FORCES AND NEWTON S LAWS OF MOTION

Transcription

1 45807_04_p1-48 6/17/05 3:31 PM Page 1 C H A P T E R 4 FORCES AND NEWTON S LAWS OF MOTION Cutnell Johnson & PHYSICS, 7e coming January 2006 As this windsurfer is propelled through the air, his motion is determined by forces due to the wind and his weight. The relationship between the forces acting on an object and the resulting motion is discussed in this chapter. (Tom King/The Image ank/getty Images) 4.1 THE CONCEPTS OF FORCE AND MASS In common usage, a force is a push or a pull, as the examples in Figure 4.1 illustrate. In basketball, a player launches a shot by pushing on the ball. The tow bar attached to a speeding boat pulls a water skier. Forces such as those that launch the basketball or pull the skier are called contact forces, because they arise from the physical contact between two objects. There are circumstances, however, in which two objects exert forces on one another even though they are not touching. Such forces are referred to as noncontact forces or action-at-a-distance forces. One example of such a noncontact force occurs when a diver is pulled toward the earth because of the force of gravity. The earth exerts this force even when it is not in direct contact with the diver. In Figure 4.1, arrows are used to represent the forces. It is appropriate to use arrows, because a force is a vector quantity and has both a magnitude and a direction. The direction of the arrow gives the direction of the force, and the length is proportional to its strength or magnitude. 1

2 45807_04_p1-48 6/20/05 7:39 AM Page 2 2 CHAPTER 4 FORCES AND NEWTON S LAWS OF MOTION F The word mass is just as familiar as the word force. A massive supertanker, for instance, is one that contains an enormous amount of mass. As we will see in the next section, it is difficult to set such a massive object into motion and difficult to bring it to a halt once it is moving. In comparison, a penny does not contain much mass. The emphasis here is on the amount of mass, and the idea of direction is of no concern. Therefore, mass is a scalar quantity. During the seventeenth century, Isaac Newton, starting with the work of Galileo, developed three important laws that deal with force and mass. Collectively they are called Newton s laws of motion and provide the basis for understanding the effect that forces have on an object. ecause of the importance of these laws, a separate section will be devoted to each one. F (a) 4.2 NEWTON S FIRST LAW OF MOTION THE FIRST LAW To gain some insight into Newton s first law, think about the game of ice hockey (Figure 4.2). If a player does not hit a stationary puck, it will remain at rest on the ice. After the puck is struck, however, it coasts on its own across the ice, slowing down only slightly because of friction. Since ice is very slippery, there is only a relatively small amount of friction to slow down the puck. In fact, if it were possible to remove all friction and wind resistance, and if the rink were infinitely large, the puck would coast forever in a straight line at a constant speed. Left on its own, the puck would lose none of the velocity imparted to it at the time it was struck. This is the essence of Newton s first law of motion: NEWTON S FIRST LAW OF MOTION An object continues in a state of rest or in a state of motion at a constant speed along a straight line, unless compelled to change that state by a net force. (b) In the first law the phrase net force is crucial. Often, several forces act simultaneously on a body, and the net force is the vector sum of all of them. Individual forces matter only to the extent that they contribute to the total. For instance, if friction and other opposing forces were absent, a car could travel forever at 30 m/s in a straight line, without using any gas after it has come up to speed. In reality gas is needed, but only so that the engine can produce the necessary force to cancel opposing forces such as friction. This cancellation ensures that there is no net force to change the state of motion of the car. When an object moves at a constant speed along a straight line, its velocity is constant. Newton s first law indicates that a state of rest (zero velocity) and a state of constant velocity are completely equivalent, in the sense that neither one requires the application of a net force to sustain it. The purpose served when a net force acts on an object is not to sustain the object s velocity, but, rather, to change it. F (c) Figure 4.1 The arrow labeled represents the force that acts on (a) the basketball, (b) the water skier, and (c) the cliff diver. (a, Nathaniel S. utler/nae/getty Images; b, P. eavismasterfile; c, Amy and Chuck Wiley/Wales/Index Stock) F Figure 4.2 The game of ice hockey can give some insight into Newton s laws of motion. ( Royalty-Free/Corbis Images)

3 45807_04_p1-48 6/17/05 3:31 PM Page NEWTON S FIRST LAW OF MOTION 3 INERTIA AND MASS A greater net force is required to change the velocity of some objects than of others. For instance, a net force that is just enough to cause a bicycle to pick up speed will cause only an imperceptible change in the motion of a freight train. In comparison to the bicycle, the train has a much greater tendency to remain at rest. Accordingly, we say that the train has more inertia than the bicycle. Quantitatively, the inertia of an object is measured by its mass. The following definition of inertia and mass indicates why Newton s first law is sometimes called the law of inertia: IN LIERTY GOD WE TRUST 1996 Penny (0.003 kg) ook (2 kg) DEFINITION OF INERTIA AND MASS Inertia is the natural tendency of an object to remain at rest or in motion at a constant speed along a straight line. The mass of an object is a quantitative measure of inertia. icycle (15 kg) SI Unit of Inertia and Mass: kilogram (kg) The SI unit for mass is the kilogram (kg), whereas the units in the CGS system and the E system are the gram (g) and the slug (sl), respectively. Conversion factors between these units are given on the page facing the inside of the front cover. Figure 4.3 gives the masses of various objects, ranging from a penny to a supertanker. The larger the mass, the greater is the inertia. Often the words mass and weight are used interchangeably, but this is incorrect. Mass and weight are different concepts, and Section 4.7 will discuss the distinction between them. Figure 4.4 shows a useful application of inertia. Automobile seat belts unwind freely when pulled gently, so they can be buckled. ut in an accident, they hold you safely in place. One seat-belt mechanism consists of a ratchet wheel, a locking bar, and a pendulum. The belt is wound around a spool mounted on the ratchet wheel. While the car is at rest or moving at a constant velocity, the pendulum hangs straight down, and the locking bar rests horizontally, as the gray part of the drawing shows. Consequently, nothing prevents the ratchet wheel from turning, and the seat belt can be pulled out easily. When the car suddenly slows down in an accident, however, the relatively massive lower part of the pendulum keeps moving forward because of its inertia. The pendulum swings on its pivot into the position shown in color and causes the locking bar to block the rotation of the ratchet wheel, thus preventing the seat belt from unwinding. Figure 4.3 The masses of various objects. The physics of seat belts. Car (2000 kg) Jetliner (1.2 x 10 5 kg) Supertanker (1.5 x 10 8 kg) AN INERTIAL REFERENCE FRAME Newton s first law (and also the second law) can appear to be invalid to certain observers. Suppose, for instance, that you are a passenger riding in a friend s car. While the car moves at a constant speed along a straight line, you do not feel the seat pushing against your back to any unusual extent. This experience is consistent with the first law, which indicates that in the absence of a net force you should move with a constant velocity. Suddenly the driver floors the gas pedal. Immediately you feel the seat pressing against your back as the car accelerates. Therefore, you sense that a force is being applied to you. The first law leads you to believe that your motion should change, and, relative to the ground outside, your motion does change. ut relative to the car, you can see that your motion does not change, because you remain stationary with respect to the car. Clearly, Newton s first law does not hold for observers who use the accelerating car as a frame of reference. As a result, such a reference frame is said to be noninertial. All accelerating reference frames are noninertial. In contrast, observers for whom the law of inertia is valid are said to be using inertial reference frames for their observations, as defined below: DEFINITION OF AN INERTIAL REFERENCE FRAME An inertial reference frame is one in which Newton s law of inertia is valid. Seat belt Ratchet wheel Locking bar Motion of car Pivots Pendulum Figure 4.4 Inertia plays a central role in one seat-belt mechanism. The gray part of the drawing applies when the car is at rest or moving at a constant velocity. The colored parts show what happens when the car suddenly slows down, as in an accident.

