 So, let us take the problems you know very quickly we will just try to go through the problems. So, now here the problem is like this it is a again a pendulum problem as we see here only thing is that a spring is attached now that spring what happens let us say assume the length of the spring is very long, so that is one of the assumptions we will make and the spring can adjust its position that means it can vertically come down or go up. So, what is this problem actually you think like this that suppose I do not have this weight if I do not have the weight then what is happening? It is said the spring is unstressed when theta equals to 30 degrees, so we are neglecting the weight of this bar which is able to rotate about this point ok, so spring is unstressed let us say at theta equals to 30 degrees then what happens? We put the weight now therefore what can happen as you see 30 degree will be somewhere here therefore the bar is going to reconfigure its position, so that position is you know said to be equals to theta ok, so at theta now our question is what value of theta will give me the equilibrium configuration that is the question here ok, so how do I now determine the potential energy of this system of rigid body, so as we see that we can easily choose our datum right here ok, so first of all potential energy has to be determined and we have to make sure that we have a datum somewhere that is a fixed point and that fixed point is definitely here, so that will give me you know height of this body and that will be the potential energy of the weight then we can get the spring force, so we can look at the how much spring is stretched by coming to the you know configuration theta, so we can take the spring stretching that way, so it is very simple as such, so just look at the potential energy function now as I said this is negative 100 force 10 cosine theta, so 10 cosine theta is the distance from the datum which is chosen at the hinge location ok and negative sign because now w is as I said below the datum line, so therefore it is a negative sign then we have the spring force the spring energy is given by half multiplied by k, now what is the stretching x is the stretching, so that stretching is going to be 5 sin theta negative 5 sin 30 degree ok, remember it should have been actually 5 sin 30 degree negative 5 sin theta, but since we have a square term it does not matter, so even if we do a plus minus you know mistake here it is not going to hamper the result, but here we have to be very very careful, when we are looking at the weight potential energy of the weight or a force then we have to be extremely careful ok, so once the potential energy is done then we can look at the equilibrium, so dv d theta is given and the higher order derivative d2b d theta2 is also given, now this part is not required unless otherwise we want to determine the stability of the problem ok, so from we can just look at this part now because I have to now find theta as a solution, remember it is a trigonometric function and theta cannot be found so easily, so in this way this one we can simply use some of the root finding methods, for example we can use Newton-Raphson method to find the root of this function, let us say this is a function of f theta then what is the root of the function, so we can use you know we can take help of numerical methods and I am not going to go to that, so basically what I said that we can really look at the Newton-Raphson method which is very commonly used and therefore I will get a solution that solution is roughly 16 degree ok, so the equilibrium configuration is determined by 16 degree and the stable equilibrium actually comes as a by-product because we know the theta and if we have the d2v d theta2 expression we just substitute theta and what it gives me is this value which is greater than 0 therefore it is a stable equilibrium clear, so let us move on to the next problem, so just take this problem very very quickly and try to understand what it is doing, so all are spring mass type of problem and we see that you know lot of practical applications that we have they are actually we can you know simplify those applications into a mass spring system, there are lot of problems you will see also those type of problems in tutorials, so here the problem is I have a 10 kg block that is attached to this rim, so it is a rim or a disc let us say and this disc is oval to rotate about 0.2, now you have another disc but that is firmly connected to the bigger disc, now on this disc small disc I have a spring that spring is attached to that disc and remember unstretched length of the spring is when theta equals to 0 that means when this spring right here you see this point that is right here that point we say that spring is unstretched, now the problem is like this that I put a weight right here if I put this weight then when I am going to achieve the equilibrium configuration just think of that way, so what at what theta that mass is just going to stabilize, so that is the basically problem, now remember in this case steps are again very simple first we have to determine the potential energy what are the forces coming into play one is the weight I have to derive the potential energy of that other one is the spring stretching right, so that is we have to find what is the spring stretching from that we have to get the potential energy, so these are two parts, so let us look at how do I get it, so remember that is the kind of free body diagram of the problem, so what is the potential energy of the weight that is simply given by weight multiplied by distance which is b cosine theta, so I have mg b cosine theta and then we have the spring stretching now spring stretching has to be defined by this arc, so that arc how much it is stretched is a multiplied by theta, so that is