 And this session and the next one would be on 3D equilibrium. In 2D equilibrium is already completed. So, from there what we have learnt that for a rigid body to be in equilibrium we need 3 equilibrium conditions that were sum of force along x equals to 0, sum of force along y equals to 0 and moment about a point was equals to 0. Now, these are 3 conditions of equilibrium for a rigid body then if you have more rigid bodies that mean interconnected rigid bodies then they will multiply. So, based on that you can actually solve for the support reactions. So, for a single rigid body in today equilibrium we discuss that we can solve 3 support reactions and if it is so then problem was statically determinate. Now, in case of 3D equilibrium remember that we are now going to get 6 scalar equations that will be required for the equilibrium and we can this you can use these 6 equilibrium conditions as you see here in scalar form that is 3 force equations and 3 moment equations they can be used to solve no more than 6 unknowns and these 6 unknowns generally represent reactions at supports or connections. But remember it is not mandatory that all the time I have to use 3 force equations and 3 moment equations. So, for example, you can actually augment one of the force equations by another moment equations. So, that is always possible now the main idea that would be actually if you think of it how do I get this scalar solutions you know a scalar rather equilibrium equations. So, we get that equilibrium equations based on the fact that I make the resultant force on the rigid body equals to 0 and the resultant moment at a particular point O right that is equals to 0. So, we have already discussed this. So, how to do R cross F? So, if you have various forces on the system. So, we have to do this vector operations. So, ultimately as you see that moment about a point ok. So, moment about any point O if you consider that equals to 0 then you are going to get 3 scalar equations and that are actually placed here. So, therefore, now what is important is that to understand that instead of doing this operation right here that is R cross F we can also do directly moment about an axis instead of moment about a point. I can also do moment about an axis by using lambda dot R cross F and we have done that in the very beginning that lambda dot R cross F that is the moment of a force about an axis and if you add the effect of all the forces that is also going to give you a equation and that equation we can use to also get the support reactions as possible ok. So, in 3D equilibrium as I said that it will be more relevant in most of the problems as we will see that attempt to use moment about an axis that can be done as you know that force times the perpendicular distance that way or if it is more complex you know system then you can also do lambda dot R cross F. So, ultimately as we have seen here that for a given rigid body I can solve 6 unknowns. Now, if you have more than 6 unknowns for a given rigid body then I would not be able to solve for all the unknowns it is not possible. But again remember therefore, what is happening is that if the problem remains statically indeterminate that is my number of unknowns are exceeding the number of equilibrium equations which are 6 in this case. Therefore, I may not be able to solve for all the unknowns but in problems in applications you will see that few unknowns can still be solved. So, we just have to keep this in mind. Now therefore, what remains for us to understand that how do we visualize the support reactions as well as the connections ok. So, that would be the basis of the discussion for a while now. So, likewise as we have done in 2D equilibrium as you have learnt that for a roller support I have one reaction that is vertical for a pin support I have two reactions that is horizontal and vertical. That means, again you have one forced unknown along with the line its line of action is also unknown. So, for a pin support you have actually 2 unknowns at support reactions and for a thick support I had 3 unknowns at support reactions that is horizontal reaction vertical reaction as well as a moment reactions. Now, how did you get to that? Remember the main idea was to understand that whenever I do not allow a movement I am going to get a reaction back. That means, how many movements are possible in a 3D body as we know that I have 3 translational motion and 3 rotational motion. So, therefore, now as we understand that therefore, whenever one of this motion is prevented that means, body is unable to do that motion I am bound to get a reaction back. So, if you just go by words this basic logic we will try to see what are the reactions that are going to come into the free body diagram when I isolate the body from the supports. So, here now you have basically we have a rigid body here and that is placed on a ball. So, what is happening now the body is unable to make a movement downward. So, we are going to get a reaction that reaction is now vertical and we also know the line of action. Therefore, my unknown is just one. So, I have just one unknown here that is the support reaction. So, this will happen on a frictionless surface or if a body is resting on a ball let us say. Similarly, let us say I have a rigid body which is connected by a cable. So, therefore, as we can see clearly that cable will induce a reaction back to the body and that line of action is known only thing that we know again unknown is the unknown is