114 Chapter The Laws of Motion Categorize This part of the problem belongs to kinematics rather than to dynamics, and Equation (3) shows that the acceleration ax is constant Therefore, you should categorize the car in this part of the problem as a particle under constant acceleration d ϭ 12axt Analyze Defining the initial position of the front bumper as xi ϭ and its final position as xf ϭ d, and recognizing that vxi ϭ 0, apply Equation 2.16, xf ϭ xi ϩ vxit ϩ 12axt 2: Solve for t : (4) Use Equation 2.17, with vxi ϭ 0, to find the final velocity of the car: E XA M P L E 2d 2d ϭ a B x B g sin u vxf2 ϭ 2axd (5) Finalize We see from Equations (4) and (5) that the time t at which the car reaches the bottom and its final speed vxf are independent of the car’s mass, as was its acceleration Notice that we have combined techniques from Chapter with new techniques from this chapter in this example As we learn more techniques in later chapters, this process of combining information from several parts of the book will occur more often In these cases, use the General Problem-Solving Strategy to help you identify what analysis models you will need tϭ vxf ϭ 22axd ϭ 22gd sin u What If? What previously solved problem does this situation become if u ϭ 90°? Answer Imagine u going to 90° in Figure 5.11 The inclined plane becomes vertical, and the car is an object in free-fall! Equation (3) becomes ax ϭ g sin u ϭ g sin 90° ϭ g which is indeed the free-fall acceleration (We find ax ϭ g rather than ax ϭϪg because we have chosen positive x to be downward in Fig 5.11.) Notice also that the condition n ϭ mg cos u gives us n ϭ mg cos 90° ϭ That is consistent with the car falling downward next to the vertical plane, in which case there is no contact force between the car and the plane One Block Pushes Another Two blocks of masses m1 and m2, with m1 Ͼ m2, are placed in contact with each other on a frictionless, horizontal surface as in Active Figure 5.12a A constant horiS zontal force F is applied to m1 as shown (A) Find the magnitude of the acceleration of the system F m1 (a) n1 n2 y P21 F SOLUTION Conceptualize Conceptualize the situation by using Active Figure 5.12a and realize that both blocks must experience the same acceleration because they are in contact with each other and remain in contact throughout the motion Categorize We categorize this problem as one involving a particle under a net force because a force is applied to a system of blocks and we are looking for the acceleration of the system Analyze First model the combination of two blocks as a single particle Apply Newton’s second law to the combination: m2 x P12 m1 m2 m 2g m 1g (b) (c) ACTIVE FIGURE 5.12 (Example 5.7) A force is applied to a block of mass m1, which pushes on a second block of mass m2 (b) The free-body diagram for m1 (c) The free-body diagram for m2 Sign in at www.thomsonedu.com and go to ThomsonNOW to study the forces involved in this two-block system a Fx ϭ F ϭ 1m1 ϩ m2 ax (1) ax ϭ F m1 ϩ m2 Section 5.7 Some Applications of Newton’s Laws 115 Finalize The acceleration given by Equation (1) is the same as that of a single object of mass m1 ϩ m2 and subject to the same force (B) Determine the magnitude of the contact force between the two blocks SOLUTION Conceptualize The contact force is internal to the system of two blocks Therefore, we cannot find this force by modeling the whole system (the two blocks) as a single particle Categorize Now consider each of the two blocks individually by categorizing each as a particle under a net force Analyze We first construct a free-body diagram for each block as shown in Active Figures 5.12b and 5.12c, where S the contact force is denoted by P From Active Figure 5.12c we see that the only horizontal force acting on m2 is the S contact force P12 (the force exerted by m1 on m2), which is directed to the right Apply Newton’s second law to m2: (2) Substitute the value of the acceleration ax given by Equation (1) into Equation (2): (3) a Fx ϭ P12 ϭ m 2ax P12 ϭ m2ax ϭ a m2 bF m1 ϩ m2 Finalize This result shows that the contact force P12 is less than the applied force F The force required to accelerate block alone must be less than the force required to produce the same acceleration for the two-block system To finalize further, let us check this expression for P12 by considering the forces acting on m1, shown in SActive FigS ure 5.12b The horizontal forces acting on m1 are the applied Sforce F to the right and the contact force P21 to the S left (the force exerted by m2 on m1) From Newton’s third law, P21 is the reaction force to P12, so P21 ϭ P12 Apply Newton’s second law to m1: (4) Solve for P12 and substitute the value of ax from Equation (1): a Fx ϭ F Ϫ P21 ϭ F Ϫ P12 ϭ m1ax P12 ϭ F Ϫ m1ax ϭ F Ϫ m1 a m2 F b ϭ a bF m1 ϩ m2 m1 ϩ m2 This result agrees with Equation (3), as it must S What If? Imagine that the force SF in Active Figure 5.12 is applied toward the left on the right-hand block of mass m2 Is the magnitude of the force P12 the same as it was when the force was applied toward the right on m1? Answer When the force is applied toward the left on m2, the contact force must accelerate m1 In the original sitS uation, the contact force accelerates m2 Because m1 Ͼ m2, more force is required, so the magnitude of P12 is greater than in the original situation E XA M P L E Weighing a Fish in an Elevator A person weighs a fish of mass m on a spring scale attached to the ceiling of an elevator as illustrated in Figure 5.13 (A) Show that if the elevator accelerates either upward or downward, the spring scale gives a reading that is different from the weight of the fish SOLUTION Conceptualize The reading on the scale is related to the extension of the spring in the scale, which is related to the force on the end of the spring as in Figure 5.2 Imagine that the fish is hanging on a string attached to the end of the spring In this case, the magnitude of the force exerted on the spring is equal to the tension T in the string 116 Chapter The Laws of Motion S a Therefore, we are looking for T The force T pulls down on the string and pulls up on the fish a Categorize We can categorize this problem by identifying the fish as a particle under a net force T T mg (a) mg (b) Figure 5.13 (Example 5.8) Apparent weight versus true weight (a) When the elevator accelerates upward, the spring scale reads a value greater than the weight of the fish (b) When the elevator accelerates downward, the spring scale reads a value less than the weight of the fish Analyze Inspect the free-body diagrams for the fish in Figure 5.13 and notice that the external forces acting Son the fish are the downward gravitaS S tional force Fg ϭ m g and the force T exerted by the string If the elevator is either at rest or moving at constant velocity, the fish is a particle in equilibrium, so ͚ Fy ϭ T Ϫ Fg ϭ or T ϭ Fg ϭ mg (Remember that the scalar mg is the weight of the fish.) Now suppose the elevator is moving with an S acceleration a relative to an observer standing outside the elevator in an inertial frame (see Fig 5.13) The fish is now a particle under a net force a Fy ϭ T Ϫ mg ϭ may Apply Newton’s second law to the fish: Solve for T : (1) T ϭ may ϩ mg ϭ mg a ay g ϩ b ϭ Fg a ay g ϩ 1b where we have chosen upward as the positive y direction We conclude from Equation (1) that the scale reading T is S S greater than the fish’s weight mg if a is upward, so ay is positive, and that the reading is less than mg if a is downward, so ay is negative (B) Evaluate the scale readings for a 40.0-N fish if the elevator moves with an acceleration ay ϭ Ϯ2.00 m/s2 S Evaluate the scale reading from Equation (1) if a is upward: S Evaluate the scale reading from Equation (1) if a is downward: T ϭ 140.0 N2 a T ϭ 140.0 N2 a 2.00 m>s2 9.80 m>s2 ϩ b ϭ 48.2 N Ϫ2.00 m>s2 9.80 m>s2 ϩ b ϭ 31.8 N Finalize Take this advice: if you buy a fish in an elevator, make sure the fish is weighed while the elevator is either at rest or accelerating downward! Furthermore, notice that from the information given here, one cannot determine the direction of motion of the elevator What If? Suppose the elevator cable breaks and the elevator and its contents are in free-fall What happens to the reading on the scale? Answer If the elevator falls freely, its acceleration is ay ϭϪg We see from Equation (1) that the scale reading T is zero in this case; that is, the fish appears to be weightless E XA M P L E The Atwood Machine When two objects of unequal mass are vertically over a frictionless pulley of negligible mass as in Active Figure 5.14a, the arrangement is called an Atwood machine The device is sometimes used in the laboratory to calculate the value of g Determine the magnitude of the acceleration of the two objects and the tension in the lightweight cord Section 5.7 Some Applications of Newton’s Laws 117 SOLUTION Conceptualize Imagine the situation pictured in Active Figure 5.14a in action: as one object moves upward, the other object moves downward Because the objects are connected by an inextensible string, their accelerations must be of equal magnitude Categorize The objects in the Atwood machine are subject to the gravitational force as well as to the forces exerted by the strings connected to them Therefore, we can categorize this problem as one involving two particles under a net force T T + m1 m1 m2 m2 + m1g Analyze The free-body diagrams for the two objects m2g are shown in Active Figure 5.14b Two forces act on S (b) (a) each object: the upward force T exerted by the string and the downward gravitational force In problems such ACTIVE FIGURE 5.14 as this one in which the pulley is modeled as massless (Example 5.9) The Atwood machine (a) Two objects connected by a massless inextensible cord over a frictionless pulley and frictionless, the tension in the string on both sides (b) The free-body diagrams for the two objects of the pulley is the same If the pulley has mass or is subSign in at www.thomsonedu.com and go to ThomsonNOW to ject to friction, the tensions on either side are not the adjust the masses of the objects on the Atwood machine and same and the situation requires techniques we will learn observe the motion