So with that, I will let you guys look over and introduce yourselves and go from here. Oh, wait, wait, wait. Sorry, sorry, sorry. One thing. I like, I'm not done yet. So this is our last talk in the Python track at 255 Eastern. So in 30 minutes from now, there is a closing ceremony talk. It's going to be the three track leaders in the bar. That's going to be over in the closing ceremony track. So there's a link to it from the agenda. If you go to the top left of your screen where you see the schedule and you see the name of the session, and you click more, you should see a closing ceremony. The platform should let me bump everyone over to it automatically. So depending on time it may happen and you still kind of put thank you and warn you on bumping you or you may all of a sudden just get unceremoniously dropped into closing ceremony depending on how things go from time. And then the event head over there for our final thoughts. And that's where we're going to wrap it up. It's not going to be in this track. It's in that specific track. But that's it. I hand it over to you. I would just start presenting the screen. Hello, everyone. My name is Farul Singh and I'm a software engineer at Red Hat. I'm sorry if there's a background noise. It's, it's, it's terrible. Like there's some construction going on my building and I can't have that. So, yeah, I am a software engineer at Red Hat. And today we are going to talk about content computing on OpenShift. With me you have a smell and Luciano from IBM. Why don't you introduce yourself? Hello, people. I am Ismail Faro. I am the technical lead in the cloud and software part of IBM Reset. Hello, I'm Luciano. I'm in a senior developer in Qiskit, our quantum SDK at IBM. So we will start with the, going to talk about how the future of computing is probably going to look like. Then we will discuss some key concepts of quantum computing. Then I will demo how to launch a development environment for implementing quantum algorithm using Qiskit and OpenShift. And the team from IBM would demo how to develop a quantum circuit using Qiskit. So the age of computing started with bits as the only unit of information. Today we have both bits and neurons at the heart of computing. But there's a newcomer and that is Q-bits around the corner that can give us unprecedented advantage to computing in the fields like biology, AI, physics, you name it. Right now, SU Summit is the most powerful supercomputer that we have and it is comprised of bits and neurons system. It can do 200,000 trillion calculations per second, but it's even difficult to say it aloud. But the question is, can it solve all the problems? And the obvious answer is no. And to give you some perspective, it is impossible to completely represent the molecular configuration of Canfine on today's most powerful supercomputer. But we could do so just using 160 Q-bits on a quantum computer. And if you want to build a classical computer that is capable of doing so, you would have to have at least 10 to power 48 bits. And to build that supercomputer, you would almost consume 1 to 10% of the total atoms present on the earth. So yes, it's impossible to build such a classical computer. So there are a lot of problems that quantum computer can help us solve and they can provide us a new path to solve some of the hardest of the most memory intensive problems in business and science. Now, we have some key concepts that whenever you are in a quantum computing group or community, you come across these words. So classical computers are represented by bits that can take value 0 or 1. Quantum bits or Q-bits can take on those values, but can also pre-represent a combination of 0th or 1 state while you're computing. So if you think as the 0th state as a north pole of the sphere and 1 state as a south pole of the sphere, one of the states can be the equatorial position where the Q-bits can have the 0th state or the 1 state with equal probability. And superposition is creating a quantum state that is a combination of both 0th and 1 state. And to do so, you pass it through or you pass a Q-bit through a Hadamard gate. Similarly, entanglement is a concept that strongly connects two or more Q-bits so that they quantum states are no longer independent and behave as a single entity. And to put them into entanglement, you pass them through a control-naught gate. And Luciano will talk about these gates later in the slides, so don't worry about it. Let me ask you about the IBM quantum thing in IBM Research. We're working from a while in this quantum idea. In fact, four years ago, we released the first quantum computer. A general quantum computer online has a lot of success. Seeing that in the beginning, like Paolo said in this moment, if you try to simulate something, we need to have one interesting number of Q-bits in the beginning when we release the first version of quantum experience, our first quantum system online, we have only five Q-bits. Nowadays, the last week, we released one device that has 65 Q-bits. And just today, we announced that in 2023, we are going to be able to release one device that can handle 1,000 Q-bits. This means the hardware is very important for us because finally it's how we can manipulate these Q-bits. But obviously, the software is the key important part in our ecosystem of tools. In 2007, I was one of the first contributors to try to release the first version of Q-Bits. After that, Q-Bits is our first