 So good afternoon and welcome back. So most of us had, if not heavy, at least a moderate lunch. So our brain cells are deprived of oxygen, I am aware of that. So what I will do is that I will start with something light and then we will move on to little bit of heavy topics. But before I begin my topic, let me just introduce myself. I was not there in the morning, I am Mandar Inamdar, I am in the civil engineering department here at IT Bombay. I have been teaching this course or not I have been teaching, I have taught this course a few times in the past. So let me just add one thing. So I have been given to understand that this is a coordinator's workshop and most of you are very experienced in teaching this course. So for example, I sometimes, I feel almost embarrassed as to what am I going to teach you all experienced people. But only thing I just want to add is that we have been very fortunate to have been taught by some of the best people, both in India and abroad. So they have taught us something, some tricks, some understanding. And in addition to that, being at IT, we are also lucky enough to teach some extremely, extremely smart students. So they ask various questions and we have to answer them. Sometimes we can't answer. So we have to go back, think. And what I feel is that after being around excellent colleagues, being taught by some excellent people and teaching some excellent students has, I believe, given us some understanding which I hope to transfer at all to you. And what I'm also hoping is that at what I know or what I'm teaching, I don't claim at all that this is the end of everything. Surely many of you may also have some other ways of discussing the same problem, understanding a different concept. As we saw in the morning, there were various people out there who had different ideas. They wanted to communicate in a different way. So what I feel is that this is not a monologue, that I'm not just going to stand on the stage and keep saying things. But what I'm going to do is that I'm going to say something, what I understand, what I have learned over the years. I hope it will be of use to you. And at the same time, if there is something that you wish to add, if you feel that something can be done in a different way, better way, then you please share to everybody so that everyone in the room will benefit. So we are all co-learners. We are here to learn this thing in a better way than we have already learned and to communicate the same to our students. So any doubts, any questions, any suggestions, please feel free to interrupt, please feel free to ask. If possible, we can take those discussions to the end of the class. But if it can't wait, there is absolutely no problem if you stop me in the middle and we take over that discussion. In fact, that's what I want. I don't want it to be a monologue where I just say something and you listen or if you are too sleepy, sleep. So let us not have it that way. So let us keep it lively so that even I don't fall asleep and same for you. So again, I want to emphasize that it's not that I'm teaching something. I have learned something and I'm conveying that information to you. And if you have something to tell me, of course I'll be willing to listen. So the topic that we'll start today is essentially at the heart of engineering mechanics. What we are going to discuss is 2D equilibrium, just to begin with what is a simple definition of equilibrium. And specifically, we will discuss 2D equilibrium and we will solve a bunch of problems. Some simple, some reasonably complex. And when I say complex, complex as far as a first-year engineering student is concerned. So let us start with the topic. So when we sit at the bodies in equilibrium, so when I teach students, so they ask me. So they, for example, at IIT Bombay, students come by giving joint entrance exam. They have done a lot of problems in rigid body dynamics. So they come and tell me, okay, please tell me that we have done so many problems in rigid body mechanics, tons and tons of them. So what are we going to learn new? So in this course, you are doing essentially only equilibrium by and large. Okay, there will be one, two lectures on rigid body dynamics. But they don't form a significant portion of this course, it is essentially equilibrium. So students are really skeptical as to what are they going to learn new? The idea is that that in your high school, when you come straight away from the writing the joint entrance exam or student come from 12th standard, they have a particular way of thinking which is very physicsy, very mathematical. But the way we are going to discuss this is in a very engineering way, okay? So to begin with, a body in equilibrium, straight forward, as Shobit discussed in the morning, we say that a body is in equilibrium when, if there are certain forces acting on the body, these single lines are the forces, double lines are the torques, okay? So if there are forces and moments acting on the body, we say that the body is in equilibrium if and only if two conditions are satisfied, that sum of all forces is 0. And sum of all the torques about any point is equal to 0. Now how do we define torque? So there are torques or moments, okay? I will replaceably sometimes use torque, sometimes use moment, depending on what time of the day, what neurons are firing in my brain. But they are equivalent, torque and moments. So the total moment about any point, here it is taken to be origin, is simply equal to sum of all the torques or the couples, plus r i cross f i, where r is the position vector, f is the force, okay? We know, so Shobit discussed that and of course you know from all your teaching you are done over the years. Now equilibrium is as straight forward as this, and rigid bodies in equilibrium is, if these two conditions are satisfied, okay? Now we come to structures, like we want to apply this simple principle, just two equations, we want to apply that for various engineering systems. So what is the, so what is the deal with the structure? So now what we know is that, that any structure. So for example, this is a structure, okay? The chair we are sitting on is a structure, the building is a structure, car is a structure, bridge is a structure, okay? So all of these are structures. And what we know that any structure is composed of various components, okay? Those components are connected by links. And for example, if you have a static structure, for example, a building, then there are supports, a bridge has supports, okay? So all these things should come together properly in a structure such that when you put it together, okay? It every structure has some particular function. Say for example, the building is housed so that everybody can sit in here, the chair is designed in such a way that we can sit. So every structure has a function and the goal of any structural design, okay? Is to design that structure in such a way that the structure does not collapse. And engineering mechanics, what we do is at the very basic level, we try to understand what are the various forces, okay? Acting between various parts of the structure. So we want a structure to maintain its integrity. And in engineering mechanics, what we learn is that how do we place the structure in a very simple manner such that the integrity of the structure is maintained and to maintain the structure and to design the structure what we need are the forces that acts on the various parts of the structure. And in engineering mechanics, what we do essentially is to have both a global equilibrium of the structure. Means for example, the building should not collapse as a whole. And similarly, we also want the internal equilibrium of the structure such that for example, that some sub-structure should not collapse in. And that's what we are going to do. And we are going to do that by solving various problems. And what we will learn in a few moments is that a structure need enough number of constraints. If there are too many constraints, we see that the structure is statically determinate. If there are not enough constraints, then the structure is unstable. This for example, simple example as you pointed out. If I give another example that if the ground is too slippery, then there is no restraint in the horizontal direction. And for example, I can topple over if I try to walk on the ground. So that is less constraint. On the other hand, if somebody just ties me too much around, that is over constraint system, we will discuss a few problems and see how to analyze these kind of problems. Now the most important, one of the most important topic here is the condition about constraints and the corresponding reactions. So what we are going to discuss here is that there is an intricate relationship between kinematics. What do I mean by kinematics? Kinematics means, say for example, let us take this elbow joint. So this elbow joint allows my arm to rotate. So it is allowing me a kinematic degree of freedom which is corresponding to this rotation. Now that is one part is kinematics. And second part is reaction. Now for example, if I