CHAPTER 8 Instruction and Intervention Support

Fluid Mechanics

1 Core Instruction

Chapter Resources

■■ The Teacher’s Edition wrap has extensive teaching support for every lesson, including Misconception Alerts, Teach from Visuals, Demonstrations, Teaching Tips, Differentiated Instruction, and Problem Solving. ■■ Several additional presentation tools are available online for every chapter, including Animated Physics features, Interactive Whiteboard activities, and Visual Concepts.

Labs and Demonstrations Section Instruction Go Online for Materials Lists and Solutions Lists.

■■ Textbook: Fluids and Buoyant Force ■■ Lab: Buoyant Vehicle (STEM) 8.1 Animated Physics: Buoyant Force ■■ Lab: Buoyancy (Open Inquiry) Visual Concepts: Fluid • Mass Density • Equation for ■■ Demonstrations: Volume of Liquids and Gases • Buoyant Mass Density • Buoyant Force • Equation for Buoyant Force • Float an Egg Force • Buoyant Force on Floating Objects Teaching Visuals: Displaced Volume of a Fluid • Buoyant Force • Densities of Some Common Substances PowerPresentations

■■ Textbook: Fluid Pressure ■■ Demonstration: Defining Pressure 8.2 Visual Concepts: Pressure • Equation for Pressure • Pascal’s Principle • Fluid Pressure as a Function of Depth • Equation for Fluid Pressure as a Function of Depth PowerPresentations

■■ Textbook: Fluids in Motion ■■ Demonstrations: Fluid Flow Around a Table-Tennis Ball • 8.3 Visual Concepts: Characteristics of an Ideal Fluid • Fluid Flow Between Two Cans Equation for Continuity • Bernoulli’s Principle • Equation for Bernoulli’s Principle Teaching Visuals: The Continuity Equation and Bernoulli’s Principle PowerPresentations

268A Chapter 8 PREMIUM Content Find all of your resources online at HMDScience.com.

2 Support and Intervention 3 Specialized Support

■■ Study Guide ■■ Chapter Summary Audio Files Concept Maps ■■ Differentiated Instruction: Below Level (TE wrap) ■■ Scientific Reasoning Skill Builder Interactive Demos Sample Problem Set I Sample Problem Set II

Where do I find it? PRINT DVD ONLINE Enrichment and Challenge Animated Physics ■ ■ Demonstrations (TE wrap) ■ ■ ■■ Differentiated Instruction: Pre-AP (TE wrap) Labs ■ ■ ■■ Take It Further: Properties of Gases (SE) Power Presentations ■ ■ ■■ Take It Further: Fluid Pressure (SE) QuickLabs ■ ■ ■ ■■ Timeline—Physics and Its World: 1690–1785 (SE) Teaching Visuals ■ ■ Textbook ■ ■ ■ Visual Concepts ■ ■

Interactive Demos ■ ■ Concept Maps ■ ■ Sample Problem Set I ■ ■ Sample Problem Set II ■ ■ Scientific Reasoning Skill Builder ■ ■ Assessment Study Guide ■ ■

■■ Section Quizzes Chapter Summary Audio Files ■ ■ ■■ Chapter Tests A and B Differentiated Instruction (TE wrap) ■ ■■ Alternative Assessment (SE) Online Assessment and Remediation ■■ ExamView Banks

Fluid Mechanics 268B CHAPTER 8

Chapter Overview Section 1 defines ideal fluids, calculates buoyant force, and explains why objects float or sink. Section 2 calculates pressures transferred by a fluid in a hydraulic lift and explains how hydrostatic pressure varies with depth. Section 3 introduces the equation of continuity and applies Bernoulli’s equation to solve problems of fluids in motion.

About the Image This kayaker is hurtling over Rainbow Falls, on the south fork of the Tuolumne River, in northern California. The Tuolumne is a favorite river among advanced kayakers and rafters, prized for its exciting rapids and waterfalls.

Kayakers know that if their weight (Fg) exceeds the upward, buoyant force

(FB) that causes them to float, they are sunk—literally! For an object, such as a kayak, that is immersed in a fluid, buoyant force equals the weight of the fluid that the object FB displaces. Buoyant force causes a Fg kayak to pop to the surface after a plunge down a waterfall. (bg) ©David Madison/Getty Images Lab 268Preview The following investigations support the Demonstrations concepts presented in this chapter. Untitled-282 268 Volume of Liquids and Gases 5/12/2011 4:28:35 AM LabS Buoyant Force Buoyant Vehicle (STEM) Float an Egg Buoyancy (Open Inquiry) Defining Pressure Fluid Flow Around a Table-Tennis Ball Fluid Flow Between Two Cans

268 Chapter 8 CHAPTER 8 SECTION 1 Fluids and Buoyant Force SECTION 2 Focus and Motivate Fluid Fluid Pressure SECTION 3 Activate Prior Fluids in Motion Knowledge Mechanics Knowledge to Review • Forces can cause changes in an object’s motion or in its shape. Why It Matters • Energy can be kinetic energy or Many kinds of hydraulic devices, such as the potential energy. brakes in a car and the • In the absence of friction, the total lifts that move heavy equipment, make use of mechanical energy of a system is the properties of fluids. constant. The total mass of a closed An understanding of system is constant. the properties of fluids is needed to design such devices. Items to Probe • Operational understanding of the concepts of area and volume: Ask students to compare the volume of containers of different shapes (with approximately the same capacity). • Ability to relate density, mass, and volume in a meaningful way: Ask students to calculate m, given ρ and V, using correct units.

ONLINE Physics HMDScience.com

ONLINE LABS Buoyant Vehicle PREMIUM Buoyancy CONTENT Physics HMDScience.com

Buoyant Force (bg) ©David Madison/Getty Images Why It Matters 269 ConnectING to History in or floated on a fluid, its weight decreased. He observed that the magnitude of decrease Untitled-282 269 The concept of buoyancy is a phenomenon 5/12/2011 4:28:46 AM that drew scholars’ attention from early days in the weight of each object is equal to the of civilization. People wondered how a large weight of the amount of the liquid that is ship could float while a small metal box would displaced. sink to the bottom of the sea. The Greek After briefly discussing this history with physicist Archimedes answered this question students, ask them how Archimedes was able for the first time in 276 BCE. He carried out to use this observation to weigh objects while numerous experiments with floating and they were immersed in water. Was he making a sinking objects. In every experiment, he direct measurement or an indirect noticed that when an object was immersed measurement?

Fluid Mechanics 269 SECTION 1 SECTION 1

Objectives Plan and Prepare Define a fluid. Fluids and Buoyant Distinguish a gas from a liquid. Preview Vocabulary Determine the magnitude of the Force Visual Vocabulary Hold up any solid buoyant force exerted on a object and ask students whether it is floating object or a submerged Key Terms object. fluid mass density buoyant force considered a fluid, and if not, why not. Then pour water from one container to Explain why some objects float and some objects sink. Defining a Fluid another. Ask students whether water is Matter is normally classified as being in one of three states—solid, liquid, considered a fluid, and if so, why? Hold or gaseous. Up to this point, this book’s discussion of motion and the up sealed bottles of other fluids, such as causes of motion has dealt primarily with the behavior of solid objects. shampoo, honey, and juice. Tilt them This chapter concerns the mechanics of liquids and gases. back and forth. Ask students what they Figure 1.1(a) is a photo of a liquid; Figure 1.1(b) shows an example of a gas. Pause for a moment and see if you can identify a common trait see happening in the bottles, and relate between them. One property they have in common is the ability to flow the behavior of the liquid to the and to alter their shape in the process. Materials that exhibit these definition of the term fluid. fluid a nonsolid state of matter in properties are called fluids. Solid objects are not considered to be fluids which the atoms or molecules are free because they cannot flow and therefore have a definite shape. to move past each other, as in a gas or a liquid Liquids have a definite volume; gases do not. Teach Even though both gases and liquids are fluids, there is a difference between them: One has a definite volume, and the other does not. Liquids, like solids, have a definite volume, but unlike solids, they do not Demonstration have a definite shape. Imagine filling the tank of a lawn mower with gasoline. The gasoline, a liquid, changes its shape from that of its original container to that of the tank. If there is a gallon of gasoline in the Volume of Liquids and Gases container before you pour, there will be a gallon in the tank after you Purpose Demonstrate that, unlike gases, pour. Gases, on the other hand, have neither a definite volume nor a liquids have a definite volume and definite shape. When a gas is poured from a smaller container into a larger one, the gas not only changes its shape to fit the new container but cannot be significantly compressed. also spreads out and changes its volume within the container. Materials large syringe, large eyedropper or turkey baster (a) (b) FIGURE 1.1 Procedure Pull the piston of the syringe to half the tube’s length, letting air into Fluids Both (a) liquids and (b) gases are considered fluids because they can the syringe. Hold your finger tightly over flow and change shape. the opening and pull the piston back as far as possible. Read the volume of air inside the syringe. Without removing your finger, push the piston in as far as possible. Have a student record the air ©Richard Megna/Fundamental York New Photographs, volume on the board for each case. Point out that no air has been let into or Differentiated270 Chapter 8 Instruction out of the syringe tube, but the same amount of air occupies all the space. Below Level Next, fill the syringe halfway with water, Untitled-36If you 270 ask students for examples of fluids, they 5/9/2011 10:05:11 AM and repeat the demonstration. are most likely to mention water, milk, juice, Cover the opening of the eyedropper and soda, all of which are liquids. They may and squeeze the bulb to show that the not think to include any gases, such as oxygen volume of air in the bulb and pipe can or carbon dioxide. Ask students what proper- be reduced. Fill the eyedropper halfway ties both gases and liquids share. Students with water. Note the water level. Hold should recognize that both gases and liquids the end sealed and squeeze the bulb to change shape as they flow. demonstrate that the air volume can be reduced but the water level cannot.

