 We were discussing earlier about the stability of floating bodies and we will continue with that. So if you recall this sketch which we were discussing in the previous class, we identified a point known as meta center and the motivation behind identifying this point is that when it is a floating body and it gets displaced and its submerged portion within the water changes its configuration, the center of buoyancy does not remain fixed at its own position. So referring to a fixed center of buoyancy might not work as a stability criterion. Now we will try to find out as what we have done for the case of totally submerged bodies that how we can find out a stability criterion for floating bodies. As you can see the presence of or the existence of this meta center is something which is equivalent to the presence or existence of the center of buoyancy for a case when the center of buoyancy remains at its own position. So the meta center somehow reflects the location of the center of buoyancy with respect to the axis of symmetry of the body. So it has some relationship with the center of buoyancy definitely but by specifying the meta center or by specifying the center of buoyancy just as their absolute locations it is not just possible to talk about the stability criterion. So we have to look into more details. To look into more details we will set up some coordinate axis. Say we have x axis like this, say we have z axis like this and the y axis is perpendicular to the plane of the figure. So the y axis is this dotted line which represents an axis in the other view. Now if we consider that how we calculate the submerged volume, the volume of the solid which is there within the fluid. Let us say that at a distance of x we take a small element of width dx and this element has a depth of z. So this element if you look into the other view will be like this which is basically located at a distance x from the y axis. Let us say that this shaded area is da. So you can say that z into da is a representation of the elemental volume of this shaded portion. Now when it is tilted, even when it is tilted we fix the x and z axis with respect to the body and we keep them as same. So we still have this as the x axis and maybe this as the z axis. These are fix relative to the body. So although the body has tilted we are using a body fitted set of axis so to say. Now let us say that the angle of tilt is theta. If the angle of tilt is theta then what happens? We can find out that what is now the displaced volume. So if you consider again say at a distance x from here some strip of width dx. Now if you consider the displaced volume, the displaced volume is of course corresponding to this part. This part is like z same as this one because it is the same axis that has got tilted. So if this is z, this displaced part is also z. But you have the additional displaced part which maybe let us mark with a different color. So this is an additional displaced volume. So this additional displaced part will correspond to a particular length along the z axis and what is this? Consider that if this angle is theta then the angle of tilt of the x axis with respect to the horizontal is also theta. So what is this dimension? You know that this is x. So that is x tan theta. So if we calculate what is the displacement in centroid of the displaced volume? That is what is the difference between the x coordinate of say the original center of buoyancy. Let us give it a name that there was an original center of buoyancy say CB and now the new center of buoyancy CB prime. So the difference in x coordinate of CB prime and CB that is what we are interested to find out. What is that? Center of buoyancy you can calculate by utilizing the formula for centroid of a volume which is here the displaced volume. So this will be integral of x dv. So x is this x. What is dv? So you have this total height as x tan theta plus z dA is the other area in the other view. So the depth times the dA is the volume. So this divided by the volume of the immersed part because of the symmetry the volume of the total immersed volume does not change. Whatever comes down the same volume goes up above the water. So the total displaced volume does not change. Minus what was the original one integral of xz dA divided by v. It may be easier if we mark the original CB and the displaced CB here. So there has been a displacement from CB to CB prime. The location of the center of buoyancy has got shifted. So this is nothing but equal to CB, CB prime that is the length that we are talking about and this is equal to integral of x square tan theta dA divided by v. Theta being like a constant parameter tan theta you can take out of the integral and this area integral is carried out over the shaded area dA. So what does it represent if you take tan theta out of the integral? Second moment of area with respect to which axis? No with respect to you see it is you refer to now this plan view integral of x square dA is the second moment of area with respect to the y axis. Now it is tan theta into Iyy by v where what is Iyy it is integral of x square dA. Now referring to this figure you can say that for a small angle theta CB, CB prime this particular small distance is a small arc of maybe a circle. So you can write this as approximately CB m into theta for small theta. It is just like s equal to r theta for a small part of an arc of a circle because this is very small we are assuming a small displacement we test stability not with a large displacement but with a small displacement and meta center m is the meta center. Now for the same smallness tan theta will be approximately equal to theta of course you are expressing theta in radian that is understood implicitly. So then if you equate the 2 parts what you will get CB m is equal to Iyy by v. Now look into this figure in this figure you see that m is above g when m is above g the couple moment created by the forces is trying to restore the body to its original configuration. If g was above m it would have been just the opposite case that you can clearly visualize. So what is important here is the location of m relative to g just like for a totally submerged body it was location