 So, today we will start with what is my really favorite topic in engineering mechanics which is friction and as we will see that there are extremely interesting problems that can be discussed in this topic. And students typically find this topic a bit weird and a simple reason to cut the long story short is that the problems in friction are actually statically indeterminate problems. So, what that thing is will become more clear as we proceed further. Let us begin with our main topic, so it is friction ok, simple plain English that is a retarding force which prevents motion. So, for example, you take two blocks placed on each other they have a contact surface. Now what happens is that assume for the time being that both blocks are stationary. Now if you start applying an horizontal force to this block what happens is that there is a friction resistance between these two blocks which comes into picture. And as a result we cannot just apply a simple force and then expect that a block to move and friction was precisely the reason why for example many many years ago way before Newton that Aristotle had thought that in order to keep a body in motion you need to apply a constant force. What he had not taken into account was the friction that acts between one surface and the surface which is which it is rolling on and this was which was recognized by Galileo and then formalized by Newton and that is what we are doing now in this course ok. It is a very old story to this picture ok, so frictions is of two types static friction kinetic friction and fluid friction. So fluid friction is something that is not the topic of this course what we will extensively discuss in today's class as well as the class that will happen next week on dynamics is static friction and kinetic friction. Sources of die friction in a very naive way we can say it is a asperities between the contacting surfaces ok that can cause the friction. But strictly speaking friction is not a very very well understood question it is still a topic of extensive modern research there is an entire field called tribology which studies sources of friction lubrication wear tear etcetera ok. So all the details ok the simple all the gory details of what happens between these two contacting surfaces is not at all a straight forward topic and it is a topic of modern research. So we will not go into the details of that ok. So now let us get back to the coefficient of friction what do we mean by that. Now the friction that we are going to discuss is called as Coulombic friction ok dry friction Coulombic friction whatever terminology you wish to use. Now let us say that we have a block placed on a surface and there is some resistance or there is some friction between the top block and the bottom block at the contacting surface. Now what happens is that just look at this structure if you draw the free body diagram for this then what happens. So when the applied force is 0 clearly there is no resisting force of friction. Now if you start increasing the applied force you will see that to maintain balance how to maintain equilibrium the resisting force will match exactly with the applied force. But then what happens that this contacting surface does not have infinite capacity to exert a resisting force. So what does Coulomb's law tell from a bunch of empirical experiments which work really well for a lot of engineering problems ok for a lot of contemporary engineering problems this kind of approximation works very well. What it says is that that the resistance between these two surfaces it depends not on the contact area and nothing else but it only depends on the normal reaction that happens that is transferred between the top surface and the bottom surface. So what it says is that that the maximum value that the maximum value of the resistance which this block gets from the underlying substrate is equal to mu which is a coefficient of static friction I will come back to that quickly what is static static friction multiplied by a normal reaction n which is exerted by the top block and the bottom block ok. So top block exerts a force on it because of the weight and bottom block exerts a reaction n. So mu times n is the maximum resistance force that can be provided by the underlying surface on the top body. Now what is another experience is that that till we reach this point the block is completely stationary and if we now increase the force just a bit beyond this maximum value then the block starts moving and when the block starts moving what is seen experimental is that that there is another kind of friction coefficient which comes into picture which is called as a coefficient of kinetic friction and the exact details of that you can have say something that now the bonds are broken and so it is easy for it to maintain the motion there can be various macroscopic reasons can be given but if you do tribology in great details then you can figure out that for many many surfaces this mu s ok is typically greater than mu k ok that mu s is typically greater than mu k and when the block starts moving then the resistance actually decreases and then when there is a complete relative slippage between the top surface and the bottom surface then the force that is exchanged the horizontal resisting force that is exchanged between these two surfaces is given by just mu k times n where mu k is the coefficient of kinetic friction ok. So this is the simple idea behind coefficient of friction that for only for impending relative motion mu is equal to mu s and for actual relative motion mu equal to mu k and you can even logically show that mu k cannot be more than mu s or mu k has to be less than or equal to mu s ok. So there is just one quick point ok which I want to emphasize the quick point I want to emphasize is this that suppose we have two blocks the top block exerts a normal reaction equal to its weight on the bottom block. Now even if both of these blocks are moving by velocity v 1 together there is no slippage between this impender between these two contacting surfaces. But the moment v 1 becomes a little bit larger than v 2 by a small amount epsilon ok this is a very small quantity then what we will immediately see is that if you draw the free body diagram then the top body has a relative motion in this direction with respect to the bottom body as a result it will get a force F which is equal to mu s times the normal reaction which is transferred from the bottom block to the top block ok. When epsilon is small there is just a little bit slippage this also is called as impending slippage however if epsilon becomes a finite quantity or a relatively large quantity then what happens is that that there is a finite slippage between this surface and this surface and as a result the force now becomes a force of kinetic friction which is equal to mu k times n ok. So, this is one thing which I wanted to emphasize that these two bodies need not be really at rest to begin with only thing that is really important is what is the relative velocity between these two surfaces. If the relative velocity between two surfaces is infinitesimal then it is a case of impending motion you have to use static friction if there is no relative slippage at all if epsilon is 0 then we cannot say anything about the friction force here, but when V1 is significant or is higher than V2 by a finite amount then in that case there is a finite relative slippage between the top surface and the bottom surface and in that case the friction transfer is equal to mu k times n ok. Now, let us get back to our slides. So, these are various values which are experimentally measured for various surfaces metal on metal, metal on wood, metal on stone and so on ok. These tables will be available in various books in various manuals. What is more important the take home message from this table is that that friction is not property of one particular material it is a property of two materials coming together or two surfaces coming together ok. Clearly it is a cliche to say ok that friction is a necessary wheel why is it a necessary wheel clearly the fact that we are comfortably sitting in our chairs we cannot sit if there is no friction between ourselves and chairs similarly if you are standing comfortably on the ground that comfortable standing becomes a nightmare if you try to stand on ice ok. Also another important use is that when for example vehicles they want to stop if there is no friction between the brakes and the wheels and between the wheels and the ground then the vehicles cannot be put to stop ok. Similarly, there are various other applications of friction where friction is actually important the drawbacks of friction clearly is that that because of all the rubbing of two surfaces there is lot of wear and tear and you lose material you