your s, so ultimately what is happening that you have half k a theta square that is the stretching of the spring clear, so now problem is really math, so we will just go to the math I am going to you know go quickly, so what is happening if you do dv d theta equals to 0 I am definitely going to get two configurations one would be theta equals to 0 as you see here and another one is defined by this which will give me theta equals to 51.7 degree, so there are two equilibrium configurations possible right, now question is if I go to the higher order which one is give me the stable solution, so again we need to do d2v d theta 2 and for theta equals to 0 you see that is what I am going to get d2v d theta 2 for theta equals to 0 it is going to be negative, so that is less than 0 therefore it is a unstable equilibrium configuration okay and for theta equals to 51.7 degree I am going to get an stable equilibrium configuration and it is very natural also because what happens spring was unstretched you put the mass it is trying to go to the you know stable condition by rotating by some amount, so that rotation amount is 51.7 degree and that is going to be the stable configuration stable configuration, so another problem this will happen in the buckling case, now what is it we have two vertical bars these two bars are attached to a single spring here remember I have a pin here okay and I have a pin here, so therefore these two bars can rotate okay and the spring is connected at this point C and one more thing importantly observed that there is a roller that is called guided roller again what it will do it will only allow the motion in the vertical direction there is no motion in the horizontal direction. Now what is the main question here is to determine the range of values of P for which the equilibrium of the system is stable in the position so on, so remember in this problem what will happen the stable equilibrium will be completely determined by the value of K how much load I can put in that will be completely dependent on the value of K and the geometry of the problem, so idea is that for some values of P the system is going to be in stable but as you go beyond that value of P then system is going to be unstable, so it is a completely buckling problem, so the value of P for the stable equilibrium is determined by the value of K and the geometry of the problem is that clear, so if you look at this way then what is happening first of all remember I have to now disturb the system by a small amount theta and try to see that disturbance theta whether it is going to give me the stable equilibrium or unstable equilibrium, in other words therefore I have to now rotate the system, so I am going to do that, so therefore what is happening first we give it a disturbance small theta let us say, so if this is theta then due to geometry let us say this is phi, now theta and phi will have again a relationship that involves the geometry of the problem, so what is that relationship we know that a sin theta must be equals to 2 a sin phi, now if theta and phi are small then phi should be equals to theta by 2 therefore now we have to determine the potential energy, so how do we determine the potential energy of this force, again this force will be treated as same as a body weight almost because as you can see this force will not change the direction, so its potential energy again will be always determined by the position of it, so it is acting almost like a body weight and the potential energy will simply be given by P multiplied by this entire distance that is y a, so this is the datum we have to choose, so therefore what I have for the spring also what is the stress configuration that is negative a sin theta, so total potential energy becomes half k multiplied by spring stretching square that is a sin theta square plus I have the potential energy of the force P. Remember body weight is ignored in this case of this bars, so I have only this force and the stretching of the spring, now as you can see if you look at this potential energy function and if you do dv by d theta what does it give, so dv by d theta will lead to a solution for theta as you can see theta equals to 0 is a solution and which is the position that is shown, so the equilibrium configuration is determined by theta equals to 0, so theta equals to 0 is a possible equilibrium configuration, so what we are asked now you have to find the P on that configuration such that it is a stable system, what should be the range of values of P such that it is stable at theta equals to 0, so that was the question, so now we just say that d2 v d theta 2 therefore we take this you know d2 v d theta 2 expression and we said greater than 0 for stable equilibrium, so therefore we can see that if we do this operation at theta equals to 0 of this expression right here then we are going to get P less than 4 by 7 kA, very nice problem to think about as I said now below this number 4 by 7 kA right then we are going to get an stable equilibrium configuration that means in buckling problem we already we always try to study you know this problem in terms of buckling if you give it a small lateral disturbance if your P is really less than 4 by 7 kA right then it is going to come back, so it is not going to go otherwise right, so that is the stable equilibrium configuration and therefore now this problem really brings in the essence of buckling okay, now if you really P equals to 4 by 7 kA that is the limit of it after that if you increase the load system is simply going to collapse okay, so it is going to get to some other equilibrium configuration, so it is going to be highly unstable it is simply going to deform sideways okay, so another very nice problem we will just study now this problem is going to be quite you know intriguing for the students as well as