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d. 1.0 × 102 m/s2 13. A girl rolls a basketball across a basketball court. The ball slowly decelerates at a rate of −0.20 m/s2. If the initial velocity was 2.0 m/s and the ball rolled to a stop at 5.0 sec after 12:00 p.m., at what time did she start the ball rolling? a. 0.1 seconds before noon b. 0.1 seconds after noon 5 seconds before noon c. 5 seconds after noon d. Performance Task 3.2 Representing Acceleration with Equations and Graphs 16. Design an experiment to measure displacement and elapsed time. Use the data to calculate final velocity, average velocity, acceleration, and acceleration. Materials • a small marble or ball bearing • a garden hose • a measuring tape • a stopwatch or stopwatch software download • a sloping driveway or lawn as long as the garden 3.2 Representing Acceleration with Equations and Graphs 14. A swan on a lake gets airborne by flapping its wings and running on top of the water. If the swan must reach a velocity of 6.00 m/s to take off and it accelerates from rest at an average rate of 0.350 m/s2, how far will it travel before becoming airborne? a. −8.60 m b. 8.60 m c. −51.4 m d. 51.4 m 15. A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of above the pool. and her takeoff point is How long are her feet in the air? a. b. c. d. e. 1.28 s hose with a level area beyond (a) How would you use the garden hose, stopwatch, marble, measuring tape, and slope to measure displacement and elapsed time? Hint—The marble is the accelerating object, and the length of the hose is total displacement. (b) How would you use the displacement and time data to calculate velocity, average velocity, and acceleration? Which kinematic equations would you use? (c) How would you use the materials, the measured and calculated data, and the flat area below the slope to determine the negative acceleration? What would you measure, and which kinematic equation would you use? 112 Chapter 3 • Test Prep TEST PREP Multiple Choice 3.1 Acceleration 17. Which variable represents displacement? a
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. a b. d c. t d. v 18. If a velocity increases from 0 to 20 m/s in 10 s, what is the average acceleration? a. 0.5 m/s2 b. 2 m/s2 10 m/s2 c. 30 m/s2 d. 3.2 Representing Acceleration with Equations and Graphs 19. For the motion of a falling object, which graphs are Short Answer 3.1 Acceleration 21. True or False—The vector for a negative acceleration points in the opposite direction when compared to the vector for a positive acceleration. a. True b. False 22. If a car decelerates from to in, what is a. b. c. d.? -5 m/s -1 m/s 1 m/s 5 m/s 23. How is the vector arrow representing an acceleration of magnitude 3 m/s2 different from the vector arrow representing a negative acceleration of magnitude 3 m/ s2? a. They point in the same direction. b. They are perpendicular, forming a 90° angle between each other. c. They point in opposite directions. d. They are perpendicular, forming a 270° angle between each other. 24. How long does it take to accelerate from 8.0 m/s to 20.0 m/s at a rate of acceleration of 3.0 m/s2? a. 0.25 s b. 4.0 s c. 9.33 s Access for free at openstax.org. straight lines? a. Acceleration versus time only b. Displacement versus time only c. Displacement versus time and acceleration versus time d. Velocity versus time and acceleration versus time 20. A bullet in a gun is accelerated from the firing chamber to the end of the barrel at an average rate of 6.30×105 m/ s2 for 8.10×10−4 s. What is the bullet’s final velocity when it leaves the barrel, commonly known as the muzzle velocity? a. b. c. d. 7.79 m/s 51.0 m/s 510 m/s 1020 m/s d. 36 s 3.2 Representing Acceleration with Equations and Graphs 25. If a plot of displacement versus time is linear, what can be said about the acceleration? a. Acceleration is 0. b. Acceleration is a non-zero constant. c. Acceleration is positive.
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d. Acceleration is negative. 26. True or False: —The image shows a velocity vs. time graph for a jet car. If you take the slope at any point on the graph, the jet car’s acceleration will be 5.0 m/s2. a. True b. False 27. When plotted on the blank plots, which answer choice would show the motion of an object that has uniformly accelerated from 2 m/s to 8 m/s in 3 s? Chapter 3 • Test Prep 113 the plot on the right shows a line from (0,2) to (3,2). d. The plot on the left shows a line from (0,8) to (3,2) while the plot on the right shows a line from (0,3) to (3,3). 28. When is a plot of velocity versus time a straight line and a. The plot on the left shows a line from (0,2) to (3,8) while the plot on the right shows a line from (0,2) to (3,2). b. The plot on the left shows a line from (0,2) to (3,8) while the plot on the right shows a line from (0,3) to (3,3). c. The plot on the left shows a line from (0,8) to (3,2) while b. c. d. when is it a curved line? a. It is a straight line when acceleration is changing and is a curved line when acceleration is constant. It is a straight line when acceleration is constant and is a curved line when acceleration is changing. It is a straight line when velocity is constant and is a curved line when velocity is changing. It is a straight line when velocity is changing and is a curved line when velocity is constant. Extended Response 3.1 Acceleration 29. A test car carrying a crash test dummy accelerates from and then crashes into a brick wall. Describe to the direction of the initial acceleration vector and compare the initial acceleration vector’s magnitude with respect to the acceleration magnitude at the moment of the crash. a. The direction of the initial acceleration vector will point towards the wall, and its magnitude will be less than the acceleration vector of the crash. b. The direction of the initial acceleration vector will point away from the wall, and its magnitude will be less than the vector of the crash. c. The direction of the initial acceleration vector
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will point towards the wall, and its magnitude will be more than the acceleration vector of the crash. d. The direction of the initial acceleration vector will point away from the wall, and its magnitude will be more than the acceleration vector of the crash. 30. A car accelerates from rest at a stop sign at a rate of 3.0 m/s2 to a speed of 21.0 m/s, and then immediately begins to decelerate to a stop at the next stop sign at a rate of 4.0 m/s2. How long did it take the car to travel from the first stop sign to the second stop sign? Show your work. a. b. c. d. 1.7 seconds 5.3 seconds 7.0 seconds 12 seconds 3.2 Representing Acceleration with Equations and Graphs 31. True or False: Consider an object moving with constant acceleration. The plot of displacement versus time for such motion is a curved line while the plot of displacement versus time squared is a straight line. a. True b. False 32. You throw a ball straight up with an initial velocity of 15.0 m/s. It passes a tree branch on the way up at a height of 7.00 m. How much additional time will pass before the ball passes the tree branch on the way back down? a. 0.574 s b. 0.956 s 1.53 s c. 1.91 s d. 114 Chapter 3 • Test Prep Access for free at openstax.org. CHAPTER 4 Forces and Newton’s Laws of Motion Figure 4.1 Newton’s laws of motion describe the motion of the dolphin’s path. (Credit: Jin Jang) Chapter Outline 4.1 Force 4.2 Newton's First Law of Motion: Inertia 4.3 Newton's Second Law of Motion 4.4 Newton's Third Law of Motion Isaac Newton (1642–1727) was a natural philosopher; a great thinker who combined science and philosophy to INTRODUCTION try to explain the workings of nature on Earth and in the universe. His laws of motion were just one part of the monumental work that has made him legendary. The development of Newton’s laws marks the transition from the Renaissance period of history to the modern era. This transition was characterized by a revolutionary change in the way people thought about the physical universe. Drawing upon earlier work by scientists Galileo Galilei and Johannes Kepler, Newton’s laws of
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motion allowed motion on Earth and in space to be predicted mathematically. In this chapter you will learn about force as well as Newton’s first, second, and third laws of motion. 116 Chapter 4 • Forces and Newton’s Laws of Motion 4.1 Force Section Learning Objectives By the end of this section, you will be able to do the following: • Differentiate between force, net force, and dynamics • Draw a free-body diagram Section Key Terms dynamics external force force free-body diagram net external force net force Defining Force and Dynamics Force is the cause of motion, and motion draws our attention. Motion itself can be beautiful, such as a dolphin jumping out of the water, the flight of a bird, or the orbit of a satellite. The study of motion is called kinematics, but kinematics describes only the way objects move—their velocity and their acceleration. Dynamics considers the forces that affect the motion of moving objects and systems. Newton’s laws of motion are the foundation of dynamics. These laws describe the way objects speed up, slow down, stay in motion, and interact with other objects. They are also universal laws: they apply everywhere on Earth as well as in space. A force pushes or pulls an object. The object being moved by a force could be an inanimate object, a table, or an animate object, a person. The pushing or pulling may be done by a person, or even the gravitational pull of Earth. Forces have different magnitudes and directions; this means that some forces are stronger than others and can act in different directions. For example, a cannon exerts a strong force on the cannonball that is launched into the air. In contrast, a mosquito landing on your arm exerts only a small force on your arm. When multiple forces act on an object, the forces combine. Adding together all of the forces acting on an object gives the total force, or net force. An external force is a force that acts on an object within the system from outsidethe system. This type of force is different than an internal force, which acts between two objects that are both within the system. The net external force combines these two definitions; it is the total combined external force. We discuss further details about net force, external force, and net external force in the coming sections. In mathematical terms, two forces acting in opposite directions have opposite signs(positive or negative). By convention, the negative sign is assigned to any movement to the left or downward
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. If two forces pushing in opposite directions are added together, the larger force will be somewhat canceled out by the smaller force pushing in the opposite direction. It is important to be consistent with your chosen coordinate system within a problem; for example, if negative values are assigned to the downward direction for velocity, then distance, force, and acceleration should also be designated as being negative in the downward direction. Free-Body Diagrams and Examples of Forces For our first example of force, consider an object hanging from a rope. This example gives us the opportunity to introduce a useful tool known as a free-body diagram. A free-body diagram represents the object being acted upon—that is, the free body—as a single point. Only the forces acting onthe body (that is, external forces) are shown and are represented by vectors (which are drawn as arrows). These forces are the only ones shown because only external forces acting on the body affect its motion. We can ignore any internal forces within the body because they cancel each other out, as explained in the section on Newton’s third law of motion. Free-body diagrams are very useful for analyzing forces acting on an object. Access for free at openstax.org. 4.1 • Force 117 Figure 4.2 An object of mass, m, is held up by the force of tension. Figure 4.2 shows the force of tension in the rope acting in the upward direction, opposite the force of gravity. The forces are indicated in the free-body diagram by an arrow pointing up, representing tension, and another arrow pointing down, representing gravity. In a free-body diagram, the lengths of the arrows show the relative magnitude (or strength) of the forces. Because forces are vectors, they add just like other vectors. Notice that the two arrows have equal lengths in Figure 4.2, which means that the forces of tension and weight are of equal magnitude. Because these forces of equal magnitude act in opposite directions, they are perfectly balanced, so they add together to give a net force of zero. Not all forces are as noticeable as when you push or pull on an object. Some forces act without physical contact, such as the pull of a magnet (in the case of magnetic force) or the gravitational pull of Earth (in the case of gravitational force). In the next three sections discussing Newton’s laws of motion, we will learn about three specific types of forces: friction, the normal force, and the gravitational force. To
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analyze situations involving forces, we will create free-body diagrams to organize the framework of the mathematics for each individual situation. TIPS FOR SUCCESS Correctly drawing and labeling a free-body diagram is an important first step for solving a problem. It will help you visualize the problem and correctly apply the mathematics to solve the problem. Check Your Understanding 1. What is kinematics? a. Kinematics is the study of motion. b. Kinematics is the study of the cause of motion. c. Kinematics is the study of dimensions. d. Kinematics is the study of atomic structures. 2. Do two bodies have to be in physical contact to exert a force upon one another? 118 Chapter 4 • Forces and Newton’s Laws of Motion a. No, the gravitational force is a field force and does not require physical contact to exert a force. b. No, the gravitational force is a contact force and does not require physical contact to exert a force. c. Yes, the gravitational force is a field force and requires physical contact to exert a force. d. Yes, the gravitational force is a contact force and requires physical contact to exert a force. 3. What kind of physical quantity is force? a. Force is a scalar quantity. b. Force is a vector quantity. c. Force is both a vector quantity and a scalar quantity. d. Force is neither a vector nor a scalar quantity. 4. Which forces can be represented in a free-body diagram? a. Internal forces b. External forces c. Both internal and external forces d. A body that is not influenced by any force 4.2 Newton's First Law of Motion: Inertia Section Learning Objectives By the end of this section, you will be able to do the following: • Describe Newton’s first law and friction, and • Discuss the relationship between mass and inertia. Section Key Terms friction inertia law of inertia mass Newton’s first law of motion system Newton’s First Law and Friction Newton’s first law of motion states the following: 1. A body at rest tends to remain at rest. 2. A body in motion tends to remain in motion at a constant velocity unless acted on by a net external force. (Recall that constant velocitymeans that the body moves in a straight line and at a constant speed.) At first glance, this law may seem to contradict your everyday experience. You have probably noticed that a moving object will
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usually slow down and stop unless some effort is made to keep it moving. The key to understanding why, for example, a sliding box slows down (seemingly on its own) is to first understand that a net external force acts on the box to make the box slow down. Without this net external force, the box would continue to slide at a constant velocity (as stated in Newton’s first law of motion). What force acts on the box to slow it down? This force is called friction. Friction is an external force that acts opposite to the direction of motion (see Figure 4.3). Think of friction as a resistance to motion that slows things down. Consider an air hockey table. When the air is turned off, the puck slides only a short distance before friction slows it to a stop. However, when the air is turned on, it lifts the puck slightly, so the puck experiences very little friction as it moves over the surface. With friction almost eliminated, the puck glides along with very little change in speed. On a frictionless surface, the puck would experience no net external force (ignoring air resistance, which is also a form of friction). Additionally, if we know enough about friction, we can accurately predict how quickly objects will slow down. Now let’s think about another example. A man pushes a box across a floor at constant velocity by applying a force of +50 N. (The positive sign indicates that, by convention, the direction of motion is to the right.) What is the force of friction that opposes the motion? The force of friction must be −50 N. Why? According to Newton’s first law of motion, any object moving at constant velocity has no net external force acting upon it, which means that the sum of the forces acting on the object must be zero. The mathematical way to say that no net external force acts on an object is N of force, then the force of friction must be −50 N for the two forces to add up to zero (that is, for the two forces to canceleach So if the man applies +50 or Access for free at openstax.org. other). Whenever you encounter the phrase at constant velocity, Newton’s first law tells you that the net external force is zero. 4.2 • Newton's First Law of Motion: Inertia 119 Figure 4.3 For a box sliding across a floor, friction acts in the direction opposite to the velocity. The force of friction depends on
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two factors: the coefficient of friction and the normal force. For any two surfaces that are in contact with one another, the coefficient of friction is a constant that depends on the nature of the surfaces. The normal force is the force exerted by a surface that pushes on an object in response to gravity pulling the object down. In equation form, the force of friction is 4.1 where μis the coefficient of friction and N is the normal force. (The coefficient of friction is discussed in more detail in another chapter, and the normal force is discussed in more detail in the section Newton's Third Law of Motion.) Recall from the section on Force that a net external force acts from outside on the object of interest. A more precise definition is that it acts on the system of interest. A system is one or more objects that you choose to study. It is important to define the system at the beginning of a problem to figure out which forces are external and need to be considered, and which are internal and can be ignored. For example, in Figure 4.4 (a), two children push a third child in a wagon at a constant velocity. The system of interest is the wagon plus the small child, as shown in part (b) of the figure. The two children behind the wagon exert external forces on this system (F1, F2). Friction facting at the axles of the wheels and at the surface where the wheels touch the ground two other external forces acting on the system. Two more external forces act on the system: the weight W of the system pulling down and the normal force N of the ground pushing up. Notice that the wagon is not accelerating vertically, so Newton’s first law tells us that the normal force balances the weight. Because the wagon is moving forward at a constant velocity, the force of friction must have the same strength as the sum of the forces applied by the two children. 120 Chapter 4 • Forces and Newton’s Laws of Motion Figure 4.4 (a) The wagon and rider form a systemthat is acted on by external forces. (b) The two children pushing the wagon and child provide two external forces. Friction acting at the wheel axles and on the surface of the tires where they touch the ground provide an external force that act against the direction of motion. The weight W and the normal force N from the ground are two more external forces acting on the system. All external forces are represented in the figure by arrows. All of the external forces acting on the
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system add together, but because the wagon moves at a constant velocity, all of the forces must add up to zero. Mass and Inertia Inertia is the tendency for an object at rest to remain at rest, or for a moving object to remain in motion in a straight line with constant speed. This key property of objects was first described by Galileo. Later, Newton incorporated the concept of inertia into his first law, which is often referred to as the law of inertia. As we know from experience, some objects have more inertia than others. For example, changing the motion of a large truck is more difficult than changing the motion of a toy truck. In fact, the inertia of an object is proportional to the mass of the object. Mass is a measure of the amount of matter (or stuff) in an object. The quantity or amount of matter in an object is determined by the number and types of atoms the object contains. Unlike weight (which changes if the gravitational force changes), mass does not depend on gravity. The mass of an object is the same on Earth, in orbit, or on the surface of the moon. In practice, it is very difficult to count and identify all of the atoms and molecules in an object, so mass is usually not determined this way. Instead, the mass of an object is determined by comparing it with the standard kilogram. Mass is therefore expressed in kilograms. TIPS FOR SUCCESS In everyday language, people often use the terms weightand massinterchangeably—but this is not correct. Weight is actually a force. (We cover this topic in more detail in the section Newton's Second Law of Motion.) WATCH PHYSICS Newton’s First Law of Motion This video contrasts the way we thought about motion and force in the time before Galileo’s concept of inertia and Newton’s first law of motion with the way we understand force and motion now. Click to view content (https://www.khanacademy.org/embed_video?v=5-ZFOhHQS68) Access for free at openstax.org. 4.2 • Newton's First Law of Motion: Inertia 121 GRASP CHECK Before we understood that objects have a tendency to maintain their velocity in a straight line unless acted upon by a net force, people thought that objects had a tendency to stop on their own. This happened because a specific force was not yet understood. What was that force? a. Gravitational force b
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. Electrostatic force c. Nuclear force d. Frictional force Virtual Physics Forces and Motion—Basics In this simulation, you will first explore net force by placing blue people on the left side of a tug-of-war rope and red people on the right side of the rope (by clicking people and dragging them with your mouse). Experiment with changing the number and size of people on each side to see how it affects the outcome of the match and the net force. Hit the "Go!" button to start the match, and the “reset all” button to start over. Next, click on the Friction tab. Try selecting different objects for the person to push. Slide the applied forcebutton to the right to apply force to the right, and to the left to apply force to the left. The force will continue to be applied as long as you hold down the button. See the arrow representing friction change in magnitude and direction, depending on how much force you apply. Try increasing or decreasing the friction force to see how this change affects the motion. Click to view content (https://phet.colorado.edu/sims/html/forces-and-motion-basics/latest/forces-and-motionbasics_en.html) GRASP CHECK Click on the tab for the Acceleration Lab and check the Sum of Forcesoption. Push the box to the right and then release. Notice which direction the sum of forces arrow points after the person stops pushing the box and lets it continue moving to the right on its own. At this point, in which direction is the net force, the sum of forces, pointing? Why? a. The net force acts to the right because the applied external force acted to the right. b. The net force acts to the left because the applied external force acted to the left. c. The net force acts to the right because the frictional force acts to the right. d. The net force acts to the left because the frictional force acts to the left. Check Your Understanding 5. What does Newton’s first law state? a. A body at rest tends to remain at rest and a body in motion tends to remain in motion at a constant acceleration unless acted on by a net external force. b. A body at rest tends to remain at rest and a body in motion tends to remain in motion at a constant velocity unless acted on by a net external force. c. The rate of change of momentum of a body is directly
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proportional to the external force applied to the body. d. The rate of change of momentum of a body is inversely proportional to the external force applied to the body. 6. According to Newton’s first law, a body in motion tends to remain in motion at a constant velocity. However, when you slide an object across a surface, the object eventually slows down and stops. Why? a. The object experiences a frictional force exerted by the surface, which opposes its motion. b. The object experiences the gravitational force exerted by Earth, which opposes its motion c. The object experiences an internal force exerted by the body itself, which opposes its motion. d. The object experiences a pseudo-force from the body in motion, which opposes its motion. 122 Chapter 4 • Forces and Newton’s Laws of Motion 7. What is inertia? a. b. c. d. Inertia is an object’s tendency to maintain its mass. Inertia is an object’s tendency to remain at rest. Inertia is an object’s tendency to remain in motion Inertia is an object’s tendency to remain at rest or, if moving, to remain in motion. 8. What is mass? What does it depend on? a. Mass is the weight of an object, and it depends on the gravitational force acting on the object. b. Mass is the weight of an object, and it depends on the number and types of atoms in the object. c. Mass is the quantity of matter contained in an object, and it depends on the gravitational force acting on the object. d. Mass is the quantity of matter contained in an object, and it depends on the number and types of atoms in the object. 4.3 Newton's Second Law of Motion Section Learning Objectives By the end of this section, you will be able to do the following: • Describe Newton’s second law, both verbally and mathematically • Use Newton’s second law to solve problems Section Key Terms freefall Newton’s second law of motion weight Describing Newton’s Second Law of Motion Newton’s first law considered bodies at rest or bodies in motion at a constant velocity. The other state of motion to consider is when an object is moving with a changing velocity, which means a change in the speed and/or the direction of motion. This type of motion is addressed by Newton’s second law of motion, which states how force causes