create lot of heat that is damaging to various applications. So, because it has positive aspects and negative aspects we emphasize that it is a necessary wheel. Now, let us get back quickly to laws of dry friction we will just remember what we had done if there is take a two take two blocks only normal force no lateral force clearly equation of equilibrium in the horizontal direction will tell us that a friction is equal to 0. Take the second case if now this P is not vertical, but it has a horizontal component in that case what happens is that that F is equal to the horizontal component is equal to P x and if this horizontal component is less than static friction times n then just F will remain equal to P s and normal reaction of course is P y plus w by equilibrium in the vertical direction. So, there is no motion whatsoever if P x or the horizontal force is less than the maximum possible force which is mu s times n. Now, when the horizontal force becomes exactly equal to mu s times n then we say that the motion is impending it is just about to start and when the motion actually happens when P x is more than when this P is more than coefficient of kinetic friction then in that case we say that F k or the resistance that comes from here is simply equal to mu k times n as simple as that and this point will become even more clear when we will solve a few problems in dynamics. Now, there is another convenient way to represent friction we represent that with what is called as angle of friction. So, note one thing. So, from the surface we can have a normal reaction we can have a friction force which is in the horizontal direction now think about it. So, a normal reaction and a horizontal force we express now this combination of two forces as their vector sum given by a resultant reaction R. Now, this resultant reaction R can make some angle phi with respect to the vertical or the normal reaction. So, this is the angle that it can make what does the friction law tell you is that this angle the maximum possible value of this angle will be decided by what is the maximum possible ratio between the horizontal force which is P x divided by the normal reaction which is n. So, which is tan of phi is equal to F by n and the maximum possible of possible value of this friction force can only be equal to mu s times n and as a result the maximum possible angle here will be equal to tan of phi will be equal to mu s or the maximum angle that the resultant reaction makes with the normal direction is equal to tan inverse of mu s. And when the motion is happening when the motion is impending then this angle phi becomes equal to phi s and when the motion is a finite motion when there is a finite slippage between these two surfaces that this surface is moving with some relative velocity with respect to the bottom surface then this angle phi will be replaced by phi of k. Now, this way of representation you may ask or student may ask that what is the big deal we are just taking this fx f and normal reaction and just replacing that with a reaction acting at a particular angle with respect to the normal reaction and the maximum possible value of that angle is equal to phi s for impending motion and phi k for a finite motion or equal to the kinetic friction angle. So, the idea is that we will see that there are a bunch of problems that if you represent this problems in this manner they becomes extremely easy and extremely transparent to solve. We will look at many such problems in the subsequent slides. So, now with this much of a preamble let us go to actual problem solving and say how to invoke laws of friction. What we had seen just a few moments ago that just because there is a friction resistance that is acting does not mean that the friction force is equal to mu s times n or mu k times n. The mu s times n only happens when there is just an impending motion and mu k times n is the friction force only when the top surface is moving with relative there is a relative sub finite motion between the top surface and the bottom surface. Otherwise you cannot immediately say what is this value of f that we have to emphasize this to the students a lot of because that is one thing which they do that moment they see f they said f is equal to mu times n that is not always true that is only true for impending motion and for finite motion. And we will see a bunch of problems in which we will see how to invoke laws of friction. So, there is a beautiful side of summary on pages 4 4 1 4 4 2 of beer and Johnston 8 tradition. It is totally worth it having a look at the summary where a lot of problems in friction are nicely categorized. So, what we have done is that we have used that as an inspiration and we have done our own categorization that how can we categorize problems of friction. So, let us say the first type of problems are problem type 1. Now, what are this problem type 1 we will see that with examples, but just as a brief preamble these are like usual equilibrium problems. So, in the last 2 lectures which I had discussed we have solved a lot of problems in equilibrium. So, these type of problems are essentially like equilibrium problems. We have to use the same concepts that we had used earlier. The only difference is that that towards the end of this question the problem will be verify if the surface can handle the load or it is capable of providing that much friction or find the minimum friction coefficient required or find out if slipping has occurred then what is the friction coefficient. So, there are various of these variations, but essentially as far as finding forces are concerned these are essentially like the problems of equilibrium that we had discussed earlier. And in these kind of problems friction law this is one of the things which I feel that we should invoke the friction law as late as possible. So, like a good batsman who commits to the shot not as early only as late as possible so that he understands what is happening. So, if you want to have a good understanding of the problem without confusing the matters then I feel that the friction law should not be invoked in the beginning it has to it should be invoked as late as possible only when it becomes mandatory that that law has to be invoked that mu k times n or mu s times n has to be invoked depending on what the problem situation is. So, let us discuss this in terms of ladder problems. So, I have seen a few example papers from Anna University, Pune University and so on and what I have seen is that this problem of ladder and friction is extremely popular. So, what we will do is that we will try to categorize our problems with respect to this humble ladder. So, this simple problem we all know what we have is that we have a ladder AB mass 10 kg for the time being without losing without any confusion we say that the mass is uniformly distributed. So, that the center of gravity or center of mass is at the center of this ladder horizontal dimension vertical dimension is given on the top of that what we are told is that there is no friction at point B and only friction at point A. Now, the question that is asked is that what is the smallest value of coefficient of static friction at A for which equilibrium can be maintained for the structure. Now, even before we go to the solution to this problem which is just two line solution what is more important is a concept and the concept is as follows. Now, let us first in all these friction problems the most important thing is to keep track of the number of unknowns in the problem and the number of equilibrium equations that we can write. It is a very key concept it is extremely important for equilibrium problems very very important for trust problems also important from frames machines everything it is important but nowhere are they as indispensable as problems in friction if you want to understand what is happening. Now, look here this surface is frictionless to the only reaction as we had seen earlier that can be exerted by the wall on this ladder is a normal reaction which acts in the horizontal direction. Look at surface A on surface A there is a possibility of having friction if required on top of that this surface penetration in this direction is not allowed. So, the surface can also exert a normal reaction in the vertical direction. Now, how many unknowns we have as far as the forces are concerned one unknown two unknowns three unknowns this force W is given to us. So, we have three unknowns we can write down three equations of equilibrium and. So, when all the forces everything is given to us all these three reactions reaction at B vertical reaction or normal reaction at A and a friction at A all can be found out immediately no need to invoke law of friction here at all but where do we need to invoke the law of friction. So, we go to