 So, now we are simply going to go to the tutorials just keep solving ourselves and again you know this problem maybe we will take a little bit longer to solve also because lot of geometries are usually involved in 3D equilibrium. So, we can just go through the steps as such when you know we are solving it. So, if you can just go through these problems quickly we will just work on the tutorial now. So, anyway I will just explain these are two bearings here and these are T shaped welded bar. So, this total body is continuous there is no breaking here. So, only supported by two bearings at A and B and we want to find out the reaction that is coming from the wall sorry about this. So, there is D is that wall ok. So, ultimately you know what is the concept here is that taking the moment about the A B axis, but before we do that you know we need to kind of convince people ok how the bearing reactions are acting why we are able to do the you know why we are able to take the moment about the A B axis that is the whole issue ok. Now, you take the moment about A B axis, but can we really try to figure out that I do have a you know moment that I do have a force from the wall that is along the x axis. So, when you take the moment about the A B axis of that force which component actually contributes to that you know moment about A B you have a D x that is acting along the x axis right. So, we need to take the component of that D x which is perpendicular to the A B that is all. So, we need to find the distance here and that distance is you know typically if we create a point here hip let us say this point is F here and we need to find this F c which is the perpendicular distance from C to the wall ok. So, that distance needs to be found out. So, that distance once we know that this problem can be solved very easily that is the hint. So, one have already you know one of you has found the this thing the distance D that I was talking about that is actually perpendicular distance from C to the wall ok. So, that distance is very important once we know that distance and also we need an angle somewhere because we have to now kind of you know because we have this D x here that D x has to be rotated such that it is again perpendicular to the line A B ok. So, once we do that right then we can simply take the moment very easily. So, 375 Newton is the answer just quickly show you the information that we need. So, the basically we are looking at this distance that distance is stated as F c ok. So, there is F and this F c should be equals to 300 that is the first thing. Second thing is that see ultimately we can see there is a theta here that theta needs to be calculated because we want to take the component of the D x right which is again parallel to the C f that is ok. So, ultimately you know we have this small D is understood that D here right that D is actually the length F c small D ok. So, this is the equation that is important is the only equation that is required to solve this problem ok. E multiplied by D over 2 because you have this D is actually F c and then we have D x sin theta. So, that sin theta component will be right parallel to the F c ok and then you multiply that by 300 to get the moment about A B. So, the final answer will be 375 Newton as one of you have answered. So, the moment about an axis here again is very important you know moment replace an important role that we have seen. So, let us move on to the next problem as such it will actually take you know all the problems will take minimum half an hour. So, let us just move on to the next problem. It is a very popular problem I would say it is given in all the you know courses that I have seen so far and it is nice to look at also because you know we play with this thing in day to day life. So, we have the tripod here and just remember some of the things that we have to keep in mind is the camera is uniformly weight of the camera is uniformly distributed ok. So, the camera weight is uniformly distributed which should pass through its mass center own mass center and also the line of action of the weight of the tripod that is said to be passing through D the D point is here. So, the vertical components of the reactions at A B and C we have to find when theta is equals to 0. Now, what is important is this camera is being moved is being rotated about the y axis ok and we want to find out that when it is going to actually tip off alright that tip off condition is very important. So, this problem is given just to you know understand that what is the tip off condition when the tip off will happen ok. So, first part is as such you know very straight forward. So, we make the theta equals to 0. So, what happens? So, all the if you look at the weight that is coming from the tripod and also the weight that is coming from the camera they should all pass through a maximum value of theta again you know these are all conversion problem. So, the maximum value of theta we keep on moving the camera right rotating the camera what is the maximum value of theta if the tripod is not to tip over. So, we look to we need to look at you know this problem from the top view. So, we just have to make sure that we draw the free body diagram in this plane all the forces again will be acting perpendicular to this plane. All the forces that we see they are acting perpendicular to the plane that is relevant to this problem ok. So, we draw the free body diagram only on that plane and these are all parallel force systems again we are looking at all parallel force system and those forces are going to be perpendicular to this. So, reactions are vertical we are going to get the vertical reactions from A, B and C that is all ok. Now, draw that free body diagram on this plane with all the forces those are actually parallel force system we have. As such it is a you know it does have some computation as if specially for this part we are going to do in a lot of computation, but this part is very straight forward that answer at least we should be able to do it. The idea is here that we are rotating the camera ok the camera is rotated about the y axis. So, as you keep doing that what happens the mass center of the camera is going to be shifted also right and what happens what is the basic concept can we imagine that there is a line A C. If you draw this line A C and think of moment about A C to topple over what will happen is that we have one force that is coming from the tri force. The other force we also coming from the mass center right. So, now these two will create a equivalent system as if and when that force that equivalent force is actually going you know on the other direction of the line A C it is going to topple. So, ultimate condition for the toppling will be what is the condition for the tip over. The reaction at