 So, just quickly now what I will do since three problems are already solved before I go to more problems, can we keep it open for 5 minutes discussion? Any question about these three problems? Sir, can you tell me the practical applications of roller and roller support in civil engineering structures actually? Usually, we use a rocker than roller actually. So, what happens in bridge? In bridge structure we tend to use the rocker which is also a roller, we assume that to be a roller. So, basically vertical movement is prevented. So, as such we tend to avoid as much as we could, but rocker is one of the examples where you tend to put those as a roller support. So, they also use as a roller support. There was a question from 1095. Question is that, can we show roller support in an invented way? Depend on the way the force is applied, it is not a usual practice actually. See what will happen? Now, I have to explain this. See depending on how the force is acting on the body, which way without absence of the roller, think of this, why do I put a support? It depends on the force, how I apply. So, if I let us say, if I take a beam, if I apply a force like this, then the motion of the body is in the direction of the force. To prevent that motion, I am applying the roller. Now, there is nothing as such called inverted roller, but let us say my force is applied from this side, then of course I will prefer to put it like this. But again it depends on the applications. See, roller has to be grounded. See, roller has to be grounded. You have always ground is always your base. On top of that you put the roller. So, you should not apply a force in the body that is resting on the roller that has tendency to lift off from the roller. So, force should not appear in that way. That is the whole issue. There is nothing called inverted roller or so on. Now, depending on the force that you are trying to apply, you could attempt to put the roller on top of the body or bottom of the body. But remember, roller has to be grounded formally. Okay? One, two, double, five. Hello sir, I have doubt about the support reaction at roller support which is in the rail. I have explained about this twice, but even though I have doubt, but it is, I have two reactions about, I mean, two reactions only there. But the bottom thing is, the rotation about y axis is also prevented. So, you do one normal and normal, correct sir? See, that is the, your question is that wheel on the rail. Okay? When you have a wheel on the rail, see what happened? Rail can also topple. So, think of this is your rail, right? And this is the wheel. On top of the wheel you have the bogie of the train. Let us say it is a train. Okay? So, now you have rail, you have wheel, you have the bogie. Now, this wheel can have tendency to topple. You see that? So, if this is the rail, it has the tendency to topple like this. Okay? So, that rotation is possible. That small rotation, if it topples, then the train can actually topple. So, therefore, I cannot have any moment reaction. Okay? But you see, what is, what it cannot do if I have this rail like this, the wheel cannot come up like this. So, it cannot slide. But smallest rotation about that axis is possible. The smallest, slightest rotation about that axis is possible. And that slight rotation is allowing the moment not to happen in that axis. Clarified? Okay? So, you can think of also, it can try to undergo some twist also. Because there is a gap, there is a very small gap between the rail and the wheel. But it can just slide off. It cannot go, you know, directly slide off from the rail. Okay? Just try to again, when you go to the railway platform, just look at the wheel. You know, when you go to the railway rail station, just look at the wheel, how it is placed on the rail. Try to observe. Although it is dangerous, but observe from the platform. Okay? 1146. Could you please explain one example related to rocker support? See, rocker support would be something like this. Okay? So, the pattern is like this. So, it could be pinned right here. But what happens on the ground? You see that this force is actually transmitted directly here. Okay? So, typically, you know, in even bridges, we tend to give this kind of rocker support. And these are just typical, you know, type of supports that we tend to use. See, in general, you know, if you roller, if you look at it, see, how do you connect, you know, you cannot just place like this. Okay? So, it is difficult to placement. So, there is a, you know, pin kind of connected here. And then we put the rocker. Okay? As such, although, you know, roller, if it is a continuous span in a bridge, then it is okay to put the roller. But if it is at the support, right, at the extreme end, then we try to put the rocker. Is that clear? Okay? So, let us move to the, any other. So, I will, I have another two problems. I will just quickly go through that. So, let us go back to the lecture a little bit. And we can definitely keep the discussions during tutorial as well. Okay? Now, the problem number four, there are three identical steel balls. So, these are three identical, each of mast M are placed in the cylindrical ring. So, there is a cylindrical ring here, which rests on a horizontal surface. And the height of this ring, if we look at it, it is slightly greater than the