 Hi, I'm Zor. Welcome to Unizor education. I would like to solve just one relatively short problem, but I think it's kind of educational and probably some exams which Also are part of this course Will be based on very similar concepts, which I'm going to talk about Right now. So we're talking about friction. This is a friction problems number three This is how you get it through mechanics and dynamics to friction and problems number three Presented on Unizor comm. That's the course called physics for teens. There is a prerequisite course, which is called math for teens and I do suggest you actually to be Rather familiar with such concepts as vectors and calculus from from any course of mathematics and if you want obviously can use math for teens on this website or All these courses are free and they don't have any advertisements or no monetization alright, so One problem for today is as follows Let's consider you have a horizontal table and an object on it Let's say this object has mass m Now obviously there is a friction. I mean the whole problem is about fruit friction, right? now there are two pulleys here and The thread goes down here and down here and there are some weights Mass of this is M1 and mass of this a smaller one is M2 So basically we understand that since this is bigger than this heavier So the weight will be greater on this side than on this side and The object on the table would move to the left in this particular case now. Let's assume that we have some kind of a Friction and new is the coefficient of the friction Which is basically the ratio between the pressure this object Exords on the table and And the force which prevents the movement so new is actually F divided by W where this is the friction force which is always against the movement so in this case the friction force is Directed this way obviously weight directed down and and the weight so Our purpose is to find This coefficient of friction if I know that this Particle object is moving to the left with acceleration a now without the friction So if new is equal to zero Well, this force goes with the force of the weight actually and this force goes also with the force of the a of the weight and It will just fall down without any Restriction so this guy will move it was exactly the same acceleration as this guy's falling down which is gravity acceleration of gravity which is g Now if there is a friction then obviously things change Now as usually we have to examine all the objects and all the forces which are acting on these objects Okay, so let's start with let's say this object. What kind of forces are acting on it? Well, there are two forces basically one force goes down and this is it's weight and Then are another is up which is tension of the thread Okay, as a result of these two forces this thing goes down now what is The movement of this thing as far as its speed acceleration, etc Well in absolute terms Not in terms of the vector but in terms of magnitude of this vector The acceleration of this one gap going down is exactly the same as acceleration of this one going to the left because they are Attached to each other and the thread obviously is assumed to be non-elastic. So as The acceleration of this down is equal to acceleration of this to the left and we know it a So what we can do right now is to completely make An equation of the second Newton's law for this particular weight the Force which goes down is w1 Well, let's assume that down is a positive direction So all these numbers all these forces will be positive in magnitude and I will use the sign to basically To tell which particular direction it goes. So I will have it as a minus t1 so The way it goes down the tension goes up and this is basically all the forces which are acting on this particular weight right and Obviously, it's equal to along according to the second Newton's law mass times its acceleration again I'm talking about absolute terms Acceleration obviously goes down as a vector, but this is just a scalar positive scalar. All right Okay, that's it for this guy now this guy Goes up obviously this is heavier So the whole system moves this way and it moves up also with the same acceleration a because this is the threat which is not elastic and And obviously pulls up with the same speed as this one goes to the left. What forces are acting on this particular Object Well the same thing down goes its weight up goes tension Which is on this side tension is different on this side. We have t1 on this side We have t2 T1 pulls it this way t2 pulls this object this way and this object obviously backwards so the tension is always Pulling things together so to speak so if for this guy it moves up the tension is directed upwards for this guy tension is directed directed this way it pulls it and And for this guy again the tension for this guy pulls it up and for this pulls it to the right so Equation will be very very similar except since we are moving up I will do it instead of weight minus tension. I will have tension minus weight because I wanted to assume that all these Numbers t1, t2, a whatever all are positive. That's why I'm using the signs since I know that we are moving to the left alright now There is also the middle object and we have to determine The coefficient of friction for this guy. Well, let's examine the forces t1 moves it to the left t2 Resists and pulls to the right and f is the friction force which is also prevents the movement since movement to the left I direct my friction force to the right These are three fun three main Forces now obviously there is a gravity, but there is a reaction of the table and they nullify each other So we can just disregard these two guys, right? all right, so in the horizontal movements is t1 to the left minus t2 and minus f is Capital M times a so these are Three equations which basically constitute the picture of everything in this case All I have to do right now is to express whatever I know In terms whatever are given so instead of w1 I will obviously put M1 Times acceleration of the free fall right because the gravity the force of gravity is equal to again according to the Second Newton's law is mass times acceleration of the free fall same thing here Instead of w2 I will put M2 times g and I also know the f my Friction force is equal to Weight times The coefficient of the friction and weight is m times g times mu and this is equal to m a Now let's look at this What's unknown t1 t2 and mu are unknown Everything else is known So three linear equations with three unknowns Now what do they have to determine in this problem? Well, I didn't forget Well, it's new coefficient of the friction and the tensions on both sides All right, so that's exactly what is unknown and that's exactly what we can determine very easily So let me just do this little exercise in solving the linear equations These are these happen to be very easy So we have from the first equation t1 already basically Determined it's m1 times g minus a right T1 and why and m1 m1 a goes to the left. So that's what we have T2 from the second one is equal to m2 Plus g plus a Now, let me just stop here for a second. Why is tension Less than the weight weight is m1 times g now. We are assuming that everything is positive, right? That's why there is a sign here. So why the tension for the left weight is Less than the weight. Well, because we are moving down, right? So if I have certain weight and I'm holding this weight and It does not move then obviously my force which I'm forcing it to stay in place is equal to the weight Right. So in this case they are equal, but if I move down, I don't have to hold it as tightly my thread Which connects to this particular object if there is a thread here the tension on this thread would be less and Actually, if I'm moving my hand With the same speed down the same acceleration down as my object is falling down the tension would be zero at all Because there is no tension. I'm moving so that's very important to understand and Obviously, that's exactly how it feels it should move the greater acceleration of the movement of the supporting block or whatever it is The obviously the less will be the tension and eventually if a is equal to G which means if I'm moving down with the same speed as this thing is falling then the thread between us would have No tension at all. It would be zero, right? On the other hand The second object we are moving against the weight, right? So we are not only not allowing it to fall down We are pulling it up Against the weight and that's exactly why we are adding this a so I just want to explain the physical Meaning it's not just algebraic manipulation. There is a very good physical meaning of subtracting here and adding here because this guy goes down and this guy goes up against the Weight and that's why the tension should compensate for both the gravity And acceleration so that's very natural Okay, and from here we obviously can determine new Which is equal to? T1 minus T2 minus m a Divided by m mg Where T1 is okay? I can just write it down, too So mu is equal to T1 which is m1 times g minus a minus m2 G plus a Minus m a use the square brackets here and we can divide it by Mg and this is my move so this is Basically the answer to the problem. We have determined the tensions Tension on the thread which is on the left would be less than the weight the tension on the right would be greater than the weight of the right object and the Coefficient of the friction in in the middle for the middle object, which is on the horizontal table is expressed in this particular Well rather long formula So I do suggest you to go to the website to mechanics dynamics friction and problem number three and read the notes for this lecture which basically the same as a textbook and Be prepared for exams that would be the next Topic in this particular category for the friction. So the friction has certain theoretical Lecture in the in the beginning then few problems and the exams So that's it that's it for today. Thank you very much and good luck