F=ma From Problems and Solutions in Introductory Mechanics (Draft version, August 2014) David Morin,

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1 Chapter 4 F=a Fro Probles and Solutions in Introductory Mechanics (Draft version, August 2014) David Morin, [email protected] 4.1 Introduction Newton s laws In the preceding two chapters, we dealt with kineatics. We took the otions of objects as given and then looked at positions, velocities, and accelerations as functions of tie. We weren t concerned with the forces that caused the objects otions. We will now deal with dynaics, where the goal is to understand why objects ove the way they do. his chapter and the following ones will therefore be concerned with force, ass, energy, oentu, etc. he otion of any object is governed by Newton s three laws: First Law: A body oves with constant velocity (which ay be zero) unless acted on by a force. If you think hard about this law, it sees a bit circular because we haven t defined what a force is. But if you think harder, there is in fact soe content there. See Section 3.1 in Morin (2008) for a discussion of this. Second Law: he rate of change of the oentu of a body equals the force acting on the body. In cases where the ass of the body doesn t change (we ll deal with the ore general case in Chapter 6), this law becoes F = a. (4.1) his is a vector equation, so it is really three equations, naely F x = a x, F y = a y, and F z = a z. hird Law: Given two bodies A and B, if A exerts a force on B, then B exerts an equal and opposite force on A. As we ll see in Chapter 6, this law basically postulates conservation of oentu. here is a great deal of physical content in this law; it says that things don t happen in isolation by agic. Instead, if an object feels a force, then there ust be another object soewhere feeling the opposite force. he second law, F = a, is the one we ll get the ost ileage out of. he unit of force is called a newton (N), and fro F = a we see that 1 N = 1 kg /s 2. A few types of forces (gravitational, tension, noral, friction, and spring) coe up again and again, so let s take a look at each of these in turn. 68

2 4.1. INRODUCION 69 Gravity he gravitational force on an object near the surface of the earth is proportional to the ass of the object. More precisely, the force is g downward, where g = 9.8 /s 2. his g force is a special case of the ore general gravitational force we will encounter in Chapter 11. Substituting g for the force F in Newton s second law quickly gives g = a = a = g. hat is, all objects fall with the sae acceleration (in the absence of air resistance). ension If you pull on a rope, the rope pulls back on you with the sae force, by Newton s third law. he agnitude of this force is called the tension in the rope. As far as the direction of the tension goes, the question, At a given interior point in the rope, which way does the tension point? can t be answered. What we can say is that the tension at the given point pulls leftward on the point just to its right, and pulls rightward on the point just to its left. Equivalently, the tension pulls leftward on soeone holding the right end of the rope, and it pulls rightward on soeone holding the left end of the rope. Noral force Whereas a tension arises fro a aterial resisting being stretched, a noral force arises fro a aterial resisting being copressed. If you push leftward on the right face of a wooden block, then the noral force fro the block pushes rightward on your hand. And siilarly, the leftward force fro your hand is itself a noral force pushing on the block. In the case where you push inward on the ends of an object shaped like a rod, people soeties say that there is a negative tension in the rod, instead of calling it a noral force at the ends. But whatever nae you want to use, the rod pushes back on you at the ends. If instead of a rigid rod we have a flexible rope, then the rope can support a tension, but not a noral force. Friction he friction force between two objects is extreely coplicated on a icroscopic scale. But fortunately we don t need to understand what is going on at that level to get a rough handle on friction forces. o a good approxiation under ost circustances, we can say the following things about kinetic friction (where two objects are oving with respect to each other) and static friction (where two objects are at rest with respect to each other). Kinetic friction: If there is slipping between two objects, then to a good approxiation under non-extree conditions, the friction force is proportional to the noral force between the objects, with the constant of proportionality (called the coefficient of kinetic friction) labeled as µ k : F k = µ k N. (4.2) he friction force is independent of the contact area and relative speed. he direction of the friction force on a given object is opposite to the direction of the velocity of that object relative to the other object. Static friction: If there is no slipping between two objects, then to a good approxiation under non-extree conditions, the axiu value of the friction force is proportional to the noral force between the objects, with the constant of proportionality (called the coefficient of static friction) labeled as µ s : F s µ s N. (4.3) As in the kinetic case, the friction force is independent of the contact area. Note well the equality in Eq. (4.2) and the inequality in Eq. (4.3). Equation (4.3) gives only an upper liit on the static friction force. If you push on an object with a force that is saller

3 70 CHAPER 4. F=MA than µ s N, then the static friction force is exactly equal and opposite to your force, and the object stays at rest.1 But if you increase your force so that it exceeds µ s N, then the axiu friction force isn t enough to keep the object at rest. So it will ove, and the friction force will abruptly drop to the kinetic value of µ k N. (It turns out that µ k is always less than or equal to µ s ; see Proble 4.1 for an explanation why.) he expressions in Eqs. (4.2) and (4.3) will of course break down under extree conditions (large noral force, high relative speed, pointy shapes). But they work fairly well under noral conditions. Note that the coefficients of kinetic and static friction, µ k and µ s, are properties of both surfaces together. A single surface doesn t have a coefficient of friction. What atters is how two surfaces interact. Spring force o a good approxiation for sall displaceents, the restoring force fro a spring is proportional to the stretching distance. hat is, F = k x, where k is the spring constant. A large value of k eans a stiff spring; a sall value eans a weak spring. he reason why the F = kx relation is a good approxiation for sall displaceents in virtually any syste is explained in Proble If x is positive then the force F is negative; and if x is negative then F is positive. So F = kx does indeed describe a restoring force, where the spring always tries to bring x back to zero. he F = k x relation, known as Hooke s law, breaks down if x is too large, but we ll assue that it holds for the setups we re concerned with. he tension and noral forces discussed above are actually just special cases of spring forces. If you stand on a floor, the floor acts like a very stiff spring. he atter in the floor copresses a tiny aount, exactly the aount that akes the upward spring force (which we call a noral force in this case) be equal to your weight. We ll always assue that our springs are assless. Massive springs can get very coplicated because the force (the tension) will in general vary throughout the spring. In a assless spring, the force is the sae everywhere in it. his follows fro the reasoning in Proble 4.3(a). Centripetal force he centripetal force is the force that keeps an object oving in a circle. Since we know fro Eq. (3.7) that the acceleration for unifor (constant speed) circular otion points radially inward with agnitude a = v 2 /r, the centripetal force likewise points radially inward with agnitude F = a = v 2 /r. his force ight be due to the tension in a string, or the friction force acting on a car s tires as it rounds a corner, etc. Since v = rω, we can also write F as rω 2. he ter centripetal isn t the sae type of ter as the above gravity, tension, etc. descriptors, because the latter ters describe the type of force, whereas centripetal siply describes the direction of the force (radially inward). he word centripetal is therefore ore like the words eastward or downward, etc. For exaple, we ight say, he downward force is due to gravity, or he centripetal force is due to the tension in a string. he centripetal force is not a agical special new kind of force. It is siply one of the standard forces (or a cobination of the) that points radially inward, and whose agnitude we know always equals v 2 /r. Free-body diagras he F in Newton s second law in Eq. (4.1) is the total force on an object, so it is iportant to deterine what all the various forces are. he best way to do this is to draw a picture. he picture of an object that shows all of the forces acting on it is called a free-body diagra. More precisely, 1he friction force certainly can t be equal to µ s N in this case, because if you push rightward with a very sall force, then the leftward (incorrect) µ s N friction force would cause there to be a nonzero net force, which would hurl the object leftward back toward you!

4 4.1. INRODUCION 71 a free-body diagra shows all of the external forces (that is, forces due to other objects) acting on a given object. here are undoubtedly also internal forces acting within the object; each ato ight be pushing or pulling on the ato next to it. But these internal forces cancel in pairs (by Newton s third law), so they don t produce any acceleration of the object. Only external forces can do that. (We ll assue we have a rigid object, so that the distance between any two given points reains fixed.) In siple cases (for exaple, ones involving only one force), you can get away with not drawing a free-body diagra. But in ore coplicated cases (for exaple, ones involving forces pointing in various directions), a diagra is absolutely critical. A proble is often hopeless without a diagra, but trivial with one. he length of a force vector is technically a easure of the agnitude of the force. But when drawing a free-body diagra, the ain point is just to indicate all the forces that exist. In general we don t yet know the relative sizes, so it s fine to give all the vectors the sae length (unless it s obvious that a certain force is larger than another). Also, the exact location of each force vector isn t critical (at least in this chapter), although the ost sensible thing to do is to draw the vector near the place where the force acts. However, when we discuss torque in Chapter 7, the location of the force will be iportant. Note that due to Newton s third law, for every force vector that appears in the free-body diagra for one object, there is an opposite force vector that appears in the free-body diagra for another object. An exaple involving two blocks on a table is shown in Fig If a person applies a force F to the left block, then the two free-body diagras are shown (assue there is no friction fro the table). Note that the force pushing the right block rightward is only the noral force between the blocks, and not the applied force F. rue, the N force wouldn t exist without the F force, but the right block feels only the N force; it doesn t care about the original cause of N. F F N 1 g 1 2 N 2 g table (picture of setup) N 1 N 2 (free-body diagras) Figure 4.1 If we happened to be concerned also with the free-body diagra for the person applying the force F, then we would draw a force F acting leftward, along with a downward g gravitational force and an upward noral force fro the ground (and also probably a rightward friction force fro the ground). Likewise, the free-body diagra for the table would involve gravity and the N 1 and N 2 noral forces pointing downward, along with upward noral forces fro the ground at the bases of the legs. But if we re concerned only with the blocks, then all of this other inforation is irrelevant. If the direction of the acceleration of an object is known, it is often helpful to draw the acceleration vector in the free-body diagra. But you should be careful to draw this vector with a dotted line or soething siilar, so that you don t istake it for a force. Reeber that although F equals a, the quantity a is not a force; see the last section of this introduction. Atwood s achines he nae Atwood s achine is the ter used for any syste of pulleys, strings, and asses. Although a subset of these systes is certainly very useful in everyday life (a block and tackle enables you to lift heavy objects; see Proble 4.4), the ain reason for all the Atwood s probles in this chapter is siply that they re good practice for drawing free-body diagras and

