 對不對, as I said the fundamental concept, these are again all the prerequisites that we talked about. Now I am really going to go through the vectors and kind of go through the scalar product, dot product, mix-triadım product. These are the three roles that we are going to follow. and how this scalar product dot product and the mixed triple product that is going to be used to solve the engineering mechanics problem related to forces. So first of all as we all know that vectors can be you know taken there are two types of vector fixed vectors and the free vectors right. So fixed vectors have well defined points of application that cannot be changed without affecting an analysis whereas free vector that can be freely moved in the space anyone what is a free vector in engineering mechanics problem weekly very good so excellent so we all know it's a couple moment so couple is a free vector right okay so we want to make this interactive again alright so that just keep that in mind equal vectors and negative vectors are there now we are going to go through the addition of vectors now addition of vectors is parallel of law that we have already defined we also have the triangle rule that we have to keep in mind so ultimately you have P and Q right and the other side should be P plus Q right so R equals to P plus Q so laws of cosines come into play if we want to get the magnitude of the resultant force R then R square equals to P square plus Q square negative 2 P Q cosine of the angle between these two vectors okay so laws of science are also there then vector addition is commutative this is very important what is happening is that does not matter how I go here that is P plus Q should be equals to Q plus P if that is not there then I cannot say that's a vector okay now addition of vectors that means what we are going to see here that you can use that rules repeatedly to you know when we have more than two vectors in place so we can just get P plus Q plus S in that way so now as I said a quantity which has magnitude and direction but does not follow parallelogram law cannot be a vector anyone any quantity finite rotation okay someone said in the beginning that we can clearly tell this with examples very simple example that finite rotation cannot be a vector okay why just take an example of a book see first this is my x y and z so I have the book place now you have both magnitude and the direction that means angle itself is the magnitude and we also have a clockwise or anti-clockwise that is the direction so if I do this operation let's say I rotate the book in this fashion that is my first operation and then I will simply move do the operation number 2 right so I get to this position this is my P plus Q but the inverse will not be true that means if I first do this operation leave the book you know first I make the rotation like this and then I make the rotation like that then it is not going to actually be in the same configuration okay so finite rotation is something that is that cannot be terms at a so the application why we are talking about these vectors now so far as such we know that basically we have you know we have several concurrent forces so we have to really solve the problems that involves the concurrent forces where all the forces meet at a particular point okay so therefore when we are solving this type of problems when forces are meeting at a particular point which we call the concurrent forces we have to use this kind of laws okay again the rectangular components of a vector so that was given in terms of the trigonometry now we can also define the vector in the rectangular coordinates so we have basically the force f that is has you know fx and fy so these are just component wise display and we have to make sure there is a unique vector so the force vector f will be taken as fxi plus fyj so addition of vectors you know when we have more than one vector that means one force we have three forces here so three forces can be also added as per our rectangular components so basically we will have r equals to p plus q plus s and all of us know that there is going to be rx that means x component of the resultant force and the y component of the resultant force so this is going to be you know then they will make an angle theta which is tan inverse ry over rx okay so let's take an example as soon as we do this we do this exercise then we are actually going to go through the problems how to deal with the problem of concurrent forces okay just to apply this knowledge that students will develop through this so the first example where we have two forces these two bars are actually bolted with a let's say bracket or a gusset plate you see the similar example is actually for a truss in case of a truss member we will see that there are multiple members that are actually connected at a particular joint so that gives a quick you know feedback of what the concurrent forces would be and the two forces are given here let's say 20 kilo newton and 30 kilo newton we are interested in find out first using triconometric rule that means the triangle rule or parallelogram law using that can I find out the resultant force of these two forces that is acting in these two members okay so ultimately the solution key is what we look at so once we give the problem student has to do some kind of brainstorming and try to come up with what would be my expected answer that is the first thing so here we can actually guess the resultant force how it would actually look like so that would be the guess that we actually try to see whether students are capable of doing it or not so as such this problem is very simple problem to begin with so we just draw the use the trigonometry so I have one force here 20 kilo newton and the other force is 30 kilo newton and the resultant is going to look like this so if I if they draw it in a scale I can immediately figure out what is the resultant force and its direction so as such we are not really going to go through all the solution but main intension will be just the demonstrate demonstrate the problem and look at can we solve it in one go or not all