 So, I will just flash the tutorials. So, some of these are routine exercise some of them has you know tricky parts with it. So, if we can figure out those tricky issues we can solve the problem in a fraction of second. So, the first problem is that three forces and the couple is shown that is applied on an angle bracket we just want to first reduce the system to a force and a couple at B and then ultimately we want to replace it by a single resultant that single resultant we want to know where it intersects the line A B and line B C. So, basically we have to find out the equation of the line and the intercept at the A B and B C. Basically we are finding out the point of action of the resultant force and the moment of the resultant force about a point is equal to some of the moments due to the components of the resultant force about a point that is basically Weigernon's theorem. Is there any specific reason as to why we are not acknowledging that name Weigernon's theorem? It is not necessary I mean that sense because as such you know here we are just telling that we want to replace it by a resultant force and the moment resultant I mean there is no you know we are not really going into that deeper as such. These are x intercept and y intercept right that is what you are saying. X is 10.24. That is great. Y is 4.3. Yes, now kind of we are hearing more answers. So, just another three minutes wrap it up. So, the answer should be x intercept that is with the line A B that is 10.3 centimeter with line B C it should be 4.36 centimeter. Resultant moments should be 1.3 to 4.4 is 49.5. This is the moment is clockwise 120 centimeter. Yes. The point is through distance it is coming around the 4.01 centimeter. The 4 centimeter. See the 4 centimeter about 4.36 will be the answer. That is what x intercept is. x intercept is 10.3. x intercept is 4.36. Yes. And it is perpendicular distance coming together is 4.01 centimeter. X is 10.34 and y is 2.28 and that resultant is coming in the second quadrant. So, what I will do since I think we got the answers from many of these. So, the answer I think I have it all this thing in word file. So, just take a look if all the answers are as you said the moment at B that is negative 120 that is clockwise and we also have the resultant force that is 29.866. So, ultimately the x intercept we want to find a similar way. So, balance the moment about this axis right here right. So, this is the final solution we have. This picture right here x intercept and y intercept x is 10.3 and y is 4.36 and this resultant force that will be 29 roughly ok. Shall we move on to the next one? It is done. So, earlier one was 4 planar system now we are in parallel force system. So, for this problem the line of action of the resultant force is actually known. So, that is the condition given and based on that we have to find out where the fourth child is going to be placed on that raft. Sir, 3.5 centimeter from NB. Along z axis. Yes. Other one I have not yet done. 3.9 centimeter. One answer was correct the z, z required is along the z direction we should have 3.55 meter. Along x axis is 19.02. Y axis it is 22.5. No. Along x axis from NB. 7.97 is something that sounds good. 7. Great. So, answer was found now. So, answer is actually along z. It should be 3.55 meter and along x it should be 7.03. So, just 7 meter roughly ok. So, along the x it should be 7 meter yeah 7.03. Now, I have the solution in pound foot I mean it is just the replacement not the conversion basically what I was given is 85 pound and 60 pound like that. And also 15 foot and like that. So, but we do not have to convert. See that is the equivalent system we are looking at. This is the first part of it. So, the equivalent system is that what is already given that resultant force must pass through the center of the raft right. So, 7.5 and 7.5 is given. Just think of meter and kg not pound and foot. So, the resultant is 330 the sum of all these forces. Then we do moment balance about the x axis. It is already given that r times z is. So, what we know is this is known the right hand side is known here. Only thing I am interested to find out what is z d that is the fourth child ok. So, right hand side here is known. So, the answer should be for z. This is the z d that is going to be 3.55 meter. And that can be found just by balancing the moment about x axis. Similarly, if we balance the moment about z axis the answer should be 7.03 meter. So, we need not to worry about sign also for this problem because everything is actually on the same side of it. So, you can either choose everything positive or everything negative. It does not matter ok. N e and that is why it was 7.7. Ok, ok. Along x axis from N e I have told and this one is given from other end sir. Ok, you told from the N d. Origin is already given right here. So, it is better always to consider x and z axis as a reference frame whenever you are doing that calculation. Yes, you can choose the origin anywhere that is not an issue. Any question on this? Shall we move to the next one? So, it is related to hydrostatic pressure now. See again there was a question from the back I think. The equilibrium approach someone said that you know what is the reaction during the barge up opening right. So, if you think of this problem also if we think about it at the barge up opening the reaction here is actually equals to 0. But we are not following equilibrium here do not forget. So, we have to think in this problem everything in terms of the resultant force its direction and so on so forth. So, water is