The Laws of Motion. chapter

Transcription

1 chpter The Lws of Motion The Concept of Force 5.2 Newton s First Lw nd Inertil Frmes 5.3 Mss 5.4 Newton s econd Lw 5.5 The Grvittionl Force nd Weight 5.6 Newton s Third Lw 5.7 Anlysis Models Using Newton s econd Lw 5.8 Forces of Friction In Chpters 2 nd 4, we described the motion of n object in terms of its position, velocity, nd ccelertion without considering wht might influence tht motion. Now we consider tht influence: Why does the motion of n object chnge? Wht might cuse one object to remin t rest nd nother object to ccelerte? Why is it generlly esier to move smll object thn lrge A person sculls on clm wterwy. The wter exerts forces on the ors to ccelerte the bot. (Tetr Imges/Getty Imges) object? The two min fctors we need to consider re the forces cting on n object nd the mss of the object. In this chpter, we begin our study of dynmics by discussing the three bsic lws of motion, which del with forces nd msses nd were formulted more thn three centuries go by Isc Newton. 5.1 The Concept of Force Everyone hs bsic understnding of the concept of force from everydy experience. When you push your empty dinner plte wy, you exert force on it. imilrly, you exert force on bll when you throw or kick it. In these exmples, the word force refers to n interction with n object by mens of musculr ctivity nd some chnge in the object s velocity. Forces do not lwys cuse motion, however. For exmple, when you re sitting, grvittionl force cts on your body nd yet _05_c05_p indd 103 6/29/09 10:34:16 AM

2 104 CHAPTER 5 The Lws of Motion Figure 5.1 ome exmples of Contct forces pplied forces. In ech cse, force is exerted on the object within the boxed re. ome gent in the environment externl to the boxed re exerts force on the object. c b Field forces m M Bridgemn-Girudon/Art Resource, NY d Isc Newton English physicist nd mthemticin ( ) Isc Newton ws one of the most brillint scientists in history. Before the ge of 30, he formulted the bsic concepts nd lws of mechnics, discovered the lw of universl grvittion, nd invented the mthemticl methods of clculus. As consequence of his theories, Newton ws ble to explin the motions of the plnets, the ebb nd flow of the tides, nd mny specil fetures of the motions of the Moon nd the Erth. He lso interpreted mny fundmentl observtions concerning the nture of light. His contributions to physicl theories dominted scientific thought for two centuries nd remin importnt tody _05_c05_p indd 104 q e Q Iron N f you remin sttionry. As second exmple, you cn push (in other words, exert force) on lrge boulder nd not be ble to move it. Wht force (if ny) cuses the Moon to orbit the Erth? Newton nswered this nd relted questions by stting tht forces re wht cuse ny chnge in the velocity of n object. The Moon s velocity chnges in direction s it moves in nerly circulr orbit round the Erth. This chnge in velocity is cused by the grvittionl force exerted by the Erth on the Moon. When coiled spring is pulled, s in Figure 5.1, the spring stretches. When sttionry crt is pulled, s in Figure 5.1b, the crt moves. When footbll is kicked, s in Figure 5.1c, it is both deformed nd set in motion. These situtions re ll exmples of clss of forces clled contct forces. Tht is, they involve physicl contct between two objects. Other exmples of contct forces re the force exerted by gs molecules on the wlls of continer nd the force exerted by your feet on the floor. Another clss of forces, known s field forces, does not involve physicl contct between two objects. These forces ct through empty spce. The grvittionl force of ttrction between two objects with mss, illustrted in Figure 5.1d, is n exmple of this clss of force. The grvittionl force keeps objects bound to the Erth nd the plnets in orbit round the un. Another common field force is the electric force tht one electric chrge exerts on nother (Fig. 5.1e), such s the chrges of n electron nd proton tht form hydrogen tom. A third exmple of field force is the force br mgnet exerts on piece of iron (Fig. 5.1f). The distinction between contct forces nd field forces is not s shrp s you my hve been led to believe by the previous discussion. When exmined t the tomic level, ll the forces we clssify s contct forces turn out to be cused by electric (field) forces of the type illustrted in Figure 5.1e. Nevertheless, in developing models for mcroscopic phenomen, it is convenient to use both clssifictions of forces. The only known fundmentl forces in nture re ll field forces: (1) grvittionl forces between objects, (2) electromgnetic forces between electric chrges, (3) strong forces between subtomic prticles, nd (4) wek forces tht rise in certin rdioctive decy processes. In clssicl physics, we re concerned only with grvittionl nd electromgnetic forces. We will discuss strong nd wek forces in Chpter 46. The Vector Nture of Force It is possible to use the deformtion of spring to mesure force. uppose verticl force is pplied to spring scle tht hs fixed upper end s shown in Figure 5.2. The spring elongtes when the force is pplied, nd pointer on the scle reds the extension of the spring. We cn clibrte the spring by defining reference force F 1 s the force tht produces pointer reding of 1.00 cm. If we now pply different downwrd force F 2 whose mgnitude is twice tht of the reference force 6/29/09 10:34:19 AM

3 5.2 Newton s First Lw nd Inertil Frmes 105 A downwrd force F 1 elongtes the spring 1.00 cm. A downwrd force F 2 elongtes the spring 2.00 cm. F 1 When nd F 2 re pplied together in the sme direction, the spring elongtes by 3.00 cm. When F 1 is downwrd nd F 2 is horizontl, the combintion of the two forces elongtes the spring by 2.24 cm F F 1 u F 2 F b F 2 c F 1 F 2 d Figure 5.2 The vector nture of force is tested with spring scle. F1 s seen in Figure 5.2b, the pointer moves to 2.00 cm. Figure 5.2c shows tht the combined effect of the two colliner forces is the sum of the effects of the individul forces. Now suppose the two forces re pplied simultneously with F1 downwrd nd F2 horizontl s illustrted in Figure 5.2d. In this cse, the pointer reds 2.24 cm. The single force F tht would produce this sme reding is the sum of the two vectors F1 nd F2 s described in Figure 5.2d. Tht is, 0 F1 0 5!F F units, nd its direction is u 5 tn 21 (20.500) Becuse forces hve been experimentlly verified to behve s vectors, you must use the rules of vector ddition to obtin the net force on n object. 5.2 Newton s First Lw nd Inertil Frmes We begin our study of forces by imgining some physicl situtions involving puck on perfectly level ir hockey tble (Fig. 5.3). You expect tht the puck will remin sttionry when it is plced gently t rest on the tble. Now imgine your ir hockey tble is locted on trin moving with constnt velocity long perfectly smooth trck. If the puck is plced on the tble, the puck gin remins where it is plced. If the trin were to ccelerte, however, the puck would strt moving long the tble opposite the direction of the trin s ccelertion, just s set of ppers on your dshbord flls onto the floor of your cr when you step on the ccelertor. As we sw in ection 4.6, moving object cn be observed from ny number of reference frmes. Newton s first lw of motion, sometimes clled the lw of inerti, defines specil set of reference frmes clled inertil frmes. This lw cn be stted s follows: Airflow Electric blower Figure 5.3 On n ir hockey tble, ir blown through holes in the surfce llows the puck to move lmost without friction. If the tble is not ccelerting, puck plced on the tble will remin t rest. If n object does not interct with other objects, it is possible to identify reference frme in which the object hs zero ccelertion. uch reference frme is clled n inertil frme of reference. When the puck is on the ir hockey tble locted on the ground, you re observing it from n inertil reference frme; there re no horizontl interctions of the puck with ny other objects, nd you observe it to hve zero ccelertion in tht direction. When you re on the Newton s first lw Inertil frme of reference 27819_05_c05_p indd 105 6/29/09 10:34:20 AM

4 106 CHAPTER 5 The Lws of Motion Pitfll Prevention 5.1 Newton s First Lw Newton s first lw does not sy wht hppens for n object with zero net force, tht is, multiple forces tht cncel; it sys wht hppens in the bsence of externl forces. This subtle but importnt difference llows us to define force s tht which cuses chnge in the motion. The description of n object under the effect of forces tht blnce is covered by Newton s second lw. Another sttement of Newton s first lw Definition of force trin moving t constnt velocity, you re lso observing the puck from n inertil reference frme. Any reference frme tht moves with constnt velocity reltive to n inertil frme is itself n inertil frme. When you nd the trin ccelerte, however, you re observing the puck from noninertil reference frme becuse the trin is ccelerting reltive to the inertil reference frme of the Erth s surfce. While the puck ppers to be ccelerting ccording to your observtions, reference frme cn be identified in which the puck hs zero ccelertion. For exmple, n observer stnding outside the trin on the ground sees the puck sliding reltive to the tble but lwys moving with the sme velocity with respect to the ground s the trin hd before it strted to ccelerte (becuse there is lmost no friction to tie the puck nd the trin together). Therefore, Newton s first lw is still stisfied even though your observtions s rider on the trin show n pprent ccelertion reltive to you. A reference frme tht moves with constnt velocity reltive to the distnt strs is the best pproximtion of n inertil frme, nd for our purposes we cn consider the Erth s being such frme. The Erth is not relly n inertil frme becuse of its orbitl motion round the un nd its rottionl motion bout its own xis, both of which involve centripetl ccelertions. These ccelertions re smll compred with g, however, nd cn often be neglected. For this reson, we model the Erth s n inertil frme, long with ny other frme ttched to it. Let us ssume we re observing n object from n inertil reference frme. (We will return to observtions mde in noninertil reference frmes in ection 6.3.) Before bout 1600, scientists believed tht the nturl stte of mtter ws the stte of rest. Observtions showed tht moving objects eventully stopped moving. Glileo ws the first to tke different pproch to motion nd the nturl stte of mtter. He devised thought experiments nd concluded tht it is not the nture of n object to stop once set in motion: rther, it is its nture to resist chnges in its motion. In his words, Any velocity once imprted to moving body will be rigidly mintined s long s the externl cuses of retrdtion re removed. For exmple, spcecrft drifting through empty spce with its engine turned off will keep moving forever. It would not seek nturl stte of rest. Given our discussion of observtions mde from inertil reference frmes, we cn pose more prcticl sttement of Newton s first lw of motion: In the bsence of externl forces nd when viewed from n inertil reference frme, n object t rest remins t rest nd n object in motion continues in motion with constnt velocity (tht is, with constnt speed in stright line). In other words, when no force cts on n object, the ccelertion of the object is zero. From the first lw, we conclude tht ny isolted object (one tht does not interct with its environment) is either t rest or moving with constnt velocity. The tendency of n object to resist ny ttempt to chnge its velocity is clled inerti. Given the sttement of the first lw bove, we cn conclude tht n object tht is ccelerting must be experiencing force. In turn, from the first lw, we cn define force s tht which cuses chnge in motion of n object. Quick Quiz 5.1 Which of the following sttements is correct? () It is possible for n object to hve motion in the bsence of forces on the object. (b) It is possible to hve forces on n object in the bsence of motion of the object. (c) Neither sttement () nor sttement (b) is correct. (d) Both sttements () nd (b) re correct. 5.3 Mss Imgine plying ctch with either bsketbll or bowling bll. Which bll is more likely to keep moving when you try to ctch it? Which bll requires more effort to throw it? The bowling bll requires more effort. In the lnguge of physics, we sy 27819_05_c05_p indd 106 6/29/09 10:34:21 AM

5 5.4 Newton s econd Lw 107 tht the bowling bll is more resistnt to chnges in its velocity thn the bsketbll. How cn we quntify this concept? Mss is tht property of n object tht specifies how much resistnce n object exhibits to chnges in its velocity, nd s we lerned in ection 1.1, the I unit of mss is the kilogrm. Experiments show tht the greter the mss of n object, the less tht object ccelertes under the ction of given pplied force. To describe mss quntittively, we conduct experiments in which we compre the ccelertions given force produces on different objects. uppose force cting on n object of mss m 1 produces chnge in motion of the object tht we cn quntify with the object s ccelertion 1, nd the sme force cting on n object of mss m 2 produces n ccelertion 2. The rtio of the two msses is defined s the inverse rtio of the mgnitudes of the ccelertions produced by the force: m 1 ; 2 (5.1) m 2 1 For exmple, if given force cting on 3-kg object produces n ccelertion of 4 m/s 2, the sme force pplied to 6-kg object produces n ccelertion of 2 m/s 2. According to huge number of similr observtions, we conclude tht the mgnitude of the ccelertion of n object is inversely proportionl to its mss when cted on by given force. If one object hs known mss, the mss of the other object cn be obtined from ccelertion mesurements. Mss is n inherent property of n object nd is independent of the object s surroundings nd of the method used to mesure it. Also, mss is sclr quntity nd thus obeys the rules of ordinry rithmetic. For exmple, if you combine 3-kg mss with 5-kg mss, the totl mss is 8 kg. This result cn be verified experimentlly by compring the ccelertion tht known force gives to severl objects seprtely with the ccelertion tht the sme force gives to the sme objects combined s one unit. Mss should not be confused with weight. Mss nd weight re two different quntities. The weight of n object is equl to the mgnitude of the grvittionl force exerted on the object nd vries with loction (see ection 5.5). For exmple, person weighing 180 lb on the Erth weighs only bout 30 lb on the Moon. On the other hnd, the mss of n object is the sme everywhere: n object hving mss of 2 kg on the Erth lso hs mss of 2 kg on the Moon. 5.4 Newton s econd Lw Definition of mss Mss nd weight re different quntities Newton s first lw explins wht hppens to n object when no forces ct on it: it either remins t rest or moves in stright line with constnt speed. Newton s second lw nswers the question of wht hppens to n object when one or more forces ct on it. Imgine performing n experiment in which you push block of mss m cross frictionless, horizontl surfce. When you exert some horizontl force F on the block, it moves with some ccelertion. If you pply force twice s gret on the sme block, experimentl results show tht the ccelertion of the block doubles; if you increse the pplied force to 3F, the ccelertion triples; nd so on. From such observtions, we conclude tht the ccelertion of n object is directly proportionl to the force cting on it: F ~. This ide ws first introduced in ection 2.4 when we discussed the direction of the ccelertion of n object. We lso know from the preceding section tht the mgnitude of the ccelertion of n object is inversely proportionl to its mss: 0 0 ~ 1/m. These experimentl observtions re summrized in Newton s second lw: Pitfll Prevention 5.2 Force Is the Cuse of Chnges in Motion An object cn hve motion in the bsence of forces s described in Newton s first lw. Therefore, don t interpret force s the cuse of motion. Force is the cuse of chnges in motion s mesured by ccelertion. When viewed from n inertil reference frme, the ccelertion of n object is directly proportionl to the net force cting on it nd inversely proportionl to its mss: ~ F m 27819_05_c05_p indd 107 6/29/09 10:34:21 AM