this slide we see we write down equation of equilibrium the x direction we see R B is equal to F A immediately equilibrium of y direction will tell us that R A should be equal to weight and taking moment about point A you will see another relation between R B is equal to 6 R B equal to 1.25 W straight forward and from these three equations we get all values of R B R A and F B. Now, where is the question coming about the coefficient of static friction the question comes because we want that the ratio F A by R A which is the friction force divided by the normal reaction this should never exceed the static friction which is present between those two points. If this exactly equal to the static friction then an impending motion just about starts here and if this quantity mu s is lesser than the ratio F A by R A then the system is not in equilibrium it has already started to slip. So, we keep that in mind and what we say is that that F A by R A should be less than or equal to mu s and from this this is the ratio and we immediately see that mu s has to be less than or equal to greater than or equal to 0.2083 and the lowest value of the friction is 0.2. Now, this problem can also be solved in another way. Now, this is the three force member clearly why because force acts at 3 points 1 weight second and third is the force acting here we know the line of action of R B we know the line of action of W these two intersect at this one point and now for this body to be in equilibrium what should we need that the total reaction should pass through point B but note one thing what is this angle? This angle is nothing but the angle made by the effective reaction with respect to the normal reaction and this angle we know cannot exceed the angle of static friction and as a result we can just say that tan phi should be less than or equal to tan of phi s which in this case is 1.25 by 6 and from this we immediately figure out what is phi s and our coefficient of friction is nothing but tan of phi s which again comes out to be 0.2083. So, this problem just says that this is very much like any equilibrium problem that we had discussed earlier that is no difference between this problem and other problems we had solved the only additional criteria that came here is that when we got all the reactions we just wanted to be sure that the friction at this point A what should be the coefficient of static friction such that this friction is sufficient to keep this body in equilibrium that is the only place where this came into picture. So, these are the category one problems or type one problems where everything is as is done in the equilibrium problems the only way you have to only place you have to invoke the friction law is towards the end note in this in one case what I have seen even when I was a student we used to have very lot of difficulties there used to be rumors floating around that friction direction cannot be chosen arbitrarily because in equilibrium problems what we had seen is that we can choose whatever direction of the reaction you want to what happens is that if our direction that we have assumed is not right we will get a negative answer that is the opposite direction but then there will be rumors floating around like during the final exam that no friction we cannot do it like this. So, what we want to emphasize here is that we want to discuss here is that that should we keep the direction does it come automatically it all depends on the problem if the problem is type one then you will see that you do not have to worry about anything the direction of the friction force will automatically come from equilibrium equations. Now we move on to this another problem which I had discussed during the coordinators workshop I will not go over the solution of this problem the solutions will be uploaded on the course website in fact it may be the slides would already be uploaded you can download them and you can see what is the solution but the principle is as follows these are self are the tongs which are operated under the under the weight which they are already lifting. So, what is done is that this is a friction tongue you have all these components put together you bring this in contact with the block you want to lift start putting some weight on the top because of its self weight this two these two pads they start to close in and then start to close in on this weight when you start lifting up they will close in even more the normal reaction increase now you can start lifting even more and ultimately you can lift this block completely up from the ground and what you are asked here is that in this problem what should be the minimum coefficient of static friction between this surface and this surface what you will realize is that that this is exactly like equilibrium problem that we had discussed earlier these are two force members this top weight should be equal to 750 you solve what is the force here then you take the free body diagram BCD there will be two reactions here two reactions here okay but what you know further is that these two reactions equal and opposite act on this block and as a result to maintain equilibrium of this block we know that the friction force here should be equal to 750 divided by 2 and you can easily find out the normal reaction by taking this free body diagram and taking moment about point C and you get the normal reaction. So nowhere have we invoke friction law and only place finally we invoke the friction law is we get the normal reaction by the moment balance for this free body diagram about point C we get the friction force which is resisting the displacement of this mass in the downward direction and f divided by n we only say at the very end at f divided by n should be less than or equal to mu s and that immediately gives us what is the minimum coefficient of static friction that is required. So exactly it is a type one problem law of friction need to be invoke only at the very end there are a bunch of other problems or you can have a look at them now with this we move on to problem type 2 now these problems can be pretty tricky now what happens in these problems is that the number of equations that you have is less than the number of unknowns we have discussed problems and then you will see what happens now the number of equations typically come from equations of equilibrium but if the number of unknowns is more than the number of equations then what happens is that this problem become what are called as statically indeterminate problems why because we had discussed that if the number of unknowns is more than the number of equations then using simple methods in engineering mechanics you cannot solve that problem this set of equations will have infinite number of solutions and what is the appropriate solution now depends what additional information you provide and in these problems the additional information is provided in terms of the laws of friction now because you need to invoke extra equations you can see that a motion of body should clearly be impending in this case okay what will become even more clear what you are typically asked to find out is some force or torque required to start the impending motion or some distance some angle some coefficient of static friction etc that is required for the impending motion okay in this case we need to use a law of friction at the impending surface and we need to invoke them quite soon and in these type of problems you will see that we have to be very very careful about what are the science of friction forces okay that we put and this will become very clear when we discuss our favorite example of ladder leaning against a wall now in our favorite example what we had done is we have added one more twist in the previous problem we had discussed in the previous problem we had discussed that that there is only a normal reaction why because there is no friction between B and the wall whereas now what we do we modify the question just a bit okay all friction problems you can just modify them as much as you want okay and to put a twist in the problem the simple thing in this problem is as follows we put friction here and we ask ourselves that everything else remains the same only thing now we are asking is that we are asking the same question that determine the smallest value of u s for which the equilibrium can be maintained so the question remains almost the same only change is that that at point B instead of having just a normal reaction you can also have some friction why because there is a coefficient of friction now there is one point which I want to emphasize which I also emphasize to the students and we all do but I just want to remind us about that is that just because friction coefficient is there between a surface and the wall doesn't always mean that the friction