given on both sides. Remember there is a weight of the gate also in this case ok. There is a weight in the gate as well as there are water levels water level difference is there. Now, pressure is acting from both sides ok. So, that we have to keep in mind. So, what is the pressure distribution in this gate A B? That is the main clue actually. What is that pressure distribution we are looking at on that gate? Weight of gate is given in submerged condition. Yes that is the weight of the body and on that condition yes. 50,000 kg when submerged. Yes. So, what would be the direction of the resultant force? See I will just give a quick hint weight is acting right downward. Just think of this problem when the height is actually same on both sides. If the height remains same then there is no pressure on the gate right. So, only thing is that we have a weight. Weight is passing through the middle of the gate. Now, you keep filling the water. So, you make the one side more. So, ultimately pressure is coming from the left hand side right and it will actually now you have a weight here weight is going down. Now, you keep filling the pressure no you keep filling the water. So, therefore, we are going to have the force right from the pressure distribution that is acted. So, ultimately you have a weight and the force that is coming from the pressure distribution these two will make a resultant right. Okay. Downward force of the water. Water force of the downstream side of the water. One is original force by the right side of the water and it is vertical force above the gate A B downward force and self weight of the gate. Three forces should be taken into account. You can look out in two ways. For example, there is a force coming from this side. It acts normal to the gate. And there that is also normal. It acts normal to the gate. Weight is vertical. Now, the deal is this you can do it in two ways. One way is that you find out the complete pressure acting from this side. Find out the complete pressure acting from this side. Take the resultant. The resultant should pass through the base in order for it. Okay. For not having any reaction there. But now note one thing. This problem can be solved in a more intelligent way. You should realize that it is just the difference between these two heights. Okay. Because this component from this will be common from this. Only thing that is different is that extra height. So, if you do the problem in that way it can be ready to just two loads. One is the differential pressure coming from this side and other is the weight. And the problem becomes a little bit simpler. Less from two errors. Is it fine? Just the differential height is what important. That pressure. The answer is given already. You can solve it in multiple ways. But the main thing would be that what would be the simplest procedure to solve this kind of problem. So, ultimately think of there will be actually two forces. When the gate is about to open is coming from the differential height of the water. That pressure, how that pressure will act that is one thing. And the weight is also there. These are the two forces that is actually contributing to a resultant. That will contribute to a resultant. The resultant must pass through AB. You can also see much easier to think from the equilibrium actually than all of this. But we have not covered equilibrium yet. So, there can be two approaches here. What is the differential force that acts on this gate? Rho g H minus 4 multiplied by the width. That is the differential force. Now that will act like this and the mg the weight is like that. So, if you apply the triangle rule ultimate conclusion is that the resultant must pass through this. That is the condition given. See, ultimately if you think of an equilibrium approach, what will happen? Moment about these two equals to zero of these two forces. One will be coming from the pressure differential pressure. Another one is the weight. If you take the moment about B, you can solve the answer. Because at the verge of opening reaction is also actually zero here. 11.18 of water level. The answer would be it is 4.5 meter. That's the H, the weight is shown. The answer will be 4.5 meter. I think I will quickly show the solution. Simple sketch can actually do the job. See, this is the sketch I need. What is the differential pressure distribution here? It is going to be same everywhere, right? So, I am going to get a uniformly distributed load. That is rho g H minus 4 because the difference in the height of the water is H minus 4. Therefore, everywhere you go in the gate is going to be same. So, that is the, you know this force is coming from the differential pressure. So, rho g H minus 4 multiplied by the width of the gate multiplied by 10. That is your total force. Pressure is rho g H minus 4, right? It is same everywhere in this case. So, if I look at the force that will be simply 10 multiplied by 4. That's the area of the gate, right? 