we can see here what is this problem state is that let us say you have a rectangular uniform solid body of mass m and height h that is resting on a fixed cylinder with a semicircular section, so this is a fixed cylinder what we have to do set up a criteria for stable and unstable equilibrium for the rectangular body in terms of h and r for the position shown, so it is completely determined by the geometry of the problem remember this body has its way which is passing through the mass center right, now if you disturb the system a little bit question is what condition should prevail between h and r such that the body can come back to its original configuration very nice problem to look at the main part here will be as I said that always how do I get the potential energy function correctly, if we get the potential energy correctly then we can solve this problem very very easily, so just look at it as I said let us assume that this is a equilibrium configuration I disturb the system by an amount theta okay, so what I will do in the next slide you can see here that body is tilted, so therefore the mass center that was at the middle of the rectangular block that is also going to be shifted to a new position alright, so that new position is now say O, the question is what is that new height of the mass center from the datum, so we have chosen the datum as dd prime, so it is again completely geometry that will control, now how I am going to get that adjusted height the final height of the you know center where the weight is passing through what is that height how do I calculate it remember there is a little kinematics involved remember as I rotate the rectangular body from its original configuration this translational component that you see here right that is equal to the arc, so this arc length is equals to that bb prime okay or ab let us say ab is equal to the ab is equals to really the arc length, so that is the only thing that we have to keep in mind okay because, so it is a curvilinear here the curved is actually curvilinear motion become a rectilinear motion here okay, so now you see here that what is the total you know height of the position now of the mass center, so we will go step by step firstly remember oa, oa is still going to be h over 2, now what is my ab, ab is equals to as I said you have to now look at the arc length, so arc length is r theta, so finally what we have the total height if we look at the vertical height, so cb cosine theta plus ab sin theta plus oa cosine theta okay, so that is the total height, so you can get the you know potential energy mg multiplied by the height, now rest is very simple, so as long as I understand how to get the potential energy for a rotation of theta then I can do the other steps in a very simple way as we have been doing, so rest is math, so dv d theta equals to 0, indeed is giving me theta equals to 0 as an equilibrium position, so theta equals to 0 was indeed an equilibrium configuration and then we do d2v d theta 2 right, so d2v d theta 2 now we have to say that for theta equals to 0 we get this expression for d2v d theta 2, that will be greater than 0 for r to be greater than h over 2, so that means the stable equilibrium condition will prevail when r is greater than h over 2 okay and for unstable equilibrium d2v d theta 2 should be less than 0, so r must be less than h over 2, so idea is depending on what we have whether r greater than h over 2 or r less than h over 2, if I disturb the rectangular body it is either going to come back to its original configuration or it is simply going to go to you know just slide down that means it is going to go to an unstable mode, so either it will be stable equilibrium or unstable depending on the relationship between r and h okay, so I will just close the you know lecture problems, so we have studied you know quite a few problems that actually you know encounter in real life examples and you will see that more in tutorial problems, but main idea that we attempt to you know deliver is the fact that can I get the potential energy expression correctly, can I understand the potential energy and once I have the potential energy function I can now find out what is the equilibrium configuration and whether the equilibrium configuration is stable or not okay, so I will just take the discussions now and then we can go for tutorials quickly. We have principle for minimum potential energy, what about principle any principle relating to other forms of energy? There are you know other form of energy theorem is available for example you know we can see the castigliano's theorem also okay, so you will see that when we have a deformable bodies, for the deformable bodies we can take those kind of principles also, but when it is a rigid body mostly we talk about you know kinetic energy as well as the potential energy, so here I do not have a dynamic problem, but if once the dynamic problem comes into play then we will have sum of this energy right that will be always conserved that is potential energy plus kinetic energy will be always conserved, but for the rigid bodies in this course engineering mechanics course we are only going to limit our self kinetic energy and the potential. Sir as you explained in the problems there are zero force members in the truss, in practical is it necessary to provide this type of the member or can we avoid this type of the member to save the cost or economy? Your question is again coming back to zero force member in a truss and what we tried to tell repeatedly that as a designer I design a truss remember zero force member is happening due to special loading conditions, suppose I have a set of loading on the structure some member may go to zero force, but remember if I change my loading pattern those members may not be zero force, as a designer I look at all