 So, now, we are going to study bell drives. Now, that is going to be important and what we will do? We will systematically look at all the free body diagrams. Now, we will study this based on first you know concept. So, we will take up a problem and try to see the concept can we draw the individual free body diagram and try to decide that where the impending slip will happen. See problem is like this. We have a motor. Let us say B has a motor. So that means it is a I will say this is a driver drum. This is my driver pulley and this is the driven pulley. So, ultimately motor you are applying some torque that torque is being transmitted to this pulley and in this pulley some machine tool is connected which is rotating. So, ultimately motor is helping that machine tool to rotate on the same direction do not forget. Now, what I will do? I am saying the velocity is constant. So, it is a constant speed problem. So, that means belt and the pulley both are rotating at same speed. But there is no slippage between the belt. That means there is no actual slippage between the belt and the pulley. They have some same angular velocities. Now, what happens? If you really try to draw the free body diagram very carefully. So, this is the free body diagram of the driver pulley. How do I decide on which side will be tension side? If the drum is rotating clockwise. So, here I am applying a torque. This is my motor. Now, that torque is simply balanced against the frictional resistance. You see that this is the frictional resistance direction. That frictional resistance as per the Newton's third law will be transferred to the belt. If you detach the belt try to draw the free body diagram. So, that is the direction of the friction on that belt. Therefore, which side should have a greater tension? It is this side. So, this side should have a greater tension. That means in this belt, this side should have a greater tension. Now, if you do not want to detach the belt from the body, then it is also clear that T2 minus T1 multiplied by the R should be equals to the moment applied. So, there is a clockwise torque. To balance that I need an anticlockwise torque. So, therefore, this side tension has to be greater than this side tension. Is that clear now? What happens to this? Ultimately torque is being transmitted here. How? Remember, I have this tension now is greater than this tension. So, therefore, I could say T2 minus T1 R. So, actually drum will be rotating like this clockwise fashion. No question about that, but how do I draw the free body diagram? Which is rotating the drum by the way? Which is rotating the drum? It is basically the frictional force that is acting. Now, look at the direction of the friction. This is helping the drum to rotate. Now, what is the M1 then? M1 is remember I have considered the free body diagram of the driven drum. So, for the driven pulley there is a machine tool attached. So, that machine tool will give a force back or torque back as per the Newton's third law on to the drum. So, that is shown here. So, ultimately you are definitely transmitting a torque T2 minus T1 R which is clockwise, but to show the equilibrium, I have a machine tool that is giving back the torque to the drum which is M1. But when you look at the machine tool which is connected here, that will have a clockwise torque at the immediate end. So, from this free body, you can clearly see ultimately this drum the driven pulley is rotated by the frictional resistance and the transmitted torque is nothing more than T2 minus T1 multiplied by R. Now, here is the question. What is the maximum power that I can transmit? What is the maximum torque that I can transmit? So, the maximum torque will be determined by the condition of impending slip. Impending slip means the belt is just trying to slip over one of the drum such that the maximum friction force has been attained. Maximum friction force is already being achieved and that is the capability that you have to transmit the torque. Now, question is where should I have the impending slip of the belt with respect to these two drums or these two pulleys. So, impending slip as I said it will be completely controlled by two parameters. One is the mu s that is the static coefficient of friction and the beta. For in this problem remember it is a constant velocity problem. So, we are going to use the static coefficient of friction. So, mu s and beta now if both have same mu s then ultimately wherever I have beta small that means contact angle small they are impending slip prevail the impending slip will first happen over there. So, with that now you can quickly solve a problem just look at this problem. So, a flat belt connects pulley A which drives a machine tool. So, now this is again the machine tool I have there is a machine tool connected here. This is the motor that is that is my driver this is a driver drum this is a driven drum as we said. Now, here we are really interested in finding out determine the largest torque which can be exerted by the belt on pulley A. So, what is the largest torque that can be transmitted? So, as I said it is very simple first we find out that where the impending slip will happen. So, we have to decide on the fact that