Newton's Laws of Motion - Physics Textbook PDF

? Under what circumstances does the ­barbell push on the weightlifter just as hard as she pushes on the barbell? (i) When she holds the barbell stationary; (ii) when she raises the barbell; (iii) when she lowers the barbell; (iv) two of (i), (ii), and (iii); (v) all of (i), (ii), and (iii)...

? Under what circumstances does the ­barbell push on the weightlifter just as hard as she pushes on the barbell? (i) When she holds the barbell stationary; (ii) when she raises the barbell; (iii) when she lowers the barbell; (iv) two of (i), (ii), and (iii); (v) all of (i), (ii), and (iii); (vi) none of these. 4 Newton’s Laws of Motion W LEARNING OUTCOMES e’ve seen in the last two chapters how to use kinematics to describe motion in In this chapter, you’ll learn... one, two, or three dimensions. But what causes objects to move the way that they 4.1 What the concept of force means in do? For example, why does a dropped feather fall more slowly than a dropped physics, why forces are vectors, and baseball? Why do you feel pushed backward in a car that accelerates forward? The an- the significance of the net force on an swers to such questions take us into the subject of dynamics, the relationship of motion to object. the forces that cause it. 4.2 What happens when the net external The principles of dynamics were clearly stated for the first time by Sir Isaac Newton force on an object is zero, and the sig- nificance of inertial frames of reference. (1642–1727); today we call them Newton’s laws of motion. Newton did not derive the 4.3 How the acceleration of an object is de- laws of motion, but rather deduced them from a multitude of experiments performed by termined by the net external force on the other scientists, especially Galileo Galilei (who died the year Newton was born). Newton’s object and the object’s mass. laws are the foundation of classical mechanics (also called Newtonian mechanics); using 4.4 The difference between the mass of an them, we can understand most familiar kinds of motion. Newton’s laws need modification object and its weight. 4.5 How the forces that two objects exert on only for situations involving extremely high speeds (near the speed of light) or very small each other are related. sizes (such as within the atom). 4.6 How to use a free-body diagram to help Newton’s laws are very simple to state, yet many students find these laws difficult to analyze the forces on an object. grasp and to work with. The reason is that before studying physics, you’ve spent years You’ll need to review... walking, throwing balls, pushing boxes, and doing dozens of things that involve motion. 1.7 Vectors and vector addition. Along the way, you’ve developed a set of “common sense” ideas about motion and its 1.8 Vector components. causes. But many of these “common sense” ideas don’t stand up to logical analysis. A big 2.4 Straight-line motion with constant part of the job of this chapter—and of the rest of our study of physics—is helping you rec- acceleration. ognize how “common sense” ideas can sometimes lead you astray, and how to adjust your 2.5 The motion of freely falling objects. understanding of the physical world to make it consistent with what experiments tell us. 3.2 Acceleration as a vector. 3.4 Uniform circular motion. 3.5 Relative velocity. 4.1 FORCE AND INTERACTIONS A force is a push or a pull. More precisely, a force is an interaction between two objects or between an object and its environment (Fig. 4.1). That’s why we always refer to the force that one object exerts on a second object. When you push on a car that is stuck in the snow, you exert a force on the car; a steel cable exerts a force on the beam it is hoist- ing at a construction site; and so on. As Fig. 4.1 shows, force is a vector quantity; you can push or pull an object in different directions. 100  4.1 Force and Interactions    101 When a force involves direct contact between two objects, such as a push or pull that Figure 4.1 Some properties of forces. you exert on an object with your hand, we call it a contact force. Figures 4.2a, 4.2b, A force is a push or a pull. and 4.2c show three common types of contact forces. The normal force (Fig. 4.2a) is A force is an interaction between two objects exerted on an object by any surface with which it is in contact. The adjective “normal” or between an object and its environment. A force is a vector quantity, with magnitude means that the force always acts perpendicular to the surface of contact, no matter what and direction. the angle of that surface. By contrast, the friction force (Fig. 4.2b) exerted on an object S by a surface acts parallel to the surface, in the direction that opposes sliding. The pull- F (force) ing force exerted by a stretched rope or cord on an object to which it’s attached is called S a tension force (Fig. 4.2c). When you tug on your dog’s leash, the force that pulls on her F Push collar is a tension force. Pull In addition to contact forces, there are long-range forces that act even when the objects are separated by empty space. The force between two magnets is an example of a long- range force, as is the force of gravity (Fig. 4.2d); the earth pulls a dropped object toward it even though there is no direct contact between the object and the earth. The gravitational Figure 4.2 Four common types of forces. force that the earth exerts on your S body is called your weight. S (a) Normal force n: When an object rests or To describe a force vector F, we need to describe the direction in which it acts as pushes on a surface, the surface exerts a push on well as its magnitude, the quantity that describes “how much” or “how hard” the force it that is directed perpendicular to the surface. pushes or pulls. The SI unit of the magnitude of force is the newton, abbreviated N. (We’ll S give a precise definition of the newton in Section 4.3.) Table 4.1 lists some typical force n magnitudes. S n A common instrument for measuring force magnitudes is the spring balance. It con- sists of a coil spring enclosed in a case with a pointer attached to one end. When forces are applied to the ends of the spring, it stretches by an amount that depends on the force. We can make a scale for the pointer by using a number of identical objects with weights of S (b) Friction force f: In addition to the normal exactly 1 N each. When one, two, or more of these are suspended simultaneously from the force, a surface may exert a friction force on an balance, the total force stretching the spring is 1 N, 2 N, and so on, and we can label the object, directed parallel to the surface. corresponding positions of the pointer 1 N, 2 N, and so on. Then we can use this instru- S ment to measure the magnitude of an unknown force. We can also make a similar instru- n ment that measures pushes instead of pulls. S f Figure 4.3 (next page) shows a spring balance being used to measure a pull or push that we apply to a box. In each case we draw a vector to represent the applied force. The length of the vector shows the magnitude; the longer the vector, the greater the force magnitude. S (c) Tension force T: A pulling force exerted on an object by a rope, cord, etc. Superposition of Forces When you hold a ball in your hand to throw it, at least two forces act on it: the push of S yourShand and the d­ ownward pull of gravity. Experiment shows that when two forces F1 S T and F2 act at the same time at the same point on an object S (Fig. 4.4, next page), the effect on the object’s motion is the same as if a SsingleS forceS R were acting equal to the vector sum, or resultant, of the original forces: R = F1 + F2. More generally, any number of S (d) Weight w: The pull of gravity on an object is a long-range force (a force that acts over TABLE 4.1 Typical Force Magnitudes a distance). Sun’s gravitational force on the earth 3.5 * 1022 N Weight of a large blue whale 1.9 * 106 N S w Maximum pulling force of a locomotive 8.9 * 105 N Weight of a 250 lb linebacker 1.1 * 103 N Weight of a medium apple 1N Weight of the smallest insect eggs 2 * 10-6 N Electric attraction between the proton and the electron in a hydrogen atom 8.2 * 10-8 N Weight of a very small bacterium 1 * 10-18 N Weight of a hydrogen atom 1.6 * 10-26 N Weight of an electron 8.9 * 10-30 N Gravitational attraction between the proton and the electron in a hydrogen atom 3.6 * 10-47 N 102    CHAPTER 4 Newton’s Laws of Motion Figure 4.4 Superposition of forces. Figure 4.3 Using a vector arrow to denote the force that we exert when (a) pulling a block with a S S string or (b) pushing a block with a stick. Two forces F1 and F2 acting on an object at point O have the same effect as a single force (a) A 10 N pull directed 30° above (b) A 10 N push directed 45° below S R equal to their vector sum. the horizontal the horizontal S F2 S R 45° O 10 N S 30° 10 N F1 Figure S 4.5 Fx and Fy are the components forces applied at a point on an object have the same effect as a single force equal to the of F parallel and perpendicular to the vector sum of the forces. This important principle is called superposition of forces. sloping surface of the inclined plane. Since forces are vector quantities and add like vectors, we can use all of the rules of vector We cross out a mathematics that we learned in Chapter 1 to solve problems that involve vectors. This would vector when we be a good time to review the rules for vector addition presented in Sections 1.7 and 1.8. replace it by its components. We learned in Section 1.8 thatS it’s easiest to add vectors by using components. That’s x why we often describe a force F in terms of its x- and y-components Fx and Fy. Note that y F S the x- and y-coordinate axes do not have to be horizontal and vertical, S respectively. As an example, Fig. 4.5 shows a crate being pulled up a ramp by a force F. In this situation it’s Fy u Fx most convenient to choose one axis to be parallel to the ramp and the other to be perpen- dicular to the ramp. For the case shown in Fig. 4.5, both Fx and Fy are positive; S in other O situations, depending on your choice of axes and the orientation of the force F, either Fx or Fy may be negative or zero.    CAUTION    SUsing a wiggly line in force diagrams In Fig. 4.5 we draw a wiggly line through the The x- and y-axes can have any orientation, just so they’re mutually perpendicular. force vector F to show that we have replaced it by its x- and y-components. Otherwise, the diagram would include the same force twice. We’ll draw such a wiggly line in any force diagram where a force is replaced by its components. We encourage you to do the same in your own diagrams! ❙ call this the net force acting on the object. We’ll use the Greek letter g (capital sigma, We’ll often need to find the vector sum (resultant) of all forces acting on an object. We S equivalent S S to the Roman S) as a shorthand notation for a sum. If the forces are labeled F1, F2, F3, and so on, we can write R = gF = F1 + F2 + F3 + c S S S S S The net force (4.1) acting on an object... Figure 4.6 Finding the components S of the vector sum (resultant) R of two forces... is the vector sum, or resultant, of all individual forces acting on that object. We read g F as “the vector sum of the forces” or “the net force.” The x-component of S S F1 and F2. S S The y-component of R equals the sum of S the y-S The same is true for the net force is the sum of the x-components of the individual forces, and likewise for the components of F1 and F2. the x-components. y-component (Fig. 4.6): Rx = g Fx Ry = g Fy (4.2) y S R = ΣF S Each component may be positive or negative, so be careful with signs when you evaluate F2y these sums. R = g F acting on the object. The magnitude is S F2 S OnceS we have Rx and Ry we can find the magnitude and direction of the net force Ry F1y S F1 R = 2Rx2 + Ry2 S x and the angle u between R and the +x@axis can be found from the relationship O F1x F2x tan u = Ry>Rx. The components Rx and Ry may be positive, negative, or zero, and the angle u Rx may be in any of the four quadrants.  4.2 Newton’s First Law    103 equation Rz = g Fz to Eqs. (4.2). The magnitude of the net force is then In three-dimensional problems, forces may also have z-components; then we add the R = 2Rx2 + Ry2 + Rz2 EXAMPLE 4.1 Superposition of forces: Finding the net force WITH ARIATION PROBLEMS S Three professional wrestlers are fighting over a champion’s belt. is positive. The angle between F3 and the negative x-axis is 53°, S so the Figure 4.7a shows the horizontal force each wrestler applies to the absolute value of its x-component is equal to the magnitude of F3 times belt, as viewed from above. The forces have magnitudes F1 = 50 N, the cosine of 53°. SThe absolute value of the y-component is therefore F2 = 120 N, and F3 = 250 N. Find the x- and y-components of the net the magnitude of F3 times the S sine of 53°. Keeping track of the signs, force on the belt, and find its magnitude and direction. we find the components of F3 are IDENTIFY and SET UP This is a problem in vector addition in which F3x = -(250 N) cos 53° = -150 N F3y = (250 N) sin 53° = 200 N From Eqs. (4.2) the components of the net force R = gF are the vectors happen S to represent forces. To find the x- and y-components S S of the net force R, we’ll use the component method of vectorS addition expressed by Eqs. (4.2). Once we know the components of R, we can Rx = F1x + F2x + F3x = 50 N + 0 N + 1-150 N2 = - 100 N find its magnitude and direction. S Ry = F1y + F2y + F3y = 0 N + 1- 120 N2 + 200 N = 80 N EXECUTE Figure 4.7a shows that force F1 (magnitude 50 N) points in the positive x-direction. Hence it has a positive x-component and zero The net force has a negative x-component and a