try to bend my arm beyond this, what happens is that that kinematic degree of freedom, which is the rotation, is restrained. And how is it restrained by a force at a torque that is produced at the elbow joint? So there are kinematic degrees of freedom. There are reactions. And what we are going to see here is that if a structure is such or if the connection is such that it prevents the kinematic degree of freedom, then it will give rise to the corresponding force or torque and vice versa. So let us look at it from the point of view of this simple reaction supports we have. So this one here is what, for example, we all know is a roller support. This is another roller support. Now what is happening here? That in this roller support, if you try to push this roller in the horizontal direction, there is no resistance. It will freely roll. But if you try to press it in the vertical direction, you will see that there is a restraint that is provided by the ground. And so as a result of that, roller support cannot provide any reaction in the horizontal direction. It is completely free, whereas it can provide a reaction in the vertical direction. So this is one important point is that whenever any kinematic degree of freedom is restrained, so the kinematic degree of freedom, which is the motion in the vertical direction, if it is restrained, it gives rise to a reaction. Since the horizontal motion or the horizontal kinematic degree of freedom is not restrained, there is no force or no reaction in the horizontal direction. So this is one point which I have seen that many students get confused about, and we try to emphasize that point. Similarly, there are this something called as a link. The same principle apply. If you see this link here, this is the main part of the structure. If you try to take this structure, pull in the direction of the link, the link is rigid, and it will prevent that motion. And that's why it will lead to a force in this direction along the direction of the link. But if you try to apply a force in the perpendicular direction, or you try to move this link in the perpendicular direction, there is no resistance coming from that. So the only reaction will be along the direction of the link. Is this point clear? Fine. And everybody knows this, but I just want to emphasize it in this particular language. Similarly for a slot. Now, if this is a frictionless slot, this slot is free to move along the rod, because of which it will not exert any reaction in this direction. But if you try to push this in the perpendicular direction to the rod, there will be reaction. That's why you will see that if a link is of this sort, a color on a frictionless rod, or a frictionless spin in a slot, in both case, this degree of freedom is free. So no reaction. If you try to pull in the perpendicular direction, then there will be a corresponding reaction, which is the unknown in this system. Second is frictionless spin or hinge. Now, what is free here? That there are these two degrees of freedom that if you come here, if you try to take the structure and pull it upwards, this spin will prevent that upward motion, all downward motion. So that will give rise to a vertical reaction. If you try to take it and try to push it in the horizontal direction, there is a resistance that will lead to horizontal force. But on the other hand, what you will notice is that because it's a pin, there is a free rotation because of which there is no bending moment. Ultimately, if you have a fixed support, then at this point, you are neither allowed horizontal motion, not upward motion, neither any rotation. And since all three kinematic degrees of freedom, displacement in the vertical, in the horizontal, and the rotation are constrained, this support will end up giving both forces in the vertical direction or horizontal direction, as well as a torque or a moment. These are two simple examples. Now we come to the concept of free body diagram. So this is the single most important concept in engineering mechanics. And what I have seen is that many students don't grasp that concept. They try to write equations of equilibrium on something which is not a free body diagram. So what I have seen is that we emphasize this as much as we can that anything in equilibrium, anything what you want to write equations of equilibrium, you have to always write that on a free body diagram. So what do we mean by free body diagram? As we had also briefly discussed in the morning, you have a structure with kinematic constraints. A free body diagram essentially means you remove all the kinematic constraints, replace the kinematic constraints using this rule. That whatever the kinematic constraint was that it was restraining a particular degree of freedom, we will solve many problems, it will become more clear. Then that will be replaced by corresponding force. And ultimately, yes please? This one? 90 degree force you have shown over there. It's a 90 degree, yes? Yes, but it's a direction should be towards that link, sir. No, no, no. So it all depends on what is the force that you are applying. If I am for example pushing in this direction, you will get one reaction. But that red one is the reactive force, sir. Which one? Red one is a reactive force. Right, no? The red one is a force that this support exerts on this link. Yes, so it should be towards that link. No, no, no, no. It depends on what force you are applying. If I am pulling this or pushing this, that will decide what is the reaction. So just by looking at it, you cannot say that. But, but, your rightly point, but this, I can take it to the top one, if you go to the frictionless surface. Yes, sir. I can only have a reaction in the vertical direction, there will be no pulling reaction. But in this case, because there are two contact points, that the internal slot can contact both here or here, it is capable of providing reactions in both directions. But what direction it will provide will decide, depends on the force that is being applied. Sir, here one more. Yes, please. Which in that roller? Yeah, on the cases. All the cases. Upward reaction you put. What about in the case of if I apply the load on the opposite direction? If you apply in the opposite direction, so say for example, if you see, let me come here. If you come to this structure, the first one, if you apply a reaction in the force in the upward direction, you are screwed. OK, because it cannot exert any reaction, because it will lift off. It will lift off. Yes, but if it's a guided roller, this one, then even if you try to apply, then this will be tried to get in touch with the top rail. Yeah, second one. And so it cannot be a downward reaction. So there are some rollers. So there is, for example, there is a rocker support in which, for example, in bridges, you use the rocker support. But what we know is that the typical load is only downwards, so you don't worry about it. Sir, how can we refer that reactions on such a case? Which case? Allowing the motion in the upward direction. How to refer that reaction force? When imbalance force is there, possibly it can move up or down. There are two possibilities we can expect. It can either move, it will not. Let me use the words carefully. It will try to move up or down. And it has tried to move up or down. When we are loading it, there is a possibility of moving up. There is a possibility of going down. But it depends on the load that is being applied. It depends on what direction the load is being applied. OK, we will see some examples. There should be some more clear if it is not clear. So the reaction will depend on the applied load. If it so happened that on this structure, the first roller, there is an upward force, then it cannot take the structure, cannot maintain equilibrium. Professor Rinnam, that. Yes, please. I can have a single statement to answer that question. To draw the free body diagram, we need to first of all identify the constraint. Remove the constraint. Of course, that's what I've been telling. That's what, sir. That's a single statement we can make to clear the idea as to what direction the force should be applied. Yeah, but the point is that that depends still on the load. Yes, of course. OK, that's what I'm trying to say. What I'm saying is that you remove the constraint, replace it with an appropriate reaction. But the direction, for example, the first roller, it cannot provide a support reaction in the downward direction. One simple sentence can be inserted in your statement to make it more clear. Let me make it like this. Identify constraint. Remove the constraint. Imagine the direction in which the body is going to move. Apply the reaction in the opposite direction. But that will depend on the loading. That's what we have to come to. Yes. OK, it will all depend on loading. All the questions can be answered, sir. Of course. Yes. Yeah. Thank you. Of course, yeah. But the point is that the loading should be known. OK, let us go to the actual problems. Let us get the actual loading. And then we will see what can be done. Your point is very valid. But everything ultimately