 So, now what we are going to do, we are going to start the next set line on a screw jack. Square threaded screws, I have not gone to the any other type of screws, I have only taking square threaded screws, because for other screws mechanics is more complex. We are only going to typically consider you know where it is being used, it is used in as you can see here the jacks, you know house jacks press, then you have clamps. So, all of these we are going to use the screw jacks, so we call it typically we call it you know square threaded screws being used in jacks, if it is so then we are going to say screw jack. But remember there are other applications also, for example it is used for clamp. So, as we all know what is the purpose of a screw, what is a screw? A screw converts a rotational motion into a translational motion, if you think of the motion, similarly it converts therefore, it converts a torque into an axial force, that is the basic concept. Remember what would be important here is that what is the pitch, pitch means that is the distance between the two threads and what is the lead, lead means if I make a perfect one revolution then how much vertical distance it will move, so that is the lead. So, these are two important aspect, so with that now how do I come to the mechanics of it, how do I draw the free body diagram and try to solve it. Remember that here is the screw, the screw is on a base, there is a cap and let us assume that the friction between the cap and screw is ignored for the time being. So, I have load, so I have to really either raise the load, so it is mostly used to raise the load. However, depending on the amount of friction you have, we can also make it self locking, we can hold the load and so on. So, it is again a question of raising the load or holding the load that we have to keep in mind. So, the question that comes into play that I am trying to let us say for the time being raising that load, remember the contact between screw and base, so here is the contact that is the main contact I am talking about and this is happening along a portion of their threads. So, you have a base thread, you have a screw thread, so this interface is going to be critical that is what the delivering the friction. Remember I have to apply a load P that means in the handle a torque is being you know a load is being applied, this force P is applied such that it produces a torque and that torque will be converted to an axial force mostly, so because it is going up and down ok. So, ultimately what is happening, if you think of it we can simply unwrap this inclined plane, let us say you think of this base, there is a thread on the base, how about if I unwrap it, it makes it inclined, how it is going to be inclined, but based on the and the lead going to be inclined, so 2 pi r if I open the amount by 2 pi r vertical distance moved is L that is the lead. So, the angle will be tan inverse L by 2 pi r ok, so you will have a lead and you will have a 2 pi r that is the revolution, where r is the mean diameter of the screw ok. So, remember therefore, once I unwrap the base ultimately what I can do, I can take the screw like the load coming on to the screw as a you know on a block, say block as if a block is on the inclined surface, remember this torque is now we have to take this torque as a equivalent force that is coming horizontal. So, if we go to the next slide we will see that what is happening exactly, so the unwrapping the thread with a lead angle of theta right. So, this angle will be called theta, now how I have unwrapped it, remember this is going to be 2 pi r and if I unwrap the thread then the vertical distance moved is L right, remember L and pitch is not same just be careful lead and pitch will only be same if it is a single threaded screw. So, if you have single threaded screw the lead L is exactly equals to the pitch, where pitch is the distance measured between 2 consecutive threads. Therefore, now as I was telling that what is this equivalent force Q that is trying to raise the load of the inclined plane. So, therefore, this entire friction force as we have learnt that friction force is independent of contact area although it is for a simple friction that is the assumption we are deriving if we skip the tribology of it. So, therefore what is happening you can clearly see that I have now during the impending motion upward that means, when you are trying to raise the load. Therefore, when we are trying to look at the motion of this block up that means, we are trying to raise the load screw is trying to advance on the upward direction what is happening during the impending state. We can clearly see that this coefficient of friction from that we can calculate the angle of friction right, why this angle of friction is coming into play remember impending motion is upward. If impending motion is upward the friction force is downward. So, what is this resultant the resultant between the friction force maximum friction force that is mu s and the n, n is the normal force. So, ultimately you can construct that force triangle I have a n here and I have a friction force here. So, n goes like this right and friction force comes like this and so, friction force and n that gives me resultant and q is the equivalent force remember that is coming due to the torque the q equals to a q multiplied by r. So, at the interface if you look at it I have a horizontal force right. So, that horizontal force multiplied by r is actually equals to p times a. So, that torque is being converted to this horizontal force that is as if helping the load means the load that I am trying to lift it is really moving on an inclined plane where this inclined plane is determined by what? Determined by the thread angle right the lead angle rather ok. So, ultimate idea all the time will be for a given w how do I find out the q clear. So, for impending motion upward this is my going to be free body diagram this was