4 45807_04_p1-48 6/17/05 3:31 PM Page 4 4 CHAPTER 4 FORCES AND NEWTON S LAWS OF MOTION The acceleration of an inertial reference frame is zero, so it moves with a constant velocity. All of Newton s laws of motion are valid in inertial reference frames, and when we apply these laws, we will be assuming such a reference frame. In particular, the earth itself is a good approximation of an inertial reference frame. 4.3 NEWTON S SECOND LAW OF MOTION Newton s first law indicates that if no net force acts on an object, then the velocity of the object remains unchanged. The second law deals with what happens when a net force does act. Consider a hockey puck once again. When a player strikes a stationary puck, he causes the velocity of the puck to change. In other words, he makes the puck accelerate. The cause of the acceleration is the force that the hockey stick applies. As long as this force acts, the velocity increases, and the puck accelerates. Now, suppose another player strikes the puck and applies twice as much force as the first player does. The greater force produces a greater acceleration. In fact, if the friction between the puck and the ice is negligible, and if there is no wind resistance, the acceleration of the puck is directly proportional to the force. Twice the force produces twice the acceleration. Moreover, the acceleration is a vector quantity, just as the force is, and points in the same direction as the force. Often, several forces act on an object simultaneously. Friction and wind resistance, for instance, do have some effect on a hockey puck. In such cases, it is the net force, or the vector sum of all the forces acting, that is important. Mathematically, the net force is written as F, where the Greek capital letter (sigma) denotes the vector sum. Newton s second law states that the acceleration is proportional to the net force acting on the object. In Newton s second law, the net force is only one of two factors that determine the acceleration. The other is the inertia or mass of the object. After all, the same net force that imparts an appreciable acceleration to a hockey puck (small mass) will impart very little acceleration to a semitrailer truck (large mass). Newton s second law states that for a given net force, the magnitude of the acceleration is inversely proportional to the mass. Twice the mass means one-half the acceleration, if the same net force acts on both objects. Thus, the second law shows how the acceleration depends on both the net force and the mass, as given in Equation 4.1. NEWTON S SECOND LAW OF MOTION When a net external force acts on an object of mass m, the acceleration a that results is directly proportional to the net force and has a magnitude that is inversely proportional to the mass. The direction of the acceleration is the same as the direction of the net force. a F (4.1) m or F ma SI Unit of Force: kg m/s 2 newton (N) F Note that the net force in Equation 4.1 includes only the forces that the environment exerts on the object of interest. Such forces are called external forces. In contrast, internal forces are forces that one part of an object exerts on another part of the object and are not included in Equation 4.1. According to Equation 4.1, the SI unit for force is the unit for mass (kg) times the unit for acceleration (m/s 2 ), or SI unit for force (kg) m kg m s2 s 2 The combination of kg m/s 2 is called a newton (N) and is a derived SI unit, not a base unit; 1 newton 1 N 1 kg m/s 2.

5 45807_04_p1-48 6/17/05 3:31 PM Page NEWTON S SECOND LAW OF MOTION 5 Table 4.1 Units for Mass, Acceleration, and Force System Mass Acceleration Force SI kilogram (kg) meter/second 2 (m/s 2 ) newton (N) CGS gram (g) centimeter/second 2 (cm/s 2 ) dyne (dyn) E slug (sl) foot/second 2 (ft/s 2 ) pound (lb) In the CGS system, the procedure for establishing the unit of force is the same as with SI units, except that mass is expressed in grams (g) and acceleration in cm/s 2. The resulting unit for force is the dyne; 1 dyne 1 g cm/s 2. In the E system, the unit for force is defined to be the pound (lb),* and the unit for acceleration is ft/s 2. With this procedure, Newton s second law can then be used to obtain the unit for mass: E unit for force lb (unit for mass) ft s 2 Unit for mass lb s2 ft The combination of lb s 2 /ft is the unit for mass in the E system and is called the slug (sl); 1 slug 1 sl 1 lb s 2 /ft. Table 4.1 summarizes the various units for mass, acceleration, and force. Conversion factors between force units from different systems are provided on the page facing the inside of the front cover. When using the second law to calculate the acceleration, it is necessary to determine the net force that acts on the object. In this determination a free-body diagram helps enormously. A free-body diagram is a diagram that represents the object and the forces that act on it. Only the forces that act on the object appear in a free-body diagram. Forces that the object exerts on its environment are not included. Example 1 illustrates the use of a free-body diagram. Example 1 Pushing a Stalled Car Two people are pushing a stalled car, as Figure 4.5a indicates. The mass of the car is 1850 kg. One person applies a force of 275 N to the car, while the other applies a force of 395 N. oth forces act in the same direction. A third force of 560 N also acts on the car, but in a direction opposite to that in which the people are pushing. This force arises because of friction and the extent to which the pavement opposes the motion of the tires. Find the acceleration of the car. Reasoning According to Newton s second law, the acceleration is the net force divided by the mass of the car. To determine the net force, we use the free-body diagram in Figure 4.5b. In this diagram, the car is represented as a dot, and its motion is along the x axis. The diagram makes it clear that the forces all act along one direction. Therefore, they can be added as colinear vectors to obtain the net force. Solution From Equation 4.1, the acceleration is a ( F)/m. The net force is F 275 N 395 N 560 N 110 N Problem solving insight A free-body diagram is very helpful when applying Newton s second law. Always start a problem by drawing the free-body diagram. +y 275 N 275 N 395 N 560 N 395 N Figure 4.5 (a) Two people push a stalled car, in opposition to a force created by friction and the pavement. Opposing force = 560 N (b) A free-body diagram that shows the (a) (b) Free-body diagram of the car horizontal forces acting on the car. +x * We refer here to the gravitational version of the E system, in which a force of one pound is defined to be the pull of the earth on a certain standard body at a location where the acceleration due to gravity is ft/s 2.

