 Now, what we are going to do? We are simply going to talk about equivalent systems. Now remember in the previous session, we already learnt how to take the resultant of a force, just a vector addition. Similarly, how to do the resultant of moment, again a vector addition. So, when we discuss about equivalent system, we should know that let us say I have a rigid body. Then in the rigid body, if I have a number of forces acting, ultimately we know that the resultant force system that we will get, that means the resultant force as a vector and the moment resultant that itself is going to be an equivalent system. So, physical significance would be as follows. Let us say we talk about two system of forces, they are equivalent if they can be reduced to the same force couple system. That means they should have the same force resultant and same moment resultant. Therefore, what will happen? The system must produce same push or pull. So, push or pull is coming from the fact that both the systems are having the same resultant vector r. Similarly, both the system should have the same turning action about a axis. That means if in order to that to happen, then we should have the same resultant couple vector m r. So, as we understand now that once we let us say have a system and we resolve it to a resultant force and a resultant moment, then we actually have an equivalent system corresponding to the original system. So, if we look at from the two system perspective, the story is same that means two system must be resolved to a resultant force and resultant couple vector. So, in order to do that we need to adopt certain procedure. So, let us say I have a rigid body and in that rigid body I apply a force at a distance r from the origin o. So, origin is chosen by me. So, my objective is to find out what is the resultant force couple system right here at origin o. So, how do I do that? So, what I am going to do now? I am simply going to add a force f and add a force negative f at point o. So, remember adding this force will not change the action of the body because they are equal to each other and they are along the their line of action remains same, but one is positive and one is negative. Now, in the process what we have eventually done is that this f that is my originally applied load f force f and this one negative f will now create a couple and we know already from the previous session that this couple will now give me a moment that moment will be simply r cross f which will be perpendicular to the plane containing f and minus f. So, therefore what essentially happened is that we have translated this force f that is my original force f that is being translated to o simply by f as well as by a moment vector. So, therefore we call this a force couple system. Now, if you look at this system and if you look at that system the original system they are simply equivalent in nature because they will create same turning action as well as same push or pull action. So, now think of that instead of having the one force here if I have multiple forces on the body then we can simply take all of these forces at o and simply apply Varignan's theorem that we have studied to get the moment resultant of those forces that are applied on the body. So, our next slide we will consider this see ultimately I have now number of forces acting on this rigid body. If I have number of forces and I want to resolve all of these forces at particular point o that means I am trying to establish equivalent force couple system. So, the idea would be that I resolve all of these forces at o and I also get the corresponding moment component. So, therefore I must get a resultant force here and the resultant moment at the point o. So, to do this again we perform the similar operation as we have done before. So, all of this force if I consider force f 1. So, f 1 you know has gone to point o with f 1 as well as there is a corresponding moment component which is m 1. So, this moment is again the couple moment effect because I have to produce a positive force as well as negative force at point o. So, ultimately now what happens we see clearly that is at point o I have set of concurrent forces. If I resolve this way I have set of concurrent forces and concurrent moment also at this point o. Remember these are all free moments. So, they can be added in any case, but I am more concerned about the forces that forces are now becoming concurrent forces. So, the concurrent forces will give me the resultant force r. Similarly, this moment can be added and this moment can be added to give me the moment resultant. So, finally remember in this problem I have only concentrated forces. If in this problem I also have some couple moment that is couple is already present let us say some force f here and negative f here then I will simply add that that couple will be simply coming as a couple moment which is a free vector on the body. So, ultimately I have a resultant force which is simply some of the all forces vectorial sum as well as I have a moment resultant at point o that is simply given done by r cross f. So, we have put a summation sign here. So, that goes from i equals to 1 to 3 and if there are extra couple then we have to just add the couple moment additionally. So, therefore this becomes an equivalent force couple system. Now remember this system can be further taken to some other point o prime. So, how do I do that? That again is simple. So, ultimately this r should come here as an r, but there has to be a moment component also that will be s cross r. So, s cross r that will produce an additional moment and then you have this already m naught r at o. So, therefore your net moment will become m naught r prime that is result at o prime. So, that should be m naught r plus s cross r. So, therefore this system can be moved anywhere and all the time what we are looking at then again this system and this system becomes equivalent system. So, now the question