 Hi, I'm Zor. Welcome to Unizor Education. Equilibrium of solids, that's what we will discuss today. Now, this lecture is part of the course called Physics for Teens, presented on Unizor.com. I do suggest you to watch this lecture from the website, which is, by the way, completely free, and no advertising, because if you found it on YouTube, for instance, searching on something, well, it contains only video. The website, Unizor.com, contains not only this course, which is called Physics for Teens, but also mass for teens, which is prerequisite for this course and some other thing. And also, every lecture has a textual notes to this, which basically serves as a textbook. And there are even exams for those who would like some challenge. So, equilibrium of solids. Well, first of all, let me just remind you that before, in the previous lecture, actually, on this topic, we were talking about equilibrium of points, and this is kind of an easy thing, because it's actually a direct consequence of the first Newton's law, which says that if all the forces are balanced, then the point, the point object, if you wish, which contains, you know, it has certain mass, etc. The point object is either at rest or in the uniform motion along the straight line in some kind of inertial frame reference, frame of reference. In other words, you can say that if the frame of reference is associated with this point object, with all the forces balanced, then this frame of reference is inertial. So, if you have a point and you have certain forces which are acting on this point, where i is an index from whatever to whatever, then the vector sum of all the forces should be equal to null vector, and this is a condition of the equilibrium of a point. And that's what we were covering in the previous lecture. Now, in this lecture, I would like to talk about solids. And the first obvious consequence from the fact that this is a solid is that this is not a sufficient condition for an object to be in equilibrium. And the obvious example is if you have some kind of a disc or whatever, and you have one force in this direction and another force in this direction, then they are, if they are of the same magnitude and they are opposite in direction, this condition is satisfied, but obviously this disc will rotate and rotation is definitely not an inertial reference frame. So, this is insufficient. All right, so we have to prevent also the rotation. Now, we did study rotation separately in the rotational dynamics topic, and we were talking about a very important concept related to rotation. And this rotation and rotation is actually related to this topic in exactly the same fashion as a translational movement relates to the force. So, what is this? Well, this is torque. Torque for rotation plays exactly the same role as force for translational movement along the straight line trajectory. And whatever in a translational movement is an acceleration caused by the force. In rotation, it's the angular acceleration. So, force causes acceleration of the point object, torque causes the angular acceleration for objects which are rotating around one particular axis. Okay, now, what if we have a very complex movement in the three-dimensional space of a solid? For instance, it can move around its own axis and then around some other axis, etc. Well, you can always imagine any kind of a complex movement of the object as the separate movement around each of the three-coordinate axes. In exactly similar manner as any translational movement along a trajectory can actually be always represented as movement along the x-axis, y-axis and z-axis. So, three functions, x as a function of the time t, y as the function of the t and z as the function of the t basically describe the movement in the three-dimensional space of a point. Now, in this case of the solid, we are talking about rotation. So, we can imagine rotation along one axis and then another axis and then the third axis. And these are kind of independent rotation. And every rotation can be basically represented as these three different rotations. Like, first along one axis, then along another and then along another. And as a result, you will have exactly the same kind of rotation which you are dealing with. So, that's why I'm not going to talk about a complicated rotation of the three-dimensional solid. I will just talk about rotation of a solid around one particular axis. One, one out of three. Doesn't really matter which one. And the torque, obviously, can be explained in the terms of rotation around one particular axis. And again, if we have a very complicated movement, then we have basically three different torques, if you wish. One rotates around this axis, another torque rotates this and another this. So, every force is causing some kind of a torque. And this force can be really considered as a separate cause of rotation around each of the coagulate axes. So, let me remind you what exactly torque is if you have a force and you have basically a position of the axis. Alright, so, here it is. If you have an axis and you have some kind of an object which is... I'm talking about the point object which is rotating, but it can be actually any solid because every little piece of that solid is basically a point object. And obviously, we are talking about basically integration of the entire solid from all these