has to act it just depends on the loading and it depends on the it depends on the loading that is provided on the problem and it depends on what is asked in the problem okay so friction only tells you that the surface is capable of exerting a friction force as opposed to just the normal force okay now how many unknowns normal reaction friction normal reaction friction 4 unknowns and additionally okay so 4 unknowns equations of equilibrium okay I am not finished with the number of unknowns 4 force unknowns and the number of equations of equilibrium we can write is only 3 so done we can cannot use simple equations of equilibrium and can get all these 4 reactions but then there is a twist the question is asking us that determine the smallest value of mu s okay determine the smallest value of mu s for which equilibrium can be maintained or in other words what you are telling is that that there is a minimum value mu s at which the motion will be impending if the friction coefficient is less than that value then gone equilibrium is already gone and if friction coefficient is more than that value well you are good so what we are asked essentially here is that we are asked here that what is the minimum value of mu s which equivalently means that find out that mu s for which the motion will be impending now let us ask ourselves okay what kind of impending motion impending motion is just some infinite similar motion that the structure can have let us ask ourselves is that that under the action of the applied load what can be the possible impending motion of this thing so let us briefly discuss is very important okay it is straight forward but it is very important because for many other problems this will become pretty mandatory to understand how the impending motion happens now let us look at our ladder this is our ladder this is the vertical wall this is the horizontal wall and this is the weight of gravity which is acting on the ladder now though it is not rigorous okay this is a very important statement though it is not it cannot be rigorously shown whatever procedure I am following now it is a rule of thumb and can be used exceedingly well in many problems okay so solving friction problems at some stage is always big almost becomes some kind of an art that you have solved enough problems you have enough insight and there is no exact algorithm to solve them but by and large there are these rules of thumbs that if you use properly then most of the problems in friction can be solved in a very efficient manner now let us say what happens if there is no friction between B and A in that case these surfaces can exert only two normal reactions and these two normal reactions pass about they intersect at this point which we will learn later is what is called as a instantaneous center of rotation okay the name is not important we will discuss that when we study dynamics in the next week but these two reactions pass about this point now note one thing that what is this weight doing the weight can exert a torque about point O what kind of torque it will exert it can exert a clockwise torque and as a result this body will have a clockwise rotation and the impending motion will look something like this why clearly there is a clockwise rotation so what happens that the point B will slide downwards relative to the wall point A slide in the sideward left direction relative to the wall okay delta 1 and delta 2 what is delta 1 and delta 2 we will see when we do virtual work tomorrow we will do the kinematics of that but what we see is that these are the displacements that are happening now when there is a relative slide in the downward direction what is the friction force that will ask when it slides slides downwards it will exert a friction in the downward direction on the wall so as far as the ladder is concerned at point B the friction force will be exerted in the vertical direction this is normal reaction R B this is the friction force F B this is the normal reaction R A okay the point is sliding in the horizontal direction so it will exert a force in the left direction on the surface but then by Newton's third law it will get an equal and opposite force coming from the wall F A so we have four force unknowns and one unknown is mu s that is also an unknown so we have five unknowns we have three equations of equilibrium of equilibrium but we have five unknown so we need two extra equations so note how beautifully our kinematics is telling us that impending motion will happen in such a way that slip happens at two surfaces and then clearly those are the two extra equations we have that F A because there is an impending motion is equal to mu s times R A and F B is equal to mu s times R B and note one thing here is that why is the direction so important because you may it may easily be taken if you don't under if you don't realize that the impending motion can happen like this I can easily change the direction of the friction force in the opposite direction this can be put opposite or have various combinations and that will not be right so when for example the rumor floats around that in friction problems you have to choose a direction of the friction force it depends on the problem it depends on the question that is asked and the way the question is asked here okay the way the problem is this is the impending motion that is happening here now with this impending motion and with these two extra equations our five unknowns are four forces the coefficient of friction which itself is an unknown because we need to find out what is that minimum value okay for which the equilibrium is maintained for this assembly okay with this much discussion in mind okay the problem is very clear now this is the friction force here this is the reaction here friction force reaction what we do now we can immediately write so for example as I had mentioned to you earlier okay that we don't want to commit to this friction law as much as possible in the beginning of the problem but now we have read the mind of the problem we know that this is a short pitch delivery okay so we can just go and cream through the covers okay whatever so in this case we know exactly what the problem is we have anticipated everything we have all the premeditation all the thought had gone into it now with that as our with that at our disposal we just say that we clearly say that the friction force will be fb is equal to mu b rb f a is equal to mu a r a and we can just write various equations of equilibrium that are convenient in this case fx equal to 0 would imply that rb is equal to mu s r a which is equal to f a m about point a equal to 0 will immediately give you what is rb okay in terms of the weight f y equal to 0 will immediately give you okay a third equation which will tell you what is r a in terms of the weight and now with this much idea we have we use these 3 equations what we will get is we will get a equation okay in terms of r a but r a is common everywhere now it is a quadratic equation in coefficient of static friction which we can solve it will give us two values one is minus 2.4 okay one is 0.2 so we neglect that minus 2.4 value okay strictly speaking okay this minus 2.4 value is not really a bad choice it only means that the direction of friction happening here we will discuss more about this later is actually not outwards but with inwards what that means okay we will discuss a little bit later okay but what we take here is that we take the positive value of mu s which is equal to 0.2 we forget about the negative value we say that mu s equal to 0.2 and when the coefficient of static friction is equal to 0.2 the motion is just impending when this will be less than 0.2 no equilibrium done the system is out of equilibrium and when mu s mean is equal to some value which is much which is more than 0.2 then perfect the equilibrium will be maintained but only thing that you do not know will be what are the forces then only thing you can say for sure is that the equilibrium will be maintained when mu s is greater than 0.2 but what exactly that will be it remains unknown why because this is actually a statically determinate problem and the best we can say is that that what is the mu s value for which the equilibrium can be maintained okay so with this much of a preamble so there is another problem for example that these are two blocks one over the other there is a string that is connecting okay this is the problem that we had discussed this is a problem that we had discussed earlier in the coordinators workshop they are asked to find out that for this assembly what is the minimum force required for which the system will be in equilibrium and what we can see here is that as a number of unknowns here as that 1 2 3 4 4 4 unknowns p also is an unknown so 5 unknowns are present in