 So welcome back, it's a very good morning. I hope it stays the same way for the next couple of hours. So what we are going to do today is one of the very important topics also conceptually it is supposed to be from whatever feedback I have obtained from the students and even from my own student days this is one of the more involved topics in engineering mechanics. Not that it is extremely difficult or anything but essentially systems or engineering mechanical systems which have frictions between their contacts because so far when we did 2D equilibrium when Shobhik taught 3D equilibrium we did trusses there was no mention of any friction. Everywhere we were said that assume that the contacts are frictionless, assume that the surface is smooth and so their friction was almost thrown away. But what we know is that in real life friction is always there. Sometime it's create havoc, the way it create havoc is for example if you have friction in your car engines, friction in fans, friction in vehicles, friction in all moving objects okay or on your shoes what they do is that friction causes damage, damage in the form of deterioration, extreme heat and so on. But at the same time we also know that friction is extremely important. For example the fact that I could walk here on this stage is because there is a friction between my sandals and the stage. Similarly you can sit so nicely on these chairs clearly because of the friction. Just try wearing silk and sitting on a wooden chair okay then you will realize what I am trying to say. Similarly okay you are holding your pens. The fact that you could hold your pens clearly is because of friction okay try holding a soap when it is wet okay you will see okay it doesn't stay in your hands. So it is essentially friction. What it does is that even okay it causes lot of damage, lot of deterioration, lot of heat okay wastage okay at the same time we need it okay essentially to survive okay otherwise any day to day activity becomes extremely difficult. For example if you are training some students here you will see that for example there is a strategy like there's a question okay it's a very pet question of physicists they ask you that if for example you are trapped on a frozen lake okay those frozen lakes we don't see in most parts of the country but in North India especially Kashmir okay you can see that for example lakes are frozen and if you are trapped in the middle of a frozen lake how do you get out? There is no friction to get out then they say okay there are some rocks and pebbles around then the strategy is that because you don't have friction you throw them in the opposite direction and try to move forward okay but the idea the point that I'm trying to make is the presence of friction which we take for granted makes our life extremely easy. So what we are going to do in this is that we are going to discuss a few problems okay that if an engineering mechanical system has frictions either internally or at the contacts then what happens how do we attempt those problems? Can we say anything about the stability of the system okay can we say when the system will collapse when it will mechanically not be stable and so on those will be the topics and there are a few strategies that reasonably easy strategies and especially I remember that when I was a student we always used to have this query that there used to be this rumors flying around okay when I was rumors is because the day before the exam all students will be studying one problem or the other and like one so why I solve it in this way then somebody say I solved it in this way then there is a rumor that I heard that apparently friction cosine should not be taken in some certain direction okay so what I realize what I'm just trying to say is that it can cause lot of solving those problems can be a bit challenging but what I'm trying to say is that at over the last few years talking to various people getting feedback from the students we have some kind of a reasonable strategy in fact almost some kind of classification as to how to solve those problems and what I'm going to do today is share those ideas with you of course as an engineering mechanics sharing of ideas is essentially done by solving problems so we'll solve problems share ideas okay and towards the end of the class if you have some additional ideas or feedbacks you can come forward okay and discuss them okay with the entire community only thing is that a request is only questions in the when I'm talking okay when I'm done okay then you can come forward and give your suggestion okay this is just in the interest because we have limited time okay and everyone wants to be on board okay so if you have any suggestions only towards the end of the class any questions always welcome even in the middle okay so with this preamble let me briefly discuss what is friction okay so what is friction simply is what we know is if you have two surfaces that are contacted each other because by definition friction only means that there is one surface there is another surface can come in from any source and when those two surface try to move past each other okay or actually move past each other so note the choice of my words okay they try to move just trying to move or they are fully moving okay so these are two different categories and as a few moments pass we will see that what those categories mean but what happens is that at the point of contact even on day one okay somebody had asked me a question okay you had asked me a question that if you have a point and a cylinder in contact with each other then what is the line of action and we saw that if there is no friction then it is normal to the surface of the cylinder but if there is friction in small forces that we have been looking at till now what we also have friction force is a bandwidth to the surface now this type of friction okay is called as braille friction okay the what we are going to discuss so while the friction can look under an entire branch called as tribology okay which is dedicated to understand how surfaces interact with each other okay and make more and more detailed studies but what we are going to do in this class in this lecture is a very very simple version of this friction which is called as coulombic friction or dry friction now in this case what we see is that that there are two types of friction static friction and kinetic friction what that is is let me briefly discuss here okay I will not take too long okay within five minutes we will move on to the problems which is essentially of interest to all of us so force of frictions okay so just look at this block on this block suppose there is a weight acting on the the block has some weight which will produce a normal reaction from the surface in addition suppose we are trying to pull the block with an applied force FF now what we see is that just look at this this is force of friction this is force R if applied force is 0 then if we draw the free body diagram of this block you will immediately see that the resultant horizontal force is also 0 clearly now you start increasing F applied and you will see that to balance or to keep this block in equilibrium F will be equal to F applied till some certain value and when it reaches this