positive y-component, y-component: as Fig. 4.7b shows. S The magnitude of R is F1x = 50 N F1y = 0 N S R = 2Rx2 + Ry2 = 21- 100 N2 2 + 180 N2 2 = 128 N Force F2 points in the negative y-direction and so has zero x-­component and a negative y-component: To find the angle between the net force and the +x-axis, we use F2x = 0 N F2y = - 120 N Eq. (1.7): S Force F3 doesn’t point along either the x-direction or the y-direction: Ry 80 N u = arctan = arctan a b = arctan 1- 0.802 Figure 4.7a shows that its x-component is negative and its y-component Rx - 100 N The arctangent of -0.80 is - 39°, but Fig. 4.7b shows that the net force R = gF and its components. Figure S 4.7 S (a) Three forces acting on a belt. (b) The net force lies in the second quadrant. Hence the correct solution is u = - 39° + 180° = 141°. (a) y (b) y EVALUATE The net force is not zero. Wrestler 3 exerts the greatest force Net force S F3y S R = ΣF S on the belt, F3 = 250 N, and will walk away with it when the struggle F3 x- and y-components ends. S Ry S S S of F3 You S should check the direction of R by adding Sthe vectors S S F1, FS2, 53° S u = 141° and F3 graphically. Does your drawing show that R = F1 + F2 + F3 F1 x x points in the second quadrant as we found? S F3x F1 has zero Rx S F2 has zero S y-component. KEYCONCEPT The net force is the vector sum of all of the individ- x-component. F2 ual forces that act on the object. It can be specified by its components or by its magnitude and direction. S TEST YOUR UNDERSTANDING OF SECTION 4.1 Figure 4.5 shows a force F acting on a crate. (a) With the x- and y-axes shown in the figure, is the x-component of the gravitational force that the earth exerts on the crate (the crate’s weight) positive, negative, or zero? (b) What about the y-component? the gravitational force has both an x-component and a y-component, and both are negative. ANSWER the x-axis points up and to the right, and the y-axis points up and to the left. With this choice of axes, ❙ (a) negative, (b) negative. The gravitational force on the crate points straight downward. In Fig. 4.5 4.2 NEWTON’S FIRST LAW How do the forces acting on an object affect that object’s motion? Let’s first note that it’s impossible for an object to affect its own motion by exerting a force on itself. If that were possible, you could lift yourself to the ceiling by pulling up on your belt! The forces that affect an object’s motion are external forces, those forces exerted on the object by other 104    CHAPTER 4 Newton’s Laws of Motion Figure 4.8 The slicker the surface, the objects in its environment. So the question we must answer is this: How do the external farther a puck slides after being given an forces that act on an object affect its motion? initial velocity. On an air-hockey table (c) To begin to answer this question, let’s first consider what happens when the net exter- the friction force is practically zero, so the puck continues with almost constant nal force on an object is zero. You would almost certainly agree that if an object is at rest, velocity. and if no net external force acts on it (that is, no net push or pull from other objects), that (a) Table: puck stops short. object will remain at rest. But what if there is zero net external force acting on an object in motion? To see what happens in this case, suppose you slide a hockey puck along a horizontal tabletop, applying a horizontal force to it with your hand (Fig. 4.8a). After you stop push- ing, the puck does not continue to move indefinitely; it slows down and stops. To keep it moving, you have to keep pushing (that is, applying a force). You might come to the “com- mon sense” conclusion that objects in motion naturally come to rest and that a force is required to sustain motion. But now imagine pushing the puck across a smooth surface of ice (Fig. 4.8b). After you quit pushing, the puck will slide a lot farther before it stops. Put it on an air-hockey table, where it floats on a thin cushion of air, and it moves still farther (Fig. 4.8c). In each case, what slows the puck down is friction, an interaction between the lower surface of the puck and the surface on which it slides. Each surface exerts a friction force on the puck that (b) Ice: puck slides farther. resists the puck’s motion; the difference in the three cases is the magnitude of the fric- tion force. The ice exerts less friction than the tabletop, so the puck travels farther. The gas molecules of the air-hockey table exert the least friction of all. If we could eliminate friction completely, the puck would never slow down, and we would need no force at all to keep the puck moving once it had been started. Thus the “common sense” idea that a force is required to sustain motion is incorrect. Experiments like the ones we’ve just described show that when no net external force acts on an object, the object either remains at rest or moves with constant velocity in a straight line. Once an object has been set in motion, no net external force is needed to keep it moving. We call this observation Newton’s first law of motion: NEWTON’S FIRST LAW OF MOTION An object acted on by no net external force (c) Air-hockey table: puck slides even farther. has a constant velocity (which may be zero) and zero acceleration. The tendency of an object to keep moving once it is set in motion is called inertia. You use inertia when you try to get ketchup out of a bottle by shaking it. First you start the bottle (and the ketchup inside) moving forward; when you jerk the bottle back, the ketchup................. tends to keep moving forward and, you hope, ends up on your burger. Inertia is also the................ tendency of an object at rest to remain at rest. You may have seen a tablecloth yanked out.............................. from under a table setting without breaking anything. The force on the table setting isn’t............................ great enough to make it move appreciably during the short time it takes to pull the table-........... cloth away........... It’s important to note that the net external force is what matters in Newton’s first law. For example, a physics book at rest on a horizontal tabletop has two forces acting on it: an upward supporting force, or normal force, exerted by the tabletop (see Fig. 4.2a) and the downward force of the earth’s gravity (see Fig. 4.2d). The upward push of the surface is just as great as the downward pull of gravity, so the net external force acting on the book (that is, the vector sum of the two forces) is zero. In agreement with Newton’s first law, if the book is at rest on the tabletop, it remains at rest. The same principle applies to a hockey puck sliding on a horizontal, frictionless surface: The vector sum of the upward push of the surface and the downward pull of gravity is zero. Once the puck is in motion, it continues to move with constant velocity because the net external force acting on it is zero. Here’s another example. Suppose a hockey puck rests on a horizontal surface with neg- ligible friction, such as an air-hockey S table or a slab of wet ice. If the puck is initially at rest and a single horizontal force F1 acts on it (Fig. 4.9a), the puck starts to move. If the puck is in motion to begin with, the force changes its speed, its direction, orS both, depending on the ­direction of the force. In this case the net external force is equal to F1, which is not zero. (There are also two vertical forces: the earth’s gravitational attraction and the upward normal force exerted by the surface. But as we mentioned earlier, these two forces cancel.)  