the engineering mechanics problem will be like that only so ultimate answer would be as we see that this one students would be able to get a visualization okay my resultant force is going to look like this next now here we have more than two forces actually so we have three forces and it is an example of a polar now what it is in a simple sense we use the umbrella every day basically you have a hollow cylinder that is gliding against a rod or a bar that is a collar now remember in that collar all the spokes are coming and meeting at that collar okay so here the problem is defined in such a way that determine the value of angle for which the resultant of the three forces is horizontal so what condition is given that the resultant forces is going to be horizontal the corresponding magnitude of the resultant so therefore what would be my solution that student has to immediately tell that okay that means you are saying that r y equals to 0 that condition is going to give me the angle that is unknown that angle unknown angle will be and then what is the magnitude of the resultant that is going to be r equals to simply sum up all the forces along x direction right because already we said r y equals to 0 okay so if we look at the solution since r is horizontal therefore r y equals to 0 so we just take the component of all these forces along the y axis and set it equals to 0 so we get the solution for the alpha and then we look at the next part of it that is part b so what is the resultant force now so resultant force is of course going to be horizontal so the resultant force will be simply sum up all the forces along the x direction now we can also tell them about the rectangular component in the space but logic remains same as such so we do not have to go through here but what is most important that we tend to say is the unit vector lambda how they calculate that unit vector that is very very important that we should highlight again and again so that calculation actually plays a very important role when we solve this kind of problem in the space now the scalar product of two vectors as we know that if we have two vectors p and q then p dot q should be equals to p cosine theta remember that p dot q can be also written down in rectangular components and the components will simply look like this now what is the main purpose of doing the scalar product what we can do in terms of forces with the scalar product anyone what would be the application when it comes to the forces nice so the answer is done so it is actually projection of the force onto a line and that is what we use the scalar product okay so that is the first thing we have to keep in mind so it is used for two different purpose one is that we can definitely calculate the angle between the two forces if it is not right so if we have the force vector here and force vector here we should be able to get the angle between these two and the second one is the very important that what is the projection of this force onto this line lambda so that can only be done by lambda dot product with p so the projection of the force onto a line that is bringing in the essence of scalar product okay and that we will be doing many times in this course next is the vector product so as such that we have p and q two vectors so the cross product will be perpendicular to the plane of p and q and the magnitude of this cross product which is giving me v should be pq sin theta and make sure that direction of v should form a right hand rule right so we have to apply the right hand rule to get the direction of the force that is in the third direction then p and q so in this case what would be the quick application of this again when it comes to the forces where do I use p cross q exactly so the ultimately what we have is the calculation of the moment so that therefore let's say we have a rigid body and the force is applied at f which is located as at a distance way so that is my position vector r so r cross f should give the moment which is actually perpendicular to the plane defining r and f and make sure we are also using the right hand rule okay remember what is important here is to look at what is the magnitude of this moment that magnitude of the moment should be f times d where d is the shortest distance between the line force and the origin o right now once we do this we can see a large number of problems that deals with a moment and moment is probably the most important concept in engineering mechanics because what we find that student has real difficulty getting the moment probably you know when they do it r cross f they may get the numbers correctly but the visualization that is going to be really really difficult for them that which direction it is going to actually act okay so that type of things we have to kind of throw them so as such when we want to use the you know vector in vector mechanics so here we have moment is r cross f right and this can be to get the moment we can write it down in this fashion so it is going to be basically we are going to take at the determinant of this matrix and that is going to get in this form now we can also do component wise that means if we break fx fy and fz each of these forces which of these components is actually going to give me the moment about let say o that is the origin and remember for example if I want to calculate moment about the y axis then actually two forces will affect the moment about the y axis they are going to be fx the component fx will affect as well as the component fz okay so like that also we can tell students that in complex problem okay you may want to just use the r cross f directly but in simplified problem it is always better to look at component wise so instead of taking the moment at the origin let say I take the moment of this force which is applied at a with respect to point b so moment is taken about b therefore what is necessary at the first place to calculate the position vector from b to a so we will get the rba basically and we know the f so again we are going to use the r cross f and the ultimately it will be put together like this to find the