 There are 4 problems, so let us take 15 minutes for problem, I will briefly discuss the problem here, it is a very straight forward, I think it is a very one of the more popular problems. So what is happening here is that we have a step of size h, the coefficient of friction between all points of contact is given to be equal to mu s, you are applying a force p at the top and the weight of the cylinder is 25 kgs, radius of the cylinder is given to be 3 meters and you are asked that if the coefficient of friction is equal to 0.3, I want to lift this cylinder which means that the cylinder while lifting will lose contact with this point and completely be hinged about point A and you are asked to find out that what is the maximum value of h such that the cylinder will not slip at point A, okay. It is a problem clear, it is a well known problem I think, so a couple of people had asked me this problem but any question about the problem, okay, just have a look at this, okay, it will not take you more than 10 minutes, okay, it is a 3 force member, okay, you realize that 3 force member and all the ideas we had used about the tipping toppling will all be applicable here, okay, so shall we move on or are there any common questions? So the typical strategy which I think many of you have figured out of course is that while the cylinder is being lift off, it has to lose contact with the ground because of which now it has 3 forces p w which pass through point A and the third force has to point through past A otherwise it will create a torque about point A and that simply will not be an equilibrium. Now that immediately tells you that as a function of height, okay, if this angle is phi, this angle is phi, note that this angle phi is angle made by the effective force with respect to the normal, okay, from this point this is the normal, this phi is the angle made by the total reaction at point A with respect to the normal and so phi can maximum value be equal to phi s or coefficient of friction tan inverse of mu. This is phi, this is phi, this will be 2 phi and what we know that the height will be given by r times, this r is the radius 1 minus cos 2 phi, note one thing that h max will be possible if this quantity is as small as possible when phi becomes as large as possible but since phi max is equal to tan inverse mu, you will at least put that in and you get what is the maximum value of the height, okay. Now note one question for you, if for example when you barely lift off the cylinder, okay pull on it, the cylinder is barely lifted from the ground and when it is barely lifted from the ground, there is no slippage at A but your ultimate goal is to take the cylinder completely up, do you think that if you can achieve the goal about lifting the cylinder just about the ground without slipping at A, can you lift the complete cylinder afterwards? Is that point clear that your ultimate goal is what to take the cylinder completely upwards? You start that process by putting a force such that it loses contact with the ground completely pinned at A but the question is that now if you can manage to successfully do that, will you be completely guaranteed that you can lift it up fully now, yes or no? No, yes, okay if you will be guaranteed why because think about it, if you take the cylinder it is just losing contact, now if you take it upward and upward what will happen is that this point will keep going up and up, what is effectively happening that the step height, apparent step height will keep decreasing and the apparent step height that the cylinder will see because if this point keep moving upward this will rotate about point A and the apparent height it will see will keep decreasing because how much it gets lift up by that will cut off from that height edge and since we know that larger height which will cause a problem that effective height because it is decreasing once you are assured that this lift off like the initial lift off is completely risk free then you can just take the cylinder upwards, okay so that is the overall logic of this problem, any question? Because otherwise like if you lift it up and then while trying to lift it again falls down so what is the point you lift fall, lift slip, lift slip so that is not but here you are ensured that if you can manage to lift it just off the ground we can just take it off fully, okay? Because if point A there is no impending slip at point A in this case there will be no impending slip afterwards, why? Because the effective height is decreasing, height decreasing. In all the problems it happens. Where? In all the problems it happens. No it does not happen that's why so I did not explicitly mention you think about it, you think about it and you will realize that when we get that limit within that limit you cannot find the forces but what you know is that outside that limit it will slip, outside that limit it will slip but within that and how do you confirm? One way to confirm is this that if you are within some limit, okay some force is there where there is no impending slipage. What you do is this you assume that what will be the coefficient of friction that is required to cause an impending slipage and if that coefficient of friction comes out to be lower than the given value then you know that the system is stable. You see the point like