6 45807_04_p1-48 6/17/05 3:31 PM Page 6 6 CHAPTER 4 FORCES AND NEWTON S LAWS OF MOTION Problem solving insight The direction of the acceleration is always the same as the direction of the net force. The acceleration can now be obtained: a F m/s 2 (4.1) m 110 N 1850 kg The plus sign indicates that the acceleration points along the x axis, in the same direction as the net force. 4.4 THE VECTOR NATURE OF NEWTON S SECOND LAW OF MOTION When a football player throws a pass, the direction of the force he applies to the ball is important. oth the force and the resulting acceleration of the ball are vector quantities, as are all forces and accelerations. The directions of these vectors can be taken into account in two dimensions by using x and y components. The net force F in Newton s second law has components F x and F y, while the acceleration a has components a x and a y. Consequently, Newton s second law, as expressed in Equation 4.1, can be written in an equivalent form as two equations, one for the x components and one for the y components: F x ma x (4.2a) F y ma y (4.2b) This procedure is similar to that employed in Chapter 3 for the equations of two-dimensional kinematics (see Table 3.1). The components in Equations 4.2a and 4.2b are scalar components and will be either positive or negative numbers, depending on whether they point along the positive or negative x or y axis. The remainder of this section deals with examples that show how these equations are used. Example 2 Applying Newton s Second Law Using Components A man is stranded on a raft (mass of man and raft 1300 kg), as shown in Figure 4.6a. y paddling, he causes an average force P of 17 N to be applied to the raft in a direction due east N W S E A sin 67 +y A +y A = 15 N A cos 67 +x 67 +x P = 17 N Free-body diagram of the raft Figure 4.6 (a) A man is paddling a raft, as in Examples 2 and 3. (b) The freebody diagram shows the forces P and A that act on the raft. Forces acting on the raft in a direction perpendicular to the surface of the water play no role in the examples and are omitted for clarity. (c) The raft s acceleration components a x and a y. (d) In 65 s, the components of the raft s displacement are x 48 m and y 23 m. a y +y a x (a) (c) +x +y y = 23 m (b) (d ) +x x = 48 m

7 45807_04_p1-48 6/21/05 4:29 PM Page THE VECTOR NATURE OF NEWTON S SECOND LAW OF MOTION 7 (the x direction). The wind also exerts a force on the raft. This force has a magnitude of 15 N and points 67 north of east. Ignoring any resistance from the water, find the x and y components of the raft s acceleration. Reasoning Since the mass of the man and the raft is known, Newton s second law can be used to determine the acceleration components from the given forces. According to the form of the second law in Equations 4.2a and 4.2b, the acceleration component in a given direction is the component of the net force in that direction divided by the mass. As an aid in determining the components F x and F y of the net force, we use the free-body diagram in Figure 4.6b. In this diagram, the directions due east and due north are the x and y directions, respectively. Solution Figure 4.6b shows the force components: Force x Component y Component P 17 N 0 N A (15 N) cos 67 6 N (15 N) sin N A F x 17 N 6 N 23 N F y 14 N The plus signs indicate that F x points in the direction of the x axis and F y points in the direction of the y axis. The x and y components of the acceleration point in the directions of F x and F y, respectively, and can now be calculated: These acceleration components are shown in Figure 4.6c. Example 3 The Displacement of a Raft (4.2a) (4.2b) At the moment the forces P and begin acting on the raft in Example 2, the velocity of the raft is 0.15 m/s, in a direction due east (the x direction). Assuming that the forces are maintained for 65 s, find the x and y components of the raft s displacement during this time interval. Reasoning Once the net force acting on an object and the object s mass have been used in Newton s second law to determine the acceleration, it becomes possible to use the equations of kinematics to describe the resulting motion. We know from Example 2 that the acceleration components are a x m/s 2 and a y m/s 2, and it is given here that the initial velocity components are v 0x 0.15 m/s and v 0y 0 m/s. Thus, Equation 3.5a (x and Equation 3.5b (y can be used with t 65 s to determine the x and y components of the raft s displacement. 0x t 1 2 a xt 2 ) a x F x m 23 N 1300 kg a y F y m 14 N 1300 kg Solution According to Equations 3.5a and 3.5b, the x and y components of the displacement are x v 0x t 1 a 48 m 2 xt 2 (0.15 m/s)(65 s) 1 (0.018 m/s 2 2 )(65 s) 2 y v 0y t 1 2 a yt 2 (0 m/s)(65 s) 1 2 (0.011 m/s2 )(65 s) 2 Figure 4.6d shows the final location of the raft. A 0yt 1 2 a yt 2 ) m/s m/s 2 23 m CHECK YOUR UNDERSTANDING 1 All of the following, except one, cause the acceleration of an object to double. Which one is it? (a) All forces acting on the object double. (b) The net force acting on the object doubles. (c) oth the net force acting on the object and the mass of the object double. (d) The mass of the object is reduced by a factor of two. (The answer is given at the end of the book.) ackground: This problem depends on the concepts of force, net force, mass, and acceleration, because Newton s second law of motion deals with them. For similar questions (including calculational counterparts), consult Self-Assessment Test 4.1. This test is described at the end of Section 4.5. Problem solving insight Applications of Newton s second law always involve the net external force, which is the vector sum of all the external forces that act on an object. Each component of the net force leads to a corresponding component of the acceleration. Need more practice? Interactive LearningWare 4.1 A catapult on an aircraft carrier is capable of accelerating a kg plane from 0 to 56.0 m/s in a distance of 80.0 m. Find the net force, assumed constant, that the jet s engine and the catapult exert on the plane. Related Homework: Problems 4, 6 For an interactive solution, go to

8 45807_04_p1-48 6/17/05 3:31 PM Page 8 8 CHAPTER 4 FORCES AND NEWTON S LAWS OF MOTION 4.5 NEWTON S THIRD LAW OF MOTION Imagine you are in a football game. You line up facing your opponent, the ball is snapped, and the two of you crash together. No doubt, you feel a force. ut think about your opponent. He too feels something, for while he is applying a force to you, you are applying a force to him. In other words, there isn t just one force on the line of scrimmage; there is a pair of forces. Newton was the first to realize that all forces occur in pairs and there is no such thing as an isolated force, existing all by itself. His third law of motion deals with this fundamental characteristic of forces. These two wapiti (elk) exert action and reaction forces on each other. (First Light/Corbis Images) NEWTON S THIRD LAW OF MOTION Whenever one body exerts a force on a second body, the second body exerts an oppositely directed force of equal magnitude on the first body. Problem solving insight Even though the magnitudes of the action and reaction forces are always equal, these forces do not necessarily produce accelerations that have equal magnitudes, since each force acts on a different object that may have a different mass. The third law is often called the action reaction law, because it is sometimes quoted as follows: For every action (force) there is an equal, but opposite, reaction. Figure 4.7 illustrates how the third law applies to an astronaut who is drifting just outside a spacecraft and who pushes on the spacecraft with a force P. According to the third law, the spacecraft pushes back on the astronaut with a force P that is equal in magnitude but opposite in direction. In Example 4, we examine the accelerations produced by each of these forces. Example 4 The Accelerations Produced by Action and Reaction Forces Suppose that the mass of the spacecraft in Figure 4.7 is m S kg and that the mass of the astronaut is m A 92 kg. In addition, assume that the astronaut exerts a force of P 36 N on the spacecraft. Find the accelerations of the spacecraft and the astronaut. Reasoning According to Newton s third law, when the astronaut applies the force 36 N to the spacecraft, the spacecraft applies a reaction force P 36 N to the astronaut. As a result, the spacecraft and the astronaut accelerate in opposite directions. Although the action and reaction forces have the same magnitude, they do not create accelerations of the same magnitude, because the spacecraft and the astronaut have different masses. According to Newton s second law, the astronaut, having a much smaller mass, will experience a much larger acceleration. In applying the second law, we note that the net force acting on the spacecraft is P, while the net force acting on the astronaut is P. F Solution Using the second law, we find that the acceleration of the spacecraft is a S P 36 N m S kg The acceleration of the astronaut is a A P m A 36 N 92 kg F m/s m/s 2 P P +P Figure 4.7 The astronaut pushes on the spacecraft with a force P. According to Newton s third law, the spacecraft simultaneously pushes back on the astronaut with a force P.

