 So, we are now going to start actually 3D equilibrium. So, once we have covered the 2D equilibrium, then it will be easy to understand the 3D equilibrium. The only point is that we have to keep in mind is the supports and connections. So, those you know how to realistically model the supports and connections, what are the forces and moments are coming as a reactions from the supports. So, that is what we are going to discuss briefly. And then we are simply going to solve again many problems because as such you know 2D and 3D equilibrium conceptually it is very simple as you have seen, but we have to solve a large number of problems to give a flavor to the students of different nitty-gritty. So, the 3D equilibrium as we know that now we are going to have the 6 scalar equation instead of 3 scalar equations that means we said that resultant force equals to 0 and moment resultant is 0. So, each one will produce 3 equilibrium conditions right. So, total we are going to have 6 equilibrium conditions to explore and for a given rigid body therefore, we can actually solve for 6 unknown constants that are coming from as a support reactions let us say. So, we can solve for this 6 reactions support reactions and we can say that problem is statically determinate as long as we have exactly 6 unknowns. So, 6 support reactions. Now, it is important to understand that there could be problem where we actually have more than 6 unknowns as support reactions, but it is still ok it is not necessary that we have to always solve a statically determinate problem. The problem can be statically indeterminate, but some of the desired forces can be solved ok. So, let us just quickly go through various types of supports and reactions that we see. First of all the most simplified version would be just a ball and on top of that ball we have this member or let us say a member is just resting on a frictionless surface. So, when we are looking at the equilibrium of this body we are trying to look at the free body diagram rather then we are actually going to say that there is one support reaction is coming into play. So, the concept of support reaction will be very similar to that of the 2D equilibrium. In one sentence if I have to say how do I determine the support reaction? See in 3D equilibrium we have 6 degrees of freedom right, 3 translational component, 3 rotational component. That means if a degrees of freedom is prevented in that degrees of freedom therefore, we are going to get the reaction ok. So, that is in one sentence we can just state like that if a degrees of freedom is prevented there we are going to get the reaction back ok. So, if we look at this way then all the you know support reactions can be very easily understood. So, in this case we can clearly say see that the support you know the movement of the body is actually restricted in this direction right and therefore, we are going to get this support reaction just one support reaction. Similarly, for a cable which is attached to the body if we want to draw the free body diagram again the motion is only prevented along the direction of the cable and therefore, we are going to get this as a reaction on to the body. So, let us move on to the next one roller on a rough surface. So, if the surface is rough that means we are actually going to have friction rather. So, we can say the roller is free to rotate that is not an issue, but we are going to get a friction force which is perpendicular to the you know on the horizontal direction in the plane we are going to get a reaction this way right because there is a friction force that is coming into play and that is actually going to prevent the motion on that direction. So, that degrees of freedom will be also prevented. So, we are actually going to see two reactions here and both reactions are actually force reactions right. Remember, the rotation is allowed in all three directions right also the roller can actually translate in this direction which we can say x axis. So, therefore, there is no force as a reaction there are no moments as a reaction ok. So, let us go to one more step ahead now we are talking about a ball and socket joint. So, what is this basically we have the you know member which has a ball head and that is put in a cup ok. So, that is the representation of a ball and socket joint. So, ultimately the translational motion is prevented for this body in all three directions. Therefore, I am going to get the support reactions in all these three translational directions. So, I will have three translational forces three concentrated forces as the reaction. However, the motion when we look at the rotational motion that is allowed in every direction that means about any axis it can actually rotate. So, I should not have any kind of moment reaction in this case. So, the best example is actually our hip joint this is just a representation of a ball and socket joint. The other one that I have presented here is that of a solar roof panel on a frame structure you can see that ball is going into this cup. Now, let us look at the other you know type of connections that are used. For example, we have the universal joint. Now, universal joint is you know used in many cases specially when we see the you know gears rotational gears and so on so forth. So, in this case actually what it is that let us say we have two small members that are welded together. So, you have these cross bars, but they are actually hinged at the two ends ok. So, they are actually hinged at the two ends. Therefore, what is happening now the body can actually the motion that will be the rotational motion that is prevented is actually about its own axis. When we are looking at the free body of the member itself. So, what the member cannot do is basically go like this about its own axis. So, therefore, I should have a movement as a reaction which is about its own axis ok. So, that movement is coming into play, but remember what can happen since here and here we have the hinge or pin connection. So, the rotation will be allowed about the z axis as well as rotation will be allowed about the y axis. So, we are not going to have any moment