 Let us continue what we were discussing in the last class. We were looking at the forces that act on a vehicle as the vehicle goes over a ramp or a slope. One of the key things that we discussed in the last class was about the rolling resistance of the tire. There are a number of questions after the class. I would like to explain that again so that you are clear about how we have converted the rolling resistance, the phenomena of rolling resistance into a force. In fact, this is the key question that was asked after the class. Let me explain that again. We talked about equilibrium with the pressure, inflation pressure. We will come to that later. The only point which I wanted to make is that every part of the tire has to be under equilibrium with the lower contact pressure that is acting as well as the inflation pressure. So, at the right of the center when there is nothing else is happening, then the contact pressure should be equilibrated by the inflation pressure. We will talk about that when we come to the tire. I just wanted to make a passing comment. We will explain it later. Now, let us come back to this topic of rolling resistance. We said that the rolling resistance is due to viscoelasticity and we said that that is also that also results in an unsymmetric pressure distribution. We said that if the tire is stationary standing, then it has to be a symmetric pressure distribution where a number of tread blocks are involved and as it when it is rolling, when we follow a tread block, the pressure distribution due to a tread block will not be symmetric because the loading and the unloading curves of a viscoelastic materials are different and so it has to be unsymmetric and we said loading curves being higher than the unloading curves. We said it has to be like this and that this results in an unsymmetric force, the force which is the force that is exerted by the ground onto the wheel. That force is not going to be symmetric right at the center but will be away from the center. We will call this later. We will give specific names to these things. We will not do it right now but let us understand first that there is a shift. There is a trail. Now, this force is the reality because of viscoelasticity. So, if I now want to shift the force to another place right at the center, I will have to do it with the force as well as a moment that is active. So, if this is the direction of rotation and hence the driving torque's direction, then this force would give rise to a torque which opposes this driving torque and so will be in that direction. It opposes the driving torque. The question that was asked is how did I now replace this force by this force like this? Before that, how did I replace the driving torque force by a force like this? By a force at the ground. This was the question that was asked. Please, I want to take you back to the free body diagram. It is a very simple thing. So, we will explain this with a very simple example so that you understand small niceties of drawing a free body diagram. Suppose let us take that I have a tractor pulling a trailer, a horse drawing a cart which is a very early example. Now, there is a link and now this tractor is pulling this trailer which means that this tractor applies a force in this direction to the trailer. Now, if you want to draw the free body diagram of this tractor, what will you do? Apart from other forces, we will just neglect the other forces. So, I have to draw a free body diagram. What will be the direction of the force? It will be in this direction. So, this will be the direction in which I will draw this free body diagram. So, in other words, I will look at the action of the trailer onto the tractor. So, what does the trailer do to the tractor? It is what I will look at and apply this. I will neither apply two forces in this direction. That means it cancels. That is ridiculous. And I want to apply a force in this direction that in that case it looks like trailer is pulling, pushing it. There is no resistance. So, I will apply it obviously only in this direction. So, that is the reaction that you apply. In the same fashion, when the tire rolls, it actually pushes the ground. You can imagine that it pushes the ground. You would have seen this when there is water over the sand. It pushes the ground and so the force that it exerts on the ground will be in this direction. Like the force that is exerted here is in this direction. So, when I want to put the free body diagram now with the force that has to be acting on the tire, which I want to represent. Now, in other words, the effect of road like we have effect of trailer and tractor, effect of road on to the vehicle, I have to put a force in the opposite direction. So, hence this force now becomes this. Clear? It force the force becomes like that. So, this is the now the traction force and that is due to the friction. We will see how friction holds this. That is the first one. Now, since the traction force is in this direction, I have substituted it with the force like this. Because of the effect of the ground on to the wheel, I have substituted it like that. I do the same thing