 Hi, I'm Zor. Welcome to Unisor Education. Today I would like to solve a little bit more difficult problems, just a little bit, related to the superposition of the forces. It's the series of the problems called Problems 2. It's part of the Physics 14s on Unisor.com. That's the website where this course is presented. On the same side, there is a prerequisite course which is called Math Proteins, which is relatively comprehensive course of mathematics for high school level students. And I do recommend you to watch this lecture and every other lecture actually from Unisor.com rather than from YouTube or any other source, because every lecture on the Unisor.com has very detailed notes. It has exams for those who would like to take them. And it also has certain educational functionality if you would like to study under the supervision of the parent or a teacher or a flipped classroom or something. Okay, so the problems. Problem number one, you have an incline, a slope with a block on the top. Well, this is basically a pulley. There is a little wheel which basically rotating without friction obviously. Now here we have one point object with mass M1. Then there is a thread around the pulley. It goes down and this is M2. So this is the vertical line. This is an incline and there is an angle, let's say alpha. Now, what's interesting is that, well, let's just consider that M2 is greater than M1, which means that my thread moving this way down. Now obviously it's all happening on Earth, so the weight is equal to mass times acceleration of the free falling, which is 9.8 meters per second square. Alright, so what's interesting is that as my combination of these two masses is moving around the pulley, it exerts certain pressure on the pulley. I need to know the value of this pressure. Well, let's just think about it from this way. First we have to analyze the forces which are acting. Now obviously we have the weight. Also, since this is an incline, there is a pressure which this object exerts on the incline and the reaction force back. So these are two forces, the weight, and there is a reaction force. Is that it? Obviously not, because these two forces are the only forces which are acting. Obviously the mass will go downhill. It goes uphill because there is a tension, which is the result of this guy going down and it pulls the thread. The thread we are considering to be non-stretchable, so there is no friction, no stretching. So there is a tension on this particular thread which basically pulls this particular object upwards. And obviously there should be the balance here. So this is the resultant of these two forces. This is R1, this is W1, this is weight, this is reaction from the incline. And their resultant is this way. And obviously the tension should go along the same line, along the thread basically, because our object is moving that way. Right? Now, what kind of forces are acting on this guy? Again, it's weight, W2, and again the tension. Tension just slows it down. Basically it still goes up, but the tension slows it down. Now, so what we have right now is these tension forces, they are the same because it's the same thread, right? And their resultant, one is going this way, another is going that way. So the resultant would be going upwards. Now, obviously there should be some kind of a force, which is the pressure which the thread exhorts on the pulling on the support mechanism here, which basically keeps the pulley intact, right? So the combination of these two forces and this pulley is zero, and that's why we have this pulley standing basically on the same place without moving. And this is the pressure which we would like actually to find out. Well, obviously it's supposed to be related to these two forces, the tensions, right? So unless we find the tension, we cannot find the pressure on the pulley. Alright, so let's start with the tension. What can we say about this particular problem? Well, now this is angle alpha, so this is angle alpha, right? Because it's too perpendicular to these two, which means we can define this force. Let's call it F1, because it's the first object, right? So F1 is equal to the W1 times sine of alpha, or M1g sine of alpha. Okay, got that. Now, what can we say about moving of these two things? The linear moving along this incline or this downward, these movements are supposed to have certain acceleration which is exactly the same, because they are connected by the same thread. So this going this way and this going this way, they're supposed to be movements with exactly the same acceleration, speed and acceleration. So what's the acceleration using the second law of Newton of this guy? Well, acceleration, let's call it A, it's equal to T minus F1, right? Because T goes along the movement and F1 goes against the movement and that should be divided by M1, the mass. So that's the acceleration of this guy. Now this guy has acceleration of also the same A. It would be W2 down along minus T goes up prevents divided by M2. They should be equal to each other and that's how we will resolve it for T, right? So let's write it down here. C minus F1, well I can actually substitute instead of F1, I can substitute the whole value. M1g sin A divided by M1 equals 2. W2 is M2g minus T divided by M2. So we have to resolve it for T and that's our first step to find out the pressure. Alright, so obviously we have to multiply by M1, M2, so we will have T times M2 minus M1, M2g sin alpha equals M1, M2g minus M1, T. T goes this way, G goes that way, so we will have T times M1 plus M2 equals M1, M2g 1 plus sin alpha. Am I right? Hope I'm right. Okay, so we found T. Now how can I find the pressure if I have the T? Well, basically if you have these two forces, let's call it T, this is from this one so it's T2 and this is from T1. These are continuation of these but they are the same actually so I don't have to really have an index since it's all the same thread. So I have this particular story. This is T, this is T. Now what's the angle between them? This angle is, this is the right triangle so it's the same as this one which is pi over 2 minus alpha, right? And I have to find a force which is basically equivalent to them which means it should be on the bisector, right? So the combination of these two is, well I can find this piece, this is the rhombus by the way, right? Because these two are the same and it's parallelogram. So I have half of this angle which is pi over 4 minus alpha divided by 2. I know this T so I can find this one, it's this times cosine. So it's T times cosine of pi over 4 minus, that's the half and if I multiply by 2 it will be the whole line, right? Because