 Another problem of that same type. So what we have here is this. We have a cylinder on which a rod AB is resting. You may say that these are more theoretical problems but towards the end we will also come with some reasonable practical examples, I will emphasize them. But these are very good conceptual problems to clear that how to solve these various type of problems. Now a rod AB which is pinned at point A is resting on a cylinder. What we are given is that at a weight, now this pounds because I took this problem from Bear and Johnston. It is in pounds multiplied by what 2.2, I can convert it to kg but it is a huge pain. This is just bear with me. Pounds kg, it does not matter so long as you stick with the units. So the weight or the mass of this rod is 12 pounds. The dimensions are given in inches. The mass of this cylinder is 36 pounds. And what we are told in this problem that determine the largest magnitude of force P for which the assembly is stable or in equilibrium. And we are given that what is the point and line of action of force P. So we want to find out that if the force acts at this position then what is the largest magnitude of force for which the assembly is in equilibrium. Any force larger than that the equilibrium will be disturbed. What does that mean? That the slippage will happen at one or more surfaces and it will not be impending. It will be a finite slippage and then the entire assembly will just be destroyed. So can you tell me how do we attempt this problem? How many unknowns are there? How many equations do we have? And how do we take care of the slippages? How many unknowns do we have? At point A we have 2, point of contact between the rod and the cylinder we have 2, bottom we have 2 and load P is the answer. So we have 7 unknowns. How many equations do we have? We have 6 equations but we have 7 unknowns. So we need 1 slippage. How many surfaces do we have to slip? 2. Okay. So for example if I give this problem to a reasonably average student who is doing this problem for the first time the first gut reaction they will have is that they say oh the rod will slip out like this. So there is a slippage here and there is a slippage here. But is that the only way to create slippage? Okay. For example now let us think kinematically this cylinder if the entire assembly is to be disturbed by sliding this cylinder the naive way or the simplest or the most bone headed way you can think of is you just move this cylinder outside okay just like this so that there is a slippage here, there is slippage here. But is that the only way in which the slippage can happen? Cylinder may rotate. So the cylinder may roll about this point okay what is rolling essentially is rolling is essentially rotation plus translation okay the cylinder may roll about this point if it rolls about the bottom point slippage happens at the top if the cylinder rolls about the top point slippage will happen at the bottom and what we need to know now that we understand the kinematics but we want to figure out now that is the slippage happen at the top or will it happen at the bottom and once we decide that what is the effective force that is acting on the body is the point clear fine now note one thing can we find out the normal reaction at the point of contact can we yes easily because we can take the top free body diagram take moment about point A the friction does not contribute because it pass through point A so we will get the normal reaction can we get the normal reaction at the bottom point yes because that top normal reaction plus the way it gives the bottom normal reaction now what we want to know is that see I have the general cautious policy that I do not want to commit to the direction of the friction force as much as possible may I want to wait till I am sure that the direction because direction is such a dicey thing in this friction problems that you really want to be sure that to the best of our abilities if you can figure out the directions by looking at simple equilibrium and then only say that friction at wherever slippage equal to mu times n so now by doing some basic mechanical analysis can you tell me that for the bottom free body diagram what should be the direction of friction should it be in this direction or should it be in the other direction this or this will it be in the direction of P or will it be in the opposite to direction of P opposite how do we know because we take the bottom free body and take torque about the top contact point we immediately know that for that to be balanced okay the friction direction has to be opposite to P what about the top point will the direction of friction at a top point will be in the direction of P or opposite to the direction of P again opposite because now we take torque for this free body diagram about the bottom contact point and we immediately know that for moment equilibrium or torque equilibrium the direction should be opposite so one thing is good so we know the direction that is one good thing now only thing that remains is to figure out if the slippage happens at the top point or it happens at the bottom point is the only thing we need to figure out at top point you think why center if we take then it will be anti clockwise means it will rotate that wheel at point of contact B in anticlockwise direction no but you cannot say that immediately right because what you want to know suppose think about it this way if I give you a problem where the friction coefficient at the bottom was very very low as compared to the top point it may slip from the bottom only visible force right now is P no but but but you draw the free body diagrams all forces become visible after that right so what you are saying the approach is fine you want to think kinematically but a final we only need to find out what is the ratio friction divided by normal reaction at the top compare with the friction by the normal reaction at the bottom and what you want is that that they should be less than or equal to the coefficient of static friction for the assembly to be stable or it to be impending stability so what can we do we know the directions of the frictions you can just do one simple thing okay so what we can do is that for this second free body diagram taking torque about point C we can obtain what is F2 exactly in terms of P of course okay similarly by sum of equilibrium of F1 and F2 should be equal to P so we immediately know what is F1 now we have already obtained okay previously that what are the normal reactions so we have F1 by N1 and F2 by N2 okay so we get our ratios immediately F1 by N1 and F2 by N2 and you can see that F1 by N1 ratio okay 0.0453 okay so larger the ratio okay more chance of slippage okay so what we say is that we want our P in such a way that this quantity is equal to mu s and that will give us the maximum value of P that can be applied so just equate these two that this top ratio is equal to mu s so P will be equal to 7.72 pounds okay this ratio is larger so if ratio is larger what do we need that for stability both should be less than or equal to mu s but this is anyway if this is less than or equal to mu s this will automatically be less than or equal to mu s so this is the critical point and that critical point we invoke that this ratio is equal to mu s and P is equal to 7.72 so we tried our best to get all the directions properly and do it in such a way that friction is invoked as late as possible in this entire equation sometimes we have to invoke it early like in the previous problems but try to push it down as much as possible is the idea clear force is acting at a point D here yes okay here also yeah force is acting at point P yes and that is the only force which causes the slip yes of course if there is no P no slip yes coefficient of friction at both the contact surfaces are same right then moment produced by the force P at that bar at a point B is greater than moment produced by the same force at the contact surfaces so due to that we can say slip is there