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Welcome to Edvantage Science: AP Physics 1
Before diving into yiour courrse, we invite you to read this introduction. It outlines key ideas from modern cognitive science research—ideas that shaped how this textbook was designed to help you learn more effectively and remember for longer.Many students who succeed in high school struggle in university—not because they’re less capable, but because they weren’t taught how to learn effectively. We want to change that. What you’re about to read is not extra; it’s essential preparation that will help you avoid asking the question many college students eventually do: “Why wasn’t I taught this earlier?”
What Is Learning?
Let’s start with a research-based definition from Make It Stick: The Science of Successful Learning: “Learning is acquiring knowledge and skills and having them readily available from memory so you can make sense of future problems and opportunities.”
Brown, Roediger, & McDaniel (2014), p. 2
This means that real learning isn’t just about reading or listening. It’s about getting ideas and skills into your long-term memory so that you can apply them when you face new challenges. That’s what this course—and this textbook—are designed to help you do.
How Your Brain Learns: A Simplified Model
The diagram you’ll see below is a simplified version of the Cognitive Theory of Multimedia Learning (Mayer, 2005), which integrates decades of research on how people learn from words, images, and experiences.
Reproduced with permission from Failsafe Strategies for Science and Literacy
The diagram highlights three key memory systems:
• Sensory memory: Information enters through your senses (especially eyes and ears).
• Working memory: Where your brain holds and organizes new information.
• Long-term memory: Where learning is stored permanently as it’s integrated into your prior knowledge. Understanding how these systems interact is critical to you learning effectively.
Attention: The Gatekeeper of Learning
You can’t learn what you don’t notice. Your sensory memory only processes what you give attention to. If you're distracted—by your phone, for example—you won’t retain what your teacher is saying, no matter how good the explanation.
That’s why attention is the first essential skill for learning. This textbook is structured to help you focus on what matters most in each lesson.
Cognitive Load: Why Less Is More
Your working memory can only hold about 4 to 7 pieces of new information at once—and often fewer when the material is unfamiliar. If too much is thrown at you at once, your brain gets overwhelmed and can’t process it deeply enough to store in long-term memory.
That’s called cognitive overload, and it’s a major barrier to learning.
To address this, Edvantage Science: AP Physics 1 is carefully designed to reduce unnecessary cognitive load so your brain can focus on understanding core ideas. Clear diagrams, structured problem-solving prompts, and progressive practice are just a few of the strategies we’ve built in to help.
The Importance of Forgetting (and Remembering)
Forgetting is normal—it happens to everyone. In fact, it begins as soon as we learn something. This is known as the Ebbinghaus Forgetting Curve and is illustrated in the graph below. But here’s the good news: forgetting can be slowed—and even reversed—by reviewing information at spaced intervals as shown in the graph.
This process is called retrieval practice, and it’s one of the most powerful learning strategies discovered in cognitive science. You’ll find many activities in this book designed to help you remember what you learn—when you need it most.
This Book Supports How You Learn
Partially Worked Solutions
Every sample problem includes a QR code. After thinking through the problem using guiding prompts, you can scan the code to reveal a partially worked solution. This helps you compare your thinking to the partial solution, without overloading your working memory with too many steps at once.
What’s My Mark?
At the end of each unit, you become the teacher. You'll mark a sample student’s test—allowing you to retrieve, evaluate, and apply what you’ve learned. This kind of elaborative practice strengthens long-term memory and makes your learning more durable.
Lab@Desk Activities
These short, focused paper-based labs let you connect abstract concepts to concrete experiences. This type of activity helps reduce cognitive load and build deeper understanding.
Remembering
Ebbinghaus
Periodically, you’ll encounter questions that revisit earlier topics—even if they’re not part of your current unit. This deliberate spacing and retrieval strengthens memory and prepares you to use physics concepts flexibly, not just repeat them on a test.
Final Thoughts
If you’ve read this far, congratulations—you’ve already taken the first step in developing strategies that help you learn. This textbook is more than a source of information; it’s a tool to help you become a better learner.
Good luck on your journey through AP Physics 1—and don’t forget to visit edvantagescience.com for additional resources and support.
References and Suggested Reading
Ashman, G. (2022). Little Guide for Teachers: Cognitive Load Theory
Brown, P. C., Roediger, H. L., & McDaniel, M. A. (2014). Make It Stick: The Science of Successful Learning.
Christodoulou, D. (2014). Seven Myths About Education
Kirschner, P., Hendrick, C. (2020). How Learning Happens: Seminal Works in Educationl Psychology and What They Mean in Practice
Ebbinghaus, H. (1885/1913). Memory: A Contribution to Experimental Psychology.
Mayer, R. E. (2005). The Cambridge Handbook of Multimedia Learning. Mirabelli, S. Sandner, L. (2023) Failsafe Strategies for Science and Literacy
8 Fluids
This unit focuses on Fluids from the AP Physics 1 CED and will cover the following ideas.
Think about feeling the air around you or watching water pour; fluids are all around us, and in this unit, we’ll explore their unique properties like pressure and flow. You’ll discover how fluid pressure increases with depth, which directly explains the crucial upward buoyant force acting on any object immersed in a fluid. We’ll use Archimedes’ Principle as a tool to figure out exactly how strong that buoyant force is and predict whether something will float or sink. Finally, we’ll investigate the principles governing fluid motion, like the continuity equation, to understand how moving fluids behave in different situations.
By the end of this chapter, you should know the meaning of these key terms:
• absolute pressure
• Bernoulli’s Principle
• buoyant force
• Continuity equation
• density
• guage pressure
• pressure
• specific gravity
By the end of this chapter, you should be able to use and know when to use the following formulae:
“Why does a hot air balloon float? This unit will explore fluid concepts necessary to answer this question and more
8.1 Internal Structure and Density
Warm Up
Scan the QR code and watch the video. In your own words make at least three observations and one inference to explain what is happening in the jar.
Matter most commonly exists in one of several states: solid, liquid, gas, or plasma. These different states are called the common phases of matter. Solids maintain a definite shape and a specific volume. Liquids, while possessing a definite volume, adapt their shape to that of their container. Gases, in contrast, lack both a definite shape and a specific volume, as their molecules move to fill the entire container they occupy. Plasmas also share this characteristic of having neither a definite shape nor volume. It is important to note that liquids, gases, and plasmas are classified as fluids because they deform under shearing forces, whereas solids resist such deformation. The ease and speed with which fluids deform, and thus flow, is determined by a property known as viscosity. The phases of matter and the distinction between fluids and solids can be understood by examining the interatomic forces that govern the behavior of matter in each phase.
8.1.1
(a) Atoms in a solid always have the same neighbors, held near home by forces represented here by springs. These atoms are essentially in contact with one another. A rock is an example of a solid. This rock retains its shape because of the forces holding its atoms together.
(b) Atoms in a liquid are also in close contact but can slide over one another. Forces between them strongly resist attempts to push them closer together and also hold them in close contact. Water is an example of a liquid. Water can flow, but it also remains in an open container because of the forces between its atoms.
(c) Atoms in a gas are separated by distances that are considerably larger than the size of the atoms themselves, and they move about freely. A gas must be held in a closed container to prevent it from moving out freely.
(d) A plasma is composed of electrons, protons, and ions that, like gases, are spaced far apart and move about freely.
Phases of Matter
Figure
Density
In solids, atoms are in close contact, with forces between them that allow the atoms to vibrate but not to change positions with neighboring atoms. These forces can be conceptualized as springs that can be stretched or compressed, but not easily broken. Consequently, a solid resists all types of stress. A solid cannot be easily deformed because its constituent atoms are not able to move about freely. Solids also resist compression, because their atoms form part of a lattice structure in which the atoms maintain a relatively fixed distance apart. Under compression, the atoms would be forced into one another. Most of the examples we have studied so far have involved solid objects which deform very little when stressed.
In contrast, liquids deform easily when stressed and do not return to their original shape once the force is removed. This is because the atoms are free to slide about and change neighbors, a property we define as flow. Thus, liquids are classified as a type of fluid, with the molecules held together by mutual attraction. When a liquid is placed in a container with no lid, it remains in the container, provided the container has no openings below the liquid’s surface. Because the atoms are closely packed, liquids, like solids, resist compression.
