 In this lecture we will integral form of the conservation of linear momentum. To do that we will start with the Reynolds transport theorem general expression which is valid for any conservation. Now if we want to conserve the linear momentum then what should be this n? N is the mass into the velocity that is the linear momentum. So we can write the left hand side as ddT of mv of the system okay. So when we write ddT of mv for the system by Newton's second law of motion this is the resultant force which is acting on the system. We have to keep in mind that we have derived this expression with a limit as delta t tends to 0. In the limit as delta t tends to 0 system tends to the control volume that we have to keep in mind and therefore in that limit this also tends to the resultant force acting on the control volume. So the left hand side essentially becomes the resultant force which is acting on the control volume and the right hand side we can express by writing what is small n? Small n is capital N per unit mass. So this is V and this is also V. Very important observation this V in a reference frame which is stationary. We have considered a stationary reference frame here. We will see later on that it is not necessary to have a stationary reference frame. We may have a moving reference frame not only that a reference frame moving arbitrarily with arbitrary rotation as well. So a non-inertial reference frame what happens to the statement of the Reynolds transport theorem in a non-inertial reference frame that we will also see. But here we are considering inertial and a special case of inertial which is stationary. So here this V is the absolute velocity but even in this particular term we are not trying to disturb this particular term because this particular term is going to be always this irrespective of the control volume moving or stationary or whatever because this is something which is giving rise to a net flow of mass across the control surface and that depends only on the relative velocity not the absolute velocity. So here it will be absolute velocity, here it will be relative velocity and that we will keep in mind but when it is a moving reference frame this will also change to relative velocity that we will see. So the statement of the Newton's second law here boils down to like a expression which is the resultant force acting on the control volume and see the power of the Reynolds transport theorem we never committed that n should be a scalar vector or whatever. So we have applied it for a vector and there is no restriction towards that. With this understanding on the form of the conservation of linear momentum for a stationary reference frame let us try to work out a problem. Let us say that you have a water jet which comes out of a tube or maybe a small pipe say there is a plate which is frictionless and oriented inclined to the jet. So this is a plate with a frictionless surface. So what happens to the water? The water extreme stream lines come like that and water leaves across the ends of the plate in this way. We have to find out if the volume flow rate here is q0 and here the volume flow rate is q1 and if the volume flow rate is q2 and let us say that the angle made by the plate with the vertical is theta which is the angle between the plate and the vertical. When we say vertical let us neglect the change in height between various points. So just to simplify the situation let us say that it is like vertical means a vertical line drawn in this plane the entire thing may be located in a horizontal arrangement or even if it is a vertical arrangement the plate height is so small that the change in elevation that the change in potential energy between various points is insignificant as compared to the other forms of energy like the kinetic energy. Now let us say that this area of cross section of this extreme stream line so these are like stream tubes these are called. So if you see that this is not a pipe basically. So this is a free water jet exposed to the atmosphere but the water jet takes a form like a tube. Its outer periphery takes the form of a tube and outer ones are the extreme stream lines. So the figure that is drawn here the lines represent the envelope of all the stream lines. So this is like this is called as a stream tube which engulfs all the stream lines the same is there for the exit ones. So what are the considerations that you now could apply for finding out so what is our objective? The objective is to find out q1 and q2 as a function of q0 and theta that is our objective. To do that one important thing that we should not forget that mass conservation is always applicable and we should try to apply that not that whenever we are trying to apply a momentum conservation mass conservation is insignificant not like that. So no matter whatever conservation we are applying again first identify a control volume. Let us say that this is the boundary or the surface of the control volume that we are looking for. You will see that like whenever we draw the boundary of a control volume or control surface we just draw an arbitrary dotted line which is not coincident with the surface. This is just for the clarity in representation. This is very very symbolic. It is not that if I have drawn it a bit far that means the air here is involved as a part of the control volume. So do not take it literally. Now when this control volume is drawn we are interested to find out write an expression first for the mass balance conservation of mass. Assume that the density of the fluid is a constant. Let us say that the areas of flow here are like here it is a2 and here it is a1. These are the parameters which we do not know because these are not within our direct control. What is in our direct control may be this area because this depends on the ejected jet from the tube or the pipe whatever nozzle but when it comes on the plate and moves this depends on the many other things. So directly we cannot say but one thing we can say is that