at that contact upper one no no see you can't because it see there are so many variables here I will tell you one simple example that if we decrease if we if we make this P go lower and lower the friction at the bottom end becomes larger and larger okay so what happens is that if you make it go to the exact bottom yes that's why I'm telling you exact bottom then what is the particular top perpendicular distance is going to increase therefore more moment is going to produce at the contact surfaces at B at therefore more slip is there at B as compared to no no no is exactly the other way round if you make this point P okay keep going down and down okay so what you will see is that at a friction at point B will keep decreasing because you note that if point if force P passes through the bottom then by taking this equilibrium what do we see that there is no friction at the top so there is no question of slippage no no friction is there at both the contact surfaces but is it's not that it is there it can exist and what is that value will depend on P okay we are going to find out where more slip is there not where more slip is there we are going to find out where is the possibility of impending slippage it is not a question of more or less it's a question of will it happen here before or will it happen here before yes is that's why I'm telling now that is its answer so yeah okay miss if you use the word more to mean that then I agree with you sir yes in this problem we are applying P at eccentrically at what eccentric at not in the center yes whether the slip will happen or rotation of the cylinder will happen sir see rotate okay you can you clarify what do you mean that slip will happen or rotation what do you mean the slip means that the both the bottom point and top point moving in the same direction slip so there is a there is a there is a simple idea that it may so happen that both point at the top and the bottom may slip at the same time but this this is again a rule of thumb that if the system can be displaced from an equilibrium position by slipping at minimum surfaces it will do that first the slippage will try to happen as as few surfaces are possible which is compatible with the overall kinematics of the problem and then you will see that mathematically also the same thing will be true that a slippage at one surface most likely there will be no slippage other surfaces only one surface either top or the bottom not both and there may be certain values of weights and frictions where all the numbers conspired together in such a way that slippage is happening at both places but a priory you should never assume that it will happen at both places we look mathematically what is happening mathematically I know that I need only one equation extra so I need to have slippage only at one point kinematically speaking also you will realize that to get this assembly away from equilibrium you don't need to move that complete cylinder you need to just either move the top surface or the bottom surface if both happens it's a very special case but a priory it is not nice you cannot assume that like it is happening at a two places you do everything and it may turn out that some numbers conspired in such a way that it all happened like this also sleep this also sleep we can always find some numbers you can do the back calculation such that the weight you can back calculate the weight such that the two ratios f1 by n1 is equal to m2 by n2 but that's a very special case more often than not it will not happen no no no no no no it can't be because take this if this is an equilibrium take torque about point d this force is creating a torque what in the anticlockwise direction because it's eccentric load there because of the p it has to rotate now instead of slipping then f1 should be right word no no no no no no no no okay okay if you take torque about point d you got me confused okay I got really scared for a second look at talk about point d talk about point d is in the clockwise direction so to balance that torque f2 has to be in the anticlockwise direction so the direction has to be this note take torque about point d what is the torque about point d I kept on thinking about the center and I got really scared okay take a talk about point d what is the torque it's in the clockwise direction take torque of this force over point d it has to be the opposite direction so this is to the direction okay if it rotates about the center it can do two things it can both rotate and it can translate so if I want to have slippage at point c and not at point d what will you do you will rotate it like this sorry you will rotate it like this okay then what is happening is that he this point is going here okay moment is in the anticlockwise direction then wait a second let what is the question again what is the question okay I lost the question I lost no no you don't answer this question I will answer the question okay if I can't then you answer no but what say again again that's a very good point that's a very good point what is friction providing friction providing a pseudo hinge but it's a hinge where for example there are limits in a in a hinge which we discussed in engine in the last couple of classes that hinge can provide any reactions in principle but here this is a very special hinge where it can provide both horizontal and vertical reaction but only when f by n is less than or equal to mu s so it is a hinge of course it is acting like a hinge but the hinge has special properties okay so you don't answer each other's question okay let us go it through me okay any other questions this is the directions are this only there is there is no way you can have different directions and in some problems you will see that the directions are not a priori clear but you can't help in those cases but here you know and if we know then we should make sure that the directions are appropriately put sir this is a sliding friction or a rolling friction no no no no see rolling friction the if you look at any standard textbook you will see that rolling friction cut technical definition is very different different yes so these are all sliding frictions let's put it this way so they're rolling rolling is only what may happen rolling is for example rolling is a very lame and term what is rolling means is that there is no slippage at a point of contact but rolling friction is a loaded term okay so we cannot use rolling friction you can say that it is rolling but ultimately friction is due to impending sliding yes so we are applying concept of a sliding friction so there is a chance of rotation no opposite direction then so you are true sir yes this is sliding friction only there is no rolling friction has a very different connotation yes which which no no no ultimately say ultimately what happens see what we are trying to do is this we are trying to figure out that our mathematics tell us that we need only one equation so slippage has to happen only at one point and then we are trying to clarify our own thinking we try to think that what can be the possible impending motion now see look here if this is the direction of friction okay point C has to slide only in this direction okay and how can that slide but but point D there is no impending sliding how can you achieve that you can achieve that but having a rotation in the clockwise direction what will that do about the center this point C will move here listen this point C will move here this point D will move here but we don't want any sliding at D so then we translate the whole assembly so it's a combination of rotation and translation which will make this point slip here and this point D not go anywhere and this point D slipping here there is an impending slippage like this which is consistent with the direction of friction that we have given there so this kinematics is only for our understanding that okay that our mathematics is telling us something is our kinematics of the problem consistent with the mathematics of what