individual pieces. And obviously, whenever a force is applied to this particular point, like in case of our disc for instance, we have one force applied to this point and this to another. Each one has its own torque, basically. And since we are talking about rotation around the axis which is probably perpendicular to this board, then we can talk about the torque of this one, torque of that one, and then add them together to cause the angular acceleration. So, again, talking about one particular point, this is an axis and this is the trajectory around this axis. So, you can always have a vector which is r, which basically is from the axis to the point. And you have the force which is also the vector. And torque, by definition, is a vector product of the radius to this particular point which is rotating around this axis and the force. So, if you don't really remember this, I would suggest you to go to rotational dynamics where I explain at length what exactly the torque is and why we are multiplying our force by the radius vector and why this is a vector product. Now, being a vector product, in a simple case like this one, so you have an axis which is perpendicular to the board. These are two forces and these are two radiuses, right? This is one and this is another. Radius, radius, this is force and this is force. That's all vectors. So, the vector product is perpendicular to the components in this particular case, right? So, the vector product of r times f would be along this axis in this case and along this axis in this case. And if I'm using the right-hand rule, then it would be either towards the board or outside of the board. And obviously, in this particular case from r to r, so I think it's this way. I think it's this way. But it doesn't really matter. What does matter is it's along the axis. So, the vector product of the force which is applied to this point and the radius vector of this is supposed to be perpendicular to both of them and it's within, it's along the axis of rotation, right? So, it's also a vector. This is tau. Now, after we have basically recalled what's the definition of torque and we have also recalled that the torque is the cause of angular acceleration, well, that means that the necessary condition for a solid object not to rotate is basically that sum of all torques which are acting on this solid must be equal to zero null vector. These are all vectors. They're all directed along the axis so we can basically add them. Some of them will be with plus, some of them will be with minus. Now, in this particular case, both are in one direction. But if I will change the direction of this force to this, then if I will calculate the vector product of these two, it will have certain magnitude which is the product of magnitude of r and magnitude of f and it will be directed into one direction. The product of these two vector product would be with an opposite sign because this r, well, if this is r, this is minus r actually, right? Because it's supposed to have some direction and so the product will be negative and it will be towards another end of this axis which is perpendicular to the board. So this is a necessary condition, sum of all torques to be equal to zero for a solid to be in equilibrium. So we have basically two different things. We have to prohibit translational movement along some kind of a trajectory and this is the condition. These are the multiple forces which are acting, must be equal to zero and we have to prevent the rotation to be in equilibrium, right? To have this particular solid in equilibrium in some kind of a system. So that's another necessary condition. Now, one more very interesting point and to tell you the truth, I don't know how to really very rigorously prove it but it feels right and I would like actually to refer it to you for a judgment. So if all the forces are equal to zero, forget about the torques, then there must be a point inside the solid which is actually not moving. Well, we are talking about some kind of inertial system and we are considering our solid within this inertial system and again, if sum of all forces equals to zero, it can rotate because I'm not talking about the torques which these forces have but if they just equal in magnitude and direction in such a way that their sum is equal to zero, null vector, then there must be one particular point, well at least one particular point, maybe many points if it's just rotating, the whole axis will be at one particular position in the space, not moving, right? So again, if the forces are summarized and give you null vector, then there must be at least one point in the solid object which is not moving. That's an interesting kind of a concept. I didn't really think thoroughly about how to prove it but it seems right, right? So only the torques basically are causing some rotation but whenever you have a real rotation, you always have some kind of a point which is not moving at all, right? In the center of rotation or on the axis of rotation. Okay, now back to forces and torques. So three forces along three dimensions and three torques, sum of all the torques along every one of the three dimensions, co-ordinate dimensions, x, y and z. These represent, as I mentioned in the previous lecture, six degrees of freedom. So you have, for a solid, you have six degrees of freedom. It can move along either of