The Laws of Motion. chapter

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1 chpter 5 The Lws of Motion 5.1 The Concept of Force 5.2 Newton s First Lw nd Inertil Fres 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 otion of n object in ters of its position, velocity, nd ccelertion without considering wht ight influence tht otion. Now we consider tht influence: Why does the otion of n object chnge? Wht ight cuse one object to rein t rest nd nother object to ccelerte? Why is A person sculls on cl wterwy. The wter exerts forces on the it generlly esier to ove sll object thn lrge ors to ccelerte the bot. (Tetr Iges/Getty Iges) object? The two in fctors we need to consider re the forces cting on n object nd the ss of the object. In this chpter, we begin our study of dynics by discussing the three bsic lws of otion, which del with forces nd sses nd were forulted ore thn three centuries go by Isc Newton. 5.1 The Concept of Force Everyone hs bsic understnding of the concept of force fro everydy experience. When you push your epty dinner plte wy, you exert force on it. iilrly, you exert force on bll when you throw or kick it. In these exples, the word force refers to n interction with n object by ens of usculr ctivity nd soe chnge in the object s velocity. Forces do not lwys cuse otion, however. For exple, when you re sitting, grvittionl force cts on your body nd yet 103

2 104 CHAPTER 5 The Lws of Motion Figure 5.1 oe exples of pplied forces. In ech cse, force is exerted on the object within the boxed re. oe gent in the environent externl to the boxed re exerts force on the object. Contct forces b c Field forces M q Q Iron N d e f Bridgen-Girudon/Art Resource, NY Isc Newton English physicist nd theticin ( ) Isc Newton ws one of the ost brillint scientists in history. Before the ge of 30, he forulted the bsic concepts nd lws of echnics, discovered the lw of universl grvittion, nd invented the theticl ethods of clculus. As consequence of his theories, Newton ws ble to explin the otions of the plnets, the ebb nd flow of the tides, nd ny specil fetures of the otions of the Moon nd the Erth. He lso interpreted ny fundentl observtions concerning the nture of light. His contributions to physicl theories dointed scientific thought for two centuries nd rein iportnt tody. you rein sttionry. As second exple, you cn push (in other words, exert force) on lrge boulder nd not be ble to ove 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 oves 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 oves. When footbll is kicked, s in Figure 5.1c, it is both defored nd set in otion. These situtions re ll exples of clss of forces clled contct forces. Tht is, they involve physicl contct between two objects. Other exples of contct forces re the force exerted by gs olecules 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 epty spce. The grvittionl force of ttrction between two objects with ss, illustrted in Figure 5.1d, is n exple of this clss of force. The grvittionl force keeps objects bound to the Erth nd the plnets in orbit round the un. Another coon 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 for hydrogen to. A third exple of field force is the force br gnet exerts on piece of iron (Fig. 5.1f). The distinction between contct forces nd field forces is not s shrp s you y hve been led to believe by the previous discussion. When exined t the toic 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 odels for croscopic phenoen, it is convenient to use both clssifictions of forces. The only known fundentl forces in nture re ll field forces: (1) grvittionl forces between objects, (2) electrognetic forces between electric chrges, (3) strong forces between subtoic prticles, nd (4) wek forces tht rise in certin rdioctive decy processes. In clssicl physics, we re concerned only with grvittionl nd electrognetic forces. We will discuss strong nd wek forces in Chpter 46. The Vector Nture of Force It is possible to use the defortion of spring to esure 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 c. If we now pply different downwrd force F 2 whose gnitude is twice tht of the reference force

3 5.2 Newton s First Lw nd Inertil Fres 105 A downwrd force F 1 elongtes the spring 1.00 c. A downwrd force F 2 elongtes the spring 2.00 c. F 1 When nd F 2 re pplied together in the se direction, the spring elongtes by 3.00 c. When F 1 is downwrd nd F 2 is horizontl, the cobintion of the two forces elongtes the spring by 2.24 c 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 oves to 2.00 c. Figure 5.2c shows tht the cobined effect of the two colliner forces is the su of the effects of the individul forces. Now suppose the two forces re pplied siultneously with F1 downwrd nd F2 horizontl s illustrted in Figure 5.2d. In this cse, the pointer reds 2.24 c. The single force F tht would produce this se reding is the su 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 experientlly verified to behve s vectors, you ust use the rules of vector ddition to obtin the net force on n object. 5.2 Newton s First Lw nd Inertil Fres We begin our study of forces by igining soe physicl situtions involving puck on perfectly level ir hockey tble (Fig. 5.3). You expect tht the puck will rein sttionry when it is plced gently t rest on the tble. Now igine your ir hockey tble is locted on trin oving with constnt velocity long perfectly sooth trck. If the puck is plced on the tble, the puck gin reins where it is plced. If the trin were to ccelerte, however, the puck would strt oving 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, oving object cn be observed fro ny nuber of reference fres. Newton s first lw of otion, soeties clled the lw of inerti, defines specil set of reference fres clled inertil fres. 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 ove lost without friction. If the tble is not ccelerting, puck plced on the tble will rein t rest. If n object does not interct with other objects, it is possible to identify reference fre in which the object hs zero ccelertion. uch reference fre is clled n inertil fre of reference. When the puck is on the ir hockey tble locted on the ground, you re observing it fro n inertil reference fre; 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 fre of reference