which will go undergo the impending slip first. Remember we already know which side has the higher tension. So, it is given already maximum allowable tension in the belt is 600 Newton. So, if the allowable tension is given maximum then which side that will happen? It has to be on the top side. So, this belt here you will have the higher tension and we have already shown that based on the logic because this is a driven drum. So, here I have the maximum tension and then we decide on the contact angle since mu s remains same then from the contact angle I can determine which one will undergo impending slip. So therefore, now if I find out the impending slip where it happens first in this case you will be able to show that B will slip first because there you have a less contact angle then we have to then we can impose the friction lock condition that is T2 by T1 equals to eta mu s B and find out the other torque in the belt and once that is found then we can say that what is the torque transmitted to the pulley A is that clear? Now, you can further show that there will be no slip in this case if the impending slip is at B then there is no slip at the at pulley A by saying that T2 by T1 is less than e to the power mu s beta or the mu s if you calculate based on the relationship of T2 and T1 if you calculate the mu s that should be less than the mu s max. Mu s max is actually what is the static coefficient of friction. So, now we can easily look into this problem and we do not want to get into more detail as I said first main target will be to find out the contact angle look at the contact angles carefully. So, here I have total contact angle equals to 240 degree here it has to be 30 plus 90 that is 120 degree. So, in radiance we measure the contact angles and as I said where the impending slip will happen wherever I have low contact angle. So, it is like a friction force also mu s n remember f max equals to mu s n that was in the simple friction wherever you had a value of less n right they are friction you know they are they are impending slip was happening. So, it is always controlled by mu s and n in this case it is controlled by mu s and beta now you have two parameters. So, beta was less here impending slip happens here. So, therefore, I have one unknown in this problem which is T1 and I know this is going to be the higher tension side. So, T2 by T1 equals to e to the power mu s beta and I can solve for T1. So, now what we have to do we can check that impending slip is actually not happening here. So, how do I check it I already know the value of tensions on the both sides. So, we can check that the check would be we simply calculate the mu s that mu s should be less than 0.25 ok. The torque transmitted can be found the torque transmitted is really T2 minus T1 multiplied by R, but that is a very simple operation T2 minus T1 multiplied by R is what the torque is being transmitted. Remember this MA this is given by the machine tool to the drum I again repeat this MA is something that is given by the machine tool that is connected here to the drum ok. And this part is simply a verification that impending slip is not happening at A this is one way of doing it. Otherwise if you just do T2 by T1 that has to be less than e to the power mu s the mu s that is given mu x beta of this pulley clear. So, can we have a quick discussion on this problem specially any question on this problem ok. Just to clarification there is a question why the rotation of the driven pulley is in the opposite direction that is not true at all what is shown is the free body diagram there driven pulley will be on the same direction as that of the driver pulley. So, what is given is that is the torque if we go back to the slide this is just the torque ok this is a driven pulley torque is actually being applied if you can see this torque negative this torque multiplied by R that will of course rotate the drum in this direction. It is a constant velocity problem do not forget we have always assumed that it is a you know constant velocity is achieved. So, as per Newton's first law the torque transmitted is T2 minus T1 R. Now, this is shown as per the free body diagram because this torque is given by the machine tool to the drum. So, the machine tool that is connected will see the torque in the same direction. Direction means clockwise direction machine tool will see the torque on the clockwise direction, but drum will see the torque on counter clockwise direction, but pulley is rotating clockwise do not forget because you have T2 minus T1 multiplied by R ok. Any question just go visit? So, we are it is a camshaft. So, let us say you have a camshaft then what happens ok. So, question is it becomes complex see friction itself is a complex aspect and if you try to now in the very first class tell students that what happens in a camshaft I think it is not going anywhere. We have to be very you know kind of specialized on telling them how to draw the free body diagram. Let us not try to get into complex problems. In fact, you know camshaft is itself is in a research oriented field. Can we draw the free body diagram? Have we understood the free body diagram? Have you understand the constant motion problem? That