Atoms in gases and charged particles in plasmas are separated by distances that are large compared with the size of the particles. The forces between the particles are therefore very weak, except during collisions. As a result, gases and plasmas not only flow and are therefore considered to be fluids, but are also relatively easy to compress because of the significant space and minimal force between the particles. When placed in an open container, gases, unlike liquids, will escape. The major distinction is that gases are easily compressed, whereas liquids are not. Plasmas are difficult to contain due to their high energy content. When discussing how substances flow, we shall generally refer to both gases and liquids as fluids, distinguishing between them only when their behavior differs.
To further simplify the study of fluid statics and dynamics, we often introduce the concept of an ideal fluid. An ideal fluid is a theoretical construct that is assumed to be both incompressible and non-viscous. Incompressibility means that the fluid’s density remains constant, even under pressure changes. Non-viscous means that the fluid has no internal friction, and therefore experiences no resistance to flow. While no real fluid is truly ideal, this model allows us to apply simplified equations and principles to analyze fluid behavior in many practical situations.
Which weighs more, a ton of feathers or a ton of bricks? This old riddle plays with the distinction between mass and density. A ton is a ton, of course; but bricks have much greater density than feathers, and so we are tempted to think of them as heavier.
Density, as you will see, is an important characteristic of substances. For example, in determining whether an object sinks or floats in a fluid. Density is the mass per unit volume of a substance or object. In equation form, density is defined as
where the Greek letter ρ (rho) is the symbol for density, m is the mass, and V is the volume occupied by the substance.
In the riddle regarding the feathers and bricks, the masses are the same, but the volume occupied by the feathers is much greater, since their density is much lower. The SI unit of density is kg/m3, representative values are given in Table 8.1.1. The metric system was originally devised so that water would have a density of , equivalent to 1 gm/cm3, equivalent to 103 kg/m3. Thus the basic mass unit, the kilogram, was first devised to be the mass of 1000 mL (or 1 litre) of water, which has a volume of 1000 cm3.
Ideal Fluid
As you can see by examining Table 8.1.1, the density of an object may help identify its composition. The density of gold, for example, is about 2.5 times the density of iron, which is about 2.5 times the density of aluminum. Density also reveals something about the phase of the matter and its substructure. Notice that the densities of liquids and solids are roughly comparable, consistent with the fact that their atoms are in close contact. The densities of gases are much less than those of liquids and solids, because the atoms in gases are separated by large amounts of empty space.
Sample Problem — Calculating the Mass of a Reservoir From Its Volume
A reservoir has a surface area of 50.0 km2 and an average depth of 40.0 m. What mass of water is held behind the dam?
What to Think About
1. Solve for m.
2. To find volume we need to multiple the surface area times it’s average depth.
How to Do It
3. Find the density of water from Table 8.1.1.
4. Solve for m
8.1 Review Questions
1. The image above shows a glass of ice water filled to the brim. Will the water overflow when the ice melts? Explain your answer.
5. Suppose you have a coffee mug with a circular cross section and vertical sides (uniform radius). What is its inside radius if it holds 375 g of coffee when filled to a depth of 7.50 cm? Assume coffee has the same density as water.
2. Gold is sold by the troy ounce (31.103 g). What is the volume of 1 troy ounce of pure gold?
6. A trash compactor can reduce the volume of its contents to 0.350 their original value. Neglecting the mass of air expelled, by what factor is the density of the rubbish increased?
3. Mercury is commonly supplied in flasks containing 34.5 kg (about 76 lb). What is the volume in liters of this much mercur y?
4. a) What is the mass of a deep breath of air having a volume of 2.00 L?
b)Discuss the effec t taking such a breath has on your body’s volume and density.
7. A 2.50-kg steel gasoline can holds 20.0 L of gasoline when full. What is the average density of the full gas can, taking into account the volume occupied by steel as well as by gasoline?
Remembering Ebbinghaus
1.A student compresses a spring with a spring constant of 600 N/m by 0.1 m and uses it to launch a 1.5 kg cart on a horizontal frictionless surface.
(a) What is the speed of the cart when it leaves the spring?
(b) What type of energy transformation takes place?
Lab@Desk
Four spheres are hung from a variety of different springs. The table below describes the characteristics of both the spheres and the springs from which they are hung. Use this information to rank the density of each sphere from least to greatest. Show work supporting your ranking.
8.2 Pressure
Warm Up
Scan the QR code to watch a short video demonstrating an example of pressure. Why do you think the water flow in the horizontal straw changes as the vertical straw depth in the bottle changes?
Defining Pressure
You’ve likely encountered the concept of pressure in various contexts, from discussions about blood pressure (whether it’s high or low) to weather reports detailing high- and low-pressure systems. These everyday examples highlight the importance of pressure in understanding the behavior of fluids.
Pressure (P), a scalar quantity, is defined as the force (F) applied perpendicularly to a given area (A): This relationship reveals a crucial concept: a specific force can produce vastly different effects depending on the area over which it’s exerted. For instance, in Figure 8.2.1 a gentle finger poke exerts a force that causes only minor irritation. However, the same force, when applied across the minuscule area of a needlepoint, can easily penetrate the skin. This demonstrates that pressure is not simply force, but force concentrated over a specific area.
The SI unit for pressure is the pascal (Pa), equivalent to one Newton per square meter (1 Pa = 1 N/m²). The unit pascal is named after French scientist, Blaise Pascal (1623-1662). However, pressure is also commonly expressed in other units. Meteorologists often use millibars (mb) to describe atmospheric pressure (100 mb = 1 x 104 Pa), while tire pressure gauges may display readings in pounds per square inch (psi). In the medical field, blood pressure is still frequently measured in millimeters of mercury (mm Hg). While pressure is a property of all states of matter, it plays a particularly vital role in understanding the behavior of fluids, which are substances that flow.
8.2.1 (a) While the person being poked with the finger might be irritated, the force has little lasting effect. (b) In contrast, the sameforce applied to an area the size of the sharp end of a needle is great enough to break the skin.
FIGURE
Atmospheric Pressure
The force exerted on the end of a tank by a static or stationary fluid is perpendicular to its inside surface. This direction is inherent because fluids cannot withstand shearing (sideways) forces; they can only exert forces perpendicular to a surface. Fluid pressure itself has no intrinsic direction, as it is a scalar quantity. However, the forces due to pressure have well-defined directions: they are always exerted perpendicular to any surface. For example, consider the force exerted by the air pressure inside a tire as shown by Figure 8.2.2; this force acts perpendicularly to the tire's inner surface at every point.
FIGURE 8.2.2 Pressure inside this tire exerts forces perpendicular to all surfaces it contacts. The arrows give representative directions and magnitudes of the forces exerted at various points. Note that static fluids do not exert shearing forces
Atmospheric pressure is an example of pressure due to the weight of a fluid, in this case due to the weight of air above a given height. The atmospheric pressure at the Earth’s surface varies a little due to the large-scale flow of the atmosphere induced by the Earth’s rotation (this creates weather “highs” and “lows”). However, the average pressure at sea level is given by the standard atmospheric pressure Patm, measured to be
This relationship as illustrated in Figure 8.2.3 means that, on average, at sea level, a column of air above 1.00 m2 of the Earth’s surface has a weight of 1.01 × 105 N equivalent to 1 atm.
8.2.3 Atmospheric pressure at sea level averages 1.01 × 105 Pa (equivalent to 1 atm), since the column of air over this 1 m2, extending to the top of the atmosphere, weighs 1.01 × 105 N
Gauge and Absolute Pressure
It is also important to distinguish between gauge pressure and absolute pressure. Gauge pressure is the pressure relative to atmospheric pressure. For example, a tire pressure reading of 30 psi is gauge pressure, meaning the pressure inside the tire is 30 psi greater than the atmospheric pressure outside. Absolute pressure, on the other hand, is the total pressure, including atmospheric pressure. To obtain absolute pressure, you would add atmospheric pressure to the gauge pressure. P = P guage + Patm
Sample Problem — Calculating Force Exerted by the Air: What Force Does a Pressure Exert?
An astronaut is working outside the International Space Station where the atmospheric pressure is essentially zero. The pressure gauge on her air tank reads 6.90 × 106 Pa. What force does the air inside the tank exert on the flat end of the cylindrical tank, a disk 0.150 m in diameter?