done also from yesterday's lecture right. So, we have once we find the q I should be able to know that what is the force p remember I am lifting the w at various strokes. So, that is assumed giving some stroke it is simply trying to lift the load. So, I am trying to determine that what is the force required such that there is no slippage at the point threads ok when I am trying to. So, this is for impending motion upward. Now, remember now you can easily think of what happens when the impending motion is downward. Downward means that load is just trying to come down ok this is verge of coming down. So, I am just applying a small force again remember when I want to you know lift the object or you know then I have a larger load when it is trying to come down that is lower the load then I have to apply a very small force p. So, when the impending motion is downward as we can all see the two situations can happen remember when impending motion is downward what happen in the process my friction force is reversed which direction is the friction force on the upward direction. So, if you will think of the force triangle where is my resultant now. So, depending on the coefficient of friction right I have the friction angle already defined. So, if I put here f max where f max is nothing but mu s times n n would be the normal force. So, the angle between the friction force and the normal force that is my friction angle right that is phi x. Now depending on the value of the friction force this phi s can be or rather depending on the coefficient of friction this phi s can be smaller than theta or greater than theta. Now here what we are saying it is smaller than theta. So, if it is smaller than theta then remember again I will be able to solve for a q and that q will be ultimately give me the value of p and that p will be much lower than what we have done in the previous case when you are trying to raise the load. So, when we are trying to raise the load I have to apply more force than compared to when I am trying to lower the load. Now what is self locking you can see that you may have to apply if the friction coefficient is really large such that your friction angle is greater than theta right. So, resultant of the friction force and the normal force as you can see that will make the resultant which is actually on the left hand side of this vertical line. So, in other words phi s is greater than theta. So, coefficient of friction is large ok. So, in that way what we have to do now you have to maintain the equilibrium of the body I have to reverse the q. So, that means instead of applying the torque as we are doing in the previous case we are applying the torque how counter clockwise right. The torque was applied counter clockwise. So, force was this way perpendicular to the handle right then you have to really give a clockwise torque. So, we have to give it a clockwise torque to actually lower the load ok. So, now can you tell me when we do not need the q when q goes to 0 q goes to 0 now that is very critical. So, q goes to 0 when r is collinear with the w that means phi s equals to theta now that really brings in one very good point. What does it state that when q equals to 0 r is actually collinear with w. In other words phi s is equals to theta that is the critical angle of self locking that means I do not need to apply any load on the handle and I can always find the minimum load that will be sustained by the screw. The screw is not going to come down automatically. So, it is completely determined by the coefficient of friction and when that angle of friction becoming theta I can say that that is the critical angle for self locking as if I have to apply no load q right and the body will be able to sustain some load is that clear now. So, three situations as I explained here that first one is the raising the load just to raise the load what is happening remember friction force is this way right now downward friction force is downward. Therefore, resultant is moved this way between the friction force and the normal force. The other case is impending motion downward now that can lead to two situation when you are just trying to hold the load and in other case the friction angle is very high. Therefore, you really have to apply a force right that is from the other direction in order to lower the load. Now, can you have a brief discussion is that concept is clear now. If you just buy this simple mechanics logic now you can approximately solve some of the problems really challenging problem using the concept of threaded screws yes. So, just you know few questions I can take for discussions on only the threaded screws square threaded screws 1 0 6 0 go ahead for your question. Sir, I doubt in belt friction sir. Ok, just go ahead quickly. Ok sir, the sliding friction will act when the tension is maximum sir. So, what is the direction of sliding friction and the tension is maximum the sliding friction is seen on the maximum. The friction force right in the impending state whenever you have an impending state friction force has achieved this maximum value is that clear. Friction force will achieve a maximum value when it is at the impending state that means the body is just trying to slip out. Now, how do you determine the direction of the friction force is that what your question? Think of it in the belt friction I have this drum just look at my hand. If I put a cable right and let us say cable is trying to tilt this way. So, from your direction it will be counter clockwise right then your friction force has to be on the resisting that motion right it is trying to resist that. Therefore, tension side will be always this side higher tension side will be this side lower tension side will be this side because that negative friction force should be equals to the other one. It is equilibrium, but you cannot take just you know sum of force along x equals to 0 in this case. Think this way that this tension