is what we need to tell the students. As a teacher is fine we can always gain our knowledge you know through different research aspect, but my request to all of you would be keep the topic to a simple as simple as it can and try to analyze the free body diagrams. Can you see the slide now? Rotation is happening on the same direction again what I have drawn this is the free body of the forces nothing to do with the kinematics of the problem. There is a difference between kinetics and kinematics kinetics deals with the force. So, what I have shown here it is all the forces. Forces is a general term that also include the movement. So, now tell me what is missing here what we do not to rotate this drum to rotate the motor I need a torque that is being applied. Everything is rotating clockwise that is the kinematics of it fine. Now, I am trying to understand is that the torque has to be applied applied torque is this. What balances the torque? If you really take this drum out give it a thought it is balanced by the frictional resistance that is coming into play it is a constant velocity problem. So, therefore, something has to balance that torque that torque from the body is going to the pulley sorry the body from the pulley it is going to the belt. So, on the belt you see the direction is reversed that is Newton's third law. Now you can clearly see if my frictional direction is this way. So, free body of the belt if I do then this side has to have the larger tension because the resistance is in this way clockwise. So, this is my T2 greater than T1 is that clear now. Similarly, in this one this torque is a consequence of the machine tool giving back to the drum. If you are detaching isolating the drum this is the torque coming from the machine tool to the drum because everything has to be in equilibrium it is a equilibrium problem constant velocity Newton's first law. So, therefore, this is the torque, but it does not mean it is rotating the drum in this direction it is just the torque that is coming into play this torque is balanced by T2 minus T1 multiplied by r because this is a higher torque. So, this is going to give it a clockwise torque. So, what the purpose of saying that this is very helpful to understand why this thing is rotating what is making it to rotate it is simply rotated by the frictional resistance you see that this friction force will always rotate that drum clockwise it cannot rotate counter clockwise clear minimum contact angle that will be determined by what you tell me now see your question is that what is the minimum contact angle you tell me give the answer tension it is determined by tension. So, it depends the question should be posed based on the problem framework see if your question is what is the I want to drive this much of maximum torque let us say your question is like that I want to drive this much of maximum torque what is the contact angle necessary that makes more relevant minimum contact angle that is desired okay thank you yeah that is a you know. So, your question is how do you draw the free body diagram of this pulley see here first of all you have a T2 and T1 right. So, based on that remember this is a fixed this is not a hinge right. So, do you want me to draw the free body diagram quickly I will draw it for you okay I am now drawing the free body diagram okay both the pulley and the belt okay. So, in this problem now let us say first of all understand we said the kinematics of the problem is that it is slipping in this direction right okay. So, this is the slip let us draw step by step everything is clear what I will do I will try to detach the belt first. So, I have detached the belt if I detach the belt what kind of friction force it is going to see since the slip direction is this way the friction force is going to be is that clear if this is my friction force direction then which direction the tension will be higher is that clear now okay now consider the drum drum is fixed means it can take a reaction as a torque do not forget that if I have to fix it that means it can take a reaction as a torque okay. So, now you transfer this external resistance in here. So, that means this is the you know complete detached free body diagram. So, here you have this contact angle right that contact angle was there. So, this area we have the friction force right then you have the reactions reactions and how the moment should come into play moment should come into play like this way this is the moment reaction. Now if you combine these two together if I really combine these two then all you see is do you do not see the interface friction force friction force at the interface. So, your free body diagram becomes okay. Now you tell me what is the problem to solve for these three reactions if I know T2 and T1 is there any problem I take sum of force along x0 sum of force along y0 and moment about O right sum of forces equals to 0. So, remember this m0 will therefore be simply equals to T2 minus T1 multiplied by R. So, remember the entire thing here is it is a fixed drum if it is not fixed that means you make a pin here it is going to rotate okay then you have a acceleration or deceleration depending on what you are doing on the system. So, again