of b relative to or the center of buoyancy relative to g. So what we are interested to find out is what is the height gm. So gm you can write it as CB m-Cbg. So Iyy by v-Cbg see why we have written the formula in this way in the right hand side you have terms which are independent of the deformed configuration or deflected configuration. So if you see here Iyy v these can be calculated based on what is the original configuration and CB is the location of the original center of buoyancy relative to the body g is the location of the center of gravity relative to the body and the distance between those 2 that does not change with deflection. So we are able to express something which should be a function of deflection in terms of certain things which are not functions of deflection. Therefore this is also not a function of deflection and m remains sort of fixed but you have to keep in mind that there are certain assumptions that go behind this. What are the important assumptions? Theta is small. So this analysis is valid only when you are having a small theta not only that we have implicitly assumed that the z axis is an axis of symmetry for calculating or for coming up with this expression. So we are essentially dealing with a simplified case with symmetric bodies that we have to also keep in mind. Let us look into an example where we illustrate the use of this expression for finding out the stability criterion. So what is the stability criterion here? Let us summarize if m is above g it is stable. If m is below g it is unstable and obviously if m coincides with g it is neutral. So you can clearly understand that m plays the role of equivalent role of center of buoyancy in this case but it is not exactly the center of buoyancy. We consider a very simple example and we will illustrate the use of this through that example. Let us say that you have a body of this shape rectangular parallel pipette shape and you can give dimension say let us say a maybe this is b and let us say this is c. Say this is partially immersed in a fluid. So it is like a floating body. Now let us try to see that how we can apply this stability criterion here and what are the important issues. First important issue is that when you apply this criterion what should be the y axis with respect to which you are considering the second moment of area. If you look into this problem very critically you will see that there are certain non-trivial issues like you can consider maybe this as one of the frontal surfaces you might consider this as one of the frontal surfaces and even the third one. So accordingly it is not very straightforward to say so when we drew the picture of the boat see it could be this front part it could be this side part both are exposed to the water I mean both are within the water. So if you could say that what should be the corresponding axis that you need to consider or thus the choice of the axis change if you shift your attention from the frontal area to the side area. At the end what you are bothered with you are bothered with the plan view. So when you have the plan view which is like the top view of this one what is the plan view it is the intersection of the body with the free water surface or free fluid surface. So that particular view it boils down to the same when you consider the this surface or this surface. But again when you look from the side and when you look from this one maybe in one case you are considering this as the y axis in another case you are considering this as the y axis both axis are relative to the top view. So question is with respect to which axis you should evaluate this. See this is what this is like evaluation of a safety criteria that it will be stable. So when you want that it should be stable what should be the guideline the guideline should be that like you want this to be positive right. That means this height is greater than this height then like m is above g. So this the way in which this is written the right hand side has to be positive. So more positive means so to say it is like it is more stable so to say. So if we take the least of this one least possible value of this one and still find that the right hand side is positive then for all other cases it should be positive. So it is like a safety design. So you look into the most adverse condition make sure that it is positive even for that. So for better conditions it would always be better. So out of these two axis for one axis it will be what will be the i it will be b a cube by 12 another will be a b cube by 12. So you take the smaller one that means if b is smaller than a as an example you take this one. So then when you substitute that in this expression and still you calculate this as positive you are assured that it is safe because with respect to other axis when you calculate the metacentric height it will definitely be greater. This is known as metacentric height and really when you calculate different metacentric heights based on different views these represent different types of angular motions like rolling, pitching these are different technical names based on with respect to what type of axis it is tilting. So but just for design simple design you can consider the smaller one and you divide it by the volume the volume not the total volume but the volume of the submerged part. So that you can easily calculate based on what part is submerged. Let us say that d is the submerged depth it is easy to calculate this because you can use the equilibrium that for equilibrium the buoyancy force must be equal to the weight. So from that if you know the density of the fluid and density of the solid body you can come up with what should be the equilibrium d just by very simple equating of the two forces. So you can calculate what is the volume which is immersed in the fluid center of buoyancy location you can get from the centroid of the displaced volume center of gravity also has a fixed position with respect to the body. So you can clearly substitute these dimensions and find out what is the metacentric height. So this effectively requires only the calculation of the immersed volume the calculation of the second moment of area with respect to