What to Think About
1. Solve for F
2. Determine surfacer area of end of cylinder. Which is a circle.
3. Solve for F
4.Consider your answer
Practice Problems — Calculating Pressure
How to Do It
1. As a woman walks, her entire weight is momentarily paced on the high-heeled shoes. Calculate the pressure exerted on the floor by the heel if it has an area of 1.50 cm2 and the woman’s mass is 55.0 kg. Express the pressure in Pa. (In the early days of commercial flight, women were not allowed to wear highheeled shoes because aircraft floors were too thin to withstand such large pressures.)
12. The pressure exerted by a phonograph needle on a record is surprisingly large. If the equivalent of 1.00 g is supported by a needle, the tip of which is a circle 0.200 mm in radius, what pressure is exerted on the record in N/m2?
13.Nail tips exert tremendous pressures when they are hit by hammers because they exert a large force over a small area. What force must be exerted on a nail with a circular tip of 1.00 mm diameter to create a pressure of 3.00 × 109 N/m2 (This high pressure is possible because the hammer striking the nail is brought to rest in such a short distance.)
Pressure and Depth
The pressure within a fluid also varies with depth. You've likely experienced this phenomenon if your ears have ever popped on a plane flight or ached during a deep dive in a swimming pool. At the Earth's surface, the air pressure you experience is due to the weight of the air column above you. This pressure decreases as you ascend in altitude, as the weight of the air above diminishes. Conversely, underwater, the pressure increases with increasing depth. The pressure exerted on you underwater results from the combined weight of the water column above you and the atmosphere pressing down on the water's surface. You might notice a subtle air pressure change during a rapid elevator ride, such as in a tall skyscraper. However, the pressure change is far more noticeable when descending even a meter or so below the surface of a swimming pool. This disparity arises from the fact that water is significantly denser than air—approximately 775 times denser.
We can derive a relationship to quantify this pressure variation with depth. Consider a container filled with a fluid as in Figure 8.2.4. The fluid at the bottom of the container supports the weight of the fluid above it. The pressure (P) exerted on the bottom is therefore the weight of the fluid (mg) divided by the supporting area (A), which is the area of the container's bottom:
P = mg/A
We can express the fluid's mass (m) in terms of its volume (V) and density (ρ):
m = ρV
The volume of the fluid is related to the container's dimensions:
V = Ah
where A is the cross-sectional area and h is the depth of the fluid. Combining these equations, we get
m = ρAh
Substituting this into the pressure equation yields:
P = (ρAhg)/A.
The area cancels out, leaving us with:
FIGURE 8.2.4 The bottom of this container supports the entire weight of the fluid in it. The vertical sides cannot exert an upward force on the fluid (since it cannot withstand a shearing force), and so the bottom must support it all.
P = ρgh
This equation holds true beyond the specific scenario we used to derive it. Even without the container, the surrounding fluid exerts this pressure, maintaining static equilibrium. Therefore, the equation P = ρgh represents the pressure due to the weight of any fluid with an average density ρ at a depth h below its surface. For liquids, which are nearly incompressible, this equation remains valid to great depths. For gases, which are compressible, the equation is applicable as long as density changes are relatively small over the depth considered. The next Sample Problem illustrates this situtation.
The total pressure within a fluid also depends on the external pressure ( P o ) exerted on it’s surface by the atmosphere. In general the total pressure at a depth h in a fluid of density ρ is:
P = P o + ρgh
Sample Problem — Calculating Average Density: How Dense Is a Column Air?
Calculate the average density of the atmosphere, given that it extends to an altitude of 120 km. Compare this density with that of air listed in Table 8.1.1.
What to Think About
1. Solve for ρ
2. Enter known values and solve for ρ.
4. Consider your answer
How to Do It
Practice Problems — Calculating Pressure and Depth
1. Calculate the depth below the surface of water at which the pressure due to the weight of the water equals 1.00 atm.
2. What depth of mercury creates a pressure of 1.00 atm?
3. The greatest ocean depths on the Earth are found in the Marianas Trench near the Philippines. Calculate the pressure due to the ocean at the bottom of this trench, given its depth is 11.0 km and assuming the density of seawater is constant all the way down.
Fluid Shear and Its Consequences
A shearing force is a force applied parallel to the surface of an object or fluid layer, causing it to deform by sliding rather than by being compressed or stretched. Unlike solids, which can resist such forces and maintain their shape, fluids offer no permanent resistance to shearing forces, continuously deforming under their influence. The inability of fluids to withstand shearing forces leads to several important consequences:
• First, fluids conform to the shape of their containers. Unlike solids, which have a fixed shape, fluids flow and adapt to the geometry of any vessel they occupy.
• Second, pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and to the walls of the container. This principle, known as Pascal’s Law, is fundamental to the operation of hydraulic systems, such as those found in car brakes and construction equipment.
• Third, fluids exert pressure perpendicular to any surface they contact, as previously discussed. This characteristic is crucial for understanding phenomena like buoyancy, where the upward force on an object submerged in a fluid is a result of the increasing pressure with depth.
Remembering Ebbinghaus
1. A mass on a spring is oscillating back and forth horizontally with no friction. At what point is its speed greatest, and how is that related to concepts of acceleration and force?
2. Two disks of the same mass and radius roll down the same incline. Disk A is solid; Disk B is a thin ring. Which one reaches the bottom first, and why?
3. You whirl a bucket of water in a vertical circle fast enough so the water doesn’t fall out at the top. What force keeps the water in the bucket at the top of the path?
4.A physical pendulum and a mass-spring oscillator both have the same period. If you move both to the Moon (where g is smaller), which system’s period changes more, and why?
8.2 Review Questions
1. A submarine is stranded on the bottom of the ocean with its hatch 25.0 m below the surface. Calculate the force needed to open the hatch from the inside, given it is circular and 0.450 m in diameter. Air pressure inside the submarine is 1.00 atm.
2. How much force is exerted on one side of an 8.50cm by 11.0 cm sheet of paper by the atmosphere? How can the paper withstand such a force?
3. What pressure is exerted on the bottom of a 0.500-m-wide by 0.900-m-long gas tank that can hold 50.0 kg of gasoline by the weight of the gasoline in it when it is full?
4. Calculate the average pressure exerted on the palm of a shot-putter’s hand by the shot if the area of contact is and he exerts a force of 800 N on it. Express the pressure in and compare it with the pressures sometimes encountered in the skeletal system.
8.3 Fluids and Newton’s Laws
Warm Up
Take a piece of foil, roll it up into a ball and drop it into water. Does it sink? Why or why not? Can you make it sink? How would you explain your results when the density of aluminum is 2.7 times the density of water (Table 8.1.1).
Forces and Fluids
When we dive into mechanics, we often start by analyzing the motion of solid objects – blocks sliding, balls falling, and so on. But the fundamental rules governing motion, Newton’s laws, apply universally, extending their reach to the vast numbers of tiny particles that constitute fluids, whether we’re talking about liquids or gases. Each individual atom or molecule within a fluid follows these same basic laws. The intricate movements we see in fluids, from gentle currents to violent turbulence, are the result of the combined actions of these countless microscopic components.
Let’s think about a single particle within a fluid. According to Newton’s First Law – the principle of inertia – this particle will maintain a constant velocity (meaning constant speed in a constant direction) unless it experiences a net external force. So, a particle in a still fluid remains in its state of motion or rest unless forces like gravity or pressure from its surroundings push it. If it’s part of a steady flow, it will continue with that velocity unless a force changes things.
Newton’s Second Law, stated as Fnet =ma, tells us precisely how the velocity of a fluid particle changes. If there’s a net force acting on a tiny bit of fluid – perhaps due to differences in pressure pushing harder on one side than another, or the pull of gravity, or viscosity from neighboring layers – that net force causes the fluid particle to accelerate. The resulting change in the particle’s velocity is directly proportional to the strength of that net force and happens in the same direction as the force, while being inversely proportional to the particle’s mass.
Newton’s Third Law, the law of action and reaction, plays a vital role in the internal dynamics of fluids. When fluid particles interact, such as bumping into one another or pushing against a boundary, they exert equal-magnitude, oppositely-directed forces on each other. These mutual forces between particles are what allow pressure and momentum to be transmitted throughout the fluid, enabling it to push on submerged objects or flow in response to external influences.