has to be higher because resistance is going this way. 1 0 2 6 go ahead for your question. Good afternoon sir. So, regarding your presentation in the first position first problem to lower the load or to keep equilibrium the force record is 181.15 Newton. At the same time to lift the load upward it is 132811 Newton. There is a drastic change in load record for lower tension side. So, considering the load as well as lifting load. Is there anything wrong in the answer or that much of force is required to lift the load. So, your question is just do not think about the number just think of physically that why do I need a large force when I am trying to lift an object up and why do I when do I need a very small force when I am trying to hold when do I need a very large force. When I am trying to pull the object up because there is a friction force that is resisting right think of physically. In one case the impending motion is upward right another case impending motion is downward. You think of in simple sense if I have to lift the object up I will always need a higher load. If I have to hold the object which is trying to going down I will always need a very small load. This is intuitively think it out that number that was put in the first problem is absolutely correct because it is controlled by the you know friction that is coming into play. There are so many wraps and all that. So, those wraps are taking care of those friction, but point is that when you are trying to raise the load we have to apply really large force to overcome all of the friction effects right. But when I am just trying to hold it that means this mass is trying to coming down I just need a very small force. So, just physically thought it out with the free body diagram. Thank you. Screw jack problem is clear because I am not getting screw jack questions. The screws concept is clear it is a block on an inclined plane that is all we said 1, 2, 0, 5, Sir my question is during relationship of T2 by T1 is equals to e to the power mu theta we choose the fixed pulley instead of having the motion between the pulley and belt and the same is regarding belt drive we are having both motion between belt and pulley. Your question is during the derivation we have used fixed belt on a pulley. Does not matter suppose I have the pulley and I have the belt right the way I have solved is that this belt the belt is trying to slide in some direction. It is still not slipping, but it is verge of slipping. So, I tried to find out which direction my tension will be maximum because there is a friction force if it tends to go this way then friction force is this way right ok. Now, based on that logic I derived the formula considering the impending motion. Now, remember the same was done during the simple friction problem also f max equals to mu s n how did you go to that. So, what happens I will just again explain it why it is like this. See the basic concept is as follow you are increasing this value of p. So, p value is increased right. So, you will have a friction force friction force. So, you are trying to break this barrier of friction force there is a maximum friction force that will be controlled always by this n, n is equals to w in this case let us say ok. So, we are really going here keep on increasing the p friction force will gradually increase. There is a barrier point we call this that is the maximum friction force that the body can take. What happens afterwards? It simply drops right and then we say ok I have a kinetic friction coming into play because when you have you know motion is already created then I need actually very small force to move this block forward. But see my point here is I am only interested in ok. Now, this part will result the result the block under motion. So, up to this one I always have the equilibrium and this one I am going to have the dynamic that means body will be either in acceleration right then I have to really equate p minus f should be equals to let us say mass of this block multiplied by acceleration ok. So, you always derive the formula at the impending state at the impending motion I will always derive the formula and I will simply can say that the same can be used when we are talking about the kinetic coefficient of friction. Remember kinetic coefficient of friction comes into play also for constant motion problem ok. So, that is the whole idea that we first get the impending motion and then they just substitute the kinetic coefficient of friction to get the you know same formula. But remember kinetic coefficient of friction when we are using the kinetic coefficient friction two things can happen one is the constant velocity problem right one is the sleeping actual sleeping that is taking place. So, when the actual sleeping is happening we are always going to use the kinetic coefficient of friction when there is no relative motion then we are always going to use the static coefficient of friction ok. Ok sir thank you yeah go ahead. I have one doubt related to the belt and pulley system at our domestic floor mill sometimes foreign particles are entered you know between the belt and pulley. So, the effect of efficiency of the system decreases. So, can you suggest me some considerations to increase the belt from the system efficiency? Your question is how do I increase the efficiency of that system? Do you remember that problem we talked about regarding the idler pulley introduction of idler pulley? See you can increase the contact angle that is one of the options. How do you increase the efficiency? I have only choice is to increase the contact angle. So, that will increase the torque transmission capability. Now there are other ways I am not getting into that. So, that requires more conceptual you know understanding, but simple way would be increase the contact angle ok. Yes sir thank you sir. See there are problems here, but I think this problem once we understand the you know block on an