 Okay, we're now going to work the second half of our example problem involving a control volume moving at constant velocity. And so if you recall, we were looking at a case where we had a water jet impinging upon a cart with a blade on it and a blade was rotating or turning the angle of the jet of water impinging on it. We came up with a control volume that was moving. And what we did in the last segment at the end, we ended with the continuity equation and we expressed it in this format here and kind of left area one, area two undefined because we weren't sure how the area of the fluid crossing the control surface was changing. This being A1 here and this being A2 there. So what we're going to do now is we're going to use our second equation for control volume analysis and that is the linear momentum equation. However, here we have linear momentum with a control volume moving at constant velocity. So that's what we're going to do in this segment. We'll begin with the x momentum equation. Okay, so we have that for the x momentum equation. A couple of things we can note right off the beginning. First of all, we were told to neglect body forces. That goes away. And we were also told that the flow is steady and consequently that term goes away. Do we have a surface force? Yeah, sure we do. That's the force that is causing the control volume to move at constant velocity. So we do have a surface force in the x. So let's go on and continue working with that. Okay, so that's what we get applying linear momentum. Remember I've used the convention of the signs depending upon the way that the flow is going across the control surface. For a control surface, one, we have the area vector this way, the velocity that way, so we have a negative. For a control surface, two, we have the area in that direction and the velocity so it's a positive. So that is dealing with essentially the mass flux term coming across. And I've also put in here the u component of velocity because that will be the x component of velocity, both the inlet control surface one and the x control surface two. So what we need to do now, we need to figure out what those velocities are for the fluid coming into and out of our control surface. So for that, what we'll do is we will write out the magnitude of the velocity. Okay, so when we go through this, what we find is that the magnitude of the velocity on the inlet and on the exit control surface is v-u. The significance of that, if you recall back to the end of the last segment, we were dealing with continuity and we came up with this expression here. So if we plug those values into this expression, what we find is A1 is equal to A2. So the surface area one is equal to surface area two for our control surfaces. So we can then write A1 equals A2 from continuity. And we will use that again as we move on through the analysis. So let's come back to this equation for the reaction force. And here what I'm going to do, I'm going to make the substitution of A1 for A2. So we get that for a reaction force in the x direction. We can now plug in values and solve for it. So the reaction force in the x direction comes out at 600 Newtons to the left. Now let's look at the y component of the linear momentum equation in order to solve for Ry. Now with this again, steady flow, neglect body forces. So notice that we do not have a component on surface control surface one and that is because their control surface one is in this direction and consequently there is no component in the y direction and that's why we don't have to worry about it. So we can plug in the values that we computed earlier. Again the mass flux term is going to be positive and that could be either A2 or A1 if they're interchangeable. So what we can do now, we can plug in the values and we get 1039 Newtons in the vertical direction. So with that what we can do, we can conclude that in order to maintain a constant velocity on the bucket or the blade that's on wheels, you need to exert a force of the following and this is in Newton. So when you exert that kind of a force onto your bucket where we had the problem that we had this coming, that this was our blade and then we had this and it was on wheels and it was moving and the jet was going up in that way. You need to put a force that you can see. It's actually a force that is going, let's see, we have a component that is in the x direction but it's going in this direction so like that and then we have a positive for the y so it's like that. So kind of a force in this direction I guess that's kind of a crude way of doing it but if you do that, that would give us the positive y and the negative x and so that would be rx and ry. If you put a force like that on your bucket it would then move at a constant velocity given the conditions that we have for this particular problem. So that gives you an example of how to apply control volume analysis to a moving blade and it's moving at constant velocity. What we're going to do in the remainder of this lecture is we're going to take a look at a case of where the control volume is accelerating but it's accelerating just through pure translation but it's a little more complex than what we've looked at here but that will be in the next couple of segments. We'll be looking at an accelerating control volume.