possible combinations of the forces that make the structure safe right all combination of external loading that is going to make the structure safe, in that way when I am really designing the problem I do not have any zero force member okay, so all the members needs to be put. For that particular loading state you may not be willing to put those members fine enough, but remember if some other loading comes into play then the structure will collapse right because I do not have actually those members in the truss, so therefore I have to put those members on the truss okay, so zero force member is just depending on how the load is acting on the truss, is that clear now and no matter what you can always send your question by email, you have my personal email send the question I am able to respond to many people who are sending me questions okay, in fact the zero force member I also answered by email okay, there is one question coming up related to the non-conservative force although that is not a point of discussion but remember for the non-conservative force I said the friction force is one form of non-conservative force and it is true actually because you know that friction problem okay, the friction force can change its direction as the body goes from one point to the other point because the trajectory is very important friction force is always going to be tangent to the path right, it is always going to be tangent to the path, therefore what happens the force is actually changing its direction as it is moving from one point to the other point therefore the definition of potential as we started this session the definition of potential comes from the fact when the work done is independent of the path followed, now if the force is following the path that is it is keep on changing the direction then work done for each and every you know distance it travels going to be different okay, so we cannot define potential for such forces is that clear now triple one two sir it is said the potential energy depends on the position of the body that means it so if a body weighs mg and if it is at a height of h the energy that will be applied it will be mg h that is correct that is what we have done am I correct sir yes what are the things that there should be considered in resolving a problem see the what we are considering mostly the sign of the potential energy mg h is the potential energy of the body with respect to the weight weight means gravitational force, so gravitational force is what is the force there what is important to understand so just think this way that I have a datum somewhere here okay and I lift this pen so body weight of this pen is acting downward right and height is h so potential energy is positive now if I put the pen down here and this is my datum still then what is the potential energy it is negative so only thing that we have to be careful is the sign of it nothing else only thing we will be caring about sign of it okay depending on where we choose the datum okay one eight two okay the question is how to choose the datum line okay the question that question is more appropriate and I am going to discuss this okay how to choose the datum line I said during the problem the datum line should be something that is fixed on the space that means if the body is moving datum line should not move that is the first thing okay so now let me just try to explain it quickly okay remember let us say this is the bar l it has its weight at its mass center okay as the body moves okay now when we are talking about this fixed datum why I am choosing right here remember this height if I choose the datum here then y can be expressed in terms of theta right let us say this is l l is the total height let us say this is l over 2 so this is l over 2 here and this is also l over 2 so ultimately y should be equals to in this case l over 2 cosine theta now remember suppose suppose I choose somewhere here does not matter and I know this height is h so in the potential energy expression in this case what I have w equal sorry v equals to w y so w l over 2 cosine theta whereas in this case y equals to l over 2 cosine theta plus h okay so v equals to w l over 2 cosine theta plus h now remember as I says potential energy is not an absolute parameter this is relative so relative to the datum that you choose so this is datum 1 this is datum 2 but I am only concerned about getting the equilibrium configuration therefore I am to make dv by d theta equals to 0 remember when I do dv d theta h is independent of theta right so therefore dv d theta term will not have anything to do with the h it will simply disappear because that is a constant term so in this type of problems you can choose datum anywhere you want to but remember the position of this mass position of this you know force gravitational force is always defined in terms of theta or in a some other problem it is x okay so as long as that is true it does not matter if you just add a height h which is thus a constant term but this datum has to be always fixed axis it should not change its position as the body moves or as the body tilts okay so h can be you know that h is a constant height so that will not factor into your solution that you are trying to establish so your question is going back to theory that we have explained why do we need to take derivative double derivative to make sure it is stable or unstable why I am doing it because if you look at it what is dv d theta what is dv d theta in physical sense just tell me that first it is the slope right remember you have a potential energy function right potential energy function you have drawn it has a slope wherever slope is 0 that is the equilibrium configuration right and then we are going to one order up why because now we are concerned about curvature so ideally think of a ball see if