9 45807_04_p1-48 6/17/05 3:31 PM Page TYPES OF FORCES: AN OVERVIEW 9 Mechanism for actuating trailer brakes Figure 4.8 Some rental trailers include an automatic brake-actuating mechanism. There is a clever application of Newton s third law in some rental trailers. As Figure 4.8 illustrates, the tow bar connecting the trailer to the rear bumper of a car contains a mechanism that can automatically actuate brakes on the trailer wheels. This mechanism works without the need for electrical connections between the car and the trailer. When the driver applies the car brakes, the car slows down. ecause of inertia, however, the trailer continues to roll forward and begins pushing against the bumper. In reaction, the bumper pushes back on the tow bar. The reaction force is used by the mechanism in the tow bar to push the brake pedal for the trailer. The physics of automatic trailer brakes. SELF-ASSESSMENT TEST Test your understanding of the material in Sections : Newton s First Law Newton s Second Law Newton s Third Law 4.6 TYPES OF FORCES: AN OVERVIEW CONCEPTS AT A GLANCE Newton s three laws of motion make it clear that forces play a central role in determining the motion of an object. In the next four sections some common forces will be discussed: the gravitational force (Section 4.7), the normal force (Section 4.8), frictional forces (Section 4.9), and the tension force (Section 4.10). In later chapters, we will encounter still others, such as electric and magnetic forces. It is important to realize that Newton s second law is always valid, regardless of which of these forces may act on an object. One does not have a different law for every type of common force. Thus, we need only to determine what forces are acting on an object, add them together to form the net force, and then use Newton s second law to determine the object s acceleration. The Concepts-at-a-Glance chart in Figure 4.9 illustrates this important idea. Figure 4.9 CONCEPTS AT A GLANCE When any of the four external forces listed here act on an object, they are included as part of the net force F in any application of Newton's second law. Each of the four external forces acts on this (very) reluctant bull as the farmers join efforts to pull it aboard the boat. (Diether Endlicher/ AP/Wide World Photos) CONCEPTS AT A GLANCE External Forces 1. Gravitational Force (Section 4.7) 2. Normal Force (Section 4.8) 3. Frictional Forces (Section 4.9) 4. Tension Force (Section 4.10) Newton s Second Law ΣF = ma

10 45807_04_p1-48 6/17/05 3:31 PM Page CHAPTER 4 FORCES AND NEWTON S LAWS OF MOTION In nature there are two general types of forces, fundamental and nonfundamental. Fundamental forces are the ones that are truly unique, in the sense that all other forces can be explained in terms of them. Only three fundamental forces have been discovered: 1. Gravitational force 2. Strong nuclear force 3. Electroweak force The gravitational force is discussed in the next section. The strong nuclear force plays a primary role in the stability of the nucleus of the atom (see Section 31.2). The electroweak force is a single force that manifests itself in two ways (see Section 32.6). One manifestation is the electromagnetic force that electrically charged particles exert on one another (see Sections 18.5, 21.2, and 21.8). The other manifestation is the so-called weak nuclear force that plays a role in the radioactive disintegration of certain nuclei (see Section 31.5). Except for the gravitational force, all of the forces discussed in this chapter are nonfundamental, because they are related to the electromagnetic force. They arise from the interactions between the electrically charged particles that comprise atoms and molecules. Our understanding of which forces are fundamental, however, is continually evolving. For instance, in the 1860s and 1870s James Clerk Maxwell showed that the electric force and the magnetic force could be explained as manifestations of a single electromagnetic force. Then, in the 1970s, Sheldon Glashow (1932 ), Abdus Salam ( ), and Steven Weinberg (1933 ) presented the theory that explains how the electromagnetic force and the weak nuclear force are related to the electroweak force. They received a Nobel prize in 1979 for their achievement. Today, efforts continue that have the goal of further reducing the number of fundamental forces. 4.7 THE GRAVITATIONAL FORCE NEWTON S LAW OF UNIVERSAL GRAVITATION Objects fall downward because of gravity, and Chapters 2 and 3 discuss how to describe the effects of gravity by using a value of g 9.80 m/s 2 for the downward acceleration it causes. However, nothing has been said about why g is 9.80 m/s 2. The reason is fascinating, as we will now see. The acceleration due to gravity is like any other acceleration, and Newton s second law indicates that it must be caused by a net force. In addition to his famous three laws of motion, Newton also provided a coherent understanding of the gravitational force. His law of universal gravitation is stated as follows: m 1 m 2 +F F Figure 4.10 The two particles, whose masses are m 1 and m 2, are attracted by gravitational forces and F. r F NEWTON S LAW OF UNIVERSAL GRAVITATION Every particle in the universe exerts an attractive force on every other particle. A particle is a piece of matter, small enough in size to be regarded as a mathematical point. For two particles that have masses m 1 and m 2 and are separated by a distance r, the force that each exerts on the other is directed along the line joining the particles (see Figure 4.10) and has a magnitude given by F G m 1m 2 (4.3) r 2 The symbol G denotes the universal gravitational constant, whose value is found experimentally to be G N m 2 /kg 2 The constant G that appears in Equation 4.3 is called the universal gravitational constant, because it has the same value for all pairs of particles anywhere in the universe, no matter what their separation. The value for G was first measured in an experiment by