reaction in these cases ok. So, ultimately we are going to have four reactions in this case three concentrated forces and one moment reaction. Now, hinge and bearing in case of a bearing see bearings are usually used for you know where we have to have the radial load right. So, in case of a radial load we actually use these bearings ok. Now, in this case if this is so, then therefore, the body is only about how to rotate about again its own axis. Therefore, I should not have any moment reaction, but body should not rotate about the y axis and about the z axis. Therefore, we should have now two moments in this case as a reaction ok. And also note that in this case the body can actually also translate along its own axis. So, this is not a thrust bearing if it is a thrust bearing then what happens you can actually apply an axial load also and to prevent you know the motion on the axial direction I must have a support reaction back. So, that is actually represented in the next case. So, there are two typical cases one is that without a thrust bearing and another one is with a thrust bearing ok. So, in one case I do not have the axial load in its direction in its axis in the other case I have the axial you know reaction ok. And typical example of a hinge very good examples we see there today life is that of a door let us say. So, what the door can do actually it can rotate about the axis of the hinge. So, that is the only motion that is allowed. Therefore, I should not have moment reaction about its own axis, but all other reactions should be present in the free body ok. So, hinge connections is a typically you know given a lot when we are trying to look at various types of problems also the bearing will come into place and the ball socket joint will come into place more often ok. So, this with this we can directly you know try to solve the problems. So, I will just take an example here. So, let us say this window is temporarily held open in this position by a wooden prop. So, let us say you know there is some problem with the cranking mechanism and I just want to keep the window in some position. So, I use a wooden prop to support that window ok. And remember that we also have two hinges here at A and B ok it has its own weight. So, the idea is here that we want to calculate what is the compressive force in that wooden prop. Problem is simple, but there are special tricks that we are going to use. So, remember in 3D equilibrium we are never you know we do not really recommend the students to really work on lambda dot r cross sorry the you know take a moment about a point. We really try to work on lambda dot r cross f that is moment of a force about an axis. So, every time we try to take the moment about an axis to solve the problem ok. So, in this case therefore, what would be my solution key? First of all let us say that how many unknowns do I have in this problem? If I have hinge here and hinge here. So, hinge should have 5 unknowns. So, I have 2 hinges therefore, I have 10 unknowns here as reactions also I should have the force as an unknown. So, there are total 11 unknowns. So, not everything can be solved. So, in the question just quickly it is asked how can I solve this force in member CD. So, now you we look at this way that there is no moment reaction about line A B right. So, once this is said that there is no moment reaction about the line A B then we can actually take the equilibrium about that line A B. So, I take the moment of the forces that are acting in the window about the axis A B. So, that is the only conclusion we can make and I can get the directly I can solve for the force directly. So, now here is the key this force FCD has an infinite line of action. So, what we are looking at we are typically looking at the free body diagram of the window right. So, I will just jump on to the next slide. So, this is the free body diagram I have the little force FCD here and I also have the weight at the mass center of the window. Now, once I want to take the moment about the line A B what would be the best way to do it? Resolve into components, but where someone said that it is going to be D. Now, if I resolve it into D then what happens how many components are actually giving out the moment about A B how many components? Two components right one would be the FCD along x and another one is the FCD along y. So, two components of that force FCD will give the moment about A B. So, naturally we have to take both into account to get the moment about line A B, but remember I can also resolve this force C D at point C without violating any equilibrium. So, once I resolve it at C then what will happen that FCD y that will pass through A B. So, y component of the force C D will not come into picture when we are taking the moment about A B only component that will contribute is the x component and that is the simplest way to solve this problem. So, this is one of the you know beautiful thing when we are trying to look at 3D equilibrium where should I resolve the force? I am sure in 2D equilibrium probably you have seen similar scenarios that where should I resolve the force such that I can take you know a minimum contribution from the force components are going to come into the moment equally. So, as such the problem remains very simple except for the fact that one has to calculate the force FCD that means we have to now find out what are the components of the FCD along x direction along y direction and along z direction although we should not be bothered about these two. So, all I am interested in finding out what is the component of C D along x and that is point 828 FCD. So, once we do that now it is very simple because weight is actually acting at the mass center and it has its distance that distance can be easily found right the perpendicular distance and we also know there is a FCD x and the perpendicular distance is simply 0.8 meter. So, through this we can very quickly solve the problem just one moment equation and one component of the force is coming into from the reaction FCD one component of the force is just balancing the weight ok the moment due to the weight so the answer will be 227 