0:00 Welcome to lesson 3a, Submerged Vertical Plate. In this lesson, we'll begin a discussion about hydrostatic forces on submerged surfaces. In this lesson, we'll look at the resultant force on a submerged vertical plate, which is the simplest case. Consider some rectangular container with vertical walls, perhaps the wall of a swimming pool. The container is filled up with water. Note the different notation here. In these problems, y is down from the surface parallel to the plate in question. Since our plate is vertical here, y is straight down. Later we'll look at inclined plates and y will be parallel to the plate. Here's the pressure distribution in absolute pressure, p atmosphere at the top, and then increasing linearly as we go down. If we subtract off atmospheric pressure on both sides, we get the gauge pressure distribution. So p gauge is zero on the right side of the
0:48 plate, and starts from zero and increases linearly to the bottom. It's much easier to work with gauge pressure in these types of problems. The notation in the textbook is that P0 is the pressure at the surface of the water. In terms of absolute pressure, P0 is P atmospheric in this diagram. In terms of gauge pressure, P0 is zero here. This turns out to be a great simplification later on. Let's define the centroid of the submerged portion of this plate. If the plate is rectangular into the page, the centroid will be right in the middle. Note this is the middle of the submerged portion only. We're not concerned about this portion above the surface. Our goal is to find the resultant force acting on the plate, which we'll call Fr. Notice that I drew it acting lower than the centroid. Why? Well, if you think about
1:34 it, there's a larger pressure force on the bottom portion of the plate than on the top portion, so the force must act at some distance below the centroid, below halfway here. Mathematically, using gauge pressure, the resultant force is the integral over the wetted portion of this plate, which we call A, of P gauge dA. We can integrate for any shape plate at any depth. You would just integrate P gauge, which varies with area on the plate, over this whole area. I also want to point out that we'll let the x-axis be into the page. in these kinds of problems. Gravity, of course, is acting down. This point where the resultant force acts, we'll call the center of pressure, or Cp. Let's define this technically. Cp is the center of pressure, which is the location on the plate where the resultant force acts. As we
2:24 often do in Emet class, imagine replacing force distribution by a single resultant force. This force acts at the center of pressure. For our case of a vertical plate here, in a single liquid, the pressure increases linearly with depth. So if you do this integration for a vertical plate like this, you'll find that FR is P gauge average times A, where A is the area of the plate on one side, of course, just the submerged side. and the average gauge pressure is the pressure halfway down. Again, this is because of the linear nature of the pressure distribution. In other words, the pressure at the centroid, we'll call P gauge C, is equal to P gauge average, where again, C is the centroid of the plate. This resultant force acts
3:09 perpendicular to the plate. Let's quickly review what a centroid is. A centroid is the mathematical center of the surface area the submerged portion of the surface of the plate. Let's redraw the submerged portion of our vertical plate. The gauge pressure is linear. The height of the plate is h. The centroid is halfway down, h over 2. And the center of pressure is somewhere further down. Recall that the resultant force acts at the center of pressure, and P gauge comma C is the gauge pressure at the centroid. Let's look at the plate straight on from the left. We're talking about a rectangular plate, so it looks like this from the left. We're looking at the submerged surface of the plate. Since everything is symmetric right and left and top and bottom on the plate, the centroid is right in the middle. The
3:57 pressure distribution is not symmetric up and down, so the center of pressure is lower than the centroid, as we've already mentioned. Recall that the resultant force is p gauge average, which is p gauge at the centroid, times the area. So we know what this resultant force is, its magnitude. What remains is to find the location of this center of pressure. This simplest case was with the plate extending all the way up to the surface of the water. We'll come back to this simple case, but for now let's look at general case. The goal is how to calculate the location of the center of pressure. Let's do the general analysis of a submerged vertical plate. Here's the surface of the water or any other liquid. Let P0 be the pressure at the surface. As I mentioned before, this is atmospheric pressure if we work with absolute
4:45 pressures. But P0 is zero for gauge pressures. Let's let our vertical plate be some distance s from the surface to the top of the plate. If this is a rectangular plate, the centroid will be right at the center. But our analysis here will actually work for a plate of any shape where the centroid may or may not be in the center of the plate. Using our notation that y is the distance from the surface, we'll let yc be the distance to centroid c. The center of pressure is somewhat lower than that. We'll call that yp. yp is the depth to the center of pressure, and yc is the depth to the centroid. of the plate. Let's add some dimensions to our plate. Again, if it's a rectangular plate, let A be the horizontal dimension, and we'll let B equal the vertical dimension. B is equal