here really the magnitude of that reaction. Similarly, if you go to the more complex supports for example, let us look at this the rigid body is connected to a wheel and that wheel is on a rail. So, now what happens in this case we have to look at the three translational motion and the three rotational motion and we have to identify that what are the motions are not possible. If the motions are not possible then I am going to get support reaction. Therefore, what is happening as you see here that basically the rigid body cannot move downward. Therefore, I have a reaction that is upward. So, f y is one reaction. Similarly, since it is on a wheel that the rail is there therefore, wheel is unable to move also horizontally this way and therefore, I am going to get a reaction. Remember the other motion this rigid body can do all other motion. For example, it can rotate about y axis. Similarly, if you consider this is x axis which is along the rail it can rotate very slightly about that axis as well. Similarly, since there is a gap possible it can also rotate a little bit about a y axis. Therefore, I am not going to get moment as reactions in this case as well as remember there is no reaction coming from the x axis as well because wheel can move therefore, body is able to move in this direction. Next would be the ball and socket joint. Now, in a ball and socket joint that is very much used as you can see very perfect example would be basically our you know famous system if you look at the bones our hip joint for example. Now, as you can see here that one you know this socket is actually you know two bones are connected by the ball and socket. So, it can actually rotate about any axis, but what it cannot do it cannot actually allow the translational motion. Therefore, what is happening the translational motion is prevented in the three directions and if it is so therefore, we are going to get again support reactions in the three directions translational directions. So, therefore, we have three support reactions as unknowns. So, if you go to more a complicated system I will just talk about the fixed support first because in 2D you also have a fixed support and for 2D we have already seen that for a 2D body x y and z if you choose then along the x the body cannot move along the y it cannot move and it cannot rotate about the y axis let us say therefore, you have moment about it cannot rotate about the z axis I am sorry. So, therefore, you have moment about the z axis. So, you have basically f x f y and m z was the three support reactions. So, in the fixed case in the 3D equilibrium case now you can think of what is happening basically motion is prevented in all you know possible motions are prevented. So, all translational motion as well as rotational motions are prevented therefore, I have six reactions I have three translational reactions and I have three moment as reactions ok. So, three force reactions and three moment reactions. So, now if you look at some other joints that are very much used this is called a universal joint. Let us look at the mechanism what we have is basically see you have basically you know kind of a cross section here. So, you see here that these two are perfectly welded here you see that these are two perfectly welded at the joint ok, but this pin right here this member right here is connected by a pin to the rigid body. So, these two ends here they have screws you can think of they have just screws ok. Similarly, these two ends of this member has screws here ok. So, what will happen eventually therefore, if you look at the x and y this body the body can rotate about this axis ok. So, if you look at this it will be able to rotate about this axis this body and that body as well. So, if they are rotating about this axis then there will be no moment. So, this axis is actually z axis. So, there is no moment about this axis. So, there is no reaction similarly, these two bodies can also rotate about this axis right here which is our y axis ok. So, if it can able to rotate about the y axis then again I do not have a moment as a reaction, but what this bodies cannot do they cannot rotate about their own axis. Own axis means that is x axis ok. So, if they cannot do that therefore, I have moment as a reaction. So, again I will repeat that this is a interconnected mechanism and these two small members. So, this joint basically has two small members perfectly welded and they are basically screwed at their ends with the connected bodies that is the basic concept and therefore, what is happening now you can see none of the rigid bodies will make any translational motion. So, translational motion will be prevented in all directions therefore, we are going to have three reactions, four reactions, but they will be able to rotate about the y axis they will be able to rotate about y axis. So, I do not have moment as a reactions. However, these two bodies cannot rotate about their own axis. So, therefore, I am going to get a moment reaction clear. So, next look at some of the problems you know involving bearings. Now, bearings are typically used in lot of mechanical problems as you have seen journal bearings, thrust bearings and so on and so forth. What happens in that case? So, remember this is a bearing. So, actual photograph is given that is your bearing and an axial shaft will go into this bearing. So, we have this axial shaft. So, our intention would be to understand what is the free body of this axle or shaft that is on the