value which is given by mu s mu is the mu s is the coefficient of static friction s clearly for static n is the normal reaction which in this case will be simply weight and only at that point what we say is that that the friction bearing capacity is up to the maximum that the surface can provide you a reaction okay and horizontal reaction to balance the applied force but beyond that applied force okay the six the surface says that okay I'm done this is the max I can do and the body goes into an impending motion impending motion when I say impending motion it means the body just tries to move or there is just a bare minimum relative sliding between the two surfaces and now we know okay then you push even more then actually what is seen is that that the body cannot take too much force okay when the body actually starts to slide not just impending again I use the word impending to mean that the body is just about to slide the body moves into actual sliding and in that case the law of dry friction again let me emphasize that this is not a universal law like Newton's law there are many many many exceptions for this law this is a simple model and what is seen is that that there are many engineering problems for which this model works reasonably well and that's the only reason that we solve many problems of this type otherwise nothing okay this is not a universal law but as far as this course is concerned we can say that there is an ample quantity of engineering problems where these properties are reasonably obeyed so force matches up with the applied force maximum force mu s times n and is a normal reaction then there is a dip if you apply if you try to push even more and then if there's a relative there is a finite sliding what is a finite sliding that the body is actually relatively moving with respect to the second body then the force acting on it is mu k is a kinetic friction times n and you may ask me this is a topic of dynamics that if I apply larger force which is more than mu k n then what happens what happens is that the body accelerates okay that's the only thing that happens but let's not worry about that for the time being what it just means is that if the body is moving at a finite velocity a constant velocity then the force no matter what the force of kinetic friction acting on it will be equal to mu k times n note one thing that by and large okay again this is not a law of nature but for many engineering problems it has been seen that mu s is less than mu k sorry mu s is more than mu k pardon me that a kinetic friction coefficient is typically smaller than the static friction coefficient do you have any practical example okay I will tell you one very funny example that how do we know or where does this manifest itself that mu k is less than mu s bicycle okay the point is with a very good example that if you want to start the bicycle the initial rush for example you need to apply larger load okay and once it starts moving it is easier to keep it in that place okay that's one thing the second very funny example yes what has to apply cut horse pulling the card yeah perfectly the same thing okay so horse is a physical in a cycle you are doing the function of the horse okay you're perfectly right okay it is what is happening now slipping down from manana okay but there is a kinetic static I don't know okay I think the time period of that is so short okay I don't know if anybody has made an experiment to figure out that is actually the impending friction okay is more than the static which already happens so suddenly let's hope it doesn't happen to anyone yes yes so that's the example I was talking about so now it is okay all of us are distinguished people we are pampered by the government of India we go by flights okay but in our olden days if you remember traveling back those very bad coaches you will see that the fan you put on the switch fan doesn't get on so what do you do okay males they take their comb and start rotating it and the point is that the moment it starts rotating okay then after it reaches some speed it suddenly starts moving it's almost like a miracle okay but now we know okay not now you also know but now we are also expressing that it is not a miracle it is just that kinetic friction is less than static friction okay so with this preamble okay so let us move a little bit forward so these are yes please in the static friction so that reason now for example like it's it's very difficult because friction I told you that tribology is this topic we study friction one simple example what people a very hand-waving reason is that initially there is some interlocking okay for example there is a if you see but again let me emphasize that this is a very very crude example if you look at this you will see that there are asperities and there is an interlocking of the asperities so the initial impending friction is to break those asperities but when you start moving then the point is that the asperities are broken and a different scale of asperities are now coming inside okay means this is a larger scale of asperities which you break and then while moving there will be another set of asperities okay finer set okay which will intermingle with inter mesh with the other finer scale of asperities okay again I'm no expert on tribology okay okay let me put I don't know but what I know is for example there is one thing I know for sure this was a paper in nature a few years ago that we use this term called smooth for surfaces and say surface is smooth no friction but actually it has been seen that if you make the surface super smooth at a nano scale then there are the so-called van der Waals forces which come into play and they actually increase the friction and the friction for example in this simple kinetic friction we see that friction is independent of velocity it just depends on the normal reaction but it has been seen at many nano scale experiments that the friction is proportional to velocity with some power law so there are all these things happen it's a very deep field okay and I maybe know maybe one by millionth of that field okay but a point is that for engineering problem see engineers for example we don't really care what is happening we just want some working law we use our safety factors of two okay so everything is good but what we need is some ballpark idea of what things are happening and for most engineering problems for the scales of interest this works reasonably well okay but if you are interested okay we can talk okay you can teach me or we can look upon various sources and see what this is okay but one simple answer that detail mechanism I don't know which are having the relative motion yes and cannot make the component like a highly finished so highly highly finished so while manufacturing the component so peaks and valleys are there yeah there is some locking interlocking between the peaks and valleys that's what when we when we pull the object so that interlocking has been broken out so once it has been broken out so it will move along the direction of the applied force okay if this if this applied force exceed the limited condition then it will continue along the applied for directions okay so that's what we are discussing but okay thanks for repeating it