5 72 CHAPER 4. F=MA applying F = a. An additional ingredient in solving any Atwood s proble is the so-called conservation of string relation. his is the condition that the length of any given string doesn t change. his constrains the otion of the various asses and pulleys. A few useful Atwood s facts that coe up again and again are derived in Proble 4.3. In this chapter we will assue that all strings and pulleys are assless. he four forces Having discussed any of the forces we see in everyday life, we should ake at least a brief ention of where all these forces actually coe fro. here are four known fundaental forces in nature: Gravitational: Any two asses attract each other gravitationally. We are quite failiar with the gravitational force due to the earth. he gravitational force between everydaysized objects is too sall to observe without sensitive equipent. But on the planetary scale and larger, the gravitational force doinates the other three forces. Electroagnetic: he single word electroagnetic is indeed the proper word to use here, because the electric and agnetic forces are two aspects of the sae underlying force. (In soe sense, the agnetic force can be viewed as a result of cobining the electric force with special relativity.) Virtually all everyday forces have their origin in the electric force. For exaple, a tension in a string is due to the electric forces holding the olecules together in the string. Weak: he weak force is responsible for various nuclear processes; it isn t too iportant in everyday life. Strong: he strong force is responsible for holding the protons and neutrons together in a nucleus. Without the strong force, atter as we know it wouldn t exist. But taking the existence of atter for granted, the strong force doesn t show up uch in everyday life. a is not a force! Newton s second law is F equals a, which says that a equals a force. Does this iply that a is a force? Absolutely not. What the law says is this: Write down the su of all the forces on an object, and also write down the ass ties the acceleration of the object. he law then says that these two quantities have the sae value. his is what a physical law does; it says to take two things that aren t obviously related, and then deand that they are equal. In a siple freefall setup, the F = a equation is g = a, which tells us that a equals g. But just because a = g, this doesn t ean that the two side of g = a represent the sae type of thing. he left side is a force, the right side is a ass ties an acceleration. So when drawing a free-body diagra, you should not include a as one of the forces. If you do, you will end up double counting things. However, as entioned above, it is often helpful to indicate the acceleration of the object in the free-body diagra. Just be careful to distinguish this fro the forces by drawing it with a dotted line. 4.2 Multiple-choice questions 4.1. wo people pull on opposite ends of a rope, each with a force F. he tension in the rope is (a) F/2 (b) F (c) 2F

6 4.2. MULIPLE-CHOICE QUESIONS You accelerate the two blocks in Fig. 4.2 by pushing on the botto block with a force F. he top block oves along with the botto block. What force directly causes the top block to accelerate? (a) the noral force between the blocks (b) the friction force between the blocks (c) the gravitational force on the top block (d) the force you apply to the botto block 4.3. hree boxes are pushed with a force F across a frictionless table, as shown in Fig Let N 1 be the noral force between the left two boxes, and let N 2 be the noral force between the right two boxes. hen F M Figure 4.2 F 2 3 Figure 4.3 (a) F = N 1 = N 2 (b) F + N 1 = N 2 (c) F > N 1 = N 2 (d) F < N 1 < N 2 (e) F > N 1 > N wo blocks with asses 2 kg and 1 kg lie on a frictionless table. A force of 3 N is applied as shown in Fig What is the noral force between the blocks? (a) 0 (b) 0.5 N (c) 1 N (d) 2 N (e) 3 N 4.5. In the syste shown in Fig. 4.5, the ground is frictionless, the blocks have ass and 2, and the string connecting the is assless. If you accelerate the syste to the right, as shown, the tension is the sae everywhere throughout the string connecting the asses because (a) the string is assless (b) the ground is frictionless (c) the ratio of the asses is 2 to 1 (d) the acceleration of the syste is nonzero (e) he tension is the sae throughout any string; no conditions are necessary. 3 N 2 kg 1 kg Figure Figure You are in a plane accelerating down a runway during takeoff, and you are holding a pendulu (say, a shoe hanging fro a shoelace). he string of the pendulu (a) hangs straight downward (b) hangs downward and forward, because the net force on the pendulu ust be zero (c) hangs downward and forward, because the net force ust be nonzero (d) hangs downward and backward, because the net force ust be zero (e) hangs downward and backward, because the net force ust be nonzero 4.7. When you stand at rest on a floor, you exert a downward noral force on the floor. Does this force cause the earth to accelerate in the downward direction? (a) Yes, but the earth is very assive, so you don t notice the otion. (b) Yes, but you accelerate along with the earth, so you don t notice the otion. (c) No, because the noral force isn t a real force. (d) No, because you are also pulling on the earth gravitationally. (e) No, because there is also friction at your feet.

7 74 CHAPER 4. F=MA 4.8. If you stand at rest on a bench, the bench exerts a noral force on you, equal and opposite to your weight. Which force is related to this noral force by Newton s third law? (a) the gravitational force fro the earth on you (b) the gravitational force fro you on the earth (c) the noral force fro you on the bench (d) none of the above 4.9. he driver of a car steps on the gas, and the car accelerates with acceleration a. When writing down the horizontal F = a equation for the car, the F acting on the car is (a) the noral force between the tires and the ground (b) the friction force between the tires and the ground (c) the force between the driver s foot and the pedal (d) the energy obtained by burning the gasoline (e) the backward friction force that balances the forward a force he static friction force between a car s tires and the ground can do all of the following except (a) speed the car up (b) slow the car down (c) change the car s direction (d) It can do all of the above things A car is traveling forward along a road. he driver wants to arrange for the car s acceleration to point diagonally backward and leftward. he driver should F f g N (a) turn right and accelerate (b) turn right and brake (c) turn left and accelerate (d) turn left and brake A block is at rest on a plane inclined at angle. he forces on it are the gravitational, noral, and friction forces, as shown in Fig hese are not drawn to scale. Which of the following stateents is always true, for any? (a) g N and g F f Figure 4.6 (b) g N and g F f (c) F f = N (d) F f + N = g (e) F f > N if µ s > 1

8 4.2. MULIPLE-CHOICE QUESIONS A block sits on a plane, and there is friction between the block and the plane. he plane is accelerated to the right. If the block reains at the sae position on the plane, which of the following pictures ight show the free-body diagra for the block? (All of the vectors shown are forces.) (a) F f N (b) N (c) F f N a a g g g (d) F f (e) (zero force) g A block with ass sits on a frictionless plane inclined at angle, as shown in Fig If the plane is accelerated to the right with the proper acceleration that causes the block to reain at the sae position with respect to the plane, what is the noral force between the block and the plane? (a) g (b) g sin (c) g/ sin (d) g cos (e) g/ cos a Figure A bead is arranged to ove with constant speed around a hoop that lies in a vertical plane. he agnitude of the net force on the bead is (a) largest at the botto (b) largest at the top (c) largest at the side points (d) the sae at all points A toy race car travels through a loop-the-loop (a circle in a vertical plane) on a track. Assuing that the speed at the top of the loop is above the threshold to reain in contact with the track, the car s acceleration at the top is (a) downward and larger than g (b) downward and saller than g (c) zero (d) upward and saller than g (e) upward and larger than g

9 76 CHAPER 4. F=MA A plane in a holding pattern is flying in a horizontal circle at constant speed. Which of the following free-body diagras best illustrates the various forces acting on the plane at the instant shown? (See Proble 4.20 for a quantitative treatent of this setup.) (a) (b) (c) (d) (e) A ass hangs fro a spring and oscillates vertically. he top end of the spring is attached to the top of a box, and the box is placed on a scale, as shown in Fig he reading on the scale is largest when the ass is (a) at its axiu height scale (b) at its iniu height Figure 4.8 (c) at the idpoint of its otion (d) All points give the sae reading A spring with spring constant k hangs vertically fro a ceiling, initially at its relaxed length. You attach a ass to the end and bring it down to a position that is 3g/k below the initial position. You then let go. What is the upward acceleration of the ass right after you let go? k k (a) 0 (b) g (c) 2g (d) 3g (e) 4g l Figure 4.9 l wo springs both have spring constant k and relaxed length zero. hey are each stretched to a length l and then attached to two asses and a wall, as shown in Fig he asses are siultaneously released. Iediately afterward, the agnitudes of the accelerations of the left and right asses are, respectively, (a) 2kl/ and kl/ (b) kl/ and 2kl/ (c) kl/ and kl/ (d) 0 and 2kl/ (e) 0 and kl/ A ass 2 suspended fro a given spring causes it to stretch relative to its relaxed length. he ass and the spring are then each cut into two identical pieces and connected as shown in Fig Is the botto of the lower ass higher than, lower than, or at the sae height as the botto of the original ass? (his one takes a little thought.) not drawn to scale Figure 4.10 (a) higher (b) lower (c) sae height

10 4.3. PROBLEMS What is the conservation-of-string relation for the Atwood s achine shown in Fig. 4.11? All of the accelerations are defined to be positive upward. (a) a 3 = 2(a 1 + a 2 ) (b) a 3 = (a 1 + a 2 ) (c) a 3 = (a 1 + a 2 )/2 (d) a 3 = (a 1 + a 2 )/4 (e) a 3 = 2a What is the conservation-of-string relation for the Atwood s achine shown in Fig. 4.12? All of the accelerations are defined to be positive upward. a 1 a 2 Figure 4.11 a 3 (a) a 3 = a 1 a 2 (b) a 3 = 2a 1 2a 2 (c) a 3 = 4a 2 (d) 2a 3 = a 1 a 2 (e) 4a 3 = a 1 a 2 a 1 a 2 a 3 Figure What is the conservation-of-string relation for the Atwood s achine shown in Fig. 4.13? All of the accelerations are defined to be positive upward. (a) a 1 = a 3 (b) a 1 = a 3 (c) a 2 = (a 1 + a 3 )/2 (d) a 2 = (a 1 + a 3 ) (e) a 2 = 2(a 1 + a 3 ) a 1 a 2 a 3 Figure Probles he first five probles are foundational probles Coefficients of friction Explain why the coefficient of static friction, µ s, ust always be at least as large as the coefficient of kinetic friction, µ k Cutting a spring in half A spring has spring constant k. If it is cut in half, what is the spring constant of each of the resulting shorter springs? 4.3. Useful Atwood s facts In the Atwood s achine shown in Fig. 4.14(a), the pulleys and strings are assless (as we will assue in all of the Atwood s probles in this chapter). Explain why (a) the tension is the sae throughout the long string, as indicated, (b) the tension in the botto string is twice the tension in the long string, as indicated, and (c) the acceleration of the right ass is negative twice the acceleration of the left ass. Also, in Fig. 4.14(b), explain why (d) the acceleration of the left ass equals negative the average of the accelerations of the right two asses.

11 78 CHAPER 4. F=MA (a) (b) 2 Figure Block and tackle (a) What force on the rope ust be exerted by the person in Fig. 4.15(a) in order to hold up the block, or equivalently to ove it upward at constant speed? he rope wraps twice around the top of the top pulley and the botto of the botto pulley. (Assue that the segent of rope attached to the center of the top pulley is essentially vertical.) (b) Now consider the case where the person (with ass ) stands on the block, as shown in Fig. 4.15(b). What force is now required? (a) (b) M M Figure Figure Equivalent ass In Fig you support the pulley syste, with your hand at rest. If you have your eyes closed and think that you are instead supporting a single ass M at rest, what is M in ters of 1 and 2? Is M siply equal to 1 + 2? 1 2 Figure 4.17 M he following eight probles involve Atwood s achines. his large nuber of Atwood s probles shouldn t be taken to iply that they re terribly iportant in physics (they re not). Rather, they are included here because they provide good practice with F = a Atwood s 1 Consider the Atwood s achine shown in Fig he asses are held at rest and then released. In ters of 1 and 2, what should M be so that 1 doesn t ove? What relation ust hold between 1 and 2 so that such an M exists?