this system the equations of equilibrium we can write 2 for this 2 for this 2 to 4 but there are 5 equations now what do we do so 4 unknowns 5 equations what we do is that we have to invoke friction law if there is an impending motion what will happen this will try to slip in this direction because the string is inextensible this will move upwards so automatically you will see that this motion creates slippages at 2 places so impending motion automatically implies 2 slippages but 5 unknowns 3 equations from equilibrium plus 2 extra equations we get 5 equations and now we can solve this problem to find out what is the minimum value of p what is the max what is the value of force p for the motion of this block is impending and any value of p over that this system will no longer be in equilibrium so you see that a combination of mathematics and understanding how the system can move in impending motion will give you perfect insights into any friction problem now problem type 3 okay so typically this is a third category problem now these type of problems are very very similar to problem type 2 what that the number of equations will be will be less than the number of unknowns that the number of force equations that you can get from the equilibrium of the rigid bodies will not be will be less than the total number of unknown present unknowns present in the problem okay but in this case there can be multiple ways in which the slipping can happen now in this problem of ladder for example slipping can happen only in one way B slips A slips okay end of the story nothing else can happen similarly here this slips this has to slip okay end of the story okay there cannot be any other variation but there are some type of problems in which the modes of slippage can be more than one okay now which particular contact the impending slippage happens has now to be decided by trial and error inspection okay physical intuition now there is no hard and fast rule for that it depends on the problem and it just depends on how our intuition will work okay if you solve enough problems if you solve enough problems with various types then we will build some physical intuition which will quickly tell us that what will be the surface at which the impending slippage can happen but ultimately when we do everything has the consistency has to be checked that all possible surfaces the ratio or the friction force should be less than mu h times n okay where mu is the coefficient of friction and n is the particular normal reaction at that surface but in this case also we may have to be careful about the direction of the forces it all depends on the problem and how it will become clear as we proceed further now this is another ladder like problem okay so since we have decided today that ladder is the theme of today's class okay so let us discuss all problems okay of that type now in this case what happens is that we have a modified ladder so our ladder AB is present but the ladder AB is now leaning against a mass okay which in turn is attached is not attached is it has contact with the wall okay the mass this particular object has a mass m knot this ladder or a tilted rod has a mass m what is told to us is that that at this surface and at this surface the coefficient of static friction is mu s and what we are asked to find out is what is the range of theta for which this entire assembly will be in equilibrium now let us think about it how many equations are present how many unknowns are present that is the first mathematical thing we first think about that how many unknowns do we have we have two unknown reactions here okay the line of action we do not worry okay it is an extended object it has an extended contact so the line of action exactly is not clear so we do not worry about it for the time being let us say that we are two unknowns one is the normal reaction one is a friction force how many unknowns here this is a roller so only one unknown which is the horizontal normal reaction point B one normal reaction possibility of a friction so we have five force unknowns of 1 2 3 4 5 we have 5 force unknowns in addition we are also asked to find out what is this angle theta okay what is the range of this angle theta for the equilibrium is maintained so we have 5 force unknowns sorry 1 2 3 4 5 force unknowns and 1 angle unknown how many equations of equilibrium we can write two equations of equilibrium for this okay 1 2 3 equations for this so we can write 5 equations of equilibrium so we need one slippage because we need one extra equation and that one extra equation has to come from laws of coulomb friction or laws of dry friction now what is that that the now one surf so one extra equation can come from a slippage happening here or here now what happened in the previous problem in the previous problem if slippage happened here then compulsorily the slippage also has to happen here there is no way you can you can have slippage here and not here but look at this problem you can independently slip here and without this slippage you can independently slip here the kinematics of the problem is precisely allowing you to do that that this can remain perfectly stable this can slip or this can remain perfectly stable and this can slip so these are two modes in which the slippage can happen and depending on if the slippage happen here or the slippage happen here you will get one theta and another theta so one theta will tell you when the slippage happens here another theta will tell you when the slippage happens here so in this case even though this is a problem in which the number of unknowns present is less is more than the number of force equilibrium equations we have. The mode of slippage is drastically different here. There is no compulsion on us here that a slippage happens both here and here and that is a major difference between this problem and a problem that we had discussed earlier. Now very simple once we recognize this let us understand that why does this problem put this limitation that there should be a range of theta for which this system will be in equilibrium. The idea is as follows that if theta is equal to 0 our we immediately see that there can be no normal reaction. If there is no normal reaction here then can be no normal reaction for the equilibrium of this block and as a result when there is no normal reaction but the vertical friction force should be equal to m naught times g. So done no normal reaction there is no way this friction that this surface can provide a friction force if this angle is 0. So what does that mean that we have to keep increasing this angle such that the normal reaction increases such that at a critical angle the motion of this block is impending. What does the impending motion means that the friction force should be equal to m naught g and that should be equal to mu times the reaction that is acting here. So that we clearly know what is a what is the logic behind having a some lower limit on theta. Now let us say we keep on increasing theta what happens we will see that if we keep increasing theta the normal reaction here increases. So this mass stays happily in equilibrium but when that increases this force also increases okay this force also increases but the normal reaction we will see will be simply equal to weight and the friction force keep increasing if you increase theta and add some critical value the normal reaction times the friction force will exceed the friction force and the body will no longer be in equilibrium and that is the reason why we have a range of thetas for which the system will be in equilibrium and the problem is nicely discussed here. We can see here that for the top free body diagram this is the reaction Ra what is this Ra we can immediately see that will be equal to mg by 2 and theta where mg is the weight of this and we see from the free body diagram from this that the normal reaction is simply equal to Ra and f2 by n2 okay this is f2 for the friction force here is again equal to Ra by equilibrium for this free body diagram in the horizontal direction the vertical normal reaction at this point B is simply equal to mg n2 is equal to mg so we know that n1 is equal to Ra is equal to mg tan theta f1 is equal to m0g for this equilibrium so f1 by n1 will be equal to this ratio which has to be less than or equal to mu s that will tell you why should you have some limit on theta and ultimately what do we have finals the second thing what you should what we we need to know that f2 by n2 will become equal to tan theta by 2 should be less than or equal to mu s that will tell you that there should be an upper limit on theta because if theta is very small this will sleep if theta is very large okay then this will sleep and for intermediate thetas where both these conditions are satisfied that f1 by n1 is