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changes in motion. Newton’s second law of motion is used to calculate what happens in situations involving forces and motion, and it shows the mathematical relationship between force, mass, and acceleration. Mathematically, the second law is most often written as 4.2 where Fnet (or ∑F) is the net external force, mis the mass of the system, and a is the acceleration. Note that Fnet and ∑F are the same because the net external force is the sum of all the external forces acting on the system. First, what do we mean by a change in motion? A change in motion is simply a change in velocity: the speed of an object can become slower or faster, the direction in which the object is moving can change, or both of these variables may change. A change in velocity means, by definition, that an acceleration has occurred. Newton’s first law says that only a nonzero net external force can cause a change in motion, so a net external force must cause an acceleration. Note that acceleration can refer to slowing down or to speeding up. Acceleration can also refer to a change in the direction of motion with no change in speed, because acceleration is the change in velocity divided by the time it takes for that change to occur, andvelocity is defined by speed and direction. From the equation we see that force is directly proportional to both mass and acceleration, which makes sense. To accelerate two objects from rest to the same velocity, you would expect more force to be required to accelerate the more massive object. Likewise, for two objects of the same mass, applying a greater force to one would accelerate it to a greater velocity. Now, let’s rearrange Newton’s second law to solve for acceleration. We get In this form, we can see that acceleration is directly proportional to force, which we write as 4.3 4.4 where the symbol means proportional to. This proportionality mathematically states what we just said in words: acceleration is directly proportional to the net external Access for free at openstax.org. 4.3 • Newton's Second Law of Motion 123 force. When two variables are directly proportional to each other, then if one variable doubles, the other variable must double. Likewise, if one variable is reduced by half, the other variable must also be reduced by half. In general, when one variable is multiplied by a number, the other variable is also multiplied by the same number. It seems reasonable that the acceleration of a
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system should be directly proportional to and in the same direction as the net external force acting on the system. An object experiences greater acceleration when acted on by a greater force. It is also clear from the equation that acceleration is inversely proportional to mass, which we write as 4.5 Inversely proportionalmeans that if one variable is multiplied by a number, the other variable must be dividedby the same number. Now, it also seems reasonable that acceleration should be inversely proportional to the mass of the system. In other words, the larger the mass (the inertia), the smaller the acceleration produced by a given force. This relationship is illustrated in Figure 4.5, which shows that a given net external force applied to a basketball produces a much greater acceleration than when applied to a car. Figure 4.5 The same force exerted on systems of different masses produces different accelerations. (a) A boy pushes a basketball to make a pass. The effect of gravity on the ball is ignored. (b) The same boy pushing with identical force on a stalled car produces a far smaller acceleration (friction is negligible). Note that the free-body diagrams for the ball and for the car are identical, which allows us to compare the two situations. Applying Newton’s Second Law Before putting Newton’s second law into action, it is important to consider units. The equation units of force in terms of the three basic units of mass, length, and time (recall that acceleration has units of length divided by time squared). The SI unit of force is called the newton (abbreviated N) and is the force needed to accelerate a 1-kg system at the rate of 1 m/s2. That is, because is used to define the we have One of the most important applications of Newton’s second law is to calculate weight (also known as the gravitational force), which is usually represented mathematically as W. When people talk about gravity, they don’t always realize that it is an acceleration. When an object is dropped, it accelerates toward the center of Earth. Newton’s second law states that the net external force acting on an object is responsible for the acceleration of the object. If air resistance is negligible, the net external force on a falling object is only the gravitational force (i.e., the weight of the object). Weight can be represented by a vector because it has a direction. Down is defined as the direction in which gravity pulls, so weight is
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normally considered a downward force. By using Newton’s second law, we can figure out the equation for weight. 4.6 Consider an object with mass mfalling toward Earth. It experiences only the force of gravity (i.e., the gravitational force or weight), which is represented by W. Newton’s second law states that the gravitational force, we have Substituting these two expressions into Newton’s second law gives We know that the acceleration of an object due to gravity is g, so we have Because the only force acting on the object is 124 Chapter 4 • Forces and Newton’s Laws of Motion This is the equation for weight—the gravitational force on a mass m. On Earth, now the direction of the weight) of a 1.0-kg object on Earth is 4.7 so the weight (disregarding for 4.8 Although most of the world uses newtons as the unit of force, in the United States the most familiar unit of force is the pound (lb), where 1 N = 0.225 lb. Recall that although gravity acts downward, it can be assigned a positive or negative value, depending on what the positive direction is in your chosen coordinate system. Be sure to take this into consideration when solving problems with weight. When the downward direction is taken to be negative, as is often the case, acceleration due to gravity becomes g = −9.8 m/s2. When the net external force on an object is its weight, we say that it is in freefall. In this case, the only force acting on the object is the force of gravity. On the surface of Earth, when objects fall downward toward Earth, they are never truly in freefall because there is always some upward force due to air resistance that acts on the object (and there is also the buoyancy force of air, which is similar to the buoyancy force in water that keeps boats afloat). Gravity varies slightly over the surface of Earth, so the weight of an object depends very slightly on its location on Earth. Weight varies dramatically away from Earth’s surface. On the moon, for example, the acceleration due to gravity is only 1.67 m/s2. Because weight depends on the force of gravity, a 1.0-kg mass weighs 9.8 N on Earth and only about 1.7 N on the moon. It is important to remember that weight and mass are very different, although they are closely related. Mass is the quantity of matter (
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how much stuff) in an object and does not vary, but weight is the gravitational force on an object and is proportional to the force of gravity. It is easy to confuse the two, because our experience is confined to Earth, and the weight of an object is essentially the same no matter where you are on Earth. Adding to the confusion, the terms mass and weight are often used interchangeably in everyday language; for example, our medical records often show our weight in kilograms, but never in the correct unit of newtons. Snap Lab Mass and Weight In this activity, you will use a scale to investigate mass and weight. • • 1 bathroom scale 1 table 1. What do bathroom scales measure? 2. When you stand on a bathroom scale, what happens to the scale? It depresses slightly. The scale contains springs that compress in proportion to your weight—similar to rubber bands expanding when pulled. 3. The springs provide a measure of your weight (provided you are not accelerating). This is a force in newtons (or pounds). In most countries, the measurement is now divided by 9.80 to give a reading in kilograms, which is a of mass. The scale detects weight but is calibrated to display mass. If you went to the moon and stood on your scale, would it detect the same massas it did on Earth? 4. GRASP CHECK While standing on a bathroom scale, push down on a table next to you. What happens to the reading? Why? a. The reading increases because part of your weight is applied to the table and the table exerts a matching force on you that acts in the direction of your weight. b. The reading increases because part of your weight is applied to the table and the table exerts a matching force on you that acts in the direction opposite to your weight. c. The reading decreases because part of your weight is applied to the table and the table exerts a matching force on you that acts in the direction of your weight. d. The reading decreases because part of your weight is applied to the table and the table exerts a matching force on Access for free at openstax.org. you that acts in the direction opposite to your weight. 4.3 • Newton's Second Law of Motion 125 TIPS FOR SUCCESS Only net external forceimpacts the acceleration of an object. If more than one force acts on an object and you calculate the acceleration by using only one of these forces, you will not get the correct acceleration for that object
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. WATCH PHYSICS Newton’s Second Law of Motion This video reviews Newton’s second law of motion and how net external force and acceleration relate to one another and to mass. It also covers units of force, mass, and acceleration, and reviews a worked-out example. Click to view content (https://www.khanacademy.org/embed_video?v=ou9YMWlJgkE) GRASP CHECK True or False—If you want to reduce the acceleration of an object to half its original value, then you would need to reduce the net external force by half. a. True b. False WORKED EXAMPLE What Acceleration Can a Person Produce when Pushing a Lawn Mower? Suppose that the net external force (push minus friction) exerted on a lawn mower is 51 N parallel to the ground. The mass of the mower is 240 kg. What is its acceleration? Figure 4.6 Strategy Because Fnet and mare given, the acceleration can be calculated directly from Newton’s second law: Fnet = ma. Solution Solving Newton’s second law for the acceleration, we find that the magnitude of the acceleration, a, is given values for net external force and mass gives Entering the 4.9 126 Chapter 4 • Forces and Newton’s Laws of Motion Inserting the units for N yields 4.10 Discussion The acceleration is in the same direction as the net external force, which is parallel to the ground and to the right. There is no information given in this example about the individual external forces acting on the system, but we can say something about their relative magnitudes. For example, the force exerted by the person pushing the mower must be greater than the friction opposing the motion, because we are given that the net external force is in the direction in which the person pushes. Also, the vertical forces must cancel if there is no acceleration in the vertical direction (the mower is moving only horizontally). The acceleration found is reasonable for a person pushing a mower; the mower’s speed must increase by 0.21 m/s every second, which is possible. The time during which the mower accelerates would not be very long because the person’s top speed would soon be reached. At this point, the person could push a little less hard, because he only has to overcome friction. WORKED EXAMPLE What Rocket Thrust Accelerates This Sled?
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Prior to manned space flights, rocket sleds were used to test aircraft, missile equipment, and physiological effects on humans at high accelerations. Rocket sleds consisted of a platform mounted on one or two rails and propelled by several rockets. Calculate the magnitude of force exerted by each rocket, called its thrust, T, for the four-rocket propulsion system shown below. The sled’s initial acceleration is the mass of the system is 2,100 kg, and the force of friction opposing the motion is 650 N. Figure 4.7 Strategy The system of interest is the rocket sled. Although forces act vertically on the system, they must cancel because the system does not accelerate vertically. This leaves us with only horizontal forces to consider. We’ll assign the direction to the right as the positive direction. See the free-body diagram in Figure 4.8. Solution We start with Newton’s second law and look for ways to find the thrust T of the engines. Because all forces and acceleration are along a line, we need only consider the magnitudes of these quantities in the calculations. We begin with 4.11 is the net external force in the horizontal direction. We can see from Figure 4.8 that the engine thrusts are in the where same direction (which we call the positive direction), whereas friction opposes the thrust. In equation form, the net external force is Access for free at openstax.org. Newton’s second law tells us that Fnet= ma, so we get After a little algebra, we solve for the total thrust 4T: which means that the individual thrust is Inserting the known values yields 4.3 • Newton's Second Law of Motion 127 4.12 4.13 4.14 4.15 4.16 Discussion The numbers are quite large, so the result might surprise you. Experiments such as this were performed in the early 1960s to test the limits of human endurance and to test the apparatus designed to protect fighter pilots in emergency ejections. Speeds of 1000 km/h were obtained, with accelerations of 45 g. (Recall that g, the acceleration due to gravity, is of 45 g is ) Living subjects are no longer used, and land speeds of 10,000 km/h have now been obtained with rocket sleds. In this example, as in the preceding example, the system of interest is clear. We will see in later examples that choosing the system of interest is crucial—and that the choice is not always obvious.
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which is approximately An acceleration Practice Problems 9. If 1 N is equal to 0.225 lb, how many pounds is 5 N of force? a. 0.045 lb b. 1.125 lb c. 2.025 lb 5.000 lb d. 10. How much force needs to be applied to a 5-kg object for it to accelerate at 20 m/s2? a. b. c. d. 1 N 10 N 100 N 1,000 N Check Your Understanding 11. What is the mathematical statement for Newton’s second law of motion? a. F = ma b. F = 2ma c. d. F = ma2 12. Newton’s second law describes the relationship between which quantities? a. Force, mass, and time b. Force, mass, and displacement c. Force, mass, and velocity d. Force, mass, and acceleration 13. What is acceleration? a. Acceleration is the rate at which displacement changes. b. Acceleration is the rate at which force changes. c. Acceleration is the rate at which velocity changes. 128 Chapter 4 • Forces and Newton’s Laws of Motion d. Acceleration is the rate at which mass changes. 4.4 Newton's Third Law of Motion Section Learning Objectives By the end of this section, you will be able to do the following: • Describe Newton’s third law, both verbally and mathematically • Use Newton’s third law to solve problems Section Key Terms Newton’s third law of motion normal force tension thrust Describing Newton’s Third Law of Motion If you have ever stubbed your toe, you have noticed that although your toe initiates the impact, the surface that you stub it on exerts a force back on your toe. Although the first thought that crosses your mind is probably “ouch, that hurt” rather than “this is a great example of Newton’s third law,” both statements are true. This is exactly what happens whenever one object exerts a force on another—each object experiences a force that is the same strength as the force acting on the other object but that acts in the opposite direction. Everyday experiences, such as stubbing a toe or throwing a ball, are all perfect examples of Newton’s third law in action. Newton’s third law of motion states that whenever a first object exerts a force on a second object, the first object experiences a force equal in magnitude but
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opposite in direction to the force that it exerts. Newton’s third law of motion tells us that forces always occur in pairs, and one object cannot exert a force on another without experiencing the same strength force in return. We sometimes refer to these force pairs as action-reactionpairs, where the force exerted is the action, and the force experienced in return is the reaction (although which is which depends on your point of view). Newton’s third law is useful for figuring out which forces are external to a system. Recall that identifying external forces is important when setting up a problem, because the external forces must be added together to find the net force. We can see Newton’s third law at work by looking at how people move about. Consider a swimmer pushing off from the side of a pool, as illustrated in Figure 4.8. She pushes against the pool wall with her feet and accelerates in the direction opposite to her push. The wall has thus exerted on the swimmer a force of equal magnitude but in the direction opposite that of her push. You might think that two forces of equal magnitude but that act in opposite directions would cancel, but they do not because they act on different systems. In this case, there are two different systems that we could choose to investigate: the swimmer or the wall. If we choose the swimmer to be the system of interest, as in the figure, then motion. Because acceleration is in the same direction as the net external force, the swimmer moves in the direction of is an external force on the swimmer and affects her Because the swimmer is our system (or object of interest) and not the wall, we do not need to consider the force because it originates fromthe swimmer rather than acting onthe swimmer. Therefore, does not Note that the swimmer pushes in the direction directly affect the motion of the system and does not cancel opposite to the direction in which she wants to move. Access for free at openstax.org. 4.4 • Newton's Third Law of Motion 129 Figure 4.8 When the swimmer exerts a force on the wall, she accelerates in the direction opposite to that of her push. This means that the net external force on her is in the direction opposite to This opposition is the result of Newton’s third law of motion, which dictates that the wall exerts a force on the swimmer that is equal in magnitude but that acts in the direction opposite to the force that
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the swimmer exerts on the wall. Other examples of Newton’s third law are easy to find. As a teacher paces in front of a whiteboard, he exerts a force backward on the floor. The floor exerts a reaction force in the forward direction on the teacher that causes him to accelerate forward. Similarly, a car accelerates because the ground pushes forward on the car's wheels in reaction to the car's wheels pushing backward on the ground. You can see evidence of the wheels pushing backward when tires spin on a gravel road and throw rocks backward. Another example is the force of a baseball as it makes contact with the bat. Helicopters create lift by pushing air down, creating an upward reaction force. Birds fly by exerting force on air in the direction opposite that in which they wish to fly. For example, the wings of a bird force air downward and backward in order to get lift and move forward. An octopus propels itself forward in the water by ejecting water backward through a funnel in its body, which is similar to how a jet ski is propelled. In these examples, the octopus or jet ski push the water backward, and the water, in turn, pushes the octopus or jet ski forward. Applying Newton’s Third Law Forces are classified and given names based on their source, how they are transmitted, or their effects. In previous sections, we discussed the forces called push, weight, and friction. In this section, applying Newton’s third law of motion will allow us to explore three more forces: the normal force, tension, and thrust. However, because we haven’t yet covered vectors in depth, we’ll only consider one-dimensional situations in this chapter. Another chapter will consider forces acting in two dimensions. The gravitational force (or weight) acts on objects at all times and everywhere on Earth. We know from Newton’s second law that a net force produces an acceleration; so, why is everything not in a constant state of freefall toward the center of Earth? The answer is the normal force. The normal force is the force that a surface applies to an object to support the weight of that object; it acts perpendicular to the surface upon which the object rests. If an object on a flat surface is not accelerating, the net external force is zero, and the normal force has the same magnitude as the weight of the system but acts in the opposite direction. In equation form, we write that Note that this equation
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is only true for a horizontal surface. The word tensioncomes from the Latin word meaning to stretch. Tension is the force along the length of a flexible connector, such as a string, rope, chain, or cable. Regardless of the type of connector attached to the object of interest, one must remember that the connector can only pull (or exert tension) in the direction parallelto its length. Tension is a pull that acts parallel to the connector, and that acts in opposite directions at the two ends of the connector. This is possible because a flexible connector is simply a long series of action-reaction forces, except at the two ends where outside objects provide one member of the actionreaction forces. Consider a person holding a mass on a rope, as shown in Figure 4.9. 4.17 130 Chapter 4 • Forces and Newton’s Laws of Motion Figure 4.9 When a perfectly flexible connector (one requiring no force to bend it) such as a rope transmits a force T, this force must be parallel to the length of the rope, as shown. The pull that such a flexible connector exerts is a tension. Note that the rope pulls with equal magnitude force but in opposite directions to the hand and to the mass (neglecting the weight of the rope). This is an example of Newton’s third law. The rope is the medium that transmits forces of equal magnitude between the two objects but that act in opposite directions. Tension in the rope must equal the weight of the supported mass, as we can prove by using Newton’s second law. If the 5.00 kg mass in the figure is stationary, then its acceleration is zero, so weight W and the tension T supplied by the rope. Summing the external forces to find the net force, we obtain The only external forces acting on the mass are its 4.18 where T and W are the magnitudes of the tension and weight, respectively, and their signs indicate direction, with up being positive. By substituting mg for Fnet and rearranging the equation, the tension equals the weight of the supported mass, just as you would expect For a 5.00-kg mass (neglecting the mass of the rope), we see that 4.19 4.20 Another example of Newton’s third law in action is thrust. Rockets move forward by expelling gas backward at a high velocity. This means that the rocket exerts a large force backward on the gas in the rocket combustion chamber,
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and the gas, in turn, exerts a large force forward on the rocket in response. This reaction force is called thrust. TIPS FOR SUCCESS A common misconception is that rockets propel themselves by pushing on the ground or on the air behind them. They actually work better in a vacuum, where they can expel exhaust gases more easily. LINKS TO PHYSICS Math: Problem-Solving Strategy for Newton’s Laws of Motion The basics of problem solving, presented earlier in this text, are followed here with specific strategies for applying Newton’s laws of motion. These techniques also reinforce concepts that are useful in many other areas of physics. First, identify the physical principles involved. If the problem involves forces, then Newton’s laws of motion are involved, and it Access for free at openstax.org. is important to draw a careful sketch of the situation. An example of a sketch is shown in Figure 4.10. Next, as in Figure 4.10, use vectors to represent all forces. Label the forces carefully, and make sure that their lengths are proportional to the magnitude of the forces and that the arrows point in the direction in which the forces act. 4.4 • Newton's Third Law of Motion 131 Figure 4.10 (a) A sketch of Tarzan hanging motionless from a vine. (b) Arrows are used to represent all forces. T is the tension exerted on Tarzan by the vine, is the force exerted on the vine by Tarzan, and W is Tarzan’s weight (i.e., the force exerted on Tarzan by Earth’s gravity). All other forces, such as a nudge of a breeze, are assumed to be negligible. (c) Suppose we are given Tarzan’s mass and asked to find the tension in the vine. We define the system of interest as shown and draw a free-body diagram, as shown in (d). is no longer shown because it does not act on the system of interest; rather, acts on the outside world. (d) The free-body diagram shows only the external forces acting on Tarzan. For these to sum to zero, we must have Next, make a list of knowns and unknowns and assign variable names to the quantities given in the problem. Figure out which variables need to be calculated; these are the unknowns. Now carefully define the system: which objects are of interest for the problem. This decision is important, because Newton’
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s second law involves only external forces. Once the system is identified, it’s possible to see which forces are external and which are internal (see Figure 4.10). If the system acts on an object outside the system, then you know that the outside object exerts a force of equal magnitude but in the opposite direction on the system. A diagram showing the system of interest and all the external forces acting on it is called a free-body diagram. Only external forces are shown on free-body diagrams, not acceleration or velocity. Figure 4.10 shows a free-body diagram for the system of interest. After drawing a free-body diagram, apply Newton’s second law to solve the problem. This is done in Figure 4.10 for the case of Tarzan hanging from a vine. When external forces are clearly identified in the free-body diagram, translate the forces into equation form and solve for the unknowns. Note that forces acting in opposite directions have opposite signs. By convention, forces acting downward or to the left are usually negative. GRASP CHECK If a problem has more than one system of interest, more than one free-body diagram is required to describe the external forces acting on the different systems. a. True b. False 132 Chapter 4 • Forces and Newton’s Laws of Motion WATCH PHYSICS Newton’s Third Law of Motion This video explains Newton’s third law of motion through examples involving push, normal force, and thrust (the force that propels a rocket or a jet). Click to view content (https://www.openstax.org/l/astronaut) GRASP CHECK If the astronaut in the video wanted to move upward, in which direction should he throw the object? Why? a. He should throw the object upward because according to Newton’s third law, the object will then exert a force on him in the same direction (i.e., upward). b. He should throw the object upward because according to Newton’s third law, the object will then exert a force on him in the opposite direction (i.e., downward). c. He should throw the object downward because according to Newton’s third law, the object will then exert a force on him in the opposite direction (i.e., upward). d. He should throw the object downward because according to Newton’s third law, the object will then exert a force on him in the same direction (i.