for rolling resistance torque, which is produced by this force. So, when I shifted it here, I had a force in this direction and I also had that moment and now that is very similar to the traction force. So, I replace it with another force that is acting like this. So, it is equivalent, same thing, same torque which is that is the F rolling resistance and that is the F traction. Clear? That is it. So, that is why we have two of them, the front traction force and the rolling resistance force, which opposes the motion. Clear? Any questions? Yes. So, what is friction? What is friction coefficient? We will come to that in a minute. How does the friction act? We will come to that in a minute. Let us understand the forces that are acting. So, now that is the rolling resistance force. Yes, that is what we explained last class that we will have a shift because of the viscoelastic effect and that the loading, when tread block goes through loading and unloading, then the path for load are different. One is loading and another is unloading. So, for the same displacements or the same strain, the forces that are given, sigma are different and hence, there cannot be the same for the same, whether it is loading or unloading will decide that the stresses are higher or lower because of which forces are higher or lower and that is the reason why it cannot be the same. So, that is why this is shift. Now, having come to this, we will go back to our equilibrium equations and write down the equilibrium equations for these things. The direction in which you draw that rolling resistance of the... Yes. So, what decides, what is this one? So, what is this force? What is the direction of this force? What is the direction of this force? That is the question. Obviously, when I am giving a driving torque, the force is in this direction. If I am giving a braking torque, it will be in the opposite direction. Obviously, that is the thing. On the other hand, rolling resistance depends upon the direction of rolling. I am rolling it in... I am rolling it like this. The rolling resistance does not care whether you are actually accelerating, decelerating or doing neither of the two or either of the two and traveling at the same velocity. It does not matter to it. As long as there is compression and relaxation, we are going into the viscoelastic effect and that will give the rolling resistance. So, in either case, you will have that force here. So, interestingly what really happens, which we pointed out in the last class, that the rolling resistance is something like braking and so helps braking when the vehicle brakes and it decelerates the accelerating vehicle. In other words, it is essentially say a bad force. It is acting. Either way, since it opposes the motion, we are to compensate for this energy lost from the engine and so it is a gas gussler. It consumes energy, so it is a gas gussler. So, the best thing, it is a very, very important force now. Especially if you look at passenger, whether it is a passenger car in commercial vehicles, especially in passenger car today, everyone wants very, very low rolling resistance. And when rolling resistance have to be low, you also get into one more problem. Because I said that it acts in the same direction of the braking force, so the braking force takes a beating, total force takes a beating. So, you have to be careful. You will see later that the tire itself, tire design itself is like a spider diagram. You cannot improve everything. When we improve something, there has to be a sacrifice on something else. So, we will continue now. We will write down the equations here. Note that I had, I am up the hill, so I have W, this is W that is acting and that is the theta s, which is the slope. So, I resolve that. So, this force becomes here, it becomes W cos theta s and that becomes W sin theta s and so on. One of the assumptions that we have made, obviously is that cos theta is approximately equal to 1 in order to write this. Strictly speaking, I have to now resolve it and put it, but I am just, I am going to write that. In other words, I said that I will replace the rolling resistance force by means of a rolling coefficient or rolling resistance coefficient F r into W. Actually, the W has to be in this direction and so I have to actually resolve it and put it as W cos theta s. But I am assuming that that cos theta is very small so cos theta is equal to 1. So, I am just replacing this r r f as F r into W f and so on. Now, again, the rolling resistance coefficient, we are assuming that is the same in the front and the rear. So, these are small assumptions. Now, we will write down, we will write down the equilibrium equations. Yes, this is normal force of course, exactly. So, that is what I just pointed out that we had a force which is away. When I want to now shift it to the center, then we will have a force, normal force will still be there plus a moment and the moment becomes the horizontal force and the normal force is still there. So, that is a reaction force has to be there. That is a reaction force. Let us continue now. Let us write down two things we are going to do. One