in Chapter 10 We must be very careful with signs in problems such as this In Active Figure 5.14a, notice that if object accelerates upward, object accelerates downward Therefore, for consistency with signs, if we define the upward direction as positive for object 1, we must define the downward direction as positive for object With this sign convention, both objects accelerate in the same direction as defined by the choice of sign Furthermore, according to this sign convention, the y component of the net force exerted on object is T Ϫ m1g, and the y component of the net force exerted on object is m2g Ϫ T Apply Newton’s second law to object 1: (1) a Fy ϭ T Ϫ m1g ϭ m1ay Apply Newton’s second law to object 2: (2) a Fy ϭ m2g Ϫ T ϭ m2ay Ϫm1g ϩ m2g ϭ m1ay ϩ m2ay Add Equation (2) to Equation (1), noticing that T cancels: Solve for the acceleration: Substitute Equation (3) into Equation (1) to find T: (3) (4) ay ϭ a m2 Ϫ m1 bg m1 ϩ m2 T ϭ m1 1g ϩ ay ϭ a 2m1m2 bg m1 ϩ m2 Finalize The acceleration given by Equation (3) can be interpreted as the ratio of the magnitude of the unbalanced force on the system (m2 Ϫ m1)g to the total mass of the system (m1 ϩ m2), as expected from Newton’s second law Notice that the sign of the acceleration depends on the relative masses of the two objects What If? Describe the motion of the system if the objects have equal masses, that is, m1 ϭ m2 Answer If we have the same mass on both sides, the system is balanced and should not accelerate Mathematically, we see that if m1 ϭ m2, Equation (3) gives us ay ϭ What If? What if one of the masses is much larger than the other: m1 ϾϾ m2? Answer In the case in which one mass is infinitely larger than the other, we can ignore the effect of the smaller mass Therefore, the larger mass should simply fall as if the smaller mass were not there We see that if m1 ϾϾ m2, Equation (3) gives us ay ϭ –g 118 Chapter E XA M P L E The Laws of Motion Acceleration of Two Objects Connected by a Cord A ball of mass m1 and a block of mass m2 are attached by a lightweight cord that passes over a frictionless pulley of negligible mass as in Figure 5.15a The block lies on a frictionless incline of angle u Find the magnitude of the acceleration of the two objects and the tension in the cord y a T m2 m1 a m 1g u SOLUTION (a) Conceptualize Imagine the objects in Figure 5.15 in motion If m2 moves down the incline, m1 moves upward Because the objects are connected by a cord (which we assume does not stretch), their accelerations have the same magnitude (b) yЈ n T Categorize We can identify forces on each of the two objects and we are looking for an acceleration, so we categorize the objects as particles under a net force m2g sin u u Analyze Consider the free-body diagrams shown in Figures 5.15b and 5.15c x m1 xЈ m 2g cos u m 2g (c) Figure 5.15 (Example 5.10) (a) Two objects connected by a lightweight cord strung over a frictionless pulley (b) The free-body diagram for the ball (c) The free-body diagram for the block (The incline is frictionless.) Apply Newton’s second law in component form to the ball, choosing the upward direction as positive: (1) a Fx ϭ (2) a Fy ϭ T Ϫ m1g ϭ m1ay ϭ m1a For the ball to accelerate upward, it is necessary that T Ͼ m1g In Equation (2), we replaced ay with a because the acceleration has only a y component For the block it is convenient to choose the positive xЈ axis along the incline as in Figure 5.15c For consistency with our choice for the ball, we choose the positive direction to be down the incline Apply Newton’s second law in component form to the block: (3) a Fx¿ ϭ m2g sin u Ϫ T ϭ m2ax¿ ϭ m2a (4) a Fy¿ ϭ n Ϫ m2g cos u ϭ In Equation (3), we replaced axЈ with a because the two objects have accelerations of equal magnitude a Solve Equation (2) for T: Substitute this expression for T into Equation (3): Solve for a: Substitute this expression for a into Equation (5) to find T: (5) T ϭ m1 1g ϩ a2 m2g sin u Ϫ m1 1g ϩ a2 ϭ m2a (6) (7) aϭ Tϭ m2g sin u Ϫ m1g m1 ϩ m2 m 1m 2g 1sin u ϩ 12 m1 ϩ m2 Section 5.8 Forces of Friction 119 Finalize The block accelerates down the incline only if m2 sin u Ͼ m1 If m1 Ͼ m2 sin u, the acceleration is up the incline for the block and downward for the ball Also notice that the result for the acceleration, Equation (6), can be interpreted as the magnitude of the net external force acting on the ball–block system divided by the total mass of the system; this result is consistent with Newton’s second law What If? What happens in this situation if u ϭ 90°? Answer If u ϭ 90°, the inclined plane becomes vertical and there is no interaction between its surface and m2 Therefore, this problem becomes the Atwood machine of Example 5.9 Letting u S 90° in Equations (6) and (7) causes them to reduce to Equations (3) and (4) of Example 5.9! What If? What if m1 ϭ 0? Answer If m1 ϭ 0, then m2 is simply sliding down an inclined plane without interacting with m1 through the string Therefore, this problem becomes the sliding car problem in Example 5.6 Letting m1 S in Equation (6) causes it to reduce to Equation (3) of Example 5.6! 5.8 Forces of Friction When an object is in motion either on a surface or in a viscous medium such as air or water, there is resistance to the motion because the object interacts with its surroundings We call such resistance a force of friction Forces of friction are very important in our everyday lives They allow us to walk or run and are necessary for the motion of wheeled vehicles Imagine that you are working in your garden and have filled a trash can with yard clippings You then try to drag the trash can across the surface of your concrete patio as in Active Figure 5.16a This surface is real, not an idealized, friction- n n Motion F F fs fk mg (a) mg (b) |f| fs,max fs =F fk = mk n O F Static region Kinetic region (c) ACTIVE FIGURE 5.16 S When pulling on a trash can, the direction of the Sforce of friction f between the can and a rough surface is opposite the direction of the applied force F Because both surfaces are rough, contact is made only at a few points as illustrated in the “magnified” view (a) For small applied forces, the magnitude of the force of static friction equals the magnitude of the applied force (b) When the magnitude of the applied force exceeds the magnitude of the maximum force of static friction, the trash can breaks free The applied force is now larger than the force of kinetic friction, and the trash can accelerates to the right (c) A graph of friction force versus applied force Notice that fs, max Ͼ fk Sign in at www.thomsonedu.com and go to ThomsonNOW to vary the applied force on the trash can and practice sliding it on surfaces of varying roughness Notice the effect on the trash can’s motion and the corresponding behavior of the graph in (c) 120 Chapter The Laws of Motion S Force of static friction ᮣ Force of kinetic friction ᮣ PITFALL PREVENTION 5.9 The Equal Sign Is Used in Limited Situations In Equation 5.9, the equal sign is used only in the case in which the surfaces are just about to break free and begin sliding Do not fall into the common trap of using fs ϭ ms n inany static situation F to the trash can, acting to less surface If we apply an external horizontal force S F the right, the trash can remains stationary when is small The force on the trash S can that counteracts F andSkeeps it from moving acts toward the left and is called fs As long as the trash canS is not moving, the force of static friction fs ϭ F ThereS S S fore, if F is increased, fs also increases Likewise, if F decreases, fs also decreases Experiments show that the friction force arises from the nature of the two surfaces: because of their roughness, contact is made only at a few locations where peaks of the material touch, as shown in the magnified view of the surface in Active Figure 5.16a At these locations, the friction force arises in part because one peak physically blocks the motion of a peak from the opposing surface and in part from chemical bonding (“spot welds”) of opposing peaks as they come into contact Although the details of friction are quite complex at the atomic level, this force ultimately involves an electrical interaction between atoms or molecules S If we increase the magnitude of F as in Active Figure 5.16b, the trash can eventually slips When the trash can is on the verge of slipping, fs has its maximum value fs,max as shown in Active Figure 5.16c When F exceeds fs,max, the trash can moves and accelerates to the right.S We call the friction force for an object in motion the force of kinetic friction f k When the trash can is in motion, the force of kinetic friction on the can is less than fs,max (Active Fig 5.16c) The net force F Ϫ fk in the x direction produces an acceleration to the right, according to Newand the trash can moves to the ton’s second law If F ϭ fk , the acceleration is zero S right with constantS speed If the applied force F is removed from the moving can, the friction force f k acting to the left provides an acceleration of the trash can in the Ϫx direction and eventually brings it to rest, again consistent with Newton’s second law Experimentally, we find that, to a good approximation, both fs,max and fk are proportional to the magnitude of the normal force exerted on an object by the surface The following descriptions of the force of friction are based on experimental observations and serve as the model we shall use for forces of friction in problem solving: ■ fs Յ m sn PITFALL PREVENTION 5.10 Friction Equations Equations 5.9 and 5.10 are not vector equations They are relationships between the magnitudes of the vectors representing the friction and normal forces Because the friction and normal forces are perpendicular to each other, the vectors cannot be related by a multiplicative constant PITFALL PREVENTION 5.11 The Direction of the Friction Force Sometimes, an incorrect statement about the friction force between an object and a surface is made—”the friction force on an object is opposite to its motion or impending motion”—rather than the correct phrasing, “the friction force on an object is opposite to its motion or impending motion relative to the surface.” The magnitude