 Okay, welcome back, there is one small point about this problem which a few students asked me a question and I think I should have emphasized that properly which I did not, just note one thing that we had implicitly taken that the platform is perfectly horizontal and there is no tilt in the platform and the reason clearly being is that this spacing D e is the same as this spacing AB okay 300, what does that mean that the horizontal distance between E and B is the same as the horizontal distance between D and A because of which that the vertical displacement of point D will be the same as vertical virtual displacement of point D and the platform will not tilt. If the top distance were different than 300 then you will see that there will be tilt in the platform and you have to add that component also when you find out what is the work done on the mass which is work done by the mass which is resting on the platform. Okay, so this one point I forgot to mention but it is very important that the top distance 300 is the same as the bottom distance 300 because of which both point D and point E have the same vertical displacement, is it clear? Okay, so let us move on to this problem okay which is very similar to one of the problems we had done in class and what we will see is that with principle of virtual work okay the entire problem becomes extremely easy. The mechanism is as follows, we have a platform okay a platform which on one side has a slot okay there is a frictionless slot and a platform can move up and down that slot okay so this is a vertical slot okay this is as a figure I drew okay so I could not exactly draw a vertical slot but it is a vertical slot and a platform is free to slide up and down through this slot okay now this platform can be hoisted up or lowered down by this mechanism. There are three members 1, 2, 3 which are connected to roller and these rollers can slide inside the platform such that if you increase this angle theta okay this entire assembly goes up these rollers can slide and the platform can appropriately move up if this angle theta is lowered then the wheels will slide like this the platform will come down okay and the entire mechanism is lowered. Now this entire assembly is supported by a hydraulic cylinder all the dimensions are given to us okay this distance this epsilon everything is known so in principle we know what is the orientation of this hydraulic cylinder AB with respect to the horizontal. What we are asked is that given all the dimensions of this problem and when theta is equal to 60 what should be the force that should be provided by this hydraulic cylinder to so as to keep this assembly in equilibrium. Now as far as the specifics of this problem is concerned we have mentioned that what is the diameter of this thing okay and so that if you know the diameter we need to find out what is the internal pressure that will give us p times the area will be equal to the force acting at point A. So as far as the mechanics of the problem is concerned this hydraulic cylinder is exerting a force at point A along this direction and keeping this mechanism in equilibrium you increase this boom the assembly goes up you decrease the size of this boom the assembly goes down and when theta is equal to 60 all the configurations are given to you all the angles are known and can we use principle of virtual work to find out in terms of the force in the hydraulic cylinder and all the other dimensions that what is the force required okay to keep this weight okay which is given 5000 Newton in equilibrium. Is the problem clear? Problem statement it is very similar to the problem we had done earlier only thing is that if you recall the problem we had done in 2D equilibrium there was a two force member connecting one link to the platform to resist any horizontal forces whereas in this problem that job is done by this vertical slot that any horizontal moment is prevented by the vertical slot the platform can only move up or down and the internal wheels can slide to raise it up or down. Is the mechanism clear? Now this problem if you want to solve it using equilibrium okay we can do it the same way we had done the problem earlier towards the end of the class in 2D equilibrium but what I'm asking is that what is the easiest way okay or what is the what are the virtual displacement that we can provide to this system that we can find out the virtual work done by the the mass on the top here and the horizontal cylinder such that we can say that the system to be in equilibrium the virtual work should be equal to zero and we can find out what is the force in the hydraulic cylinder okay so what are the virtual displacements that we should apply our virtual displacement rotation combination whatever yes so angular rotation note that these three rods are all parallel to each other so if you apply a small rotation theta here delta theta we want to apply the same rotation to all the three rods such that all of them remain as a parallel gram this distance does not change okay so far so good we apply a small rotation delta theta to all these inclined rods now if we apply a small rotation delta theta to any of the rods what will be the vertical motion of this entire assembly how much will it move up by 2L 2L cos theta delta theta will the entire assembly move upwards by so the work done by the gravity will be minus W okay or if you reduce the theta then it will be plus W increase the theta it will be minus W so we know what is the vertical displacement of the platform if this rod is rotated by small amount delta theta then this entire assembly will go up by delta theta into this horizontal distance which is 2L cos theta fine that point is clear now how do we find out what is the work done by the hydraulic cylinder if we hydraulic cylinder is a two