4.2 Newton’s First Law    105 (a) A puck on a frictionless surface (b) This puck is acted on by two Figure 4.9 (a) A hockey puck acceleratesS accelerates when acted on by a horizontal forces whose vector sum in the direction of a net applied force F1. single horizontal force. is zero. The puck behaves as though (b) When the net external force is zero, no external forces act on it. the acceleration is zero, and the puck is in S equilibrium. ΣF = 0 S S a a=0 S S S F1 F1 F2 APPLICATION Sledding with Newton’s First Law The downward force S S Now suppose we apply a second force, F2 (Fig. 4.9b), equal Sin magnitude to F1 but op- of gravity acting on the child and sled S is balanced by an upward normal force posite in direction. The two forces are negatives of each other, F2 = −F1, and their vector exerted by the ground. The adult’s foot sum is zero: g F = F1 + F2 = F1 + 1 −F1 2 = 0 exerts a forward force that balances the S S S S S backward force of friction on the sled. Hence there is no net external force on Again, we find that if the object is at rest at the start, it remains at rest; if it is initially the child and sled, and they slide with a constant velocity. moving, it continues to move in the same direction with constant speed. These results show that in Newton’s first law, zero net external force is equivalent to no external force at all. This is just the principle of superposition of forces that we saw in Section 4.1. When an object is either at rest or moving with constant velocity (in a straight line with constant speed), we say that the object is in equilibrium. For an object to be in equilib- rium, it must be acted on by no forces, or by several forces such that their vector sum— that is, the net external force—is zero: gF = 0 Newton’s first law: S... must be zero if the object Net external force on an object... is in equilibrium. (4.3) We’re assuming that the object can be represented adequately as a point particle. When the object has finite size, we also have to consider where on the object the forces are ap- plied. We’ll return to this point in Chapter 11. CONCEPTUAL EXAMPLE 4.2 Using Newton’s first law I In the classic 1950 science-fiction film Rocketship X-M, a spaceship is to move in a straight line with constant speed. Some science-fiction moving in the vacuum of outer space, far from any star or planet, when movies are based on accurate science; this is not one of them. its engine dies. As a result, the spaceship slows down and stops. What does Newton’s first law say about this scene? KEYCONCEPT If the net external force on an object is zero, the object either remains at rest or keeps moving at a constant velocity. SOLUTION No external forces act on the spaceship after the engine dies, so according to Newton’s first law it will not stop but will continue CONCEPTUAL EXAMPLE 4.3 Using Newton’s first law II You are driving a Porsche 918 Spyder on a straight testing track at a your Porsche than it does on the Volkswagen. But a backward force also constant speed of 250 km>h. You pass a 1971 Volkswagen Beetle doing acts on each car due to road friction and air resistance. When the car a constant 75 km>h. On which car is the net external force greater? is traveling with constant velocity, the vector sum of the forward and backward forces is zero. There is a greater backward force on the fast- SOLUTION The key word in this question is “net.” Both cars are in moving Porsche than on the slow-moving Volkswagen, which is why the equilibrium because their velocities are constant; Newton’s first law Porsche’s engine must be more powerful than that of the Volkswagen. therefore says that the net external force on each car is zero. This seems to contradict the “common sense” idea that the faster car KEYCONCEPT If an object either remains at rest or keeps moving at must have a greater force pushing it. Thanks to your Porsche’s high- a constant velocity, the net external force on the object is zero. power engine, it’s true that the track exerts a greater forward force on 106    CHAPTER 4 Newton’s Laws of Motion Inertial Frames of Reference In discussing relative velocity in Section 3.5, we introduced the concept of frame of refer- ence. This concept is central to Newton’s laws of motion. Suppose you are in a bus that is traveling on a straight road and speeding up. If you could stand in the aisle on roller skates, you would start moving backward relative to the bus as the bus gains speed. If instead the bus was slowing to a stop, you would start moving forward down the aisle. In either case, it looks as though Newton’s first law is not obeyed; there is no net external force acting on you, yet your velocity changes. What’s wrong? The point is that the bus is accelerating with respect to the earth and is not a suitable frame of reference for Newton’s first law. This law is valid in some frames of reference and not valid in others. A frame of reference in which Newton’s first law is valid is called an inertial frame of reference. The earth is at least approximately an inertial frame of reference, but the bus is not. (The earth is not a completely inertial frame, owing to the ac- celeration associated with its rotation and its motion around the sun. These effects are quite small, however; see Exercises 3.25 and 3.34.) Because Newton’s first law is used to define what we mean by an inertial frame of reference, it is sometimes called the law of inertia. Figure 4.10 helps us understand what you experience when riding in a vehicle that’s accelerating. In Fig. 4.10a, a vehicle is initially at rest and then begins to accelerate to the right. A passenger standing on roller skates (which nearly eliminate the effects of friction) has virtually no net external force acting on her, so she tends to remain at rest relative to the inertial frame of the earth. As the vehicle accelerates around her, she moves backward relative to the vehicle. In the same way, a passenger in a vehicle that is slowing down tends to continue moving with constant velocity relative to the earth, and so moves forward rela- tive to the vehicle (Fig. 4.10b). A vehicle is also accelerating if it moves at a