we know that we need slippage only at one point can we have a motion or not I should not say motion but an infinite similar impending motion in such a way that it is consistent with the mathematics of the problem and I will always find that it will happen suppose the same problem yes only the line of action of p is given huh that Russia is not given okay that's a good point how to tackle yes then the entire as entire thing you just turn everything the other way that should be assumed both directions and but even in that case what you will see is that that directions of frictions we just sleep excuse me sir yeah where is this I have a question this side this side yeah please sir did there is a point of this force P at D okay what is it D what as I think yeah as I think it should not sleep it should rotate only because there is a torque because of no no no if it is acting at sea then okay you are correct but if there is any torque so it's a and under the surfaces are rough or there is a friction between all the surfaces then there must be a torque but when you say these things what is the logic that you have in mind when you say such things okay because you said it okay but how do you know that that is right or how do you convince yourself that what you're saying is actually right and one more thing is here if it is sleeping then directions in free body diagram 2 it is correct but in free body diagram 1 it may not be f1 in this direction it should be opposite if f1 is this f1 on the other one it should be opposite there is no other choice there is no other choice sir if it is if it is sleeping only without rolling I don't think that there should be any problem in deciding the direction of the forces it will be always opposite why we are going for so much pain what wait a second okay what is the pain and what we are going for okay can you clarify again I am I am saying that if it is sleeping only without rolling but but you don't know that our prairie he's saying something and you are assuming what he's saying is right okay it is not right so sir you said that we are okay sorry okay if I said that okay let me look at the video and then clarified what I did not to the best of my knowledge okay I did not I said it will not it will not slip in this problem at D only slippage will happen at C and the way that can be achieved kinematically is that there is a overall rotation about the center followed by a translation such that this point it doesn't go anywhere but point C slips in this direction that's what I said I have a small doubt sir okay I think one last question okay sir if we apply the force P at point C exactly okay point C exactly problem with the top and bottom surfaces will have equal chance of slipping not equal chances why because the friction forces will be the same at top and bottom but the normal reactions are different sir sir sir as well as the beam AB is concerned no what the baby is concerned that is the top point yes that is contact point C center center C oh oh oh I made this contact point C in this okay I screwed up okay let us refer to that the normal load which is coming is only R1 is only R1 R1 yeah yeah yeah and the that is the frictional forces F1 okay and you were say the torque which is applied that is P into the eccentricity what you have okay that is also in the anticlockwise direction yeah and what is the anticlockwise direction about which point but we have to talk about at which point about C okay anticlockwise yeah therefore the that is rotating like this I know I know I know I know I have to clarify I have to step in here it is not rotating the in there can be an impending rotation okay there's a big difference yes yeah the possible thing is yeah infinite symbol impending rotation yeah yeah and thereby the normal reaction is that the frictional force is that and this moment now is also adding to that frictional rotation okay okay whereas when okay when you're coming to the bottom contact point here the the total normal load is which is acting as the R1 plus the weight of the disc yes the normal reaction is much more yes therefore the frictional resistance which is generated at the bottom point is higher higher yes when compared to the frictional force which is developed only because of the weight of the beam yes that is one point to be considered yes and in addition to that you are rotating it when you rotate it the frictional resistance may come down at the bottom but the weights which suggests they are less yes that is the point I'm making that but the numbers are conspired in such a way a prairie without putting in the numbers you cannot say that the normal reaction is larger at the bottom so it is stable there no because this is down so the friction also is larger here so it has to be a ratio that you should look at okay so I think we should no no no any any questions any doubts after this okay so I have to finish one last problem okay I'm not after this okay let me quickly finish this so what we have here is a very well known problem okay which we face in our day-to-day life is that we have a cupboard or we have for example this which I I don't know why I would want to slide this but suppose if I wish to okay so what we have here is that so we have an extended body okay of weight w it's in contact with the ground and you're asked to find out that if the width of this is B then what should be the maximum height at which I can apply it okay a load P okay what is the maximum height as we can apply the load P such that this thing this entire assembly okay will not topple because generally what happens is that if you you realize that for example you're trying to move a cupboard okay or somebody is trying to move a cupboard you're trying to watch or help him whatever way you see that if you push on the top the thing tends to topple okay so many times what you do is that you like try to push it from the bottom okay or like lie down and push with your legs so the question asked here is that that given all these dimensions okay it's a very simple problem how do you find out what is the maximum height at which this load should act such that this system does not topple okay now when we say something topples okay visually arise at what so suppose suppose suppose friction at what at a corner point so what we look at is we want that something is about to topple okay so two things can happen perfect the total reaction the point at which the reaction acts is what we are trying to find out for example many times we made a point at the reaction point will adjust such that torque balance is obeyed but in some cases especially when the bodies are big that need not be always achievable okay so in this case what we want is that that the line of action will pass through some point if you increase the height at which the load is applied it will keep shifting and when it reaches the edge okay is when the thing is about to topple but you will also realize that if the friction is very low okay the friction is very low then before toppling it will actually slide okay so the competition is between toppling and sliding so it may for example go all the way okay on the edge but still when you are pushing on it it need not topple it may start to slide okay the moment you apply any load okay so it will it may start to slide and what we are asked here is that that given the coefficient of friction between the bottom and discovered or what are on this block what is the maximum height such that it topples before it slips is the question is the point clear that even when it is when you think that it is about to topple it may just start sliding okay so what is the so how do you solve this problem it can be solved in actually one line if you just realize that this is a three body member okay which is a much maligned three body member that this is a weight acting to the center line all the reactions are concentrated at the edge because it is about to topple and what we want to understand is will it slip will it topple before