moment about b so let us look at one more example so this is actually to do deal with a shower head and let say mechanic came and try to tighten the shower head with a wrench so once we see this problem so the force is applied look at the force how it is actually applied on the shower head it will definitely have three components right and these three components we can find out what are these components actually so just briefly explain this problem this force that is being applied at sea if you look at this shower head shower head sea here that is the range where the force is applied then this is actually 15 millimeter down from the z axis or in other words you have the x z plane it is actually 15 millimeter down from the x z plane okay it is also oriented with respect to the with respect to an axis which is parallel to the z axis so this is actually 12 degree right so first of all you know procedure is very straight forward question is here can we ask the students what kind of direction I am looking at for the ultimate moment vector the moment that is going to be produced at a the direction of that vector how it should look like just tell me component wise anyone you can tell me okay it is along the x axis or maybe negative x axis one component will be like this another component will be like that anyone weekly so just think of r cross f that plane just draw that plane so how that plane is going to look like we have r ac right that is my r okay and when we do r cross f the moment should be perpendicular to that plane that means the answer that I am actually looking at not about z s one component yes just go ahead anyone positive y that is good one component will be positive y component another component negative z okay that is that is what I was actually looking at see you are going to get ultimately two components those are going to be positive x and y but one component that is actually z component that is going to be negative right so we need to define this plane here that plane will look like something like this so this will be my plane and it is going to be kind of you know within the wall of that shower head where it is attached okay so ultimately you know that do this process is again simple but breaking of the component into forces would be very important so you can see that there are three components of forces the force component x how do I get this so you have basically first we bring it on to this this plane so I have cosine of 45 degree right then we move to a parallel to the x axis so that will be sine 12 degree so I first bring it to here and then move sine of 12 degree so that is my x component make sure that is negative similarly if I look at the z component let's say so z component will also be negative so ultimately it is going to point like this so I have x component y component and z component remember y component is relatively easy because y component is just f sine 45 degree which is parallel to the y axis and it is pointing downward so difficulty is getting the x component of the force and the z component of the force correctly now when we want to do it in a component wise let's say I want to see the moment about x axis what are the forces actually going to contribute to the moment about x axis then it's a bit difficult in this problem so maybe student will just prefer to do this one r cross f okay once they do r cross f but as I said that intuitively we have to make sure they get the final answer correctly and as some of you have said that I am expecting really negative answer for the z so z component of moment should be negative and x and y should be positive just look at the direction of it see ultimately if you really break it into components so you are coming to first of all f cosine 45 degree that brings you to this line and it is directed towards negative z axis not that axis actually there is a inclined axis here so ultimately now you are going to take this one into sine so ultimately you see this force is going to be like that with this tip should be heading towards the negative x so the arrow should be along the negative x axis so as I said moment is always that gives travel to students I mean this is what we have to know this is just a coordinator workshop I can't look at a number of problems but ultimately what will happen in the main workshop we are actually going to give multiple problems to the students just to get a feeling of how the moment should look like they can simply sketch the r cross f plane and show it by picture that ok this figure in a moment is going to be like this direction now mixed triple product so this involves three forces as if looks like so pq and s and the mixed triple product is going to be s dot p cross q right ultimate outcome is that I am going to get a scalar quantity and that scalar quantity represent the volume of parallel so because p cross q that will simply give me the area right the magnitude of it right and when you are doing s dot p cross q so s dot p cross q actually going to give the height here right so height multiplied by area so that is giving me the volume of the parallel pi now what is the application basic application of s dot p cross q in engineering mechanics anyone related to moment projection of moment on an axis so it is actually related to moment of a force about an arbitrary axis on the space so let us say that I have a force applied at a what is that moment of this force about an axis well let us say this line is passing through the origin o then simply we can say the moment is going to be this is again a scalar quantity moment about that axis o l is simply going to be lambda dot r cross f where lambda is the unit vector of that line now remember if the axis is not passing through o let us say you have some arbitrary axis but the unit vector is defined we can still calculate the moment of the force about any arbitrary axis so for example I have here this axis which is not passing through o but this axis let us say I have points here b and c so what is the moment of this force about this axis simply lambda dot mb now what is my mb mb is r cross f right so to get