 Good morning to all of you and welcome to the fifth day of this workshop. If you look at the schedule for the morning sessions, we are mostly going to deal with complex friction and basically as we have you know considered during the coordinator workshop, we are going to look into belt friction as well as square threaded screws that is screw jacks more or less. So, just to start off with as you have seen that friction you know we are approximating the solution on the basis of Coulomb friction or dry friction we also sometime call that point friction. Again that is a very rough assumption and as we all know that friction will depend on the tribology between the contacting surfaces. But more or less what we have studied that if you for the time being just consider the point friction then the analysis could be very simple. Only thing that we have to careful is with the direction of the friction force, how it is going to act that is the first one. Secondly as we have all seen that when we try to look at problem involving impending motion that means most of the problem that we look at without slippage problem. That means what is the for example if we look at a problem that I am trying to really push a body up on an incline then my question would be let us say what is the load required to just push the block up on the incline without slippage. So, I am really therefore trying to you know enforce the condition of impending motion. And what it does basically it does you know and one thing that you have noticed that it makes the number of unknowns equals to number of you know conditions that are required to solve the problem. So, remember the friction you know law therefore that actually tells me or allows me to solve the problem in a statically determinate way. So, therefore, when we have multi body interaction if we have looked at it carefully we have said that in a multi body interaction problem let us say we have you know two contact surfaces may participate. In that case what was the main you know principle that we have used first we try to identify the number of unknowns. So, we have identified the number of unknown forces right then we have limited number of equilibrium equations right. So, we have always seen that number of unknowns exceed the number of equilibrium equations and therefore the balance unknown ok. So, that is the total number of unknowns minus the number of equilibrium equation that is my balance unknown and what we do is basically if the problem is self consistent we are going to say that I should have that many contact surfaces where I would induce the friction loss. So, it is as simple as that in word however remember we are always enforcing a problem which is statically determinate in nature and you have seen this from the very beginning that no matter what topic I choose my problem is always statically determinate otherwise what happens in the process that to if the number of forces exceed the number of equations then we need to invoke the you know deflection that means kinematics as well deformation of the body as well to solve for the additional unknown forces. With that what I will now try to look at some of the complex friction now why do I call it complex friction. Remember as you see in case of a belt friction what is happening I have illustrated few different examples here. For example, here you have a rope right that is passing over a bollard. What is the objective the person can actually apply a very small amount of force to hold a very large load. You can see this very often if you go to jet ease for anchoring of boards for anchoring of ships and so on so forth. You can see that by applying a small force I can hold a large amount of load. Similarly now what is happening that there is a large area of contact coming into play however remember friction force is distributed between the rope and the surface. However we are going to introduce the fact that we can still say that each and every point on the contacting surface should have simple friction law that means Coulomb friction and we are going to derive the relationship between the tension on the both sides based on that. So ultimately if there is no friction for example between the pulley and the rope then we say the tension in the both sides are same which is perfectly fine. But when the friction exists then we should understand that that force that is the tension in one side of the belt will differ on the other side. So again this problem will help us to you know study applications that involve raising or lowering the loads. One thing that instantly come to our mind is as follows. I would always require small force to hold a load. Suppose I put a 20 kg block here I would probably need 5 kg to just hold the both block. Two conditions are coming into play. In one case in the first case the block is trying to go down that is impending motion is downward. So we are trying to invoke the impending motion condition when it is simply trying to go down and we are trying to find out what is the load required to hold this body right here. Similarly in the second case if I consider the impending motion upward right then what will happen I have to apply a large amount of load and that is all possible because remember the friction direction is always being changed here it is being reversed as you look at the problem involving loading you know holding the load or the problem involving just raising the load. So friction will change the direction and tension side will always be on end side that we will try to oppose the friction force the larger tension side ok. So another application as you see here that is you know we see in the transmission of torque problem right. So you have somewhere a motor that motor may be connected to some machine tool. So we have again pulley or drum and belt problem ok. Remember this problem however