 So, let us move on now, I have still two more problems, do it now, do this one, since we are so fresh, why do not we do this problem. Let us solve this, any questions you please ask, do it using calculus approach, much easier. Now any doubts, any questions will automatically get clarified. I think most of you are pretty comfortable with this approach, let me briefly discuss this and let us move on. So what we have here, we write down the Y coordinate of P, XY is set up like this at point A, Y, we write coordinate for Y, for this one, for this one, X coordinate for point D and X coordinate for point E and then just differentiate, that is delta or this position, I should write it H, this is my error, this is not pH. So delta YH, delta YG, delta XD and delta XE, all components are there, here P will be minus P, minus 2B, Q, FEX and then just put this, this multiplied by delta, minus P multiplied by delta YP, minus 2P multiplied by delta YG, do not change sign again, same sign, okay, delta XD into Q, positive and delta XE into FEX, solve it and you immediately get this is the answer for FEX, just note one thing, if Q is 0, that is what we saw, if Q is 0, what will tend to happen, this point E will try to go inside, so only way to prevent it from going inside is to apply reaction which is outside, so FEX is proportional to P, not minus P, we are good, now to find out what is the vertical reaction at E, now that is a mundane problem, so this is an exaggerated displacement, we give it, just tell me if I give it an exaggerated displacement like this, okay, what have we done, rotated the entire assembly rigidly about point A, take this entire assembly, release the support, rotate it about point A, now what is the horizontal displacement at point E, when you do the full rotation, 0, 0, because it will just move upward, just note that this is the line, displacement will be perpendicular to that, yes, okay, so FEX will not do work in that case, what is the horizontal displacement at point Q, rotation is about point A, what do you do, you draw a line parallel, horizontal direction draw a perpendicular from A, what is the displacement, so A sin theta into delta theta, what is the vertical displacement at point P, A cos theta into delta theta, vertical displacement at point G, 3A cos theta delta theta, just put in all those values here, equal to 0, you will immediately find out what is the value of FEY, and the same thing will happen, if you draw the complete free body diagram and take torque or moment about point O, you will get the same answer, because these two processes are equivalent. Excuse me sir, yes please, sir, previous slide sir, previous slide, yes, sir actually what I have shown to you, in the point E I have taken the direction of reaction in X, that is in opposite, okay, you take inverse, okay, so according to my assumption, so I am getting the reverse sign in minus inverse, there is no problem, absolutely not, because whatever answer you get, like for example, if I have taken this convention, you have taken this convention, we should get exactly opposite answers. That is why I am showing the service. Yeah, exactly opposite answers is what we will. That will be correct sir. Of course, yes. Thank you sir. It is fine, no questions, can we move on? Okay, why do not we do this problem, okay, I did not intend it as a tutorial problem, but this is a bit easier, why do not we do this? So use principle of virtual work to find force in the member CD, okay, this is another, it will be a good experience, that find out using principle of virtual work, what is the force in member CD, okay? The idea is this, simple idea, and you will work on this, that CD is a two-force member, we know, so like Shobik I discussed earlier, CD is a two-force member, what will it do? If I assume that CD is in tension, then I remove this member CD and at point C I put one force, which is pulling, equal and opposite force at point B, which is pulling, so that is an unknown, and I want to find that unknown using principle of virtual work, you can do it using method of joint section or whatever, but let us do it with principle of virtual work just for demonstration. Is the idea clear? You remove CD, assume CD is in tension for example, and replace force at FCD, pulling force at C, equal and opposite pulling force at D, an unknown, and what you want is that you want to give this entire assembly a virtual displacement, virtual rotation, combination, whatever in such a way that only those forces and the applied force W do the work, and then in one shot we can find out what is the force in that member? It is an equilateral triangle, okay, sorry, sorry, everything, all the lengths are equal, so all angles are 60 degrees, okay, good point, sorry, so all the angles are 60 degrees, the absolute value of lengths are immaterial. The second answer is W by 2 L sin theta. L will not come, because the dimensions are wrong, L will not come, yeah, W by 2 sin theta. W by 2 sin theta. L sin theta. L will not come, L will not come, because look the dimensionality