it's slipping or not so then what we do is this okay this is n this is the effective reaction this is the angle between n and f this is the line of action of force P H what we want is that for that body to be in equilibrium we want all of this this complete reaction weight and P should meet at one point now what we see is this that if we increase this theta okay if we increase this if it decrease this theta then height keeps increasing right if you keep increasing this theta then this height will also keep increasing but what is the maximum value of the what is the maximum value of theta that you can have no the corner reaction can't be zero otherwise there will be no balance because we know that f should be equal to P n should be equal to w okay so it will be just what we just find out is that that tan theta will be equal to by definition okay by geometry is equal to B by 2 H okay but what we also realize is that that there is a limitation on this that tan theta come maximum value can be mu because we want to increase this H as much as possible okay that we want that for example that what is the maximum height we can go to such that before slipping it tips okay and so that what we do is that the tan theta will be equal to mu and so H is equal to B by 2 mu anytime you increase the height above that value then it may tend to topple but it will actually slide okay it may actually tend to topple but it will really slide okay so that is the entire idea and if the if the value of height okay is less than that value then it will topple first before sliding and the only thing what what we what we realize is that that if the height H okay is suppose the height H is somewhere here okay somewhere lower at the through the center okay then what do we what do we realize is that that then the impending slippage okay that impending slippage will guarantee that this is theta is equal to phi s and we will see that the corresponding line of contact or the point of point where this force acts will be simply given by mu H by 2 by simple geometry theta is remaining same because we are looking at the impending slippage also the theta will be equal to phi s so the problem is question is this that we are trying to move this so when we are trying to move this there has to be we are trying to do two things it can either topple or it can have an impending slippage so what we are just saying is that if the height is less than the critical height then there will be no tip there will be no tipping or there will be no toppling but we still want to in a exert enough force that there is still impending slippage because you want to slide this entire thing but then we turn the question around and ask ourselves that in this case can you tell me when the impending slippage is about to happen what is the point at which the effective reaction acts and that you get from here that when the impending slippage is about to happen so problem number 8 is a reasonably interesting problem it seems very formidable but actually it's very straightforward and the idea is as follows so if you have already read the statement of the problem okay then maybe this is a repetition but what we have here is that that there is a central hinge and we have this device where the central wheel can rotate in the clockwise direction okay by an internal torque this is connected to some motor for example which is providing it a torque in the clockwise direction and this device okay is for example is any torque that is being applied to it this two small spheres or these two small cylinders they are preventing rotation of this when that torque M is applied okay this two small these two small rollers have a contact point here and here a contact point here and here so this entire assembly has a 180 degree rotational symmetry if you take this and rotate by 180 degrees you will get exactly the same device and what we are asked to find out that if there is a common coefficient of static friction between this point this point this point and this point there are four contact points then what is the range of D if you can see what is D D is the distance between the wall and the center okay such that the device will operate okay as desired what does that mean is that the braking mechanism can be provided by this we want the range of values of D when this can happen now when you think about it we have put our problems in various categories okay category 1 2 3 4 this is a very peculiar problem how many unknowns do you think are present in this problem so there is if I look at the lower cylinder then there are two unknowns here two unknowns here 2 2 4 okay 5 6 7 8 at this point how many 9 10 2 reactions plus the unknown torque there is also a torque that is present okay 11 so we have 11 unknowns and how many free world diagrams can we draw 3 3 3 so 3 free world diagrams so 9 so we still need two extra equations okay but a point is this you will realize something funny very soon okay so we need two extra equations which means that we need slippage at two points okay but can you tell me now is there anything special about this cylinder what is the last day is that neglect the weight of these two rollers okay roller or cylinder neglect the weight of this neglect the weight of this so if we neglect the weight of this one and this one can we simplify the problem by this assumption or let me put it this way that the forces that are involved in this problem are much larger the contact forces will be much larger compared to the weight so we neglected so this fact that we can neglect the weight of the two rollers can that simplify our problem or what can we say about the rollers then or their reactions symmetry though is there so whatever happens at the top same thing will happen at the bottom okay that symmetry is present there is a 180 degree symmetry you wrote it the entire assembly by 180 you get the same thing okay so you immediately know that that symmetry is present but if we neglect the weight of the rollers then what can we say about the reactions at this point and at this point there will be normal there will be friction but just think about it is there is the reactions are happening where only at the ends the reactions are happening or are acting only at the at the edges two points the contact point there are two contact points that the reactions are acting exactly at two contact points what can you say about those two the two force members okay so there are two force members and if there are two force members then what do we know we know that if we have a line joining this point with this point that will be the line of action of the reaction do you agree with me okay so that will be the line of action of this reaction same for this but then can you tell me that why do why should we why do we need to have a requirement on D that why can't D become smaller than a particular value what is the maximum possible value of D can you tell me R plus capital R plus 2 R that is the maximum value of D you can have but why is there a requirement for a minimum value why do you think so we what we figured out is this that it is a seemingly complicated problem but we are asked to find out what is the range of D we could kind of guess okay that the largest value of D by geometry can be capital R plus two times small R but then the question is demanding that there is also a small there is a lower limit on D and lower limit of D is a geometry quantity we know that the line of action of this force is purely governed by the geometry this point and this point you join okay so in this case what will happen is if you draw this free body diagram so this is the point of contact okay this is the point of contact then you will see that no way two force member so this should be the line of action of the two reactions now if you just if I zoom this in what will I see look at the top cylinder here just look at the top cylinder I zoom here this is the larger cylinder the larger the larger wheel this is the smaller cylinder line of