Ultimately, the observable, macroscopic properties of a fluid – its flow, pressure, resistance to motion (viscosity), and how it interacts with objects – are emergent phenomena. They arise from the collective behavior of all those particles, each following Newton’s laws, interacting with each other and responding to any external forces applied to the fluid body as a whole. Consider the pressure at the bottom of a swimming pool: it’s the cumulative effect of the gravitational forces on trillions upon trillions of water molecules above that point, with their weight transmitted downwards through the repulsive forces between molecules as described by Newton’s Third Law.
A Deeper Dive into the Buoyant Force
You may have noticed it’s easier to lift a large rock or other heavy object in a swimming pool than out of it. Or a beach ball held underwater requires a significant push to keep it down. These experiences highlight an upward force exerted by fluids. Objects like boats float because of this upward push, and even objects that sink still experience it as a partial support. For instance, an anchor feels heavier when lifted out of the water than it does when it’s submerged. This upward force is always present on objects within a fluid, regardless of whether they float, sink, or remain suspended. This net upward force exerted by a fluid that opposes the weight of an immersed object is called the buoyant force
To understand the magnitude of this buoyant force, consider what happens when a submerged object is removed from a fluid. The space it occupied is then filled by the surrounding fluid, which has a specific weight. This weight of the displaced fluid is supported by the surrounding fluid. Therefore, the buoyant force must be equal to the weight of the fluid that was displaced by the object.
This principle is a testament to the insight of the Greek mathematician and inventor Archimedes (c. 287–212 BCE), who articulated it long before the formal concepts of force were established. Archimedes’ principle states: The buoyant force on an object equals the weight of the fluid it displaces. In equation form, Archimedes’ principle is:
Fb = Wfluid displaced
Where Fb is the buoyant force and Wfluid displaced is the weight of the fluid displaced by the object. Archimedes’ principle holds true in general, for any object in any fluid, whether the object is partially or totally submerged (Figure 8.3.1).
FIGURE 8.3.1 The force of gravity on the metal block in (a) (that is, its weight) is 20 N. When the block is placed in water the volume of water displaces a volume of water equal to its own volume of 300 mL.
The buoyant force is a direct consequence of how fluid pressure changes with depth. As discussed earlier, the pressure in a fluid increases as you go deeper due to the weight of the fluid above. When an object is submerged, the upward pressure on its bottom surface is greater than the downward pressure on its top surface. This pressure difference results in a net upward force on the object—the buoyant force.
Imagine a simple cube submerged in water. The water pressure pushing up on the bottom of the cube is stronger than the water pressure pushing down on the top. This imbalance in pressure creates a net upward force on the cube.
The Archimedes’ principle provides a way to calculate the magnitude of this buoyant force. It tells us that the buoyant force doesn’t depend on the object’s weight or composition, but only on the weight of the fluid it displaces.
Here’s a more intuitive way to think about it:
When an object is submerged, it pushes aside (displaces) a volume of the surrounding fluid. This volume of fluid has a certain weight. The buoyant force is equal to that weight. Essentially, the surrounding fluid is trying to reclaim the space occupied by the submerged object. The force it uses to do so is the buoyant force.
Density and Archimedes Principle
Density plays a crucial role in Archimedes’ principle. The average density of an object is what ultimately determines whether it floats. If its average density is less than that of the surrounding fluid, it will float. This is because the fluid, having a higher density, contains more mass and hence more weight in the same volume. The buoyant force, which equals the weight of the fluid displaced, is thus greater than the weight of the object. Likewise, an object denser than the fluid will sink.
The extent to which a floating object is submerged depends on how the object’s density is related to that of the fluid. In For example, an unloaded ship has a lower density and less of it is submerged compared with the same ship loaded. We can derive a quantitative expression for the fraction submerged by considering:
The volume submerged equals the volume of fluid displaced, which we call Vfl. Now we can obtain the relationship between the densities by substituting into the expression ρ = m/V. This gives
Specific Gravity
Where ρobj is the average density of the object and ρfl is the density of the fluid. Since the object floats, its mass and that of the displaced fluid are equal, and so they cancel from the equation, leaving We use this last relationship to measure densities. This is done by measuring the fraction of a floating object that is submerged—for example, with a hydrometer. It is useful to define the ratio of the density of an object to a fluid (usually water) as specific gravity:
where is the average density of the object or substance and ρw is the density of water at 4.00 0C. Specific gravity is dimensionless, independent of whatever units are used for. If an objec t floats, its specific gravity is less than one.
If it sinks, its specific gravity is greater than one. Moreover, the fraction of a floating object that is submerged equals its specific gravity. If an object’s specific gravity is exactly 1, then it will remain suspended in the fluid, neither sinking nor floating. Scuba divers try to obtain this state so that they can hover in the water. We measure the specific gravity of fluids, such as battery acid, radiator fluid, and urine, as an indicator of their condition.
FIGURE 8.3.2 This hydrometer is floating in a fluid of specific gravity 0.87. The glass hydrometer is filled with air and weighted with lead at the bottom. It floats highest in the densest fluids and has been calibrated and labeled so that specific gravity can be read from it directly.
Sample Problem — Submerged or Not Icebergs?
An iceberg, composed of freshwater ice, is observed floating motionless in the salty seawater of the North Atlantic. Calculate the proportion (expressed as both a fraction and a percentage) of the iceberg’s total volume that lies below the surface of the seawater. See Table 8.1.1 for densities
What to Think About
1. Consider an object floats in static equilibrium when the upward buoyant force exerted by the fluid exactly balances the downward force of the object’s weight.
2. Find the buoyant force. Use Archimedes’ principle. The buoyant force equals the weight of the fluid displaced, which depends on the fluid density and the submerged volume.
3. Express the object’s weight. The weight of the iceberg depends on its total volume and its own density.
4. Set up the equilibrium equation: Substitute the expressions for buoyant force and weight into the equilibrium condition from Step 1.
5. Solve for the fraction submerged.
6. Calculate the numerical value: Substitute the given densities into the ratio to find the fraction submerged.
7. Convert to percentage. The proportion of the iceberg’s volume submerged is approximately 0.895, or 89.5%
More Density Measurements
How to Do It
8.3.3 (a) A coin is weighed in air. (b) The apparent weight of the coin is determined while it is completely submerged in a fluid ofknown density. These two measurements are used to calculate the density of the coin.
One of the most common techniques for determining density is shown below in Figure 8.3.3.
FIGURE
An object, here a coin, is weighed in air and then weighed again while submerged in a liquid. The density of the coin, an indication of its authenticity, can be calculated if the fluid density is known. This same technique can also be used to determine the density of the fluid if the density of the coin is known. All of these calculations are based on Archimedes’ principle. Archimedes’ principle states that the buoyant force on the object equals the weight of the fluid displaced. This, in turn, means that the object appears to weigh less when submerged; we call this measurement the object’s apparent weight. The object suffers an apparent weight loss equal to the weight of the fluid displaced. Alternatively, on balances that measure mass, the object suffers an apparent mass loss equal to the mass of fluid displaced. That is:
apparent weigth loss = weight of fluid displaced or
apparent mass loss = mass of fluid displaced
The next Sample Problem illustrates the use of this technique.
Sample Problem — Is the Coin Authentic?
The mass of an ancient Greek coin is determined in air to be 8.630 g. When the coin is submerged in water as shown in Figure 8.3.1, its apparent mass is 7.800 g. Calculate its density, given that water has a density of and that effects caused by the wire suspending the coin are negligible.
What to Think About
1. To calculate the coin’s density, we need its mass and its volume. Remember the mass of water displaced equals the apparent mass loss
2. The volume of the coin equals the volume of water displaced.
3. The volume of water displaced Vw can be found by using the equation for density.
4. Determine denisty of coin.
How to Do It
8.3 Review Questions
1. Draw three free body diagrams ilustrating and object floating, sinking and rising. Your diagrams should show the magnitude and direction of the objects weight and buoyant force.
2. For each diagram in question 1, indicate the specific gravity of the object in each situation.
5. If your body has a density of 9.95 kg/m3, what fraction of you will be submerged when floating gently in: (a)freshwater? (b)salt water?