again okay and so let us okay so let us let us not go into this direction okay what I am saying this we are we say again the course we are doing is engineering mechanics we are not doing tribology so let us not get into details of that okay if you have any things again we will come at the end of the class and discuss okay I'm not saying I'm not saying that we should not do this do it at the end of the class okay these are some okay but thanks for pointing it out will be more yeah no no so again again I'm saying okay I think I made it amply clear should I say in other language also we will not go into details of this okay I don't see who that person is but okay who that is okay universal message we are not going to details of micro mechanics of friction okay that's not the goal of this course yes so that is viscosity is another type of friction and the simplest viscosity for example if you look at the dampers dole dampers what we know is that in dole dampers or if you have done a basic course and dynamics we say that the resisting force is proportional to velocity that is a major difference whereas in this dry friction the force is independent of velocity if there's a finite velocity then the force is given by strictly speaking mu k velocity divided by mod of velocity so it is only the direction of velocity that is important the magnitude is not important whereas in viscous friction it is a magnitude c times v is the actual friction force okay so I think we will move on to the problem so there are many problems that we have to solve so this is what for example one point which I want to emphasize again that just because there is a friction there is a possibility of friction or there's a friction coefficient between two surfaces doesn't always mean that a friction will act okay this for example on this block if the horizontal force is zero then friction force is also zero even though the two surfaces are not smooth so friction what I want to say is that that friction is a reactive force that you apply some force and friction is a reaction to that force and that reaction will only be invoked as and when it is necessary okay so one of the common mistakes which I have seen which happens okay not that this is what we emphasize to the students that it doesn't mean that if there's a weight acting then the frictional force is always mu times w that's not true okay if there's a horizontal force then you will immediately say that if px is less than fm what is fm mu s times n okay then no motion and impending motion only when px is a horizontal force is equal to fm okay the maximum friction force which is mu s times n okay another very important point that just because there is a friction okay doesn't mean that it needs to slide just because I have like a lot of money okay I don't have but if I have a lot of money I just mean that I need to go out and spend it okay so it just as per the requirement it's a reactive force okay it acts depending on what is the applied forces on the system okay and when there's a finite motion impending motion is about to slide finite motion is that it is actually finitely sliding okay then friction is equal to okay f times mu k times k and if px is more than that friction then the body will accelerate and not stay in a constant velocity that's the only thing there is one very important concept which I want to say is very important because we will see that in many problems by just turning the argument around and representing the friction in this way problems become very simple okay even initially for example when I had just started like just learning this course teaching this course in the very beginning what I really let's see what why do you take this extra representation okay so what is this representation before I say that why we do this what is that representation that at the point of contact there will be a normal reaction there will be a friction and as we had discussed in the first few classes that and we all know that the simple parallelogram law will tell that the resultant will be R in some direction and let's say that the resultant reaction R makes an angle phi with respect to the normal reaction this is a normal this is R so this has to be phi and what we know is that immediately that then this friction force okay is simply equal to okay simply equal to this reaction times sin phi and normal reaction is simply equal to n is equal to R times cos phi and what we know is because the ratio of friction and normal reaction cannot become more than mu s this angle of friction maximum value will be such that tan of phi s is equal to mu s okay that is phi will have a maximum value of phi s which is given by tan inverse of mu s very simple point okay and only when the motion is impending will this phi become equal to phi s okay so even I used to have many misgivings that okay why this representation okay you just say friction in the horizontal friction force which is in a tangential direction normal force in the normal direction why this representation we will see that there are many problems that if we have a representation like this okay you can drastically simplify the problem so far so good okay so we have essentially finished our theory we move on to the problems but any questions before we move on to solving the problems actually yeah fine let us move on in this case the point of application of the normal reaction is shifted according to the figure so why it's so yeah it's because a good question the point of application of the normal reaction in a block typically see as opposed to a cylinder which has a perfect point of contact so the reaction can only act to that point but if you have an extended object like a block as shown here then the normal effective reaction the statically equivalent reaction can have any action and the action the line of action will be such okay it will adjust in such a way that the torque or the moment which is created by all these forces get ultimately balanced so the line of reaction can shift depending on what is a height at which you apply we'll solve a problem depending on how much is the magnitude the line of reaction can clearly shift so that's why for example when you are solving equilibrium of just 2d of bodies which I have a finite contact not a point contact we don't write the three equations of equilibrium we write only two equations of equilibrium because third equation will only give you the line of action we'll discuss many such problems here okay it's good question that the line of contact will change why because depending on the applied force the line of action of the force the moment has to be balanced and the line of reaction will shift okay in order to balance that moment fine yeah okay some jokes okay time later on okay so what we'll do is that the friction problems okay by and large okay again say this is not for example a taxonomical categorization okay it's not a hard and fast that like this will only go into this section other will only go into this section but typically that friction problems we can classify it into some simple cases okay we have some three or four classification okay their empirical classification you may either add or subtract from them at the end of this class so after like discussing with many