 So, just make a note here that coefficient of static friction is different between the rope and the horizontal pipe and rope and the vertical pipe. And what we have to again do that we are trying to find out what is the P for which equilibrium is maintained, range of values of P. So, again we are trying to you know figure out that P is going down that means impending motion downward and impending motion upward ok. So, these are the two situation we have to look at. We just need some number at least the one part let us say we do the impending motion upward that is just to let us say hold the 100 pound load. If we just want to hold the 100 pound load that means P is trying to go up ok. So, can we just solve that part. So, it is clear that that answer again it is the similar problem we have solved that when it is trying to when I am trying to hold the load P will be less and when I am trying to pull it down P will be large ok. Contact angles are very easy in this case one would be just pi over 2 that is the horizontal contact angle and vertical contact angle is with the vertical pipe you know pipe is pi. So, one is pi over 2 one is pi and make sure the friction you know tension will change. So, you take the horizontal pipe right immediately once it is crossing the horizontal pipe there is a change in the tension. Then again when it is coming back right rounding up and coming back there again we have a change in the tension right. Sir, 156 lb sir. No, I am not getting the correct answer yet. 156 lb? No, so many answers, but nothing is actually correct so far. One answer is correct that is 411 is the upper range that is one answer that means that is actually impending motion downward. When you have impending motion downward we expect the P to be the largest ok 24.3 that is the lower end answer ok. So, two answers are 24.3 and 411. In any case so this free diagram will solve the problem in any case. So, you have P and Q right then again you are looking at Q and R and then R and 100. So, if I do impending motion downward then which one is greater which one is T2? T2 is going to be P right and then the lowest will be 100. So, it goes sequentially like this. So, impending motion downward answer is 411 another is 24.3 is that clear? This is basically the free body diagram that will control everything and we just have to understand that for impending motion downward P should be greater than Q, Q should be greater than R and R should be greater than 100. Then we reverse this when the impending motion is upward ok. So, let us move on to the next problem. The second problem is that of a bell drive again, but what is done here that let us say one side of it gets slacked ok. Then we are going to use a idler pull it is very often commonly used actually to tighten the bell back because it will increase the contact angle and once the contact angle is increased therefore we are going to get the torque carrying capacity will also be increased. But remember the idler pulley that is a very small mass first of all and it is free to rotate. So, you make it frictionless ok. So, therefore now you can start thinking about it. If it is free to rotate or the frictionless surface then the tension in the both sides will be same for that particular pulley idler pulley ok. Now that is also used for various different applications because once you want to you know kind of channelize the bell in different directions we just keep using different types of you know idler pulley in various places ok. So, in serpentine belt it is more or less used. So, just see what is given here it is said that the slack side is 1000 Newton that is the tension. So, slack side is going to be upward right the upper one is the slack side and this is actually the tension side is the bottom one. And remember if the you know all the balancing is happening correctly or not that also you check because I have applied torque right the torque is shown here that arrow right. So, that is my driver pulley answer again. So, what was the answer I am on side from the back. So, answer is 234.9 Newton meter that is the torque transmitted from A to B. So, which one is under impending slip in this case impending slip condition where we have to invoke which one has the smaller one right ok that has lesser contact angle. So, answer is 234 Newton meter I repeat 234 roughly Newton meter rather 235 actually is that clear. So, this is the larger tension side T and that is the slack side that is already given 1000 ultimately one impending surface we have to find where the slip happens that surface is actually the lower you know smaller pulley right because contact angle is less over there. We just apply T over let us say 1000 equals to e to the power mu s of beta let us say that pulley is B right smaller pulley is B then beta B ok. So, that is how we solve the torque tension on that side in the larger tension side and then we can just do the T minus 1000 times r ok. The important thing to understand that impending slip happens on B ok that is the driven pulley any question on this ok this one is actually combined point friction as well as belt friction problem whatever we have you know learnt in the morning session that also coming into play here. So, I will just explain the problem. So, this crate is being lowered and we can see that arrangement. So, there is a pulley that pulley is free to rotate and then it is going to that rope is going to a cup stand here ok. Now, once it is being lowered down observe that there is a weight that is the mass center ok and this gap is given very small. So, that gap is very small gap ok. So, ultimately what is happening if you look at