bearing when I isolate this from the bearing. Therefore, I have to understand that what are the support reactions coming from the bearing to the body that body is the shaft or the axle. Now, look at carefully what can happen? What is being said here is that hinge and bearing supporting radial load only. So, that means the type of load that I am going to see is only the moment about its own axis. So, it is carrying the radial load. So, whatever load I apply they are radial in nature. So, in that way remember what will happen in the process you may have a vertical load you may attach one perpendicular bar put some loading here and here. So, in the process what is happening as we can see what the body cannot do basically it is not allowed to you know move up like this it is not allowed to move out of the bearing like that right. So, y and z motion is prevented translational therefore, I have two support reactions here. The body can however since it is not carrying any axial load that means I do not need the axial you know reaction as well. So, that axial reaction is also missing. Remember what it can do it is actually can rotate about its own axis that is the whole idea. So, if it can rotate about its own axis that means that rotation is allowed therefore, I should not have the moment as a reaction. So, what I do not have basically in this problem mostly I do not have the moment as a reaction about its own axis of this body of this shaft or of this you know axle. We have all other things present that means I have moment about this axis I have moment about this axis as well as reactions are present. Only thing I do not have a axial reaction the reason being it is not taking any axial load also. So, I am not putting any load like this or maybe any inclined load on this body which is my shaft or a axle. Now, just take this and try to just understand what happens if I now apply an inclined load or maybe just if I apply an axial load then in order to prevent the motion in the axial direction all I have to do I have to somehow manage to block the motion you know in the axial direction therefore, thrust bearings are used. So, in the thrust bearing you will also see that I will have a reaction along the axial direction. So, this is a force reaction along the axial direction very common example we see day to day life is the hinge on a door or on a window. We see this right on a door I can only rotate the door about this hinge axis the door is unable to make any other motion about any other axis. So, only thing that is allowed to do is rotate about its own axis which is the x axis in this case therefore, as we see that I have no moment as a reaction. So, typically speaking when the problem comes most often we are going to do we are going to see this and this one this will be typical problems as well as this one fixed support and if I go back a little bit we are also going to see lot of ball and socket joint as well as this is very common. So, ultimately problem will involve when we discuss to the you know come to the discussions of different problems we are going to see more often this one ball and socket joint or may be a roller and so on. So, before I proceed to you know different problems now you can only understand this by you know taking many problems and try to see how we can you know do the simplification. So, basically as I said the reaction is one part that we first understand then what we really have to do we have to think that how to get this unknowns to solve for the unknowns we will always try to attempt to do the moment about an axis. We will try to avoid as much as we can doing the r cross f that means, moment about a point O and create three equations out of that. Typically speaking if we keep doing the moment about an axis we will be able to solve some unknowns very very easily and remember moment about an axis can be solved by force times perpendicular distance from that axis. So, we will can keep maybe let us say 5 minutes just for the sake of discussions if there is any you know doubts regarding the support reactions that we have discussed so far. Sir why there is no moment in x direction about which one which one which reaction are you talking about. Sir last one last one that door the example of a door that is hinge and bearing support yes. Radial load ok. So, think of the so the question is in case of a door the hinge look at the you know hinge properly. So, there we can clearly see that what see to understand the support reaction important thing to understand that what are the degrees of freedom is actually doing the motion right. I have 3 degrees 6 degrees of freedom 3 translational and 3 rotational. Now, can you see clearly that which degrees of freedom is actually doing the you know some motion is there. The only motion that is allowed that is about the x axis that is the rotation about the own axis of the hinge because when you pull the door right you are able to make the movement only about its own axis right. So, I have a rotational motion that is about its own axis therefore, we are not going to have any support reaction there we are not going to get any movement reaction there. So, wherever whenever I have a motion allowed I should not have a support reaction. If the support is allowing the motion then I should not have a support reaction is that clear ok. I have another question popped up again I will just explain the wheel on a rail. So, there was a question on wheel on a rail just go back there. So, wheel on a rail