is of course, the well known F is equal to ma. We will write down F is equal to ma equation and the other one which we are going to write down, that is the first set of equations. So, let me write the other way ma is equal to the forces that are acting in the, I said that this would be the x direction. So, we are writing this F is equal to ma in the x direction. So, what are the forces that are acting? The two forces that are going to aid the vehicle to accelerate are the traction forces F F and F R. So, these are the traction forces and then the forces that are going to oppose the motion are the aerodynamic forces. F forces as we are going to see now or later is that it is proportional to V square, velocity square. So, it is a several parabolic distribution. So, at higher speeds they are going to be important. So, there is a drag coefficient C D and you are going to look at the projected area A and you are going to look at V square. So, half rho A, C D A V square. We will see that bit later. So, we are now dumping everything as R A. Then I have the rolling resistance force front and rear. I am combining them and writing them as the rolling resistance force. So, what is the other force that is acting against it? It is W sin theta s, which we will call as the gravitational force, a resistance due to gravitational force. You can call that as R G if you want and the other force which opposes is a drawbar pull. We will neglect that. We can add it if you want. We will later neglect it. Drawbar pull is due to the force is applied if you have a trailer attached to it and so that is the drawbar pull. The next assumption which we are going to make is that H which is the centre of gravity height is equal to HA is equal to HD. It is quite valid for a car, passenger car. So, we are going to make an assumption like that. Now, my next job is to find out W F and W R and W F and W R are determined by taking moments about that point A and this point B. That is a drawbar pull. In other words, if you have a trailer attached to it, then that is the force. So, I am neglecting that. What is, how do you calculate W F and W R? Just take the moment about A. Let us get that. I hope you do not miss out anything. So, W F, let us see, look at this. W F into L in the clockwise direction plus RA into HY plus that is also in the clockwise direction. Look at the direction if I make a mistake, substitute a point A. The onus is on U. So then, what are the other forces? Rg, W sin theta that is also in the clockwise direction into H. Now, I am going to replace this MA by means of that is this force by means of a D'Alembert's force, which is W by g into A. So, when I replace that, then that is also in the clockwise direction. So, g into A into H. I also have this force, which is W cos theta, that force and that force acts in the anticlockwise direction. So, I got to write that as, because this is in the anticlockwise direction, write this as W cos theta S into multiplied by L2 is equal to 0. Rearrange it, but only a comment which I am going to make and which will be valid for the next class, this class and next class is that. You have to be careful with this sign here. It would vary whether you are going uphill or downhill. So, you have to be careful in that. So, right now, we are going uphill and so it does the same. So, it opposes. So, it is the same direction as that of Ra. If it is downhill, then the direction will be different. It will be opposite to Ra. So, good thing to do is to look at Ra and where we stand. What is the direction? So, write down and other thing which we said is that we will make this to be equal to 1. Sin theta will retain it, so that we will get W1 by L into WL2 minus the rest of the quantity. Why not you write that? Minus Ra into H minus W sin theta into H minus W by Ga into H. So, I am just going to put minus or plus here to take into account whether it is going uphill or downhill. Write down for now WR. I will give a minute. Write it down for WR. How will it be? Just take the moment about B now. 1 by L will have WL1 into cos theta. Ra will become now plus Ra into H. So, all the directions will be different plus minus W sin theta H plus W by Ga into H. So, let me call this equation as 1. That equation is 2 equals 3. It is simple equations. Nothing, all of you can derive it very easily. Let us rearrange this equation in a fashion that will be easier for us to interpret. So, I will write down that as WLF is equal to I will take the first term L2 by L, the first term. So, I will put a minus sign there minus H by L into, let me write down the rest of them, Ra plus. So, how should it be? Just the same thing, but so that the sum has to be, we will consistently use just plus or minus so that you will know when uphill, downhill you will accordingly, you will do that so that minus plus plus minus there will be a confusion. You can very easily find out which is plus which is minus. So, I will consistently use plus minus. According to the situation you will put plus and minus, sin theta S plus W by G into A. Now, let me call this as 4 and 5. I will simplify the 4 and 5 equation by using equation 