the mass will be conserved. So if you write the conservation of mass by this time we have got habituated in writing for different cases. So we will not write the full Reynolds transport theorem form. We know which terms are important and which terms are not. Left hand side first term is 0 always because the system has a fixed mass. The second term if you have a control volume which is of a fixed volume and density has a constant that will be 0. So it will be only the rate of mass outflow-inflow. So what is the rate of mass outflow? So that is rho into q1 plus q2 is outflow-q0. So that is the outflow-inflow. So you can straight away write q1 plus q2 is equal to q0. Let us make another assumption that the velocity is uniform over each section. So you can write this as good as a1 v1 plus a2 v2 is equal to a0 v0 or q0 whatever equal to q0. Now there is a relationship between v1, v2 and v0. It is possible to have a relationship between v1, v2 and v0 but before coming into that relationship let us complete the exercise of looking into the linear momentum conservation. What linear momentum conservation gives us? See one very important thing is it is considered to be a frictionless surface. That means there is no shear force between the fluid and the plate in the direction tangential to the plate. So if you write a linear momentum conservation with the components say along x, x is like say tangential to the plate or maybe let us call it s to indicate that it is a tangential to the plate and maybe a direction n which is like normal to the plate but we are bothered about for the tangential to the plate. So tangential to the plate what is the situation? The situation is that there should not be any force on the water which is there in the control volume because it is a frictionless thing. No matter whether it is tangential or normal so let us write this f for the control volume as let us write it in the general vector form. So let us write the conservation of linear momentum f as fs into unit vector along x s plus fn into unit vector along n. This is the resultant force on the control volume. General form with we know that fs is 0 from the physical considerations of a frictionless plate. This is equal to for the first term we are assuming that it is a steady flow velocity is not a function of time. We are assuming a constant density, density is not a function of time and the volume of the control volume is not also a function of time. So all these 3 considerations lead to the fact that this term has to be 0. So when this term is 0 then you come to the second term. So what will be the second term? So for the second term you have so the integral is not important because the velocities are uniform over each areas. So we have to write this for the 3 surfaces 0, 1 and 2 just like that individually. Integral will eventually not have to be evaluated because of the uniformity of the velocity. So let us write this for the section 0. So for the section 0 you see what is v. So maybe let us again make a chart what we did for solving one of our earlier problems just for your convenience to begin with. So we have a section say 0 at which we are interested to find out what is v and what is the normal vector. The normal vector will be insignificant because eventually we are dealing with the flow rates. So if you consider the v to be uniform and rho to be uniform you see rho into v you can take out of the integral. If you take rho into v out of the integral then what remains within the integral is integral of vr.n da. That is what? That is the total volume flow rate over the section that you are considering. So it is as good as rho into velocity vector into the volume flow rate over the section with the proper sign plus or minus. So keeping that in mind we will not write this vector not that it is not existing but it is not important for solving our problem. It will just give a redundant exercise. We will put too much of effort which is not necessary but the velocity vector is important. So the velocity vector for this one like let us say that we have our original axis like this is x and this is y and the s axis is like this which makes an angle theta with y. Maybe you can also choose a n axis which is normal to that. So what is v for this one? v is say v0 i cap. Maybe instead of writing this normal vector let us write what is vr.n a or more fundamentally integral of vr.n da. So what is there that for the surface 0? So this is see vr and or v and this unit vector they are oriented in the same sense or opposite sense. This is the direction of v. This is the direction of the unit vector of this surface. So this will be what? Minus of q0. Then for the surface one so you are having a flow for the surface one in this direction which is having a velocity of v1. So that may be resolved along x and y. So v1 will have components what? So v1 sin theta i plus v1 cos theta j. What is v.n da here? It is q1 because the flow direction and the unit vector out of the area they are oriented in the same sense. For the surface 2 what is v? Magnitude is v2. So the velocity is like this now. This is v2. So it will have some component along x and some component along y. What is the component along x? Minus v2 sin theta i and minus v2 cos theta j. What is this one? q2. So this is very straight forward but I want to emphasize on this because this is a place where students make mistake many times because by default it is a common thing that the outflow boundary gives rise to a positive flow rate. But the velocity that comes with it it has nothing to do with the flow rate. So that you have to write in proper vector sense. So this term plus or minus has no conflict with the flow rate term that you should keep in mind. Otherwise just by inertia of writing you might write this in a similar sense as we have written this. So let us now substitute that. So if you substitute that let us say that we are interested about writing the expression of the flow rate term. In terms of the tangential and the