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, ultiple forces tht cncel; it sys wht hppens in the bsence of externl forces. This subtle but iportnt difference llows us to define force s tht which cuses chnge in the otion. The description of n object under the effect of forces tht blnce is covered by Newton s second lw. Another stteent of Newton s first lw Definition of force trin oving t constnt velocity, you re lso observing the puck fro n inertil reference fre. Any reference fre tht oves with constnt velocity reltive to n inertil fre is itself n inertil fre. When you nd the trin ccelerte, however, you re observing the puck fro noninertil reference fre becuse the trin is ccelerting reltive to the inertil reference fre of the Erth s surfce. While the puck ppers to be ccelerting ccording to your observtions, reference fre cn be identified in which the puck hs zero ccelertion. For exple, n observer stnding outside the trin on the ground sees the puck sliding reltive to the tble but lwys oving with the se velocity with respect to the ground s the trin hd before it strted to ccelerte (becuse there is lost 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 fre tht oves with constnt velocity reltive to the distnt strs is the best pproxition of n inertil fre, nd for our purposes we cn consider the Erth s being such fre. The Erth is not relly n inertil fre becuse of its orbitl otion round the un nd its rottionl otion bout its own xis, both of which involve centripetl ccelertions. These ccelertions re sll copred with g, however, nd cn often be neglected. For this reson, we odel the Erth s n inertil fre, long with ny other fre ttched to it. Let us ssue we re observing n object fro n inertil reference fre. (We will return to observtions de in noninertil reference fres in ection 6.3.) Before bout 1600, scientists believed tht the nturl stte of tter ws the stte of rest. Observtions showed tht oving objects eventully stopped oving. Glileo ws the first to tke different pproch to otion nd the nturl stte of tter. He devised thought experients nd concluded tht it is not the nture of n object to stop once set in otion: rther, it is its nture to resist chnges in its otion. In his words, Any velocity once iprted to oving body will be rigidly intined s long s the externl cuses of retrdtion re reoved. For exple, spcecrft drifting through epty spce with its engine turned off will keep oving forever. It would not seek nturl stte of rest. Given our discussion of observtions de fro inertil reference fres, we cn pose ore prcticl stteent of Newton s first lw of otion: In the bsence of externl forces nd when viewed fro n inertil reference fre, n object t rest reins t rest nd n object in otion continues in otion 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. Fro the first lw, we conclude tht ny isolted object (one tht does not interct with its environent) is either t rest or oving with constnt velocity. The tendency of n object to resist ny ttept to chnge its velocity is clled inerti. Given the stteent of the first lw bove, we cn conclude tht n object tht is ccelerting ust be experiencing force. In turn, fro the first lw, we cn define force s tht which cuses chnge in otion of n object. Quick Quiz 5.1 Which of the following stteents is correct? () It is possible for n object to hve otion in the bsence of forces on the object. (b) It is possible to hve forces on n object in the bsence of otion of the object. (c) Neither stteent () nor stteent (b) is correct. (d) Both stteents () nd (b) re correct. 5.3 Mss Igine plying ctch with either bsketbll or bowling bll. Which bll is ore likely to keep oving when you try to ctch it? Which bll requires ore effort to throw it? The bowling bll requires ore effort. In the lnguge of physics, we sy