force member let us replace that by an equivalent force so you tell me yes how do we get the force in the hydraulic cylinder only y component x component what do we do why no but look at this hydraulic cylinder this hydraulic cylinder AB you can replace by a force along AB let us say there is a compressive force in the hydraulic cylinder then this can be replaced by a force acting towards a along direction AB we want to find out what is the component of displacement along that right now let us say if this angle is alpha this angle the cylinder makes is alpha what will be the horizontal component that this of the virtual displacement if you rotate this assembly a will have a component in this direction how do we find out that component so what we do is that we just extend this line from this point drop a perpendicular and what we need to know is what is the perpendicular distance of a line which is perpendicular from this point to the cylinder AB now if this angle is alpha the total angle is what theta plus alpha so simply the displacement along the hydraulic cylinder will be equal to L sin theta plus alpha into delta theta so if I show this in this figure what you will see is this if this is alpha then this angle will be theta plus alpha what we want is just this vertical distance or from point a drop a vertical line okay along this direction what is the distance of that it will be L sin theta plus alpha so when we provide a virtual rotation like this delta theta at this point where the force is acting on the hydraulic cylinder the displacement in this direction will be simply given by L sin theta plus alpha into delta theta done and alpha direction you can easily find out why because all the dimensions are given to you we can find out what is the value of alpha is a is a question clear is a is a is the approach clear is there any doubt any question about this please ask so what we want is this okay if I repeat again if I repeat again then let us look at this central rod this force acting by the hydraulic cylinder we are replacing by fh now if I zoom onto this rod we are rotating this rod by a small virtual displacement delta theta about point a and we want to find out that for that small rotation delta theta about point a what will be the displacement at point o along direction of fh and if you just do the geometry what do we need to do from a you drop a perpendicular line to o and find out what is that distance multiplied by delta theta will be the corresponding displacement but this angle if this is alpha this is theta this angle will be theta plus alpha and l sin theta plus alpha will be that distance of a and a perpendicular line to this okay yes sir if I think in that way that from the ground to the point a the length is a from what sir think about okay from the ground to point a the length is a along the yes yeah now if we give a small displacement delta theta so from perpendicular along this line the displacement is a delta theta that we can say perpendicular to which line perpendicular to that incline line perpendicular to this line right and then you can do that thing also you can take that and the component of that in that but fine that case the distance coming is different now it will come out to be the same obviously that is a delta theta and the cost component of that but what happens is that when you look at a perpendicular distance that cost becomes sine okay because that angle you are to look at okay it becomes the vertical angle there okay so this and this so simple thing is for example that let's look at it this way let me move to this slide it will come out to be exactly the same okay what you are saying is fine if you look at this line this angle is theta this is L delta theta this is theta okay this will be theta right so because of which how much is this distance will be delta cos theta which will be L cos theta delta theta will be which will be the displacement in the vertical direction which is the same as doing this okay so whatever way you do you will get the same answer yes I was telling the same thing if we take the components of P fine that will be a I think it will be a better method than this I will tell you one thing okay so better method lesser method again it is a idiosyncratic thing okay you may find that comfortable so it is fine okay so I'm not saying that better or not better okay no no there's nothing to be sorry about okay because some are more comfortable with that okay some some people may find that other one more comfortable but why was this why looking at the perpendicular distance and going for it why is that beneficial that if you look here you will see especially this problem you will see that all of them should have the same distance it's just the horizontal distance times delta theta if you come to this problem I don't need to find out any angles nothing if you really want to use the method that you are suggesting I have to draw this line find the angle take component and do data theta but in this case for finding out vertical displacement I just need this distance easier easier okay it's not a question easier depends on the problem and depends on for example what state of like what you are comfortable with but if you look here this is like a general approach where you know that all the points along that it only distance on this distance depends on this distance find it but you are right like there will be some other problems in which what you are suggesting may be beneficial but both are equivalent thank you sir okay so as somebody what they suggested is that that at point O which is equivalent you know that the displacement is perpendicular to this line so then just take that displacement and find out what is the component along this both are the same thing whatever you are comfortable with you