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e., downward). WORKED EXAMPLE An Accelerating Subway Train A physics teacher pushes a cart of demonstration equipment to a classroom, as in Figure 4.11. Her mass is 65.0 kg, the cart’s mass is 12.0 kg, and the equipment’s mass is 7.0 kg. To push the cart forward, the teacher’s foot applies a force of 150 N in the opposite direction (backward) on the floor. Calculate the acceleration produced by the teacher. The force of friction, which opposes the motion, is 24.0 N. Figure 4.11 Strategy Because they accelerate together, we define the system to be the teacher, the cart, and the equipment. The teacher pushes backward with a force system. Because all motion is horizontal, we can assume that no net force acts in the vertical direction, and the problem becomes one dimensional. As noted in the figure, the friction f opposes the motion and therefore acts opposite the direction of of 150 N. According to Newton’s third law, the floor exerts a forward force of 150 N on the Access for free at openstax.org. 4.4 • Newton's Third Law of Motion 133 We should not include the forces because these are exerted bythe system, not onthe system. We find the net external force by adding together the external forces acting on the system (see the free-body diagram in the figure) and then use Newton’s second law to find the acceleration., or, Solution Newton’s second law is The net external force on the system is the sum of the external forces: the force of the floor acting on the teacher, cart, and equipment (in the horizontal direction) and the force of friction. Because friction acts in the opposite direction, we assign it a negative value. Thus, for the net force, we obtain 4.21 The mass of the system is the sum of the mass of the teacher, cart, and equipment. Insert these values of net F and minto Newton’s second law to obtain the acceleration of the system. 4.22 4.23 4.24 4.25 Discussion None of the forces between components of the system, such as between the teacher’s hands and the cart, contribute to the net external force because they are internal to the system. Another way to look at this is to note that the forces between components of a system cancel because they are equal in magnitude and opposite in direction. For example
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, the force exerted by the teacher on the cart is of equal magnitude but in the opposite direction of the force exerted by the cart on the teacher. In this case, both forces act on the same system, so they cancel. Defining the system was crucial to solving this problem. Practice Problems 14. What is the equation for the normal force for a body with mass mthat is at rest on a horizontal surface? a. N = m b. N = mg c. N = mv d. N = g 15. An object with mass mis at rest on the floor. What is the magnitude and direction of the normal force acting on it? a. N = mvin upward direction b. N = mgin upward direction c. N = mvin downward direction d. N = mgin downward direction Check Your Understanding 16. What is Newton’s third law of motion? a. Whenever a first body exerts a force on a second body, the first body experiences a force that is twice the magnitude and acts in the direction of the applied force. b. Whenever a first body exerts a force on a second body, the first body experiences a force that is equal in magnitude and acts in the direction of the applied force. c. Whenever a first body exerts a force on a second body, the first body experiences a force that is twice the magnitude but acts in the direction opposite the direction of the applied force. d. Whenever a first body exerts a force on a second body, the first body experiences a force that is equal in magnitude but 134 Chapter 4 • Forces and Newton’s Laws of Motion acts in the direction opposite the direction of the applied force. 17. Considering Newton’s third law, why don’t two equal and opposite forces cancel out each other? a. Because the two forces act in the same direction b. Because the two forces have different magnitudes c. Because the two forces act on different systems d. Because the two forces act in perpendicular directions Access for free at openstax.org. Chapter 4 • Key Terms 135 KEY TERMS dynamics the study of how forces affect the motion of objects and systems external force a force acting on an object or system that originates outside of the object or system force a push or pull on an object with a specific magnitude and direction; can be represented by vectors; can be expressed as a multiple of a standard force free-body diagram a diagram showing all external forces acting on a body Newton’s second law
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of motion the net external force, on an object is proportional to and in the same direction as the acceleration of the object, a, and also proportional to the object’s mass, m; defined mathematically as or Newton’s third law of motion when one body exerts a force on a second body, the first body experiences a force that is equal in magnitude and opposite in direction to the force that it exerts freefall a situation in which the only force acting on an normal force the force that a surface applies to an object; object is the force of gravity friction an external force that acts in the direction opposite to the direction of motion inertia the tendency of an object at rest to remain at rest, or for a moving object to remain in motion in a straight line and at a constant speed law of inertia Newton’s first law of motion: a body at rest remains at rest or, if in motion, remains in motion at a constant speed in a straight line, unless acted on by a net external force; also known as the law of inertia mass the quantity of matter in a substance; measured in kilograms net external force the sum of all external forces acting on an object or system net force the sum of all forces acting on an object or system Newton’s first law of motion a body at rest remains at rest or, if in motion, remains in motion at a constant speed in a straight line, unless acted on by a net external force; also known as the law of inertia SECTION SUMMARY 4.1 Force • Dynamics is the study of how forces affect the motion of objects and systems. • Force is a push or pull that can be defined in terms of various standards. It is a vector and so has both magnitude and direction. • External forces are any forces outside of a body that act on the body. A free-body diagram is a drawing of all external forces acting on a body. 4.2 Newton's First Law of Motion: Inertia • Newton’s first law states that a body at rest remains at rest or, if moving, remains in motion in a straight line at a constant speed, unless acted on by a net external force. This law is also known as the law of inertia. Inertia is the tendency of an object at rest to remain at rest or, if moving, to remain in motion at constant velocity. Inertia is related to an object’s mass. • acts perpendicular and away from the surface with which the
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object is in contact system one or more objects of interest for which only the forces acting on them from the outside are considered, but not the forces acting between them or inside them tension a pulling force that acts along a connecting medium, especially a stretched flexible connector, such as a rope or cable; when a rope supports the weight of an object, the force exerted on the object by the rope is called tension thrust a force that pushes an object forward in response to the backward ejection of mass by the object; rockets and airplanes are pushed forward by a thrust reaction force in response to ejecting gases backward weight the force of gravity, W, acting on an object of mass m; defined mathematically as W = mg, where g is the magnitude and direction of the acceleration due to gravity • Friction is a force that opposes motion and causes an object or system to slow down. • Mass is the quantity of matter in a substance. 4.3 Newton's Second Law of Motion • Acceleration is a change in velocity, meaning a change in speed, direction, or both. • An external force acts on a system from outside the system, as opposed to internal forces, which act between components within the system. • Newton’s second law of motion states that the acceleration of a system is directly proportional to and in the same direction as the net external force acting on the system, and inversely proportional to the system’s mass. In equation form, Newton’s second law of motion is • or or. This is sometimes written as • The weight of an object of mass mis the force of gravity that acts on it. From Newton’s second law, weight is 136 Chapter 4 • Key Equations • given by If the only force acting on an object is its weight, then the object is in freefall. 4.4 Newton's Third Law of Motion • Newton’s third law of motion states that when one body exerts a force on a second body, the first body experiences a force that is equal in magnitude and opposite in direction to the force that it exerts. • When an object rests on a surface, the surface applies a force on the object that opposes the weight of the object. KEY EQUATIONS 4.2 Newton's First Law of Motion: Inertia Newton's first law of motion or This force acts perpendicular to the surface and is called the normal force. • The pulling force that acts along a stretched flexible connector, such as a rope or cable, is called tension. When
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a rope supports the weight of an object at rest, the tension in the rope is equal to the weight of the object. • Thrust is a force that pushes an object forward in response to the backward ejection of mass by the object. Rockets and airplanes are pushed forward by thrust. Newton’s second law of motion to solve weight 4.4 Newton's Third Law of Motion 4.3 Newton's Second Law of Motion Newton’s second law of motion or normal force for a nonaccelerating horizontal surface tension for an object at rest Newton’s second law of motion to solve acceleration CHAPTER REVIEW Concept Items 4.1 Force 1. What is dynamics? a. Dynamics is the study of internal forces. b. Dynamics is the study of forces and their effect on motion. c. Dynamics describes the motion of points, bodies, and systems without consideration of the cause of motion. d. Dynamics describes the effect of forces on each other. 2. Two forces acting on an object are perpendicular to one another. How would you draw these in a free-body diagram? a. The two force arrows will be drawn at a right angle to one another. b. The two force arrows will be pointing in opposite directions. c. The two force arrows will be at a 45° angle to one another. Access for free at openstax.org. d. The two force arrows will be at a 180° angle to one another. 3. A free-body diagram shows the forces acting on an object. How is that object represented in the diagram? a. A single point b. A square box c. A unit circle d. The object as it is 4.2 Newton's First Law of Motion: Inertia 4. A ball rolls along the ground, moving from north to south. What direction is the frictional force that acts on the ball? a. North to south b. South to north c. West to east d. East to west 5. The tires you choose to drive over icy roads will create more friction with the road than your summer tires. Give another example where more friction is desirable. a. Children’s slide b. Air hockey table Ice-skating rink c. Jogging track d. 6. How do you express, mathematically, that no external force is acting on a body? a. Fnet = −1 b. Fnet = 0 c. Fnet = 1 d. Fnet = ∞ 4.3 Newton's Second Law
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of Motion 7. What does it mean for two quantities to be inversely proportional to each other? a. When one variable increases, the other variable decreases by a greater amount. b. When one variable increases, the other variable also increases. c. When one variable increases, the other variable decreases by the same factor. d. When one variable increases, the other variable also increases by the same factor. Chapter 4 • Chapter Review 137 8. True or False: Newton's second law can be interpreted based on Newton's first law. a. True b. False 4.4 Newton's Third Law of Motion 9. Which forces cause changes in the motion of a system? a. internal forces b. external forces c. both internal and external forces d. neither internal nor external forces 10. True or False—Newton’s third law applies to the external forces acting on a system of interest. a. True b. False 11. A ball is dropped and hits the floor. What is the direction of the force exerted by the floor on the ball? a. Upward b. Downward c. Right d. Left Critical Thinking Items direction. 4.1 Force 12. Only two forces are acting on an object: force A to the left and force B to the right. If force B is greater than force A, in which direction will the object move? a. To the right b. To the left c. Upward d. The object does not move 13. In a free-body diagram, the arrows representing tension and weight have the same length but point away from one another. What does this indicate? a. They are equal in magnitude and act in the same direction. c. The magnitude is xand points in the downward direction. d. The magnitude is 2xand points in the downward direction. 15. Three forces, A, B, and C, are acting on the same object with magnitudes a, b, and c, respectively. Force A acts to the right, force B acts to the left, and force C acts downward. What is a necessary condition for the object to move straight down? a. The magnitude of force A must be greater than the magnitude of force B, so a > b. b. The magnitude of force A must be equal to the magnitude of force B, so a = b. c. The magnitude of force A must be greater than the b. They are equal in magnitude and act in opposite magnitude of force C, so A > C.
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directions. c. They are unequal in magnitude and act in the same d. The magnitude of force C must be greater than the magnitude of forces A or B, so A < C > B. direction. d. They are unequal in magnitude and act in opposite 4.2 Newton's First Law of Motion: Inertia directions. 14. An object is at rest. Two forces, X and Y, are acting on it. Force X has a magnitude of xand acts in the downward direction. What is the magnitude and direction of Y? a. The magnitude is xand points in the upward direction. 16. Two people push a cart on a horizontal surface by applying forces F1 and F2 in the same direction. Is the magnitude of the net force acting on the cart, Fnet, equal to, greater than, or less than F1 + F2? Why? a. Fnet < F1 + F2 because the net force will not include the frictional force. b. The magnitude is 2xand points in the upward b. Fnet = F1 + F2 because the net force will not include 138 Chapter 4 • Chapter Review the frictional force in acceleration and mass. c. Fnet < F1 + F2 because the net force will include the component of frictional force 4.4 Newton's Third Law of Motion d. Fnet = F1 + F2 because the net force will include the frictional force 17. True or False: A book placed on a balance scale is balanced by a standard 1-kg iron weight placed on the opposite side of the balance. If these objects are taken to the moon and a similar exercise is performed, the balance is still level because gravity is uniform on the moon’s surface as it is on Earth’s surface. a. True b. False 4.3 Newton's Second Law of Motion 18. From the equation for Newton’s second law, we see that Fnet is directly proportional to a and that the constant of proportionality is m. What does this mean in a practical sense? a. An increase in applied force will cause an increase in acceleration if the mass is constant. b. An increase in applied force will cause a decrease in acceleration if the mass is constant. c. An increase in applied force will cause an increase in acceleration, even if the mass varies. d. An increase in applied force will cause an increase 19. True or False: A person accelerates while walking on the ground
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by exerting force. The ground in turn exerts force F2 on the person. F1 and F2 are equal in magnitude but act in opposite directions. The person is able to walk because the two forces act on the different systems and the net force acting on the person is nonzero. a. True b. False 20. A helicopter pushes air down, which, in turn, pushes the helicopter up. Which force affects the helicopter’s motion? Why? a. Air pushing upward affects the helicopter’s motion because it is an internal force that acts on the helicopter. b. Air pushing upward affects the helicopter’s motion because it is an external force that acts on the helicopter. c. The downward force applied by the blades of the helicopter affects its motion because it is an internal force that acts on the helicopter. d. The downward force applied by the blades of the helicopter affects its motion because it is an external force that acts on the helicopter. Problems 4.3 Newton's Second Law of Motion 55.0 kg c. d. 91.9 kg on Earth. What is its 4.4 Newton's Third Law of Motion 21. An object has a mass of weight on the moon? a. b. c. d. 22. A bathroom scale shows your mass as 55 kg. What will it read on the moon? a. 9.4 kg 10.5 kg b. Performance Task 4.4 Newton's Third Law of Motion 24. A car weighs 2,000 kg. It moves along a road by applying a force on the road with a parallel component of 560 N. There are two passengers in the car, each weighing 55 kg. If the magnitude of the force of friction Access for free at openstax.org. 23. A person pushes an object of mass 5.0 kg along the floor by applying a force. If the object experiences a friction force of 10 N and accelerates at 18 m/s2, what is the magnitude of the force exerted by the person? a. −90 N b. −80 N c. 90 N 100 N d. experienced by the car is 45 N, what is the acceleration of the car? a. 0.244 m/s2 b. 0.265 m/s2 c. 4.00 m/s2 d. 4.10 m/s2 TEST PREP Multiple Choice 4.1 Force 25. Which of the following is a physical quantity that can be described by dynamics but not
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by kinematics? a. Velocity b. Acceleration c. Force 26. Which of the following is used to represent an object in a free-body diagram? a. A point b. A line c. A vector 4.2 Newton's First Law of Motion: Inertia 27. What kind of force is friction? a. External force b. Internal force c. Net force 28. What is another name for Newton’s first law? a. Law of infinite motion b. Law of inertia c. Law of friction 29. True or False—A rocket is launched into space and escapes Earth’s gravitational pull. It will continue to travel in a straight line until it is acted on by another force. a. True b. False 30. A 2,000-kg car is sitting at rest in a parking lot. A bike and rider with a total mass of 60 kg are traveling along a road at 10 km/h. Which system has more inertia? Why? a. The car has more inertia, as its mass is greater than the mass of the bike. b. The bike has more inertia, as its mass is greater than the mass of the car. c. The car has more inertia, as its mass is less than the mass of the bike. d. The bike has more inertia, as its mass is less than the mass of the car. 4.3 Newton's Second Law of Motion 31. In the equation for Newton’s second law, what does Fnet stand for? Internal force a. b. Net external force c. Frictional force 32. What is the SI unit of force? Chapter 4 • Test Prep 139 a. Kg b. dyn c. N 33. What is the net external force on an object in freefall on Earth if you were to neglect the effect of air? a. The net force is zero. b. The net force is upward with magnitude mg. c. The net force is downward with magnitude mg. d. The net force is downward with magnitude 9.8 N. 34. Two people push a 2,000-kg car to get it started. An acceleration of at least 5.0 m/s2 is required to start the car. Assuming both people apply the same magnitude force, how much force will each need to apply if friction between the car and the road is 300 N? a. 4850 N 5150 N b. c. 97000 N 10300 N d. 4.4 Newton
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's Third Law of Motion 35. One object exerts a force of magnitude F1 on another object and experiences a force of magnitude F2 in return. What is true for F1 and F2? a. F1 > F2 b. F1 < F2 c. F1 = F2 36. A weight is suspended with a rope and hangs freely. In what direction is the tension on the rope? a. parallel to the rope b. perpendicular to the rope 37. A person weighing 55 kg walks by applying 160 N of force on the ground, while pushing a 10-kg object. If the person accelerates at 2 m/s2, what is the force of friction experienced by the system consisting of the person and the object? 30 N a. b. 50 N c. 270 N d. 290 N 38. A 65-kg swimmer pushes on the pool wall and accelerates at 6 m/s2. The friction experienced by the swimmer is 100 N. What is the magnitude of the force that the swimmer applies on the wall? a. −490 N b. −290 N c. 290 N d. 490 N 140 Chapter 4 • Test Prep Short Answer 4.1 Force 39. True or False—An external force is defined as a force generated outside the system of interest that acts on an object inside the system. a. True b. False 40. By convention, which sign is assigned to an object moving downward? a. A positive sign ( b. A negative sign ( c. Either a positive or negative sign ( d. No sign is assigned ) ) ) 41. A body is pushed downward by a force of 5 units and upward by a force of 2 units. How would you draw a free-body diagram to represent this? a. Two force vectors acting at a point, both pointing up with lengths of 5 units and 2 units b. Two force vectors acting at a point, both pointing down with lengths of 5 units and 2 units c. Two force vectors acting at a point, one pointing up with a length of 5 units and the other pointing down with a length of 2 units d. Two force vectors acting at a point, one pointing down with a length of 5 units and the other pointing up with a length of 2 units 42. A body is pushed eastward by a force of four units and southward by a force of three units. How would you draw a free-body diagram to represent this? a. Two force vectors acting at a
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point, one pointing left with a length of 4 units and the other pointing down with a length of 3 units b. Two force vectors acting at a point, one pointing left with a length of 4 units and the other pointing up with a length of 3 units c. Two force vectors acting at a point, one pointing right with a length of 4 units and the other pointing down with a length of 3 units d. Two force vectors acting at a point, one pointing right with a length of 4 units and the other pointing up with a length of 3 units 4.2 Newton's First Law of Motion: Inertia 43. A body with mass m is pushed along a horizontal surface by a force F and is opposed by a frictional force f. How would you draw a free-body diagram to represent this situation? a. A dot with an arrow pointing right, labeled F, and an arrow pointing left, labeled f, that is of equal length or shorter than F Access for free at openstax.org. b. A dot with an arrow pointing right, labeled F, and an arrow pointing right, labeled f, that is of equal length or shorter than F c. A dot with an arrow pointing right, labeled F, and a smaller arrow pointing up, labeled f, that is of equal length or longer than F d. A dot with an arrow pointing right, labeled F, and a smaller arrow pointing down, labeled f, that is of equal length or longer than F 44. Two objects rest on a uniform surface. A person pushes both with equal force. If the first object starts to move faster than the second, what can be said about their masses? a. The mass of the first object is less than that of the second object. b. The mass of the first object is equal to the mass of the second object. c. The mass of the first object is greater than that of the second object. d. No inference can be made because mass and force are not related to each other. 45. Two similar boxes rest on a table. One is empty and the other is filled with pebbles. Without opening or lifting either, how can you tell which box is full? Why? a. By applying an internal force; whichever box accelerates faster is lighter and so must be empty b. By applying an internal force; whichever box accelerates faster is heavier and so the other box must be empty c. By applying an external force; whichever box accelerates faster is lighter and so must
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be empty d. By applying an external force; whichever box accelerates faster is heavier and so the other box must be empty 46. True or False—An external force is required to set a stationary object in motion in outer space away from all gravitational influences and atmospheric friction. a. True b. False 4.3 Newton's Second Law of Motion 47. A steadily rolling ball is pushed in the direction from east to west, which causes the ball to move faster in the same direction. What is the direction of the acceleration? a. North to south b. South to north c. East to west d. West to east 48. A ball travels from north to south at 60 km/h. After being hit by a bat, it travels from west to east at 60 km/ Chapter 4 • Test Prep 141 h. Is there a change in velocity? a. Yes, because velocity is a scalar. b. Yes, because velocity is a vector. c. No, because velocity is a scalar. d. No, because velocity is a vector 49. What is the weight of a 5-kg object on Earth and on the moon? a. On Earth the weight is 1.67 N, and on the moon the weight is 1.67 N. b. On Earth the weight is 5 N, and on the moon the weight is 5 N. b. F c. 2F 30F d. 52. A fish pushes water backward with its fins. How does this propel the fish forward? a. The water exerts an internal force on the fish in the opposite direction, pushing the fish forward. b. The water exerts an external force on the fish in the opposite direction, pushing the fish forward. c. The water exerts an internal force on the fish in the same direction, pushing the fish forward. c. On Earth the weight is 49 N, and on the moon the d. The water exerts an external force on the fish in the weight is 8.35 N. same direction, pushing the fish forward. d. On Earth the weight is 8.35 N, and on the moon the weight is 49 N. 50. An object weighs 294 N on Earth. What is its weight on the moon? 50.1 N a. b. 30.0 N c. 249 N 1461 N d. 4.4 Newton's Third Law of Motion 51. A large truck with mass 30 m crashes into a small sedan with mass m. If the truck exerts a
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force F on the sedan, what force will the sedan exert on the truck? a. Extended Response 4.1 Force 55. True or False—When two unequal forces act on a body, the body will not move in the direction of the weaker force. a. True b. False 56. In the figure given, what is Frestore? What is its magnitude? a. Frestore is the force exerted by the hand on the spring, and it pulls to the right. b. Frestore is the force exerted by the spring on the hand, and it pulls to the left. 53. True or False—Tension is the result of opposite forces in a connector, such as a string, rope, chain or cable, that pulls each point of the connector apart in the direction parallel to the length of the connector. At the ends of the connector, the tension pulls toward the center of the connector. a. True b. False 54. True or False—Normal reaction is the force that opposes the force of gravity and acts in the direction of the force of gravity. a. True b. False c. Frestore is the force exerted by the hand on the spring, and it pulls to the left. d. Frestore is the force exerted by the spring on the hand, and it pulls to the right. 4.2 Newton's First Law of Motion: Inertia 57. Two people apply the same force to throw two identical balls in the air. Will the balls necessarily travel the same distance? Why or why not? a. No, the balls will not necessarily travel the same distance because the gravitational force acting on them is different. b. No, the balls will not necessarily travel the same distance because the angle at which they are thrown may differ. c. Yes, the balls will travel the same distance because the gravitational force acting on them is the same. d. Yes, the balls will travel the same distance because the angle at which they are thrown may differ. 58. A person pushes a box from left to right and then lets the box slide freely across the floor. The box slows down as it slides across the floor. When the box is sliding 142 Chapter 4 • Test Prep freely, what is the direction of the net external force? a. The net external force acts from left to right. b. The net external force acts from right to left. c. The net external force acts upward. d. The net external force acts downward. 4.3