constant speed but is turning (Fig. 4.10c). In this case a passenger tends to continue moving relative to the earth at constant speed in a straight line; relative to the vehicle, the passenger moves to the side of the vehicle on the outside of the turn. In each case shown in Fig. 4.10, an observer in the vehicle’s frame of reference might be tempted to conclude that there is a net external force acting on the passenger, since the passenger’s velocity relative to the vehicle changes in each case. This conclusion is simply Figure 4.10 Riding in an accelerating vehicle. (a) Initially, you and the (b) Initially, you and the (c) The vehicle rounds a turn vehicle are at rest. vehicle are in motion. at constant speed. S S v=0 v t = 0 t = 0 t = 0 S S a S a a S v S v t = ∆t t = ∆t S S a a t = ∆t S S v v S a t = 2∆t S t = 2∆t a S a S S v v t = 3∆t t = 3∆t t = 2∆t S S a a S a You tend to remain at rest as the You tend to continue moving vehicle accelerates around you. with constant velocity as the vehicle slows down around you. S v You tend to continue moving in a straight line as the vehicle turns.  4.3 Newton’s Second Law    107 wrong; the net external force on the passenger is indeed zero. The vehicle observer’s mis- Figure 4.11 From the frame of reference take is in trying to apply Newton’s first law in the vehicle’s frame of reference, which is of the car, it seems as though a force is pushing the crash test dummies forward as not an inertial frame and in which Newton’s first law isn’t valid (Fig. 4.11). In this book the car comes to a sudden stop. But there we’ll use only inertial frames of reference. is really no such force: As the car stops, We’ve mentioned only one (approximately) inertial frame of reference: the earth’s sur- the dummies keep moving forward as a face. But there are many inertial frames. If we have an inertial frame of reference A, in consequence of Newton’s first law. which Newton’s first law is obeyed, then any second frame of reference B will also be S inertial if it moves relative to A with constant velocity vB>A. We can prove this by using the relative-velocity relationship Eq. (3.35) from Section 3.5: S S S vP>A = vP>B + vB>A S Suppose that P is an object that moves with constant velocity vP>A with respect to an inertial frame A. By Newton’s first law the net external force on this object is zero. The S S S velocity of P relative to another frame B has a different value, vP>B =vP>A − vB>A. But if S S the relative velocity vB>A of the two frames is constant, then vP>B is constant as well. Thus B is also an inertial frame; the velocity of P in this frame is constant, and the net external force on P is zero, so Newton’s first law is obeyed in B. Observers in frames A and B will disagree about the velocity of P, but they will agree that P has a constant velocity (zero acceleration) and has zero net external force acting on it. There is no single inertial frame of reference that is preferred over all others for formu- lating Newton’s laws. If one frame is inertial, then every other frame moving relative to it with constant velocity is also inertial. Viewed in this light, the state of rest and the state of motion with constant velocity are not very different; both occur when the vector sum of forces acting on the object is zero. TEST YOUR UNDERSTANDING OF SECTION 4.2 In which of the following situations is there zero net external force on the object? (i) An airplane flying due north at a steady 120 m>s and at a constant altitude; (ii) a car driving straight up a hill with a 3° slope at a constant 90 km>h; (iii) a hawk circling at a constant 20 km>h at a constant height of 15 m above an open field; (iv) a box with slick, frictionless surfaces in the back of a truck as the truck accelerates forward on a level road at 5 m>s2. ANSWER In (iii), the hawk is moving in a circle; hence it is accelerating and is not in equilibrium. ­reference frame of the ground as the truck accelerates forward, like the person on skates in Fig. 4.10a.] object is zero. [In (iv), if the truck starts from rest, the box remains stationary as seen in the inertial ❙ (i), (ii), and (iv) In (i), (ii), and (iv) the object is not accelerating, so the net external force on the 4.3 NEWTON’S SECOND LAW Newton’s first law tells us that when an object is acted on by zero net external force, the object moves with constant velocity and zero acceleration. In Fig. 4.12a (next page), a hockey puck is sliding to the right on wet ice. There is negligible friction, so there are no mal force exerted by the ice surface sum to zero. So the net external force g F acting on horizontal forces acting on the puck; the downward force of gravity and the upward S nor- the puck is zero, the puck has zero acceleration, and its velocity is constant. But what happens when the net external force is not zero? In Fig. 4.12b we apply a con- g F is constant and in the same horizontal direction as v. We find that during the time the stant S horizontal force to a sliding puck in the same direction that the puck is moving. Then S force is acting, the velocity of the puck changes at a constant rate; that is, the puck moves same direction as v and g F. S with constant acceleration. SThe speed of the puck increases, so the acceleration a is in the In Fig. 4.12c we reverse the direction of the force on the puck so that g F acts opposite to S S g S v. In this case as well, the puck has an acceleration; the puck moves more and more slowly to previous case, experiment shows that the acceleration is constant if g F is constant. S S the right. The acceleration a in this case is to the left, in the same direction S as F. As in the We conclude that a net external force acting on an object causes the object to accelerate in the same direction as the net external force. If the magnitude of the net external force is constant, as in Fig. 4.12b and Fig. 4.12c, then so is the magnitude of the acceleration. 