6 108 CHAPTER 5 The Lws of Motion If we choose proportionlity constnt of 1, we cn relte mss, ccelertion, nd force through the following mthemticl sttement of Newton s second lw: 1 Newton s second lw Newton s second lw: component form Pitfll Prevention 5.3 m Is Not Force Eqution 5.2 does not sy tht the product m is force. All forces on n object re dded vectorilly to generte the net force on the left side of the eqution. This net force is then equted to the product of the mss of the object nd the ccelertion tht results from the net force. Do not include n m force in your nlysis of the forces on n object. Definition of the newton Exmple 5.1 F 5 m (5.2) In both the textul nd mthemticl sttements of Newton s second lw, we hve indicted tht the ccelertion is due to the net force g F cting on n object. The net force on n object is the vector sum of ll forces cting on the object. (We sometimes refer to the net force s the totl force, the resultnt force, or the unblnced force.) In solving problem using Newton s second lw, it is impertive to determine the correct net force on n object. Mny forces my be cting on n object, but there is only one ccelertion. Eqution 5.2 is vector expression nd hence is equivlent to three component equtions: F x 5 m x F y 5 m y F z 5 m z (5.3) Quick Quiz 5.2 An object experiences no ccelertion. Which of the following cnnot be true for the object? () A single force cts on the object. (b) No forces ct on the object. (c) Forces ct on the object, but the forces cncel. Quick Quiz 5.3 You push n object, initilly t rest, cross frictionless floor with constnt force for time intervl Dt, resulting in finl speed of v for the object. You then repet the experiment, but with force tht is twice s lrge. Wht time intervl is now required to rech the sme finl speed v? () 4 Dt (b) 2 Dt (c) Dt (d) Dt/2 (e) Dt/4 The I unit of force is the newton (N). A force of 1 N is the force tht, when cting on n object of mss 1 kg, produces n ccelertion of 1 m/s 2. From this definition nd Newton s second lw, we see tht the newton cn be expressed in terms of the following fundmentl units of mss, length, nd time: 1 N ; 1 kg? m/s 2 (5.4) In the U.. customry system, the unit of force is the pound (lb). A force of 1 lb is the force tht, when cting on 1-slug mss, 2 produces n ccelertion of 1 ft/s 2 : 1 lb ; 1 slug? ft/s 2 (5.5) A convenient pproximtion is 1 N < 1 4 lb. An Accelerting Hockey Puck A hockey puck hving mss of 0.30 kg slides on the frictionless, horizontl surfce of n ice rink. Two hockey sticks strike the puck simultneously, exerting the forces on the puck shown in Figure 5.4. The force F1 hs mgnitude of 5.0 N, nd the force F2 hs mgnitude of 8.0 N. Determine both the mgnitude nd the direction of the puck s ccelertion. y F 2 F 1 = 5.0 N F 2 = 8.0 N OLUTION Conceptulize tudy Figure 5.4. Using your expertise in vector ddition from Chpter 3, predict the pproximte direction of the net force vector on the puck. The ccelertion of the puck will be in the sme direction. Figure 5.4 (Exmple 5.1) A hockey puck moving on frictionless surfce is subject to two forces F 1 nd F F 1 x 1 Eqution 5.2 is vlid only when the speed of the object is much less thn the speed of light. We tret the reltivistic sitution in Chpter The slug is the unit of mss in the U.. customry system nd is tht system s counterprt of the I unit the kilogrm. Becuse most of the clcultions in our study of clssicl mechnics re in I units, the slug is seldom used in this text _05_c05_p indd 108 6/29/09 10:34:22 AM

7 5.5 The Grvittionl Force nd Weight cont. Ctegorize Becuse we cn determine net force nd we wnt n ccelertion, this problem is ctegorized s one tht my be solved using Newton s second lw. Anlyze Find the component of the net force cting on the puck in the x direction: Find the component of the net force cting on the puck in the y direction: Use Newton s second lw in component form (Eq. 5.3) to find the x nd y components of the puck s ccelertion: F x 5 F 1x 1 F 2x 5 F 1 cos F 2 cos N N N F y 5 F 1y 1 F 2y 5 F 1 sin F 2 sin N N N x 5 F x m y 5 F y m N 5 29 m/s kg N 5 17 m/s kg Find the mgnitude of the ccelertion: 5 "129 m/s m/s m/s 2 Find the direction of the ccelertion reltive to the positive x xis: u5tn 21 y x b 5 tn b 5 31 Finlize The vectors in Figure 5.4 cn be dded grphiclly to check the resonbleness of our nswer. Becuse the ccelertion vector is long the direction of the resultnt force, drwing showing the resultnt force vector helps us check the vlidity of the nswer. (Try it!) WHAT IF? uppose three hockey sticks strike the puck simultneously, with two of them exerting the forces shown in Figure 5.4. The result of the three forces is tht the hockey puck shows no ccelertion. Wht must be the components of the third force? Answer If there is zero ccelertion, the net force cting on the puck must be zero. Therefore, the three forces must cncel. We hve found the components of the combintion of the first two forces. The components of the third force must be of equl mgnitude nd opposite sign so tht ll the components dd to zero. Therefore, F 3x N nd F 3y N. 5.5 The Grvittionl Force nd Weight All objects re ttrcted to the Erth. The ttrctive force exerted by the Erth on n object is clled the grvittionl force F g. This force is directed towrd the center of the Erth, 3 nd its mgnitude is clled the weight of the object. We sw in ection 2.6 tht freely flling object experiences n ccelertion g cting towrd the center of the Erth. Applying Newton s second lw g F 5 m to freely flling object of mss m, with 5 g nd g F 5 F g, gives g F 5 mg Therefore, the weight of n object, being defined s the mgnitude of Fg, is equl to mg: F g = mg (5.6) Becuse it depends on g, weight vries with geogrphic loction. Becuse g decreses with incresing distnce from the center of the Erth, objects weigh less t higher ltitudes thn t se level. For exmple, kg pllet of bricks used in the construction of the Empire tte Building in New York City weighed N t street level, but weighed bout 1 N less by the time it ws lifted from sidewlk 3 This sttement ignores tht the mss distribution of the Erth is not perfectly sphericl. Pitfll Prevention 5.4 Weight of n Object We re fmilir with the everydy phrse, the weight of n object. Weight, however, is not n inherent property of n object; rther, it is mesure of the grvittionl force between the object nd the Erth (or other plnet). Therefore, weight is property of system of items: the object nd the Erth. Pitfll Prevention 5.5 Kilogrm Is Not Unit of Weight You my hve seen the conversion 1 kg lb. Despite populr sttements of weights expressed in kilogrms, the kilogrm is not unit of weight, it is unit of mss. The conversion sttement is not n equlity; it is n equivlence tht is vlid only on the Erth s surfce _05_c05_p indd 109 6/29/09 10:34:23 AM

8 110 CHAPTER 5 The Lws of Motion NAA/Eugene Cernn The life-support unit strpped to the bck of stronut Hrrison chmitt weighed 300 lb on the Erth nd hd mss of 136 kg. During his trining, 50-lb mock-up with mss of 23 kg ws used. Although this strtegy effectively simulted the reduced weight the unit would hve on the Moon, it did not correctly mimic the unchnging mss. It ws more difficult to ccelerte the 136-kg unit (perhps by jumping or twisting suddenly) on the Moon thn it ws to ccelerte the 23-kg unit on the Erth. level to the top of the building. As nother exmple, suppose student hs mss of 70.0 kg. The student s weight in loction where g m/s 2 is 686 N (bout 150 lb). At the top of mountin, however, where g m/s 2, the student s weight is only 684 N. Therefore, if you wnt to lose weight without going on diet, climb mountin or weigh yourself t ft during n irplne flight! Eqution 5.6 quntifies the grvittionl force on the object, but notice tht this eqution does not require the object to be moving. Even for sttionry object or for n object on which severl forces ct, Eqution 5.6 cn be used to clculte the mgnitude of the grvittionl force. The result is subtle shift in the interprettion of m in the eqution. The mss m in Eqution 5.6 determines the strength of the grvittionl ttrction between the object nd the Erth. This role is completely different from tht previously described for mss, tht of mesuring the resistnce to chnges in motion in response to n externl force. In tht role, mss is lso clled inertil mss. We cll m in Eqution 5.6 the grvittionl mss. Even though this quntity is different in behvior from inertil mss, it is one of the experimentl conclusions in Newtonin dynmics tht grvittionl mss nd inertil mss hve the sme vlue. Although this discussion hs focused on the grvittionl force on n object due to the Erth, the concept is generlly vlid on ny plnet. The vlue of g will vry from one plnet to the next, but the mgnitude of the grvittionl force will lwys be given by the vlue of mg. Quick Quiz 5.4 uppose you re tlking by interplnetry telephone to friend who lives on the Moon. He tells you tht he hs just won newton of gold in contest. Excitedly, you tell him tht you entered the Erth version of the sme contest nd lso won newton of gold! Who is richer? () You re. (b) Your friend is. (c) You re eqully rich. Conceptul Exmple 5.2 How Much Do You Weigh in n Elevtor? You hve most likely been in n elevtor tht ccelertes upwrd s it moves towrd higher floor. In this cse, you feel hevier. In fct, if you re stnding on bthroom scle t the time, the scle mesures force hving mgnitude tht is greter thn your weight. Therefore, you hve tctile nd mesured evidence tht leds you to believe you re hevier in this sitution. Are you hevier? OLUTION No; your weight is unchnged. Your experiences re due to your being in noninertil reference frme. To provide the ccelertion upwrd, the floor or scle must exert on your feet n upwrd force tht is greter in mgnitude thn your weight. It is this greter force you feel, which you interpret s feeling hevier. The scle reds this upwrd force, not your weight, nd so its reding increses. 5.6 Newton s Third Lw If you press ginst corner of this textbook with your fingertip, the book pushes bck nd mkes smll dent in your skin. If you push hrder, the book does the sme nd the dent in your skin is little lrger. This simple ctivity illustrtes tht forces re interctions between two objects: when your finger pushes on the book, the book pushes bck on your finger. This importnt principle is known s Newton s third lw: Newton s third lw If two objects interct, the force F12 exerted by object 1 on object 2 is equl in mgnitude nd opposite in direction to the force F21 exerted by object 2 on object 1: F12 52F 21 (5.7) 27819_05_c05_p indd 110 6/29/09 10:34:26 AM

9 5.6 Newton s Third Lw 111 When it is importnt to designte forces s interctions between two objects, we will use this subscript nottion, where Fb mens the force exerted by on b. The third lw is illustrted in Figure 5.5. The force tht object 1 exerts on object 2 is populrly clled the ction force, nd the force of object 2 on object 1 is clled the rection force. These itlicized terms re not scientific terms; furthermore, either force cn be lbeled the ction or rection force. We will use these terms for convenience. In ll cses, the ction nd rection forces ct on different objects nd must be of the sme type (grvittionl, electricl, etc.). For exmple, the force cting on freely flling projectile is the grvittionl force exerted by the Erth on the projectile Fg 5 FEp (E 5 Erth, p 5 projectile), nd the mgnitude of this force is mg. The rection to this force is the grvittionl force exerted by the projectile on the Erth FpE 52F Ep. The rection force FpE must ccelerte the Erth towrd the projectile just s the ction force FEp ccelertes the projectile towrd the Erth. Becuse the Erth hs such lrge mss, however, its ccelertion due to this rection force is negligibly smll. Consider computer monitor t rest on tble s in Figure 5.6. The rection force to the grvittionl force Fg 5 FEm on the monitor is the force FmE 52F Em exerted by the monitor on the Erth. The monitor does not ccelerte becuse it is held up by the tble. The tble exerts on the monitor n upwrd force n 5 Ftm, clled the norml force. (Norml in this context mens perpendiculr.) This force, which prevents the monitor from flling through the tble, cn hve ny vlue needed, up to the point of breking the tble. Becuse the monitor hs zero ccelertion, Newton s second lw pplied to the monitor gives us g F 5 n 1 mg 5 0, so n j^ 2 mg j^ 5 0, or n 5 mg. The norml force blnces the grvittionl force on the monitor, so the net force on the monitor is zero. The rection force to n is the force exerted by the monitor downwrd on the tble, Fmt 52F tm 52n. Notice tht the forces cting on the monitor re Fg nd n s shown in Figure 5.6b. The two forces FmE nd Fmt re exerted on objects other thn the monitor. Figure 5.6 illustrtes n extremely importnt step in solving problems involving forces. Figure 5.6 shows mny of the forces in the sitution: those cting on the monitor, one cting on the tble, nd one cting on the Erth. Figure 5.6b, by contrst, shows only the forces cting on one object, the monitor, nd is clled force digrm or digrm showing the forces on the object. The importnt pictoril representtion in Figure 5.6c is clled free-body digrm. In free-body digrm, the prticle model is used by representing the object s dot nd showing the forces tht ct on the object s being pplied to the dot. When nlyzing n object subject to forces, we re interested in the net force cting on one object, which we will model s prticle. Therefore, free-body digrm helps us isolte only those forces on the object nd eliminte the other forces from our nlysis. 2 F 12 F 12 F 21 F 21 Figure 5.5 Newton s third lw. The force F 12 exerted by object 1 on object 2 is equl in mgnitude nd opposite in direction to the force F 21 exerted by object 2 on object 1. Pitfll Prevention 5.6 n Does Not Alwys Equl mg In the sitution shown in Figure 5.6 nd in mny others, we find tht n 5 Norml mg (the norml force force hs the sme mgnitude s the grvittionl force). This result, however, is not generlly true. If n object is on n incline, if there re pplied forces with verticl components, or if there is verticl ccelertion of the system, then n? mg. Alwys pply Newton s second lw to find the reltionship between n nd mg. Pitfll Prevention 5.7 Newton s Third Lw Remember tht Newton s third-lw ction nd rection forces ct on different objects. For exmple, in Figure 5.6, n 5 Ftm 52mg 52F Em. The forces n nd mg re equl in mgnitude nd opposite in direction, but they do not represent n ction rection pir becuse both forces ct on the sme object, the monitor. 1 n F tm n F tm n F tm F mt F g F me F Em F g F Em b c F g F Em Figure 5.6 () When computer monitor is t rest on tble, the forces cting on the monitor re the norml force n nd the grvittionl force F g. The rection to n is the force F mt exerted by the monitor on the tble. The rection to F g is the force F me exerted by the monitor on the Erth. (b) A digrm showing the forces on the monitor. (c) A freebody digrm shows the monitor s blck dot with the forces cting on it _05_c05_p indd 111 6/29/09 10:34:27 AM