12 4.3. PROBLEMS Atwood s 2 Consider the Atwood s achine shown in Fig Masses of and 2 lie on a frictionless table, connected by a string that passes around a pulley. he pulley is connected to another ass of 2 that hangs down over another pulley, as shown. Find the accelerations of all three asses. (top view) (side view) 2 2 Figure Atwood s 3 Consider the Atwood s achine shown in Fig. 4.19, with asses,, and 2. Find the acceleration of the ass Atwood s 4 In the Atwood s achine shown in Fig. 4.20, both asses are. Find their accelerations Atwood s 5 Consider the triple Atwood s achine shown in Fig What is the acceleration of the rightost ass? Note: he ath isn t as bad as it ight see at first. You should take advantage of the fact that any of your F = a equations look very siilar Figure 4.19 Figure 4.20 Figure Atwood s 6 Consider the Atwood s achine shown in Fig he iddle ass is glued to the long string. Find the accelerations of all three asses, and also the tension everywhere in the long string Atwood s 7 In the Atwood s achine shown in Fig. 4.23, both asses are. Find their accelerations Atwood s 8 In the Atwood s achine shown in Fig. 4.24, both asses are. he two strings that touch the center of the left pulley are both attached to its axle. Find the accelerations of the asses.

13 80 CHAPER 4. F=MA Figure 4.22 Figure 4.23 Figure M a (side view) Figure No relative otion All of the surfaces in the setup in Fig are frictionless. You push on the large block and give it an acceleration a. For what value of a is there no relative otion aong the asses? Slipping blocks A block with ass sits on top of a block with ass 2 which sits on a table. he coefficients of friction (both static and kinetic) between all surfaces are µ s = µ k = 1. A string is connected to each ass and wraps halfway around a pulley, as shown in Fig You pull on the pulley with a force of 6g. (a) Explain why the botto block ust slip with respect to the table. Hint: Assue that it doesn t slip, and show that this leads to a contradiction. (b) Explain why the top block ust slip with respect to the botto block. (Sae hint.) (c) What is the acceleration of your hand? µ=1 2 F = 6g Figure 4.26 M Figure Block and wedge A block with ass M rests on a wedge with ass and angle, which lies on a table, as shown in Fig All surfaces are frictionless. he block is constrained to ove vertically by eans of a wall on its left side. What is the acceleration of the wedge? Up and down a plane A block with ass is projected up along the surface of a plane inclined at angle. he initial speed is v 0, and the coefficients of both static and kinetic friction are equal to 1. he block reaches a highest point and then slides back down to the starting point. (a) Show that in order for the block to in fact slide back down (instead of reaining at rest at the highest point), ust be greater than 45. (b) Assuing that > 45, find the ties of the up and down otions. (c) Assuing that > 45, is the total up and down tie longer or shorter than the total tie it would take (with the sae initial v 0 ) if the plane were frictionless? Or does the answer to this question depend on what is? (he solution to this gets a little essy.)

14 4.3. PROBLEMS Rope in a tube A rope is free to slide frictionlessly inside a circular tube that lies flat on a horizontal table. In part (a) of Fig. 4.28, the rope oves at constant speed. In part (b) of the figure, the rope is at rest, and you pull on its right end to give it a tangential acceleration. What is the direction of the net force on the rope in each case? (a) v (b) at rest pull rope (top views) Figure Circling bucket You hold the handle of a bucket of water and swing it around in a vertical circle, keeping your ar straight. If you swing it around fast enough, the water will stay inside the bucket, even at the highest point where the bucket is upside down. What, roughly, does fast enough here ean? (You can specify the axiu tie of each revolution.) Make whatever reasonable assuptions you want to ake for the various paraeters involved. You can work in the approxiation where the speed of the bucket is roughly constant throughout the otion Banking an airplane A plane in a holding patter flies at speed v in a horizontal circle of radius R. At what angle should the plane be banked so that you don t feel like you are getting flung to the side in your seat? At this angle, what is your apparent weight (that is, what is the noral force fro the seat)? Breaking and turning You are driving along a horizontal straight road that has a coefficient of static friction µ with your tires. If you step on the brakes, what is your axiu possible deceleration? What is it if you are instead traveling with speed v around a bend with radius of curvature R? Circle of rope A circular loop of rope with radius R and ass density λ (kg/) lies on a frictionless table and rotates around its center, with all points oving at speed v. What is the tension in the rope? Hint: Consider the net force on a sall piece of rope that subtends an angle d Cutting the string A ass is connected to the end of a assless string of length l. he top end of the string is attached to a ceiling that is a distance l above the floor. Initial conditions have been set up so that the ass swings around in a horizontal circle, with the string always aking an angle with respect to the vertical, as shown in Fig If the string is cut, what horizontal distance does the ass cover between the tie the string is cut and the tie the ass hits the floor? l Figure 4.29 l

15 82 CHAPER 4. F=MA l v Figure 4.30 R ω Circling around a cone A ass is attached by a assless string of length l to the tip of a frictionless cone, as shown in Fig he half-angle at the vertex of the cone is. If the ass oves around in a horizontal circle at speed v on the cone, find (a) the tension in the string, (b) the noral force fro the cone, and (c) the axiu speed v for which the ass stays in contact with the cone Penny in a dryer Consider a clothes dryer with radius R, which spins with angular frequency ω. (In other words, consider a cylinder that spins with angular frequency ω around its axis, which is oriented horizontally.) A sall object, such as a penny, is in the dryer. he penny rotates along with the dryer and gets carried upward, but eventually loses contact with the dryer and sails through the air (as clothes do in a dryer), eventually coing in contact with the dryer again. Assue for siplicity that the coefficient of friction is very large, so that the penny doesn t slip with respect to the dryer as long as the noral force is nonzero. Let s assue that you want the trajectory of the penny to look like the one shown in Fig. 4.31, starting and ending at diaetrically opposite points. In order for this to happen, where ust the penny lose contact with the dryer? (Give the angle with respect to the vertical.) What ust ω be, in ters of g and R? (side view) Figure Multiple-choice answers 4.1. b he tension is siply F. A coon error is to double F (because there are two people pulling) and say that the tension is 2F. his is incorrect, because every piece of the rope pulls on the piece to its right with a force F, and also on the piece to its left with a force F. his coon value is the tension. Reark: If you try to change the setup by replacing one person with a wall, then the wall still pulls on the rope with a force F (assuing that the other person still does), so the setup hasn t actually changed at all. If you instead reove one person (say, the left one) and replace her with nothing, then the setup has certainly changed. he reaining (right) person will accelerate the rope rightward, and the tension will vary over the length (assuing that the rope has ass). Points closer to the left end don t have as uch ass to their left that they need to accelerate, so the tension is saller there b Friction is the horizontal force that acts on the top block. Reark: he friction force wouldn t exist without the noral force between the blocks, which in turn wouldn t exist without the gravitational force on the top block. But these forces are vertical and therefore can t directly cause the horizontal acceleration of the top block. Likewise, the friction force wouldn t exist if you weren t pushing on the botto block. But your force isn t what directly causes the acceleration of the top block; if the surfaces are greased down, then you can push on the botto block all you want, and the top block won t ove e he three boxes all have the sae acceleration; call it a. hen the force F equals F = (6)a because this is the force that accelerates all three boxes, which have a total ass of 6. Siilarly, N 1 = (5)a because N 1 is the force that accelerates the right two boxes. And N 2 = (3)a because N 2 is the force that accelerates only the right box. herefore, F > N 1 > N 2. As a double check, the net force on the iddle block is N 1 N 2 = 5a 3a = 2a, which is correctly (2)a c he acceleration of the syste is a = F/ = (3 N)/(3 kg) = 1 /s 2. he noral force N on the 1 kg block is what causes this block to accelerate at 1 /s 2, so N ust be given by N = a = (1 kg)(1 /s 2 ) = 1 N a he assless nature of the string iplies that the tension is the sae everywhere throughout it, because if the tension varied along the length, then there would exist a

16 4.4. MULIPLE-CHOICE ANSWERS 83 assless piece that had a net force acting on it, yielding infinite acceleration. Conversely, if the string had ass, then the tension would have to vary along it, so that there would be a net force on each little assive piece, to cause the acceleration (assuing the acceleration is nonzero). So none of the other answers can be the reason why the tension is the sae everywhere along the string e he pendulu is accelerating forward (as is everything else in the plane), so there ust be a forward net force on it. If the pendulu hangs downward and backward, then the tension force on the pendulu s ass is upward and forward. he upward coponent cancels the gravitational force (the weight), and the forward (uncanceled) coponent is what causes the forward acceleration. Reark: If the pendulu s ass is, then fro the above reasoning, the vertical coponent of the tension is g, and the horizontal coponent is a (where a is the acceleration of the plane and everything in it). So if the string akes an angle with the vertical, then tan = a/g = a = g tan. You can use this relation to deduce your acceleration fro a easureent of the angle. Going in the other direction, a typical plane ight have a takeoff acceleration of around 2.5 /s 2, in which case we can deduce what is: tan = a/g 1/4 = 15. For coparison, a typical car ight be able to go fro 0 to 60 ph (27 /s) in 7 seconds, which iplies an acceleration of about 4 /s 2 and an angle of = 22. A ore extree case is a fighter jet taking off fro an aircraft carrier. With the help of a catapult, the acceleration can be as large as 3g 30 /s. A pendulu in the jet would therefore hang at an angle of tan 1 (3) 72, which is ore horizontal than vertical. In the accelerating reference frae of the jet (accelerating fraes are the subject of Chapter 12, so we re getting ahead of ourselves here), the direction of the hanging pendulu defines downward. So the pilot effectively lives in a world where gravity points diagonally downward and backward at an angle of 72 with respect to the vertical. he direction of the jet s forward otion along the runway is therefore nearly opposite to this downward direction (as opposed to being roughly perpendicular to downward in the case of a passenger airplane). he jet pilot will therefore have the sensation that he is flying upward, even though he is actually oving horizontally along the runway. A possible dangerous consequence of this sensation is that if it is nighttie and there are inial visual cues, the pilot ay istakenly try to correct this error by turning downward, causing the jet to crash into the ocean d he noral force fro you on the earth is equal and opposite to the gravitational force fro you on the earth. (Yes, you pull on the earth, just as it pulls on you.) So the net force on the earth is zero, and it therefore doesn t accelerate. Reark: Siilarly, you also don t accelerate, because the noral force fro the earth on you is equal and opposite to the gravitational force fro the earth on you. So the net force on you is zero. his wouldn t be the case if instead of standing at rest, you jup upward. You are now accelerating upward (while your feet are in contact with the ground). And consistent with this, the upward noral force fro the earth on you is larger than the downward gravitational force fro the earth on you. So the net force on you is upward. In contrast, after you leave the ground the noral force drops to zero, so the downward gravitational force is all there is, and you accelerate downward c Newton s third law says that the forces that two bodies exert on each other are equal in agnitude and opposite in direction. he given force is the (upward) noral force fro the bench on you. he two objects here are the bench and you, so the force that is related by the third law ust be the (downward) noral force fro you on the bench. Siilarly, choices (a) and (b) are a third-law pair. Reark: he earth isn t relevant at all in the third-law stateent concerning you and the bench. Of course, the gravitational force fro the earth on you (that is, your weight) is related to the given noral force fro the bench on you, but this relation does not involve the third law. It involves the second law. More precisely, the second law says that since you are at rest (and hence not accelerating), the total force on you ust be zero. So the downward force fro the earth on you (your weight) ust be equal and opposite to the upward noral force fro the bench on you. Note that since three objects (earth, you, bench) were entioned in the preceding stateent, there is no way