students our colleagues reading beer and Johnston we we realize okay that typically we can invoke laws of friction in different ways so problem type one okay let us call it one for the lack of any fancy term it's like a usual equilibrium problem whatever we have done in the class till now so we have to just solve those problems okay like we have done all these problems okay within view without even thinking about friction and then what we do is the problem statement is such that you will be asked that verify if a particular surface is capable of handling that load what does that mean is that at that particular surface the friction divided a normal reaction should be less than or equal to mu okay or find out the minimum coefficient of friction required or given the slipping of occurs at any surface find the friction coefficient so in these problems the important thing is that that a friction law okay that any invocation of the Coulomb's friction law that the impending motion implies that the friction is equal to mu times n needs to be invoked at the very end till then the problem is like any normal problem what we have done in the class two days ago and we'll do that okay with a simple example so what we have in this example is these are what are called as friction tongs okay so in this friction tongs what is being done is that suppose there is a there is a block which is resting on the ground and you want to lift it up okay using the tongs suppose for example this is a hot metal block you don't want to touch it so these tongs are used to lift up that block now how does this mechanism work the mechanism works is that that you bring this tongs from your house okay if you have it okay of course to bring it there will be some weight for that tongue okay now you go and put these jaws at point D and D prime okay now on the top try to lift it because of the self-weight of the tongue this jaws will try to close and when they close they exert a normal reaction now when you try to lift it the normal reaction will increase friction will increase and when you try it lifting more and more okay it's a self-feeding mechanism where the normal reaction keeps increasing friction keeps increasing till you lift that block up completely from the ground okay so that's the overall mechanism it's a self-locking mechanism that you just put that in the weight will do the initial job keep lifting lifting lifting okay it will clamp on it and ultimately lift that block now what we are asked in this problem that all the dimensions are given to us okay all the dimensions of the problems are given to us all the weights okay this 750 is much larger than the supposed weight of the tongs and what we are asked to do is we are asked to find out that at there are two contact surfaces of the tongs with this block at D and D prime and we are asked to find out that's what is the smallest allowable value of coefficient of static friction between the casting this and the blocks D and D prime okay that what is the minimum coefficient of friction that is required here and at this point such that we can lift it off without the block falling down problem statement clear the mechanism approximate mechanism clear okay now tell me how will you solve this problem can we solve this problem first of all what we need essentially what we are asked we are asked to find out what is the minimum coefficient of friction at D and D prime what does that mean that at the point of contact f the friction force divided by the normal reaction generated okay whatever those values are Unka ratio will be the minimum coefficient of friction now the question we ask is that that without invoking the friction lock can we get those quantities beforehand just think about it can be or can't be if you live this if you live this block what will be the tension in this string if you're lifting this by a string at the top what will be the force you need to exert on the string equal to weight equal to weight and for moment equilibrium that weight and the string should exactly be in the same line like we did the problem a couple of days ago so one question is solved now if I know what is a vertical force acting at joint a what are BA and AB BA and AB prime what kind of members are they they are two force members can we find out what is a force in those two members immediately we can take joint a okay great you also got the answer okay so we immediately know that what are the forces in them by taking joint equilibrium of a in the vertical direction by symmetry or by equilibrium in the horizontal direction both the both the forces are safe now we get what is a force at B we also know the direction now how do we get what is a normal reaction at D and what is a friction at D can you tell me what will be the friction at D 375 perfect so how do we get that 375 if you take the equilibrium of the small block here then by symmetry the friction force which is acting upwards now friction force has to act upward here will be equal to how much this the 750 divided by 2 so we also get a friction force here now what we need is a normal reaction how do we get a normal reaction we will take just this free body diagram say BCD okay yeah and take moment about see will immediately get the normal reaction so we have not used friction till now at all what we have done is that we have just simple done simple 2d equilibrium and only at the end we get the friction force also which is 375 we can find out the normal reaction and what we say is that the minimum coefficient of friction will be just f by n and any friction coefficient which is higher than that the tongs will be functional point clear so nowhere are we using these are the type of problems where friction comes at the very end is the idea clear fine okay so this is a solution you can solve it and this is a coefficient of friction comes out to be 0.19 all the procedure exactly as many of you suggested let us do this problem okay suppose in this problem what is given is that a bicyclist pedals the bike up a slope so there is a small slope which is given to you 5 by 100 gradient the bicyclist is moving at a constant speed the speed is constant and what we are given is the combined weight of the bicyclist and the cycle it passes through point g whose coordinate is also given it's 460 perpendicular along the direction of the slope the wheel dimensions are given the spacing between the center of the wheels is also given and what we are asked to do here that if the rear wheel this wheel at the back is on the verge of slipping determine the coefficient of friction mu s between the rear tire and the road that is the first part of the question the second part of the question is is the coefficient of friction is doubled what will be the force acting on the rear wheel okay now can we so now the only difference in this problem is that that the bicyclist is moving at a constant speed but what we know is that even if an object is moving at a constant speed okay all the forces all the torque should be balanced on that okay we know that also okay because we just change our reference frame okay and we see that a bike and we will see that a bicyclist is stationary in that inertial frame so all the laws which we have used till now are equally valid now