 Okay, so welcome back to the last session for the day after the tea break. So straight going into the action, so what we have here, this is one example which I want to demonstrate where there is no way you can get any solution quicker than using to principle of minimum potential energy. So what we are asked here is something like this, a determiner angle of inclination of each linkage in this figure. Okay, so there are three linkages, W1L1, W2L2, W3L3, these are pinned at these two points and this is a frictionless surface on which these wheels can roll. And what you are asked to find out is that what is the angle of inclination or in other words what is the final configuration for the system when the system ultimately is in equilibrium and to say is that configuration stable or unstable and let me give you a hint you do not need to do a single calculation in order to find out what is the final configuration. Not a single calculation, if you try to do this with equilibrium or principle of virtual work it will be a huge pain, but with principle of minimum potential energy okay, just you can just do it, is the problem clear to everyone, yeah, though my one hint is that that no equation required, can you just stare at this problem and like just figure out what is the configuration okay, final configuration that this system can have. Note one thing that we want to minimize the potential energy okay, so minima is the best any time when you can find a minima that will be the absolute global minima and the system will be stable. For this to go into minima what do we need, each of the rods should be the center of mass should be as low as possible. So yes so the, so L1 and L3 vertical okay and the line joining the two ends okay, so that will be one minimum configuration straight away okay, so this is one of the interesting that these things you can do that if L1 plus L2 is greater than L3. Suppose for the time being assume that L3 is greater than L1, the one L3 is greater than L1 without losing any generality, if you think that okay no L2 is later less than L1 just go to the other side and see okay, L1 will become L2, L2 will become L3 okay, so without losing generality let's say L3 is greater than L1 then there are two conditions, one is when L1 plus L2 the two links is greater than L3 then that configuration shown at the top will be the minima energy configuration. Any configuration you try to perturb it from this core confirmation whatever way you will see that the center of mass will only rise, are you convinced okay, so the portion is that that without losing generality we said that L3 is greater than L1. Now look at this to minimize the energy what do we want, we want the center of mass of each rod to go as low as possible okay, now take this confirmation what we are saying is that this is as low as possible for one will be when one is completely vertical, center of mass as low as possible for two if two is perfectly vertical okay and what is the center of third particle which will be nothing but L1 plus L2 by 2 will be the position of the center for third one, so if you make L1 and L2 okay the centers of the two chains as large as possible okay as down as possible automatically the center for the third chain okay for the third link will get as down as possible, so this is possible but in this case what you require is that L1 plus L2 should be greater than L3, now in case suppose L1 plus L2 is less than L3 then the second confirmation which I have shown okay that will be the best confirmation what you will see is that in this case L1 plus L2 is less than L3, L1 is as low as possible, L2 is as low as possible and you will see that how do we figure out what is this angle of this third chain third link is just from this point draw an arc of the circle given by L3 such that it intersects this point and that will be the final confirmation and angle is immediately known that if that angle is nothing but cos theta is equal to L1 plus L2 divided by L3, so there are two possible cases without losing generality L3 is greater than L1 and then afterwards L1 plus L2 greater than L3 is the top one and L1 plus L2 less than L3 okay is the bottom one and just try perturbing the system from this point you will see that your energy will always be more than whatever the energy you have now because the center of mass will always go up if you depart from this confirmation makes sense. In the second diagram the last two terms are W3 into L1 plus L2 by 2 yeah because just note here assume that for example the center of mass for this this is L1 this is L2 assume that uniform mass okay I am assuming that all the mass is uniform at center of gravity is at the center okay so where will the center of this will be where will the vertical height vertical height is the same as the center of this which is L1 plus L2 by 2 thank you sir okay fine so this is this is for example one example where you use equilibrium you use principle of virtual work there is no way you are going to get this cleaner answer okay and of course this is the global minima and this will be a stable configuration oh oh it will be L1 plus L2 by 2 sorry it is L1 plus L2 by 2 not the common my mistake okay it is L1 plus L2 by 2 because center of mass will be what this distance plus the center so it is L1 plus L2 by 2 good sorry for the mistake okay so let us move on just note that okay because these slides I will give to the person in charge to just note that if you look at the solution okay first of all I cannot vouch that all solutions are perfectly accurate I tried to be as good as possible but sometimes like clearly I have made mistakes okay so please pardon me for that and be on the guard for some mistakes if you are referring to those slides now let me do this problem actually okay because the gentleman there here suggested that we use small angles approximation so let us do one problem where we actually do the problem with small angle approximation so what is given to us is a mechanism as follows okay so I will discuss this problem and then we will solve two or three more problems in the tutorial okay you will solve and at 515 I will give a small quiz okay and then we will depart so let me just discuss this problem briefly what we are given is this that we have a mechanism where we have a mass m resting on this two force member which is pinned here at the top and at the bottom similarly it is resting on this another two force member of length B which is pinned here and pinned here now this mechanism if there were no springs in between then