action is from here to here suppose this angle is phi this angle also has to be phi and this angle will be 2 phi then what will be the distance distance will be small r plus capital R cos 2 phi plus smaller do you agree with this statement that small that the distance okay this distance is nothing but join the is nothing but take this line join center to center okay so it is r plus r this is phi phi 2 phi r plus r cos 2 phi plus smaller so this horizontal distance is capital R plus small r cos 2 phi plus smaller do you agree with me if I say that this is the distance D yes now the question is that how why is there a minimum requirement on D if phi increases what happens to D if this angle phi you see do you realize what is this angle phi this angle phi is nothing okay but the angle made with respect to horizontal by the point joining this contact and this contact by the line joining these two contact points what happens if phi increases phi increases cos phi decreases D decreases okay if we make D very small okay very very very small what is the implication on this cos 2 phi no no it is not a question of normal reaction here just think in terms terms of this angle just think of this angle that if you decrease D what happens to this angle this angle keeps increasing okay this angle phi keeps increasing because larger the phi smaller is cos 2 phi and smaller will be it so if you keep decreasing D angle phi will increase but we know the coefficient of static friction between took all the contacts so what is the maximum value you can have of that angle phi tan inverse mu tan inverse mu is the maximum value you can have for that angle so what does that mean that any value of angle when it becomes larger D becomes smaller but you cannot exceed that angle while at the same time keeping the system in equilibrium so that immediately tells you that there has to be a lower limit and your D cannot become smaller than that because if your D becomes smaller than that your phi increases and it increases beyond tan inverse of mu which will make the system not be in equilibrium so that immediately gives you phi maximum value is phi s you substitute that and you will immediately see that D mean will come out to be this simple expression and D max is good it is 2r plus small r and that phi requirement is automatically satisfied because phi is equal to 0 which is less than tan inverse of mu so you are good so the maximum limit is when D is equal to capital R plus 2r and the small limit is governed by this requirement when this angle phi becomes equal to phi s so what we have done is that this seemingly formidable problem just by recognizing that these two wheels are two force members and by invoking angle of friction we could solve this problem in essentially three lines okay so in the beginning I had said that I had my own doubts about like what is the need for that extra angle of friction okay we have mu we have f we have normal reaction let us work with it but what we see here is that the geometry when used appropriately can be extremely helpful as is seen in this problem any question you have with this problem any question with the solution there is one interesting point that I am to make after this but do you have any other any other question or any doubt which two rollers okay so you are talking about this friction this and this in between how do you have they are not in contact with each other right there is one roller here one roller here for the wheel and a roller okay so wheel and a roller there is clearly friction now just look here this r is making an angle with respect to normal reaction just look here this is r this is the normal the line joining the centers of these two will be normal okay so this r will have some angles so clearly there is a friction force right r times sin phi will be the friction force because this angle is phi this angle is also phi so r sin phi will be this friction force and note one thing what is what is the question of this problem that this device is used to prevent clockwise are the friction directions consistent with that think about it are the directions of friction consistent with this what is the reaction we are what is the reaction that will act here on this wheel like this okay this is the reaction which is provided by the wheel on the roller which is outwards so what is the reaction that the roller provides on the wheel is in this direction okay what is the direction of friction the reaction of friction will be upwards what kind of torque it will provide counter clockwise torque same for the bottom one and there will be two counter clockwise torque that will provided by the friction force and they will resist the clockwise rotation and note that the geometry of this assembly is such that the friction can only be provided in the clockwise direction sorry in the anticlockwise direction so if you try to rotate this assembly okay in the other direction or in the anticlockwise direction it will fail it will not do that so in order to do we need two wheels provided here and here okay so because we want a rotation in this direction it is that is why we have chosen that this should be the orientation of the wheels is the point clear this one now another question you solve this problem fully you will realize that there is no way in the world you can obtain the normal reaction you can only get that all the ratios are fine but the actual values of normal reaction and a friction will not be obtainable from this problem is not possible so what do you think may be happen you solve this problem you you will never be able to say radically indeterminate problem where you know that this assembly is good that it can provide me the requisite friction but what it doesn't tell you that what is the maximum torque that it can balance all it tells you is that that with that distances it is good to go but what it doesn't tell you is what is the moment carrying capacity why because you solve this problem the normal reactions will never come so what may be happening here how do you decide what is the torque carrying capacity because we have not talked about that how much moment you try to rotate it within the clockwise direction will be balanced by this assembly how do you decide that any guess typically what will happen in this case is that these two wheels they will be pre-stressed they will be compressed and then put inside and so they will try to expand and depending on how much pre-stress you have given in those wheels that normal reaction will appropriately increase and so that becomes a problem in solid mechanics now in conjunction with the simple engineering mechanics and the amount of pre-stress the stiffness of the wheels so the stiffness of the cylinders will decide what is the normal reaction that will be generated and that in turn will decide that what is the maximum moment or what is the maximum torque that this assembly can provide a counter balance to is the idea clear okay so that problem is statically indeterminate it is not a topic of humane but note that the question that is asked that what is the maximum value or minimum value of D that is still solvable by using simple arguments but we cannot solve everything here so that's why I said that there were categories 1 2 3 4 but always those are rules of thumb you can always have funny problems okay which doesn't fit any category is the idea clear behind this very simple mechanism but a geometry once you look at a geometry okay that immediately simplifies the problem so that you can like this formidable looking problem so called can even be solved in just three or four lines in doubts any questions okay it will be symmetric to the top one just look now there's a 180 degree symmetry what bias huh when I pencil 5 maximum we have considered tan inverse maybe same thing will happen at the bottom because this assembly is maintained in a symmetric way that that is true sir I agree yeah only my if suppose one more roller has been added so there is no effect