there is a know my unknown is really the tension in that cable I have to find out the tension in that cable that is the target ok, but what is happening in the process can anyone tell what is happening in this case Cg is not matching. So, therefore, what happens it is going to rest on this wall as well as this one now what are the contacting points here the contacting points are going to be a and b right a and b are going to be the contacting points. So, it is going to now we are just assuming this gap is very very small ok. So, now you start your problem. So, this is you know from the morning session how many contact how many unknowns you have once you draw the free body diagram of the crate how many unknowns do I have how many equilibrium conditions are there what are the number of contacting surfaces involved and how many you know law of friction I have to use. So, first let us try to just solve the tension in the cable from the session that was done in the morning there is no bell friction to do that ok. So, very nice problem to look at remember there is a you know we have already said that kinetic coefficient is going to be taken into account why because why it is so because this is always going to be slipped over the wall right and remember that everything is happening at a very slow speed that is the key ok. So, we are again going to use the static conditions everywhere. So, as long as we solve for this key right here just give me that number. So, let us this is another tutorial from the morning session and then we can just use the bell friction on the top. So, how many unknowns 5 right if I just take the equilibrium of this is it 5. So, I have a tension here that is 1 then I am going to have a normal reaction here and the friction force which direction friction force upward similarly here normal reaction plus friction force. So, total unknowns are 5 how many contact surfaces I have 2 contact surfaces right. So, friction law now they are remember the kinematics of the problem suggest that I have to invoke the friction law at both surfaces because they are not independent surfaces right because it is just going like this. So, both surfaces slip has to take place simultaneously. In any case in this problem we are just you know going to use the e to the power mu k times 5 right. So, 3 equilibrium conditions are already coming into play from this and we can use the friction law here as well as here right. So, total 5 equations 5 unknowns and solve for the t. So, I need the number is that clear the final answer will be 10 kilo Newton ok. The final answer means the t that we are trying to find out. If you want the intermediate steps that means what is the tension in this cable right here that is link to the crate anyone got this answer ultimate answer is 10 kilo Newton anyone got 434.78 yeah that is 435 let us say ok and that tension down there will be 10 kilo Newton once we apply the belt friction it is wrapped around 2 times. So, look at the free body diagram here. So, again I repeat there are 5 unknowns there are 3 equilibrium conditions for this crate there is a slippage surface is only 2 slippage surface has to be there. So, invoke the friction law. So, here we said already mu k r b mu k r a ok. So, friction law is taken care of now in this body you have actually 3 unknowns to solve for e r r a and r b this needs to be solved. So, ultimately we solve for the t 1 here that is 434.78 kilo Newton and then we go to the belt friction right here. So, 2 terms where they are beta equals to 2 times 2 pi t 1 must be equals to t 2 ok. So, we apply that and we get the value 10 kilo Newton. Any question on this because gap is very small that was the condition. Gap is very it is just slightly I mean you can think of it in finite similarly small you know it is just trying to go down, but there is a you know slippage here that is not it that was already given in the problem it is just you know very small gap it is trying to adjust ok it is little bit tilted, but. . Tension is given in the right side already right side when you are going to lift with the help of the fourth applied at d it is eccentric now it will be touches surfaces at c and also the d no no there is a geometric center no it is tilted no so it will be tilted like this only no which way you have the no which way you have the carpal look at the carpal no we are applying tension in the upward direction no if it is freely suspend that is the case so when you are applying the tension at the top side the whole body will be tilting towards this side only ok look at this there is a t 1 this way and 500 this way so tendency of tilting will be counter clockwise that it is shown here there is a couple here right we can't say it's a couple actually because they are actually different value because we have friction so the moment here the whole body is tilting counter clockwise done whole body is tilting counter clockwise like this no no that diagram is done ok this this tilt line you are saying it is ok these are all straight line assume all straight lines ok vertical straight lines that is not an issue but ultimately this is shown no tension t 1 is I think lesser in comparison the tension t because the block is moving in the upward direction so the ratio t upon t 1 will be e to the mute again you are lowering it's a lowering you are trying to lower it down no you need to give a very small force compared to your 500 kilo Newton you are lowering down everything why do you once once they say lower it down how can tell me how can t can be greater than 500 kilo Newton are you expecting that are you expecting that once are you expecting that tell me no ok