depends on what the load is in the system. Is it appropriate to set it as follows? It's potato-potato, the way they say it, tomato-tomato. We can say torque or moment. It's fine. But if you feel more comfortable not saying torque, I will not force you to do that. So these are some examples. So let us go to the problems. So as you rightly pointed, we have said that we have to replace the kinematic constraints by the corresponding reactions. How? We'll solve a few problems. Simple examples of free body diagrams. So let us take the example of a beam. Beam, this is a typical representation of a pin support, a triangle and a pin. A triangle and a pin. This is a pin support or a hinge support. Now note one thing that if I want to draw a free body diagram for this structure. If I want to draw a free body diagram for this structure, what can we do? We remove this kinematic constraint A, remove this kinematic constraint B. But when we remove a kinematic constraint, we cannot just remove it. That kinematic constraint was there to do something. And by removing that, we have to replace that by a corresponding force. Now just look here. This reaction is a pin support. So the horizontal motion is prevented by the pin support. Vultical moment is prevented by the pin support. So the moment we replace this kinematic constraint at A, we have to replace that by two unknown reactions, A, X, and A, Y. Same goes for reaction at B. If you take this second free body diagram, this triangle with some gap, either you put two rollers here or you have a gap here. This is saying that this is a roller support. This is a clamp support. Now what is happening here? If I remove this kinematic degree of constraint, then at this point, translation is prevented. Vertical translation is prevented. Rotation is prevented. So we replace them with a vertical reaction, a horizontal reaction, and a moment or a torque. At this roller, yes, please? The magnitude of A, X, R equal to 0, no? They may be, but you have to still put them. If they are 0 or not, they will only come after you apply equations of equilibrium. You cannot, a priori, just dismiss that because what happens is that the bookkeeping will really get destroyed. Because you may think there are three equations of equilibrium, but you already made it 0 and somehow everything will get topsy-turvy. So we put it and then make it 0 later on. The better way is to make that force F inclined. Yeah, you can do it that way. F inclined at an angle. But two unknown reactions, if you look here, now again, see, better or worse, OK, see, again, that depends on the personal taste, personal idiosyncrasy. But if you look here, you can have a vertical reaction at an angle, OK? But vertical reaction, so you can have a force at an angle that is equivalent to saying that there are two unknowns, Fx and Fy. So whatever is convenient, and you are totally right, that there will be some problems in which representation in the way you are saying will be more beneficial, but there are some other problems, even if you have Fx and Fy, it will be better, OK? We will see those problems. Yes, sir, at this stage, I feel. Yes, please. Degrees of freedom also has to be introduced. Which? Degrees of freedom of a particular type of support, and then degrees of constraint. But that's what I introduced, right? So just correct me, is that not what I introduced? I mean, that hinge, actually, it has only one degree of freedom, that is rotation. It has two degrees of constraint, one is horizontal movement and the other is vertical movement. That's what I said. And if that force is made inclined, then Ax will also not be equal to 0. Yeah, but inclined force at an angle is equivalent to saying there is Fx and there is Fy, OK? That's two ways of saying the same thing. Depending on what problem you are doing, you may have to do it one way or the other, OK? That's what essentially, what's what I'm saying, right? You replace the kinematic constraints, which are the degrees of freedom, but you cannot replace them just like that. You have to introduce forces, and how are the forces that depend on? As in this case, only degree of freedom, that is free is the rotation. This displacement, this displacement are constraint, so you replace that with two reactions, unknown reactions. Same here, these three degrees of freedom, all of them are constraint, so you get Ax Ay at a bending moment. At a roller, there is only one degree of freedom that is constraint, which is the motion, or the degree of freedom in the vertical direction, so you replace that by reaction in the vertical direction. These are some simple examples of free body diagram, that in this case, there is a hinge support, there is a roller support, we replace that, OK? With two reactions, one vertical reaction, a clamp support replaced by three unknown reactions. This, OK, if this is a frictionless support, smooth surface at contacted mass m, only reaction can be in the normal direction, this is n. This is a pin support, two degrees of freedom are constraint, correspondingly, two forces, and so on. So far, so good, I think everybody is on the same page. In fact, people are going ahead of this topic already. So should I move on to the new topic, or are there any other outstanding questions about what we have done till now? Any outstanding questions? So the idea is that, when you draw a free body diagram, you release kinematic degree of freedom, what is kinematic? Yx, Y, rotation. So you release kinematic degree of constraint, and replace that by corresponding forces. In case of such a small support, yeah, two direction moving along x and along y. No, not along x and along y. Along y, this motion is restrained. If you say that this is y, this is x. We are extracting along the z axis only. But the mobility along x and y is there, no? So let's see. We are talking about currently 2D supports, OK? We are not going to 3D. Shobik is going to discuss that today afternoon. Right now, these are all 2D supports. And what a 2D problem is? I will come back to that in a few moments. But these we are discussing only two dimensional systems. That there are only three degrees of freedom, x displacement in the y direction, in the x direction, and rotation. 2D systems of equilibrium, OK? 3D, there will be extra degrees of freedom. There will be this, there will be this, and there will be z motion also. That Shobik will discuss when we discuss this 3D equilibrium in tomorrow afternoon, OK? But these are 2D equilibrium problems right now. And what a 2D equilibrium problem is also that I will come to that in a few moments, OK? Now, this is one thing, OK, which I have emphasized our students a lot, that in 2D equilibrium or a planar equilibrium, you cannot write more than three equations of equilibrium, OK? Any time you write some students, write four equations, that's a straight penalty, OK? If you write more than three equations for a planar equilibrium problem or a 2D equilibrium problem, that is not legal, OK? Not allowed. Now, we can write equations in so many different ways. You can write equation equilibrium in the x direction, force is balanced, sigma fx equal to 0, sigma fy, some of all forces in the y direction is 0, and sigma ma is rotation of all the torques by taking torque about point a to be equal to 0, for example. We can also use equations like that only sigma fx equal to 0, sigma ma equal to 0, sigma mb equal to 0. So one force, two moments. And the most general case is if we take a point c such that abc are not collinear or not in a straight line, then all these three equations can be written which are equivalent to these other two equations, sigma ma equal to 0, sigma mb equal to 0, and sigma mc equal to 0, OK? So general way of writing down equations of equilibrium for a 2D or planar body. So we'll discuss one very simple problem, OK? So what we are asked to do in this problem is we have a rigid link, ACD, which is pinned at point c. On this link, we are applying a load of 150 newtons at one point, and there is a string that goes around abc, which is where this is a roller or this is a hinge, whatever the way you want to look at it. And what you're asked to find out is determine the tension in the cable and support reactions at c. So assuming that the friction is less pulley. Puli, when you say frictionless pulley, do you mean the friction at what point? That's a good point that you raise, OK? So let me ask you friction. Contact friction only over the point. Contact friction did not be 0. It did not be 0. Then there will be the difference in tension on either side, sir. Only thing you need is that at this pin, it is well oiled. That is free to rotate. That is the only thing that you need. Means same tension will develop. There's the same tension will develop. And that's what we'll discuss in the next free body diagram. Is that this is one point, for example, where students ask that should this pulley, let me show it here, is one thing, one point of confusion which I've seen many times, that is this pulley frictionless or not frictionless? And when I say that, which means that is