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Newton's Second Law of Motion 59. A 55-kg lady stands on a bathroom scale inside an elevator. The scale reads 70 kg. What do you know about the motion of the elevator? a. The elevator must be accelerating upward. b. The elevator must be accelerating downward. c. The elevator must be moving upward with a constant velocity. d. The elevator must be moving downward with a constant velocity. 60. True or False—A skydiver initially accelerates in his jump. Later, he achieves a state of constant velocity called terminal velocity. Does this mean the skydiver becomes weightless? a. Yes b. No 4.4 Newton's Third Law of Motion 61. How do rockets propel themselves in space? a. Rockets expel gas in the forward direction at high velocity, and the gas, which provides an internal force, pushes the rockets forward. b. Rockets expel gas in the forward direction at high velocity, and the gas, which provides an external force, pushes the rockets forward. c. Rockets expel gas in the backward direction at high velocity, and the gas, which is an internal force, pushes the rockets forward. d. Rockets expel gas in the backward direction at high velocity, and the gas, which provides an external force, pushes the rockets forward. 62. Are rockets more efficient in Earth’s atmosphere or in outer space? Why? a. Rockets are more efficient in Earth’s atmosphere than in outer space because the air in Earth’s atmosphere helps to provide thrust for the rocket, and Earth has more air friction than outer space. b. Rockets are more efficient in Earth’s atmosphere than in outer space because the air in Earth’s atmosphere helps to provide thrust to the rocket, and Earth has less air friction than the outer space. c. Rockets are more efficient in outer space than in Earth’s atmosphere because the air in Earth’s atmosphere does not provide thrust but does create more air friction than in outer space. d. Rockets are more efficient in outer space than in Earth’s atmosphere because the air in Earth’s atmosphere does not provide thrust but does create less air friction than in outer space. Access for free at openstax.org. CHAPTER 5 Motion in Two Dimensions Figure 5.1 Billiard balls on a pool table are in motion after being hit with a cue stick. (Popperipopp, Wikimedia Commons) Chapter Outline 5.1 Vector Addition and Subtraction: Graphical
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Methods 5.2 Vector Addition and Subtraction: Analytical Methods 5.3 Projectile Motion 5.4 Inclined Planes 5.5 Simple Harmonic Motion In Chapter 2, we learned to distinguish between vectors and scalars; the difference being that a vector has INTRODUCTION magnitude and direction, whereas a scalar has only magnitude. We learned how to deal with vectors in physics by working straightforward one-dimensional vector problems, which may be treated mathematically in the same as scalars. In this chapter, we’ll use vectors to expand our understanding of forces and motion into two dimensions. Most real-world physics problems (such as with the game of pool pictured here) are, after all, either two- or three-dimensional problems and physics is most useful when applied to real physical scenarios. We start by learning the practical skills of graphically adding and subtracting vectors (by using drawings) and analytically (with math). Once we’re able to work with two-dimensional vectors, we apply these skills to problems of projectile motion, inclined planes, and harmonic motion. 144 Chapter 5 • Motion in Two Dimensions 5.1 Vector Addition and Subtraction: Graphical Methods Section Learning Objectives By the end of this section, you will be able to do the following: • Describe the graphical method of vector addition and subtraction • Use the graphical method of vector addition and subtraction to solve physics problems Section Key Terms graphical method head (of a vector) head-to-tail method resultant resultant vector tail vector addition vector subtraction The Graphical Method of Vector Addition and Subtraction Recall that a vector is a quantity that has magnitude and direction. For example, displacement, velocity, acceleration, and force are all vectors. In one-dimensional or straight-line motion, the direction of a vector can be given simply by a plus or minus sign. Motion that is forward, to the right, or upward is usually considered to be positive(+); and motion that is backward, to the left, or downward is usually considered to be negative(−). In two dimensions, a vector describes motion in two perpendicular directions, such as vertical and horizontal. For vertical and horizontal motion, each vector is made up of vertical and horizontal components. In a one-dimensional problem, one of the components simply has a value of zero. For two-dimensional vectors, we work with vectors by using a frame of reference such as a coordinate system. Just as with one-dimensional vectors
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, we graphically represent vectors with an arrow having a length proportional to the vector’s magnitude and pointing in the direction that the vector points. Figure 5.2 shows a graphical representation of a vector; the total displacement for a person walking in a city. The person first walks nine blocks east and then five blocks north. Her total displacement does not match her path to her final destination. The displacement simply connects her starting point with her ending point using a straight line, which is the shortest distance. We use the notation that a boldface symbol, such as D, stands for a vector. Its magnitude is represented by the symbol in italics, D, and its direction is given by an angle represented by the symbol Note that her displacement would be the same if she had begun by first walking five blocks north and then walking nine blocks east. TIPS FOR SUCCESS In this text, we represent a vector with a boldface variable. For example, we represent a force with the vector F, which has both magnitude and direction. The magnitude of the vector is represented by the variable in italics, F, and the direction of the variable is given by the angle Figure 5.2 A person walks nine blocks east and five blocks north. The displacement is 10.3 blocks at an angle north of east. The head-to-tail method is a graphical way to add vectors. The tail of the vector is the starting point of the vector, and the head (or tip) of a vector is the pointed end of the arrow. The following steps describe how to use the head-to-tail method for graphical vector addition. Access for free at openstax.org. 1. Let the x-axis represent the east-west direction. Using a ruler and protractor, draw an arrow to represent the first vector (nine blocks to the east), as shown in Figure 5.3(a). 5.1 • Vector Addition and Subtraction: Graphical Methods 145 Figure 5.3 The diagram shows a vector with a magnitude of nine units and a direction of 0°. 2. Let the y-axis represent the north-south direction. Draw an arrow to represent the second vector (five blocks to the north). Place the tail of the second vector at the head of the first vector, as shown in Figure 5.4(b). Figure 5.4 A vertical vector is added. 3. If there are more than two vectors, continue to add the vectors head-to-tail as described in
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step 2. In this example, we have only two vectors, so we have finished placing arrows tip to tail. 4. Draw an arrow from the tail of the first vector to the head of the last vector, as shown in Figure 5.5(c). This is the resultant, or the sum, of the vectors. 146 Chapter 5 • Motion in Two Dimensions Figure 5.5 The diagram shows the resultant vector, a ruler, and protractor. 5. To find the magnitude of the resultant, measure its length with a ruler. When we deal with vectors analytically in the next section, the magnitude will be calculated by using the Pythagorean theorem. 6. To find the direction of the resultant, use a protractor to measure the angle it makes with the reference direction (in this case, the x-axis). When we deal with vectors analytically in the next section, the direction will be calculated by using trigonometry to find the angle. WATCH PHYSICS Visualizing Vector Addition Examples This video shows four graphical representations of vector addition and matches them to the correct vector addition formula. Click to view content (https://openstax.org/l/02addvector) GRASP CHECK There are two vectors and. The head of vector touches the tail of vector. The addition of vectors and gives a resultant vector. Can the addition of these two vectors can be represented by the following two equations? ; a. Yes, if we add the same two vectors in a different order it will still give the same resultant vector. b. No, the resultant vector will change if we add the same vectors in a different order. Vector subtraction is done in the same way as vector addition with one small change. We add the first vector to the negative of the vector that needs to be subtracted. A negative vector has the same magnitude as the original vector, but points in the opposite direction (as shown in Figure 5.6). Subtracting the vector B from the vector A, which is written as A − B, is the same as A + (−B). Since it does not matter in what order vectors are added, A − B is also equal to (−B) + A. This is true for scalars as well as vectors. For example, 5 – 2 = 5 + (−2) = (−2) + 5. Access for free at openstax.org. 5.1 • Vector Addition and Subtraction: Graphical Methods 147 Figure
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5.6 The diagram shows a vector, B, and the negative of this vector, –B. Global angles are calculated in the counterclockwise direction. The clockwise direction is considered negative. For example, an from the positive x-axis. angle of south of west is the same as the global angle which can also be expressed as Using the Graphical Method of Vector Addition and Subtraction to Solve Physics Problems Now that we have the skills to work with vectors in two dimensions, we can apply vector addition to graphically determine the resultant vector, which represents the total force. Consider an example of force involving two ice skaters pushing a third as seen in Figure 5.7. Figure 5.7 Part (a) shows an overhead view of two ice skaters pushing on a third. Forces are vectors and add like vectors, so the total force on the third skater is in the direction shown. In part (b), we see a free-body diagram representing the forces acting on the third skater. In problems where variables such as force are already known, the forces can be represented by making the length of the vectors proportional to the magnitudes of the forces. For this, you need to create a scale. For example, each centimeter of vector length could represent 50 N worth of force. Once you have the initial vectors drawn to scale, you can then use the head-to-tail method to draw the resultant vector. The length of the resultant can then be measured and converted back to the original units using the scale you created. You can tell by looking at the vectors in the free-body diagram in Figure 5.7 that the two skaters are pushing on the third skater with equal-magnitude forces, since the length of their force vectors are the same. Note, however, that the forces are not equal because they act in different directions. If, for example, each force had a magnitude of 400 N, then we would find the magnitude of the total external force acting on the third skater by finding the magnitude of the resultant vector. Since the forces act at a right angle to one another, we can use the Pythagorean theorem. For a triangle with sides a, b, and c, the Pythagorean theorem tells us that Applying this theorem to the triangle made by F1, F2, and Ftot in Figure 5.7, we get 148 Chapter 5 • Motion in Two Dimensions or Note that, if the vectors were not at a right angle
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to each other Pythagorean theorem to find the magnitude of the resultant vector. Another scenario where adding two-dimensional vectors is necessary is for velocity, where the direction may not be purely east-west or north-south, but some combination of these two directions. In the next section, we cover how to solve this type of problem analytically. For now let’s consider the problem graphically. to one another), we would not be able to use the WORKED EXAMPLE Adding Vectors Graphically by Using the Head-to-Tail Method: A Woman Takes a Walk Use the graphical technique for adding vectors to find the total displacement of a person who walks the following three paths north of (displacements) on a flat field. First, he walks 25 m in a direction east. Finally, he turns and walks 32 m in a direction Strategy Graphically represent each displacement vector with an arrow, labeling the first A, the second B, and the third C. Make the lengths proportional to the distance of the given displacement and orient the arrows as specified relative to an east-west line. Use the head-to-tail method outlined above to determine the magnitude and direction of the resultant displacement, which we’ll call R. north of east. Then, he walks 23 m heading south of east. Solution (1) Draw the three displacement vectors, creating a convenient scale (such as 1 cm of vector length on paper equals 1 m in the problem), as shown in Figure 5.8. (2) Place the vectors head to tail, making sure not to change their magnitude or direction, as shown in Figure 5.9. Figure 5.8 The three displacement vectors are drawn first. (3) Draw the resultant vector R from the tail of the first vector to the head of the last vector, as shown in Figure 5.10. Figure 5.9 Next, the vectors are placed head to tail. Access for free at openstax.org. 5.1 • Vector Addition and Subtraction: Graphical Methods 149 Figure 5.10 The resultant vector is drawn. (4) Use a ruler to measure the magnitude of R, remembering to convert back to the units of meters using the scale. Use a protractor to measure the direction of R. While the direction of the vector can be specified in many ways, the easiest way is to measure the angle between the vector and the nearest horizontal or vertical axis. Since R is south of the eastward pointing axis (
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the x-axis), we flip the protractor upside down and measure the angle between the eastward axis and the vector, as illustrated in Figure 5.11. Figure 5.11 A ruler is used to measure the magnitude of R, and a protractor is used to measure the direction of R. In this case, the total displacement R has a magnitude of 50 m and points this vector can be expressed as south of east. Using its magnitude and direction, and 5.1 5.2 Discussion The head-to-tail graphical method of vector addition works for any number of vectors. It is also important to note that it does not matter in what order the vectors are added. Changing the order does not change the resultant. For example, we could add the vectors as shown in Figure 5.12, and we would still get the same solution. 150 Chapter 5 • Motion in Two Dimensions Figure 5.12 Vectors can be added in any order to get the same result. WORKED EXAMPLE Subtracting Vectors Graphically: A Woman Sailing a Boat A woman sailing a boat at night is following directions to a dock. The instructions read to first sail 27.5 m in a direction west of north). If the north of east from her current location, and then travel 30.0 m in a direction woman makes a mistake and travels in the oppositedirection for the second leg of the trip, where will she end up? The two legs of the woman’s trip are illustrated in Figure 5.13. north of east (or Figure 5.13 In the diagram, the first leg of the trip is represented by vector A and the second leg is represented by vector B. Strategy We can represent the first leg of the trip with a vector A, and the second leg of the trip that she was supposed totake with a vector B. Since the woman mistakenly travels in the opposite direction for the second leg of the journey, the vector for second leg of the trip she actuallytakes is −B. Therefore, she will end up at a location A + (−B), or A − B. Note that −B has the same south of east, as illustrated in Figure 5.14. magnitude as B (30.0 m), but is in the opposite direction, Figure 5.14 Vector –B represents traveling in the opposite direction of vector B. We use graphical vector addition to find where the woman arrives A + (−B). Access for free at openstax.org.
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5.1 • Vector Addition and Subtraction: Graphical Methods 151 Solution (1) To determine the location at which the woman arrives by accident, draw vectors A and −B. (2) Place the vectors head to tail. (3) Draw the resultant vector R. (4) Use a ruler and protractor to measure the magnitude and direction of R. These steps are demonstrated in Figure 5.15. Figure 5.15 The vectors are placed head to tail. In this case and 5.3 5.4 Discussion Because subtraction of a vector is the same as addition of the same vector with the opposite direction, the graphical method for subtracting vectors works the same as for adding vectors. WORKED EXAMPLE Adding Velocities: A Boat on a River A boat attempts to travel straight across a river at a speed of 3.8 m/s. The river current flows at a speed vriver of 6.1 m/s to the right. What is the total velocity and direction of the boat? You can represent each meter per second of velocity as one centimeter of vector length in your drawing. Strategy We start by choosing a coordinate system with its x-axis parallel to the velocity of the river. Because the boat is directed straight toward the other shore, its velocity is perpendicular to the velocity of the river. We draw the two vectors, vboat and vriver, as shown in Figure 5.16. Using the head-to-tail method, we draw the resulting total velocity vector from the tail of vboat to the head of vriver. 152 Chapter 5 • Motion in Two Dimensions Figure 5.16 A boat attempts to travel across a river. What is the total velocity and direction of the boat? Solution By using a ruler, we find that the length of the resultant vector is 7.2 cm, which means that the magnitude of the total velocity is By using a protractor to measure the angle, we find Discussion If the velocity of the boat and river were equal, then the direction of the total velocity would have been 45°. However, since the velocity of the river is greater than that of the boat, the direction is less than 45° with respect to the shore, or xaxis. 5.5 Practice Problems 1. Vector, having magnitude, pointing south of east and vector having magnitude, pointing north of east are added. What is the magnitude of the resultant vector? a. b. c. d. 2. A person walks north of west
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for and east of south for. What is the magnitude of his displacement? a. b. c. d. Virtual Physics Vector Addition In this simulation (https://archive.cnx.org/specials/d218bf9b-e50e-4d50-9a6c-b3db4dad0816/vector-addition/), you will experiment with adding vectors graphically. Click and drag the red vectors from the Grab One basket onto the graph in the middle of the screen. These red vectors can be rotated, stretched, or repositioned by clicking and dragging with your mouse. Check the Show Sum box to display the resultant vector (in green), which is the sum of all of the red vectors placed on the Access for free at openstax.org. 5.2 • Vector Addition and Subtraction: Analytical Methods 153 graph. To remove a red vector, drag it to the trash or click the Clear All button if you wish to start over. Notice that, if you click on any of the vectors, the component, and Ryis its vertical component. You can check the resultant by lining up the vectors so that the head of the first vector touches the tail of the second. Continue until all of the vectors are aligned together head-to-tail. You will see that the resultant magnitude and angle is the same as the arrow drawn from the tail of the first vector to the head of the last vector. Rearrange the vectors in any order head-to-tail and compare. The resultant will always be the same. is its direction with respect to the positive x-axis, Rx is its horizontal is its magnitude, Click to view content (https://archive.cnx.org/specials/d218bf9b-e50e-4d50-9a6c-b3db4dad0816/vector-addition/) GRASP CHECK True or False—The more long, red vectors you put on the graph, rotated in any direction, the greater the magnitude of the resultant green vector. a. True b. False Check Your Understanding 3. While there is no single correct choice for the sign of axes, which of the following are conventionally considered positive? a. backward and to the left b. backward and to the right forward and to the right c. forward and to the left d. 4. True or False—A person walks 2 blocks east and 5 blocks north. Another person walks 5
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blocks north and then two blocks east. The displacement of the first person will be more than the displacement of the second person. a. True b. False 5.2 Vector Addition and Subtraction: Analytical Methods Section Learning Objectives By the end of this section, you will be able to do the following: • Define components of vectors • Describe the analytical method of vector addition and subtraction • Use the analytical method of vector addition and subtraction to solve problems Section Key Terms analytical method component (of a two-dimensional vector) Components of Vectors For the analytical method of vector addition and subtraction, we use some simple geometry and trigonometry, instead of using a ruler and protractor as we did for graphical methods. However, the graphical method will still come in handy to visualize the problem by drawing vectors using the head-to-tail method. The analytical method is more accurate than the graphical method, which is limited by the precision of the drawing. For a refresher on the definitions of the sine, cosine, and tangent of an angle, see Figure 5.17. 154 Chapter 5 • Motion in Two Dimensions Figure 5.17 For a right triangle, the sine, cosine, and tangent of θare defined in terms of the adjacent side, the opposite side, or the hypotenuse. In this figure, xis the adjacent side, yis the opposite side, and his the hypotenuse. Since, by definition, length of yby using, we can find the length xif we know hand by using. Similarly, we can find the. These trigonometric relationships are useful for adding vectors. When a vector acts in more than one dimension, it is useful to break it down into its x and y components. For a two-dimensional vector, a component is a piece of a vector that points in either the x- or y-direction. Every 2-d vector can be expressed as a sum of its x and y components. For example, given a vector like produce it. In this example, common situation in physics and happens to be the least complicated situation trigonometrically., add to form a right triangle, meaning that the angle between them is 90 degrees. This is a in Figure 5.18, we may want to find what two perpendicular vectors, and and Figure 5.18 The vector, with its tail at the origin of an x- y-coordinate system, is shown together with its x-
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and y-components, and These vectors form a right triangle. and triangle. are defined to be the components of along the x- and y-axes. The three vectors,,, and, form a right If the vector and y-components, we use the following relationships for a right triangle: is known, then its magnitude (its length) and its angle (its direction) are known. To find and, its x- and is the magnitude of A in the x-direction, where resultant with respect to the x-axis, as shown in Figure 5.19. is the magnitude of A in the y-direction, and is the angle of the Access for free at openstax.org. 5.2 • Vector Addition and Subtraction: Analytical Methods 155 Figure 5.19 The magnitudes of the vector components and can be related to the resultant vector and the angle with trigonometric identities. Here we see that and Suppose, for example, that Figure 5.20. is the vector representing the total displacement of the person walking in a city, as illustrated in Figure 5.20 We can use the relationships and to determine the magnitude of the horizontal and vertical component vectors in this example. Then A = 10.3 blocks and, so that This magnitude indicates that the walker has traveled 9 blocks to the east—in other words, a 9-block eastward displacement. Similarly, 5.6 5.7 indicating that the walker has traveled 5 blocks to the north—a 5-block northward displacement. 156 Chapter 5 • Motion in Two Dimensions Analytical Method of Vector Addition and Subtraction Calculating a resultant vector (or vector addition) is the reverse of breaking the resultant down into its components. If the perpendicular components definition, are known, then we can find analytically. How do we do this? Since, by of a vector and we solve for to find the direction of the resultant. Since this is a right triangle, the Pythagorean theorem (x2 + y2 = h2) for finding the hypotenuse applies. In this case, it becomes Solving for A gives In summary, to find the magnitude in Figure 5.21, we use the following relationships: and direction of a vector from its perpendicular components and, as illustrated Figure 5.21 The magnitude and direction of the resultant vector can be determined once the horizontal components and have been determined. Sometimes, the vectors added are not perfectly perpendicular to one another. An
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example of this is the case below, where the vectors are added to produce the resultant as illustrated in Figure 5.22. and Access for free at openstax.org. 5.2 • Vector Addition and Subtraction: Analytical Methods 157 Figure 5.22 Vectors and are two legs of a walk, and is the resultant or total displacement. You can use analytical methods to determine the magnitude and direction of. represent two legs of a walk (two displacements), then If and up at the tip of x-direction and then in the y-direction. Those paths are the x- and y-components of the resultant,. There are many ways to arrive at the same point. The person could have walked straight ahead first in the is the total displacement. The person taking the walk ends and If we know and, we can find and using the equations and. 1. Draw in the xand ycomponents of each vector (including the resultant) with a dashed line. Use the equations and angle of by vector A). to find the components. In Figure 5.23, these components are, and with its own x-axis (which is slightly above the x-axis used Vector makes an,, with the x-axis, and vector makes and angle of Figure 5.23 To add vectors and first determine the horizontal and vertical components of each vector. These are the dotted vectors shown in the image. 2. Find the xcomponent of the resultant by adding the xcomponent of the vectors and find the ycomponent of the resultant (as illustrated in Figure 5.24) by adding the ycomponent of the vectors. 158 Chapter 5 • Motion in Two Dimensions Figure 5.24 The vectors and add to give the magnitude of the resultant vector in the horizontal direction, Similarly, the vectors and add to give the magnitude of the resultant vector in the vertical direction, Now that we know the components of we can find its magnitude and direction. 3. To get the magnitude of the resultant R, use the Pythagorean theorem. 4. To get the direction of the resultant WATCH PHYSICS Classifying Vectors and Quantities Example This video contrasts and compares three vectors in terms of their magnitudes, positions, and directions. Click to view content (https://www.youtube.com/embed/Yp0EhcVBxNU) GRASP CHECK. Vector points in the northwest. If the vectors,, and were added points to the northeast. Vector points to the
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, and, have the same magnitude of. Vector Three vectors,, southwest exactly opposite to vector together, what would be the magnitude of the resultant vector? Why? a. b. c. d.. All of them will cancel each other out.. Two of them will cancel each other out. units. All of them will add together to give the resultant.. Two of them will add together to give the resultant. TIPS FOR SUCCESS In the video, the vectors were represented with an arrow above them rather than in bold. This is a common notation in math classes. Access for free at openstax.org. 5.2 • Vector Addition and Subtraction: Analytical Methods 159 Using the Analytical Method of Vector Addition and Subtraction to Solve Problems Figure 5.25 uses the analytical method to add vectors. WORKED EXAMPLE An Accelerating Subway Train Add the vector the y-axis is along the north–south directions. A person first walks vector The person then walks in a direction to the vector shown in Figure 5.25, using the steps above. The x-axis is along the east–west direction, and north of east, represented by in a direction north of east, represented by vector Figure 5.25 You can use analytical models to add vectors. Strategy The components of We will solve for these components and then add them in the x-direction and y-direction to find the resultant. along the x- and y-axes represent walking due east and due north to get to the same ending point. and Solution First, we find the components of and along the x- and y-axes. From the problem, we know that =, and. We find the x-components by using, which gives and Similarly, the y-components are found using and The x- and y-components of the resultant are and 160 Chapter 5 • Motion in Two Dimensions Now we can find the magnitude of the resultant by using the Pythagorean theorem 5.8 so that Finally, we find the direction of the resultant This is Discussion This example shows vector addition using the analytical method. Vector subtraction using the analytical method is very similar. It is just the addition of a negative vector. That is, components of. Therefore, the x- and y-components of the resultant. The components of – are the negatives of the are and and the rest of the method outlined above is identical to that for addition. Practice Problems 5.