17 84 CHAPER 4. F=MA that it can be a third-law stateent, because such a stateent ust involve only two objects. Just because two forces are equal and opposite, this doesn t ean they are related by the third law b he friction force is the horizontal force that akes the car accelerate; you won t go anywhere on ice. he other choices are incorrect because: (a) the noral force is vertical, (c) the force applied to the pedal is an internal force within the car, and only external forces appear in F = a (and besides, the force on the pedal is far saller than the friction between the tires and the ground), (d) energy isn t a force, and (e) the friction force doesn t point backward, and a isn t a force! (See the discussion on page 72.) d When you step on the gas, the friction force speeds you up (if you are on ice, you won t go anywhere). When you hit the brakes, the friction force slows you down (if you are on ice, you won t slow down). And when you turn the steering wheel, the friction force is the centripetal force that causes you to ove in the arc of a circle as you change your direction (if you are on ice, your direction won t change). Reark: Note that unless you are skidding (which rarely happens in everyday driving), the friction force between the tires and the ground is static and not kinetic. he point on a tire that is instantaneously in contact with the ground is instantaneously at rest. (he path traced out by a point on a rolling wheel is known as a cycloid, and the speed of the point is zero where it touches the ground.) If this weren t the case (that is, if we were perpetually skidding in our cars), then we would need to buy new tires every week, and we would be listening constantly to the sound of screeching tires d Braking will yield a backward coponent of the acceleration, and turning left will yield a leftward coponent. his leftward coponent is the centripetal acceleration for the circular arc (at least locally) that the car is now traveling in. F f N Reark: Depending on what you do with the gas pedal, brake, and steering wheel, the total acceleration vector (or equivalently, the total force vector) can point in any horizontal direction. he acceleration can have a forward or backward coponent, depending on whether you are stepping on the gas or the brake. And it can have a coponent to either side if the car is turning. he relative size of these coponents is arbitrary, so the total acceleration vector can point in any horizontal direction. g Figure b he total acceleration (and hence total force) is zero. In Fig we have broken the g force into its coponents parallel are perpendicular to the plane, and we have drawn the forces to scale. Zero net force perpendicular to the plane gives N = g cos, which iplies g N. And zero net force along the plane gives F f = g sin, which iplies g F f. he 0 liit gives a counterexaple to choices (a), (c), and (e), because F f 0 when 0. Choice (d) would be alost true if we were talking about vectors (the correct stateent would be F f + N + g = 0). But we re dealing with agnitudes here, so choice (d) is equivalent to sin + cos = 1, which isn t true c he gravitational force exists, as does the noral force (to keep the block fro falling though the plane). he friction force ight or ight not exist (and it ight point in either direction). So the correct answer ust be (a), (b), or (c). But a isn t a force, so we re left with choice (c) as the only possibility. If you want to draw the acceleration vector in a free-body diagra, you ust draw it differently (say, with a dotted line) to signify that it isn t a force! Reark: Since we are assuing that the block reains at the sae position on the plane, it has a nonzero a x but a zero a y. Given a x, the (positive) horizontal coponent of N plus the (positive or negative or zero) horizontal coponent of F f ust equal a x. (F f will point down along the plane if a x is larger than a certain value; as an exercise, you can deterine this value.) And zero a y eans that the upward coponents of N and F f ust balance the downward g force.

18 4.4. MULIPLE-CHOICE ANSWERS e Since the plane is frictionless, the only forces acting on the block are gravity and the noral force, as shown in Fig Since the block s acceleration is horizontal, the vertical coponent of the noral force ust equal g, to yield zero net vertical force. he right triangle then iplies that N = g/ cos. N g Liits: If = 0 then N = g. his akes sense, because the plane is horizontal and the noral force siply needs to balance the weight g. If 90 then N. his akes sense, because a needs to be huge to keep the block fro falling. g Reark: Consider instead the case where there is friction between the block and the plane, and where the entire syste is static. hen the noral force takes on the standard value of N = g cos. his is ost easily derived fro the fact that there is no acceleration perpendicular to the plane, which then iplies that N is equal to the coponent of gravity perpendicular to the plane; see Fig his should be contrasted with the original setup, where g was equal to a coponent (the vertical coponent) of N. In any case, the noral force doesn t just agically turn out to be g cos or g/ cos or whatever. It (or any other force) is deterined by applying F = a d he agnitude of the radial acceleration is v 2 /R, which is the sae at all points because v is constant. And the tangential acceleration is always zero because again v is constant. So the net force always points radially inward with constant agnitude v 2 /R. he vertical orientation of the hoop is irrelevant in this question, given that the speed is constant. he answer would be the sae if the hoop were horizontal. Figure 4.33 F friction N g Figure 4.34 Reark: Be careful not to confuse the total force on the bead with the noral force fro the hoop. he noral force does depend on the position of the bead. It is largest at the botto of the hoop, because there the radial F = a equation (with inward taken to be positive for both N and a) is N bot g = v2 R = N bot = v2 R + g. (4.4) At the top of the hoop, the radial F = a equation (with inward again taken to be positive for both N and a) is N top + g = v2 = N top = v2 g. (4.5) R R If this is negative (if v is sall), it just eans that the noral force actually points radially outward (that is, upward) a In the threshold case where the car barely doesn t stay in contact with the track at the top, the noral force N is zero, so the car is in freefall (while oving sideways). he downward acceleration is therefore g. If the speed is above the threshold value, then N is nonzero. he total downward force, F = g + N, is therefore larger than the g due to gravity. he downward acceleration, which is F/ = g + N/, is therefore larger than g. Reark: It isn t necessary to ention the v 2 /R expression for the centripetal acceleration in this proble. But if you want to write down the radial F = a equation at the top of the loop, you can show as an exercise that the iniu speed required to barely aintain contact with the track at the top of the loop (that is, to ake N 0) is v = gr c Gravity acts downward. he force fro the air ust have an upward coponent to balance gravity (because there is zero vertical acceleration), and also a radially inward coponent to provide the nonzero centripetal acceleration. So the net force fro the air points in a diagonal direction, upward and leftward. he gravitational and air forces are the only two forces, so the correct answer is (c). Reark: Choices (d) and (e) are incorrect because although they have the correct gravitational and air forces, they have an incorrect additional horizontal force. Reeber that a = v 2 /r is not a force (see the discussion on page 72), so (d) can t be correct. Choice (e) would be correct if we were working in an accelerating frae and using fictitious forces. But we won t touch accelerating fraes until Chapter 12.

19 86 CHAPER 4. F=MA b At the botto of the otion, the upward force fro the spring on the ass is axiu (because the spring is stretched axially there), which eans that the downward force fro the spring on the box is axiu (because the spring exerts equal and opposite forces at its ends). his in turn eans that the upward force fro the scale on the box is axiu (because the net force on the box is always zero, because it isn t accelerating). And this force is the reading on the scale c he forces on the ass are the spring force, which is k y = k(3g/k) = 3g upward, and the gravitational force, which is g downward. he net force is therefore 2g upward, so the acceleration is 2g upward e he left ass feels forces fro both springs, so it feels equal and opposite forces of kl. Its acceleration is therefore zero. he right ass feels only the one force of kl fro the right spring (directed leftward). Its acceleration is therefore kl/ leftward. he aount that the left spring is stretched is copletely irrelevant as far as the right ass goes. Also, as far as the left ass goes, it is irrelevant that the far ends of the springs are attached to different things (the iovable wall and the ovable right ass), at least right at the start a Let the stretching, relative to the equilibriu position, of the spring in the original scenario be 2l. hen the given inforation tells us that half of the spring stretches by l when its tension is 2g (because the whole spring stretches by 2l when its tension is 2g, and the tension is the sae throughout). herefore, the top half of the spring in the second scenario is stretched by l, because it is holding up a total ass of 2 below it (this ass is split into two pieces, but that is irrelevant as far as the top spring is concerned). But the botto half of the spring is stretched by only l/2, because it is holding up only a ass below it (and x F by Hooke s law). he total stretch is therefore l + l/2 = 3l/2. his is less than 2l, so the desired answer is higher. Reark: his answer of higher can be ade a little ore believable by looking at soe liiting cases. Consider a ore general version of the second scenario, where we still cut the spring into equal pieces, but we now allow for the two asses to be unequal (although they ust still add up to the original ass). Let the top and botto asses be labeled t and b, and let the original ass be M. In the liit where t = 0 and b = M, we siply have the original setup, so the answer is sae height. In the liit where t = M and b = 0, we have a ass M hanging fro a shorter spring. But a shorter spring has a larger spring constant (see Proble 4.2), which eans that it stretches less. So the answer is higher. herefore, since the t = M/2 and b = M/2 case presented in the proble lies between the preceding two cases with answers of sae height and higher, it is reasonable to expect that the answer to the original proble is higher. You can also ake a siilar arguent by splitting the original ass into two equal pieces, but now allowing the spring to be cut into unequal pieces b (a 1 + a 2 )/2 is the acceleration of the botto pulley, because the average height of the botto two asses always stays the sae distance below the botto pulley. And the acceleration of the botto pulley equals the acceleration of the iddle pulley, which in turn equals a 3 /2. his is true because if the iddle pulley goes up by d, then 2d of string disappears above it, which ust therefore appear above 3 ; so 3 goes down by 2d. Putting all this together yields a 3 = (a 1 + a 2 ). See the solution to Proble 4.3 for ore discussion of these concepts b If the left ass goes up by y 1, there is 2y 1 less string in the segents above it. Likewise, if the iddle ass goes up by y 2, there is 2y 2 less string in the segents above it. All of this issing string ust appear above the right ass, which therefore goes down by 2y 1 + 2y 2. So y 3 = 2y 1 2y 2. aking two tie derivatives gives choice (b) d If the left ass goes up by y 1, then 2y 1 worth of string disappears fro the left region. Siilarly, if the right ass goes up by y 3, then 2y 3 worth of string disappears fro the right region. his 2y 1 + 2y 3 worth of string ust appear in the iddle region. It gets