the three co-ordinate axes and it can rotate around either of the three co-ordinate axes and that constitutes basically six degrees of freedom which means that some of the projections of the forces to each axis should be equal to zero and some of the torques along each of the axes, if you represent the rotation as a combination of rotation around one axis and another axis and the third axis. So on each axis it should be equal to zero. So we have basically like six equations which must be satisfied to have the system, to have the solid in the state of equilibrium. Okay, now we are talking about a couple of examples. So let me just represent whatever I'm just talking about in pictures and you will see what I mean. So I have two very simple examples. One is the following. So consider you have a thread, a rod and the distance is r here and here and you have the same weight w here and w here. So let's just examine the equilibrium of this system. Now obviously it must be an equilibrium, right? Just intuitive without this. But let's just think about it how exactly it works. So what kind of forces we are acting on this rod? Well the first one obviously is tension of the thread and it's directed upwards. Now obviously it's fixed somehow. Now then this gives you one vector which is w and this gives you one vector which is w. These are equal in magnitude and direction. This one is opposite. So all the vectors must be, some of all the vectors must be equal to zero if we are talking about equilibrium. So if this point is not moving then obviously t plus w plus w is supposed to be equal to zero, null vector which means t is equal to minus 2w. Well obviously minus is here because if this is the positive direction of the, well let's say z-axis or whatever the axis we are choosing, then this is a negative for vice versa. So that's why we have this negative. It's obvious solution to this vector equation. Now let's talk about torques. Well this force does not create any torques because it's in the center. We are talking about equilibrium of rotation relative to this center. Why? Well because again radius is equal to zero, tau is equal to radius vector product force, right? So the force is the tension but the radius is zero in this particular case. Now in this case the force is w and the radius is r. So if we consider this direction as a positive and this is x whatever, then obviously this is minus r times w in absolute terms, right? We are talking about absolute terms and this is vectors. Now this is tau 1. Now tau 2 which is this one is equal to plus r, not w. And obviously sum of these three, this, this and this. Now this is equal to zero, zero times t. So this is zero, this is the torque of this force of the tension. This is the torque which weight creates, radius times w. And this is the torque which this weight creates, this radius times w. And these two vectors obviously are opposite to each other. That's why I put minus in one case and plus in another case. So this is 1 and this is tau 2. And sum as you see is zero so that's what actually makes this system to be in equilibrium. Now the next example is very similar except that I will do it with different radiuses. Let's consider this is 2w and this is 2r and w. So this length is twice as big as this one but this weight is twice as big as this one, right? So again this is zero. So tau is equal to zero times t is zero vector. And tau 1 which is this one is minus r times 2w. And tau 2 is equal to positive 2r times w. And from the properties of the vector product this coefficient can be actually taken outside of the vector product as well as this one. And in both cases we will have minus 2r times w. And in this case we will have 2r times w. And the sum will be equal to zero. So we have increased the radius of rotation by the factor of 2 in one case leaving the weight as it is. But in another case we increased the weight. And the system will still be in equilibrium. And obviously this is the way how people are pulling with a lever a heavy stone or something. If you have some kind of a stone here, very heavy. And you would like to, this is the person. And this person is pushing down this part. So this part will go up. And if this length r and this length let's say 10r then the effort to lift should be one tenth of the weight of this stone. Exactly the same thing as this one. Well, is that the way how Egyptians build the pyramids? Maybe, I don't know. But in any case, this is basically an illustration. Two examples which illustrate how solids are in equilibrium as far as translational motion. And which is much more important for solids as far as their rotation. In one case we need all the forces to sum up to zero along each axis. And in another case for the solids we need not only the forces but for rotation to be in equilibrium, not to rotate basically. That means we have to have all torques not to be in balance along each axis of rotation. Okay, that's it. I can refer you back to Unisor.com and read the comments to this lecture. It can serve you as a textbook. And I also recommend you to be relatively fluent in the course called mass proteins. Because everything, whichever we are discussing in physics depends very much on the mass. Especially the vector algebra calculus. So you have to be really comfortable in these areas. That's it. Thank you very much and good luck.