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What is the magnitude of a vector whose x-component is 4 cm and whose y-component is 3 cm? 1 cm a. 5 cm b. c. 7 cm d. 25 cm 6. What is the magnitude of a vector that makes an angle of 30° to the horizontal and whose x-component is 3 units? a. 2.61 units 3.00 units b. c. 3.46 units d. 6.00 units Access for free at openstax.org. 5.2 • Vector Addition and Subtraction: Analytical Methods 161 LINKS TO PHYSICS Atmospheric Science Figure 5.26 This picture shows Bert Foord during a television Weather Forecast from the Meteorological Office in 1963. (BBC TV) Atmospheric science is a physical science, meaning that it is a science based heavily on physics. Atmospheric science includes meteorology (the study of weather) and climatology (the study of climate). Climate is basically the average weather over a longer time scale. Weather changes quickly over time, whereas the climate changes more gradually. The movement of air, water and heat is vitally important to climatology and meteorology. Since motion is such a major factor in weather and climate, this field uses vectors for much of its math. Vectors are used to represent currents in the ocean, wind velocity and forces acting on a parcel of air. You have probably seen a weather map using vectors to show the strength (magnitude) and direction of the wind. Vectors used in atmospheric science are often three-dimensional. We won’t cover three-dimensional motion in this text, but to go from two-dimensions to three-dimensions, you simply add a third vector component. Three-dimensional motion is represented as a combination of x-, y- and zcomponents, where zis the altitude. Vector calculus combines vector math with calculus, and is often used to find the rates of change in temperature, pressure or wind speed over time or distance. This is useful information, since atmospheric motion is driven by changes in pressure or temperature. The greater the variation in pressure over a given distance, the stronger the wind to try to correct that imbalance. Cold air tends to be more dense and therefore has higher pressure than warm air. Higher pressure air rushes into a region of lower pressure and gets deflected by the spinning of the Earth, and friction slows the wind at Earth’s surface. Finding how wind changes over distance and multiplying vectors lets meteorologists,
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like the one shown in Figure 5.26, figure out how much rotation (spin) there is in the atmosphere at any given time and location. This is an important tool for tornado prediction. Conditions with greater rotation are more likely to produce tornadoes. GRASP CHECK Why are vectors used so frequently in atmospheric science? a. Vectors have magnitude as well as direction and can be quickly solved through scalar algebraic operations. b. Vectors have magnitude but no direction, so it becomes easy to express physical quantities involved in the atmospheric science. c. Vectors can be solved very accurately through geometry, which helps to make better predictions in atmospheric science. d. Vectors have magnitude as well as direction and are used in equations that describe the three dimensional motion of the atmosphere. Check Your Understanding 7. Between the analytical and graphical methods of vector additions, which is more accurate? Why? a. The analytical method is less accurate than the graphical method, because the former involves geometry and 162 Chapter 5 • Motion in Two Dimensions trigonometry. b. The analytical method is more accurate than the graphical method, because the latter involves some extensive calculations. c. The analytical method is less accurate than the graphical method, because the former includes drawing all figures to the right scale. d. The analytical method is more accurate than the graphical method, because the latter is limited by the precision of the drawing. 8. What is a component of a two dimensional vector? a. A component is a piece of a vector that points in either the xor ydirection. b. A component is a piece of a vector that has half of the magnitude of the original vector. c. A component is a piece of a vector that points in the direction opposite to the original vector. d. A component is a piece of a vector that points in the same direction as original vector but with double of its magnitude. 9. How can we determine the global angle (measured counter-clockwise from positive ) if we know and? a. b. c. d. 10. How can we determine the magnitude of a vector if we know the magnitudes of its components? a. b. c. d. 5.3 Projectile Motion Section Learning Objectives By the end of this section, you will be able to do the following: • Describe the properties of projectile motion • Apply kinematic equations and vectors to solve problems involving projectile motion Section Key Terms air resistance maximum height (of a projectile) projectile projectile motion
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range trajectory Properties of Projectile Motion Projectile motion is the motion of an object thrown (projected) into the air. After the initial force that launches the object, it only experiences the force of gravity. The object is called a projectile, and its path is called its trajectory. As an object travels through the air, it encounters a frictional force that slows its motion called air resistance. Air resistance does significantly alter trajectory motion, but due to the difficulty in calculation, it is ignored in introductory physics. The most important concept in projectile motion is that horizontal and vertical motions areindependent, meaning that they don’t influence one another. Figure 5.27 compares a cannonball in free fall (in blue) to a cannonball launched horizontally in projectile motion (in red). You can see that the cannonball in free fall falls at the same rate as the cannonball in projectile motion. Keep in mind that if the cannon launched the ball with any vertical component to the velocity, the vertical displacements would not line up perfectly. Access for free at openstax.org. Since vertical and horizontal motions are independent, we can analyze them separately, along perpendicular axes. To do this, we separate projectile motion into the two components of its motion, one along the horizontal axis and the other along the vertical. 5.3 • Projectile Motion 163 Figure 5.27 The diagram shows the projectile motion of a cannonball shot at a horizontal angle versus one dropped with no horizontal velocity. Note that both cannonballs have the same vertical position over time. We’ll call the horizontal axis the x-axis and the vertical axis the y-axis. For notation, d is the total displacement, and x and y are its components along the horizontal and vertical axes. The magnitudes of these vectors are xand y, as illustrated in Figure 5.28. Figure 5.28 A boy kicks a ball at angle θ, and it is displaced a distance of s along its trajectory. As usual, we use velocity, acceleration, and displacement to describe motion. We must also find the components of these variables along the x- and y-axes. The components of acceleration are then very simple ay = –g = –9.80 m/s2. Note that this definition defines the upwards direction as positive. Because gravity is vertical, ax = 0. Both accelerations are constant, so we can use the kinematic equations. For review, the kinematic equations from a previous chapter are summarized in Table 5.
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1. (when ) (when ) Table 5.1 Summary of Kinematic Equations (constant a) Where x is position, x0 is initial position, v is velocity, vavg is average velocity, tis time and a is acceleration. 164 Chapter 5 • Motion in Two Dimensions Solve Problems Involving Projectile Motion The following steps are used to analyze projectile motion: 1. Separate the motion into horizontal and vertical components along the x- and y-axes. These axes are perpendicular, so and are used. The magnitudes of the displacement along x- and y-axes are called and The magnitudes of the components of the velocity are and, where is the magnitude of the velocity and is its direction. Initial values are denoted with a subscript 0. 2. Treat the motion as two independent one-dimensional motions, one horizontal and the other vertical. The kinematic equations for horizontal and vertical motion take the following forms Vertical motion (assuming positive is up ) 3. Solve for the unknowns in the two separate motions (one horizontal and one vertical). Note that the only common variable between the motions is time. The problem solving procedures here are the same as for one-dimensional kinematics.. We can use the analytical method of vector 4. Recombine the two motions to find the total displacement and velocity addition, which uses displacement and velocity. and to find the magnitude and direction of the total is the direction of the displacement, and is the direction of the velocity. (See Figure 5.29 Access for free at openstax.org. 5.3 • Projectile Motion 165 Figure 5.29 (a) We analyze two-dimensional projectile motion by breaking it into two independent one-dimensional motions along the vertical and horizontal axes. (b) The horizontal motion is simple, because and is thus constant. (c) The velocity in the vertical direction begins to decrease as the object rises; at its highest point, the vertical velocity is zero. As the object falls towards the Earth again, the vertical velocity increases again in magnitude but points in the opposite direction to the initial vertical velocity. (d) The x- and y-motions are recombined to give the total velocity at any given point on the trajectory. TIPS FOR SUCCESS For problems of projectile motion, it is important to set up a coordinate system. The first step is to choose an initial position for and. Usually, it is simplest to set the initial position of the object
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so that and. 166 Chapter 5 • Motion in Two Dimensions WATCH PHYSICS Projectile at an Angle This video presents an example of finding the displacement (or range) of a projectile launched at an angle. It also reviews basic trigonometry for finding the sine, cosine and tangent of an angle. Click to view content (https://www.khanacademy.org/embed_video?v=ZZ39o1rAZWY) GRASP CHECK Assume the ground is uniformly level. If the horizontal component a projectile's velocity is doubled, but the vertical component is unchanged, what is the effect on the time of flight? a. The time to reach the ground would remain the same since the vertical component is unchanged. b. The time to reach the ground would remain the same since the vertical component of the velocity also gets doubled. c. The time to reach the ground would be halved since the horizontal component of the velocity is doubled. d. The time to reach the ground would be doubled since the horizontal component of the velocity is doubled. WORKED EXAMPLE A Fireworks Projectile Explodes High and Away During a fireworks display like the one illustrated in Figure 5.30, a shell is shot into the air with an initial speed of 70.0 m/s at an angle of 75° above the horizontal. The fuse is timed to ignite the shell just as it reaches its highest point above the ground. (a) Calculate the height at which the shell explodes. (b) How much time passed between the launch of the shell and the explosion? (c) What is the horizontal displacement of the shell when it explodes? Figure 5.30 The diagram shows the trajectory of a fireworks shell. Strategy The motion can be broken into horizontal and vertical motions in which be zero and solve for the maximum height. and. We can then define and to Solution for (a) By height we mean the altitude or vertical position above the starting point. The highest point in any trajectory, the maximum height, is reached when ; this is the moment when the vertical velocity switches from positive (upwards) to negative (downwards). Since we know the initial velocity, initial position, and the value of vy when the firework reaches its maximum height, we use the following equation to find Access for free at openstax.org. Because and are both zero, the equation simplifies to Solving for gives Now we must find initial velocity of 70.0 m/
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s, and is the initial angle. Thus,, the component of the initial velocity in the y-direction. It is given by 5.3 • Projectile Motion 167, where is the and is so that Discussion for (a) Since up is positive, the initial velocity and maximum height are positive, but the acceleration due to gravity is negative. The maximum height depends only on the vertical component of the initial velocity. The numbers in this example are reasonable for large fireworks displays, the shells of which do reach such heights before exploding. Solution for (b) There is more than one way to solve for the time to the highest point. In this case, the easiest method is to use is zero, this equation reduces to. Because Note that the final vertical velocity,, at the highest point is zero. Therefore, Discussion for (b) This time is also reasonable for large fireworks. When you are able to see the launch of fireworks, you will notice several seconds pass before the shell explodes. Another way of finding the time is by using equation for., and solving the quadratic Solution for (c) Because air resistance is negligible, velocity multiplied by time as given by and the horizontal velocity is constant. The horizontal displacement is horizontal, where is equal to zero where is the x-component of the velocity, which is given by Now, The time for both motions is the same, and so is Discussion for (c) The horizontal motion is a constant velocity in the absence of air resistance. The horizontal displacement found here could be useful in keeping the fireworks fragments from falling on spectators. Once the shell explodes, air resistance has a major effect, and many fragments will land directly below, while some of the fragments may now have a velocity in the –x direction due to the 168 Chapter 5 • Motion in Two Dimensions forces of the explosion. The expression we found for while solving part (a) of the previous problem works for any projectile motion problem where air resistance is negligible. Call the maximum height ; then, This equation defines the maximum height of a projectile. The maximum height depends only on the vertical component of the initial velocity. WORKED EXAMPLE Calculating Projectile Motion: Hot Rock Projectile Suppose a large rock is ejected from a volcano, as illustrated in Figure 5.31, with a speed of the horizontal. The rock strikes the side of the volcano at an altitude 20.0 m lower than its starting point. (a) Calculate the time it takes the rock to follow this path. and at an angle
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above Figure 5.31 The diagram shows the projectile motion of a large rock from a volcano. Strategy Breaking this two-dimensional motion into two independent one-dimensional motions will allow us to solve for the time. The time a projectile is in the air depends only on its vertical motion. Solution While the rock is in the air, it rises and then falls to a final position 20.0 m lower than its starting altitude. We can find the time for this by using If we take the initial position vertical component of the initial velocity, found from to be zero, then the final position is Now the initial vertical velocity is the 5.9 Substituting known values yields Rearranging terms gives a quadratic equation in This expression is a quadratic equation of the form –20.0. Its solutions are given by the quadratic formula, where the constants are a= 4.90, b= –14.3, and c= Access for free at openstax.org. 5.3 • Projectile Motion 169 This equation yields two solutions t= 3.96 and t= –1.03. You may verify these solutions as an exercise. The time is t = 3.96 s or –1.03 s. The negative value of time implies an event before the start of motion, so we discard it. Therefore, Discussion The time for projectile motion is completely determined by the vertical motion. So any projectile that has an initial vertical velocity of and lands 20.0 m below its starting altitude will spend 3.96 s in the air. Practice Problems 11. If an object is thrown horizontally, travels with an average x-component of its velocity equal to, and does not hit the ground, what will be the x-component of the displacement after a. b. c. d.? 12. If a ball is thrown straight up with an initial velocity of upward, what is the maximum height it will reach? a. b. c. d. The fact that vertical and horizontal motions are independent of each other lets us predict the range of a projectile. The range is the horizontal distance R traveled by a projectile on level ground, as illustrated in Figure 5.32. Throughout history, people have been interested in finding the range of projectiles for practical purposes, such as aiming cannons. 170 Chapter 5 • Motion in Two Dimensions Figure 5.32 Trajectories of projectiles on level ground. (a) The greater the initial speed, the greater the range for a given initial angle. (b) The
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effect of initial angle on the range of a projectile with a given initial speed. Note that any combination of trajectories that add to 90 degrees will have the same range in the absence of air resistance, although the maximum heights of those paths are different. How does the initial velocity of a projectile affect its range? Obviously, the greater the initial speed shown in the figure above. The initial angle range of a projectile on level groundis also has a dramatic effect on the range. When air resistance is negligible, the, the greater the range, as is the initial speed and where apply to problems where the initial and final y position are different, or to cases where the object is launched perfectly horizontally. is the initial angle relative to the horizontal. It is important to note that the range doesn’t Virtual Physics Projectile Motion In this simulation you will learn about projectile motion by blasting objects out of a cannon. You can choose between objects such as a tank shell, a golf ball or even a Buick. Experiment with changing the angle, initial speed, and mass, and adding in air resistance. Make a game out of this simulation by trying to hit the target. Click to view content (https://archive.cnx.org/specials/317dbd00-8e61-4065-b3eb-f2b80db9b7ed/projectile-motion/) Access for free at openstax.org. 5.4 • Inclined Planes 171 GRASP CHECK If a projectile is launched on level ground, what launch angle maximizes the range of the projectile? a. b. c. d. Check Your Understanding 13. What is projectile motion? a. Projectile motion is the motion of an object projected into the air, which moves under the influence of gravity. b. Projectile motion is the motion of an object projected into the air which moves independently of gravity. c. Projectile motion is the motion of an object projected vertically upward into the air which moves under the influence of gravity. d. Projectile motion is the motion of an object projected horizontally into the air which moves independently of gravity. 14. What is the force experienced by a projectile after the initial force that launched it into the air in the absence of air resistance? a. The nuclear force b. The gravitational force c. The electromagnetic force d. The contact force 5.4 Inclined Planes Section Learning Objectives By the end of this section, you will be able to do the
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following: • Distinguish between static friction and kinetic friction • Solve problems involving inclined planes Section Key Terms kinetic friction static friction Static Friction and Kinetic Friction Recall from the previous chapter that friction is a force that opposes motion, and is around us all the time. Friction allows us to move, which you have discovered if you have ever tried to walk on ice. There are different types of friction—kinetic and static. Kinetic friction acts on an object in motion, while static friction acts on an object or system at rest. The maximum static friction is usually greater than the kinetic friction between the objects. Imagine, for example, trying to slide a heavy crate across a concrete floor. You may push harder and harder on the crate and not move it at all. This means that the static friction responds to what you do—it increases to be equal to and in the opposite direction of your push. But if you finally push hard enough, the crate seems to slip suddenly and starts to move. Once in motion, it is easier to keep it in motion than it was to get it started because the kinetic friction force is less than the static friction force. If you were to add mass to the crate, (for example, by placing a box on top of it) you would need to push even harder to get it started and also to keep it moving. If, on the other hand, you oiled the concrete you would find it easier to get the crate started and keep it going. Figure 5.33 shows how friction occurs at the interface between two objects. Magnifying these surfaces shows that they are rough on the microscopic level. So when you push to get an object moving (in this case, a crate), you must raise the object until it can skip along with just the tips of the surface hitting, break off the points, or do both. The harder the surfaces are pushed together (such as if another box is placed on the crate), the more force is needed to move them. 172 Chapter 5 • Motion in Two Dimensions Figure 5.33 Frictional forces, such as f, always oppose motion or attempted motion between objects in contact. Friction arises in part because of the roughness of the surfaces in contact, as seen in the expanded view. The magnitude of the frictional force has two forms: one for static friction, the other for kinetic friction. When there is no motion between the objects, the magnitude of static friction fs is is the coefficient of static friction and N is the magnitude of the
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normal force. Recall that the normal force opposes the where force of gravity and acts perpendicular to the surface in this example, but not always. Since the symbol means less than or equal to, this equation says that static friction can have a maximum value of That is, Static friction is a responsive force that increases to be equal and opposite to whatever force is exerted, up to its maximum limit. Once the applied force exceeds fs(max), the object will move. Once an object is moving, the magnitude of kinetic friction fk is given by where is the coefficient of kinetic friction. Friction varies from surface to surface because different substances are rougher than others. Table 5.2 compares values of static and kinetic friction for different surfaces. The coefficient of the friction depends on the two surfaces that are in contact. System Static Friction Kinetic Friction Rubber on dry concrete Rubber on wet concrete Wood on wood 1.0 0.7 0.5 Waxed wood on wet snow 0.14 Metal on wood Steel on steel (dry) Steel on steel (oiled) Teflon on steel 0.5 0.6 0.05 0.04 Bone lubricated by synovial fluid 0.016 Shoes on wood 0.9 Table 5.2 Coefficients of Static and Kinetic Friction 0.7 0.5 0.3 0.1 0.3 0.3 0.03 0.04 0.015 0.7 Access for free at openstax.org. 5.4 • Inclined Planes 173 System Static Friction Kinetic Friction Shoes on ice Ice on ice Steel on ice 0.1 0.1 0.4 0.05 0.03 0.02 Table 5.2 Coefficients of Static and Kinetic Friction Since the direction of friction is always opposite to the direction of motion, friction runs parallel to the surface between objects and perpendicular to the normal force. For example, if the crate you try to push (with a force parallel to the floor) has a mass of 100 kg, then the normal force would be equal to its weight perpendicular to the floor. If the coefficient of static friction is 0.45, you would have to exert a force parallel to the floor greater than to move the crate. Once there is motion, friction is less and the coefficient of kinetic friction might be 0.30, so that a force of only 290 N would keep it moving at a constant speed. If the floor were lubricated, both coefficients would be much smaller than they
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would be without lubrication. The coefficient of friction is unitless and is a number usually between 0 and 1.0. Working with Inclined Planes We discussed previously that when an object rests on a horizontal surface, there is a normal force supporting it equal in magnitude to its weight. Up until now, we dealt only with normal force in one dimension, with gravity and normal force acting perpendicular to the surface in opposing directions (gravity downward, and normal force upward). Now that you have the skills to work with forces in two dimensions, we can explore what happens to weight and the normal force on a tilted surface such as an inclined plane. For inclined plane problems, it is easier breaking down the forces into their components if we rotate the coordinate system, as illustrated in Figure 5.34. The first step when setting up the problem is to break down the force of weight into components. Figure 5.34 The diagram shows perpendicular and horizontal components of weight on an inclined plane. When an object rests on an incline that makes an angle with the horizontal, the force of gravity acting on the object is divided into two components: A force acting perpendicular to the plane, perpendicular force of weight, acting parallel to the plane, the object, so it acts upward along the plane., is typically equal in magnitude and opposite in direction to the normal force, The force, opposes the motion of, causes the object to accelerate down the incline. The force of friction,, and a force acting parallel to the plane,. The 174 Chapter 5 • Motion in Two Dimensions It is important to be careful when resolving the weight of the object into components. If the angle of the incline is at an angle to the horizontal, then the magnitudes of the weight components are Instead of memorizing these equations, it is helpful to be able to determine them from reason. To do this, draw the right triangle formed by the three weight vectors. Notice that the angle of the incline is the same as the angle formed between and. Knowing this property, you can use trigonometry to determine the magnitude of the weight components WATCH PHYSICS Inclined Plane Force Components This video (https://www.khanacademy.org/embed_video?v=TC23wD34C7k) shows how the weight of an object on an inclined plane is broken down into components perpendicular and parallel to the surface of the plane. It explains the geometry for finding the angle in more detail. GRASP CHECK Click to view content
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(https://www.youtube.com/embed/TC23wD34C7k) This video shows how the weight of an object on an inclined plane is broken down into components perpendicular and parallel to the surface of the plane. It explains the geometry for finding the angle in more detail. When the surface is flat, you could say that one of the components of the gravitational force is zero; Which one? As the angle of the incline gets larger, what happens to the magnitudes of the perpendicular and parallel components of gravitational force? a. When the angle is zero, the parallel component is zero and the perpendicular component is at a maximum. As the angle increases, the parallel component decreases and the perpendicular component increases. This is because the cosine of the angle shrinks while the sine of the angle increases. b. When the angle is zero, the parallel component is zero and the perpendicular component is at a maximum. As the angle increases, the parallel component decreases and the perpendicular component increases. This is because the cosine of the angle increases while the sine of the angle shrinks. c. When the angle is zero, the parallel component is zero and the perpendicular component is at a maximum. As the angle increases, the parallel component increases and the perpendicular component decreases. This is because the cosine of the angle shrinks while the sine of the angle increases. d. When the angle is zero, the parallel component is zero and the perpendicular component is at a maximum. As the angle increases, the parallel component increases and the perpendicular component decreases. This is because the cosine of the angle increases while the sine of the angle shrinks. TIPS FOR SUCCESS Normal force is represented by the variable This should not be confused with the symbol for the newton, which is also represented by the letter N. It is important to tell apart these symbols, especially since the units for normal force ( to be newtons (N). For example, the normal force, difference is that normal force is a vector, while the newton is simply a unit. Be careful not to confuse these letters in your calculations!, that the floor exerts on a chair might be One important ) happen To review, the process for solving inclined plane problems is as follows: Access for free at openstax.org. 5.4 • Inclined Planes 175 1. Draw a sketch of the problem. 2. 3. Draw a free-body diagram (which is a sketch showing all of the forces acting
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on an object) with the coordinate system Identify known and unknown quantities, and identify the system of interest. rotated at the same angle as the inclined plane. Resolve the vectors into horizontal and vertical components and draw them on the free-body diagram. 4. Write Newton’s second law in the horizontal and vertical directions and add the forces acting on the object. If the object does not accelerate in a particular direction (for example, the x-direction) then Fnet x= 0. If the object does accelerate in that direction, Fnet x= ma. 5. Check your answer. Is the answer reasonable? Are the units correct? WORKED EXAMPLE Finding the Coefficient of Kinetic Friction on an Inclined Plane A skier, illustrated in Figure 5.35(a), with a mass of 62 kg is sliding down a snowy slope at an angle of 25 degrees. Find the coefficient of kinetic friction for the skier if friction is known to be 45.0 N. Figure 5.35 Use the diagram to help find the coefficient of kinetic friction for the skier. Strategy. Therefore, The magnitude of kinetic friction was given as 45.0 N. Kinetic friction is related to the normal force N as we can find the coefficient of kinetic friction by first finding the normal force of the skier on a slope. The normal force is always perpendicular to the surface, and since there is no motion perpendicular to the surface, the normal force should equal the component of the skier’s weight perpendicular to the slope. That is, Substituting this into our expression for kinetic friction, we get which can now be solved for the coefficient of kinetic friction μk. Solution Solving for gives Substituting known values on the right-hand side of the equation, Discussion This result is a little smaller than the coefficient listed in Table 5.2 for waxed wood on snow, but it is still reasonable since values 176 Chapter 5 • Motion in Two Dimensions of the coefficients of friction can vary greatly. In situations like this, where an object of mass mslides down a slope that makes an angle θwith the horizontal, friction is given by WORKED EXAMPLE Weight on an Incline, a Two-Dimensional Problem The skier’s mass, including equipment, is 60.0 kg. (See Figure 5.36(b).) (a) What is her acceleration if friction is negligible? (b) What is her acceleration if the frictional
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force is 45.0 N? Figure 5.36 Now use the diagram to help find the skier's acceleration if friction is negligible and if the frictional force is 45.0 N. Strategy The most convenient coordinate system for motion on an incline is one that has one coordinate parallel to the slope and one perpendicular to the slope. Remember that motions along perpendicular axes are independent. We use the symbol perpendicular, and to mean parallel. to mean, and in the free-body diagram. The only external forces acting on the system are the skier’s weight, friction, and the normal force exerted by the ski slope, labeled, direction of either axis, so we must break it down into components along the chosen axes. We define weight parallel to the slope and two separate problems of forces parallel to the slope and forces perpendicular to the slope. to be the component of the component of weight perpendicular to the slope. Once this is done, we can consider the is always perpendicular to the slope and is parallel to it. But is not in the Solution The magnitude of the component of the weight parallel to the slope is the component of the weight perpendicular to the slope is, and the magnitude of (a) Neglecting friction: Since the acceleration is parallel to the slope, we only need to consider forces parallel to the slope. Forces perpendicular to the slope add to zero, since there is no acceleration in that direction. The forces parallel to the slope are the amount of the skier’s weight parallel to the slope acceleration parallel to the slope is and friction. Assuming no friction, by Newton’s second law the Where the net force parallel to the slope, so that is the acceleration. (b) Including friction: Here we now have a given value for friction, and we know its direction is parallel to the slope and it opposes motion between surfaces in contact. So the net external force is now Access for free at openstax.org. 5.4 • Inclined Planes 177 and substituting this into Newton’s second law, gives We substitute known values to get or which is the acceleration parallel to the incline when there is 45 N opposing friction. Discussion Since friction always opposes motion between surfaces, the acceleration is smaller when there is friction than when there is not. Practice Problems 15. When an object sits on an inclined plane that makes an angle θwith the horizontal, what is the expression for the component of the objects weight force that is parallel to the incline? a. b.