270 Chapter 8 Density and Buoyant Force Have you ever felt confined in a crowded elevator? You probably felt that way because there were too many people in the elevator for the amount Demonstration of space available. In other words, the density of people was too high. In general, density is a measure of a quantity in a given space. The quantity can be anything from people or trees to mass or energy. Buoyant Force Purpose Show the relationship between Mass density is mass per unit volume of a substance. buoyant force and submerged volume. When the word density is used to describe a fluid, what is really being Materials large spring scale, several measured is the fluid’s mass density. Mass density is the mass per unit mass density the concentration of cylinders of the same size but of different volume of a substance. It is often represented by the Greek letter ρ (rho). matter of an object, measured as the mass per unit volume of a substance materials (preferably of low density), a

the upward force clear container partially filled with water Mass Density buoyant force m exerted by a liquid on an object ρ = _ immersed in or floating on the liquid Procedure Point out that all of the V mass cylinders have the same volume. Measure mass density = _ volume this volume using a graduated cylinder or by the overflow method. Hang a cylinder 3 The SI unit of mass density is kilograms per cubic meter (kg/m ). on the scale, read the weight, and slowly In this book we will follow the convention of using the word density to FIGURE 1.2 refer to mass density. Figure 1.2 lists the densities of some fluids and a few DENSITIES OF SOME lower the cylinder into the water. important solids. COMMON SUBSTANCES* Students should observe that the scale’s Solids and liquids tend to be almost incompressible, meaning that Substance ρ (kg/m3) reading drops continuously as more of their density changes very little with changes in pressure. Thus, the the cylinder is submerged. Explain that the densities listed in Figure 1.2 for solids and liquids are approximately Hydrogen 0.0899 independent of pressure. Gases, on the other hand, are compressible and water is exerting an upward force on the can have densities over a wide range of values. Thus, there is not a Helium 0.179 cylinder. Have a student record the scale’s standard density for a gas, as there is for solids and liquids. The densities Steam (100°C) 0.598 reading before the cylinder’s immersion, listed for gases in Figure 1.2 are the values of the density at a stated temperature and pressure. For deviations of temperature and pressure Air 1.29 midway through its immersion, and when from these values, the density of the gas will vary significantly. Oxygen 1.43 it is completely submerged. Repeat with the other cylinders. Examining all of the Carbon dioxide 1.98 Buoyant forces can keep objects afloat. data will show that for the same volume Ethanol 0.806 103 Have you ever wondered why things feel lighter underwater than they do × submerged, the buoyant force is the in air? The reason is that a fluid exerts an upward force on objects that are Ice 0.917 × 103 partially or completely submerged in it. This upward force is called a same, regardless of the cylinder’s weight. 3 buoyant force. If you have ever rested on an air mattress in a swimming Fresh water (4°C) 1.00 × 10 pool, you have experienced a buoyant force. The buoyant force kept you Sea water (15°C) 1.025 × 103 and the mattress afloat. 3 Because the buoyant force acts in a direction opposite the force of Iron 7.86 × 10 gravity, the net force acting on an object submerged in a fluid, such as Mercury 13.6 × 103 water, is smaller than the object’s weight. Thus, the object appears to 3 weigh less in water than it does in air. The weight of an object immersed Gold 19.3 × 10 in a fluid is the object’s apparent weight. In the case of a heavy object, * All densities are measured at 0°C such as a brick, its apparent weight is less in water than its actual weight and 1 atm unless otherwise is in air, but it may still sink in water because the buoyant force is not noted. enough to keep it afloat. ©Richard Megna/Fundamental York New Photographs,

Fluid Mechanics 271

Pre-AP the cube will not move from side to side. The forces on the bottom face and top face of the Untitled-36 271 Explain to students that all the principles in 5/9/2011 10:05:11 AM physics work together consistently. Draw a jar cube, however, must be unbalanced, and the of water with an immersed cube and ask difference between these forces results in the students to analyze the forces on the cube. cube’s buoyancy. Explain that the liquid exerts a force on all sides of the cube, and ask what they can deduce about these forces. Students should conclude that the forces exerted on the side faces of the cube are equal and balanced, as

Fluid Mechanics 271 FIGURE 1.3

Archimedes’ Principle (a) A brick Teach continued is being lowered into a container of water. (b) The brick displaces water, causing the water to flow into a smaller container. The Language of (c) When the brick is completely submerged, the volume of the displaced Physics water (d) is equal to the volume of (a) (b) (c) (d) Students may need help interpreting the brick. all the symbols in the equations and relating them to prior knowledge. It may Archimedes’ principle describesHRW • Holt the Physics magnitude of a buoyant force. PH99PE-C09-002-001-A be helpful to remind students that g Imagine that you submerge a brick in a container of water, as shown in is free-fall acceleration, with a value of Figure 1.3. A spout on the side of the container at the water’s surface allows water to flow out of the container. As the brick sinks, the water level rises 2 9.81 m/s , which allows them to find an and water flows through the spout into a smaller container. The total object’s weight in newtons when its volume of water that collects in the smaller container is the displaced mass is known in kilograms. volume of water from the large container. The displaced volume of water is equal to the volume of the portion of the brick that is underwater. The magnitude of the buoyant force acting on the brick at any given time can be calculated by using a rule known as Archimedes’ principle. Misconception Alert! This principle can be stated as follows: Any object completely or partially submerged in a fluid experiences an upward buoyant force equal in Students may wonder why the buoyant magnitude to the weight of the fluid displaced by the object. Most people have experienced Archimedes’ principle. For example, recall that it is force, FB, is treated like weight (mg) even though it pushes upward. By carefully relatively easy to lift someone if you are both standing in a swimming pool, even if lifting that same person on dry land would be difficult. reading the statement of Archimedes’ Using mf to represent the mass of the displaced fluid, Archimedes’ principle, they may realize that the principle can be written symbolically as follows: magnitude of that force equals the weight of the fluid that would otherwise Did YOU Know? Buoyant Force Archimedes was a Greek mathematician occupy the space taken up by the FB = Fg (displaced fluid) = mf g submerged object. who was born in Syracuse, a city on the island of Sicily. According to legend, magnitude of buoyant force = weight of fluid displaced the king of Syracuse suspected that a certain golden crown was not pure gold. While bathing, Archimedes figured out Whether an object will float or sink depends on the net force acting how to test the crown’s authenticity on it. This net force is the object’s apparent weight and can be when he discovered the buoyancy principle. He is reported to have then calculated as follows: exclaimed, “Eureka!” meaning “I’ve Fnet = FB − Fg (object) found it!” Now we can apply Archimedes’ principle, using mo to represent the mass of the submerged object.

Fnet = mf g − mo g Remember that m = ρV, so the expression can be rewritten as follows:

Fnet = ( ρfVf − ρoVo)g Note that in this expression, the fluid quantities refer to the displaced fluid.

Problem272 Chapter Solving 8 Deconstructing Problems Untitled-36Show 272 students the following steps to explain 5/9/2011 10:05:12 AM how the expression is rewritten to obtain the formula for the net force. Replace the following formulas:

mf = ρfVf

m0 = ρ0V0

in the formula Fnet = mf g - m0 g and simplify:

Fnet = ρfVf g - ρ0V0 g

Fnet = (ρfVf - ρ0V0)g

272 Chapter 8 For a floating object, the buoyant force equals the object’s weight. FIGURE 1.4 Imagine a cargo-filled raft floating on a lake. There are two forces acting Floating The raft and cargo are on the raft and its cargo: the downward force of gravity and the upward floating because their weight and the Demonstration buoyant force of the water. Because the raft is floating in the water, the raft buoyant force are balanced. is in equilibrium and the two forces are balanced, as shown in Figure 1.4. For floating objects, the buoyant force and the weight of the object are Float an Egg F equal in magnitude. B Purpose Demonstrate the effect of liquid density on the buoyant force. Buoyant Force on Floating Objects F = F (object) = m g Materials uncooked egg, demonstra- B g o tion scale, glass jar, water, table salt, buoyant force = weight of floating object measuring spoon, stirring stick Procedure Measure the egg’s weight Notice that Archimedes’ principle is not required to find the buoyant force on a floating object if the weight of the object is known. Fg and record it on the board. Fill the glass jar about two-thirds full with water. The apparent weight of a submerged object depends on density. Demonstrate that the egg sinks in the water. Is a buoyant force acting on it? Imagine that a hole is accidentally punched in the raft shown in Figure 1.4 FIGURE 1.5 and that the raft begins to sink. The cargo and raft eventually sink below PHYSICSYes, but less than the egg’s weight Sinking The raft and cargo sink the water’s surface, as shown in Figure 1.5. The net force on the raft and Spec. ExplainNumber PH that99 PE C09-001-006-Aadding salt to the water will cargo is the vector sum of the buoyant force and the weight of the raft because their density is greater than Boston Graphics, Inc. and cargo. As the volume of the raft decreases, the volume of water the density of water. 617.523.1333increase the density of the water. Have displaced by the raft and cargo also decreases, as does the magnitude the class watch the egg as you stir one of the buoyant force. This can be written by using the expression for the FB spoonful of salt at a time into the water. net force: The egg will start to float but will remain F = ( ρ V − ρ V )g net f f o o submerged. Ask students to estimate Because the raft and cargo are completely submerged, V and V are equal: f o the buoyant force. same as the egg’s F = ( ρ − ρ )Vg net f o weight Keep adding salt until the egg Notice that both the direction and the magnitude of the net force floats partially above water. Again ask Fg depend on the difference between the density of the object and the students to estimate the buoyant force. density of the fluid in which it is immersed. If the object’s density is greater than the fluid density, the net force is negative (downward) and same as the egg’s weight Explain that the the object sinks. If the object’s density is less than the fluid density, the buoyant force balances the egg’s weight net force is positive (upward) and the object rises to the surface and in both cases, but as the density of the floats. If the densities are the same, the object hangs suspended PHYSICSwater increases, the volume of the egg underwater. Spec. Number PH 99 PE C09-001-007-A A simple relationship between the weight of a submerged object and Bostonthat Graphics, is immersed Inc. decreases. the buoyant force on the object can be found by considering their ratio 617.523.1333 as follows:

Fg (object) ρ Vg _ = _o FB ρfVg

F (object) ρ _g _o = ρ FB f This last expression is often useful in solving buoyancy problems.

Fluid Mechanics 273

Take it Further Divide both sides of this equation by V0 ρf and simplify. Untitled-36 273 Explain to students that, for a floating object, 5/9/2011 10:05:13 AM ρ V ρ V you can use the relative densities of the object _​ 0 0 ​ = _​ f f ​ V ρ V ρ and liquid to determine what volume of the 0 f 0 f ρ V object is submerged: _​ 0 ​ = _​ f ​ ρ V For a floating object, the buoyant force f 0

equals the weight of the object. Therefore, In this equation, V0 is the total volume of the

mo = mf , and we can conclude that floating object and Vf is the volume of the displaced liquid, and therefore the submerged ρ0V0 = ρfVf portion of the object.

Fluid Mechanics 273 PREMIUM CONTENT

Buoyant Force Interactive Demo HMDScience.com Sample Problem A A bargain hunter purchases a “gold” Teach continued crown at a flea market. After she gets home, she hangs the crown from a scale and finds its weight to be 7.84 N. She then weighs the crown while it is immersed in water, and the scale reads 6.86 N. Classroom Practice Is the crown made of pure gold? Explain. Buoyant Force Given: Calculate the actual weight, the buoyant ANALYZE Fg = 7.84 N apparent weight = 6.86 N ρ = ρ = 1.00 × 103 kg/m3 force, and the apparent weight of a f water Unknown: ρ = ? 5.00 × 10−5 m3 iron ball floating at rest in o mercury. Diagram: Answer: 3.86 N, 3.86 N, 0.00 N Tips and Tricks 7.84 N 6.86 N The use of a diagram can help clarify a problem and How much of the ball’s volume is FT,1 the variables involved. In FB FT,2 immersed in mercury? this diagram, FT, 1 equals −5 3 the actual weight of the Answer: 2.89 × 10 m F F crown, and FT, 2 is the g g apparent weight of the crown when immersed In air In water in water.