let us discuss that how do we solve this problem how do we get that coefficient of friction how many unknowns do we have and can we get all those unknowns on the front wheel how many unknowns will we have traction force means driving force okay sir driving force is not needed yes so what is the conclusion rolling only excuse me it is under pure rolling that is fine but my question is simple my question is that is there a friction force on the front wheel or not yes definitely sir okay then somebody said no I want to hear from that person somebody said no right from this side I heard no friction is under rear wheel but can you tell me the justification that while the front wheel there will be no friction I agree with you okay in fact I have to tell that I agree with this person here impending case the motion is impending in the rear wheel only suppose if it is not impending you have some something to that I want to hear that logic I could not disturb it clearly sir it is rolling it is wait a second you think it is like that okay that's your okay now now you tell somebody interrupt it is rolling sir actually to roll a body there must be a torque and for that this T which is torque it must be balanced by the force so there must be friction over the rear wheel okay one last okay then I will move on to the what my version of the events is without friction it cannot roll okay then I will move on to my version sir the wheel is yes so you think will there be a friction on the front wheel or not no okay so the idea is this if we assume that the axles are well lubricated again as usual okay if the axles are rusty then all bits are off but the axles are well lubricated and note that this entire bike is moving at a constant speed which also implies that if there is which also implies that the wheels okay are also moving at a constant angular velocity okay now if you draw the free body diagram of the front wheel as many of you pointed out that there is no driving torque front wheel is a free wheeling it's a free wheeling cylinder because of which what is happening is that and there is no because the axle is well lubricated as we saw in the case of pulley there can be only two reactions horizontal reaction and vertical reaction but no torque if it is not well lubricated then again this is wrong but assuming it is well lubricated okay which we hope we keep our bicycles then there is no torque here then if we assume that there is some friction which is acting here then all these three forces they meet at the center so the friction will create a torque at the center which means that the wheel cannot rotate at a constant angular velocity which is contradicting the given statement of the problem which means that for example for that particular state where everything is moving at a constant speed there cannot be torque or there cannot be friction on the front wheel because it will create a torque at the center and it contradicts what we are seeing okay or what is given to us in the problem so if the condition is maintained like this then there is no friction force on the front wheel no no but then no no no because we are doing an approximation here that it's a point friction but what we see clearly is that the wheel is not a point contact it's a very simple approximation wheel if you see it always flattens out it is not really remaining a point okay so then you have to use different ideas in that but it's a simple course to the first order okay to the first correction to the most basic degree of understanding okay the friction fields force will not be as dominant on the front wheel as on the back wheel let me put it this way yes of course there will be because the wheel will not have a point contact those are called a rolling friction yeah now okay so front wheel doesn't have the front wheel only has a normal reaction now at the back wheel there can be a friction why there can be a friction because the chain is connected to a wheel at the back okay I don't recall the name for that and we are okay yeah good okay so that is connected to the rear wheel and when we pedal we are exerting a torque and the rear wheel and that torque has to be balanced by the friction that is provided at the rear wheel so there has to be friction at the rear wheel and now if you take the full free body diagram then what you realize is that that is the biker is moving up there is a slope so there will be a component of the reaction along the plane so that friction will balance the component of the of the weight along the slope so what we see here is that so essentially that is also telling us that when we are biking we are creating that friction by the pedaling and that friction in turn is essentially making sure that for example it is balancing the the reaction of the weight along the direction of the slope now essentially what do we have we just have three reactions three unknowns we can immediately find out what is the normal reaction by taking the torque for the full cycle about point a how do we know RBX by balancing the react forces along the slope so we know RBX we know RBY and we take the ratio of them at the very end of the problem okay and we say that because it is given that the wheel is on the verge of slipping that ratio should be equal to the coefficient of friction between the rear wheel and the surface and now this entire solution procedure immediately makes it clear that if you double the friction what happens anything changes if you double the friction now does anything in the problem changes only change that will happen is that now the wheel will not have impending slippage that is the only thing that will happen okay so now these are the point of problems where for example what we do is we keep track of how many unknowns are there in the system how many equations we can have from the number of free body diagram and if they to match then only at the end for example will be asked that is the motion impending what is the minimum friction and so on so these are essentially simple problems like what we had done earlier only friction is invoked at the very end of solving the problem is the overall idea clear fine okay because what happens is that typically I have seen that for example that in many students there is a quick tendency that friction okay suddenly just put everywhere mu s times n okay so that's not the idea that we just have to figure out what is happening what are the unknowns acting at every point can we get all those unknowns without invoking the friction law we have we should invoke the friction that mu s times n only at the very end if we can help it that is the experience because otherwise if you just put it somewhere then there is no way to track that because if you put all those arrows put mu s times n and at the end okay you realize okay that it was not sliding there but then all the job is done you will completely lose track of the bookkeeping and the problem will be a mess so as far as possible one of the rules of thumb is that you try to invoke the friction law only as late as possible okay so that is one of the observations so can we move on to the next type or the next class of problems okay very simple problem here I have to this problem is a very simple problem is there for a reason okay