 So, now so far see what we have discussed you know since friction is covered see usually we say that if there is no friction we are taking a pulley if it is a friction less then tension in one side is equals to the tension in other side right that has to be there because we want to balance the moment about the rotational axis right. So, now this belt friction application what we are going to see actually is how the tension will differ on the both sides if we take into account the friction on the body where it is going to be roping ok. So, many applications are there. So, first example is a capstone or you can think of a you know bullet we call it. For example, you know when we are going to anchor a boat or anchor a ship when it comes to the let us say to the you know to this parking area then the person who is actually going to hold it he has to apply a very small force in order to stabilize the large boats or ships ok. So, tension on the both side will differ in that case similarly we can take a you know example of a pulley that is raising or lowering the loads right. So, I can actually apply a very small force to either hold the you know weight that is on the other side or I can even apply a very large force to really bring that weight up. So, again you can kind of guess that there are two conditions here going on impending motion the weight that is on the other side can actually tend to go down or tend to come up. So, same logic as that of point friction that we have discussed in the morning is going to apply in this case ok. Similarly, we can also think about some cases where the torque is being transmitted. So, we have two pulleys or let us say one is connected to a motor where we are actually going to give the applied torque. So, motor is giving the torque right and that torque is going to be transmitted to a tool that is connected with a with the other pulley. So, that means here actually let us say there is another machine tool that is going to be rotated by the applied torque that is coming into play here right, but in this case the type of problem that we are going to study is always going to be either constant velocity let us say that means uniform motion such that we can apply the statics ok. So, we are not getting into any kind of situation where there will be acceleration or deceleration into the body ok just keep that in mind. Similarly, we have you know break problem band break problem where there is a drum here and then drum is actually connected to again some kind of component and there is a band on top of that that is the belt ok. And let us say this drum is moving at a constant speed which can be attained let us say by applying a certain amount of torque and then we are going to see that when we are you know this belt when we say that there are tensions you know both sides are going to be unequal that has to balance that torque. So, we are actually eventually going to apply certain amount of torque in order to bring the drum into equilibrium in a constant speed. So, think like that way ok. So, in all of these cases we have to first find out that what relationship exist between the tension in one side of the belt with the other if we take into account the friction in these problems ok. So, remember now the point of discussion here would be is that let us say I will just assume that this drum is fixed and there is a belt on the top and we have friction in between. Now, what happens let us say this belt is about to slip to the right that means, that is the impending slip is going to happen. If the impending slip is going to happen which side will have the higher tension right side why because if the belt is trying to slip that way the friction should try to oppose that right. So, there is a opposing force that means, this tension in the right side should be higher than the tension in the left side. So, T 2 should be always greater than T 1 as long as the belt is trying to slip towards the right. So, first we have said that it is a impending condition is already established right ok. Now, how I am going to solve this problem relate this you know T 2 and T 1 is as follows. We have already said in the point friction what is the maximum friction force possible mu s n right that is the maximum friction force during the impending slip condition right. So, I am going to apply the same rule here, but for a differential element that means, for a small element I am going to say that that small element is under impending slip. And then I will simply integrate the effect over the contact angle that contact angle is actually beta ok, because friction is happening everywhere. So, therefore, now it is also assumed if it is a rotating problem let us say this drum is actually rotating along with the belt the speed is reasonably low such that there is no force the centrifugal force is not coming into play which is acting radially ok. So, with this assumption just look at it is not that simple to derive a formula that actually gives me the tension between T 2 and T 1 what relationship exists in terms of what in terms of friction coefficient of friction static coefficient of friction mu s and the contact angle beta this is my governing you know these are the two things again going to govern the entire friction law ok. So, let us look at this differential element right here which is actually at a distance theta and which has an angle of delta theta ok. So, if I assume the tension in this side is T therefore, on the other side