you see here clearly see what it can do now when you place the wheel on this rail ok. You can think of this wheel can little bit tilt about the x axis why there is no moment as a reaction because wheel can little bit tilt about x axis or on the same logic it can tilt about any other axis as well that means x axis y axis as well as z axis. So, it can little bit of rotate. So, that rotation the little movement is allowed. So, it is therefore, not taking any moment as reactions ok. So, only thing that it cannot do it cannot slide from the rail the wheel cannot slide from the rail and therefore, I have a force as a reaction ok that will prevent the you know sliding of the rail from the wheel from the rail as well as we cannot do any vertical movement because it has to be properly grounded on the rail. So, therefore, I am going to get this support reaction it can of course make a translational motion along the x axis right therefore, I do not have any reaction over there ok clear now. So, now what I will do you know I will just try to go to the problem. So, I am going to take different problems and try to look at in a very simplified way that how we solve this problems ok. So, the problem involves here is that of a window and we can see here that window is hinged at a and b ok. So, that means, the only motion that can it can do if I attempt to you know if I have to really look at what are the support reactions here you can clearly see that if I choose my z axis right here if this is my z axis let us say. So, the you know movement is allowed that mean rotation is allowed about the z axis right because if I try to close the window or if I try to open the window it should rotate about the z axis. So, therefore, what I do not have I just do not have the moment reaction about the z axis ok I do not have moment reaction about the z axis. So, this is a hinge problem again just keep in mind now what is being asked here that there was a cranking type opening mechanism that somehow failed. Now, what we have done we have basically provided a wooden prop. So, CD is a wooden prop right. So, to weld the you know this one in position in certain configuration I just put a prop that prop is CD. Remember the window itself has a weight right that weight is given. So, it has its own mass center which is at the middle and that weight is given by 50 kg or 500 Newton ok. Now, we are supposed to find out the compressive force CD again why it should be compressive just get a physical feeling that you know it is trying to push this you know prop due to its weight ok. So, therefore, this is going to be compressive on the prop CD or the wooden bar CD. Now, what I have to find out I have to find out for this compressive force. So, what would be the solution to this problem how do I attack this problem if I look at it carefully I have hinge here I have hinge here 3D you know problem. So, total reactions as you see here at A and B how many reactions are there total for each one of this as we said there will be 5 reactions. So, 5, 5, 10 reactions are there ok that means, 10 unknowns plus I have one more unknown coming here that is 11 unknowns are there. Now, I would not be able to solve for all of these unknowns because if I isolate this rigid body that is my window if I really isolate this then I am going to get only 6 equilibrium equations now, the problem demands that I have to only solve for CD. So, I am not interested in finding out all other reactions also. So, therefore, the problem remains statically indeterminate. However, I can still solve for this force the question is how I am going to solve for this force remember if I take the moment about the A B axis. Since, there is no moment reaction about A B right and all other reactions are passing through you know will not come into play to the when I am going to take the moment of this force and the weight about the A B axis no other reactions are going to occupy the equilibrium equation no other you know reactions are actually going to come into the equilibrium equations. The only forces that I am going to see in the equilibrium equations would be the FCD right and as well as the weight. So, these are the two forces FCD as well as the weight they are going to contribute to the moment equation. So, that is quite interesting. So, what now if you really try to do this problem let us say r cross F you choose at about point A let us say you take moment about point A it is not going to go anywhere because if you take moment about point A then other forces from the other reactions are going to you know come to your momentum equation because you have to do r cross F everywhere. So, the point here is then instead of doing r cross F I am really going to lambda dot r cross F. Now, even that lambda dot r cross F operation I may not need to do the if I try to find out what is the perpendicular distance of different forces from the line A B. So, if I am taking moment about A B all I would be interested to find out hey what is the perpendicular distance of this weight from let us say this line A B. Now, perpendicular distance means how I am going to do. So, really I have to look at this plane. So, from plane I have also force here what is the distance remember this weight will have infinite line of action and therefore, I just have to get the shortest path what is possible shortest distance between the line of action of the load and the line A B. So, now there is another trick here the trick is as I said just now I said that