should arise that once I resolve the system of forces at a particular point o by a resultant force and by a resultant moment can I further reduce it? The simplest form should be what? So, if we look at it now I have r and I have m naught r at a point o. Remember now this m naught r that vector which is a couple moment that can be decomposed parallel to r as well as perpendicular to r. So, now I am going to decompose it into two components one is m1, m1 is parallel to r that means it is producing a turning action about the line of action of r and I also have a m2 which is turning action will be like this. Now remember this system effectively can be further reduced. How? I can simply take into account this moment by shifting this resultant force associated with that couple moment to some point a. So, what will happen in this process is that r multiplied by some distance d which is perpendicular from the line of action of m2 should produce the m2. So, r cross d will become now m2. So, ultimately the entire system is now resolved to a one resultant force and along with a moment m1 which is again creating a turning action along the line of axis of r. So, we have a resultant force as well as a moment. Now if you look at the application of this if we just think of a screwdriver. This screwdriver will now have a turning action about its own axis as well as when we are tightening the screw it is being pushed. So, r is creating that pushing you know the push and the m1 that is creating the turning action about its own axis. So, the ultimate idea is that any three dimensional force system can be resolved to a that of a screwdriver or a range such that it has the line of action of force is along the direction of the screwdriver axis of the screwdriver and the moment should be about the axis of the screwdriver. So, although we will we are not going to you know take up any problem on the three dimensional force system we are only going to be you know use the planar force system or parallel force system. So, we can see that the resultant force couple system for a system of forces will be mutually perpendicular. That means now we are talking about if I look at a co-planar force system what is the resultant force resultant force should be in the plane xy plane. But the moment if you look at the moment component that is going to be perpendicular to the xy plane that means that is going to be about the z axis. So, it is trying to rotate the body about the z axis which is perpendicular to the plane. So, therefore, we can say that resultant force and the resultant moment is going to be perpendicular if and only if the system is co-planar similarly if we look at here in the parallel force system. So, remember the resultant force will be along the y direction, but all the moment the resultant moment will be on the xy plane. So, again we are talking about perpendicular system that means resultant and the moment resultant they are perpendicular to each other. So, we are going to study these two special cases that is co-planar and parallel force system where the resultant and the moment resultant are perpendicular to each other. Now, what happens in this case as we can see that once we resolve the resultant and the moment resultant at point O they can be further reduced to a single force such that we have to bring that resultant force to some place and its moment is going to take care of the moment resultant because they are perpendicular to each other resultant and the moment resultant. That means the co-planar system can be reduced to a single force similarly a parallel force system can be also reduced to a single force. So, we will take one at a time. So, let us choose the co-planar force system and how to reduce to a single force. So, I have three forces applied, but there can be any number of forces applied on this rigid body. Again these are all parallel force system because all the forces are applied on the x y plane. So, I can calculate the resultant force at O and the resultant moment at O. Remember the resultant force and the resultant moment they are going to perpendicular to each other and the operation that I have to do is very simple. We know already how to get the resultant force, how to get the resultant moment by simple vector operation or it can be even done component wise. So, next part would be how to reduce this to a single force system. So, the effective idea would be that we move the resultant force to some distance d such that d times r should be equals to m naught r. Now mathematically to find the d is not that simple. So, we are going to take help of algebraic operations to get the intercept with the x axis as well as intercept with the y axis of this resultant force. So, if we look at carefully what is happening I have the resultant force and I have the moment resultant. Now let us say the resultant has been moved such that it has an intercept with the x axis. So, we are interested in finding out the interception of the resultant force with the x axis. So, we split the you know break the resultant force at point B into two components one is r x the other one is r y. Remember the r y sorry r x is not going to contribute to the moment about o because it is already passing through o. So, therefore we can say that x times r y should be equals to m naught r. So, therefore I can get the x intercept of the resultant force. Similarly how do I get the y intercept of the resultant force. So, again I resolve the line of action that is the you know resultant force at point C into two components r x and r y. Remember r y will not contribute to the moment. So, r x times y must be equals to m naught r. However we must be careful with the science. So, remember whatever shown here is a positive notation because we have chosen the resultant force in such a way that both components should be positive. As well as moment is chosen in such a way that it is also a positive moment we can see from the right hand