Choose an equation or situation: PLAN PHYSICS Because the object is completely submerged, consider the ratioSpec. of theNumber PH 99 PE C09-001-008-A weight to the buoyant force. Boston Graphics, Inc. 617.523.1333 FB − Fg = apparent weight F ρ _g _o = ρ FB f Rearrange the equation to isolate the unknown:

FB = Fg − (apparent weight) F _g ρo = ρf FB

SOLVE Substitute the values into the equation and solve:

FB = 7.84 N − 6.86 N = 0.98 N F _g _7.84 N 3 3 ρo = ρf = (1.00 × 10 kg/m ) FB 0.98 N 3 3 ρo = 8.0 × 10 kg/m

From Figure 1.2, the density of gold is 19.3 × 103 kg/m3. Because 8.0 × 103 kg/m3 < 19.3 × 103 kg/m3, the crown cannot be pure gold.

Continued Problem274 Chapter Solving 8 Alternative Approaches This volume of gold should weigh 18.9 N (0.00010 m3 × 9.81 m/s2 × density of gold), Untitled-36Describe 274 the following method as an 5/9/2011 10:05:14 AM alternative to the given solution: which is much greater than the 7.84 N weight given in the problem. FB = Factual - Fapparent

FB = 7.84 N - 6.86 N = 0.98 N

FB = ρfluidVg 0.98 N V = ​ ___ ​ (1.00 × 103 kg/m3)(9.81 m/s2) V = 0.00010 m3

274 Chapter 8 Buoyant Force (continued) PROBLEM guide A 1. A piece of metal weighs 50.0 N in air, 36.0 N in water, and 41.0 N in an unknown Use this guide to assign problems. liquid. Find the densities of the following: SE = Student Edition Textbook a. the metal b. the unknown liquid PW = Sample Problem Set I (online) PB = Sample Problem Set II (online) 2. A 2.8 kg rectangular air mattress is 2.00 m long, 0.500 m wide, and 0.100 m thick. What mass can it support in water before sinking? Solving for:

3. A ferry boat is 4.0 m wide and 6.0 m long. When a truck pulls onto it, the boat sinks ρ SE Sample, 1; 4.00 cm in the water. What is the weight of the truck? Ch. Rvw. 8–9 PW 6–8 4. An empty rubber balloon has a mass of 0.0120 kg. The balloon is filled with helium at 0°C, 1 atm pressure, and a density of 0.179 kg/m3. The filled balloon has a radius PB 5–7 of 0.500 m. m SE 2, 3; Ch. Rvw. 23 a. What is the magnitude of the buoyant force acting on the balloon? (Hint: See Figure 1.2 for the density of air.) PW Sample, 1–3 b. What is the magnitude of the net force acting on the balloon? PB 8–10

FB SSE 4; Ch. Rvw. 24*, 25*, 26*, 29*, 30*, 32*, 34*, 35* SECTION 1 FORMATIVE ASSESSMENT PW 4–5 Reviewing Main Ideas PB Sample, 1–4 1. What is the difference between a solid and a fluid? What is the difference between a gas and a liquid? *Challenging Problem 2. Which of the following objects will float in a tub of mercury? a. a solid gold bead b. an ice cube c. an iron bolt Answers d. 5 mL of water Practice A 3. A 650 kg weather balloon is designed to lift a 4600 kg package. What 1. a. 3.57 × 103 kg/m3 volume should the balloon have after being inflated with helium at 0°C and 1 atm pressure to lift the total load? (Hint: Use the density values in b. 6.4 × 102 kg/m3 Figure 1.2.) 2. 97 kg 4. A submerged submarine alters its buoyancy so that it initially accelerates 3 upward at 0.325 m/s2. What is the submarine’s average density at this 3. 9.4 × 10 N time? (Hint: the density of sea water is 1.025 × 103 kg/m3.) 4. a. 6.63 N

Critical Thinking b. 5.59 N 5. Many kayaks are made of plastics and other composite materials that are denser than water. How are such kayaks able to float in water? Assess and Reteach

Assess Use the Formative Assessment on this page to evaluate student Fluid Mechanics 275 Answers to Section Assessment mastery of the section. 1. A solid has a definite shape, while a fluid Reteach For students who need additional instruction, download the Untitled-36 275 does not; a liquid has a definite volume, 5/9/2011 10:05:15 AM while a gas does not. Section Study Guide. 2. b, c, and d Response to Intervention To reassess 3. 4.7 × 103 m3 students’ mastery, use the Section Quiz, 4. 9.92 × 102 kg/m3 available to print or to take directly online at HMDScience.com. 5. The kayak’s effective density includes the material of the kayak itself and the air within the kayak. Taken together, these are less than the density of the water.

Fluid Mechanics 275

Plan andPrepare Demonstration 276 definition ofpressure as how thisdemonstration relates to the observe adeeperdeformation. Discuss the same questions. Students will pad with its smaller face down and raise of contact Now place thebrickon centimeters isthisforce distributed? area brick’s weight Over how many square the brickexerts onthepad.same as the mation ofthepad.Askhow muchforce down. Have studentsnotice thedefor- Place the brick on the pad, large side weight ofthebrickanditsdimensions. Procedure box-shaped demonstration object, scale Materials on whichaforce isexerted. Purpose Definng Pressure area, creating agreater pressure. the same force is applied over a smaller should realize that without snowshoes, only whenwearing snowshoes. Students person can walk safely across the snow happens. Besure they explainhow the on thesnow. Have studentsexplainwhat removes thesnowshoes andtriesto walk after aheavy snowfall. Theperson then imagine aperson walking onsnowshoes Visual Vocabulary Preview Vocabulary SECTION Teach Chapter 8 Relate pressure andthearea flat foam pad,brickorother Measure and record the 2 Ask students to P =​ __ A F

​. Differentiated Instruction Untitled-37 276 1 kPa = use kilopascals (kPa) for measuring pressure. a dollarbilllyingontable.Scientists generally small, about the amount of pressure exerted by students that apressure of1Pa very is actually standard unit of air pressure. Point out to text, meteorologists use theatmosphere asthe divided by any unit of area. As mentioned in the Pressure may be measured in any unit of force Below L evel 276 pressur on asurfaceperunitarea FIGURE 2.1 up to 610 m. the fluid in the ocean at depths of to withstand the pressure ex erted by Atmospheric diving suits allow divers Protection from Pressure with depth in a fluid. Calculate how pressure varies by a fluid. Calculate the pressure ex erted 1000Pa. SECTION 2 e Chapter 8 themagnitudeofforce Objectives Fluid Pressure Pressure pressure Key Term is about 1.01 1.01 about is level sea at atmosphere the of pressure pressure.The of unit small a is pascal In theexamples above, thefluidsexert Pressure is force per unit area. 2.1 Figure in shown one the like suits diving atmospheric wear explorers Deep-sea noted this fact in what is now called called now is what in fact this noted (1623–1662) Pascal Blaise amount. same the exactly by container the inside points all at increases pressure the tire), the of valve the at as (such container result, thepressure increases by amount anequal throughout thetire. not only against thepump butagainst also thewalls ofthetire. Asa turn aforce exerts ontheairinside thetire. The airresponds by pushing youWhen tire, pump abicycle you apply aforce onthepump that in Applied pressure is transmitted equally throughout a fluid. It canasfollows: written be Pressure isameasure ofhow much force isapplied over agiven area. the bottom of a swimming pool, drive up a mountain, or ride in an airplane. an in ride or mountain, a up drive pool, swimming a of bottom the to dive you when ears your on forces similar of effects the Youexperience pressure inside a typical automobile tire is about 3 3 about is tire automobile typical a inside pressure calculating pressure,10 calculating the unit, another for basis the is equally to every point of the fluid and to the walls of the container. The SI unit of pressure is the the is pressure of unit SI The In general, if the pressure in a fluid is increased at any point in a a in point any at increased is fluid a in pressure the if general, In Pascal’s Principle Pressure Pressure applied to a fluid in a closed container is transmitted

to resist the forces exerted by water in the depths of the ocean. × 10 5 Pa. This amount of air pressure under normal conditions conditions normal under pressure air of amount Pa. This 5 Pa is about the same as 1 atm. The absolute air air absolute The atm. 1 as same the about Pa is pressure pressure atmosphere pascal P = Pascal’sprinciple _ A F (Pa), which is equal to 1 N/m 1 to equal is which (Pa), = pressure

_ force area (atm). For the purpose of of purpose Forthe (atm).

onyour eardrums. × 10 (or 5 Pa, or 3 atm. Pa,3 or Pascal’slaw 2 ): . The The . 5/9/2011 10:05:43AM

©Alexis Rosenfeld/Photo Researchers, Inc. A hydraulic lift, such as the one shown in FIGURE 2.2 Figure 2.2, makes use of Pascal’s principle. A small Hydraulic Lift The pressure is the same on both force F1 applied to a small piston of area A1 causes Misconception Alert! a pressure increase in a fluid, such as oil. sides of the enclosed fluid, allowing a small force to lift a According to Pascal’s principle, this increase in heavy object. Students may confuse the pressure

pressure, Pinc , is transmitted to a larger piston of increase in Pascal’s principle with the area A2 and the fluid exerts a force F2 on this pressure of the fluid itself. While an piston. Applying Pascal’s principle and the defini- tion of pressure gives the following equation: increase in pressure is transmitted equally throughout a fluid, the total F F P = _ 1 = _ 2 F pressure at different points in the fluid inc A A 1 1 2 may vary. For example, pressure varies Rearranging this equation to solve for F2 produces with depth. the following: A1 A2 F2 A F = _ 2 F 2 A 1 1 Classroom Practice This second equation shows that the output Pressure force, F2, is larger than the input force, F1, by a factor equal to the ratio of the areas of the two In a hydraulic lift, a 620 N force is pistons. However, the input force must be applied PHYSICS exerted on a 0.20 m2 piston in order Spec. Number PH 99 PE C09-002-006-A over a longer distance; the work required to lift the Boston Graphics, Inc.to support a weight that is placed on a truck is not reduced by the use of a hydraulic lift. 617.523.1333 2.0 m2 piston. How much pressure is

PREMIUM CONTENT exerted on the narrow piston? How much weight can the wide piston lift? Pressure Interactive Demo HMDScience.com Answer: 3.1 × 103 Pa; 6.2 × 103 N Sample Problem B The small piston of a hydraulic lift has an area of 0.20 m2. A car weighing 1.20 × 104 N sits on a rack mounted on the large piston. The large piston has an area of 0.90 m2. How large a force must be applied to the small piston to support the car?

2 2 ANALYZE Given: A 1 = 0.20 m A2 = 0.90 m

4 F2 = 1.20 × 10 N

Unknown: F1 = ?