a force will have a infinite line of action. Now, when I isolate this body let us try to do the free body diagram I isolate this body. So, if we look at the free body diagram by isolating the body remember I have 5 reactions here it is shown here 5 reactions here. So, here A x A y A z moment about x and moment about y I do not have moment about z that is the most important thing. Once I isolate this body and I look at f c d now what I can do here this f c d instead of decomposing f c d at d I can decompose f c d also at c. So, therefore, you can clearly see now f c d will have component in this case right in all directions. So, what is going to happen you are going to say f c d x f c d y f c d z. So, therefore, when you take moment about line A B your both f c d x that means force along the x direction as well as force along the y direction both are going to contribute to the moment. So, if you bypass that and how do I bypass that if I resolve it at c because force at line infinite line of action if I resolve at c then you see if I decompose here only the x component of the force c d is going to contribute to the moment about A B. The other force that means the y component is already passing through A B. So, that will not contribute to the moment. Now, only nitty-gritty about the 3 d equilibrium is that you have no choice you have to find out the position vector you have to decompose the force into several components. So, those we have already studied in the vector mechanics when we discussed in the first session. So, therefore, as we can see you do have to find out what is my f c d that is a vector. So, that will have you know we have to find those and to do that I have to first find out what is the unit vector along c d. Once I know the unit vector along c d I just multiply that by the magnitude of the force f c d to get the force as a vector. And if you clearly see here that what will happen that force c d again we will have 3 components remember it is going to have 3 components these 3 components are given here. So, this is along x this is along y this is along z. Now I am not really going to talk about here numbers how you are getting those numbers those can be given to the students you know as an just you know in that just to solve by themselves. So, here we are only going to be talking about the concept. So, as such we have found the f c d. Now only thing that I said is that now look at carefully if I resolve the f c d right at point d then what are the components will participate in the moment about a b. I have f c d x multiplied by the perpendicular distance that perpendicular distance will be 0.8 cosine 50 degree right that is the moment about line a b is not that and for the f c d y if you just you know look at the infinite line of action what is the contribution f c d y multiplied by 0.8 sine 50. So, if I decompose here two you know part are going to contribute to the moment. So, if you try to avoid that calculation I can decompose right here I can still choose f c d x. So, ultimate solution would be I have f c d x multiplied by 0.8. See in this case it was f c d x multiplied by 0.8 cosine 50, but when you decompose at c then I have only f c d x multiplied by 0.8. So, y component of the force is passing through a b line a b therefore it will not contribute to the moment. What else do I have? I have also weight the weight of the body which is 50 kg. So, how does weight contributes to the moment about line a b? It is simply going to be as we can see from the distances that 0.4 right get the perpendicular distance. So, that will be we have to figure it out. So, that is 0.4 sine of 50 degree. So, this is the moment of the weight about the a b. So, therefore, what I am basically doing I am neither taking help of r cross f neither I am doing a operation lambda dot r cross f. All I am saying that there are problems where you can simply get the moment by considering force multiplied by the perpendicular distance from the line. So, if we look at now the solution. So, ultimately I have explained all of this, but just spend a few minutes here. So, what I basically said that we take the moment of the forces about the line a b that line is passing through the z. So, we have chosen the z axis here does not matter. So, we are taking the moment about this line and as we can see here clearly. So, the first part is the weight. So, we have the weight here I have taken 50 multiplied by 10 let us say 9.81. So, 50 multiplied by that is the force here right and we have said already what is the perpendicular distance it is simply going to be 0.4 sin 50. Now, if I choose this one this one will be negative if I assume that counter clockwise moment is positive anti clockwise moment is positive. Then we can simply say this is negative then the other one if you think of it right. Then this one is going to give f c d x multiplied by 0.8. So, this is f c d x this component right here 0.828 f c d is f c d x x component of f c d multiplied by 0.8 that is equals to 0. So, I get the solution very very easily. So, neither I have taken help from r cross f neither lambda dot r cross f we have also made use of a very you know special thing that we tend to do that force has an infinite line of action and I can decompose it anywhere when I take the moment and I have taken the moment by decomposing the force at c not at d. So, this we have to keep in mind when we try to solve you know 3D problems. So, only thing is that yes we agree that we do have to get the you