it should be T plus delta T ok. Now, we simply take the equilibrium of this remember here I have the normal force delta n and I have applied I have already invoked the friction law here which is mu s times delta n ok. So, if I try to take the equilibrium of this so, I do sum of force along x 0 and sum of force along y 0 ok. Ultimately we are going to get this one when we eliminate the delta n from this. So, ultimate objective will be to just eliminate delta n and get a relationship with delta T T and the other parameter that is delta theta and mu s so on ok. Now, when delta theta goes to 0 then this one straight goes to 1 and this parameter right here that goes to 1 right. So, therefore, ultimately I have delta T by delta theta right negative mu s T remember when delta theta goes to 0 the delta T is also going to go to 0 right. So, therefore, I have d theta d you know delta T divide by delta theta equals to let us say mu s delta T mu s T sorry ok. So, therefore, now we are going to simply apply the separation of variables that means, we are now going to say that d T by T should be equals to mu s d theta ok and therefore, now you integrate it. So, we integrate it from theta 0 to beta and therefore, the integration limit on the other side will be T 1 to T 2. So, T goes from T 1 to T 2 and theta goes from 0 to beta we have made already separation of variables. Therefore, ultimately what I have is T 2 by T 2 by T 1 should be e to the power mu s beta very simple relationship we establish, but that has you know that is going to do unique things for us. So, what is it first of all can I compare it with a point friction can I compare it with a point friction it is see there is no definite maximum force for the belt friction. That means, there is nothing I can call mu s time cell or e to the power mu s times beta, but what I can say that at the impending slip the maximum friction force that is developed is given by the ratio of the torque T 2 by T 1 should be equals to e to the power mu s beta. So, therefore, for impending slip T 2 by T 1 equals to e to the power mu s beta if there is no slip that means, T 2 by T 1 should be less than e to the power mu s beta period. So, ultimately we just take this concept into account and therefore, now interesting facts will come into play we all know the tension in the belt that pulls that should be T 2 and tension in the belt that resets that is T 1 that means, for this belt to slip towards the right T 2 should be greater than T 1. Now, if actually this belt slips over the drum then I cannot say T 2 by T 1 equals to e to the power mu s beta again compare it with the point friction. So, I have to say T 2 by T 1 equals to e to the power mu k beta that is the kinetic coefficient of friction. So, if there is a relative slip actually taking place between the drum and the belt. Not necessarily velocity for the constant speed problem we will try to come to that, but in this case you know do not think about velocity now whatever problem we are going to solve they should be always statically determinate. See in friction case what happens point friction you always create an unbalance there will be imbalance in the system as soon as the barrier is broken the maximum force is broken that means you have a imbalance in the system. Then you can think of that there is a acceleration to the system, but here remember all the problems so far you have seen never we have cast upon that line that there is actually you know imbalance force equals to mass times acceleration we have never said that. So, the component either that we have solved they are statically determinate and that they can be solved by the static equilibrium conditions. So, number of problems can be actually solved flat belt passing over a fixed cylindrical drum ropes or belt wrapped around a post or capstan or pulleys or belt drives as I showed that all of this problem can be solved by this single you know friction law which is now T 2 by T 1 equals to e to the power mu s beta at the during the impending slip that means the belt is just trying to slip over the drum rope or you know drum or whatever it is connected. Now, these problems can be best understood by solving you know a multiple number of problems with different types of contacts. So, again how we solve the friction problem we always try to understand how many number of unknowns I have and how many number of static equilibrium conditions I have how many slippage condition I have to improve at how many surfaces is in that that is that was the basis of solving all the friction problem. Count the number of unknowns figure out how many static equilibrium condition I have therefore, the rest balance amount that means whatever extra unknown I have that many contact surfaces have to have to solve this problem is in that the basis. So, if you think of the force from the force perspective it is very very simple, but if you take it to the kinematic deformation that where actually I have to you know invoke the friction law that is a bit complex right. So, now let us look at this problem. So, this problem says that a chord that is a cable let us say y you know it is wrapped around twice at a and wrapped around three times at b and then it is going to go over a barrel shaped you know thing right here that is the half barrel and supports a mass of 500 kg. The question is what is the tension T required to maintain this load a coefficient of static friction 0.1 for all surfaces of contact. See in this