you have a you know surface like this let us say this is a arc okay this is really v theta function now what is this what we have discussed really is that the slope of it if you keep on calculating slope everywhere you get slope equals to 0 right at this point so that is definitely going to be equilibrium question is why it is stable or unstable think of putting a ball now if I move this ball right here can I come back to its original configuration that's the question if it is so if the ball can come back to its original configuration then it says stable configuration now how do I decide this it will be decided based on the curvature that the v right that is taking place now you are concerned about what is the curvature rho the curvature is defined by what rho equals to always d2v d theta 2 so ultimately this brings in the concept of curvature if the curvature is greater than 0 then it will be stable so that means this is a math problem that means that will define the minimum potential energy right if the curvature is less than 0 that will define the maximum potential energy okay so this this will be always minimum potential energy and this state here will be maximum potential okay is that clear now so it is completely determined by whether this shape is concave or whether this shape is convex nothing else okay and it is a complete mathematical you know model that we use even class 12 standard has the same you know thing but you talk about in math but it is more in physical sense now 1179 suppose if you are considering a motion of a body in a inclined surface without any rolling how can you find out this minimum and maximum condition whether it is stable or unstable if it is rolling it will be always no it that depends on what kind of forces it has right actually we do have a problem in the tutorial but i will give more that as a homework your question is if the body is moving on an inclined surface okay and that is a more complex problem to discuss but for the problem that i will try to give it in the homework would be as follows suppose i have a ball okay now this ball has its own weight w now i put another small mass here so that mass is let us say m can you tell me whether this equilibrium is stable or unstable so what condition should prevail between this w and m such that i can say the you know equilibrium is stable or unstable this kind of problem can be studied okay so this will will give it in the homework and we can just brainstorm that when should i get the stable or unstable configuration yeah 1 2 6 3 when we are choosing the problem uh treated as the potential energy or the work energy method your question is when do we use potential energy when we use work energy two are completely different see potential energy is being used for the rigid body problems but work energy principle that we see right remember it can also go get into the deformable bodies when the body is actually deforming okay now in rigid body aspects work energy principle also comes into play but mostly work energy principle when you see it you will see it in terms of deformable bodies because deformable body has to now do work in terms of strain energy so we call that strain energy okay so then we say work energy principle is coming into that but here everything is rigid body so that is not coming at all okay sir i want to know that castigliano's theorem or strain energy maybe uh at all be thought as an adverse version of this virtual work theory or minimum potential energy theory absolutely correct you are absolutely correct your question is castigliano's theorem whether castigliano's theorem can be considered as more advanced version but tell me where when do we get the castigliano's theorem again you are going back to deformable bodies when the body is actually undergoing a deformation for example see this is again not the scope of work but the question came like this castigliano's theorem see remember typically what we do let's say you have a fixed beam i apply the load now what happens the body actually deform the body will actually deform so what will happen now all the you know if you look at all the particles that particles are actually doing a work by stretching themselves right okay so we write therefore what happens once you look apply the load the strain energy is stored on the body in terms of the deformation of the fibers or particles okay so castigliano's theorem if you look at it you will see that we define a strain energy okay that will be integral of half there will be e i term coming into play but see remember what happens in these cases this is completely departure from the rigid body mechanics why it is so because at the first hand i am assuming the body is actually deforming which is not the case in case of a rigid body so therefore in terms of rigid body this body has only potential in terms of this force applied it does not have the strain energy that is coming into play when the body is actually deforming is that clear now in more advanced mechanics we actually look at hamilton's principle in hamilton's principle we say that the total energy which will be composed of strain energy plus potential energy plus kinetic energy there will be three terms okay so all of these three terms now i have to minimize that function so we will choose a you know hamilton's principle is more general that is the most general concept and i will just give you these tasks to look at at homework okay this is most general but remember this concept is introduced when we look at mostly deformable bodies a deformable bodies should have strain energy stored okay because this body is doing some work right this force is doing some work due to the application of p and that will store the you know energy additional energy in terms of strain energy is that clear now but it is yes it is more advanced