5 5.4 Newton s econd Lw 107 tht the bowling bll is ore 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 uch resistnce n object exhibits to chnges in its velocity, nd s we lerned in ection 1.1, the I unit of ss is the kilogr. Experients show tht the greter the ss of n object, the less tht object ccelertes under the ction of given pplied force. To describe ss quntittively, we conduct experients in which we copre the ccelertions given force produces on different objects. uppose force cting on n object of ss 1 produces chnge in otion of the object tht we cn quntify with the object s ccelertion 1, nd the se force cting on n object of ss 2 produces n ccelertion 2. The rtio of the two sses is defined s the inverse rtio of the gnitudes of the ccelertions produced by the force: 1 ; 2 (5.1) 2 1 For exple, if given force cting on 3-kg object produces n ccelertion of 4 /s 2, the se force pplied to 6-kg object produces n ccelertion of 2 /s 2. According to huge nuber of siilr observtions, we conclude tht the gnitude of the ccelertion of n object is inversely proportionl to its ss when cted on by given force. If one object hs known ss, the ss of the other object cn be obtined fro ccelertion esureents. Mss is n inherent property of n object nd is independent of the object s surroundings nd of the ethod used to esure it. Also, ss is sclr quntity nd thus obeys the rules of ordinry rithetic. For exple, if you cobine 3-kg ss with 5-kg ss, the totl ss is 8 kg. This result cn be verified experientlly by copring the ccelertion tht known force gives to severl objects seprtely with the ccelertion tht the se force gives to the se objects cobined 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 gnitude of the grvittionl force exerted on the object nd vries with loction (see ection 5.5). For exple, person weighing 180 lb on the Erth weighs only bout 30 lb on the Moon. On the other hnd, the ss of n object is the se everywhere: n object hving ss of 2 kg on the Erth lso hs ss of 2 kg on the Moon. 5.4 Newton s econd Lw Definition of ss Mss nd weight re different quntities Newton s first lw explins wht hppens to n object when no forces ct on it: it either reins t rest or oves in stright line with constnt speed. Newton s second lw nswers the question of wht hppens to n object when one or ore forces ct on it. Igine perforing n experient in which you push block of ss cross frictionless, horizontl surfce. When you exert soe horizontl force F on the block, it oves with soe ccelertion. If you pply force twice s gret on the se block, experientl results show tht the ccelertion of the block doubles; if you increse the pplied force to 3F, the ccelertion triples; nd so on. Fro 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 fro the preceding section tht the gnitude of the ccelertion of n object is inversely proportionl to its ss: 0 0 ~ 1/. These experientl observtions re surized in Newton s second lw: Pitfll Prevention 5.2 Force Is the Cuse of Chnges in Motion An object cn hve otion in the bsence of forces s described in Newton s first lw. Therefore, don t interpret force s the cuse of otion. Force is the cuse of chnges in otion s esured by ccelertion. When viewed fro n inertil reference fre, the ccelertion of n object is directly proportionl to the net force cting on it nd inversely proportionl to its ss: ~ F

6 108 CHAPTER 5 The Lws of Motion If we choose proportionlity constnt of 1, we cn relte ss, ccelertion, nd force through the following theticl stteent of Newton s second lw: 1 Newton s second lw Newton s second lw: coponent for Pitfll Prevention 5.3 Is Not Force Eqution 5.2 does not sy tht the product 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 ss of the object nd the ccelertion tht results fro the net force. Do not include n force in your nlysis of the forces on n object. Definition of the newton Exple 5.1 F 5 (5.2) In both the textul nd theticl stteents 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 su of ll forces cting on the object. (We soeties refer to the net force s the totl force, the resultnt force, or the unblnced force.) In solving proble using Newton s second lw, it is ipertive to deterine the correct net force on n object. Mny forces y be cting on n object, but there is only one ccelertion. Eqution 5.2 is vector expression nd hence is equivlent to three coponent equtions: F x 5 x F y 5 y F z 5 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 tie intervl Dt, resulting in finl speed of v for the object. You then repet the experient, but with force tht is twice s lrge. Wht tie intervl is now required to rech the se 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 ss 1 kg, produces n ccelertion of 1 /s 2. Fro this definition nd Newton s second lw, we see tht the newton cn be expressed in ters of the following fundentl units of ss, length, nd tie: 1 N ; 1 kg? /s 2 (5.4) In the U.. custory syste, the unit of force is the pound (lb). A force of 1 lb is the force tht, when cting on 1-slug ss, 2 produces n ccelertion of 1 ft/s 2 : 1 lb ; 1 slug? ft/s 2 (5.5) A convenient pproxition is 1 N < 1 4 lb. An Accelerting Hockey Puck A hockey puck hving ss of 0.30 kg slides on the frictionless, horizontl surfce of n ice rink. Two hockey sticks strike the puck siultneously, exerting the forces on the puck shown in Figure 5.4. The force F1 hs gnitude of 5.0 N, nd the force F2 hs gnitude of 8.0 N. Deterine both the gnitude 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 fro Chpter 3, predict the pproxite direction of the net force vector on the puck. The ccelertion of the puck will be in the se direction. Figure 5.4 (Exple 5.1) A hockey puck oving 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 uch less thn the speed of light. We tret the reltivistic sitution in Chpter The slug is the unit of ss in the U.. custory syste nd is tht syste s counterprt of the I unit the kilogr. Becuse ost of the clcultions in our study of clssicl echnics re in I units, the slug is seldo used in this text.