SOLVE Use the equation for pressure and apply Pascal’s principle. F F _ 1 = _ 2 A1 A2 A 2 F = _ 1 F = 0.20_ m (1.20 × 104 N) 1 2 2 ( A2 ) ( 0.90 m )

3 F1 = 2.7 × 10 N ©Alexis Rosenfeld/Photo Inc. Researchers, Continued Problem Solving Fluid Mechanics 277 Take it Further Untitled-37 277 The formula used in this problem takes on 5/9/2011 10:05:44 AM another form, where the areas of pistons and the distances they move upward and down-

ward are proportional. That is, if d1 and d2 are the distances the pistons move upward and downward, then we have the following formula: d d _​ 1 ​ = _​ 2 ​ A2 A1

Fluid Mechanics 277 Pressure (continued) Teach continued 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. PROBLEM guide B a. How large a force must the compressed air exert to lift a 1.33 × 104 N car? Use this guide to assign problems. b. What pressure produces this force? Neglect the weight of the pistons. SE = Student Edition Textbook 2. A 1.5 m wide by 2.5 m long water bed weighs 1025 N. Find the pressure that the PW = Sample Problem Set I (online) water bed exerts on the floor. Assume that the entire lower surface of the bed PB = Sample Problem Set II (online) makes contact with the floor. Solving for: 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. F SE Sample, 1a, 3b; The radius of each eardrum is 0.40 cm. The pressure of the atmosphere drops from Ch. Rvw. 14–16, 21 1.010 × 105 Pa at the bottom of the lift to 0.998 × 105 Pa at the top. PW Sample, 1–2 a. What is the pressure difference between the inner and outer ear at the top of the mountain? PB 5–7 b. What is the magnitude of the net force on each eardrum? P SE 1b, 2, 31a; Ch. Rvw. 22, 27, 33* PW 6–7 PB 8–10 Pressure varies with depth in a fluid. A PW 3–5 As a submarine dives deeper in the water, the pressure of the water against the hull of the submarine increases, so the hull must be strong PB Sample, 1–4 enough to withstand large pressures. Water pressure increases with depth *Challenging Problem 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- Answers sectional area of the column and h is its height. Hence the mass of this Practice B column of water is m = ρV = ρAh. Using the definitions of density and 1. a. 1.48 × 103 N pressure, the pressure at this depth due to the weight of the column of water can be calculated as follows: b. 1.88 × 105 Pa mg ρVg ρAhg 2 P = _F = _ = _ = _ = ρhg 2. 2.7 × 10 Pa A A A A 3. a. 1.2 × 103 Pa This equation is valid only if the density is the same throughout the fluid. −2 b. 6.0 × 10 N 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 P0 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 ρ as follows:

Differentiated278 Chapter 8 Instruction Pre-AP Untitled-37Ask students 278 why the roof of a building does 5/9/2011 10:05:44 AM not collapse under the tremendous pressure exerted by our atmosphere. The pressure inside the building is approximately equal to the pressure outside the building. This may be easier for students to conceptualize in regards to a sports arena with an inflated roof.

278 Chapter 8 Fluid Pressure as a Function of Depth TEACH FROM VISUALS P = P + ρgh 0 FIGURE 2.3 Point out that the fluid absolute pressure = exerts force in all directions because of atmospheric pressure + (density × free-fall acceleration × depth) the pressure. Pressure against the sides FIGURE 2.3 of the box also increases with depth. This expression for pressure in a fluid can be used to help understand Pressure and Depth The fluid buoyant forces. Consider a rectangular box submerged in a container of pressure at the bottom of the box is Ask Suppose the box in the diagram is water, as shown in Figure 2.3. The water pressure at the top of the box is greater than the fluid pressure at the top immersed in water and L = 2.0 m. How P + ρgh , and the water pressure at the bottom of the box is P + ρgh . The of the box. 0 1 0 2 does pressure on the box’s sides 1.0 m downward force on the box is in the negative direction and given by AP0 from its top compare with pressure on −A(P0 + ρgh1), where A is the area of the top of the box. The upward force on the box is in the positive direction and given by A(P0 + ρgh2). The net the sides near the top and near the force on the box is the sum of these two forces. A(P0 + ρgh1 ) bottom?

Fnet = A(P0 + ρgh2) − A(P0 + ρgh1) = ρg(h2 − h1)A = ρgV = mf g L Answer: ∆P = ρg∆h = (1.00 × 103 kg/m3)(9.81 m/s2) Note that this is an expression of Archimedes’ principle. In general, A(P0 + ρgh2 ) we can say that buoyant forces arise from the differences in fluid (1.0 m) = 9800 Pa. This means that the pressure between the top and the bottom of an immersed object. pressure on the middle area of the sides of the box is 9800 Pa higher than the HRW • Holt Physics PH99PE-C09-002-011-A pressure near the top and 9800 Pa lower SECTION 2 FORMATIVE ASSESSMENT than the pressure near the bottom of the box. Reviewing Main Ideas 1. Which of the following exerts the most pressure while resting on a floor? a. a 25 N cube with 1.5 m sides b. a 15 N cylinder with a base radius of 1.0 m Assess and Reteach c. a 25 N cube with 2.0 m sides d. a 25 N cylinder with a base radius of 1.0 m Assess Use the Formative Assessment 2. Water is to be pumped to the top of the Empire State Building, which is on this page to evaluate student 366 m high. What gauge pressure is needed in the water line at the base of mastery of the section. the building to raise the water to this height? (Hint: See Figure 1.2 for the density of water.) Reteach For students who need 3. When a submarine dives to a depth of 5.0 × 102 m, how much pressure, in Pa, must its hull be able to withstand? How many times larger is this additional instruction, download the pressure than the pressure at the surface? (Hint: See Figure 1.2 for the Section Study Guide. density of sea water.) Response to Intervention To reassess Critical Thinking students’ mastery, use the Section Quiz, 4. Calculate the depth in the ocean at which the pressure is three times atmospheric pressure. (Hint: Use the value for the density of sea water available to print or to take directly given in Figure 1.2.) online at HMDScience.com.

Answers to Section Assessment Fluid Mechanics 279

1. a 6 Untitled-37 279 2. 3.59 × 10 Pa 5/9/2011 10:05:45 AM 3. 5.0 × 106 Pa; 5.0 × 101 4. 20.1 m

Fluid Mechanics 279

Plan andPrepare Demonstration Demonstration 280 them anddrawing themtogether. cans, lowering the pressure between that theairmoves faster between the cans will move toward each other. Explain volunteer to blow between thecans.The between the two cans. Ask a student would happen if you were to blow air or other flat surface. Ask students what sides approximately 2 cm apart on a table Procedure Materials flowof fluid andBernoulli’s principle. Purpose Fluid loween T Betw faucet, theballmoves toward thewater. from thewater. Whenyou turnonthe centimeters below thefaucet butaway string so that the ball is suspended a few things around them. Holdtheendof inmotion“suckingena offluids in” section covers thesurprisingphenom- Procedure water faucet end ofapiece oflight-weight string, Materials and Bernoulli’s principle. of theinteresting effects flow offluid Purpose T Fluid low Ar sticky nature. which means having anadhesive and derived from the Latin word “viscositas,” Latin Word Origins Preview Vocabulary SECTION Teach able-Tennis Ball Chapter 8 Further demonstrate theeffects Introduce studentsto some two empty soda cans table-tennis ballgluedto the Lay thesoda cansontheir Tell theclass that this ound a 3 The word wo Cans viscosity is Differentiated Instruction Untitled-38 280 it important to compare viscosity at the same list them in order of increasing viscosity. Why is Ask students to identify five to ten liquids and This greater friction results in a greater viscosity. the greater between thefriction themolecules. force varies. The greater the attractive force is, The strength of this intramolecular attractive always an attractive force between molecules. viscosity. They shouldunderstandthat there is Have studentsexplore themolecularbasisof Pre-AP incompressible friction orviscosityandis ideal fluid 280 FIGURE 3.1 Fluid Flow exhibits laminar flow and turbulent flow. fluid motion. Bernoulli’s principle on Recognize the effects of using the continuity equation. Examine the motion of a fluid SECTION 3 Chapter 8 Laminar flow afluidthathasnointernal Objectives The water flowing around this rock Turbulent flow Fluid Flow ideal fluid Key Term Fluids inMotion of the fluid always remains constant. were practically incompressible. A fluid is incompressible if the density density and buoyancy, we assumed all of the fluids used in problems of real fluids, so the model is a useful tool for analysis. While discussing an ideal fluid, the ideal fluid model does help explain many properties been rocksbeen ordramatic that bends created foamy whitewater rapids. or to simply paddle along. At otherplaces intheriver, there may have thatnoticed of theflowed part river smoothly, allowing you to float calmly Have you evercanoeing gone orrafting Ifso, downariver? you may have behavior of an Many features of fluid motion can be understood by considering the The ideal fluid model simplifies fluid-flow analysis. the moving fluid. nonturbulentfluid isalso , which means that there are currents noeddy in and pressure at each point inthefluidare constant. Ideal flow ofanideal characterizedalso by flow asteady . In otherwords, thevelocity,density, flow. asthey Ideal nokinetic energy dueto friction lose they so fluidsare slidingparticles past each other. Ideal fluidsare considered, nonviscous fluid istransformed into internal ofthefluid energy dueto thefriction a low Asaviscousfluidflows, viscosity. ofthekinetic energy ofthe part ahighflows viscosity fluid. Afluidwith more slowly than afluidwith does The term viscosity The term currents, are characteristic ofturbulent flow. sharp turns inariver. Irregular motionsofthefluid,called eddy changesabrupt invelocity,such as where there are obstacles or above velocityorunderconditionsthat a certain can cause laminar flow. that point earlier. The stretches smooth are ofariver regions of along thesamepath smooth traveled by that theparticles passed laminar characterizedbe ways. in oneoftwo The flow issaid to be chaotic andunpredictable. becauseto model itispredictable. Turbulent flow isextremely laminar flow andtheturbulent flow. Laminar flow ismuch easier Notice thedramatic difference inflow patterns the between In contrast, theflow irregular, ofafluidbecomes orturbulent, afluid,suchwater,When asriver isinmotion,theflow can Figure 3.1 decreasing temperature increases viscosity. Increasing temperature reduces viscosity, while motor oil. Temperature affects viscosity. answers: water, milk, honey, vegetable oil, temperature for each substance? Sample ideal fluid. if every particle that particle point aparticular passes moves ifevery shows aphotograph ofwater flowing past arock. refers to theamount ofinternal a within friction Although no real fluid has all the properties of 5/9/2011 10:06:11AM

©Richard Megna/Fundamental Photographs, NY

(br) ©Oote Boe Photography 2/Alamy Principles of Fluid Flow Fluid behavior is often very complex. Several general principles describing the flow of fluids can be derived relatively easily from basic physical laws. Teaching Tip The viscosity index on motor oil cans The continuity equation results from mass conservation. consists of a Society of Automotive Imagine that an ideal fluid flows into one end of a pipe and out the other FIGURE 3.2 Engineers (SAE) number (typically from end, as shown in Figure 3.2. The diameter of the pipe is different at each 5 to 50) followed by the letter W. The end. How does the speed of fluid flow change as the fluid passes through Mass Conservation in a Pipe The mass flowing into the pipe must equal higher the SAE number, the more viscous the pipe? the mass flowing out of the pipe in the the oil is. Low-viscosity oils are meant for Because mass is conserved and because the fluid is incompressible, we same time interval. v2 use in severe winter climates because know that the mass flowing into the bottom of the pipe, m1, must equal the mass flowing out of the top of the pipe, m2, during any given time interval: A2 low-viscosity oils flow more easily in cold temperatures. For hot or high-speed m1 = m2 v1 driving conditions, a high-viscosity oil can This simple equation can be expanded by recalling that m =ρV and by A1 be used because the excessive heat using the formula for the volume of a cylinder, V = A∆x. effectively thins the oil.