we will be tackling based on the fact that both are rotating at a constant speed. If both rotates at a constant speed then we will try to understand that when the belt is trying to slip over the drum what relationship exist between the tensions of these two. We should first try to understand also that where the slip would happen first whether it is going to be in the motor or whether it is going to be on the machine tool that is connected right that is being driven by the motor. The other example will be the band break problem. So you can see a disc and you have the you know band that is on top of the disc. Now you need some special arrangement here there may be a connected to a lever arm where we are going to apply a force. So once you apply the force this band is actually going to tighten on the disc and it will you know reduce the speed of this disc. So again our idea would be that can I find the force required such that a constant velocity is maintained here ok. So depending on the amount of force you can actually control the speed of the disc that is rotating. Remember here band is fixed on top of the disc or in other words there is a constant slippage between the disc and the belt or the band ok. Therefore this will actually involve kinetic coefficient of friction than that of static coefficient friction. So whenever we see that we have relative motion exist between the two contacting surfaces then we are going to say that we have to induce kinetic coefficient of friction. So let us try to just look at how do I derive the theory involving the belt friction problem. So what is the question here? The question is relate T1 and T2 as you see the T1 and T2 are the two tensions on the both sides relate T1 and T2 when belt is about to slide to the right. That means I am invoking the impending motion condition towards the right. So if the belt is just about to slide over this drum or let us say disc towards the right then which should be the direction of the friction. The friction will oppose the motion of the belt therefore friction will be on the opposite direction and we can clearly see therefore that T2 tension on this side has to be greater than tension on this side. Now remember as I said that there is a contact angle beta that plays an enormous role. So therefore we can easily get a feeling that whatever formula I derived T2 by T2 T1 relationship that will be completely controlled by this angle beta as well as the coefficient of friction between the belt and the drum. So now how do I get into the theoretical part? So again as I said it is a complex friction because friction is actually applied throughout. Now what we are going to enforce I am going to simply take a very small segment of the belt and I am going to apply the concept of point friction or simple friction there and I would assume that is everywhere and finally I have to get an integrated effect from P2 to P1. So that is the basic concept. Therefore let us just detach a segment PP prime. Suppose I detach a segment PP prime of the belt and I try to understand in this differential element that I have chosen PP prime what are the forces that have been induced. Since as I said that T2 has to be greater than T1 so I can see here that here I have T plus delta T okay and here you have only T and from the geometry we also understand that this angle is delta theta therefore this is delta theta over 2 delta theta over 2. Now here is the important part. The important point here is that I should have a normal force delta N that is given by the drum to the belt. I should have a friction force delta F. Now when the impending motion prevails then we say that delta F equals to mu s delta N that is our friction law. Now therefore you can see clearly that I have two unknowns in this problem one is the delta T and delta N right. So T and N these will be more or less my unknowns. So therefore now I will treat this body as a particle and I will try to understand that what is sum of force along x direction 0 and sum of force along y direction 0. So if we do sum of force along x direction 0 we see that T plus delta T cosine delta theta by 2 negative T minus cosine delta theta by 2 negative mu s delta N 0 that is one equilibrium equation and second equilibrium equation can always be written like this. Now we have to find out now I am only interested in finding out what relationship exists between T mu and beta let us say. So my objective here would be to somehow eliminate delta N. Therefore what I will do I will simply take this equation remember in this equation this term and this term gets cancelled. So delta N is delta T sin delta theta by 2 which I will simply substitute back here and therefore I am going to get this equation now this equation remember I have also divide everything by delta A delta theta because delta theta cannot be 0. So now what happens as you see now the equation has been revised I have one equation to play with and that equation becomes this ok. Now in the limit as delta theta goes to 0 what happens remember these terms will become 1 delta theta goes to 0 10 to 0. So this is 0 that is one what else happen when really delta theta tends to 0 we can also say the delta theta this one delta T right that is also tends to 0 ok. So therefore what I eventually get I eventually get a very simple equation which looks like dT by d theta equals to negative mu st that is equals to 0 ok. Now remember that we have to separate you know we have to use the separation of variables so we have dT by T must be equals to mu s d theta. So once I use separation of variables dT by T equals to mu s d theta and I will integrate both sides. So one side will go from T1 to T2 other side I will have 0 to beta. So theta should go from 0 to beta and T should grow from T1 to T2. So these are the two integration limit that I am going to use after separation of