 Hi, I'm Zor. Welcome to Unisor Education. I'd like to solve a couple of problems related to the state of equilibrium. We are talking about statics right now. Now this lecture is part of the course called Physics for Teens. It's presented on Unisor.com. So if you found this lecture on YouTube, for instance, then you will definitely have this lecture only, but if you will go through the website, you will get also well, the whole course basically, but for every lecture you will have notes. They're very detailed like textbook. Then there are certain exams if you wish to take them. It's completely, the site is completely free and results of the exams are not really going anywhere, so you can take as many times as you want. Plus, this website contains the prerequisite course of math for teens. Especially something like vectors and calculus, you really have to be very comfortable with to listen to these lectures. Anyway, let's talk about problems on equilibrium. So the first problem is really very easy. You have a weightless rod and you have two masses and M and M. This is bigger than this one, obviously. Now the rod is weightless, so the weight is concentrated only in these two weights. You can consider them to be basically point objects. Now my problem is I have to put a thread somewhere here, let's say. So the rod is horizontal. Now, obviously, we're talking about the whole thing is happening on earth. So there is a weight here, which is equal to Mg and weight here, which is equal to capital Mg. So what I have to do is find the point where exactly my rod should be hanging from to maintain the equilibrium. Well, let's say the whole length is capital L. So I need to know this length and this length. Well, obviously, this problem is related to equality of the momentum in this particular case. Now to be more rigorous, I have to really have all the forces which are acting on this system to nullify each other and all the moments to nullify each other. Well, as far as the forces, that's basically easy because you have one force and another force and here you have a tension. So the tension should be equal to some of these two forces, but that's not what I'm asking. I'm actually asking where exactly the point should be the hanging point should be located. And as far as the forces, these weights and the tension, they're not related. Wherever the point of hanging is, these forces will still be balancing to each other, so that's irrelevant. The moments are different. So the moment of this guy, so let's say this is x and this is y. Well, obviously x plus y equals to L. I think I have to change my So, let's write it again, x plus y equals L. Okay, that's obvious. Now, what I have to do right now is I have to equate the moments. Well, if this is horizontal, this is 90 degrees and this is 90 degrees. So basically my vector, now you remember that momentum is vector product of radius times the force. So the radius is this way. In this case, and in this case, the radius is opposite, right? Now, the force is mg down. So if I multiply this by this and then this by that, that should be equal, but with opposite signs, and that's why the sum is equal to zero, right? Opposite signs because the forces are directed the same way, but the radius are opposite. So one radius is positive, another is negative. Whatever it is, doesn't matter which one is which, but obviously in one case we will have the positive, in another case we will be having negative. So our purpose is to equate their absolute value, which is actually x times mg, should be equal to y times capital mg. From which, obviously, g is of no interest. We have x divided by y is equal to m divided by m, and considering that, let's say, x is equal to l minus y, so we have l minus y divided by y is equal to m divided by m. So l divided by y minus 1 equals to m divided by m. So this goes to to the right. This is capital M, obviously. So that's m plus m divided by m. So y is equal to m divided by m plus m times l, and x is equal to capital N divided by m plus ml. So this is the answer. Well, obviously, if capital M is greater than lowercase m, then the x is greater than y, and that's how it's supposed to be. The weight is less, so the radius should be greater to equalize the moment with this guy. So that's it for this first problem. Let's go to the next one. Okay, the next one is the following. We have a horizontal surface and an object of weight w. Now the coefficient of static friction is mu. Static is very important in this case. It's static friction. Now we have a rope which makes an angle phi with the horizon. And question is, what is the force F, which is required, the minimum force F, required to start moving. Okay, let's just think about this. Again, we have to equalize the balance. We have to balance all the forces and all the moments. Well, in this case, we don't have any rotation, so it's only to equalize the forces. Now, what forces are acting on this particular object? Well, obviously the weight, definitely. Another is the reaction of the floor. Right? And then we have two components of the F force. Let's call it Fy and Fx. Well, if this is an X and this is a Y, that's why I called it this way. So these are two components. Now force Fx acts horizontally and it's supposed to move the body, move this object forward. Force Fy makes the object lighter, so to speak, not to press on the ground as much. Right? So what we have to do now, we have to equalize the vertical forces and the horizontal forces. Now, what's the horizontal force? Well, this is basically the force of friction and we're talking about the minimum force F, so it's just the beginning whenever you start moving and that's exactly what static friction actually is. This is the resistance to start the move, to start the motion. So basically we have to equalize all these forces. Now, again, let's start from the vertical position, let's say. Well, vertical position is n plus Fy should be equal to w. Now, this is fine and but this is not really very interesting because it's vertical balance. We know that n would be with balance because the force F is not strong enough to start the object flying. Although it might actually be the case, but we are not talking about this. We're