case what happens first of all understand like this I can either attempt to just hold the load that means this can simply try to come down this is trying to come down I am simply trying to hold the load or I want to really pull that load up that means my impending slip condition is upward. So, there is impending slip condition downward impending slip condition upward because remember why I am going to impending slip because that gives me the maximum numbers. So, I have to always assume that I want to break the barrier that barrier is the friction force right ok that is at the contact. Now, think of it how many unknowns I have let us say this is T 1 this is T 2 this is T that means I have three unknowns. So, now there is no equilibrium in this case what it is basically I have to have three contacting surfaces. So, I have to apply the friction law three times to solve this problem and that is true that because indeed I have three contacting surfaces we cannot see equilibrium. So, T 2 by T 1 equals to me to the power mu s beta is not equilibrium condition it is like mu s n in your friction problem it is a friction law ok right. So, now what happens in this case remember if I just want to hold the load in this position I will I will have to apply small force if I have to pull it up I have to apply more force. So, that you know has to be kept in mind and we have to whatever answer we get at the end must follow that logic it is just by logic simple logic ok in one case I am just trying to hold. So, I do not need any much force, but in other case I am just trying to pull it up. So, I need a large amount of force. So, now I can definitely go to this with this you know part that I have said that in this case when the upward impending motion of mass in that case T must be larger than that of the weight and when impending motion is downward that means you are trying to hold the load then T should be less than mg ok right. So, this problem would be simple as long as we can calculate all the contact angles angle of contact needs to be calculated very fast. So, what are the angle of contacts? So, in A it was wrapped around twice. So, it is 4 pi in radian in this case it is 6 pi and for this small half barrel it is actually 0.5 pi over 2 90 degree those are my contact angles. What are the free bodies? First I have a weight right. So, let us say this is T 2 then force will again change at T 1 and this is what I want to find 3 unknowns are there. So, I will repeatedly apply the friction law right in all the contact surfaces to establish the relationship between T 2 and T 1. So, just make sure first I assume that impending motion is downward for this mass. If the impending motion is downward then these logic has to follow right. Then I can say if the impending motion is downward then mg must be T 2 because this one is going to give me the maximum tension right here in the cable right and this will be the smallest ok right. So, therefore, I should say that mg must be greater than T 2 greater than T 1 greater than T. Then I am going to yes any question is that clear because I am just trying to hold the load. Therefore, my weight has to be the maximum tension in the cable right and the load will be the minimum ok that is one case of it right. When I count the impending motion downward because what is the friction direction in this case when it is trying to come down right then it is always going to be upward. So, that means this tension has to be greater than this tension ok. So, like that therefore, what I have done here I have simply applied the friction law. So, first right here I have said mg divided by T 2 must be equals to e to the power 0.1 pi over 2 just for this particular half barrel ok. So, this is now my T 2 actually and this is my T 1 this is larger this is smaller. So, that thing we have to always examine which one gives me the greater tension that is always going to be T 2 in the formula ok. So, what is shown here this T 2 is not actually the T 2 in the formula just think about that because that is not the larger tension ok clear. So, probably I should have put a different notation here instead of T 2 and T 1 we could have taken T 1 and T 2 that would have been much you know useful ok just symbol wise not to confuse. So, now again now in this capstone here I have T 2 by T 1. So, this will be the larger side again this will be the smaller side. So, I have T 2 by T 1 equals to this and right here I have T 1 by T equals to this ok. So, now I have established this relationship therefore, I can simply you know multiply all of these three I will get a relationship between mg and T ok. So, the final answer will come out to be 181 Newton. Now, let us look at impending motion is just upward. So, now it is upward impending motion if it is upward impending motion then what I have the tension has to be maximum here and minimum here and I will go by that rule again. So, now I have to reciprocate everything. So, in other words I will just say that instead of saying T by mg you know if you equals to e to the power negative that I will just say mg by T equals to e to the power minus that one and that will give me the solution. We can go to the same steps logically it is going to you know we can follow the same steps. So, ultimately you will see the final outcome would be that T mg by T is equals to e to the power negative 1.055 ultimately ok. So, with that therefore, I get the tension value. So, look at the tension value right here it is so