the mb first of all I get the moment about a point b right and then I take the projection of that moment on to that line so lambda dot r cross f where r is simply this vector ba now instead of taking ba can I actually take ca rca answer is yes so it does not matter it is independent where that point is it could be a it could be b I should get the same answer so this is another very important point okay and you can clearly see that if I want to prove that my answer would be same does not matter whether I choose a b and c right so I can simply say that lambda dot r cross f I had this now r can be broken that rba can be broken into two parts one is the bc and another is the ca right so now once you break it into two parts let's say rba and rbc one component will be simply zero because that cannot make the volume of the parallel pipe that means this rbc lambda dot rbc cross f will have no make sense that has to be zero okay the other component will give you the moment so ultimately we can show that mb should be equals to mc okay now we can just come to a quick example of this we have a frame acd this frame is actually hinged at a and d probably student at this time does not have any background that whether you know he or she understands the hinge or not does not matter 11 but this problem is stated like this that let's say I have a frame acd and it is supported by a cable that passes through a ring at b and is attached to hooks at g and h what is given it actually that the tension in the cable is 1125 Newton so determine the moment about the diagonal ad of the force exerted on the frame by portion bh of the cable so I am really interested in that what moment this force on this cable bh will produce with respect to this moment about ad so this is again a moment of a force about an axis and that axis is ad okay so I do have here tension in this cable is let's say bh so that I have to first show this is my force vector and I am really taking the moment of this force vector about the ad so first let us try to understand this way can we get the direction correctly what would be my direction of the moment remember I am going to get an scalar quantity right yes it is along no it is not along a b I am trying to get the moment about ad right and let's say I define my lambda as lambda ad let's say I define lambda as lambda ad that means tip of the arrow is the towards d what answer I am expecting positive or negative if I say my lambda is from a to d right then it is going to you just look at the hand so I have this force force distance so ultimately my thumb is going not from a to d but rather from d to a so ultimately I am expecting a negative answer okay so once we give this feedback to the students or let them first thing that what is the expected answer then others can be done actually others are just mathematical process as such so they just find out okay first I have to choose a point on this line ad let's say that point is a right they can also choose the point d does not matter define the lambda ad so what is the lambda ad then I define the position vector from a to b which is r a b right and once these are defined I also know what is tbh in vector form so basically I need lambda I need r and I need t then I can calculate the moment as lambda dot r cross f which is again a scalar quantity and the answer will be negative 180 Newton meter so basically that's how you know we tend to kind of impart this you know at the beginning of the problem that can you guess the answer just the direction direction absolutely plays a very very important I'll just take five minutes to finish it off so couple moment again this is a very simple logic and ideally we have two forces f and negative wave having the same magnitude parallel lines of action and opposite sense they should form a couple right so we can actually think of a steering very easy example okay so remember at the very beginning of the you know this session we said that this is going to be a free vector okay and basically f times d should be the moment that we are that I am looking at that's the magnitude of the moment and that is going to be perpendicular to the plane where the couple lies so I am going to get this moment vector which is actually perpendicular to this plane now without affecting the equilibrium this moment can be transferred anywhere when I am going to take the equilibrium right so it does not depend on the position of that moment so two couples will have equal moments if and only if these three criteria are satisfied that means let's say I am talking about a couple f1 here and f2 here these two can only be same if and only if f1 multiplied by d1 equals to f2 multiplied by d2 that means they are magnitude the moment magnitude has to be same two couples must lie in parallel planes right they have to be parallel to each other and two couples have same sense or tendency to cause the rotation in the same direction like moment we can add since it's a vector couple can also be added similar methodology will follow so ultimately we have let's say here actually two planes are given one is having the couple f1 you can see f1 and negative f1 that is on the plane e1 the other couple is on the plane e2 and in a similar way we can simply say that the resultant moment of these two couples should be just some of the individual couple moment which is m1 plus m2 so this m1 vector will act perpendicular to p1 m2 vector will act perpendicular to the p2 and we can simply add this to get the moment okay so in summary a couple can always be represented by a vector with magnitude and direction equal to the moment of the couple that is f multiplied by d couple vectors also obey the law of addition of vectors I can always add two couples and they are free vectors and their point of application is not significant and couple vectors can also be resolved in component wise like the moment so their performance is same as moment but we do not have a definite point of application of this moment that is coming now okay now we can just take a t break and we can start our next topic that will be on equivalent force couple system