talking about much smaller F, so the object doesn't fly anywhere. It's just on the surface and it's supposed to go horizontally. So let's talk about horizontal movement. So what exactly are the forces? So Fx is equal to F times cos 5, right? Now, what is the friction force? Well, friction force is basically the result of weight minus Fy because that's actually the normal reaction of the of the floor times mu, which is the coefficient of static friction. So again, w goes down, Fy goes up. The difference between them is exactly the reaction of the floor, normal perpendicular to the floor and that's what's supposed to be multiplied by the coefficient of friction to get the force which resists the motion and it's supposed to be equal to Fx, which means F times cos 5. Now, Fy is equal to F times sin 5. So what do we have now? Mu times w minus mu times F times sin 5 equals F times cos 5 from which F is equal to, so F goes to this way, so on the top we will have mu w and on the bottom we will have cos sin 5 plus mu sin 5. So that's the answer. This is the force which we have to apply to the rope to start moving this particular object if weight is given and the coefficient of static friction is given. That's it. That's the solution. Number three. Okay, number three also has no rotation. So no moments are involved. We just have to carefully examine the forces which are moving. So let's say we have a platform and on the platform we have a person. This is the person and there is a rope. Which goes through some kind of a pulley and it connects here. I think it's not really a good thing. I'll do it much bigger than that. Let me start from the rope. So we have a pulley, a rope, platform, another rope, another pulley and here we have a person. Okay, now what do we have? We have the weight of the person, W. We have the weight of the platform P and we know that the system is in equilibrium because the person is pulling this particular rope down. Now as he pulls it down, obviously this rope supports the platform from this side and from this side because this is another rope. So what we have to find out is what is the force F. This poor person should pull down the rope to stabilize his position not to fall down and not to go up, basically. So he is in equilibrium. He and the platform are in equilibrium. Let's just think what kind of forces are acting here. Well, obviously there is a tension force here. Well, the tension force here is F. That's the force actually which we have to find out. Now, what also is important? Well, obviously there is a weight and there is a reaction of the floor which actually goes this way. So, now let's consider only the person and all the forces which are acting upon this particular person. Now, since he is in equilibrium, all forces which are acting on the guy must be equal to each other. So what are those forces? Well, since he pulls the rope with the force F, everything is in equilibrium, right? That means that the rope pushes him up or pulls him up, whatever, with the same force F. That's the third law of Newton, right? So, on this person we have the force F which goes up. Reaction of the floor also goes up. His weight goes down. So, basically the weight should be equal to N plus F. To maintain him in balance, that's what's necessary. Now, let's talk about the platform. Now, the platform is also in equilibrium. What acts on the platform? Well, first of all, the platform has its own weight, right? Now, the platform, between the platform and the person, we have this reaction force. Reaction force pushes him up, but at the same time he is pushing the platform down. So, the weight is P and the reaction N is also down. So, these are all down things. Now, what's up? Up is tension of the rope here and tension of the rope here. Well, this is, I don't know what it is, let's put it T. But this tension is equal to force F because that's exactly the effort we are pulling this rope down. So, that's what the tension is, since it's all in equilibrium. Now, finally, this particular pulley is in equilibrium. So, what pulls it down is this force F and the same tension here, right? So, it's 2F, it's unstretchable rope. But the up is the same tension as in this side. So, they are equal. So, these are three conditions to balance our platform and the person and the pulley to balance the whole system. What's necessary to find out is, well, W and P are given and we have to find basically the force F. So, how do we do it? Well, very easily. Let's substitute 2F instead of this T, we will have P plus N equals 3F, right? Now, N from here is equal to W minus F, so we can put P plus W minus F instead of N equals to 3F. So, F goes here, it's 4F, so F is equal to P plus W divided by 4. So, that's the answer. And by the way, you see, divided by 4, we use two pulleys and that allows us to pull with a significantly smaller efforts than the weights which are P and W. Let's say the platform is weightless, right? Then, if the person wants to just pull himself up, it will be just one quarter of his weight, W divided by 4. So, in any case, it's easier to use these system of pulleys to pull something up. And that's exactly how people in practice do, how they use these systems of pulleys. Okay. The last problem also has certain practical implications. Okay, so let's consider you have the wall and the floor and there is a ladder. Well, let's assume that the ladder has the length D and we also have to assume that there is certain static coefficient of friction, which is new. Now, as you understand, if we put this ladder very close to the wall, it will stand very easily. But if you would like to move the base of the ladder out from the wall, the further we put it from the wall, the more chances are that it will start slipping and the friction between the ladder and the floor would not be sufficient to keep the ladder in this position. And at some point, the whole thing might actually start slipping away. So, my question is that given the coefficient of static friction, what would be the angle, the minimum angle, obviously. The maximum is 90 degrees, but the minimum angle will be where the ladder can still actually stand in the equilibrium position without slipping to the right. Okay, so