for example apparent height whatever you are talking now that apparent height now what you do is we solve the problem again assuming impending slipage at A, okay but now you will get a coefficient of friction now because H is given you will back a coefficient of friction but that coefficient of friction now will turn out to be lesser than the value there which means that for impending slip to happen your friction should be lower but you actually have higher slip friction so no impending slipage will happen. You are turning the problem around think about it. For example to convince yourself that this will not happen that you just say suppose impending slipage happen back track what should be the coefficient of friction that coefficient will turn out to be lower than what is given you are solving it in the other way round, okay shall we move on now. Sir the solution can be done by using Lambert's theorem here. What I don't know this great theorem. Lambert's theorem that is sin rule P by sin alpha. 3 because 3 you can always use sin rule. Yes sir. If there is no friction at the contact in surface the reaction is possible to it of course and as there is friction there the reaction force will be making an angle of 5. It will making 90 minus 5. 90 minus 5. That 5 is an unknown. 5 is an unknown. You are given the value of 5 mu. But only at that maximum height H when the slipage is about to happen 5 will become equal to 5s. Yes sir that is for this equilibrium you can take it as equilibrium. Impending equilibrium. Impending equilibrium when there is an impending slipage at point there. It is also equilibrium. Yeah but at impending equilibrium that is a special point any height lower than the critical height you cannot say you can only say that the direction is given by the geometry. Yes. But that will not be equal to 5s. Okay so essentially that having all those intersect at the same point is equivalent to the Lambert's theorem or sin rule okay they are all the same thing so whatever you find convenient okay you can definitely do that. Which one? P. P is the top right. Where reaction? Center K weight is acting through center right. Weight is assumed it is a nice uniform so I should have said it explicitly nice cylinder all load acting through the center. So two lines intersect at the top so third has to go there otherwise it will create a torque about the top point okay. Yes please. How it is coming? 2 pi. 2 pi? 2 pi. So 2 pi. Oh that is 5 that is 5. So there is an isosceles triangle so if this is 5 other angle other angle is also 5, so 5 plus 5 is 2 5 there, external angle, yes the external angle right, so it is 5 5 is 2 5, okay, fine, so let us move on, little bit of a difficult problem, not too much, some thinking will be required, so what we have here is we have a wedge, it is a wedge problem, so what I have tried to do that, I have tried to cover as many problems of different categories as possible, so we have a wedge, we have a cylinder, okay, the wedge angle is 5 degrees, this cylinder is resting on the wedge, what is told is a coefficient of friction between all 3 points of contacts, 1, 2 and 3 is 1 by 4, and you are asked to find out that if that is given, the weight of the cylinder is 1000 pounds, weight of the wedge is taken to be negligible, you are asked to find out what is the force P required to move the wedge, and clearly from the direction, what is the force P required to move the wedge internally, okay, you want to move it inwards, okay, and when I say P what is the minimum force P required, of course you apply larger force it will move, what is the minimum force P required to move this wedge in the inward direction, is the problem clear, your statement is really straight forward like most problems in friction, some thinking not much, how many unknowns do we have, 4 unknowns on the cylinder, okay, 1, 2, 3, 4, bottom how many unknowns, 2 unknowns, line is also unknown, but forget about that, equations and P, P is also an unknown, so we have the number of unknowns is 4 plus 2, 6 plus 1, 7, 7, how many equations do we have, 3 for the cylinder, 2 for the wedge, so we need, we have 7 unknowns, 5 equations, we need 2 extra equations, so slip should happen at 2 places, okay, how many surfaces do we have, 3, okay, so there are 3 possible combinations in which the slip can happen, and now in this problem it will be very beneficial that if we can have an idea about how the motion, how the impending motion can happen, so that you can immediately fix what are the directions, if you know the impending, if you can visualize what is the impending motion, the directions will automatically get fixed now, so 2 surfaces should, you have slip edge, one thing, okay, any question, one thing you can realize is it is not possible, sir, yes very well, in the previous question what is the role of W, role of W, that was given as 25, the W is the wheel and here, the W is making your life difficult, okay, right, the W is making your life difficult, if W is 0, what, this problem is not there, you just come and take it up, I think that has not been used in any of the equation, I think, no, no, but it depends on what we are asked, right, we are asked to find out H max, if you are asked to find out P, then it will come, no, and P depends on H, so if you are asked to find out what is P for H