 So, let us go to this problem, second problem what we have is this. So, we have a mechanism like this A B C. So, A B is a two force member, B C is a two force member. Now, at point C we have a piston which is free to slide in the horizontal direction, no friction between this surface and this surface. The piston is free to slide on this mechanism at point B, we apply some weight. Now, if we do not provide any pressure or any resistance from here, what will happen we know is that this point C will tend to slide in this direction and this will remain a mechanism and not be a balanced system in equilibrium. So, in order to prevent that what we are doing is we are applying some pressure which acts in the piston, the diameter of the piston is given and the end result is that there will be a force in the horizontal direction generated at point C, which will resist the weight. So, what we are asked is that at the configuration shown all the angles are given. So, given the pressure, given the diameter of the cylinder which means we are given what is the horizontal force acting at point C and with this information we are asked that for the system to be in equilibrium at theta equal to 10 degrees, what is the load that we should apply at point B? Is the problem clear? Simple problem. You can do this in like three lines using two force members, it is not an issue at all. But the idea is that how do we do this with principle of virtual work and we will see that there are next few problems which can be easily solved after figuring this out. Problem clear? Can you tell me now what kind of virtual displacements should we put to this system because note this system, this system is A, B, C, two rigid rods combined with a hinge at point B. If there were no, this point C has a force F applied in the horizontal direction, we know that force. There is a vertical reaction which comes from here which we do not know. What is the horizontal force strictly speaking we know because it is a two force member, but let us say that this reaction can exist. A, B force we do not know, we do not know W. So what we want is that we want to provide virtual displacements or virtual rotations or a combination to this system in such a way that only W does work and this force does work and all the other components do not do work at all. If I displace what system, the complete system, I can take this point C and slide it horizontally. But then what will happen? Is it a pure sliding? Is it a pure translation or is it a combination of translation and rotation? So point B will come down. Point B will come down, point C will move sideways. But then how do we decide that what, but the point is that now the virtual work will be done because of point B coming down by delta Y and point C sliding by delta X. But now delta Y and delta X are two quantities. There has to be a relation between these two otherwise we cannot eliminate them. How do we get the relation between delta Y and delta X is the question. That we can get from the geometry? Yes, we can get from geometry but how? Yes, you get the point, right? You are totally right, I totally agree with you. We want virtual displacement in such a way that point C slides in the horizontal direction. Then automatically point B will move down in the vertical direction. But there are two displacements, they will do work. W into delta Y, F into delta X. But delta Y and delta X has to be a relation between them otherwise we cannot solve this problem. And how do we get that relation? No differentiation for the time being, okay? Differences in later, okay? Right now, let's stick to the more physical approach. So delta Y by delta X is equal to 10 to thetas? Yes, it is related to thetas, of course, yes. Yes, yes, it is fine 10 theta but you are totally right, okay? You are almost there, okay? So what we do is this, okay? A very simple approach. I think combined I am almost sure that if you are given 10 more minutes you will figure out what to do. But because time is precious here, let me do this thing. Let us look at this rod as a ladder leaning against the imaginary vertical wall. Second rod as a ladder leaning against this imaginary vertical wall. Now suppose there were two ladders leaning, I let them slide the way we did previously. What is the issue? What will happen is that if these two are the ladder, we let them slide. There were two like this, both of them slide. But then what happens? This point moves sideways, but this point is also moving sideways. But note what is happening here? Point A is a hinge. We don't want no work to happen here. Then what we do? Both sideways, we give a combined translation to the entire thing. Such that we get back to point A to where it came from. And that is the resultant virtual displacement that we have. So if you look here, suppose we have a system like this. This is angle theta 1 BC, AB angle theta 2. If we think of AB as a ladder and BC also as a ladder, then giving the virtual displacement the same way as we did here, what do we get? B will move to B1, but A will also move to A1 and C will also move to C1. What do we know? That if this displacement is delta y, what is delta x? As one professor here pointed out that delta x2 by delta y should be equal to tan theta 2. You remember this problem. Delta x by delta y is equal to tan theta. By similar logic, delta y by delta x2 is equal to delta x2 by delta y equal to tan theta 2, delta x1 by delta y1 is equal to tan theta 1. But now we know that there is a problem by giving a virtual displacement like this y because point A is supposed to be fixed. It is not supposed to move anywhere. So what we do? We take this entire assembly and translate it by how much distance? By distance of delta x2. So just note then point C1, C had originally gone to C1. A full translation will make it go to C prime and the total displacement of point C from C to C prime will be delta x1 plus delta x2. But we already know that delta