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c. d. rests on a plane inclined from horizontal. What is the component of the weight force that 16. An object with a mass of is parallel to the incline? a. b. c. d. Snap Lab Friction at an Angle: Sliding a Coin An object will slide down an inclined plane at a constant velocity if the net force on the object is zero. We can use this fact to measure the coefficient of kinetic friction between two objects. As shown in the first Worked Example, the kinetic friction on a slope opposite directions, so when they have equal magnitude, the acceleration is zero. Writing these out, and the component of the weight down the slope is equal to. These forces act in Solving for, since we find that • • • 1 coin 1 book 1 protractor 1. Put a coin flat on a book and tilt it until the coin slides at a constant velocity down the book. You might need to tap the book lightly to get the coin to move. 5.10 178 Chapter 5 • Motion in Two Dimensions 2. Measure the angle of tilt relative to the horizontal and find. GRASP CHECK True or False—If only the angles of two vectors are known, we can find the angle of their resultant addition vector. a. True b. False Check Your Understanding 17. What is friction? a. Friction is an internal force that opposes the relative motion of an object. b. Friction is an internal force that accelerates an object’s relative motion. c. Friction is an external force that opposes the relative motion of an object. d. Friction is an external force that increases the velocity of the relative motion of an object. 18. What are the two varieties of friction? What does each one act upon? a. Kinetic and static friction both act on an object in motion. b. Kinetic friction acts on an object in motion, while static friction acts on an object at rest. c. Kinetic friction acts on an object at rest, while static friction acts on an object in motion. d. Kinetic and static friction both act on an object at rest. 19. Between static and kinetic friction between two surfaces, which has a greater value? Why? a. The kinetic friction has a greater value because the friction between the two surfaces is more when the two surfaces are in relative motion. b. The static friction has a greater value because the friction between the two surfaces is more when the two surfaces are in relative motion. c.
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The kinetic friction has a greater value because the friction between the two surfaces is less when the two surfaces are in relative motion. d. The static friction has a greater value because the friction between the two surfaces is less when the two surfaces are in relative motion. 5.5 Simple Harmonic Motion Section Learning Objectives By the end of this section, you will be able to do the following: • Describe Hooke’s law and Simple Harmonic Motion • Describe periodic motion, oscillations, amplitude, frequency, and period • Solve problems in simple harmonic motion involving springs and pendulums Section Key Terms amplitude deformation equilibrium position frequency Hooke’s law oscillate period periodic motion restoring force simple harmonic motion simple pendulum Hooke’s Law and Simple Harmonic Motion Imagine a car parked against a wall. If a bulldozer pushes the car into the wall, the car will not move but it will noticeably change shape. A change in shape due to the application of a force is a deformation. Even very small forces are known to cause some deformation. For small deformations, two important things can happen. First, unlike the car and bulldozer example, the object returns to its original shape when the force is removed. Second, the size of the deformation is proportional to the force. This Access for free at openstax.org. 5.5 • Simple Harmonic Motion 179 second property is known as Hooke’s law. In equation form, Hooke’s law is where x is the amount of deformation (the change in length, for example) produced by the restoring force F, and k is a constant that depends on the shape and composition of the object. The restoring force is the force that brings the object back to its equilibrium position; the minus sign is there because the restoring force acts in the direction opposite to the displacement. Note that the restoring force is proportional to the deformation x. The deformation can also be thought of as a displacement from equilibrium. It is a change in position due to a force. In the absence of force, the object would rest at its equilibrium position. The force constant k is related to the stiffness of a system. The larger the force constant, the stiffer the system. A stiffer system is more difficult to deform and requires a greater restoring force. The units of k are newtons per meter (N/m). One of the most common uses of Hooke’s law is solving problems involving
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springs and pendulums, which we will cover at the end of this section. Oscillations and Periodic Motion What do an ocean buoy, a child in a swing, a guitar, and the beating of hearts all have in common? They all oscillate. That is, they move back and forth between two points, like the ruler illustrated in Figure 5.37. All oscillations involve force. For example, you push a child in a swing to get the motion started. Figure 5.37 A ruler is displaced from its equilibrium position. Newton’s first law implies that an object oscillating back and forth is experiencing forces. Without force, the object would move in a straight line at a constant speed rather than oscillate. Consider, for example, plucking a plastic ruler to the left as shown in Figure 5.38. The deformation of the ruler creates a force in the opposite direction, known as a restoring force. Once released, the restoring force causes the ruler to move back toward its stable equilibrium position, where the net force on it is zero. However, by the time the ruler gets there, it gains momentum and continues to move to the right, producing the opposite deformation. It is then forced to the left, back through equilibrium, and the process is repeated until it gradually loses all of its energy. The simplest oscillations occur when the restoring force is directly proportional to displacement. Recall that Hooke’s law describes this situation with the equation F = −kx. Therefore, Hooke’s law describes and applies to the simplest case of oscillation, known as simple harmonic motion. Figure 5.38 (a) The plastic ruler has been released, and the restoring force is returning the ruler to its equilibrium position. (b) The net force is zero at the equilibrium position, but the ruler has momentum and continues to move to the right. (c) The restoring force is in the opposite direction. It stops the ruler and moves it back toward equilibrium again. (d) Now the ruler has momentum to the left. (e) In the absence of damping (caused by frictional forces), the ruler reaches its original position. From there, the motion will repeat itself. 180 Chapter 5 • Motion in Two Dimensions When you pluck a guitar string, the resulting sound has a steady tone and lasts a long time. Each vibration of the string takes the same time as the previous one. Periodic motion is a motion that repeats itself at regular time intervals,
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such as with an object bobbing up and down on a spring or a pendulum swinging back and forth. The time to complete one oscillation (a complete cycle of motion) remains constant and is called the period T. Its units are usually seconds. Frequency fis the number of oscillations per unit time. The SI unit for frequency is the hertz (Hz), defined as the number of oscillations per second. The relationship between frequency and period is As you can see from the equation, frequency and period are different ways of expressing the same concept. For example, if you get a paycheck twice a month, you could say that the frequency of payment is two per month, or that the period between checks is half a month. If there is no friction to slow it down, then an object in simple motion will oscillate forever with equal displacement on either side of the equilibrium position. The equilibrium position is where the object would naturally rest in the absence of force. The maximum displacement from equilibrium is called the amplitude X. The units for amplitude and displacement are the same, but depend on the type of oscillation. For the object on the spring, shown in Figure 5.39, the units of amplitude and displacement are meters. Figure 5.39 An object attached to a spring sliding on a frictionless surface is a simple harmonic oscillator. When displaced from equilibrium, the object performs simple harmonic motion that has an amplitude X and a period T. The object’s maximum speed occurs as it passes through equilibrium. The stiffer the spring is, the smaller the period T. The greater the mass of the object is, the greater the period T. The mass mand the force constant k are the onlyfactors that affect the period and frequency of simple harmonic motion. The period of a simple harmonic oscillator is given by and, because f= 1/T, the frequency of a simple harmonic oscillator is Access for free at openstax.org. 5.5 • Simple Harmonic Motion 181 WATCH PHYSICS Introduction to Harmonic Motion This video shows how to graph the displacement of a spring in the x-direction over time, based on the period. Watch the first 10 minutes of the video (you can stop when the narrator begins to cover calculus). Click to view content (https://www.khanacademy.org/embed_video?v=Nk2q-_jkJVs) GRASP CHECK If the amplitude of the displacement of a spring were larger, how would
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this affect the graph of displacement over time? What would happen to the graph if the period was longer? a. Larger amplitude would result in taller peaks and troughs and a longer period would result in greater separation in time between peaks. b. Larger amplitude would result in smaller peaks and troughs and a longer period would result in greater distance between peaks. c. Larger amplitude would result in taller peaks and troughs and a longer period would result in shorter distance between peaks. d. Larger amplitude would result in smaller peaks and troughs and a longer period would result in shorter distance between peaks. Solving Spring and Pendulum Problems with Simple Harmonic Motion Before solving problems with springs and pendulums, it is important to first get an understanding of how a pendulum works. Figure 5.40 provides a useful illustration of a simple pendulum. Figure 5.40 A simple pendulum has a small-diameter bob and a string that has a very small mass but is strong enough not to stretch. The linear displacement from equilibrium is s, the length of the arc. Also shown are the forces on the bob, which result in a net force of −mgsinθ toward the equilibrium position—that is, a restoring force. Everyday examples of pendulums include old-fashioned clocks, a child’s swing, or the sinker on a fishing line. For small displacements of less than 15 degrees, a pendulum experiences simple harmonic oscillation, meaning that its restoring force is directly proportional to its displacement. A pendulum in simple harmonic motion is called a simple pendulum. A pendulum has an object with a small mass, also known as the pendulum bob, which hangs from a light wire or string. The equilibrium position for a pendulum is where the angle is zero (that is, when the pendulum is hanging straight down). It makes sense that without any force applied, this is where the pendulum bob would rest. The displacement of the pendulum bob is the arc length s. The weight mg has components mg cos along the string and mg sin tangent to the arc. Tension in the string exactly cancels the component mg cos parallel to the string. This leaves a net restoring force back toward the equilibrium position that runs tangent to the arc and equals −mg sin. 182 Chapter 5 • Motion in Two Dimensions For a simple pendulum, The period is The only things that affect the period of a simple pendulum are its length and the acceleration due to gravity.
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The period is completely independent of other factors, such as mass or amplitude. However, note that Tdoes depend on g. This means that if we know the length of a pendulum, we can actually use it to measure gravity! This will come in useful in Figure 5.40. TIPS FOR SUCCESS Tension is represented by the variable T, and period is represented by the variable T. It is important not to confuse the two, since tension is a force and period is a length of time. WORKED EXAMPLE Measuring Acceleration due to Gravity: The Period of a Pendulum What is the acceleration due to gravity in a region where a simple pendulum having a length 75.000 cm has a period of 1.7357 s? Strategy We are asked to find g given the period Tand the length Lof a pendulum. We can solve for g, assuming that the angle of deflection is less than 15 degrees. Recall that when the angle of deflection is less than 15 degrees, the pendulum is considered to be in simple harmonic motion, allowing us to use this equation. Solution 1. Square and solve for g. 2. Substitute known values into the new equation. 3. Calculate to find g. Discussion This method for determining g can be very accurate. This is why length and period are given to five digits in this example. WORKED EXAMPLE Hooke’s Law: How Stiff Are Car Springs? What is the force constant for the suspension system of a car, like that shown in Figure 5.41, that settles 1.20 cm when an 80.0-kg person gets in? Access for free at openstax.org. 5.5 • Simple Harmonic Motion 183 Figure 5.41 A car in a parking lot. (exfordy, Flickr) Strategy Consider the car to be in its equilibrium position x = 0 before the person gets in. The car then settles down 1.20 cm, which means it is displaced to a position x = −1.20×10−2 m. At that point, the springs supply a restoring force F equal to the person’s weight w = mg = (80.0 kg)(9.80 m/s2) = 784 N. We take this force to be F in Hooke’s law. Knowing F and x, we can then solve for the force constant k. Solution Solve Hooke’s law, F = −k
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x, for k. Substitute known values and solve for k. Discussion Note that F and x have opposite signs because they are in opposite directions—the restoring force is up, and the displacement is down. Also, note that the car would oscillate up and down when the person got in, if it were not for the shock absorbers. Bouncing cars are a sure sign of bad shock absorbers. Practice Problems applied to a spring causes it to be displaced by. What is the force constant of the spring? 20. A force of a. b. c. d. 21. What is the force constant for the suspension system of a car that settles when a person gets in? a. b. 184 Chapter 5 • Motion in Two Dimensions c. d. Snap Lab Finding Gravity Using a Simple Pendulum Use a simple pendulum to find the acceleration due to gravity g in your home or classroom. • • • 1 string 1 stopwatch 1 small dense object 1. Cut a piece of a string or dental floss so that it is about 1 m long. 2. Attach a small object of high density to the end of the string (for example, a metal nut or a car key). 3. Starting at an angle of less than 10 degrees, allow the pendulum to swing and measure the pendulum’s period for 10 oscillations using a stopwatch. 4. Calculate g. GRASP CHECK How accurate is this measurement for g? How might it be improved? a. Accuracy for value of gwill increase with an increase in the mass of a dense object. b. Accuracy for the value of gwill increase with increase in the length of the pendulum. c. The value of gwill be more accurate if the angle of deflection is more than 15°. d. The value of gwill be more accurate if it maintains simple harmonic motion. Check Your Understanding 22. What is deformation? a. Deformation is the magnitude of the restoring force. b. Deformation is the change in shape due to the application of force. c. Deformation is the maximum force that can be applied on a spring. d. Deformation is regaining the original shape upon the removal of an external force. 23. According to Hooke’s law, what is deformation proportional to? a. Force b. Velocity c. Displacement d. Force constant 24. What are oscillations? a. Motion resulting in small displacements b. Motion which repeats
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itself periodically c. Periodic, repetitive motion between two points d. motion that is the opposite to the direction of the restoring force 25. True or False—Oscillations can occur without force. a. True b. False Access for free at openstax.org. Chapter 5 • Key Terms 185 KEY TERMS air resistance a frictional force that slows the motion of objects as they travel through the air; when solving basic physics problems, air resistance is assumed to be zero maximum height (of a projectile) the highest altitude, or maximum displacement in the vertical position reached in the path of a projectile amplitude the maximum displacement from the oscillate moving back and forth regularly between two equilibrium position of an object oscillating around the equilibrium position analytical method the method of determining the magnitude and direction of a resultant vector using the Pythagorean theorem and trigonometric identities component (of a 2-dimensional vector) a piece of a vector that points in either the vertical or the horizontal direction; every 2-d vector can be expressed as a sum of two vertical and horizontal vector components points period time it takes to complete one oscillation periodic motion motion that repeats itself at regular time intervals projectile an object that travels through the air and experiences only acceleration due to gravity projectile motion the motion of an object that is subject only to the acceleration of gravity range the maximum horizontal distance that a projectile deformation displacement from equilibrium, or change in travels shape due to the application of force restoring force force acting in opposition to the force equilibrium position where an object would naturally rest caused by a deformation in the absence of force frequency number of events per unit of time graphical method drawing vectors on a graph to add them using the head-to-tail method head (of a vector) the end point of a vector; the location of the sum of the a collection of vectors resultant resultant vector simple harmonic motion the oscillatory motion in a the vector sum of two or more vectors system where the net force can be described by Hooke’s law the vector’s arrow; also referred to as the tip simple pendulum an object with a small mass suspended head-to-tail method a method of adding vectors in which from a light wire or string the tail of each vector is placed at the head of the previous vector Hooke’s law proportional relationship between the force F static friction a force that opposes the motion of two systems that are in contact and are not moving relative to one another on a material and the deformation it causes, tail the starting point of a
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vector; the point opposite to the kinetic friction a force that opposes the motion of two systems that are in contact and moving relative to one another SECTION SUMMARY 5.1 Vector Addition and Subtraction: Graphical Methods • The graphical method of adding vectors and involves drawing vectors on a graph and adding them by using the head-to-tail method. The resultant vector is defined such that A + B = R. The magnitude and direction of protractor. are then determined with a ruler and • The graphical method of subtracting vectors A and B involves adding the opposite of vector B, which is defined as −B. In this case, Next, use the head-totail method as for vector addition to obtain the resultant vector. • Addition of vectors is independent of the order in which they are added; A + B = B + A. • The head-to-tail method of adding vectors involves head or tip of the arrow trajectory the path of a projectile through the air vector addition adding together two or more vectors drawing the first vector on a graph and then placing the tail of each subsequent vector at the head of the previous vector. The resultant vector is then drawn from the tail of the first vector to the head of the final vector. • Variables in physics problems, such as force or velocity, can be represented with vectors by making the length of the vector proportional to the magnitude of the force or velocity. • Problems involving displacement, force, or velocity may be solved graphically by measuring the resultant vector’s magnitude with a ruler and measuring the direction with a protractor. 5.2 Vector Addition and Subtraction: Analytical Methods • The analytical method of vector addition and subtraction uses the Pythagorean theorem and trigonometric identities to determine the magnitude 186 Chapter 5 • Key Equations and direction of a resultant vector. • The steps to add vectors method are as follows: and using the analytical 1. Determine the coordinate system for the vectors. Then, determine the horizontal and vertical components of each vector using the equations and 2. Add the horizontal and vertical components of each vector to determine the components resultant vector, and of the and 3. Use the Pythagorean theorem to determine the magnitude,, of the resultant vector • To solve projectile problems: choose a coordinate system; analyze the motion in the vertical and horizontal direction separately; then, recombine the horizontal and vertical components using vector addition equations. 5.4 Inclined Planes • Friction is a contact force between systems
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that opposes the motion or attempted motion between them. Simple friction is proportional to the normal force N pushing the systems together. A normal force is always perpendicular to the contact surface between systems. Friction depends on both of the materials involved. • µs is the coefficient of static friction, which depends on both of the materials. • µk is the coefficient of kinetic friction, which also depends on both materials. • When objects rest on an inclined plane that makes an angle with the horizontal surface, the weight of the object can be broken into components that act perpendicular the plane. and parallel ( ) to the surface of 4. Use a trigonometric identity to determine the direction, • An oscillation is a back and forth motion of an object 5.5 Simple Harmonic Motion, of 5.3 Projectile Motion • Projectile motion is the motion of an object through the air that is subject only to the acceleration of gravity. • Projectile motion in the horizontal and vertical directions are independent of one another. • The maximum height of an projectile is the highest altitude, or maximum displacement in the vertical position reached in the path of a projectile. • The range is the maximum horizontal distance traveled by a projectile. KEY EQUATIONS 5.2 Vector Addition and Subtraction: Analytical Methods resultant magnitude resultant direction x-component of a vector A (when an angle is given relative to the horizontal) Access for free at openstax.org. between two points of deformation. • An oscillation may create a wave, which is a disturbance that propagates from where it was created. • The simplest type of oscillations are related to systems that can be described by Hooke’s law. • Periodic motion is a repetitious oscillation. • The time for one oscillation is the period T. • The number of oscillations per unit time is the frequency • A mass msuspended by a wire of length Lis a simple pendulum and undergoes simple harmonic motion for amplitudes less than about 15 degrees. y-component of a vector A (when an angle is given relative to the horizontal) addition of vectors 5.3 Projectile Motion angle of displacement velocity angle of velocity maximum height range Chapter 5 • Chapter Review 187 perpendicular component of weight on an inclined plane parallel component of weight on an inclined plane 5.4 Inclined Planes Hooke’s law 5.5 Simple Harmonic Motion force of static friction force of kinetic friction period in simple harmonic motion frequency in simple harmonic motion period of a simple
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pendulum CHAPTER REVIEW Concept Items 5.1 Vector Addition and Subtraction: Graphical Methods d. 5.2 Vector Addition and Subtraction: Analytical Methods 1. There is a vector, with magnitude 5 units pointing 4. What is the angle between the x and y components of a, with magnitude 3 units, towards west and vector pointing towards south. Using vector addition, calculate the magnitude of the resultant vector. a. 4.0 b. 5.8 c. 6.3 d. 8.0 2. If you draw two vectors using the head-to-tail method, how can you then draw the resultant vector? a. By joining the head of the first vector to the head of the last vector? a. b. c. d. 5. Two vectors are equal in magnitude and opposite in direction. What is the magnitude of their resultant vector? a. The magnitude of the resultant vector will be zero. b. The magnitude of resultant vector will be twice the magnitude of the original vector. b. By joining the head of the first vector with the tail of c. The magnitude of resultant vector will be same as the last magnitude of the original vector. c. By joining the tail of the first vector to the head of d. The magnitude of resultant vector will be half the the last magnitude of the original vector. d. By joining the tail of the first vector with the tail of the last 3. What is the global angle of south of west? a. b. c. 6. How can we express the x and y-components of a vector, and direction, global angle in terms of its magnitude,? a. b. c. 188 Chapter 5 • Chapter Review d. 7. True or False—Every 2-D vector can be expressed as the c. d. 12. What equation gives the magnitude of kinetic friction? a. b. c. d. 5.5 Simple Harmonic Motion 13. Why is there a negative sign in the equation for Hooke’s law? a. The negative sign indicates that displacement decreases with increasing force. b. The negative sign indicates that the direction of the applied force is opposite to that of displacement. c. The negative sign indicates that the direction of the restoring force is opposite to that of displacement. d. The negative sign indicates that the force constant must be negative. 14. With reference to simple harmonic motion, what is the equilibrium position? a. The position where velocity is
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the minimum b. The position where the displacement is maximum c. The position where the restoring force is the maximum d. The position where the object rests in the absence of force 15. What is Hooke’s law? a. Restoring force is directly proportional to the displacement from the mean position and acts in the the opposite direction of the displacement. b. Restoring force is directly proportional to the displacement from the mean position and acts in the same direction as the displacement. c. Restoring force is directly proportional to the square of the displacement from the mean position and acts in the opposite direction of the displacement. d. Restoring force is directly proportional to the square of the displacement from the mean position and acts in the same direction as the displacement. displacement will be the same as it would have been had he followed directions correctly. a. True b. False product of its x and y-components. a. True b. False 5.3 Projectile Motion 8. Horizontal and vertical motions of a projectile are independent of each other. What is meant by this? a. Any object in projectile motion falls at the same rate as an object in freefall, regardless of its horizontal velocity. b. All objects in projectile motion fall at different rates, regardless of their initial horizontal velocities. c. Any object in projectile motion falls at the same rate as its initial vertical velocity, regardless of its initial horizontal velocity. d. All objects in projectile motion fall at different rates and the rate of fall of the object is independent of the initial velocity. 9. Using the conventional choice for positive and negative axes described in the text, what is the y-component of the acceleration of an object experiencing projectile motion? a. b. c. d. 5.4 Inclined Planes 10. True or False—Kinetic friction is less than the limiting static friction because once an object is moving, there are fewer points of contact, and the friction is reduced. For this reason, more force is needed to start moving an object than to keep it in motion. a. True b. False 11. When there is no motion between objects, what is the relationship between the magnitude of the static friction and the normal force? a. b. Critical Thinking Items 5.1 Vector Addition and Subtraction: Graphical Methods 16. True or False—A person is following a set of directions. He has to walk 2 km east and then 1 km north. He takes a wrong turn and walks in the opposite direction
