 We will start our next session which is on equivalent system. Now what is equivalent system? Mathematically two systems of forces are equivalent if they can be reduced to the same force couple system. So, see so far what we have done we learnt that there is a force result and there is a moment result and to you know every system. So, here what we are going to mostly focus is that we do have a force system let us say in a body we have multiple number of forces or couples and we want to reduce this force system to a equivalent force couple system. So, ultimately the main point here is that if we look at two force system they must produce the same push or pull. In physical term they must produce same push or pull in the rigid body that means they should have the same resultant force R. Each system must produce also same turning action that is in physical term about any point in the rigid body that is they should have the same resultant moment couple vector M R. So, to do this now how it is done actually suppose I have a rigid body and I apply a force F can I translate this force to a point O. So, the ideally if I want to do this then I should add a positive force and a negative force and they should have the same direction as that the line of action should be same as that of F. So, ultimately now I have a couple F and negative F that will create a couple moment M naught along with the force which is now translated at O. So, in other words all of these systems we look at they are actually equivalent in nature because they must produce the same push or pull or same turning action about the point O. So, we can calculate what is that moment vector M naught through simply R cross F. Now, this is for a simple single rigid body with a force F. So, one force is applied on the rigid body now we want to extend it to multiple number of forces, but the operation will be same. So, if we just go through these figures I have multiple number of forces F 1, F 2 and F 3 all of these therefore, can be brought to a point O just by adding a couple. So, all of these forces will be ultimately translated to point O with the force itself plus a couple moment. So, for example, F 1 is translated at O with F 1 plus a moment that is M 1. Similarly, for F 2 and F 3 now you see these force systems and the moment system they are concurrent in nature. So, therefore, I can add this based on the concept of concurrent forces and I can effectively get a resultant force R and a resultant moment M naught. So, the procedure is same ultimately what we are doing is what is my resultant force simply sum of all the forces vectorially and what is the moment resultant? The moment resultant is simply R cross F and that is done over different forces. I have added an extra term C. So, this C is actually if there is existing couple on the body see here we are just talking about in terms of forces, but remember I can also have a couple that is already applied on the body. So, that is a free vector and I can simply add that free vector couple vector C on top of this moment resultant to get my final moment resultant vector. This can be further you know if I want to take the this system which is now resultant force and resultant moment here to let us say point O prime. So, I want to bring it from O to O prime then I will do the same operation that means this R will be translated with a added moment component again ok. So, that will be now you have a new moment resultant M naught R that new moment resultant will be simply the previous moment resultant plus S cross R. So, S cross R is coming just for the translation of the R to the point O prime ok. Now, if we really look at pictorially all of these whatever is drawn here they are actually all equivalent system ok. So, this is now very interesting that actually what happens in a three dimensional system of forces what would be my ultimate reduction of that force system. It can be shown that a three dimensional force system can be reduced to just a simply range that means just a screwdriver. It will be application of screwdriver how it is suppose I resolved that R and M naught R right that is resultant force and resultant moment. Now, this can be further reduced how remember I can resolve the moment resultant along the axis of the resultant and perpendicular to the axis of the resultant of the force. So, M naught can be decomposed in two components let us say one is M 1 and another one is M 2. M 1 is that vector is parallel to R along the axis of R and M 2 is perpendicular to the R that can always be done. Therefore, now what I can do I can simply move this R along with M 1 to another point A such that R times let us say O A cross R that must be equals to M 2 ok. So, O A cross R that should be equals to M 2. So, ultimately now if we look at the what it is actually doing in the body should be simply represented by a screwdriver. There is a push right as well as there is a rotation, but their direction remains same vectorially ok. So, any three dimensional force system can be reduced to a range. Remember we can further simplify this depending on the type of forces that a body has ok. So, for example, I can have a coplanar force system and I can have a parallel force system. Remember in both cases let us say consider the coplanar force system where all the forces are applied on the x y plane. So, the consequence is that my moment resultant of this system is along z direction that means perpendicular to the plane right ok. Similarly, in this