x1 by delta y1 is tan theta 1, delta x2 by delta y will be equal to tan theta 2 and so delta xc is nothing but tan theta 1 plus tan theta 2 into delta y. So we immediately know that if this is the delta y we provide where the weight will do work then the corresponding work done by the force acting in this direction will be nothing but f times delta xc which is again related to delta y. Is the idea clear? Very simple idea. So now we have a relation between delta y, the virtual displacement of point B on the top and the horizontal displacement of point C on the side. And we know that w times delta y should now weight is downwards delta y is upwards. So minus w times delta y plus f times delta xc which we have a relation for is equal to 0. So we know that w should be equal to f times tan theta 1 plus tan theta 2. And in the problem which we have just shown theta 1 and theta 2 both are same and equal to 10 degrees. Is the idea clear? So two problems. So first we solved a very elementary problem of just a ladder leaning against a wall. Second two ladders leaning against a wall but such that one point is fixed. So two basic problems and using these two okay let us solve okay a third problem okay which is a combination now. Now you will see that what is the importance of having a simple approach like this before. So let us move on to this problem. It looks kind of painful but it is not really okay it's really straight forward. What we have is this. We have a mechanism with three rods AB two force member BC two force member, BD two force member, C and D are sliders. They can slide without friction on this bar. So note that at point C there's a perfect freedom for sliding horizontally point D perfect freedom for sliding horizontally and this entire mechanism is connected to a spring. And what we know is that that we want this configuration okay to be in equilibrium when theta is equal to 25 degrees. So everything is given to us. All the distances are given to us. The configuration is given to us. The spring constant is given to us and what we are told is that the constant of the spring is 1.6 kilo Newton meter and the spring is unstretched when member BD is horizontal. What does that mean that when that entire assembly is completely flat. It is when the spring was at rest length when that assembly moved up the spring got stretched because of which it extended a force at point D in the horizontal direction in this direction okay. So that force is also known to us. How do we know that force? Just note here that 3L if this entire assembly is horizontal how much is the distance between A and D? This will be if this is 2L BC and BD are 150 so it is half so it is 300 plus 150. So 450 is the entire horizontal distance and now what is the distance? The distance is if you say that this angle say that this angle is 5 okay this angle then 300 cos 5 plus 150 sin theta or sin theta is equal to 25 degrees sorry cos 25 degrees will be the horizontal length and this P okay. Is the force in the spring? So now what this problem boils down to okay there is one error that I made this P I have replaced by Q and this Q I have replaced by P in this diagram okay because there are some older version of the book where it was Q and P okay and I had solved it like this. So one smaller just note one thing that this P I have written as Q in the solution and this force I have written as P and this P can be easily obtained because why we know this theta equal to 25 degree we all know the dimensions and from geometry we can also find what this phi is but without going into now for example like these are all the peripherals but going to the crux of the problem what we essentially have is we have this assembly at equilibrium when theta is equal to 25 degrees and we want to find out that what is the magnitude of P required to keep the system in equilibrium given that there is sorry what is the magnitude of Q here required to keep the system in equilibrium given that there is a P already acting given by this particular value is the problem clear or no not clear is the problem clear or no yes. So this force is given at this point and we want this entire assembly to be in equilibrium in this configuration where theta is equal to 25 degrees and what we are asking is that to maintain this configuration in equilibrium what force P we should apply here such that this system stays the way it is. So can we solve this problem very easily using the ideas developed before what are the virtual displacements I need to give to this system the only work should be done by the force at D and by the force at C that's the only work that should be done P and D okay so if that perfect okay so if you say that AB rotates about point O then okay yes come back this fine you are what you are saying is fine that what the approach you are trying to take is the valid approach that you assume that if all the members were completely independent of other members now you give it all virtual displacement rotation and now what you do is that now you translate them and rotate them in such a way that the connections come back together and that will be the ultimate assembly that we want and in that final virtual displacement okay the work will only be done by P and at by this force at C and a force at D but my question to you is this can we use the ideas which we have developed in the last two problems to get an answer to this in one short even without now without much thinking can we map this problem to a combination of problems which we had done in the last few minutes and solve this problem in just like three lines BC and BD can be taken as ladders okay so two ladder problem perfect BC BD so the tool and the second the piston problem okay is essentially AB and BD so what we know is that if B moves upwards by delta y this is phi this is theta so we know that the horizontal displacement here delta x will be related to the vertical displacement by tan theta plus tan