do it and you will see that whatever you do the answer will come out to be only this it will be L sin theta plus alpha into delta theta will be the virtual displacement at point O because of this rotation and now you know what is the virtual displacement along the hydraulic cylinder virtual displacement for the way just do the virtual work principle and you will immediately get the answer that the force will be given by W cos theta divided by sin of theta plus alpha just write down one equation you will see that a force in the hydraulic cylinder is W cos theta by sin theta plus alpha how do we get alpha all the dimensions are known from here okay you can immediately find what is tan of alpha from that you will get alpha theta is already given plug in all the values you will find out force is I think 2.08 in the whatever units they are given 2.08 Pascal's if I am not mistaken is the numerical value of the force but the formula is just this any questions so what you see is that that by using principle of virtual work if you try to solve this problem using conventionally equilibrium first of all this is the statically indeterminate system because there are three rods there so you cannot find out force in all the rods but as far as finding the force in the hydraulic cylinder is concerned it is immaterial okay so that is the power of principle of virtual work that you immediately can figure out that if a mechanism can be generated with respect to a particular force you are done excuse me sir yes please work done is equals to force into displacement force into displacement in that direction so our aim is to find out displacement in the direction along the direction of the force that is our goal yes so we have to concentrate over the displacement in the direction of force only that's what we did yes what we are doing over here first of all we are finding the displacement along the line which is perpendicular to that one but then we are resolving so it becomes complicated if you find out the displacement like this directly then you automatically what is the displacement along this because this is what this distance is nothing but from point a I have drawn a line perpendicular to O that distance into delta theta is nothing but the virtual displacement along this straight away but we request in y direction only because we don't need y direction we need from the top which we already got even at point a also displacement displacement in y direction why because the only force acting at point a yes is the force from the hydraulic cylinder which is inclined at an angle alpha and that is y component only no because there is no displacement in x direction no no no the force is acting in an inclined direction right okay the force is inclined it is not like this so work done so displacement now you can think about it this way x and y directions what the x and y take is my choice if I take x in this direction y perpendicular to this then what I want I don't care about what is the displacement in the y direction I only want force in the x direction where my x and y axis are oriented such that this is x and this is y you can do it either ways okay whatever way you do you want what is the component of virtual displacement along the direction of the force that's all you want and that's what we have done here in that some sir through point a we can drop perpendicular on the ground you drop perpendicular to the ground yes fine that becomes a right angle triangle yes hypotenuse is known angle is known yeah we can find out height of the triangle yeah so we can find out dy in that direction dy is equals to y first of all I will give you dx the horizontal projection which is L cost that I will give you the displacement in the y direction but dx is not required now sir because there is no displacement along x axis because forces what is happening here is that we are deciding that this is not required when what is not required is like the entire discussion about dx and dy really is a moot point because what we only need is a displacement along this direction okay and if you are still not convinced okay we can continue this after the after the lecture okay sir in this second free body diagram the body line not a free body diagram again in this active force diagram okay it's not a free body diagram in the second I have not removed all the forces right all the the two lines at oh and at a it is both are inclined lines at alpha angle yeah so this line is parallel to this okay actually I should have done a much sensible thing just take this point one dropped a perpendicular line so these was these diagrams were drawn in my younger days okay when for example I was not experienced okay but what you know is that from a you drop a perpendicular line okay that would have been much more convenient it too much work again to do everything okay this was already there okay but this distance is the same as if I from a point a I drop a perpendicular to oh why because this line and this line I have drawn to be parallel to each other okay so easiest will be you take point a drop a perpendicular to this okay and straight away you are done okay this is circuitous we are doing it okay but now I know better yes any other questions okay so let us move on okay simple problem much simpler problem but of a different type what we are asked okay but you note that the last problem principle of virtual work for example like the beauty and the power of that method is apparent when we solve the last two problems if you try doing those problems using standard approach it is not at all clear what free body diagrams it's not difficult not impossible but here by just knowing how the mechanism can happen we impose that particular virtual displacement and you are done literally in one equation okay now let's come