10 112 CHAPTER 5 The Lws of Motion Pitfll Prevention 5.8 Free-Body Digrms The most importnt step in solving problem using Newton s lws is to drw proper sketch, the free-body digrm. Be sure to drw only those forces tht ct on the object you re isolting. Be sure to drw ll forces cting on the object, including ny field forces, such s the grvittionl force. Quick Quiz 5.5 (i) If fly collides with the windshield of fst-moving bus, which experiences n impct force with lrger mgnitude? () The fly. (b) The bus. (c) The sme force is experienced by both. (ii) Which experiences the greter ccelertion? () The fly. (b) The bus. (c) The sme ccelertion is experienced by both. Conceptul Exmple 5.3 You Push Me nd I ll Push You A lrge mn nd smll boy stnd fcing ech other on frictionless ice. They put their hnds together nd push ginst ech other so tht they move prt. (A) Who moves wy with the higher speed? OLUTION This sitution is similr to wht we sw in Quick Quiz 5.5. According to Newton s third lw, the force exerted by the mn on the boy nd the force exerted by the boy on the mn re third-lw pir of forces, so they must be equl in mgnitude. (A bthroom scle plced between their hnds would red the sme, regrdless of which wy it fced.) Therefore, the boy, hving the smller mss, experiences the greter ccelertion. Both individuls ccelerte for the sme mount of time, but the greter ccelertion of the boy over this time intervl results in his moving wy from the interction with the higher speed. (B) Who moves frther while their hnds re in contct? OLUTION Becuse the boy hs the greter ccelertion nd therefore the greter verge velocity, he moves frther thn the mn during the time intervl during which their hnds re in contct. 5.7 Anlysis Models Using Newton s econd Lw In this section, we discuss two nlysis models for solving problems in which objects re either in equilibrium or ccelerting long stright line under the ction of constnt externl forces. Remember tht when Newton s lws re pplied to n object, we re interested only in externl forces tht ct on the object. If the objects re modeled s prticles, we need not worry bout rottionl motion. For now, we lso neglect the effects of friction in those problems involving motion, which is equivlent to stting tht the surfces re frictionless. (The friction force is discussed in ection 5.8.) We usully neglect the mss of ny ropes, strings, or cbles involved. In this pproximtion, the mgnitude of the force exerted by ny element of the rope on the djcent element is the sme for ll elements long the rope. In problem sttements, the synonymous terms light nd of negligible mss re used to indicte tht mss is to be ignored when you work the problems. When rope ttched to n object is pulling on the object, the rope exerts force on the object in direction wy from the object, prllel to the rope. The mgnitude T of tht force is clled the tension in the rope. Becuse it is the mgnitude of vector quntity, tension is sclr quntity _05_c05_p indd 112 6/29/09 10:34:30 AM

11 5.7 Anlysis Models Using Newton s econd Lw 113 Anlysis Model: The Prticle in Equilibrium If the ccelertion of n object modeled s prticle is zero, the object is treted with the prticle in equilibrium model. In this model, the net force on the object is zero: F 5 0 (5.8) Consider lmp suspended from light chin fstened to the ceiling s in Figure 5.7. The force digrm for the lmp (Fig. 5.7b) shows tht the forces cting on the lmp re the downwrd grvittionl force F g nd the upwrd force T exerted by the chin. Becuse there re no forces in the x direction, o F x 5 0 provides no helpful informtion. The condition o F y 5 0 gives o F y 5 T 2 F g 5 0 or T 5 F g Agin, notice tht T nd F g re not n ction rection pir becuse they ct on the sme object, the lmp. The rection force to T is downwrd force exerted by the lmp on the chin. F g Figure 5.7 () A lmp suspended from ceiling by chin of negligible mss. (b) The forces cting on the lmp re the grvittionl force F g nd the force T exerted by the chin. b T Anlysis Model: The Prticle Under Net Force If n object experiences n ccelertion, its motion cn be nlyzed with the prticle under net force model. The pproprite eqution for this model is Newton s second lw, Eqution 5.2: F 5 m (5.2) Consider crte being pulled to the right on frictionless, horizontl floor s in Figure 5.8. Of course, the floor directly under the boy must hve friction; otherwise, his feet would simply slip when he tries to pull on the crte! uppose you wish to find the ccelertion of the crte nd the force the floor exerts on it. The forces cting on the crte re illustrted in the free-body digrm in Figure 5.8b. Notice tht the horizontl force T being pplied to the crte cts through the rope. The mgnitude of T is equl to the tension in the rope. In ddition to the force T, the free-body digrm for the crte includes the grvittionl force F g nd the norml force n exerted by the floor on the crte. We cn now pply Newton s second lw in component form to the crte. The only force cting in the x direction is T. Applying o F x 5 m x to the horizontl motion gives F x 5 T 5 m x or x 5 T m No ccelertion occurs in the y direction becuse the crte moves only horizontlly. Therefore, we use the prticle in equilibrium model in the y direction. Applying the y component of Eqution 5.8 yields o F y 5 n 1 (2F g ) 5 0 or n 5 F g Tht is, the norml force hs the sme mgnitude s the grvittionl force but cts in the opposite direction. If T is constnt force, the ccelertion x 5 T/m lso is constnt. Hence, the crte is lso modeled s prticle under constnt ccelertion in the x direction, nd the equtions of kinemtics from Chpter 2 cn be used to obtin the crte s position x nd velocity v x s functions of time. In the sitution just described, the mgnitude of the norml force n is equl to the mgnitude of F g, but tht is not lwys the cse, s noted in Pitfll Prevention 5.6. For exmple, suppose book is lying on tble nd you push down on the book with force F s in Figure 5.9. Becuse the book is t rest nd therefore not ccelerting, o F y 5 0, which gives n 2 F g 2 F 5 0, or n 5 F g 1 F 5 mg 1 F. In this sitution, the norml force is greter thn the grvittionl force. Other exmples in which n? F g re presented lter. b n F g Figure 5.8 () A crte being pulled to the right on frictionless floor. (b) The free-body digrm representing the externl forces cting on the crte. T F Physics F g n Figure 5.9 When force F pushes verticlly downwrd on nother object, the norml force n on the object is greter thn the grvittionl force: n 5 F g 1 F. y x 27819_05_c05_p indd 113 6/29/09 10:34:30 AM

12 114 CHAPTER 5 The Lws of Motion Problem-olving trtegy APPLYING NEWTON LAW The following procedure is recommended when deling with problems involving Newton s lws: 1. Conceptulize. Drw simple, net digrm of the system. The digrm helps estblish the mentl representtion. Estblish convenient coordinte xes for ech object in the system. 2. Ctegorize. If n ccelertion component for n object is zero, the object is modeled s prticle in equilibrium in this direction nd o F 5 0. If not, the object is modeled s prticle under net force in this direction nd o F 5 m. 3. Anlyze. Isolte the object whose motion is being nlyzed. Drw free-body digrm for this object. For systems contining more thn one object, drw seprte freebody digrms for ech object. Do not include in the free-body digrm forces exerted by the object on its surroundings. Find the components of the forces long the coordinte xes. Apply the pproprite model from the Ctegorize step for ech direction. Check your dimensions to mke sure tht ll terms hve units of force. olve the component equtions for the unknowns. Remember tht you generlly must hve s mny independent equtions s you hve unknowns to obtin complete solution. 4. Finlize. Mke sure your results re consistent with the free-body digrm. Also check the predictions of your solutions for extreme vlues of the vribles. By doing so, you cn often detect errors in your results. Exmple 5.4 A Trffic Light t Rest A trffic light weighing 122 N hngs from cble tied to two other cbles fstened to support s in Figure The upper cbles mke ngles of 37.0 nd 53.0 with the horizontl. These upper cbles re not s strong s the verticl cble nd will brek if the tension in them exceeds 100 N. Does the trffic light remin hnging in this sitution, or will one of the cbles brek? 37.0 T T 3 T 2 T 3 T y 53.0 T 2 x OLUTION Conceptulize Inspect the drwing in Figure Let us ssume the cbles do not brek nd nothing is moving. Ctegorize If nothing is moving, no prt of the system is ccelerting. We cn now model the light s prticle in equilibrium on which the net force is zero. imilrly, the net force on the knot (Fig. 5.10c) is zero. F g b c Figure 5.10 (Exmple 5.4) () A trffic light suspended by cbles. (b) The forces cting on the trffic light. (c) The free-body digrm for the knot where the three cbles re joined. T 3 Anlyze We construct digrm of the forces cting on the trffic light, shown in Figure 5.10b, nd free-body digrm for the knot tht holds the three cbles together, shown in Figure 5.10c. This knot is convenient object to choose becuse ll the forces of interest ct long lines pssing through the knot. Apply Eqution 5.8 for the trffic light in the y direction: o F y 5 0 T 3 2 F g 5 0 T 3 5 F g N 27819_05_c05_p indd 114 6/29/09 10:34:32 AM

13 5.7 Anlysis Models Using Newton s econd Lw cont. Choose the coordinte xes s shown in Figure 5.10c nd resolve the forces cting on the knot into their components: Force x Component y Component T 1 2T 1 cos 37.0 T 1 sin 37.0 T 2 T 2 cos 53.0 T 2 sin 53.0 T N Apply the prticle in equilibrium model to the knot: (1) o F x 5 2T 1 cos T 2 cos (2) o F y 5 T 1 sin T 2 sin (2122 N) 5 0 Eqution (1) shows tht the horizontl components of T 1 nd T 2 must be equl in mgnitude, nd Eqution (2) shows tht the sum of the verticl components of T 1 nd T 2 must blnce the downwrd force T 3, which is equl in mgnitude to the weight of the light. cos 37.0 olve Eqution (1) for T 2 in terms of T 1 : (3) T 2 5 T 1 cos 53.0 b T 1 ubstitute this vlue for T 2 into Eqution (2): T 1 sin (1.33T 1 )(sin 53.0 ) N 5 0 T N T T N Both vlues re less thn 100 N ( just brely for T 2 ), so the cbles will not brek. Finlize Let us finlize this problem by imgining chnge in the system, s in the following Wht If? WHAT IF? uppose the two ngles in Figure 5.10 re equl. Wht would be the reltionship between T 1 nd T 2? Answer We cn rgue from the symmetry of the problem tht the two tensions T 1 nd T 2 would be equl to ech other. Mthemticlly, if the equl ngles re clled u, Eqution (3) becomes T 2 5 T 1 cos u cos u b 5 T 1 which lso tells us tht the tensions re equl. Without knowing the specific vlue of u, we cnnot find the vlues of T 1 nd T 2. The tensions will be equl to ech other, however, regrdless of the vlue of u. Conceptul Exmple 5.5 Forces Between Crs in Trin Trin crs re connected by couplers, which re under tension s the locomotive pulls the trin. Imgine you re on trin speeding up with constnt ccelertion. As you move through the trin from the locomotive to the lst cr, mesuring the tension in ech set of couplers, does the tension increse, decrese, or sty the sme? When the engineer pplies the brkes, the couplers re under compression. How does this compression force vry from the locomotive to the lst cr? (Assume only the brkes on the wheels of the engine re pplied.) OLUTION While the trin is speeding up, tension decreses from the front of the trin to the bck. The coupler between the locomotive nd the first cr must pply enough force to ccelerte the rest of the crs. As you move bck long the trin, ech coupler is ccelerting less mss behind it. The lst coupler hs to ccelerte only the lst cr, nd so it is under the lest tension. When the brkes re pplied, the force gin decreses from front to bck. The coupler connecting the locomotive to the first cr must pply lrge force to slow down the rest of the crs, but the finl coupler must pply force lrge enough to slow down only the lst cr _05_c05_p indd 115 7/13/09 11:11:28 AM

14 116 CHAPTER 5 The Lws of Motion Exmple 5.6 The Runwy Cr A cr of mss m is on n icy drivewy inclined t n ngle u s in Figure y (A) Find the ccelertion of the cr, ssuming the drivewy is frictionless. OLUTION Conceptulize Use Figure 5.11 to conceptulize the sitution. From everydy experience, we know tht cr on n icy incline will ccelerte down the incline. (The sme thing hppens to cr on hill with its brkes not set.) Ctegorize We ctegorize the cr s prticle under net force becuse it ccelertes. Furthermore, this exmple belongs to very common ctegory of problems in which n object moves under the influence of grvity on n inclined plne. u x b mg cos u u mg sin u g = m g Figure 5.11 (Exmple 5.6) () A cr on frictionless incline. (b) The freebody digrm for the cr. The blck dot represents the position of the center of mss of the cr. We will lern bout center of mss in Chpter 9. n F x Anlyze Figure 5.11b shows the free-body digrm for the cr. The only forces cting on the cr re the norml force n exerted by the inclined plne, which cts perpendiculr to the plne, nd the grvittionl force Fg 5 mg, which cts verticlly downwrd. For problems involving inclined plnes, it is convenient to choose the coordinte xes with x long the incline nd y perpendiculr to it s in Figure 5.11b. With these xes, we represent the grvittionl force by component of mgnitude mg sin u long the positive x xis nd one of mgnitude mg cos u long the negtive y xis. Our choice of xes results in the cr being modeled s prticle under net force in the x direction nd prticle in equilibrium in the y direction. Apply these models to the cr: olve Eqution (1) for x : (1) o F x 5 mg sin u 5 m x (2) o F y 5 n 2 mg cos u 5 0 (3) x 5 g sin u Finlize Note tht the ccelertion component x is independent of the mss of the cr! It depends only on the ngle of inclintion nd on g. From Eqution (2), we conclude tht the component of F g perpendiculr to the incline is blnced by the norml force; tht is, n 5 mg cos u. This sitution is nother cse in which the norml force is not equl in mgnitude to the weight of the object. It is possible, lthough inconvenient, to solve the problem with stndrd horizontl nd verticl xes. You my wnt to try it, just for prctice. (B) uppose the cr is relesed from rest t the top of the incline nd the distnce from the cr s front bumper to the bottom of the incline is d. How long does it tke the front bumper to rech the bottom of the hill, nd wht is the cr s speed s it rrives there? OLUTION Conceptulize Imgine tht the cr is sliding down the hill nd you use stopwtch to mesure the entire time intervl until it reches the bottom. Ctegorize This prt of the problem belongs to kinemtics rther thn to dynmics, nd Eqution (3) shows tht the ccelertion x is constnt. Therefore, you should ctegorize the cr in this prt of the problem s prticle under constnt ccelertion _05_c05_p indd 116 6/29/09 10:34:34 AM

15 5.7 Anlysis Models Using Newton s econd Lw cont. Anlyze Defining the initil position of the front bumper s x i 5 0 nd its finl position s x f 5 d, nd recognizing tht v xi 5 0, pply Eqution 2.16, x f 5 x i 1 v xi t x t 2 : olve for t: Use Eqution 2.17, with v xi 5 0, to find the finl velocity of the cr: Finlize We see from Equtions (4) nd (5) tht the time t t which the cr reches the bottom nd its finl speed v xf re independent of the cr s mss, s ws its ccelertion. Notice tht we hve combined techniques from Chpter 2 with new techniques from this chpter in this exmple. As we lern more techniques in lter chpters, this process of combining nlysis models nd informtion from severl prts of the book will occur more often. In these cses, use the Generl Problem-olving trtegy to help you identify wht nlysis models you will need. d x t 2 2d 2d (4) t 5 5 Å x Å g sin u v xf x d (5) v xf 5 "2 x d 5 "2gd sin u WHAT IF? Wht previously solved problem does this sitution become if u 5 90? Answer Imgine u going to 90 in Figure The inclined plne becomes verticl, nd the cr is n object in free fll! Eqution (3) becomes x 5 g sin u 5 g sin 90 5 g which is indeed the free-fll ccelertion. (We find x 5 g rther thn x 5 2g becuse we hve chosen positive x to be downwrd in Fig ) Notice lso tht the condition n 5 mg cos u gives us n 5 mg cos Tht is consistent with the cr flling downwrd next to the verticl plne, in which cse there is no contct force between the cr nd the plne. Exmple 5.7 One Block Pushes Another Two blocks of msses m 1 nd m 2, with m 1. m 2, re plced in contct with ech other on frictionless, horizontl surfce s in Active Figure A constnt horizontl force F is pplied to m 1 s shown. (A) Find the mgnitude of the ccelertion of the system. ACTIVE FIGURE 5.12 (Exmple 5.7) () A force is OLUTION pplied to block of mss m 1, which pushes on second block Conceptulize Conceptulize the sitution by using Active Figure 5.12 nd ing on m 1. (c) The forces cting of mss m 2. (b) The forces ct- relize tht both blocks must experience on m 2. the sme ccelertion becuse they re in contct with ech other nd remin in contct throughout the motion. Ctegorize We ctegorize this problem s one involving prticle under net force becuse force is pplied to system of blocks nd we re looking for the ccelertion of the system. y x P 21 b F n1 F m 1 m 2 n2 m m 2 1 g c P 12 g Anlyze First model the combintion of two blocks s single prticle under net force. Apply Newton s second lw to the combintion in the x direction to find the ccelertion: o F x 5 F 5 (m 1 1 m 2 ) x F (1) x 5 m 1 1 m 2 continued 27819_05_c05_p indd 117 6/29/09 10:34:34 AM