variables ok. So and that will lead to this because dT by T that is simply log of T right. So therefore we have log T2 by T1 equals to mu s beta because 0 to theta that integral will become beta. So ultimately what is the relationship T2 by T1 equals to e to the power mu s beta. So therefore what we have obtained that during the impending slip when the belt is just about to slide over the drum I have this relationship T2 by T1 equals to e to the power mu s beta. Remember there is no unique relationship that exists between you know we cannot find both T2 and T1 from just one equation. So what it tells me that I have a ratio that ratio is controlled by e to the power mu s beta where mu s is the static coefficient of friction and beta is the contact angle that should be measured in radian. So therefore to avoid the impending slip what formula shall I use T2 by T1 should be less than e to the power mu s beta. If there is no slip between the before the impending slip I should always use T2 by T1 less than e to the power mu s beta and if there is an actual slip that means there is a relative motion between the belt and the drum I should use T2 by T1 equals to e to the power mu k beta. So if there is a relative motion that exists then I should use e to the power mu k beta. So therefore again if we have to summarize we can summarize like this that when the belt is about to slide towards the right during that impending motion T2 by T1 equals to e to the power mu s beta. So it is completely controlled by the coefficient of static friction and the contact angle which is measured in radian we should first understand that T2 has to be greater than T1 if the belt is about to slide on this direction because all the friction forces will try to oppose that motion okay and remember that if there is no slip then T2 by T1 has to be less than e to the power mu s beta. If there is an actual slip that means there is a relative slip between the belt and the drum then T2 by T1 should be equals to e to the power mu k beta. Therefore I can actually you know take this problem to solve a large class of you know application we can consider using this concept. What are those problems? As I said the formula can be applied for impending slip condition for problems involving those are the applications I have shown in the beginning that flat belt passing over a fixed cylindrical drum okay. So that means your drum is fixed and belt is passing over that drum let us say you are trying to lower a object you know or you are trying to raise a object or you are trying to lower an object. So in those kind of problems it is going to come into play. Similarly ropes, belts you know around a poster, capstan or pulis we have gone through that also and belt drives. So this is going to be very important remember belt drives problem we have both the drum and the belt they rotate together at a constant velocity. So the question will be when there will be a slip between the drum and the belt. So now when it involves the motion problem that means when it is rotating at a very fast speed you can take one thing you could ask me that okay is there a centrifugal force. So did I consider that in that equation? So therefore now you can see the only thing is that we are always trying to understand here the main assumption is belt is always in contact with the drum. It should not try to you know just separate away from the drum. So in other words if the speed is really large we may have to include the rho omega square r effect in this formula. However as long as the mass of this belt remains small we can actually ignore that term as well. So remember in engineering mechanics problem we are always trying to approximate the actual problem and we are trying to get an answer which is as close as that of the actual solution. So we can always have this plus minus 10 percent error as we say okay in as a thumb rule in any of these problems. Is that clear now? So now we are going to take few problem a cord is wrapped around, wrapped twice around a capstan at A and three times around second capstan at B. Finally cord goes over a half barrel section so that is the half barrel section and supports a mass m of 500 kg okay. Now the question is what is the tension T required to maintain this load? Now only thing is that what is being asked what is the tension T required to maintain this load or what is you can also ask what is the range of values of T for which equilibrium is maintained okay. Now to study this problem we should understand that there are two scenarios possible what are these two scenarios? One is the upward impending motion of mass that means we are actually trying to lift the mass up therefore I have to first understand the T must be greater than the weight of the body here. The final T that you get should be greater than the weight of the body here. So this will be always your large tension side okay. So now to get the minimum value of a tension that means we are simply trying to hold the load that means load is trying to come down now okay. So in that case my impending motion of the mass will be downward therefore we can say that in this case the tension in this side will be greater than the tension that you need here. So that means mg has to be greater than that of T. So remember now how do you understand the problem in terms of the force concept whether do I have a statically determinate problem or not that is the first key. Remember here the assumption was always that the belt that is the rope is always going to be a rigid body okay. Now we try to understand this like this way that if I have a tension here my tension will be different here then I have another tension here I have another tension here. So how many problems how many unknowns I have the number of unknowns I have 1, 2, 3 that means tension here, here and here remember tension here is already