ρ1V1 = ρ2V2

ρ1A1∆x1 = ρ2 A2 ∆x2 TEACH FROM VISUALS The length of the cylinder, ∆x, is also the distance the fluid travels, which is equal to the speed of flow multiplied by the time interval FIGURE 3.2 Point out that the volumes (∆x = v∆t). of the two shaded sections of the pipe ρ A v ∆t = ρ A v ∆t FIGURE 3.3 1 1 1 2 2 2 must be equal in order for the continuity Narrowing of Falling Water The time interval and, for an ideal fluid, the density are the same on The width of a stream of water narrows equation to apply. each side of the equation, so they cancel each other out. The resulting as the water falls and speed up. equation is called the continuity equation: Ask Assume A1 is smaller than pictured

in Figure 3.2. Would ∆x1 need to be Continuity Equation longer or shorter for the mass of liquid

A1v1 = A2v2 in each section to still be equal?

area × speed in region 1 = area × speed in region 2 Answer: longer

The speed of fluid flow depends on cross-sectional area.

Note in the continuity equation that A1 and A2 can represent any two different cross-sectional areas of the pipe, not just the ends. This equation implies that the fluid speed is faster where the pipe is narrow and slower where the pipe is wide. The product Av, which has units of volume per unit time, is called the flow rate. The flow rate is constant throughout the pipe. The continuity equation explains an effect you may have observed as water flows slowly from a faucet, as shown in Figure 3.3. Because the water speeds up due to gravity as it falls, the stream narrows, satisfying the continuity equation. The continuity equation also explains why a river tends to flow more rapidly in places where the river is shallow or narrow

©Richard Megna/Fundamental NY Photographs, (br) ©Oote Boe Photography 2/Alamy than in places where the river is deep and wide. Problem Solving Fluid Mechanics 281 Take it Further Untitled-38 281 Point out that the continuity equation can be 5/9/2011 10:06:13 AM applied to any type of fluid flow, including the circulation of blood in living organisms. As an example, have students find the average flow velocity of blood in a vessel with radius 6 mm that takes blood away from the heart. Assume that a normal heart pumps blood at a rate of 83 mL/s. 13..84 mL/s

Fluid Mechanics 281 FIGURE 3.4 The pressure in a fluid is related to the speed of flow. Suppose there is a water-logged leaf carried along by the water in a Pressure and Speed A leaf speeds drainage pipe, as shown in Figure 3.4. The continuity equation shows that Teach continued up as it passes into a constriction in a drainage pipe. The water pressure on the the water moves faster through the narrow part of the tube than through right is less than the pressure on the left. the wider part of the tube. Therefore, as the water carries the leaf into the constriction, the leaf speeds up. Misconception Alert! P1 If the water and the leaf are accelerating as they enter the constriction, One might think that the water coming a P2 an unbalanced force must be causing the acceleration, according to out of the hose is at a higher pressure Newton’s second law. This unbalanced force is a result of the fact that the than the water in the hose, but the water pressure in front of the leaf is less than the water pressure behind the leaf. The pressure difference causes the leaf and the water around it opposite is true. The pressure outside the v 2 to accelerate as it enters the narrow part of the tube. This behavior v hose is atmospheric pressure, while the 1 illustrates a general principle known as Bernoulli’s principle, which can pressure inside the hose is higher than be stated as follows: FIGURE 3.5 atmospheric pressure. In fact, the water flows out of the hose because the Bernoulli’s Principle As air flows Bernoulli’s Principle around an airplane wing, air above moves greater pressure in the hose results in a faster than air below, producing lift. The pressure in a fluid decreases as the fluid’s velocity increases. net force on the water at the end of the F hose, which pushes the water out. The lift on an airplane wing can be explained, in part, with Bernoulli’s principle. As an airplane flies, air flows around the wings and body of the plane, as shown in Figure 3.5. Airplane wings are designed to direct the flow of air so that the air speed above the wing is greater than the air speed below the wing. Therefore, the air pressure above the wing is less Assess and Reteach than the pressure below, and there is a net upward force on the wing, called lift. The tilt of an airplane wing also adds to the lift on the plane. Assess Use the Formative Assessment The front of the wing is tilted upward so that air striking the bottom of the on this page to evaluate student wing is deflected downward. mastery of the section. Reteach For students who need SECTION 3 FORMATIVE ASSESSMENT additional instruction, download the Section Study Guide. Reviewing Main Ideas Response to Intervention To reassess 1. Water at a pressure of 3.00 × 105 Pa flows through a horizontal pipe at a speed of 1.00 m/s. The pipe narrows to one-fourth its original diameter. students’ mastery, use the Section Quiz, What is the speed of the flow in the narrow section? available to print or to take directly 2. A 2.0 cm diameter faucet tap fills a 2.5 × 10−2 m3 container in 30.0 s. online at HMDScience.com. What is the speed at which the water leaves the faucet?

Critical Thinking 3. The time required to fill a glass with water from a large container with a spigot is 30.0 s. If you replace the spigot with a smaller one so that the speed of the water leaving the nozzle doubles, how long does it take to fill the glass?

Interpreting Graphics 4. For this problem, refer back to Figure 3.2. Assume that the cross-sectional area, A2, in the tube is increased. Would the length, ∆x2, need to be longer or shorter for the mass of liquid in both sections to still be equal? Answers282 Chapter to S ection8 Assessment

1. 16.0 m/s

Untitled-38 2. 2.7 282 m/s 5/9/2011 10:06:14 AM 3. 30.0 s, because the volume rate of flow is constant 4. shorter

282 Chapter 8 Untitled-40 283

©Richard Megna/Fundamental Photographs, New York The Ideal Gas Law equations influidmechanicsrelates thesethree quantitiestoeachother. describe themacroscopic stateofanidealgas.Onethemostimportant ideal gas. plasma. Theseobservationshaveledscientiststodeveloptheconceptofan of temperatures andpressures, provided theirparticlesare notcharged,asina is agoodapproximation forthebehaviorofmanyreal gasesoverawiderange of thegastendtoberelated tooneanotherinafairlysimpleway. Thisrelationship When thedensityofagasissufficiently low, thepressure, volume,andtemperature Properties ofGases TAKE ITFURTHER Volume, pressure, andtemperature are thethree variablesthatcompletely the initial andfinal states ofthegas are related asfollows: form. useful Ifthenumberparticularly inthegas ofparticles isconstant, (or any combination theideal gas ofthese), law can expressed be ina In thissense, theideal gas law isuniversally applicable to allgases. gas. The gas could oxygen, be particles carbon dioxide, orany othergas. scale.) Also, theideal gas law makes nomention ofthecomposition ofthe in theKelvin scale. thechapter (See “Heat” to learn about theKelvin that whenapplying theideal gas law, you must express thetemperature from T flask sits over theburner, thetemperature oftheairinside itincreases a heat source, over placed aballoon with theopening oftheflask. Asthe room temperature (T experiment, a flask filled with air(V experiment, with aflask filled temperature ofagas. This relationship canasfollows: written be lawThe isanexpression ideal gas that relates thevolume, pressure, and experimentally determined to be approximately to be determined experimentally 1.38×10 If agas undergoes achange involume, pressure, ortemperature The symbol k This relation isillustrated shownin intheexperiment Ideal GasLaw number of gas particles particles gas of number 1 to T 2 . B represents Boltzmann’s constant. Its value has been 1 ) andatmospheric pressure (P pressure pressure × PV _ P Boltzmann’s constant constant Boltzmann’s T 1 V N 1 =Nk

1 1

× = = 1 equals thevolume equals oftheflask) at volume volume N _ P B T 2 2 V T 2

2

= 1 =P × temperature Figure 1. 0 -23 ) is placed over) isplaced J/K. Note J/K. In this FIGURE 1 Volume Temperature, Pressure, and the air inside are both increasing. because the volume and pressure of Fluid Mechanics The balloon is inflated 5/9/2011 10:06:47AM 283 Demonstration return to itsoriginalsize. the flask, to avoid breakage) and watch it Place theballoonunder cold water (not amount asbefore, butinalarger volume together.in theballoonandflask same fewer moleculesAskthemhow muchis now.air isintheflask volume, same but approximate size.Askthemhow much estimate andrecord theballoon’s balloon expands.Have students Heatof theflask untilthe theflask this closed system. aboutequalto that Ask studentswhat theairvolume isin ortheballoon. in oroutoftheflask balloon. Emphasizethat nogascanflow Procedure heated glassware. burner, anduse tongs whenhandlingthe at alltimeswhenoperating theBunsen Caution Bunsen burner, goggles Materials at ahighertemperature. amount ofgasoccupies alarger volume Purpose and Volume Tempera Take itFur temperature. behaves asitadjuststo room and have studentsobserve how it After 15minutes, remove theballoon a helium-filledballooninrefrigerator. For analternative demonstration, place Teaching Tip Be sure to wear safety goggles Demonstrate that thesame ressure, Pressure, ture, heat-resistant balloon, flask, Cap the flask withthe Cap theflask Fluid Mechanics ther 283 Take it Further

According to the ideal gas law, when the temperature increases, either Key Models and the pressure or the volume—or both—must also increase. Thus, the air inside the flask exerts a pressure (P2) on the balloon that serves to inflate Analogies MATERIALS the balloon. Because the balloon is expandable, the air expands to a larger volume (V ) to fill the balloon. When the flask is taken off the Graphs offer a convenient way to • 1 plastic 1 L bottle 2 • 1 quarter burner, the pressure, volume, and temperature of the air inside will slowly represent the relationship between return to their initial states. temperature, volume, and pressure for IDEAL GAS LAW Another alternative form of the ideal gas law indicates the law’s the following special cases of the ideal Make sure the bottle is empty, dependence on mass density. Assuming each particle in the gas has a gas law: and remove the cap. Place the mass m, the total mass of the gas is N × m = M. The ideal gas law can bottle in the freezer for at then be written as follows: least10 min. Wet the quarter P with water, and place the Mk T = _B quarter over the bottle’s PV = NkBT m opening as you take the bottle Constant T out of the freezer. Set the Mk T k T ρk T P = _ B = _M _ B = _ B bottle on a nearby tabletop; mV ( V ) m m then observe the bottle and quarter while the air in the V bottle warms up. As the air A Real Gas V inside the bottle begins to return to room temperature, An ideal gas is defined as a gas whose behavior is accurately described by the the quarter begins to jiggle ideal gas law. Although no real gas obeys the ideal gas law exactly for all Constant P around on top of the bottle. temperatures and pressures, the ideal gas law holds for a broad range of What does this movement tell physical conditions for all gases. The behavior of real gases departs from the you about the pressure inside behavior of an ideal gas at high pressures or low temperatures, conditions the bottle? What causes this under which the gas nearly liquefies. However, when a real gas has a relatively T change in pressure? high temperature and a relatively low pressure, such as at room temperature P Hypothesize as to why you and atmospheric pressure, its behavior approximates that of an ideal gas. need to wet the quarter before For problems involving the motion of fluids, we have assumed that all placing it on top of the bottle. Constant V gases and liquids are ideal fluids. An ideal fluid is a liquid or gas that is assumed to be incompressible. This is usually a good assumption because it is difficult to compress a fluid—even a gas—when it is not confined to a container. A fluid will tend to flow under the action of a force, changing its shape while maintaining a constant volume, rather than compress. T This feature, however, considers confined gases whose pressure, volume, and temperature may change. For example, when a force is applied to a piston, the gas inside the cylinder below the piston is compressed. Even though an ideal gas behaves like an ideal fluid in many situations, it cannot QuickLab be treated as incompressible when confined to a container.