of the force of static friction between any two surfaces in contact can have the values ■ where the dimensionless constant ms is called the coefficient of static friction and n is the magnitude of the normal force exerted by one surface on the other The equality in Equation 5.9 holds when the surfaces are on the verge of slipping, that is, when fs ϭ fs,max ϵ msn This situation is called impending motion The inequality holds when the surfaces are not on the verge of slipping The magnitude of the force of kinetic friction acting between two surfaces is fk ϭ m kn ■ ■ ■ (5.9) (5.10) where mk is the coefficient of kinetic friction Although the coefficient of kinetic friction can vary with speed, we shall usually neglect any such variations in this text The values of mk and ms depend on the nature of the surfaces, but mk is generally less than ms Typical values range from around 0.03 to 1.0 Table 5.1 lists some reported values The direction of the friction force on an object is parallel to the surface with which the object is in contact and opposite to the actual motion (kinetic friction) or the impending motion (static friction) of the object relative to the surface The coefficients of friction are nearly independent of the area of contact between the surfaces We might expect that placing an object on the side having the most area might increase the friction force Although this method provides more points in contact as in Active Figure 5.16a, the weight of the Section 5.8 121 Forces of Friction TABLE 5.1 Coefficients of Friction Rubber on concrete Steel on steel Aluminum on steel Glass on glass Copper on steel Wood on wood Waxed wood on wet snow Waxed wood on dry snow Metal on metal (lubricated) Teflon on Teflon Ice on ice Synovial joints in humans ms mk 1.0 0.74 0.61 0.94 0.53 0.25–0.5 0.14 — 0.15 0.04 0.1 0.01 0.8 0.57 0.47 0.4 0.36 0.2 0.1 0.04 0.06 0.04 0.03 0.003 Note: All values are approximate In some cases, the coefficient of friction can exceed 1.0 30Њ object is spread out over a larger area and the individual points are not pressed together as tightly Because these effects approximately compensate for each other, the friction force is independent of the area F (a) Quick Quiz 5.6 You press your physics textbook flat against a vertical wall with your hand What is the direction of the friction force exerted by the wall on the book? (a) downward (b) upward (c) out from the wall (d) into the wall F 30Њ Quick Quiz 5.7 You are playing with your daughter in the snow She sits on a sled and asks you to slide her across a flat, horizontal field You have a choice of (a) pushing her from behind by applying a force downward on her shoulders at 30° below the horizontal (Fig 5.17a) or (b) attaching a rope to the front of the sled and pulling with a force at 30° above the horizontal (Fig 5.17b) Which would be easier for you and why? E XA M P L E 1 (b) Figure 5.17 (Quick Quiz 5.7) A father slides his daughter on a sled either by (a) pushing down on her shoulders or (b) pulling up on a rope Experimental Determination of Ms and Mk The following is a simple method of measuring coefficients of friction Suppose a block is placed on a rough surface inclined relative to the horizontal as shown in Active Figure 5.18 The incline angle is increased until the block starts to move Show that you can obtain ms by measuring the critical angle uc at which this slipping just occurs y n fs mg sin u mg cos u u u SOLUTION Conceptualize Consider the free-body diagram in Active Figure 5.18 and imagine that the block tends to slide down the incline due to the gravitational force To simulate the situation, place a coin on this book’s cover and tilt the book until the coin begins to slide Categorize The block is subject to various forces Because we are raising the plane to the angle at which the block is just ready to begin to move but is not moving, we categorize the block as a particle in equilibrium mg x ACTIVE FIGURE 5.18 (Example 5.11) The external forces exerted on a block lying on a rough incline are the gravitational S S forceSm g, the normal force n, and the force of friction f s For convenience, the gravitational force is resolved into a component mg sin u along the incline and a component mg cos u perpendicular to the incline Sign in at www.thomsonedu.com and go to ThomsonNOW to investigate this situation further 122 Chapter The Laws of Motion S S AnalyzeS The forces acting on the block are the gravitational force mg, the normal force n, and the force of static friction f s We choose x to be parallel to the plane and y perpendicular to it Apply Equation 5.8 to the block: Substitute mg ϭ n/cos u from Equation (2) into Equation (1): (3) (1) a Fx ϭ mg sin u Ϫ fs ϭ (2) a Fy ϭ n Ϫ mg cos u ϭ fs ϭ mg sin u ϭ a When the incline angle is increased until the block is on the verge of slipping, the force of static friction has reached its maximum value msn The angle u in this situation is the critical angle uc Make these substitutions in Equation (3): n b sin u ϭ n tan u cos u m sn ϭ n tan uc m s ϭ tan uc For example, if the block just slips at uc ϭ 20.0°, we find that ms ϭ tan 20.0° ϭ 0.364 Finalize Once the block starts to move at u Ն uc , it accelerates down the incline and the force of friction is fk ϭ mkn If u is reduced to a value less than uc , however, it may be possible to find an angle ucЈ such that the block moves down the incline with constant speed as a particle in equilibrium again (ax ϭ 0) In this case, use Equations (1) and (2) with fs replaced by fk to find mk: mk ϭ tan u¿c where u¿c u c E XA M P L E The Sliding Hockey Puck n A hockey puck on a frozen pond is given an initial speed of 20.0 m/s If the puck always remains on the ice and slides 115 m before coming to rest, determine the coefficient of kinetic friction between the puck and ice Motion fk SOLUTION Conceptualize Imagine that the puck in Figure 5.19 slides to the right and eventually comes to rest due to the force of kinetic friction mg Categorize The forces acting on the puck are identified in Figure 5.19, but the text of the problem provides kinematic variables Therefore, we categorize the problem in two ways First, the problem involves a particle under a net force: kinetic friction causes the puck to accelerate And, because we model the force of kinetic friction as independent of speed, the acceleration of the puck is constant So, we can also categorize this problem as one involving a particle under constant acceleration Figure 5.19 (Example 5.12) After the puck is given an initial velocity to the right, the only external forces acting on it are the gravitational force S S mg, the normal force n, and the force S of kinetic friction f k Analyze First, we find the acceleration algebraically in terms of the coefficient of kinetic friction, using Newton’s second law Once we know the acceleration of the puck and the distance it travels, the equations of kinematics can be used to find the numerical value of the coefficient of kinetic friction Apply the particle under a net force model in the x direction to the puck: (1) a Fx ϭ Ϫfk ϭ max Apply the particle in equilibrium model in the y direction to the puck: (2) a Fy ϭ n Ϫ mg ϭ Section 5.8 Substitute n ϭ mg from Equation (2) and fk ϭ mkn into Equation (1): 123 Forces of Friction Ϫ m kn ϭ Ϫ m kmg ϭ max ax ϭ Ϫ m k g The negative sign means the acceleration is to the left in Figure 5.19 Because the velocity of the puck is to the right, the puck is slowing down The acceleration is independent of the mass of the puck and is constant because we assume that mk remains constant ϭ vxi2 ϩ 2ax xf ϭ vxi2 Ϫ 2m k gxf Apply the particle under constant acceleration model to the puck, using Equation 2.17, vxf2 ϭ vxi2 ϩ 2ax 1xf Ϫ xi , with xi ϭ and vf ϭ 0: mk ϭ mk ϭ Finalize on ice vxi2 2gxf 120.0 m>s2 2 19.80 m>s2 1115 m2 ϭ 0.117 Notice that mk is dimensionless, as it should be, and that it has a low value, consistent with an object sliding E XA M P L E Acceleration of Two Connected Objects When Friction Is Present A block of mass m1 on a rough, horizontal surface is connected to a ball of mass m2 by a lightweight cord over a lightweight, frictionless pulley as shown in Figure 5.20a A force of magnitude F at an angle u with the horizontal is applied to the block as shown and the block slides to the right The coefficient of kinetic friction between the block and surface is mk Determine the magnitude of the acceleration of the two objects y a m1 F sin u x u n F T F u T F cos u fk m2 a m 1g m 2g m2 (a) (c) (b) S SOLUTION S Conceptualize Imagine what happens as F is applied to S the block Assuming F is not large enough to lift the block, the block slides to the right and the ball rises Figure 5.20 (Example 5.13) (a) The external force F applied as shown can cause the block to accelerate to the right (b, c) The free-body diagrams assuming the block accelerates to the right and the ball accelerates upward The magnitude of the force of kinetic friction in this case is given by fk ϭ mkn ϭ mk (m1g Ϫ F sin u) Categorize We can identify forces and we want an acceleration, so we categorize this problem as one involving two particles under a net force, the ball and the block S Analyze First draw free-body diagrams for the two objects as shown in Figures 5.20b and 5.20c The applied force F has x and y components F cos u and F sin u, respectively Because the two objects are connected, we can equate the magnitudes of the x component of the acceleration of the block and the y component of the acceleration of the ball and call them both a Let us assume the motion of the block is to the right Apply the particle under a net force model to the block in the horizontal direction: (1) a Fx ϭ F cos u Ϫ fk Ϫ T ϭ m1ax ϭ m1a Apply the particle in equilibrium model to the block in the vertical direction: (2) a Fy ϭ n ϩ F sin u Ϫ m1g ϭ Apply the particle under a net force model to the ball in the vertical direction: (3) a Fy ϭ T Ϫ m2g ϭ m2ay ϭ m2a Problems At one moment, the pole makes an angle of 35.0° with the vertical and the water exerts a horizontal drag force of 47.5 N on the boat, opposite to its forward velocity of magnitude 0.857 m/s The mass of the boat including its cargo and the worker is 370 kg (a) The water exerts a buoyant force vertically upward on the boat Find the magnitude of this force (b) Model the forces as constant over a short interval of time to find the velocity of the boat 0.450 s after the moment described 129 exerted by the wind on the sail) and for n (the force exerted by the water on the keel) (b) Choose the x direction as 40.0° north of east and the y direction as 40.0° west of north Write Newton’s second law as two component equations and solve for n and P (c) Compare your solutions Do the results agree? Is one calculation significantly easier? 