normal components. Why we want to do that? Because we are interested about equating this tangential force to 0. So that we want to exploit. So in that case these coordinates are not good coordinates for us. This I just wrote for a practice of showing that how to write different terms. But more convenient will be to use the s and the n coordinate. So in terms of the s and n coordinate if you want to write the good thing is the flow rate does not change. It is independent of the coordinate type. But what changes is now the expressions for the velocity vectors that you are looking for. So let us write these expressions now in terms of the s and n vector which will be useful for solving this problem. So in terms of the velocity at section 0. So you have a velocity at section 0 like this you have v0. You have a coordinate s like this and a coordinate perpendicular to that as n and with this one it makes an angle 90 degree-theta. So you can resolve v0 into components along s and n. So typically how the resolution will look like maybe you have this along s this along n. So the resultant is this one. So what will be the component along s? So v0 sin theta-v0 cos theta epsilon n. This is for the surface 0. Then for the surface 1 v1 epsilon s for the surface 2-v2 epsilon s okay. Let us put that here. So if you put that then it will become first we put for the surface 1. So rho into v, v is v1-v1-v1-v1-v1-v1-v1. So v0 sin theta epsilon s-v0 cos theta epsilon n into-q0. Then for the surface 1-rho v1 epsilon s-plus q1. For the next one-rho v2 epsilon s-multiplied. So if you extract the tangential component and keep in mind that because of the frictionless case this is equal to 0. You are left with that. So if you equate the epsilon s components you will get 0 equal to rho-rho v0 q0 sin theta-plus rho q1-v1-v1-v1-v1-v1-v1-v1-v1. So again see the deceiving thing as if this minus sign will is giving inflow but this is a minus but this is outflow. So the origin of the plus and minus has to be clear. It is not because of outflow or inflow the plus or minus that has come as a combination of outflow and the direction of the velocity vector. So rho being a constant you can write like from that equation q1 v1-q2 v2 is equal to q0 v0 sin theta. Now the question is what is the relationship between say v0 v1 and v2. Let us say somehow you could measure both a1 and a2. So somehow you could measure. Now let us try to see that how that measurement could give us a good picture. Let us say you know that what is v0 you are interested to find out what is v1. So what you will be tempted to do? If you know what is v0 your objective is to know what is v1 from that. What intuitively you try to use? Let us consider a streamline that goes from the section 0 to the section 1 because it is a uniform velocity like if you take a point 0 here and a point 1 here. These are representatives of the velocities as good as at section entire section 0 and entire section 1 because it is a uniform velocity at each section that we have assumed. Now if you apply the Bernoulli's equation between any 2 points on the streamline. So let us try to apply that and see that what new information we get. So the first information that we get is first of all let us write it and then the information will be clear. We assume that both 0 and 1 are atmospheric and it is true. It is not an assumption because it is a free jet in the atmosphere. This is also a free liquid film being passed on to the atmosphere. So in both places sections 0 and 1 it is exposed to atmospheric pressure and because the height difference is not that large atmospheric pressure will not change substantially. So these 2 are equal and because the height difference is not appreciable as we assumed earlier. So the g0 and g1 are approximately the same or the difference is negligible. So from this we can conclude that at least in an approximate sense v0 is same as v1 and with a similar consideration you can say that v0 similarly you can say v0 is approximately same as v2. So in this expression where you have v1, v2 and v0 you can cancel that from both sides. So what you get? You get q1-q2 is equal to q0 sin theta. So let us see we get 2 expressions involving q1 and q2 in terms of q0 and theta. So you can solve for q1 and q2 individually as a function of theta from these 2 very simple. Important thing is it does not require any information on what are the functions of the areas of these liquid films at 1 and 2. So it does not require what is a1 and what is a2. There could be a very interesting observation. Now you find out what is the normal component of the force. Normal component of the force definitely is not 0 and the normal component of force will come out if you see that because of the term rho v0 q0 cos theta. So because of the normal component of the force, this is the force on what? This we have to keep in mind. Let us write it. So fn is –rho v0 sorry not –rho v0 q0 cos theta. q0 you can write in terms of a0 and v0. So it is like rho a0 v0 square cos theta. So this is the force on what? These are very subtle points. See always it is not here the answer ends. We have found out the normal component of force. We must ask ourselves that is force on what? Force exerted by whom on whom? If we are not clear about that then what is the good in finding out a mathematical expression. So this is we have to keep in mind that this is the force on the control volume. So if you have drawn a free body diagram of the control volume if you would have done then what would have been the forces acting on that control volume that would have shown in the free body diagram. Let us say that this could be your control volume. So you could have a force in the normal direction say fs you could have a force sorry in the tangential direction fs and normal direction fn. Assume that the same thing is occurring almost in a horizontal plane. So the weight you are not considering in this plane. Not that weight is not there but in this plane of forces weight is not