max, W will play a role, but just note one thing that even though there are two combinations, so slip can happen, combinations, what combinations, here and here, here and bottom, here and bottom, but the first assumption that slip happening here and here is not possible because we want the wedge to go in, and just by having slip here simultaneously it is not going to rotate, so this mode of motion, impending motion where slip happens here and here is not, you can immediately throw that out, so essentially the competition between does the slip happen here and here or here and here, and think about the wedge is trying to move inside, so we need to find out what is the impending motion, so we can immediately decide what is the direction of friction, think about it, any questions asked? So let me just briefly discuss the problem, okay, the idea is this, what is given, okay, let us ask the question, we are asked to find out that this wedge, what force minimum P we should apply such that the wedge move inwards, minimum force, so what do we know is that the impending motion of this wedge should be inverse, now just think about it, how many unknowns we had, we had just looked at 1, 2, 3, 4, 5, 6, 7 unknowns, we had 3 equations from this free body, 2 from this, so 5 equations, 7 unknowns, we need 2 slippage, now the 2 slippage can happen in different ways, it can like for example this center cylinder can have an infinitesimal rotation giving slippage at 1 and 2, but this is useless, why? Because first of all a cylinder cannot simultaneously start to rotate, on top of that in this mechanism wedge does not go anywhere which is what we want, so this is out, first one is completely out, come to the second one, if the wedge moves inwards there is clearly a slippage here at the bottom point, but when the wedge move in just note that this cylinder center has to go up, because in here height is this, if you push the wedge in the cylinder has to go up little bit, you agree with me? Cylinder has to briefly go up, but then what happens that if you think about it, if the cylinder center just translates upwards then you will have a slip at this point also and at this point also, both at the bottom and the top point, we do not want that, so if we want to have slippage only here for example and not here then what we can do is that, we give it a clockwise rotation, such that this point in addition it will move even more upwards little bit, whereas this bottom point whatever distance it had gone slipped up, it will cover that distance, so this is one mode of slipping, but now look at other mode of slipping, if the slippage does not happen here, but here, center has to move up, so this motion is mandatory, but this point will also slip, in addition if you provide a rotation in the anticlockwise direction of very small rotation, then you can cover up, this moved up, but we rotated little bit, so that we covered that distance down and so no slip here, but this point moved up, so there has to be slippage here and no matter what you do, if this mode of slippage happen, this point has ultimately slide upwards with respect to the wall, so the friction will be down, if this point is slipping, then this point is going to slide upwards relative to the wedge, so the friction will be downwards and ultimately the free body diagram will always look like this, this is F, for this mode this is F, but torque balance about points will immediately guarantee that this also has to be F, if the slippage happen here, then this point moved relatively upwards, so the slippage is upwards, so the friction force should be down and that also automatically guarantees for moment equilibrium of this free body about point C, that the friction at top point should be in the downward direction, so this has to be the free body diagram for the top cylinder, F, F, these are the directions, normal reaction 1, normal reaction 2 equal and opposite on the wedge, this is P and there is a slippage here N3 and because there is a slippage F3 will be equal to mu as N3, so in the free body diagram clear that this is how the free body diagram will look like from the considerations of the problem, both kinematic and equilibrium considerations and now we have to process this further, any questions about this? The kinematics now just visualize, so for example tomorrow we are going to do principle of virtual work and when we solve principle of virtual problem you will realize that how to take care of infinitesimal rotations and see that like how different components when they move what is their motion, it will become more clear but if you think about it you will realize that that is the only way for this problem you can have the directions of all the frictional forces, yes unambiguously, somebody had a question? Think just little bit like and it's not difficult, just think how that infinitesimal motion can happen and this is the only way you can have all the friction forces, there is no other possible way which is consistent with the definition of the problem that the wedge goes inwards and that all the sub parts and the full system is in equilibrium, any question about this? And once you know this you are good, now second idea is to figure out will the slip happen, is this mode going to happen or this mode going to happen because we know by problem