high. So, that means, to just move this you know 500 Newton load upward I need a very high amount of force whereas, just to hold it I need a very small amount of force clear. So, what is the you know condition? So, what is the range of T that will give me the equilibrium all the time satisfied that will be 181 Newton to let us say 132 kilo Newton. So, that is the range in which my equilibrium is always going to be maintained. If I am outside let us say if I do not apply this force if I do not apply this tension it is suddenly going to downward ok. So, very simple problem let us any question on this clear. So, that you know kind of forms the basis and I am really comparing with now with that point friction that was studied ok. So, in this problem we have three pulleys pulleys e pulleys d and pulleys e. Now, what is being said that two of the pulleys d and e they are locked to prevent rotation. So, they are actually locked they are fixed they cannot rotate however, c is slowly rotated this one is slowly rotated. So, what we want to find out determine the largest weight which can be raised if pulleys rotated of course, it has to be clockwise ok. So, remember what is happening that means, the belt and the pulleys e these are actually going together in this case. If we look at the motion of the pulleys e the belt is actually in always contact in touch right it is just rotating like this. Whereas, since d and e is locked what is happening belt is actually constantly slipping over those two pulleys. They will take mucus grade. So, everyone got it. So, this is the key feature here. So, what is happening if we look at the impending slip at c then that is the max condition I am trying to achieve right. I am just assuming this is my impending slip condition right and here it is going to actually go over the pulleys. So, that means, there is relative slip between the pulleys and the belt ok. So, with that we can quickly draw all the you know free body diagrams that are necessary. What are the free body diagrams necessary? Just look at this w a t 2 these are the two tensions in the cable. So, here I have t 2 t 1 and then you have t 1 16 ok. Now, when I am invoking the friction law at c right. So, I am assuming in order for to find out the largest weight right. That means, it has to simply try to break the barrier here the friction force that is in the contact right. So, I have to invoke the friction law there. So, remember now in this case therefore, w a must be greater than t. So, tension in this cable should be greater than tension in this cable. Why that is happening? By the way can is the equilibrium is violated? If I do so, how the torque is being applied? See I have to always balance the torque that I applied do not forget I am applying a clockwise torque. So, in this problem in part c I have a clockwise torque I have to balance that. What is balancing? w a minus t 2 multiplied by r. So, one is clockwise another is counter clockwise. So, pull is in under equilibrium. We have already said we are moving it a in a very slow speed right. So, in that way it is maintaining the equilibrium over there ok. So, once I do this therefore, w a is my greater tension side. So, w a by t 2 is equals to e to the power mu s. Now, we have to find this contact angle beta c which can also be found out right. What is this contact angle? Just look at here beta c is going to be 2 pi over 3 ok. Now, in the other cases these two pull is they are actually prevented to do rotation right. That means, what is this? This will be kind of not hinged here, but it is kind of fixed here. There is a fixed support it cannot rotate right. So, that means that that since it is fixed it can actually take the moment as a reaction that will balance the torque that is given by the tension in the cable on the both sides ok right. So, therefore, in this case now t 1 is actually greater than t 2 because what is happening in this case? There is a relative slip relative slip of the belt with respect to the pulley. So, which way it is slipping? It is slipping this way which way is the slip? t 1 direction if the slip is in t 1 direction therefore, t 1 has to be greater than that of t 2. That side tension has to be larger is not that and also since there is a relative slip we are going to say that it is e to the power mu k not mu s ok right. So, we get that. So, therefore, t 1 by t 2 equals to e to the power mu k beta d. Similarly, in this case we can also find out which way this is actually coming down right. So, this is coming down therefore, friction force is opposing. So, this has to be my larger side. So, 16 is my larger side now. So, this is t 2 and this is t 1 in turn. So, this is e to the power mu k beta d. Now, this problem is both you have kinetic coefficient of friction as well as static coefficient friction and in every wire equilibrium is maintained no issue. Static equilibrium is maintained each and every wire ok. So, with that we can just solve again from these three equations we can simply solve w a any question. Now, there is no see in these problems there is again as I said in the when I assumed this now when I made the assumption I said that this is always going to be low speed low speed problem in high speed there may be a centrifugal force that will actually try to throw the bolt you know throw the belt away from the pulley. We cannot define like that, but there will be an error still you can try to solve this problem using this approach. There will be only an