important. Then any other force on this? Yes. Think of the fundamental considerations of forces in continuum mechanics that we discussed. You have surface force and body force. So body force we have considered the weight just not shown in this plane because it is not significantly there in this plane. Surface force. Pressure distribution is there. So there is a surface force from all sides normal to the areas whatever. Have we considered the contribution of this? We have not considered. But magically we are expecting that it is giving us the correct information on forces. And this magical thing is possible because if you have a closed contour over which you have a uniform pressure distribution then irrespective of the shape of the contour the resultant force is 0. Because of a very special nature that pressure locally is always normal to the direction of the surface. So no matter how bad the shape of the contour of the surface is so long as it is closed and if you have a uniform pressure throughout then the resultant force due to that will be 0. This is a very important thing and you should be able to derive it by yourself. If you are not able to do that you let me know later on I will try to help you out but you should be able to do it. So if that is the case then if there is a variation of pressure only that will matter to a resultant force. But if throughout it is atmospheric then it does not matter. Therefore our negligence of the pressure in this case has not mattered because it is a uniform pressure. That is why you will see that what is important for us is not what is the atmospheric pressure but the deviation from the atmospheric pressure that only can give rise to a local force. So always when we consider forces here due to pressure we consider the gauge pressures at different sections because the deviation from atmospheric pressure is what is interesting for us. Atmospheric pressure if it was the sole feature it would have existed throughout equally and it would have given rise to a 0 net force. So in terms of giving rise to a net force any deviation from atmospheric pressure is important. So the gauge pressure is the important quantity. Now here of course the pressure contribution does not come so from fs and fn these two are there. Question is that what should be fs and sorry on which this fs is acting and it is exerted by whom on whom and on which fn is acting. fs is 0 we have seen and that gives a clue that it is between whom and whom right. When we said fs equal to 0 we were thinking about an interaction between the plate and the So, fn also should be a consequence of interaction between the plate and the water. So it has 2 possibilities, what force exerted by plate on the water or exerted by water on the plate, yes, there should be same but same in magnitude not same in sense. So whatever answer we have got here, this has a particular sense. So it represents only a unique thing. So when we have said fn epsilon a, it has a proper sense of the vector epsilon a. So this is force exerted by water on the plate or plate on the water. It is force exerted by plate on the water because it is force exerted on the control volume and what is there in the control volume is not plate but water. So this is force exerted by the plate on the water present in the control volume. By Newton's third law, you will be having the force exerted by the water on the plate as minus of this. So the force exerted by water on the plate will be opposite to the positive n direction. So force exerted by water on the plate will be in this direction. This is you see, this is in the positive n direction that we have considered. So opposite to that is the force exerted by water on the plate and that is quite obvious. Logically if you see just jet strikes on the plate like this, it should have a force in this direction not the opposite direction. So this there is a resultant force which is exerted on the plate in its normal direction. So this force might be good enough to make the plate move and the plate should be adequately supported to prevent such movement. In general if that jet is falling on the plate, the plate will also move because of this normal force and later on we will encounter such problems. We will try to see that if the object on which the jet is striking or which is interacting with the jet in any form, if that itself is moving then how we adapt our analysis of momentum conservation. So this is like a very simple problem but it gives a lot of important concepts that we should keep in mind. Let us try to solve another problem. So in the next problem we will revisit the case of a boundary layer on a flat plate that we discussed quite some time back when we were discussing about viscosity. So what is the situation that you have a flat plate like this? Fluid is coming from far stream with a uniform velocity say u infinity. This fluid is falling on the plate and because of its viscous interactions the effect of viscosity within the plate. So the within the fluid the first thing is the plate tries to slow down the fluid which is in contact with that and that effect is propagated to the outer fluid through the viscosity of the fluid. So there is a thin layer close to the wall where this effect of the plate is felt and the velocity gradient is created and when you go outside that layer the velocity gradient is no more there so outside that layer the fluid does not actually feel the effect of the plate so to say and that thin layer is known as the boundary layer. So if you have a boundary layer growing on the plate so you have a velocity profile at different layers like this. This we have discussed in some physical details without going into some mathematics but at least the physical details when we were discussing about the no slip boundary condition and the viscosity. So let us say that at the edge of the plate you are given the velocity profile. What is the velocity profile that is given? Let us say that locally the thickness of the boundary layer is