inclined plane these problems can be tackled. So, I had one problem in the lecture slide and all of you should have the lecture slide in the moodle ok. So, in this problem you see we are using the clamp to hold two pieces. Now the clamp only thing is that we have a double square thread of mean diameter equals to 10 mm with a pitch of 2 mm ok. Now if a maximum torque of 14 Newton meter is applied in tightening the clamp determine the force exerted on the pieces of wood. So, we are interested in finding out what force is going to this wood as I apply a torque of 14 Newton meter. The another one is the torque required to loosen the clamp. So, maximum torque is one part the another one is the torque required to loosen the clamp. See one in one case we can think once we are tightening then you are trying to actually raise the load that is the first part. So, we are trying to raise the load in one part in other one we are simply trying to hold the load right. So, loosen the clamp. So, in loosen the clamp will be again you can think of that whether my phi s that means the angle of friction is greater than theta lead angle or less than theta that is less than lead angle ok. So, the concept is similar to what we have studied. So, step by step if you want to go ahead go ahead calculate the lead angle ok. And as I said that lead will be twice the pitch for a double straight double threaded screw and then we can simply use the concept of block on inclined plane ok. So, that will solve for the first problem first part and the second part we can again use the concept of you know impending motion downward and that will solve it. So, I will try to flash it very quickly the steps that are required because as such answers are very simple these are very straight forward problem. So, first find the theta tan theta equals to L by 2 pi r and here you can see L equals to 4 mm why the 4 because it is 2 times the pitch ok. Then we can calculate the friction angle also angle of friction was calculated angle of lead was calculated right. Then ultimately I had to find out this q remember q multiplied by r that should always be equals to the torque. So, what is the value of q? q times r mean diameter that is equals to the torque. So, I get the value of q ok. So, now therefore, what we have is in this case if you draw the triangle rule you know just draw the use the triangle rule this is theta plus phi s right we can get the value of w. Similarly, when the impending motion is downward remember you want to loosen the clamp and remember to start the problem first we have found out that phi s phi s is greater than theta. So, phi s is greater than theta will create a self locking mechanism that we have already discussed. So, therefore, if you look at the resultant the resultant will be shifted this way and q will come from the other direction. So, you have to really apply the torque on the other direction ok. So, with this then you can again calculate by based on the triangle rule we can find out what is the torque. Remember every time you are doing this problem it is a body with three forces right and they should be concurrent ok. So, that is the concept. So, there are three forces which are concurrent in nature and you can also do sum of force along x 0 sum of force along y 0 that is also the possibility ok. So, but triangle rule would be more appropriate to solve this class of problems. The other problem I am not going to go through it once we understand this concept you can also think of you know more practical applications as you see this kind of problem. So, the automobile jack is being lowered or it is being lifted we are applying a torque here right. So, we have to find out the torque to raise the automobile let us say we have a automobile here. So, that is 4 kilo Newton you know load is coming out. So, we want to find out the magnitude of couple required to if we want to just raise the try to raise the automobile. So, again impending motion condition comes into play. Now, remember the thread direction is different in this case in one side I have this side right handed thread this side left handed thread ok. So, ultimately as you apply the torque a and c point will try to come close to each other. If you apply a counter clockwise torque then it will try to go away from each other. So, in one case you will raise the load in another case we are going to lower the load that is all. Remember what would be the first step in this problem to get the force transmitted in the screw and that you can do because these are all 2 force members connected. So, if you use the 2 force member concept we can find out what is the load that is coming in the screw. So, this 4 kilo Newton will be converted to an axial load along the screw that is the whole idea and once we get that you can draw the free body diagram from this free body diagrams one can get what is the load coming on the screw ok. So, once you know this load. So, this is going to be now applied. So, this is going to be the applied on that block. So, now you are going to the block analogy. So, in that block this is now my FAC ok and then you can see that to you know raise the load we can again try to find out the lead angle is found here. So, that is done right then you have the phi s phi s is the angle of friction ok. So, based on that we will be able to find out the value of q right once q is found we should be able to get the moment. So, only thing this moment this torque is one side. So, remember you have to multiplied by 2 times because you have taken only FAC. So, we have solved the problem from one side ok. You have to the final torque that is required is this ok. So, point the time trying to make. So, this is going to be same as that of our simple friction concept that we have developed yesterday and based on that I can always you know calculate let us say what is the torque required to raise a particular load or to lower a particular load on a screw jack ok.