108    CHAPTER 4 Newton’s Laws of Motion S Figure 4.12 Using a hockey puck on (a) If there is zero net external force on the puck, so ΣF = 0,... a frictionless surface to explore the force gF on an object and the resulting relationship S between the net external S S S S S S v v v v v acceleration a of the object. S S... the puck has zero acceleration (a = 0) and its velocity v is constant. S (b) If a constant net external force ΣF acts on the puck in the direction of its motion... S S S S S ΣF ΣF ΣF ΣF ΣF S S S S S a a a a a S S S S S v v v v v S... the puck has a constant acceleration a in the same direction as the net force. S (c) If a constant net external force ΣF acts on the puck opposite to the direction of its motion... S S S S S ΣF ΣF ΣF ΣF ΣF Figure 4.13 A top view of a hockey S S S S S puck in uniform circular motion on a a a a a a frictionless horizontal surface. Puck moves at constant speed S S S S S around circle. v v v v v S S v... the puck has a constant acceleration a in the same direction as the net force. S S ΣF a S v S a ΣF S These conclusions about net external force and acceleration also apply to an object Rope moving along a curved path. For example, Fig. 4.13 shows a hockey puck moving in a horizontal circle on an ice surface of negligible friction. A rope is attached to the puck and ΣF S S a to a stick in the ice, and this rope exerts an inward tension force of constant magnitude on S the puck. The net external force and acceleration are both constant in magnitude and di- v rected toward the center of the circle. The speed of the puck is constant, so this is uniform S At all points, theSacceleration a and the net circular motion (see Section 3.4). external force ΣF point in the same direction— always toward the center of the circle. Figure 4.14a shows another experiment involving acceleration and net external force. We apply a constant horizontal force to a puck on a frictionless horizontal surface, using the spring balance described in Section 4.1 with the spring stretched a constant amount. Figure 4.14 The magnitude of an object’s As in Figs. 4.12b and Figs. 4.12c, this horizontal force equals the net external force on the S acceleration a is directly proportional to puck. If we change the magnitude of the net external force, the acceleration changes in gF acting on the object of mass m. theSmagnitude of the net external force the same proportion. Doubling the net external force doubles the acceleration (Fig. 4.14b), halving the net external force halves the acceleration (Fig. 4.14c), and so on. Many such (a) A constant net external force ΣF S experiments show that for any given object, the magnitude of the acceleration is directly S causes a constant acceleration a. proportional to the magnitude of the net external force acting on the object. S a Mass and Force Our results mean that for a given object, the ratio of the magnitude 0 g F 0 of the net exter- x m S S S ΣF = F1 nal force to the magnitude a = 0 a 0 of the acceleration is constant, regardless of the mag- S (b) Doubling the net external force nitude of the net external force. We call this ratio the inertial mass, or simply the mass, of 0 gF 0 0 gF 0 doubles the acceleration. the object and denote it by m. That is,   or   0 g F 0 = ma   or   a = S 2a S S S x m =  (4.4) m S S a m ΣF = 2F1 Mass is a quantitative measure of inertia, which we discussed in Section 4.2. The last of the equations in Eqs. (4.4) says that the greater an object’s mass, the more the object “re- (c) Halving the net external force halves the acceleration. sists” being accelerated. When you hold a piece of fruit in your hand at the supermarket 1S and move it slightly up and down to estimate its heft, you’re applying a force and seeing 2a how much the fruit accelerates up and down in response. If a force causes a large accelera- x tion, the fruit has a small mass; if the same force causes only a small acceleration, the fruit m S S ΣF = 12 F1 has a large mass. In the same way, if you hit a table-tennis ball and then a basketball with  4.3 Newton’s Second Law    109 force gF acting an object, the the same force, the basketball has much smaller acceleration because it has much greater Figure 4.15 S For a given net external mass. acceleration is inversely proportional The SI unit of mass is the kilogram. We mentioned in Section 1.3 that the kilo- to the mass of the object. Masses gram is officially defined in terms of the definitions of the second and the meter, as add like ordinary scalars. well as the value of a fundamental quantity called Planck’s constant. We can use this S definition: (a) A known net external force ΣF causes an object with mass m1 to have S an acceleration a1.  One newton is the amount of net external force that gives an acceleration of S 1 meter per second squared to an object with a mass of 1 kilogram. a1 S ΣF x m1 This definition allows us to calibrate the spring balances and other instruments used to measure forces. Because of the way we have defined the newton, it is related to the units of mass, length, and time. For Eqs. (4.4) to be dimensionally consistent, it must be true that S 1 newton = 11 kilogram211 meter per second squared2 (b) Applying the same net external force ΣF to a second object and noting the acceleration or 1 N = 1 kg # m>s2 allow us to measure the mass. S a2 Here’s an application of Eqs. (4.4). Suppose we apply a constant net external force g F S We’ll use this relationship many times in the next few chapters, so keep it in mind. ΣF S x m2 to an object of known mass m1 and we find an acceleration of magnitude a1 (Fig. 4.15a). We then apply the same force to another object of unknown mass m2 , and we find an ac- celeration of magnitude a2 (Fig. 4.15b). Then, according to Eqs. (4.4), (c) When the two objects are fastened together, the same method shows that m1 a1 = m2 a2 their composite mass is the sum of their m2 a1 individual masses. =    (same net external force) (4.5) S a3 m1 a2 For the same net external force, the ratio of the masses of two objects is the inverse of the ΣF S ratio of their accelerations. In principle we could use Eq. (4.5) to measure an unknown m1 + m2 x mass m2 , but it is usually easier to determine mass indirectly by measuring the object’s weight. We’ll return to this point in Section 4.4. When two objects with masses m1 and m2 are fastened together, we find that the mass of the composite object is always m1 + m2 (Fig. 4.15c). This additive property of mass may seem obvious, but it has to be verified experimentally. Ultimately, the mass of an object is related to the number of protons, electrons, and neutrons it contains. This wouldn’t be a good way to define mass because there is no practical way to count these particles. But the con- cept of mass is the most fundamental way to characterize the quantity of matter in an object. Stating Newton’s Second Law Experiment shows that the net externalS force S S on an object is what causes that object to ac- celerate. If a combination of forces F1 , F2 , F3 , and so on is applied to an object, the object S will have the same acceleration vectorS a asSwhenS only a single force is applied, if that single force is equal to the vector sum F1 + F2 + F3 + P. In other words, the principle of superposition of forces (see Fig. 4.4) also holds true when the net external force is not zero and the object is accelerating. Equations (4.4) relate the magnitude of the net external force on an object to the mag- nitude of the acceleration that it produces. We have also seen that the direction of the net external force is the same as the direction of the acceleration, whether the object’s path is straight or curved. Finally, we’ve seen that the forces that affect an object’s motion are external forces, those exerted on the object by other objects in its environment. Newton wrapped up all these results into a single concise statement that we now call Newton’s second law of motion: NEWTON’S SECOND LAW OF MOTION If a net external force acts on an object, the object accelerates. The direction of acceleration is the same as the direction of the net external force. The mass of the object times the acceleration vector of the object equals the net external force vector. 