thumb rule because moment vector is applied perpendicular to the plane x y. So, therefore we are eventually going to get a positive x intercept in this case and negative y intercept in the this case. So, the line of equation if we say the intercept is a this one is a and intercept this one is b then we can just write the line of action of the resultant force should be x over a plus y over b equals to 1. So, that will give me the line of action of the resultant force. Now this is going to be very useful to calculate the x intercept and y intercept. So, once we get the intercepts of these two then my resultant force is fixed and entire system is replaced by a single resultant force. So now we can discuss a quick problem you will see nice example this is a warship. So, this warship when it is you know going to park here is very difficult when it comes to shore and we are trying to park this you know warship. So, what happens then we are going to take help of tug boards. So, what this tug boards are doing basically they are applying the force in such a way such that there is a resultant force and in the direction of the resultant force this vehicle is going to move and being parked in the appropriate or desired location. So, again it is a good example of parallel force system as we can see that this tug boards are going to apply parallel you know forces. So, forces are always going to be in plane and ultimately we are going to get a resultant force that resultant force direction is going to take this ship to the location desired. So, we will take an example of this and we will go step by step. So, again let us consider this is a boat or ship that is the you know and we are trying to take this ship to a certain direction. So, ultimately what is happening again this tug boards there are four tug boards that are applying the forces. So, how much forces are applied each tug boards are applying 100 kilo Newton push remember the forces are applied on the body of this ship. So, what we want to do we want to replace these four tug boards by a more powerful tug board such that that powerful tug board that is single tug board is going to create the same effect as that of these four original tug boards. So, we are therefore interested in finding out the total push by that single tug board and the direction that it should be applied and where it should be applied. So, we are interested to find out is it going to be applied here or is it going to be applied here that means where in the body the single force should be applied. Now if we think of this problem let us try to think of intuitively can we find out a answer quickly. So, as I said ultimately what are the steps the steps will involve let us say we have to choose a origin we have to define the coordinate axis. So, coordinate axis will be at o let us say x and y and then we are simply going to take the moment of all of these forces about this point and sum them up. So, therefore we are going to get the resultant moment similarly we are going to get the resultant force at this point o. So, once the resultant force and resultant moment is defined we can simply move the resultant force to a distance such that the moment due to that resultant force will be equals to the moment resultant as we have done the in the previous study. So, effectively I have to again find the x intercept and the y intercept and from that I have to figure out where should be the location of the resultant force or it can be right here or it can be here ok. So, if we just try to think little bit it is very clear since I have 3 forces here. So, the direction of the resultant force has to be down note. So, that means it has to be applied from this side now which way it is going to tilt is it going to be tilted this way or is it going to be tilted that way. So, that we have to decide if we just look at effect of these 2 forces it can be easily seen that the resultant force the single force has to be the direction it is shown here. So, our ultimately objective is what is the value of the r and where does it touch the body of this ship. So, the step one would be simply to get the resultant force again we can see here that all the forces these are all 100 kilo Newton force tugboat 1 2 3 and 4. So, we can simply add them vectorially as it is being done here. So, resolve the components of this. So, these are again slope is given 3 over 4. So, we can resolve into x and y components keep on adding. So, therefore we get the resultant force. So, I know how the resultant force would look like and if I look at the sign of it right along the x axis I have a component and along the y axis I have a negative components. So, therefore the resultant if we look at it it has to be inclined the way I have shown in the previous slide. So, we get the m naught r m naught r can be obtained using the vector operation again. So, we have to define r cross f for each one of this instead of doing that we can also look at component wise. For example, these force if we look at it. So, this will be simply 100 multiplied by 24 that is the moment that is produced at o. Remember that sign will be negative because it is clockwise counter clockwise should have been positive as per the right hand thumb rule. Similarly, if we look at the tug board number 2 we can break it into two components right here. So, the x component will produce a moment which is equals to x component multiplied by this distance that perpendicular distance is 24 right. Similarly, the y component will produce also moment. So that would be the y component multiplied by the perpendicular distance this perpendicular distance is 144. So, you see two values here one is 144 that is the distance the other one is the 24. So, this is coming from the x component multiplied by the perpendicular distance this is coming from the y component multiplied by the perpendicular distance clear. So, like this way we can get the moment resultant at o which will be this