that's the problem. We have to find phi based on mu. Actually, the result will not depend on the D, by the way, but we will come to this later on. Okay, so let's just think about what kind of forces are involved here. Well, there is obviously the friction which is directed this way. Now, there is a reaction of the floor and there is a reaction of the wall. Let's say this is N1, this is N2, and there is a force which is the weight. I put it in the middle. Well, you might say that, look, the weight is not really in the middle. The weight is concentrated basically at every infinitesimally small segment of the ladder. Strictly speaking, you are right. And if you will start doing more precise calculations, we will basically come up with exactly the same result. In cases like this, you can always assume that the weight is concentrated in the center of mass. And obviously, in this particular case, center is in the middle. So the whole body actually behaves as if the weight is concentrated in the center of the mass. So I assume that you agree with me on this particular point, and I hope you do. I put this particular weight here, thinking that the whole weight is here and there is no other forces involved. Now, okay, fine. So let's think about this fact. Now what is the force of the friction? The force of the friction should be equal to the reaction of the floor, which is n, 2 times coefficient of the static friction. I mean, whenever you have this type of contact between the body and the floor or whatever, it's the reaction of this floor which is supposed to be multiplied by the coefficient of the friction to get the force of the friction. So that's all. Basically, that's all horizontal movement, which we are talking about. So if we want this to be an equilibrium, this must be true. At the same time, this is a much more complex problem. There are also other forces which are involved. What are the forces? Well, the forces are which are rotating. You see, when this ladder would shift, the bottom would shift to the right, the ladder would rotate. So we have to really talk about the momentum of rotation. So that's what's very important. So what are the momentum of rotation here? Well, we have two main forces. Well, first of all, we have to choose the center of rotation. We have sufficiently a large number of forces. So we have to really choose the momentum which basically simplifies our calculations. And I suggest we choose as a center of momentum this point because at least two forces already can be completely ignored because their radius is zero. So momentum would be zero. But the momentum of N1 and W is not zero, obviously, relative to this point. So what are the momentum? Well, you always have to represent the force as the sum of two forces. One is along the radius and the other is perpendicular to the radius. That's the easiest way. So if I will do this, so this would be the force which I have to just multiply by the radius. And in this case, again, I will do the perpendicular. That would be my force which I have to multiply by the radius because that's what actual vector product is. So instead of doing sine of whatever, we can always do this. It's kind of easier. So what's the projection? Now if this is phi, then this is phi and this is phi. So this is N1 times sine phi. And the radius is equal to the length of the letter which is D. Now, and that actually turns the whole letter this way. And it should be balanced with another force which is trying to rotate it this way. So if this is phi, this is phi, this is phi. So if this is W, this is, let me see. No, this is phi. No, this is not phi. This is 9 to the degree minus phi. So this is phi. So it's W times cosine. So it should be equal W times cosine phi. And the radius is equal to D divided by 2. So this is the equation of balancing of the momentum. And lo and behold, we are not dependent on the length of the okay. Now, so what do we have here right now? What do we have here right now? Horizontally we have this is the force which goes to the left and this is the force which goes to the right. So that should be equal to N1. Okay, vertically we have N2 here and we have W down. So N2 is supposed to be equal to W. So we have basically balanced, we don't really need this anymore. So we have balanced all the forces. This is horizontal balancing. This is vertical balancing. And this is momentum balancing. Well, that's it. We should really be able to find out phi. Okay, so let's do it this way. So from this N1 is equal to W divided by mu. N1 is equal to N2 which is not multiplied by mu. Sorry. So we will substitute it here. So this is W times mu. And lo and behold, it doesn't depend on the weight of the ladder. And now we have a very simple thing. Mu is equal to cosine phi divided by 2 sine phi. Or tangent phi which is sine over cosine. And 2 goes here so it will be equal to 1 over 2 mu. Right? So phi is equal to arc tangent of 1 over 2 mu. And that's the answer. Well, I was trying to put the problems in increasing difficulty kind of things. But again, there is a common approach. All forces and all momentum should be balanced. And if we are talking about the forces, the best way is to go vertically, separately from the horizontally. So the vertical forces must be equal to each other, balanced to each other. And horizontal forces must balance each other. And obviously momentum should balance each other. So balance means that some of them should be equal to 0 basically. That's what the force which goes in one direction should be equal by magnitude to the force which goes to another direction. Okay, that's it. I suggest you to go to the website unizor.com and read the notes for this particular lecture. I think the first two problems which are relatively easy, I've given just answers. So you can just try to basically solve the same problems just by yourself and check the answer. The third and the fourth problems as a little bit more difficult. I do have solutions, but why don't you just try not to read the solution, solve the problem yourself again, check the answer. And if something is not right, read the solution presented in written format as notes to this lecture. That's it for today. Thank you very much and good luck.