20 A bag of cement of weight 325 N hangs in equilibrium from three wires as shown in Figure P5.20 Two of the wires make angles u1 ϭ 60.0° and u2 ϭ 25.0° with the horizontal Assuming the system is in equilibrium, find the tensions T1, T2, and T3 in the wires u1 u2 © Tony Arruza/CORBIS T1 T2 T3 w Figure P5.15 16 A 3.00-kg object is moving in a plane, with its x and y coordinates given by x ϭ 5t Ϫ and y ϭ 3t ϩ 2, where x and y are in meters and t is in seconds Find the magnitude of the net force acting on this object at t ϭ 2.00 s 17 The distance between two telephone poles is 50.0 m When a 1.00-kg bird lands on the telephone wire midway between the poles, the wire sags 0.200 m Draw a freebody diagram of the bird How much tension does the bird produce in the wire? Ignore the weight of the wire 18 An iron bolt of mass 65.0 g hangs from a string 35.7 cm long The top end of the string is fixed Without touching it, a magnet attracts the bolt so that it remains stationary, displaced horizontally 28.0 cm to the right from the previously vertical line of the string (a) Draw a free-body diagram of the bolt (b) Find the tension in the string (c) Find the magnetic force on the bolt 19 ⅷ Figure P5.19 shows the horizontal forces acting on a sailboat moving north at constant velocity, seen from a point straight above its mast At its particular speed, the water exerts a 220-N drag force on the sailboat’s hull (a) Choose the x direction as east and the y direction as north Write two component equations representing Newton’s second law Solve the equations for P (the force Figure P5.20 Problems 20 and 21 21 A bag of cement of weight Fg hangs in equilibrium from three wires as shown in Figure P5.20 Two of the wires make angles u1 and u2 with the horizontal Assuming the system is in equilibrium, show that the tension in the lefthand wire is T1 ϭ Fg cos u sin 1u ϩ u 2 22 ⅷ You are a judge in a children’s kite-flying contest, and two children will win prizes, one for the kite that pulls the most strongly on its string and one for the kite that pulls the least strongly on its string To measure string tensions, you borrow a mass hanger, some slotted masses, and a protractor from your physics teacher, and you use the following protocol, illustrated in Figure P5.22 Wait for a child to get her kite well controlled, hook the hanger onto the kite string about 30 cm from her hand, pile on slotted masses until that section of string is horizontal, record the mass required, and record the angle between the horizontal and the string running up to the kite (a) Explain how this method works As you construct your explanation, imagine that the children’s parents ask you about your method, that they might make false assumptions about your ability P 40.0Њ n N W E S 220 N Figure P5.22 Figure P5.19 = intermediate; = challenging; Ⅺ = SSM/SG; ᮡ = ThomsonNow; Ⅵ = symbolic reasoning; ⅷ = qualitative reasoning 130 Chapter The Laws of Motion 27 Figure P5.27 shows the speed of a person’s body as he does a chin-up Assume the motion is vertical and the mass of the person’s body is 64.0 kg Determine the force exerted by the chin-up bar on his body at (a) time zero, (b) time 0.5 s, (c) time 1.1 s, and (d) time 1.6 s 30 speed (cm/s) without concrete evidence, and that your explanation is an opportunity to give them confidence in your evaluation technique (b) Find the string tension if the mass is 132 g and the angle of the kite string is 46.3° 23 The systems shown in Figure P5.23 are in equilibrium If the spring scales are calibrated in newtons, what they read? Ignore the masses of the pulleys and strings, and assume the pulleys and the incline in part (d) are frictionless 5.00 kg 5.00 kg 20 10 5.00 kg (a) (b) 0.5 1.5 2.0 Figure P5.27 5.00 kg 30.0Њ 5.00 kg 1.0 time (s) 5.00 kg (d) (c) Figure P5.23 24 Draw a free-body diagram of a block that slides down a frictionless plane having an inclination of u ϭ 15.0° The block starts from rest at the top, and the length of the incline is 2.00 m Find (a) the acceleration of the block and (b) its speed when it reaches the bottom of the incline 25 ᮡ A 1.00-kg object is observed to have an acceleration of 10.0 m/s2 in a direction 60.0° east of north (Fig P5.25) S The force F2 exerted on the object has a magnitude of 5.00 N and is directed north Determine the magnitude S and direction of the force F1 acting on the object 28 Two objects are connected by a light string that passes over a frictionless pulley as shown in Figure P5.28 Draw free-body diagrams of both objects Assuming the incline is frictionless, m1 ϭ 2.00 kg, m2 ϭ 6.00 kg, and u ϭ 55.0°, find (a) the accelerations of the objects, (b) the tension in the string, and (c) the speed of each object 2.00 s after they are released from rest m1 m2 u 60.0Њ F2 aϭ 10 /s m Figure P5.28 1.00 kg F1 Figure P5.25 26 A 5.00-kg object placed on a frictionless, horizontal table is connected to a string that passes over a pulley and then is fastened to a hanging 9.00-kg object as shown in Figure P5.26 Draw free-body diagrams of both objects Find the acceleration of the two objects and the tension in the string ᮡ A block is given an initial velocity of 5.00 m/s up a frictionless 20.0° incline How far up the incline does the block slide before coming to rest? 30 In Figure P5.30, the man and the platform together weigh 950 N The pulley can be modeled as frictionless Determine how hard the man has to pull on the rope to lift himself steadily upward above the ground (Or is it impossible? If so, explain why.) 29 5.00 kg 9.00 kg Figure P5.26 = intermediate; Figure P5.30 Problems 26 and 41 = challenging; Ⅺ = SSM/SG; ᮡ = ThomsonNow; Ⅵ = symbolic reasoning; ⅷ = qualitative reasoning Problems 131 S 31 In the system shown in Figure P5.31, a horizontal force Fx acts on the 8.00-kg object The horizontal surface is frictionless Consider the acceleration of the sliding object as a function of Fx (a) For what values of Fx does the 2.00-kg object accelerate upward? (b) For what values of Fx is the tension in the cord zero? (c) Plot the acceleration of the 8.00-kg object versus Fx Include values of Fx from Ϫ100 N to ϩ100 N 8.00 kg Fx 2.00 kg Figure P5.31 32 An object of mass m1 on a frictionless horizontal table is connected to an object of mass m2 through a very light pulley P1 and a light fixed pulley P2 as shown in Figure P5.32 (a) If a1 and a2 are the accelerations of m1 and m2, respectively, what is the relation between these accelerations? Express (b) the tensions in the strings and (c) the accelerations a1 and a2 in terms of g and of the masses m1 and m2 P1 36 A 25.0-kg block is initially at rest on a horizontal surface A horizontal force of 75.0 N is required to set the block in motion, after which a horizontal force of 60.0 N is required to keep the block moving with constant speed Find the coefficients of static and kinetic friction from this information 37 Your 3.80-kg physics book is next to you on the horizontal seat of your car The coefficient of static friction between the book and the seat is 0.650, and the coefficient of kinetic friction is 0.550 Suppose you are traveling at 72.0 km/h ϭ 20.0 m/s and brake to a stop over a distance of 45.0 m (a) Will the book start to slide over the seat? (b) What force does the seat exert on the book in this process? 38 ⅷ Before 1960, it was believed that the maximum attainable coefficient of static friction for an automobile tire was less than Then, around 1962, three companies independently developed racing tires with coefficients of 1.6 Since then, tires have improved, as illustrated in this problem According to the 1990 Guinness Book of Records, the fastest time interval for a piston-engine car initially at rest to cover a distance of one-quarter mile is 4.96 s Shirley Muldowney set this record in September 1989 (a) Assume the rear wheels lifted the front wheels off the pavement as shown in Figure P5.38 What minimum value of ms is necessary to achieve the record time interval? (b) Suppose Muldowney were able to double her engine power, keeping other things equal How would this change affect the time interval? P2 Jamie Squire/Allsport/Getty Images m1 m2 Figure P5.32 Figure P5.38 33 A 72.0-kg man stands on a spring scale in an elevator Starting from rest, the elevator ascends, attaining its maximum speed of 1.20 m/s in 0.800 s It travels with this constant speed for the next 5.00 s The elevator then undergoes a uniform acceleration in the negative y direction for 1.50 s and comes to rest What does the spring scale register (a) before the elevator starts to move, (b) during the first 0.800 s, (c) while the elevator is traveling at constant speed, and (d) during the time interval it is slowing down? 34 In the Atwood machine shown in Figure 5.14a, m1 ϭ 2.00 kg and m2 ϭ 7.00 kg The masses of the pulley and string are negligible by comparison The pulley turns without friction and the string does not stretch The lighter object is released with a sharp push that sets it into motion at vi ϭ 2.40 m/s downward (a) How far will m1 descend below its initial level? (b) Find the velocity of m1 after 1.80 seconds ᮡ A 3.00-kg block starts from rest at the top of a 30.0° incline and slides a distance of 2.00 m down the incline in 1.50 s Find (a) the magnitude of the acceleration of the block, (b) the coefficient of kinetic friction between block and plane, (c) the friction force acting on the block, and (d) the speed of the block after it has slid 2.00 m 40 A woman at an airport is towing her 20.0-kg suitcase at constant speed by pulling on a strap at an angle u above the horizontal (Fig P5.40) She pulls on the strap with a 35.0-N force The friction force on the suitcase is 20.0 N Draw a free-body diagram of the suitcase (a) What angle does the strap make with the horizontal? (b) What normal force does the ground exert on the suitcase? 