the B will be equals to 0 that is the condition we have to use. So, at the verge of tipping R B equals to 0. So, the question was it is not to tip over that means R B or B Y let us say. So, B Y should be greater than as long as B Y is greater. So, I will quickly you know show you the free body diagram I think just a hint to this problem. I think ultimately how do you draw the free body diagram? See this is the part A free body diagram ok. So, this is the you know point D through which the weight of the tripod the weight of the tripod is 0.44 pound right and the camera right. So, camera weight that is going to be acting at a distance of 1.4 inch because if we look at the mass center of the camera right. So, that is going to be 1.4 inch away from the D. Why? Because see ultimately it is 4.8 divided by 2. So, 2.4 negative 1 right. So, that should be the distance of the mass center from D 2.4 negative 1 ok. So, that is how we draw the free body diagram which is very very important and the rest of the procedure will easily follow now. So, now how do I find out A in one go? The A Y can be found out by taking the moment about the line B C and someone already said the answer is going to be 0.65 pound ok. Similarly, what we can do? We can try to find out either B or C by taking the moment about x axis right. So, B and C that equation of what is the relationship between B and C? So, that they are equal in this case ok. I was in this case it is fine, but ultimately we have also we have to also make use of the vertical equilibrium condition. So, A Y plus B Y plus C Y equals to the force. So, from there we can actually solve for the B Y and C Y. So, in the first case yes they are equal ok B Y and C Y. So, in the second problem what is basically happening? I will again show the free body diagram. So, procedure is going to be same. So, this is the free body diagram ultimately this theta needs to be found. So, I have rotated the camera. So, mass center has been moved by this theta. So, remember this 1.4 inch is still there right. So, only concern is this theta which we have to find and the condition is that not to tip over B Y should always be greater than 0. So, ultimately find the expression for B Y in terms of theta that is what we have to do and once we say that expression is established. So, B Y just greater than equals to 0 we will solve for the theta ok. There is also an important part here as I said that one can also think of this problem like this if we join AC ok there is a line. See ultimately what is happening if we study this problem very carefully these two makes an equivalent system that is located somewhere between let us say D and this point as we keep moving the as we keep rotating the camera right ok. And ultimately what will happen that equivalent force as soon as it actually cuts the AC line as soon as it cuts the AC line then immediately B Y will be equals to 0 because if we take the moment of everything about the line AC ok. So, that procedure can also be adopted that also we can. I mean it is a bit lengthy problem and definitely it is not going to be an exam problem I would not say, but it is going to be more or less a homework problem. But condition of keeping is something that we have to you know impart to the student very carefully. So, the answer if we want to find out we have to go through the process ultimately setting up expression for B Y that we can do now. How do we set up the expression for B Y? One is that we take the moment about the x axis that is already done. So, you get a relationship between C Y and B Y now ok. Other thing that we should do take the moment about an axis that is parallel to Z, but passing through A. So, let us say axis A Z. If we take the moment about this then again we can have a equation that is relating B Y and C Y. So, from these two equation we solve for B Y in terms of theta and then say B Y is greater than 0. So, these are again so far we had basically you know single rigid body action. Now, we just study also the multi body actions ok. So, all the connections are here ball and socket joint. There are three members everything is a ball and socket joint. And the question is posed very very carefully. Use only three equilibrium conditions to find x component of reactions at C B and A. So, first part let us just try to do that and then others can follow. We can try to find out what are the number you know how many unknowns are there, how many equations we have. Just a hint because we have already started to the equilibrium. Now, in 3 D is there in two force member in this body in this frame? Is there any two force member? B E is a two force member why because they are connected by pins it is almost pins because ball and socket joint now it forms a pin. So, to hold the equilibrium the net reaction has to be equal and opposite ok. So, therefore, at the very far beginning we have to understand that B Y equals to 0 and E Y equals to 0 because it is a two force it becomes a two force member. You have only axial load in B E ok and due to that we should not have any B Y and E Y. Now, we can start solving I mean now the issue will be very very simple. So, that is the only concept that is required here because no lateral load is applied in B E it is simply ball and socket joint at the two ends right and therefore, it will only carry the axial load in this problem. Now, this is a typical you know final exam problem actually. So, C X is 2 that is correct A X is the answer the minus plus is a issue definitely how we have chosen, but magnitude is not correct A X should be 0.6 and B X is 2.6. So, how do you do that? So, can you take a moment about B Y C X is 2 right and C X can be found moment about Y axis passing through point. So, it has to be you know you can find it very easily from this freeway diagram D Y take moment about D Y see one in one go you can find it from that freeway only. So, in one go if you really want to find out what is C X I should take a moment about D Y of this particular bar ok I take a moment about Y axis of this bar that will solve for the C X very easily. So, C X is at the top take the moment about D Y then we come to that member A B C in the member A B C if we take a moment about D Y that will solve for A X and if we do the sum of force along X direction equals to 0 that will solve for B X ok. So, thing will follow like that. So, that is why the question was spoke was that only use three equilibrium conditions to solve for A X B X and C X, but see here I do not know what the main intension was to find out that there is a two force member and that really you know solve the problem very quickly. The rest of the you know things we can just keep taking the various moment and force conditions that we have from various free bodies, but these are all the answers is that ok.