1, substitute that from the first equation so that WF is equal to L2 by L into W minus H by L. When I want to substitute that so that I have F and when I substitute that, what would be that there? For Ra, W and Rd, of course, we have made it 0 and Rg so it will be, this will be F so it will simply be, what will that be? When I substitute there in that expression, I will have to bring all those rest of it to the third three terms to the right hand side. So, F minus Rr, which I will replace by Fr into, obviously Wr will be L2 by L W plus H by L into F plus sorry F minus. Obviously, when I add these two equations or these two equations 6 and 7 or 5 and 6, you would notice happy that you are very active, lunch time has not dampened your enthusiasm, I am very happy. So, that is L1. So, now obvious that when I sum the map WF plus WL has to be W and that is what has happened here. So, you see a minus sign here and a plus sign here, which we experience every day when you do an acceleration that there is your weight actually goes back or there is a redistribution of weights in the vehicle and so the front wheel loses some of the reactions in the rear gains this reaction. So, sum of them is equal to 0. Simple, I do not think there will be any questions on this, we will go to the next step. Now, as we have written acceleration, you see in every advertisement of cars that 0 to 16, 6 seconds or 0 to 19, 8 seconds and so on and so forth. So, what are the limitations of this driving force or F? Can I have unlimited F? That is our first question or is it limited or is that a F max? Of course, the attractive force as it is called is controlled by two things. One is the ability of your power plant or the engine to generate the kind of torque, which is applicable or which is available at the wheel. That is the first thing. Maybe your car would not be able to develop it, that is the first thing. Another important thing is the ability of the tire to take that force, the FF without you without what would happen if it is more, you would have seen it that for example, if it is a wet road or an ice when you have traction, the wheel starts rotating slips, major slip. We have to be very careful in using these terms. Slip has another meaning as far as the tire is concerned. Right now, we say that we use a word English word called slip, but we will then refine it when we study tires. So, the ability of the tire to support that force, traction force or in other words the tire without as you said slipping. So, the tire is the next reason for it. So, we can write that the F max in an analogy to the coulomb friction. We can write this to be mu into analogy to coulomb friction. We can write it as mu into W. Is it coulomb friction that acts? Is this mu very similar to what I know? Some box sliding and a friction coefficient mu acting. These debates are very important and interesting. We will go into those debates when we look at tire mechanics. So, when I say mu into W, mu is a sort of a dumped quantity. So, mu is equal to or equivalent, I would say equivalent coulomb friction. So, that law brings us to the difference between a front wheel drive and a rear wheel drive. So, obviously when it is a front wheel drive or both, it can also be four wheel drive. Let us now look at only the front wheel and the rear wheel drive. So, mu is equal to W F which is the front wheel drive, maximum force and if it happens to be a rear wheel drive, then mu W R will be the maximum force. So, what I am now going to do? I am going to do is to substitute for F. It is a front wheel drive or a rear wheel drive. Mu W F is equal to F max. I will substitute for F, there here with some front wheel and the rear wheel drive and then rearrange these quantities. So, that I will know what is the F max or A by G. Very simple now. The first step is I will just substitute in this equation. Let me call that as 8 and 9. Substitute equation 6 there and I will rearrange the terms for F max because there is an F there. Rearrange the term F max. So, I will just write down what is F max. You can do that as substituting or using I will say using 6 and 6 into 8, 6 and 7 into 8 and 9 if you want. So, I can say F max. Maybe you can quickly do that. Thus for a front wheel drive is equal to mu sorry mu W into L2 is there. So, W have taken it out. So, L2 plus FR into H, there is a L there divided by W. Since I had rearranged the terms, I will get divided by 1 plus mu H by L. So, this is the maximum force that is developed if it happens to be a front wheel drive. If it happens to be a rear wheel drive, I will do the same thing with the second equation and I will get something like this for the rear wheel drive. The maximum F max is equal to mu W into L1 minus FR into H divided by L whole thing divided by 1 minus mu H. Verify this, I hope this is correct L minus no L1 sorry mu W into minus FR into H divided by L. No, we had substituted, please note we have substituted RR by FR into W. So, the W is what we are taking it out. So, that is why we get a W here. If I put RR then it would not come. That is L1. Now, this is the front wheel and the rear wheel drive, the maximum