given by delta which is a function of x. So here you have let us say the value of this delta is the boundary layer thickness at this location where the x is l so x axis is oriented along the free stream direction. Let us say this value of delta is delta 0 and the velocity profile is given. Let us say it is given by u by u infinity is equal to y by delta-1.5. Let us say this is given what is the great sanctity of these types of velocity profiles or whether it could be different or not that we will discuss in details later on in one of our chapters known as boundary layer theory. But right now let us say that this is something which is given to you. Now what is your interest? Your interest is to find out what is the net drag force exerted by the plate on the fluid. This is a physically interesting parameter because of the viscous effect there is a drag force that is there that is the plate tries to slow down the fluid and our objective is to find out what is that net force acting over this length l. So to find out the force we may understand that we might require a linear momentum conservation because the force is involved and when we have a linear momentum conservation the mass part of that should also be conserved so we should also be consistent with the mass conservation. So it depends like we should choose a control volume. Let us first choose a control volume which is intuitive and try to solve the problem and then see we will try to investigate whether that intuitive choice of the control volume is good or bad. So let us say that we choose a so this black line what is there it is the edge of the boundary layer what is the edge of the boundary layer we know that within this layer only the velocity profile is there beyond this the velocity is uniform equal to u infinity almost. So if you want to take a control volume let us say we choose a control volume like this just engulfing the edge of the boundary layer control volume. With respect to this control volume let us write the first the conservation of mass let us give some names of the edges of the control volume let us say a b and then this c. So if you write the conservation of mass what we are getting integral form left hand side first term is 0 because we are talking about a identified system of fixed mass. Right hand side first term let us assume again it is a rho equal to constant and this is not a deformable control volume. So right hand side first term is 0. Then the next thing is basically what basically the net mass flow rate. So you have 3 surfaces here clearly across a b there is no flow because of what no penetration. So because with across a b there is no flow but across b c there is some flow what is the flow across b c so integral of so rho is coming out of the integral now we will directly write the integral because we have experienced how to write that in earlier examples. So rho this is the velocity profile so the velocity and the outward normal is in the same direction so dot product will give something positive. So we just write the scalar form so if we take at a distance y from the bottom if we take a strip of width dy what is the dA dA is dy into the width of the plate let us say b is the width of the plate. So rho into u into dy into b y is integrated from 0 to delta not okay so and u is a function of y and then there is a flow through ac right. So when there is a flow through ac let us not try to write it in a complicated way let us say let us call it m dot ac okay. So m dot ac this is algebraic so it may be plus or minus here you can see that m dot ac is equal to minus. So what it appears that see this is such a control volume there is nothing in the left so to say across one surface something is leaving to make a balance of that as if some mass is coming from outside across the edge of the boundary layer to compensate for that and that is given by this one. So it is like a m dot entering the control volume. Now conservation of linear momentum simply conservation of momentum till now we have only talked about linear momentum so let us write what is the x component of the force because see the drag force should be oriented along the direction of the relative velocity between the free stream and the plate so that is oriented along x so the force might have some component but let us write only the x component so fx is equal to the right hand side the time derivative term is 0 we are writing the Reynolds transport theorem for linear momentum conservation. Then next term look into that next term basically we are writing this. So the next term so that has contributions again for bc and ac so for bc let us first write what is for bc so for bc you see we are writing only the x component so we will just keep in mind that we will write only the x component of the velocity vector. So when we write the x component of the velocity vector what should be here first we are writing for the surface bc so integral of rho see this v dot n this is like a scalar it does not give a directionality so the directionality the direction for which you are looking for the force should come from the component of the velocity vector that you are taking. So we should take u and here u is anyway the only component so rho not that u is the only component u is the only important component because you can clearly see that this delta is a function of x this delta is varying with x so if you differentiate u partially with respect to x you will get that as not equal to 0 because it appears not to be a function of x but implicitly it is a function of x because delta is a function of x this being not equal to 0 the continuity equation for incompressible flow the corresponding the other term this is also not equal to 0 and there because v is 0 at y equal to 0 and it is gradient is not equal to 0 so you have a v not that you do not have a y component of velocity from the continuity equation given this you can find out what is the y component of velocity how it is varying but it is it is much smaller than