20 4.5. PROBLEM SOLUIONS 87 divided evenly between the two segents there, so the iddle ass goes down by y 1 + y 3. Hence y 2 = (y 1 + y 3 ). aking two tie derivatives gives choice (d). 4.5 Proble solutions 4.1. Coefficients of friction Assue, in search of a contradiction, that µ k is larger than µ s. Iagine pushing a block that rests on a surface. If the applied force is saller than µ s N (such as the force indicated by the point A on the scale in Fig. 4.35), then nothing happens. he static friction exactly cancels the applied force, and the block just sits there. However, if the applied force lies between µ s N and µ k N (such as the force indicated by the point B), then we have proble. On one hand, the applied force exceeds µ s N (which is the axiu static friction force, by definition), so the block should ove. But on the other hand, the applied force is saller than the kinetic friction force, µ k N, so the block shouldn t ove. Basically, as soon as the block oves even the slightest infinitesial aount (at which point the kinetic friction force becoes the relevant force), it will decelerate and stop because the kinetic friction force wins out over the applied force. So it actually never oves at all, even for the applied force represented by point B. his eans, by the definition of µ s, that µ s N is actually located higher than point B. he above contradiction (where the block both does ove and doesn t ove) will arise unless µ s N µ k N. So we conclude that it ust be the case that µ s µ k. here then exists no point B below µ k N and above µ s N. In the real world, µ s is rarely larger than twice µ k. In soe cases the two are essentially equal, with µ s being a hair larger. B A F µ k N µ s N his situation with µ k > µ s leads to a contradiction Figure Cutting a spring in half ake the half-spring and stretch it a distance x. Our goal is to find the force F that it exerts on soething attached to an end; the spring constant is then given by k 1/2 = F/x (ignoring the inus sign in Hooke s law; we ll just deal with the agnitude). Iagine taking two half-springs that are each stretched by x, lying along the sae line, and attaching the right end of one to the left end of the other. We are now back to our original spring with spring constant k. And it is stretched by a distance 2x, so the force it exerts is k(2x). But this is also the force F that each of the half-springs exerts, because we didn t change anything about these springs when we attached the together. So the desired value of k 1/2 is k 1/2 = F/x = k(2x)/x = 2k. Reark: he sae type of reasoning shows (as you can verify) that if we have a spring and then cut off a piece with a length that is a factor f ties the original (so f = 1/2 in the above case), then the spring constant of the new piece is 1/ f ties the spring constant of the original. Actually, the reasoning works only in the case of rational nubers f. (It works with f of the for f = 1/N, where N is an integer. And you can show with siilar reasoning that it also works with f of the for f = N, which corresponds to attaching N springs together in a line. Cobining these two results yields all of the rational nubers.) But any real nuber is arbitrarily close to a rational nuber, so the result is true for any factor f. Basically, a shorter spring is a stiffer spring, because for a given total aount of stretching, a centieter of a shorter spring ust stretch ore than a centieter of a longer spring. So the tension in the forer centieter (which is the sae as the tension throughout the entire shorter spring) is larger than the tension in the latter centieter Useful Atwood s facts (a) Assue, in search of a contradiction, that the tension varies throughout the string. hen there exists a segent of the string for which the tension is different at the two ends. his eans that there is a nonzero net force on the segent. But the string

21 88 CHAPER 4. F=MA is assless, so the acceleration of this segent ust be infinite. Since this can t be the case, the tension ust in fact be the sae throughout the string. In short, any assless object ust always have zero net force acting on it. (Ignoring photons and such!) Reark: If there is friction between a string and a pulley, and if the pulley has a nonzero oent of inertia (a topic covered in Chapter 7), then the tension in the string will vary, even if the string is assless. But the stateent, Any assless object ust always have zero net force acting on it, still holds; there is now a nonzero friction force fro the pulley acting on a segent of the string touching it. So the net force on the segent is still zero. (b) he reasoning here is siilar to the reasoning in part (a). Since the left pulley is assless, the net force on it ust be zero. If we draw a free-body diagra as shown in Fig. 4.36, then we see that two s protrude fro the top of the box. So the downward tension fro the botto string ust be 2, to ake the net force be zero. (By the sae reasoning, the tension in the short string above the right pulley is also 2.) We ll use this result any ties in the Atwood s probles in this chapter. 2 Figure Figure 4.37 d d Figure d Reark: In the above free-body diagra, the top string does indeed get counted twice, as far as the forces go. In a odified scenario where we have two people pulling upward on the ends of the string, each with a force, as shown in Fig. 4.37, their total upward force is of course 2. And the pulley can t tell the difference between this scenario and the original one. (c) Iagine that the left ass (and hence left pulley) goes up by a distance d. hen a length d of string disappears fro each of the two segents above the pulley, as shown in Fig So a total length 2d of string disappears fro these two segents. his string has to go soewhere, so it appears above the right ass. he right ass therefore goes down by 2d, as shown. So the downward displaceent of the right ass is always twice the upward displaceent of the left ass. aking two derivatives of this relation tells us that downward acceleration of the right ass is always twice the upward acceleration of the left ass. (Equivalently, the general relation d = at 2 /2 holds, so the ratio of the accelerations ust be the sae as the ratio of the displaceents.) his is the so-called conservation of string stateent for this setup. he sae reasoning applies if the left ass instead goes down. In any case, the sign of the right ass s acceleration is the negative of the sign of the left ass s acceleration. Rearks: When thinking about conservation-of-string stateents, it is often helpful to iagine cutting out pieces of string in soe parts of the setup and then splicing the into other parts. his isn t what actually happens, of course; the string just slides around like a snake. But if you take a photo at two different ties during the otion, the splicing photo will look the sae as the actual photo. In short, while it is often hard to visualize what is happening as the string oves, it is generally uch easier to iagine the string as having oved. Conservation of string by itself doesn t deterine the otion of the asses. We still need to apply F = a to find out how the asses actually ove. If you grab the asses and ove the around in an arbitrary anner while always aking sure that the strings stay taut, then the conservation-of-string condition will be satisfied. But the otion will undoubtedly not be the sae as the otion where the asses are acted on by only gravity. here is an infinite nuber of possible otions consistent with conservation of string, but only one of these otions is also consistent with all of the F = a equations. And conversely, the F = a equations alone don t deterine the otion; the conservation-of-string relation is required. If the strings aren t present, the otion will certainly be different; all the asses will be in freefall! (d) he average height of the right two asses always reains a constant distance below the right pulley (because the right string keeps the sae length). So y p = (y 2 + y 3 )/2 + C. aking two derivatives of this relation gives a p = (a 2 + a 3 )/2. But the downward (or upward) acceleration of the right pulley (which we just showed equals the average of the accelerations of the right two asses) equals the upward

22 4.5. PROBLEM SOLUIONS 89 (or downward) acceleration of the left ass (because the left string keeps the sae length), as desired. A few of the Atwood s-achine probles in this chapter contain soe unusual conservation-of-string relations, but they all involve the types of reasoning in parts (c) and (d) of this proble Block and tackle (a) Let be the tension in the long rope (the tension is the sae throughout the rope, fro the reasoning in Proble 4.3(a)). Since the botto pulley is assless, the reasoning in Proble 4.3(b) tells us that the tension in the short rope attached to the block is 4 (there would now be four s protruding fro the top of the dashed box in Fig. 4.36). We want this 4 tension to balance the Mg weight of the block, so the person ust pull on the rope with a force = Mg/4. (he angle of the rope to the person doesn t atter.) (b) he free-body diagra for the botto part of the setup is shown in Fig Five tensions protrude fro the top of the dashed box, and the two weights Mg and g protrude fro the botto. he net force ust be zero if the setup is at rest (or oving with constant speed), so the tension ust equal = (M + )g/5. his is the desired downward force exerted by the person on the rope. Liits: In the case where M = 0 (so the person is standing on a assless platfor), he ust pull down on the rope with a force equal to one fifth his weight, if he is to hoist hiself up. In this case, his g weight is balanced by an upward g/5 tension force fro the rope he is holding, plus an upward 4g/5 noral force fro the platfor (which basically coes fro the 4g/5 tension in the short rope attached to the platfor) Equivalent ass With the various paraeters defined as in Fig. 4.40, the two F = a equations are (with upward taken as positive for the left ass, and downward positive for the right) 1 g = 1 a and 2 g = 2 a. (4.6) M Mg Figure g We have used the fact that since your hand (and hence the pulley) is held at constant height, the accelerations of the asses are equal in agnitude and opposite in direction. We can solve for by ultiplying the first equation by 2, the second by 1, and then subtracting the. his eliinates the acceleration a, and we obtain = g/( ). he upward force you apply equals the tension 2 in the upper string; see Proble 4.3(b) for the explanation of the 2. his force of 2 equals the weight Mg of a single ass M if M = (4.7) his result is not equal to the su of the asses, In the special case where the asses are equal ( 1 = 2 ), the equivalent ass does siply equal 2. But in general, M isn t equal to the su. Liits: In the liit where 1 is very sall, we can ignore the 1 in the denoinator of Eq. (4.7) (but not in the nuerator; see the discussion in Section 1.1.3), which yields M 4 1. So the cobination of a arble and a bowling ball looks basically like four arbles (this liit akes it clear that M can t be equal to in general). his can be seen fairly intuitively: the bowling ball is essentially in freefall downward, so the arble accelerates upward at g. he tension ust therefore be equal to 2 1 g to ake the net upward force on the arble be 1 g. he tension 2 in the upper string is then 2(2 1 g), which yields M = 4 1. If we want to solve for a in Eq. (4.6), we can siply add the equations. he result is a = g (4.8) a 1 2 Figure 4.40 a

23 90 CHAPER 4. F=MA his acceleration akes sense, because a net force of 2 g 1 g pulls down on the right side, and this force accelerates the total ass of Various liits check correctly: if 2 = 1 then a = 0; if 2 1 then a g; and if 2 1 then a g (in Fig we defined positive a to be upward for 1 ) Atwood s 1 If 1 is at rest, the tension in the string supporting it ust be 1 g. Fro the reasoning in Proble 4.3(b), the tension in the lower string is then 1 g/2. he F = a equations for the two lower asses are therefore (with upward taken to be positive for 2 and downward positive for M) 2 : M : 1 g 2 2g = 2 a, Mg 1g = Ma. (4.9) 2 We have used the fact that the accelerations of 2 and M have the sae agnitude, because the botto pulley doesn t ove if 1 is at rest. Equating the two resulting expressions for a fro the above two equations gives 1 g g = g 1g 2 2 2M = 1 4 = 1 2 M = M = (4.10) In order for a physical M to exist, we need the denoinator of M to be positive. So we need 2 > 1 /4. Alternatively, you can solve this proble by solving for 2 in Eq. (4.7) in the solution to Proble 4.5 and then relabeling the asses appropriately. Reark: If 2 is saller than 1 /4, then even an infinitely large M won t keep 1 fro falling. he reason for this is that the best-case scenario is where M is so large that it is essentially in freefall. So 2 gets yanked upward with acceleration g, which eans that the tension in the string pulling on it is 2 2 g (so that the net upward force is 2 2 g 2 g = 2 g). he tension in the upper string is then 4 2 g. (We ve basically just repeated the reasoning for the sall- 1 liit discussed in the solution to Proble 4.5.) his is the largest the tension can be, so if 4 2 g is saller than 1 g (that is, if 2 is saller than 1 /4), then 1 will fall downward. Note that if 2 (with 2 finite), then Eq. (4.10) says that M = 1 /4, which is consistent with the preceding reasoning. You can show, as you ight intuitively expect, that the sallest su of 2 and M that supports a given 1 is achieved when 2 and M are both equal to 1 / Atwood s 2 Fro the reasoning in Proble 4.3(b), the tensions in the two strings are and 2, as shown in Fig he F = a equations are therefore = a 1, = (2)a 2, (2)g 2 = (2)a 3. (4.11) We have three equations but four unknowns here: a 1, a 2, a 3, and. So we need one ore equation the conservation-of-string relation. he average position of the left two asses reains the sae distance behind the left pulley, which oves the sae distance as the right pulley, and hence right ass. So the conservation-of-string relation is a 3 = (a 1 + a 2 )/2. his is the sae reasoning as in Proble 4.3(d). he first two of the above F = a equations quickly give a 1 = 2a 2. Plugging this into the conservation-of-string relation gives a 3 = 3a 2 /2. he second two F = a equations are