39 Section 5.8 Forces of Friction 35 A car is traveling at 50.0 mi/h on a horizontal highway (a) If the coefficient of static friction between road and tires on a rainy day is 0.100, what is the minimum distance in which the car will stop? (b) What is the stopping distance when the surface is dry and ms ϭ 0.600? = intermediate; = challenging; Ⅺ = SSM/SG; ᮡ = ThomsonNow; u Figure P5.40 Ⅵ = symbolic reasoning; ⅷ = qualitative reasoning 132 Chapter The Laws of Motion 41 A 9.00-kg hanging object is connected, by a light, inextensible cord over a light, frictionless pulley, to a 5.00-kg block that is sliding on a flat table (Fig P5.26) Taking the coefficient of kinetic friction as 0.200, find the tension in the string 42 Three objects are connected on a table as shown in Figure P5.42 The rough table has a coefficient of kinetic friction of 0.350 The objects have masses of 4.00 kg, 1.00 kg, and 2.00 kg, as shown, and the pulleys are frictionless Draw a free-body diagram for each object (a) Determine the acceleration of each object and their directions (b) Determine the tensions in the two cords and downward as shown in Figure P5.45 Assume the force is applied at an angle of 37.0° below the horizontal (a) Find the acceleration of the block as a function of P (b) If P ϭ 5.00 N, find the acceleration and the friction force exerted on the block (c) If P ϭ 10.0 N, find the acceleration and the friction force exerted on the block (d) Describe in words how the acceleration depends on P Is there a definite minimum acceleration for the block? If so, what is it? Is there a definite maximum? P 1.00 kg Figure P5.45 4.00 kg 2.00 kg Figure P5.42 43 Two blocks connected by a rope of negligible mass are being dragged by a horizontal force (Fig P5.43) Suppose F ϭ 68.0 N, m1 ϭ 12.0 kg, m2 ϭ 18.0 kg, and the coefficient of kinetic friction between each block and the surface is 0.100 (a) Draw a free-body diagram for each block (b) Determine the tension T and the magnitude of the acceleration of the system T m1 m2 46 Review problem One side of the roof of a building slopes up at 37.0° A student throws a Frisbee onto the roof It strikes with a speed of 15.0 m/s, does not bounce, and then slides straight up the incline The coefficient of kinetic friction between the plastic and the roof is 0.400 The Frisbee slides 10.0 m up the roof to its peak, where it goes into free fall, following a parabolic trajectory with negligible air resistance Determine the maximum height the Frisbee reaches above the point where it struck the roof 47 The board sandwiched between two other boards in Figure P5.47 weighs 95.5 N If the coefficient of friction between the boards is 0.663, what must be the magnitude of the compression forces (assumed horizontal) acting on both sides of the center board to keep it from slipping? F Figure P5.43 44 ⅷ A block of mass 3.00 kg is pushed against a wall by a S force P that makes a u ϭ 50.0° angle with the horizontal as shown in Figure P5.44 The coefficient of static friction between the block and the wall is 0.250.S (a) Determine the possible values for the magnitude of P that allow the block to remain stationary (b) Describe what happens if S P has a larger value and what happens if it is smaller (c) Repeat parts (a) and (b) assuming the force makes an angle of u ϭ 13.0° with the horizontal u P Figure P5.44 45 ⅷ A 420-g block is at rest on a horizontal surface The coefficient of static friction between the block and the surface is 0.720, and the coefficient of kinetic friction is 0.340 A force of magnitude P pushes the block forward = intermediate; = challenging; Ⅺ = SSM/SG; ᮡ Figure P5.47 48 A magician pulls a tablecloth from under a 200-g mug located 30.0 cm from the edge of the cloth The cloth exerts a friction force of 0.100 N on the mug, and the cloth is pulled with a constant acceleration of 3.00 m/s2 How far does the mug move relative to the horizontal tabletop before the cloth is completely out from under it? Note that the cloth must move more than 30 cm relative to the tabletop during the process 49 ⅷ A package of dishes (mass 60.0 kg) sits on the flatbed of a pickup truck with an open tailgate The coefficient of static friction between the package and the truck’s flatbed is 0.300, and the coefficient of kinetic friction is 0.250 (a) The truck accelerates forward on level ground What is the maximum acceleration the truck can have so that the package does not slide relative to the truck bed? (b) The truck barely exceeds this acceleration and then moves with constant acceleration, with the package sliding along its bed What is the acceleration of the package relative to the ground? (c) The driver cleans up the frag- = ThomsonNow; Ⅵ = symbolic reasoning; ⅷ = qualitative reasoning Problems ments of dishes and starts over again with an identical package at rest in the truck The truck accelerates up a hill inclined at 10.0° with the horizontal Now what is the maximum acceleration the truck can have such that the package does not slide relative to the flatbed? (d) When the truck exceeds this acceleration, what is the acceleration of the package relative to the ground? (e) For the truck parked at rest on a hill, what is the maximum slope the hill can have such that the package does not slide? (f) Is any piece of data unnecessary for the solution in all the parts of this problem? Explain Additional Problems 50 The following equations describe the motion of a system of two objects: ϩn Ϫ 16.50 kg2 19.80 m>s2 cos 13.0° ϭ fk ϭ 0.360n ϩT ϩ 16.50 kg2 19.80 m>s2 sin 13.0° Ϫ fk ϭ 16.50 kg2a ϪT ϩ 13.80 kg2 19.80 m>s2 ϭ 13.80 kg2 a (a) Solve the equations for a and T (b) Describe a situation to which these equations apply Draw free-body diagrams for both objects 51 An inventive child named Pat wants to reach an apple in a tree without climbing the tree Sitting in a chair connected to a rope that passes over a frictionless pulley (Fig P5.51), Pat pulls on the loose end of the rope with such a force that the spring scale reads 250 N Pat’s true weight is 320 N, and the chair weighs 160 N (a) Draw free-body diagrams for Pat and the chair considered as separate systems, and another diagram for Pat and the chair considered as one system (b) Show that the acceleration of the system is upward and find its magnitude (c) Find the force Pat exerts on the chair Figure P5.51 Problems 51 and 52 52 ⅷ In the situation described in Problem 51 and Figure P5.51, the masses of the rope, spring balance, and pulley are negligible Pat’s feet are not touching the ground (a) Assume Pat is momentarily at rest when he stops pulling down on the rope and passes the end of the rope to another child, of weight 440 N, who is standing on the ground next to him The rope does not break Describe the ensuing motion (b) Instead, assume Pat is momentar2 = intermediate; = challenging; Ⅺ = SSM/SG; ᮡ 133 ily at rest when he ties the end of the rope to a strong hook projecting from the tree trunk Explain why this action can make the rope break S 53 A time-dependent force, F ϭ 18.00ˆi Ϫ 4.00t ˆj N, where t is in seconds, is exerted on a 2.00-kg object initially at rest (a) At what time will the object be moving with a speed of 15.0 m/s? (b) How far is the object from its initial position when its speed is 15.0 m/s? (c) Through what total displacement has the object traveled at this moment? 54 ⅷ Three blocks are in contact with one another on a frictionless, horizontal surface as shown in Figure P5.54 A S horizontal force F is applied to m1 Take m1 ϭ 2.00 kg, m2 ϭ 3.00 kg, m3 ϭ 4.00 kg, and F ϭ 18.0 N Draw a separate free-body diagram for each block and find (a) the acceleration of the blocks, (b) the resultant force on each block, and (c) the magnitudes of the contact forces between the blocks (d) You are working on a construction project A coworker is nailing plasterboard on one side of a light partition, and you are on the opposite side, providing “backing” by leaning against the wall with your back pushing on it Every hammer blow makes your back sting The supervisor helps you to put a heavy block of wood between the wall and your back Using the situation analyzed in parts (a), (b), and (c) as a model, explain how this change works to make your job more comfortable F m1 m2 m3 Figure P5.54 55 ⅷ A rope with mass m1 is attached to the bottom front edge of a block with mass 4.00 kg Both the rope and the block rest on a horizontal frictionless surface The rope does not stretch The free end of the rope is pulled with a horizontal force of 12.0 N (a) Find the acceleration of the system, as it depends on m1 (b) Find the magnitude of the force the rope exerts on the block, as it depends on m1 (c) Evaluate the acceleration and the force on the block for m1 ϭ 0.800 kg Suggestion: You may find it easier to part (c) before parts (a) and (b) What If? (d) What happens to the force on the block as the rope’s mass grows beyond all bounds? (e) What happens to the force on the block as the rope’s mass approaches zero? (f) What theorem can you state about the tension in a light cord joining a pair of moving objects? 