16 118 CHAPTER 5 The Lws of Motion 5.7 cont. Finlize The ccelertion given by Eqution (1) is the sme s tht of single object of mss m 1 1 m 2 nd subject to the sme force. (B) Determine the mgnitude of the contct force between the two blocks. OLUTION Conceptulize The contct force is internl to the system of two blocks. Therefore, we cnnot find this force by modeling the whole system (the two blocks) s single prticle. Ctegorize Now consider ech of the two blocks individully by ctegorizing ech s prticle under net force. Anlyze We construct digrm of forces cting on the object for ech block s shown in Active Figures 5.12b nd 5.12c, where the contct force is denoted by P. From Active Figure 5.12c, we see tht the only horizontl force cting on m 2 is the contct force P 12 (the force exerted by m 1 on m 2 ), which is directed to the right. Apply Newton s second lw to m 2 : (2) o F x 5 P 12 5 m 2 x ubstitute the vlue of the ccelertion x given by Eqution (1) into Eqution (2): m 2 (3) P 12 5 m 2 x 5 bf m 1 1 m 2 Finlize This result shows tht the contct force P 12 is less thn the pplied force F. The force required to ccelerte block 2 lone must be less thn the force required to produce the sme ccelertion for the two-block system. To finlize further, let us check this expression for P 12 by considering the forces cting on m 1, shown in Active Figure 5.12b. The horizontl forces cting on m 1 re the pplied force F to the right nd the contct force P 21 to the left (the force exerted by m 2 on m 1 ). From Newton s third lw, P 21 is the rection force to P 12, so P 21 5 P 12. Apply Newton s second lw to m 1 : olve for P 12 nd substitute the vlue of x from Eqution (1): (4) o F x 5 F 2 P 21 5 F 2 P 12 5 m 1 x F P 12 5 F 2 m 1 x 5 F 2 m 1 b 5 bf m 1 1 m 2 m 1 1 m 2 m 2 This result grees with Eqution (3), s it must. WHAT IF? Imgine tht the force F in Active Figure 5.12 is pplied towrd the left on the right-hnd block of mss m 2. Is the mgnitude of the force P 12 the sme s it ws when the force ws pplied towrd the right on m 1? Answer When the force is pplied towrd the left on m 2, the contct force must ccelerte m 1. In the originl sitution, the contct force ccelertes m 2. Becuse m 1. m 2, more force is required, so the mgnitude of P 12 is greter thn in the originl sitution. Exmple 5.8 Weighing Fish in n Elevtor A person weighs fish of mss m on spring scle ttched to the ceiling of n elevtor s illustrted in Figure (A) how tht if the elevtor ccelertes either upwrd or downwrd, the spring scle gives reding tht is different from the weight of the fish. OLUTION Conceptulize The reding on the scle is relted to the extension of the spring in the scle, which is relted to the force on the end of the spring s in Figure 5.2. Imgine tht the fish is hnging on string ttched to the end of the spring _05_c05_p indd 118 6/29/09 10:34:35 AM

17 5.7 Anlysis Models Using Newton s econd Lw cont. In this cse, the mgnitude of the force exerted on the spring is equl to the tension T in the string. Therefore, we re looking for T. The force T pulls down on the string nd pulls up on the fish. Ctegorize We cn ctegorize this problem by identifying the fish s prticle under net force. When the elevtor ccelertes upwrd, the spring scle reds vlue greter thn the weight of the fish. When the elevtor ccelertes downwrd, the spring scle reds vlue less thn the weight of the fish. Anlyze Inspect the digrms of the forces cting on the fish in Figure 5.13 nd notice tht the externl forces cting on the fish re the downwrd grvittionl force F g 5 mg nd the force T exerted by the string. If the elevtor is either t rest or moving t constnt velocity, the fish is prticle in equilibrium, so o F y 5 T 2 F g 5 0 or T 5 F g 5 mg. (Remember tht the sclr mg is the weight of the fish.) Now suppose the elevtor is moving with n ccelertion reltive to n observer stnding outside the elevtor in n inertil frme. The fish is now prticle under net force. T T mg Figure 5.13 (Exmple 5.8) A fish is weighed on spring scle in n ccelerting elevtor cr. b mg Apply Newton s second lw to the fish: olve for T: o F y 5 T 2 mg 5 m y (1) T 5 m y 1 mg 5 mg y g 1 1b 5 F g y g 1 1b where we hve chosen upwrd s the positive y direction. We conclude from Eqution (1) tht the scle reding T is greter thn the fish s weight mg if is upwrd, so y is positive (Fig. 5.13), nd tht the reding is less thn mg if is downwrd, so y is negtive (Fig. 5.13b). (B) Evlute the scle redings for 40.0-N fish if the elevtor moves with n ccelertion y m/s 2. OLUTION Evlute the scle reding from Eqution (1) if is upwrd: 2.00 m/s2 T N2 1 1b N m/s Evlute the scle reding from Eqution (1) if is downwrd: m/s2 T N2 1 1b N 9.80 m/s 2 Finlize Tke this dvice: if you buy fish in n elevtor, mke sure the fish is weighed while the elevtor is either t rest or ccelerting downwrd! Furthermore, notice tht from the informtion given here, one cnnot determine the direction of motion of the elevtor. WHAT IF? uppose the elevtor cble breks nd the elevtor nd its contents re in free fll. Wht hppens to the reding on the scle? Answer If the elevtor flls freely, its ccelertion is y 5 2g. We see from Eqution (1) tht the scle reding T is zero in this cse; tht is, the fish ppers to be weightless _05_c05_p indd 119 6/29/09 10:34:37 AM

18 120 CHAPTER 5 The Lws of Motion Exmple 5.9 The Atwood Mchine When two objects of unequl mss re hung verticlly over frictionless pulley of negligible mss s in Active Figure 5.14, the rrngement is clled n Atwood mchine. The device is sometimes used in the lbortory to determine the vlue of g. Determine the mgnitude of the ccelertion of the two objects nd the tension in the lightweight cord. OLUTION Conceptulize Imgine the sitution pictured in Active Figure 5.14 in ction: s one object moves upwrd, the other object moves downwrd. Becuse the objects re connected by n inextensible string, their ccelertions must be of equl mgnitude. Ctegorize The objects in the Atwood mchine re subject to the grvittionl force s well s to the forces exerted by the strings connected to them. Therefore, we cn ctegorize this problem s one involving two prticles under net force. Anlyze The free-body digrms for the two objects re shown in Active Figure 5.14b. Two forces ct on ech object: the upwrd force T exerted by the string nd the downwrd grvittionl force. In problems such s this one in which the pulley is modeled s mssless nd frictionless, the tension in the string on both sides of the pulley is the sme. If the pulley hs mss or is subject to friction, the tensions on either side re not the sme nd the sitution requires techniques we will lern in Chpter 10. ACTIVE FIGURE 5.14 (Exmple 5.9) The Atwood mchine. () Two objects connected by mssless inextensible cord over frictionless pulley. (b) The freebody digrms for the two objects. We must be very creful with signs in problems such s this one. In Active Figure 5.14, notice tht if object 1 ccelertes upwrd, object 2 ccelertes downwrd. Therefore, for consistency with signs, if we define the upwrd direction s positive for object 1, we must define the downwrd direction s positive for object 2. With this sign convention, both objects ccelerte in the sme direction s defined by the choice of sign. Furthermore, ccording to this sign convention, the y component of the net force exerted on object 1 is T 2 m 1 g, nd the y component of the net force exerted on object 2 is m 2 g 2 T. + m 1 m 2 + T m 1 m 1 g b T m 2 m 2 g Apply Newton s second lw to object 1: Apply Newton s second lw to object 2: Add Eqution (2) to Eqution (1), noticing tht T cncels: (1) o F y 5 T 2 m 1 g 5 m 1 y (2) o F y 5 m 2 g 2 T 5 m 2 y 2 m 1 g 1 m 2 g 5 m 1 y 1 m 2 y olve for the ccelertion: (3) y 5 m 2 2 m 1 m 1 1 m 2 bg ubstitute Eqution (3) into Eqution (1) to find T: (4) T 5 m 1 (g 1 y ) 5 2m 1m 2 m 1 1 m 2 bg Finlize The ccelertion given by Eqution (3) cn be interpreted s the rtio of the mgnitude of the unblnced force on the system (m 2 2 m 1 )g to the totl mss of the system (m 1 1 m 2 ), s expected from Newton s second lw. Notice tht the sign of the ccelertion depends on the reltive msses of the two objects. WHAT IF? Describe the motion of the system if the objects hve equl msses, tht is, m 1 5 m 2. Answer If we hve the sme mss on both sides, the system is blnced nd should not ccelerte. Mthemticlly, we see tht if m 1 5 m 2, Eqution (3) gives us y 5 0. WHAT IF? Wht if one of the msses is much lrger thn the other: m 1.. m 2? Answer In the cse in which one mss is infinitely lrger thn the other, we cn ignore the effect of the smller mss. Therefore, the lrger mss should simply fll s if the smller mss were not there. We see tht if m 1.. m 2, Eqution (3) gives us y 5 2g _05_c05_p indd 120 6/29/09 10:34:38 AM

19 5.7 Anlysis Models Using Newton s econd Lw 121 Exmple 5.10 Accelertion of Two Objects Connected by Cord A bll of mss m 1 nd block of mss m 2 re ttched by lightweight cord tht psses over frictionless pulley of negligible mss s in Figure The block lies on frictionless incline of ngle u. Find the mgnitude of the ccelertion of the two objects nd the tension in the cord. m 1 m 2 u y T x m 1 g OLUTION b Conceptulize Imgine the objects in Figure 5.15 in motion. If m 2 moves down the incline, then m 1 moves upwrd. Becuse the objects re connected by cord (which we ssume does not stretch), their ccelertions hve the sme mgnitude. Ctegorize We cn identify forces on ech of the two objects nd we re looking for n ccelertion, so we ctegorize the objects s prticles under net force. Figure 5.15 (Exmple 5.10) () Two objects connected by lightweight cord strung over frictionless pulley. (b) The freebody digrm for the bll. (c) The free-body digrm for the block. (The incline is frictionless.) c T n y m 2 g sin u 2 g cos u x m u m 2 g Anlyze Consider the free-body digrms shown in Figures 5.15b nd 5.15c. Apply Newton s second lw in component form to the bll, choosing the upwrd direction s positive: (1) o F x 5 0 (2) o F y 5 T 2 m 1 g 5 m 1 y 5 m 1 For the bll to ccelerte upwrd, it is necessry tht T. m 1 g. In Eqution (2), we replced y with becuse the ccelertion hs only y component. For the block, it is convenient to choose the positive x9 xis long the incline s in Figure 5.15c. For consistency with our choice for the bll, we choose the positive direction to be down the incline. Apply Newton s second lw in component form to the block: (3) o F x9 5 m 2 g sin u 2 T 5 m 2 x9 5 m 2 (4) o F y9 5 n 2 m 2 g cos u 5 0 In Eqution (3), we replced x9 with becuse the two objects hve ccelertions of equl mgnitude. olve Eqution (2) for T : (5) T 5 m 1 (g 1 ) ubstitute this expression for T into Eqution (3): m 2 g sin u 2 m 1 (g 1 ) 5 m 2 olve for : (6) 5 m 2 sin u2m 1 bg m 1 1 m 2 ubstitute this expression for into Eqution (5) to find T : (7) T 5 m 1m 2 1sin u112 bg m 1 1 m 2 Finlize The block ccelertes down the incline only if m 2 sin u. m 1. If m 1. m 2 sin u, the ccelertion is up the incline for the block nd downwrd for the bll. Also notice tht the result for the ccelertion, Eqution (6), cn be interpreted s the mgnitude of the net externl force cting on the bll block system divided by the totl mss of the system; this result is consistent with Newton s second lw. WHAT IF? Wht hppens in this sitution if u 5 90? continued 27819_05_c05_p indd 121 6/29/09 10:34:39 AM

20 122 CHAPTER 5 The Lws of Motion 5.10 cont. Answer If u 5 90, the inclined plne becomes verticl nd there is no interction between its surfce nd m 2. Therefore, this problem becomes the Atwood mchine of Exmple 5.9. Letting u 90 in Equtions (6) nd (7) cuses them to reduce to Equtions (3) nd (4) of Exmple 5.9! WHAT IF? Wht if m 1 5 0? Answer If m 1 5 0, then m 2 is simply sliding down n inclined plne without intercting with m 1 through the string. Therefore, this problem becomes the sliding cr problem in Exmple 5.6. Letting m 1 0 in Eqution (6) cuses it to reduce to Eqution (3) of Exmple 5.6! 5.8 Forces of Friction When n object is in motion either on surfce or in viscous medium such s ir or wter, there is resistnce to the motion becuse the object intercts with its surroundings. We cll such resistnce force of friction. Forces of friction re very importnt in our everydy lives. They llow us to wlk or run nd re necessry for the motion of wheeled vehicles. Imgine tht you re working in your grden nd hve filled trsh cn with yrd clippings. You then try to drg the trsh cn cross the surfce of your concrete ptio s in Active Figure This surfce is rel, not n idelized, frictionless surfce. If we pply n externl horizontl force F to the trsh cn, cting to the right, the trsh cn remins sttionry when F is smll. The force on the trsh cn tht countercts F nd keeps it from moving cts towrd the left nd is clled For smll pplied forces, the mgnitude of the force of sttic friction equls the mgnitude of the pplied force. When the mgnitude of the pplied force exceeds the mgnitude of the mximum force of sttic friction, the trsh cn breks free nd ccelertes to the right. n n Motion fs F fk F mg mg f b f s,mx ACTIVE FIGURE 5.16 () nd (b) When pulling on trsh cn, the direction of the force of friction f between the cn nd rough surfce is opposite the direction of the pplied force F. (c) A grph of friction force versus pplied force. Notice tht f s,mx. f k. c O f s F ttic region f k m k n Kinetic region F 27819_05_c05_p indd 122 6/29/09 10:34:40 AM