to a simpler problem this is much simpler very simple assembly you have an inclined plane which is taken to be frictionless and this this weight is supported by a string passing over the system of pulleys okay string goes up there is weight for this pulley there is another weight for this pulley which is providing a counter that there is a tension in the string and ultimately there is a load queue that I apply at the bottom and what we are asking is that determine queue for the equilibrium of the systems if the if the pulleys are frictionless okay this slope is also frictionless and all the masses are given use principle of virtual work okay because we are doing virtual work you can easily do this even with equilibrium but just for the purposes of demonstration how do we do this by using principle of virtual work quantities W3, W1, W2 are given we need to find a queue in terms of those quantities no friction at any point of contact or any surface of contact okay so you realize that for example if I give it a virtual displacement to this block upwards a small delta displacement upwards what will move we don't want to extend the string we want the length of the string to remain the same because if we extend the string then the internal force in the string should also do well so we don't want to do that so we want the length of the string to remain the same just move this block upward by small distance delta what will happen this entire string this will go down W2 will go down by how much same delta but now suppose there will no support here then what will happen this will penetrate inside by delta Q will move down by delta but now this cannot penetrate inside so I have to leave this entire assembly by an amount delta but because of that that part of string will go to this side so delta plus delta 2 delta will how much the Q will move down by and so if this moves up by delta this will move down by 2 delta so what are the virtual work done upwards delta so W3 sin theta okay minus into delta W2 okay downwards displacement of how much this same delta is in here so delta into W2 plus 2 delta into Q should be equal to 0 delta cancels out you get a relation between W1 W2 and Q the W1 at the top doesn't do any work because there is no displacement as far as this scheme is concerned idea clear okay so we will solve a very small problem so what do we have is we have a simple mechanism composed entirely here of two force members so two force member no no no sorry sorry so one two force member second two force member a whole rod another second second complete rigid body pinned at this point again these two two force members and in these two members we apply a force of P and P in this direction all the dimensions are given to you L1 L1 L1 L2 L2 L2 and we are asked to find out what should be the force Q that is required to keep this assembly in equilibrium or I can turn the argument around and say that if this Q is given what should be the two forces required to keep this assembly in equilibrium at this given theta now this problem is very easily solved using the calculus approach okay that what you can do is take your coordinate system at this point say this is a fixed point because in the calculus approach what you need is a you need a point which is fixed which is not moving so at this fixed point have a coordinate system x y and then write down the coordinates of these two points now what are we interested we are interested in the vertical displacement at this point vertical displacement of this point so we write down the y coordinate of these two and at this point we are interested in the horizontal displacement why because the work is force is horizontal so the work will only be done for the horizontal displacement so we can write down in terms of the angle theta What is the x coordinate of this end point? What are the y coordinates of these two points? And then we use the differentiation approach, take the derivative, then as a function of delta theta we immediately know what is the vertical displacement of point P, what is the vertical displacement here, what is the horizontal displacement here, as a function of small delta theta and then we can apply the principle of virtual world. So we can write it this way and we will immediately see that y of this point is L1 sin theta clearly. Now before there is one point which I did not emphasize, I have used a degree of freedom which is theta. Is this a one degree of freedom problem or is it a multiple degree of freedom problem? If I change that theta, will all, can I get the configuration of the entire assembly by just knowing one angle? Perfect. Okay, so just one theta will immediately ensure that I know the entire configuration. So it is a one degree of freedom problem. So I can write down all the coordinates in terms of theta, ya L1 sin theta. So delta ya will be L1 cos theta delta theta. Note it is consistent that if you increase delta theta, which means delta theta is positive, this will go up, delta yi is positive. Yb will be equal to minus L1 sin theta, so delta yb equal to minus L1 cos theta delta theta. So if delta theta increases, this will go down. Now know that in this approach, you do not have to worry about like if something is going down or up. What you just say that my x coordinate, y coordinate is up, my x coordinate is sideways. This delta ya is only upwards or only sideways. If it is negative, it just means that it is not in this direction, it is in the opposite direction. But you do not need to explicitly take care of the signs. It will be automatically here. So in that case, this is minus p, this is plus p. What is the x coordinate of point c? L1 cos theta plus L1 cos theta plus L2 cos theta plus L2 cos theta. So it will be 2L1 plus L2 cos theta. So delta xc will be minus 2L1 plus L2 sin