known because that is simply going to be the weight. So therefore I have to repeatedly invoke the friction laws three times my friction law now becomes T2 by T1 equals to e to the power mu s beta. So I will apply that repeatedly between this tension, tension on the left side of the capstan A with the tension in the right hand side of the capstan A. Similarly left side of the capstan B with the right side of the capstan B and left side of the barrel with the right side of the barrel okay. So I have to really derive these three equations in order to get the ultimate tension that is required. So as such it looks very simple to us now so what we are going to do firstly remember that we must find out what is the angle of contact as we have seen that rope here the rope here rope is wrapped twice here right. So the angle of contact will be 4 pi here rope is wrapped three times so the angle of contact will be 6 pi okay and what is the angle of contact in the barrel it is simply pi over 2. So first we understand what are the angle of contacts of the belt with the respective objects. Now we try to draw the free body diagram as I have told that you can see I am I am interested really to find out what is T here right. So T1 and T so I have one unknown the tension here tension here this is two second unknown third unknown is T2 okay. So I have three unknowns in this problem and I am going to solve it. So how do I solve it now? First assume that impending downward motion of the mass so that means you are simply trying to hold this mass the mass can try to go down. So we are simply trying to hold this mass by applying the tension T. So therefore we can use this relationship one at a time so in this case therefore what is happening this tension has to be greater than this tension. So mg by T2 should be e to the power mu s multiplied by beta of the barrel of the pulley okay. So ultimately what you have this tension is greater than this tension okay. So this will now in the formula becomes T2 and this is actually T1 so mg by T2 then this tension has to be greater than this tension. So here T2 by T1 so this is higher tension side so T2 by T1 is equals to e to the power again mu s and beta what is the beta? Beta is 6 pi because the rope is wrapped around three times. Similarly this T1 must be greater than the T so we will use this one then only thing I have to do basically multiply all of this so that I get a relationship between mass and the tension that I am trying to find out. So the final tension will be this. So if I do this problem by saying now I am trying to pull the object up okay that means now I have to understand that impending upward motion of M that means now friction force will change the direction. So now friction force will all be you can tell like in clockwise in this case okay. So therefore I can clearly tell that this side will have lower tension than this side. So now this side is always going to have higher tension so now I have to just rearrange all of this. So as I said the above relations should be reciprocated. So here we already have relationship T equals to Mg to the power negative the total this thing 1.05 pi so we will just reciprocate this relationship why? Because now I have T greater than Mg okay. Now on other words you can actually flip all of this equations just flip all of this reverse all of this equation same answer same result so what is happening now since we have reversed it so you are going to get very large value of the T you can see that it is so high compared to when you are just simply trying to hold the load you are simply trying to hold the you know mass such that it does not go down. But when you are trying to raise the load right when I am trying to raise the load upward then I need a very large amount of tension in the row okay. So is there any question so far? So if this concept is clear then we can go ahead any discussion on this issue short discussions. Hello 1.17 yeah go ahead ma'am. Sir actually I wanted to ask about the coefficient of static friction and coefficient of kinetic friction there is a general notion that the coefficient of kinetic friction is greater than the static friction. So are there any conditions where we can get the both the coefficients equal or the static coefficient of static friction is greater than the kinetic friction? No no there is no such rule actually see that is not always true actually what you are saying that kinetic coefficient of friction is greater than static coefficient friction I looked actually that is always hold goods okay. So that we need to clarify that first okay that may not be hold true first of all. So secondly see why this static coefficient of friction and kinetic coefficient of friction in one case there is no relative slip between the surfaces right that is participating when we say static coefficient of friction there is no actual relative slip but kinetic coefficient of friction when there is actual slip. Sir there is one more question from today's lecture. So what is the difference between V belt and flat belt friction coefficient? So okay now your question relates to that is a very natural question comes into mind that what is the difference between a V belt and the flat belt right. Now I will try to you know everyone understood the question the question was a flat belt and a V belt. Now V belt will only you know introduce one more you know unknown to the solution. So one more force to the solution rather to solve the problem okay. Now how that will happen? So remember I have to now draw it okay so in the V belt so first let us say when we are discussing flat belt seen flat belt now you are really looking at the cross section of the belt okay. Now in the flat belt case if you look at let us say cross section and try to look at from this side the cross section is rectangular in shape right okay. Now