case also if I do R cross F the moment resultant that direction will again be along or on the x z plane right. That means, ultimately your resultant force and the moment resultant they are perpendicular to each other. If that is the case if the resultant and the moment resultant they are perpendicular to each other I can simply reduce it by a single resultant force. So, operation is quite simple the idea here would be I have this coplanar force system first I resolve at point O right by a resultant and by a moment resultant. Remember this moment resultant is shown in this way, but actually if you look at the vector it is moment about the z axis. So, vector wise it is along the z axis ok. Now, idea is how do I replace this moment resultant I want to take the effect of this moment resultant. Therefore, I will simply move this resultant force to a distance d such that magnitude wise R multiplied by d must be equals to m naught R ok. So, as we see here if I want to do it in rectangular coordinate system as we preferred we have to do it in a manner that is consistent. So, idea is let us say that there is a line of action of R and that line of action of R intercept the x axis as well as y axis. If I want to find the x intercept of it that means the point B what I will do I will simply resolve this resultant R into two component R y and R x R x will not contribute to the moment right where R y will contribute to the moment. So, the x intercept will be simply x equals to m naught R divide by R y whereas, the y intercept will be m naught R divide by R x if I do the same way. So, here I have resolved the forces at C ok. So, therefore, the equation of line can be established through this relationship which is x R y negative y R x equals to m naught R. Therefore, my ultimately this force system is now reduced to a single resultant force whose line of action is known and I also know the interception with the x axis and y axis. So, logically it is very you know same. So, let us look at a problem this is a very nice example of a coplanar force system whereby a warship this is US John Kennedy this is a warship that is being actually adjacent near the dock. So, how do you do it when it comes to the shore or let us say parking area then we are actually going to you know help it moving by using the tugboats. So, tugboats are going to apply forces in such a way. So, that this whole ship will be moving to a particular direction. So, that means ultimately there will be a resultant force of these 3 forces that are coming from the tugboat that will help the ship to move towards the shore or parking area ok. So, let us just try to see this example itself this is my ocean liner or the ship and there are 3 tugboats that are giving the forces on the body of that ocean liner or ship. So, body is called the hull. So, we want to replace these 4 tugboats by a more powerful single tugboat such that it will create the same effect as that of 4 original tugboat. So, that means we are really interested in find out how do I replace these 4 forces that are coming from the tugboat by a single resultant force that will be applied on to the body, but it will have the same effect as that of the 4 tugboat. So, again the process will be simple the process will be really fast find out the resultant force and the moment about O of these 4 forces. So, we will just find the resultant force at O and the resultant moment and then we will simply move the resultant force such that resultant force multiplied by the distance equals to m naught r moment resultant. So, we want to find out the line of action of that resultant force which is creating the same effect as that of the 4 tugboat. So, first what is done here say again what we you know try to do can we guess the answer I have shown the answer ultimately you can see there are 4 there are 3 forces on the top of the body right and just one force here ok. So, if you look at the direction of the moment resultant how the moment resultant should look like moment resultant should be clockwise at O right ok. Therefore, I always have to move it to the right hand side of O that single resultant force also there are 2 forces the action is perpendicular to let us say x axis, but there is one force like this and one force like that tugboat 3 and 4. So, look at what would be the direction of the single resultant force that will really come into effect based on this 2 tugboats their magnitude and their direction. So, you can kind of visualize the answer that I am looking at ok I am going to have a resultant force which may look like this and that will replace the all tugboats if we go by just the vector algebra. So, we have the capital R that is the resultant force. So, resultant force is simply sum of all these 4 forces and the resultant moment is done again at O that is through r cross f which you have to calculate. Now, you can either follow the r cross f or you can go component wise for example, what is the moment produced by the tugboat at O simply equals to the force here multiplied by this distance, but we have to put a sign should be negative sign ok. Similarly, we can break this one into 2 components x component and y component and we can look at what is the moment is being produced at O. So, x component should give me x component multiplied by this distance which is 24 meter there I have to add a negative sign and the y component will also give the moment which will be negative in nature ok. So, like that once we get the r and m naught r