see you know dynamic problem if say constant motion problem those things can be analyzed based on the fact or very slow motion can be analyzed based on static problem whatever static problem we have done based on those concept it can always be solved is that clear just one more problem of that of a break okay band break we have discussed. So, let us say I have a band break now again remember we are posing the problem in a such a way I can solve it based on equilibrium and I have thoroughly told you that each and every free body diagram can be drawn free body diagram can be drawn and equilibrium can be taken no question about that okay. Now in this problem I have a differential band break what it does it is rotating at a constant speed. So, there is let us say some axle is there and it is just rotating like this does not matter okay it is pin about this point it is rotate what we are doing essentially that this band okay this belt is allowing it to maintain a constant speed again how it is maintaining a constant speed because remember for a given value of force I have some tension induced here therefore I have friction force that friction force will be you know fixed for a given value of P in other words for a given value of P I am going to get some given tension here and therefore the torque can be controlled. So, to maintain a constant speed I need basically let us say Ta minus Tb equals Ta minus Tb multiplied by R the problem is clear now what is being asked determine the magnitude of the force P that is exerted on end A of the lever arm when the drum is rotating clockwise and the drum is rotating counter clockwise remember it is rotating at a constant speed and for that we are transferring a couple of magnitude 125 pound-foot that means a torque is already being said to be there and this torque is going to keep the speed constant. So, the only question is if the torque is clockwise if the torque is counter clockwise then what is the value of the force remember this band break this band will be constantly slipping over this drum that means there is always a relative slip between the band and the drum. So, as I said there is a relative slip between belt and drum what we have to do moment equilibrium of drum about center of drum will give the tension generated in both sides of the drum again moment equilibrium of lever that means this is the lever ABC ABD about point B will give the force P how many unknowns do I have in this problem see I have actually if you look at it carefully very very carefully this is my unknown P tension in B tension in A three unknowns plus there are two reactions if I just look at this right. Now, how many equations now for this lever how many equilibrium equations I have three that means I have two extra unknowns those two extra unknowns how I am going to solve basically I have to invoke the friction law that means there has to be a relationship between T A and T B as well as remember we are also given with a condition here that the magnitude of the couple is given that means T B minus T 1 multiplied by R is also given. So, to look at the problem in this way we can clearly draw all the free body diagrams. So, first problem was it is rotating clockwise that means a clockwise torque is also given you see that clockwise torque if a clockwise torque is given that way then which side of the tension should be greater T A T A should be greater than T C has to happen to balance that torque. Now therefore, from the first part I just solve the moment is given that torque applied torque is given. So, I write down that T A minus T C multiplied by R that must be equals to the torque remember it is just converted by this tool do not look at the numbers here T A minus T C multiplied by R equals to 125 and it is converted in inch. So, this is pound inch. So, I get a relationship of T A and T C what is my second condition for slip in belt actual slip relative slip. I have T A by T C that should be equals to e to the power mu k multiplied by the contact angle. So, T A and T C can be solved clear. So, similarly you can now think of once I know the T A and T C I can now take the equilibrium of this arm lever arm. So, this lever equilibrium suggests that I should be able to solve for the force P since T A and T C is already known I take the moment about point B to solve. So, ultimately we are going to get the answer that is the P required equals to this. That means, to maintain that constant speed at that value of torque and the drum is rotating clockwise I have to apply a force P equals to 146 pound. Now, the problem can just be reversed. Reversed mean now it is rotating counter clockwise. So, if it rotates counter clockwise in the anticlockwise direction then what happens then I have T C greater than T A that is the only thing. So, I will just reverse the problem T C and T A is reversed this will be exactly same. So, you are going to get an answer of 9.41. So, that means, to rotate it clockwise I need larger force than to rotate it counter clockwise. Now, I leave it to for you in homework then why to rotate in clockwise I need a larger force than that of this when I am rotating counter clockwise why do I need a smaller force. So, let us take that as