I found only the half of it. From this times cosine of this angle it's half of the resultant. Okay, so that's the answer and it corresponds to whatever I just said. Because right now I can say that this is equal to T is this divided by this sum so it's m1, m2, g1 plus sin alpha divided by m1 plus m2. This one and times cosine of pi over 4 minus alpha over 2. That's the answer. So that's the pressure which is applied by the whole system of these objects and pulley and all this. The pressure applied on the base of the pulley basically. I scratched my m. Okay, next. Next is you have an object on the thread. I have an angle 5 and the length of the thread. And now this object I'm basically launching to do the circular movement. So it goes this way. So in the horizontal plane this thing is making a circular trajectory. Now what do I have to find out? I have to find out if I have some time T I would like to find number of rotations per time T. Okay, now again let's see what kind of forces are acting. Obviously this is the force which is w1. Well w actually have only one object which is mg. And I have obviously a tension here. And also I had a force which initial push around the circle. It's no longer participating basically in other movement. So as long as my speed is constant it just continues indefinitely circulating considering there is no friction etc. So this force which is along the trajectory which was in the very beginning to basically start the movement is no longer participating because it's no longer applied. It was just initial push and that's it. So there is no such force anymore. There are only two forces down and along the thread, the tension. Now but since my object is moving in a circle obviously there supposed to be some kind of force which keeps it on the circle. And obviously this is the force which is the combination of these two. So this is the resultant force and this is the centripetal force which keeps it on the orbit. Now we know that the force is equal to mass, linear speed divided by r or mass r omega divided, multiplied by r omega square. So this is angular speed, this is linear speed. Well let's concentrate on this one. So this is the force which is the resultant of these two forces. Now if this is phi, this is phi and this is phi. Now the weight I know, well I don't know the mass but I probably don't need it. So how can I find this one? Well if this divided by this is equal to tangent, so it's the weight multiplied by tangent equals to my force, let's call it f, which keeps it tangent of what? Tangent of phi. This is the force which keeps it on the trajectory which is supposed to be this one. Oh by the way I didn't know the r, r obviously is calculated from l. So r is equal to l times sine of phi, right? l times sine of phi times mass times omega square, right? So and the weight is equal to mg, so I can put mg here. Now what we have from here is basically all we need is omega because this angular speed will obviously allow us to define how many rotations per unit of time we have. So we obviously can cancel t, I will have g is equal to l times sine of phi times omega square. So omega is equal to square root of g divided by l sine of phi. Okay we've got omega. So this is angular speed which is number of radians per, let's say second, right? Number of radians per second. What we need is number of full rotations per time t. Well if this is number of radians per second then within the time period t we will make number of radians would be obviously t times greater which is t times square root of g over l sine phi. So that's number of radians. Now if you would like to change it to a number of rotations and each rotation is 2 pi radians I have to divide it by 2 pi. And this is my answer. This is number of rotations per unit of time t, rather per t units of time. Say correctly. During the time t. Okay, next. Next is the following. So you have a wedge. This is phi and this is phi. Wage is going down. Now here are two similar masses, point masses, point objects obviously m and m. Now the wedge has mass m. Now it's all on the table. There is no friction. So as wedge goes down it pushes aside both objects, right? Now my task is to find out the acceleration. These two objects will be pushed aside, right? All right, so again what we start with obviously the forces. Now what kind of forces actually are involved here? Well, the weight here and weight here are insignificant because they are completely balanced by the reaction of the table. So that's not important. So what is important is, let me just scratch this. So we know that this is angle phi and phi. What is important is the reaction from this particular movement. So it's always perpendicular. Now these are point objects, right? So the reaction will be perpendicular to the sides of the wedge. Let's call it n. n here and n there. So the wedge goes down and it pushes with certain force our blocks aside. And the blocks, these two blocks are reacting by these forces back to the wedge. Now the wedge goes down, let's say with acceleration a w. And these two are going left and right correspondingly with acceleration a. Now how can I transform these forces into the forces which are acting along the direction of the movement? Now direction of the movement of the wedge is down, direction of the movement of these is horizontal, left and right. So I have to convert all the forces into the forces which are parallel to these two directions. Now in case of the wedge, I have the weight, it goes down and it's equal to mg, right? And I have the force which is perpendicular to this one and perpendicular to this one. Now I would like to convert this force into horizontal and vertical components. Now horizontally they will act against each other because they will act along this line here and here. Vertically they will go up here and up here. So I would like to find out these two because they are acting opposite to my weight, right? So let me just do a bigger picture and you will see, okay, this is my force and I need to know the vertical component and horizontal component. So this is my vertical component. Now this is angle phi. So this is angle phi, right? Because this is perpendicular to this and this is perpendicular to this. So if I have this which is n, unknown obviously, then the vertical is equal to n times sine of phi. And I have from another side exactly the same thing. So they are both acting that way and they are acting with a minus sign because they are going directed upwards and the weight goes down. So mg. That's my total force acting vertically on this particular guy. And that's why the acceleration is equal to this. This is