in case of a V belt now where that is applied if you look at the you know the cross section of it. So the cross section will look at so it is on a groove actually the belt will now be on a groove and this is the cross section okay. Now what happens due to that let us say I have an angle V here because it is a V shape right that angle first I have to define so that angle let us say alpha. So what will happen there will be additional reaction that is coming into play let us say delta R from this side and this side okay right. So what will happen now this has a resultant these two if you take the equilibrium you can show like this so ultimately what is happening you have delta R 2 delta R is in that. So what happens now we discuss this problem now so now in now you go back to your flat belt case so where we have looked at only this now here I had a perfect delta N right I have T 1 I had T 2 sorry this will be T plus delta T that is the differential element I am choosing. So basically that effect will come into play so this is T this is T plus delta T. Now in addition to that what is happening this force will come into play which is given by the cross section if you look at this which is given by the groove onto the belt and that will come in your perpendicular equation of equilibrium okay. So in addition to that I am also going to get another force that is coming from the V belt clear. So problem can be solved still but we are not going to take that for the time being but is that concept clear that how this force is coming into the free body and it can be solved okay. So you will just going to get another additional alpha in the equation. So therefore this alpha will come now into the equation okay it will come I think in terms of sin alpha alright. Any other discussion? Sir is there a difference between the friction coefficient of V belt and flat belt? See coefficient of friction is determined by the two contacting surfaces right it all depends on there is nothing your question is there a difference between the coefficient of friction of flat belt and V belt the answer is no coefficient of friction is not but what usually is written in the book is as follows as I said in the equation I cannot really recall the equation correctly but for V belt I think it is something sin alpha by 2 that comes into play okay. Now what we tend to say that this multiplied by the this. So basically mu s sin alpha by 2 that is for the mu s you know of the V belt I think that is what you are trying to ask is there a change in the coefficient of friction between the flat belt and V belt. My general answer is there is nothing change in coefficient of friction but it is how you are writing it. So basically I am now saying mu s sin alpha that becomes the effective coefficient of friction in the V belt okay. Thank you sir. Yeah go ahead sir. In linear motion we consider f is equal to m into a but in rotation we consider tau is equal to moment of inertia into alpha just few days before we have thought that moment of inertia is a task of resistance but in linear motion we consider mass is a task of resistance. Now you are going to a completely different aspect remember so your question is how the inertia comes into play your question is mass times acceleration and I times alpha. Answer is very simple I said I am taking only problems involving constant velocity okay. Even if my drum and the belt are rotating they are rotating at the same angular speed okay. So when it is a constant velocity problem then again you can go back to Newton's first law. So therefore you can again use static equilibrium equation. I am not taking any problem that will involve dynamic but think this way, think this way just simple experimental demonstration okay. I am rotating a disc, a disc is being rotated so I apply some couple right to rotate the disc. I ask someone just put a belt on top of that to decelerate the motion what happens how do I solve that equation of motion. So now what it comes now the belt that you are really trying to decelerate the system belt you are holding tightly right to decelerate the system and therefore now the angular acceleration right that will become a decelerating body. Now if you take equilibrium of that there you are going to see I alpha. So I alpha that is going to come into play and then you have t2 minus t1 multiplied by the distance that is the torque. So that torque is actually going to equals to be equals to I times alpha. Now you are coming to moment equals to I times alpha. So now you solve for that alpha is that clear okay. So problem is very simple whether it you are solving a dynamic problem or whether you are solving a static problem. Remember constant motion problem is itself is a static problem you can actually define it as a static problem okay. Sir actually F is equal to m into a it means Newton's formula is valid only for inertial frame of reference. But we consider in a professional but in professional there is a dynamic in nature then how can I apply Newton's formula. What is here Newton why cannot you see if you are in a non inertial frame then you will simply use the D'Alembert's principle and that I am going to discuss when we talk about it. In non inertial frame means that means you are actually inside that rotating you know you are actually inside that moving body. Then what happen your resultant force still becomes 0 okay. But in a inertial frame how you write it you write force equals to mass times acceleration. But in a non inertial frame you will write force negative mass times acceleration equals to 0. That means there is no relative motion again if you are in a non inertial frame okay. Thank you sir. 1, 1, 4, 5 go ahead for your question. Is there any friction between the center screw of the pulley? In the width with respect to what? Sir in the center of the pulley there is a screw no. Is