then we can easily calculate what is the x intercept x intercept is nothing, but m naught r divide by r x right and y intercept is similarly m naught r divide by r y. So, once we get this line of action then we can actually point out that ok what where it is going to intercept at x and where it is going to intercept what that did not solve the problem yet it is just the line of action. I want to find out where on the body where it is going to apply. Now, to do that remember there are 2 cases can happen one is that r can actually act right here right on that horizontal line of the body or r can act also at the inclined portion of the body. Therefore, now I have to say ok what is the equation of line for this and what is the equation of line for that and I have to find the intercept of this line with those lines right. So, equation of hull can have 2 choices one is the inclined portion that means y equals to x m x and the other one is the horizontal line where y equals to point before. So, I will try this. So, I will simply take this and substitute this y in this formula right to get what is my x. So, if I do this exercise x is going to be equals to 82.58 meter if I substitute this equation of if I give this y. So, that tells this is greater than 64 meter ok. So, that answer is not possible because it is greater than 40. So, therefore, try with y equals to 24 now I can get a value of x that x is equals to 88 meter ok. So, ultimately the point of application of the R will be not on the inclined portion, but on the horizontal portion of the body. So, let us move to the next part of it. So, that is how we do the coplanar force system let us come to the parallel force system now as we said parallel force system can also be reduced to a single force. Now, how we do that again simple logic that again I have taken everything here in a positive sense. So, let us assume that resultant and the moment resultant you know we look like this. The idea is that if resultant is this and if I have to move this resultant on the positive x and z axis then r times resultant multiplied by the distance that should give me a moment which is actually pointing this way ok. But we have to be careful with the actual sign when it comes to the problem that we are solving. So, the idea is that first we get the resultant forces if someone wants to do it in vectorial form then just find the resultant forces. So, that will come with some number multiplied by with a unit vector which is actually j right and similarly for the moment resultant we will also get right the you know component will be actually i and k component of the moments ok. So, ultimate idea is that to get the resultant force it is line of action we have to find this x g and z g right where it is actually going to be placed and that can be simply done by equating the moments of these two sits. So, I have to simply say the moment about x axis of this system should be equals to moment about x axis of that system right. Similarly, the moment about the z axis of this system should be moment about z axis of the original system. So, just be careful with the signs that we are going to use. So, we have to tell repeatedly just be very careful with the sign of the moment that we are dealing with. Again a problem of that of a concrete foundation here we can easily see that all the forces are actually you know parallel force system that is parallel to the y axis and it is actually a matte foundation in concrete. So, it is a said that the geometry of the foundation is a regular hexagon. Remember if I want to replace this by a resultant force procedure is there just moment about x and moment about z that is what we have to do that is how we equate it. Can anyone guess the answer? At least let us try to find out what would be whether it is in the positive coordinate negative coordinate and so on so forth for the x and z. Someone said negative z anything else positive x. Let us see if we are ok with that. So, negative z and positive x is the answer ok and very quickly that is a feasibility because the higher concentration of load if we look at it that is actually in the positive x direction right. So, therefore, this has to be positive x and now these two loads are along the x axis right. Therefore, in the when we talk about moment about you know x is x axis right these two are not coming into play ultimately moment about x axis will be contributed by these two loads and higher concentration on 60 kilo Newton side right. So, it is negative z absolutely answer is correct, but you know the process has to be done the way we do it the answer is actually you know everything take in a positive sense, but ultimately answer is coming as a negative z ok. So, this is going that way and positive x is the answer for the resultant. So, ultimately it is going to be placed somewhere in this quadrant. So, now we have completed two important you know systems those are coplanar and parallel force system most of the engineering problems actually takes that into account. Now, we are going to go through the distributed force and how to reduce that to a single force remember how this distributed force can come into play distributed force can come into play because simply the weight of the beam itself right because it has a density and density multiplied by the area can give me the line load. So, that will be have the unit of Newton per meter. So, distributed load is nothing more than a line load which has a unit Newton per meter distributed