what happens is that it is an unstable mechanism if you just give it a little bit of a nudge then this rectangle okay will become a parallel gram and this entire assembly will collapse so what is being done here is that in order to keep this system in equilibrium what we have done is that we have taken two springs we have pre-stressed we have given a stretch of delta and then put inside this assembly so spring whose approximate length is this length little bit of a stretch of delta put it in second spring take that length put little bit of delta put it inside and now we have this assembly where this internal spring has a little bit has a pre-stress of delta pre-stretch of delta it is not rest within this spring there is a tiny extension of delta within this spring there is a tiny extension of delta we have taken this assembly this mass is m and what we are asked to find out is determine the minimum stiffness k of each spring which will ensure stability of the platform in the position shown each spring has a tensile preset deflection of data see the question that this position should be in stable equilibrium and what we want is that given all these dimensions what should be the stiffness of the spring such that this configuration is actually a stable equilibrium conformation and now we will use for example this small theta approximation because why we are giving a small perturbation about the position where the theta which we can take to be the angle of the top links or the vertical links to be equal to 0 what do you think should we do if you perturb this system okay this is a rectangle let us put a bit by little bit let us make it a parologram so this link let us put a bit by small amount such that it makes an angle theta with respect to the vertical now if you put up that link okay the other link will also put up there is no other choice so this angle becomes theta other angle will also become a small theta now what will happen because of that if we make this link okay move by small angle theta what will happen to that spring it will get compressed okay why because there is a component of the displacement along that it will get compressed what will happen to the second spring it will get stretched okay can we express that compression and stretch in terms of theta very easily because all the dimensions are given to us if we say that the vertical that angle if this angle okay so if I say that this top angle is alpha here okay if this top angle is alpha then what do we know that if this small rotation is theta this top angle is alpha how much is the compression in the spring extra draw a perpendicular from here this is alpha 90-alpha okay so this also will be alpha so because alpha into data theta or the small perturbation theta will be how much there will be an extra compression in the spring you agree if alpha I will show it in the solution also if alpha is the angle on the top then draw a line from here perpendicular to the spring if this is alpha this is 90-alpha so B sin 90-alpha or B cos alpha is the perpendicular distance so B sin alpha sorry B cos alpha into theta will be the compression in that in this spring and correspondingly B cos alpha into theta for small theta will be the extra extension in this spring will agree with me okay fine now what is the potential energy contribution for a small theta for the top mass if I take base okay as my datum what is the potential energy of the top mass as a function of this theta the B cos theta if what is theta the way I am defining theta is B rotates by small angle theta so theta is the new angle with respect to vertical so B cos theta no vertical vertical vertical height cos theta B cos theta okay B cos theta then you are putting we have to then take the datum top level as a datum then it will be minus mg B 1 minus cos theta which will come out to be the same plus some additional constant everybody agrees or there is some some descent or there there's somebody doesn't agree okay it is not by intimidation okay if you don't agree just we will we will we will discuss this just look here this angle is theta this is this vertical arm it was position was here what is this distance B cos theta and B cos theta as we had discussed earlier for small theta can also be rewritten as 1-theta square by 2 okay because we want to do everything without cos and signs we can say that B cos theta can be written as 1-theta square by 2 does everybody agree that for small theta cos theta can be written as 1-theta square by 2 yes okay so we'll go by consensus now what do we do V gravity is what mg B cos theta here I have used cos theta but I say that even if you replace this cos theta by 1-theta square by 2 it still will be fine only thing what we want we want to know is for stability purposes this should at least be quadratic okay at least be quadratic now what is the extension in this spring initial extension was delta this spring got compressed so what is the final extension delta minus B theta cos alpha you agree with me what is the extension of this spring delta plus B theta cos alpha so my total potential energy will be mg B cos theta plus half k delta minus B theta cos alpha square plus half k theta minus B theta cos alpha square just expand it out what you will see is that your potential energy just looks like this now you take the derivative d V by d theta you will see that at theta equal to 0 this indeed is 0 which means that theta equal to 0 as expected was an equilibrium position and d 2 V by d theta square will just come up to be 2 k B square cos square alpha minus mg B just simple algebra and you want for stability purposes that this should be greater than 0 that will immediately tell you that minimum spring constant will be mg to be cos square alpha and you will realize that what is cos square alpha cos alpha what is cos alpha cos alpha is nothing but L divided by L square plus B square you agree with me cos alpha will be L divided by square root of L square plus B square so cos square alpha will be L square divided by L square plus B square just put in all the values and you will see that the minimum stiffness will be mg by 2B 1 minus B square by L square and if your stiffness is at least this much the assembly okay or stiffness should be more than the strictly speaking because exactly at that stiffness what happens we don't know but if your stiffness is more than that then your assembly will be in stable equilibrium is the problem clear okay any doubt any doubt so can we move on to solve some three tutorial problems and then go on with the quiz