 Consider a moving plate moving parallel to a stationary plate at a distance of 20 mm with SAE 30-weight oil at 20°C, which is a viscous fluid between them. If the moving plate has a velocity of 3 m per second, determine the drag force acting on 0.2 m2. So for us to talk about the drag force, we are going to be describing the shear stress and recognizing that that is going to be the force acting per unit area. Therefore, to answer the question, we are going to need to calculate the shear stress on the fluid and multiply it by the area, which is 0.2 m2. If we assume that the fluid is a Newtonian fluid, then we can say that the shear stress is going to vary linearly with shear rate. And the slope of that linear line we call dynamic viscosity. Therefore, our force is going to be the dynamic viscosity multiplied by the area, multiplied by the derivative of the x-component of the velocity of the fluid with respect to y. So note in this diagram that y is positive going up from our perspective and the x-component of velocity, u, is moving to the right. So u as a function of y is referring to how this velocity changes the further up you go. I am observing that it is a linear line that is 0 at a y-position of 0 and it is v at a y-position of h. So at 0, the u-component of velocity is 0 and at h it is v. Why v? Why couldn't it have been something else to try to limit our confusion with these variables? I don't know. And then because it's linear, I can say that the velocity at any point is going to be the proportion of the height between 0 and h. You are multiplied by v. So y over h multiplied by v. So if you're halfway up, that is half of h over h, then your velocity is half of v. Does that make sense? So this is the function we are going to need to take the derivative of down here. So I'm saying u as a function of y is y over h times big v. And for convenience, I can write that as v over h times y. Therefore, du dy would just be the constant in front of y, which is v over h. So my drag force here is going to be viscosity multiplied by area, multiplied by velocity over h. I know v, that was given in the problem. It is 3 meters per second. I know h, it was 20 millimeters. I know a, that was 0.2 square meters. So the only unknown is the dynamic viscosity of SAE 30 weight oil at 20 degrees Celsius. And for that, we are going to go into our appendices. So we want table A3, which is going to have properties of some common liquids at 20 degrees Celsius conveniently. And for SAE 30 weight oil, the dynamic viscosity is 2.9 times 10 to the negative first kilograms per meter second. Now it's time to just plug everything in. The problem does not specify a unit for the force that it wants. So let's just default to Newtons. A Newton is a kilogram meter per second squared, starting at my destination and working backwards into primary dimensions. So far, seconds and seconds cancel second squared, kilogram cancels kilogram, meter squared cancels meters and meters. So to finish up, I just need to convert millimeters to meters and then I can cancel millimeters and meters. That will give me an answer in Newtons. So if our calculator will deign to help us out, you can say 2.9 e to the negative first, multiplied by 0.2, multiplied by 3, multiplied by 1000, multiplied by 1 over 20. Why I'm not just dividing by 20? I don't know. Really, I'm gonna be different. Anyway, the force is 8.7 Newtons. So there are 8.7 Newtons of force acting on a 0.2 square meter section of the plate. And note that that's the drag force on the moving plate. Also with this analysis, we could figure out what velocity a certain pulling force would achieve. If you were tugging on this plate with a 15 Newton force and its surface area in contact with the fluid was actually only 0.2 square meters, then we could figure out what velocity would yield equilibrium. That is, the force going forward equalling the force going backwards.