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for the second leg of the trip. The magnitude of his total Access for free at openstax.org. 5.2 Vector Addition and Subtraction: Analytical Methods 17. What is the magnitude of a vector whose x-component and whose angle is? is a. b. c. d. 18. Vectors and are equal in magnitude and opposite in direction. Does have the same direction as vector or? a. b. Chapter 5 • Chapter Review 189 force pushes against it and it starts moving when is applied to it. What can be said about the coefficient of kinetic friction between the box and the floor? a. b. c. d. 22. The component of the weight parallel to an inclined with the horizontal is plane of an object resting on an incline that makes an angle of the object’s mass? a. b. c. d.. What is 5.3 Projectile Motion 5.5 Simple Harmonic Motion 19. Two identical items, object 1 and object 2, are dropped building. Object 1 is dropped, while object 2 is thrown. from the top of a with an initial velocity of straight downward with an initial velocity of What is the difference in time, in seconds rounded to the nearest tenth, between when the two objects hit the ground? after object 2. a. Object 1 will hit the ground after object 2. b. Object 1 will hit the ground c. Object 1 will hit the ground at the same time as object 2. d. Object 1 will hit the ground after object 2. 20. An object is launched into the air. If the y-component of its acceleration is 9.8 m/s2, which direction is defined as positive? a. Vertically upward in the coordinate system b. Vertically downward in the coordinate system c. Horizontally to the right side of the coordinate system d. Horizontally to the left side of the coordinate system 5.4 Inclined Planes 21. A box weighing is at rest on the floor. A person Problems 5.1 Vector Addition and Subtraction: Graphical Methods 25. A person attempts to cross a river in a straight line by navigating a boat at. If the river flows at from his left to right, what would be the magnitude of the boat’s resultant velocity? In what direction would the boat go, relative to the straight line 23. Two springs are attached to two hooks. Spring A has a greater force constant than spring B. Equal weights
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are suspended from both. Which of the following statements is true? a. Spring A will have more extension than spring B. b. Spring B will have more extension than spring A. c. Both springs will have equal extension. d. Both springs are equally stiff. 24. Two simple harmonic oscillators are constructed by attaching similar objects to two different springs. The force constant of the spring on the left is that of the spring on the right is force is applied to both, which of the following statements is true? a. The spring on the left will oscillate faster than and. If the same spring on the right. b. The spring on the right will oscillate faster than the spring on the left. c. Both the springs will oscillate at the same rate. d. The rate of oscillation is independent of the force constant. across it? a. The resultant velocity of the boat will be The boat will go toward his right at an angle of to a line drawn across the river. b. The resultant velocity of the boat will be The boat will go toward his left at an angle of to a line drawn across the river. c. The resultant velocity of the boat will be The boat will go toward his right at an angle of... 190 Chapter 5 • Chapter Review to a line drawn across the river. d. The resultant velocity of the boat will be. The boat will go toward his left at an angle of to a line drawn across the river. 26. A river flows in a direction from south west to north east. A boat captain wants to cross at a velocity of this river to reach a point on the opposite shore due east of the boat’s current position. The boat moves at. Which direction should it head towards if the resultant velocity is a. b. c. d. It should head in a direction It should head in a direction It should head in a direction It should head in a direction east of south. south of east. east of south. south of east.? 5.2 Vector Addition and Subtraction: Analytical Methods 30. A person wants to fire a water balloon cannon such that it hits a target 100 m away. If the cannon can only be launched at 45° above the horizontal, what should be the initial speed at which it is launched? 31.3 m/s a. 37.2 m/s b. c. 980.0 m/s 1,385.9 m/s d. 5.4 Incl
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ined Planes 31. A coin is sliding down an inclined plane at constant to the horizontal, velocity. If the angle of the plane is what is the coefficient of kinetic friction? a. b. c. d. 27. A person walks 10.0 m north and then 2.00 m east. 32. A skier with a mass of 55 kg is skiing down a snowy slope Solving analytically, what is the resultant displacement of the person? a. b. c. d. = 10.2 m, θ = 78.7º east of north = 10.2 m, θ = 78.7º north of east = 12.0 m, θ = 78.7º east of north = 12.0 m, θ = 78.7º north of east that has an incline of 30°. Find the coefficient of kinetic friction for the skier if friction is known to be 25 N. a. b. c. d. 28. A person walks north of west for and 5.5 Simple Harmonic Motion south of west for. What is the magnitude 33. What is the time period of a long pendulum on of his displacement? Solve analytically. a. b. c. d. earth? a. b. c. d. 34. A simple harmonic oscillator has time period. If the, what is the force constant of mass of the system is the spring used? a. b. c. d. period 2 seconds). 5.3 Projectile Motion 29. A water balloon cannon is fired at at an angle of above the horizontal. How far away will it fall? a. b. c. d. Performance Task 5.5 Simple Harmonic Motion 35. Construct a seconds pendulum (pendulum with time Access for free at openstax.org. TEST PREP Multiple Choice 5.1 Vector Addition and Subtraction: Graphical Methods 36. True or False—We can use Pythagorean theorem to calculate the length of the resultant vector obtained from the addition of two vectors which are at right angles to each other. a. True b. False 37. True or False—The direction of the resultant vector depends on both the magnitude and direction of added vectors. a. True b. False with a headwind blowing. What is the resultant velocity 38. A plane flies north at from the north at of the plane? a. b. c. d. north south north south 39. Two hikers
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take different routes to reach the same spot. southeast, then turns and goes The first one goes at south of east. The second hiker goes south. How far and in which direction must the second hiker travel now, in order to reach the first hiker's location destination? a. b. c. d. east south east south 5.2 Vector Addition and Subtraction: Analytical Methods 40. When will the x-component of a vector with angle be greater than its y-component? a. b. c. d. 41. The resultant vector of the addition of vectors and. The magnitudes of is are, respectively. Which of the following is true? a., and,,, and b. c. Chapter 5 • Test Prep 191 d. 42. What is the dimensionality of vectors used in the study of atmospheric sciences? a. One-dimensional b. Two-dimensional c. Three-dimensional 5.3 Projectile Motion 43. After a projectile is launched in the air, in which direction does it experience constant, non-zero acceleration, ignoring air resistance? a. The x direction b. The y direction c. Both the x and y directions d. Neither direction 44. Which is true when the height of a projectile is at its maximum? a. b. c. 45. A ball is thrown in the air at an angle of 40°. If the maximum height it reaches is 10 m, what must be its initial speed? a. 17.46 m/s b. 21.78 m/s 304.92 m/s c. d. 474.37 m/s 46. A large rock is ejected from a volcano with a speed of above the horizontal. The and at an angle rock strikes the side of the volcano at an altitude of lower than its starting point. Calculate the horizontal displacement of the rock. a. b. c. d. 5.4 Inclined Planes 47. For objects of identical masses but made of different materials, which of the following experiences the most static friction? a. Shoes on ice b. Metal on wood c. Teflon on steel 48. If an object sits on an inclined plane and no other object makes contact with the object, what is typically equal in magnitude to the component of the weight perpendicular to the plane? 192 Chapter 5 • Test Prep a. The normal force b. The total weight c. The parallel force of weight 49. A 5 kg box is at rest on the
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floor. The coefficient of static friction between the box and the floor is 0.4. A horizontal force of 50 N is applied to the box. Will it move? a. No, because the applied force is less than the maximum limiting static friction. b. No, because the applied force is more than the maximum limiting static friction. c. Yes, because the applied force is less than the maximum limiting static friction. d. Yes, because the applied force is more than the maximum limiting static friction. 50. A skier with a mass of 67 kg is skiing down a snowy slope with an incline of 37°. Find the friction if the coefficient of kinetic friction is 0.07. a. 27.66 N 34.70 N b. 36.71 N c. d. 45.96 N 5.5 Simple Harmonic Motion 51. A change in which of the following is an example of deformation? a. Velocity b. Length c. Mass d. Weight 52. The units of amplitude are the same as those for which of the following measurements? a. Speed b. Displacement c. Acceleration d. Force 53. Up to approximately what angle is simple harmonic motion a good model for a pendulum? a. b. c. d. 54. How would simple harmonic motion be different in the absence of friction? a. Oscillation will not happen in the absence of friction. b. Oscillation will continue forever in the absence of friction. c. Oscillation will have changing amplitude in the absence of friction. d. Oscillation will cease after a certain amount of time in the absence of friction. 55. What mass needs to be attached to a spring with a force in order to make a simple harmonic constant of oscillator oscillate with a time period of a. b. c. d.? Short Answer 5.1 Vector Addition and Subtraction: Graphical Methods 56. Find for the following vectors: the addition of these vectors? a. Zero b. Six c. Eight d. Twelve 108 cm, a. 108 cm, b. c. 206 cm, d. 206 cm, 57. Find for the following vectors: 108 cm, a. b. 108 cm, c. 232 cm, d. 232 cm, 59. Two people pull on ropes tied to a trolley, each applying 44 N of force. The angle the ropes form with each other is 39.5°. What is the magnitude of the net force exerted
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on the trolley? a. 0.0 N b. 79.6 N c. 82.8 N d. 88.0 N 5.2 Vector Addition and Subtraction: Analytical Methods 60. True or False—A vector can form the shape of a right 58. Consider six vectors of 2 cm each, joined from head to tail making a hexagon. What would be the magnitude of angle triangle with its x and y components. a. True Access for free at openstax.org. b. False 61. True or False—All vectors have positive x and y a. True b. False Chapter 5 • Test Prep 193 69. For what angle of a projectile is its range equal to zero?. What is in terms of and a. b. c. d. or or or or 5.4 Inclined Planes 70. What are the units of the coefficient of friction?. What is in terms of and a. b. c. d. unitless 71. Two surfaces in contact are moving slowly past each components. a. True b. False 62. Consider? a. b. c. d. 63. Consider? a. b. c. d. 64. When a three dimensional vector is used in the study of atmospheric sciences, what is z? a. Altitude b. Heat c. Temperature d. Wind speed 65. Which method is not an application of vector calculus? a. To find the rate of change in atmospheric temperature b. To study changes in wind speed and direction c. To predict changes in atmospheric pressure d. To measure changes in average rainfall 5.3 Projectile Motion 66. How can you express the velocity, terms of its initial velocity, time,? a. b. c. d. 67. In the equation for the maximum height of a projectile, what does stand for? Initial velocity in the x direction a. b. Initial velocity in the y direction c. Final velocity in the x direction d. Final velocity in the y direction other. As the relative speed between the two surfaces in contact increases, what happens to the magnitude of their coefficient of kinetic friction? a. It increases with the increase in the relative motion. It decreases with the increase in the relative motion. It remains constant and is independent of the relative motion. b. c. 72. When will an object slide down an inclined plane at constant velocity? a. When the magnitude of the component of the weight along the slope is equal to the magnitude of the
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frictional force. b. When the magnitude of the component of the weight along the slope is greater than the magnitude of the frictional force. c. When the magnitude of the component of the weight perpendicular to the slope is less than the magnitude of the frictional force. 73. A box is sitting on an inclined plane. At what angle of incline is the perpendicular component of the box's weight at its maximum? a. b. c. d. 5.5 Simple Harmonic Motion 74. What is the term used for changes in shape due to the, of a projectile in, acceleration,, and d. When the magnitude of the component of the weight perpendicular to the slope is equal to the magnitude of the frictional force. 68. True or False—Range is defined as the maximum vertical distance travelled by a projectile. application of force? a. Amplitude 194 Chapter 5 • Test Prep b. Deformation c. Displacement d. Restoring force 75. What is the restoring force? a. The normal force on the surface of an object b. The weight of a mass attached to an object c. Force which is applied to deform an object from its original shape d. Force which brings an object back to its equilibrium position 76. For a given oscillator, what are the factors that affect its period and frequency? a. Mass only b. Force constant only c. Applied force and mass d. Mass and force constant 77. For an object in simple harmonic motion, when does the Extended Response 5.1 Vector Addition and Subtraction: Graphical Methods 80. True or False—For vectors the order of addition is important. a. True b. False 81. Consider five vectors a, b, c, d,and e.Is it true or false that their addition always results in a vector with a greater magnitude than if only two of the vectors were added? a. True b. False 5.2 Vector Addition and Subtraction: Analytical Methods 82. For what angle of a vector is it possible that its magnitude will be equal to its y-component? a. b. c. d. 83. True or False—If only the angles of two vectors are known, we can find the angle of their resultant addition vector. a. True b. False 84. True or false—We can find the magnitude and direction of the resultant vector if we know the angles of two vectors and the magnitude of one. Access for free at openstax.org.
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maximum speed occur? a. At the extreme positions b. At the equilibrium position c. At the moment when the applied force is removed d. Midway between the extreme and equilibrium positions 78. What is the equilibrium position of a pendulum? a. When the tension in the string is zero b. When the pendulum is hanging straight down c. When the tension in the string is maximum d. When the weight of the mass attached is minimum 79. If a pendulum is displaced by an angle θ, what is the net restoring force it experiences? a. mgsinθ b. mgcosθ c. –mgsinθ d. –mgcosθ a. True b. False 5.3 Projectile Motion 85. Ignoring drag, what is the x-component of the acceleration of a projectile? Why? a. The x-component of the acceleration of a projectile is because acceleration of a projectile is due to gravity, which acts in the y direction. b. The x component of the acceleration of a projectile is because acceleration of a projectile is due to gravity, which acts in the x direction. c. The x-component of the acceleration of a projectile is because acceleration of a projectile is due to gravity, which acts in the x direction. d. The x-component of the acceleration of a projectile is because acceleration of a projectile is due to gravity, which acts in the y direction. 86. What is the optimum angle at which a projectile should be launched in order to cover the maximum distance? a. b. c. d. 5.4 Inclined Planes 87. True or False—Friction varies from surface to surface because different substances have different degrees of roughness or smoothness. a. True b. False 88. As the angle of the incline gets larger, what happens to the magnitudes of the perpendicular and parallel components of gravitational force? a. Both the perpendicular and the parallel component d. The force constant kis related to the friction in the system: The larger the force constant, the lower the friction in the system. Chapter 5 • Test Prep 195 will decrease. b. The perpendicular component will decrease and the parallel component will increase. c. The perpendicular component will increase and the parallel component will decrease. d. Both the perpendicular and the parallel component will increase. 5.5 Simple Harmonic Motion 89. What physical characteristic of a system is its force constant related to? a. The force constant kis related to the stiffness of a system
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: The larger the force constant, the stiffer the system. b. The force constant kis related to the stiffness of a system: The larger the force constant, the looser the system. c. The force constant kis related to the friction in the system: The larger the force constant, the greater the friction in the system. 90. How or why does a pendulum oscillate? a. A pendulum oscillates due to applied force. b. A pendulum oscillates due to the elastic nature of the string. c. A pendulum oscillates due to restoring force arising from gravity. d. A pendulum oscillates due to restoring force arising from tension in the string. 91. If a pendulum from earth is taken to the moon, will its frequency increase or decrease? Why? a. It will increase because on the Moon is less than on Earth. It will decrease because on the Moon is less than b. c. d. on Earth. It will increase because on the Moon is greater than on Earth. It will decrease because on the Moon is greater than on Earth. 196 Chapter 5 • Test Prep Access for free at openstax.org. CHAPTER 6 Circular and Rotational Motion Figure 6.1 This Australian Grand Prix Formula 1 race car moves in a circular path as it makes the turn. Its wheels also spin rapidly. The same physical principles are involved in both of these motions. (Richard Munckton). Chapter Outline 6.1 Angle of Rotation and Angular Velocity 6.2 Uniform Circular Motion 6.3 Rotational Motion You may recall learning about various aspects of motion along a straight line: kinematics (where we learned INTRODUCTION about displacement, velocity, and acceleration), projectile motion (a special case of two-dimensional kinematics), force, and Newton’s laws of motion. In some ways, this chapter is a continuation of Newton’s laws of motion. Recall that Newton’s first law tells us that objects move along a straight line at constant speed unless a net external force acts on them. Therefore, if an object moves along a circular path, such as the car in the photo, it must be experiencing an external force. In this chapter, we explore both circular motion and rotational motion. 198 Chapter 6 • Circular and Rotational Motion 6.1 Angle of Rotation and Angular Velocity Section Learning Objectives By the end of this section, you will be able to do the following: • Desc
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ribe the angle of rotation and relate it to its linear counterpart • Describe angular velocity and relate it to its linear counterpart • Solve problems involving angle of rotation and angular velocity Section Key Terms angle of rotation angular velocity arc length circular motion radius of curvature rotational motion spin tangential velocity Angle of Rotation What exactly do we mean by circular motionor rotation? Rotational motion is the circular motion of an object about an axis of rotation. We will discuss specifically circular motion and spin. Circular motion is when an object moves in a circular path. Examples of circular motion include a race car speeding around a circular curve, a toy attached to a string swinging in a circle around your head, or the circular loop-the-loopon a roller coaster. Spin is rotation about an axis that goes through the center of mass of the object, such as Earth rotating on its axis, a wheel turning on its axle, the spin of a tornado on its path of destruction, or a figure skater spinning during a performance at the Olympics. Sometimes, objects will be spinning while in circular motion, like the Earth spinning on its axis while revolving around the Sun, but we will focus on these two motions separately. When solving problems involving rotational motion, we use variables that are similar to linear variables (distance, velocity, acceleration, and force) but take into account the curvature or rotation of the motion. Here, we define the angle of rotation, which is the angular equivalence of distance; and angular velocity, which is the angular equivalence of linear velocity. When objects rotate about some axis—for example, when the CD in Figure 6.2 rotates about its center—each point in the object follows a circular path. Figure 6.2 All points on a CD travel in circular paths. The pits (dots) along a line from the center to the edge all move through the same angle in time. The arc length,, is the distance traveled along a circular path. The radius of curvature, r, is the radius of the circular path. Both are shown in Figure 6.3. Access for free at openstax.org. 6.1 • Angle of Rotation and Angular Velocity 199 Figure 6.3 The radius (r) of a circle is rotated through an angle. The arc length,, is the distance covered along the circumference. Consider a line from the center of the CD to its edge. In a given time, each pit(used to record information) on this line moves through the same
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angle. The angle of rotation is the amount of rotation and is the angular analog of distance. The angle of rotation is the arc length divided by the radius of curvature. The angle of rotation is often measured by using a unit called the radian. (Radians are actually dimensionless, because a radian is defined as the ratio of two distances, radius and arc length.) A revolution is one complete rotation, where every point on the radians (or 360 degrees), and therefore has an angle of rotation circle returns to its original position. One revolution covers of revolutions, and degrees using the relationship radians, and an arc length that is the same as the circumference of the circle. We can convert between radians, 1 revolution = rad = 360°. See Table 6.1 for the conversion of degrees to radians for some common angles. Degree Measures Radian Measures 6.1 Table 6.1 Commonly Used Angles in Terms of Degrees and Radians Angular Velocity How fast is an object rotating? We can answer this question by using the concept of angular velocity. Consider first the angular speed is the rate at which the angle of rotation changes. In equation form, the angular speed is 6.2 which means that an angular rotation given time, it has a greater angular speed. The units for angular speed are radians per second (rad/s). occurs in a time,. If an object rotates through a greater angle of rotation in a Now let’s consider the direction of the angular speed, which means we now must call it the angular velocity. The direction of the 200 Chapter 6 • Circular and Rotational Motion angular velocity is along the axis of rotation. For an object rotating clockwise, the angular velocity points away from you along the axis of rotation. For an object rotating counterclockwise, the angular velocity points toward you along the axis of rotation. Angular velocity (ω) is the angular version of linear velocity v. Tangential velocity is the instantaneous linear velocity of an object in rotational motion. To get the precise relationship between angular velocity and tangential velocity, consider again a pit on the rotating CD. This pit moves through an arc length so its tangential speedis in a shorttime From the definition of the angle of rotation,, we see that. Substituting this into the expression for vgives 6.3 says that the tangential speed vis proportional to the distance rfrom the center of rotation. Consequently, The equation tangential speed is greater for
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a point on the outer edge of the CD (with larger r) than for a point closer to the center of the CD (with smaller r). This makes sense because a point farther out from the center has to cover a longer arc length in the same amount of time as a point closer to the center. Note that both points will still have the same angular speed, regardless of their distance from the center of rotation. See Figure 6.4. Figure 6.4 Points 1 and 2 rotate through the same angle ( ), but point 2 moves through a greater arc length ( ) because it is farther from the center of rotation. means large vbecause Now, consider another example: the tire of a moving car (see Figure 6.5). The faster the tire spins, the faster the car moves—large, will produce a greater linear (tangential) velocity, v, for the car. This is because a larger radius means a longer arc length must contact the road, so the car must move farther in the same amount of time.. Similarly, a larger-radius tire rotating at the same angular velocity, Access for free at openstax.org. 6.1 • Angle of Rotation and Angular Velocity 201 Figure 6.5 A car moving at a velocity, v, to the right has a tire rotating with angular velocity. The speed of the tread of the tire relative to the axle is v, the same as if the car were jacked up and the wheels spinning without touching the road. Directly below the axle, where the tire touches the road, the tire tread moves backward with respect to the axle with tangential velocity, where ris the tire radius. Because the road is stationary with respect to this point of the tire, the car must move forward at the linear velocity v. A larger angular velocity for the tire means a greater linear velocity for the car. However, there are cases where linear velocity and tangential velocity are not equivalent, such as a car spinning its tires on ice. In this case, the linear velocity will be less than the tangential velocity. Due to the lack of friction under the tires of a car on ice, the arc length through which the tire treads move is greater than the linear distance through which the car moves. It’s similar to running on a treadmill or pedaling a stationary bike; you are literally going nowhere fast. TIPS FOR SUCCESS Angular velocity ω and tangential velocity v are vectors, so we must include magnitude and direction. The