21 5.8 Forces of Friction 123 the force of sttic friction f s. As long s the trsh cn is not moving, f s 5 F. Therefore, if F is incresed, f s lso increses. Likewise, if F decreses, f s lso decreses. Experiments show tht the friction force rises from the nture of the two surfces: becuse of their roughness, contct is mde only t few loctions where peks of the mteril touch. At these loctions, the friction force rises in prt becuse one pek physiclly blocks the motion of pek from the opposing surfce nd in prt from chemicl bonding ( spot welds ) of opposing peks s they come into contct. Although the detils of friction re quite complex t the tomic level, this force ultimtely involves n electricl interction between toms or molecules. If we increse the mgnitude of F s in Active Figure 5.16b, the trsh cn eventully slips. When the trsh cn is on the verge of slipping, f s hs its mximum vlue f s,mx s shown in Active Figure 5.16c. When F exceeds f s,mx, the trsh cn moves nd ccelertes to the right. We cll the friction force for n object in motion the force of kinetic friction f k. When the trsh cn is in motion, the force of kinetic friction on the cn is less thn f s,mx (Active Fig. 5.16c). The net force F 2 f k in the x direction produces n ccelertion to the right, ccording to Newton s second lw. If F 5 f k, the ccelertion is zero nd the trsh cn moves to the right with constnt speed. If the pplied force F is removed from the moving cn, the friction force f k cting to the left provides n ccelertion of the trsh cn in the 2x direction nd eventully brings it to rest, gin consistent with Newton s second lw. Experimentlly, we find tht, to good pproximtion, both f s,mx nd f k re proportionl to the mgnitude of the norml force exerted on n object by the surfce. The following descriptions of the force of friction re bsed on experimentl observtions nd serve s the model we shll use for forces of friction in problem solving: The mgnitude of the force of sttic friction between ny two surfces in contct cn hve the vlues f s # m s n (5.9) where the dimensionless constnt m s is clled the coefficient of sttic friction nd n is the mgnitude of the norml force exerted by one surfce on the other. The equlity in Eqution 5.9 holds when the surfces re on the verge of slipping, tht is, when f s 5 f s,mx 5 m s n. This sitution is clled impending motion. The inequlity holds when the surfces re not on the verge of slipping. The mgnitude of the force of kinetic friction cting between two surfces is f k 5 m k n (5.10) where m k is the coefficient of kinetic friction. Although the coefficient of kinetic friction cn vry with speed, we shll usully neglect ny such vritions in this text. The vlues of m k nd m s depend on the nture of the surfces, but m k is generlly less thn m s. Typicl vlues rnge from round 0.03 to 1.0. Tble 5.1 (pge 124) lists some reported vlues. The direction of the friction force on n object is prllel to the surfce with which the object is in contct nd opposite to the ctul motion (kinetic friction) or the impending motion (sttic friction) of the object reltive to the surfce. The coefficients of friction re nerly independent of the re of contct between the surfces. We might expect tht plcing n object on the side hving the most re might increse the friction force. Although this method provides more points in contct, the weight of the object is spred out over lrger re nd the individul points re not pressed together s tightly. Becuse these effects pproximtely compenste for ech other, the friction force is independent of the re. Force of sttic friction Force of kinetic friction Pitfll Prevention 5.9 The Equl ign Is Used in Limited itutions In Eqution 5.9, the equl sign is used only in the cse in which the surfces re just bout to brek free nd begin sliding. Do not fll into the common trp of using f s 5 m s n in ny sttic sitution. Pitfll Prevention 5.10 Friction Equtions Equtions 5.9 nd 5.10 re not vector equtions. They re reltionships between the mgnitudes of the vectors representing the friction nd norml forces. Becuse the friction nd norml forces re perpendiculr to ech other, the vectors cnnot be relted by multiplictive constnt. Pitfll Prevention 5.11 The Direction of the Friction Force ometimes, n incorrect sttement bout the friction force between n object nd surfce is mde the friction force on n object is opposite to its motion or impending motion rther thn the correct phrsing, the friction force on n object is opposite to its motion or impending motion reltive to the surfce _05_c05_p indd 123 6/29/09 10:34:40 AM

22 124 CHAPTER 5 The Lws of Motion 30 F TABLE 5.1 Coefficients of Friction M s M k Rubber on concrete teel on steel Aluminum on steel Glss on glss Copper on steel Wood on wood Wxed wood on wet snow Wxed wood on dry snow 0.04 Metl on metl (lubricted) Teflon on Teflon Ice on ice ynovil joints in humns Note: All vlues re pproximte. In some cses, the coefficient of friction cn exceed 1.0. b F 30 Figure 5.17 (Quick Quiz 5.7) A fther slides his dughter on sled either by () pushing down on her shoulders or (b) pulling up on rope. Quick Quiz 5.6 You press your physics textbook flt ginst verticl wll with your hnd. Wht is the direction of the friction force exerted by the wll on the book? () downwrd (b) upwrd (c) out from the wll (d) into the wll Quick Quiz 5.7 You re plying with your dughter in the snow. he sits on sled nd sks you to slide her cross flt, horizontl field. You hve choice of () pushing her from behind by pplying force downwrd on her shoulders t 30 below the horizontl (Fig. 5.17) or (b) ttching rope to the front of the sled nd pulling with force t 30 bove the horizontl (Fig. 5.17b). Which would be esier for you nd why? Exmple 5.11 Experimentl Determintion of M s nd M k The following is simple method of mesuring coefficients of friction. uppose block is plced on rough surfce inclined reltive to the horizontl s shown in Active Figure The incline ngle is incresed until the block strts to move. how tht you cn obtin m s by mesuring the criticl ngle u c t which this slipping just occurs. OLUTION Conceptulize Consider Active Figure 5.18 nd imgine tht the block tends to slide down the incline due to the grvittionl force. To simulte the sitution, plce coin on this book s cover nd tilt the book until the coin begins to slide. Notice how this exmple differs from Exmple 5.6. When there is no friction on n incline, ny ngle of the incline will cuse sttionry object to begin moving. When there is friction, however, there is no movement of the object for ngles less thn the criticl ngle. Ctegorize The block is subject to vrious forces. Becuse we re rising the plne to the ngle t which the block is just redy to begin to move but is not moving, we ctegorize the block s prticle in equilibrium. Anlyze The digrm in Active Figure 5.18 shows the forces on the block: the grvittionl force mg, the norml force n, nd the force of sttic friction f s. We choose x to be prllel to the plne nd y perpendiculr to it. fs mg cos u u mg n y mg sin u ACTIVE FIGURE 5.18 (Exmple 5.11) The externl forces exerted on block lying on rough incline re the grvittionl force mg, the norml force n, nd the force of friction f s. For convenience, the grvittionl force is resolved into component mg sin u long the incline nd component mg cos u perpendiculr to the incline. u x 27819_05_c05_p indd 124 6/29/09 10:34:41 AM

23 5.8 Forces of Friction cont. Apply Eqution 5.8 to the block in both the x nd y directions: (1) o F x 5 mg sin u 2 f s 5 0 (2) o F y 5 n 2 mg cos u 5 0 ubstitute mg 5 n/cos u from Eqution (2) into Eqution (1): When the incline ngle is incresed until the block is on the verge of slipping, the force of sttic friction hs reched its mximum vlue m s n. The ngle u in this sitution is the criticl ngle u c. Mke these substitutions in Eqution (3): For exmple, if the block just slips t u c , we find tht m s 5 tn (3) f s 5 mg sin u5 n b sin u5n tn u cos u m s n 5 n tn u c m s 5 tn u c Finlize Once the block strts to move t u $ u c, it ccelertes down the incline nd the force of friction is f k 5 m k n. If u is reduced to vlue less thn u c, however, it my be possible to find n ngle u9 c such tht the block moves down the incline with constnt speed s prticle in equilibrium gin ( x 5 0). In this cse, use Equtions (1) nd (2) with f s replced by f k to find m k : m k 5 tn u9 c, where u9 c, u c. Exmple 5.12 The liding Hockey Puck A hockey puck on frozen pond is given n initil speed of 20.0 m/s. If the puck lwys remins on the ice nd slides 115 m before coming to rest, determine the coefficient of kinetic friction between the puck nd ice. n Motion OLUTION Conceptulize Imgine tht the puck in Figure 5.19 slides to the right nd eventully comes to rest due to the force of kinetic friction. Ctegorize The forces cting on the puck re identified in Figure 5.19, but the text of the problem provides kinemtic vribles. Therefore, we ctegorize the problem in two wys. First, it involves prticle under net force: kinetic friction cuses the puck to ccelerte. Furthermore, becuse we model the force of kinetic friction s independent of speed, the ccelertion of the puck is constnt. o, we cn lso ctegorize this problem s one involving prticle under constnt ccelertion. fk mg Figure 5.19 (Exmple 5.12) After the puck is given n initil velocity to the right, the only externl forces cting on it re the grvittionl force mg, the norml force n, nd the force of kinetic friction f k. Anlyze First, let s find the ccelertion lgebriclly in terms of the coefficient of kinetic friction, using Newton s second lw. Once we know the ccelertion of the puck nd the distnce it trvels, the equtions of kinemtics cn be used to find the numericl vlue of the coefficient of kinetic friction. The digrm in Figure 5.19 shows the forces on the puck. Apply the prticle under net force model in the x direction to the puck: (1) o F x 5 2 f k 5 m x Apply the prticle in equilibrium model in the y direction to the puck: ubstitute n 5 mg from Eqution (2) nd f k 5 m k n into Eqution (1): (2) o F y 5 n 2 mg m k n 5 2 m k mg 5 m x x 5 2 m k g The negtive sign mens the ccelertion is to the left in Figure Becuse the velocity of the puck is to the right, the puck is slowing down. The ccelertion is independent of the mss of the puck nd is constnt becuse we ssume m k remins constnt. continued 27819_05_c05_p indd 125 6/30/09 3:06:08 PM

24 126 CHAPTER 5 The Lws of Motion 5.12 cont. Apply the prticle under constnt ccelertion model to the puck, using Eqution 2.17, v xf 2 5 v xi x (x f 2 x i ), with x i 5 0 nd v f 5 0: olve for the coefficient of kinetic friction: m k 5 v xi 2 ubstitute the numericl vlues: m k v xi x x f 5 v xi 2 2 2m k gx f 2gx f m/s m/s m Finlize Notice tht m k is dimensionless, s it should be, nd tht it hs low vlue, consistent with n object sliding on ice. Exmple 5.13 Accelertion of Two Connected Objects When Friction Is Present A block of mss m 2 on rough, horizontl surfce is connected to bll of mss m 1 by lightweight cord over lightweight, frictionless pulley s shown in Figure A force of mgnitude F t n ngle u with the horizontl is pplied to the block s shown, nd the block slides to the right. The coefficient of kinetic friction between the block nd surfce is m k. Determine the mgnitude of the ccelertion of the two objects. OLUTION Conceptulize Imgine wht hppens s F is pplied to the block. Assuming F is not lrge enough to lift the block, the block slides to the right nd the bll rises. Ctegorize We cn identify forces nd we wnt n ccelertion, so we ctegorize this problem s one involving two prticles under net force, the bll nd the block. m 1 m 2 u F y m 1 m 1 g x fk b c T T n F sin u m 2 u m 2 g F F cos u Figure 5.20 (Exmple 5.13) () The externl force F pplied s shown cn cuse the block to ccelerte to the right. (b, c) Digrms showing the forces on the two objects, ssuming the block ccelertes to the right nd the bll ccelertes upwrd. Anlyze First drw force digrms for the two objects s shown in Figures 5.20b nd 5.20c. Notice tht the string exerts force of mgnitude T on both objects. The pplied force F hs x nd y components F cos u nd F sin u, respectively. Becuse the two objects re connected, we cn equte the mgnitudes of the x component of the ccelertion of the block nd the y component of the ccelertion of the bll nd cll them both. Let us ssume the motion of the block is to the right. Apply the prticle under net force model to the block in the (1) o F x 5 F cos u 2 f k 2 T 5 m 2 x 5 m 2 horizontl direction: Becuse the block moves only horizontlly, pply the prticle in equilibrium model to the block in the verticl direction: Apply the prticle under net force model to the bll in the verticl direction: olve Eqution (2) for n: (2) o F y 5 n 1 F sin u 2 m 2 g 5 0 (3) o F y 5 T 2 m 1 g 5 m 1 y 5 m 1 n 5 m 2 g 2 F sin u ubstitute n into f k 5 m k n from Eqution 5.10: (4) f k 5 m k (m 2 g 2 F sin u) ubstitute Eqution (4) nd the vlue of T from Eqution (3) into Eqution (1): olve for : F cos u 2 m k (m 2 g 2 F sin u) 2 m 1 ( 1 g) 5 m 2 (5) 5 F 1cos u 1m k sin u 2 2 1m 1 1m k m 2 2g m 1 1 m _05_c05_p indd 126 6/29/09 10:34:43 AM

25 ummry cont. Finlize The ccelertion of the block cn be either to the right or to the left depending on the sign of the numertor in Eqution (5). If the motion is to the left, we must reverse the sign of f k in Eqution (1) becuse the force of kinetic friction must oppose the motion of the block reltive to the surfce. In this cse, the vlue of is the sme s in Eqution (5), with the two plus signs in the numertor chnged to minus signs. Wht does Eqution (5) reduce to if the force F is removed nd the surfce becomes frictionless? Cll this expression Eqution (6). Does this lgebric expression mtch your intuition bout the physicl sitution in this cse? Now go bck to Exmple 5.10 nd let ngle u go to zero in Eqution (6) of tht exmple. How does the resulting eqution compre with your Eqution (6) here in Exmple 5.13? hould the lgebric expressions compre in this wy bsed on the physicl situtions? Definitions ummry An inertil frme of reference is frme in which n object tht does not interct with other objects experiences zero ccelertion. Any frme moving with constnt velocity reltive to n inertil frme is lso n inertil frme. We define force s tht which cuses chnge in motion of n object. Concepts nd Principles Newton s first lw sttes tht it is possible to find n inertil frme in which n object tht does not interct with other objects experiences zero ccelertion, or, equivlently, in the bsence of n externl force, when viewed from n inertil frme, n object t rest remins t rest nd n object in uniform motion in stright line mintins tht motion. Newton s second lw sttes tht the ccelertion of n object is directly proportionl to the net force cting on it nd inversely proportionl to its mss. Newton s third lw sttes tht if two objects interct, the force exerted by object 1 on object 2 is equl in mgnitude nd opposite in direction to the force exerted by object 2 on object 1. The grvittionl force exerted on n object is equl to the product of its mss ( sclr quntity) nd the freefll ccelertion: F g 5 mg. The weight of n object is the mgnitude of the grvittionl force cting on the object. The mximum force of sttic friction f s,mx between n object nd surfce is proportionl to the norml force cting on the object. In generl, f s # m s n, where m s is the coefficient of sttic friction nd n is the mgnitude of the norml force. When n object slides over surfce, the mgnitude of the force of kinetic friction f k is given by f k 5 m k n, where m k is the coefficient of kinetic friction. continued 27819_05_c05_p indd 127 6/29/09 10:34:44 AM