Teacher’s Notes Did YOU Know? This activity is meant to demonstrate A third way of writing the ideal gas law may be familiar to you from your study of chemistry: that pressure increases with temperature PV = nRT when the volume of a gas is constant. In this equation, n is the number of moles of gas (one mole is equal to 6.02 × 1023 Eventually, the pressure becomes great particles). The quantity R is a number called the molar (universal) gas constant and has a value of 8.31 J/(mol�K). enough to overcome the weight of the quarter, and it jumps up slightly, allowing air to escape. 284 Chapter 8 Homework Options This QuickLab can easily be performed outside of the physics lab room. Untitled-40 284 5/9/2011 10:06:49 AM

284 Chapter 8 Take it Further TAKE IT FURTHER Fluid Pressure Demonstration Hydrostatic Pressure This feature discusses some topics related to fluid pressure, including atmospheric Purpose Demonstrate that pressure pressure and the kinetic theory of gases. It also covers Bernoulli’s equation, which is a more general form of Bernoulli’s principle. increases with depth. Materials 32-oz or 2-L plastic soda Atmospheric Pressure bottle, tape, water, bucket; optional: FIGURE 1 The weight of the air in the upper portion of Earth’s atmosphere exerts plastic straw or small pieces of glass or pressure on the layers of air below. This pressure is called atmospheric Mercury Barometer The metal tubing and modeling clay or pressure. The force that atmospheric pressure exerts on our bodies is height of the mercury in the tube of a 2 extremely large. (Assuming a body area of 2 m , this force is on the order barometer indicates the atmospheric silicone putty of 200 000 N, or 40 000 lb.) How can we exist under such tremendous pressure. C09-002-011-A(This illustration is not forces without our bodies collapsing? The answer is that our body cavities drawn to scale.) Procedure Drill three holes at different and tissues are permeated with fluids and gases that are pushing outward heights along the side of the bottle, with a pressure equal to that of the atmosphere. Consequently, our Empty each about 10 cm above the other with bodies are in equilibrium—the force of the atmosphere pushing in equals the internal force pushing out. the lowest hole near the bottom of the An instrument that is commonly used to measure atmospheric bottle. Stagger the holes horizontally pressure is the mercury barometer. Figure 1 shows a very simple mercury about 1 cm apart so that streams of barometer. A long tube that is open at one end and closed at the other is water flowing through them do not filled with mercury and then inverted into a dish of mercury. Once the tube is inverted, the mercury does not empty into the bowl. Instead, collide. If you are using straws or tubing, the atmosphere exerts a pressure on the mercury in the bowl. This insert short segments of tubing tightly atmospheric pressure pushes the mercury in the tube to some height into the holes to improve the flow. above the bowl. In this way, the force exerted on the bowl of mercury by the atmosphere is equal to the weight of the column of mercury in Cover the holes or open end of the the tube. Any change in the height of the column of mercury means tubing with tape, clay, or putty and fill Mercury that the atmosphere’s pressure has changed. the bottle with water. Place the bottle Kinetic Theory of Gases at the edge of a table. Place it high enough above the bucket so that the Many models of a gas have been developed over the years. Almost all of these models attempt to explain the macroscopic properties of a gas, such effects of depth on the range of each as pressure, in terms of events occurring in the gas on a microscopic stream will be easily observed. Ask scale. The most successful model by far is the kinetic theory of gases. students to predict how the water In kinetic theory, gas particles are likened to a collection of billiard streams will compare. Quickly pull the balls that constantly collide with one another. This simple model is successful in explaining many of the macroscopic properties of a gas. tape, clay, or putty away from all three For instance, as these particles strike a wall of a container, they transfer holes and have students observe the some of their momentum during the collision. The rate of transfer of shape of each stream. Water shooting momentum to the container wall is equal to the force exerted by the out from near the bottom of the bottle gas on the container wall, in accordance with the impulse-momentum theorem. This force per unit area is the gas pressure. exits the bottle at a higher speed than water shooting out from near the top does. The reason is that the pressure in the bottle increases with depth. Fluid Mechanics 285

Untitled-40 285 5/9/2011 10:06:51 AM

Fluid Mechanics 285 Take it Further

Bernoulli’s Equation Classroom Practice Imagine a fluid moving through a pipe of varying cross-sectional area and Bernoulli’s Equation elevation, as shown in Figure 2. When the cross-sectional area changes, A camper creates a shower by attaching the pressure and speed of the fluid can change. This change in kinetic energy may be compensated for by a change in gravitational potential a tube to the bottom of a hanging energy or by a change in pressure (so energy is still conserved). The bucket that is open to the atmosphere expression for the conservation of energy in fluids is called Bernoulli’s on top. If the water level in the bucket equation. Bernoulli’s equation is expressed mathematically as follows: is 3.15 m above the end of the tube Bernoulli’s Equation (the shower head), then what is the 1 P + __ ρv2 + ρgh = constant speed of the water exiting the tube? 2 Answer: 7.86 m/s pressure + kinetic energy per unit volume + gravitational potential energy per unit volume = constant along a given streamline A horizontal pipe narrows from a cross section of 2.0 m2 to 0.30 m2. If the speed FIGURE 2 Bernoulli’s equation differs slightly from the law of of the water flowing through the wider Conservation of Energy ∆x2 conservation of energy. For example, two of the terms on area of the pipe is 8.0 m/s, what is the As a fluid flows through this pipe, the left side of the equation look like the terms for kinetic P A speed of the water flowing through the it may change velocity, pressure, 2 2 energy and gravitational potential energy, but they and elevation. contain density, ρ, instead of mass, m. The reason is that narrow part? the conserved quantity in Bernoulli’s equation is energy v2 Answer: 53 m/s per unit volume, not just energy. This statement of the ∆x1 conservation of energy in fluids also includes an addi- y2 Assuming incompressible flow, what is P1A1 tional term: pressure, P. If you wish to compare the the change in pressure as the pipe energy in a given volume of fluid at two different points, y1 Bernoulli’s equation takes the following equivalent form: v1 narrows __1 2 __1 2 P1 + ρv 1 + ρgh1 = P2 + ρv 2 + ρgh2 Answer: 1.4 × 106 Pa 2 2 Comparing Bernoulli’s Principle and Equation Two special cases of Bernoulli’s equation are worth mentioning here. First, if the fluid is at rest, then both speeds are zero. This case is a static situation, such as a column of water in a cylinder. If the height at the top

of the column, h1, is defined as zero and h2 is the depth, then Bernoulli’s equation reduces to the equation for pressure as a function of depth, introduced in the chapter on fluids:

P1 = P2 + ρgh2 (static fluid) Second, imagine again a fluid flowing through a horizontal pipe with a constriction. Because the height of the fluid is constant, the gravitational potential energy does not change. Bernoulli’s equation then reduces to: 1 2 1 2 __ __ P1 + 2 ρv 1 = P2 + 2 ρv 2 (horizontal pipe)

This equation suggests that if v1 is greater than v2 at two different points in the flow, then P1 must be less than P2. In other words, the pressure decreases as speed increases—Bernoulli’s principle. Thus, Bernoulli’s principle is a special case of Bernoulli’s equation and is strictly true only when elevation is constant.

286 Chapter 8

Untitled-40 286 5/9/2011 10:06:53 AM

286 Chapter 8 Chapter summary CHAPTER 8 Summary Teaching Tip SECTION 1 Fluids and Buoyant Force KEY TERMS Ask students to prepare a concept map • Force is a vector quantity that causes changes in motion. fluid of the chapter. The concept map should • A fluid is a material that can flow, and thus it has no definite shape. Both mass density include most of the vocabulary terms, gases and liquids are fluids. buoyant force along with other integral terms or • Buoyant force is an upward force exerted by a fluid on an object floating on or submerged in the fluid. concepts. • The magnitude of a buoyant force for a submerged object is determined by Archimedes’ principle and is equal to the weight of the displaced fluid. • The magnitude of a buoyant force for a floating object is equal to the weight of the object because the object is in equilibrium.

SECTION 2 Fluid Pressure KEY TERM

• Pressure is a measure of how much force is exerted over a given area. pressure • According to Pascal’s principle, pressure applied to a fluid in a closed container is transmitted equally to every point of the fluid and to the walls of the container. • The pressure in a fluid increases with depth.

SECTION 3 Fluids in Motion KEY TERM

• Moving fluids can exhibit laminar (smooth) flow or turbulent flow. ideal fluid • An ideal fluid is incompressible, nonviscous, and, when undergoing ideal flow, nonturbulent. • The continuity equation is derived from the fact that the amount of fluid leaving a pipe during some time interval equals the amount entering the pipe during that same time interval. • According to Bernoulli’s principle, swift-moving fluids exert less pressure than slower-moving fluids.

VARIABLE SYMBOLS

Quantities Units Conversions Problem Solving

ρ density kg/m3 kilogram per meter3 = 10−3 g/cm3 See Appendix D: Equations for a summary of the equations introduced in this chapter. = N/m2 If you need more problem-solving practice, P pressure Pa pascal = 10−5 atm see Appendix I: Additional Problems.

