 In this session we are also involving some fluid kinematics, that is when the fluid container is accelerated in the horizontal or vertical direction. Let us see what are the outcomes. At the end of this session, learner will be able to calculate force on the curved surface, analyze liquid subjected to constant horizontal acceleration, analyze liquid subjected to constant vertical acceleration, and analyze liquid subjected to combined acceleration. Now these are various important applications wherein we have to apply the equations of fluid mechanics or fluid kinematics that we are going to see in detail. Now in the last session we have seen that when we calculate the force exerted on the plate which is kept vertical, horizontal or inclined. And we have made it clear that the force is always perpendicular to the surface because in fluid statics normal force is the pressure per unit area. Now here many applications are there in which we have a curved surface. Now let us see the geometry. Here we have a curved surface and if I consider this particular part of the curved surface where I want to find out what is the force exerted. This may be a gate or this may be some tank section in which the liquid is present. Then what happens? We know that if the liquid level is up to this point then the total force exerted will be continuously increasing from top to bottom because pressure increases gradually. Now the question is how to find out exactly the force exerted on this curved portion. So there is very simple tool that we employ. We know that suppose I assume that this part as a curve, this part as a curve. Then what I do? Whatever the fluid is present in front of this, in front of this. So I will have some pressure along this line and if I take the projection of this, so this is the projection of this surface. So what we are interested in? We are interested in finding out the force exerted on this projected area, on this projected area represents the x component. So this represents fx and the force represented on this component which is the projected area of this part. This represents the fy component. Now this fx and fy, they are always mutually perpendicular, fx and fy are mutually perpendicular. So I can find the resultant force coming is fx square plus fy square. And if I want to find out at what angle it will be operating, I know that it is tan theta is equal to fy by fx. So it is a fairly simple calculation. So what we require? We must know that how to find out the pressure that is gradually increasing as we move. I will just illustrate a simple case for you. Suppose I have got this container in which I put a plate like this and plate is touching the water surface here. So pressure here is atmospheric. Then gradually I move down. The pressure goes on increasing. So pressure here is p0 and here it is p0 plus rho gh. Forget about p0. So what has happened? Pressure has changed from 0 to rho gh. So what is the average pressure? It is rho gh by 2. So the average pressure is this much. And this pressure will act at the center of pressure. Now this average pressure which is acting, I must evaluate this pressure for each geometry. For this particular I find out what is the pressure on this particular part? What is the pressure on this particular part? So by doing this I can calculate fx and fy. I am not talking about the width available for this channel in the other direction. You see in the assignments and numerical examples during our model courses. Right now we must know that the pressure at the top is 0. Pressure at the bottom that is at the height h is equal to rho gh. Pressure goes on increasing from 0 to rho gh. And if I find out the height of this, it is 1 half base into height. So base is h and height is 1 half rho gh. So it is rho gh square by 2. This is my total pressure force that is coming. Now this is the thing that we use in our calculations. Now we go to the next part that is when fluid is uniformly accelerated. Now first of all we will see exactly what is the need of this particular type. There are many containers. Suppose this is a container and which is moving with certain acceleration in this direction. Now as a simple example, I know that you know that suppose this is a liquid present here and a smooth table is present. If you push it, if you push the glass, glass moves in the forward direction and the liquid level becomes like this. If there is more acceleration, it may happen that water may spill down. So in many situations we have to find out that exactly how much height of the container should be so that if the vehicle is accelerated, water will not come out from the other edges. And for that we must know that what is the relation between the acceleration, between the gravity and between the theta that the angle that it makes. And finally this will be determined by this particular angle theta it makes. Now one important thing about this geometry is that when we have in this situation g is here, in this situation g effective is here, g is still here, acceleration is here because whenever we have acceleration in the forward direction even if I want to apply Newtonian mechanics and when the frame is accelerating or when I am working in accelerating say frame, I cannot apply Newton's law directly. I have to use D'Alembert's construction or I have to include the pseudo force that is acting opposite to the direction of the motion of the body. So if the body is accelerating in this direction, I have to show the acceleration here on the other side. This is my g and this is my g effective. So you will see that g effective is the direction which is always perpendicular to the fluid surface. Means whenever I see any container suppose it is moving in the right direction and suppose it is like this then immediately I can guess that g effective is in this direction. Means g is here and I can find out the acceleration here so that net resultant is going to be this one. So the g effective will decide the level or the angle made by the fluid surface with the horizontal. Now this is same as the problem that we encounter in our mechanics. Suppose in a car or in a vehicle we have a pendulum. Now what happens when the pendulum is there when we move in the forward