anas m.

Advertisement

456 Chapter 6: Applications of Definite Integrals Fluid Pressures and Forces We make dams thicker at the bottom than at the top (Figure 6.64) because the pressure against them increases with depth. The pressure at any point on a dam depends only on how far below the surface the point is and not on how much the surface of the dam happens to be tilted at that point. The pressure, in pounds per square foot at a point h feet below the surface, is always 62.4h. The number 62.4 is the weight-density of water in pounds per cubic foot. The pressure h feet below the surface of any fluid is the fluid?s weight-density times h. 6.7 The Pressure-Depth Equation In a fluid that is standing still, the pressure p at depth h is the fluid?s weight- density w times h: (1)p = wh. In this section we use the equation to derive a formula for the total force ex- erted by a fluid against all or part of a vertical or horizontal containing wall. The Constant-Depth Formula for Fluid Force In a container of fluid with a flat horizontal base, the total force exerted by the fluid against the base can be calculated by multiplying the area of the base by the pressure at the base. We can do this because total force equals force per unit area (pressure) times area. (See Figure 6.65.) If F, p, and A are the total force, pressure, and area, then = whA. = pressure * area = pA F = total force = force per unit area * area p = wh Weight-density A fluid?s weight-density is its weight per unit volume. Typical values are Gasoline 42 Mercury 849 Milk 64.5 Molasses 100 Olive oil 57 Seawater 64 Water 62.4 slb>ft 3 d from Eq. (1) p = wh Fluid Force on a Constant-Depth Surface (2)F = pA = whA For example, the weight-density of water is so the fluid force at the bottom of a rectangular swimming pool 3 ft deep is For a flat plate submerged horizontally, like the bottom of the swimming pool just dis- cussed, the downward force acting on its upper face due to liquid pressure is given by Equation (2). If the plate is submerged vertically, however, then the pressure against it will be different at different depths and Equation (2) no longer is usable in that form (because h varies). By dividing the plate into many narrow horizontal bands or strips, we can create a Riemann sum whose limit is the fluid force against the side of the submerged vertical plate. Here is the procedure. = 37,440 lb. F = whA = s62.4 lb>ft 3 ds3 ftds10 # 20 ft 2 d 10 ft * 20 ft 62.4 lb>ft 3 , FIGURE 6.64 To withstand the increasing pressure, dams are built thicker as they go down. h FIGURE 6.65 These containers are filled with water to the same depth and have the same base area. The total force is therefore the same on the bottom of each container. The containers? shapes do not matter here. 4100 AWL/Thomas_ch06p396-465 9/1/04 12:56 PM Page 456 6.7 Fluid Pressures and Forces 457 The Variable-Depth Formula Suppose we want to know the force exerted by a fluid against one side of a vertical plate submerged in a fluid of weight-density w. To find it, we model the plate as a region ex- tending from to in the xy-plane (Figure 6.66). We partition [a, b] in the usual way and imagine the region to be cut into thin horizontal strips by planes perpendicular to the y-axis at the partition points. The typical strip from y to is units wide by L(y) units long. We assume L(y) to be a continuous function of y. The pressure varies across the strip from top to bottom. If the strip is narrow enough, however, the pressure will remain close to its bottom-edge value of The force exerted by the fluid against one side of the strip will be about Assume there are n strips associated with the partition of and that is the bot- tom edge of the k th strip having length and width The force against the entire plate is approximated by summing the forces against each strip, giving the Riemann sum (3) The sum in Equation (3) is a Riemann sum for a continuous function on [a, b], and we ex- pect the approximations to improve as the norm of the partition goes to zero. The force against the plate is the limit of these sums. F L a n k = 1 sw # sstrip depthd k # Ls y k dd ¢y k . ¢y k .Lsy k d y k a ? y ? b = w # sstrip depthd # Ls yd ¢y. ¢F = spressure along bottom edged * saread w * sstrip depthd. ¢yy +¢y y = by = a y Surface of fluid Strip length at level y Submerged vertical plate b y a H9004y Strip depth L(y) FIGURE 6.66 The force exerted by a fluid against one side of a thin, flat horizontal strip is about w * sstrip depthd * Lsyd ¢y. ¢F = pressure * area = The Integral for Fluid Force Against a Vertical Flat Plate Suppose that a plate submerged vertically in fluid of weight-density w runs from to on the y-axis. Let L(y) be the length of the horizontal strip meas- ured from left to right along the surface of the plate at level y. Then the force ex- erted by the fluid against one side of the plate is (4)F = L b a w # sstrip depthd # Lsyd dy. y = by = a EXAMPLE 1 Applying the Integral for Fluid Force A flat isosceles right triangular plate with base 6 ft and height 3 ft is submerged vertically, base up, 2 ft below the surface of a swimming pool. Find the force exerted by the water against one side of the plate. Solution We establish a coordinate system to work in by placing the origin at the plate?s bottom vertex and running the y-axis upward along the plate?s axis of symmetry (Figure 6.67). The surface of the pool lies along the line and the plate?s top edge along the line The plate?s right-hand edge lies along the line with the upper right vertex at (3, 3). The length of a thin strip at level y is Ls yd = 2x = 2y. y = x,y = 3. y = 5 x (ft) 0 Pool surface at Depth: 5 H11002 y y (3, 3) H9004y y H11005 5 y H11005 3 y H11005 x or x H11005 y y (ft) (x, x) H11005 (y, y) x H11005 y FIGURE 6.67 To find the force on one side of the submerged plate in Example 1, we can use a coordinate system like the one here. 4100 AWL/Thomas_ch06p396-465 9/1/04 12:56 PM Page 457 The depth of the strip beneath the surface is The force exerted by the water against one side of the plate is therefore Eq. (4) Fluid Forces and Centroids If we know the location of the centroid of a submerged flat vertical plate (Figure 6.68), we can take a shortcut to find the force against one side of the plate. From Equation (4), = w * sdepth of plate?s centroidd * sarea of plated. = w * smoment about surface level line of region occupied by plated = w L b a sstrip depthd * Ls yd dy F = L b a w * sstrip depthd * Ls yd dy = 124.8c 5 2 y 2 - y 3 3 d 0 3 = 1684.8 lb. = 124.8 L 3 0 s5y - y 2 d dy = L 3 0 62.4s5 - yd2y dy F = L b a w # a strip depth b # Ls yd dy s5 - yd. 458 Chapter 6: Applications of Definite Integrals Surface level of fluid h H11005 centroid depth Plate centroid FIGURE 6.68 The force against one side of the plate is area.w # h # plate Fluid Forces and Centroids The force of a fluid of weight-density w against one side of a submerged flat ver- tical plate is the product of w, the distance from the plate?s centroid to the fluid surface, and the plate?s area: (5)F = whA. h EXAMPLE 2 Finding Fluid Force Using Equation (5) Use Equation (5) to find the force in Example 1. Solution The centroid of the triangle (Figure 6.67) lies on the y-axis, one-third of the way from the base to the vertex, so The triangle?s area is Hence, = 1684.8 lb. F = whA = s62.4ds3ds9d = 1 2 s6ds3d = 9. A = 1 2 sbasedsheightd h = 3. 4100 AWL/Thomas_ch06p396-465 9/1/04 12:56 PM Page 458 Commercial_CD 4100 AWL/Thomas_ch06p396-465

Advertisement

"The semester I found StudyBlue, I went from a 2.8 to a 3.8, and graduated with honors!"

Jennifer Colorado School of Mines
StudyBlue is not sponsored or endorsed by any college, university, or instructor.

© 2015 StudyBlue Inc. All rights reserved.

© 2015 StudyBlue Inc. All rights reserved.