So in this video, let's work on a special application. This is actually a little bit different from what we have just covered so far, like applying the Newton's law, F equals ma. This special cases are the case where you cannot really directly apply F equals ma because m is bearing. So two typical examples are steady mass flow and the variable mass cases. Like a steady flow, like a fluid system, with a control volume when we defining the control volume, and also the variable masses, a case where you skip losing the weight, losing the mass, or gaining the mass like a rocket and the chain problem. So instead of applying just F equals ma relationship, we would like to take a derivative of the linear momentum mv over t to obtain the corresponding relationship. Okay. As I said that earlier, like I'd like to apply Newton's second law to the fluidic system, such as turbine or pumps or the nozzle and so on. Since those flow keep changing its best flows, how can I formulate them? To do so, I'm going to define the control volume. Control volume is the volume where you are handling the inlet of the mass coming in and coming out in all those environment, that's control volume. Then when the mass of the coming in and coming out are balanced, which means the net mass of the control volume does not change anymore, that's why we call it as a steady flow. So we are going to apply the Newton's second law to describe the motion of the steady flow. So first, let free the body. What are the forces applied to it? So maybe to maintain this control volume, keep holding all those flow coming in and all those fluid coming out. You should actually hold this. So there should be external force, should be applied, and there is a inlet outlet static pressure and the gravity resist. Then we apply all the net forces are generating acceleration, mass times acceleration. Instead of applying mass times acceleration, we would use the linear momentum change. Linear momentum by definition is delta mv, and we handled the case where m is a finite, like a particle, but in that case it's a fluidic motion. So there are some bests flows in and out. So that's what I would call as a delta m, multiply by the v2 outlet and v1 inlet. So time derivative of the linear momentum will be formulated like this. So here if I defining delta m over delta t time derivative as m prime, which call it as a mass flow rate, my Newton's second law of formulation is written as m prime delta v. So instead of having F equals ma, I would have m prime, delta v. All the time derivative components are applied to the mass instead of the v. Similar to the linear motion, linear momentum case, angular momentum case, I would apply the similar manner. So angular momentum is r cross mv, but instead of constant as the mass, I would have it as a variable like a deviation, delta m, so that I have a r2, v2 and r1, v1 cross-product. If I have a time derivative of H angular momentum to equate with the moment applied, I would have m prime multiplied by the r1, v1 and r2, v2. Well, let's solve the problem. There's a high velocity water jet where the abrasive mixture is cutting the metal. The metal has a thickness and the nozzle has an area A, and the flow rate is Q, given Q, and the outlet speed of the fluid is v naught. The density of the mixture is rho. Then when it cuts the metal, the released water velocity at the bottom is going to be reduced because it loses the energy. It's going to be the v. In that case, what will be the horizontal force that is required to make this happen, cutting happen? Because it's a really high pressure water jet, without holding it, you can't cut the metal. So that's the problem asks us to find. Since the velocity changes over the cutting process is given and you are supposed to find the force, I guess maybe applying the so F equals ma would solve the problem. So let's recall how we can formulate the F equals ma, Newton's Second Law in steady-flow. First, you should define the control volume and there are the mass flow coming in and out, keep coming in and out, that's delta m, all the forces applied on it is the friction force, normal force, gravity, and the external forces, and those static pressure forces could be ignored. Then your net force will generate m prime delta v. What's m prime here? M prime is delta m over delta t, which is rho A, area of the nozzle in the distance traveled and divide by the t, that's the velocity. Initially, the components given, like you have a flow rate is given which is Av naught, and since you know the density of the mixture, you can calculate what's going to be the m prime and also v naught. So you can obtain the equations of motion in x direction, horizontal, and the vertical direction, all the net forces on the left-hand side and the momentum change, now expressed by the m prime v2 minus v1, in the horizontal and vertical components. By solving this, you could find out what's going to be the velocity at the outlet and you can figure that out, what's going to be the force, horizontal force, holding force required for the cutting. So in this video we learned how we could apply the F equals ma to the special case like a steady mass flow. In this case, we define the control volume. So instead of applying ma, we had a mass flow rate m prime and formulated in terms of that term. Next time, we are going to study about the variable mass.