radius of the ring. The diameter of the ring is such that the balls are virtually touching one another. So, balls are basically virtually touching one another. So, these three balls, first you put these three balls, they are virtually touching one another. What it otherwise means that there is no contact force that is being transmitted from one ball to the other ball horizontally. Okay? Now, also understand that this ring is provided in such a way that its height is just slightly greater than the radius of the balls. Okay? Now, what we are going to do? We are going to put a fourth ball on top of the three balls. Now, we try to understand that determine the force P exerted by the ring on each of the lower three balls. So, we have to find out what is the force that is exerted by the ring on each of the three lower balls. Let us assume there are no friction in this problem. Okay? So, how do I think about it? So, to do this problem, remember one thing we should keep in mind how the force are transferred from one ball to the other ball. So, the top ball if you look at the contact between the balls, the force are being transmitted right through the contact. So, now think of this. So, can we just think of just connect the centers, connect the force centers of the ball then what we will get? If I really connect the centers of this ball I am going to get a tetrahedron. Okay? So, one way to visualize this problem in 3D just first draw a tetrahedron and if I draw a tetrahedron then actually on the edges the forces are going to be transferred. So, these are contact forces that is going from one ball to the other ball and then what is happening remember this ball has a weight does the other balls as well but this weight has to be balanced by the contact forces that is coming into play due to the contact of this ball with the other balls and those will be the edges of the tetrahedron. The line of action will be the edges of the tetrahedron. Next thing to look at once I get those forces now how do I get that I will come in a minute but ultimately it is a symmetric problem in one way. Okay? So, then we will simply try to take the lower ball and try to understand that how this lower ball is pushed against the ring. Okay? So, it is a two step procedure. The one step would be the equilibrium of the top ball. Remember three equal contact reactions balances the weight and why they are equal as we can see that reactions are equal because of the axisymmetry. Then we take the equilibrium of one of the lower balls to find the force exerted by the ring. Now, this is a very appealing problem although you know these are kind of little bit of not on the application side but quite funny to play with I mean because it has the geometrical aspect that comes into play. Now, I am having another problem. So, now you see first of all to draw this tetrahedron. So, the tetrahedron is drawn here and you can see this edges here. So, if you kind of think of this way what is going to happen I am going to have the reactions along the edges. So, ultimately if we get a top view and if you try to take a section along away and look at the side view. So, these are the two things coming into play one is the top view and one is the side view by taking the section through away. So, that means if I can just cut like this and try to look at this either cut like this and try to look at this I am going to get this picture by taking the top ball and the lower ball and we can see the forces. So, here it is easy you have R, R and R these are the three reactions that will equate the mg. Now, only thing as I said that it will involve some geometrical calculations to get the angles and all that basically this theta needs to be found and that is already done. So, let us say we know the theta. So, ultimately what will happen that 3R cosine theta equals to mg. So, therefore, we can get the R. Now, when we look at the lower ball if I draw the free body diagram of the lower ball that lower ball will simply have that reaction that is coming from the top ball through this edges of the tetragon. So, that R is still there that reaction then we are going to get a reaction from the bottom and then we are going to get a reaction here which is exerted by the ring to the ball and weight of this ball is also there. So, ultimately the lower ball is going to see four different forces what are these forces one is the contact force that is coming from the top ball to the lower ball that is R another is the weight another is the normal force and another is the P this P I am interested in finding out and that can be very easily solved if you just take the fx equals to 0 because R is already found from the previous step. So, I just take P equals to R sin theta now and therefore, what we have the solution for the P. So, again in this case what we have seen that there is no moment coming into play at all and either I have to take the you know moment about any axis. So, this is totally a problem involved within the effect of forces and the last problem that I am going to discuss quickly. Now, I am going to just know this one we can take it as homework we do not have to solve everything in this case I have the solution with me, but problem involves like this. We will just take this problem as a homework problem and we are not going to you know check on the