Chapter Summary 287

Untitled-34 287 5/9/2011 10:03:43 AM

Fluid Mechanics 287 CHAPTER REVIEW CHAPTER 8 Review Answers 1. Buoyant force opposes and so Density and Buoyancy Pressure reduces the effect of weight. REVIEWING MAIN IDEAS REVIEWING MAIN IDEAS

2. the object’s weight 1. How is weight affected by buoyant force? 10. Is a large amount of pressure always caused by a large force? Explain your answer. 3. The weight of the water displaced is 2. Buoyant force equals what for any floating object? greater than the weight of the ball. 11. What is the SI unit of pressure? What is it equal to, in CONCEPTUAL QUESTIONS terms of other SI units? 4. The level falls because ρice < ρwater. 5. in the ocean; Because 3. If an inflated beach ball is placed beneath the surface CONCEPTUAL QUESTIONS of a pool of water and released, the ball shoots ρsea water > ρfresh water , less sea water upward. Why? 12. After a long class, a physics teacher stretches out for a nap on a bed of nails. How is this possible? must be displaced. 4. An ice cube is submerged in a glass of water. What 6. because the average density of the happens to the level of the water as the ice melts? 13. When drinking through a straw, you reduce the pressure in your mouth and the atmosphere 5. Will a ship ride higher in an inland freshwater lake or boat, including air inside the hollow moves the liquid. Could you use a straw to drink in the ocean? Why? hull, is less than the density of the on the moon? water 6. Steel is much denser than water. How, then, do steel boats float? PRACTICE PROBLEMS 7. The amount of the block that is submerged will decrease. The 7. A small piece of steel is tied to a block of wood. When For problems 14–16, see Sample Problem B. the wood is placed in a tub of water with the steel on 14. volume of water displaced remains top, half of the block is submerged. If the block is The four tires of an automobile are inflated to an × 5 the same when the block is inverted. inverted so that the steel is underwater, will the absolute pressure of 2.0 10 Pa. Each tire has an area of 0.024 m2 in contact with the ground. When the block is inverted, the steel amount of the wooden block that is submerged increase, decrease, or remain the same? Determine the weight of the automobile. occupies some of the displaced 15. A pipe contains water at 5.00 × 105 Pa above atmo- volume. Thus, the amount of wood PRACTICE PROBLEMS spheric pressure. If you patch a 4.00 mm diameter hole in the pipe with a piece of bubble gum, how submerged is less than half the block. For problems 8–9, see Sample Problem A. much force must the gum be able to withstand? 8. a. 6.3 × 103 kg/m3 8. An object weighs 315 N in air. When tied to a string, 16. A piston, A, as shown at 500.0 N b. 9.2 × 102 kg/m3 connected to a balance, and immersed in water, it right, has a diameter of weighs 265 N. When it is immersed in oil, it weighs 0.64 cm. A second piston, F 9. 2.1 × 103 kg/m3 B 269 N. Find the following: B, has a diameter of 3.8 cm. a. the density of the object 10. no; A force on a small area can Determine the force, F, b. the density of the oil A produce a large pressure. necessary to support the 9. A sample of an unknown material weighs 300.0 N in 500.0 N weight in the 2 11. the pascal; 1 pascal (1 Pa) = 1 N/m air and 200.0 N when submerged in an alcohol absence of friction. 3 3 12. The force opposing F is spread out solution with a density of 0.70 × 10 kg/m . What is g the density of the material? over a large number of nails, so no HRW • Holt Physics PH99PE-C09-CHR-004-A single nail exerts very much pressure. 13. No, there would be no way to reduce the pressure in your mouth below the zero atmospheric pressure outside the liquid. 4 14. 1.9 × 10 N 288 Chapter 8 15. 6.28 N 16. 14 N downward

Untitled-34 288 5/9/2011 10:03:44 AM

288 Chapter 8 CHAPTER REVIEW

24. When a load of 1.0 × 106 N is placed on a battleship, Fluid Flow the ship sinks only 2.5 cm in the water. Estimate the 17. The air flow causes the pressure over cross-sectional area of the ship at water level. the entrance in the mound to be CONCEPTUAL QUESTIONS (Hint: See Figure 1.2 for the density of sea water.) lower than the pressure over the 17. Prairie dogs live in underground burrows with at least 25. A 1.0 kg beaker containing other entrance. Thus, air is pushed two entrances. They ventilate their burrows by 2.0 kg of oil with a density of building a mound around one entrance, which is 916 kg/m3 rests on a scale. through the mound by the higher- open to a stream of air. A second entrance at ground A 2.0 kg block of iron is sus- pressure area. level is open to almost stagnant air. Use Bernoulli’s pended from a spring scale principle to explain how this construction creates air and completely submerged in 18. The water at the ground floor has flow through the burrow. the oil, as shown at right. Find less potential energy relative to the the equilibrium readings of 18. Municipal water supplies are often provided by reservoir than the water at the both scales. (Hint: See reservoirs built on high ground. Why does water from Figure 1.2 for the density of higher floor does. Because energy is such a reservoir flow more rapidly out of a faucet on iron.) the ground floor of a building than out of an identical conserved, the water flowing to the faucet on a higher floor? 26. A raft is constructed of wood having a density of lower faucet has greater kinetic 600.0 kg/m3. The surface area of the bottom of the raft 19. If air from a hair dryer is blown over the top of a table- PHYSICSenergy and thus greater velocity than is 5.7 m2, and the volume of the raft is 0.60 m3. When tennis ball, the ball can be suspended in air. Explain Spec. Number PH 99 PE C09-CHR-006-A the raft is placed in fresh water having a density of Bostonthe Graphics, water Inc.flowing to the higher how this suspension is possible. 1.0 × 103 kg/m3, how deep is the bottom of the raft 617.523.1333faucet does. below water level? 19. The moving air above the ball creates Mixed Review 27. A physics book has a height of 26 cm, a width of a low pressure area so that the air 21 cm, and a thickness of 3.5 cm. REVIEWING MAIN IDEAS a. What is the density of the physics book if it below the ball exerts a force that is 20. An engineer weighs a sample of mercury weighs 19 N? equal and opposite Fg. 3 3 b. Find the pressure that the physics book exerts on a (ρ = 13.6 × 10 kg/m ) and finds that the weight of −5 3 the sample is 4.5 N. What is the sample’s volume? desktop when the book lies face up. 20. 3.4 × 10 m c. Find the pressure that the physics book exerts on 21. 1.01 × 1011 N 21. About how much force is exerted by the atmosphere the surface of a desktop when the book is balanced 2 5 on 1.00 km of land at sea level? on its spine. 22. 5.9 × 10 Pa −1 22. A 70.0 kg man sits in a 5.0 kg chair so that his weight 28. A natural-gas pipeline with a diameter of 0.250 m 23. 6.11 × 10 kg is evenly distributed on the legs of the chair. Assume 3 delivers 1.55 m of gas per second. What is the flow 24. 4.0 × 103 m2 that each leg makes contact with the floor over a speed of the gas? circular area with a radius of 1.0 cm. What is the 25. 17 N, 31 N pressure exerted on the floor by each leg? 29. A 2.0 cm thick bar of soap is floating in water, with 1.5 cm of the bar underwater. Bath oil with a density 26. 6.3 × 10−2 m 23. A frog in a hemispherical bowl, as shown below, just of 900.0 kg/m3 is added and floats on top of the water. 3 3 3 3 floats in a fluid with a density of 1.35 × 10 kg/m . How high on the side of the bar will the oil reach 27. a. 1.0 × 10 kg/m If the bowl has a radius of 6.00 cm and negligible when the soap is floating in only the oil? b. 3.5 × 102 Pa mass, what is the mass of the frog? 3 30. Which dam must be stronger, one that holds back c. 2.1 × 10 Pa 5 3 1.0 × 10 m of water 10 m deep or one that holds 28. 31.6 m/s back 1.0 × 103 m3 of water 20 m deep? 29. 1.7 × 10−2 m 30. the one that holds back water 20 m deep, because pressure increases with increasing depth

Chapter Review 289

Untitled-34 289 5/9/2011 10:03:45 AM

Fluid Mechanics 289 CHAPTER REVIEW CHAPTER REVIEW

31. A light spring with a spring constant of 90.0 N/m 35. A thin, rigid, spherical shell with a mass of 4.00 kg 31. 0.605 m rests vertically on a table, as shown in (a) below. and diameter of 0.200 m is filled with helium (adding 32. a. 84 g/s A 2.00 g balloon is filled with helium (0°C and 1 atm negligible mass) at 0°C and 1 atm pressure. It is then pressure) to a volume of 5.00 m3 and connected to released from rest on the bottom of a pool of water b. 2.8 × 10−2 cm/s the spring, causing the spring to stretch, as shown that is 4.00 m deep. in (b). How much does the spring stretch when the a. Determine the upward acceleration of the shell. 33. 6.3 m system is in equilibrium? (Hint: See Figure 1.2 for the b. How long will it take for the top of the shell to 34. 21 Pa density of helium. The magnitude of the spring force reach the surface? Disregard frictional effects. equals k∆x.) 35. a. 0.48 m/s2 36. A student claims that if the strength of Earth’s gravity doubled, people would be unable to float on water. b. 4.0 s Do you agree or disagree with this statement? Why?

36. disagree; because the buoyant force 37. A light spring with a spring constant of 16.0 N/m rests is also proportional to g vertically on the bottom of a large beaker of water, as shown in ( ) below. A 5.00 × 10−3 kg block of wood −3 a 37. 1.7 × 10 m with a density of 650.0 kg/m3 is connected to the spring, and the mass-spring system is allowed to 38. The buoyant force causes the net k come to static equilibrium, as shown in (b) below. force on the astronaut to be close to k How much does the spring stretch? zero. In space, the astronauts (a) (b) accelerate at the same rate as the craft, so they feel as if they have no 32. The aorta in an average adult has a cross-sectional m area of 2.0 cm2. net force acting on them. a. Calculate the flow rate (in grams per second) of PHYSICS blood (ρ = 1.0 g/cm3) in the aorta if the flow speed 39. Helium is less dense than air and Spec. Number PH 99 PE C09-CHR-011-A k is 42 cm/s. Boston Graphics, Inc. k therefore floats in air. b. Assume that the aorta branches to form a large 617.523.1333 number of capillaries with a combined cross-sec- (a) (b) tional area of 3.0 × 103 cm2. What is the flow speed in the capillaries? 38. Astronauts sometimes train underwater to simulate conditions in space. Explain why. 33. A 1.0 kg hollow ball with a radius of 0.10 m is filled PHYSICS with air and is released from rest at the bottom of a 39. Explain why balloonists use helium instead of air Spec. Number PH 99 PE C09-CHR-012-A 2.0 m deep pool of water. How high above the surface in balloons. Boston Graphics, Inc. of the water does the ball rise? Disregard friction and 617.523.1333 the ball’s motion when the ball is only partially submerged.

34. In testing a new material for shielding spacecraft, 150 ball bearings each moving at a supersonic speed of 400.0 m/s collide head-on and elastically with the material during a 1.00 min interval. If the ball bearings each have a mass of 8.0 g and the area of the tested material is 0.75 m2, what is the pressure exerted on the material?

290 Chapter 8

Untitled-34 290 5/9/2011 10:03:46 AM

290 Chapter 8 CHAPTER REVIEW CHAPTER REVIEW

ALTERNATIVE ASSESSMENT as quickly as possible or as slowly as possible? Design experiments to determine the answer by Alternative 1. Build a hydrometer from a long test tube with some using model motor boats in a swimming pool. sand at the bottom and a stopper. Adjust the amount Indicate exactly what you will measure and how. Assessment Answers of sand as needed so that the tube floats in most If your teacher approves your plan and you are liquids. Calibrate it, and place a label with markings able to locate the necessary equipment, perform 1. Students should realize that the on the tube. Measure the densities of the following the experiment. higher the tube floats (more buoyant liquid foods: skim milk, whole milk, vegetable oil, 3. Record any examples of pumps in the tools, force), the greater the fluid’s density. pancake syrup, and molasses. Summarize your machines, and appliances you encounter in one findings in a chart or table. week, and briefly describe the appearance and 2. Student plans should be safe and 2. The owner of a fleet of tractor-trailers has contacted function of each pump. Research how one of these should measure force or acceleration you after a series of accidents involving tractor-trailers pumps works, and evaluate the explanation of the perpendicular to the motion. When passing each other on the highway. The owner wants pump’s operation for strengths and weaknesses. to know how drivers can minimize the pull exerted as Share your findings in a group meeting and create boats or trucks pass, the speed of one tractor-trailer passes another going in the same a presentation, model, or diagram that summarizes the fluid between them increases the group’s findings. direction. Should the passing tractor-trailer try to pass (as in a narrow pipe) and pressure drops. 3. Student answers should recognize that pumps use pressure differences to move fluids. Possible answers include air pump for tires, gas pump, or vacuum cleaner.