error in this formula t 2 by t 1 equals to e to the power mu s b. We can solve the same problem by you considering the centrifugal tension also. Yeah I am not saying it cannot be done, but this problem listen this problem the objective is that everything is again under static equilibrium. That is that considering centrifugal tension then and then only it will be in the equilibrium condition also then only it will be. But equilibrium that that type condition will be. That is ok you know you will have a vertical equilibrium will be maintained that you have to figure out that will give you extra tension in the cables. No yeah. That is about that is yes. So, that is ok. So, we are not discussing those problems here. Yeah. Because that will be you know little bit on advance side. Hello sir. Yes. In this equation the third equation in that equation for when delta theta is very small why we are neglecting delta t is 0 because delta t is the main thing that we are. No no because now this is a differential right t plus delta t over 2. So, what is contributing more is t not delta t that is a very small components small fraction. I have to when I am ignoring the term I have to simply ignore the term that is associated with the delta t multiplied by things no because that term is contributing very less to the formulation ok. In any case theta equals to 0 if you just say right then actually this matches right t and delta t. So, there is no delta t ok. And regarding your question you can do the centrifugal force issue here. So, you can put a centrifugal force on the belt and that will simply change that relationship ok. Now, the important problem that we will discuss now is belt drives ok. So, in a belt drive again remember they are actually belt along with the two pulleys or two drums they are actually rotating at a same speed. So, everything is a constant speed problem ok. And here I have a motor let us say this is a my driver pulley. So, where a motor is applied. So, motor is giving a torque that is a externally applied torque ok. And once I give this torque remember somehow I already you know said that transient state is gone both the pulleys are now at a constant velocity. Therefore, I am not analyzing that transient state it is already in the you know Newton's first law ok. So, it is in uniform motion right. So, I will use the static equilibrium condition. Now, what happens if I give a torque here which side of the belt will give me higher tension? Sure. Sure. If my applied torque is clockwise the upward one right. And you can do that very easily using the you know equilibrium position to balance this torque I must have T on this side greater than this. Because I need to have an counter clockwise torque to balance that clockwise torque right ok. Now, remember both are rotating at a constant speed and I am going to invoke this friction law T 2 by T 1 equals to e to the power mu s beta. If I want to transmit the maximum torque to the driven pulley right. So, to to transmit the torque into the driven pulley which is my bigger pulley I have to invoke the friction law. Because that ratio T 2 by T 1 equals to e to the power mu s beta that is my impending slip condition which states that friction force is now somewhere maximum ok. Now, there are two different bodies here. How do I decide that where the friction force will reach maximum first? Not only angle of wrap, but also mu s. You have two terms mu s and beta they are actually like mu s and n. You have to look at both it is like your you know almost like your Coulomb friction ok. We have two parameters mu s as well as n. So, we have to see that multiplication of these two mu s times n mu s times beta in this case wherever that is less right. They are the impending slip will happen fast is that clear ok. So, now let us draw all the free bodies and that will give a very good perception as well that how the equilibrium in means melt and both at the belt as well as at the drum at the pulleys. So, therefore, what I am going to do I am simply going to take the belt out. So, what is shown here belt is taken out and I have shown the friction force here. So, this remember this side is a higher tension side right. This is my higher tension side this is the driver pulleys where the motor is connected. So, there is a applied torque that applied torque is shown by m that is a clockwise torque given ok. To balance that I must say t 2 minus t 1 multiplied by r should be equals to this m right. So, I must get a counter clockwise torque. Now how the friction is happening internally it will simply oppose this tension. So, I have shown the friction here right on the belt and on the pulleys it is simply going to be reversed as per Newton's third law. Just look at the equilibrium of the pulleys itself how the pulleys driven actually. I am applying a torque right that torque is simply balanced by the friction force right. Torque due to the friction force is balancing the applied torque. Now when I glue these two belt as well as you know the pulleys when I glue that then I do not see the friction force at the counter. So, I say ok t 2 minus t 1 times r equals to that applied torque. Similarly, for the driver pulleys I have the same situation for the driven pulleys that is being driven. How it is driven it is actually driven by the friction force again same situation right ok. So, we have t 2 here because this tension has to be same tension. So, t 2 t 1 now this t 2 t 1 is going to get this kind of friction