is not right, W is force, okay, okay, memberation is force, so W I L. L will cancel, W by 2 sin theta. Yes, and it is tension or compression, you have to be very careful about us members, tension or compression, you have to specify that explicitly, compression, compression, okay, so I think most of you got the answer, so quickly let me discuss, so if this is the active force diagram, we have FCD, FCD, we have assumed tension, now what do we do, what kind of virtual displacement we need to give? We give this entire assembly a virtual rotation about point O, this triangle, now note that point line A is horizontal, so this point will move exactly downwards, okay, no horizontal displacement. Now second point also, second triangle also we want to give some rotation, but how much should we give? If we give it a different rotation, then point E which is common to this triangle and this triangle there will be split, so in order to prevent that split, both of this triangle should have the same rotation, and then what is the corresponding horizontal rotation at C, it will be simply this vertical displacement, this vertical distance A sin theta into delta theta, and straight away here also A sin theta into delta theta, so by this delta theta work done will be positive here, also positive here, what is the vertical displacement here? Simply A which is the length of the any arm into delta theta, just add up all the virtual works, you will see that 2 FCD sin theta plus W should be equal to 0 or FCD is equal to minus W by 2 sin theta, now this approach, why is it so good? What you realize is this, that in the presence of W, and again this is also very relevant to what Shobik had taught for example the adequacy of the constraints, suppose FCD were not there, then what will happen? Your assembly will precisely deform like this, okay, this will go into a mechanism like this, so principle of virtual work has so many different advantages that not only can we understand how does a mechanism, like for example if you remove some constraint, how will it go into a mechanism, but we can also find out correspondingly what the forces are, and also solve more complicated problems, okay, so it gives you intuition at various levels, is the entire idea clear? And since the answer is negative, this sign means that there was a pulling force at joint C and D, we had assumed this to be in tension, but the sign coming out to be negative automatically implies that this was actually compression, magnitude is W by 2 sin 60 or W by root 3, fine, fine, okay, let us solve this problem, okay, so we are given this assembly, this is a roller, a guided roller, one whole member BAC, one member AO, this is a hinge, there is a mass which is attached here, okay, whose center is this at point C, M naught, mass of this member BC is 2M, whose center of mass will be at A, and mass of this member AO is M, and whose center of gravity is at the center, now for this assembly we are asked that to keep this in equilibrium, what should be the moment that should be applied at this point such that this entire assembly can stay in equilibrium, whose principle of virtual work, okay, answer will come in like literally three or four lines, is the problem clear? Any ambiguity about the problem or the mechanism? So note that and you will also immediately realize that if there are no moment here, this will go into a mechanism which means that like you can freely deform it without changing any lengths of the members, so what is this M providing, is providing a counterbalance to that, fine, is the mechanism clear? At what? Location B, the reaction will only be vertical, because it is a free roller, it is a free roller, so only reaction will be vertical, at point O is the pin, you can have two reactions, so what it means is that we want to give virtual displacement in such a way that those reactions don't do work or you don't open any linkages, for example point A is common to two members, we don't want to give virtual displacement such that that opens up or it deforms any members, okay, so let us briefly discuss the solution, okay, so what we can see here is something like this, the virtual displacement, suppose we rotate this bar by some angle delta theta, okay, then what happens if you rotate this by delta theta, point A will move down by how much? Point A will move down by, this is L, okay, this is theta, this is also theta, so L cos theta delta theta will this move down by, okay, now the question is that we want to apply some other displacement here, okay, because this is a second point, so what is the displacement that you should, or what is the rotation that you should apply here, if this we apply delta theta, okay, same rotation, yes or no? Why? Because if you don't apply same rotation, what will happen? This point A is common to both AO and BC, so if this rotation is delta theta, then the vertical displacement of point A will be L cos theta delta theta and if we apply delta phi for this one, then the vertical displacement of this point will be L cos theta