3. Suppose a 60.0-kg woman floats in freshwater with of her volume submerged when her lungs are full of air. What is her average density?
6.A rock with a mass of 540 g in air is found to have an apparent mass of 342 g when submerged in water. (a) What mass of water is displaced?
4. Find the density of a fluid in which a hydrometer having a density of 0.750 g/mL, floats with 20% of its volume submerged.
(b) What is the volume of the rock?
(c) What is its average density? Is this consistent with the value for granite (2.7 g/cm3)?
8.4 Fluids and Conservation Laws
Warm Up
Scan the QR code to watch a video showing a phenomenon you will investigate in this topic. Describe what you observe and suggest a possible explanation. You may wish to review the headings of this unit for potential clues.
Describing Flow: Volume Flow Rate
Have you ever watched a river flow, seen smoke rise from a chimney, or felt the wind on your face? These are all examples of fluids in motion. Recall a fluid is any substance that can flow and take the shape of its container – primarily liquids and gases. While the study of stationary fluids, called fluid statics, deals with pressure and buoyancy, fluid dynamics explores the more complex world of fluids in motion. Understanding how fluids move is essential for designing everything from airplanes and pipelines to artificial hearts.
One of the most fundamental questions we can ask is: why do fluids flow in the first place? Imagine water sitting in a perfectly level pipe, closed at both ends. It doesn’t move. Now, imagine tilting the pipe, or perhaps using a pump to increase the pressure at one end. The water begins to flow. The driving force behind most fluid motion is a difference in pressure. A difference in pressure between two locations causes a fluid to flow. Fluids naturally move from regions of higher pressure to regions of lower pressure, much like a ball rolls downhill from higher gravitational potential energy to lower potential energy. This pressure difference provides the net force needed to accelerate the fluid and overcome any resistive forces like viscosity (internal friction in the fluid) or friction with the container walls.
When a fluid moves through a pipe, tube, or channel, we often want to quantify how much fluid is passing through a particular point over some time. Think about filling a bucket with a garden hose. The rate at which the bucket fills depends on how much water comes out of the hose each second. This concept is formalized as the volume flow rate, often symbolized as Q or simply V/t It represents the volume (V) of fluid that passes through a specific cross-section of the pipe per unit time (t).
How does this volume flow rate relate to the properties of the flow itself? Consider a segment of fluid moving through a pipe with a uniform cross-sectional area A. In a time interval t, the fluid particles at the leading edge of this segment will travel a distance d. If the fluid is moving with an average speed v, then this distance is given by d=vt. The volume of fluid that has passed through the area A in time t is the volume of a cylinder (or other shape depending on the pipe shape) with base area A and length d. So, the volume is V = A × d = A × (vt)
To find the volume flow rate, we divide this volume by the time interval t:
Fluid Flow
Conservation of Mass and the Continuity Equation
The time t cancels out, leaving us with a fundamental relationship:
This equation tells us that the rate at which matter flows into a location is proportional to the cross-sectional area of the flow and the speed at which the fluid flows. A wider pipe (larger A) or a faster flow (larger v) will result in a greater volume of fluid passing a point per second. The units for volume flow rate are typically cubic meters per second (m3/s) in the SI system.
Now, let’s consider a fluid flowing steadily through a pipe that might change in size. Imagine a pipe that is wide in one section and narrow in another. We’ll focus on a specific type of fluid called an incompressible fluid. An incompressible fluid is one whose density (ρ) remains constant regardless of changes in pressure. Most liquids, like water or oil, are excellent approximations of incompressible fluids under typical conditions. Gases are generally compressible, but can sometimes be treated as incompressible if the pressure changes are small and the flow speeds are relatively low compared to the speed of sound.
Consider Figure 8.4.1, an incompressible fluid is flowing through a pipe that expands. Let the first section have a cross-sectional area A1 and the fluid speed there be v1. Let the second, narrower section have a cross-sectional area A2 and the fluid speed there be v2.
FIGURE 8.4.1 An incompressible fluid is flowing through a pipe that increases in diameter. The lines in the diagram are called streamlines Streamlines are imaginary lines drawn in a fluid that represent the path a massless fluid particle would follow as the fluid flows.
A fundamental principle in physics is the conservation of mass. Mass cannot simply appear or disappear. If the flow is steady (meaning the flow pattern isn’t changing over time) and there are no leaks or sources/sinks of fluid between section 1 and section 2, then the mass of fluid entering section 1 per unit time must equal the mass of fluid exiting section 2 in the same unit time. The rate at which matter enters a fluid-filled tube open at both ends must equal the rate at which matter exits the tube.
We can express the mass flow rate (mass per unit time or m/t) using the volume flow rate (V/t=Av) and the density (ρ). Since density is mass per unit volume (ρ=m/V), mass is density times volume (m=ρV). Therefore, the mass flow rate is:
Applying the principle of mass conservation, the mass flow rate at section 1 must equal the mass flow rate at section 2:
Implications of the Continuity Equation
This equation applies generally. However, we made the crucial assumption that the fluid is incompressible. For an incompressible fluid, the density is constant throughout: ρ1=ρ2. Since the density is the same and non-zero, we can divide both sides of the equation by ρ:
This elegantly simple result is known as the continuity equation for incompressible fluid flow. It mathematically expresses the conservation of mass flow rate in incompressible fluids. It tells us that the product of the cross-sectional area and the fluid speed must remain constant along a pipe of varying size, as long as the fluid is incompressible and there are no leaks or additions of fluid.
The continuity equation A1v1=A2v2 has direct and observable consequences. It shows an inverse relationship between the cross-sectional area of the flow channel and the speed of the fluid. If the area A decreases, the speed v must increase proportionally to keep the product Av constant. Conversely, if the area A increases, the speed v must decrease.
You experience this phenomenon every time you use a garden hose. Water flows relatively slowly through the main body of the hose (larger area A1). If you partially cover the opening with your thumb or attach a nozzle, you decrease the exit area (A2<A1). According to the continuity equation, the water must speed up (v2>v1) as it exits, allowing you to spray the water much farther.
Another example is a river. In wide, deep sections (large A), the water flows slowly and gently. Where the river narrows or becomes shallow (small A), the water speeds up, often forming rapids. The volume of water passing through any cross-section of the river per second (the volume flow rate Av) remains essentially constant (assuming no tributaries or significant evaporation/seepage), demonstrating the principle of continuity.
Quick Check
In the section ‘Implications of the Continuity Equation’, two scenarios are described. Draw a picture of one of the scenarios and label A1, A2, v1 and v2. Caption the diagram by explaining how the Continuity Equation applies.
Sample Problem — Changing River Water Flows
A river flows through a changing channel. At a point where the river is 15.0 meters wide and averages 2.50 meters deep, the water flows at an average speed of 0.800 m/s. Further downstream, the channel narrows to a width of 10.0 meters, and the average depth decreases to 2.00 meters. Assuming the water is incompressible and the flow is steady, what is the average speed of the water in this narrower section of the channel?
What to Think About
1. This problem requires the continuity equation
2. Find the two areas of water flow
3.Solve for water speed at second point
How to Do It
The result shows that the water speed increases in the narrower and shallower section, which is expected based on the continuity equation and conservation of mass for an incompressible fluid
Practice Problems —
1. What are the limitations of modeling real-world fluid flow using only the concepts of incompressible fluids and the continuity equation presented here? What other factors might influence fluid behavior?
2.An incompressible fluid flows through a pipe. At point A, the pipe’s cross-sectional area is $0.12 \text{ m}^2$ and the fluid’s speed is 3.0 m/s. At point B, the pipe narrows, and the fluid’s speed is measured to be 5.0 m/s. What is the crosssectional area of the pipe at point B?
3.A main water pipe with a diameter of 20 cm branches into two smaller pipes, each with a diameter of 10 cm. If the water in the main pipe flows at a speed of 1.2 m/s, what is the speed of the water in each of the smaller pipes? (Assume the flow splits equally and smoothly).
4.A fluid flows through a pipe that expands in diameter. At a point where the pipe’s diameter is 8.0 cm, the fluid’s speed is unknown. Further downstream, the pipe widens to a diameter of 12.0 cm, and the fluid’s speed is measured to be 3.6 m/s. What was the speed of the fluid in the narrower section?
