 So, we will just start our next topics which is review of vector algebra and its applications how we are going to use that in solving the mechanics problem. Now, what are the basic things we are going to look at in this topics or as follows. We will be looking at addition of vectors that is one of the important thing because you know there could be several forces that are acting on a body or maybe at a particle let us say and we have to find the resultant of the force. So, that is one aspect. So, vector addition and subtraction then we can see the scalar product why do I use the scalar product and how it can be applied for the solution of mechanics problem. So, that is another consideration second would be you know the cross product. So, cross product is going to play an important role because basically through the cross product we are going to get the moment at a particular point and the lastly we will be discussing about a triple cross product. So, through the triple cross product what we get is the moment of a force about an arbitrary axis. So, as such we are really going to use the slide. So, we can control our timing depending on our need, but if you have any queries please use your module and we have also TAs sitting around. So, they can collect your questions and we can discuss that at a separate discussion session. So, vectors they are classified in two groups one is the fixed or bound vectors and we cannot change the point of application of the fixed or bound vectors without affecting an analysis and we will see that most often in all the problems and there is also a free vector. So, free vectors are those that can be freely moved in space without changing their effect on an analysis. So, what would be a free vector we will see the later on that couple is going to be a free vector. So, equal and negative vectors we all know this next is the addition of vectors. So, remember as we said that we are going to use the parallelogram rule for vector addition and those are illustrated here. So, we have P and Q. So, if they form two sides of the parallelogram then the resultant would be along the diagonal of it. Remember vector can be added based on the triangle rule as well. So, as we can see P and Q are two vectors therefore, we have P plus Q as the resultant of these two vectors. So, once we adopt the laws of science then we can clearly see that resultant which is nothing but P plus Q and the magnitude can be represented in terms of magnitude of the P magnitude of Q and the angle that is made by this two vectors. So, we have R square equals to P square plus Q square negative 2 P cosine of B where B is the angle made by two vector. Similarly, laws of science can be applied. So, sine A divided by Q must be sine B divided by R should be equals to sine C divided by E. So, remember this is very important vector addition is commutative that means we can add it either by P plus Q or Q plus P it would not change the result and vector subtraction is also illustrated in this diagram. So, we have P negative Q now gives me the resultant E minus Q. So, when we go to the addition of three or more vectors then we can either apply the triangle rule repeatedly or we can use the polygon rule. So, as we see here that in this case we have really taken the triangle rule repeatedly. So, as the vectors are commutative in nature their addition. So, first we can simply do Q plus S and then P is added on to Q plus S to form the P plus Q plus S. Similarly, for polygon rule we can always get the polygon and the resultant will be defined by the P plus Q plus S. So, remember in this case it is associative that means P plus Q plus S can be considered first P plus Q then add on to the S or we can first consider the P and then added with Q plus S. So, now there is a simple question here as we said that a quantity which has magnitude and direction and also follows the parallelogram law should be a vector, but can we define a quantity which has the magnitude and the direction, but does not follow the parallelogram law. So, I just want to hear from at least one of the remote centers if we can get that answer that means I am looking at a quantity which has both the magnitude and direction, but they are not following the parallelogram law. So, they are therefore it cannot be a vector. Can I get an answer quickly from one of the remote centers? Just type it on chat. Now I see one of the answers probably from one of the centers. Let us just there are so many actually. Now we cannot go through all of this, but I see one of the answers that is coming up which is simple to illustrate that would be the finite rotation. So, we will just quickly take this. So, the answer that I am getting from some of the chats are you know the finite rotation. How do I explain this? So, if we just go quickly here we can see just take an example of a book actually. So, you can see that a finite rotation we can define the magnitude of the rotation as well as its direction that is clockwise or anticlockwise. So, if we take this example let us say I have a book. So, as we see here that we are having a you know counter clockwise rotation sorry yeah let us say counter clockwise with respect to x then clockwise with respect to z. So, I do this operation first this is my operation number 1 then the operation number 2 then the book is actually placed right here. Now if I reverse it that means I am going to adopt the commutative property. So, what I am doing here first I am going to do the operation number 1 which would be rotation about the z axis and the rotation about the x axis I can clearly see the book is not at the same position. Therefore, we can clearly say that p plus q is not equals to p plus p and therefore finite rotation cannot be a vector. Now, next come to the you know one of the important