force which is actually rotating the drum. This friction force is going to rotate the drum and what is this m 1 shown m 1 is something that is coming from the machine tool back to the pulleys that is as per again Newton's third law. Because pulleys is going to give a clockwise torque to the machine tool that is attached and machine tool is simply going to back give that back to the pulleys that is the equilibrium you know of everything in this problem ok. So, ultimately the torque that is being transmitted is going to be equals to t 2 minus t 1 multiplied by the radius of the big pulleys. See the problem is very simple, but when you want to draw the free body diagram as I said in the very first day that every time we should tell student you think of it in terms of each and every free body diagram that presents in that particular problem and everything will be clear crystal clear there is no ambiguity here yes. That radius of the pulleys given is 35 centimeter. Just do not worry about this I mean I am not looking at number this 0.2 is coming from that pulleys that was the driven pulleys ok. 0.2 meter is given do not look at the number just say r ok do not worry all right. It is 2 whatever that is ok yes, yes anything. The direction of this will be I think friction would be clockwise. Which one? This pulleys is rotating in a kilo this is not rotation there is nothing called rotation in statics. So, there is always force what I have shown here is a force it is a moment I have written here. So, the direction will be in the direction direction will be clockwise it can rotate clockwise, but what is happening that machine tool is giving the torque back to the pulley that is how we show the free body diagram it is Newton's third law. So, the pulley the pulley is giving the torque this way clockwise as you see to the tool, but when we apply the Newton's third law then reaction is coming back to the pulley ok. This is the effect of that. Yes. So, this is the effect of that. So, that tells me that everything is under static equilibrium it is a constant speed problem do not forget ok. Now, here is the question if there is a slip then it is in dynamic problem then I have to bringing one is accelerating one is decelerating. Think of this way this is a very simple problem. Now, question is why I am invoking the friction law in one surface what was the deal in the morning session? Why I am invoking the friction law in one surface? In this case I said that I am going to invoke the friction law in one surface why did I do that? It is a multi body problem right why both invoked friction law is not invoked in both the pulleys. That is in terms of force is very easy as I said you can count the number of unknowns. So, the problem will be given like this look I want to find out T 2 and T 1 right. Let us say I give you applied torque I have two unknowns I need two equations right period. What are my two equations? One is the static equilibrium equation that is coming from the torque balance. What is the other equation? Friction law T 2 by T 1 equals to e to the power mu s beta right. Now, that can only be applied at one surface not both surfaces simultaneously is that clear? Remember I have two unknowns here I need two conditions I have only one static you know equilibrium condition to solve for in this case because both pulleys will do the same thing right. Now, what happens I am applying it only the friction law at one contacting surfaces although there are two contacting surface here right. Now, why it is like that can you think of it what happens when actually the you know belt tends to slip kinematically what happens? Not really then one will see what happens system will be disturbed one of these remember what happens in this case for example, the motor the pulley that is connected with the motor that is actually going to let us say because beta is let us say mu s is same let us say mu s is same beta is different. So, this one has a lesser beta therefore, the impending slip will happen at that pulley right impending slip will happen at which direction pulley is trying to pull slip that will be along the t2 right ok. Now, as soon as that friction barrier is broken what happens really this one is going to be disturbed totally disturbed there will be unbalance there will be a change in speed suddenly is not that because remember this one is still intact this is still intact this one is going to get the imbalance this is actually going into dynamic. So, this will either you know you can think of that there is a infinitesimally small change in the angular speed. So, now, you have a relative change in the speed between the two now you think of this problem this way I have one fixed pulley let us say I have one fixed pulley and one of this is slowly rotated where the belt is actually intact with the pulley what happens think of it one is fixed I am rotating one slowly that means this is actually now what I am looking at relative difference between this two because I have said that there is always relative change is going to happen relative change in the speed right. So, I said there is a delta omega somewhere now I am locking one of this that means I made the you know rotation is not as if happening here that is 0 omega is 0 another is simply delta omega where you know that for what is happening one is actually rotating with the belt in other one it is actually slipping against see the point that means there is a actual slip is taking place in one