110    CHAPTER 4 Newton’s Laws of Motion Figure 4.16 The design of high- In symbols, performance motorcycles depends gF = ma fundamentally on Newton’s second law. Newton’s second law: S S... the object accelerates in To maximize the forward acceleration, the If there is a net external the same direction as the (4.6) designer makes the motorcycle as light as force on an object... Mass of object net external force. possible (that is, minimizes the mass) and uses the most powerful engine possible (thus maximizing the forward force). An alternative statement is that the acceleration of an object is equal to the net external force acting on the object divided by the object’s mass: gF S S a= Lightweight m object (small m) Newton’s second law is a fundamental law of nature, the basic relationship between force and motion. Most of the remainder of this chapter and all of the next are devoted to learning how to apply this principle in various situations. Equation (4.6) has many practical applications (Fig. 4.16). You’ve actually been using it all your life to measure your body’s acceleration. In your inner ear, microscopic hair cells are attached to a gelatinous substance that holds tiny crystals of calcium carbonate called otoliths. When your body accelerates, the hair cells pull the otoliths along with the rest of your body and sense the magnitude and direction of the force that they exert. By Newton’s second law, the acceleration of the otoliths—and hence that of your body as a whole—is proportional to this force and has the same direction. In this way, you can sense Powerful engine (large F) the magnitude and direction of your acceleration even with your eyes closed! Using Newton’s Second Law At least four aspects of Newton’s second law deserve special attention. First, Eq. (4.6) is a vector equation. Usually we’ll use it in component form, with a separate equation for each component of force and the corresponding component of acceleration: Newton’s second law: Each component of the net external force on an object... gFx = max gFy = may gFz = maz (4.7)... equals the object’s mass times the corresponding acceleration component. This set of component equations is equivalent to the single vector Eq. (4.6). Second, the statement of Newton’s second law refers to external forces. As an exam- pieces of the ball together. That’s why only external forces are included in the sum g F in ple, how a kicked soccer ball moves isn’t affected by the internal forces that holdSthe Eqs. (4.6) and (4.7). Third, Eqs. (4.6) and (4.7) are valid only when the mass m is constant. It’s easy to think of systems whose masses change, such as a leaking tank truck or a moving railroad car being loaded with coal. Such systems are better handled by using the concept of momen- tum; we’ll get to that in Chapter 8. Finally, Newton’s second law is valid in inertial frames of reference only, just like the first law. It’s not valid in the reference frame of any of the accelerating vehicles in Fig. 4.10; relative to any of these frames, the passenger accelerates even though the net external force on the passenger is zero. We’ll usually treat the earth as an adequate ap- proximation to an inertial frame, although because of its rotation and orbital motion it is not precisely inertial.    CAUTION    ma is not a force Even though the vector ma is equal to the vector sum gF of all the S S S S forces acting on the object, the vector ma is not a force. Acceleration is the result of the net external force; it is not a force itself. It’s “common sense” to think that a “force of acceleration” pushes you back into your seat when your car accelerates forward from rest. But there is no such force; instead, your inertia causes you to tend to stay at rest relative to the earth, and the car accelerates around you (see Fig. 4.10a). The “common sense” confusion arises from trying to apply Newton’s second law where it isn’t valid—in the noninertial reference frame of an accelerating car. We’ll always examine motion relative to inertial frames of reference only, and we strongly recommend that you do the same in solving problems. ❙  4.3 Newton’s Second Law    111 In learning how to use Newton’s second law, we’ll begin in this chapter with examples of straight-line motion. Then in Chapter 5 we’ll consider more general kinds of motion and develop more detailed problem-solving strategies. EXAMPLE 4.4 Newton’s second law I: Determining acceleration from force WITH ARIATION PROBLEMS A worker applies a constant horizontal force with magnitude 20 N to a Just as we did for the forces in Example 4.1 (Section 4.1), we’ll find the box with mass 40 kg resting on a level, freshly waxed floor with negli- vector sum of these external forces using components. That’s why the sec- gible friction. What is the acceleration of the box? ond step in any problem involving forces is choosing a coordinate system IDENTIFY and SET UP This problem involves force and acceleration, for finding vector components. It’s usually convenient to take one axis either so we’ll use Newton’s second law. That means we’ll have to find the net along or opposite the direction of the object’s acceleration, which in this case external force acting on the box and set it equal to the mass of the box is horizontal. Hence we take the +x-axis to be in the direction of the applied multiplied by its acceleration. In this example, the acceleration is our horizontal force (which is the direction in which the box accelerates) and the target variable. +y-axis to be upward. In most force problems that you’ll encounter (includ- In any problem involving forces, to find the net external force we must ing this one), the force vectors all lie in a plane, so the z-axis isn’t used. first identify all of the individual external forces that act on the object The box doesn’t move vertically, so the y-acceleration is zero: in question. (Remember that the net external force is the vector sum of ay = 0. Our target variable is the x-acceleration, ax. We’ll find it by these individual forces.) To identify these forces, we’ll use the idea that using Newton’s second law in component form, Eqs. (4.7). S two broad categories of forces act on an object like the box in Fig. 4.17: EXECUTE The force F exerted by the worker has a positive x-­component S the weight of the object w—that is, the downward gravitational force ex- and zero y-component (so Fx = F = 20 N, Fy = 0); the normal force erted