Newton. Next problem is that of a sign board and this sign board is actually held in position by two cables as well as there are some supports that supports are properly defined here. So, support at D is that of a ball and socket joint. So, we should have three reactions here force reactions right in x y and z direction and what is also said that at point C oh sorry the point C was said to be ball and socket joint and point D here we have actually you know in y direction there is only a support ok. So, there is a ball and socket joint at C and there is just a y direction right force in D. So, now, how I am going to solve this problem. So, is it a first of all statically determinate problem how many reactions do I have I have three reactions here all force reaction sorry all force reactions at C right three forces I just have one force here. So, total four and then these two therefore, total I have six unknowns. Now, this problem is perfectly statically determinate. Now, one important thing we can observe here is that if we just look at the x axis we can see that a number of forces are passing through this x axis that is T 1, T 2 there is D y and only thing is that except for the C y nothing contributes to the moment about x axis that is a very important you know point here except for C y which is coming from the ball and socket joint no other force is contributing to the moment about x axis. Therefore, what is C y first of all C y has to be equals to. So, now, to find this unknowns first thing is that we can simply take a moment about the x axis and just make sure that C y equals to 0. Then other five unknowns can be solved very systematically for example, if I just take a moment about the z axis let us say if I take moment about z axis then I can find a relationship between T 1 and T 2 that is done. Then another important point here is the let us take a moment about a b if I take a moment about a b what do I find remember this C is shown as a resultant of the three forces C x, C y and C z. So, actually we have already said that C x C y is going to be equals to 0 that is what we have said there is still C x and C z. Now, once I take a moment about a b I can easily find the force C x y because T 1 and T 2 is already passing through the a b they do not contribute. So, we just take a moment about the C x C z is also passing through a b and this r that is also parallel to y axis. So, does not contribute in the moment about a b. So, only thing that is contributing moment about a b is the C x and the moment of this weight. So, we balance that take a moment about a b equals to 0 and therefore, we can find the C x. So, once that is found then we can actually explore the you know other equilibrium conditions. So, we take a moment about z axis and that I as I said that moment about z axis will actually give a relationship between T 1 and T 2. Let us say that is that equation I have in my hand and then we will just simply say that sum of force along x equals to 0. So, sum of force along x equals to 0 we will now have T 1 and T 2 components. So, x components of T 1 and T 2 will come into play and as well as C x. Now, C x is already solved C x was solved from the previous step. So, therefore, we can now say what is T 1 and T 2. So, through this process we eventually solved actually 4 you know unknown so far and then we can do the you know explore the other equation. So, sum of force along y direction equals to 0 and sum of force along z equals to 0 this will solve for the other 2 remaining unknowns that is R and the C z. So, now, we move on to problem number 3. So, far we have just talked about a single rigid body. So, single rigid body I may or may not have 6 unknowns let us say to solve, but problem can you know some unknowns can always be found. Now, we are going to talk about multibody that means more than one body is connected then how I am going to solve for the unknowns. So, in this problem again what is being stated is that there are 2 bars A B and O D they are pinned together at C. So, there is a pins connection here. So, it is almost like a you know it is like a C z and we have ball and socket joints at O and A and also there is a cable that is cable is B a vertical load is being applied. So, find the components of reactions at A and O. So, 2 bodies are connected and we are going to find out that what are the reactions at A and O. Now, try to just understand that how many unknowns we have in this problem and you can see that if you look at globally without detaching the bodies there are 2 rigid bodies. So, ultimately we can actually say that you know if it is statically determinate then I have to show that there are 12 unknowns indeed and from each rigid body there are 6 equilibrium equations. So, 12 equilibrium equations can solve the problem ok. Now, let us try to find out if there are actually 12 unknowns here or not ok. So, we can see here the since A and O are ball and socket joint therefore, I have 3 reactions here force reaction 3 force reactions at O total 6 plus 1 from the cable 7 right and then right here we also have a hinge. Now, that hinge what it can actually do the only thing that it can do is move about an axis that is perpendicular to its plane right. So, there is no moment reaction about that axis that is you know axis that is passing through C parallel to Y C Y axis let us say. So, there should not be just any moment, but in all other directions we are going to get the moment. So, we draw the free body diagram and just solve step by step in this case. So, what can happen actually even if we just look at the without detaching the bars let us say I just want to you know solve the unknowns. So, few unknowns can be solved not all the unknowns can be solved, but few unknowns can be solved still. So, what we will do we will go step by step. So, as we can see that let us take a moment about an axis that is about the x axis here in this case. So, this is x axis is passing through O here. So, we have assumed this is our origin. So, in this case we clearly see that ultimately what we are going to get the moment contribution is given by A y and P. So, we can find the relationship