that for example in many friction problems for example you may be asked that what should be the maximum what should be the inclination what is the maximum value of x so as is given in this problem so what we are asked here is we have a vertical mass which is guided by rollers in this slot and we want to keep this mass in equilibrium by putting a stick AB such that there is a friction between B and the surface and the same and friction between A and the surface what is given is that the coefficient of static friction is 0.3 at the upper end of the bar and 0.4 at the lower end of the bar okay find the friction force acting on each end for x equal to 75 millimeter also find the maximum value of x for which the bar will not slip so two parts of the question so can you tell me how do we solve this problem is the problem statement clear first problem is that that a coefficient of friction is 0.3 at the upper end of the bar and 0.4 at the lower end of the bar and if this x value okay the length is 300 this x is horizontal this is given okay find the friction force acting on each end of the bar at B and at A and second part is find the maximum value of x for which the bar will not slip how do we solve this problem if we neglect the weight of this bar AB okay suppose because this is 60 kg hopefully the weight of this will be far low okay what is AB it's a two-force member what will be the direction along which the reaction the total reaction at A will act along AB yes two-force member it's a two-force member okay what about the reaction at B it will be along BA so it has to be a compressive force clearly in this case so the reactions will act like this okay now what do we know about the normal reaction that is provided at B no at point B what will be the vertical reaction 16 to 15 to 15 to G or just you can say 50 kg if you want to stick with kg units so 50 G or if take G equal to 10 it will be 500 Newton will be the force which is normal reaction which will act at B now can you immediately find out what is friction at what is the friction force here just no no no no no no no no no that's what I'm saying 0.3 is the value of friction but in this case we know that is a two-force member we also know that for that two-force member at point B what is the vertical component of the free of the force is equal to 50 into 10 or 500 kilo Newton and if you know the direction then immediately we can find out what is the horizontal force yes we don't need to invoke friction here you see the point okay most of you but if not okay please see that point you can't say 0.3 into that but what you do now is you find out the value of friction okay how do you know that it will slip or not slip is it's the entire assembly stable or not stable 500 sin theta that's fine where theta is this no not find it find a finite tan theta okay no no what are we talking here now member force is fine but we are asked to find a friction force okay fine so you do that but now my question to you is this that you can find out the friction force why because this force line of action is along BA normal is 500 Newton so friction force will be just this so the ratio of friction by normal will be just equal to how much the angle this angle is how much this angle will be 75 divided by 300 sin inverse of that okay so tan so 500 into tan of that angle will be the friction force fine great but now what I'm asking is okay good what I'm asking is that keep we also have to verify that the for that system to be in equilibrium it should provide that much force but the coefficient of frictions are given if the coefficient of friction is 0 at point B this assembly is not stable right this assembly is not stable but on the other hand we are given that a coefficient of static friction is 0.3 at the upper end 0.4 at the lower end so what should we how what should we compare AB if I if I just draw it here let me just go here this is the reaction that is provided okay this alpha or this alpha cannot be more than the phi max or phi s which is tan inverse of mu at this contact and similarly for this contact see the point yes no just tell me yes or no fine so then we just verify that if that angle this resulting angle okay if this angle because the minimum friction now here is what 0.3 and the angle is same both at A and B so we have to verify that this angle is less than tan inverse of 0.3 if it is less you are good the assembly is stable if not the assembly is unstable and the second part of the question is that find the maximum value of x for which the bar will not slip so what do we do in that case we see that if we increase this angle then x will keep increasing but we cannot increase this angle beyond tan inverse of 0.3 so we take this angle to be tan inverse of 0.3 and x will be equal to 300 sign of that angle okay you immediately get what is the value so what we have done is that strictly speaking this system you don't need to worry about a friction problem only at the end to check the stability of the assembly okay we need to invoke friction saying that at all the forces we could obtain all the forces but we just need to verify that the forces are such that the angle of friction is less than 5s or f divided by n is less than us both are equivalent this idea fine okay now we come to another interesting category of problem that in these type of problems what we know is that the number of equations okay if we write all the equations for the free body diagrams is less than the number of unknowns in the system okay now in this case this problem is a statically indeterminate problem so the problems in these category are defined in such a way that you are asked to find out that what is the range of force that I should apply since the entire assembly is in equilibrium or what should be the range of angles or what should be the range of distances over which the system should be spread such that the entire assembly is in equilibrium okay so now these category of problems okay we have to be very careful okay because now we need to invoke the law of friction mu s times n very early as compared to the category one problems where the friction has to be invoked only at the very end otherwise it was like a 2d equilibrium problem okay so the idea is always to keep track of the number of unknowns that are present in the system and the number of equations that are present in the system so let us move on to this problem okay what we have in this case is 2 inclined blocks okay 15 degree angle block a block b the coefficient of static friction between this is 0.4 static friction 0.4 between this now these two blocks are connected by a string a b and look carefully what is the definition of this problem the definition of this problem states that what is the force p that you should apply to this system for which the motion of the 30 kg block okay this block is impending or sometimes you may rephrase the problem in a different way that what is a minimum pulling force you should apply to this block so that this entire assembly is no longer in equilibrium okay you can rephrase it in another way now note that in this case how many unknowns are present 1 2 string there are 2 unknown reactions here 3 4 2 unknown reactions here okay so 4 5 6 so there are 6 unknowns and how many equations