aspect as I said that why we are doing this you know addition of vectors because we want to get the resultant of several concurrent forces. So, in many applications we see that number of forces are being applied at point a let us say of a body. So, now we are going to talk about p q s these are actually three forces and we have to get the resultant of this and as we know that we can either use the triangle rule repeated triangle rule or polygonal rule to get this. We can also adopt the rectangular components of a vector that means now what we are saying that if we have a force F it can be decomposed into two components one is the F x and one is the F y and we define the unit vector that is i and j and therefore in terms of the rectangular components we can define the force must be equal to F x i plus F y j and remember F x is equals to F cosine theta and F y equals to F sin theta. So, we can apply this again to look at addition of concurrent forces that means once I define in the rectangular components I can take that up for the addition of the concurrent forces and the way to do it is that every force is being now broken into two components that means let us say consider force p it will be broken into p x and p y similarly s and q therefore what we are going to get we are going to get the resultant r that resultant r will have two components one is the r x and one is the r y. So, r x should have the sum of x component of all the forces and r y should have sum of y component of all the forces therefore I can get the resultant r the magnitude of resultant r should be defined by square root of r x square plus r y square and the direction cosine should be obtained through tan inverse r y by r x. So, now we will take simple examples to look at this. So, let us look at this example very carefully. So, it is a example of a bracket where two members are attached to the bracket. So, the member A is subject to a force that force is 20 kilo Newton and it is compressive in nature and member B is subject to another force 30 kilo Newton which is also compressive in nature. Now, the question is determine using trigonometry. So, only we are asking that using the trigonometry that we have studied before the magnitude and direction of the resultant of the forces applied to the bracket by members A and B. So, we are interested to get the resultant of these two forces. So, you can also see a demo you know example that of a truss member where we see that several you know truss members are connected at a particular joints and this is also in the example of a bracket that is also used for different supports. So, how do I get the solution? Steps are very simple first construct the force triangle and apply the sine and cosine rules to get the resultant of these two forces. So, if I just adopt the triangle rule. So, again I have the 20 kilo Newton force and then we are going to add the 30 kilo Newton force on top of that if I just go back quickly remember the 20 kilo Newton load was 45 degree angle and the 30 kilo Newton load was 25 degree angle with respect to the vertical axis. So, what we have done here we have taken first 20 kilo Newton and then the 30 kilo Newton and resultant is pretty obvious. So, which is going to look like this. So, the answer that I am trying to get here really is the angle that the resultant makes with respect to the x axis that means the direction cosine and the magnitude of the r. So, these are my requirements. So, if we try to solve this problem let us say. So, what are the known information I have I know these angles 25 degree and 45 degree. Therefore, the angle that is made between the 20 kilo Newton and 30 kilo Newton can be obtained. So, that angle let us say it is a gamma. So, gamma is 110 degree ok. Now therefore, we can get the resultant r. So, resultant r can be obtained since I know the magnitude 30 and 20 and the angle that is made between these 2 forces. Therefore, we apply the cosine rule and therefore, we say that r should be equals to 41.357 kilo Newton. So, how do I get the phi to get the phi I can simply apply the sin rule. So, I am trying to now get the answer for alpha right here. Once I know the alpha then I should be able to calculate the phi also. So, alpha can be obtained by simply using the sin rule. So, 30 divided by sin of alpha should be equals to r divided by sin of gamma and therefore, I could get the value of alpha through this operation. So, alpha is known therefore, phi should be known. So, phi should become alpha plus 45 degree. So, I get the value of phi to be roughly 88 degrees. So, we can display the final result. So, that will be my resultant and which will make 88 degree angle with respect to the horizontal axis. So, remember this can also be obtained through the rectangular components. So, we will leave that as a homework exercise let us say that can we get the same answer if I use the rectangular components of forces. So, another problem now this will be done through the rectangular components of forces. So, now this is a example of a collar. The collar are typically used as you see that there is a vertical rod and this collar can actually glide against this vertical rod. So, you can see this picture actually in case of an umbrella. So, we could clearly see that when we are trying to open the umbrella we have this spokes this small members these are actually transmitting the forces. So, very similar example here taken now what is in question here that three forces are being applied one is 90 pound one is 70 pound and one is 130 pound. So, determine part A the value of the angle for which the resultant of the three forces is horizontal. So, you want to find out what should be this angle such that resultant of the three forces is horizontal and the corresponding magnitude of the resultant. So, if