7 5.5 The Grvittionl Force nd Weight cont. Ctegorize Becuse we cn deterine net force nd we wnt n ccelertion, this proble is ctegorized s one tht y be solved using Newton s second lw. Anlyze Find the coponent of the net force cting on the puck in the x direction: Find the coponent of the net force cting on the puck in the y direction: Use Newton s second lw in coponent for (Eq. 5.3) to find the x nd y coponents 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 y 5 F y N 5 29 /s kg N 5 17 /s kg Find the gnitude of the ccelertion: 5 "129 /s /s /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 siultneously, with two of the exerting the forces shown in Figure 5.4. The result of the three forces is tht the hockey puck shows no ccelertion. Wht ust be the coponents of the third force? Answer If there is zero ccelertion, the net force cting on the puck ust be zero. Therefore, the three forces ust cncel. We hve found the coponents of the cobintion of the first two forces. The coponents of the third force ust be of equl gnitude nd opposite sign so tht ll the coponents 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 Fg. This force is directed towrd the center of the Erth, 3 nd its gnitude 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 to freely flling object of ss, with 5 g nd g F 5 Fg, gives g F 5 g Therefore, the weight of n object, being defined s the gnitude of Fg, is equl to g: F g = g (5.6) Becuse it depends on g, weight vries with geogrphic loction. Becuse g decreses with incresing distnce fro the center of the Erth, objects weigh less t higher ltitudes thn t se level. For exple, kg pllet of bricks used in the construction of the Epire tte Building in New York City weighed N t street level, but weighed bout 1 N less by the tie it ws lifted fro sidewlk 3 This stteent ignores tht the ss distribution of the Erth is not perfectly sphericl. Pitfll Prevention 5.4 Weight of n Object We re filir with the everydy phrse, the weight of n object. Weight, however, is not n inherent property of n object; rther, it is esure of the grvittionl force between the object nd the Erth (or other plnet). Therefore, weight is property of syste of ites: the object nd the Erth. Pitfll Prevention 5.5 Kilogr Is Not Unit of Weight You y hve seen the conversion 1 kg lb. Despite populr stteents of weights expressed in kilogrs, the kilogr is not unit of weight, it is unit of ss. The conversion stteent is not n equlity; it is n equivlence tht is vlid only on the Erth s surfce.

8 110 CHAPTER 5 The Lws of Motion NAA/Eugene Cernn The life-support unit strpped to the bck of stronut Hrrison chitt weighed 300 lb on the Erth nd hd ss of 136 kg. During his trining, 50-lb ock-up with ss of 23 kg ws used. Although this strtegy effectively siulted the reduced weight the unit would hve on the Moon, it did not correctly iic the unchnging ss. It ws ore difficult to ccelerte the 136-kg unit (perhps by juping 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 exple, suppose student hs ss of 70.0 kg. The student s weight in loction where g /s 2 is 686 N (bout 150 lb). At the top of ountin, however, where g /s 2, the student s weight is only 684 N. Therefore, if you wnt to lose weight without going on diet, clib ountin 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 oving. Even for sttionry object or for n object on which severl forces ct, Eqution 5.6 cn be used to clculte the gnitude of the grvittionl force. The result is subtle shift in the interprettion of in the eqution. The ss in Eqution 5.6 deterines the strength of the grvittionl ttrction between the object nd the Erth. This role is copletely different fro tht previously described for ss, tht of esuring the resistnce to chnges in otion in response to n externl force. In tht role, ss is lso clled inertil ss. We cll in Eqution 5.6 the grvittionl ss. Even though this quntity is different in behvior fro inertil ss, it is one of the experientl conclusions in Newtonin dynics tht grvittionl ss nd inertil ss hve the se 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 fro one plnet to the next, but the gnitude of the grvittionl force will lwys be given by the vlue of g. 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 hi tht you entered the Erth version of the se contest nd lso won newton of gold! Who is richer? () You re. (b) Your friend is. (c) You re eqully rich. Conceptul Exple 5.2 How Much Do You Weigh in n Elevtor? You hve ost likely been in n elevtor tht ccelertes upwrd s it oves towrd higher floor. In this cse, you feel hevier. In fct, if you re stnding on bthroo scle t the tie, the scle esures force hving gnitude tht is greter thn your weight. Therefore, you hve tctile nd esured 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 fre. To provide the ccelertion upwrd, the floor or scle ust exert on your feet n upwrd force tht is greter in gnitude 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 kes sll dent in your skin. If you push hrder, the book does the se nd the dent in your skin is little lrger. This siple ctivity illustrtes tht forces re interctions between two objects: when your finger pushes on the book, the book pushes bck on your finger. This iportnt 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 gnitude nd opposite in direction to the force F21 exerted by object 2 on object 1: F12 52F 21 (5.7)