added okay then the point is that then the problem becomes it is it won't be solvable by simple emic ideas it will still be solvable okay only thing what will happen is that it is nothing to do with the but in that case what will happen is that because now here because of the symmetry this d you decide that is also automatically decide but if you add one more then it's a funny thing the point is that you add a redundant roller because these are good enough both of them what the redundant roller you have to just decide key what is the top roller will give you some d bottom roller will give you some d okay depending on what roller you want to be functional you may have to choose between maximum or minimum okay but adding one more roller you can still get the angles but there is no unambiguous answer for d because there is a d coming from the top roller d coming from the bottom roller so the problem becomes a bit ill conceived okay because there is a there are two lengths now that will be involved if you put a roller on the top also okay some other third roller if you add then the number of lengths that are involved or more then it will all boil down to what length you are interested in my concern is about the frictional force so frictional force will still come but if you put a roller if the number of rollers are gone increasing that total force will be more opposite force or a frictional force yeah so so here is it that tan inverse members will be holds good if no no no see this problem is just telling you that there will be some angle phi which depends on this distance and what we only want is that that phi should not exist exceed phi s that is the only requirement we need but we also saw that the distance d is also related to phi and so we say that for example that we don't want phi to go beyond a certain value which automatically implies that d should not go below a certain value if you add one more roller some other length will come but this phi arguments everything will remain the same because all of them still will be two force members but then the point is that what is the definition of d like do you look at d of the third roller so it it it is little bit of a what do you say ill condition problem but what this is telling you that this assembly is sufficient and at least as far as the minimum distance is concerned we can tell from simple engineering mechanics principle that there is a minimum distance and the largest distance governed simple by the 2r plus r the overall dimensions of the assembly yes so let us move on so this problem okay this is a very interesting problem okay so it will involve lot of like some geometrical thinking and at the same time okay like how how do we decide what are the contact surfaces where the slippage is occur how to do that analysis most of the things will be taken care of in this problem and in the tea break some some of you came to me asking like some clarifications and I realized that I made some statements okay which were not wrong but they have the potential to be interpreted in such a way that that can create further confusion so let me clarify those points also at the same time while we are solving the problem so problem definition is straight forward like most problem in engineering mechanics what we have is two cylinders okay are on an inclined plane with angle 15 degrees the coefficient of friction at all the contacting surfaces 1 2 and 3 is 0.2 weight of each of the cylinder is 30 pounds okay pounds I took it from beer and johnson old book everything is in pounds not beer and johnson this is a merriam problem 1972 edition everything is in pounds so 30 pounds what we are asked to find out is minimum force p required to move the two identical cylinders up the incline so it is like for example you have seen that there are like people are carrying barrels up the slope okay it's a small slope but if you are carrying two barrels at the same time pushing them up then given the dimensions given the frictional properties between all contact points you're asked to find out what is the minimum force p required so that the assembly just starts to move upwards is the problem statement clear fine so now how many unknowns are present in this problem before you even go to the the solution slide what are the number of unknowns so seven unknowns okay so 1 2 3 4 5 6 p p is also an unknown because we need to find out that p okay if there is no p what happens p is 0 entire assembly rolls down so what we need to do is to we need to apply some bare minimum force p to prevent it from coming down but in addition we need to provide even extra force p so that the assembly starts to move up okay so here the problem definition is very important that we want that p minimum such that the assembly starts impending motion upwards so there are seven equations how many free body diagrams two so six equations sorry six equations seven unknowns so we need slip at exactly one plane okay but there are three surfaces now till now all the problems we had discussed okay the slippage like there were two competing surfaces and we decided okay it will happen here or here but in this problem there are three competing surfaces now how do we decide where the slip will happen to do that first we need to figure out when those two cylinders are trying to move up how do you think they will like how will the impending motion be now this is one point I want to emphasize again that while solving that rod and a cylinder problem I kept on saying that a cylinder can roll and slip so some of you kind of misinterpreted saying that rolling means rolling friction so there is no rolling friction here if you go and look at any standard textbook rolling friction has a very very different connotation no rolling friction we are all talking about slips okay that if this going to slip other going to slip all those kinematic ideas when I say kinematics that how the system is going to be disturbed by a small motion by impending a small impending motion we are just trying to visualize that what can be a small geometric motion that is compatible okay with the given statement of the problem is what we are trying to do okay so when I say all those diagram that rotation and translation it is not like huge translation and rotation it is an impending kinematics that when the system is now is in equilibrium it is just disturbed from the equilibrium position then what is the kind of motion like tiny infinitesimal motions that can happen to the system so that they are compatible or they are consistent with the overall definition of the problem now as you will see in this problem that visualizing how this system can go up can make our life very easy to understand what will be the directions of friction if the problem for the cylinder problem just by doing simple moment equilibrium we could find direction of friction immediately okay without making any assumptions we could get the direction of friction and it could be only this but in this problem okay because there are so many surfaces we want to first understand what do we mean when this assembly like what what may happen kinematically when this entire assembly is trying to go up the slope okay some thoughts so how do you think this assembly when you push up how do you think will this go up the slope in a very infinitesimal sense okay not like just rolling up very infinitesimal motion just about to start how will it happen some thoughts both how will they move what do you think will they will they rotate or will they translate what will happen without any rotation both cylinders will translate along that surface okay now if both of them move upwards slip is happening where slip is happening both for the bottom point and the top point right bottom no no if the both cylinders move upwards slip is happening at the top also at content surface between two cylinders there will be no no no no no no you are saying that both cylinders without any any rotation they can just move just translate upwards yes but