24 4.5. PROBLEM SOLUIONS 91 2 a a 2 a 3 2 Figure 4.41 then = (2)a 2, (2)g 2 = (2)(3a 2 /2). (4.12) he second equation plus twice the first gives 2g = 7a 2 = a 2 = 2g/7. We then have a 1 = 2a 2 = 4g/7, and a 3 = 3a 2 /2 = 3g/7. As a check, a 3 is indeed the average of a 1 and a 2. Additionally, the tension is = a 1 = 4g/7. Reark: When solving any proble, especially an Atwood s proble, it is iportant to (1) identify all of the unknowns and (2) ake sure that you have as any equations as unknowns, as we did above Atwood s 3 If is the tension in the lowest string, then fro Proble 4.3(b) the tensions in the other strings are shown in Fig Let all of the accelerations be defined with upward being positive. hen the three F = a equations are g = a 1, g = a 2, he first two of these equations quickly give a 1 = a 2. (2)g = (2)a 3. (4.13) Now for the conservation-of-string stateent. Let a p be the acceleration of the upper right pulley (which is the sae as the acceleration of the lower right pulley). he average height of the two right asses always reains the sae distance below this pulley. herefore a p = (a 2 + a 3 )/2. But we also have a 1 = 2a p, because if the pulley goes up a distance d, then a length d of string disappears fro both segents above the pulley, so 2d of string appears above the left ass. his eans that it goes down by 2d; hence a 1 = 2a p. (We ve just redone Proble 4.3(c) and (d) here.) Cobining this with the a p = (a 2 +a 3 )/2 relation gives ( a2 + a ) 3 a 1 = 2 = a 1 + a 2 + a 3 = 0. (4.14) 2 Since we know fro above that a 1 = a 2, we obtain a 3 = 2a 2. he last two of the above F = a equations are then a 1 2 a p a 2 a 3 2 Figure 4.42 g = a 2, (2)g = (2)( 2a 2 ). (4.15) aking the difference of these equations yields g = 5a 2, so a 2 = g/5. (And this is also a 1.) he desired acceleration of the ass 2 is then a 3 = 2a 2 = 2g/5. his is negative, so the ass 2 goes downward, which akes sense. (If all three asses are equal to, you can quickly show that = g and all three accelerations are zero. Increasing the right ass to 2 therefore akes it go downward.)

25 92 CHAPER 4. F=MA 4.9. Atwood s a 4a Let be the tension in the string connected to the right ass. hen fro the reasoning in Proble 4.3(b), the tensions in the other strings are 2 and 4, as shown in Fig he conservation-of-string relation tells us that the accelerations are a and 4a, as shown. his is true because if the botto pulley goes up by d, then the iddle pulley goes up by 2d, fro the reasoning in Proble 4.3(c). Fro the sae reasoning, the right ass then goes down by 4d. So the ratio of the distances oved is 4. aking two tie derivatives of this relation tells us that the (agnitudes of the) accelerations are in the sae ratio. he F = a equations are therefore Figure g = a, g = (4a). (4.16) he first equation plus four ties the second gives 3g = 17a = a = 3g/17. his is the upward acceleration of the left ass. he downward acceleration of the right ass is then 4a = 12g/17. Reark: Consider the ore general case where we have a tower of n ovable pulleys extending down to the left (so the given proble has n = 2). If we define N 2 n, then as an exercise you can show that the upward acceleration of the left ass and the downward acceleration of the right ass are a left = g N 1 N and a right = g N2 N N (4.17) hese expressions correctly reproduce the above results when n = 2 = N = 4. If N, we have a left 0 and a right g. You can think physically about what is going on here; in soe sense the syste behaves like a lever, where forces and distances are agnified. he above expressions for the a s are also valid when n = 0 = N = 1, in which case both accelerations are zero; we just have two asses hanging over the fixed pulley connected to the ceiling. a 1 a a a b a 3 2 a 4 Figure Atwood s 5 Let be the tension in the botto string. hen fro the reasoning in Proble 4.3(b), the tensions in the other strings are 2 and 4, as shown in Fig If all of the accelerations are defined with upward being positive, the four F = a equations are 4 4g = 4a 1, 2 2g = 2a 2, g = a 3, he first three of these equations quickly give a 1 = a 2 = a 3. 2g = 2a 4. (4.18) We ust now deterine the conservation-of-string relation. Let a and a b be the accelerations of the iddle and botto pulleys (with upward being positive). he average position of the botto two asses stays the sae distance below the botto pulley, so a b = (a 3 + a 4 )/2. Siilarly, a = (a 2 + a b )/2. Substituting the value of a b fro the first of these relations into the second, and using the fact that a 1 = a, we have ( ) a2 + (a 3 + a 4 )/2 a 1 =. (4.19) 2 Using a 1 = a 2 = a 3, this becoes ( ) a3 + (a 3 + a 4 )/2 a 3 = 2 he last two of the F = a equations in Eq. (4.18) are then a = a 3, = a 4 = 7a 3. (4.20) 2g = 2( 7a 3 ). (4.21)

26 4.5. PROBLEM SOLUIONS 93 Subtracting these equations eliinates, and we find a 3 = g/15. he acceleration of the rightost ass is then a 4 = 7a 3 = 7g/15. his is negative, so this ass goes down. he other three asses all ove upward with acceleration g/ Atwood s 6 he iportant thing to note in this proble is that the tension in the long string is different above and below the iddle ass. he standard fact that we ordinarily use (that the tension is the sae everywhere throughout a assless string; see Proble 4.3(a)) holds only if the string is assless. If there is a ass attached to the string, then the tensions on either side of the ass will be different (unless the ass is in freefall, as we see fro the second F = a equation below). Let these tensions be 1 and 2, as shown in Fig If we label the accelerations as a 1, a 2, and a 3 fro left to right (with upward defined to be positive for all), then the F = a equations are 1 g = a 1, 1 2 g = a 2, 2 2 g = a 3. (4.22) Conservation of string quickly gives a 2 = a 1. And it also gives a 3 = a 1 /2, fro the reasoning in Proble 4.3(c). he above F = a equations then becoe Figure g = a 1, 1 2 g = ( a 1 ), 2 2 g = ( a 1 /2). (4.23) We now have three equations in three unknowns ( 1, 2, and a 1 ). Solving these by your ethod of choice gives a 1 = 2g/9. Hence a 2 = 2g/9 and a 3 = g/9. he tensions turn out to be 1 = 11g/9 and 2 = 4g/ Atwood s 7 Let the tension in the upper left part of the string in Fig be. hen the two other upper parts also have tension, as shown. Because the left pulley is assless, the net force on it ust be zero, so the string below it ust have tension 2. his then iplies the other 2 shown. Finally, the tension in the botto string is 4 because the net force on the botto pulley ust be zero. Note that the tension in the long string is allowed to change fro to 2 at the ass, because the assless string reasoning in Proble 4.3(a) doesn t hold here. With the accelerations defined as shown in the figure, the F = a equations for the botto and top asses are g 4 = a b, g + 2 = a t. (4.24) a b Figure 4.46 a t We ust now deterine the conservation-of-string relation. Fro the reasoning in Proble 4.3(c), the acceleration of the top ass is twice the acceleration of the left pulley (with the opposite sign). Additionally, if the left pulley goes up by d, then an extra length d of string appears below the dotted line in Fig his is true because d disappears fro each of the two parts of the string above the left pulley, but d ust also be inserted right below the pulley. So a net length d is left over, and this appears below the dotted line. It gets divided evenly between the two parts of the string touching the botto pulley, so the botto pulley (and hence the botto ass) goes down by d/2. So the acceleration of the botto ass is half the acceleration of the left pulley (with the opposite sign). But fro the previous paragraph, the acceleration of the left pulley is half the acceleration of the top ass (with the opposite

27 94 CHAPER 4. F=MA sign). Putting this all together gives a b = ( a t /2)/2, or a t = 4a b. he above F = a equations therefore becoe g 4 = a b, g + 2 = (4a b ). (4.25) he first equation plus four ties the second gives 5g = 17a b = a b = 5g/17. And then a t = 4a b = 20g/17. It is perfectly fine that a t > g. In the liit where the top ass is zero, the botto ass is in freefall, so we have a b = g and a t = 4g a Figure a Atwood s 8 Let be the tension in the string connected to the right ass. hen fro Proble 4.3(b), zero net force on the right pulley tells us that the tension in the other long string is 2, as shown in Fig And then again fro Proble 4.3(b), zero net force on the left pulley tells us that the tension in the botto short string is 4. Let a be the upward acceleration of the left ass. We clai that the conservation-ofstring relation says that the downward acceleration of the right ass is 4a. his is true for the following reason. If the left pulley (and hence left ass) oves up by a distance d, then a length d of string disappears fro each of the two segents of string above it that touch it tangentially. Likewise, if the right pulley oves down by a distance d (and it does indeed ove the sae distance as the left pulley, because the centers of the two pulleys are connected by a string), then a length d of string disappears fro each of the two segents of string below it (assuing for a oent that the right ass doesn t ove). A total of 4d of string has therefore disappeared (teporarily) fro the long string touching the right ass. his 4d ust end up being inserted above the right ass. So this ass goes down by 4d, which is four ties the distance the left ass goes up, hence the accelerations of a and 4a shown in the figure. he F = a equations for the left and right asses are therefore 4 g = a, g = (4a). (4.26) he first equation plus four ties the second gives 3g = 17g = a = 3g/17. his is the upward acceleration of the left ass. he downward acceleration of the right ass is then 4a = 12g/17. hese accelerations happen to be the sae as the ones in Proble No relative otion First note that by the following continuity arguent, there ust exist a value of a for which there is no relative otion aong the asses. If a is zero, then 2 falls and 1 oves to the right. On the other hand, if a is very large, then intuitively 1 will drift backward with respect to M, and 2 will rise (reeber that there is no friction anywhere). So for soe interediate value of a, there will be no relative otion. Now let s find the desired value of a. If there is no relative otion aong the asses, then 2 stays at the sae height, which eans that the net vertical force on it ust be zero. he tension in the string connecting 1 to 2 ust therefore be = 2 g. he horizontal F = a equation on 1 is then = 1 a = 2 g = 1 a = a = ( 2 / 1 )g. his is the acceleration of the syste, and hence the desired acceleration of M. Liits: a is sall if 2 is sall (ore precisely, if 2 1 ) and large if 2 is large ( 2 1 ). hese results ake intuitive sense.