phi one ladder problem not at the second ladder problem it's again a second ladder problem but this is theta this is theta but according to convention we have used theta is now minus theta as far as the two ladder problem is concerned so we immediately see from here that if we give it a virtual displacement point B will move upwards by delta y but then delta x will be nothing but tan phi plus tan theta into delta y will be the horizontal displacement of this and the horizontal displacement of this point okay will be equal to it will it will be in this direction will be given by delta y multiplied by tan phi into minus tan theta because according to ladder convention okay that was plus theta this becomes minus theta and you can convince yourself that this will work okay the two of them here so theta is just replaced by minus theta displacement goes in the inward direction so if you do this you'll immediately see that there's two equations delta x 1 delta x 2 and what is the total virtual work total virtual work done will be q delta x 1 q delta x 2 okay and the problem gets solved okay the final will answer will be that q will be equal to P tan theta plus tan phi divided by tan theta minus tan phi what is the relation between theta and phi we know that the vertical displacement is the same so 300 sin phi will be equal to 150 sin theta and so you know that sin phi is equal to sin theta by 2 given theta you can find out phi and you immediately know what is the relation between this force and this force is the procedure clear idea clear yes no yes do we move on okay now let us solve this problem okay this gets solved so easily and so beautifully in principle of virtual work immediately in one line if you use 2d equilibrium it will like take a few moments here to draw appropriate free word diagrams and it may look very complicated but in virtual work this gets done in literally one equation so what we have here is an assembly okay this is back of a truck so we have a member we have this platform this is the back of a car or a truck and this flat this platform is connected to the back by two members a b e b and d a okay yeah by e b and d a and it is used to support some weight and now this entire assembly is supported by a hydraulic cylinder cf okay and what we are asked to find out is what should be the force in the hydraulic cylinder such that this assembly is in equilibrium is the question that we are asked problem clear now if we go back to our equilibrium approach okay now what I want to connect here let us do this problem first using virtual work and then I want want to collect that how we had motivated our problems in equilibrium approach okay and then we will go from there now you tell me how do we do this problem what are the two what is the quantity that we are interested in we are interested in finding out force in the hydraulic cylinder so work done by that hydraulic cylinder and work done by gravity on the weight are the only two quantities okay that we are interested in now how do we provide this system a virtual displacement or a virtual rotation or a combination whatever you want to call such that we can provide that virtual displacement that virtual rotation and the work done will be only by the weight and hydraulic cylinder so that we just get one relation between them is that doable and if yes how yes of course yes how to give the angular rotation about point a perfect you can you give angular rotation about point a yes if you give angular rotation about point of delta theta suppose that there is a it's a it's a member which is crooked but it doesn't concern us okay you give it an angular rotation the same angular rotation has also to be provided to be because if that is not provided what will happen let us think about it if I give dca an angular rotation of delta theta 1 and e b an angular rotation of delta theta 2 if we do that what is what what may happen or why is it not a right thing to do platform will tell but big deal platform because it is connected by the same piston piston is not there so piston doesn't do anything what is common d e what is common 300 dimension is common for both of them yes yes that is fine but what I'm asking is that if given these are the dimension that these two links are exactly identical okay both dca and e b are identical what my question is that and both of one is fixed at d other is fixed at e there is a geometric consequence of what will happen if e c b a and e b are given different rotations because virtual rotation is in our control suppose I decide okay one fine day I get up say no I don't want to give them same rotation but what is happen nature will not like me okay it has some own set of rules I do something it will react back what is that backlash or what is that reaction which mechanics will give me if I provide both of them different rotations note one thing point d and point a e have a spacing between them okay some constant spacing tell me if I give this assembly a c d or rotation of delta theta what will be the horizontal displacement of point d horizontal displacement of point d note what we saw horizontal displacement of point d what do we do horizontal line who's go I draw a perpendicular from here and that distance multiplied by delta theta will be the horizontal displacement of d yes what is that displacement 650-300 350 no forgets no no no it is it is six 650 so 650 is that distance so 650 into delta theta will be the horizontal displacement of d but if I give a different rotation for b how much will be the displacement of e it will be equal to 650 into delta theta prime what does that mean d moves horizontally with some distance e moves with different distance so we are stretching that distance so we are increasing that distance between d and e and if we increase the distance we have to pay the penalty that the internal forces will do work which we don't want because those are unknown what is delta x no no it is not