force that is developed. So, beyond this, the wheel will not develop a force and we will not be able to reach the kind of deceleration that we want. Now, what we are going to do is to change track and look at braking. So, this is the maximum force and braking becomes very important and we will change it and look at braking. The forces that are acting, all other forces are the same for braking except that there is not going to be a traction force but there is going to be a braking force. So, the braking force will be in the opposite direction. So, that will be the braking force FB front and that will be the braking force. Is that the only difference? One more difference is that we are going to decelerate. So, I will replace though it is not the right thing to do. It is not theoretically very correct. I should have A and put it as minus A. That is what I should be doing but just for ease of understanding, every time you have to look at A then you have to say that actually I am in minus A when I am braking plus A when I am accelerating and so on. I will replace A minus A by D. So, I would replace A with D where D is nothing but deceleration. That makes my job much simpler. So, when I am decelerating what happens to my D'Alembert's force then it will be in the switch. This will switch. Now, let us write down same thing. I am going to follow the same procedure. Nothing very difficult or okay. So, let us write down the forces. Why not do that in a minute? Now you can very easily look at this. So, the only thing I am going to write the first equation is W by G into D I will write is equal to rolling resistance force R A. Then what is that it is supposing? I will put this as plus or minus W sin theta. You know the physics. So, that would be the second opposing force and rolling resistance force is now opposing the motion and of course we have the braking force, the front braking force plus the force of front braking force here. I am neglecting the drawbar pull and all this because I did not want to add some more on to this. No. This is that is the accelerating direction. I am now decelerating. I am decelerating. So, I am breaking. So, I assume that I am decelerating. Assuming that I am of course decelerating. So, the decelerating, the deceleration is written like this. Actually, how should I write this? I should have written like this. W by G into A is equal to minus R A minus and so on, R R and so on. So, this is the so what I have done is actually I should have written it as when I write F is equal to M A. If I write the directions here, I am explicitly writing the directions here. Obviously, A will become negative. Hence, I am replacing that negative A with D. So, in other words I am replacing all the negative forces by plus forces. Just to understand this, that is all. No, no, no, no. Please note that minus A is what I am replacing with D. I said I would not use the symbol A, but I will use it D. D is equal to minus A. That is what I said. If you want minus A is equal to D. The symbol D indicates deceleration. Obviously, it is minus A. There is no questions about it. Do the same thing. Take the moment about A, moment about B in order to find out the W F and W R. I am not going into those details again. I think I hope I am sure all of you can do that. It is not a problem. So, I will only write the final result for W F and this will be 1 by L into what will be the case, W into L2 minus R plus. It has to be plus because the force transfer will be such that there will be more load to the front when you decelerate. So, this will be plus H into FB. I am putting this as a total breaking force plus H into FB. This is F R into W. Once I write this, you can write W R. I am skipping steps here. This is on breaking. So, I am calling that as 1. I will call this as 2. I will call this as 3. Strictly speaking, I have to split this. I am going to do that as FB front plus FB the maximum breaking force, which I would call as FB F max or F max breaking force in the front and the maximum breaking force of the rear are similarly controlled by mu because it is that quantity called friction, which gives the ability for the breaking force to be developed. You know this all the time. So, that is equal to mu into W F and maximum breaking force at the rear is equal to mu into F. Now, this is an interesting factor here. This is an interesting factor here. In the previous case, we looked at the vehicle and said it is a front wheel drive or a rear wheel drive. Now, we cannot do that in breaking because breaking is applied to both the wheels. So, if I say that total force, breaking force is FB, a fraction of the total breaking force is applied to the front and a fraction of the breaking force is applied to the rear. Let me call that fraction as front to be KBF into FB and the fraction at the rear to be FBR. So, KBR into FB, obviously KBR has to be 1 minus. So, the key question which we are going to answer in the next class is how should the breaking force be distributed between the front and the rear? That is the question which we are going to answer in the next class. Stop here and we will continue in the next class.