much. So, you can see it is going to apply clockwise. So, it is going to be applied clockwise here. Now, what do you do? Now, we know we can have this equation. So, ultimately I will be interested in finding out what is the intercept of r with the x and y let us say. So, to do that I can easily do this because I know the magnitude of the moment. So, magnitude divide by the r y that will give me the x intercept whereas, magnitude of the moment divide by r x that will give me the y intercept. So, what we can finally find out that x intercept is going to be equals to this 106.67 and y intercept is going to be 137.14. So, for sure I know the line of action of the ultimate resultant force. So, that is the more powerful single tugboat. So, that is how the line of action should be. However, in the problem it was asked at the body where it should be applied. Now, that is that can be clarified based on this. See remember although I have the x and y intercept. So, it makes sense that it will be somewhere here, but if I want to argue we can do as follows. Can it be applied here? Let us try to find out can it be applied on this inclined portion or not. So, what do I do? So, I know the equation of line of this. The equation of line of this is simply going to be y equals to mx where m is defined by the slope of this. So, once I look at this now if I take this y, if I take this y and substitute in the equation of action of the force then what I am going to get? I am going to get a value of 82.58 meter. So that means it is greater than 64. So, if it is greater than 64 now 64 is actually what is where the inclined portion is. So, if the now we are coming up with a fictitious answer because it is not between you know it is not for x less than 64 therefore, it cannot be applied at this location in the inclined portion. So, if you take y equals to 24 now which is the line of equation of the body right here and substitute here then what is my x. So, that x will give me the value equals to 88 meter that means the location of this point on the body has to be equals to 88 meter from the y axis. So, we have covered or completed a problem on the coplanar force system. Now, let us move on to the parallel force system. So, remember in the parallel force system we have again whole bunch of forces. So, these are going to be replaced at the origin first. So, once I take the resultant force and the moment resultant of all of these forces at the point o then that is how it looks like. So, we have a resultant force that will be simply sum of all of these forces let us say 1 to sum number n p is varying from 1 to n that many forces. Similarly, I could get the moment resultant about o and that is very simple. So, moment resultant can be obtained also at o just by using r cross f for each and every force. Now, the point is I have to bring this resultant at some location r. So, I have to replace this force and couple system by a single resultant force. So, therefore, we can clearly see that how I am going to do it that means this one and that one is going to give me the same moment balance and same force balance. So, if I look at the new position g of the resultant then I can say that r g. So, r g cross r that should be equals to what I get from this. So, in other words what we can do more easily what we can do actually is as follows. So, why do not we just simply balance the moment about x axis from this system with the moment about x axis of all the forces from this system that will give me the location of the single resultant force. So, I will get the value of z g if I equate the moment about x axis. Similarly, if I equate the moment about z axis I will get the value of the x g. So, therefore, I can clearly define the location of the resultant force at point g. Now, this system and that system is r therefore equivalent to each other because they will have the same resultant force as well as resultant moment about any point. So, again let us consider a problem. So, ideally speaking what is given here is a concrete foundation and we have as you know that in the you know civil structures we have the foundation and all the columns are going to transfer mostly vertical loads. So, we have whole bunch of columns, but there are only four forces acting on this column. So, what we want to effectively do determine the magnitude and the point of application of the resultant four loads. So, that means resultant of this four loads should be found out and its location should be found out. So, in other words simply replace all of this column by a single column that will be the equivalent system of the original problem. So, again if we look at the approach it is very clear that what we are essentially trying to do I have to do a moment equate the moment about x axis of this system with the equivalent resultant system. So, basically the moment has to be equated as well as about the x axis as well as about the z axis. Now remember can I get a quick answer here that where should be the resultant force is it possible to find out where should be the resultant force quickly. See I am interested in knowing just intuitively that which way it should be in which quadrant let us say quickly what should be the presentation of the single resultant force just intuitively in which quadrant or rather I just want to know I know whether it is positive x or negative x positive y or negative y like that. Yes, so that is great. So, we are actually going to it is going to be positive x side that is clear any comment on the z axis which side it should be positive z or negative z negative z axis. So, perfect. So, we got the answers and quick explanation can be always given because if you try to look at it see ultimately overall picture is like this. If we look at the moment balance about the z axis then we have three forces here. So, effectively that will produce more moment. So, to balance this I need a single resultant force which should be on the positive x side