 Hi, I'm Zor. Welcome to Unisor Education. I would like to solve a couple of problems, relatively easy problems, related to friction. And it's part of the course called Physics 14 presented on Unisor.com. I recommend you to use the website to watch this and all other lectures. There are a few reasons for this. One of the reasons, for instance, this is the second course. The first one, which is a prerequisite, is Math 14. And all the lectures are organized in some kind of menus, etc. So it's very easy to use as a course, basically. Also, every lecture has detailed textual explanation. It's like a textbook, so you have the advantage of watching the lecture and reading the textbook at the same time. Plus, there are exams on that website. So, the website, by the way, is free for all, no advertisement. So, the problems. My first problem is the follows. You have an inclined and there is an object here of mass m. So, you know that inclined plane has an angle phi. And you also know that there is a friction between an object and the plane itself. The mu is the coefficient of kinetic friction. So, you remember that there is a static friction and kinetic. Static is when the object is actually on the surface and little bumps are basically kind of deeply embedded between the object and the surface. But when it moves, they are not as deep inserted in between each other. So, it's easier to move when it's already moving. And that's why the coefficient of the friction, the kinetic, is less than static. Alright. And again, as a reminder, the coefficient of friction basically characterizes the force between the force of resistance of the motion. So, if you have a surface and you have certain pressure, whatever the pressure is, then the resistance is this friction. Then the resistance is this force of pressure. Let's put it P here. Times certain coefficient, which depends basically on the quality of the surfaces of the object lying on the surface and the surface itself. All these little bumps, it all depends on how they're structured. The more polished, obviously, the surfaces are the less the coefficient of friction is and the less the force of the resistance to the motion will be. Now, these are not vectors. These are absolute values, magnitudes, because the vectors of pressure goes perpendicularly to the surface. Always perpendicular, by the way. And the resistance goes parallel to the surface. Alright, so, we know that. It's just a small repetition. So, what I know also about this, that I need the force F up to move this object uphill with a constant velocity, no acceleration. What I have to determine is what would be an acceleration of this object if it just goes by itself under its own weight. So, there is a weight here. And as we know, the weight is the force basically going down. It's a gravity, which depends on the mass as this. Where G is the acceleration of the free fall. So, this mass, this is acceleration. This must be the force, right? And this is a known constant on the surface of the earth. It's like 9.8 meters per second squared. Alright, so, M known, G obviously is a constant known, phi known. The force which is needed to pull the object up is known. What is not known is acceleration of this object as it goes down. Now, let's just think about it. If I need certain force to move something with constant velocity, no acceleration, it means that the forces which are working against it, against this movement are exactly equal and balanced basically by this force F up. So, let's just think about what kind of forces are restricting my motion, are directed against the motion. Well, with this purpose, we obviously, as in all, basically in all problems related to movements on the inclined plane, we have to represent the force which is acting in a direction not really related to the inclined to two components. One component is perpendicular to inclined. And that would be the source of the pressure and the friction. And the force which goes along the plane. That would be the force which basically is moving the object downhill. Now, so this is actually somewhere here. And this is somewhere here. So these forces, let's call it force of pressure. And this would be force of downhill movement. So any weight can be represented as a sum of these two vectors. This is parallelogram, not just a parallelogram, it's a rectangle because we are talking about one side to be perpendicular to this line and another parallel. So this is the right angle. And if this is the right angle, and this is phi, obviously this angle and this angle are also phi because the sides of this angle are correspondingly perpendicular to sides of this angle. Now, we can actually think about what forces are actually acting upon our object. Well, obviously this direction is force which is known, f up. Now, what represents the force which is going to opposite direction is number one, the weight component which is fd. And number two is since there is a pressure, there is a friction. So pressure times the coefficient of friction gives me the force directed against the motion. Motion is that way, right? The force is against the motion which represents this restricting force which is restrict the motion forward. Forward means up in this case. Now, since my object as a result of application of this force moves with a constant velocity, that means that these forces should be in magnitude equal. Now, this is just a plain addition because this force is directed this way and this force is directed this way. This force is directed opposite, right? So we're talking about vectors which are all collinear and this is the direction along this plane. One directed downhill, another directed uphill. That's why we can actually just add their magnitude if we want to compare them, right? So the magnitudes are supposed to be exactly the same because this is directed downhill and this is directed uphill. Okay, fine. We actually need the acceleration of the object as it goes just by itself, right? So let's just think about it. If there is no force f up, if it goes down by itself, in this case, what kind of forces are acting upon our object? Well, obviously this component of the weight fd would be directed downhill, right? But now, when the object goes down, the friction works against this motion. So the same friction would actually be towards the uphill direction, preventing as much as possible movement downhill. So in this case, the force which is acting this direction is this one. We should put the minus sign here because the direction of this friction would be in this case exactly the same in magnitude, but directed against the motion now is downhill. And this would be, obviously, since we know the force which goes down, we know the acceleration would be related with the second Newton's law, right? Well, basically that's it. That's all we need in this particular case because if we will add them up, we will have 2 fd equals 2 f up plus ma. Now, 2 fd. What is fd? This is the downhill component of the weight, which is obviously equal to weight, which is mg times sin of this angle, right? Of phi. So this is 2 mg sin phi, from which we can find out the A. A is equal to 2 mg sin phi minus f up divided by m, right? This is ma is this minus f up. So acceleration is this, which is equal to 2g sin phi minus f up divided by m. So that's the answer. By the way, it's independent of mu. That's what's interesting, right? So even if the coefficient of this will be high or low, doesn't really matter, whatever it is, it's independent and that will be acceleration. If we know the force needed to pull the object with a constant velocity uphill, and this phi is basically the angle of the hill, then this would be acceleration of its free sliding downhill with the friction, whatever