definition that slip will always happen at the bottom portion, there is no choice because the problem is telling us that the wedge should go inwards but now the question is will the slip happen here at the vertical wall or between the cylinder and the wedge and to answer that question what we realize is this, draw this line two tangents ok along the direction of friction, just note here if I draw the free body diagram for this what you will see is that that N1 times D plus W times R is equal to N2 times D for moment equilibrium, so immediately it becomes clear from here that N2 is more than N1 and since the friction forces are the same the coefficient of frictions are also the same, you immediately know that N2 more than N1 implies that there will be no slippage at this point, only slippage at this point and when the slippage happens at the vertical point you automatically know that F will be equal to mu s nu N1 as far as this problem is concerned, now what we do? We have N1 mu s N1, take this free body diagram, take talk about point B, you will immediately find out what is the value of N1, you get the N1 will be equal to 120.3 approximately, now what you do is that you know N1 you know F, so it is a very simple way to do this problem is you look at the full free body diagram 1 plus 2 put together, now full system is also in equilibrium, we want a normal reaction, what is normal reaction? F plus W will be equal to N3 for equilibrium of this complete system but we already know F which is equal to mu s N1, so we immediately know what is N3, now once you know N3 there is a slippage at the bottom point, so F3 is equal to mu s N3, so we know also the friction, now the only unknown in the system is P is what we want for this complete free body diagram take equilibrium in the x direction, so N1 plus Fs or plus F3 will be equal to P and you immediately see that the force P required to move the block inwards will be approximately equal to 377.8 or 378 pounds, so by kinematic considerations by free body equilibrium taking appropriate free body diagrams you could see that the problem so called, it looks reasonably formidable to begin with but by doing it in some way, you can see that like by just 3 equations which are not too complicated, we could immediately solve the full answer that we decided, N2, you don't need N2 really because, okay if you want to get N2, okay then just come here, look at this free body diagram, if F is known, first of all do you think F is known or not, if F is known or if F is equal to mu times N2 assuming slippage is here, then you just talk about point that you will get N2 and if you want to solve what is N2 for this diagram, okay what you can do is take equilibrium of this free body along this direction, there are so many different ways in which you can get N2, yes please, what 5 equations, I can write 5 possible equations, right, what are the possible equations, you know 3 equations for this free body diagram, moment x direction, force x, force y, another 2, about 3 points which are not collinear and for the bottom way Fx and Fy, equilibrium, so 3 plus 2, 5, so 6 unknowns, what is the 6 unknowns, 1, 2, 3, 4, 5, 6 plus 1, so 6 plus 1, 7, UK unknowns, 5 equations, so we need a difference which is 2, we need 2 extra equations to solve this problem, 2 slips, okay because the most naive thing to do is for example assume that like slipping happening everywhere, but then that problem becomes over constrained and you will see that you will not get a consistent answer which satisfy all the equilibrium conditions. W cos theta, okay, okay if you take W cos theta is equal to no it is not because just look now W cos theta vertical component, there is also a F which is acting in the W cos theta is what, only this force, now onwards, but now for the top free body diagram, we also have vertical, the friction cos is acting vertical, even here, n took a vertical component, this friction will also have a vertical component, which will be equal to F sin theta. It will be because of the friction, in this case more, in this case more, F plus W will be equal to n3. W prime is 0 here, okay the weight of this wedge is very small, we are taking it to be 0. That is 0, okay the whole weight of the sphere will not directly transfer to this wedge, yes, so the bottom n3 is equal to W plus F, no we do not need to why, listen because what we can do is we can draw the complete free body diagram, like for example do not expose n2 at all, the cylinder and the wedge is my free body, who is stopping me from doing that, because if the system is in equilibrium, not only the whole body is in equilibrium, any small part of the system and combinations, all are in equilibrium, so I can draw the complete free body, means like just merge these two together, so this will cancel each other out, no contribution equal opposite equal opposite gone, so the only forces which are not canceling each other out is this, this, this and the weight, full free body diagram and why, you may say why, so at some point you have to use your judgment, you could have as well done for example like found out n2 and then transferred at n2, but that is an extra step and I do not need n2, I even need it you do it, but in this problem we do not need it, all the time we cannot do like this, all the time you cannot, of course