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direction of the angular velocity is along the axis of rotation, and points away from you for an object rotating clockwise, and toward you for an object rotating counterclockwise. In mathematics this is described by the right-hand rule. Tangential velocity is usually described as up, down, left, right, north, south, east, or west, as shown in Figure 6.6. Figure 6.6 As the fly on the edge of an old-fashioned vinyl record moves in a circle, its instantaneous velocity is always at a tangent to the circle. The direction of the angular velocity is into the page this case. 202 Chapter 6 • Circular and Rotational Motion WATCH PHYSICS Relationship between Angular Velocity and Speed This video reviews the definition and units of angular velocity and relates it to linear speed. It also shows how to convert between revolutions and radians. Click to view content (https://www.youtube.com/embed/zAx61CO5mDw) GRASP CHECK For an object traveling in a circular path at a constant angular speed, would the linear speed of the object change if the radius of the path increases? a. Yes, because tangential speed is independent of the radius. b. Yes, because tangential speed depends on the radius. c. No, because tangential speed is independent of the radius. d. No, because tangential speed depends on the radius. Solving Problems Involving Angle of Rotation and Angular Velocity Snap Lab Measuring Angular Speed In this activity, you will create and measure uniform circular motion and then contrast it with circular motions with different radii. • One string (1 m long) • One object (two-hole rubber stopper) to tie to the end • One timer Procedure 1. Tie an object to the end of a string. 2. Swing the object around in a horizontal circle above your head (swing from your wrist). It is important that the circle be horizontal! 3. Maintain the object at uniform speed as it swings. 4. Measure the angular speed of the object in this manner. Measure the time it takes in seconds for the object to travel 10 revolutions. Divide that time by 10 to get the angular speed in revolutions per second, which you can convert to radians per second. 5. What is the approximate linear speed of the object? 6. Move your hand up the string so that the length of the string is 90 cm. Repeat steps 2–5. 7. Move your hand up the string so
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that its length is 80 cm. Repeat steps 2–5. 8. Move your hand up the string so that its length is 70 cm. Repeat steps 2–5. 9. Move your hand up the string so that its length is 60 cm. Repeat steps 2–5 10. Move your hand up the string so that its length is 50 cm. Repeat steps 2–5 11. Make graphs of angular speed vs. radius (i.e. string length) and linear speed vs. radius. Describe what each graph looks like. GRASP CHECK If you swing an object slowly, it may rotate at less than one revolution per second. What would be the revolutions per second for an object that makes one revolution in five seconds? What would be its angular speed in radians per second?. The angular speed of the object would be a. The object would spin at. b. The object would spin at. The angular speed of the object would be. Access for free at openstax.org. c. The object would spin at d. The object would spin at. The angular speed of the object would be. The angular speed of the object would be.. 6.1 • Angle of Rotation and Angular Velocity 203 Now that we have an understanding of the concepts of angle of rotation and angular velocity, we’ll apply them to the real-world situations of a clock tower and a spinning tire. WORKED EXAMPLE Angle of rotation at a Clock Tower The clock on a clock tower has a radius of 1.0 m. (a) What angle of rotation does the hour hand of the clock travel through when it moves from 12 p.m. to 3 p.m.? (b) What’s the arc length along the outermost edge of the clock between the hour hand at these two times? Strategy We can figure out the angle of rotation by multiplying a full revolution ( hour hand in going from 12 to 3. Once we have the angle of rotation, we can solve for the arc length by rearranging the equation radians) by the fraction of the 12 hours covered by the since the radius is given. Solution to (a) In going from 12 to 3, the hour hand covers 1/4 of the 12 hours needed to make a complete revolution. Therefore, the angle between the hour hand at 12 and at 3 is (i.e., 90 degrees). Solution to (b) Rearranging the equation we get Inserting the known values gives an arc length
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of 6.4 6.5 6.6 Discussion We were able to drop the radians from the final solution to part (b) because radians are actually dimensionless. This is because the radian is defined as the ratio of two distances (radius and arc length). Thus, the formula gives an answer in units of meters, as expected for an arc length. WORKED EXAMPLE How Fast Does a Car Tire Spin? Calculate the angular speed of a 0.300 m radius car tire when the car travels at 15.0 m/s (about 54 km/h). See Figure 6.5. Strategy In this case, the speed of the tire tread with respect to the tire axle is the same as the speed of the car with respect to the road, so we have v= 15.0 m/s. The radius of the tire is r= 0.300 m. Since we know vand r, we can rearrange the equation, to get and find the angular speed. Solution To find the angular speed, we use the relationship:. 204 Chapter 6 • Circular and Rotational Motion Inserting the known quantities gives Discussion When we cancel units in the above calculation, we get 50.0/s (i.e., 50.0 per second, which is usually written as 50.0 s−1). But the angular speed must have units of rad/s. Because radians are dimensionless, we can insert them into the answer for the angular speed because we know that the motion is circular. Also note that, if an earth mover with much larger tires, say 1.20 m in radius, were moving at the same speed of 15.0 m/s, its tires would rotate more slowly. They would have an angular speed of 6.8 6.7 Practice Problems 1. What is the angle in degrees between the hour hand and the minute hand of a clock showing 9:00 a.m.? a. 0° b. 90° 180° c. 360° d. 2. What is the approximate value of the arc length between the hour hand and the minute hand of a clock showing 10:00 a.m if the radius of the clock is 0.2 m? a. 0.1 m b. 0.2 m c. 0.3 m d. 0.6 m Check Your Understanding 3. What is circular motion? a. Circular motion is the motion of an object when it follows a linear path. b. Circular
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motion is the motion of an object when it follows a zigzag path. c. Circular motion is the motion of an object when it follows a circular path. d. Circular motion is the movement of an object along the circumference of a circle or rotation along a circular path. 4. What is meant by radius of curvature when describing rotational motion? a. The radius of curvature is the radius of a circular path. b. The radius of curvature is the diameter of a circular path. c. The radius of curvature is the circumference of a circular path. d. The radius of curvature is the area of a circular path. 5. What is angular velocity? a. Angular velocity is the rate of change of the diameter of the circular path. b. Angular velocity is the rate of change of the angle subtended by the circular path. c. Angular velocity is the rate of change of the area of the circular path. d. Angular velocity is the rate of change of the radius of the circular path. 6. What equation defines angular velocity,? Take that is the radius of curvature, is the angle, and is time. a. b. c. d. 7. Identify three examples of an object in circular motion. Access for free at openstax.org. 6.2 • Uniform Circular Motion 205 a. an artificial satellite orbiting the Earth, a race car moving in the circular race track, and a top spinning on its axis b. an artificial satellite orbiting the Earth, a race car moving in the circular race track, and a ball tied to a string being swung in a circle around a person's head c. Earth spinning on its own axis, a race car moving in the circular race track, and a ball tied to a string being swung in a circle around a person's head d. Earth spinning on its own axis, blades of a working ceiling fan, and a top spinning on its own axis 8. What is the relative orientation of the radius and tangential velocity vectors of an object in uniform circular motion? a. Tangential velocity vector is always parallel to the radius of the circular path along which the object moves. b. Tangential velocity vector is always perpendicular to the radius of the circular path along which the object moves. c. Tangential velocity vector is always at an acute angle to the radius of the circular path along which the object moves. d. Tangential velocity vector is always at an obtuse angle to the radius of the circular
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path along which the object moves. 6.2 Uniform Circular Motion Section Learning Objectives By the end of this section, you will be able to do the following: • Describe centripetal acceleration and relate it to linear acceleration • Describe centripetal force and relate it to linear force • Solve problems involving centripetal acceleration and centripetal force Section Key Terms centrifugal force centripetal acceleration centripetal force uniform circular motion Centripetal Acceleration In the previous section, we defined circular motion. The simplest case of circular motion is uniform circular motion, where an object travels a circular path at a constant speed. Note that, unlike speed, the linear velocity of an object in circular motion is constantly changing because it is always changing direction. We know from kinematics that acceleration is a change in velocity, either in magnitude or in direction or both. Therefore, an object undergoing uniform circular motion is always accelerating, even though the magnitude of its velocity is constant. You experience this acceleration yourself every time you ride in a car while it turns a corner. If you hold the steering wheel steady during the turn and move at a constant speed, you are executing uniform circular motion. What you notice is a feeling of sliding (or being flung, depending on the speed) away from the center of the turn. This isn’t an actual force that is acting on you—it only happens because your body wants to continue moving in a straight line (as per Newton’s first law) whereas the car is turning off this straight-line path. Inside the car it appears as if you are forced away from the center of the turn. This fictitious force is known as the centrifugal force. The sharper the curve and the greater your speed, the more noticeable this effect becomes. Figure 6.7 shows an object moving in a circular path at constant speed. The direction of the instantaneous tangential velocity is shown at two points along the path. Acceleration is in the direction of the change in velocity; in this case it points roughly toward the center of rotation. (The center of rotation is at the center of the circular path). If we imagine smaller, then the acceleration would point exactlytoward the center of rotation, but this case is hard to draw. We call the acceleration of an object moving in uniform circular motion the centripetal acceleration ac because centripetal means center seeking. becoming smaller and 206 Chapter 6 • Circular and Rotational Motion Figure 6
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.7 The directions of the velocity of an object at two different points are shown, and the change in velocity is seen to point approximately toward the center of curvature (see small inset). For an extremely small value of, points exactly toward the center of the circle (but this is hard to draw). Because, the acceleration is also toward the center, so ac is called centripetal acceleration. Now that we know that the direction of centripetal acceleration is toward the center of rotation, let’s discuss the magnitude of centripetal acceleration. For an object traveling at speed vin a circular path with radius r, the magnitude of centripetal acceleration is Centripetal acceleration is greater at high speeds and in sharp curves (smaller radius), as you may have noticed when driving a car, because the car actually pushes you toward the center of the turn. But it is a bit surprising that ac is proportional to the speed squared. This means, for example, that the acceleration is four times greater when you take a curve at 100 km/h than at 50 km/h. We can also express ac in terms of the magnitude of angular velocity. Substituting into the equation above, we get. Therefore, the magnitude of centripetal acceleration in terms of the magnitude of angular velocity is 6.9 TIPS FOR SUCCESS The equation expressed in the form ac = rω2 is useful for solving problems where you know the angular velocity rather than the tangential velocity. Virtual Physics Ladybug Motion in 2D In this simulation, you experiment with the position, velocity, and acceleration of a ladybug in circular and elliptical motion. Switch the type of motion from linear to circular and observe the velocity and acceleration vectors. Next, try elliptical motion and notice how the velocity and acceleration vectors differ from those in circular motion. Click to view content (https://archive.cnx.org/specials/317a2b1e-2fbd-11e5-99b5-e38ffb545fe6/ladybug-motion/) Access for free at openstax.org. 6.2 • Uniform Circular Motion 207 GRASP CHECK In uniform circular motion, what is the angle between the acceleration and the velocity? What type of acceleration does a body experience in the uniform circular motion? a. The angle between acceleration and velocity is 0°, and the body experiences linear acceleration. b. The angle between acceleration and velocity
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is 0°, and the body experiences centripetal acceleration. c. The angle between acceleration and velocity is 90°, and the body experiences linear acceleration. d. The angle between acceleration and velocity is 90°, and the body experiences centripetal acceleration. Centripetal Force Because an object in uniform circular motion undergoes constant acceleration (by changing direction), we know from Newton’s second law of motion that there must be a constant net external force acting on the object. Any force or combination of forces can cause a centripetal acceleration. Just a few examples are the tension in the rope on a tether ball, the force of Earth’s gravity on the Moon, the friction between a road and the tires of a car as it goes around a curve, or the normal force of a roller coaster track on the cart during a loop-the-loop. Any net force causing uniform circular motion is called a centripetal force. The direction of a centripetal force is toward the center of rotation, the same as for centripetal acceleration. According to Newton’s second law of motion, a net force causes the acceleration of mass according to Fnet = ma. For uniform circular motion, the acceleration is centripetal acceleration: a = ac. Therefore, the magnitude of centripetal force, Fc, is. By using the two different forms of the equation for the magnitude of centripetal acceleration, get two expressions involving the magnitude of the centripetal force Fc. The first expression is in terms of tangential speed, the second is in terms of angular speed:, we and and. Both forms of the equation depend on mass, velocity, and the radius of the circular path. You may use whichever expression for centripetal force is more convenient. Newton’s second law also states that the object will accelerate in the same direction as the net force. By definition, the centripetal force is directed towards the center of rotation, so the object will also accelerate towards the center. A straight line drawn from the circular path to the center of the circle will always be perpendicular to the tangential velocity. Note that, if you solve the first expression for r, you get From this expression, we see that, for a given mass and velocity, a large centripetal force causes a small radius of curvature—that is, a tight curve. 208 Chapter 6 • Circular and Rotational Motion
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Figure 6.8 In this figure, the frictional force fserves as the centripetal force Fc. Centripetal force is perpendicular to tangential velocity and causes uniform circular motion. The larger the centripetal force Fc, the smaller is the radius of curvature rand the sharper is the curve. The lower curve has the same velocity v, but a larger centripetal force Fc produces a smaller radius. WATCH PHYSICS Centripetal Force and Acceleration Intuition This video explains why a centripetal force creates centripetal acceleration and uniform circular motion. It also covers the difference between speed and velocity and shows examples of uniform circular motion. Click to view content (https://www.youtube.com/embed/vZOk8NnjILg) GRASP CHECK Imagine that you are swinging a yoyo in a vertical clockwise circle in front of you, perpendicular to the direction you are facing. Now, imagine that the string breaks just as the yoyo reaches its bottommost position, nearest the floor. Which of the following describes the path of the yoyo after the string breaks? a. The yoyo will fly upward in the direction of the centripetal force. b. The yoyo will fly downward in the direction of the centripetal force. Access for free at openstax.org. 6.2 • Uniform Circular Motion 209 c. The yoyo will fly to the left in the direction of the tangential velocity. d. The yoyo will fly to the right in the direction of the tangential velocity. Solving Centripetal Acceleration and Centripetal Force Problems To get a feel for the typical magnitudes of centripetal acceleration, we’ll do a lab estimating the centripetal acceleration of a tennis racket and then, in our first Worked Example, compare the centripetal acceleration of a car rounding a curve to gravitational acceleration. For the second Worked Example, we’ll calculate the force required to make a car round a curve. Snap Lab Estimating Centripetal Acceleration In this activity, you will measure the swing of a golf club or tennis racket to estimate the centripetal acceleration of the end of the club or racket. You may choose to do this in slow motion. Recall that the equation for centripetal acceleration is or. • One tennis racket
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or golf club • One timer • One ruler or tape measure Procedure 1. Work with a partner. Stand a safe distance away from your partner as he or she swings the golf club or tennis racket. 2. Describe the motion of the swing—is this uniform circular motion? Why or why not? 3. Try to get the swing as close to uniform circular motion as possible. What adjustments did your partner need to make? 4. Measure the radius of curvature. What did you physically measure? 5. By using the timer, find either the linear or angular velocity, depending on which equation you decide to use. 6. What is the approximate centripetal acceleration based on these measurements? How accurate do you think they are? Why? How might you and your partner make these measurements more accurate? GRASP CHECK Was it more useful to use the equation or in this activity? Why? a. It should be simpler to use because measuring angular velocity through observation would be easier. b. c. It should be simpler to use because measuring tangential velocity through observation would be easier. It should be simpler to use because measuring angular velocity through observation would be difficult. d. It should be simpler to use because measuring tangential velocity through observation would be difficult. WORKED EXAMPLE Comparing Centripetal Acceleration of a Car Rounding a Curve with Acceleration Due to Gravity A car follows a curve of radius 500 m at a speed of 25.0 m/s (about 90 km/h). What is the magnitude of the car’s centripetal acceleration? Compare the centripetal acceleration for this fairly gentle curve taken at highway speed with acceleration due to gravity (g). 210 Chapter 6 • Circular and Rotational Motion Strategy Because linear rather than angular speed is given, it is most convenient to use the expression the centripetal acceleration. to find the magnitude of Solution Entering the given values of v= 25.0 m/s and r= 500 m into the expression for ac gives Discussion To compare this with the acceleration due to gravity (g= 9.80 m/s2), we take the ratio. Therefore,, which means that the centripetal acceleration is about one tenth the acceleration due to gravity. WORKED EXAMPLE Frictional Force on Car Tires Rounding a Curve a. Calculate the centripetal force exerted on a 900 kg car that rounds a 600-m-radius curve on horizontal ground at 25.0
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m/s. b. Static friction prevents the car from slipping. Find the magnitude of the frictional force between the tires and the road that allows the car to round the curve without sliding off in a straight line. Access for free at openstax.org. 6.2 • Uniform Circular Motion 211 Strategy and Solution for (a) We know that. Therefore, Strategy and Solution for (b) The image above shows the forces acting on the car while rounding the curve. In this diagram, the car is traveling into the page as shown and is turning to the left. Friction acts toward the left, accelerating the car toward the center of the curve. Because friction is the only horizontal force acting on the car, it provides all of the centripetal force in this case. Therefore, the force of friction is the centripetal force in this situation and points toward the center of the curve. Discussion Since we found the force of friction in part (b), we could also solve for the coefficient of friction, since. Practice Problems 9. What is the centripetal acceleration of an object with speed going along a path of radius? a. b. c. d. 10. Calculate the centripetal acceleration of an object following a path with a radius of a curvature of 0.2 m and at an angular velocity of 5 rad/s. a. b. c. d. 1 m/s 5 m/s 1 m/s2 5 m/s2 Check Your Understanding 11. What is uniform circular motion? 212 Chapter 6 • Circular and Rotational Motion a. Uniform circular motion is when an object accelerates on a circular path at a constantly increasing velocity. b. Uniform circular motion is when an object travels on a circular path at a variable acceleration. c. Uniform circular motion is when an object travels on a circular path at a constant speed. d. Uniform circular motion is when an object travels on a circular path at a variable speed. 12. What is centripetal acceleration? a. The acceleration of an object moving in a circular path and directed radially toward the center of the circular orbit b. The acceleration of an object moving in a circular path and directed tangentially along the circular path c. The acceleration of an object moving in a linear path and directed in the direction of motion of the object d. The acceleration of an object moving in a linear path and directed in the direction opposite to the motion of the object 13. Is there a
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net force acting on an object in uniform circular motion? a. Yes, the object is accelerating, so a net force must be acting on it. b. Yes, because there is no acceleration. c. No, because there is acceleration. d. No, because there is no acceleration. 14. Identify two examples of forces that can cause centripetal acceleration. a. The force of Earth’s gravity on the moon and the normal force b. The force of Earth’s gravity on the moon and the tension in the rope on an orbiting tetherball c. The normal force and the force of friction acting on a moving car d. The normal force and the tension in the rope on a tetherball 6.3 Rotational Motion Section Learning Objectives By the end of this section, you will be able to do the following: • Describe rotational kinematic variables and equations and relate them to their linear counterparts • Describe torque and lever arm • Solve problems involving torque and rotational kinematics Section Key Terms angular acceleration kinematics of rotational motion lever arm tangential acceleration torque Rotational Kinematics In the section on uniform circular motion, we discussed motion in a circle at constant speed and, therefore, constant angular velocity. However, there are times when angular velocity is not constant—rotational motion can speed up, slow down, or reverse directions. Angular velocity is not constant when a spinning skater pulls in her arms, when a child pushes a merry-go-round to make it rotate, or when a CD slows to a halt when switched off. In all these cases, angular acceleration occurs because the angular velocity rate of change of angular velocity. In equation form, angular acceleration is changes. The faster the change occurs, the greater is the angular acceleration. Angular acceleration is the increases, then is the change in angular velocity and is the change in time. The units of angular acceleration are (rad/s)/s, or rad/ where s2. If is negative. Keep in mind that, by convention, counterclockwise is the positive direction and clockwise is the negative direction. For example, the skater in Figure 6.9 is rotating counterclockwise as seen from above, so her angular velocity is positive. Acceleration would be negative, for example, when an object that is rotating counterclockwise slows down. It would be positive when an object that is rotating counterclockwise speeds up. decreases, then is positive. If Access for free at open
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stax.org. 6.3 • Rotational Motion 213 Figure 6.9 A figure skater spins in the counterclockwise direction, so her angular velocity is normally considered to be positive. (Luu, Wikimedia Commons) The relationship between the magnitudes of tangential acceleration, a, and angular acceleration, 6.10 These equations mean that the magnitudes of tangential acceleration and angular acceleration are directly proportional to each other. The greater the angular acceleration, the larger the change in tangential acceleration, and vice versa. For example, consider riders in their pods on a Ferris wheel at rest. A Ferris wheel with greater angular acceleration will give the riders greater tangential acceleration because, as the Ferris wheel increases its rate of spinning, it also increases its tangential velocity. Note that the radius of the spinning object also matters. For example, for a given angular acceleration, a smaller Ferris wheel leads to a smaller tangential acceleration for the riders. TIPS FOR SUCCESS Tangential acceleration is sometimes denoted at. It is a linear acceleration in a direction tangent to the circle at the point of interest in circular or rotational motion. Remember that tangential acceleration is parallel to the tangential velocity (either in the same direction or in the opposite direction.) Centripetal acceleration is always perpendicular to the tangential velocity. So far, we have defined three rotational variables: Table 6.2 shows how they are related.,, and. These are the angular versions of the linear variables x, v, and a. Rotational Linear Relationship x Table 6.2 Rotational and Linear Variables 214 Chapter 6 • Circular and Rotational Motion Rotational Linear Relationship v a Table 6.2 Rotational and Linear Variables, and We can now begin to see how rotational quantities like, that starts at rest has a large angular acceleration for a fairly long time, it ends up spinning rapidly and rotates through many revolutions. Putting this in terms of the variables, if the wheel’s angular acceleration the final angular velocity undergoes a large linear acceleration, then it has a large final velocity and will have traveled a large distance. and angle of rotation are large. In the case of linear motion, if an object starts at rest and are related to each other. For example, if a motorcycle wheel is large for a long period of time t, then The kinematics of rotational motion describes the relationships between the angle of rotation, angular velocity, angular acceleration, and time. It
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