we look at the solution key the basic step that is required. So, what is being said here actually that r y should be equals to 0 because the resultant of the three forces is horizontal. Therefore, the y component of the resultant that is sum of the vertical component of the forces they should be equals to 0. So, if we apply this condition then definitely I am going to get the value of alpha because now I have one equation one condition with one unknown that unknown is alpha. So, we look at the solution carefully. So, here we set r y equals to 0 r y is nothing, but sum of all the forces along the y direction should be equals to 0. So, we do that we have 90 plus 70 sin of alpha and 130 cosine alpha remember we have to add this algebraically. So, this will get the positive sign whereas this will get the negative sign we set this to equals to 0 and try to solve for the alpha we will end up getting a quadratic equation here. So, one of the answer should be alpha that should be equals to 24.1 degree. So, once we know the alpha then we can get the second part was what is the actual resultant. So, actual resultant should then be equals to simply r equals to r x. So, we can also get the r x. So, r x should be just sum of all the forces along x direction. Therefore, we get the resultant force which is horizontal in this case is equals to 117. So, now we have studied so far two dimensional problem now we can extend it to suddenly to the three dimensional problem. So, by applying the vector mechanics we can clearly see that a force in a three dimensional space x y and z can again be represented in rectangular components. So, that will be f x plus f y plus f z and remember that f x should be equals to f cosine theta x. So, theta x is the direction cosine that is the angle between the force f and the x axis similarly for f y and f z. So, ultimately what we see that what is most important here is to get an unit vector along O A. So, once we get the unit vector along O A which is determined by the direction cosines then we can define the force f should be equals to lambda multiplied by the magnitude of the force. So, remember in application we may have you know the magnitude of the force vector is f and the direction of the force is defined by the locations of two points. So, force is being applied let us say somewhere at point m and we know basically the direction of the force that can be determined by the direction you know the two points through which the force is actually passing. So, what we have to do in this case as we know that we have to get the unit vector. So, how to get the unit vector by knowing the two points in the space. So, we get the unit vector by this operation as illustrated here. So, basically we will get the unit vector lambda and therefore, the force can be represented. So, force will be equals to magnitude of the force and then the unit vector. So, the main point here is the getting the unit vector is going to play an important role to solve the problems. So, we can quickly study a problem again that relates to three dimensional force. So, what is given in this problem basically you will see this one in real life that lot of antennas must must of antennas you know they are actually very unstable system. So, you can think of a transmission tower itself. So, they are to stabilize the system we need guy wire. So, guy wires are connected in various directions such that we can stabilize the system. Now, let us say that the tension in the guy wire is 2.5 kilo Newton. Therefore, what we have to find is the components of Fx, Fy, Fz of the force acting on the bolt at A. So, we want to find out what are the components of forces at A and the direction cosines of the force. So, now to start as we really looked at this problem theoretically. So, ultimately what we need to find out we really need to find out the unit vector that unit vector should be from A to B. So, we will be interested to finding out what is the lambda. So, if the lambda AB is found then the magnitude times the lambda AB should give me the force vector and the direction cosines can also be found. So, we look at the solution of this. So, how do I get the lambda AB or lambda unit vector along AB which is pointing from A to B. So, we first get the position vector AB. So, that can be obtained. So, we have to get to get this we can either do you know to freeze the coordinate system first. So, coordinate system is chosen here as origin. So, we have R B minus R A that will give me the position vector AB or we can simply march from A to B in various direction that is x y and z to get the R AB. So, the position vector AB can be obtained as we see here. Therefore, the unit vector which is the position vector divided by the magnitude of the position vector. So, unit vector can also be obtained. So, therefore, we determine the components of the force. So, which is simply magnitude of the force which was given as 2.5 kilo Newton multiplied by the unit vector. So, the second part was how to get the direction cosines theta x theta y and theta z that can also be obtained because we already got the unit vector and unit vector is nothing but the direction cosines multiplied by the unit vector in its directions. So, therefore, we can get this angles for the direction cosines. So, there they will be you know like this as displayed. Now, suppose we ask what are the components of the force in the wire then at point B. So, instead of saying the component of the forces at A can we get the components of forces at the B. Remember in this problem we already said that it was in tension and we have used the fact that from lambda unit vector direction was from A to B. Now, the question is the unit vector direction is from B to A. So, we can certainly ask the student can we give me the FBA instead of