these axis and specification of the locations of the center of buoyancy and center of gravity with respect to the axis fixed on the body. So that is a sufficient information to calculate the metacentric height. So what we can clearly see is that if m is above g that makes it stable that means lower the location of the center of gravity it is having a greater chance that it will be stable because then it is a greater chance that m is above g lower the location of g. So location of g if it goes higher and higher it might make a previously stable system or convert it into an unstable one. Let us look into an animated example to consider this case. So in this animated example what we will see? We will see that how the stability may be disturbed because of the shifting of the distribution of weight of the body. So just look into it carefully. So there is a body which is given a slight displacement and you will see that for small displacement it oscillates like a pendulum. We will easily derive what is the time period of that. Now you see that the distribution of the weight is being altered. So the center of gravity is being shifted higher and higher by putting the load more and more towards the top and you see that it topples. So obviously this is clear illustration of this concept that we have learnt in this example that if you have metacenter above the center of gravity it makes it more stable and otherwise it is not so. So now for small oscillations you could see that for small deflection you could see that it oscillates like a pendulum. So when it oscillates like a pendulum it has a time period and that may be calculated by calculating the moment of the resultant force with respect to the axis. So if you have a tilted axis like this and if you have force Fb which is the buoyancy force this force has a moment with respect to the axis of the body. So what is the moment of this force with respect to the axis of the body you can just find out the perpendicular distance of this force from the axis of the body okay. So it will be just what will be the perpendicular distance of this buoyancy force line of action of the buoyancy force I mean you can see that it eventually passes through the axis of the body right. So it is not this force individually which is important it is the couple moment that you are having the so called the restoring couple. So you have also the w and it is basically the couple moment of this forces that you need to consider. So the perpendicular distance between these two in terms of the metacentric height could you express this. So if you have this as the metacenter m so we are interested about this distance. So it is possible to express it in terms of the metacentric height. So this angle being theta it is what is this perpendicular distance gm sin theta. So the moment of this force is w gm sin theta and the resultant moment of all the forces is nothing but if it is like rotation of a rigid body with respect to a fixed axis. It is a restoring moment so you should have a minus sign associated with one that means whatever is the chosen positive direction of theta dot theta double dot this has a direction opposite to that to bring it to its original configuration. So you can write this as so I is what now I is the mass moment of inertia it is not the same I that we were talking about it is the real mass moment of inertia with respect to the axis of the body relative to which it is tilting. So this plus for small theta again this is approximately equal to theta that is equal to 0. So it is just like equation of a spring mass system mx double dot plus kx equal to 0. So what is the natural frequency of oscillation of the system root over equivalent stiffness by equivalent mass. So w into gm divided by i that is the natural frequency of oscillation of the system it is an angular oscillation not a linear one. So you can clearly see that greater the meta centric height greater will be the frequency of oscillation. So if it is a ship it will be more uncomfortable to the passenger. So these are two conflicting designs. See greater the meta centric height you expect it to be a safe in terms of stability but it will have more oscillation within that stability regime that means for a passenger it may be quite uncomfortable. At the same time if it is for a used for a particular say critical purpose like warfare and so on. So their stability of the ship is more important than the comfort of the crew and there obviously it is the stability that should be the driving factor for design. So when a ship is design you have two conflicting things one is the comfort another is the stability and comfort factor comes from this high frequency of oscillation and the stability comes from the meta centric height or location of the meta centre relative to the centre of gravity. Now that we have seen this stability criterion so on we will come to our final topic in fluid statics which ironically is not fluid statics but fluid with rigid body motion. And as we discussed earlier that we are going to address this issue within the purview of fluid statics for a very simple reason that when you have fluid under rigid body motion it is still fluid element without any shear. So when it is without any shear it is just the normal force which is acting on the surface. So the distribution of force in terms of pressure on the surface remains unaltered no matter whether it is at rest or under rigid body motion. When the fluid elements are deforming then only you have the shear. So let us take a simple example for fluid under rigid body motion to begin with. Let us say that we have a tank partly filled up with water and the tank is accelerating towards the right with an acceleration of a x along x fixed acceleration and neglect the deformation of the water that is there in the tank. In reality that deformation is there. So whatever analysis that we are presenting here is not perfectly correct because in reality there will be shear and deformation and so on but just as an idealization. Just like many times we discussed about frictionless surfaces not that they are there but through this idealization we learn certain concepts. So let us say that there is