there a possibility of friction between this rope and a pulley, this rope and a pulley? The point is that there can be friction. Means the coefficient of friction between the rope and the pulley need not be 0. Only thing you need is that at this point, the pulley is rotating about something, right? It is rotating on a screw. That should be well lubricated. There should be no bearing or general friction between that pulley and the support. That is the only requirement you have. That's OK. Then the tension will be acting through the point, sir. It is required for what? To say that the tensions are equal on two sides, OK? Of course it is not required. Even then tensions will be equal. Of course not. No, because there will be no rotation of the pulley because it is in equilibrium. There is no, it is not a question of there is rotation or no rotation. There is a question of that can the pulley, if you let it rotate, can there be a friction between this axle? Let me tell you the complete thing, what I am thinking. OK, fine. If it is not rotating, then we can find out the moment of the center of the pulley of the set, this is T1, this is T2. T1 into R minus T2 is equal to I into alpha. If alpha is 0, then T1 equal to T. No, let me do this, OK? Since you are asking this, let me do this. Suppose this is the free body diagram of the pulley, OK? What will you draw here? This is the center, which is rotating. There is a hinge, so AX, AY. OK? So only thing that you need, but what if, for example, I don't lubricate the hinge? If the hinge is really fixed, then that degree of freedom is constrained. So it can develop, in principle, this torque. It can develop because of friction. For example, if you have a rusty pulley, you don't lubricate it for months together. It will get jammed on the axle, and it will not rotate freely. That will act, for example, that will prevent the rotation in the direction, in this rotational direction. Of course. What about that hinge? My question is, if I draw the free body diagram of the pulley? Yeah, but when you draw free body, I am taking about three, but a moment of the center? No, no, no. But the point is that when you draw free body diagram, what are the support reactions at the hinge? What are the support reactions? If you tell me, only when that support is well oiled, can you say that the only support reaction is that the translation is prevented? So exit reaction, vertical translation is prevented, so why reaction? But this rotation is free, so no torque or no moment coming from the axle. Excuse me, sir. I would like to explain his question. Sure. OK. No, you don't answer. You talk to me. Yes, yes. When frictionless pulley is there, when oil is there, that will act as a hinge support. Yes, perfect. Yes, and wherever hinge support is there, two reactions are there, HA and VA. Perfect. And when that is a rush pulley, according to you, it is not going to rotate, then it will act as a fixed support. Yes. That's why three reactions are horizontal, vertical, as well as moment. Yeah, and he got it. OK, thank you. OK, so thanks. OK, that's the only bottom line. So the point was that at these points, there can be friction, the coefficient of friction between this rope and this pulley need not be 0. Only thing that you need is that this movement should not come. And this movement will only not come when that it's a proper free joint. Is that fine? And when that happens, now you take this free body diagram, T, T, take torque about this point. This is A, this is A, this is T. So for equilibrium, for moment about this point, it can only happen if this T1 and this T2 are both equal to T. So that's the only requirement. The friction can still be there between the rope and the pulley. So now we ask ourselves the question first. Now the idea is that what is the tension in the pulley doing? We ask ourselves this question. Before even solving this problem, as the professor there also mentioned, that we try to see in what direction or what happens to the, if you remove a support or you remove something, what will happen to the motion? So we will think in that direction only, okay? So now if you remove the string ABD, suppose you say that there were no string, there were no ABD, what will happen to this structure ACD? So it will try to rotate counterclockwise about point C. So that, what does it immediately tell you? That immediately tells us that what is this tension doing? That for the free body diagram for ACD, this tension will try to prevent rotation of this in the anticlockwise direction, okay? So immediately we know what is our free body diagram and what is our equation of equilibrium? So what we do here immediately is that for this free body diagram, we replace the hinge support at C with two reactions, FCX, FCY, we replace this pulley with this string with tension T, with tension T. And we saw just a few moments ago that by drawing the free body diagram of the pulley, we see that the two tensions are equal. Because the torque has to be balanced or there should be moment equilibrium for this pulley about the center. Now what do we do? We just saw that in the absence of tension, there can be rotation of this free body about point C. So we just draw sigma MC or the moment of all the forces about point C to be equal to zero and that will immediately give us what is the tension? Straight forward, okay? So what we have done is that that if you want to find tension, we remove that the string which is providing the tension and see what happens in the absence of that. In the absence of that, we'll try to get the rotation, the entire assembly will try to rotate about point C. So we immediately know that for that particular free body diagram, we need to write down the equation of equilibrium by taking torque of all the forces including the tension about point C. Professor, is this the free body diagram of the pulley? That's the free body diagram of pulley. So there should be another force, balancing these two T. Yeah, so that's the two forces. Yes, another force should be there which will be the resultant of these two forces opposite to the direction. Yeah, so the point is this, is those directions, I have chosen some direction that may turn out to be different. So the sign will automatically flip. Those two reactions which I've drawn at point B, those are the two. FBH and FBY. Yeah, that is FBY. I have not written FBY because we don't concern about them. I don't want to needlessly complicate my free body diagram. So I don't draw, write FBX, FBY. Because the point is that those reactions, I don't care about them. They are not asked in the problem. So I don't write FBX, FBY. So it is there, it's a proper free body diagram. So this is just a brief category of what are different categories of equilibrium in 2D. Straight forward. If all the forces, and I think Shobit discussed briefly in the morning, let me just quickly gloss over it. If all the forces are collinear, then the two equations, which is the equilibrium in the y direction and the torque becomes redundant. And the only equation you need is sigma FX where this is the x direction is equal to zero. If the forces are concurrent, then the moment balance is again a redundant quantity. Only thing you need is sigma FX equal to zero, sigma FI equal to zero. If the forces are parallel, then along y direction, the equation is previally satisfied, equation of equilibrium. So we only need equation for sigma FX equal to zero and moment about the z direction is equal to zero. And the most general case is what we had seen a few moments ago. Now, this is a very important thing. Is how do we know that for a given rigid body, if the constraints that we are providing are they adequate or are they not adequate? So let us come to the first system here. What do we have? We have this blue rigid body, which is supported by a link here, another vertical link, another third link. Do you think this body is properly constrained? Each link can provide reaction in the horizontal direction as we had seen. Vertical reaction can provide a force in a vertical direction. The second link can provide a force in a vertical direction. Are they support adequate? What do you think? Now, suppose we remove this link and put it at another point, such that the line of this link or the line at which the reaction will be obeyed or reaction will be provided will come and meet at point A. Is this structure adequately constrained or not? If it is not constrained, can you tell me why? If you apply any force, which is not passing through point A, there will be a torque about point A or moment about point A that none of the supports will be able to balance and the system cannot be in equilibrium. So this is inadequately constrained system. Suppose I do this, put three links like this. Is the system adequately constrained? No, why? Because all these reactions are only horizontal, so they are all parallel to each other. So if you apply force in a perpendicular direction, it cannot be balanced. So this is not adequate. And lastly, if I apply one extra link, this was already fully constrained. So