10 multiplied by 4 is the area of the gate. Multiply it by rho g H minus 4, okay? Just think this way, what is happening? As you, you know, keep feeling this. So, Fp is gradually increasing. Fp is gradually increasing. Therefore, your resultant, see, initially let's say, you know, difference is very small between the water level. Ultimately, your R will be somewhere towards the, you know, not along the AB, but it will be just somewhat toward the left of the gate such that gate is always closed. But you keep feeling the water. So, Fp magnitude will keep on increasing as per the triangle rule. So, Fp magnitude will keep on increasing to an extent when the R is parallel to the gate. So, it can be done using simple triangle rule. As long as I know Fp equals to what? And Mg equals to what? And those are actually given. Mg is given. Fp we have to find out that is coming from the force, pressure. That is coming from the pressure. That is the total force. So, that's the small part we have to work on, okay? If you think of equilibrium, we are not doing, but you can think of equilibrium also. Moment about B should be equals to 0 from these two forces again, okay? And that will also give you the 4.5 meters. So, that check can be done. Now, this can be given as an equilibrium problem as well as, you know, with this setting. So, ultimately this angle is what is known to us, this theta, that angle is known, okay? That is the condition when the resultant is parallel to the ad. Resultant coincides with ad. Okay, I think let's move on to the next one. Next one is, you know, would require some kind of calculations. So, in the previous problem, there was no integration required. So, no integration was required. In this case, we have to do a little bit, okay? So, what is the concept in just one sentence? Just one sentence. Inside the steel cone is equal to atmospheric pressure. Weight of. So, the upward force that is generated by the pressure should be equals to the weight of the cone. So, all we have to do, we are just telling students, okay, get me the upward force that is coming from the pressure distribution, okay? And then we have to adopt the integration approach in this case. Now, you have to take a strip circumferential, you know, in the circumference of the cone and we have to integrate it out. 15.7, no. The answer is 41.8. I am just giving you the answer. 41.8, 41.8 kilo Newton. If you want the answer in terms of algebraic quantities, it is 4 pi divided by 3. 4 pi divided by 3 multiplied by rho g. Rho is the density and g is the gravitational. So, anyone could explain how do you approach for this problem just quickly? Hello. The approach will be, we will be getting the center of gravity of this cone will be 1 fourth of the height from the base, h by 4. And the total weight of the water that is volume into density, that will be, we will take moment about that edge where the slant height is meeting that circle. So, we will balance it and we will get the height, we will get this mg. And don't think of in terms of equilibrium though. No sir, area of the base multiplied by the water force that is a buoyant force which we will be having now. That should be equal to the weight of the cone, that is area of the beam. But how the pressure is acting on the wall? Pardon? How the pressure is acting on the wall? Everything is, everything will be upward only. Pressure acting on the wall. It is always perpendicular to the surface. What you have to do, take the vertical cone. That will compensate, whatever is there from the top and bottom, now that will compensate each other. Only at the base I am taking, base. Hello sir. Sir, if we consider a very small strip at the surface, suppose at a distance r from the top end or the bottom end, that we can consider at a small distance which is a variable. And that surface area will be 2 pi r into d r in that form. And the pressure is acting that is normal to that surface. We have to take the vertical component only. And then we have to integrate it from the, considering a variable height from the bottom to that particular height. And that we have to get with the weight of the cone. Absolutely correct. So that's the answer. It is exactly the same way as we do the, you know, when we set up the problem of a distributed load. Okay, so what we do here? First you create a strip, that means you take a, so you take a distance z, okay, and in that strip first we make sure what is the pressure and therefore what is the force. So you can think of this strip on the, you know, periphery of that cone on the surface, right. Now there are issues here. The main issue is that there is a slanted height and there is a vertical height. But they are linked. The slanted height, the inclined distance and the vertical height has to be linked because we have to do the integration. And that integration we can go from on the height or on the slanted surface. We have two choices. So in this case it will be appropriate if we consider a, let's say distance z from the top of the cone. So let us find out what is the small force, dr. Now that small force dr will act perpendicular to this surface. So that is equals to rho gz, that is the pressure coming for this height z and then 2 pi r, that's the radius of the cone at that particular height z. Rz is the radius of the cone and 2 pi r is the perimeter multiplied by this distance ds. So that gives the surface area as if of that strip. So I have a small force dr which is actually acting perpendicular to that ds. And then we take the component of this. So you have to take the component upward. That is fv, vertical force. So that vertical force whatever we get is the answer for the weight. So finally remember we do have to change, there is a s, s is the slanted height as I said and there is a z but there are relationship exist between z and s because of this angle. So these are more complex problem and it is difficult for the student also to absorb that at the beginning but ultimately it is a systematic procedure that we are following here. It is the same as that of how we calculate the distributed load if we have what is the centroid and what is the area of that distributed load.