less than or equal to mu s and f2 by n2 is equal is less than or equal to mu s this entire assembly will be in equilibrium at this end point okay at this end point the slippage happens here at this end point the slippage happens here okay and in between the system will be in equilibrium okay so that is the logic behind this problem there are a few other problems also so we do not have enough time let me see let me let us see how we how we are doing on time if you get enough time I will discuss this problems but this was discussed extensively in the coordinators workshop last type of problem okay that we have to ask if there is an impending slipping or tipping now just note that impending slippage happen when f equal to mu n and tipping happens okay when the reactions at all points other than the point of tipping are equal to 0 okay so this is one special case we will quickly look at one small problem and then we will we will take questions so what we have here is we have a block of weight w of weight b you can think that this block is for example our armilla Almira that we want to move we want to move the Almira okay but this block is resting on the floor and there is a coefficient of static friction mu s between this block and the floor now what we do is that we start pushing against this block at height h with some force peak now we know clearly that if you want to start pushing this block there has to be some impending motion without that impending motion the block cannot move so what we know from here is that our P or the force that we apply should be equal to the normal reaction at the bottom surface multiplied by the coefficient of friction in this case clearly the normal reaction will be will be equal to w so w times mu is simply equal to the P that we need to apply but we know that it is not as straight forward as that we see that if we try to apply a load which is a bit high then this tries to topple okay and so what we do the trick we do is that many a times with the if there is lot of friction between this and the surface we get down and try to apply load at a lower level and then we see that this starts to move now what is happening here is that what happens here is we will look here okay so what is asked in this problem is that that what is the critical height h okay such that the block will start sliding now start sliding means there will be just an impending motion okay without the tipping now what we do is that that we see that we keep on increasing the height h what happens now clearly the effective reaction that exist between this bottom surface okay that the floor exerts on the bottom surface will be some force and what we know is that these two forces pass through point B straight away okay this is the geometry of the problem it applied at height h this goes through the center the weight so P and W meet at point B now for the equilibrium of this block to be maintained what we should have is that the total reaction that should act from here should pass through B now what we are asked is this let us take this particular case where we are applying a load through the center now what we will see that when the load is applied through the center okay and the and another thing that the load is applied through the center and the block is about to slide because that is our purpose then what happens is that that this friction force the resultant friction force should pass through this point okay where this line of application of force P and a weight meet why because if not we saw yesterday that we can take torque moment balance about this point and there is nothing to balance so if the block is in equilibrium then we should have this effective reaction passing through this point now you will see that this angle theta will be equal to phi s what phi phi s because this angle theta is the angle made by the friction by the resulting force with respect to the normal reaction at slippage so this is equal to phi s or the angle of friction and what we see is that that when the load is at the center this angle will be somewhere here now if we start moving this upwards what you will see is that that that this will start moving out because this angle theta is equal to phi s we keep moving this up our geometry tells you that this will start moving up and ultimately when does tipping happen tipping happens when the entire reaction is concentrated at one edge now what does that mean that the raw that the almirah is about to tip or this top block is about to tip the only line of contact it has is at point a but it is also about to slide so this angle theta with the effective reaction makes with the perpendicular the normal direction is equal to phi s or tan inverse of mu or the angle of friction now what do we say how do we find out what is the critical height is that the critical height is when okay that the rod that the block is just about to tip before sliding okay so sliding and tipping are just about to happen together okay in this case that is a critical point this angle is equal to phi s and then pure geometry tells us that this is b by 2 is the width this is height h and simple geometry tell us that tan theta is equal to phi s is equal to is equal to b by 2 h or the maximum height will be b by 2 than theta which because we want a sliding to start happening okay impending sliding will just become b by 2 mu okay so as simple as that so in this problem what we are asked is that that we want to start that moving but without the tipping because if it tips then you are done so we want to slowly start sliding without tipping and what this is telling you is that that what should be the critical height and note again here it is a very interesting point is if the width is larger b then of course the height can be larger which is our experience that if you are trying to move a long cupboard okay it is more difficult to move but if this is short t-point then it is easier to move okay then this does not arise or if the coefficient of friction is large then this height decreases which also makes sense that when the coefficient of friction is large okay then the sliding becomes more difficult but tipping becomes more easy okay so this is the basic logic behind this problem and with this much we have solved all typically all kind of categories of problem that we encounter in friction and now what we will have is that we will have a discussion session for the next half an hour okay so there are a few general questions okay so which we will try to answer okay via the chat okay so during the tutorial when I have ample amount of time I will answer those questions okay but there are some general questions so first question is by center number 1068 that how can we justify that frictional force is directly proportional to normal reaction there is no direct justification okay so this is a field of tribology and in fact there are many many cases in which this law is not true but for a lot of engineering problems it is a decent first approximation that is all we can say okay to go beyond this is to go into deep waters okay if you are interested what I can suggest is that I will write down the name of one book okay these things are explained in somewhat details in that book it is a reasonably advanced book on okay it is a book by the authors are Andy Ruina okay and Rudra Pratap this is a freely available book you do Google okay search for Ruina and Cornell you will reach his website okay and the book is available just for download at his website you take this book okay in that look upon this look in this chapter that what are the limitations of this friction approximation it is very nicely given we will move on to the next question it is by center 1068 again okay why the coefficient of correction is less than 1 again look at that book 1170 how can we justify that friction force is directly proportional to normal reaction same question again okay same book what is limiting friction the limiting friction as we had discussed that as far as this problems are concerned this friction problems are concerned limiting friction is mu s times n where what is n we look at two surfaces this is the normal reaction that is exchange between these two surfaces and the friction force by definition is tangential to this f and limiting friction is mu s s is the coefficient of static friction times n what does that mean that if this value if somehow you apply external forces in such a way that this force has to exceed this value okay or it is just equal to this value then these two surfaces they start barely slipping with respect to each other this is what I think is meant by limiting friction next if the body it is a center 1130 again with the body is resting on a table in the absence of external forces and what does that mean it means that body is resting on