FAB. So, the answer will be very simple. We can simply say that FBA must be equals to F negative of FAB because the force in the wire must be same throughout its length and the force at B and acting towards A must be the same magnitude but opposite in direction to the force at A. So, far we have completed the rectangular components of the forces. Now, we will do the you know scalar product and the cross product of the two vectors and we will try to find out what are the applications how it is coming into the force applications. So, as we all know the scalar product of two vectors P and Q they can be simply given by P dot Q which is equals to P Q cosine theta. Remember that there are special characteristics of P dot Q. So, they are commutative in nature. So, that means P dot Q should be equals to Q dot P distributed as well. So, P dot Q 1 plus Q 2 should be P dot Q 1 plus P dot Q 2 but do not have the associated feature that means that is undefined. So, as it is explained here. So, ultimately what we know that when we take the dot product between P and Q in the rectangular Cartesian coordinates we will get a scalar quantity out of it which is simply the multiplication of the individual components. So, P x, Q x plus P y, Q y, P z, Q z. So, if we are talking about the same vector then it will be simply the magnitude square. So, how do I apply this in the force? So, what are the things we are looking at mostly if we look at the two forces two important things we want to know from the P dot Q operation the scalar product operation if the P and Q are given then we should be able to find out what is the angle between these two. So, one application is to get the angle between the two force vector. What is the other application? The other application would be which is very important is the projection of a force onto a line. So, suppose in this case I want to find out the projection of this force P onto an arbitrary line O L assume the vector in this direction is Q just for the time being. So, we know that P dot Q equals to P Q cosine theta. Therefore, P cosine theta should be the projection of P onto the line of the line O L which is nothing, but P dot Q divided by the Q. So, what is now Q divided by you know vector Q divided by Q? We already know that is nothing, but the unit vector along the axis from O to L. So, what it comes out to be if we want to get the projection of the force P onto a line O L which is defined by the unit vector lambda then we just do this scalar product operation which is P dot lambda to get the projection of the force onto the line O L which will be like this. Now, vector product of two vectors. So, now we are talking about the cross products of the two vectors. Now that is very important and we have to be extremely careful when we talk about this. So, as we know that P cross Q is going to give me three important information. It will result in a new vector V that V is perpendicular to plane containing P cross Q. The magnitude of the V is P Q sin theta and the direction of the V has to be obtained from the right hand thumb rule. So, we have three important things going on that one is the new vector that we get should be perpendicular to the plane where P and Q lies and the direction must be determined by the right hand thumb rule and the magnitude has to be equals to P Q sin theta. Remember if we look at the characteristics so, we have already looked at the scalar product. Now, in case of a vector products they are not commutative, but they are distributed. So, P cross Q 1 plus Q 2 must be equal to P cross Q 1 plus P cross Q 2 and they are not associative. So, that means P cross Q cross S should not be equals to P cross Q cross S. So, how do we make this operation in rectangular components? So, I have a force let us say P. So, P is defined in rectangular components and Q is also defined in the rectangular components. So, therefore, what is P cross Q that will give me the vector V? We know it has to be done by taking the determinant of this matrix. So, the first row of this matrix will be the unit vector second row should have the components of the P vector and the third row should have the components of the Q vector. Now, how it is vector product? Why it is important in mechanics application and so, it will lead to taking the moment of a force about a point O. So, as it is explained here that what it will do that moment of a force will produce a turning action on a rigid body. So, when we look at it suppose that I have a rigid body as displayed here and the point of application of the force is at A. So, we have the force applied F and the position of the point A is given by the vector R. So, therefore, if I want to take the moment of the force about the point O then we should get R cross F. So, R cross F will give me the moment which is perpendicular to the plane of R vector as well as F vector. Remember its direction must be determined by the thumb rule as it is shown here. So, what we can clearly see the magnitude of M naught that is simply equals to R F sin theta. So, R F sin theta from this if we look at the geometric carefully what is R sin theta R sin theta should be equals to D. So, therefore, to get the moment we can also say the magnitude of the moment should be determined by the F multiplied by the perpendicular distance of the force from the position O which is the origin let us say. So, the main idea here that is explained that it is trying to rotate that body about an axis and that is what we mean that axis is defined by the direction of the moment vector itself. So, in case of a 2 dimensional structure since the depth can be ignored. So, we can clearly say that if we apply a force then the moment of that force about O can be obtained through F multiplied by the D. However, we have to be very careful with the