between this. So, A y can be solved just by taking the moment about the x axis. Let us explore also moment about z axis what that will give that will directly solve the the force in the cable right because we take the component of this force that is T sin of 45 degree and multiply by the distance and that should be equals to P times A. So, we get the cable force T points about y x line. What is the distance perpendicular to the brunner distance of force P from y x line? Yes, I think it is A it is all A. So, now we do since you know some of the forces are known for example, T we have already solved. So, we can just go like this some of force along y ultimately what will happen that I will be able to find actually 5 unknowns from this. If we keep doing this kind of operations, but then you know we would not be able to find the all unknowns. So, now what we have to do we just have to detach one of the member and expose the you know unknowns at C and from there we can just take the equilibrium of the detached body and try to solve for the unknown force ok. So, ultimately remember in this case we have solved A x, A y that is solved and O x, O y that is solved also T is solved. So, what we do not know yet is that A z and O z. Now, once we detach one of the bars so this bar is being detached there is actually a small mistake here. So, the way to you know look at it. So, if I really have to draw it there is a moment reactions missing in this problem ok. So, remember there should not be any moment about this axis as a reaction, but we should have the moment reaction about the C x axis and about the C z axis ok. So, that was just a simple you know mistake it has got, but ultimately what will happen now since there is no moment reaction about the C y axis. So, we can take a you know moment about this axis equals to 0 and what it will give it will immediately give a relationship between O x and O z. So, O x and O z that once we know that then remember we had already solved for the O x that was equals to P. So, you know O z will be also equals to P. So, here the idea was unless otherwise I detach this body I would not be able to solve for one of the you know unknowns that was left over ok. So, ultimately all of the unknowns can be found through this process just by taking the equilibrium about you know different axis and the force and the force equally. So, the next problem will again be a multi body problem. So, what is being done here actually we have three identical balls each of mass m are placed in the cylindrical ring which rests on a horizontal surface and the height of this ring is slightly greater than the radius of the ball. Diameter of the ring is such that the balls are virtually touching one another that is a important point to observe. And then these three balls are actually virtually touching to each other and then we are going to put a fourth ball on the top. So, what is being asked is that determine the force P exerted by the ring on each of the three lower balls. So, once we do that once we dump that ball on the top basically you know there will be tendency to spread over and that that is why we are going to get the reaction back from the ring. So, now how do I approach for this problem? Remember that there are point contacts between the balls right and this point contact will give the reactions. So, you are going to get basically contact forces here and that contact forces line of action will be from center to center ok. So, we can basically think of a tetrahedron if we join the you know centers of four balls then we are going to get a tetrahedron. So, how this reaction is happening is basically through the age of this tetrahedron ok. So, now once we take the equilibrium of the top ball remember all these reaction three reactions forces will be identical right and we can simply take the vertical equilibrium of the top ball to find the reaction first. So, one that reaction can be found that way and then we will simply go to the lower ball for the equilibrium right because lower ball will also have a force that is exerted by the ring on the ball also it will have a reaction from the bottom ok. So, these are the two equilibrium that we are going to explore. As I said just concept would be simply to think about a tetrahedron right here which is connecting the you know centers of this four balls that is the only thing we have to tell the students probably ok. And then they will get to know ok then I can actually look the look at the problem from the top view. So, what I have created here is a top view and then another view that we have taken is that of a let us say if we pass a plane that is you know to the AO axis right going like this we take a slice actually and that you know view can also be explored ok. So, these are the two views I required to solve this problem. So, in the first case remember the reactions are passing through the edges if we look at the equilibrium of the top ball ok. And therefore, what is necessary to find out just this angle that little angle theta is what that is controlling everything ok. That is the you know angle that we have to find out and once we do the equilibrium of the top ball we should be able to find out what is the R that reaction ok. And that reaction is now came back to the lower ball when we are taking the equilibrium of the lower ball then that reaction is here along with a force P which we want to find out as well as there is a normal reaction, but I am not interested in the normal reaction. So, therefore, we can just set sum of force equals the sum of force along x direction equals to 0 and we get the result. So, the only issue here would be the calculation of theta and that have to be done using the you know geometry of the problem. So, as such as I said that we are really going to work on multiple problems. So, I just demonstrated few problems where we have seen you know difficulty specially these problems students have some difficulty in visualization and also that multi body you know problem where you really have to detach the bar and how the you know conditions of equilibrium will be used about the various axis that needs to be understood.