know position vectors we do have to get the force vector that thing cannot be avoided and those are coming mostly based on the geometry of the problem clear. So, we will move to the next problem as we see in day to day life. So, the next problem is that of a rectangular sign board. So, you see this sign board very often. Now, just understand that how this sign board is supported what are the different supports I have we can clearly see it is connected by 2 cables on the wall. So, there is a cable that will give reactions force reaction right. So, I will have tension in the cables what is also said that the support against the wall at point c that is a ball and socket joint. So, at point c I have ball and socket joint and at point d right here only support is provided along the y direction that means motion is not allowed along the y direction. So, therefore, there is a reaction along the y direction. So, how many unknowns do I have in this problem I have exactly 6 unknowns because this is my one unknown second unknown 3 unknowns here because ball and socket joint how many unknowns I have 3 translational reaction force reactions right. So, c x c y and c z and here I have another unknown that unknown is simply d y. So, total 6 unknowns are there and I have one single rigid body that means I have 6 equilibrium equations and therefore, I can I should be able to solve it all the reactions can be solved. So, it is a statically determinate problem. Now, here is the you know main point as I said now if you want to make the thing more of a vector mechanics approach you can simply take you know moment about any point any point moment about any point you can take up all of this forces that means you do r cross f for all the forces equate it to 0 right. Therefore, you are going to get a vector equation which will have 3 you know i j k. So, you can have 3 scalar equations out of that that is the moment right. Similarly, you could simply do sum up force along x 0 y 0 and z 0 you again have 3 equations. Now, I am not going to do either of this the reason being I want to you know visually inspect it and try to see if I can find out unknowns one at a time. How do I find out unknowns one at a time? See observe one thing let us say I talk about x axis I can clearly see there are so many reactions passing through x axis. So, many forces t 1 t 2 both pass through x axis similarly y pass through x axis right that force right here will pass through the x axis. So, what can happen similarly you see that c z right and c x c x is parallel to x axis and c y is also passing through x axis. So, therefore, c z and c x will not participate in the moment about x axis. So, only thing will happen is that you have c y right if you take the moment about x axis will be surprised that we are only going to get c y multiplied by 1 equals to 0. So, that means what I am saying here that if you do a little bit calculations in this case if you start looking at that first of all I am not going through this process because I said that we are can bypass these operations you have to find out what is t 1 and what is t 2 right that will come from the geometry of the problem. So, t 1 is the force vector we have found in terms of magnitude of t 1 t 2 force vector we have found in terms of magnitude of t 2. So, these studies we have to anyway do only thing let at look at the moment equations how I am doing it basically taking the moment about x axis as I said t 1 t 2 pass through x axis you have this r right that is the only reaction here and now c will have 3 components remember c x c y and c z, but c x is parallel to x axis therefore, it cannot contribute moment about x axis c z is passing through x axis it will not contribute either. So, therefore, what will happen that only c y will contribute and if I say c y multiplied by 1 that should be equals to simply 0 right because weight is also passing through x axis. So, c y becomes 0. So, one unknown is already solved by doing just visually inspect the problem what else I can do think of it I can solve for c y is already solved right. Now c x let us look at can I solve it by taking moment about 1 axis we have to interrogate we have to introspect that can I find any of the reaction by taking moment about any axis. If I just choose c x right and attempt to take the moment about a b. So, what is happening in this case moment of a b will only be contributed by c x and the w, w will give a moment you can see w multiplied by 2 ok that is the perpendicular distance right you have this plane here and you can see the perpendicular distance if you project this infinite line of action the perpendicular distance is this right. So, we have w multiplied by 2 that should be equals to c x multiplied by 3.5 all other forces will not participate in the moment about a b axis is that clear. So, therefore, I can easily solve for c x. So, now c y and c x is solved at least I was able to solve for 2 unknowns by taking moment about different axis. Let us move on to the what else I can do can you tell me quickly I can get a very nice relation about t 1 and t 2 how do I get a relationship between t 1 and t 2. I can simply take moment about the z axis remember no other forces will contribute to the moment about z axis except for t 1 and t 2. So, therefore, if I do that the first what I have done take the moment about the z axis right. So, what are the components are participating of t 1 and t 2 basically we can see the x component will pass through the z