26 128 CHAPTER 5 The Lws of Motion Anlysis Models for Problem olving Prticle Under Net Force If prticle of mss m experiences nonzero net force, its ccelertion is relted to the net force by Newton s second lw: F 5 m (5.2) m F Prticle in Equilibrium If prticle mintins constnt velocity (so tht 5 0), which could include velocity of zero, the forces on the prticle blnce nd Newton s second lw reduces to F 5 0 (5.8) 0 m F 0 Objective Questions 1. An experiment is performed on puck on level ir hockey tble, where friction is negligible. A constnt horizontl force is pplied to the puck, nd the puck s ccelertion is mesured. Now the sme puck is trnsported fr into outer spce, where both friction nd grvity re negligible. The sme constnt force is pplied to the puck (through spring scle tht stretches the sme mount), nd the puck s ccelertion (reltive to the distnt strs) is mesured. Wht is the puck s ccelertion in outer spce? () It is somewht greter thn its ccelertion on the Erth. (b) It is the sme s its ccelertion on the Erth. (c) It is less thn its ccelertion on the Erth. (d) It is infinite becuse neither friction nor grvity constrins it. (e) It is very lrge becuse ccelertion is inversely proportionl to weight nd the puck s weight is very smll but not zero. 2. In Figure OQ5.2, locomotive hs broken through the wll of trin sttion. During the collision, wht cn be sid bout the force exerted by the locomotive on the wll? () The force exerted by the locomotive on the wll ws lrger thn the force the wll could exert on the locomotive. (b) The force exerted by the locomotive on the wll denotes nswer vilble in tudent olutions Mnul/tudy Guide ws the sme in mgnitude s the force exerted by the wll on the locomotive. (c) The force exerted by the locomotive on the wll ws less thn the force exerted by the wll on the locomotive. (d) The wll cnnot be sid to exert force; fter ll, it broke. 3. The third grders re on one side of schoolyrd, nd the fourth grders re on the other. They re throwing snowblls t ech other. Between them, snowblls of vrious msses re moving with different velocities s shown in Figure OQ5.3. Rnk the snowblls () through (e) ccording to the mgnitude of the totl force exerted on ech one. Ignore ir resistnce. If two snowblls rnk together, mke tht fct cler. 9 m/s 400 g 400 g 300 g 12 m/s b 8 m/s 10 m/s c 500 g 12 m/s 200 g e d tudio Lévy nd ons Figure OQ5.2 Figure OQ The mnger of deprtment store is pushing horizontlly with force of mgnitude 200 N on box of shirts. The box is sliding cross the horizontl floor with forwrd ccelertion. Nothing else touches the box. Wht must be true bout the mgnitude of the force of kinetic friction cting on the box (choose one)? () It is greter thn 200 N. (b) It is less thn 200 N. (c) It is equl to 200 N. (d) None of those sttements is necessrily true. 5. The driver of speeding empty truck slms on the brkes nd skids to stop through distnce d. On second tril, the truck crries lod tht doubles its mss. Wht 27819_05_c05_p indd 128 6/29/09 10:34:46 AM

27 Conceptul Questions 129 will now be the truck s skidding distnce? () 4d (b) 2d (c)!2d (d) d (e) d/2 6. The driver of speeding truck slms on the brkes nd skids to stop through distnce d. On nother tril, the initil speed of the truck is hlf s lrge. Wht now will be the truck s skidding distnce? () 2d (b)!2d (c) d (d) d/2 (e) d/4 7. An object of mss m moves with ccelertion down rough incline. Which of the following forces should pper in free-body digrm of the object? Choose ll correct nswers. () the grvittionl force exerted by the plnet (b) m in the direction of motion (c) the norml force exerted by the incline (d) the friction force exerted by the incline (e) the force exerted by the object on the incline 8. A lrge crte of mss m is plce on the fltbed of truck but not tied down. As the truck ccelertes forwrd with ccelertion, the crte remins t rest reltive to the truck. Wht force cuses the crte to ccelerte? () the norml force (b) the grvittionl force (c) the friction force (d) the m force exerted by the crte (e) No force is required. 9. A crte remins sttionry fter it hs been plced on rmp inclined t n ngle with the horizontl. Which of the following sttements is or re correct bout the mgnitude of the friction force tht cts on the crte? Choose ll tht re true. () It is lrger thn the weight of the crte. (b) It is equl to m s n. (c) It is greter thn the component of the grvittionl force cting down the rmp. (d) It is equl to the component of the grvittionl force cting down the rmp. (e) It is less thn the component of the grvittionl force cting down the rmp. 10. An object of mss m is sliding with speed v i t some instnt cross level tbletop, with which its coefficient of kinetic friction is m. It then moves through distnce d nd comes to rest. Which of the following equtions for the speed v i is resonble? () v i 5!22mmgd (b) v i 5!2mmgd (c) v i 5!22mgd (d) v i 5!2mgd (e) v i 5!2md 11. If n object is in equilibrium, which of the following sttements is not true? () The speed of the object remins constnt. (b) The ccelertion of the object is zero. (c) The net force cting on the object is zero. (d) The object must be t rest. (e) There re t lest two forces cting on the object. 12. A truck loded with snd ccelertes long highwy. The driving force on the truck remins constnt. Wht hppens to the ccelertion of the truck if its triler leks snd t constnt rte through hole in its bottom? () It decreses t stedy rte. (b) It increses t stedy rte. (c) It increses nd then decreses. (d) It decreses nd then increses. (e) It remins constnt. 13. Two objects re connected by string tht psses over frictionless pulley s in Active Figure 5.14, where m 1, m 2 nd 1 nd 2 re the mgnitudes of the respective ccelertions. Which mthemticl sttement is true regrding the mgnitude of the ccelertion 2 of the mss m 2? () 2, g (b) 2. g (c) 2 5 g (d) 2, 1 (e) 2. 1 Conceptul Questions 1. A person holds bll in her hnd. () Identify ll the externl forces cting on the bll nd the Newton s third-lw rection force to ech one. (b) If the bll is dropped, wht force is exerted on it while it is flling? Identify the rection force in this cse. (Ignore ir resistnce.) 2. If cr is trveling due westwrd with constnt speed of 20 m/s, wht is the resultnt force cting on it? 3. In the motion picture It Hppened One Night (Columbi Pictures, 1934), Clrk Gble is stnding inside sttionry bus in front of Cludette Colbert, who is seted. The bus suddenly strts moving forwrd nd Clrk flls into Cludette s lp. Why did this hppen? 4. Your hnds re wet, nd the restroom towel dispenser is empty. Wht do you do to get drops of wter off your hnds? How does the motion of the drops exemplify one of Newton s lws? Which one? 5. A pssenger sitting in the rer of bus clims tht she ws injured when the driver slmmed on the brkes, cusing suitcse to come flying towrd her from the front of the bus. If you were the judge in this cse, wht disposition would you mke? Why? denotes nswer vilble in tudent olutions Mnul/tudy Guide 6. A sphericl rubber blloon inflted with ir is held sttionry, with its opening, on the west side, pinched shut. () Describe the forces exerted by the ir inside nd outside the blloon on sections of the rubber. (b) After the blloon is relesed, it tkes off towrd the est, gining speed rpidly. Explin this motion in terms of the forces now cting on the rubber. (c) Account for the motion of skyrocket tking off from its lunch pd. 7. If you hold horizontl metl br severl centimeters bove the ground nd move it through grss, ech lef of grss bends out of the wy. If you increse the speed of the br, ech lef of grss will bend more quickly. How then does rotry power lwn mower mnge to cut grss? How cn it exert enough force on lef of grss to sher it off? 8. A child tosses bll stright up. he sys tht the bll is moving wy from her hnd becuse the bll feels n upwrd force of the throw s well s the grvittionl force. () Cn the force of the throw exceed the grvittionl force? How would the bll move if it did? (b) Cn the force of the throw be equl in mgnitude to the grvittionl force? Explin. (c) Wht strength cn ccurtely be ttributed to the force of the throw? Explin. (d) Why does the bll move wy from the child s hnd? 27819_05_c05_p indd 129 6/29/09 10:34:57 AM

28 130 CHAPTER 5 The Lws of Motion 9. A rubber bll is dropped onto the floor. Wht force cuses the bll to bounce? 10. The myor of city reprimnds some city employees becuse they will not remove the obvious sgs from the cbles tht support the city trffic lights. Wht explntion cn the employees give? How do you think the cse will be settled in medition? 11. Blncing crefully, three boys inch out onto horizontl tree brnch bove pond, ech plnning to dive in seprtely. The third boy in line notices tht the brnch is brely strong enough to support them. He decides to jump stright up nd lnd bck on the brnch to brek it, spilling ll three into the pond. When he strts to crry out his pln, t wht precise moment does the brnch brek? Explin. uggestion: Pretend to be the third boy nd imitte wht he does in slow motion. If you re still unsure, stnd on bthroom scle nd repet the suggestion. 12. When you push on box with 200-N force insted of 50-N force, you cn feel tht you re mking greter effort. When tble exerts 200-N norml force insted of one of smller mgnitude, is the tble relly doing nything differently? 13. A weightlifter stnds on bthroom scle. He pumps brbell up nd down. Wht hppens to the reding on the scle s he does so? Wht If? Wht if he is strong enough to ctully throw the brbell upwrd? How does the reding on the scle vry now? 14. Give resons for the nswers to ech of the following questions: () Cn norml force be horizontl? (b) Cn norml force be directed verticlly downwrd? (c) Consider tennis bll in contct with sttionry floor nd with nothing else. Cn the norml force be different in mgnitude from the grvittionl force exerted on the bll? (d) Cn the force exerted by the floor on the bll be different in mgnitude from the force the bll exerts on the floor? 15. A cr is moving forwrd slowly nd is speeding up. A student clims tht the cr exerts force on itself or tht the cr s engine exerts force on the cr. () Argue tht this ide cnnot be ccurte nd tht friction exerted by the rod is the propulsive force on the cr. Mke your evidence nd resoning s persusive s possible. (b) Is it sttic or kinetic friction? uggestions: Consider rod covered with light grvel. Consider shrp print of the tire tred on n sphlt rod, obtined by coting the tred with dust. 16. In Figure CQ5.16, the light, B A tut, unstretchble cord B joins block 1 nd the lrger-mss 2 1 block 2. Cord A exerts force on block 1 to mke it ccelerte Figure CQ5.16 forwrd. () How does the mgnitude of the force exerted by cord A on block 1 compre with the mgnitude of the force exerted by cord B on block 2? Is it lrger, smller, or equl? (b) How does the ccelertion of block 1 compre with the ccelertion (if ny) of block 2? (c) Does cord B exert force on block 1? If so, is it forwrd or bckwrd? Is it lrger, smller, or equl in mgnitude to the force exerted by cord B on block 2? 17. Identify ction rection pirs in the following situtions: () mn tkes step (b) snowbll hits girl in the bck (c) bsebll plyer ctches bll (d) gust of wind strikes window 18. Twenty people prticipte in tug-of-wr. The two tems of ten people re so evenly mtched tht neither tem wins. After the gme they notice tht cr is stuck in the mud. They ttch the tug-of-wr rope to the bumper of the cr, nd ll the people pull on the rope. The hevy cr hs just moved couple of decimeters when the rope breks. Why did the rope brek in this sitution when it did not brek when the sme twenty people pulled on it in tug-of-wr? 19. An thlete grips light rope tht psses over low-friction pulley ttched to the ceiling of gym. A sck of snd precisely equl in weight to the thlete is tied to the other end of the rope. Both the snd nd the thlete re initilly t rest. The thlete climbs the rope, sometimes speeding up nd slowing down s he does so. Wht hppens to the sck of snd? Explin. 20. Cn n object exert force on itself? Argue for your nswer. 21. Describe two exmples in which the force of friction exerted on n object is in the direction of motion of the object. 22. As shown in Figure CQ5.22, student A, 55-kg girl, sits on one chir with metl runners, t rest on clssroom floor. tudent B, n 80-kg boy, sits on n identicl chir. Both students keep their feet off the floor. A rope runs from student A s hnds round light pulley nd then over her shoulder to the hnds of techer stnding on the floor behind her. The low-friction xle of the pulley is ttched to second rope held by student B. All ropes run prllel to the chir runners. () If student A pulls on her end of the rope, will her chir or will B s chir slide on the floor? Explin why. (b) If insted the techer pulls on his rope end, which chir slides? Why this one? (c) If student B pulls on his rope, which chir slides? Why? (d) Now the techer ties his end of the rope to student A s chir. tudent A pulls on the end of the rope in her hnds. Which chir slides nd why? tudent B tudent A Figure CQ5.22 Techer 23. uppose you re driving clssic cr. Why should you void slmming on your brkes when you wnt to stop in the shortest possible distnce? (Mny modern crs hve ntilock brkes tht void this problem.) 27819_05_c05_p indd 130 6/29/09 10:34:58 AM

29 Problems 131 Problems The problems found in this chpter my be ssigned online in Enhnced WebAssign 1. denotes strightforwrd problem; 2. denotes intermedite problem; 3. denotes chllenging problem 1. full solution vilble in the tudent olutions Mnul/tudy Guide 1. denotes problems most often ssigned in Enhnced WebAssign; these provide students with trgeted feedbck nd either Mster It tutoril or Wtch It solution video. ections 5.1 through A 3.00-kg object undergoes n ccelertion given by i^ j^ 2 m/s 2. Find () the resultnt force cting on the object nd (b) the mgnitude of the resultnt force. 2. The verge speed of nitrogen molecule in ir is bout m/s, nd its mss is kg. () If it tkes s for nitrogen molecule to hit wll nd rebound with the sme speed but moving in the opposite direction, wht is the verge ccelertion of the molecule during this time intervl? (b) Wht verge force does the molecule exert on the wll? 3. A toy rocket engine is securely fstened to lrge puck tht cn glide with negligible friction over horizontl surfce, tken s the xy plne. The 4.00-kg puck hs velocity of 3.00 i^ m/s t one instnt. Eight seconds lter, its velocity is 18 i^ 1 10 j^ 2 m/s. Assuming the rocket engine exerts constnt horizontl force, find () the components of the force nd (b) its mgnitude. 4. A certin orthodontist uses wire brce to lign ptient s crooked tooth s in Figure P5.4. The tension in the wire is djusted to hve mgnitude of 18.0 N. Find the mgnitude of the net force exerted by the wire on the crooked tooth. 14 T y Figure P5.4 x Review. The grvittionl force exerted on bsebll is 2.21 N down. A pitcher throws the bll horizontlly with velocity 18.0 m/s by uniformly ccelerting it long stright horizontl line for time intervl of 170 ms. The bll strts from rest. () Through wht distnce does it move before its relese? (b) Wht re the mgnitude nd direction of the force the pitcher exerts on the bll? 6. Review. The grvittionl force exerted on bsebll is 2F g j^. A pitcher throws the bll with velocity v i^ by uniformly ccelerting it long stright horizontl line T denotes sking for quntittive nd conceptul resoning denotes symbolic resoning problem denotes Mster It tutoril vilble in Enhnced WebAssign denotes guided problem shded denotes pired problems tht develop resoning with symbols nd numericl vlues for time intervl of Dt 5 t t. () trting from rest, through wht distnce does the bll move before its relese? (b) Wht force does the pitcher exert on the bll? 7. Review. An electron of mss kg hs n initil speed of m/s. It trvels in stright line, nd its speed increses to m/s in distnce of 5.00 cm. Assuming its ccelertion is constnt, () determine the mgnitude of the force exerted on the electron nd (b) compre this force with the weight of the electron, which we ignored. 8. Besides the grvittionl force, 2.80-kg object is subjected to one other constnt force. The object strts from rest nd in 1.20 s experiences displcement of i^ j^ 2 m, where the direction of j^ is the upwrd verticl direction. Determine the other force. 9. One or more externl forces, lrge enough to be esily mesured, re exerted on ech object enclosed in dshed box shown in Figure 5.1. Identify the rection to ech of these forces. 10. A brick of mss M hs been plced on rubber cushion of mss m. Together they re sliding to the right t constnt velocity on n ice-covered prking lot. () Drw free-body digrm of the brick nd identify ech force cting on it. (b) Drw free-body digrm of the cushion nd identify ech force cting on it. (c) Identify ll of the ction rection pirs of forces in the brick cushion plnet system. 11. An object of mss m is dropped t t 5 0 from the roof of building of height h. While the object is flling, wind blowing prllel to the fce of the building exerts constnt horizontl force F on the object. () At wht time t does the object strike the ground? Express t in terms of g nd h. (b) Find n expression in terms of m nd F for the ccelertion x of the object in the horizontl direction (tken s the positive x direction). (c) How fr is the object displced horizontlly before hitting the ground? Answer in terms of m, g, F, nd h. (d) Find the mgnitude of the object s ccelertion while it is flling, using the vribles F, m, nd g. 12. A force F pplied to n object of mss m 1 produces n ccelertion of 3.00 m/s 2. The sme force pplied to second object of mss m 2 produces n ccelertion of 1.00 m/s 2. () Wht is the vlue of the rtio m 1 /m 2? (b) If m 1 nd m 2 re combined into one object, find its ccelertion under the ction of the force F _05_c05_p indd 131 6/29/09 10:34:59 AM