not, it all depends what is the requirement of the problem, how the problem is, so it is a judgment after sometime it becomes more of an art like than science actually, so definitely we should move by judging by determining the value of n2 then, you can find it, but in this problem you do not need n2, is all I am trying to say, if you feel comfortable finding n2 and going further, fine not an issue at all, there is nothing wrong with it, but what I am trying to say that as far as this problem is concerned, what is asked, we do not need n2. Sir, interrupt taking moment, if you take free body of each, each okay, I have taken each, first and second, interrupt taking moment, taking moment for what free body and about what point, for both bodies, no, no, just we will, if you go for second body moment is irrelevant, why because the line of action is not for first body also for fx0 and fx0, we can calculate n1 and n2, we are not taking any moments, we have taken moment there, if we take 10 knots, only if fx and ff will not give the answer, because there are 3 unknowns, sir we will get answer, we will not, if you get answer, that is not right answer, because if you take only fx and ff, just note, this is not a joke, what I am saying is that, suppose f is equal to mu n1, how many unknowns are there, one unknown, here how many unknowns, 2, 3, okay, but if f is equal to that, you can get it, yes, we will get it, you will get it, only thing is, without taking moment, you will have to solve two simultaneous equations, simultaneous equation, but what we are trying to do is, getting rid of simultaneous equation, we want to solve everything in one shot, we have to convert, you are right, you can do simultaneous equation, you can get it, but by taking torque about point B, we got everything in one shot, simultaneous equation, the risk is that you make mistake in one, you make mistake in the other. Sir excuse me, sir here, sir this side, sir for FVD 1, you have taken the moment about B, sir as you are mentioning that the B point at which this wage and roller, they are rolling over each other, no they are not, see, see, you have to kind of, see there is a kinematics aspect, there is a equilibrium aspect, the kinematics aspect is only for you to get an idea of how the friction or what is happening, how is the slippage happening, it is not that they are always rolling, but the point is that, what I have shown here is that, that these are the possible modes in which that sliding can happen, and that visualisation is for you to understand that if this moves in like this, and if I assume the slippage here, then what should be the direction of friction? Sir you are explaining it is very true sir, but I have one doubt also sir, that while they are rolling over each other, they are not rolling over each other, they say infinite symbol roll, whether the friction come there, at B point you have shown one friction, friction can come, they are equal and opposite for both FVD, but only thing you know that what we have seen here is that at N1, sorry, N1 is less than N2, so the impending slip is happening at 1 and not 2, okay, it is happening at 1 and not at B, what does that mean? That the slippage is happening like this, which means that there is no relative slip at these two points, but no relative slip only means that the friction at that point is not equal to mu times the reaction, normal reaction at that point, okay, it does not mean that the friction is not there, friction can be there, but we cannot say that the friction is equal to mu times the normal reaction. But it will change numerical to numerical, these conditions. Depend on problem to problem, we have to change, okay, but you need not have impending slippage, but there can still be friction. Okay, thank you. But what it means is that F by N is not equal to mu s, it is less than mu s. Thank you, thank you sir. Sir, it means we are neglecting the friction that. No, we are not neglecting, we are just saying that there is no impending friction there, so if the normal reaction there is N2, F is not equal to mu times N2, is all we are saying. Okay, it is some value. It is some value. So what is the value? That is what we get, because what we saw is that the impending friction can happen at 1, so F is equal to mu times N1. By torque equilibrium or moment equilibrium or point C, you see that this F and this F are to be the same, so if I can find out N1, I know F and that is the value of F that will come. Hello sir, simply by using the concept of friction is opposite to the direction of motion. Impending motion, okay. Impending motion for static friction, finite motion for kinetic friction. Okay, okay. And using the sign rule or Lame's theorem, the problem becomes so simple. How will you use Lame's theorem here? Because you know the line, don't know the line of. There are only three forces. For the top one you can get it, yes. You can get it. You are not for the wage, wage. Even also for the wage, there are three forces. There are three forces. You can't use Lame's theorem because the point is that what is the line of action you have to figure out from and I have not even given you what is the height of this P. I am not giving you what is height of P. Without knowing the height of P, how can you use Lame's theorem? Hello sir. Sir. Myself