delta phi and they should match, otherwise A will split apart, so which means that both two delta theta's should be equal to same, so we immediately know that delta yc for this point will be L by 2 because the center of mass is at the center, so L by 2 cos theta delta theta, for point A it will be L cos theta delta theta and for point c will be 2L cos theta delta theta, but now the deal is that that point A because of rotation from this rod will move to this side whereas point A because of rotation from this rod will move to this side, so there will be a split, how to create the, how to get rid of that split? We have this entire assembly which is on roller, so I can move this entire assembly on the other side and close that split, okay, so ultimately that's how the final displacement diagram will look like for the virtual displacements and we know what is delta yc, delta ya, delta yc, ultimately what is the work done by the moment? If you apply a rotation of delta theta in this way the work done will be minus m into delta theta because the rotation is anticlockwise on this rod but the moment is clockwise, so to add up all the virtual works and you will get a simple equation for m, is the idea clear? Any questions please ask? Explain which point? Okay, let me explain everything again maybe, first portion, which portion? This equation? This one, okay, just note you agree with me that the rotation, the virtual rotation that we should provide to this rod should be the same as the virtual rotation we provide to this rod, you agree with me, so if this is delta theta this also should be delta theta, now note that the mass, center of mass, okay, because the weight will act through the center of mass, okay, so weight is acting through the center, so we need to find out what is the vertical displacement of the center, so it will be just half because the displacement of this point is somewhere, this distance is l cos theta, this entire distance, so this is l by 2 cos theta and l by 2 cos theta into delta theta will be the vertical displacement of the center, the point clear? Yes please, first one is c, oh it should be, okay, wrong, okay, my mistake, this is c is this point, okay, I should like this is enough, like the way we use in computer, the same operator we keep changing its values, okay, so this c is first applied here, then you replace it by c here, okay, sorry, this c the way I read, I should write some point b o here and this is the vertical virtual displacement of the center of mass, center of a, yes of this point, okay, sorry, it is not c, okay, so last problem before the quiz, okay, solve this problem, I have used one method to solve this which I use long time ago which is not good, but there is a easier way of solving this problem, okay, what we have is we have this assembly mass m, okay, which is supported, there are two links here, okay, 1, 2, 3, 4 and two hydraulic cylinders which are supporting these two links and what we are asked is that, that in this configuration where this angle is theta, what should be the force in the hydraulic cylinder such that the system is in equilibrium in this configuration, okay, so my suggestion will be use delta theta, okay, use delta theta of this member, this member and relate how that delta theta gives rise to vertical displacement here and that corresponding delta theta gives rise to displacement along the direction of the hydraulic cylinder, so the problem becomes very easy. This can also be done in the calculus language by setting up a coordinate system here or here, but it is much more, and you can only take one assembly because everything is symmetric, but it is much more satisfactory if you can have a mechanical, physical picture of how the virtual displacements happen, direction you can find out from the geometry, okay, all the dimensions are given, you can figure out all the dimensions from the geometry, note that this is also b, this is b, okay, so this becomes an isosceles triangle, okay, you can find out the dimensions again because those two angles will equal the third angle you can automatically get. Mg cos theta upon sin theta by 2, yes, okay, so let us discuss briefly, so what we realize is this, okay, so this rod, the way you want to give the virtual displacement, first of all you note that this angle is b, this angle is b, so this angle here and this angle will be the same, so this angle will be 90-theta by 2, this angle will be 90-theta by 2, okay, that much is clear, right, this 90-theta by 2, 90-theta by 2, now what we do is that we want to give this entire system some virtual displacement, how do we give? We rotate this rod, okay, about this point by a small angle delta theta, so how much will point b move up by, sorry, so we don't know, we rotate it downwards, let's suppose let us rotate it downwards, okay, we give it a clockwise rotation of delta theta, anticlockwise also fine, we give, suppose we give it an anticlockwise small rotation of delta theta, how much will point b move up by? b cos theta delta theta, okay, same thing we do, okay, let's