direction. So, in this case it should tend to rotate the body about O in a counter clockwise fashion. So, in this case what we see that we can take this as a positive. So, remember moment is actually you know the direction of the moment is applied perpendicular to the axis which is outward to the slide. Similarly, in the next case if we apply the force in the other direction. So, it will try to rotate the body clockwise about the point O. So, therefore, in this case we can say the magnitude should be taken as negative as we see because it is going into the plane of the slide or plane of the board as we see here. So, we can quickly once we introduce this concept in 2D and to make sure that the students are actually had a good grasp how to take the moment now we can simply take a very small problem. So, these are various applications of how moments are applied in real life examples as we can see here. So, all of these will actually demonstrate how the moment is applied and what would be the direction of the moment vector. So, one important aspect is the Varygnon theorem. So, what is this is basically it is stated that moment about a given point O of the resultant of several concurrent forces is equal to the sum of the moments of the various forces about the same point O. So, that means, if I do have a resultant of these set of forces whatever moment that gives about O should be equals to if I simply go component wise and take the moment of each and every component of the forces about O. Now, that is going to play an enormous role because what we now learn is that that determination of moment of a force F by the moments of 2 or more components of forces of F. Now, it is nothing but nothing more than we are basically using the distributed property of the vector. If we can clearly see the distributed property of the vector is resulting on to the Varygnon theorem, but it will make understanding much easier if we get the moment component wise and then algebraically sum this out. So, the next would be you know getting the moments about point O of an force which is applied at point A. So, this is a very standard operation. So, what we are doing we have set up the origin here. So, we are trying to take the moment of the force about the origin. So, which could which should be simply given by r cross F and r is already known to us. So, that is given by x plus y plus z as we see and the component of forces are also known. So, we will ultimately get the moment component individual moment components that are acting about the x axis, y axis and z axis. Now, if you want to take the moment about some other point instead of let us say origin O we can do that as well. So, instead of taking the moment about point O I am now taking the moment of the force that is applied at A with respect to point B. So, that can also be done by saying that I have r B A. So, r B A is my now vector that is the position vector and then we have the force. So, again we apply the vector mechanics r cross F. So, therefore, we get the m B. Now, as for the Varygnan's theorem what we discussed that moment can be taken in component wise as well it is demonstrated here for a two dimensional structure. Let us say I want to take the moment about point O. So, how do I do this simply if the force is given by two components. So, F x and F y. So, therefore, we can clearly see the moment about point O should be simply be equals to x multiplied by F y. So, that is counter clockwise. So, that is positive then we have to take the y multiplied by F x, but that is going to create a clockwise action. Therefore, we should take that negative y F x. Similarly, instead of taking the moment about point O if I want to take the moment about point B of this two dimensional force system then we can clearly see that same operation I am doing except for that x is being replaced by x A minus x B and the y is replaced by y A minus y B. So, ultimately this is what plays an important role that what would be the direction of the individual component of the moment sorry individual component of the forces when we take the moment about point O what are their direction and how we are going to algebraically at those component of moment. So, we take a very you know simple illustrative example the question here is that we have a bar let us say rigid bar O A and a vertical force is applied at the tip of it at point A which is 500 Newton. We want to find out the following the moment about the O the horizontal force at A which creates the same moment then the smallest force at A which produces the same moment and the location for a 1200 Newton vertical force to produce the same moment. So, first let us just study this part which is very simple as we can clearly see that what would be the moment due to this 500 Newton. So, if we look at the component wise. So, we need to find out the perpendicular distance from O which is easy in this case. So, it will be simply 0.6 cosine of 60 degree. So, 500 multiplied by 0.6 cosine of 60 degree that will give me the moment about O. Now, it will try to turn the body about O in a clockwise manner. Remember it is rotating about an axis which is perpendicular to the slide passing through O. So, the first part is already solved. So, we can simply do this operation that F multiplied by D that should be equals to M naught that is 150 Newton meter direction is shown that is clockwise. The second problem was part B the horizontal force at A. So, we now need to find out what is the horizontal force at A which produces the same moment. So, therefore, we apply the horizontal force. So, same moment that means what we found from the part A. So, we can clearly see now we are talking about F multiplied by the vertical distance D and we are interested in finding out what is the magnitude of F such that we get