numbers, but we can just think of how to solve this problem is it possible to solve by taking moment about an axis if I want to do that how complicated that would be. So, the problem involves basically a uniform steel plate. So, this is a plate right here the dimension is given this is 0.5 by 0.75 it has a mass 40 kg in and as in a and is attached to ball and socket joints at A and B. So, it is attached at ball and socket joints at A and B knowing that the plate leans against a frictionless vertical wall at D. So, this is a vertical wall which is also frictionless it is just leaning against this wall and only point of contact that it has it at D. Now, determine first of all the location of D and then the reaction at D. So, that is the problem now why it is asking you to find the location of D remember only thing that is given is the length of this one. Now, think of it that I go to you know some shop and just buy you know some kind of mirror or something just a rectangular plate and I am trying to lean against one wall. So, it is just lean against one wall, but there is a ball and socket joint at the base. So, ultimately what we see here that ball and socket joints are going to give me the reactions. There are three reactions here, three reactions here and then I am going to get a reaction from the wall. So, that reaction will be parallel to the x axis which is at the D. Now, what will happen in this case I can clearly see that reaction can be obtained how I am going to obtain the reaction. I am going to simply take the moment about A B right if I take moment about A B it can be solved very easily. So, the first part now what will happen if you really try to attempt to find out how to get the moment of the forces about A B it is not going to be an easy task because the DX is going to come then remember DX has to be taken perpendicular to this plane also right and then only I can take the moment of that about A B. So, as long as I have the DX I can find out what is the perpendicular component of the DX on this plane then I can take the moment about A B. Now, that is not going to be that simple and therefore, what I have said that vector mechanics could be easier. So, the point to you know bring in this problem is that that we can also use vector mechanics to solve some class of problems. So, I will just show you the solution the first part involve geometry. So, from the geometry I can find the location of D. So, let us say like this. So, therefore, what we have this is the complete free body diagram you see. So, you can see this N D which is again parallel to the X axis that is the you know reaction from the wall it is a frictionless wall as we said now how do you find the location of D very simple. So, all I had to remember the dot product of this equals to 0 because they are two perpendicular axis A D and A B they are perpendicular axis therefore, the dot product will be 0. So, if I get R A D and if I get R A B. So, R A D dot R A B equals to 0. So, this will lead to some kind of equation you can see here this will lead to some kind of equation where the unknowns will be the coordinates. So, let us say this coordinate is defined by Y and Z right. So, I am going to get some equations. So, I will get the Z out of it then I have to still find the Y how do I find the Y I know the another equation is required and that equation will be simply the length. So, length of R A D that vector position vector right. So, that is going to give me another equation I know the length 0.75 that is going to give Y. Now, I am simply going to do the vector operation here. So, what I will do I will simply try to use lambda A B because I want to take the moment about A B. So, lambda dot R cross F right. So, I need somehow lambda A B. So, in the next slide what we have shown how to find the lambda A B right. So, I am going to use lambda dot R cross F now to take the moment of all the forces individual forces two forces are anyway going to give the moment one is the N D right. So, I will write N D I because that is the I component and another one would be the W that is another component ok right. So, all of these are found now remember how do I get the R R cross F needs to be done. So, here I can simply say R A D that is not a problem I already found, but to take the moment of W what I need to take the moment of W I need R B G or R A G. So, therefore, R B G or R A G needs to be found. So, that we can do again from the geometry are not an issue. So, I have here written since we have already you know we can find the R B D it is already found here that is R B D and R G B R B G therefore, B to G is nothing but 0.5 half of R B D ok. So, now you can see I am going to simply take moment about A B. So, that is lambda A B dot R A D cross N D plus lambda A B dot R B G cos W that is equals to 0. So, what is my unknown unknown will be the magnitude of the reaction that is N D as you see here. So, again here I am really going to use that mixed triple product for moment of a force about an axis that we have discussed in the very first class ok. So, in that way I can solve this problem to get the magnitude of the N D ok. Therefore, the vector N D the force is simply magnitude I ok because that is the x component. So, I think we have studied variety of problems now. So, what will happen we will just take a short break.