a homework and we can discuss later on. So, 2 minute discussion 2 3 minutes anyone has any question regarding this everything as I am saying that as long as static is concerned we can also include in static constant velocity constant motion problem no issue and we can simply try to draw the free body diagrams to get the desired forces. Remember every time I am solving a statically determinate problem in engineering mechanics I cannot deliver a knowledge on statically indeterminate. I need to go to the advanced level of studies where the statically indeterminate problem is solved based on the concept of kinematics that means, actual deformation of the body. Yes 1 0 8 8. Starting problem number 2 why the weight of block A is greater than tension T 2. Your question is in this problem why the tension in this side is greater than this side. Look at the problem statement who is direction rotating it is rotating clockwise that means, you are actually applying a torque. So, if I apply a clockwise torque to balance that what needs to be happened what we have to do I need a counter clockwise torque. How do I generate a counter clockwise torque W A minus this multiplied by R is in that clear therefore, W A has to be greater than this clear now good question. Sir why the weight of belt is not considered in the derivation of belt friction what will happen if I even consider that we are not physicists we are engineering we do not care about 10 percent error in our solution is that answer to your question usually tension will be much higher than if I have to include that effect of load 1 double to 4. Sir I have the answer for second problem you asked that because the weight of the AB should be lifted up that is why for rotating clockwise it is more difficult than rotating anti clockwise. 1 0 3 7 How the friction between belt and pulley can be reduced if we use oil or grease then well we will start slipping how to reduce friction between belt and pulley. Your question is how to reduce the friction between belt and pulley why do I want to do that I want to transmit the maximum torque possible then why should I try to reduce it I should try to maximize it right. In case of fluid friction sir when to take dynamic viscosity and when to take kinematic we are not talking about fluid mechanics here. So, we are limiting our set to only static coefficient of friction and kinetic coefficient 1 0 4 3 can you rear me yeah just go ahead please. Yes sir you already explained in the problem number 2 that WA the weight WA is more it is more than tension in the rope but still we are not getting it. See what we are trying to look at the problem statement what is being said in the problem number 2 right I am applying the you know pulley is rotated very slowly clockwise. See I am trying to raise the load remember I have one load here on that pulley and then there are other connections through that there is another load 16 pound right. So, I am slowly rotating it to raise the load. So, rotating direction remember is clockwise now how do you rotate a body first of all I have to apply a torque that is not given but I know that applied torque has to be balanced by the tension produced. So, now you think of it I am applying a clockwise torque which is not given in the problem that is not necessary. Now therefore, to balance that clockwise torque I need a counter clockwise torque. So, now how to counter clockwise torque will come into play WA multiplied by R negative tension on the other side multiplied by R because therefore WA has to be greater than the tension on the other side okay clear. You can also look at the way I have shown the friction right again if you think if you detach the belt if you detach the belt way belt then what is actually causing that rotation what is helping this rotation friction right the friction is causing that rotation to happen is not that. So, from that also you can get this but simple way to look at is that you are rotating it clockwise but I need a counter clockwise torque to balance it clear okay. I have another question sir if what happened if rope is wound spirally on the capstone rope is what will be the effect on wound spirally sir spiral not horizontally in the circular form spirally fine no that it is the. So, your question is if the rope is spiral in the sense what that means you are trying to tie it many times like the Bollard problem bouncing spirally sir many times yeah. So, that is a wrap right how many wrap you have in a let us say Bollard or a wrap in spirally yes yes. So, in that way it is all depends on the contact angle nothing else we just have to figure out what is the contact angle and we have solved that problem in a simple way right that Bollard problem you remember when you go to the jetty no when you go to a fish jetty you see the boats are coming and someone suddenly jumps from the boat and try to put the you know rope around the that Bollard or you know. So, that is a wrapping problem practically they are wrapping like that only but what we studied in that they must be wrapped closely. Okay I mean yeah. So, that will make it more complex yes yes that will make it more complex nothing else I mean okay yeah I think we can now take a break I think.