Fluid Mechanics with teacher's notes

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$Holt Math Minnesota Test Prep Workbook for Grade 9$

$Math Skills: More Practice$

PROBLEM SOLUTION Pressure SAMPLE PROBLEM 9B The small piston of a hydraulic lift has an area of 0.20 m 2 . A car weighing 1.20 10 4 N sits on a rack mounted on the large piston. The large piston has an area of 0.90 m 2 . How large a force must be applied to the small piston to support the car? Given: A 1 = 0.20 m 2 Unknown: F 1 = ? F 2 = 1.20 × 10 4 N A 2 = 0.90 m 2 Use the equation for pressure from page 325. Pressure Copyright © by Holt, Rinehart and Winston. All rights reserved. F1 F2 ⎯⎯ = ⎯⎯ A1 A2 F1 = ⎯A 1 A2 ⎯ F 2 = ⎯0 F 1 = 2.7 × 10 3 N 2 . 20 m ⎯ 2 0. 90 m (1.20 × 104 N) 1. In a car lift, compressed air exerts a force on a piston with a radius of 5.00 cm. This pressure is transmitted to a second piston with a radius of 15.0 cm. a. How large a force must the compressed air exert to lift a 1.33 × 10 4 N car? b. What pressure produces this force? Neglect the weight of the pistons. 2. A 1.5 m wide by 2.5 m long water bed weighs 1025 N. Find the pressure that the water bed exerts on the floor. Assume that the entire lower surface of the bed makes contact with the floor. 3. A person rides up a lift to a mountaintop, but the person’s ears fail to “pop”—that is, the pressure of the inner ear does not equalize with the outside atmosphere. The radius of each eardrum is 0.40 cm. The pressure of the atmosphere drops from 1.010 × 10 5 Pa at the bottom of the lift to 0.998 × 10 5 Pa at the top. a. What is the pressure on the inner ear at the top of the mountain? b. What is the magnitude of the net force on each eardrum? PRACTICE 9B Fluid Mechanics 327 SECTION 9-2 Classroom Practice The following may be used as a teamwork exercise or for demonstration at the chalkboard or on an overhead projector. PROBLEM Pressure In a hydraulic lift, a 620 N force is exerted on a 0.20 m 2 piston in order to support a weight that is placed on a 2.0 m 2 piston. How much pressure is exerted on the narrow piston? How much weight can the wide piston lift? Answer 3.1 × 10 3 Pa; 6.2 × 10 3 N PRACTICE GUIDE 9B Solving for: F PE Sample, 1a, 3b; Ch. Rvw. 16–18, 19b*, 32, 50* PW Sample, 1–2 PB 5–7 P PE 1b, 2, 3a; Ch. Rvw. 33, 37, 38b*, 47, 55*, 67* PW 6–7 PB 8–10 A PW 3–5 PB Sample, 1–4 ANSWERS TO Practice 9B Pressure 1. a. 1.48 × 10 3 N b. 1.88 × 10 5 Pa 2. 2.7 × 10 2 Pa 3. a. 1.2 × 10 3 Pa b. 6.0 × 10 −2 N 327
SECTION 9-2 Visual Strategy Figure 9-8 Point out that the fluid exerts pressure in all directions. Pressure against the sides of the box also increases with depth. Suppose the box in the dia- Q gram is immersed in water and L = 2.0 m. How does pressure on the box’s sides 1.0 m from its top compare with pressure on the sides near the top and near the bottom? ∆P = rg∆h = (1.00 × A 328 10 3 kg/m 3 )(9.81 m/s 2 ) (1.0 m) = 9800 Pa. This means that the pressure on the middle area of the sides of the box is 9800 Pa higher than the pressure near the top and 9800 Pa lower than the pressure near the bottom of the box. ANSWERS TO Conceptual Challenge 1. The pressure inside the building is approximately equal to the pressure outside the building. 2. The small piston must be pushed a long distance to move the object a short distance. 3. With snowshoes, her weight is applied over a larger area, so the pressure is small. Without snowshoes, the force is applied over a smaller area, so the pressure is large. L 328 Chapter 9 P0 P0 + ρgh1 P0 + ρgh2 Figure 9-8 The fluid pressure at the bottom of the box is greater than the fluid pressure at the top of the box. Pressure varies with depth in a fluid 1. Atmospheric pressure Why doesn’t the roof of a building collapse under the tremendous pressure exerted by our atmosphere? 2. Force and work In a hydraulic lift, which of the two pistons (large or small) moves through a longer distance while the hydraulic lift is lifting an object? (Hint: Remember that for an ideal machine, the input work equals the output work.) As a submarine dives deeper in the water, the pressure of the water against the hull of the submarine increases, and the resistance of the hull must be strong enough to withstand large pressures. Water pressure increases with depth because the water at a given depth must support the weight of the water above it. Imagine a small area on the hull of a submarine. The weight of the entire column of water above that area exerts a force on the area. The column of water has a volume equal to Ah, where A is the cross-sectional area of the column and h is its height. Hence the mass of this column of water is m = rV = rAh. Using the definitions of density and pressure, the pressure at this depth due to the weight of the column of water can be calculated as follows: F mg P = ⎯⎯ = ⎯⎯ = ⎯ A A rVg rAhg ⎯ = ⎯⎯ = rhg A A Note that this equation is valid only if the density is the same throughout the fluid. The pressure in the equation above is referred to as gauge pressure. It is not the total pressure at this depth because the atmosphere itself also exerts a pressure at the surface. Thus, the gauge pressure is actually the total pressure minus the atmospheric pressure. By using the symbol P 0 for the atmospheric pressure at the surface, we can express the total pressure, or absolute pressure, at a given depth in a fluid of uniform density r as follows: FLUID PRESSURE AS A FUNCTION OF DEPTH P = P 0 + rgh absolute pressure = atmospheric pressure + (density × free-fall acceleration × depth) This expression for pressure in a fluid can be used to help understand buoyant forces. Consider a rectangular box submerged in a container of water. The water pressure pushing down on the top of the box is −(P 0 + rgh 1), and 3. Snowshoes A woman wearing snowshoes stands safely in the snow. If she removes her snowshoes, she quickly begins to sink. Explain what happens in terms of force and pressure. Copyright © by Holt, Rinehart and Winston. All rights reserved.
Page 1 and 2: Chapter 9 Overview Section 9-1 defi
Page 3 and 4: Section 9-1 Demonstration 1 Volume
Page 5 and 6: SECTION 9-1 The Language of Physics
Page 7 and 8: SECTION 9-1 Demonstration 4 Lifting
Page 9 and 10: SECTION 9-1 PRACTICE GUIDE 9A Solvi
Page 11: SECTION 9-2 STOP 326 Misconception
Page 15 and 16: SECTION 9-2 Classroom Practice The
Page 17 and 18: Section 9-3 Demonstration 7 Fluid f
Page 19 and 20: SECTION 9-3 Misconception STOP Aler
Page 23 and 24: Section 9-4 Demonstration 10 Temper
Page 27 and 28: Chapter 9 Summary Teaching Tip Ask
Page 29 and 30: 9 REVIEW & ASSESS 12. because the p
Page 31 and 32: 9 REVIEW & ASSESS 42. 7.1 m 43. 17
Page 33 and 34: 9 REVIEW & ASSESS 63. 60.9 m/s 2 64
Page 35 and 36: Chapter 9 Laboratory Exercise NOTE
Page 37 and 38: CHAPTER 9 LAB Boyle’s Law Apparat

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