11 45807_04_p1-48 6/17/05 3:31 PM Page 11 the English scientist Henry Cavendish ( ), more than a century after Newton proposed his law of universal gravitation. To see the main features of Newton s law of universal gravitation, look at the two particles in Figure They have masses m 1 and m 2 and are separated by a distance r. In the picture, it is assumed that a force pointing to the right is positive. The gravitational forces point along the line joining the particles and are F, the gravitational force exerted on particle 1 by particle 2 F, the gravitational force exerted on particle 2 by particle 1 These two forces have equal magnitudes and opposite directions. They act on different bodies, causing them to be mutually attracted. In fact, these forces are an action reaction pair, as required by Newton s third law. Example 5 shows that the magnitude of the gravitational force is extremely small for ordinary values of the masses and the distance between them. Example 5 Gravitational Attraction What is the magnitude of the gravitational force that acts on each particle in Figure 4.10, assuming m 1 12 kg (approximately the mass of a bicycle), m 2 25 kg, and r 1.2 m? Reasoning and Solution The magnitude of the gravitational force can be found using Equation 4.3: F G m 1m 2 r 2 ( N m 2 /kg 2 ) (12 kg)(25 kg) (1.2 m) N For comparison, you exert a force of about 1 N when pushing a doorbell, so that the gravitational force is exceedingly small in circumstances such as those here. This result is due to the fact that G itself is very small. However, if one of the bodies has a large mass, like that of the earth ( kg), the gravitational force can be large. 4.7 THE GRAVITATIONAL FORCE 11 As expressed by Equation 4.3, Newton s law of gravitation applies only to particles. However, most familiar objects are too large to be considered particles. Nevertheless, the law of universal gravitation can be applied to such objects with the aid of calculus. Newton was able to prove that an object of finite size can be considered to be a particle for purposes of using the gravitation law, provided the mass of the object is distributed with spherical symmetry about its center. Thus, Equation 4.3 can be applied when each object is a sphere whose mass is spread uniformly over its entire volume. Figure 4.11 shows this kind of application, assuming that the earth and the moon are such uniform spheres of matter. In this case, r is the distance between the centers of the spheres and not the distance between the outer surfaces. The gravitational forces that the spheres exert on each other are the same as if the entire mass of each were concentrated at its center. Even if the objects are not uniform spheres, Equation 4.3 can be used to a good degree of approximation if the sizes of the objects are small relative to the distance of separation r. Moon M M +F F Earth M E WEIGHT r The weight of an object arises because of the gravitational pull of the earth. M M +F F M E DEFINITION OF WEIGHT The weight of an object on or above the earth is the gravitational force that the earth exerts on the object. The weight always acts downward, toward the center of the earth. On or above another astronomical body, the weight is the gravitational force exerted on the object by that body. SI Unit of Weight: newton (N) Figure 4.11 The gravitational force that each uniform sphere of matter exerts on the other is the same as if each sphere were a particle with its mass concentrated at its center. The earth (mass M E ) and the moon (mass M M ) approximate such uniform spheres.

12 45807_04_p1-48 6/20/05 7:39 AM Page CHAPTER 4 FORCES AND NEWTON S LAWS OF MOTION R E Object of mass m W Mass of earth = M E Figure 4.12 On or above the earth, the o weight W of an object is the gravitational force exerted on the object by the earth. Problem solving insight When applying Newton s gravitation law to uniform spheres of matter, remember that the distance r is between the centers of the spheres, not between the surfaces. r Using W for the magnitude of the weight,* m for the mass of the object, and M E for the mass of the earth, it follows from Equation 4.3 that W G M Em (4.4) r 2 Equation 4.4 and Figure 4.12 both emphasize that an object has weight whether or not it is resting on the earth s surface, because the gravitational force is acting even when the distance r is not equal to the radius R E of the earth. However, the gravitational force becomes weaker as r increases, since r is in the denominator of Equation 4.4. Figure 4.13, for example, shows how the weight of the Hubble Space Telescope becomes smaller as the distance r from the center of the earth increases. In Example 6 the telescope s weight is determined when it is on earth and in orbit. Example 6 The Hubble Space Telescope The mass of the Hubble Space Telescope is kg. Determine the weight of the telescope (a) when it was resting on the earth and (b) as it is in its orbit 598 km above the earth s surface. Reasoning The weight of the Hubble Space Telescope is the gravitational force exerted on it by the earth. According to Equation 4.4, the weight varies inversely as the square of the radial distance r. Thus, we expect the telescope s weight on the earth s surface (r smaller) to be greater than its weight in orbit (r larger). Solution (a) On the earth s surface, the weight is given by Equation 4.4 with r m (the earth s radius): W G M E m r 2 W N ( N m 2 /kg 2 )( kg)( kg) ( m) 2 (b) When the telescope is 598 km above the surface, its distance from the center of the earth is r m m m The weight now can be calculated as in part (a), except that the new value of r must be used: W N. As expected, the weight is less in orbit. The space age has forced us to broaden our ideas about weight. For instance, an astronaut weighs only about one-sixth as much on the moon as on the earth. To obtain his weight on the moon from Equation 4.4, it is only necessary to replace M E by M M (the mass of the moon) and let r R M (the radius of the moon). W ( 10 5 N) R E = 6.38 x 10 6 m r ( 10 6 m) Figure 4.13 The weight of the Hubble Space Telescope decreases as the telescope gets farther from the earth. The distance from the center of the earth to the telescope is r. RELATION ETWEEN MASS AND WEIGHT Although massive objects weigh a lot on the earth, mass and weight are not the same quantity. As Section 4.2 discusses, mass is a quantitative measure of inertia. As such, mass is an intrinsic property of matter and does not change as an object is moved from one location to another. Weight, in contrast, is the gravitational force acting on the object and can vary, depending on how far the object is above the earth s surface or whether it is located near another body such as the moon. The relation between weight W and mass m can be written in two ways: m G M E W (4.4) r 2 p W m g (4.5) * Often, the word weight and the phrase magnitude of the weight are used interchangeably, even though weight is a vector. Generally, the context makes it clear when the direction of the weight vector must be taken into account.

13 45807_04_p1-48 6/17/05 3:31 PM Page THE GRAVITATIONAL FORCE 13 Equation 4.4 is Newton s law of universal gravitation, and Equation 4.5 is Newton s second law (net force equals mass times acceleration) incorporating the acceleration g due to gravity. These expressions make the distinction between mass and weight stand out. The weight of an object whose mass is m depends on the values for the universal gravitational constant G, the mass M E of the earth, and the distance r. These three parameters together determine the acceleration g due to gravity. The specific value of g 9.80 m/s 2 applies only when r equals the radius R E of the earth. For larger values of r, as would be the case on top of a mountain, the effective value of g is less than 9.80 m/s 2. The fact that g decreases as the distance r increases means that the weight likewise decreases. The mass of the object, however, does not depend on these effects and does not change. Conceptual Example 7 further explores the difference between mass and weight. Problem solving insight Mass and weight are different quantities. They cannot be interchanged when solving problems. Conceptual Example 7 Mass Versus Weight A vehicle is being designed for use in exploring the moon s surface and is being tested on earth, where it weighs roughly six times more than it will on the moon. In one test, the acceleration of the vehicle along the ground is measured. To achieve the same acceleration on the moon, will the net force acting on the vehicle be greater than, less than, or the same as that required on earth? Reasoning and Solution The net force F required to accelerate the vehicle is specified by Newton s second law as F The Lunar Roving Vehicle that m a, where m is the vehicle s mass and a is the acceleration astronaut Eugene Cernan is driving on along the ground. For a given acceleration, the net force depends only on the mass. ut the the moon and the Lunar Excursion mass is an intrinsic property of the vehicle and is the same on the moon as it is on the earth. Module (behind the Roving Vehicle) Therefore, the same net force would be required for a given acceleration on the moon as on have the same mass that they have on the earth. Do not be misled by the fact that the vehicle weighs more on earth. The greater the earth. However, their weight is weight occurs only because the earth s mass and radius are different than the moon s. In any different on the moon than on the earth, event, in Newton s second law, the net force is proportional to the vehicle s mass, not its as Conceptual Example 7 discusses. weight. (NASA/Johnson Space Center) Related Homework: Problems 22, 87 CHECK YOUR UNDERSTANDING 2 One object has a mass m 1, and a second object has a mass m 2, which is greater than m 1. The two are separated by a distance 2d. A third object has a mass m 3. All three objects are located on the same straight line. The net gravitational force acting on the third object is zero. Which of the drawings correctly represents the locations of the objects? The answer is given at the end of the book.) m1 m3 m 2 m1 m3 m 2 m1 m 2 m3 d d d d d d (a) (b) (c) m1 m3 m 2 d (d) d ackground: The gravitational force and Newton s law of universal gravitation are the focus of this problem. For similar questions (including calculational counterparts), consult Self-Assessment Test 4.2. This test is described at the end of Section 4.10.