but what I am saying is that but to do that before what I am saying is that we want our mechanism we want our virtual display in such a way we don't need this displacement I agree with you but what a point I am trying to make is that we have the choice because there are two links we have to give this some rotation this some rotation what I am asking is that why should both of them have the same rotation it's because if you don't give them the same rotation then d e distance we have to change to be changed okay so that's the reason why both of them should have the same rotation is the point clear okay both of them should have the same rotation now the question okay so now we know both should have the same rotation good now with this fact can you tell me how do you use principle of virtual work the only virtual displacement we can give is give both the links rotation about a and b respectively suppose I give a d and b e small rotation of delta theta about point a and b respectively delta theta now you immediately tell me what are the what is the work done by gravity or what is the work done by the mass or what is the work done by the weight of the body with respect to this virtual displacement we have to find out how much it goes down by how much will that mass go down by you tell me delta y yes what is delta y if delta theta is given to you what will be delta y what we had seen just 500 into delta theta just 500 into delta theta you agree with me so 500 into delta theta is the vertical displacement of the top mass what the look at tell me what is the force which is which will be provided by the hydraulic cylinder it's a two force member force will provide in the horizontal direction what will be the work done by the hydraulic cylinder if I provide it a delta theta like this just this distance into delta theta what is that 650 minus 300 which is 350 okay so the work done will be 350 into the force in the hydraulic cylinder whatever direction if you choose a direction which is outwards okay what is that which is the compression so then it's done so 650 500 into delta theta into weight plus force in the hydraulic cylinder into 350 is equal to 0 that's what principle of virtual work will demand and you're done one small equation answer is out you see the point the answer is out in just one small equation any question any doubt about how we give the virtual displacement what were the requirements why were the virtual displacement desired in such a way okay we discussed this point but still if there are any questions please feel free to ask any doubt any question yeah it's clear so you saw the power of this approach right that just by being able to visualize what is happening we immediately figure out this will happen now you tell me if you have the ability to visualize how some mechanism can happen suppose this the cylinder were not there suppose if we had discussed some problems like this in 2d equilibrium what we saw that the entire system will collapse but now we know how will it collapse that this will tend to rotate about point a this will tend to rotate about point b in the same same way delta theta 1 and delta theta 1 okay the angles will be the same and because of that what happens point c moves in the outward direction and so we know that it starts to move in the outward direction so most likely the force in this hydraulic cylinder will be tensile okay because it without this it will try to move like this but in the presence of this that motion will be prevented so that's another check for us that whatever answer we saw even using free body diagrams and 2d equilibrium the final answer should be consistent with all the mechanisms of the problem if that body or that hydraulic cylinder were not there so the moment we can visualize how things can move we immediately have an idea that is that problem solvable okay now not another thing the beauty of virtual work is as follows many times what happens is that sometimes you open up a problem in a textbook there may be some error for example some dimension may not be mentioned they forget to mention some dimension or sometimes they mention some 50 dimensions but not mention the dimension which is actually of importance if you can understand how the mechanism can happen and if you can figure out that these dimensions we decide what is a relative displacement of one and other so we know that a problem for example given in a textbook is right or not right so we can be our own authors that way okay that once you understand how things move what are the relative motions then everything becomes crystal clear even then for example you will realize if instead of having three rods suppose I have five rods it's a big assembly three four five will the answer change instead of two if I have five or ten as far as the force in the hydraulic cylinder is concerned I answer will not change okay so that's what you immediately know by knowing that how for example the system displaces when we provide some virtual displacements and rotations and what are the virtual words that are created whereas if you have a very naive approach about counting the number of equation number of unknowns for a 2d equilibrium problem you will realize that the problem is statically indeterminate and you may feel oh you may not be able to find out the force in the cylinder but by knowing the approach like this you know that okay the system is statically indeterminate as far as the other forces are concerned but as far as the force in the cylinder is concerned it doesn't matter okay so that's the power of this approach that it will give you multiple ways in how to think even the 2d equilibrium problem that once you understand that how if you remove one component how the mechanism happened you immediately know okay that how to solve the corresponding 2d equilibrium problem is the idea clear the motivation for this entire procedure everything is this fine okay so we do we take a break now