open-source solution to allow people to create algorithms and circuits and create all the software on top of this hardware to allow to the people in one moment to create or build some kind of application that try to connect the real world with this quantum computer era. Additionally, we work and have some kind of services to teach people and to help people to understand how work these quantum computers and obviously to have access to these quantum computers. This thing connects with the last pillar for us that is the quantum community. We try to pay a lot of attention and teach people how work these quantum computing phenomena and how you can develop something using Q-Bits on top of all of these layers of the abstraction that we are going to show to you. Additionally, we work with researchers, developers and business companies to try to figure out what are going to be the next things to continue to address or define what are going to be the next challenge and try to cover that using our hardware and our software. This video is a video that showed to you all the executions that happened in the last four years. In the beginning, obviously we have more simulators than quantum devices thinking that in the beginning we have only one quantum device and we have several simulators on the cloud but nowadays we have, we delivered in the last four years 29 quantum devices and you can see, I don't know, it's easy to see. In this moment the blue line is the execution in the real chips and the red line is the execution in the simulators. In this moment we have more than 250,000 users registered in our platform and you can see that the execution in the last two years is most in our quantum real devices. That helped us to pass from the research to the production. And obviously to try to achieve all of these executions and allow the people to use our quantum devices like we mentioned it before, we have this Qiskie that is our open source SDK and Luciano are going to show to you in detail how you can develop your first quantum hello wall. But for half a very high view about what is this Qiskie SDK or what is the element that try to cover, we have these like levels of the abstraction. On the top, the applications and programs obviously is the most easy way to allow to the people to enter and start to interact with our quantum devices and our quantum algorithm. One example like Paru said is like example, in applications we have or we try to support chemistry simulation, machine learning or initiation problems. This means some experts in this kind of areas can use the quantum devices without need to go in deep and understand what happens in these quantum devices. The next level is the patterns and the circuit. The circuit is finally our building blocks and Luciano are going to show to you in detail how you can develop this or use these elements to develop your first quantum program. After that we have this transpiler that in the general world for us is like a compiler because it helps us to optimize the quantum programs on top of the specific quantum devices. Thinking that we in this moment use some kind of technology that allows us to manipulate these quantum devices or qubits using microwave pulses. The next level is the experimental tools. The layer that our researchers use to improve our qubits and all of the hardware in low-level assets. After that we have the simulators to allow to the people to simulate in your local machine or in the cloud, make simulations, quantum simulations in a very easy way. After that use the quantum device to finalize or to try to compare if your program runs in the same way or not. And the last layer that is the IBM provider is the connector that connects your quits in local or in the cloud with our quantum hardware. Next slide. So I work at Red Hat in the emerging technologies in the office of CTO and right now what we are doing is exploring mechanism to connect the classical and the quantum world together. And I emphasize we are not productizing this right now. We are just doing exploratory work. With OpenShift we have one of the best of the classical computing environments and IBM Research has developed quantum computers based on superconducting qubits. So we are prototyping to define the best practices for running heterogeneous workflow in a co-processor model using OpenShift and IBM Quantum Services. Like Pavel mentioned, our idea is to have this hybrid cloud between the user, the cloud and obviously our real devices, real quantum devices. And from our side the strategy is very simple. It continues to double down in the open source part that is our Qiskie framework. And you can see in this diagram that in the cloud we use obviously all this knowledge and all this development in the Qiskie side like part of our services because finally the transpilation, all the demonstrations can happen in local or can happen in the cloud. And in the same way that when the people need to use this kind of quantum algorithms on a very high level, a structural level in our solar stack we need to have some kind of ability to execute this like part of our cloud services on our cloud stack. In our collaboration with Rehab the idea again is going in the same direction like trying to make open source all these elements that we are working on and the OpenShift Qiskie operator is one example. For us the OpenShift Qiskie operator is one of the key elements to allow someone to deploy in a very simple