56 A black aluminum glider floats on a film of air above a level aluminum air track Aluminum feels essentially no force in a magnetic field, and air resistance is negligible A strong magnet is attached to the top of the glider, forming a total mass of 240 g A piece of scrap iron attached to one end stop on the track attracts the magnet with a force of 0.823 N when the iron and the magnet are separated by 2.50 cm (a) Find the acceleration of the glider at this instant (b) The scrap iron is now attached to another green glider, forming a total mass of 120 g Find the acceleration of each glider when they are simultaneously released at 2.50-cm separation = ThomsonNow; Ⅵ = symbolic reasoning; ⅷ = qualitative reasoning 134 57 Chapter The Laws of Motion ᮡ An Sobject of mass M is held in place by an applied force F and a pulley system as shown in Figure P5.57 The pulleys are massless and frictionless Find (a) the tension in each sectionS of rope, T1, T2, T3, T4, and T5 and (b) the magnitude of F Suggestion: Draw a free-body diagram for each pulley T4 on the section of cable between the cars? What velocity you predict for it 0.01 s into the future? Explain the motion of this section of cable in cause-and-effect terms 60 A 2.00-kg aluminum block and a 6.00-kg copper block are connected by a light string over a frictionless pulley They sit on a steel surface as shown in Figure P5.60, where u ϭ 30.0° When they are released from rest, will they start to move? If so, determine (a) their acceleration and (b) the tension in the string If not, determine the sum of the magnitudes of the forces of friction acting on the blocks Aluminum T1 Copper m1 T2 T m2 Steel T5 u M F Figure P5.60 S 61 A crate of weight Fg is pushed by a force P on a horizontal S floor (a) The coefficient of static friction is ms, and P is directed at angle u below the horizontal Show that the minimum value of P that will move the crate is given by Figure P5.57 58 ⅷ A block of mass 2.20 kg is accelerated across a rough surface by a light cord passing over a small pulley as shown in Figure P5.58 The tension T in the cord is maintained at 10.0 N, and the pulley is 0.100 m above the top of the block The coefficient of kinetic friction is 0.400 (a) Determine the acceleration of the block when x ϭ 0.400 m (b) Describe the general behavior of the acceleration as the block slides from a location where x is large to x ϭ (c) Find the maximum value of the acceleration and the position x for which it occurs (d) Find the value of x for which the acceleration is zero Pϭ Ϫ m s tan u (b) Find the minimum value of P that can produce motion when ms ϭ 0.400, Fg ϭ 100 N, and u ϭ 0°, 15.0°, 30.0°, 45.0°, and 60.0° 62 Review problem A block of mass m ϭ 2.00 kg is released from rest at h ϭ 0.500 m above the surface of a table, at the top of a u ϭ 30.0° incline as shown in Figure P5.62 The frictionless incline is fixed on a table of height H ϭ 2.00 m (a) Determine the acceleration of the block as it slides down the incline (b) What is the velocity of the block as it leaves the incline? (c) How far from the table will the block hit the floor? (d) What time interval elapses between when the block is released and when it hits the floor? (e) Does the mass of the block affect any of the above calculations? T M ms Fg sec u m x h u Figure P5.58 H 59 ⅷ Physics students from San Diego have come in first and second in a contest and are down at the docks, watching their prizes being unloaded from a freighter On a single light vertical cable that does not stretch, a crane is lifting a 207-kg Ferrari and, below it, a 461-kg red BMW Z8 The Ferrari is moving upward with speed 3.50 m/s and acceleration 1.25 m/s2 (a) How the velocity and acceleration of the BMW compare with those of the Ferrari? (b) Find the tension in the cable between the BMW and the Ferrari (c) Find the tension in the cable above the Ferrari (d) In our model, what is the total force exerted = intermediate; = challenging; Ⅺ = SSM/SG; ᮡ R Figure P5.62 Problems 62 and 68 63 ⅷ A couch cushion of mass m is released from rest at the top of a building having height h A wind blowing along the side of the building exerts a constant horizontal force of magnitude F on the cushion as it drops as shown in Figure P5.63 The air exerts no vertical force (a) Show that the path of the cushion is a straight line (b) Does = ThomsonNow; Ⅵ = symbolic reasoning; ⅷ = qualitative reasoning Problems Ꭽ Cushion Wind force h R Figure P5.63 the cushion fall with constant velocity? Explain (c) If m ϭ 1.20 kg, h ϭ 8.00 m, and F ϭ 2.40 N, how far from the building will the cushion hit the level ground? What If? (d) If the cushion is thrown downward with a nonzero speed at the top of the building, what will be the shape of its trajectory? Explain 64 A student is asked to measure the acceleration of a cart on a “frictionless” inclined plane as shown in Figure 5.11, using an air track, a stopwatch, and a meter stick The height of the incline is measured to be 1.774 cm, and the total length of the incline is measured to be d ϭ 127.1 cm Hence, the angle of inclination u is determined from the relation sin u ϭ 1.774/127.1 The cart is released from rest at the top of the incline, and its position x along the incline is measured as a function of time, where x ϭ refers to the cart’s initial position For x values of 10.0 cm, 20.0 cm, 35.0 cm, 50.0 cm, 75.0 cm, and 100 cm, the measured times at which these positions are reached (averaged over five runs) are 1.02 s, 1.53 s, 2.01 s, 2.64 s, 3.30 s, and 3.75 s, respectively Construct a graph of x versus t 2, and perform a linear least-squares fit to the data Determine the acceleration of the cart from the slope of this graph, and compare it with the value you would get using a ϭ g sin u, where g ϭ 9.80 m/s2 65 A 1.30-kg toaster is not plugged in The coefficient of static friction between the toaster and a horizontal countertop is 0.350 To make the toaster start moving, you carelessly pull on its electric cord (a) For the cord tension to be as small as possible, you should pull at what angle above the horizontal? (b) With this angle, how large must the tension be? 66 ⅷ In Figure P5.66, the pulleys and the cords are light, all surfaces are frictionless, and the cords not stretch (a) How does the acceleration of block compare with the acceleration of block 2? Explain your reasoning (b) The mass of block is 1.30 kg Find its acceleration as it depends on the mass m1 of block (c) Evaluate your 135 answer for m1 ϭ 0.550 kg Suggestion: You may find it easier to part (c) before part (b) What If? (d) What does the result of part (b) predict if m1 is very much less than 1.30 kg? (e) What does the result of part (b) predict if m1 approaches infinity? (f) What is the tension in the long cord in this last case? (g) Could you anticipate the answers (d), (e), and (f) without first doing part (b)? Explain 67 What horizontal force must be applied to the cart shown in Figure P5.67 so that the blocks remain stationary relative to the cart? Assume all surfaces, wheels, and pulley are frictionless Notice that the force exerted by the string accelerates m1 m1 m2 M F Figure P5.67 68 In Figure P5.62, the incline has mass M and is fastened to the stationary horizontal tabletop The block of mass m is placed near the bottom of the incline and is released with a quick push that sets it sliding upward The block stops near the top of the incline, as shown in the figure, and then slides down again, always without friction Find the force that the tabletop exerts on the incline throughout this motion 69 A van accelerates down a hill (Fig P5.69), going from rest to 30.0 m/s in 6.00 s During the acceleration, a toy (m ϭ 0.100 kg) hangs by a string from the van’s ceiling The acceleration is such that the string remains perpendicular to the ceiling Determine (a) the angle u and (b) the tension in the string u u Figure P5.69 An 8.40-kg object slides down a fixed, frictionless inclined plane Use a computer to determine and tabulate the normal force exerted on the object and its acceleration for a series of incline angles (measured from the horizontal) ranging from 0° to 90° in 5° increments Plot a graph of the normal force and the acceleration as functions of the incline angle In the limiting cases of 0° and 90°, are your results consistent with the known behavior? 71 A mobile is formed by supporting four metal butterflies of equal mass m from a string of length L The points of support are evenly spaced a distance ᐉ apart as shown in 70 m1 m2 Figure P5.66 = intermediate; = challenging; Ⅺ = SSM/SG; ᮡ = ThomsonNow; Ⅵ = symbolic reasoning; ⅷ = qualitative reasoning 136 Chapter The Laws of Motion Figure P5.71 The string forms an angle u1 with the ceiling at each endpoint The center section of string is horizontal (a) Find the tension in each section of string in terms of u1, m, and g (b) Find the angle u2, in terms of u1, that the sections of string between the outside butterflies and the inside butterflies form with the horizontal (c) Show that the distance D between the endpoints of the string is Dϭ D ᐉ u1 u2 ᐉ L 12 cos u ϩ cos 3tanϪ1 12 tan u ϩ u1 u2 ᐉ ᐉ ᐉ m L ϭ 5ᐉ m m m Figure P5.71 Answers to Quick Quizzes 5.1 (d) Choice (a) is true Newton’s first law tells us that motion requires no force: an object in motion continues to move at constant velocity in the absence of external forces Choice (b) is also true A stationary object can have several forces acting on it, but if the vector sum of all these external forces is zero, there is no net force and the object remains stationary 5.2 (a) If a single force acts, this force constitutes the net force and there is an acceleration according to Newton’s second law 5.3 (d) With twice the force, the object will experience twice the acceleration Because the force is constant, the acceleration is constant, and the speed of the object (starting from rest) is given by v ϭat With twice the acceleration, the object will arrive at speed v at half the time 5.4 (b) Because the value of g is smaller on the Moon than on the Earth, more mass of gold would be required to represent newton of weight on the Moon Therefore, your friend on the Moon is richer, by about a factor of 6! = intermediate; = challenging; Ⅺ = SSM/SG; ᮡ 5.5 (i), (c) In accordance with Newton’s third law, the fly and bus experience forces that are equal in magnitude but opposite in direction (ii), (a) Because the fly has such a small mass, Newton’s second law tells us that it undergoes a very large acceleration The large mass of the bus means that it more effectively resists any change in its motion and exhibits a small acceleration 5.6 (b) The friction force acts opposite to the gravitational force on the book to keep the book in equilibrium Because the gravitational force is downward, the friction force must be upward 5.7 (b) When pulling with the rope, there is a component of your applied force that is upward, which reduces the normal force between the sled and the snow In turn, the friction force between