9 5.6 Newton s Third Lw 111 When it is iportnt to designte forces s interctions between two objects, we will use this subscript nottion, where Fb ens 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 ters re not scientific ters; furtherore, either force cn be lbeled the ction or rection force. We will use these ters for convenience. In ll cses, the ction nd rection forces ct on different objects nd ust be of the se type (grvittionl, electricl, etc.). For exple, 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 gnitude of this force is g. The rection to this force is the grvittionl force exerted by the projectile on the Erth FpE 52F Ep. The rection force FpE ust ccelerte the Erth towrd the projectile just s the ction force FEp ccelertes the projectile towrd the Erth. Becuse the Erth hs such lrge ss, however, its ccelertion due to this rection force is negligibly sll. Consider coputer onitor t rest on tble s in Figure 5.6. The rection force to the grvittionl force Fg 5 FE on the onitor is the force FE 52F E exerted by the onitor on the Erth. The onitor does not ccelerte becuse it is held up by the tble. The tble exerts on the onitor n upwrd force n 5 Ft, clled the norl force. (Norl in this context ens perpendiculr.) This force, which prevents the onitor fro flling through the tble, cn hve ny vlue needed, up to the point of breking the tble. Becuse the onitor hs zero ccelertion, Newton s second lw pplied to the onitor gives us g F 5 n 1 g 5 0, so n j^ 2 g j^ 5 0, or n 5 g. The norl force blnces the grvittionl force on the onitor, so the net force on the onitor is zero. The rection force to n is the force exerted by the onitor downwrd on the tble, Ft 52F t 52n. Notice tht the forces cting on the onitor re Fg nd n s shown in Figure 5.6b. The two forces FE nd Ft re exerted on objects other thn the onitor. Figure 5.6 illustrtes n extreely iportnt step in solving probles involving forces. Figure 5.6 shows ny of the forces in the sitution: those cting on the onitor, 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 onitor, nd is clled force digr or digr showing the forces on the object. The iportnt pictoril representtion in Figure 5.6c is clled free-body digr. In free-body digr, the prticle odel 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 odel s prticle. Therefore, free-body digr helps us isolte only those forces on the object nd eliinte the other forces fro 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 gnitude 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 g In the sitution shown in Figure 5.6 nd in ny others, we find tht n 5 Norl g (the norl force force hs the se gnitude 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 coponents, or if there is verticl ccelertion of the syste, then n? g. Alwys pply Newton s second lw to find the reltionship between n nd g. Pitfll Prevention 5.7 Newton s Third Lw Reeber tht Newton s third-lw ction nd rection forces ct on different objects. For exple, in Figure 5.6, n 5 Ft 52g 52F E. The forces n nd g re equl in gnitude nd opposite in direction, but they do not represent n ction rection pir becuse both forces ct on the se object, the onitor. 1 n F t n F t n F t F t F g F E F E F g F E b c F g F E Figure 5.6 () When coputer onitor is t rest on tble, the forces cting on the onitor re the norl force n nd the grvittionl force F g. The rection to n is the force F t exerted by the onitor on the tble. The rection to F g is the force F E exerted by the onitor on the Erth. (b) A digr showing the forces on the onitor. (c) A freebody digr shows the onitor s blck dot with the forces cting on it.

10 112 CHAPTER 5 The Lws of Motion Pitfll Prevention 5.8 Free-Body Digrs The ost iportnt step in solving proble using Newton s lws is to drw proper sketch, the free-body digr. 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-oving bus, which experiences n ipct force with lrger gnitude? () The fly. (b) The bus. (c) The se force is experienced by both. (ii) Which experiences the greter ccelertion? () The fly. (b) The bus. (c) The se ccelertion is experienced by both. Conceptul Exple 5.3 You Push Me nd I ll Push You A lrge n nd sll boy stnd fcing ech other on frictionless ice. They put their hnds together nd push ginst ech other so tht they ove prt. (A) Who oves wy with the higher speed? OLUTION This sitution is siilr to wht we sw in Quick Quiz 5.5. According to Newton s third lw, the force exerted by the n on the boy nd the force exerted by the boy on the n re third-lw pir of forces, so they ust be equl in gnitude. (A bthroo scle plced between their hnds would red the se, regrdless of which wy it fced.) Therefore, the boy, hving the sller ss, experiences the greter ccelertion. Both individuls ccelerte for the se ount of tie, but the greter ccelertion of the boy over this tie intervl results in his oving wy fro the interction with the higher speed. (B) Who oves frther while their hnds re in contct? OLUTION Becuse the boy hs the greter ccelertion nd therefore the greter verge velocity, he oves frther thn the n during the tie intervl during which their hnds re in contct. 5.7 Anlysis Models Using Newton s econd Lw In this section, we discuss two nlysis odels for solving probles in which objects re either in equilibriu or ccelerting long stright line under the ction of constnt externl forces. Reeber 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 odeled s prticles, we need not worry bout rottionl otion. For now, we lso neglect the effects of friction in those probles involving otion, which is equivlent to stting tht the surfces re frictionless. (The friction force is discussed in ection 5.8.) We usully neglect the ss of ny ropes, strings, or cbles involved. In this pproxition, the gnitude of the force exerted by ny eleent of the rope on the djcent eleent is the se for ll eleents long the rope. In proble stteents, the synonyous ters light nd of negligible ss re used to indicte tht ss is to be ignored when you work the probles. When rope ttched to n object is pulling on the object, the rope exerts force on the object in direction wy fro the object, prllel to the rope. The gnitude T of tht force is clled the tension in the rope. Becuse it is the gnitude of vector quntity, tension is sclr quntity.