way the Qiskie and all the Qiskie tools in one environment in one click that for us is perfect. From the other side in our cloud architecture obviously we use OpenShift to try to improve all of our deployment from all of our micro services that we have on the cloud and we are working to use this operator like part of our boom time to improve our way to store and execute our quantum progress on the cloud. Next slide. So now I can demo you how the OpenShift Qiskie operator works. So basically what it does that it launches Jupyter server and you can access Jupyter Notebook and this Jupyter Notebook will have all the dependencies and the package of Qiskit for like if you are working for chemistry then a specialized optimized package for Qiskit and chemistry that works together would be available and the developer who is mostly called quantum scientist. He is an expert in writing or implementing quantum algorithm so it's not his he need not care about the configuration or the matching of the dependencies. All he needs to do is like with the hit of the button install the operator. So here's a demo you need to go to the operator hub of your OpenShift environment and from there just search for the operator by the keyword Qiskit and that you can see that the OpenShift Qiskit operator is available. You install it to a specific namespace or to the entire cluster it's up to you. I have running it into a particular namespace which is Qiskit op. Once it is installed you have to create an instance of the CRD and basically what the CRD is a Jupyter server. Go to your project to see the status you can see that the pods are running and they are ready and once they are ready it exposes the Jupyter server using a route and everyone in the team can just access and create new instance of the Jupyter Notebook using the routes. So with this operator we have given two demo Jupyter Notebook and as you can see you don't need to do any pip install. The Jupyter Notebook already comes with all the necessary dependencies. And the last slides before entering the code is again to try to bring to you some minimal explanations about what is this bidding blocks and how this looks for the next slide. When the quantum science people or developers talk about the pulsates, gates and circuits this finally is the minimum element that you are going to have in this moment that is to develop your quantum programs and for that you have these horizontal lines that is the qubit, that is the Q0, Q1, Q2 and on top of that thinking that these lines are like your timeline you can apply some kind of modifiers, some kind of operations that is these rectangle boxes or these circles. These operations we call gates and with these gates we manipulate the state of each of these qubits. And at the end, obviously depending on the technology that you use at the end you need to convert this into something that you can send to the real qubit and manipulate the state of this qubit. And that is the thing that happens in the right side that is these pulses. This means all of these gates that are on top of these lines that finally is each of these qubits is something that we translate the way pulsates that we send to each of our qubits to manipulate and make operations in that case. Next slide. Okay, so let's see how this qubit works at the end, right? I can see myself, what we knew into the quantum world and as a consequence I always tried to connect classical concepts into the quantum concepts. For example, this gate that we have here on top is the XOR gate with two bits and returns the result in another bit after applying an exclusive OR operation. Similarly, there is a quantum gate doing something similar to that. She's called it C0. It's here in the bottom. Yeah, it works a bit like an if statement, right? The gate controls from the state of the wire in X and if the wire is X, sorry, if the wire is 1 it applies a node gate into the wire in the bottom. And that makes it similar to the XOR gate that we have here in the top. Notice that unlike the classical XOR, the quantum synod is reversible. That means that you can reconstruct the input if you know the output. So in order to increase the similarity between these two gates we have to add this extra wire here on top. Now they are even more similar. Again, they are not exactly the same but for the sake of connecting concepts I think I will make it. And now that we know how to the synod gate works we're going to connect that synod gate with the Hadamard gate that we know from before. It's Hadamard gate that puts qubits into superposition. So the first thing that we're going to do is to import Qiskit to the IBM Quantum Computing SDK and we're going to create a quantum resistor of length 2 here in this line, the quantum resistor and another classical resistor to put the result of our computation also length 2 and then we have those two resistors into a quantum circuit. Then we apply the Hadamard gate on the qubit 0 and then on the qubit 0 and 1 we're going to apply the synod gate represented here as CX and then we're going to measure the results. The circuit roughly, well not roughly, it looks exactly like this, right? And this is usually known as the hello world of the quantum world. It's known as the bell state. So now let's try running this circuit 1,000 times into a local simulator that is installed in our computer. You can see that roughly the result says that we most of the time, or half of the