direction pendulum starts moving in the backward direction because of the pseudo force that is acting on it and we can find out the effective acceleration there. That effective acceleration is going to give you the angle. So this tan theta is going to be defined by tan theta is equal to a by g. Now we will see how it comes as a by g. Now let us see these two figures combined together you will get it. Now suppose I consider say element mass where f1 is acting on the left side f2 is acting on the right side. Element length is given as L. Then what happens f1 minus f2 the net force f1 minus f2 is equal to mass into acceleration. This is my condition. Then height of this part is y1 height of this part is y2. So when this is y1 and this is y2 then what happens force p1 into area minus p2 into area is equal to m into a. Now after all this p1 is nothing but it is rho g into y1 this is rho g into y2 is equal to m a. So I will get y1 minus y2 upon L is equal to rho g instead of rho g I will write here a L rho a. So this L will get cancelled it will not be present. So I will get y1 minus y2 is equal to this one and finally I will get by this is equal to a by g. But from the geometry you can see y1 minus y2 is equal to upon L is equal to tan theta. So I will get tan theta is equal to a by g. Now this is the geometry that we use for doing the calculations. Now if you see it is like this. Theoretically speaking my liquid surface will behave like a simple pendulum. Now where g is in this direction g is in this direction and the fluid is moving in the forward direction. Now next important thing about the fluid is when fluid is subjected to constant vertical acceleration. Now this is the mass of the element that is the fluid that I have considered and acceleration is in the upward direction. Now what happens this is the pressure p1 into area this is pressure p2 into area this is the force mg is acting in normal direction and ma is in the upward direction. So I will get p2 minus p1 is equal to rho into g effective into h. So finally what has happened? Formulas remain same as we do in mechanics when there is a weightlessness and weight gain feeling. When we come down weightlessness when we come go up we feel that we are gaining the weight in the lift. So the same thing happens and g effective we have to evaluate. Now what is this g effective? g effective may be g plus a or g minus a depending upon whether we are moving up and down. So if we are moving up it is g plus a when we are coming down it is g minus a which is going to be g effective and it is g effective. Now many times what happens we have both the accelerations either fluid is moving in the horizontal direction or fluid may be moving in the vertical direction. Now in this situation if I say that ax and ay are the two components of acceleration in x and y direction g is going to be there and g is always going to be vertical. So if I see my geometry I will see that g and ay are going to be downward. So g plus ay are going to be in the downward direction. So as it is ax in this direction to compensate for that I have to say I must get some effective gravity that is g effective. So this is my ax. So when it is moving in this direction I take care of a pseudo acceleration or pseudo force and I get this g effective. So for this geometry tan theta will be equal to ax upon g plus ay. So if you use this equation whenever your object is moving in a horizontal direction with acceleration ax at the same time it is moving in the vertical direction with acceleration ay and acceleration g is going to be there. The angle made by this will be given by ax upon g plus ay. If there is no vertical motion it will be ax upon g already we have proved in the previous situation. Now another case whenever I have a container completely filled with liquid means this container is closed. There is no open end it is closed. It is a closed container and in that container suppose I want to find out what is the pressure at this point. Now you know that when the vehicle is accelerated in the forward direction here the liquid will try to make its shape perpendicular to g effective. So there is some point here where the pressure is going to be zero. You can visualize this suppose you completely fill completely fill this particular with liquid and give some acceleration. What happens the fluid will compress here and it will form the layers like this. So here the pressure is going to be zero. Now if I want to find out the pressure at this particular point I can calculate this pressure plus this pressure plus this pressure. Now how to evaluate this? Now this height is h this is l. Now this height is h and it is in the direction of g. So pressure will increase in the direction of g as I told you. So pressure here is zero it will increase in the direction of g up to this level so that is h. So pressure will be zero plus rho g h. Now if you look at this point for this particular direction from a if I view my vehicle is moving in this direction. What is my acceleration in the opposite direction? So for me the acceleration is in this direction and what is acceleration it is a. So when the acceleration is a in the downward direction that is in the left direction. I will say that as by my rule force pressure is equal to rho g h. So I will go for rho g into a rho g into g instead of g I will use a and instead of h I will use l. So my pressure will be rho l a rho g h plus zero is equal to pa. So by this we can calculate the pressure at a is equal to this plus this pressure. And this is a very classic example of calculating the total pressure of fluid at this one. So with this major part of the fluid statics we have covered. In next sessions if I prepare it we will discuss about Archimedes principle and fluid kinematics and Bernoulli's equation. Now meanwhile some assignments will be put on the model and you can work on them. The reference book is once again by white by TMH publication that is fluid mechanics is title. Or you can go for cook that is also a good book that is to be referred. And to some students if they are interested in seeing the numerical examples or some conventional proofs they can say go for Bunsen. So with this I stop over here. Thank you.