11 5.7 Anlysis Models Using Newton s econd Lw 113 Anlysis Model: The Prticle in Equilibriu If the ccelertion of n object odeled s prticle is zero, the object is treted with the prticle in equilibriu odel. In this odel, the net force on the object is zero: F 5 0 (5.8) Consider lp suspended fro light chin fstened to the ceiling s in Figure 5.7. The force digr for the lp (Fig. 5.7b) shows tht the forces cting on the lp re the downwrd grvittionl force Fg 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 infortion. 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 Fg re not n ction rection pir becuse they ct on the se object, the lp. The rection force to T is downwrd force exerted by the lp on the chin. F g Figure 5.7 () A lp suspended fro ceiling by chin of negligible ss. (b) The forces cting on the lp 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 otion cn be nlyzed with the prticle under net force odel. The pproprite eqution for this odel is Newton s second lw, Eqution 5.2: F 5 (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 ust hve friction; otherwise, his feet would siply 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 digr in Figure 5.8b. Notice tht the horizontl force T being pplied to the crte cts through the rope. The gnitude of T is equl to the tension in the rope. In ddition to the force T, the free-body digr for the crte includes the grvittionl force F g nd the norl force n exerted by the floor on the crte. We cn now pply Newton s second lw in coponent for to the crte. The only force cting in the x direction is T. Applying o F x 5 x to the horizontl otion gives F x 5 T 5 x or x 5 T No ccelertion occurs in the y direction becuse the crte oves only horizontlly. Therefore, we use the prticle in equilibriu odel in the y direction. Applying the y coponent of Eqution 5.8 yields o F y 5 n 1 (2F g ) 5 0 or n 5 F g Tht is, the norl force hs the se gnitude s the grvittionl force but cts in the opposite direction. If T is constnt force, the ccelertion x 5 T/ lso is constnt. Hence, the crte is lso odeled s prticle under constnt ccelertion in the x direction, nd the equtions of kinetics fro Chpter 2 cn be used to obtin the crte s position x nd velocity v x s functions of tie. In the sitution just described, the gnitude of the norl force n is equl to the gnitude of Fg, but tht is not lwys the cse, s noted in Pitfll Prevention 5.6. For exple, 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 g 1 F. In this sitution, the norl force is greter thn the grvittionl force. Other exples 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 digr 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 norl force n on the object is greter thn the grvittionl force: n 5 F g 1 F. y x

12 114 CHAPTER 5 The Lws of Motion Proble-olving trtegy APPLYING NEWTON LAW The following procedure is recoended when deling with probles involving Newton s lws: 1. Conceptulize. Drw siple, net digr of the syste. The digr helps estblish the entl representtion. Estblish convenient coordinte xes for ech object in the syste. 2. Ctegorize. If n ccelertion coponent for n object is zero, the object is odeled s prticle in equilibriu in this direction nd o F 5 0. If not, the object is odeled s prticle under net force in this direction nd o F Anlyze. Isolte the object whose otion is being nlyzed. Drw free-body digr for this object. For systes contining ore thn one object, drw seprte freebody digrs for ech object. Do not include in the free-body digr forces exerted by the object on its surroundings. Find the coponents of the forces long the coordinte xes. Apply the pproprite odel fro the Ctegorize step for ech direction. Check your diensions to ke sure tht ll ters hve units of force. olve the coponent equtions for the unknowns. Reeber tht you generlly ust hve s ny independent equtions s you hve unknowns to obtin coplete solution. 4. Finlize. Mke sure your results re consistent with the free-body digr. Also check the predictions of your solutions for extree vlues of the vribles. By doing so, you cn often detect errors in your results. Exple 5.4 A Trffic Light t Rest A trffic light weighing 122 N hngs fro cble tied to two other cbles fstened to support s in Figure The upper cbles ke 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 the exceeds 100 N. Does the trffic light rein 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 ssue the cbles do not brek nd nothing is oving. Ctegorize If nothing is oving, no prt of the syste is ccelerting. We cn now odel the light s prticle in equilibriu on which the net force is zero. iilrly, the net force on the knot (Fig. 5.10c) is zero. F g b c Figure 5.10 (Exple 5.4) () A trffic light suspended by cbles. (b) The forces cting on the trffic light. (c) The free-body digr for the knot where the three cbles re joined. T 3 Anlyze We construct digr of the forces cting on the trffic light, shown in Figure 5.10b, nd free-body digr 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