14 45807_04_p1-48 6/17/05 3:31 PM Page CHAPTER 4 FORCES AND NEWTON S LAWS OF MOTION F N W Figure 4.14 Two forces act on the block, its weight o W and the normal force F N exerted by the surface of the table. 4.8 THE NORMAL FORCE THE DEFINITION AND INTERPRETATION OF THE NORMAL FORCE In many situations, an object is in contact with a surface, such as a tabletop. ecause of the contact, there is a force acting on the object. The present section discusses only one component of this force, the component that acts perpendicular to the surface. The next section discusses the component that acts parallel to the surface. The perpendicular component is called the normal force. DEFINITION OF THE NORMAL FORCE The normal force F N is one component of the force that a surface exerts on an object with which it is in contact namely, the component that is perpendicular to the surface. 11 N 11 N F N = 26 N W = 15 N (a) F N = 4 N W = 15 N (b) Figure 4.15 (a) The normal force N is greater than the weight of the box, because the box is being pressed downward with an 11-N force. (b) The normal force is smaller than the weight, because the rope supplies an upward force of 11 N that partially supports the box. F Figure 4.14 shows a block resting on a horizontal table and identifies the two forces o that act on the block, the weight W and the normal force F N. To understand how an inanimate object, such as a tabletop, can exert a normal force, think about what happens when you sit on a mattress. Your weight causes the springs in the mattress to compress. As a result, the compressed springs exert an upward force (the normal force) on you. In a similar manner, the weight of the block causes invisible atomic springs in the surface of the table to compress, thus producing a normal force on the block. Newton s third law plays an important role in connection with the normal force. In Figure 4.14, for instance, the block exerts a force on the table by pressing down on it. Consistent with the third law, the table exerts an oppositely directed force of equal magnitude on the block. This reaction force is the normal force. The magnitude of the normal force indicates how hard the two objects press against each other. If an object is resting on a horizontal surface and there are no vertically acting forces except the object s weight and the normal force, the magnitudes of these two forces are equal; that is, F N W. This is the situation in Figure The weight must be balanced by the normal force for the object to remain at rest on the table. If the magnitudes of these forces were not equal, there would be a net force acting on the block, and the block would accelerate either upward or downward, in accord with Newton s second law. o If other forces in addition to W and F N act in the vertical direction, the magnitudes of the normal force and the weight are no longer equal. In Figure 4.15a, for instance, a box whose weight is 15 N is being pushed downward against a table. The pushing force has a magnitude of 11 N. Thus, the total downward force exerted on the box is 26 N, and this must be balanced by the upward-acting normal force if the box is to remain at rest. In this situation, then, the normal force is 26 N, which is considerably larger than the weight of the box. Figure 4.15b illustrates a different situation. Here, the box is being pulled upward by a rope that applies a force of 11 N. The net force acting on the box due to its weight and the rope is only 4 N, downward. To balance this force, the normal force needs to be only 4 N. It is not hard to imagine what would happen if the force applied by the rope were increased to 15 N exactly equal to the weight of the box. In this situation, the normal force would become zero. In fact, the table could be removed, since the block would be supported entirely by the rope. The situations in Figure 4.15 are consistent with the idea that the magnitude of the normal force indicates how hard two objects press against each other. Clearly, the box and the table press against each other harder in part a of the picture than in part b. Like the box and the table in Figure 4.15, various parts of the human body press against one another and exert normal forces. Example 8 illustrates the remarkable ability of the human skeleton to withstand a wide range of normal forces.

15 45807_04_p1-48 6/21/05 9:39 AM Page THE NORMAL FORCE 15 Free-body diagram +y Free-body diagram +y F N F N F N F N +x +x Seventh cervical vertebra 50 N 50 N 50 N 490 N 50 N (a) (b) (c) 490 N Example 8 A alancing Act In a circus balancing act, a woman performs a headstand on top of a standing performer s head, as Figure 4.16a illustrates. The woman weighs 490 N, and the standing performer s head and neck weigh 50 N. It is primarily the seventh cervical vertebra in the spine that supports all the weight above the shoulders. What is the normal force that this vertebra exerts on the neck and head of the standing performer (a) before the act and (b) during the act? Reasoning To begin, we draw a free-body diagram for the neck and head of the standing performer. efore the act, there are only two forces, the weight of the standing performer s head and neck, and the normal force. During the act, an additional force is present due to the woman s weight. In both cases, the upward and downward forces must balance for the head and neck to remain at rest. This condition of balance will lead us to values for the normal force. Solution (a) Figure 4.16b shows the free-body diagram for the standing performer s head and neck before the act. The only forces acting are the normal force F N and the 50-N weight. These two forces must balance for the standing performer s head and neck to remain at rest. Therefore, the seventh cervical vertebra exerts a normal force of F N 50 N. (b) Figure 4.16c shows the free-body diagram that applies during the act. Now, the total downward force exerted on the standing performer s head and neck is 50 N 490 N 540 N, which must be balanced by the upward normal force, so that F N 540 N. Figure 4.16 (a) A young woman keeps her balance during a performance by China s Sincuan Acrobatic group. A free-body diagram is shown (above the shoulders) for the standing performer (b) before the act and (c) during the act. For convenience, the scales used for the vectors in parts b and c are different. ( SUPRI/Reuters/Landov LLC) The physics of the human skeleton. In summary, the normal force does not necessarily have the same magnitude as the weight of the object. The value of the normal force depends on what other forces are present. It also depends on whether the objects in contact are accelerating. In one situation that involves accelerating objects, the magnitude of the normal force can be regarded as a kind of apparent weight, as we will now see. APPARENT WEIGHT Usually, the weight of an object can be determined with the aid of a scale. However, even though a scale is working properly, there are situations in which it does not give the correct weight. In such situations, the reading on the scale gives only the apparent weight, rather than the gravitational force or true weight. The apparent weight is the force that the object exerts on the scale with which it is in contact.