the same value as that of part A. So, basically now we have to take the vertical distance which is 0.6 sine 60 degree and therefore, we get the value of F which should be towards the right. Remember why it is towards the right? Why not left? Because if we take the left then it is going to be opposite to what that is being applied. So, in order to maintain the clockwise moment we have to apply towards the right. The third part is important. Now what was the question? The question was what is the minimum component of the force that will produce the same moment? So, what does that mean? Remember I have different directions that force can be applied, but I am trying to find out when it leads to the minimum force. So, ultimately to obtain the minimum force the requirement would be perpendicular distance between the force and the point O has to be maximum. So, the perpendicular distance between the force and the line of action of the F we have to get that maximum. So, maximum means that means I have to achieve the maximum perpendicular distance should be 0.6 that means force has to be applied perpendicular to O A. So, in that way now this is a very important concept because as we can see if we take any other direction then one component of the moment will be always lost. So, in order to maximize the force I should get the force should be perpendicular to O A such that I can maximize the perpendicular distance. So, therefore I want to find out the force F. So, the force F as we see here must be applied at a 30 degree angle which is nothing but defines it from the horizontal axis which is perpendicular to O A. So, the last part was to determine the point of application of a 1200 Newton force to produce the same moment. So, now we have to create the same action as that of moment that means we are always going to create the clockwise action. Therefore, we choose the distance D which is unknown to us and apply the 1200 Newton force. So, therefore what we get? We get the distance O B from this operation. So, ultimately we found the distance D that can be obtained very easily and from the distance D from the geometry we can find the point of application of this load 1200 Newton. So, that is O B. Now, we will take this up to you know exercise more complex problems. So, this is an example of an shower head. So, you can see the shower head here just understand the geometry of this problem a little bit. So, we have the shower head and that shower head if you look at it then the point B here we have the shower head it is on the Y Z plane, but the point B is 50 millimeter down from point A, but it is resting on the Y Z plane. Now, what is being essentially done? We are applying a force through a range. So, we have the range here where the force is applied. Now, how the force is being applied? First of all look at it carefully the force has a 45 degree direction. Now, that 45 degree direction how it is maintained if we choose a line which is parallel to the Z axis. So, that makes the 45 degree angle from the force on the Y Z plane and then we have another 12 degree angle that is made from the line this line right here to the axis this axis. So, this is parallel to the Z axis. So, now it involves two processes one is that we first take the component of the force on this plane and then we have to also rotate it by the 12 degree on the horizontal plane in order for this force to be perpendicular or you know when you look at the component wise basically what we are going to get we are going to get three components of moments, but remember we can do this individually. We can do this individually that means taking the component wise moment of forces resolved into different components or we can simply adopt the vector approach that is r cross f we can do that way also. So, now intuitively if we want to think out that what would be my final moment how does it look like. So, if we think of it seriously not the numbers, but let us say which direction it is going to act one could easily find out that my moment is always going to be perpendicular to a plane which will be defined by a c and f. So, that means I can always draw a plane which is where the position vector a c and the force lies and if you draw that plane what happens I can clearly see that the moment vector has to be perpendicular to that plane. That means if I examine component wise it should be behind this wall that means ultimate answer that I am going to get that I must expect that I must expect my answers such that x and y component of the moment should be positive and the z component of the moment should be negative. So, in this case we can use the vector mechanics access. So, we find the r a to c that can be always obtained from the geometry of the problem. So, we get r a c and then we look at the components of the forces that means what is my force vector what are the components just look at it carefully. So, x component of the force if I want to get the x component of the force first I take the projection on to this plane which is my x z plane. So, that is f cosine 45 degree and then the direction would be pointing towards c and then you take the components. So, along the x axis if I have to look at it. So, it should be 36 cosine 45 degree multiplied by sin 12 degree. Now, remember it is going to point again towards the c that means which is negative from the x axis. So, we get the components of the force x similarly we can clearly see what is my f y f y is nothing but f sin of 45 degree that will be parallel to y, but again it is pointing downwards that is negative y axis. So, we get that and f z can also be obtained. So, we get the force vector therefore, we get the moment vector by the operation r cross f to the vector mechanics and as I said before look at the now components. So, I have 3 components