 Yeah, we will start the tutorial session now. So, tutorial session on 3D equilibrium we are starting just now. The first problem is already displayed. So, what I will be you know interested in as we have been doing from the beginning just take 10 to 15 minutes maybe 10 minutes. Try to see if you can get the answer. If you cannot get the answer it is ok, but at least one should understand the concept that how to get you know how to solve for the unknowns. So, the first problem is displayed for the tutorial. So, the it involves as you see here the two rods these two rods let us say this point is D. So, let us say C D and A B two rods are welded together ok. And it is leaning against a frictionless walls at D. So, it is just resting on a frictionless wall at D. Now, a vertical force P of magnitude 600 Newton is applied at the midpoint E ok. Determine the reaction at D. So, just D should be here. So, D is the point where this rod is leaning against the wall. So, we are interested in finding out that reaction at D. So, D is the point with the wall ok. Just try to study the problem again and if you have any queries let us me know if there are any discussions as such you know it is supported on bearing. So, we know how many reactions are there and we can clearly see what needs to be done to find out the reaction at D. So, I can give you some hint that reaction that is coming from this wall that is along the x axis. So, that is D x. So, this point is D here. So, we are just trying to find out what is D x. So, as you all see that this is a bearing and if you think of it that there will be no movement about the axis AB. There is no support reaction that is acting about AB ok. So, therefore, what we can do we can take the moment about AB. So, answer will be just it is a one line problem. However, it involves some geometrical calculation in order to take the moment of forces about AB. So, there is a D x here and this load is applied. So, ultimately you see that I need to find the perpendicular distances. So, if one you know some of the remote centers, once you get the answers you can keep on posting what is D x that is the reaction at point D. If there are any doubts please let me know some of you have started posting the solutions looks like keep getting the solution here. The answer is 375 Newton 375 Newton and I am getting that answer from many remote centers look like. So, what I will do I will quickly display the solution now and just look at the free body diagram. This problem is just a one line problem to solve, but it involves some geometrical calculations. So, now the free body diagram is displayed. You see the free body diagram here. So, this is the reaction D x it is leaning against the wall. So, we have to find out the D x which is parallel to the x axis. Now, remember if I want to take the now what will happen as we see here. Now, what we do we join a line. So, this is a perpendicular line. So, this is a perpendicular line going from this a b. So, this imagine this line C f which is perpendicular to a b that will cut the wall at f. Now, if you take the P the P will cut here at a distance D over 2 from the a b axis. From the a b axis it will be at D over 2. So, moment of P is simply P times D over 2. Now, what we have to find out what is this D the small D needs to be found out that distance which is f c. So, that we can do based on the geometry of the problem and remember the D x right. If you really project this D x right here. So, if you are trying to look at the component of D x perpendicular to a b because that is what we are interested in. So, that will be if you say this is theta. So, this will be D x sin theta. The D x sin theta component that will be perpendicular to a b. So, D x sin theta multiplied by this vertical height that is 300. So, ultimately we have to find this theta and this small D. These are the two things that needs to be found. In the video the solution is displayed in the video. Can you see it? So, ultimately the answer will be 375 Newton and here is the equilibrium equation. That is the moment about a b. So, we need to do this D x sin theta times 300 equals to negative p small d over 2 and all the solutions will be posted also on the module. So, we can look at it afterwards also. But conceptually what it means is that this reaction from the wall that component we are taking which is perpendicular to the axis a b. So, we are taking the component of this D x perpendicular to the a b and then multiply by the distance which is this distance 300 and equate that moment is equated with p the load that is applied multiplied by small d over 2. So, we will move to the next problem now. So, it is a space frame, but remember in this problem I have multiple bodies connected. So, I have three bars and the loads are also applied all of these are taken at ball and socket joints. So, all are ball and socket joints at a b c d e. So, first thing that we should do is get the free body diagrams of all the members that is important. Remember what we have asked here use only three equilibrium conditions to find x component of reactions at c b and a. So, you have to find out these three reactions that is the x component of all these three reactions using only three equilibrium conditions. So, there are three unknowns c x b x and a x that needs to be find using three equilibrium conditions. Then the other components needs to be found for those reactions. The only trick in this problem I will again give you a hint can you identify the two force member can all of you identify the two force member is there any two force member that was done in the morning session. Once you get that then it will be very simple problem to solve then one of the bar will be gone from the picture. So, I will just say that two force member you see here it is connected by ball and socket here it is connected by ball and socket here there is no intermediate load. So, b e if you look at it the reaction should be along the axis right. So, it is a two force member that is how it is defined. So, here you will get a r here you will get a r to show the equilibrium in order for equilibrium of this b e since there is no intermediate load and it is connected by pin supports you can say because you will have b x b y and b z e x e y and e z. Now what will happen in the process the net reaction should be equal and opposite at b and e. So, that is the clue now what I want I think if you can just give me c x b x and a x I think it is solved. So, just try to give c x b x and a x again detach the bodies. So, the reactions clearly by saying that this is a two force member b e is