16 45807_04_p1-48 6/17/05 3:31 PM Page CHAPTER 4 FORCES AND NEWTON S LAWS OF MOTION a = g a a W = 700 N W = 700 N W = 700 N W = 700 N (a) No acceleration (v = constant) (b) Upward acceleration (c) Downward acceleration (d) Free-fall Figure 4.17 (a) When the elevator is not accelerating, the scale registers the true weight (W 700 N) of the person. (b) When the elevator accelerates upward, the apparent weight (1000 N) exceeds the true weight. (c) When the elevator accelerates downward, the apparent weight (400 N) is less than the true weight. (d) The apparent weight is zero if the elevator falls freely that is, if it falls with the acceleration due to gravity. To see the discrepancies that can arise between true weight and apparent weight, consider the scale in the elevator in Figure The reasons for the discrepancies will be explained shortly. A person whose true weight is 700 N steps on the scale. If the elevator is at rest or moving with a constant velocity (either upward or downward), the scale registers the true weight, as Figure 4.17a illustrates. If the elevator is accelerating, the apparent weight and the true weight are not equal. When the elevator accelerates upward, the apparent weight is greater than the true weight, as Figure 4.17b shows. Conversely, if the elevator accelerates downward, as in part c, the apparent weight is less than the true weight. In fact, if the elevator falls freely, so its acceleration is equal to the acceleration due to gravity, the apparent weight becomes zero, as part d indicates. In a situation such as this, where the apparent weight is zero, the person is said to be weightless. The apparent weight, then, does not equal the true weight if the scale and the person on it are accelerating. The discrepancies between true weight and apparent weight can be understood with the aid of Newton s second law. Figure 4.18 shows a free-body diagram of the person in o the elevator. The two forces that act on him are the true weight W mg and the normal force FN exerted by the platform of the scale. Applying Newton s second law in the vertical direction gives F y F N mg ma W = mg +y F N Figure 4.18 A free-body diagram showing the forces acting on the person riding in the elevator of Figure o W is the true weight, and F N is the normal force exerted on the person by the platform of the scale. +x where a is the acceleration of the elevator and person. In this result, the symbol g stands for the magnitude of the acceleration due to gravity and can never be a negative quantity. However, the acceleration a may be either positive or negative, depending on whether the elevator is accelerating upward ( ) or downward ( ). Solving for the normal force F N shows that F N mg ma (4.6) Apparent True weight weight In Equation 4.6, F N is the magnitude of the normal force exerted on the person by the scale. ut in accord with Newton s third law, F N is also the magnitude of the downward force that the person exerts on the scale namely, the apparent weight. Equation 4.6 contains all the features shown in Figure If the elevator is not accelerating, a 0 m/s 2, and the apparent weight equals the true weight. If the elevator accelerates upward, a is positive, and the equation shows that the apparent weight is greater than the true weight. If the elevator accelerates downward, a is negative, and the apparent weight is less than the true weight. If the elevator falls freely, a g, and the apparent weight is zero. The apparent weight is zero because when both the person and the scale fall freely, they cannot push against one another. In this text, when the weight is given, it is assumed to be the true weight, unless stated otherwise.

17 45807_04_p1-48 6/17/05 3:31 PM Page STATIC AND KINETIC FRICTIONAL FORCES 17 Figure 4.19 This photo, shot from underneath a transparent surface, shows a tire rolling under wet conditions. The channels in the tire collect and divert water away from the regions where the tire contacts the surface, thus providing better traction. (Courtesy Goodyear Tire & Rubber Co.) Microscopic contact points Figure 4.20 Even when two highly polished surfaces are in contact, they touch only at relatively few points. 4.9 STATIC AND KINETIC FRICTIONAL FORCES When an object is in contact with a surface, there is a force acting on the object. The previous section discusses the component of this force that is perpendicular to the surface, which is called the normal force. When the object moves or attempts to move along the surface, there is also a component of the force that is parallel to the surface. This parallel force component is called the frictional force, or simply friction. In many situations considerable engineering effort is expended trying to reduce friction. For example, oil is used to reduce the friction that causes wear and tear in the pistons and cylinder walls of an automobile engine. Sometimes, however, friction is absolutely necessary. Without friction, car tires could not provide the traction needed to move the car. In fact, the raised tread on a tire is designed to maintain friction. On a wet road, the spaces in the tread pattern (see Figure 4.19) provide channels for the water to collect and be diverted away. Thus, these channels largely prevent the water from coming between the tire surface and the road surface, where it would reduce friction and allow the tire to skid. Surfaces that appear to be highly polished can actually look quite rough when examined under a microscope. Such an examination reveals that two surfaces in contact touch only at relatively few spots, as Figure 4.20 illustrates. The microscopic area of contact for these spots is substantially less than the apparent macroscopic area of contact between the surfaces perhaps thousands of times less. At these contact points the molecules of the different bodies are close enough together to exert strong attractive intermolecular forces on one another, leading to what are known as cold welds. Frictional forces are associated with these welded spots, but the exact details of how frictional forces arise are not well understood. However, some empirical relations have been developed that make it possible to account for the effects of friction. Figure 4.21 helps to explain the main features of the type of friction known as static friction. The block in this drawing is initially at rest on a table, and as long as there is no attempt to move the block, there is no static frictional force. Then, a horizontal force F is applied to the block by means of a rope. If F is small, as in part a, experience tells us that the block still does not move. Why? It does not move because the static frictional force f s exactly cancels the effect of the applied force. The direction of f s is opposite to that of F, and the magnitude of f s equals the magnitude of the applied force, f s F. Increasing the applied force in Figure 4.21 by a small amount still does not cause the block to move. There is no movement because the static frictional force also increases by an amount that cancels out the increase in the applied force (see part b of the drawing). If the applied f s f s f s No movement (a) No movement (b) When movement just begins (c) Figure 4.21 Applying a small force to the block, as in parts a and b, produces no movement, because the static frictional force f s exactly balances the applied force. (c) The block just begins to move when the applied force is slightly greater than the maximum static frictional force f s. F F F F