there any friction between that screw and the pulley? Good question actually yes so your question is is there any friction between the screw or the hinge rather right hinge and the pulley what we have done so far yes no we have not taken that into account okay. Yes sir. So we will see that in journal bearing problem do not worry okay it will come into play. 1, 3, 1, 4 go ahead for your question. Sir will you please explain the concept that flat bulk can be used for a very long center design whereas V bulk is limited to 1 meter only why is it. Yes so your question is what would be the specific applications of the flat belt and the V belt yes absolutely correct. Related to. So it relates to the applications we are talking about okay usually you have a flat drum you know and you will always prefer to have a flat belt on top of that but let us say you have a groove right on the drum also. Yes. Then we are going to use the V belt right so V belt is basically put in a groove. So drum will have to have a groove okay or the rotating system has some threads or grooves there we are putting the V belt. So there will be additional force that is coming from those you know thread or groove okay. So the next question is if I use the V belt in a flat pulley what will be the reaction. No there will be no reaction because there is no nothing there no see in a flat pulley so it is basically a 3D visualization you have a flat pulley and I have this belt now if the flat if there is nothing on these sides right of the pulley okay then I am not getting any force from that side okay. So if the V belt is used in a flat pulley it makes the same thing we are going to solve the same problem T2 by T1 equals to e to the power mu s beta. So problem will remain same okay. Thank you sir thank you. 1, 3, 3, 0 go ahead for your question. My question is how to solve questions in which anything about the motion of the body is unknown. When we do not know whether body is impending having an impending motion or not at all or it is moving on the on the other surface whether anything about the motion is unknown. How to solve questions on that? How can that be you are applying the load right see your question is actually I have to restate that if raise the question so your question would be I am applying a load on a body how do I determine that whether body is under static or under dynamic condition right in a friction problem see remember friction force is a barrier it is a barrier right you are trying to break that barrier okay. So you keep think of it I have a you know a block let us say I have a block on an inclined okay. So I am trying to apply so first I start applying the load small small load I am applying right at some point of time you know when the load is small equilibrium will be maintained that means you can actually use the Newton's first law that resultant force equals to 0 right. Remember friction force is also you know if you plot that that was done yesterday itself right so you are going to also gradually you know friction force is also increasing but friction force has a limit that is mu s multiplied by n right. So as soon as your actual force that exceeds that mu s n then the body is in imbalance condition okay. So you have broken the barrier of the friction so friction barrier is always mu s times n remember that is based on the concept of point friction now when we say it is a dynamic that means you have broken the barrier now we do not know you know body is under some acceleration or deceleration depending on the problem then you have to use the dynamic equation force equals to mass times acceleration or force equals to or moment equals to I times alpha okay. So to decide whether it is under static or dynamic right it is given by the whether the resultant force on the body was 0 or non-zero okay if the resultant force is non-zero then of course it is dynamic that is all simple. Sir we have one more question yes sir why we take angle of contact in radian. Why do because that is a mathematical problem no you are saying your question is why beta is taken in the radian because you are e to the power mu s beta no beta has to be expressed in radian otherwise it is does not make any sense okay it is a mathematical problem. Sir is there any difference between angle of contact and angle of lap or it is the same thing what was that angle of contact and angle of lap I do not know personally I would not be able to answer it okay I do not know which problem you are talking about but that is irrelevant I think. Sir my question is what is the role of kinetic friction in torque transmission see it is of course important see why your question is what is the role of kinetic friction in torque transmission think of simple way think of you know a thrust bearing end bearing problem right if we have a end bearing let us say you are applying a thrust but I also need a couple right if it is rotating body I need a couple okay. So what is happening is that we have resistance from the end right that is coming through the kinetic coefficient of friction because resistance to that you think of disc brake problem I just gave an example disc brake in the disc brake right that brake problem it is under the motion right. So ultimately it is determined that what is the kinetic coefficient between the belt and the disc that will able to either decelerate the body right. So deceleration will depend on what is the kinetic coefficient of friction but not only that remember the contact angle here also plays an important role what is the contact angle that is also determined and we will come to that problem the question that you are asking the kinetic coefficient of friction if there is an actual slip between the belt and the drum right then how that torque is actually changing okay. So it helps it helps the you know changing the torque definitely