the table no forces okay nothing else will happen to that body now it is centered 1210 is the impact of contact surface area on frictional resistance in this dry friction or Coulomb friction we assume okay it is a model okay all these are models nothing is reality one model is more useful than the other model this Coulombic friction model works reasonably well for a lot of problems that is the only reason we use it it does not mean it explains all possible problems under the sun okay it does not but if you use that and in that particular model okay and there are a few experiments and it is not that we are just pulling it out of thin air in that particular model what we are saying is that that the normal reaction okay we do some simple x experiments you would be also doing in your labs you take a stack of books keep it horizontally keep them vertically and we see typically that it is dependent the limiting friction force depends and use the use of the word the limiting friction force okay is dependent only on the surfaces and the normal reaction and not within the area but it is a model and it not be always right there is another question that is asked will there be any change in kinetic friction with change in increasing velocity between the sliding surface is a very good question okay it is it has been shown that for example for dry friction between two surfaces okay that for low velocities this coefficient of static friction right this is not constant okay it is a good question but again that is beyond the topic of this particular course in engineering mechanics that force is some function the friction force okay this is a friction force is some function of the relative velocity now this v is not the absolute velocity it is a relative velocity between the surfaces and this in the simplest approximation the velocities are small you can approximate that to be equal to eta times v but this is equivalent to what is called as fluid friction but there are many many complicated laws also so again this Coulombic friction law is not a gospel okay it is not the ultimate truth but it works for a lot of problems okay x1037 explain instantaneous center of rotation and how to rotate the order clockwise perfect this is a good question let me go over through details of that although this point will become more clear tomorrow when we discuss virtual world let me still discuss that okay so think about it let us look at our ladder problem now think about it this point if it has to move what is the only way it can move it cannot move outwards why because if it moved outwards it will lose contact with the surface it cannot move inwards because then it will penetrate the surface so only direction it can have for its infinitesimal movement okay a tiny movement will be this now think about it what about this surface what is the direction in which it can move okay it can of course it can move upwards also it can move downwards also but because we have we want to discuss about clockwise motion let us take for the time being clockwise motion what about this thing it can move either in this direction or it can move in this direction but it cannot move up why because it will lose contact with the surface either can it move down why because if it moves down it will penetrate the wall which is not allowed so the only way the motion can happen is that this point moves up this point moves down this point moves left this can move right now what happens is that if we draw this is the direction in which the motion can possibly happen instantaneously we draw a tangent a normal to this okay a normal to this normal to this by definition they will intersect at point O now this point O is called as the instantaneous center of rotation why it is called as an instantaneous center of rotation is that because if you imagine a body like this okay imagine a rigid body like this okay this ladder just say it is like a rigid body now if you try to rotate this about point O what do you mean by about point O that pin this here and then try to rotate this about this point in the clockwise direction what way the motion will happen the motion will happen the instantaneous velocity or the instantaneous infinitesimal displacement will be perpendicular to this and again it will be in this direction at this point so that is the instantaneous rotation why because when you take this try to instantaneously just give a small rotation then this point will tend to come down this step point will tend to come to the other side and this was the purported direction the purported motion that the ladder will undergo if there were no friction and note that this is clearly a clockwise motion that this try to move clockwise this tries to move down this tries to move sideways there is another way in which you can interpret this motion what you can say is this I take this ladder rotate about this point A how will the resultant motion will be it will look like this perpendicular this will become more clear when we do virtual work yesterday but now what has happened now what happened is that now this is a this is a different this is another mode this is a motion in the anticlockwise direction now it has loose contact here that loss in the contact is not allowed why because then the system is not in equilibrium you do not get a reaction so what we do is that we shift this entire assembly to the left and so the final motion will look like this so this has gone up this has moved sideways and resultant rotation that happened was in the clockwise direction so there are different ways of looking at the same problem but this was the instantaneous rotation where we know which direction this point will tend to move which more direction this point will tend to move or move infinitesimally join those two reaction that point will be the infinitesimal center of rotation of rotation this point will become even more clear when in the next week we discuss that what is the instantaneous center of rotation for kinematics velocity kinetics or velocity kinematics of rigid bodies okay so it is how does the coefficient of kinetic friction becomes less than the so there is another question by center 1 2 2 7 is there a relation between kinematic and static coefficient of friction the only thing we can definitely sit definitely say is that that mu s has to be more than or equal to mu k you can never have mu k less than mu s and the reason is simple okay because that is another question which is asked by center 1 2 2 4 okay how does the coefficient of kinetic friction become less than the coefficient of static friction so microscopic answer I will not give I will give a logical answer think about it what is coefficient of static friction tells you that on this block okay this is the normal reaction this is weight so there will be a normal reaction that the surface will provide it we apply a force F and find out what is the force of friction that is applied here now what we see is that that this friction force will try to be equal to force F till you reach mu s times n now suppose okay let us take the other case where kinetic friction coefficient is actually more than static friction coefficient then what happens after you increase the force just by a little bit this body starts moving now kinetic friction comes into play but because the kinetic friction is now assumed okay for the purposes of argument to be more than equal to mu s the body will stop again so unless and until you exert a force which is equal to mu k times n the body will not move at all because if you say that coefficient of static friction is less than kinetic friction apply force more than that body starts moving but because coefficient of static friction is more that we have assumed less than the kinetic sorry coefficient of kinetic friction is assumed for argument sake to be more than the coefficient of static friction the body will stop so unless you exert a force which is more than n times mu k this body will never move and then afterwards it will have a friction force which gives which is given like this but then if this happens then this is as well saying that mu s is equal to mu k because we can never distinguish static and kinetic friction so that is simple logic to tell that why kinetic of static friction can never be less than kinetic of sorry than the coefficient of kinetic friction and if it is more suppose for argument sake then the moment the body tries to move it will stop tries to move it will stop only when the force becomes more than whatever value of mu k is okay you will see a motion like this it is a logical answer okay think about it if you have any questions we can discuss in more details any relation will link limiting friction and angle of repose okay the angle of repose essentially phi s instant center 1 to 2 for us what