5:30 to H in this case, but in general we'll use B for the height of the plate. From the left we see our dimensions A and B, our centroid in the middle, and our center of pressure somewhere further down. For this rectangular plate, C is at the center, so YC is S plus B over 2. What about YP, the depth to the center of pressure? We will analyze the general case in the next lesson. I'm just going to give you the result here. YP is YC plus I sub XXC divided by YCA, where I XXC is the second moment of area about the axis into the page. which here is the x-axis, hence the xx notation here. Ixxc is also called the centroidal
6:16 moment of inertia. Now let's apply this equation to our rectangular plate. For a rectangle it turns out that Ixxc, the centroidal moment of inertia, is ab cubed over 12. So yp from this equation is yc, which we know is s plus b over 2 in our case. So YP is S plus B over 2 plus IXXC divided by YCA. YC is S plus B over 2 again, and the area is simply A times B. A little bit of simplification gives us YP equal S plus B over 2 plus B squared over 12 S plus B over 2. Notice that little a cancels out. That's because it doesn't matter how wide the plate is. The pressure distribution is the same at any portion of this plate width wise. So the
7:04 locations of C and CP do not depend on A. So this is the CP depth for a vertical submerged rectangular plate. Now let's return to our swimming pool example, our container. That is the case where the vertical wall extends all the way up to or even through the surface of the water. We call the depth h. So yc is halfway down the plate, and yp is given by this expression above. Recall that the resultant force acts at the center of pressure. Here are some equations then. yc is h over 2. Fr is p gauge average times A, where this is the pressure acting at the centroid. And since we're a distance h over 2 down, the gauge pressure is just rho times g times that depth, which is h over 2. The area, of course, is just A times h. So Fr is equal
7:53 to rho g times h over 2 times Ah, or simplifying, Fr, the resultant force, is rho gA times h squared over 2. Thinking in terms of dimensions, rho g times a length is a pressure, and h squared is an area. So this force is a pressure times an area. So that makes sense. Yp, sometimes I'll call this Ycp, the depth to the center of pressure, is given by this expression. here. So let's do that for this simplified case, noting that b equals h here. So yp is s plus h over 2 plus h squared over 12, quantity s plus h over 2. In our case here, s, which is the depth from the surface to the top of the plate, that is the top of the submerged plate, so s equals 0 here. So s drops out, and so yp becomes h over
8:42 2 plus h squared over 6h, or 2h over 3. So yp is 2h over 3 for this simplified case of a vertical plate extending up through the surface. Caution here, this does not hold for cases where s does not equal zero. In that case you would have to go back to this more general equation. Now let's do an example problem. This is example 3.8 of our textbook edition 4. It's a very practical problem. Suppose your car plunges into a lake. We'll treat the door as a rectangular plate. Inside the car the pressure is still atmospheric, but outside the car pressure is higher. Let's calculate the hydrostatic force on the door and the location of the center of pressure. I'll label some variables. This is S, 8 meters. C is at
9:27 the center of the door, and Cp is somewhat below C, which we have to calculate. We're approximating the door as a rectangular plate, which is vertical, of dimensions A and B. Of course, we'll use gauge pressure here for convenience. Yc, the depth to the centroid, which is S plus B over 2. By the way, in our diagram, b is 1.2 meters, and a is 1.0 meters. The average gauge pressure on the door is rho g times yc, and the resultant force is that average gauge pressure times the area of the door. So fr is rho g, yc, which is s plus b over 2, times the area, which is ab, and I'll plug in the numbers. Density of water, gravitational constant, 8 plus 1.2 over 2, those are meters, 1.0 meters is a, and 1.2 meters is b. We need only one unity conversion factor.
10:16 A kilonewton is a thousand kilogram meters per second squared. You can verify that all the units cancel except kilonewtons. I get 101.208 kilonewtons, which I round to three digits as our answer. 101 kilonewtons. The location of this resultant force is at the center of pressure. Using our equation from above, this is yc plus ixxc over yca. yc is s plus b over 2. Ixxc is ab cubed over 12, yc on the bottom is s plus b over 2 again, and then again the area is ab, which simplifies to s plus b over 2 plus b squared over 12, quantity s plus b over 2. We plug in the numbers to get the depth to the center of pressure. We get 8.61395 meters to three significant digits.
11:02 Our answer is 8.61. Let's compare that to yc. Plugging in the numbers we get Yc is 8.60 meters. Comparing these two, we see that Yp, the depth to the center of pressure, is just a little bit bigger than the centroid depth. That's because the car is submerged a pretty significant amount. Finally, let's look at how much force this is. A really strong person can lift about 100 kilograms, or 981 newtons, 1 kilonewton. If you put your feet on the door from the inside, At the farthest point from the hinge, your maximum moment is about 1 kilonewton meter. But the hydrostatic moment is about 50 kilonewton meters, approximately 50 times the moment the driver can possibly generate. So you cannot open the door. So what can you do? Well, as we say here, if you roll down the window a little bit, or just wait for
11:49 the car to flood a little bit, right before the car fills with water, pressure on both sides of the door will equalize and then you should be able to open the door fairly easily. So not only are you learning fluid mechanics but you have a practical example of how you might be able to save your life if this ever happens to you.