Expanding Ideal
Forms of Energy in a Fluid
In our previous discussion, we saw how the principle of mass conservation, expressed through the continuity equation (A1 v1 = A2 v2 ), helps us understand how fluid speed changes in response to changes in the flow area for incompressible fluids. Now, we turn our attention to another fundamental conservation principle: the conservation of energy. How does the energy of a fluid element change as it moves through regions of different pressure, speed, or height?
To simplify our analysis, remember we consider all fluids discussed as an ideal fluid. For the purposes of our exploration of energy conservation, an ideal fluid is primarily one that is incompressible (constant density, ρ) and has no viscosity. Viscosity is essentially fluid friction; neglecting it means we assume no mechanical energy is lost or dissipated as heat due to internal friction within the fluid or friction between the fluid and the pipe walls. We also assume the flow is steady (velocity, pressure, and density at any point don't change with time) and irrotational (no swirling eddies). These assumptions allow us to apply the principle of conservation of mechanical energy directly.
Consider a small element of fluid with mass m and volume V moving within a larger flow. What forms of mechanical energy does it possess?
1. Kinetic Energy: Like any moving object, the fluid element has kinetic energy due to its speed v, given by KE= 1⁄₂mv2. It’s often more convenient to discuss energy per unit volume. Dividing by volume V, and recalling that density ρ=m/V, we get the kinetic energy density:
2. Gravitational Potential Energy: If the fluid element is at a height h relative to some reference level, it possesses gravitational potential energy PE=mgh. The gravitational potential energy density is:
3. Energy Associated with Pressure (Flow Energy): This form is less obvious. Imagine pushing a volume V of fluid into an adjacent region where the pressure is P. The force required is F=PA, where A is the cross-sectional area. To move the entire volume V, this force must act over a distance d such that V=Ad. The work done on the fluid volume is W=Fd=(PA)d=P(Ad)=PV. This work represents energy transferred to the volume V due to the pressure of the surrounding fluid. The energy per unit volume associated with this pressure is:
So, the pressure P itself can be interpreted as a measure of energy per unit volume (specifically, the work done by pressure forces, sometimes called “flow energy” density).
Bernoulli’s Principle
For an ideal fluid in steady flow, the total mechanical energy is conserved along its path. This means the sum of the kinetic energy density, gravitational potential energy density, and the energy density associated with pressure remains constant. This fundamental concept is known as Bernoulli's Principle
Mathematically, we can express this as:
ρ constant
This elegant equation, named after Swiss scientist Daniel Bernoulli, relates the pressure, speed, and height of a fluid at any point along its flow path. More commonly, we apply it to compare two different points (point 1 and point 2) along the flow:
Here, P is the absolute pressure (not gauge pressure), ρ is the fluid density, v is the fluid speed, g is the acceleration due to gravity, and y is the vertical height relative to a chosen zero level. Bernoulli's equation powerfully describes how energy transforms between these three forms as a fluid moves. A difference in gravitational potential energies between two locations in a fluid will result in a difference in kinetic energy and pressure between those two locations that is described by conservation laws like Bernoulli's principle. For instance, if a fluid flows downhill (y1 > y2 ), its potential energy density (ρgy) decreases. This lost potential energy must be converted into either kinetic energy density (1⁄₂ ρv2, meaning the fluid speeds up) or pressure energy (P, meaning the pressure increases), or a combination of both, such that the total sum remains constant.
Sample Problem — Pressure Changes Due to Changing Diameter of a Pipe
Water flows smoothly through a horizontal pipe at point 1. The pipe has an inner diameter of 4.0 cm, and the water’s speed is 2.0 m/s. The pressure at this point is 150,000 Pa. The pipe then narrows and rises to a height of 3.0 meters above point 1, where the inner diameter is 2.0 cm (point 2). Assuming water is an ideal, incompressible fluid and the flow is steady, what is the pressure at point 2?
What to Think About
1. This problem involves a fluid flowing through a pipe where the speed, height, and pressure change between two points. This requires the application of the full Bernoulli’s principle.
2. Find V2 using continuity principle
3. Find area of pipe for both points. Watch units
4. Solve for v2 using continuity principle
How to Do It
Continued on next page
What to Think About
5. Now that we have v2, we can return to Bernoulli’s equation. Substitute all known values (P1, v1, y1, ρ, g, v2, y2) into the equation and solve for the unknown pressure$P2. Point 1 is horizontal, so we can set its height as the reference level, h1 = 0. Point 2 is 3.0 m above point 1, so y2 = 3.0 m.
6. Solve for unknown pressure P2.
Let’s summarize why the answer (P2=90,600 Pa, which is lower than P1=150,000 Pa) makes sense for the Bernoulli’s principle pipe problem.
Bernoulli’s principle is essentially a statement of energy conservation for an ideal fluid flowing along a streamline. It says that the sum of the pressure energy per unit volume (P), the kinetic energy per unit volume , and the gravitational potential energy per unit volume (ρgy) is constant:
In the problem, as the water flows from Point 1 to Point 2:
The pipe narrows: According to the continuity equation (A1v1=A2v2), if the cross-sectional area (A) decreases, the fluid speed (v) must increase to maintain a constant flow rate. In this problem, the pipe narrowed, and the speed increased significantly from 2.0 m/s to 8.0 m/s. An increase in speed means an increase in the kinetic energy term .
The pipe rises: The fluid moves to a higher elevation (from y1=0 m to y2=3.0 m). An increase in height means an increase in the gravitational potential energy term (ρgh).
Since the total energy per unit volume (P+½ρv2+ρgy) must remain constant according to Bernoulli’s principle, and we observed that both the kinetic energy term and the gravitational potential energy term (ρgy) increased from Point 1 to Point 2, the pressure term (P) must decrease to compensate for these increases and keep the total constant.
Therefore, it makes physical sense that the calculated pressure at Point 2 (90,600 Pa) is lower than the pressure at Point 1 (150,000 Pa), because some of the initial pressure energy has been converted into increased kinetic and potential energy as the fluid moved to a higher elevation and faster speed.
A Bernoulli bar chart is a visual tool used to represent the three energy terms in Bernoulli’s equation:
Each bar represents the relative contribution of one term:
• Pressure energy (P)
• Kinetic energy per volume ( ½ρv2)
• Gravitational potential energy per volume (ρgy)
Figure 8.4.2 is a Bernoulli bar chart allowing you to compare two points in a flowing fluid (like inside a pipe or inside and at a hole in a container). Note how energy shifts between pressure energy, kinetci energy (speed), and potential energy (height) — while maintaining total energy balance.
Bernoulli Bar Chart
in both graphs the total relative energy units stay
Practice Problems — Applying Bernoulli’s Principle
Relating Pressure and Speed: A fluid flows horizontally through a pipe that narrows. According to the principle of continuity, the fluid’s speed increases in the narrower section. Based on Bernoulli’s principle, how does the pressure of the fluid in the narrower, faster-moving section compare to the pressure in the wider, slower-moving section, assuming the height remains constant? Explain your reasoning using the terms in Bernoulli’s equation.
Relating Height and Speed: Consider an ideal fluid flowing horizontally in an open channel (like a river) where the pressure at the surface is constant (atmospheric). If the channel narrows significantly, causing the fluid’s speed to increase, how would you expect the height (depth) of the fluid in the narrower, faster-moving section to change compared to the wider, slower section? Explain using Bernoulli’s principle and considering which terms are constant or changing.
FIGURE 8.4.2 A Bernoulli bar chart comparing two points in a leaking container. Point A: Inside the container (high pressure, high height, no speed) Point B: Outside the hole near the bottom of the container (atmospheric pressure, lower height, high speed). Notice that
the same.
Implications and Applications of Bernoulli’s Principle
Bernoulli's equation reveals an important relationship, especially when height changes are negligible (y1 = y2 ). In this case, the equation simplifies to ρ constant . This implies that for horizontal flow, regions where the fluid speed (v) is high must have lower pressure (P), and regions where the speed is low must have higher pressure. This inverse relationship between speed and pressure is responsible for many familiar phenomena:
• Airplane Lift: An airplane wing (airfoil) is shaped so that air tends to flow faster over the curved upper surface than over the flatter bottom surface. According to Bernoulli’s principle, regions of faster airflow correspond to regions of lower pressure. This means the pressure above the wing is lower than below, resulting in a net upward force known as lift. While this pressure difference contributes to lift, it’s important to note that lift also arises from how the wing deflects air downward. According to Newton’s third law, the downward momentum given to the air results in an upward force on the wing. Both Bernoulli’s principle and Newton’s laws help explain how airplanes fly.