of equilibrium we have we have 4 why because as one gentleman asked us before that there is a line of action which is also an unknown so we don't take care of that so because it's an extended contact node is an extended contact so the line of the effective force is also an unknown but we don't worry about that why because for this problem for finding p is concerned that line of action is immaterial okay what we assume is that the body is not toppling and that reaction line of reaction will come somewhere we only want to find out that what are the forces so as far as the top block is concerned we have 2 equations bottom block we have 2 equations so we have 4 equations and 6 unknowns so we are in trouble 4 equations 6 unknowns we need 2 extra equations and those 2 extra equations are going to come from impending friction where we are going to invoke that a friction force is given go is equal to mu s times n okay then if you invoke those 2 equations then the number of equations number of unknowns will become the same but now how do we invoke them okay suppose this string were not present okay if the string were not present then what happens is that assembly in equilibrium no string present no p present just 2 blocks lying one on top of each other is that entire assembly in equilibrium why not may or may not be that's a very good question okay that's a very good reply may or may not be okay how do you verify that it's in equilibrium or not in equilibrium that's a very good point okay may or may not be but how do we decide that it is not or yes angle of angle so what we need to know is that tan 15 okay should be should be less than mu s in this case it will be true tan of 15 degrees you can quickly type in a calculator you will see that it is less than 0.4 so just those 2 are in equilibrium okay now let us come back to our original problem okay let's say that if there is a string that is attached now how do we create an impending motion for this block now note that's the duty of this kinematics and dynamics if there is an impending motion of this block B in the right direction okay what happens to block A will it just remain there or will it also move it will move up or down up or down because now what is happening that this string is of constant length when you pull this block in this side there is extra string that is pulled from the top so block A will move up block me will move sideways so what is happening is that that impending motion of block A is immediately causing impending motion of block B so it's a consistent that for example we needed two extra equations and the kinematics of the problem is telling us that when block B is impending block A is impending sorry when block B is impending A also becomes impending which is automatic now the only thing that you need to do is to figure out what are the directions of friction forces which are now very important which were to invoke very at the very beginning not at the very end as we did before now if the motion of block B is impending okay in the right direction what is that what is the direction of friction of acting on block B in the opposite direction okay right left okay okay opposite okay because like I see from this side okay so the impending motion of block B is this way the friction will act in the opposite direction okay on one surface now at surface A now here in these problems you need to be very careful that the kinematics of the problem should dictate what is the sign okay as opposed to for example the other problems in equilibrium where sign you take whatever and you are still good because automatically the negative sign will come but here there is a interplay between the kinematics and the forces so because of that we have to be very careful about what the sign is so if the block B moves upwards there is a relative sliding like this so what should be the direction of friction acting on block A downwards downwards and and the friction which is what block A acts on block B will be again how much upwards equal and opposite that's it you are done so now the number of unknowns are normal reaction friction is mu s times n normal reaction friction of mu s times n just draw the appropriate free body diagrams and you will immediately find out okay what is p or for example a solution procedure is briefly given here okay you can just follow that solution procedure or you can think about it that what is the easiest thing to do is as far as the top block is concerned you can have a x and y axis which is along the direction of the plane as far as the block 2 is concerned you can have x and y axis which are in the normal direction okay like horizontal and vertical just solve these two and we think like few lines you can find out what is the value of p is the procedure clear see how beautiful it is that you track the number of equations number of unknowns and the kinematics of the problem will also conspire in such a way that whatever your mathematical requirements will automatically be satisfied by the kinematics that you needed two extra sleepages to solve this problem and kinematics guaranteed that you got two slippages okay so is this fine so now note that this problem is very different and the problem we solved a few moments ago where we had to invoke friction at the very end and in fact even if you screw the direction of friction you can automatically adjust it afterwards in those problems but here you cannot screw the direction of friction the physical meaning of the problem just completely changes and the direction of frictions will be purely governed by how the impending motions can be seen is this point clear okay any doubts any questions fine okay so this is category type 2 problem okay one we will solve one more problem of this type in the tutorial okay we will you will solve it and I will just super wise problem number 3 is again the rule of thumb and the most important rule of thumb okay is that it's these type of problems are very similar to problem type 2 okay very similar but what is different is that the total number of extra equations that you need means for example you get a certain number of equations certain number of free bodies which will provide you a certain number of equations there will be a certain number of unknowns and if those unknowns in this case also will be more than the number of equations so you need to invoke friction but what will happen in these kind of problems is that the contact surfaces will be more than the number of slippages that you require so because of that there can be different ways in which the body can go into non equilibrium or for example it can lose this mechanical equilibrium so we have to now decide that for these kind of problems where will the slippage occur okay you cannot just immediately say that a slippage I can set all contact points okay and one type of problem okay we'll discuss here so what we have here okay and whatever I said if it was not already clear will become clear when we solve this problem so what we have here is a block AB there is a coefficient of friction mu s at the point of contact between A and B note that this is a point contact so there is a small cylinder so there is a point contact here or a contact or a very small area so we know immediately