Section 6.5 Reading Assignment: Work and Fluid Forces

Answer Only Exercise 1, 2, and 3 by using a screenshot provided Calculus Pearson textbook. Make sure you read these three questions very carefully and see on what it is asking for.

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Section 6.5 Reading Assignment: Work and Fluid Forces Instructions. Read through this assignment and complete the three exercises below by reading the appropriate passages of the textbook. The notion of work from physics gives a good example of how Riemann sums are typically used in physics and engineering. You may have learned that work is equal to force times displacement, but this formula assumes that the force is constant.' How can we compute work when the force is changing? Note the definition boxed on p. 400. The textbook gives a more technical explanation, but the Riemann sum argument here is that for a small enough displacement, the force can be treated as constant. Over each small displacement, the work is F(c ) Ax. Adding these up give the Riemann sum and taking the limit as the displacement goes to zero gives the integral formula. This idea that over a small enough displacement or period of time, a quantity can be treated as though it were constant is a typical argument used when formulas involve integrals in physics. Exercise 1. Read the subsection "Hooke's Law for Springs: F=kx" (p. 400- 401). Explain how the integral formula for work applies to springs. If you've seen Hooke's law before, you probably learned that the key issue is with the spring constant k. The same is true here, except with the addition of the integral formula. Exercise 2. Read Example 4 (p. 401). Describe what each factor in the integral, that is 0.08, (20- *), and 9.8 represent in the physical situation. Describe the meanings of all the terms separately. Note that the entire reason we have to consider an integral here is because of the rope. The force due to the bucket is constant. Since the example skips over some important steps, it's worth slowing down and thinking through how this problem is set up. We conclude our look into physical applications with the work used to pump liquids up. The justification for the formula for finding the work for pumping liquids is actually quite different from the formula we've considered thus far, so it's best to consider it from scratch. Exercise 3. Read Example 5 (p. 401 - 402). Summarize the Riemann sum argument in this example and how it gives the total work to pump the oil in the container. Do this by describing the physical meaning of AW and each factor of the expression for AW Notice that the Ay, which becomes dy in the limit, in this example, is part of the units in the end result. We would not get ft-lb if not for multiplying that Ay. This is also true for the integral formula for work. Integrating Newtons (N) gives us Joules () because of the differential dx, which is in meters. So, force (Newton) times dx (meters) gives work (Joules). Thinking of that differential as a literal multiplication helps to track the units and makes it clearer how the physical units are affected by calculus operations. We will not be covering any material from the subsection "Fluid Pressures and Forces" due to time. That material will not appear for any assignments in this class. Another issue with this formula is that the force and the displacement must be in the same direction. We can deal with both these issues at once in Calculus ill.Chapter 6 Applications of Definite Integrals 400 Chapter 6 Applications of Definite Integrals DEFINITION The work done by a variable force F(x) in moving an object along the x-axis from x = a to x = b is (2) The units of the integral are joules if F is in newtons and x is in meters, and foot-pounds if F is in pounds and & is in feet. So the work done by a force of F(x) = 1/x- newtons in moving an object along the x-axis from x = 1 m to x = 10m is W = / dx = 10 + 1 = 0.9J.Hooke's Law for Springs: F = kx One calculation for work arises in finding the work required to stretch or compress a Compressed spring. Hooke's Law says that the force required to hold a stretched or compressed spring F WWWWWW x units from its natural (unstressed) length is proportional to x. In symbols, F = KX. (3) WWW.. The constant k, measured in force units per unit length, is a characteristic of the Uncompressed spring, called the force constant (or spring constant) of the spring. Hooke's Law, Equa- tion (3), gives good results as long as the force doesn't distort the metal in the spring. We assume that the forces in this section are too small to do that. EXAMPLE 2 Find the work required to compress a spring from its natural length of Force (lb) F = 16x I ft to a length of 0.75 ft if the force constant is & = 16 1b/ft. 4 Work done by F from r = 0 to r = 0.25 Solution We picture the uncompressed spring laid out along the x-axis with its movable end at the origin and its fixed end at x = 1 ft (Figure 6.36). This enables us to describe the + x (1) 0.25 force required to compress the spring from 0 to a with the formula F = 16x. To compress Amount compressed the spring from 0 to 0.25 ft. the force must increase from (b) FO) = 16:0 = 01b to F(0.25) = 16 + 0.25 = 4 1b. FIGURE 6.36 The force F needed to The work done by F over this interval is hold a spring under compression increases 0.25 10.25 Eq. (2) with linearly as the spring is compressed W = 16x dx = 8x] = 0.5 ft-Ib. - 0, b - 0.25, (Example 2). Jo F(x) = 16rEXAMPLE 3 A spring has a natural length of 1 m. A force of 