depending on the coefficient of kinetic friction we have. Another question is which types of precautions should be in mind before designing the right flat belt or when you are designing a belt let us say the two things are very important coming into play of course as we have seen here one is the coefficient of friction and that was your in fact earlier question okay kinetic coefficient of friction. So that one thing another is definitely it is determined by the contact angle okay. So these are the two aspects that we should look at when we are looking at the force I am not getting into the other part of it. So if your question is related to more of a design then I am not answering to that but more or less if it is related to force right these are the two things controlling the force okay transmission of torque if you say like that then these are the two things basically controlling okay. One three zero five go ahead. My question is the problem with your just all on belt friction there you have taken that wrap round and three wrap round and two wrap round the angle of lap you have taken six pie and four pie can you elaborate separately by drawing that particular freeway diagram and can you justify how it has come six pie and four pie. Yes so the question that we have received from one three zero five that contact angle that we are saying one is wrapped around two times another is wrapped around three times. So if I have a two wrap and I simply explain it I do not see one let us say one time let us say one time wrap then what that should be if it is a one time wrap that means let me draw it here okay. So what does that mean that is it that is a one wrap so you go yeah that is two pie right no one wrap will be two pie one wrap means we are going to make full revolution right okay so that means that is it okay so that is one wrap so I go like this and then come back like this okay then two wrap put this one actually okay so you go around then one more then another okay so two full revolution right remember at the look at the problem there you will see that going like this and then exiting like that right so that makes it one wrap if you have two wrap then you have to wrap it around two times okay so just make two full circles that is all okay so one full wrap will be two pie two wrap will be four pie and three wrap will be six pie. Sir what happens if I change the geometry from circular to rectangular? Your question is you want to change it from circular to rectangular but why why why do you have that question in mind? If I take the rectangular material rectangular geometrical shape nothing will happen it is just going to get more complex because now you are going to see that you have three sides your question is if I have a rectangular segment let us say let us say not proper rectangular you have some kind of you know fillet shape here okay so now you have a belt now it just you see that what is happening as I said friction is always going to you know gets complicated because now you have to look at this part that part and this part three parts has to be separated and look into a different manner and problem gets more complicated okay because remember what we have done in case of a flat belt on a drum right we have always considered this small segment okay and we try to establish the force was delta f equals to mu delta n right so we have again applied the concept of simple friction to go to the complex friction we have to integrate in this case it is even more complex but remember this you will not see any physical problems a physical problem will always try to address a nice circular you know a drum or a pulley again it will determine it will be determined by the free body diagrams and how you take into account can you explain it again why angle of contact taken in radian angles are measured in radian you have log log value log t2 by t1 equals to mu s that is going back to simple math class you have log of t2 by t1 equals to mu s beta right now you are saying t2 by t1 equals to e to the power mu s beta so therefore beta should be measured in radian okay 1, 2, 9, 3 go ahead why you are not consider the centrifugal tension in the flat belt and v belt your question was why the centrifugal force was ignored I did get that answer when I was discussing two ways one is the center of you know the mass of the belt could be small that is one thing second one was the answer could be the rotational speed is not that great or in other words the belt will not detach from the drum or the pulley okay we can just think of that way if the if the rotation speed is very high then we may have to incorporate that. Sir one more question sir in the v belt there another term is cosmic beta is coming if someone book are given cosmic beta yes 1 over sin alpha that is what it is okay 1 over sin alpha that term I tried to explain that before also no sir I am asking to you that alpha for the lab angle right and beta for the v angle that is what it is that is what I am saying so you have e to the power mu s beta so beta is the contact angle okay lap angle right and then there is another term which is sin alpha by 2 I think 1 over sin alpha by 2 that comes from the v belt that is from the cross sectional direction that is coming into play okay so there are two terms okay that is coming into play yes we will now start little bit more you know getting into problem specially you know we should try to look at belt drives type of problem how to you know look at belt drives okay now this is another interesting problem we are going to study that involves both static coefficient of friction and kinetic coefficient of friction okay so now how this is happening a cable passes around 3 2 inch radius pulleys and support two blocks as shown so in this problem there are three pulleys remember two of the pulleys d and e are locked to prevent