sir. Here. Yeah, and even if you use Lame's theorem, okay, good. If you can, I think you can't use it here, but you can use it, good. But the point is that this procedure, I feel comfortable with this one. Sir, what are the limiting conditions to use the Lame's theorem? What are the limiting conditions to use the Lame's theorem? You should know the directions of all the forces. First of all. They are known, suppose. Okay, all the directions are known. Then the point is that they form a triangle, right? Three forces. So the directions are known here, no? For this one, no? Yeah. So P is going to the left. Okay, good, good for you. Okay, you can use Lame's theorem. Okay, yes. You can use Lame's theorem, great. It's fine. In equation summation moment about B equals to 0, can it be n1r cos theta minus mu n1r 1 plus sin of theta? Instead of 1 plus phi. Which one? Second one. No, this is, no, but this, this one. That equation, sir. Which equation are you talking? Summation moment about B equal to 0, n1r cos theta minus. Okay, so n1, did I make a mistake there? n1r cos theta. Can it be mu n1r 1 plus sin of theta? R sin of theta. S is sin theta because that paper ran off. I wrote just S. It is sin theta. Sir, hello, sir. Phi went to the edge. Sir, for this second wedge, we can't use Lame's theorem. For Lame's theorem, we require only three forces. There are three forces, no? Wait, W prime is 0. If you know what is N2 and F. Okay, then you know the direction of this. Ah, that. Okay, so you know 3. You can do it. But the point is that in this case, no? This simple equilibrium in x, y direction is so easy. So you can use it, okay? But the point is that it's completely in the same way. If you wish to use it, I completely agree that you can use it here. Hello, sir. Because they are forming a triangle, you can use it. S is sin theta. I ran out of paper, okay? So it was towards the end. So I just wrote S. S is sin theta. Sir, sorry. For in this kind of problem, multi-surface contact is there to find out where impending will occur or not, for that we are checking that number of equations, we have a number of unknowns accordingly. Yes. But how do I beforehand know that the problem will be statically determinate or not? No, no, but to make it statically determinate, what is saying is that if you apply that much force such that you are about to start that motion, now you tell me how much force I need to require. So you are making that problem statically determinate. If it is not statically determinate, I totally agree. Of course, it is in your hand. Every problem is like we have to decide what is the question we are asking. You are right. But if I ask a weird question, not a weird question. If I have a different question saying if I apply some P2, then what are the forces? You cannot get the answer. It is indeterminate problem. Sir. But what you can tell is that if that P is less than that P max, then this will be in stable equilibrium is what you can say. All you can say. In a particular structure, what are the different constraints and accordingly how do I check that beforehand we have to decide that whether it is a statically determinate or not. No, but fiction problems are essentially statically indeterminate problem, essentially. But the way we make them pseudo statically determinate is by putting this condition. So I do not ask that if these are the forces that I am applying to the system, what are the reactions and so on? We do not put them that way. But we put in this way, essentially. And for example, any real life structure, we want that structure to be stable. So what we say is that if I apply so and so load, if this is a geometry, tell us some range where the structure is in equilibrium. So we pose the problems like that and simplify our life. But you are right that in general, fiction problems are statically indeterminate problem. Only thing we can say is that for those given coefficient of frictions, for the given loads, given geometries, is the assembly stable or not stable? This is a question we can very nicely answer. But we cannot find out all the forces for any arbitrary P or anything. Why? Because number of equations, more or less than number of ones. So impending slippage, only to tell that at what force will that become unstable? Okay, then we go, okay, if the force is lesser. Because you can always ask a question if the P is in the other direction. Then at some P, you will see that it will start slipping in this direction. But in between that, it will be stable. Sir, in the first free body diagram. I think we are done, I think if you have any question. Hello sir, in the first free body diagram, at point A, at point B also, at point A, point B. The friction force is the same. Yes, yes, yes. Is it correct, sir? Yes, yes, yes. How it is justified, sir? Because you take- N2 is greater than N1. No, no, no, but at F point, so just note how are they equal. If you take torque for the top free body diagram about point C, N1, N2 are passing through the center. Weight is passing through the center. So only friction is that this is F1, F2, they should be equal. But what is happening is that that F is equal to mu times N1. But F is not equal to mu times N2. Bottom here. Yes, that's how we are justifying it. It is F, but it is not equal to mu times N2. It is still equal to mu times N1.