 A square 15 centimeter piling, those wooden pillar things near docks and shores that break the waves, is acted on by a water flow of 2 meters per second, which is 5 meters deep. You'll notice I'm not asking us to do anything here, I appear to have left that part out. But let's calculate the drag force on the beam. We are assuming a steady flow of seawater at 20 degrees Celsius, and I give you the density and dynamic viscosity for that substance. So this is a situation where our primary source of drag is the pressure drag on the body, not the friction drag that we've been using flat plate theory for, and one of the questions that we have to answer is, what is the orientation of the square piling relative to the fluid? And for this purpose, let's maybe further clarify our requirements by saying the worst case. So I don't just want to know the drag force, I want to know what the drag force is if this fluid is coming from the least advantageous angle as possible. So in one of the situations, we're going to consider a square piling where the fluid is coming directly at one of the sides. In another situation, we can consider that same piling at an orientation where one of the corners is the leading edge for the drag. We could consider any orientation between the two as well, but we recognize that these will be our boundary cases and as a result one of the two of them is going to be the worst case scenario. Actually while we're here, take a minute and think about, which one do you think is going to have more drag? I mean this one seems like it's a dragnier surface, whereas this one has a leading edge that's allowing water to split around it, but on the other hand, this one's wider than this one is, so more area of effect. Which one is worse? I guess we'll just have to find out. Our drag force calculation is going to be the same one from flat plate theory. We have, let's just go with one half times density times velocity squared times area times the coefficient of drag. We know the density of sea water, I told you a velocity. We can figure out the area of effect depending on the orientation, and the coefficient of drag for this body is going to come from our tables. Chapter seven gives us two dimensional shapes and some three dimensional shapes. Which do you think we should use? We're going to use this one. So on the two dimensional shapes table, we want the square cylinder. So this is referring to the fact that we have a square that is the same in the z direction everywhere, as opposed to a cube. Does that distinction make sense? It's a square, but only viewed from the top. It extends across the entire five meter height. So it's a two dimensional body. Okay, our coefficient of drag is 2.1 if it's dead on and 1.6 if it's coming from one of the corners. So I will call this orientation square on, and I will call this orientation diagonal. So for the square on orientation, our drag force is going to be one half times the density of seawater, which was 1,025 kilograms per cubic meter, times our free stream velocity, which was two meters per second, and that's squared, so two squared meters squared per second squared. What is our area of effect? Well keep in mind that from the perspective of the fluid, it's seeing a rectangle that is 15 centimeters across and 5 meters tall. Therefore my area is 15 centimeters times 5 meters, and I can convert that to meters while we're here, and then I have a coefficient of drag of 2.1. So again, when we're talking about pressure drag, we are looking at the body from the perspective of the fluid flow. So if we're looking at the front end of a sedan, it's probably going to be approximately a rectangular cross section. If we're looking at the front end of an aircraft, we're going to have rather thin areas of effect for our wings, larger circular or ellipsoid shape for the fuselage. For the square piling, it's a rectangle that is 5 meters tall and either 15 centimeters wide for the square on case, because this dimension is 15 centimeters, or 1.4 times 15 centimeters for this case. I mean while we're here, actually let's not call that x, let's call that y, so as to not get confused with the boundary layer thickness. So I'm saying 15 squared plus 15 squared is equal to y squared, therefore y is equal to the square root of 2 times 15 squared. So the square root of 2 times 15, which is approximately 1.4 times 15. But you know, in the interest of precision, we can say square root of 2 times 15 squared and we get 15 times the square root of 2. Thank you calculator. 21.2132 and that's a number of centimeters, a little bit of a tangent there. So our drag force in Newton, in Newton is a kilogram meter per second squared, centimeters cancel centimeters, square meters, meters and meters, cancels meters and cubic meters. Second squared cancels second squared, kilograms cancels kilograms. So 0.5 times 1025 times 2 squared times 15 times 5 times 2.1 and we are dividing by 100. We get a drag force of 3228.8 Newtons. And we can repeat the calculation for the diagonal case, which is going to start the same. The only difference is that we use a coefficient of drag of 1.6 and a cross-sectional area of 5 meters times 21.21 centimeters. So with the diagonal orientation, we get 3478.97 Newtons. So let's call that 3479. Interesting right? The diagonal orientation has a higher drag force even though it has a smaller coefficient of drag. The drag force is a function of coefficient of drag and also cross-sectional area. We have to bear in mind how the relative position of our object affects the drag force in addition to its coefficient of drag.