a two force member and take moment about axis once after you detach it. If you have the answer you can just give me c x a x and b x I think one of the answers coming up c x looks like correct 2 kilo Newton the another one answer came up already that is a x that is also correct. How about b x anyone has solved b x just solve the first part part a rest you can we can do in other time also not an issue others will come automatically some of the remote centers have posted the correct answers I think c x the x component of the reaction at c that is 2 a x reaction at a along the x axis is 0.667 kilo Newton and b x is 2.67 kilo Newton. So, what I will do quickly we will just show you the first part that is part a of the tutorial and we can carry on the later part as well. So, I will just flash the solution any discussion on this 1, 3, 1, 2. Hello sir what are the difference between theorem of movement and the Bergen and Stearns. The question is if I am repeating the question are you saying what is what are the differences between Baruchnon's theorem and the theorem of movements what is the theorem of movements that I do not know personally. So, I need to look into that, but Bergen and Stearns theorem we have discussed Bergen and Stearns theorem what we have discussed that if you have a concurrent force system resultant force system if you have right if I have a resultant of a force then you take the moment of that resultant force right that is equals to some of the moments of the individual forces. So, if you think of that I have a point okay I have on a body I have a concurrent force system. So, f1 plus you know f2, f3, f4 like that okay then we had there is a resultant there or now you could do simply r cross f right that is the distance what is small r the small r is the position vector of that point where the force is applied right. So, I could easily do small r cross capital R capital R is the resultant of the forces right. Now, what it is that capital R is simply f1 plus f2 plus f3 plus f4. So, Bergen's theorem is not as such great anything about it it is just the distributive property of a vector okay because resultant vector is now decomposed into forces f1 plus f2 plus f3. So, I cannot give you answer for theorem of moments I do not know what it is okay. So, I look into that alright. Sir, in the earlier 3D equilibrium you discussed a problem second problem regarding rectangular sign of restore. Sir, you during your equilibrium equations of enthricular question the second problem actually a ball and socket joint is provided the question they are told that a ball and socket joint is being provided for point C. Yes. So, in that case there should be in point C there should be 2 forces acting. The ball and socket joint your question is in problem number 2 for the sign board right that sign board problem ball and socket joint was provided at C okay. Now, if we have a ball and socket joint then there are 3 reactions we have to actually there are 2 forces acting because automatically one force is going to 0 if you take the moment about the x axis now take the moment about the x axis the C y will be automatically 0. See it will become a planar problem in an way right but remember there is also a reaction on the top the D point D right there is a reaction component along y axis okay. See ball and socket joint will have 3 force reactions okay. So, that is C x C y and C z now what we have done we have taken the moment about x axis already that will give me C y to be equals to 0 if you take the moment about x axis that will be 0. So, what I will do we will just flag the third problem. So, I am giving the problem number 3 which will be very interesting. So, problem 3 is now displayed tutorial problem number 3 is now displayed the problem involves a camera on a tripod. So, this is the tripod and you have camera here tripod has its own weight and the camera has its own weight now remember what is being assumed here the weight of the camera is uniformly distributed. So, if you look at it the camera will have its own mass center. So, that will be if you take it as a rectangular you know this step. So, it is passing through the centroid of that okay it has a mass center for tripod that is also weighing 0.44 pound okay and the weight is passing through the point D weight of the tripod is passing through point D. Now, we are trying to rotate the camera by an angle theta you see that now as you keep rotating there will be a situation when the tripod will try to tip over. So, the condition that is being asked the maximum value of theta we have to find the maximum value of theta if the tripod is not to tip over. So, the first part is simply theta equals to 0 right and the second problem we are trying to find out a condition when this tripod is trying to tip over. So, at what value of theta? So, we are just rotating the camera. So, theta equals to 0 means camera and tripod they are parallel to each other. Now, there are reactions that is coming from the ground okay. So, what we need to do just quick hint you need to draw a free body diagram of the tripod just on this plane which is at the ground. So, all the forces that means the weight weights are going to you know you should take the draw the free body diagram the planner free body diagram on this plane let us say A B C that is the plane on that plane we have to show the weights as well as the reaction. So, what it will become it is a basically a parallel force system. So, parallel force system that is what we are going to have. So, let us try to draw the free body diagram of that when theta equals to 0 and then as you rotate the camera you can see that the weight if you consider this line A D that line A D right when theta equals to 0 the weight of the camera will pass through the line A D is in that and now as you rotate by theta it will start deviating right. So, the camera weight will try to also rotate. So, as if the A D axis is rotating there will be a situation when one of the reactions will go to 0 when the reaction remember as you turn it to the clockwise direction right then the condition would be that reaction at B goes to 0 and for that reaction at B goes to 0 we have to find the value of theta. So, one condition one unknown. So, the condition for tip over is the reaction goes to 0 and as you see here the reaction will go to 0 at B. Now, it is a challenging problem however the first part can be solved just by you know you can see that there are three reactions will come into play reaction at A reaction at B reaction at C that is coming from the ground and you could easily solve for them you need