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 coponents: Force x Coponent y Coponent 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 equilibriu odel 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 coponents of T 1 nd T 2 ust be equl in gnitude, nd Eqution (2) shows tht the su of the verticl coponents of T 1 nd T 2 ust blnce the downwrd force T 3, which is equl in gnitude to the weight of the light. cos 37.0 olve Eqution (1) for T 2 in ters 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 proble by igining chnge in the syste, 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 fro the syetry of the proble tht the two tensions T 1 nd T 2 would be equl to ech other. Mtheticlly, if the equl ngles re clled u, Eqution (3) becoes 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 Exple 5.5 Forces Between Crs in Trin Trin crs re connected by couplers, which re under tension s the locootive pulls the trin. Igine you re on trin speeding up with constnt ccelertion. As you ove through the trin fro the locootive to the lst cr, esuring the tension in ech set of couplers, does the tension increse, decrese, or sty the se? When the engineer pplies the brkes, the couplers re under copression. How does this copression force vry fro the locootive to the lst cr? (Assue only the brkes on the wheels of the engine re pplied.) OLUTION While the trin is speeding up, tension decreses fro the front of the trin to the bck. The coupler between the locootive nd the first cr ust pply enough force to ccelerte the rest of the crs. As you ove bck long the trin, ech coupler is ccelerting less ss 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 fro front to bck. The coupler connecting the locootive to the first cr ust pply lrge force to slow down the rest of the crs, but the finl coupler ust pply force lrge enough to slow down only the lst cr.

14 116 CHAPTER 5 The Lws of Motion Exple 5.6 The Runwy Cr A cr of ss is on n icy drivewy inclined t n ngle u s in Figure y (A) Find the ccelertion of the cr, ssuing the drivewy is frictionless. OLUTION Conceptulize Use Figure 5.11 to conceptulize the sitution. Fro everydy experience, we know tht cr on n icy incline will ccelerte down the incline. (The se thing hppens to cr on hill with its brkes not set.) Ctegorize We ctegorize the cr s prticle under net force becuse it ccelertes. Furtherore, this exple belongs to very coon ctegory of probles in which n object oves under the influence of grvity on n inclined plne. u x g cos u b Figure 5.11 (Exple 5.6) () A cr on frictionless incline. (b) The freebody digr for the cr. The blck dot represents the position of the center of ss of the cr. We will lern bout center of ss in Chpter 9. u n F g sin u g = g x Anlyze Figure 5.11b shows the free-body digr for the cr. The only forces cting on the cr re the norl force n exerted by the inclined plne, which cts perpendiculr to the plne, nd the grvittionl force Fg 5 g, which cts verticlly downwrd. For probles 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 coponent of gnitude g sin u long the positive x xis nd one of gnitude g cos u long the negtive y xis. Our choice of xes results in the cr being odeled s prticle under net force in the x direction nd prticle in equilibriu in the y direction. Apply these odels to the cr: olve Eqution (1) for x : (1) o F x 5 g sin u 5 x (2) o F y 5 n 2 g cos u 5 0 (3) x 5 g sin u Finlize Note tht the ccelertion coponent x is independent of the ss of the cr! It depends only on the ngle of inclintion nd on g. Fro Eqution (2), we conclude tht the coponent of F g perpendiculr to the incline is blnced by the norl force; tht is, n 5 g cos u. This sitution is nother cse in which the norl force is not equl in gnitude to the weight of the object. It is possible, lthough inconvenient, to solve the proble with stndrd horizontl nd verticl xes. You y wnt to try it, just for prctice. (B) uppose the cr is relesed fro rest t the top of the incline nd the distnce fro the cr s front buper to the botto of the incline is d. How long does it tke the front buper to rech the botto of the hill, nd wht is the cr s speed s it rrives there? OLUTION Conceptulize Igine tht the cr is sliding down the hill nd you use stopwtch to esure the entire tie intervl until it reches the botto. Ctegorize This prt of the proble belongs to kinetics rther thn to dynics, nd Eqution (3) shows tht the ccelertion x is constnt. Therefore, you should ctegorize the cr in this prt of the proble s prticle under constnt ccelertion.