by the earth—and contact forces, which are forces exerted by other S n has zero x-component and an upward, positive y-component (so objects that the object in question is touching. Two objects are touching S nx = 0, ny = n); and the weight w has zero x-component and a down- the box—the worker’s hands and the floor—and both exertS contact forces ward, negative y-component (so wx = 0, wy = - w). From Newton’s on the box. The worker’s hands exert a horizontal force F of magnitude second law, Eqs. (4.7), gFx = F + 0 + 0 = F = 20 N = max 20 N. The floor exerts an upward supporting force; as in Section 4.1, we S gFy = 0 + n - w = may = 0 call this a normal force n because it acts perpendicular to the surface of contact. (Remember that “normal” is a synonym for “perpendicular.” It does not mean the opposite of “abnormal”!) If friction were present, the From the first equation, the x-component of acceleration is gFx floor would also exert a friction force on the box; we’ll ignore this here, since we’re told that friction is negligible. Figure 4.17 shows these three 20 N 20 kg # m>s2 external forces that act on the box. ax = = = = 0.50 m>s2 m 40 kg 40 kg Figure 4.17 Our sketch for this problem. EVALUATE The net external force is constant, so the acceleration in the The vertical components of the external forces + x-direction is also constant. If we know the initial position and veloc- on the box sum to zero, and the box has no ity of the box, we can find its position and velocity at any later time vertical acceleration. from the constant-acceleration equations of Chapter 2. ond law from Eqs. (4.7), gFy = may. Can you use this equation to To determine ax, we didn’t need the y-component of Newton’s sec- show that the magnitude n of the normal force in this situation is equal The horizontal components of to the weight of the box? the external forces on the box do not add to zero, so the box KEYCONCEPT In problems involving forces and acceleration, first has a horizontal acceleration. identify all of the external forces acting on an object, then choose a co- ordinate system. Find the vector sum of the external forces, and then set it equal to the mass of the object times the acceleration. EXAMPLE 4.5 Newton’s second law II: Determining force from acceleration WITH ARIATION PROBLEMS A waitress shoves a ketchup bottle with mass 0.45 kg to her right along Figure 4.18 Our sketch for this problem. a smooth, level lunch counter. The bottle leaves her hand moving at We draw one diagram showing the forces on the bottle 2.0 m>s, then slows down as it slides because of a constant horizontal and another one showing the bottle’s motion. friction force exerted on it by the countertop. It slides for 1.0 m before coming to rest. What are the magnitude and direction of the friction force acting on the bottle? IDENTIFY and SET UP This problem involves forces and acceleration (the slowing of the ketchup bottle), so we’ll use Newton’s second law to solve it. As in Example 4.4, we identify the external forces acting on the bottle and choose a coordinate system (Fig. 4.18). And as in 112    CHAPTER 4 Newton’s Laws of Motion S Example 4.4, we have a downward gravitational force w and an up- The negative sign shows that the net external force on the bottle is to- S ward normal force n exerted by the countertop. The countertop also ward the left. The magnitude of the friction force is f = 0.90 N. S exerts a friction force f ; this slows the bottle down, so its direction the +x-axis to the left in Fig. 4.18. You’ll find that gFx is equal to EVALUATE As a check on the result, try repeating the calculation with must be opposite the direction of the bottle’s velocity (see Fig. 4.12c). We choose the + x-axis to be in the direction that the bottle slides, and take the origin to be where the bottle leaves the waitress’s hand. +f = +0.90 N (because the friction force is now in the + x-direction), Our target variable is the magnitude f of the friction force. We’ll and again you’ll find f = 0.90 N. The answers for the magnitudes of find it by using the x-component of Newton’s second law from Eqs. forces don’t depend on the choice of coordinate axes! (4.7). We aren’t told the x-component of the bottle’s acceleration, ax , We didn’t write the y-component of Newton’s second law in but we know that it’s constant because the friction force that causes the this example. You should do this and show that it’s the same as in acceleration is constant. Hence we can use a constant-acceleration for- Example 4.4, so again the normal force and the weight have the same mula from Section 2.4 to calculate ax. We know the bottle’s initial and magnitude. final x-coordinates 1x0 = 0 and x = 1.0 m2 and its initial and final x-velocity 1v0x = 2.0 m>s and vx = 02, so the easiest equation to use is    CAUTION    Normal force and weight don’t always have the same Eq. (2.13), vx2 = v0x 2 + 2ax1x - x02. magnitude Be careful that you never assume automatically that the S S normal force n and the weight w have the same magnitude! Although EXECUTE We solve Eq. (2.13) for ax: that is the case in this example and the preceding one, we’ll see many vx 2 - v0x2 10 m>s2 2 - 12.0 m>s2 2 examples in Chapter 5 and later where the magnitude of the normal ax = = = - 2.0 m>s2 force is not equal to the weight. ❙ 21x - x02 211.0 m - 0 m2 The negative sign means that the bottle’s acceleration is toward the left KEYCONCEPT In problems involving forces in which you’re in Fig. 4.18, opposite to its velocity; this makes sense because the bottle given velocity, time, and>or displacement data, you’ll need to use the is slowing down. As in Example 4.4, neither the normal force nor the equations for motion with constant acceleration as well as Newton’s weight has an x-component. That means the net external force in the second law. x-direction is just the x-component -f of the friction force: gFx = - f = max = 10.45 kg21- 2.0 m>s22 = - 0.90 kg # m>s2 = -0.90 N Some Notes on Units A few words about units are in order. In the cgs metric system (not used in this book), the unit of mass is the gram, equal to 10-3 kg, and the unit of distance is the centimeter, equal Figure 4.19 Despite its name, the British to 10-2 m. The cgs unit of force is called the dyne: 1 dyne = 1 g # cm>s2 = 10-5 N unit of mass has nothing to do with the type of slug shown here. A common garden slug has a mass of about 15 grams, or about 10-3 slug. In the British system, the unit of force is the pound (or pound-force) and the unit of mass is the slug (Fig. 4.19). The unit of acceleration is 1 foot per second squared, so 1 pound = 1 slug # ft>s2 The official definition of the pound is 1 pound = 4.448221615260 newtons It is handy to remember that a pound is about 4.4 N and a newton is about 0.22 pound. Another useful fact: An object with a mass of 1 kg has a weight of about 2.2 lb at the earth?