my acceleration AW. Now let's talk about this guy. Now this is the same n by absolute weight, just differently directed. Now I need to know its horizontal component. So again let me just do it a little bit bigger. This is my n. This is my big block. I need this component, this one. So now this is perpendicular to this and this one is perpendicular to this. So this is phi. So it's n times cosine, okay? So my object has n times cosine phi as the force and divided by m is acceleration. That's different acceleration obviously. Now how many unknowns we have? m is unknown, a is unknown, AW is unknown. So we have three unknowns with only two equations. Not good, right? But let's consider it this way. What if you have this wedge and in the beginning the point where it actually touches of this block is here in the very corner. Then the wedge goes down and this thing is sliding to the right. As the wedge goes all the way down, this thing will move to the right by this piece, right? So whenever wedge goes by this piece down, my block will be moved to the right by this piece. So their ratio is actually the ratio of the displacement and ratio of the velocity and the ratio of accelerations obviously because they're all proportional. So if the movement down is proportional to the movement to the right, then acceleration will be also proportional. So that's how I can establish my third one. So how do they are related? If this is phi, right? Then this is acceleration w would be to acceleration, no, sorry. This will be acceleration and this will be acceleration w. This is tangent of phi, right? Now these are three linear equations with three different unknowns and we need to know the a, which is actually very easy because again these are linear equations. And the result is in my notes and I can give you the result here. A is equal to G tangent phi divided by 1 plus 2 m over m times tangent square phi. So this is basically easy exercise on system of linear equations. From here you get n, you substitute n into this and you will have only a and a w. And then from this you can get a w and substitute the result and that would be the solution. Very easy. So I'm skipping all these technicalities there through them. Notes to this lecture have a little bit more details of this trivial calculations. We don't want to stick on it. And the last problem is the following. Let's imagine the person who would like to make a long jump. And it's not exactly the way how long jump is made but this is a good kind of a physical problem which is related to this. So let's consider him accelerating to a certain speed. At this point he jumps up. So what happens? Well while he is jumping up and then falls down he continues this horizontal speed and that would be the lengths where he lands. So my purpose is to find out the lengths if I know that he was accelerating during time t with acceleration 0 from 0 to some kind of a v, some kind of a speed. And then he jumps up and I know the height. So by knowing the height and these parameters the time and acceleration I would like to find out how long he will jump. Well look at it this way. First of all let's find out the v is equal to a t. During the time t with acceleration a from velocity equal to 0 that's my maximum velocity. And this is the velocity or speed actually because it's one dimensional as he would fly forward during his jump up, right? Now so all we need to know if we know the horizontal speed we have to know the time during which it happens. So the time is as he jumps up I know the h. So from h I have to find the time. Well that's actually easy because if he goes up first it means he is pushing himself with certain vertical velocity, right? And then during the time it goes up let's say it's time t vertical, no it's better to say t up. What is his parameters? As he goes up he has a negative acceleration of the free falling which is g and at the very top his velocity is 0, right? So if I will subtract from the initial speed as he goes up this I will have 0, right? That was initial speed. At the end of the time period speed is 0 so basically that's what it is. So v vertical is equal to g times t up. Now how about going down? Down you go from the place where your initial speed is 0, your initial acceleration is basically only speed actually is only horizontal, but then you go down with acceleration g. So the h would be equal to gt, sorry, yes, acceleration and speed divided by 2, right? That's the formula because the initial speed is equal to 0. It's only horizontal so vertical speed is 0 so this is my going down. So this is down. Now considering my time to go up, time to go up should be equal to time to go down, right? So basically if I am calculating only this or I can actually prove it I guess because the same h is equal to v vertical times t up minus gt up squared divided by 2 minus because g is slowing me down, right? This is my initial speed up times the time. That's my formula for uniform acceleration and acceleration is equal to g, right? So that's basically two things where I can find out my t up and t down. They are supposed to be equal and from this we can say that v is equal to gt up, right? And because v is the beginning speed, 0 is the end of the speed. So v vertical minus gt up should be equal to 0 from which we derive this which means that in this particular case we have h is equal to gt squared gt times t minus gt squared this is up divided by 2 which is gt squared up divided by 2. As you see it's exactly the same as t down, right? So that's the proof basically that whatever the time you need to go up is exactly the same the time to go down. And from here obviously you can find time if you know the h. Time is t up plus t down but they are exactly the same so you can double the time up. So the time of the entire up and down would be equal to 2 times 2h divided by g square root. 2h divided by g. So my time is known. My speed, horizontal speed is known. So their product which is 2at square root of 2h divided by g is the length of the long jump. That's it. So what you have to do now and I strongly recommend you to do it. Go to the website, go to this particular problems number two in superposition of the forces. This is how you go. You go from the website, you have a physics for teens, then you go to mechanics. From mechanics you go to superposition of forces and there you have the problems two lecture. Now the nodes contain all these problems and answers so do it yourself. Try to solve all these problems, check against the answers and that would be great exercise if you can do everything just by yourself. Alright that's it. Thank you very much and good luck.