• Perfume Atomizer / Paint Sprayer: When the bulb is squeezed, air is forced quickly through a narrow nozzle at the top of a vertical tube connected to a reservoir of liquid. According to Bernoulli’s principle, the high-speed airflow has lower pressure compared to the stationary air in the bottle. This pressure difference causes the higher atmospheric pressure on the liquid in the reservoir to push the liquid up the tube and into the airstream. As the liquid enters the fast-moving air, it breaks into fine droplets and sprays outward.
• The Curveball in Baseball: When a pitcher throws a curveball, they impart spin to the ball. This spin alters the airflow around the ball — the side spinning in the same direction as the ball’s motion drags air along, while the opposite side resists it. This interaction causes the airflow to deflect asymmetrically, resulting in a pressure difference across the ball. The side with deflected, slower air develops higher pressure, while the side where air is carried along experiences lower pressure. This pressure difference produces a force perpendicular to the direction of motion — a phenomenon known as the Magnus effect — causing the ball to curve in flight.
• Bunsen Burner: In a Bunsen burner, natural gas is forced through a narrow nozzle at the base of the barrel. This gas accelerates as it exits the nozzle, forming a highspeed jet. According to Bernoulli’s principle, this rapid motion creates a region of low pressure along the gas stream inside the barrel. The surrounding air, at higher atmospheric pressure, is then pushed into the burner through the adjustable air holes. This mixing of air (oxygen) with fuel before ignition is essential for producing a hotter, cleaner flame. This mechanism of drawing in air due to a fast gas jet is called entrainment and is a valid application of Bernoulli’s principle.
FIGURE 8.4.3 Air flow over an airplane wing
FIGURE 8.4.4 A Perfume Atomizer
FIGURE 8.4.5 Air flow around a curveball
FIGURE 8.4.6 Air flow in a bunsen burner
Torricelli’s Theorem:
Efflux Speed from a Tank
A specific and useful application of Bernoulli's equation is determining the speed at which a liquid flows out of a small hole in a large open tank. Consider a tank filled with liquid of density ρ to a height y above a small opening in the side. We want to find the efflux speed, vout , at the opening. Let point 1 be the top surface of the liquid in the tank, and point 2 be the opening. We apply Bernoulli's equation: ρ ρ ρ ρ
We make the following reasonable assumptions:
1. The tank is open to the atmosphere, so the pressure at the top surface is atmospheric pressure: P1 = Patm
2. The liquid exits into the atmosphere, so the pressure at the opening is also atmospheric pressure: P2 = Patm .
3. The tank's cross-sectional area is much larger than the area of the opening. By the continuity equation (Av=constant), if A1 >> A2 , then v1 << v2 . We can approximate the speed of the top surface as negligible: v1 ≈ 0.
4. Let's set the reference level for height at the opening: y2 = 0. Then the height of the top surface is y1 = y.
Substituting these into Bernoulli's equation:
The atmospheric pressure terms cancel out:
Assuming the density ρ is not zero, we can cancel it:
Solving for v2 (which is the efflux speed, let's call it vefflux ):
This result is known as Torricelli’s Theorem. It states that the speed of a fluid exiting an opening is related to the difference in height (h) between the opening and the top surface of the fluid (EK 8.4.B.3). Remarkably, this is the same speed an object would acquire if it fell freely from rest through a vertical distance y (vf 2 =vi 2 +2ay = if vi =0).
This underscores that Torricelli’s theorem can be derived from conservation of energy principles, as it is a direct consequence of Bernoulli’s equation under specific conditions.
Understanding Bernoulli’s principle and Torricelli’s theorem allows us to analyze and predict the behavior of fluids based on the fundamental principle of energy conservation, adding another powerful tool to our study of fluid dynamics.
Lab@Desk
Recall Bernoulli’s Equation states that for an ideal, incompressible fluid: You will use this relationship to construct Bernoulli bar charts comparing two different points in various fluid scenarios.
Task A: Horizontal Pipe with a Narrowing
• Point 1: In the wider section
• Point 1: In the narrower section
1. Draw a Bernoulli bar chart for both points.
2. Which point has greater fluid speed? How do you know?
3. Compare the pressures at both points using Bernoulli’s equation.
4. Explain how the energy is conserved between Point 1 and Point 2.
Task B: Wind Over a Roof
• Point 1 Air inside the house
• Point 2: Wind flowing over the roof
1. Draw a Bernoulli bar chart for both points.
2. Which point has faster moving air?
3. Which point has higher pressure?
4. How can this pressure difference affect the roof?
Conclusion
For one of the situations above, write a short paragraph defending your pressure comparison using your bar chart as evidence. Use the sentence frame: In Task ___ the pressure at Point ___ is greater than at Point ___ because…
8.4 Review Questions
1. A kitchen faucet has an inner diameter of 1.5 cm. If water flows from the faucet at a speed of 0.80 m/s, what is the volume flow rate of the water in liters per minute?
4. Air flows horizontally over the top surface of an airplane wing at a speed of 80 m/s and under the bottom surface at a speed of 60 m/s. If the density of the air is 1.225 kg/ m$^3$, what is the pressure difference between the bottom and top surfaces of the wing (Pbottom−P top)? (Assume the height difference between the top and bottom surfaces is negligible).
2. What is the average inner radius of a major artery if blood flows through it at an average speed of 20 cm/s and the volume flow rate of blood in cubic centimeters per second is 15.7 cm3/s?
5.: A large cylindrical tank is filled with water to a height of 5.0 meters above a small drain plug at the bottom. If the plug is removed, allowing water to flow out, what is the initial speed of the water exiting the hole? (Assume the tank is open to the atmosphere, the hole is small compared to the tank diameter, and g=9.8 m/s2).
3.When pouring a liquid into a funnel that has a narrow stem, the speed at which the liquid level drops in the wide upper part of the funnel changes as the funnel narrows. A liquid flows out of the narrow stem of a funnel at a constant volume flow rate of 0.15 Liters per second. At a specific moment, the liquid level in the wide upper part of the funnel has a circular surface with a diameter of 12 cm. At a lower point in the funnel, where the liquid surface has a diameter of 8.0 cm, what is the speed at which the liquid level is dropping at that specific moment?
Chapter 8 Conceputal Review Questions
1. Consider a small block of wood and a small block of lead that have exactly the same volume. Which block has the greater density? Explain your reasoning using the definition of density.
2. Two identical containers are filled with water to the same depth. One is tall and narrow, the other is short and wide. Which container exerts more pressure at the bottom?
3. A diver reads a pressure gauge at 10 meters underwater. The gauge shows 98,000 Pa. Is this the total pressure the diver experiences?
4. A heavy steel ship floats, but a small steel ball sinks. Why?
5. Imagine designing a ventilation system for a large building. How would the principles of volume flow rate and the continuity equation be important considerations for the ductwork design?
6. Consider an ideal fluid flowing steadily downhill through a pipe of diameter. According to Bernoulli’s principle, what must happen to the fluid’s pressure? Explain your reasoning.
Chapter 8 Review Questions
1. The primary difference between a fluid and a solid relates to one of the following: density, temperature, ability to maintain a fixed shape against shear stress or electrical conductivity? Defend your answer.
4. A c ylindrical drum of radius 0.5 m is used to hold 400 liters of petroleum ether (density = .68 g/mL or 680 kg/ m3). (Note: 1 liter = 0.001 m3)
a. Determine the amount of pressure applied to the walls of the drum if the petroleum ether fills the drum to its top.
2. Your teacher shows you an under-inflated volleyball is pumped full of air so that its radius increases by 10%. She asked the following question: ignoring the mass of the air inserted into the ball, what will happen to the volleyball’s density? Your group comes up with five different answers. Which one is correct and why?
a. The density of the volleyball will increase by approximately 25%.
b. The density of the volleyball will increase by approximately 10%.
c. The density of the volleyball will decrease by approximately 10%.
d. The density of the volleyball will decrease by approximately 17%.
e. The density of the volleyball will decrease by approximately 25%.