load can also come into play when we are talking about a hydrostatic pressure right. So, because there will be a variation as we go down we will have a triangular distribution of pressure which can also be converted to a line load by multiplying by the width of the body where it is being applied. So, ideal is the similar methodology will follow. So, ultimately the if you look at these two as equivalent systems then the moment resultant and the force resultant must be same. So, what is the moment resultant of the distributed load? We take a small strip or area that is d a right and then we take the moment of this area about point o right and we integrate it similarly moment of this we take that is w times x bar and that is how we get the x bar and we get the net force resultant. So, it is same as the way we do the centroid calculations as it. So, remember the net force will be applied at the centroid of the load distribution ok. So, therefore, if I let us say I have a triangular distribution then we know the area of that triangle right that is my net resultant force and the that will be applied at the centroid of that triangle ok. Then you have to think about what kind of distribution you are getting actually, but that can be a function. For example, if we consider a tapered beam you are saying density is changing within if it is a composite material. No, but it cannot I mean it cannot change definitely, but logically you know you have to do that integration mechanism again. That means, in that case it will be through the center of mass not centroid. Yes, in that sense yes it is true ok. Now, one more example we will take is that of a beam with a distributed load again now here we have a uniformly distributed load and here you have a triangular distribution. So, ultimately we want to replace all of this by a resultant force make sure there is a couple moment also right. So, that couple moment is not shown how it is coming, but let us say there is a free vector which can be added anywhere. So, this is really a couple ok. So, the first step to do this problem is that we break it into two parts. Let us try to just find out what is the resultant of this uniformly distributed load and the triangular load. So, we break it into two parts and we find where the resultant force of these two systems is going to be. So, ultimately the distributed load is replaced by a single resultant force that resultant force is 180 Newton and it is going to be placed at 7.33 meter right and then we also have the 100 Newton and we also have the couple moment. Now, we are looking at a system which has three two concentrated forces and a concentrated moment. I would not say it is a concentrated moment rather it is a free moment applied anywhere in the beam. So, again just now we want to create the equivalent system. So, get the resultant force and resultant moment about point O and therefore, we can get the centroid x bar which is at 4.7 meter. So, that is the application of the total resultant force that we are getting from this kind of force system. So, this and that now is equivalent system. A hydrostatic pressure. So, once we look at problem involving you know dams and also in mechanical we have valve of different shapes and sizes. So, there we can see the pressure distribution is going to be triangular and make sure we tell also that atmospheric pressure let us say equals to 0 that we assume and we have this pressure distribution. The pressure distribution will now be converted to a line load by multiplying the pressure with the with B of this dam. Let us think of a plate just a rectangular plate. So, basically our w x in this case will be rho g d minus x that is the pressure multiplied by the width. Now, where it is going to be applied the resultant force that we all know now it is simply the centroid of this triangle and the resultant is simply going to be the area of the triangle. One can also adopt the integration procedure, but that is not necessary. We will assume at this point that students are familiar with you know what is the center and so on. So, now we are going to just exercise this problem of a hydrostatic pressure where by we have an automatic valve this valve consists of a rectangular sorry square plate 225 in mm by 225 square plate that plate is hinged here and we keep filling the water what is being said determine the depth of water d for which the valve will open. So, we want to find out what is the depth d for which the valve is open. Remember here we are not studying equilibrium yet. So, everything has to be in terms of where the resultant force is going to be. So, if there is a hinge here then the resultant force should pass through the hinge it has to be just slightly above the hinge. So, in limiting condition what we know is that condition is that resultant force must pass through the hinge. Now, the problem is not you know simple yet because I have to really look at the solution systematically. Now, what is necessary to understand that as I keep filling the water d three situations can happen. One is that d is less than 225 millimeter. One is that d is equals to 225 millimeter. Another one is d is greater than 225 millimeter. Now, first we say let us say d equals to 225 millimeter. So, through that what resultant I am getting the resultant will be now if I say the water depth is 225 millimeter then the resultant must be 225 divided by three which is less than 100 mm. So, that cannot open the gate right. So, therefore, water depth must