if that happens then you have a slip at the bottom cylinder also at the top cylinder also we need only one slip but then it can't happen you can't just have like slide both of them upwards and only slip happens at the bottom it can't happen it's not feasible right it is not feasible that's the reason why we want to think little bit in terms of like how that impending motion can happen one way it happens that both of them slide upwards but if they both of them slide upwards you have two frictions okay that is not compatible with the mathematics we need only one slip now one motion can happen is this both of them okay can do this okay so let me open this new slide so both going I see that is this thing which for example you tell students immediately say okay both of them slide upwards but then you are getting two slips you see my point and but the mathematics is telling you that I have seven seven unknowns I have six equations I need only one slip okay and there is a law a very law of nature okay which I have found out okay it's it's a very hand waving thumb okay no law but if the motion can happen with as few slips as possible as opposed to having multiple slips more slips it will try to follow that course where the number of slips will be as less as possible now you can think here yes excuse me yes okay sir so we are applying torque on only on rear wheel so right now there is no and at that time we have discussed that there is no there's there would be no friction at in first means front wheel but the reason for that was that the front wheel was rotating at a uniform angular velocity about a axle which we believed was well lubricated so it could not provide an internal torque and that was the reason which guaranteed us okay not going into details about like I has one professor pointed out that the point contact there is some deformation okay assuming that like the cycle very high pressure in the cycle and there is very little it's a point contact okay so that problem and this problem first of all let me emphasize they are drastically different okay they are drastically different why because there were hinges internally okay those conditions also had to be satisfied second of all there the bicycle was moving uniformly upwards at a constant speed here it's stationary we want to apply enough force such that there is an impending motion just happening just an impending motion okay so they are not the same at all now what we are trying to understand okay is how that impending motion can happen and what I am saying is that it is not mandatory but if you can think about the impending motion in some different ways we can really really get it deeper inside into the problem okay and can also for example like as we can see in this problem we can immediately fix what is the direction of the friction forces so if I come here okay so let us have a look at this what is the motion that can happen that the first wheel okay first wheel infinitesimal rotation about the center plus translation infinitesimal rotation of the second wheel about the center plus translation how much rotation how much translation such that 0.1 and 0.3 there is no slip pure translation what will it do it will create a slip here but if you add a small rotation then what happens is that there is no slip here because we can compensate for that distance move think about it fine and again this is only infinitesimal and this now where can the slip happen there look at this point this point will move downward coming from the first cylinder will move upwards coming from the other cylinder so how will that motion be this will try to go down this will try to go up okay it's like gear it is going to go up and down so there is a slip there this is the first mode in which the slip can happen second mode what happening here is that that the first cylinder has a tiny infinitesimal translation followed by rotation in the anticlockwise direction now what does that imply this point will slip clearly but this point if it has a same amount of rotation in the opposite direction followed by translation then note that this point we can still make this entire cylinder go up okay this two assembly okay note here that with this with this motion the entire assembly can go up with this motion also very infinitesimal motion the entire assembly can move a little bit upwards but the slip now cannot happen here because both of them rotate in the same way okay same same magnitude so no slip so only slip cannot happen here finally the third mode in which that entire assembly can move upwards is that that this one has translation and rotation so that no slip here equal and opposite rotation so no slip here but a combination of sliding and translation or sliding and some rotation which will provide a slip here so these are the three different ways in which this entire assembly can infinitesimally move upwards while having slipping only at one surface okay and note one thing immediately that with all of this all of these motions friction direction can only be this if this slippage happen here how can friction be this is trying to go down going up so friction on this one can only be down and the other one can only be up think about it it is trying to do this friction on this is up friction on the other cylinder is down okay and once this is up and down if you look at this cylinder taking talk about the center taking moment about the center if this is a f has to be only f at this point can have this has to have the same magnitude and this direction there is no other choice same here if this is f this friction also has to be f and the direction has to be downward there is no choice mechanics will give you no choice it will fix that and no matter what mode there's a body slip okay what mode does the body go upwards by you will see that you will put all the friction forces for example if the friction if the body slides upwards in this mode okay you will see that the friction only can go downwards because this point is going upwards so whatever you do you will immediately see that this has to be a free body diagram with these as the forces and all the friction forces have to be equal so the idea behind all this thinking is that that we could figure out what is the direction of friction what is the magnitude also like we say that all the friction should be equal and this becomes our simplified free body diagram if you don't do this kind of analysis then how do we interpret what do we mean by two of the cylinder going up okay what is the direction of slippage okay where can it happen but with this analysis we immediately know that the assembly going up it can go up by slipping exactly at one surface and no matter in what mode it slips the friction the free body diagram will only like this with the direction of frictions only like that there is no other choice okay so that's the reason for example in this case it was helpful to understand what does it mean that assembly is going up because we could fix the direction of all the frictions okay now it is straightforward once we know that these are the directions if this is f taking torque about or taking moment about point w we know that this has to be f in this direction because this is creating a anticlockwise torque clockwise torque ff this is a create for this free body diagram all other forces pass through the center but what we know is that if this is f this is f okay moment if this is f it is creating a torque in the anticlockwise direction so the second f has to create a torque in this direction okay and both of them will be equal then what you can do is that for the second free body diagram now now we have to figure out that frictions at all the points a b and c are the same now how do we decide where there's a slip occur we don't know still but what we know is that that a coefficient of friction at all the points is same and equal to point two so the point where the slippage will happen the first will be the point where the normal reaction is the lowest