of moment my x component is positive y component is positive and z component is negative. So, the point that I am trying to make so intuitively we had thought about it that if I really draw a plane that contains f and r then the moment vector should be perpendicular to that plane such that x and y component should be positive and z component should be negative. So, we ultimately got that. So, finally, we are going to look at the mixed triple product of the 3 vectors. So, in this case as I said this will be very important when we are going to get the moment of a force about an arbitrary axis. So, how that is coming into play let us look at it. So, I have p and q and s these are 3 vectors. So, what is s dot p cross q that is simply going to give me the scalar result. So, I get a scalar component out of that what that is going to define that is going to give me the volume of the parallel pipe that is drawn here. So, I am going to get the volume of this. Now, there are some properties of it that how it is going to act. So, we can always use the thumb rule and try to see if these are correct, but our ultimate goal is how to evaluate the final scalar quantity that comes from s dot p cross q and we can see the determinant of this matrix that is going to give me the final triple product which is the scalar quantity there is no i j k. So, remember the concept here is that we get the area through p cross q magnitude and then when we are doing the dot product. So, p cross q dotted with s we are getting the height. So, area multiplied by height is going to give me the volume of the parallel pipe. So, now how that can be applied. So, we are surely going to get the moment of this force that is acting at A through the line O L. So, now that line in this case it is assumed that passing through O. So, we know the unit vector lambda. So, unit vector lambda is known along the O L we know the R that is the position where the force is applied position vector and we also know the force vector. So, we want to get the moment of this force about this line O L. So, first we take the moment of this force about 0.0. So, that can be simply done by r cross f. Now, what r cross f will give r cross f will give me the moment vector which is perpendicular to this plane where r and f lies and then I just have to take the component of this moment on to this line that will give me the moment that is about this axis O L. So, therefore, I just have to dot product the lambda that is lambda dot m naught. So, lambda dot m naught that is going to give me lambda dot r cross f and you can clearly see that this is my mixed triple product. So, finally, what we have learnt that now the body will have a turning action about the O L. However, as we know this is going to be a scalar quantity here which is rotating the body about the O L. Now, here is the main question now when we are taking the moment of the force about an arbitrary line on the space. So, what is demonstrated here that first we choose let us say a point B on that axis. So, therefore, I can always get the r B A that is my position vector. So, r cross f will give me the moment vector and I take the projection of that moment on to this line. So, that will give me the moment of this force about the axis. So, I will get a scalar quantity. Now, question is when I am trying to get the moment m naught does it matter whether I choose the point B or point C. So, it can be shown it can be proven that it is independent of point B or point C it does not matter where I take the moment it should give me the same result and reason is very simple let us say m B is equals to lambda dot r cross f. Now, we are talking about r B A. Now, if I choose a point C r B A can be decomposed into 2 components one is r B C and other is r C A. So, r B A should be equals to r B C and r C A. So, remember now if you apply the distributive property one component that means lambda dot r B C cross f lambda dot r lambda dot r B C cross f that will simply give me 0. Why because lambda is now parallel to B C. So, that cannot constitute a volume of the parallel pi. So, therefore, ultimately what I have I have r C A cross f lambda dot r C A cross f which is equals to lambda dot r A B r B A cross f. So, it does not matter whether I choose B or whether I choose C I should get the same answer. So, we will quickly take a problem. So, it is a problem involving a frame A C D which is hinged at A and D. That means it can rotate about this point and we can clearly see that if I do not have a cable then this is going to simply collapse because A and D's they are hinged. So, therefore, what is being asked now that it is supported by a cable that passes through a ring at B and is attached to hooks at G and H. So, this cable is passing through a hook let us say it is frictionless also at this point of time. Now, the tension in the cable is given which is 125 Newton. So, we want to find out the movement about the diagonal A D of the force exerted on the frame by portion B H of the cable. So, I have B H here. So, the force is 1125, 1125 Newton that is tensile in nature. So, pointing from B to H. So, what is the movement produced that by that force on to the diagonal A D. So, we have to think out at the very beginning what answer I am expecting. Remember firstly I have to take the movement of this force B H about a point A and then dot product with the unit vector along the A D. Now, the unit vector can be chosen from A to D or D to A. So, let us say I choose lambda A D as my unit vector. So, if the lambda A D is the unit vector then can we interrogate that what is the movement that is going to be applied on the A D that means what is the direction of that movement that I am looking at. Remember since A D is on that in this direction and I have T B H. So, therefore, the final