is the instantaneous center of rotation we have discussed 1 1 0 2 does any change takes place due to frictional force at support okay this question is not clear okay what what is trying to be implied okay we will have more about this we will try to address this maybe it will become more clear okay a little bit when you solve a more a few more problems huh this is a very good question by center 1 to 2 phi please indicate how to take direction of friction in case of rolling circular bodies okay there is a problem in the tutorial we will solve that problem and then it will become clear how to take that also there is a solved problem okay which we did not have the time to go through have a look at that also okay this particular problem I am talking about okay the only difference okay let me quickly mention okay is when you have supports like this okay when you have supports like this these are extended supports okay these are the sorry so these are bodies with extended contacts so we do not know what is the line of effective line of action of the force whereas in this problem the friction force will act parallel tangential to this the normal force will act perpendicular to this so that is the only difference and also what happens is that that in this problem you will see that the number of equations we will have for equilibrium will be 3 okay there is a problem in the tutorial okay we will go over that in more details okay when we proceed to the tutorials okay 1 1 0 to ask what is the difference between angle of repose and angle of internal friction so typically angle of repose comes in geotechnical engineering that if you have sand for example and you put sand okay and dump it on the ground you see that it forms a cone and at angle of cone they call it as angle of repose and typically that is related to the internal friction of the grains okay so that is a very detailed topic granular materials okay are a subject of extensive research in engineering and sciences it is beyond the course but typically that is the angle of repose and it has some relation with what is the angle of friction internal angle of friction between various grains but this is really beyond that beyond the scope of this course lastly the another question from the same center to the frictional force is perpendicular reaction forces in all type of supports yes because that is definition that the normal reaction how do we get a normal reaction that any two surfaces okay any two surfaces they can be like this and like this or like this and like this or this and like this or like this and like this so these are the typical ways okay or so in all these cases what we do is that in this case there is a common normal to both of them this is the common normal so the normal reaction acts like this and by definition the frictional force is tangential to the normal direction or in other words it is it is perpendicular to the normal reaction in this case the normal reactions can only be like this because this is the normal here this is the tangent and the friction force can act in the tangential plane similarly here now this is a sharp corner but a sharp corner you can approximate that with something with a very small radius of curvature okay let me draw it again here we can say that this as a very small radius of curvature and in any case this will be the common normal so this will be the tangential plane the normal reaction will act like this tangential reaction will act like this two surfaces perfect normals are only like this and frictional force only act perpendicular to them and ultimately if you have a flat surface in connection with this then what happens is that that the normal reaction can only be this friction can be this and only ambiguous problem comes when you have two corners coming into contact with each other okay so these are ill posed problem where the normals cannot be uniquely defined okay so we do not worry about these problems but all other problems okay these are the normals okay and these are the corresponding friction forces okay so with this we can now visit a few centers and have a brief discussion with them 1146 kindly explain about coefficient of friction and how do they are calculated for different materials forward to you they are calculated for different materials using various experiments okay so the simplest experiment we do in our in our colleges for example right that you take a tilted material so the simplest experiment is you keep a material material one material two and you start inclining this and at a critical angle theta you see when the impending slippage happened this is the simplest okay I am not even talking about industry standard impending slippage and this particular angle theta we all know is equal to phi s and a coefficient of friction is equal to tan of phi s okay nothing that you do not know just this okay so until you reach phi equal to for theta equal to phi s this block will be perfectly fine why because the forces acting against it if this is the weight then this is mg cos theta mg sin theta friction force f normal reaction mg cos theta and we know that force should be balanced by mg cos sin theta and also force divided by the normal reaction is equal to mg cos theta which is equal to tan theta and since tan theta okay has to be less than or equal to mu s why because the surface cannot provide any more friction than the value given by mu s times n we see that the critical value becomes tan inverse of mu s okay I hope that answers the question okay okay so those those are not the dry friction problems okay but those friction problems typically are what are called as viscous friction or fluid friction or fluid friction in that case the the resisting force okay and what we have done here is was the dry friction currently and friction frictional force is in that case it is proportional to velocity it is equal to some viscous coefficient times v that is the simplest modeling for those kind of damping materials okay so these are called as linear viscoelastic materials or linear viscous materials 1068 okay sir I have two I have two questions right now okay as in what is the practical application of cone of friction that is the first question and the second question is what will be the condition when the coefficient of static friction will be less than 1 and what will be the condition when the coefficient of static friction will be more than 1 so cone of the friction right this yeah I get your question okay I get your question so cone of friction is nothing okay it is not a big deal only thing is that if you are dealing with 2d problems then what we know is that that this is a normal force and a friction force can act in this direction or in this direction and the maximum possible value of the frictional force will be equal to mu times n and so this particular thing okay this is in 2d that the reactions can only stay within this within this cone or within this particular triangle or within this within this angle and if you look at this from a 3d point of view okay that if you have a 3d structure with a friction acting on it then in that case this angle will just become a cone so as simple as that and depending on for example this phi s is equal to tan inverse of mu so if your mu is less than 1 then phi s has to be less than 45 degrees and if your mu s is more than 1 then this angle can become more than 45 degrees whatever value it wants to but mu less than 1 or equal to 1 only tells you that the angle of friction or the angle of this cone can be less than or equal to pi by 4 or 45 degrees I think that is the the best answer I can give to this question is that okay yeah I am just asking about what is the application of a cone of friction significance or significance it is telling you that is a limiting value that your reaction okay whatever application now there can be 100 applications okay what you want is that if you have certain forces acting on the body then those you do not want to have reactions coming from the surface okay the only single because there are hundreds and thousands of whatever the friction comes into play what you can say is this that if you have some forces acting on the body then you better make sure that if your body should be in equilibrium okay the forces are not such that the resulting reaction lie outside that cone because if they lie outside that cone then the body will not be in equilibrium okay so that is the that is the engineering application that I can think of that if you lie outside the cone okay mu s more than 1 mu s less than 1 does not matter it can be any value but your reactions if they are such that they have to lie outside this cone your body will not be in equilibrium that is a simple thing I can give you is that okay okay okay okay thanks we will have a tea break now right