30 132 CHAPTER 5 The Lws of Motion 13. Two forces F 1 nd F 2 ct on 5.00-kg object. Tking F N nd F N, find the ccelertions of the object for the configurtions of forces shown in prts () nd (b) of Figure P5.13. F2 m 90.0 F1 Figure P5.13 m b F F1 14. You stnd on the set of chir nd then hop off. () During the time intervl you re in flight down to the floor, the Erth moves towrd you with n ccelertion of wht order of mgnitude? In your solution, explin your logic. Model the Erth s perfectly solid object. (b) The Erth moves towrd you through distnce of wht order of mgnitude? 15. A 15.0-lb block rests on the floor. () Wht force does the floor exert on the block? (b) A rope is tied to the block nd is run verticlly over pulley. The other end is ttched to free-hnging 10.0-lb object. Wht now is the force exerted by the floor on the 15.0-lb block? (c) If the 10.0-lb object in prt (b) is replced with 20.0-lb object, wht is the force exerted by the floor on the 15.0-lb block? 16. Review. Three forces cting on n object re given by F i^ j^ 2 N, F i^ j^ 2 N, nd F i^ 2 N. The object experiences n ccelertion of mgnitude 3.75 m/s 2. () Wht is the direction of the ccelertion? (b) Wht is the mss of the object? (c) If the object is initilly t rest, wht is its speed fter 10.0 s? (d) Wht re the velocity components of the object fter 10.0 s? ection 5.7 ome Applictions of Newton s Lws 17. Review. Figure P5.17 shows worker poling bot very efficient mode of trnsporttion cross shllow lke. He pushes prllel to the length of the light pole, exerting force of mgnitude 240 N on the bottom of the lke. Assume the pole lies in the verticl plne contining the keel of the bot. At one moment, the pole mkes n ngle of 35.0 with Figure P5.17 the verticl nd the wter exerts horizontl drg force of 47.5 N on the bot, opposite to its forwrd velocity of mgnitude m/s. The mss of the bot including its crgo nd the worker is 370 kg. () The wter exerts buoynt force verticlly upwrd on the bot. Find the mgnitude of this force. (b) Model the forces s constnt over short intervl of time to find the velocity of the bot s fter the moment described. 18. An iron bolt of mss 65.0 g hngs from string 35.7 cm long. The top end of the string is fixed. Without touching AP Imges it, mgnet ttrcts the bolt so tht it remins sttionry, but is displced horizontlly 28.0 cm to the right from the previously verticl line of the string. The mgnet is locted to the right of the bolt nd on the sme verticl level s the bolt in the finl configurtion. () Drw free-body digrm of the bolt. (b) Find the tension in the string. (c) Find the mgnetic force on the bolt. 19. Figure P5.19 shows the horizontl forces cting on silbot moving north t constnt velocity, seen from point stright bove its mst. At the prticulr speed of the silbot, the wter exerts 220-N drg force on its hull nd u For ech of the situtions () nd (b) described below, write two component equtions representing Newton s second lw. Then solve the equtions for P (the force exerted by the wind on the sil) nd for n (the force exerted by the wter on the keel). () Choose the x direction s est nd the y direction s north. (b) Now choose the x direction s u north of est nd the y direction s u west of north. (c) Compre your solutions to prts () nd (b). Do the results gree? Is one method significntly esier? 20. The systems shown in Figure P5.20 re in equilibrium. If the spring scles re clibrted in newtons, wht do they red? Ignore the msses of the pulleys nd strings nd ssume the pulleys nd the incline in Figure P5.20d re frictionless kg c 5.00 kg 5.00 kg W n N 5.00 kg 5.00 kg Figure P5.20 b 5.00 kg E 220 N Figure P A block slides down frictionless plne hving n inclintion of u The block strts from rest t the top, nd the length of the incline is 2.00 m. () Drw free-body digrm of the block. Find (b) the ccelertion of the block nd (c) its speed when it reches the bottom of the incline. 22. A 3.00-kg object is moving in plne, with its x nd y coordintes given by x 5 5t nd y 5 3t 3 1 2, where x nd y re in meters nd t is in seconds. Find the mgnitude of the net force cting on this object t t s. d P u 27819_05_c05_p indd 132 6/29/09 10:35:00 AM

31 Problems The distnce between two telephone poles is 50.0 m. When 1.00-kg bird lnds on the telephone wire midwy between the poles, the wire sgs m. () Drw free-body digrm of the bird. (b) How much tension does the bird produce in the wire? Ignore the weight of the wire. 24. A bg of cement weighing 325 N hngs in equilibrium from three wires s suggested in Figure P5.24. Two of the wires mke ngles u nd u with the horizontl. Assuming the system is in equilibrium, find the tensions T 1, T 2, nd T 3 in the wires. 25. A bg of cement whose weight is F g hngs in equilibrium from three wires s shown in Figure P5.24. Two of the wires mke ngles u 1 nd u 2 with the horizontl. Assuming the system is in equilibrium, show tht the tension in the left-hnd wire is T 1 5 F g cos u 2 sin 1u 1 1u 2 2 T 1 T A setup similr to the one shown in Figure P5.26 is often used in hospitls to support nd pply horizontl trction force to n injured leg. () Determine the force of tension in the rope supporting the leg. (b) Wht is the trction force exerted to the right on the leg? Figure P kg u 1 CEMENT T 3 u 2 F g Figure P5.24 Problems 24 nd An object of mss m kg is observed to hve n ccelertion F 2 with mgnitude of m/s 2 in direction 60.0 est of north. Figure P5.27 shows m view of the object from bove. The force F 1 F2 cting on the object hs mgnitude of 5.00 N nd is Figure P5.27 directed north. Determine the mgnitude nd direction of the one other horizontl force F1 cting on the object. 28. An object of mss m kg m 1 plced on frictionless, horizontl tble is connected to string tht psses over pulley nd then is fstened to hnging object of mss m 2 m kg s shown in Figure P5.28. () Drw free-body digrms Figure P5.28 of both objects. Find (b) the mgnitude of the ccelertion of the Problems 28 nd 45. objects nd (c) the tension in the string. 29. Figure P5.29 shows the speed of person s body s he does chin-up. Assume the motion is verticl nd the mss of the person s body is 64.0 kg. Determine the force exerted by the chin-up br on his body t () t 5 0, (b) t s, (c) t s, nd (d) t s. speed (cm/s) time (s) Figure P Two objects re connected by light string tht psses over frictionless pulley s shown in Figure P5.30. Assume the incline is frictionless nd tke m kg, m kg, nd u () Drw free-body digrms of both objects. Find (b) the mgnitude of the ccelertion of the objects, (c) the tension in the string, nd (d) the speed of ech object 2.00 s fter it is relesed from rest. 31. Two blocks, ech of mss m kg, re hung from the ceiling of n elevtor s in Figure P5.31. () If the elevtor moves with n upwrd cceler tion of mgnitude 1.60 m/s 2, find the tensions T 1 nd T 2 in the upper nd lower strings. (b) If the strings cn withstnd mximum tension of 85.0 N, wht mximum ccelertion cn the elevtor hve before string breks? 32. Two blocks, ech of mss m, re hung from the ceiling of n elevtor s in Figure P5.31. The elevtor m 1 m 2 hs n upwrd ccelertion. The strings hve negligible mss. () Find the tensions T 1 nd T 2 in the upper nd lower strings in terms of m,, nd g. (b) Compre the two tensions nd determine which string would brek first if is mde sufficiently lrge. (c) Wht re the tensions if the cble supporting the elevtor breks? 33. In the system shown in Figure P5.33, horizontl force Fx cts on n object of mss m kg. The horizontl m 1 m 2 Figure P5.33 u Figure P5.30 m m T 1 T 2 Figure P5.31 Problems 31 nd 32. F x 27819_05_c05_p indd 133 6/29/09 10:35:01 AM

32 134 CHAPTER 5 The Lws of Motion surfce is frictionless. Consider the ccelertion of the sliding object s function of F x. () For wht vlues of F x does the object of mss m kg ccelerte upwrd? (b) For wht vlues of F x is the tension in the cord zero? (c) Plot the ccelertion of the m 2 object versus F x. Include vlues of F x from 2100 N to 1100 N. 34. An object of mss m 1 hngs from string tht psses over very light fixed pulley P 1 s shown in Figure P5.34. The string connects to second very light pulley P 2. A second string psses round this pulley with one end ttched to wll nd the other to n object of mss m 2 on frictionless, horizontl tble. () If 1 nd 2 re the ccelertions of m 1 nd m 2, respectively, wht is the reltion between these ccelertions? Find expressions for (b) the tensions in the strings nd (c) the ccelertions 1 nd 2 in terms of the msses m 1 nd m 2, nd g. m 2 P 2 P 1 Figure P In Exmple 5.8, we investigted the pprent weight of fish in n elevtor. Now consider 72.0-kg mn stnding on spring scle in n elevtor. trting from rest, the elevtor scends, ttining its mximum speed of 1.20 m/s in s. It trvels with this constnt speed for the next 5.00 s. The elevtor then undergoes uniform ccelertion in the negtive y direction for 1.50 s nd comes to rest. Wht does the spring scle register () before the elevtor strts to move, (b) during the first s, (c) while the elevtor is trveling t constnt speed, nd (d) during the time intervl it is slowing down? 36. In the Atwood mchine discussed in Exmple 5.9 nd shown in Active Figure 5.14, m kg nd m kg. The msses of the pulley nd string re negligible by comprison. The pulley turns without friction, nd the string does not stretch. The lighter object is relesed with shrp push tht sets it into motion t v i m/s downwrd. () How fr will m 1 descend below its initil level? (b) Find the velocity of m 1 fter 1.80 s. ection 5.8 Forces of Friction 37. Review. A rifle bullet with mss of 12.0 g trveling towrd the right t 260 m/s strikes lrge bg of snd nd penetrtes it to depth of 23.0 cm. Determine the mgnitude nd direction of the friction force (ssumed constnt) tht cts on the bullet. 38. Review. A cr is trveling t 50.0 mi/h on horizontl highwy. () If the coefficient of sttic friction between rod nd tires on riny dy is 0.100, wht is the minimum distnce in which the cr will stop? (b) Wht is the stopping distnce when the surfce is dry nd m s ? m A 25.0-kg block is initilly t rest on horizontl surfce. A horizontl force of 75.0 N is required to set the block in motion, fter which horizontl force of 60.0 N is required to keep the block moving with constnt speed. Find () the coefficient of sttic friction nd (b) the coefficient of kinetic friction between the block nd the surfce. 40. Why is the following sitution impossible? Your 3.80-kg physics book is plced next to you on the horizontl set of your cr. The coefficient of sttic friction between the book nd the set is 0.650, nd the coefficient of kinetic friction is You re trveling forwrd t 72.0 km/h nd brke to stop with constnt ccelertion over distnce of 30.0 m. Your physics book remins on the set rther thn sliding forwrd onto the floor. 41. To meet U.. Postl ervice requirement, employees footwer must hve coefficient of sttic friction of 0.5 or more on specified tile surfce. A typicl thletic shoe hs coefficient of sttic friction of In n emergency, wht is the minimum time intervl in which person strting from rest cn move 3.00 m on the tile surfce if she is wering () footwer meeting the Postl ervice minimum nd (b) typicl thletic shoe? 42. Before 1960, people believed tht the mximum ttinble coefficient of sttic friction for n utomobile tire on rodwy ws m s 5 1. Around 1962, three compnies independently developed rcing tires with coefficients of 1.6. This problem shows tht tires hve improved further since then. The shortest time intervl in which piston-engine cr initilly t rest hs covered distnce of one-qurter mile is bout 4.43 s. () Assume the cr s rer wheels lift the front wheels off the pvement s shown in Figure P5.42. Wht minimum vlue of m s is necessry to chieve the record time? (b) uppose the driver were ble to increse his or her engine power, keeping other things equl. How would this chnge ffect the elpsed time? Jmie quire/allsport/getty Imges Figure P Review. A 3.00-kg block strts from rest t the top of 30.0 incline nd slides distnce of 2.00 m down the incline in 1.50 s. Find () the mgnitude of the ccelertion of the block, (b) the coefficient of kinetic friction between block nd plne, (c) the friction force cting on the block, nd (d) the speed of the block fter it hs slid 2.00 m. 44. A womn t n irport is towing her 20.0-kg suitcse t constnt speed by pulling on strp t n ngle u bove the horizontl (Fig. P5.44). he pulls on the strp with 35.0-N force, nd the friction force on the suitcse is 20.0 N. () Drw free-body digrm of the suitcse. (b) Wht ngle does u Figure P _05_c05_p indd 134 6/29/09 10:35:04 AM