the rotation okay while the third pulleys rotated slowly at a constant speed just pay attention to this that they are rotated at a very you know this third pulleys e is rotating at a constant speed so we know the coefficient of friction between the cable and the pulleys determine the largest weight which can be raised if pulleys c is rotated clockwise that means we are interested in finding out the largest weight that can be raised when this pulleys rotated very slowly clockwise again this is rotating at a constant speed therefore remember the belt and the pulleys they are intact so belt is also have the same angular rotation as that of the pulleys okay now in this here belt is actually constantly slipping over these two pulleys so the problem involves therefore we can clearly see the two determine the largest weight that is possible two days we need to invoke the impending motion condition at c right so for the impending motion you can clearly see the largest weight that is possible will be based on when this belt will try to actually slip over the c so when the belt is trying to actually slip okay just about to slip over this pulleys then we will say that impending motion is reached and therefore I will be able to obtain my largest weight that can be lifted up now again it will involve drawing the free body diagram of each and every pulleys remember there is a kinetic coefficient of friction is coming into play in drum D and drum E should have the kinetic coefficient of friction because the belt is constantly slipping over this pulleys so ultimately you see that to decide this problem we first make sure that where the impending slip has to prevail because it relates to the largest weight see the largest load that is possible if we ask always this kind of question then impending slip logic automatically comes into play the other drum the belt is constantly slipping right so no problem here we can simply use the kinetic coefficient of friction in this problem so now we have to really look at the free bodies specially we will be concerned with the belts so you can see the tension on the belt side we have to be careful that how the tension starts to turn which side would be greater and which side would be lower one more thing to understand the contact angle that is very important so the contact angle that we can find from the geometry of the problem so first we decide on the contact angle and we express that in terms of radian again so all the contact angles are shown here now remember the C is rotated clockwise as I said therefore which side the tension would be higher in this case if it is rotated clockwise and if I try to look at the free body diagram of the drum itself how the torque has to be transmitted see you are applying a torque which is not given actually a torque is applied clockwise to balance that this side of the tension has to be greater than the other side of the tension so w a must be greater than this side so tension in this side has to be greater than the right hand side so therefore now this is becoming T2 and this is becoming T1 so we say that w a is now T2 and T2 itself becomes T1 okay so this is the higher tension side this is the lower tension side and it is very easy to understand also because I am rotating it clockwise and to balance that torque ultimately it has to be balanced by the opposite torque that is given so w a minus T2 multiplied by r that will try to balance that torque okay similarly so when I express this relationship then I go to the next one the next one will be now we decide which side the tension would be higher now remember the belt is constantly slipping over the pulley so the slip direction is already defined slip direction in this drum D and drum E is already defined therefore I know the resistance direction resistance friction will always be opposing the motion so friction force direction is already known if I know the friction force direction which will be clockwise in this case therefore this side tension on the right hand side tension will be higher than the left hand side tension so now you see in the formula I have to see say that T1 is actually higher than T2 and I have done that based on the fact that which direction the belt is slipping there is an actual slip between the belt and the drum remember drum is fixed belt is constantly slipping at a very slow speed so therefore I am going to use T1 by T2 equals to e to the power mu k beta so I put that this is another expression and the last expression as we can see here that again the same logic will be applied this side tension will be greater than this side tension because there is a slip here right and frictional resistance is coming on the other direction counter clockwise in this case right so this tension will be greater than this center so I use this formula 16 by T1 so 16 by T1 should be equals to e to the power mu k beta therefore now what I have I can now solve these three equations right to get a relationship between ultimate objective is to find out the WA which can be achieved by the similar process we have done so the value of WA the largest weight that can be raised is 11 pound okay so now this problem involves using both the two points both static coefficient of friction as well as kinetic coefficient of friction remember problem can come in variety of ways but here again we are trying to study from the point of view of equilibrium and we are trying to invoke the friction law as required so here we have a constant velocity problem constant velocity means a couple is applied right and there are four we are trying to see when this slip is trying to you know when this belt is simply trying to slip what is that condition that will be my impending slip condition right then we looked at also when slip is actually happening that is there is relative slip between the belt and the pool.