axis. So, that cannot participate. So, only component that will participate as we all see here that the component of t 2 along the y axis right that will participate in the moment about z axis. Similarly, the component of t 1 about the you know y axis along the y axis. So, these are the 2 forces that is the y component of t 1 and y component of t 2 that are going to participate when I take the moment about z axis. So, we have clearly identified that and what we can see that we are going to get a relationship alright that relationship will come out nicely. So, here there is a 0. So, just make a note of this this is 0 right here. So, we have taken the moment of the y component of the t 2 and t 1 about the z axis. So, I get a relationship with the t 1 and t 2. Now, what else I could do I could now remember c x is already solved right I have solved c x from the previous slide c x was how much 0.561 c y was 0. Now, if I do force along x direction what are the components going to come in the force along x direction. Basically the x component of t 1 x component of t 2 and c x. So, the force along x if I equate to 0 then I am again going to get a relationship between t 1 and t 2. Therefore, what happens in the process I have now 2 equations to solve for t 1 and t 2. So, that is how it is solved. So, so far we have more or less solved all of it except for I think r is not solved. So, we will try to solve that as well. So, if I just take now some of force along y direction equals to 0 I have to now do that because r is along the y axis right. So, if I solve the force along y axis since all other unknowns are already solved then I should be get the value of r. So, therefore, r is now solved remember in the process I have actually consumed 6 equilibrium equations also. If you look at it carefully I actually indeed used 6 equilibrium equations but they are absolutely in different way. I have used 2 force equations sum of force along x equals to 0 sum of force along y equals to 0. I have also used 4 moment equations as we can see this is one moment equation this is a one this is another one right. So, the point is that that I have used 6 equilibrium equations indeed to get the all unknown forces that are all unknown reactions that are possible. Now, we are going to move to the next problem just take the problems quickly. Now, we have got the idea that taking the moment about various axis can actually resolve for different type you know different unknowns. So, now, let us talk about more complex problem let us say I have multiple rigid bodies connected. So, in this problem it is almost like a scissor if you think of scissor. So, now, there you have a hinge here. So, the bar A B and bar O D they are connected by a pin at C. So, therefore, now you think of it that I have one cable also being supported there are ball and socket joints here I apply a load P here. So, a load is being applied right here I have to find out what are the components of reactions at O and A as well as I am interested in finding out what is the tension in this cable. So, now, if you look at this problem carefully I have two rigid bodies if I look at the externally what are the support reactions ball and socket joints as three support reactions. So, I have three support reactions here three support reactions here total six what are the other unknowns I have tension in the cable as unknown right. Then you see if I take a global equilibrium there are seven unknowns I cannot solve it. Therefore, what needs to be done in this problem if we think of it carefully I need to detach I have to detach O D and A B and if I try to do that then I will simply expose the reactions at this connection. How many reactions are present in this connections there will be five reactions why it is so because what this body two bodies can do they can only rotate about an axis which is parallel to y axis passing through C that means we call it C y. So, C y axis it can only rotate about this point it is possible the rotation is possible. Therefore, I do not have any moment reaction, but I have all other reaction. So, I do have total of five reactions here. So, therefore, I have three reaction here three reactions here six plus five eleven plus one twelve reactions. So, twelve unknowns sorry. So, we have twelve unknowns in this problem. Now, look at it the total number of unknowns are twelve how many rigid bodies I have I have two rigid bodies. Therefore, I should have total number of equilibrium equation should be twelve problem is perfectly statically determinant. Therefore, I should be able to solve for all reactions are all unknowns that means I should be able to solve for the ball and socket reaction here here cable reaction as well as all the reactions here, but in the question it is just asked to calculate the reaction at A reaction at O and the cable force. So, now we think how I am going to start solving it. So, first let us try to look at the free body diagram. So, that is a free body diagram and remember I am always going to get the answer in terms of the known force P. Now, this is a global equilibrium where I have not detached any bodies. So, I have just looked at the free body of the entire system. If I look at the free body of the entire system I have three reactions here three reactions here and one tension here. Now, I would not be able to as I said I would not be able to solve for all the unknowns all seven unknowns, but if I try to you know use the equilibrium equations in terms of by two