this is a statically indeterminate structure and these kind of structures are topic of structural mechanics and we are not going to discuss with stress structures in this course, okay? Find any comments, any questions? This one? This one? Okay, so note here that in this one, this link can provide reaction in the horizontal direction. This link can provide a reaction in the vertical direction. This inclined link can provide a reaction in this direction, but note that the reaction will pass through point A. So all the three reactions are intersecting at point A. Now what does that mean? That if I apply a force on this body which does not pass through point A, then it will create a moment about point A or a torque about point A which the reactions won't be able to balance and the structure will become a not in equilibrium structure. How will you know that the reaction from the link will be passing through the point A? In this, I have shown it, okay? In any problem, you have to look at the geometry and figure out that if it is going to pass or not. In this case, it is given. But for example, if you know the geometry of the problem, know the dimensions, then you can draw the lines of the reaction and figure out for ourselves there are the reactions passing through common point or not, okay? So let us briefly discuss another problem which is a single rigid body, okay? We are still talking about just one rigid body which is supported by various supports and strings and whatnot, okay? So one quick example. What we have here is this kind of garage door mechanisms typically we don't see in India but are abundant in the West, especially in the US, okay? This is a garage door. This is where the car will come from. And what does this mechanism do? That this is a door. I'm looking only from the side. There is a guide through which the door can slide in the horizontal direction at point A here. And at point B, okay, the door, okay, there is a slot through which this pin can slide. And the same mechanism is there on the other side, okay? Is the mechanism clear? Straight forward mechanism. What we have is that that if you want to open this door, how do you open it? Point B will move upwards. Point A will move sideways and the door will open. If you want to close it, point B will move downwards. A will move on to the right and the door will close. And the same mechanism is there on other side. There is one mechanism here. On the other side, there is a mechanism, okay? So even though it is really a 3D problem by symmetry, because the same mechanism is there on two sides, this becomes a 2D problem. Now how is this door operated? The door has some weight. And to close the door, okay, we release this tension. To open the door, we increase this tension. And what you are asked to find out that what is the tension that is required to maintain this door in equilibrium in this configuration is what we are asked. Now the first question is that, what if there were no tension? If there were no string pulling on the door, what will be, what will happen? Okay, the door will just close. Why? Because there is a weight acting here. What will that weight do? So look at this. Support A, what will be the direction of reaction at support A? Vertical? What is the direction of support reaction at B? Horizontal. They are meeting at one point, okay? Where are they meeting at this point here? Is the weight passing through that point? No. So there is an instantaneous rotation that can happen about that point because of which if you don't provide a third reaction that it is a mechanism and the door will simply close. So what is the string doing is it is preventing the instantaneous rotation of this door about point O. That if there were the tension where zero, then this door will start rotating about point O, keep rotating that point O will keep shifting and the door will ultimately close. So what is the string doing? It is preventing rotation. The tension is preventing rotation about point O. And so the mechanism is immediately making it clear for us that what we want is that the tension is preventing rotation of this door about point O. So we need to take for this three-body diagram, moment or torque about point O of all the forces. So reaction at A will go through point O, no torque. Reaction at point B will go through point O, no torque or no moment. Only torque or moment is provided by T and W and they are just balanced and in one shot we can obtain T. So what we have done is that we release that support, see what happens and then decide what should be our, yes please. Can you draw that three-body diagram you didn't assume what is the direction of R I? I have. That is one thing I will tell you. You should get it negative, right? Yeah. When you solve it, you are supposed to... You can get a negative answer, yes. No, in this case it has to be negative. It will be negative clearly. But you have shown it as possible. Oh, I should, that's a mistake, I should put a minus there. Okay, I am wrong there, okay, fine. Okay, so what is pointing out and rightly pointing out is that I just assume that the direction of reaction is downward, okay, without losing any generality. But in this problem, if I take equation of equilibrium in the y direction, you will see that R A plus W equal to zero and R A will come out to be negative. By mistake, I forgot the minus sign. You are totally right. The reaction should be upwards and since I have taken it downwards, the resulting answer should be negative. They are not intersecting at all. Why, they are still intersecting. The intersection need not be just the arrows pointing each other. The line of action should intersect. It is not the arrows intersecting. The line of action is what is important. Yes, please. Except for flexibly, for all kind of predictors, we can assume the reaction on both directions. You can assume any direction. If we assume to the wrong direction, we will get the negative answer. Yes, and then you can look at the support. And for example, there is one problem that in, I can discuss that problem. For example, if you have support which is on like this and if the reaction becomes negative, then it just means that the structure will become unstable. But typically, you are right that I can assume any direction, either up or down, sideways, leftwards or rightwards. And if our assumption to begin with was wrong, the answer you will get will be negative. It was a well-known fact. No rocket science and no new discovery. But I made a mistake here. So thanks for pointing out. Now comes the more important topic. We just discussed the two problems that we have solved a few moments ago, where just multiple rigid bodies, sorry, a single rigid body with end reactions. But typically, we don't have just a single rigid body, okay? There is one rigid body. We connect that to other body. Look at our body. We just are not one whole self. We have one bone connected by a link here. This connected by link, fingers connected by link. For the simple reason, is that every structure has a particular functionality. And if you make everything rigid, then that functionality is not obeyed. If all the joints were rigid, then we would just be a statue, okay? So in order for not that to happen, is that that various rigid bodies, like various rigid bones, rigid for all practical purposes, are connected by various links. And because of that, a lot of interesting behavior happens, okay? And what kind of problems can come because of that, okay? We'll discuss that, okay? In a few moments. So one very simple example, okay? So what we have here is an exercise machine, okay? So if you look, zoom in here, this is a post, this is a bracket, and this is a weight arm, okay? Now in engineering mechanics, the idea is that any real life structure is very complex. But the idea is that in order to get the forces that we want, we want to reduce that complex structure into a caricature, or into a cartoon. And then that cartoon is easy for us to analyze because instead of drawing all the complex details which are irrelevant. We just go to the crux and find out, okay, that what should be the typical structure and what is relevant for us. Now, the point that I will discuss briefly is any real life problem, okay? We live in a three dimensional world. So any problem that we will encounter will be a 3D problem. So the question that, for example, many times students ask is that, why are we solving 2D equilibrium problem? What's the deal with it? Like it is really a 3D world, why do we solve 2D problems? And the reason is briefly as follows, okay? So let us zoom on to this, okay? So this is the weight arm, okay? This is the pivot. Now, for example, sometimes showing these pictures really helps students to just see that, for example, how are various connections made and what are the possible reactions? So as you can see here, that if you zoom in here, that this is one rigid body, this is another rigid body, and those two rigid bodies, this one and two, are connected to each other through this link. So just see here that there is a hole here, okay? There's