3. A piece of aluminum foil has a known surface density of 15 g/cm2. If a 100-gram hollow cube were constructed using this foil, determine the approximate side length of this cube.
b. Determine the amount of pressure applied to the floor of the drum if the petroleum ether fills the drum to its top.
c. If the drum were redesigned to hold 800 liters of petroleum ether:
i. How would the pressure on the walls change? Would it increase, decrease, or stay the same?
ii. How would the pressure on the floor change? Would it increase, decrease, or stay the same?
5. You are pumping up a bicycle tire with a hand pump, the piston of which has a 2.00-cm radius.
(a) What force in newtons must you exert to create a pressure of 6.90 × 105 Pa.
8. Some fish have a density slightly less than that of water and must exert a force (swim) to stay submerged. What force must an 85.0-kg grouper exert to stay submerged in salt water if its body density is 1015 kg/m3?
(b) What is unreasonable about this (a) result?
(c) Which premises are unreasonable or inconsistent?
9. (a) Calculate the buoyant force on a 2.00-L helium balloon.
6. Pressure cookers have been around for more than 300 years, although their use has strongly declined in recent years (early models had a nasty habit of exploding). How much force must the latches holding the lid onto a pressure cooker be able to withstand if the circular lid is in diameter and the gauge pressure inside is 300atm? Neglect the weight of the lid.
(b) Given the mass of the rubber in the balloon is 1.50 g, what is the net vertical force on the balloon if it is let go? You can neglect the volume of the rubber
10. What fraction of an iron anchor’s weight will be supported by buoyant force when submerged in saltwater?
7. Archimedes’ principle can be used to calculate the density of a fluid as well as that of a solid. Suppose a chunk of iron with a mass of 390.0 g in air is found to have an apparent mass of 350.5 g when completely submerged in an unknown liquid.
(a) What mass of fluid does the iron displace?
11. Blood is flowing through an artery of radius 2 mm at a rate of 40 cm/s. Determine the flow rate and the volume that passes through the artery in a period of 30 s.
(b) What is the volume of iron, using its density as given in Table 8.1.1
(c) Calculate the fluid’s density and identify it.
12. If water flows steadily from a pipe with a diameter of 10 cm into a pipe with a diameter of 5 cm, how does the speed of the water in the 5 cm pipe compare to the speed in the 10 cm pipe? Explain your reasoning.
13. (a) What is the pressure drop due to the Bernoulli effect as water goes into a 3.00-cm-diameter nozzle from a 9.00-cm-diameter fire hose while carrying a flow of 40.0 L/s?
14. A large, open tank is filled with water to a height of 4.0 meters. A small spigot is located at the bottom of the tank, 4.0 meters below the surface of the water and open to the atmosphere. Assuming the water is an ideal, incompressible fluid and the tank is large enough that the water level drops very slowly, what is the speed at which the water flows out of the spigot?
(b) To what maximum height above the nozzle can this water rise? (The actual height will be significantly smaller due to air resistance.)
15. Consider the following circumstances within a fluid, and determine the answer using Bernoulli’s equation.
(a) The pressure and kinetic energy per unit volume along a fluid path increases. What must be true about the potential energy per unit volume of the fluid along the fluid path? Explain.
(b) The pressure along a fluid path increases, and the kinetic energy per unit volume remains constant. What must be true about the potential energy per unit volume of the fluid along the fluid path? Explain.
What’s My Mark?
You are the teacher and it’s time to mark the Fluids Unit test. For the multiple choice, the underlined answer is the student’s choice. What’s the mark for this student? Remember to provide feedback on questions that the student got wrong.
Part 1: Multiple Choice 1 mark each.
1. Which of the following best defines a fluid?
(A) A substance that has a fixed volume but no fixed shape.
(B) A substance that can flow and takes the shape of its container.
(C) A substance composed only of liquid particles.
(D) A substance that is easily compressible.
2. Objec t X and Object Y have the same mass. Object X takes up more space than Object Y. Which statement is true regarding their densities?
(A) Object X has a greater density than Object Y.
(B) Object Y has a greater density than Object X.
(C) Both objects have the same density.
(D) Density cannot be determined without knowing the shapes.
3. An incompressible fluid flows steadily through a pipe. At one point, the pipe widens, doubling its cross-sectional area. Compared to the speed in the narrower section, the speed in the wider section is:
(A) Doubled
(B) Halved
(C) Quadrupled
(D) Quartered
4. A main water pipe splits into three identical smaller pipes. Water flows steadily and is incompressible. If the speed in the main pipe is vmain and the speed in each smaller pipe is vsmall, and the cross-sectional area of the main pipe is Amain and each smaller pipe is Asmall, which equation correctly expresses the conservation of flow rate?
(A) Amainvmain =Asmallvsmall
(B) Amainvmain=3Asmallvsmall
(C) 3Amainvmain =Asmallvsmall
(D) Amain/vmain=3Asmall/vsmall
5. Two blocks, X and Y, made of different materials but having the same mass, are floating in freshwater. Object X is made of a less dense material than Object Y (ρX <ρY ). Which statement about the buoyant forces on the objects is true?
(A) The buoyant force on X is greater than the buoyant force on Y.
(B) The buoyant force on Y is greater than the buoyant force on X.
(C) The buoyant force on X is equal to the buoyant force on Y.
(D) The buoyant force depends on the objects’ volumes, which cannot be determined from the information given.
6. At what depth beneath the surface of the lake is the pressure in the water equal to twice atmospheric pressure?
(A) 10 m
(B) 100 m
(C) 1000 m
(D)10,000 m
7. As water falls from a faucet, the stream typically narrows. This is primarily because:
(A)Air pressure increases as the water falls.
(B) The water’s speed increases due to gravity, and volume flow rate must be conserved.
(C) The water’s density increases as it falls.
(D)Surface tension pulls the stream together more strongly as speed increases.
8. According to Bernoulli’s equation, if the pressure in a given fluid is constant and the kinetic energy per unit volume of a fluid increases, which of the following is true?
(A) The potential energy per unit volume of the fluid decreases.
(B) The potential energy per unit volume of the fluid increases.
(C) The fluid must no longer be considered incompressible.
(D) The flow rate of the fluid increases.
9. A horizontally oriented pipe has a diameter of 5.6 cm and is filled with water. The pipe draws water from a reservoir that is initially at rest. A manually operated plunger provides a force of 440 N in the pipe. Assuming that the other end of the pipe is open to the air, with what speed does the water emerge from the pipe?
(A)12 m/s
(B) 19 m/s
(C)150 m/s
(D)190 m/s
Part 2: Free Response Question: 1 mark each for a total of six marks
10.Density and Archimedes
A cube of polystyrene measuring 10 cm per side lies partially submerged in a large container of water.
a.If 90% of the polystyrene floats above the surface of the water, what is the density of the polystyrene? (Note: The density of water is 1000 kg/m3)
b.A 0.5 kg mass is placed on the block of polystyrene. What percentage of the block now remains above water?
c. The water is poured out of the container and replaced with ethyl alcohol (density = 790 kg/m3).
i. Will the block be able to remain partially submerged in this new fluid? Explain.
ii. Will the block be able to remain partially submerged in this new fluid with the 0.5 kg mass placed on top? Explain.
11. Water flow through a Pipe
Water flows through a pipe that is 10.0 meters above the ground at point A and 15.0 meters above the ground at point B.At point A, the pipe diameter is 8.0 cm, the water speed is 3.0 m/s, and the pressure is 200,000 Pa. At point B, the pipe diameter is 6.0 cm. What is the pressure at point B? (Assume water density is 1000 kg/m3 and g=9.8 m/s2 ).
Assume v a = vb
12.Ebbinghaus Remembers - System Motion and Energy
A child has two red wagons, with the rear one tied to the front by a (non-stretching) piece of metal. If the child pushes on the rear wagon, what happens to the kinetic energy of each of the wagons, and of the two-wagon system?
Only the rear wagon gains kinetic energy because it’s the wagon being pushed.
What’s my Mark? - Feedback Form
Now that you’ve marked the test, provide feedback to the student so they’re better prepared next time.