Flow Rates

Flow rate, as you learned earlier in this chapter, is described by You will use this equation to study the flow rates (Y1) for the following equation: various hose diameters (X) and flow speeds (V). The calculator flow rate = Av will produce a table of flow rates in cubic centimeters per second versus hose diameters in centimeters. Flow rate is a measure of the volume of a fluid that passes through a tube per unit time. A is the cross-sectional area of In this graphing calculator activity, you will learn how to read a the tube, and v is the flow speed of the fluid. If A has units of table on the calculator and to use that table to make predic- centimeters squared and v has units of centimeters per tions about flow rates. second, flow rate will have units of cubic centimeters per Go online to HMDScience.com to find this graphing second. calculator activity. The graphing calculator will use the following equation to determine flow rate. 2 Y1 = π * V(X/2)

Chapter Review 291

Untitled-34 291 5/9/2011 10:03:46 AM

Fluid Mechanics 291 STANDARDS-BASED ASSESSMENT Standards-Based Assessment Answers MULTIPLE CHOICE 4. A fish rests on the bottom of a bucket of water while 1. C the bucket is being weighed. When the fish begins to 2. G 1. Which of the following is the correct equation for swim around in the bucket, how does the reading the net force acting on a submerged object? on the scale change? 3. B A. Fnet = 0 F. The motion of the fish causes the scale reading B. F = (ρ − ρ )gV 4. J net object fluid object to increase. C. Fnet = (ρfluid − ρobject)gVobject G. The motion of the fish causes the scale reading 5. A D. Fnet = (ρfluid + ρobject)gVobject to decrease. H. The buoyant force on the fish is exerted down- 2. How many times greater than the lifting force must 6. J ward on the bucket, causing the scale reading the force applied to a hydraulic lift be if the ratio of to increase. the area where pressure is applied to the lifted area J. The mass of the system, and so the scale reading, is _ 1 ? 7 will remain unchanged. F. _1 49 Use the passage below to answer questions 5–6. G. _1 7 A metal block (ρ = 7900 kg/m3) is connected to a spring H. 7 scale by a string 5 cm in length. The block’s weight in air J. 49 is recorded. A second reading is recorded when the block is placed in a tank of fluid and the surface of the 3. A typical silo on a farm has many bands wrapped fluid is 3 cm below the scale. around its perimeter, as shown in the figure below. Why is the spacing between successive bands 5. If the fluid is oil (ρ < 1000 kg/m3), which of the smaller toward the bottom? following must be true? A. to provide support for the silo’s sides above them A. The first scale reading is larger than the B. to resist the increasing pressure that the grains second reading. exert with increasing depth B. The second scale reading is larger than the C. to resist the increasing pressure that the first reading. atmosphere exerts with increasing depth C. The two scale readings are identical. D. to make access to smaller quantities of grain near D. The second scale reading is zero. the ground possible 6. If the fluid is mercury (ρ = 13 600 kg/m3), which of the following must be true? F. The first scale reading is larger than the second reading. G. The second scale reading is larger than the first reading. H. The two scale readings are identical. J. The second scale reading is zero.

PHYSICS Spec. Number PH 99 PE C09-CHR-001-A Boston Graphics, Inc. 617.523.1333

292 Chapter 8

Untitled-39 292 5/9/2011 10:06:26 AM

292 Chapter 8 TEST PREP

7. B Use the passage below to answer questions 7–8. 11. A hydraulic brake system is shown below. The area of 8. H 2 the piston in the master cylinder is 6.40 cm , and the 9. mercury; because the density of Water flows through a pipe of varying width at a constant area of the piston in the brake cylinder is 1.75 cm2. mass flow rate. At point A the diameter of the pipe is dA The coefficient of friction between the brake shoe mercury is greater than that of water and at point B the diameter of the pipe is d . B and wheel drum is 0.50. What is the frictional force 10. 2.5 × 1010 capillaries 7. Which of the following equations describes the between the brake shoe and wheel drum when a force of 44 N is exerted on the pedal? 11. 6.0 N relationship between the water speed at point A, vA, and the water speed at point B, vA? Wheel drum 12. FB,oil + FB,water = Fg,block A. d v = d v A A B B Pedal (ρ - ρ )h 2 2 Brake shoe __block oil B. d v = d v 13. y = ​ ​ A A B B (ρ - ρ ) C. d d = v v water oil A B A B −2 1 2 1 2 14. 1.71 × 10 m D. __ __ 2 dA v A = 2 dB v B 2 8. If the cross-sectional area of point A is 2.5 m and Brake cylinder the cross-sectional area of point B is 5.0 m2, how Master cylinder many times faster does the water flow at point A than at point B? EXTENDED RESPONSE F. _1 Base your answers to questions 12–14 on the information below. 4 HRW • Holt Physics PH99PE-C09-CHR-003-A G. _1 Oil, which has a density of 930.0 kg/m3, floats on water. 2 A rectangular block of wood with a height, h, of 4.00 cm H. 2 and a density of 960.0 kg/m3 floats partly in the water, J. 4 and the rest floats completely under the oil layer.

SHORT RESPONSE 12. What is the balanced force equation for this situation? 9. Will an ice cube float higher in water or in mercury? 13. What is the equation that describes y, the thickness Explain your answer. of the part of the block that is submerged in water? 10. The approximate inside diameter of the aorta is 14. What is the value for y? 1.6 cm, and that of a capillary is 1.0 × 10−6 m. The average flow speed is about 1.0 m/s in the aorta and 1.0 cm/s in the capillaries. If all the blood in the aorta eventually flows through the capillaries, estimate the number of capillaries.

11 12 1 Test Tip 10 2 9 3 For problems involving several forces, 8 4 write down equations showing how the 7 6 5 forces interact.

Standards-Based Assessment 293

Untitled-39 293 5/9/2011 10:06:27 AM

Fluid Mechanics 293 294 World Physics andIts Chapter 8 Untitled-196 294 PHYSICS 294 nearly two centuries. them to control the region for effective bureaucracy enable centralized government and Ghana. The Ashanti’s strong emerges in what is now the major African kingdoms, The Ashanti empire, the last of 1698 Newcomen engine. improvements to the Watt makes significant 50 years later, steam engine. Over invents the first practical Thomas Newcomen 1712 = eff _ W Q James net h

1690 AND ITS WORLD today as the greatest Chinese novel. another writer, is widely regarded Red Chamber, attributed to him and is born. The book Chinese writer 1715 (approx.) 1700 the six Johann Sebastian Bach completes 1721 Ts’ao Hsüeh-ch’in Brandenburg Concertos. The Dream of the 1710 ship’s longitude. accurately determine a will allow navigators to four chronometers that constructs the first of John Harrison 1735 1720 1730 Empire enters a stage of decline. expands into India as the Moghul Shah, the Persian Empire Under the leadership of 1738 P is published. mechanical behavior of fluids, includes his research on the Hydrodynamics, which Daniel Bernoulli’s 1738 + __ 2 1

ρv 2

+ 1740 ρgh Nadir = constant 5/11/2011 7:27:16AM

(bc) Johann Sebastian Bach (about 1720), Johann Jakob Ihle. Oile on canvas, 82 x 66 cm. Photo ©akg-images; (tl) ©R. Sheridan/The Ancient Art & Architecture Collection LTD; (br) ©SEF/Art Resource, NY; (bl) ©Corbis; (tcl) Statue of Ts’ao Hsüeh-ch’in Photo ©Imaginechina; (tcr) Harrison’s first marine chronometer Photo ©Mary Evans Picture Library; (tr) ©Hydrodynamics by Daniel Bernoulli & Hydraulics by Johann Bernoulli, 1968 Dover Publications, Inc., New York

(tcl) Franklin’s Experiment (1876). Colored lithograph. Currier and Ives, New York. Photo ©akg-images; (tl) ©The Ancient Art & Architecture Collection, Ltd.; (tr) ©Corbis; (br), (bl) ©Bettmann/Corbis; (bcr) Caroline Lucretia Herschel (early 19th century). Engraving. Photo ©Kean Collection/Hulton Archive/Getty Images; (bcl) ©Sheila Terry/Photo Researchers, Inc.; (tcr) ©The Print Collector/Alamy 1690 – 1785 1747 1752 1770 Contrary to the favored idea that Benjamin Franklin performs the Antoine Laurent Lavoisier heat is a fluid, Russian chemist dangerous “kite experiment,” in begins his research on Mikhail V. Lomonosov which he demonstrates that lightning chemical reactions, notably publishes his hypothesis that consists of electric charge. He would oxidation and combustion. heat is the result of motion. build on the first studies of electricity Several years later, Lomonosov performed earlier in the century formulates conservation laws for by describing electricity as having mass and energy. positive and negative charge. 1775 + + + The American Revolution begins. Electrons + Holes

1740 1750 1760 1770 1780 1790

1756 1757 1772 1785 The Seven Year’s War German musician William Caroline Herschel, sister begins. British general Charles Augustin de Coulomb Herschel emigrates to of astronomer William publishes the results of experiments James Wolfe leads the England to avoid fighting in Herschel, joins her brother that will systematically and conclusively capture of Fort Louisburg, the Seven Year’s War. Over in England. She compiles prove the inverse-square law for electric in Canada, in 1758. the next 60 years, he pursues the most comprehensive force. The law has been suggested astronomy, constructing the star catalog of the era for over 30 years by other scientists, largest reflecting telescopes and discovers several such as Daniel Bernoulli, Joseph of the era and discovering new nebulae—regions of glowing Priestly, and Henry Cavendish. objects, such as binary stars gas—within our galaxy. and the planet Uranus.

q q F = k _1 2 electric C ( r2 ) (bc) Johann Sebastian Bach (about Johann 1720), Jakob Oile Ihle. on 82 canvas, x 66 Photo cm. ©akg-images; (br) Art Architecture (bl) Ancient ©SEF/Art (tl) & NY; Sheridan/The Collection ©Corbis; Resource, ©R. LTD; (tcl) Statue Ts’ao of Hsüeh-ch’in Photo ©Imaginechina; first (tcr) marine Harrison’s chronometer Photo ©Mary Evans Picture Library; (tr) ©Hydrodynamics by Daniel Bernoulli & Hydraulics by Johann Bernoulli, 1968 York Dover New Inc., Publications, Photo ©akg-images; Experiment Colored Art (1876). (tl) Currier Architecture lithograph. Ancient Ltd.; & (tcl) ©The Franklin’s York. and Collection, New (tr) Ives, ©Corbis; (bl) (br), ©Bettmann/Corbis; (bcr) Caroline Lucretia Herschel (early 19th century). Photo Engraving. ©Kean Archive/Getty Inc.; Collection/Hulton Researchers, Images;(tcr) Terry/Photo ©The (bcl) ©Sheila Print Collector/Alamy

295

Untitled-196 295 5/11/2011 7:27:19 AM

Fluid Mechanics 295