no deformation so the water which is there within the tank just get deflected in its configuration in terms of the free surface like a rigid body. So how we calculate that what should be the location new location of the free surface because of this acceleration that is what we want to see. Say initially the height of water in the tank was h0. Maybe let us specify the breadth of the tank as b and maybe the width perpendicular to the plane of the board as w. Let us say that means it is a rectangular tank. Now recall that we derived certain expressions with regard to distribution of pressure in presence of body force. So far as I remember this is the expression that we derived sometime back when we were actually starting with our discussion on fluid statics. So we will try to use this expression here. So along x we have minus because there is no body force acting along x. Let us say that the y direction so we have x and y as our chosen directions. Let us say that the y direction is the direction opposite to the acceleration due to gravity. So for y direction what you can write? So what is ay? Ay is 0. It is accelerating along x. So ay is 0. What is by-g? So you can get-rho g. Now when you have a free surface the free surface is characterized by what? The free surface has the same pressure throughout because it is exposed to the atmosphere which has same pressure throughout. So there is a pressure equilibrium between that and the atmosphere. So for the free surface you must have dp equal to 0. That should be the governing parameter for locating the free surface. So when you have dp equal to 0 remember that now p is a function of both x and y. So you can write dp as this one okay. So you can substitute in place of the partial derivative with respect to x as-rho ax dx and this one is-rho g dy is equal to 0. That means what is dy dx? That is equal to-ax by g. What does this dy dx represent? It represents the slope of the free surface. So can you tell now whether the slope of this surface will be like this or like this? 1 or 2? 1 because you can clearly see that this will represent a kind of tilt like this. So this will become the new free surface and the angle that you are considering for the slope is basically this one because this is negative ax is positive and g is positive. So this is negative so it must be an obtuse angle. So in the direction in which the tank is accelerating the liquid will be more down and in the other direction it will be more up. And if you specify this angle as say theta then that theta is nothing but 180 degree-this slope angle. So you can say that tan theta is nothing but equal to ax by g because theta is nothing but 180 degree-this angle. So you can find out that what is the extent to which the water level will rise on one side and may be fall on the other side. Let us say that this rise is delta H on one side because of symmetry it will be a fall of equivalent delta H on the other side. So it will be as if swiveling with respect to the center. So we can calculate what is delta H. So if you calculate delta H what will be that? It is nothing but delta H divided by b by 2 is tan theta that is ax by g. So from here you can calculate what is delta H ax by g into b by 2. Now let us come to a critical condition. If we are happy with this sometimes we are deceived. How? Let us say that the total height of the tank is H. If it so happens that let us say for from calculation we get delta H plus H0 it is greater than H. Practically that is not possible right because the liquid cannot occupy a height which is which is greater than that is provided by the tank. So what this will mean? This will mean some water has spilled out. So this is a condition from which you can say that it has actually spilled out. When it has spilled out it is no more this configuration. So when it has spilled out what will happen? What type of configuration you expect? So before spilling out it tried its best to climb up to the topmost level and then it spilled out. So this will be one end of the surface and maybe the other end is like this somewhere. So with spill out maybe this is the so this is with spill out. Irrespective of whether it has spilled out or not you still have this expression applicable. So now this will be the angle theta. You can find out that what should be this length let us say B1. What should be this length B1 because you can say capital H by B1 equal to tan theta which is Ax by g capital H being the height of the tank known. So from there you can find out what is B1 and therefore you can calculate what is the volume which is there now within the tank and the difference between the original volume and that volume will give you what is the volume that has got spilled. So you see that it is not just like your solution should not be driven by a magic formula but based on the numerical data given you have to come to a decision whether the water is there inside or it has got spilled and so on. Let us take a variant of this example. So the variant of the example is that now we have the tank located on an inclined plane with an angle of inclination alpha. The tank is there on the plane and it is accelerating say downwards with an acceleration of A which is a uniform acceleration may be because of a resultant force which acts along that direction. So in this case it may be more convenient if you fix up your coordinate axis relative to the inclined plane say x and y. So the similar equations will be applied and let us just do it very quickly because it is very straight forward. So you will have minus partial derivative of p with respect to x plus what will be Bx now? Bx is g sin theta. So plus rho g sin theta is equal to rho here it is not theta we have given a name alpha. So rho g sin alpha is equal to rho Ax is A. What about y? So minus this so y will be minus g cos theta. So minus rho g cos alpha is equal to 0. So from here you can find out when you have dp equal to 0 that means you can substitute the expression. So it will be let us write it at the top. The expression now becomes in place of partial derivative with respect to x you can substitute rho g sin alpha minus rho A dx plus minus rho g cos alpha dy equal to 0. So dy