okay lowest normal reaction will give you the slippage now let us look at the top free body diagram look how the free body diagram looks like note this is one tangent this is another tangent along the slope they intersect at point b now for this free body diagram if we take moment balance about this point b what do we see that normal reaction n2 times radius plus w times some d some distance will be equal to normal reaction 3 okay times r we are taking moment about this point d for the entire free body diagram and from this simple equation you will see that since d is greater than 0 just do some geometric visualization and you will see that when angle theta the slope angle is less than 45 degrees then this d will always be greater than 0 and you will see that n3 is always greater than n2 so there is no possibility of slippage at point 3 then you have to only compete between point 1 and point 2 now do we compete between point 1 and point 2 just write down the free body equation you will see that the equation for n2 okay draw the free body diagram for this equilibrium in the vertical direction will give equation for n1 equilibrium for this free body diagram okay equilibrium along x direction will give you this equation so now we still do not know that n2 n1 f w sin theta we do not know if n1 is more than n2 now at this stage when we cannot do anything further okay now like whatever analysis we do this equation still remains and there is no way to know if n2 is more than n1 now we bite the bullet finally and say let us assume that the friction happens at point 2 now you can as well assume that friction happens at point 1 but if you assume that the friction happens at point 2 then friction force f will be written as mu times n2 just substitute here solve these equations you will get that this is the value of n1 this is the value of n2 and what you will see is that n1 is equal to 0.9 w putting in all the values and n2 is equal to 0.324 w and our assumption was that the slippage is happening at point 2 was consistent why because the normal reaction at 1 is coming out to be more than the normal reaction at 2 which is what we had assumed to begin with so the friction will happen first at 2 and then at 1 substitute all the values and you will immediately see that if I draw the complete free body diagram 1 plus 2 then for balance of forces along x you will see that p which is this force will be equal to 2w sin theta plus 2 times this friction force and put friction force equal to 2 mu s times n2 you will see that what is the minimum force required to have an impending motion for these 2 cylinders up the slope so what we had done is that that we don't want to commit okay to where the slippage is happening till the very end we want to obtain the directions as unambiguously as possible and try to obtain where the slippage can happen as unambiguously as possible for example our analysis told us that n3 is more than n1 n2 so slippage cannot happen at 3 but as far as n1 and n2 are concerned we could not say anything and as you will see that for example for certain values of slopes and for certain values of coefficient of friction you plug in the values and you will actually see that n1 will turn out to be lesser than n2 and so the slippage will happen at n1 and not n2 okay so that tells you that for example there is no way by simple mechanics you can figure out where the slippage can happen at the end you just decide okay let me say that a slippage is happening at point 2 now let us see what is the reaction is n1 more than n2 or n2 more than n1 if n1 turns out to be more than n2 which it happened here then our assumption that slippage happens at 2 and not at 1 is good if not we just repeat the entire procedure in the other way around say that the slippage happens at 1 so the friction force is equal to mu times n1 solve the other problem and it will work okay so this is like a very comprehensive problem okay where for example visualization proper choice of free body diagrams proper choice of equilibrium equations will really simply for your life okay and make the solution procedure more logical and less prone to errors okay so with this if you have any questions you can ask or else we can start the tutorial yes please somebody has a question here if you take the second case that is rolling anticlockwise okay the second from the direction of friction should be in the opposite direction which which one no second case second in the middle right yeah middle one just not one thing if it is moving anticlockwise no no if it is anticlockwise look how does the point be the downward point where I have put those two lines how will that move is it going to move upwards or downwards think it will slip like this so if if just by the rotation will it move up or down the bottom point where where I have drawn those two green lines that is where the slip is happening yeah you tell me that if there is only rotation tiny rotation that point of contact will it move up or down it will move now because rotation about the center what is happening if you rotate about this this point is going to move upwards and there is in addition there's an arrow which is going upwards so the overall movement infinitesimal movement of that bottom point will be upwards so the relative sliding is upwards friction force has to act downwards I'm not talking about point B which point are you talking about second case point B point B oh friction at point B you cannot say anything from this free from this diagram you cannot say anything but you draw the free body diagram then for the balance of moment of that point because there is no impending motion now at the center point look here if both of them rotate like this this point are not slipping but impending not having an impending motion doesn't mean that friction cannot exist impending motion is the extreme case of slippage extreme case of friction is impending motion where you know that friction is equal to mu times a normal reaction so that point we cannot say anything just by looking at that diagram because there is no impending motion but if I know that the impending motion is happening downwards we know that the direction of friction is downwards and now look at that free body diagram the only direction of the friction at point B can be upwards because if you want to have moment equilibrium of free body one about the center then it is mandatory that if the bottom one is f and downwards then the top one will be f and upwards it's mandatory now mechanics guarantees that both will be f and the direction will be this no no that is now I'm talking about in the first wall point B if it is rotating anti-clockwise like this then the friction should be downward no no that's what I'm saying it is but but the point is that is there any impending motion happen at the contact point B in this case you're talking about the middle case right about this yeah is there any impending motion happening at the middle point you tell me if the two rotations are same infinite similar to the one the anti-clockwise rotation suppose it is delta omega the clockwise rotation is delta omega is there any slip that is happening at those two points no no then you cannot say anything about what is the direction of friction there immediately you cannot only when there is an impending slippage where you see that there is a relative sliding can you immediately say that the direction of friction has to be this but if there is no impending slippage up there is no way of saying that the friction is up or down because there is no relative slippage but that doesn't mean there is no friction there can be friction and the direction now will be fixed by mechanics if you have more questions you can ask me afterwards I think I made it very clear here that not having relative motion doesn't mean there is no friction and if the two for example if two bodies are moving together there can be friction only thing is that we cannot talk about the direction immediately we cannot