movement that is going to be acting on the A D that will be pointing from D to A not A to D. So, that means I am really expecting a negative answer at the end of the day. So, the scalar quantity that I will get should give me a negative answer because my chosen unit vector direction is A D not D A. One more point to observe when I am taking the movement about this line A D I can either choose A or I can either choose point D and we have proven that it does not matter which point I choose I am going to get the same answer. So, for our convenience we will choose in this case point A about which I am going to take the movement and then I am simply take the dot product with the lambda A D. So, let us look at the solution and at the end we are really going to get this answer and as I said if we intuitively think it out then the answer for the movement that is trying to turn this frame about the diagonal A D. So, its rotation should be about A D it should be negative we have already thought that out. So, ultimately now we have to work out different vectors that are involved. So, we can clearly see that what is my final answer that I am trying to get is a scalar quantity that is given by lambda A D dot R which is R A B cross F F is my T B H. So, what is lambda A D that we can calculate from the unit vector our origin is right here. So, based on that we get the lambda A D which will be this is then we have R A B R A B R A B is very simple in this case it is just going to be 0.4 I. So, that is obtained now T B H. So, T B H will again require us to calculate what is lambda B H. So, we obtain the lambda B H and multiply by that with the magnitude of the force which is 1 1 2 5. So, therefore, I will have the moment which can be obtained through the determinant of this matrix. So, we get negative 180 Newton meter. So, the next part will be the couple moment as we have talked about it before also during the introductory part that the couple moment is a free vector. So, what is a couple two forces F and minus F having the same magnitude parallel lines of action and opposite sense are set to form a couple. So, we can clearly see a good example that we are trying to take out the wheel of this car and we have the range arrangement like this. So, we have equal and forces being applied they have the same magnitude parallel line of action and opposite sense. So, therefore, if we take this let us say one force is being applied at F other force is negative F applied at B then we can take the moment of this force about O then we can clearly see that ultimate outcome would be the magnitude of the M should be simply equals to F multiplied by the perpendicular distance between these two forces. So, F time D now that is absolute a free vector because we will always get that answer no matter which point I choose to take the moment. In other words the answer is simply F D which will make it a free vector the direction of this should be perpendicular to the plane where the F and negative F lies. So, we get the direction of the moment as well. So, two couples will have equal moments if and only if as we see here there are two couples F 1 and F 2 as long as the as I can show that F 1 D 1 equals to F 2 D 2 they should lie on the parallel planes and the two couples have the same sense of the tendency to cause rotation in the same direction that means they should try to rotate the body in the same direction. So, if these three conditions are made then we can say the couples are equal. So, addition of couples can be done in a same way as we do the vector addition. So, we can clearly see here that let us say I have a couple F 1. So, you can see F 1 and negative F 1 that is on a plane P 1 I have another couple F 2 and negative F 2 that is on plane P 2. So, ultimately this couple F 1 couple will produce a moment which is perpendicular to plane P 1 and the couple F 2 will produce the moment which is perpendicular to the plane P 2. So, we can vectorially add. So, one couple moment is M 1 the other couple moment is M 2. So, once we vectorially add we can get the resultant of these two which is M and it is nothing but the main point here is that how that is acting? We can clearly see if we do the resultant of F 1 and F 2 that gives me the resultant. So, resultant becomes also a couple and it has its own direction. So, that direction will be defined by the plane where the r and minus r lies. So, you can clearly see the direction will be perpendicular to the plane where r and negative r lies. So, couple is going to be an important aspect. So, we will just summarize that a couple can be represented by a vector with magnitude and direction equal to the moment of the couple. Couple vectors obey the law of addition of vectors and if they are free vectors the point of application is not significant. So, they are not fixed vector or bound vector. So, when we do the equilibrium we can take this couple moment anywhere and we can simply add it to the equilibrium equation and couple vectors may be resolved into component vector. So, what we have finally studied is that we have introduced the mechanics that was of the first part. In the second phase we have simply reviewed the vector mechanics and the vector mechanics how it is applied to get the resultant forces, resultant moment on a system we have looked at that. Mostly we have looked at scalar product, cross product and mixed triple product and we have seen that the cross product is very important when we are talking about the moment of a force about a point and mixed triple product is going to play a dominant role when we look at the moment of a force about an arbitrary axis. In addition to that we have also looked at couple moment. So, couple moment is a free vector that is what we have discussed. So, now we are going to close this session.