the x component of velocity but here when you are writing the x component of force only the x component of velocity anyway is important so rho u and then again u v dy from y equal to 0 to delta not then you also have for ac so for ac it will be like if you take this rho as uniform sorry this v as uniform and take out of the integral then the remaining term inside over ac is the mass flow rate across ac so it is nothing but plus u over ac into mass flow rate over ac what is u over ac u infinity because ac is the edge of the boundary layer at which u becomes equal to u infinity so this is u infinity okay so you can substitute here fx yes now m dot ac you can substitute so it becomes rho rho is a constant you can take it out so 0 to delta not then again b is a constant so u square dy minus u infinity into u dy you may write it in a more compact form that is in a single integral u square minus u infinity dy from 0 to delta not let us just look into that what could be an alternative choice of the control volume may be to solve the problem a bit more elegantly not that it is too dirty but one could even like do it in a bit more elegant manner one interesting thing you observe from here that there is only x component of velocity here and y component of velocity in the free stream is 0 but there is some flow across the surface ac now could we choose a surface ac or surface of similar type such that there is no flow across it so let us try to draw a separate sketch with an alternative choice of the control volume so alternative choice of control volume so for alternative choice of the control volume let us say that again we have the velocity profile and everything you have this ab we want to choose a line from here see such that there is no flow across that line we have seen there is a flow across the edge of the boundary layer so what should be that line that we choose here so that there is no flow across it yes through what type of line you have no flow across through a streamline so if you choose a streamline which is passing through the point C this is a streamline then we know that there is no flow across it but it gives rise to a new unknown because we do not know that where that streamline is intersecting with this one let us say h0 this we do not know so now let us make a new control volume say abcd where we have tried to get rid of a problem by considering the streamline when there is no flow across it but we have a new flow boundary ad so with for that new control volume let us write the conservation of mass so let us quickly write the conservation of mass and conservation of linear momentum for the new control volume and let us try to see whether it makes us converse to the same answer so conservation of mass first you let us for ab it is 0 mass flow for bc we have seen what is that integral of rho u dyb from 0 to delta not this is for bc for cd it is 0 because it is a streamline there is no flow across it but there is something for ad so what is that for ad plus whether it is plus or minus so the normal direction is opposite to the flow direction so with a minus sign then integral of u infinity into dy into b from 0 to h0 u infinity is a constant so from here you can clearly find out what is h0 so h0 is equal to integral of u dy divided by from 0 to delta divided by u infinity then conservation of linear momentum along x fx equal to again only the last term will be there so for bc you have rho integral of u square dy into b then for cd for cd you do not have a mass flow rate so the linear momentum flow is 0 because you do not have any mass flow rate across that so if you take this v out of the integral because it is uniform you see the remaining is the mass flow rate over the area that you are considering and the mass flow rate is 0 across a streamline so for cd it will not come but again for da it will come so for da what will be that what will be that for da plus integral of rho you take out first x component of velocity that is u infinity then v dot n da that is minus u infinity dy b from this one sorry 0 to h0 whatever so what will be our fx let us just write it clearly so fx is equal to rho integral of u square dy b from 0 to delta naught minus rho b u infinity square so yes u infinity square into h0 so that if you substitute the expression for h0 it is becoming u square dy rho b you can take as common minus u infinity integral u dy 0 to this delta naught you can clearly see that you get the same expression back so 2 different choices of the control volumes are giving back the same expression and to me the alternative choice of control volume is not bad because it gives a better visualization of what is happening because this gives a direct visualization that something is entering here and something is leaving and these 2 are not participating for the case of the edge of the boundary layer it is physically not that intuitive mathematically it is not that difficult straight forward but the better physical picture is being provided by this control volume but both are fine and the remaining part is easy you may substitute u as a function of y and integrate because u as a function of y is given to find out the expression again this is the force exerted by what on what this is the force exerted by the plate on the control volume and you can you make out that whether it should be positive or negative you can clearly see that here you have u square and u is less than u infinity and here is like u into u infinity so here u is multiplied with a number which is greater than u itself so this term must be less than the second term so intuitively this should come as negative and that is what is obvious because it is a drag force so it tries to slow down the motion of the fluid so this is something which is a physical understanding that is important once you solve a problem you get a sign out of the problem and you should develop an intuition of what that sign implies let us stop here today and we will continue again in the next class thank you.