time, the result was 1,1 and half of the time was 0,0. That means there is a dependency between the first bit and the second bit that this notion is called as entanglement, this relationship between these two bits. And you're probably asking yourself, okay, what I can use this for and that's where the Doge algorithm enters. The Doge algorithm was specifically designed with the purpose of illustrating a situation where a quantum computer can do something better than a classical computer. So it sets a challenge where in a classical computer requires two executions and in a quantum computer it can be solved with only one execution. So imagine that you have a function that takes only one bit and returns only one bit. So there are only four possible functions like those ones, right? They're here as f1, f2, f3, f4 here in the top. And let's split those four functions into some of them are called valence and some of them are called constant. If we want to... So a constant function is when the result of the output does not depend on the input. And a valence function, it does depend on the input. So in a classical execution, we need to do two executions. One with the input zero, another one with the input one. Let's take the video around a bit because I think I'm getting ahead of myself. In a quantum execution, it's possible to do this with only one execution. And oracles, so this function, we're going to put it inside a box that we call an oracle. And we can ask anything to the oracle and we can see the result of the oracle, but we cannot open the oracle. We cannot see what is inside, right? And the challenge is that with one execution, can you answer the question, is this function valence or constant? So therefore we need to also interpret that oracle as a gate and therefore it's also reversible. And notice that you have extra wires in the quantum situation because we had to make sure that that function is reversible. Maybe you can fast forward a bit. So in order to do this, to represent this situation, the first thing that we have to do is to have all these functions as quantum circuits. And that's what we're going to do in the next section. I think that you can use the arrows on the back and the forward. The first situation is the constant zero. It doesn't match the input. The output will always be zero. That's one of the functions that we have. The other one is the constant one. It doesn't match the input. The output is always one. And we flip the ground situation zero into one, just by adding a not gate. Here it represented as an X. X means not in the quantum world. The next situation is the identity. Identity means that if the input is one, the output is one. And if the input is zero, the output is zero. And we know how to do that already with a C not gate. So when the input is zero, the X gate will be triggered in the bottom. And therefore the output also, sorry, when the input is one, the not gate will be triggered in the bottom. And therefore the output will be zero. And the last one is invert, which is kind of like identity, but we flip the bit at the end. When the input is zero, the output has to be one. And when the input is one, the output has to be zero. So we have all these four functions now. And now we have to run them just to see how that works. The first scenario, for example, here is the invert. We put it between these two bars. And we also try to disfunction with the input one. Zero, not, is one. And when we execute this circuit, the output should be zero. It's invert function, remember? So that means that works roughly, like as expected, right? We can try also the situation with identity. So the also input is still one. And now the output is also one, because it's identity, it's identity function. So now that we have all these four functions, we're going to put them as inside these boxes, and then inside the dodge algorithm that can answer the question, is that Oracle balance or constant? And for that, we're not going to enter in the details, but we're going to put some gates around the function. And that's the algorithm itself. And the Oracle, it's here inserted in line six. For example, in this case, we insert the Oracle one as a black box. We cannot open that box. Here we see how the circuit looks like. And that is on the execution of this program. We're going to try to answer the question, if this Oracle behaves as a balance or as a constant function. So here in this case, for example, it's the Oracle one. And when we execute it, we execute it only... Well, the answer is one, it's because it's balanced, and if the answer is zero, it's because it's constant. And notice that we are executing this only once. In line one, you can see that the shots are one. So with dodge algorithms, telling us that the Oracle one is, in this case, balanced. Let's scroll up to see if that's the case. Identity is a balance function. Therefore, dodge did it correctly in only one shot. Let's check now with another Oracle. For example, the Oracle three. And let's also execute again. And here we see that it's constant. Let's check it out if Oracle three was behaving as constant. Indeed, it's constant one. So in only one shot, we could answer the question. But not everything is so nice in quantum computing. Some things are more complicated than classical. And one of those sorts of complications is the fact that real computers are noisy. They have noise results. So let's try to run the same circuit, but