equations taking the moment about some axis then I should be able to solve for some of the unknowns. So, slowly what we are going to see that let us say I start with it the way to write it that sum of moment of all the forces about an axis x which is passing through O. So, O x is axis this is an axis which is passing through O and it is parallel to x axis. Now, in this case O itself is a origin, but we will always prefer to write like this O x that means it is indicating an axis which is parallel to the x axis passing through O. Now, if I do that what happens if I take the moment about this axis I can clearly see I will be immediately able to solve for A y. So, what is happening A y multiplied by the distance that is A times equals to P multiplied by A. So, you can clearly see A y is instantly solved similarly how I do the other let us take moment about another axis which is the z axis. So, I write O z that is z axis passing through O if it is. So, then what I am going to solve for you see these two are passing through already z axis these do not contribute none of this contribute none of this will contribute only contribution that comes from P and which component of t will contribute the y component because x component of t will pass through that z axis. So, therefore, what I have done here the t by root 2 that is the component along the y axis multiplied by A as you see that is the moment about z axis equals to P times A. So, this is solved similarly we keep going like this just keep using sum of force along y equals to 0 let us do that. Now, already I have solved for A y right. So, therefore, I should be able to solve for O y. So, through this process if I do sum of forces along the y axis equals to 0 since A y is already known right and I know P and t is going to contribute. So, here I am simply trying to solve for O y. So, O y is suddenly solved what else I can do I keep going like this how about moment about y axis also I have already taken x and z. So, why not trying the moment about y axis also if I do that then you can clearly see that A x equals to 0. So, A x will be 0 if I do the moment about y axis which is passing through O. So, similarly if I now do once A x is known let us say I do sum of force along x equals to 0 what would be the outcome the outcome would be since A x is already solved I will simply be able to solve for O x. So, O x is also solved. Now, I can still use one more equilibrium equation and you can try anything, but ultimately you will see that you are going to you are not going to solve for any other forces. So, what I have to do now now I am going to detach the body. See in the previous thing what will happen if you attempt to take any other moment equation or force equation you can use 6, but if you try to do that you will get some relationship, but nothing can be solved. So, now what I am trying to say now you have no other choice, but let us try to detach one of the bodies. So, once I detach one of the body I can clearly see that what are the different reactions in this case coming into play I have O x, O y and O z C x, C y and C z, but remember there is a moment missing in this case there is no moment about this axis which is parallel to y axis. So, C y, C y does not have any moment reaction, but we have moment reaction about the C x and C z. So, that is just a missing part please do a correction here. Then you can clearly see what I will try to do since there is no moment reaction about this axis I do not have any moment reaction I will simply take the moment about this axis. So, what will happen if I take the moment about this axis if I try to take the moment about this axis I am simply going to get a relationship between O x and O z because O x multiplied by you have to find that perpendicular distance. So, O x you know we take the component of O x so that we make a parallel to the z axis. So, O z is there so in other words if you do the moment. So, you get O x equals to O z basically and remember O x was already solved O x was already solved since O x was solved. So, I can easily get O z equals to P. So, therefore, we are able to solve for all of the unknowns so far except for there was another problem there see we still did not know I think one of the unknowns were missing that was a z I was unable to solve. See from the previous study a z I was unable to solve ok. So, a z was not solvable. So, now here we can also try to see if I can get the a z by some other means. So, the idea is set sum of force along x equals to 0 right that is from the this case. So, what we are going to get actually that in this free body a z is going to be if I do this there has to be a relations from the previous slide. So, just need the O z. So, from the previous slide actually if we solve for O z and O x is solved. So, if I go back here right here I can get a z and O z you see a z plus O z is equals to 0 right. So, from here a z and O z equals to 0. So, that equation should have come in this slide itself. So, a z plus O z equals to 0 and therefore, what we have already solved here O z is already solved. So, therefore, I can solve for a z clear. So, these equations should have gone in the previous slide by saying a z plus O z equals to 0 ok. So, a z plus O z equals to 0 now only thing I solved in the previous step is the O z equals to p. Therefore, a z equals to negative p ok. So, now more or less I have solved for all of the forces whatever in question now all of these are actually solved. So, this is the final solution.