 Now, what we are going to do now, we are going to do the tutorial. So, I am giving you three problems. So, tutorial is now displayed. It was also in the module. Quickly we will try to go through these problems and if you can find the answer, then it is great actually. So, for each problem, just take may be 5 to 7 minutes. They are not complex problems at all. The question is determine the range of values of P for which the equilibrium is maintained. That means two conditions are possible. One is that you are trying to lift the load up. That means impending motion of P is downward. The other one is that we are trying to hold the load. That means that load can actually try to go down. That means impending motion is upward on P. So, based on these two, we have to find out the largest you know the range of P. So, in one case, you will get a higher value. In other case, you will get a lower value. An equilibrium will be maintained as long as we are between these two values. So, you must give two values of P. That is the range of P between which it will not slip. There will no slippage in either of the directions. Get the contact angles very quickly. So, again you can see tension on this side. You have a horizontal pipe, vertical pipe. So, therefore tension in this side there is one tension. Then tension will change as the pipe is wrapped around and then again you have tension here. So, we have to draw three free bodies separately of that of the rope. I see one answer is coming up 411 pound. That is the larger load that is to pull it down. That means or rather I would say to lift the object up, to lift the 100 pound up. Then you need a higher value of P as you see and just to hold the load. That means it is trying to go down. Then you need a less P. So, the higher value is 411 pound. I have seen that some of you are giving the answer. Yes, so 411 pound and 24 pound. So it is the load has to be in between 24 pound, 24 and 400 and 11 pound. Pollution is being displayed. So, as I said you can draw the free body diagrams. So, there are two conditions. One is for P to impen downward. That means you are trying to lift the load up. In that case remember P has to be greater than 100. So therefore, P must be greater than Q and Q must be greater than R. R must be greater than 100. When impending motion is upward that means this 100 pound is going to go down. Then you need a less load to hold it. So therefore, 100 must be greater than R. R must be greater than Q and Q must be greater than P. So, there are two ways to solve it. Remember contact angles needs to be properly defined. With the horizontal pipe it is pi over 2 and with this it is pi. So therefore, we have for impending down P to impen downward 411 pound. For 100 to impen downward you can just reverse this relationship. So, this relationship will be reversed when we are talking about P to impen upward or 100 pound to impen downward. So, this will be the range of values for which equilibrium is maintained. Remember if the value of P is less than 24 that means what will happen? The block will go down. The 100 pound will start going down and then you are in motion actually. So, there will be a dynamic problem now. Similarly if P is greater than 411 pound therefore, the P will try to go down. Again it will become a dynamic problem. The next problem it is a problem that involves of that of an idler pulley. It has a small mass and it is free to rotate. So that means we can actually ignore the friction and therefore, we have tension on both sides to be same of that of the idler pulley. You can see the example of idler pulley why it is chosen basically. Remember during the operation the belt can be slacked. So, it can get slacked actually. So therefore, to tighten the belt we can use the idler pulley. In fact, idler pulley is used in many different purposes. For example, serpentin belt that means belt is going from let us say motor to some other machine tool in a some zigzag path. We also use serpentin belt to direct it in proper direction. So, idler pulley is used on that also. Remember what is given also the bigger pulley is a driver pulley. So, you see that torque system is rotating in this direction. So, this is a driver pulley. So, it is rotating this direction. This is also rotating in this direction. The question is if the tension in the slack side is 1000 Newton. So, the tension in this side is given. Find the maximum torque that can be transmitted by the pulleys for a coefficient of friction 0.3. So, what is the maximum torque transmitted? This is a driver pulley. So, just go back to the lecture materials. So, which side will have greater tension? Identify that. Then you decide based on the contact angle where the impending slip will happen. Is it on the driver pulley or is it on the driven pulley? So, contact angle will play a big role because friction coefficient is same. So, mu s is same. The answer is 234 Newton meter. 234, some of the remote sensors are already giving the answers. Ok, it looks like a lot of you are responding now with correct answer. So, as we have discussed that firstly for idler pulleys since it has a small mass and it is free to rotate. Therefore, tension has to be same on the both sides. That is the first thing. Now, we have a driver pulley. The driver pulley is this one. This is a driver pulley. So, some torque is applied that is not given, but point is it is rotating in this direction which means some couple is applied. So, balance that I need a counter clockwise couple and it can only be done if this side has a greater tension. So, tension on the lower side of the cable will be higher. So, higher tension on the lower side, the angle of contact at A and at B can be easily found out. We can see clearly. Then you express this in terms of radian. Now, remember since contact, since coefficient of friction remains same in both, but the contact angle is less in B. That means in the driven pulley therefore, this will undergo impending sleep. Once we know that, so we know the value of T1. So, from this if this is the tension side, so that is the only unknown I have. So, I found that and therefore, we can solve for the torque transmitted that is T minus 1000 multiplied by 0.15. So, this is 234 Newton meter. Solutions are also going to be posted on the module. So, you can also take a look later on. If there are any issues to discuss, we will discuss