18 45807_04_p1-48 6/21/05 9:39 AM Page CHAPTER 4 FORCES AND NEWTON S LAWS OF MOTION Figure 4.22 The maximum static frictional force fs would be the same, no matter which side of the block is in contact with the table. force continues to increase, however, there comes a point when the block finally breaks away and begins to slide. The force just before breakaway represents the maximum static frictional force fs that the table can exert on the block (see part c of the drawing). Any applied force that is greater than fs cannot be balanced by static friction, and the resulting net force accelerates the block to the right. Experimental evidence shows that, to a good degree of approximation, the maximum static frictional force between a pair of dry, unlubricated surfaces has two main characteristics. It is independent of the apparent macroscopic area of contact between the objects, provided that the surfaces are hard or nondeformable. For instance, in Figure 4.22 the maximum static frictional force that the surface of the table can exert on a block is the same, whether the block is resting on its largest or its smallest side. The other main characteristic of f is that its magnitude is proportional to the magnitude of the normal force F s N. As Section 4.8 points out, the magnitude of the normal force indicates how hard two surfaces are being pressed together. The harder they are pressed, the larger is f s, presumably because the number of cold-welded, microscopic contact points is increased. Equation 4.7 expresses the proportionality between f s and F N with the aid of a proportionality constant s, which is called the coefficient of static friction. STATIC FRICTIONAL FORCE The magnitude f s of the static frictional force can have any value from zero up to a maximum value of f s, depending on the applied force. In other words, f s f s, where the symbol is read as less than or equal to. The equality holds only when f s attains its maximum value, which is f s sf N (4.7) In Equation 4.7, s is the coefficient of static friction, and F N is the magnitude of the normal force. Need more practice? Interactive LearningWare 4.2 A sofa rests on the horizontal bed of a moving van. The coefficient of static friction between the sofa and van bed is The van starts from rest and accelerates for a time of 5.1 s. What is the maximum distance that the van can travel in this time period without having the sofa slide? Related Homework: Problem 82 For an interactive solution, go to It should be emphasized that Equation 4.7 relates only the magnitudes of fs and FN, not the vectors themselves. This equation does not imply that the directions of the vectors are the same. In fact, f is parallel to the surface, while F s N is perpendicular to it. The coefficient of static friction, being the ratio of the magnitudes of two forces ( s f s /F N ), has no units. Also, it depends on the type of material from which each surface is made (steel on wood, rubber on concrete, etc.), the condition of the surfaces (polished, rough, etc.), and other variables such as temperature. Table 4.2 gives some typical values of s for various surfaces. Example 9 illustrates the use of Equation 4.7 for determining the maximum static frictional force. Table 4.2 Approximate Values of the Coefficients of Friction for Various Surfaces* Coefficient of Static Coefficient of Materials Friction, s Kinetic Friction, k Glass on glass (dry) Ice on ice (clean, 0 C) Rubber on dry concrete Rubber on wet concrete Steel on ice Steel on steel (dry hard steel) Teflon on Teflon Wood on wood *The last column gives the coefficients of kinetic friction, a concept that will be discussed shortly.

19 45807_04_p1-48 6/20/05 7:39 AM Page 19 Example 9 The Force Needed To Start a Skier Moving 4.9 STATIC AND KINETIC FRICTIONAL FORCES 19 ANALYZING MULTIPLE-CONCEPT PROLEMS A skier is standing motionless on a horizontal patch of snow. She is holding onto a horizontal tow rope, which is about to pull her forward (see Figure 4.23a). The skier s mass is 59 kg, and the coefficient of static friction between the skis and snow is What is the magnitude of the maximum force that the tow rope can apply to the skier without causing her to move? Reasoning When the rope applies a relatively small force, the skier does not accelerate. The reason is that the static frictional force opposes the applied force and the two forces have the same magnitude. We can apply Newton s second law in the horizontal direction to this situation. In order for the rope to pull the skier forward, it must exert a force large enough to overcome the maximum static frictional force acting on the skis. The magnitude of the maximum static frictional force depends on the coefficient of static friction (which is known) and on the magnitude of the normal force. We can determine the magnitude of the normal force by using Newton's second law, along with the fact that the skier does not accelerate in the vertical direction. Knowns and Unknowns The data for this problem are: Description Symbol Value Mass of skier m 59 kg Coefficient of static friction s 0.14 Unknown Variable Magnitude of maximum horizontal force that tow rope can apply F? Modeling the Problem f s F N (a) +y (b) mg Figure 4.23 (a) Two horizontal forces act on the skier in the horizontal direction just before she begins to move. (b) Two vertical forces act on the skier. F +x STEP 1 Newton s Second Law (Horizontal Direction) Figure 4.23a shows the two horizontal forces that act on the skier just before she begins to move: the force F applied by the tow rope and the maximum static frictional force fs. Since the skier is standing motionless, she is not accelerating in the horizontal or x direction, so a x 0 m/s 2. Applying Newton s second law (Equation 4.2a) to this situation, we have F x ma x 0 Since the net force F x in the x direction is F x F f s, Newton s second law can be written as F 0. Thus, f s f s F f s We do not know, but its value will be determined in Steps 2 and 3. STEP 2 The Maximum Static Frictional Force The magnitude f s of the maximum static frictional force is related to the coefficient of static friction s and the magnitude F N of the normal force by Equation 4.7: f s sf N (4.7) We now substitute this result into Equation 1, as indicated in the right column. The coefficient of static friction is known, but F N is not. An expression for F N will be obtained in the next step. F f s F f s q f s sf N (1) (1) (4.7)

20 45807_04_p1-48 6/20/05 7:39 AM Page CHAPTER 4 FORCES AND NEWTON S LAWS OF MOTION STEP 3 Newton s Second Law (Vertical Direction) We can find the magnitude F N of the normal force by noting that the skier does not accelerate in the vertical or y direction, so a y 0 m/s 2. Figure 4.23b shows the two vertical forces that act on the skier: the normal force F N and her weight mg. Applying Newton s second law (Equation 4.2b) to the vertical direction gives F y ma y 0 The net force in the y direction is F y F N mg, so Newton s second law becomes F N mg 0. Thus, F N mg We now substitute this result into Equation 4.7, as shown at the right. F f s f s m s F N F N mg (1) (4.7) Solution Algebraically combining the results of the three steps, we have STEP 1 STEP 2 STEP 3 The magnitude F of the maximum force is F f s m s F N m s mg F s mg (0.14)(59 kg)(9.80 m/s 2 ) 81 N If the force exerted by the tow rope exceeds this value, the skier will begin to accelerate forward. Related Homework: Problems 100, 110, 113 The physics of rock climbing. Static friction is often essential, as it is to the rock climber in Figure 4.24, for instance. She presses outward against the walls of the rock formation with her hands and feet to create sufficiently large normal forces, so that the static frictional forces can support her weight. Once two surfaces begin sliding over one another, the static frictional force is no longer of any concern. Instead, a type of friction known as kinetic* friction comes into play. The kinetic frictional force opposes the relative sliding motion. If you have ever pushed an object across a floor, you may have noticed that it takes less force to keep the object sliding than it takes to get it going in the first place. In other words, the kinetic frictional force is usually less than the static frictional force. Experimental evidence indicates that the kinetic frictional force f k has three main characteristics, to a good degree of approximation. It is independent of the apparent area of contact between the surfaces (see Figure 4.22). It is independent of the speed of the sliding motion, if the speed is small. And lastly, the magnitude of the kinetic frictional force is proportional to the magnitude of the normal force. Equation 4.8 expresses this proportionality with the aid of a proportionality constant k, which is called the coefficient of kinetic friction. * The word kinetic is derived from the Greek word kinetikos, meaning of motion. Figure 4.24 In maneuvering her way up El Matador at Devil s Tower National Monument in Wyoming, this rock climber uses the static frictional forces between her hands and feet and the vertical rock walls to support her weight. ( Corey Rich/Coreyography)