what is the line of action of the force okay that point of action of the force it is here as opposed to this problem where this line or this point of action is not clear in this case we immediately know that the force friction or normal will only act at point B now this AB bar has a roller which is pinned at its end A and it is leaning against a mass M0 which is in contact with the vertical okay which with the vertical wall and what we are asked here is this we are asked in this problem that this mass is given okay what should be the range of theta so this angle theta that what should be the range of theta such that this entire assembly is in equilibrium okay so is the problem clear that everything is given and you are asked to find out that those theta what is the range of those such that this entire assembly remains in equilibrium problem clear or any ambiguity about the problem statement yes okay so I take it that the problem is clear so can you tell first of all before doing anything let us count the number of equations and number of unknowns for this vertical mass how many unknowns to normal reaction friction point of contact roller 3 point B 45 to unknowns friction normal reaction and angle theta so there are six unknowns in this problem how many free body diagrams to free body diagrams how many equations I can write for this for this block note extended contact line of action point I should not know so only 2 we can write only 2 equations for this and 3 for this so we have 6 unknowns and we have 5 equations so we need one extra equation and that extra equation has to come from impending slippage at some point but how many surfaces do we have now this is a frictionless surface you don't care here we have 1 and 2 we have 2 surfaces okay we have 2 surfaces but how many unknowns how many extra equations we need mathematically just 1 so the most bone headed we are doing is immediately say that there is a friction is equal to mu times n mu times n but then you will have a over determined problem so what we need is we need slip only at one surface now does that give you an inkling of why do we need a range of theta okay that a friction okay the slippage we need only at one surface but we have 2 now how do we decide okay then we come to the statement of the problem what we are asked is that what is a range of theta for which this assembly is in equilibrium so think about it like this if theta is extremely small what happens let's say theta is very small the rod is almost vertical let's say theta is 0 in the extreme case okay we can always use extremes because that gives us insight if theta is 0 what happens to the assembly there is no normal reaction that will act on this mass okay but there is a weight so friction is taking care of the weight so the top mass will topple down take second example where this AB is very horizontal okay it's like it's a this angle is this angle is very large okay it will fall down it cannot take this friction cannot be taken care of okay this friction cannot be taken care of so it will fall down okay why because a normal reaction will keep increasing with this angle okay and the friction will not be capable of taking care of that so the rod will slip down so what we know immediately from this simple analysis that very large small theta top mass will slip large theta the rod will slide so what do we want the range is required such that even this does not come down the rod does not slide so that is the range but that also immediately tells us that for small theta okay the the boundary line of equilibrium okay the boundary line that the system is in stable configuration is when the motion is impending anything beyond that the system will collapse okay so what do we do first we say that when theta is small then the we find out that for what theta we get impending motion here and then we increase theta and say that what theta we get the impending motion here and those two different mechanism that in one mechanism the system fails by this in second mechanism the system fails by this and we want this limit where the system stays in equilibrium so we look for one range here where the system is in impending motion here the second limit of theta where the system is in impending motion at the end is the logic clear okay so we don't for example it's again a problem where the number of equations is less than the number of unknowns but still we cannot directly invoke friction now at all surfaces we have to figure out that where is that slippage indeed going to happen and we use all manner of all possible logic at our disposal to figure that out okay now if you just quickly is the idea clear okay that why in this case and in this case we get a range it's a range of theta okay what is what what is happening here and the mechanism is that one theta this falls down other theta this slips out and a simple way to do it is you can take the top free body diagram normal reaction this f1 should be equal to m0g for equilibrium of this mass and n1 should be equal to ra take the second rod free body diagram what we need to know what we know is by taking torque about point B or moment about point B we immediately find out ra which will equal to mg by 2 and theta and as we are rightly suspected that when theta increases ra increases okay and n2 will be equal to mg so this is the f2 will be equal to ra so friction force is equal to this normal reaction okay is equal to mg and in the first case okay when theta is a reasonably small value we want slippage here so we immediately know that the first case of slippage is such that f1 by n1 okay so for the entire assembly to be in equilibrium we want f1 by n1 to be less than or equal to mu s from that okay this is the expression we get for f1 by n1 we can find out what is the lowest limit on tan theta or on theta from the second ratio f2 by n2 okay this is the f2 by n2 ratio which will come out to be this quantity we will get what is the largest limit on theta and your real theta should be within these limits that when your theta is between these two values your system will stay if your theta becomes smaller than this value the top block will slide if your theta become larger than this value this will slide at be is the idea clear and then you ask yourself a question okay suppose okay in a hypothetical scenario where this angle is actually more than this angle means in other words if this quantity look here this quantity is more than this quantity then what will happen it is a question for you if for example if this quantity is less than 2 mu s then we can get a nice limit on theta where the system is in equilibrium but if so happens that the dimensions of the system are such the weights of the components are such that this quantity becomes more than or equal to mu s it will never be in equilibrium this will either slip here or slip there okay you can never maintain both surfaces in equilibrium so is the logic clear for this problem like these type of problems the category type 3 or it's a category we have created you can always make it ABC or whatever whatever suits you okay but this is just for our understanding so that we can break it into various parts and then decide that what is the logic we are going to use to attack that particular problem