not to use six equilibrium conditions, but just three equilibrium conditions will be enough to solve for these reactions at A B and C. The first part is not that challenging, but the second part is what we need to focus on however it is going to be lengthy. So, just think of the second part of it and you will encounter this problem most often it appears that answers are coming up for the first part the second part we can just discuss it may be bit lengthy to solve the problem. So, the answer as I find see ultimately we need to take a moment about the line B C the steps are very simple. So, we have R A, R B and R C and we have the weight here. So, weight as at D for the tripod and we can find out the weight here also from the camera. Now, moment about B C will directly solve for R A. So, the answer for R A the reaction at A is 0.656. So, another axis you could choose probably x axis that is passing through B. So, an axis passing through B may be, but parallel to the x axis if we assume this is the x axis we had. So, we can also take the moment that way now since R A is solves R C can be solved in any case as we see here that R B should be equals to R C when theta equals to 0. So, the first we have the answers R A will be 0.656, B Y and C Y should be equal and the value would be 0.157. The second part can be done very similar way only thing is that you have to express the B Y in terms of theta. So, B Y must be expressed in terms of theta and it should be 0. So, B Y greater than 0 will be no tipping condition and B Y is just equals to 0 is the extreme case. So, that will lead to the condition for that will give you the value of theta. So, this is one way of doing it what would be the other way of doing it. So, let us just go through the solution quickly. So, for the second part B solution is now on the video just look at the video. So, that is the free body diagram for the part A and many of you have already solved it. So, again it is a parallel force system as we see this is the tripod weight passing through D and the if you calculate the camera weight it will be acting at 1.4 inch from the z axis that is passing through D. Now, in the next case so, the answers are as I said first part is simple take a moment about B C you will solve for A Y. If I take a moment about B C I solve for A Y you see that here moment about B C is 0 then what was done you choose an axis passing through B and parallel to x axis. So, B X choose an axis B X we can take the moment about that in order to solve for C Y remember B Y equals to C Y. So, B Y and C Y will get the same value you can either do this or you can do this one does not matter in both cases we should have the same answer. You can just do this operation sum of force along Y equals to 0 by saying that B Y should already be equals to C Y because again the problem is symmetric about the x axis. But, in the next case just look at the free body diagram again now the problem is as you keep rotating the theta as you keep rotating the camera. So, now theta so, therefore you see the weight of the camera is also shifted. But, does not matter I have to find out the theta for which the tripod will not keep over. So, you can see the movement more weights are actually distributed on the other side of x. So, tendency of the tilt will be on the other side of x. So, in this way if you want to do now we can do the exact same step as that of a part 1 we can follow the exact same step try to get B Y as a function of theta and the condition of not to tip would be as long as B Y is greater than 0 right at B Y equals to 0 it will try to tip. So, the condition for not tip will be B Y greater than 0 express B Y in terms of theta. So, that is a lengthy process this is one way of doing it ultimately we will get a trigonometric equation that will be inequality for B Y. So, you will get a type of equation which looks like this from that we can get theta. So, ultimately answer will be theta equals to 6 you know roughly 62 degree. So, theta equals to 62 degree would be the answer for this problem. Now, this is one way of doing it would there be another way of doing it the simplified way to understand this problem is how about if I try to see the line AC remember what will happen you have a tripod weight at D and camera weight at Q. Now, these two if you think of where is the resultant of these two only where is the resultant of W tripod and W camera. Now, whenever that resultant of these two passes through C that may AC line AC. So, you think of line AC as soon as that resultant is actually going on the other side of the line AC then it is going to tip over. So, the extreme condition would be that the resultant of these two must pass through the line AC in other words you take the moment of all the forces about line AC saying that B Y equals to 0. Now, this is completely geometrical problem when I am saying the resultant R is passing through AC when the resultant pass through the line AC. Now, it becomes a geometry problem. So, we can take the moment of the tripod and camera weight about the AC because B Y is already 0 at that tipping condition. So, the first equation you see here W 1 L 1 W tripod. So, L 1 is this distance right here. So, point P where you know intersection with the AC and the line DQ where the weights are acting. So, what I said basically here that we can think of one angle eta. So, that eta is here. So, eta equals to APQ angle APQ is eta. So, now it is a geometric problem and you take the moment about point line AC. You also know that L 1 plus L 2 equals to 1.4 inch. Then we can use some information on the triangle and we can try to see that from here basically L 1 can be solved also from the triangle property. We can get an expression from L 1 in terms of theta. So, ultimately we should get an equation on theta in this case which we need to solve for and answer would be same as that of the first one. So, solution will be given to you. Is that clear? So, the ultimately what we have now talked about in 3D equilibrium. So, before we have the quiz I again repeating now this problem is although quite computational intensive, but main objective here is to just to deliver the knowledge to students what would be the condition of tip when it is going to tip over. So, that is what we have done. As I said for the 3D equilibrium the way we have studied is actually not using r cross f or lambda dot r cross f. Although there are many problems where that may come into play, but we have systematically looked at that how to take moment about axis to get the equilibrium equations and even we can solve some challenging problems.