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