of the pulleys not in the both one pulley where belt is already attached it is rotating, but the other pulley is actually not rotating because there is a delta omega difference can you think of that way as I said kinematically if you want to think you know, but we tell that to students all the time also try to think hard in terms of kinematically because that tells you in the friction problem immediately that why I am really getting one contact surface in this case why not to what is happening ok that is that clear so therefore now this problem we can take it forward. So, look at the way problem is imposed here. So, I have a flat belt connects pulley a which is a driver pulley. So, similar problem I am really discussing this is my driver pulley this is the driven pulley what is being said that knowing that the maximum allowable tension in the belt is 600 Newton. So, one tension is given right and we are trying to find out determine the largest torque which can be exerted by the belt on pulley a that means how much torque is being transmitted at the pulley a right. So, in this case remember therefore, my unknown is what in this case what is unknown T 1 T 2 is given T 1 is unknown that means I do not get into even the business of equilibrium doing the equilibrium I just say that where the friction law needs to be invoked. So, I just make use of the friction law because I know already that all the equilibrium is taken care of. So, I have one unknown in this problem and I just invoke the friction law, but that automatically if the problem is self consistent then I must find out there is indeed one contact surface in the body in the multi in a multiple body system right. So, in this case now we can say that. So, the process will be very simple past we identify that which pulley will have the impending slip fast and we have already answered that this is going to be you know the pulley where the impending slip will happen because mu s beta is less right we can calculate that right ok. So, mu s beta if you calculate the contact angle this pulley has a lesser contact angle. So, B will slip fast that is the first thing second thing that we want to find out is the maximum torque in one side of the belt of pulley B because what is given maximum tension is already given. So, which side is that that is going to be this one that already we have discussed. So, that is also given. So, now you just invoke that friction law T 2 by T 1 equals to e to the power mu s beta problem is solved you get the T 1. Now, you prove conversely that T 2 by T 1 is indeed less than e to the power mu s beta for the other pulley I have to prove that right. If nothing is violated if my logic is correct everything is correct then I can actually show that pulley a for the pulley a T 2 by T 1 should be less than e to the power mu s beta in order for it not to slip the belt not to slip over the pulley ok. So, I solve the problem I get the T 1 right once I get T 1 problem is solved I am trying to find out the torque transmitted that is actually T 1 T 2 minus T 1 times that radius of that pulley right and I can also check. So, this is the last part is really the check where I said that mu s you can solve for that pulley because you know already T 2 and T 1 ratio right. So, there I can solve actually mu s that mu s should be less than what is given is that clear that is how you also solve the friction problem think of that way. That is what I said that friction problem you know as long as the mechanics is concerned in the first year very simple number of equation number of unknowns then you figure out how many extra unknowns I have if the problem is self consistent you must have that many contact surfaces where you have to invoke the friction law period problem is solved. Now, if you can think out which surface actually you know undergo the impending slip that we have to deal with the kinematics actual kinematics and that is not that you know simple for a first year kid to absorb they have to do lot of you know analysis. So, let us say this one is already solved I will just show you the you know final thing here. So, basically I have applied T 2 by T 1 equals to 2 to the power mu s beta and I get that T 1. So, T 1 is solved T 1 is 355 Newton. So, pulley B was under impending slip up currently at it is very good. So, moment torque you know that is transmitted that is going to be 48.92 that is the torque I am trying that is the you know torque that is being transmitted to this pulley that is the driven pulley and then we have showed that indeed there is no slip in this pulley that is the driven pulley A. There is no slip you can find out T 2 by T 1 that should be less than equals less than e to the power mu s beta for that pulley or the friction value you can calculate coefficient of friction value you can calculate that is should be less than what is given right whatever is that clear done. Now, we are going to take an example of a band break. So, as I said that the main point here is that drum is actually rotating and we are trying to you know make this drum to be rotated at a constant speed. So, the drum is constantly slipping over the belt that we have and once we adjust this P ok. So, you have to find out what is the P such that a particular torque is achieved that means there is a particular angular speed. So, that means I am controlling the angular speed based on the force that is being given the P. So, there is a relationship between P and the angular speed. So, now