not move the mass here, just give this object, okay, this member, again a rotation in the clockwise direction of delta theta, how much will b move downwards by? Same distance, b cos theta delta theta, now what is happening is that b was common, after giving the virtual rotation this moved down, this moved up, so there is a gap or there is an interpenetration between these two members by how much? 2b cos theta delta theta, so we need to move the top one by that much distance so as to get to the gap again, close the gap again, so the platform will then move up by 2b cos theta delta theta, fine, that sort of platform will move by, now let's look at this assembly, we want to know that how much is the displacement that is created at point b along this direction, okay, so just note, see here, okay, if you see here, you will realize that this angle is alpha, this is the direction in which the force from the hydraulic cylinder acts, so if we drop a perpendicular from this to here, the length of that is how much b sin alpha, so b sin alpha into delta theta will be the motion of the cylinder, the motion at point b along the direction of the hydraulic cylinder, but what is that, b delta theta sin of alpha is what sin of pi by 2 minus theta by 2, so it will be b cos theta by 2, how much is the platform gone up by 2b cos theta delta theta, so just put in all the numbers and you will immediately see, put in all the symbols, the work done by the weight will be minus w into delta y because delta y is upwards, work done by the hydraulic cylinder, if we assume this to be in compression, then it is, the sign is this, it will be fh into delta h, sum should be equal to 0, we will immediately see that the force in the hydraulic cylinder will be equal to w cos theta cos theta theta by 2, clear, fine, okay, so I think there are actually some problems, this is a reasonably difficult problem, okay, this is a very easy problem and this is also a reasonably difficult problem, so can we do that, like if you are interested we can do that after the session, okay, in the next part we can do these two problems, just one thing which I wanted to just emphasize that in the quiz problem, when you apply the virtual displacement in order to obtain what you get, so how many of you use the calculus approach to do that problem, just write down the y coordinate and differentiate with respect to theta, okay, that will give the right answer, but how many of you could get a virtual displacement diagram which looks like this, the second member, how many of you see rotation in the second member, okay, the first one there is a rotation, how many of you see a rotation also in the second member, while drawing the virtual work diagram, you got a rotation in the second member also, this one, BC, okay, because typically this was one of the problems that was given in one of the quizzes here and what we had seen is that many students could do, get the right answer even without assuming that this arm BC can rotate, just keep the arm BC straight and you could still get the right answer, but what will happen is that and even for example the calculus approach where you say that the y coordinate of point C with respect to A is A sin theta plus A and then you differentiate with respect to theta, you can get what is dy at a coordinate and then you will get delta in terms of theta, but this is extremely crucial because if you realize that if the middle member were welded, then it is not possible to have this as a mechanism, so this entire assembly can never be given any virtual displacement which will not end up deforming any members, so one thing which I should have done is if having a moment instead of having the moment here, if that moment would have been transferred to this portion, then the second part of the answer, okay, if you do not take into account this rotation would be wrong and this is one of the reasons that I will move on to the next topic, but this is one of the reasons why the physical approach where you actually draw that what are the displacements which are compatible with the constraints, what are the virtual displacements, because if you mechanically use the calculus approach, then these kind of rotations or displacements which are not immediately apparent from the calculus approach, okay, may not come and in this problem, okay, it was easy that the moment was given here, but if the moment would have been given here, then those of you who did not take into account the rotation of rod AB would have gotten the wrong answer for part two of this problem, okay, just simple point I wanted to make, I was very glad that a huge number of you got the right answer both for part one and B, if you get part one right, part two automatically comes, but I just wanted to emphasize that this rotation is not reflected in that problem and will be really reflected if this moment is transferred here and if you don't take care of this rotation, then the answer you will get will not be right, just simple point, okay, so is it, am I clear about this, okay, fine.