the friction is, doesn't really matter. All right. Second problem. Second problem is a little bit unusual in formulation, but relatively simple, obviously. Here it is. Let's imagine you have a platform, well, I don't know, maybe on wheels, doesn't really matter, platform, which is actually going with acceleration A. Now, in my problem, I actually even have numbers, but numbers are not important actually. Now, there is an object here. Now, let's assume that there is some friction. Now, what happens with this object if my platform accelerates with the certain acceleration A? Well, obviously, if it's fixed on the platform, it will also go with acceleration A, right? But what if it's not fixed on the platform? Well, if there is no friction, then there is nothing actually. If friction is equal to zero, there is nothing, no force, which acts on this particular object, because the platform will slide from underneath of this object, and if there is no friction, the object will just stay on the platform, but it will slide this direction relative to the platform, or relative to the earth, it will just stay in place on this platform as the platform goes. With certain acceleration, if there is no friction. So, if this particular object moves, then the only result of this, the only source of this movement is the friction. Again, if there is no friction, it would stay. If there is a friction which is sufficient to basically hold it in place, whatever the coefficient of friction is, it will just follow the platform with acceleration A. So, if friction is somewhere in between, maybe small one, question is what happens with this particular object if there is some friction. So, we have to really analyze this story. Now, I'm also giving you two numbers. This is the friction coefficient, this is the kinetic coefficient of friction, and this is the static. Static is greater, by the way, as you see, right? Now, first of all, why do I give this static coefficient? Well, because if the object in the beginning just stays on the platform and platform starts accelerating, it might not actually move at all. It all depends. If my platform accelerates just a little bit with a small acceleration, it might actually be fixed on the platform, the force of friction will be enough, and it will follow the platform. If my static friction is large enough or acceleration is small enough, then the object will stay in place on the platform and it will follow the platform, right? If static friction is not sufficient to hold the object, then the object will start sliding. So, the platform will go forward and the object will also go forward, but it will not be with exactly the same acceleration as the platform. It will start sliding a little bit backwards relative to the platform. Still forward towards the earth, obviously. There's nothing which pulls it backwards, so it will slide forward, but not as fast. So, what I want to say is that the friction is the only source of the movement of this object. Now, we used to think that the friction is always directed against the motion, right? But that's in case we have an external force which moves the object forward, then the friction kind of slows it down. In this case, the situation is actually reversed, because friction is the only force which acts on this particular object. I mean, there is obviously the weight, but weight is neutralized by the reaction of the platform. So, these two forces can be completely out of the equation, out of the equations. And the only force which can, in theory, move the object forward would be the force of friction. Now, let's just think about it. If I know the mass of the object and obviously the weight, then if I were multiplied by coefficient of static friction, that would be the force which is capable of pulling the object forward with the platform. So, if this force is exactly the same as the force needed to give it an acceleration, A, which is if it's equal to MA, or obviously if A is equal to G times mu static, this is exactly the moment when the object is, well, it's on the boundary. If A is slightly bigger than this, it will not maintain the position on the platform. It will start sliding down. Less than this, then that force would be basically sufficient to pull the object forward with the platform with exactly the same acceleration. So, that's why we need this static acceleration. We have to analyze how the object behaves. We need this static coefficient of friction to find out how fast the acceleration actually can be. So, the object either stays in place or starts sliding backwards relatively to the platform. Now, in this particular case, A is 10. This is 9.8 times 0.4, so it's definitely, this is smaller. So, A is greater. So, the force of the movement with the acceleration A should be greater than this. So, we can't really do it. And the object will start sliding backwards relative to the platform. Okay, as it is sliding backwards, what pulls it forward? Again, the force which is equal to mg times mu kinetic. Because it will slide, right? So, since it slides backwards relative to the platform, the platform goes forward and object will be not as fast, forward but not as fast. So, it's sliding down relative to the platform. So, it's moving relative to the platform, which means we have to use kinetic coefficient of friction since there is a movement, right? So, if there is a movement, this mg is the weight. So, this is the force which actually acts on the object. This is the friction which pulls it forward. Which means that this is exactly the force which gives it some acceleration, A OBJ, A object. Okay? So, what happens here is that we can determine what's acceleration of the object and it's equal to this. So, the object would move, I'm talking about relative to the Earth right now, the object would move with acceleration this forward, which means I have it calculated somewhere, 294 meters per second square. So, acceleration of this object will be forward 2.94 meters per second square, which is definitely slower than this one. So, this is relative to Earth since I know the force, I know the acceleration, right? And I can also determine what would be the acceleration relative to the platform. Now, obviously the platform goes forward with a 10 meters per second square with acceleration A. And my object in the same direction moves with acceleration A OBJ, then the difference between them would be acceleration relative to the platform. Now, if I'm talking about the system of reference associated with the platform, so as the platform moves forward, in this particular system the object moves backwards, so it's actually with a negative sign. So, relative to the reference frame associated with the platform, the difference would be what, minus 7.06, 10 minus 294 meters per second square. And again, negative because relative to the platform it goes backwards and the platform goes along the axis forward. Okay, this is again, in my personal view, this is slightly unusual problem in that respect that the force of movement is the friction, which in most other cases is the other way around. So, in any case, it's probably just very nice to think about friction in different terms, not always as something which prevents your motion, but also sometimes it's actually causing your motion, right? All right. Now, I usually recommend people after watching the lecture to go to the notes for this lecture on Unizor.com and basically, again, just take a look at the text of the problems. There is an answer, but before going to an answer, I don't provide the solution as here during the lecture. Try to solve the problem yourself again and see if you have the same answer that will be very educational. Okay, that's it. Thank you very much and good luck.