a hole or a slot, and this is a rigid post, and what we do here is that, that this screw will go through this hole and connect one rigid body with the other rigid body, such that this degree of freedom or the rotation is free, fine? And this, for example, if you want to go into details, say what is happening at the pin connection? At the pin connection, what is happening is that, come here, so this is the hole, this is the pin, so what happens is that, that there is a hole, there is a pin. Now, depending on what force we apply, okay, or what is the position of the weighing arm, you will see that this internal screw will either contact at this point or at this point or at this point. Now, depending upon what load you apply, the screw will keep moving and make contacts at different points, but in the end, what we say is that, that this internal screw is free to make contact at any point in the circumference, and so it can apply on an average reaction in any direction, which means that we can say that a general reaction that can be provided by the link will be a reaction in the horizontal direction and a reaction in the vertical direction, okay? So this is what the caricature will look like that at the pin and the bracket contact, the contact will happen somewhere here. Look at this one, okay? There the contact will be a line contact. Now what we will see is that, that this is the full 2D problem, full 3D problem. These are the reactions coming from the left, reaction coming from the right, this is the weight, this is the person who is exerting some load on it, coming from the left, coming from the right. Now, in a case where all these reactions are symmetric, that AL is equal to AR. For example, when you typically do exercise, you don't want to really have an imbalance. So if AL is equal to AR, or the reaction on the left is equal to reaction on right, reaction support, this force reaction here is equal to, L is equal to F of R, then this becomes a perfectly symmetric object, and you can just look at it from one side and say that this 3D problem, because of all the symmetries involved, I can just look it at one side, same like the garage door problem, and that seemingly 3D problem can be reduced to a 2D problem, and what happens in the third dimension? Actually is irrelevant because of the symmetries involved in the problem. It's fine, so far so good that what is the logic behind, for example, at why can, even though the real world is 3D, we can use 2D problems. It is because we exploit the symmetry of many structures and convert the real 3D problem into a 2D problem. Another case, even though, for example, you take the case of a scissors, okay? Even though this is not a symmetric problem, okay? It's not a symmetric problem, but if I look from the top, you will see that the third dimension is very small as compared to the two dimensions. So all the forces are approximately planar. So even in this case, you can solve this seemingly 3D problem using planar equilibrium or 2D equilibrium. Now let us come to this one simple problem. It's a planar problem. So in this case, we have this rigid rods, we have pulleys, we have string, the third dimension. The third dimension means out of the plane is much smaller compared to the other two dimensions. So it is, even though the real problem is 3D, we can use simple 2D techniques in order to solve this problem. Now in this problem, okay, what do we have? What are the different rigid bodies present? Rod AD, rod ACE, this pulley, this pulley. So this entire thing is a combination of four rigid bodies, which are pinned here, pinned here, pinned here, pinned here, and there's a string that is tied around here. So it's a combination of many rigid bodies. And what we are asked to find in this problem is determine the components of reaction at support D and support E, okay? Now the idea is that because there are so many rigid bodies. If you want to draw. We need to say that A and D are different bodies there. Yes? Which one? I should not answer. Which? A pulley, D pulley, what are they? Yeah, A pulley is a rigid pulley. It's a rigid body. This is another rigid body. Can you call it as a rigid body? Of course. Why not? What is the difference between body and particle? It's an extended body. For example, this force, okay, can create a torque about this hinge point. If it is body, can you apply the moment equation on that? Yes, of course you can. By taking the individual three-body diagram of the pulley. Yes, that's what we did a few minutes ago for the pulley. It's an extended body. It's a rigid body. Okay, means you can take the equation or in fact, that equilibrium will tell you that this tension is equal to 4.8 and this 4.8 is equal to this 4.8. Why? Because I draw the three-body diagram for this pulley. Of course, but you have not mentioned the dimensions for that pulley. Pullicle dimensions, it's a there, there's a radius of 250 millimeters there now for the pulley. You have to read the problem. Okay, the dimension is given 250 millimeters. So the idea is that there are four rigid bodies in this problem, but the only quantity that is asked is reaction at D and reaction at E. So in principle, we can just completely disassemble this problem and write down all equations of equilibrium. But is that a right thing to do in this case? If you want to just brute force it. For example, there are some students, what they do is that in the exam, like you give some problem, they will disassemble everything, who are unknowns, they will write some 10 unknowns, write down equations, not get any answer, then come to me in the end say, I written 10 equations, 10 unknowns, you can solve. So, but the point is that, is that the efficient way of solving this problem? If you only want reactions at D and E, what do you think would be a judicious choice of free body? Because these free bodies, and for example, I can take a combination of these free bodies, can also be a free body. Okay, whole system can be a free body. Okay, so now if we take the whole system as a free body, how many unknown reactions will be exposed? Because to make the whole system as a free body, we have to release this kinematic constraint at E, release kinematic constraint at D. Now, because it's a pin support, which is preventing these two degrees of freedom, two unknowns, same at D. We will release two kinematic degrees of freedom, displacement in the horizontal and vertical, two unknowns. So, we are exposing, we are getting four unknowns, when we just draw the complete free body diagram. Okay, so we cannot solve for all unknowns, because we just saw that we can write only three equations of equilibrium. So, four unknowns, one unknown will always be left. But still, can we find at least some reactions at support D and E, using the complete free body diagram? Okay, reaction at D, there are two reactions, one vertical, horizontal. Yes. The reaction at D will be in the direction of D A. In the direction of? D A. D A. No, not necessarily, that's not a two-fold. I'm not committed, it's not in the direction of D A, not necessarily at all. It will take moment of D. Moment of this full free body diagram? Then we will get one unknown. We will get horizontal reaction at E? Yes. Perfect, because what is happening is that, that if, okay, for this free body diagram, if this support were not there, what will happen, this entire structure will tend to rotate about point E. Or if I remove support E, then the entire structure will tend to rotate about point E. We will get EX. We will get EX. Okay, is that point clear? What I'm doing here, I'm not solving the full problem, I'm just trying to qualitatively discuss that there are so many different things we can do. What we are asked is to find out reactions at D and E. What may be the better way, or what is the engineering way of solving this problem? As this professor pointed out, that we can take the full free body diagram. For this full free body diagram here, free body diagram two, we can take reaction about D to be equal to zero. That will give us what is F of E? Sir, excuse me. Yes, please, where, where? Priya Slash. The full link. ABD is a straight one. ABD is a straight one. ABD is a straight one. Therefore, actually the reaction. It's a full, it's a full rigid body. The reaction component in the link, remember DA, should be along the line DA only. No, no. No, that is not true. This is if you're, sir, if you're cutting at in between B and D, if you cut that. Okay, so what free body diagram do you want me to draw? Okay, let me do it here. Okay. Only cut. Only cut? Only cut between B and D. No, only cut between B and D. Yes. It's really bad. No, no, no, sir. If you just cut it. If I cut it, yes. Only force will be along ABD. No, no, why? Because if you cut here, note that what is this rigid body doing? If I look at two sides of this, this rotation is prevented, translation is prevented. So there are all three possible reactions that will come in here. There is possibility there will be Fx, Fy and moment. And because of that moment, you will realize that the reaction need not be