24 N holds the spring stretched to a total length of 1.8 m. (a) Find the force constant k. (b) How much work will it take to stretch the spring 2 m beyond its natural length? (c) How far will a 45-N force stretch the spring? Solution (a) The force constant. We find the force constant from Equation (3). A force of 24 N maintains the spring at a position where it is stretched 0.8 m from its natural length, so 24 = k(0.8) Ey. (3) with F = 24, x = ().8 k = 24/0.8 = 30 N/m.Chapter 6 Applications of Definite Integrals 6.5 Work and Fluid Forces 401 (b) The work to stretch the spring 2 m. We imagine the unstressed spring hanging along the x-axis with its free end at x = 0 (Figure 6.37). The force required to stretch the spring x m beyond its natural length is the force required to hold the free end of the spring x units from the origin. Hooke's Law with k = 30 says that this force is F(x) = 30r. The work done by F on the spring from x = 0m to x = 2 m is 0.8 W = 30 de = 15.1- - 60 J. 24 N (c) How far will a 45-N force stretch the spring? We substitute F = 45 in the equation F = 30r to find x (m) 45 = 30x. or x = 1.5m. FIGURE 6.37 A 24-N weight stretches this spring 0.8 m beyond its unstressed A 45-N force will keep the spring stretched 1.5 m beyond its natural length. length (Example 3). Lifting Objects and Pumping Liquids from Containers The work integral is useful for calculating the work done in lifting objects whose weights vary with their elevation.EXAMPLE 4 A 5-ke bucket is lifted from the ground into the air by pulling in 20 m of rope at a constant speed (Figure 6.38). The rope weighs 0.08 kg/m. How much work 20 - was spent lifting the bucket and rope? Solution The weight of the bucket is obtained by multiplying the mass (5 kg) and the accel- eration due to gravity, approximately 9.8 m/s-. So the bucket's weight is (5)(9.8) = 49 N. and the work done lifting it alone is weight X distance = (49)(20) = 980 J. The weight of the rope varies with the bucket's elevation, because less of it is freely hanging as the bucket is raised. When the bucket is x m off the ground, the remaining por- tion of the rope still being lifted weighs (0.08)(20 - x)(9.8) N. So the work in lifting the rope is FIGURE 6.38 Lifting the bucket in -20 Example 4. Work on rope = / (0.08)(20 - x)(9.8) dx = (15.68 - 0.784x) dx 120 = 15.68x - 0.3921 = 313.6 - 156.8 = 156.8J. The total work for the bucket and rope combined is 980 + 156.8 = 1136.8 J. y = 2ror r = How much work does it take to pump all or part of the liquid from a container? Engi- neers often need to know the answer in order to design or choose the right pump, or to 10 compute the cost to transport water or some other liquid from one place to another. To find (5, 10) out how much work is required to pump the liquid, we imagine lifting the liquid out one thin horizontal slab at a time and applying the equation W = Fd to each slab. We then evaluate the integral that this leads to as the slabs become thinner and more numerous. EXAMPLE 5 The conical tank in Figure 6.39 is filled to within 2 ft of the top with olive oil weighing 57 1b/ft-. How much work does it take to pump the oil to the rim of the tank? FIGURE 6.39 The olive oil and tank in Solution We imagine the oil divided into thin slabs by planes perpendicular to the y-axis Example 5. at the points of a partition of the interval [0, 8 ].Chapter 6 Applications of Definite Integrals 402 Chapter 6 Applications of Definite Integrals The typical slab between the planes at y and y + Ay has a volume of about AV = (radius)-(thickness) = ( 2> ) Ay = XX AYR. The force Fly) required to lift this slab is equal to its weight, F(y) = 57 AV = 3/0 yl Ay lb. Weight = (weight per unit volume) X volume The distance through which F(y) must act to lift this slab to the level of the rim of the cone is about (10 - y) ft, so the work done lifting the slab is about (10 - y)y- Ay ft-lb. Assuming there are a slabs associated with the partition of [0, 8 ], and that y = y denotes the plane associated with the th slab of thickness Ay, we can approximate the work done lifting all of the slabs with the Riemann sum W = >(10 - Win An ft-lb. The work of pumping the oil to the rim is the limit of these sums as the norm of the parti- tion goes to zero and the number of slabs tends to infinity: W = lim S 4 (10 - y)y AV = 572 (10 - by2 dy 57 TT 4 ( 10 yz - 13) dy = : 57+ 10y3 4 3 = 30,561 ft-lh.Fluid Pressure and Forces 4 Dams are built thicker at the bottom than at the top (Figure 6.40) 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 hap- pens 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 freshwater in FIGURE 6.40 To withstand the increase pounds per cubic foot. The pressure h feet below the surface of any fluid is the fluid's ing pressure, dams are built thicker as they weight-density times h. go down. 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: Weight-density P = Wh. (4) A fluid's weight-density w is its weight per unit volume. Typical values (1b/ft') In a container of fluid with a flat horizontal base, the total force exerted by the fluid are listed below. against the base can be calculated by multiplying the area of the base by the pressure at the Gasoline 42 base. We can do this because total force equals force per unit area (pressure) times area. Mercury 849 (See Figure 6.41.) If F. p. and A are the total force, pressure, and area, then Milk 64.5 Molasses 100 F = total force = force per unit area X area Olive oil 57 Seawater 64 = pressure X area = pA Freshwater 62.4 = WhiA. p - wh from Eq. (4)