theta delta theta. We note clearly that if delta theta increases, point c move inwards and that is why the negative sign. They use principle of virtual work. Point a, load is what? Minus p because we are using coordinate system. Minus p delta ya, no change in any sign, plus p, p is upwards, y is upwards. So plus p delta yb minus q. Now q is opposite to the assumed direction of x. So this is minus q into delta xc. And with all the signs put in these values, and when you solve this simple equation, you will see that p is equal to q L1 plus L2 L1 tan theta. Okay, straightforward. Or if p is given to you, you want to find out q, you just change this appropriate equation and you can get q in terms of p. Now this approach, in this case for example, again this is an R that when you decide that you use the calculus approach, or you use the more physical approach that we have discussed so far, in this case the calculus approach is much simple and especially it is good to use because we have a nice fixed point here. Okay, this a is a fixed point so you can put a nice coordinate axis. And what you will see is that why this mechanism is so good that I can have many, many, many use number of triangles. Okay, for example in a Caesar jack or for example you know the toy snakes, you push on it and the snake goes out. Those kind of mechanisms, okay? Or Caesar jacks which are used to raise and lower assemblies, okay? There are so many different components if you try to use equilibrium for this, okay? You have to analyze so many free body diagrams. It's very likely to make errors, but using this kind of an approach, you will see that the answer comes in one shot and very less prone to errors. Yes, please? Which equation? This one? Force q towards left or right? No, in this approach, for example in this coordinate approach, where you write the coordinates, signs are automatically taken care of. If you decide that this is x, this is y, okay? That is okay. Automatically, if negative sign comes, means your displacement in this direction, positive means this direction and same for y. And force also we have to take it plus or minus depending on if it is consistent or opposite to the coordinate axis. Direction of motion and the direction of force is same, then work done is positive. It will come automatically here. Okay, so same thing is there, p delta y a, that is minus. Why minus? Because at point a, force is downward, so I have to take that as minus. Yes, and displacement is upward. No, upwards, according to this convention, these are all upward values. So it is not upward, it will automatically be negative. So whatever you get here, you just have to plug that here. By definition, it is upwards. If it is negative by some reason, for some reason, then it means that it was downwards. But then you don't have to do double correction. It automatically happens in this approach. That's why I'm telling sir. Yeah. Suppose that q is moving towards left, then a will move upward. Delta y is upward in the direction, force is downward in direction. So according to that, first one should be positive, second one should be negative, and q should be positive. First one should be negative by your convention, because work is downward, displacement is upward. If I increase theta, point a goes up. So point a is going up, but the load is downward. So the work should be negative in that way and that is automatically taken care of if you just follow sign convention. When theta increases, q is moving towards left. But we are talking about p, right here. We are talking about p. But sign convention has to be same for. Yes. Yes. So when c is moving towards left. Yes. So displacement of a q is dx. Is inwards. So that should be positive then. No, no, no. So that's what is happening. So xc, if delta theta is positive, xc will be negative in this direction. But note here that the work done is minus q. Why? Because q is in this direction. So minus q into minus. So work done will be positive if delta theta increases. Sir. Yes. I think there is a mistake in this. In which one? It should be plus p into delta ya. That's why I'm telling. No, no, no. It is fine. There's no mistake. Minus is fine, because just look now. No, it's okay, sir. If delta theta is positive. If delta theta is positive, delta ya is upwards, positive. But p is downward, so work done is negative. Yes, yes. Yeah. Okay, it's fine. There is no issue with this one. Sir, you use that sign convention in upper one. You are putting negative value over there. You use sign convention both for the forces. Yes, yes. As well for the virtual displacements. Yes, sir. And then you don't have to explicitly again go and decide the signs. Whatever sign comes, you have to just put them mindlessly. Inclined. Will it be in equilibrium? If q is inclined, it won't be in equilibrium. Very good point. Then, because this assembly is not supported, I need to put a roller there. Okay. Strictly speaking, I need to put a roller at C. Then, if q whatever you do, that reaction will take care of that load. But the procedure remains the same. Why? Because according to this procedure, the C cannot move upwards. But you are right. To be perfectly consistent, I should put a roller at C. If you read the second equation, it seems correct. Yeah, I read it just now. Yeah. It seems correct. It seems that q should be plus and second p should be minus. No. Because if I am using the convention, p is acting in the downward direction. Yes. So I should take that as minus p. Okay. I should take that as minus p. Because p is acting, because I should use the sign convention both for the forces as well as the displacement. Think about it. Both for forces and displacement, you use it consistently, you will get appropriate equation. Think. Okay. So why do not we now move on with the tutorial?