along direction of AB. Because the internal moment will also get exposed. If this entire structure where one BD pinned with another AB, then what you are saying is right. If there are two separate ones which are pinned, then what you are saying is that there are two force members. But this is one whole thing. If you make a cut, you will expose a bending moment. If you start from A. If you start from A. Yes, let's start from A. Yeah. This is like little bit at this pulley there will be 4.8. And opposite side that same will be the 4.8. Where? Where? And A to B. A to B will be 4.8. Yes, because this pulley is free to rotate. Okay. So at the point, at the joint A, we are having two unknowns along AB and AC. AB and AC. Now joint A, it will be more complicated. It will be more complicated. Why? Because this ABD, if you remove ABD, this rod, this beam will exert two unknowns. No. If I remove ACB, there will be two unknowns. Okay. So essentially there will be 2 plus to 4 unknowns there. Plus extra unknowns coming here and all of them will balance to form equilibrium. So removing this pulley is a very bad idea. Only the part of AB, some part of the AB we are considering. But if I remove only a part of AB. Yeah. Then what will happen? You will expose an internal bending moment because you are making a cut in the rigid body. If you make a cut in the rigid body, in the rigid body what happens? No rotation, no translation. So all three reactions can be present. So it is never a good idea to disassemble somewhere in between. Okay. Because you are exposing all three unknowns. So the good idea is only to make a cut at the joints or at the supports. Why? Because there are less possibility of having this kind of issues there. Okay. So making a cut in the rigid body is always a bad idea. You will not gain anything. Sir, excuse me. Yes, please, where? Yeah. Yes. Sir, any system, if it is in equilibrium, under the action of three forces. Action of? Under the action of three forces. They have to meet at a point. Yeah, they must be collinear. Yes, no. Concurrent. Yeah, they must be concurrent. They must be concurrent. Fine. So by applying that fundamental, if they are concurrent, then we can find out reactions at the end. You can try that. Well, the point is that the geometries will be kind of painful here. Okay. You can try that. Okay, this three force business, personally, I don't like it. There are these three force members. Okay, two force, three force. Personally, I don't like it. There are some problems where it is good, but if you are really good at geometry, now that you can really visualize where are arrows going, it is great. Okay, but that's a huge pain. Okay, so if you like it, okay, please feel free to do it, but I don't like it. Okay. Excuse me, sir. You are right. In some cases, it can be very helpful. Yes, please. Excuse me, sir. Where, where? Yes. Just open the pulley, which is at A. Okay, if I open the pulley at A. Okay, then you will get a two reaction at the center and an axle. How many reactions will you expose? Two, two, two. You will expose four reactions. No, we are going to draw a free body diagram of pulley only. Yeah, but then pulley is connected to rod AD. Yes. So, when you remove that rod AD, this one is exerting two reactions. Yes. So, you can calculate. But then you will also remove ACE. No, first of all, we are drawing free body diagram of pulley only. Just pulley, okay, just pulley. So, you will get two forces. One is 4.8. You will get four forces. No, no, sir. Four. Yes, four forces are, but only two unknowns are there. All four unknowns. No, sir. We are going to just draw the pulley. I am talking about the pulley. But at pulley, there are four unknowns. That's what I am saying. Only two reactions are there. No, that's what, like, just remove that. If you remove this ABD. Okay, see, all these three, I can think about it like this. No, sir. No, no, no, listen. There's a pulley. There's a screw coming out of the pulley. On this pulley, one member, two member. So, if you completely remove this pulley, you are removing one member. You will remove this ACE, two unknowns. You will remove ABD, two more unknowns. So, this pulley will have four unknowns when you remove. Think about it, because every time, because to draw the free body diagram of the pulley, you will have to remove this ABD. When you remove ABD, what can ABD do? It can exert a force in this and this direction because those degree of freedom are constrained. Two unknowns. In addition, I have to also remove ACE. So, two more unknowns will come. So, four unknowns will come at pulley. Sir, force exerted by ABD at the center of the pulley. At the center of pulley. Okay, ACE, also going to exert at the same point. And that center of the pulley and the axle of the pulley will act as a hinge support. So, two reactions are there, horizontal and vertical. Only from ABD, only from ABD, there will be two reactions. Only from ABD. That is a common pin, sir. But so what? That is a common hinge for both the rods. Yes, but the point is that when you remove any rod, okay, when you remove, there's a screw. Suppose you tell me one thing, suppose instead of having three. Sir, I have a point here, sir. Yes, okay, who is making a point? This side, sir. Okay. This side, sir. Okay, yeah. Now, as a matter of fact. Okay. Both of you are saying the same thing. Yeah, of course. I don't know, of course. I'm saying something, I believe he's saying the same thing. Let me intervene, let me intervene. Okay. What Professor Inamdar is trying to say is, he is cutting the member AB. I'm not cutting, I'm removing it. What other people are saying is, what other people are saying is, treat this as a truss. No, you can't treat as a truss. Just a minute, just a minute. It can be done, it can be done. Let me tell you, sir. It can be done, it's not a truss. Excuse me. Sir, let me tell you, sir. Let me complete, please. Let me complete. Okay, this is not parliament, okay. Let us have some decorum. Sir, we know, we know there are two reactions at D. We know there is one reaction at E. Right? No, no, no, no, no, no, no, no. We know two reactions at E, two reactions at E. Not one. Two. Sir, it should be made clear. It's a pin. It's a pin support. It's a pin support. No, no, agreed, sir. But once we analyze it, we realize that the vertical reaction at E would be zero. No, it won't be zero. The other way of explaining it is, The other way of explaining it is, Let me finish, let me finish, okay. Because what happens is, because everything is coming from this one system, okay, I cannot figure out who is speaking from where. Yes, let him finish, please. See, we can take the joint here. We can take the joint here. No, no, when you say take joint A, okay, okay. Find the force in the member AB and AC. No, no, that's what I'm saying, AB, only AB. And AB is not independent, sir. AB is not independent. Once again, what happens is, AB and BD. But what do you mean by force in member AB? There is no force in member AB. It changes as a function of wherever you are. No, no, one more point is, actually, this cannot be done. I'm trying to drive in the point, saying that this cannot be done as method of joints. Of course not. It cannot be done. Because this is not a trust, because trust may, all the forces are axial. And there are internal moments. Yes. Yeah, so that's the point. Okay, I agree with you. If you are saying that, then I fully agree with you. Sir. No, no, no, no. In simple word, we can say like this. D, member D, B, A, C, E. Okay, okay, okay, where does it come from? Okay, okay. Okay, okay. Okay, member D. Okay, D. D, B, A, C, E. If it is one piece. It is not one piece. Then if it is one piece. But if it is one piece, then I am Prime Minister. Okay, but it's not one piece. No, no, no. No, no, no. No, no, no. Just lightning it, okay. As it is not one piece. As it is not one piece. That's why there are four unknowns at Puli A. Yes, yes. I see. You are totally right. Totally right. I totally agree with you. What do you say? If it would have been a single piece, then there will be only two unknowns. Yes, yes. Okay, thank you. Okay, this is totally right. That's it. What have we been trying to convey? Wait, wait, wait, where? Sir, here, from here. From where? Okay. Now, simple question is that we have to calculate reaction. Simple, I hope, yeah. Simple means what I'm saying. We have to calculate only reaction at D and E. That's what is asked. Yes. Take moment about E, you will get D. What free body diagram? Take moment of what free body diagram? We will draw D, one horizontal one vertical. P is there. Free body diagram, which free body diagram are you talking about? Enter, we will complete entire figure in equilibrium. We will show weight, 4.8 kilo Newton. Then at E, two reaction will be there. At D, two reaction will be there. Horizontal vertical. We will take moment about D, we will get reaction E. We will take... Only horizontal reaction. Then we will take moment about E, then we will get reaction about D. Horizontal company. But you don't get only get some of the vertical reactions, you