actually to make a little bit more precise definition of the frame of reference. Let's consider the frame of reference is this. So, this is the z this is 0 and this is x. Okay? So, we are talking about this particular point now y coordinate is this way and it's 0 right? So, we're not talking about y point. Alright so, what can we say about the object let's say a mass m and weight w at this particular point? Well, let's examine the forces which are acting. Now, obviously there is a force which is weight. Okay? By the way if w is my absolute value of this particular force, the magnitude of this vector, the vector goes down against the z. So, basically the vector is in vector mode it's with a minus sign. So, if I would like to combine whatever forces I have using their absolute value, I have to put them with some kind of a sign. Okay? Now, what else? Well, obviously this particular motorcycle is pressing on the loop with some force, right? Now, this force is directed from the object to the loop. And again, let's put it p or whatever pressure and again the loop presses on the object according to the Newton's third law with the same force it would be minus p. The only thing I will put the letter t which means top and this will be bottom. Alright, so now the reaction force from the loop will also go down and that will be therefore minus pt. Now, at the same time we know that whenever my object is moving in a circle with constant angular or linear velocity there is a force which is supposed to keep it on the loop. So, these forces are okay. We know that they are actually acting on this particular object but the result of this should be the force which keeps the object in a circular trajectory. And we have already examined this that the force which is supposed to be keeping its mv2 divided by r, b is the linear speed or mr omega2 where omega is angular speed. So, we know from our examination of the circular motion with constant angular and linear velocities we know these formulas. It's, by the way, very easy to derive and there is a previous lecture which doesn't. Alright, so this particular force is supposed to keep our object on its orbit on its trajectory and is directed towards the center. So, this is how it is. This is how it's supposed to be. So, they must be equal, right? And I will also put it with minus sign because, again, the force is down opposite to z line. Okay, this is basically sufficient for us to find out what is the value of the pressure at the top. Well, first of all, I can always multiply by minus 1 everything. From here I will see this minus w, right? Or, considering I was just talking about before that weight and mass are related by multiplier g which is gravity constant on the surface of the earth. So, that would be r omega square minus g or m v square over r minus g because v is r times omega, right? So, v square would be r square omega square. Okay, it would be the same thing. So, this is my result. This is the pressure which my which the motorcycle exhorts on this loop and the loop exhorts on the motorcycle. Now, here we can actually talk about a very interesting situation. If v square divided by r where v is linear speed or r is the radius is equal to g, my pressure is zero. If my pressure is zero, it means the motorcycle actually feels like being basically weightless at this particular moment. So, at some very specific speed it will be weightless at the top. If it will be greater speed, the whole thing would just press upwards. So, the motorcycle will feel that his direction of his weight is up. Well, but since he is basically sitting upside down when he is on the top of the loop that kind of corresponds to normal to normality for him, right? However, again, if it's equal v square over r is equal to g, then he will feel weightless and what happens if v square divided by r is smaller than g? Well, there is no pressure over there. It's negative pressure, which means what? Which means he will fall down. He will not make the loop. That's what's important. So, whenever you are making a loop, you know the r, you know the g, doesn't really matter what your mass is, but your speed should be sufficient to make this thing positive. Otherwise, you fall down from the top. You will not make it. Okay, now let's talk about bottom. So, at the bottom again I have w down which means it's supposed to be with a minus sign. Now, my pressure from the motorcycle to the loop is down, but from the loop reaction is back, it's upwards, right? So, in this case, my pressure at the bottom is positive. And so is my centripetal force, which basically keeps my object, my motorcycle on this particular trajectory. So, that's also plus. From which we can see this bottom will be this and plus will be this and plus will be this. Now, in this case, there is no danger. He will not fall down anywhere. He will definitely go through this point. And obviously, the pressure is greater now than the weight. The weight is m times g, right? So, the pressure will be greater, which means that as he feels, as the motorcycle feels at the bottom, he will feel that the whole thing actually presses him down with the force greater than usual weight. So, it will be overweight. So, this is underweight. This is overweight. And there is absolutely no chance to get any kind of weightlessness. Weight only increases. And again, you probably felt it yourself if you are in a car and you are actually going in this particular direction. Whenever you're going down, you feel a little bit, everything goes down, which means everything is heavier, so to speak, right? So, this is why. All right. So, these are a couple of problems which basically gives you a feel how we can use certain techniques in adding or subtracting the forces, the super position of forces to solve real problems. I do suggest you to read the documentation for this lecture, the notes. It's on the website. Well, basically, that's it. I think I will spend some time solving certain problems. And that would kind of re-emphasize this principle of super position of forces. All right. That's it. Thank you very much and good luck.