28 4.5. PROBLEM SOLUIONS Slipping blocks (a) he free-body diagras are shown in Fig he noral forces N 1 (between the blocks) and N 2 (between the botto block and the table) are quickly found to be N 1 = g and N 2 = 3g. And the tension in the string is = 3g, because twice this ust equal the applied 6g force (because the net force on the assless pulley ust be zero). he friction forces F 1 (between the blocks) and F 2 (between the botto block and the table) are as yet unknown. Assue that the botto block doesn t slip with respect to the table. he axiu possible leftward static friction force fro the table acting on the botto block is µ s N 2 = 3g. his can cancel out the rightward tension = 3g acting on the block. However, there also the friction force F 1 between the blocks. (his is the kinetic friction force µ k N 1 = g, because you can quickly show that if the botto block is at rest, then the top block ust slip with respect to the botto block.) his friction force acts rightward on the botto block, which eans that the net rightward force on the botto block is nonzero. It will therefore slip with respect to the table. Hence our initial non-slipping assuption was incorrect. (b) Assue that the top block doesn t slip with respect to the botto block (which we know ust be oving, fro part (a)). hen the two blocks can be treated like a single block with ass 3. he leftward kinetic friction force fro the table is F 2 = µ k N 2 = 3g. he net force on the effective 3 block is therefore 6g 3g = 3g rightward, so the acceleration is a = g rightward. he horizontal F = a equation for the top block is then F 1 N 1 N 1 g F 1 F 2 2 2g N 2 Figure 4.48 F 1 = a = 3g F 1 = g = F 1 = 2g. (4.27) But this friction force isn t possible, because it exceeds the axiu possible static friction force between the blocks, which is µ s N 1 = g. his contradiction iplies that our initial non-slipping assuption ust have been incorrect. he blocks therefore slip with respect to each other. Reark: If the coefficient of static friction between the blocks were instead ade sufficiently large (µ s 2), then the blocks would in fact ove as one effective ass 3. he static friction force 2g between the blocks would act backward on and forward on 2, and both blocks would have acceleration g. If the friction force were then suddenly decreased to the kinetic value of µ k N 1 = g relevant to the stated proble, would accelerate faster than g (because there isn t as uch friction holding it back), and 2 would accelerate slower than g (because there isn t as uch friction pushing it forward). his is consistent with the accelerations we will find below in part (c). (c) Since we know that all surfaces slip with respect to each other, the friction forces are all kinetic friction forces. heir values are therefore F 1 = µ k N 1 = g and F 2 = µ k N 2 = 3g. he F = a equations for the two blocks are then : F 1 = a 1 = 3g g = a 1 = a 1 = 2g, 2 : + F 1 F 2 = (2)a 2 = 3g + g 3g = 2a 2 = a 2 = g/2. (4.28) By conservation of string, the average position of the two blocks stays the sae distance behind the pulley, and hence also behind your hand. So a hand = a 1 + a 2 2 = 2g + g/2 2 = 5g 4. (4.29)

29 96 CHAPER 4. F=MA M Figure 4.49 y M x Block and wedge Let N be the noral force between the block and the wedge. hen the vertical F = a equation for the block (with downward taken to be positive) is Mg N cos = Ma M. (4.30) And the horizontal F = a equation for the wedge (with rightward taken to be positive) is N sin = a. (4.31) We have two equations but three unknowns (N, a M, a ), so we need one ore equation. his equation is the constraint that the block reains on the wedge at all ties. At a later tie, let the positions of the block and wedge be indicated by the dotted lines in Fig he block has oved downward a distance y M, and the wedge has oved rightward a distance x. Fro the triangle that these lengths for (the shaded triangle in the figure), we see that y M = x tan. aking two tie derivatives of this relation gives the desired third equation, a M = a tan. (4.32) here are various ways to solve the preceding three equations. If we ultiply the first by sin and the second by cos, and then add the, the N ters cancel. If we then use a M = a tan to eliinate a M, we obtain Solving for a gives Mg sin = Ma M sin + a cos = M(a tan ) sin + a cos. (4.33) a = Mg sin cos M sin 2 + cos 2. (4.34) Liits: here are any liits we can check. You should verify that all of the following results ake sense. If 0 or 90, then a 0. If M then a 0. If M then a g/ tan = g a tan. his is siply Eq. (4.32) with a M = g, because the block is essentially in freefall. If M =, then a = g sin cos, which achieves a axiu when = 45. And since the constraining force on the left side of the block equals N sin, which in turn equals a fro Eq. (4.31), we see that = 45 necessitates the axiu force by the constraining wall (for the special case where M = ). You can check that various liits for a M (given in Eq. (4.32)) also work out correctly Up and down a plane (a) When the block stops (at least instantaneously) at its highest point, the forces along the plane are the g sin gravity coponent downward, and the static friction force F f upward. We know that F f µ s N = 1 g cos. he block will accelerate downward if the gravitational force g sin is larger than the axiu possible friction force g cos. So the block will slide back down if g sin > g cos = tan > 1 = > 45. (4.35) (b) On the way up the plane, both the gravity coponent and friction point down the plane, so the force along the plane is g sin + µ k g cos downward. herefore, since µ k = 1, the acceleration during the upward otion points down the plane and has agnitude a u = g(sin + cos ). (4.36)

30 4.5. PROBLEM SOLUIONS 97 he total tie for the upward otion is then t u = v 0 a u = v 0 g(sin + cos ). (4.37) o find the tie t d for the downward otion, we need to find the axiu distance d up the plane that the block reaches. Fro standard kineatics we have d = v 2 0 /2a u. (his can be obtained fro either v 2 f v 2 i = 2ad, or d = v 0 t at 2 /2 with t = v 0 /a.) During the downward otion, the friction force points up the plane, so the net acceleration points down the plane and has agnitude a d = g(sin cos ). (4.38) Using the value of d we just found, the relation d = a d td 2 /2 for the downward otion gives v0 2 = a dtd 2 2a u 2 = t d = v 0 au a d = v 0 g (sin + cos )(sin cos ). (4.39) (c) he total tie with friction is F = t u + t d = v 0 a u + v 0 au a d, (4.40) where a u and a d are given in Eqs. (4.36) and (4.38). Without friction, the acceleration along the plane is siply g sin downward for both directions of otion, so the total tie with no friction is no F = 2v 0 g sin. (4.41) (Equivalently, just erase the cos ters in F, since those ters cae fro the friction.) If we let s sin, c cos, and x cot c/s, then F will be larger than no F if 1 s + c + 1 (s + c)(s c) > 2 s = 1 > x x Cross ultiplying and siplifying yields = = = x + 1 > 2 1 x 2 1 (1 + 2x)2 > 1 x2 (1 + x) x > (1 + 2x)2 (1 + x). (4.42) 4x 3 2x > 0 = x > 1 2. (4.43) (here is technically also a range of negative solutions to this equation, but x is defined to be a positive nuber.) However, we also need x < 1 for Eq. (4.42) to hold. (If x > 1 then our cross ultiplication switches the order of the inequality.) So F > no F if 1 > cot > 1/ 2, or equivalently if 1 < tan < 2 = 45 < < (Of course, we already knew that > 45 fro part (a).) o suarize: If 45 < < 54.7, then F > no F. hat is, the process takes longer with friction. If 54.7, then no F F. hat is, the process takes longer without friction. he first of these results is clear in the liiting case where is only slightly larger than 45, because the block will take a very long tie to slide back down the plane, since a d 0. In the other extree where 90, we have F = no F because the friction force vanishes on the vertical plane.

31 98 CHAPER 4. F=MA Plots of F and no F for values between 45 and 90 are shown in Fig Note that F achieves a local iniu. You can show nuerically that this iniu occurs at = 1.34 radians, which is about 77. It isn t obvious that there should exist a local iniu. But what happens is that below 77, F is larger than the iniu value because the slowness of the downward otion doinates other copeting effects. And above 77, F is larger because the larger distance up the plane doinates other copeting effects (this isn t terribly obvious). Figure 4.50 F, no F (in units of v 0 /g) F (with friction) no F (without friction) Rope in a tube (a) First solution: Each little piece of the rope with tiny ass has a radially inward acceleration v 2 /R, so it feels a radially inward force of v 2 /R (applied by the outer surface of the tube). he sideways coponents of these forces cancel in pairs in Fig. 4.28(a), so we are left with a net downward force (in the plane of the page). Second solution: he center of ass (a topic in Chapter 6) of the rope is located soewhere on the y axis at the given instant. It happens to be at radius 2R/π, but that isn t iportant here. he relevant fact is that the CM travels in a circle. And since F net = a CM (see Chapter 6), we can siply iagine a point ass oving around in a circle. So the acceleration is radially inward (that is, downward in Fig. 4.28(a)) at the given instant. (b) First solution: Each little piece of the rope has a tangential acceleration a (but no radial acceleration since v is instantaneously zero), so it feels a tangential force of a. he vertical (in the plane of the page) coponents of these forces cancel in pairs, so we are left with a net rightward force in Fig. 4.28(b). Reark: Since the rope is being pulled downward, where does this rightward force coe fro? It coes fro the inner surface of the tube, with the iportant fact being that the force fro the right part of the inner circle in Fig. 4.28(b) is larger than the force fro the left part. his can be traced to the fact that the tension in the rope is larger closer to the right end that is being pulled; the tension approaches zero at the left end. Second solution: he CM is initially located on the y axis, and it is accelerated to the right at the given instant (because it will end up oving in a circle once the rope gains soe speed). Since F net = a CM, the net force is therefore rightward. If you pull on the rope with a nonzero force while it has a nonzero v, then the total force vector will have both downward and rightward coponents Circling bucket Consider a little volue of the water, with ass. Assuing that the water stays inside the bucket, then at the top of the otion the forces on the ass are both the downward

32 4.5. PROBLEM SOLUIONS 99 gravitational force g and the downward noral force N fro the other water in the bucket. So the radial F = a equation is g + N = v2 R. (4.44) (If you want, you can alternatively consider a little rock at the botto of an epty bucket; that s effectively the sae setup.) If v is large, then N is large. he cutoff case where the water barely stays in the bucket occurs when N = 0. he iniu v is therefore given by g = v2 R = v in = gr. (4.45) If we take R to be, say, 1, then this gives v in 3 /s. he tie for each revolution is then 2πR/v in 2 s. If you swing your ar around with this period of revolution, you ll probably discover that the iniu speed is a lot slower that you would have guessed. Reark: he ain idea behind this proble is that although at the top of the otion the water is certainly accelerating downward under the influence of gravity, if you accelerate the bucket downward fast enough, then the bucket will aintain contact with the water. So the requireent is that at the top of the otion,2 you ust give the bucket a centripetal (downward) acceleration of at least g. hat is, v 2 /R g, in agreeent with Eq. (4.45). If the centripetal acceleration of the bucket is larger than g, then the bucket will need to push downward on the water to keep the water oving along with the bucket. hat is, the noral force N will be positive. If the centripetal acceleration of the bucket is saller than g, then the water will accelerate downward faster than the bucket. hat is, the water will leave the bucket. It isn t necessary to use circular otion to keep the water in the bucket; linear otion works too. If you turn a glass of water upside down and iediately accelerate it straight downward with an acceleration greater than or equal to g, the water will stay in the glass. Of course, you will soon run out of roo and sash the glass into the floor! he nice thing about circular otion is that it can go on indefinitely. hat s why centrifuges involve circular otion and not linear otion Banking an airplane First solution: If you do feel like you are getting flung to the side in your seat, then you will need to counter this tendency with soe kind of force parallel to the seat, perhaps friction fro the seat or a noral force by pushing on the wall, etc. If you don t feel like you are getting flung to the side in your seat, then you could just as well be asleep on a frictionless seat, and you would reain at rest on the seat. So the goal of this proble is to deterine the banking angle that is consistent with the only forces acting on you being the gravitational force and the noral force fro the seat (that is, no friction), as shown in Fig Let the banking angle be. he vertical coponent of the noral force ust be N y = g, to ake the net vertical force be zero. his iplies that the horizontal coponent is N x = g tan. he horizontal F = a equation for the circular otion of radius R is then N x = v2 R = g tan = v2 R = tan = v2 gr. (4.46) Your apparent weight is siply the noral force, because this is the force with which a scale on the seat would have to push up on you. So your weight is g tan N g a = v 2 /R seat Figure 4.51 g N = N 2 x + N 2 y = ( ) v g 2. (4.47) R Liits: If R is very large or v is very sall (ore precisely, if v 2 gr), then 0 and N g. (Of course, v can t be too sall, or the plane won t stay up!) hese liits ake sense, because 2As an exercise, you can show that if the water stays inside the bucket at the highest point, then it will stay inside at all other points too, as you would intuitively expect.