don't get them individually. Then we will take some of it. I think that's the thing, see for example in a class, the blackboard is so convenient. I really hate writing on these things. Because with blackboard, we could just draw the free body diagram and be done with this once and for all. Okay, so let us do this. Okay, let us hope that I can draw some diagram here. Okay, so if I draw this pulley, okay, as you pointed out, if I look from the side, how does this pulley look like? It will look like this and this screw is sticking out of that pulley. Okay, there is a pulley, a screw is sticking out of the pulley. Now what is happening is that on this screw, there is one member which is pinned, okay? Which is this? Okay, on this pulley, there is another member which is pinned, which is this. So if I look from the front, what you will see is that if this is the pulley, this is the screw coming out, one member is this, other member is this. And so if I remove this member, disassemble, this is going to exert two reactions. Why? Because it is constrained both in the horizontal direction and the vertical direction. So if I remove this one, you will see this plus two reactions. In addition, if you also remove this member, then two extra reactions will come and so your pulley will have this four unknowns and whatever this force which is being exerted. So there will be four unknowns on that pulley. Is that fine? Great, okay. Yeah. Yes, fine. No, I totally agree. No, this is, for example, like we have thought about it because our students also have asked us these questions and then we have to think ways of telling them. So it is not that like I am just like pulling it out of the hat. I am not talking about it. I am just telling you. Once you have done that, it's fine. Is this fine? Okay, is this fine? Fine. Okay, so now let us move on. Okay, so let us move on now. So what we see is that exposing the pulley is a very bad idea. Okay, it's a very bad idea because you are exposing four unknowns. But we ask ourselves a question. Suppose this point E, let us ask. This is a 2D problem. This is a 2D problem. Why? Because the third dimension which is out of plane. C, C, C. No, no, no. No, no, no. You have a very valid point, but let me put it this way. Okay, that if this screw, I have now drawn it bigger for exaggeration. This screw, because your dimension are very small compared to the other dimensions. So I am taking that as a pseudo 2D problem. This is drawn so big only for our convenience. Just to demonstrate. This screw is very small compared to the other dimension. So everything is almost planar. Okay, this is exaggerated. It's a very good point, but that's a point. That why this is still a 2D problem? Why? Because the screw size is very small compared to the other dimensions which are in meters. This screw will be in some millimeters. Now let us ask ourselves one question. And the question is as follows. What if E were not a pin support, E were a roller? One reaction is lower, but is this system stable? Why is it not stable? Because if you look at this A, C, E, what is happening? This tension will rotate this rod about point A. Okay, so that's why if this E becomes a roller, the system is not in internal equilibrium. Then what is, then immediately that tells us that if I want to find out vertical reaction at E, because roller is providing the, that pin is providing the vertical reaction. So if I want to find out what is the vertical reaction, what do I need to do? Just take the free body diagram A, C, E. Okay, take free diagram A, C, E. And there will be two unknowns will be exposed at A. Okay, because we just saw that if we remove this A, C, E from this, two unknowns will come. But those two unknowns are immaterial to us. What we only need is vertical reaction at E. So if we take torque about point A, we immediately get what is vertical reaction at E. And once you get vertical reaction at E, the problem is done. Because then you can take the whole free body diagram as you rightly pointed out. We can get horizontal reaction at D at E. Vertical equilibrium for the whole free body diagram will give you what is the vertical reaction at D, done. Yes? Free body, for the entire free body diagram. The entire free body diagram if we take moment about A. It's a very bad idea. Because... Suppose, D y and E y, we will get one equation. About A. No, no, no, but about A you will get one equation. This E x will pass through A. So no component. Vertical component. But vertical component will come through. Already we have calculated a horizontal component by taking moment about E and D respectively. But nothing will happen. Then for vertical component, you tell what will happen is this, that for one free body diagram, there are four unknowns, there is nobody in the world which can give you all the four reactions from one free body diagram. Sir, already we have calculated horizontal reactions. What you will see is that, this is what I am giving as homework, that if you get D x, if you get, suppose you get D x. E x and D x we will get. Suppose you get, what I am telling you is that, that one, when you write some moment equilibrium about A, only thing you will get is that, that E y plus D y will be equal to 4.8. And summation F y. If we take summation F y is equal to 0. You will get, are you getting the same equation? Same equation, but. Not independent. Same equation. You will not get independent equation. You cannot create. Two equations and two. But those equations will be just looking similar. You will just get F x plus F y equal to 4.8 in one way. In second case, F x plus F y into 4 is equal to 4.8 into 4, something like that. It will not come. We cannot get something out of nothing. Sir, I have tried. Excuse me. Yes, please. Point D. Point D, yes. A, A D is a rigid link. A D, whole A, B, D is a whole rigid link. Am I correct? So, as you said that one, so there are two reactions, I agree. There can be. I am not saying there are. See, there is a difference. There are not. There is, there can be two reactions. So, we cannot just remove those two. Do reactions are there at D? There can be two reactions. But as it is a hinge, the moment should be zero. Yes, yes. Therefore, the resultant of those two reactions should be in the line of the A D. We still did not, because the reaction C, only if the reactions are such that the ratio of the reactions is equal to R y by R x is equal to 1.5 divided by 4, then what you are saying is fine. But that is not a priori guarantee. Okay, that is not a priori guarantee. A x A y is there, but what is the ratio? It is a ratio which will decide what is the line of action. Because it is a 2D line. It is a? It is a 2D, no. It is a 2D problem, but 2D problem doesn't mean the force can pass along. The resultant reaction at D is supposed to pass through the DA. No, it is not supposed to. That's why we will come to two-force member. In a few moments, we will come to what is typically known as a two-force member. And this is not a two-force member, so you cannot say with guarantee. I'm not saying that it cannot happen. But what I'm saying is that a priori, you should not assume that it will happen. It may so happen in the end. But that is not a priori given to you given all the constraints. It is not a two-force member. Yeah, of course, it's not a two-force. Because there is central force. That is destroying the two-force nature of that problem. Why? When we are looking at only at D. OK, so is this problem clear? So what we realize is that just by looking at that, OK, what is some support reaction doing? We can immediately recognize what is the appropriate free body. And once we get an appropriate free body diagram, we see what is the possible way, as the esteemed professors have also told us. Then we know that, for example, what is the corresponding free body diagram and the appropriate equation of equilibrium. And we are done. It's fine. This one? We move on? Yes. Now they are coming to the point of confusion was that what is called as a two-force members. So these are various examples of two-force members. Note that, for example, in each of these members, there is a member which is pinned at one end, pinned at other end. And the self-weight is very small compared to the other forces that are involved. That's the idea. Look at any of these things. This is a link, pinned here, pinned here, no forces in between, self-weight negligible compared to the other forces. Here, here, same for here, two links, pinned here, pinned here, no force in between, weight is negligible. And the most prominent, what is happening here? The most prominent are what is called as hydraulic cylinders. And we'll discuss them in a few moments. What does hydraulic cylinders are and what do they do? We have seen these on many machines. There is so much construction happening around. And we have seen these hydraulic cylinders everywhere. Now, what is the concept of two-force member? Since this point was pinned, so we have a member of any shape. The shape is not a requirement. It can be of any shape. So we'll take a 15 minutes break.