the sled and the snow is reduced, making the sled easier to move If you push from behind with a force with a downward component, the normal force is larger, the friction force is larger, and the sled is harder to move = ThomsonNow; Ⅵ = symbolic reasoning; ⅷ = qualitative reasoning 6.1 Newton’s Second Law for a Particle in Uniform Circular Motion 6.2 Nonuniform Circular Motion 6.3 Motion in Accelerated Frames 6.4 Motion in the Presence of Resistive Forces Passengers on a “corkscrew” roller coaster experience a radial force toward the center of the circular track and a downward force due to gravity (Robin Smith / Getty Images) Circular Motion and Other Applications of Newton’s Laws In the preceding chapter, we introduced Newton’s laws of motion and applied them to situations involving linear motion Now we discuss motion that is slightly more complicated For example, we shall apply Newton’s laws to objects traveling in circular paths We shall also discuss motion observed from an accelerating frame of reference and motion of an object through a viscous medium For the most part, this chapter consists of a series of examples selected to illustrate the application of Newton’s laws to a variety of circumstances 6.1 Newton’s Second Law for a Particle in Uniform Circular Motion In Section 4.4, we discussed the model of a particle in uniform circular motion, in which a particle moves with constant speed v in a circular path of radius r The particle experiences an acceleration that has a magnitude ac ϭ v2 r S The acceleration is called centripetal acceleration because ac is directed toward the S S center of the circle Furthermore, ac is always perpendicular to v (If there were a S component of acceleration parallel to v, the particle’s speed would be changing.) 137 138 Chapter Circular Motion and Other Applications of Newton’s Laws Fr m r r Fr v Figure 6.1 An overhead view of a ball moving in a circular path in a S horizontal plane A force Fr directed toward the center of the circle keeps the ball moving in its circular path ACTIVE FIGURE 6.2 An overhead view of a ball moving in a circular path in a horizontal plane When the string breaks, the ball moves in the direction tangent to the circle Sign in at www.thomsonedu.com and go to ThomsonNOW to “break” the string yourself and observe the effect on the ball’s motion Let us now incorporate the concept of force in the particle in uniform circular motion model Consider a ball of mass m that is tied to a string of length r and is being whirled at constant speed in a horizontal circular path as illustrated in Figure 6.1 Its weight is supported by a frictionless table Why does the ball move in a circle? According to Newton’s first law, the ball would move in a straight line if there were no force on it; the string, however, prevents motion along a straight S line by exerting on the ball a radial force Fr that makes it follow the circular path This force is directed along the string toward the center of the circle as shown in Figure 6.1 If Newton’s second law is applied along the radial direction, the net force causing the centripetal acceleration can be related to the acceleration as follows: Force causing centripetal acceleration PITFALL PREVENTION 6.1 Direction of Travel When the String Is Cut Study Active Figure 6.2 very carefully Many students (wrongly) think that the ball will move radially away from the center of the circle when the string is cut The velocity of the ball is tangent to the circle By Newton’s first law, the ball continues to move in the same direction in which it is moving just as the force from the string disappears E XA M P L E v2 a F ϭ mac ϭ m¬ r ᮣ (6.1) A force causing a centripetal acceleration acts toward the center of the circular path and causes a change in the direction of the velocity vector If that force should vanish, the object would no longer move in its circular path; instead, it would move along a straight-line path tangent to the circle This idea is illustrated in Active Figure 6.2 for the ball whirling at the end of a string in a horizontal plane If the string breaks at some instant, the ball moves along the straight-line path that is tangent to the circle at the position of the ball at this instant Quick Quiz 6.1 You are riding on a Ferris wheel that is rotating with constant speed The car in which you are riding always maintains its correct upward orientation; it does not invert (i) What is the direction of the normal force on you from the seat when you are at the top of the wheel? (a) upward (b) downward (c) impossible to determine (ii) From the same choices, what is the direction of the net force on you when you are at the top of the wheel? The Conical Pendulum A small ball of mass m is suspended from a string of length L The ball revolves with constant speed v in a horizontal circle of radius r as shown in Figure 6.3 (Because the string sweeps out the surface of a cone, the system is known as a conical pendulum.) Find an expression for v Section 6.1 Newton’s Second Law for a Particle in Uniform Circular Motion 139 SOLUTION Conceptualize Imagine the motion of the ball in Figure 6.3a and convince yourself that the string sweeps out a cone and that the ball moves in a circle L u T cos u T u r Categorize The ball in Figure 6.3 does not accelerate vertically Therefore, we model it as a particle in equilibrium in the vertical direction It experiences a centripetal acceleration in the horizontal direction, so it is modeled as a particle in uniform circular motion in this direction T sin u mg mg (a) (b) Analyze Let u represent the angle between the string and the vertical In S the free-body diagram shown in Figure 6.3b, the force T exerted by the string is resolved into a vertical component T cos u and a horizontal component T sin u acting toward the center of the circular path Figure 6.3 (Example 6.1) (a) A conical pendulum The path of the object is a horizontal circle (b) The free-body diagram for the object Apply the particle in equilibrium model in the vertical direction: a Fy ϭ T cos¬u Ϫ mg ϭ Use Equation 6.1 to express the force providing the centripetal acceleration in the horizontal direction: Divide Equation (2) by Equation (1) and use sin u/cos u ϭ tan u: T cos u ϭ mg (1) (2) mv a Fx ϭ T sin u ϭ mac ϭ r tan u ϭ v2 rg Solve for v: v ϭ 2rg tan u Incorporate r ϭ L sin u from the geometry in Figure 6.3a: v ϭ 2Lg sin u tan u Finalize Notice that the speed is independent of the mass of the ball Consider what happens when u goes to 90° so that the string is horizontal Because the tangent of 90° is infinite, the speed v is infinite, which tells us the string S cannot possibly be horizontal If it were, there would be no vertical component of the force T to balance the gravitational force on the ball That is why we mentioned in regard to Figure 6.1 that the ball’s weight in the figure is supported by a frictionless table E XA M P L E How Fast Can It Spin? A ball of mass 0.500 kg is attached to the end of a cord 1.50 m long The ball is whirled in a horizontal circle as shown in Figure 6.1 If the cord can withstand a maximum tension of 50.0 N, what is the maximum speed at which the ball can be whirled before the cord breaks? Assume the string remains horizontal during the motion SOLUTION Conceptualize It makes sense that the stronger the cord, the faster the ball can whirl before the cord breaks Also, we expect a more massive ball to break the cord at a lower speed (Imagine whirling a bowling ball on the cord!) Categorize Because the ball moves in a circular path, we model it as a particle in uniform circular motion Analyze Incorporate the tension and the centripetal acceleration into Newton’s second law: Solve for v : Tϭm (1) vϭ v2 r Tr Bm 140 Chapter Circular Motion and Other Applications of Newton’s Laws Find the maximum speed the ball can have, which corresponds to the maximum tension the string can withstand: v max ϭ 150.0 N2 11.50 m Tmaxr ϭ ϭ 12.2 m>s B m B 0.500 kg Finalize Equation (1) shows that v increases with T and decreases with larger m, as we expected from our conceptualization of the problem What If? Suppose the ball is whirled in a circle of larger radius at the same speed v Is the cord more likely or less likely to break? Answer The larger radius means that the change in the direction of the velocity vector will be smaller in a given time interval Therefore, the acceleration is smaller and the required tension in the string is smaller As a result, the string is less likely to break when the ball travels in a circle of larger radius E XA M P L E What Is the Maximum Speed of the Car? fs A 500-kg car moving on a flat, horizontal road negotiates a curve as shown in Figure 6.4a If the radius of the curve is 35.0 m and the coefficient of static friction between the tires and dry pavement is 0.523, find the maximum speed the car can have and still make the turn successfully SOLUTION Conceptualize Imagine that the curved roadway is part of a large circle so that the car is moving in a circular path (a) Categorize Based on the conceptualize step of the problem, we model the car as a particle in uniform circular motion in the horizontal direction The car is not accelerating vertically, so it is modeled as a particle in equilibrium in the vertical direction n Analyze The force that enables the car to remain in its circular path is the force of static friction (It is static because no slipping occurs at the point of contact between road and tires If this force of static friction were zero—for example, if the car were on an icy road—the car would continue in a straight line and slide off the road.) The maximum speed vmax the car can have around the curve is the speed at which it is on the verge of skidding outward At this point, the friction force has its maximum value fs,max ϭ msn mg (b) Figure 6.4 (Example 6.3) (a) The force of static friction directed toward the center of the curve keeps the car moving in a circular path (b) The free-body diagram for the car (1) Apply Equation 6.1 in the radial direction for the maximum speed condition: a Fy ϭ Apply the particle in equilibrium model to the car in the vertical direction: Solve Equation (1) for