33 100 CHAPER 4. F=MA N a = v 2 /R (v 2 /R) cos g Figure 4.52 v a t a r Figure 4.53 a g sin for all you know, you are essentially oving in a straight line. If R is very sall or v is very large (ore precisely, if v 2 gr), then 90 and N v 2 /R, which akes sense because gravity is inconsequential in this case. Second solution: Let s use tilted axes parallel and perpendicular to the seat. If we break the forces and the acceleration into coponents along these tilted axis, then we see that the force coponent parallel to the seat is g sin, and the acceleration coponent parallel to the seat is (v 2 /R) cos, as shown in Fig So the F = a equation for the otion parallel to the seat is F = a = g sin = v2 R cos = tan = v2 gr, (4.48) in agreeent with the result in Eq. (4.46). he F = a equation for the otion perpendicular to the seat gives us the noral force N: N g cos = v2 R sin = N = g cos + v2 R sin. (4.49) You can verify that the sin and cos values iplied by the tan = v 2 /gr relation in Eq. (4.48) ake this expression for N reduce to the one we found above in Eq. (4.47) Breaking and turning If you brake on a straight road, your acceleration vector points along the road. he friction force satisfies F f = a. (We ll just deal with agnitudes here, so both F f and a are positive quantities.) But F f µn = µ(g). herefore, a = F f µg = a µg. (4.50) So your axiu possible deceleration is µg. It akes sense that this should be zero if µ is zero. If you brake while traveling around a bend, your acceleration vector does not point along the road. It has a tangential coponent a t pointing along the road (this is the deceleration we are concerned with) and also a radial coponent a r = v 2 /R pointing perpendicular to the road. hese coponents are shown in Fig he total acceleration vector (and hence also the total friction force vector) points backward and radially inward. Since F f = a, we still have F f = a, where a is the agnitude of the a vector, which is now a = at 2 + a2 r. So the F f µn restriction on F f takes the for at 2 + ar 2 = F f µg = at 2 + (v 2 /R) 2 µg = a t (µg) 2 (v 2 /R) 2. (4.51) sin (d/2) R d (top view) Figure 4.54 Liits: his result correctly reduces to a t µg when R =, that is, when the road is straight. It also reduces to a t µg when v is very sall (ore precisely, v µgr), because in this case the radial acceleration is negligible; you are effectively traveling on a straight road. If v = µgr, then Eq. (4.51) tells us that a t ust be zero. In this case the axiu friction force µn = µg is barely large enough to provide the v 2 /R = (µgr)/r = µg force required to keep you going in a circle. Any additional acceleration caused by braking will necessitate a friction force larger than the µn liit Circle of rope he forces on a sall piece of rope subtending an angle d are the tensions at its ends. In Fig these tensions point slightly downward; they ake an angle of d/2 with respect to the horizontal. (his is true because each of the long radial sides of the pie piece in the figure akes an angle of d/2 with respect to the vertical, and the tensions

34 4.5. PROBLEM SOLUIONS 101 are perpendicular to these sides.) If is the tension throughout the rope, then fro the figure the net force on the sall piece is F = 2 sin(d/2). his force points downward (radially inward). Using the approxiation sin x x for sall x, we see that the net force is F d. his force is what causes the centripetal acceleration of the sall piece. he length of the piece is R d, so its ass is λr d. he radial F = a equation is therefore F = v2 R = d = (λr d)v2 R = = λv 2. (4.52) Rearks: Note that this result of λv 2 is independent of R. his eans that if we have an arbitrarily shaped rope (that is, the local radius of curvature ay vary), and if the rope is oving along itself (so if the rope were featureless, you couldn t tell that it was actually oving) at constant speed, then the tension equals λv 2 everywhere. If the rope is stretchable, it will stretch under the tension as it spins around. You should convince yourself that the tension s independence of R iplies that if two circles (ade of the sae aterial) with different radii have the sae v, then the stretching will cause their radii to increase by the sae factor. his λv 2 result coes up often in physics. In addition to being the answer to the present proble, λv 2 is the tension in a rope with density λ if the speed of a traveling wave is v (this is a standard result that can be found in a textbook on waves). And λv 2 is also the tension needed if you have a heap of rope and you grab one end and pull with speed v to straighten it out (see Multiple-Choice Question 6.18). All three of these results appear in a gloriously unified way in the intriguing phenoenon of the chain fountain Cutting the string We ust first find the speed of the circular otion. he free-body diagra is shown in Fig. 4.55, where we have included the acceleration for convenience. he vertical F = a equation tells us that y = g, because the net vertical force ust be zero. he horizontal coponent of the tension is therefore x = y tan = g tan. So the horizontal F = a equation is (using the fact that the radius of the circular otion is r = l sin ) x = v2 r = g tan = v2 l sin = v = gl sin tan. (4.53) After the string is cut, we siply have a projectile proble in which the initial velocity is horizontal (tangential to the circle when the string is cut). he distance down to the floor is d = l l cos, so the tie to fall to the floor is given by 1 2 gt2 = l l cos = t = he horizontal distance traveled is therefore x = vt = gl sin tan 2l(1 cos ). (4.54) g 2l(1 cos ) g = l 2 sin tan (1 cos ). (4.55) Liits: If 0 then x 0 (because both v 0 and t 0). And if 90 then x (because v ). hese liits ake intuitive sense. Units: Note that x doesn t depend on g. Intuitively, if g is large, then for a given the speed v is large (it is proportional to g). But the falling tie t is short (it is proportional to 1/ g). hese two effects exactly cancel. he independence of g also follows fro diensional analysis. he distance x ust be soe function of l,, g, and. But there can t be any dependence on g, which has units of /s 2, because there would be no way to eliinate the seconds fro the units to obtain a pure length. (Likewise for the kilogras in the ass.) 3See Understanding the chain fountain (by he Royal Society) at O0. x a = v 2 /r Figure 4.55 y g

35 102 CHAPER 4. F=MA a = v 2 /r a sin a cos g g sin Figure 4.56 N g cos Circling around a cone (a) he free-body diagra is shown in Fig he net force produces the horizontal centripetal acceleration of a = v 2 /r, where r = l sin. Let s work with axes parallel and perpendicular to the cone. Horizontal and vertical axes would work fine too, but things would be a little essier because the tension and the noral force N would each appear in both of the F = a equations, so we would have to solve a syste of equations. he F = a equation along the cone is (using a = v 2 /l sin ) ( v 2 ) g cos = sin = = g cos + v2. (4.56) l sin l Liits: If 0 then g + v 2 /l. We will find below that if contact with the cone is to be aintained, then v ust be essentially zero in this case. So we siply have g, which akes sense because the ass is hanging straight down. If π/2 then v 2 /l, which akes sense because the ass is oving in a circle on a horizontal table. (b) he F = a equation perpendicular to the cone is ( v 2 ) g sin N = cos = N = g sin v2 l sin l tan. (4.57) Liits: If 0 then N 0 v 2 /l tan. Again, we will find below that v ust be essentially zero in this case, so we have N 0, which akes sense because the cone is vertical. If π/2 then N g, which akes sense because, as above, the ass is oving in a circle on a horizontal table. (c) he ass stays in contact with the cone if N 0. Using Eq. (4.57), this iplies that Penny in a dryer g sin v2 l tan = v gl sin tan v ax. (4.58) If v equals v ax then the ass is barely in contact with the cone. If the cone were reoved in this case, the ass would aintain the sae circular otion. Liits: If 0 then v ax 0. And if π/2 then v ax. hese liits ake intuitive sense. he penny loses contact with the dryer when the noral force N becoes zero. If is easured with respect to the vertical, then the radially inward coponent of the gravitational force is g cos. he radial F = a equation is therefore N + g cos = v 2 /R, which iplies that the N = 0 condition is g cos = v2 = v 2 = gr cos. (4.59) R his equation tells us how v and are related at the point where the penny loses contact. (If v > gr, then there is no solution for, so the penny never loses contact.) o solve for and v (and hence ω = v/r) individually, we ust produce a second equation that relates v and. his equation coes fro the projectile otion and the condition that the landing point is diaetrically opposite. When the penny loses contact with the dryer, the initial position (relative to the center of the dryer) for the projectile otion is (R sin, R cos ); reeber that is easured with respect to the vertical. he initial angle of the velocity is upward to the left, as you can check. herefore, since the initial speed is v, the initial velocity coponents are x x = v cos and v y = v sin. he coordinates of the projectile otion are then x(t) = R sin (v cos )t, y(t) = R cos + (v sin )t gt 2 /2. (4.60)

36 4.5. PROBLEM SOLUIONS 103 If the penny lands at the diaetrically opposite point, then the final position is the negative of the initial position. So the final tie ust satisfy R sin (v cos )t = R sin = (v cos )t = 2R sin, (4.61) R cos + (v sin )t gt 2 /2 = R cos = gt 2 /2 (v sin )t = 2R cos. Solving for t in the first of these equations and plugging the result into the second gives g 2 ( ) 2 2R sin v sin v cos ( ) 2R sin = 2R cos v cos = gr sin2 v 2 cos 2 = sin2 cos + cos = gr sin2 v 2 cos 2 = 1 cos = v 2 = gr sin2 cos. (4.62) his is the desired second equation that relates v and. Equating this expression for v 2 (which guarantees a diaetrically opposite landing point) with the one in Eq. (4.59) (which gives the point where the penny loses contact) yields gr sin 2 cos = gr cos = tan 2 = 1 = = 45. (4.63) (And = 45, works too, if the dryer is spinning the other way. Equivalently, the corresponding value of t is negative.) here ust be a reason why the angle coes out so nice, but it eludes e. With this value of, Eq. (4.59) gives gr v = 2 = ω = v g R =. 2R (4.64) Liits: ω grows with g, as expected. he decrease with R isn t as obvious, but it does follow fro diensional analysis. Rearks: he relation in Eq. (4.62) describes any different types of trajectories. If v is large, then the corresponding is close to 90. In this case we have a very tall projectile path that generally lies outside the dryer, so it isn t physical. If v is sall, then is close to 0. In this case the penny just drops fro the top of the dryer down to the botto; the trajectory lies copletely inside the dryer. he cutoff between these two cases (that is, the case where the penny barely stays inside), turns out to be the = 45 case that solves the proble (by also satisfying Eq. (4.59)). You are encouraged to think about why this is true. As an exercise, you can also produce the second equation relating v and (Eq. (4.62), derived fro the projectile otion) by using axes that are tilted along the diaeter and perpendicular to it. he agnitudes of the accelerations in these directions are g cos and g sin, respectively, and the penny hits the diaeter a distance 2R down along it. he ath is fairly clean due to the fact that the initial velocity is perpendicular to the diaeter. So the solution turns out to be a bit quicker than the one we used above. Proble 3.18(b) involved siilar reasoning with the axis perpendicular to the diaeter/plane.