be above gate or the edge of the plate. So, the first part is that we just quickly tell that if the d is up to the edge of the plate then the resultant is never going to be 100 millimeter. The resultant force because that is going to be 225 divided by 3 which is less than 100 mm. So, therefore, my depth has to be actually higher than that of the edge of the plate. So, now what we do is as we understand that depth is actually going to be the top of the you know valve or edge of the plate. So, we can quickly go through this ultimately. So, in this valve is going to see a trapezoidal pressure distribution. See ultimately it is going to be a trapezoidal pressure distribution. If someone knows what where is the centroid of a trapezoid you or she can calculate that easily right. So, therefore, condition would be that you know we know the area of this trapezoid we get the resultant force right. We also know the centroid of the trapezoid that can be equal equals to 100 mm and we can get the answer for the depth because depth is the unknown. The other way around is that we just follow the integration approach. So, in the integration approach what we adopt is this is the condition we have right and then we have alpha equals to m by r. Now, this alpha is actually x bar. So, x bar equals to 100. So, ultimately if we do this operation with d unknown then we are going to get the solution for the d ok. So, d will be ultimately equals to 450 mm. So, in this integration when we are doing make sure that integration limit is from 0 to 225 mm because that area that we are looking at that is only happening this pressure is only acting on the valve. Now, here we have actually you know some kind of valve which is assumed that it can take any amount of load ok. So, ultimately we are interested in how the valve is open right. Take the same problem let us say we take the same problem, but in this case what we are going to do we are simply instead of taking a rectangular plate we are simply make it triangle. Once we will make it a triangle what changes remember pressure distribution is same rho g d minus x right, but what will be changing is the line load because line load is now multiplied by the width that is how you get the line load from the pressure to the width. Now, here width is varying quantity right. So, we have to find out what is b let say the width of the plate at a distance x ok. Now, since both now this is isosceles triangle so we have b must be equals to x right. So, where is the how the line load should look like. So, line load now if you really look at line load will simply be a parabola this is a very important fact that since line load is now a parabola. So, the centroid should be the at the middle ok. So, if we just tell this much that we are not now we are really not going to go through any kind of you know operations on let us say integration operation. Only thing is that we find out what is the line load distribution and from the line load distribution we can clearly say that where is the centroid. So, ultimately what a answer I am expecting for this problem anyone what would be the answer for the depth see remember my resultant has to be passing through the since the resultant has to pass through the 100 mm right and line load distribution suggest is a parabola where the resultant is going to act at the centroid of the parabola. Therefore, the actual height necessary is simply 2 times 100 that is 200 mm. Now, student can actually scratch their head and go through all the integration procedure and so on so forth, but is that really necessary that is not necessary. As long as we have a visual feedback immediately through this simple calculation that this is going to be the parabolic distribution and parabolic distribution should have a resultant at the centroid of it. So, now what is done in this one is just you know similar way that if a student cannot guess that at the beginning. So, what he or she is going to do basically try to adopt the integration procedure which is going to be very very lengthy. So, here see just make take depth of the water is below the gate level do one integration find out where is the resultant then again you take the you know depth to be higher than that find out where is the resultant that is going to create more complexity in the problem. So, simple answer can just be done based on this line load how do you convert the distributed you know how do you convert the actual pressure distribution to a line load. I think we have completed this lecture. So, it has three parts parallel force system, coplanar force system and hydrostatic distribution hydrostatic load. The zone below the hinge there will be another force I think. No, but see that is a good question, but that comes into play when we are talking about equilibrium. Now, here what we said we are trying to simply look at when the gate is trying to just open. When sir it will break the equilibrium then the gate will open. When the gate is trying to open what will be the force here the force has to be support reaction has to be equals to 0 right. So, that is the problem we are studying. But this lower force will not be. If the lower is the depth the gate cannot open because as long as the reaction is below this level right that reaction will be balanced by the force here right support reaction here. So, the gate cannot open as soon as the resultant force just goes above the point A then reaction has to be equals to 0 reaction from this particular point B.