dx will be g sin alpha minus A divided by g cos alpha. And this is now equal to the tan theta where theta is the angle relative to the original location of the free surface or like the assumed x direction. So you can see now that there is depending on the magnitude of A this may be positive or negative. So you may have a case when g sin alpha is greater than A or g sin alpha less than A. So here you cannot trivially say that whether it is 1 or 2 case 1 or case 2. Again you see that it is not a magic formula that should drive your decision. It depends on what is the physical situation that is prevailing. Let us consider a third example. Yes surface should always be normal on the resultant. See here what we are doing? Here we are implicitly applying the same condition. It is you see that we are having one very important assumption. The assumption is that you do not have an oscillation in the surface. So sometimes because of this displacement the surface oscillates. It becomes like a wavy situation and that is known as sloshing of tanks. So we are not going into that details. So we are assuming that the surface remains flat and when the surface remains flat and under these conditions when you have dp equal to 0 that is exactly the same condition. See vector analysis is not it is also mathematics. When you say that when you are dealing with mathematical analysis at film this is too mathematical. I mean I do not see any difference. Like I mean if you have vector analysis this is also vector analysis just do dealing with scalar components. So it is better to be habituated with these because again I am telling you there are situations when it will not be as straight forward as these. So you have to use these fundamental equations. So whenever you are solving a problem try to adhere to the fundamental situations. I am getting your point why you are telling this because you have been habituated in solving problems in that way through your entrance exams. But we will be encountering more challenging problems than what you have solved earlier through that type of magic situation and we will try to avoid that. So our basic intention will be that we have this basic equation. This should solely guide us for solving whatever complicated problems that we are having of this type. Let us consider another one say example 3. Now let us say you have a closed tank. Close tank partially filled with water this is closed. Again you are doing the same thing accelerating it along x. What will happen to the free surface? Now there cannot be any spilling. So when there cannot be any spilling there is even a chance that the free surface is like this. So the symmetry with respect to the central line which was there for the previous case without spilling is now broken. That symmetry is not guaranteed because it may try to escape and since it is not finding an escape route it will break the symmetry and get distributed in what way in such a way that the volume of the liquid now is conserved because it cannot go out of the tank. So if you say that this dimension is say y1 and this is x1 then you have y1 by x1 as tan theta and that is equal to ax by g and you can calculate y1 and x1 by considering that the volume which was there originally is same as the volume of water after this tilting of the interface. So this is the new interface. So this will give you another relationship involving x1 and y1 and you can solve for x1 y1. Now let us say that we are interested to find out what is the total force acting on the top surface or the lead of the container. How will you find it out? So you let us say this point is c. So you have to keep in mind that only up to ac it is in contact with the fluid and you can use this one. So you can find out the pressure distribution as a function of x from a to c. Take a small element at a distance x from a of width dx. So the force acting on that is p into dx into the third dimension. Integrate it from a to c and ac you can find out from these geometrical considerations. So that will give you the total force. Let us consider maybe a fourth variant which I will not solve but just tell you that such a variant is also possible. Say you have a tank, now it is completely filled with water and it is closed. Which p will be p0? You can you assume any one of the point say a as a reference pressure say p some reference p because always when you have pressure it is relative to some point. And so you can express the pressure distribution in terms of the pressure at the point a and then integrate it over the element. So next example is you have this tank accelerating towards the right but completely filled with water but closed. So water cannot escape. So if I tell you find out the total force on the lead AB. How will you do it? I will not do it myself leave it on you as an exercise. And I can only tell you that this is the simplest of all the cases that we have considered. But you have to keep in mind that same consideration as this also should work. So that will give you a natural pressure distribution from a to b. And similar to this you can just integrate from a to b to find out. Here you do not have the botheration of finding out what is the portion that is exposed with the liquid because since it is completely filled it should be completely exposed. Now when we consider the rigid body motion it is not always just translation it may also be rotation. So let us consider a rigid body rotation example. This type of example you have seen earlier that you have a tank filled up with water to some height h0. It is a cylindrical tank of radius r. So we are using a rz coordinate system axis symmetric coordinate system. So this is r coordinate and along this there is z coordinate. So this tank is rotated with respect to its axis with an angular velocity omega. So when it is rotated you already know it that it will come to its free surface will come to a deformed or deflected shape like this which is a parabola of revolution. Let us quickly see that how we can derive that it should be a parabola of revolution from these fundamental principles not from any magic formula okay. So let us see you start with here. So there are 2 directions r and z. So