the maximum speed and substitute for n: fs (2) v max ϭ v 2max fs,max ϭ m sn ϭ m¬ r S n Ϫ mg ϭ S n ϭ mg m smgr m snr ϭ ϭ 2ms gr m B B m ϭ 10.5232 19.80 m>s2 135.0 m ϭ 13.4 m>s Finalize This speed is equivalent to 30.0 mi/h Therefore, this roadway could benefit greatly from some banking, as in the next example! Notice that the maximum speed does not depend on the mass of the car, which is why curved highways not need multiple speed limits to cover the various masses of vehicles using the road What If? Suppose a car travels this curve on a wet day and begins to skid on the curve when its speed reaches only 8.00 m/s What can we say about the coefficient of static friction in this case? Section 6.1 Newton’s Second Law for a Particle in Uniform Circular Motion 141 Answer The coefficient of static friction between tires and a wet road should be smaller than that between tires and a dry road This expectation is consistent with experience with driving because a skid is more likely on a wet road than a dry road To check our suspicion, we can solve Equation (2) for the coefficient of static friction: v2max ms ϭ gr Substituting the numerical values gives 18.00 m>s2 v 2max ms ϭ ϭ ϭ 0.187 gr 19.80 m>s2 135.0 m which is indeed smaller than the coefficient of 0.523 for the dry road E XA M P L E The Banked Roadway A civil engineer wishes to redesign the curved roadway in Example 6.3 in such a way that a car will not have to rely on friction to round the curve without skidding In other words, a car moving at the designated speed can negotiate the curve even when the road is covered with ice Such a ramp is usually banked, which means that the roadway is tilted toward the inside of the curve Suppose the designated speed for the ramp is to be 13.4 m/s (30.0 mi/h) and the radius of the curve is 35.0 m At what angle should the curve be banked? nx u n ny SOLUTION Conceptualize The difference between this example and Example 6.3 is that the car is no longer moving on a flat roadway Figure 6.5 shows the banked roadway, with the center of the circular path of the car far to the left of the figure Notice that the horizontal component of the normal force participates in causing the car’s centripetal acceleration u Figure 6.5 (Example 6.4) A car rounding a curve on a road banked at an angle u to the horizontal When friction is neglected, the force that causes the centripetal acceleration and keeps the car moving in its circular path is the horizontal component of the normal force Categorize As in Example 6.3, the car is modeled as a particle in equilibrium in the vertical direction and a particle in uniform circular motion in the horizontal direction Analyze On a level (unbanked) road, the force that causes the centripetal acceleration is the force of static friction between car and road as we saw in the preceding example If the road is banked at an angle u as in Figure 6.5, however, the norS mal force n has a horizontal component toward the center of the curve Because the ramp is to be designed so that the force of static friction is zero, only the component nx ϭ n sin u causes the centripetal acceleration Write Newton’s second law for the car in the radial direction, which is the x direction: (1) mv2 a Fr ϭ n sin¬u ϭ r a Fy ϭ n cos¬u Ϫ mg ϭ Apply the particle in equilibrium model to the car in the vertical direction: n cos¬u ϭ mg (2) (3) Divide Equation (1) by Equation (2): Solve for the angle u: Fg u ϭ tanϪ1 a tan¬u ϭ 113.4 m>s2 v2 rg 135.0 m2 ¬19.80 m>s2 b ϭ 27.6° Finalize Equation (3) shows that the banking angle is independent of the mass of the vehicle negotiating the curve If a car rounds the curve at a speed less than 13.4 m/s, friction is needed to keep it from sliding down the bank (to the left in Fig 6.5) A driver attempting to negotiate the curve at a speed greater than 13.4 m/s has to depend on friction to keep from sliding up the bank (to the right in Fig 6.5) 142 Chapter Circular Motion and Other Applications of Newton’s Laws What If? Imagine that this same roadway were built on Mars in the future to connect different colony centers Could it be traveled at the same speed? Answer The reduced gravitational force on Mars would mean that the car is not pressed as tightly to the roadway The reduced normal force results in a smaller component of the normal force toward the center of the circle This smaller component would not be sufficient to provide the centripetal acceleration associated E XA M P L E with the original speed The centripetal acceleration must be reduced, which can be done by reducing the speed v Mathematically, notice that Equation (3) shows that the speed v is proportional to the square root of g for a roadway of fixed radius r banked at a fixed angle u Therefore, if g is smaller, as it is on Mars, the speed v with which the roadway can be safely traveled is also smaller Let’s Go Loop-the-Loop! A pilot of mass m in a jet aircraft executes a loop-theloop, as shown in Figure 6.6a In this maneuver, the aircraft moves in a vertical circle of radius 2.70 km at a constant speed of 225 m/s (A) Determine the force exerted by the seat on the pilot at the bottom of the loop Express your answer in terms of the weight of the pilot mg n bot v Top R SOLUTION v ntop Conceptualize Look carefully at Figure 6.6a Based on mg mg Bottom experiences with driving over small hills on a road or riding at the top of a Ferris wheel, you would expect to (a) (b) (c) feel lighter at the top of the path Similarly, you would Figure 6.6 (Example 6.5) (a) An aircraft executes a loop-the-loop expect to feel heavier at the bottom of the path At the maneuver as it moves in a vertical circle at constant speed (b) The bottom of the loop the normal and gravitational forces free-body diagram for the pilot at the bottom of the loop In this posion the pilot act in opposite directions, whereas at the top tion the pilot experiences an apparent weight greater than his true weight (c) The free-body diagram for the pilot at the top of the loop of the loop these two forces act in the same direction The vector sum of these two forces gives a force of constant magnitude that keeps the pilot moving in a circular path at a constant speed To yield net force vectors with the same magnitude, the normal force at the bottom must be greater than that at the top Categorize Because the speed of the aircraft is constant (how likely is that?), we can categorize this problem as one involving a particle (the pilot) in uniform circular motion, complicated by the gravitational force acting at all times on the aircraft Analyze We draw a free-body diagram for the pilot at the bottom of the loop as shown in Figure 6.6b The only S S S forces acting on him are the downward gravitational force Fg ϭ m g and the upward force nbot exerted by the seat The net upward force on the pilot that provides his centripetal acceleration has a magnitude nbot – mg Apply Newton’s second law to the pilot in the radial direction: Solve for the force exerted by the seat on the pilot: Substitute the values given for the speed and radius: v2 F ϭ n Ϫ mg ϭ m bot a r nbot ϭ mg ϩ m nbot ϭ mg a ϩ v2 v2 ϭ mg a ϩ b r rg 1225 m>s2 12.70 ϫ 103 m 19.80 m>s2 b ϭ 2.91mg S Hence, the magnitude of the force nbot exerted by the seat on the pilot is greater than the weight of the pilot by a factor of 2.91 So, the pilot experiences an apparent weight that is greater than his true weight by a factor of 2.91 Section 6.2 Nonuniform Circular Motion 143 (B) Determine the force exerted by the seat on the pilot at the top of the loop SOLUTION Analyze The free-body diagram for the pilot at the top of the loop is shown in Figure 6.6c As noted earlier, both S the gravitational force exerted by the Earth and the force ntop exerted by the seat on the pilot act downward, so the net downward force that provides the centripetal acceleration has a magnitude ntop ϩ mg Apply Newton’s second law to the pilot at this position: v2 F ϭ n ϩ mg ϭ m top a r ntop ϭ m v2 v2 Ϫ mg ϭ mg a Ϫ b r rg ntop ϭ mg a 1225 m>s2 12.70 ϫ 103 m 19.80 m>s2 Ϫ 1b ϭ 0.913mg In this case, the magnitude of the force exerted by the seat on the pilot is less than his true weight by a factor of 0.913, and the pilot feels lighter Finalize The variations in the normal force are consistent with our prediction in the conceptualize step of the problem 6.2 Nonuniform Circular Motion In Chapter 4, we found that if a particle moves with varying speed in a circular path, there is, in addition to the radial component of acceleration, a tangential component having magnitude dv>dt Therefore, the force acting on the particle must also have a tangential and a radial component Because theStotal acceleration S S S S S is a ϭ ar ϩ at , the total force exerted on the particle is ͚ F ϭ ͚ Fr ϩ ͚ Ft as shown in Active Figure 6.7 (We express the radial and tangential forces as net forces with the summation notation because each force could consist of multiple forces that S combine.) The vector ͚ Fr is directed toward the center of the circle and is S responsible for the centripetal acceleration The vector ͚ Ft tangent to the circle is responsible for the tangential acceleration, which represents a change in the particle’s speed with time Quick Quiz 6.2 A bead slides freely along a curved wire lying on a horizontal surface at constant speed as shown in Figure 6.8 (a) Draw the vectors representing the force exerted by the wire on the bead at points Ꭽ, Ꭾ, and Ꭿ (b) Suppose the bead in Figure 6.8 speeds up with constant tangential acceleration as it moves toward the right Draw the vectors representing the force on the bead at points Ꭽ, Ꭾ, and Ꭿ ACTIVE FIGURE 6.7 ⌺F ⌺ Fr ⌺ Ft When the net force acting on a particle moving in a circular path has a tangential component ͚ Ft , the particle’s speed changes The net force exerted on the particle in this case isS the vector sum of the radial force S S and the tangential force That is, ͚ F ϭ ͚ Fr ϩ ͚ Ft Sign in at www.thomsonedu.com and go to ThomsonNOW to adjust the initial position of the particle and compare the component forces acting on the particle with those for a child swinging on a swing set Ꭽ Ꭾ Ꭿ Figure 6.8 (Quick Quiz 6.2) A bead slides along a curved wire