because I am getting more moment about z axis due to these three forces than this one. So, similarly if you look at the moment about the x axis remember at point a and d these two forces does not produce any moment about the x axis. So, ultimately I am going to get a moment about x axis which will be about x axis in the let us say from z to y if I rotate then it will be on that side. So, ultimately we can see effectively from these two forces that the resultant forces has to be along the negative z axis. So, finally we got our answers. So, we can do this just by you know using the approach that we have adopted. So, I have total resultant force that is negative 280 kilo Newton. So, I am now trying to make these two equivalent. So, I am interested in finding out what is the x g and what is the z g. So, first I do the moment balance about x axis of these two system that will give me the value of z g which is obtained. So, you can clearly see that I am getting a negative value that means actually the r should be on the negative side of the z axis. Similarly, if I do the moment balance about the z axis of these two system then I should get the value of x g. So, ultimately we can see that x g is comes out to be positive. So, actually we are going to get a resultant force in this quadrant which is defined by positive x and negative z. So, we will quickly taken one more problem and this problem is to do with 3D force system, but you know just to go through how to resolve the moment at point A we are just taking this simple exercise. So, basically replace this forces with an equivalent force couple system at A. So, that is what we want to do. Now remember this point is A is located negative 30 millimeter downward from the z axis. So, effectively if I want to find out we can always go with r cross f. So, we can try to find out what is the position vector of all of these forces. Let us say if I choose c that what is r ac then I can do r ac cross f. So, that will give me the moment at A similarly for the others and therefore I can get the moment resultant and the force resultant right here. However, remember in vector mechanics could be tedious, but whoever likes the vector mechanics they can go with the vector mechanics approach. In approach number 2 what we can also do and this is what we are going to mostly do when we are solving 3D equilibrium type of point. So, there we can take look at the individual component of the moments. That means for example, if I look at it what are the forces producing the moment about y axis that is passing through A. So, I am interested in find out what are the forces that is producing moment about y axis that is passing through the A. We can clearly see the 2 components are coming into play one is this 300 Newton. So, 300 Newton multiplied by 75 millimeter that would be one component. However, that is counter clockwise and 150 Newton will always also produce a moment about the y axis which is passing through A. So, that will be 150 times 60, but that is clockwise. So, we can accordingly add them with the signs and get the moment about y axis which is passing through A. Similar thing we can do for let us say moment about the z axis which is passing through A and moment about x axis which is also passing through A. So, we have solved it based on the vector mechanics approach. Those who would like the vector mechanics they can simply add the forces and we can see clearly the force has been added to get the resultant force. So, resultant force will have negative 300, negative 240 and plus 25. Then we can get the moment. So, for the moment part we have to do r cross f for each of these forces and that is what it has been done. Remember the only thing is this 240 Newton will not see these two you can see this is actually vectorially added. So, this force right here is negative 240 and negative 125. So, there is a net force here whose vectorial form is negative 240, negative 125. This force is here 300 Newton negative 300 and this force is also here which is 150 Newton. So, ultimately I am going to get this moment which is resolved at A which has three different components. However, all of these components can be obtained individually as well as I have explained before. For example, quickly if we look at what is the z component of the moment. So, z component of the moment if we look at z axis is passing through the A. Now, z component of the moment will only be produced by the 300 Newton force. We can clearly see the 300 multiplied by 30 that is the perpendicular distance. So, that will give me a moment about z axis and if we use the right hand thumb rule also, if we just use the right hand thumb rule you can see the thumb should be pointing towards the positive z axis as well. So, therefore value that you get is 300 multiplied by 30 that is in 9000 Newton millimeter or let us say 9 Newton meter and you can see that 9 Newton meter clearly right here. So, this you know other components can also be obtained. So, likewise moment about x axis and moment about y axis just by taking the component of a moments of the individual forces. So, as I told before that you know for the students it is always a challenge to get the moment correctly. So, we have discussed now you know several problems how to resolve a force system by a equivalent force couple system at any point on the body. So, for the parallel and coplanar force system ultimately for the parallel and coplanar force system since the resultant force and the resultant moment they are perpendicular to each other. Therefore, I can always replace them by a single resultant force. However, for a general 3D force system we cannot get the single force it will always be resolved as a screwdriver whereby I have the resultant force and the moment which is also acting about its axis. However, we do not do that problem right here just because they are going to be bit lengthy. So, I think now we are going to close this session right here.