33 Problems 135 the strp mke with the horizontl? (c) Wht is the mgnitude of the norml force tht the ground exerts on the suitcse? 45. A 9.00-kg hnging object is connected by light, inextensible cord over light, frictionless pulley to 5.00-kg block tht is sliding on flt tble (Fig. P5.28). Tking the coefficient of kinetic friction s 0.200, find the tension in the string. 46. Three objects re connected on tble s shown in Figure P5.46. The coefficient of kinetic friction between the block of mss m 2 nd the tble is The objects hve msses of m kg, m kg, nd m kg, nd the pulleys re frictionless. () Drw freebody digrm of ech object. (b) Determine the ccelertion of ech object, including its direction. (c) Determine the tensions in the two cords. Wht If? (d) If the tbletop were smooth, would the tensions increse, decrese, or remin the sme? Explin. m 1 m 2 Figure P Two blocks connected by rope of negligible mss re being drgged by horizontl force (Fig. P5.47). uppose F N, m kg, m kg, nd the coefficient of kinetic friction between ech block nd the surfce is () Drw free-body digrm for ech block. Determine (b) the ccelertion of the system nd (c) the tension T in the rope. 48. A block of mss 3.00 kg is pushed up ginst wll by force P tht mkes n ngle of u with the horizontl s shown in Figure P5.48. The coefficient of sttic friction between the block nd the wll is () Determine the possible vlues for the mgnitude of P tht llow the block to remin sttionry. m 3 T m 1 m 2 F Figure P5.47 (b) Describe wht hppens if 0 P 0 hs lrger vlue nd wht hppens if it is smller. (c) Repet prts () nd (b), ssuming the force mkes n ngle of u with the horizontl. 49. Review. One side of the roof of house slopes up t A roofer kicks round, flt rock tht hs been thrown onto the roof by neighborhood child. The rock slides stright up the incline with n initil speed of 15.0 m/s. The coefficient of kinetic friction between the rock nd the roof is The rock slides 10.0 m up the roof to its pek. It crosses the ridge nd goes into free fll, following prbolic trjectory bove the fr side of the roof, with negligible ir resistnce. Determine the mximum height the rock reches bove the point where it ws kicked. u P Figure P Review. A Chinook slmon cn swim underwter t 3.58 m/s, nd it cn lso jump verticlly upwrd, leving the wter with speed of 6.26 m/s. A record slmon hs length 1.50 m nd mss 61.0 kg. Consider the fish swimming stright upwrd in the wter below the surfce of lke. The grvittionl force exerted on it is very nerly cnceled out by buoynt force exerted by the wter s we will study in Chpter 14. The fish experiences n upwrd force P exerted by the wter on its threshing til fin nd downwrd fluid friction force tht we model s cting on its front end. Assume the fluid friction force disppers s soon s the fish s hed breks the wter surfce nd ssume the force on its til is constnt. Model the grvittionl force s suddenly switching full on when hlf the length of the fish is out of the wter. Find the vlue of P. 51. Review. A mgicin pulls tblecloth from under 200-g mug locted 30.0 cm from the edge of the cloth. The cloth exerts friction force of N on the mug, nd the cloth is pulled with constnt ccelertion of 3.00 m/s 2. How fr does the mug move reltive to the horizontl tbletop before the cloth is completely out from under it? Note tht the cloth must move more thn 30 cm reltive to the tbletop during the process. Additionl Problems 52. A blck luminum glider flots on film of ir bove level luminum ir trck. Aluminum feels essentilly no force in mgnetic field, nd ir resistnce is negligible. A strong mgnet is ttched to the top of the glider, forming totl mss of 240 g. A piece of scrp iron ttched to one end stop on the trck ttrcts the mgnet with force of N when the iron nd the mgnet re seprted by 2.50 cm. () Find the ccelertion of the glider t this instnt. (b) The scrp iron is now ttched to nother green glider, forming totl mss 120 g. Find the ccelertion of ech glider when the gliders re simultneously relesed t 2.50-cm seprtion. 53. Review. A hockey puck struck by hockey stick is given n initil speed v i in the positive x direction. The coefficient of kinetic friction between the ice nd the puck is m k. () Obtin n expression for the ccelertion of the puck s it slides cross the ice. (b) Use the result of prt () to obtin n expression for the distnce d the puck slides. The nswer should be in terms of the vribles v i, m k, nd g only. 54. Why is the following sitution impossible? A book sits on n inclined plne on the surfce of the Erth. The ngle of the plne with the horizontl is The coefficient of kinetic friction between the book nd the plne is At time t 5 0, the book is relesed from rest. The book then slides through distnce of 1.00 m, mesured long the plne, in time intervl of s. 55. Two blocks of msses m 1 nd m 2 re plced on tble in contct with ech other s discussed in Exmple 5.7 nd shown in Active Figure The coefficient of kinetic friction between the block of mss m 1 nd the tble is m 1, nd tht between the block of mss m 2 nd the tble is m 2. A horizontl force of mgnitude F is pplied to the block of mss m 1. We wish to find P, the mgnitude of the contct force between the blocks. () Drw digrms showing the forces for ech block. (b) Wht is the net force on the system of two 27819_05_c05_p indd 135 6/29/09 10:35:05 AM

34 136 CHAPTER 5 The Lws of Motion blocks? (c) Wht is the net force cting on m 1? (d) Wht is the net force cting on m 2? (e) Write Newton s second lw in the x direction for ech block. (f) olve the two equtions in two unknowns for the ccelertion of the blocks in terms of the msses, the pplied force F, the coefficients of friction, nd g. (g) Find the mgnitude P of the contct force between the blocks in terms of the sme quntities. 56. A rope with mss m r m m r is ttched to block with b mss m b s in Figure P5.56. The block rests on frictionless, horizontl surfce. The rope Figure P5.56 does not stretch. The free end of the rope is pulled to the right with horizontl force F. () Drw force digrms for the rope nd the block, noting tht the tension in the rope is not uniform. (b) Find the ccelertion of the system in terms of m b, m r, nd F. (c) Find the mgnitude of the force the rope exerts on the block. (d) Wht hppens to the force on the block s the rope s mss pproches zero? Wht cn you stte bout the tension in light cord joining pir of moving objects? 57. An inventive child nmed Nick wnts to rech n pple in tree without climbing the tree. itting in chir connected to rope tht psses over frictionless pulley (Fig. P5.57), Nick pulls on the loose end of the rope with such force tht the spring scle reds 250 N. Nick s true weight is 320 N, nd the chir weighs 160 N. Nick s feet re not touching the ground. () Drw one pir of digrms showing the forces for Nick nd the chir considered s seprte systems nd nother digrm for Nick nd the chir considered s one system. (b) how tht the ccelertion of the system is upwrd nd find its mgnitude. (c) Find the force Nick exerts on the chir. F projecting from the tree trunk. Explin why this ction cn mke the rope brek. 59. In Exmple 5.7, we pushed on two blocks on tble. uppose three blocks re in contct with one nother on frictionless, horizontl surfce s shown in Figure P5.59. A horizontl force F is pplied to m 1. Tke m kg, m kg, m kg, nd F N. () Drw seprte free-body digrm for ech block. (b) Determine the ccelertion of the blocks. (c) Find the resultnt force on ech block. (d) Find the mgnitudes of the contct forces between the blocks. (e) You re working on construction project. A coworker is niling up plsterbord on one side of light prtition, nd you re on the opposite side, providing bcking by lening ginst the wll with your bck pushing on it. Every hmmer blow mkes your bck sting. The supervisor helps you put hevy block of wood between the wll nd your bck. Using the sitution nlyzed in prts () through (d) s model, explin how this chnge works to mke your job more comfortble. F m 1 m 2 m 3 Figure P On single, light, verticl cble tht does not stretch, crne is lifting kg Ferrri nd, below it, kg BMW Z8. The Ferrri is moving upwrd with speed 3.50 m/s nd ccelertion 1.25 m/s 2. () How do the velocity nd ccelertion of the BMW compre with T 4 those of the Ferrri? (b) Find the tension in the cble between the BMW nd the Ferrri. (c) Find the tension in the cble bove the Ferrri. 61. An object of mss M is held in plce by n pplied force F nd pulley system s shown in Figure P5.61. The pulleys re mssless nd frictionless. () Drw digrms showing the forces on ech pulley. Find (b) the tension in ech section of rope, T 1, T 2, T 3, T 4, nd T 5 nd (c) the mgnitude of F. F T 1 T 2 T 3 M T 5 Figure P5.61 Figure P5.57 Problems 57 nd In the sitution described in Problem 57 nd Figure P5.57, the msses of the rope, spring blnce, nd pulley re negligible. Nick s feet re not touching the ground. () Assume Nick is momentrily t rest when he stops pulling down on the rope nd psses the end of the rope to nother child, of weight 440 N, who is stnding on the ground next to him. The rope does not brek. Describe the ensuing motion. (b) Insted, ssume Nick is momentrily t rest when he ties the end of the rope to strong hook 62. An luminum block of mss Aluminum m kg nd copper m 1 Copper block of mss m kg m re connected by light 2 teel string over frictionless pulley. They sit on steel surfce u s shown in Figure P5.62, Figure P5.62 where u () When they re relesed from rest, will they strt to move? If they do, determine (b) their ccelertion nd (c) the tension in the string. If they do not move, determine (d) the sum of the mgnitudes of the forces of friction cting on the blocks _05_c05_p indd 136 6/29/09 10:35:05 AM

35 Problems A crte of weight F g is pushed by force P on horizontl floor s shown in Figure P5.63. The coefficient of sttic friction is m s, nd P is directed t ngle u below the horizontl. () how tht the minimum vlue of P tht will move the crte is given by P 5 m s F g sec u 1 2 m s tn u (b) Find the condition on u in terms of m s for which motion of the crte is impossible for ny vlue of P. 64. A student is sked to mesure the ccelertion of glider on frictionless, inclined plne, using n ir trck, stopwtch, nd meterstick. The top of the trck is mesured to be cm higher thn the bottom of the trck, nd the length of the trck is d cm. The crt is relesed from rest t the top of the incline, tken s x 5 0, nd its position x long the incline is mesured s function of time. For x vlues of 10.0 cm, 20.0 cm, 35.0 cm, 50.0 cm, 75.0 cm, nd 100 cm, the mesured times t which these positions re reched (verged over five runs) re 1.02 s, 1.53 s, 2.01 s, 2.64 s, 3.30 s, nd 3.75 s, respectively. () Construct grph of x versus t 2, with best-fit stright line to describe the dt. (b) Determine the ccelertion of the crt from the slope of this grph. (c) Explin how your nswer to prt (b) compres with the theoreticl vlue you clculte using 5 g sin u s derived in Exmple A flt cushion of mss m is relesed from rest t the corner of the roof of building, t height h. A wind blowing long the side of the building exerts constnt horizontl force of mgnitude F on the cushion s it drops s shown in Figure P5.65. The ir exerts no verticl force. () how tht the pth of the cushion is stright line. (b) Does the cushion fll with Wind force constnt velocity? Explin. (c) If m kg, h m, nd F N, how fr from the building will the cushion hit the level ground? Wht If? (d) If the cushion is thrown downwrd with nonzero speed t the top of the building, wht will be the shpe of its trjectory? Explin. 66. In Figure P5.66, the pulleys nd the cord re light, ll surfces re frictionless, nd the cord does not stretch. () How m 1 does the ccelertion of block 1 compre with the ccelertion of block 2? Explin your resoning. (b) The mss of block 2 m 2 is 1.30 kg. Find its ccelertion s it depends on the mss m 1 of Figure P5.66 block 1. (c) Wht If? Wht does the result of prt (b) predict if m 1 is very much less thn 1.30 kg? (d) Wht does the result of prt (b) predict if m 1 pproches infinity? (e) In this lst cse, wht is the tension in the cord? (f) Could you nticipte the nswers to prts (c), (d), nd (e) without first doing prt (b)? Explin. P Figure P5.63 h Figure P5.65 Cushion 67. Wht horizontl force must be pplied to lrge block of mss M shown in Figure P5.67 so tht the tn blocks remin sttionry reltive to M? Assume ll surfces nd the pulley re frictionless. Notice tht the force exerted by the string ccelertes m An 8.40-kg object slides down fixed, frictionless, inclined plne. Use computer to determine nd tbulte () the norml force exerted on the object nd (b) its ccelertion for series of incline ngles (mesured from the horizontl) rnging from 0 to 90 in 5 increments. (c) Plot grph of the norml force nd the ccelertion s functions of the incline ngle. (d) In the limiting cses of 0 nd 90, re your results consistent with the known behvior? 69. A cr ccelertes down hill (Fig. P5.69), going from rest to 30.0 m/s in 6.00 s. A toy inside the cr hngs by string u from the cr s ceiling. The bll u in the figure represents the toy, of mss kg. The ccelertion is such tht the string Figure P5.69 remins perpendiculr to the ceiling. Determine () the ngle u nd (b) the tension in the string. Chllenge Problems F M m A time-dependent force, F i^ t j^ 2, where F is in newtons nd t is in seconds, is exerted on 2.00-kg object initilly t rest. () At wht time will the object be moving with speed of 15.0 m/s? (b) How fr is the object from its initil position when its speed is 15.0 m/s? (c) Through wht totl displcement hs the object trveled t this moment? 71. The bord sndwiched between two other bords in Figure P5.71 weighs 95.5 N. If the coefficient of sttic friction between the bords is 0.663, wht must be the mgnitude of the compression forces (ssumed horizontl) cting on both sides of the center bord to keep it from slipping? Figure P5.71 m 2 Figure P Why is the following sitution impossible? A 1.30-kg toster is not plugged in. The coefficient of sttic friction between the toster nd horizontl countertop is To mke the toster strt moving, you crelessly pull on its electric cord. Unfortuntely, the cord hs become fryed from your previous similr ctions nd will brek if the tension in the 27819_05_c05_p indd 137 6/29/09 10:35:06 AM

36 138 CHAPTER 5 The Lws of Motion cord exceeds 4.00 N. By pulling on the cord t prticulr ngle, you successfully strt the toster moving without breking the cord. 73. A block of mss 2.20 kg is ccelerted cross rough surfce by light cord pssing over smll pulley s shown in Figure P5.73. The tension T in the cord T is mintined t 10.0 N, nd the pulley is m bove the top x M of the block. The coefficient of kinetic friction is () Determine the ccelertion of the block when x m. (b) Describe Figure P5.73 the generl behvior of the ccelertion s the block slides from loction where x is lrge to x 5 0. (c) Find the mximum vlue of the ccelertion nd the position x for which it occurs. (d) Find the vlue of x for which the ccelertion is zero. 74. A mobile is formed by supporting four metl butterflies of equl mss m from string of length L. The points of support re evenly spced distnce, prt s shown in Figure P5.74. The string forms n ngle u 1 with the ceiling t ech endpoint. The center section of string is horizon- tl. () Find the tension in ech section of string in terms of u 1, m, nd g. (b) In terms of u 1, find the ngle u 2 tht the sections of string between the outside butterflies nd the inside butterflies form with the horizontl. (c) how tht the distnce D between the endpoints of the string is D 5 L 5 b2 cos u cos3tn tn u r 75. Review. A block of mss m kg is relesed from rest t h m bove the surfce of tble, t the top of u incline s shown in Figure P5.75. The frictionless incline is fixed on tble of height H m. () Determine the ccelertion of the block s it slides down the incline. (b) Wht is the velocity of the block s it leves the incline? (c) How fr from the tble will the block hit the floor? (d) Wht time intervl elpses between when the block is relesed nd when it hits the floor? (e) Does the mss of the block ffect ny of the bove clcultions? H h m u D u 1 u 1 Figure P5.75 Problems 75 nd 76. R m u 2 u 2 m m Figure P5.74 m L In Figure P5.75, the incline hs mss M nd is fstened to the sttionry horizontl tbletop. The block of mss m is plced ner the bottom of the incline nd is relesed with quick push tht sets it sliding upwrd. The block stops ner the top of the incline s shown in the figure nd then slides down gin, lwys without friction. Find the force tht the tbletop exerts on the incline throughout this motion in terms of m, M, g, nd u _05_c05_p indd 138 6/29/09 10:35:08 AM