now in a real cloud. Let's first by importing the native functions. We're also going to load our IBM Q account, which used the cloud service. And we're going to ask the cloud to give us the list we see back end that has at least two qubits. Let's see if this process takes some time. Also, we want to make sure that we are not using a simulator because our IBM cloud provides us with simulators. So we're going to use a real computer with at least two qubits. And from all that offer, we want to take the one that has less jobs in the queue. The IBM Q or answer has five qubits and only four pending jobs in the queue. So it's the one that we're going to use. And we're going to run exactly the same dodge algorithm that we run before, but 1,000 times instead of only one. And instead of using the simulator, we're going to run this IBM Q or answer, which is one of the multiple computers that are available for you to use. So with execution, we take a job and now with a job that puts our circuit in the queue, we need to wait until the circuit is the queue and then process. This takes some time. That's why I'm fast forward in the video. But at some point, it will return a status that is not Q. Okay, now it's running. Now it's in the physical circuit, in the physical chip. And it's a very small circuit, so probably quickly it will be done. Now it's done. So we can analyze the results. It's exactly the same circuit that we run before locally, but now we're going to see what are the results in the real device. So remember that the real result was that the function that we were analyzing was constant. And now we have after 1,000 executions, we're going to do the same process. And notice that most of the time, the answer is the correct one, is constant. But not always. In some situations, in roughly 2% of the situations, we see that the result was balanced, was the wrong result. And this happened because of noise. Noise in quantum computers is something that is impossible to avoid and it's everywhere. It's when we set the qubits, when we read the qubits, when we operate the qubits, all those operations introduce more or more noise. Even just letting time pass introduce noise by a process called as decoherence. Because the qubits, they lose their state when they move forward. So noise is the big enemy of a quantum computer. And that's why we need to execute things more than once in order to make sure that the result that we want is the most probable one. Hopefully this demo is an invitation for you to play more with Qiskit. We're going to put a link to the notebook so you can jump into the details and you can dive into all those things that I shall quickly at the end of the presentation. We have a lot of resources online. We have the quantum computing IGN.com that allows you to learn using only the draw-and-drop without the code. And we have some kind of tutorials that allow you to learn how to work all these quantum mechanics behind scenes and what is the relevance of the use of the circuits and how you create your first circuits. And the other side, like we mentioned all the time, we have this Qiskit. And this is the two most significant things that we bring to the community to allow the people to learn and finally create new programs and algorithms and experiments. So you're free to ask this question if anybody has. So I've been keeping an eye on chat. It's looking fairly quiet in terms of questions. You guys went right up to the time so that's actually quite perfect. Thank you very much. I was super excited to have this talk in the track. Barul and I worked together a couple of months ago and getting this stuff initially kick-started into the operator. And I've been kind of keeping an eye on it as soon as we had this Python check. This is one of the first blocks I wanted to reach out to. So thank you. Thank you to you guys from IBM who I'd ever appreciate you guys coming into this. And that's all we have for the Python check. It has been a very interesting, very dense couple of hours here as we jump from operators over to MicroPipianV and talk about installation dependencies and then to AI ML and Flask and now ending on this really cool quantum concept.  I really appreciate your time and effort and all of this. Sorry, can we start over? I missed everything past the original welcome to the Python track. We just kind of do it again. So now we're going to be moving out to the closing track. I was actually wrong. I can't push people into it. The moderator in that channel will pull people over. Honestly, your best bet is going to be just going to the top under schedule. Drop by here so you can see the faces. Top next to the schedule, you'll see a more dropdown next to my name. Python track hosted by Jason Doebies. If you click more, you will see closing ceremony. If you want to hop over to that room now, we're going to be finishing up the closing ceremony there. To you three from the quantum track, if you have any links you wanted to share, the Slack room will be sticking around well past this. That's going to be the best spot for persistent links that people can follow up on that. They haven't been following or necessarily copying down in this chat. Yes, again, a lot of good feedback here. It's fascinating. So thank you to everyone. I will see you all over in the closing ceremony room in a couple of minutes. Take care, everyone. Thanks. Bye-bye.


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