you have dp dr-that plus what is the body force along r-omega square r. See these we are writing in the original form of the Newton's second law of motion not with respect to an accelerating reference frame. So no question of centrifugal okay. So you have now you tell when it is no question of centrifugal whether the body force is there or not yes or no. You have say you are standing on a platform and you are looking into it from an inertial reference frame. This is not an external force that is applied that you can see only when you have a rotating reference frame that is attached to the platform that is having angular motion. So obviously there will be no br that you have to keep in mind. This is the form of the Newton's second law in an inertial reference frame. With a non-inertial reference frame you have to use a pseudo force for the inertia force but that then you do not write equation of dynamics but equivalent to equation of static equilibrium. You convert that into an equivalent static equation through D'Alembert's principle. But here we are not talking about that. We are talking about the proper acceleration. So you have no body force but you have acceleration. What is the acceleration? This is nothing but a centripetal acceleration. So minus rho omega square r. See if you had considered it in a rotating reference frame the right hand side would be 0 because you are writing static equilibrium. This would be represented by the pseudo force. So the final equation would eventually be the same. It is just a matter of the reference frame with respect to which you are writing but this is written with respect to the inertial reference frame. Then when you come to the z direction-dpdz then let us say this is the acceleration due to gravity direction-rho g is equal to there is no acceleration it is not vertically moving. Again like whenever it is vertically moving and so on you can substitute. So start with this basic equation depending on whatever information is given in the problem you try to use it. So for the free surface you have dp equal to 0. So when you have dp equal to 0 you have this equal to 0. So this is rho omega square r. This is minus rho g. So you have dz dr is equal to omega square r by g. You can now integrate this with respect to r. So on integration what follows? Let us just complete that integrating it follows g equal to omega square r square by 2g plus a constant of integration z0. You can set up the constant of integration z0 by choosing a reference such that when r equal to 0, g equal to 0. So if we choose this as our origin of the coordinate. So this is r and this is g. So if you have at small r equal to 0, g equal to 0 then g0 equal to 0. So g will be omega square r square by 2g. This formula you have encountered earlier. So we can clearly see that it is a equation like of a parabola. So it is a paraboloid of revolution. It is a 3 dimensional situation. And you can calculate the other things just similar to what we did in the previous case with an understanding of what that again if it does not spill whatever was the volume that should be conserved. So how can you calculate the volume? So initial volume you can calculate. Final volume is pi capital R square into H0. What is the final volume? Final volume is whatever is the depression say delta H plus the volume of the shaded parabola of revolution. And I leave it on you as an exercise. If you calculate this shaded volume you will see that it will just be half of the volume of the circumscribing cylinder. That means if this height is H1 then the shaded volume will be half of pi R square H1. By simple integration you can find out this volume. So by equating the initial and the final volume you can find out totally the deflected configuration. Again you may have to check that H1 plus delta H if it becomes greater than the height of the original tank then it will spill. And spilling again may have 2 different cases. You may have in one case the cylinder is rotating in such a way that you have spilling but still it is having an interface shape like this. Again it may so happen that is rotating so fast that is spilling but only the part of the parabola of revolution is within the cylinder. So then it is an imaginary parabola of revolution even outside that you have to consider to find out what is the volume that is there inside. So it all depends on the rotational speed given dimensions and so on. So it is not just like a fixed formula but you know what is the basic principle. We have discussed enough numbers of examples to see that what is the basic principle and that basic principle should guide you to find out that what is the case. Now if it is totally closed cylinder as a final example say we have a cylinder that is totally closed and filled up with liquid and rotated with respect to its axis. What is the total force on the top lead AB? Again the basic principle is the same. You just use this dp dr formula to find out how pressure varies with r. Of course with a reference say at r equal to 0 p equal to p0 with a reference because pressure you always calculate with a reference. So you know how p varies with r by integrating this with respect to r. When you integrate with respect to r g is fixed. So then you can find out the total force by taking an element. Here an element will be 2 pi r dr. So integrating over that you can find out the total force on the top surface. Total force on the bottom surface will be that plus the weight of the fluid which is there. So we will close this discussion by seeing just one example where we will see that how this type of vortex motion is generated in practice. You see that this type of a vortex motion that is there. I mean it is not exactly a parabola of revolution but it is by rotating the fluid in a container. Why it is not exactly a parabola of revolution is because we have neglected here the viscous effects. The shear between various fluid layers. We have assumed that the fluid rotates like a rigid body. In reality that is not the case and we will look into these situations more emphatically whenever we are discussing with viscous flows. So we close our discussion on fluid statics with this. Thank you very much.