here what is said is that that angular speed is now controlled by the torque that is being applied. So, angular speed is directly taken care of by that applied torque. So, that means what would be my P such that that particular speed is achieved ok. So, that is the you know basic problem that we are going to discuss. Now, again I would like for you to kind of think about it in terms of first whether it is a statically determinate problem how many unknowns really high I am in my problem ok and how do I solve it. So, that is the most important thing is it now statically determinate if we make those assumptions. It is a rotating at a constant speed I do apply a torque right and basically I am trying to find out what the P is ok to maintain that speed. Remember there is a relative slip between the belt and the drum that is all the time slipping. So, we have to use e to the power mu k beta not mu s there is no static coefficient of friction. So, just think of the let us say we think about the number of unknowns here how many do we have total 5 unknowns right. So, can we solve all of these remember there is a pin. So, 2 reactions here these are the 2 tensions 4 and this is also unknown 5. What is my static equilibrium condition that is actually the torque produced by the tensions in the belt that is going to balance the torque that is applied right that is in equilibrium because it is in constant speed problem. So, that is the equilibrium part right. In lever lever has this lever a b c we will give how many equations equilibrium equations 3 equilibrium equations. So, total 4 right and then we have to apply the friction law here right. So, friction law is T 2 by T 1 equals to e to the power mu k beta is the approach is clear to everyone I have total 5 unknowns I have one lever that will always give me 3 equilibrium conditions right. One is coming from the equilibrium of the drum there is one static equilibrium. So, total 4 equilibrium I am having plus I am using the friction law because there is a relative slip between the belt and the drum all the time. So, if we do that way then we can now what is interesting that drum is you know there are 2 different ways we have to solve first the drum is rotated clockwise. If it is rotated clockwise then we solve for the p. Now, if the drum is rotated clockwise then we have to make sure that which side of the tension is higher. So, that will be a side. So, a side of the tension is now higher and b side is lower. So, my T 2 by T 1 may e a by T b. So, it will be T a by T b equals to e to the power mu k beta. So, such very simple problem and we can always correlate this with the you know common friction problem point friction problem that we solved. So, that is the approach actually for all the friction problems. So, we could see the you know now the free body diagrams considering that it is rotating clockwise we are applying this torque this is the applied torque on the drum right. If this is the applied torque then we have the tension T a and T c sorry this is yeah this is meeting at T c. So, there is a T c here. Now, when it is rotating clockwise T a should be greater than T c to balance that applied torque then we look at the equilibrium of the lever. So, we have two reactions B x and B y ok T a T c and T. So, simple problem as such, but make sure that we understand all the steps here applied couple torque is 125 ok. So, we can do the equilibrium. So, take moment about e for the drum. So, this is the equilibrium of the drum. So, we get a relationship between T a and T c from here right ok. Then what else if I solve T a and T c then what I can do I can simply take a I can simply come to the lever and take a moment about B. See I can solve all the unknowns not an issue, but here my target you know unknown is just to find the P, but to do so I have to find first T a and T c right ok. So, we solve that for the slip in belt which is already happening there is a relative slip between the belt and drum. So, T a by T c equals to e to the power mu k that is already given and the contact angle is actually 7 pi by 6 90 plus 30 sorry 180 plus 30 right ok. So, we get T a and T c based on these two equations ok. And once we know the T a and T c then we can move on to the equilibrium of the lever. So, we go to the equilibrium of the lever. So, we take the moment about B since T a and T c is already found. So, take the equilibrium moment about B to solve for T clear the approach is clear any question on this ok. So, T a T c is found. So, the next step that is remaining is simply we take the now we take the lever here and take the moment about B to solve for P ok. Then we do the other way around. So, now it is said if it is rotating anticlockwise then what is P. So, then we just reverse the T 2 and T 1 right. So, that has to be reversed. So, in the second case I will just flash the solution as such there will be no change as the interchange in the T a and T c ok. So, just T a and T c will interchange nothing else whatever we found from the previous one just interchange that and then again solve for the P. So, that P will be 9.41 ok. So, now I think any question on the belt friction. So, we have basically covered all the problems that I have shown in the first you know slide actually all the applications that are possible. We have you know lowering load or maybe you know take the load upward then we have belt drive we have band break ok. So, more or less that is what is covered in here in our course.