later on as well. So, let us look at the third problem. Now, third problem involves both belt friction as well as simple friction that we have studied yesterday. You see the problem. So, this crate is being lowered very slowly. So, this is important at a constant speed down an elevator shaft which has slightly. So, this gap is there is a slight gap between the crate and the elevators shaft. Now, the chord wraps around a freely turning pulley. So, again this is a freely turning pulley and then it has two rotations around a cup stand. So, we are interested in finding out the tension T needed for the operation. So, what is the T required? Tension if I have to slowly lower this crate down the elevator. Remember there is a small gap, but what happens? This is just quick hint. You can see the due to the CG of the crate through which the total weight will pass. This is offset. So, this tension that you are creating here, there is a slight offset here. The offset is 0.1 meter and due to this what happens? This crate will rotate counter clockwise. Therefore, the crate is going to be always resting at A and at B. So, it is going to slide against the wall at point A and at point B. So, now you see that whatever we have studied yesterday, this is coming to play also. Now, you have to understand that how many number of surfaces are going to participate. So, there is a point contact here. There is a point contact here. So, we have two contacting points here. Now, if you look at the kinematics of the problem that means it is coming down slowly that means the sliding has to happen at both surfaces simultaneously. That means the friction law has to be invoked right here and right here simultaneously. Remember since it is a constant motion problem, so kinetic coefficient of friction comes into play mu k. I have given mu d, but it is just mu k nothing else. So, ultimately what it boils down to finding out, try to find out the number of unknowns. So, if you draw the free body diagram, there will be a slight tilt, very slight tilt. So, now can you try to solve it? So, first try to find out the tension here by taking the free body diagram of this crate. So, take the free body diagram of the crate, try to find out the tension right here. This is exactly what we have done in the simple friction. So, here we have simple friction coming into play Coulomb friction. Then we start the belt friction talking about belt friction right here. It was just given as a mixed problem. This problem is highly intriguing and that gives you a good concept again back with the dry friction or Coulomb friction, point friction that we have studied. It is coupled with a belt friction problem right here. So, can you draw the free body diagram and give me the tension right here? I need the tension value right at the crate. Yes, some are now trying to post it. I see if the answer may be few centers are just giving the correct answer. Absolute correct answer is 10 kilo Newton. 10 kilo Newton is the absolute correct answer. So, ultimately we can clearly see if we draw the free body diagram of this. So, I am just going to go to the solution now. Please pay attention because this is also a simple friction. First we have to understand what is the free body diagram of it. Now, due to the scanning problem it is tilted actually towards the right, but that should not be it should tilt towards the left. So, it is actually tilted towards the left. There is a scanning problem here. Ultimately that has to happen because you can see the there is a imbalance T 1 and 500 right. So, that will try to create a torque. So, that torque is going to create an imbalance. So, it is trying to tilt towards the left because this force is greater than that force. So, now what happens? If you look at the free body diagram, I have one reaction here, another reaction here, then friction force is going to be up. Here it is also up. Then there is a tension here. So, how many unknowns I have? 1, 2, 3, 4, 5. If you look at general sense, the general sense suggests that I have exactly 5 unknowns, 2 reactions, 2 friction forces and 1 is the tension on the crate. Therefore, now this is a rigid body. I can only have 3 static equilibrium conditions. That means, I have now extra 2 unknowns. How do I solve that? Therefore, it is a self consistent problem. It suggests automatically that I must have 2 surfaces where I have to invoke the friction law and the problem demands that if you look at the kinematics, the motion going down. That means, at both point B and at point A, at both of these points, at point A and point B, it has to simultaneously slip. So, I apply that friction law now. So, 2 friction laws are applied, mu D multiplied by R B and mu D multiplied by R A. That means, now I have 3 unknowns and I can use the static equilibrium equation to solve the problem. Therefore, what we have here? We take the equilibrium now and through the equilibrium, first we find out the force T 1. So, T 1 will be 434. I think one Riemann center had given this before. So, T 1 is what that is being applied on the crate, on the tension on that crate. Now, as we said that there is a, now you come to the pulley and all that. So, there is a free pulley. So, that will create the same tension on the other side that is free to rotate and then you have slippage here. So, here we have to again invoke the impending condition. So, now there is a wrap. It is wrapped around, no there is no impending condition. I am sorry about that. There is a constant slippage problem again. You have the mu k again coming into play, but remember the contact angle is 2 times 2 pi because it is 2 turns. So, ultimately this side will be greater than that side because this is the direction of the slippage taking place. So, you have T 1 by T from that we can get the value of T. So, ultimate idea is that answer would be 10.02. So, what we are trying to do? We are trying to lower the load. Now, there are so many friction forces that is resisting that. So, therefore, ultimately I have to apply a very small tension in order to let that crate going down. So, that is the basic concept. Now, for 500 kilo Newton, I am just have to apply only 10 kilo Newton. So, it has reduced 50 times. So, it is a concept that is delivered based on that kind of review of what we have done in terms of let us say simple friction as well as we have a belt friction problem mixed here.