15 5.7 Anlysis Models Using Newton s econd Lw cont. Anlyze Defining the initil position of the front buper 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 fro Equtions (4) nd (5) tht the tie t t which the cr reches the botto nd its finl speed v xf re independent of the cr s ss, s ws its ccelertion. Notice tht we hve cobined techniques fro Chpter 2 with new techniques fro this chpter in this exple. As we lern ore techniques in lter chpters, this process of cobining nlysis odels nd infortion fro severl prts of the book will occur ore often. In these cses, use the Generl Proble-olving trtegy to help you identify wht nlysis odels you will need. d x t 2 2d 2d (4) t 5 5 Å x Å g sin u v 2 xf 5 2 x d (5) v xf 5 "2 x d 5 "2gd sin u WHAT IF? Wht previously solved proble does this sitution becoe if u 5 90? Answer Igine u going to 90 in Figure The inclined plne becoes verticl, nd the cr is n object in free fll! Eqution (3) becoes 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 g cos u gives us n 5 g 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. Exple 5.7 One Block Pushes Another Two blocks of sses 1 nd 2, with 1. 2, re plced in contct with ech other on frictionless, horizontl surfce s in Active Figure A constnt horizontl force F is pplied to 1 s shown. (A) Find the gnitude of the ccelertion of the syste. ACTIVE FIGURE 5.12 (Exple 5.7) () A force is OLUTION pplied to block of ss 1, which pushes on second block Conceptulize Conceptulize the sitution by using Active Figure 5.12 nd ing on 1. (c) The forces cting of ss 2. (b) The forces ct- relize tht both blocks ust experience on 2. the se ccelertion becuse they re in contct with ech other nd rein in contct throughout the otion. Ctegorize We ctegorize this proble s one involving prticle under net force becuse force is pplied to syste of blocks nd we re looking for the ccelertion of the syste. y x P 21 b F n1 F 1 2 n2 P g c g Anlyze First odel the cobintion of two blocks s single prticle under net force. Apply Newton s second lw to the cobintion in the x direction to find the ccelertion: o F x 5 F 5 ( ) x F (1) x continued

16 118 CHAPTER 5 The Lws of Motion 5.7 cont. Finlize The ccelertion given by Eqution (1) is the se s tht of single object of ss nd subject to the se force. (B) Deterine the gnitude of the contct force between the two blocks. OLUTION Conceptulize The contct force is internl to the syste of two blocks. Therefore, we cnnot find this force by odeling the whole syste (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 digr 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. Fro Active Figure 5.12c, we see tht the only horizontl force cting on 2 is the contct force P 12 (the force exerted by 1 on 2 ), which is directed to the right. Apply Newton s second lw to 2 : (2) o F x 5 P x ubstitute the vlue of the ccelertion x given by Eqution (1) into Eqution (2): 2 (3) P x 5 bf 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 ust be less thn the force required to produce the se ccelertion for the two-block syste. To finlize further, let us check this expression for P 12 by considering the forces cting on 1, shown in Active Figure 5.12b. The horizontl forces cting on 1 re the pplied force F to the right nd the contct force P 21 to the left (the force exerted by 2 on 1 ). Fro 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 1 : olve for P 12 nd substitute the vlue of x fro Eqution (1): (4) o F x 5 F 2 P 21 5 F 2 P x F P 12 5 F 2 1 x 5 F 2 1 b 5 bf This result grees with Eqution (3), s it ust. WHAT IF? Igine tht the force F in Active Figure 5.12 is pplied towrd the left on the right-hnd block of ss 2. Is the gnitude of the force P 12 the se s it ws when the force ws pplied towrd the right on 1? Answer When the force is pplied towrd the left on 2, the contct force ust ccelerte 1. In the originl sitution, the contct force ccelertes 2. Becuse 1. 2, ore force is required, so the gnitude of P 12 is greter thn in the originl sitution. Exple 5.8 Weighing Fish in n Elevtor A person weighs fish of ss 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 fro 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. Igine tht the fish is hnging on string ttched to the end of the spring.

17 5.7 Anlysis Models Using Newton s econd Lw cont. In this cse, the gnitude 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 proble 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 digrs 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 g nd the force T exerted by the string. If the elevtor is either t rest or oving t constnt velocity, the fish is prticle in equilibriu, so o F y 5 T 2 F g 5 0 or T 5 F g 5 g. (Reeber tht the sclr g is the weight of the fish.) Now suppose the elevtor is oving with n ccelertion reltive to n observer stnding outside the elevtor in n inertil fre. The fish is now prticle under net force. T T g Figure 5.13 (Exple 5.8) A fish is weighed on spring scle in n ccelerting elevtor cr. b g Apply Newton s second lw to the fish: olve for T: o F y 5 T 2 g 5 y (1) T 5 y 1 g 5 g y g 1 1b 5 F g y g 1 1b where we hve chosen upwrd s the positive y direction. We conclude fro Eqution (1) tht the scle reding T is greter thn the fish s weight g if is upwrd, so y is positive (Fig. 5.13), nd tht the reding is less thn g if is downwrd, so y is negtive (Fig. 5.13b). (B) Evlute the scle redings for 40.0-N fish if the elevtor oves with n ccelertion y /s 2. OLUTION Evlute the scle reding fro Eqution (1) if is upwrd: 2.00 /s2 T N2 1 1b N /s Evlute the scle reding fro Eqution (1) if is downwrd: /s2 T N2 1 1b N 9.80 /s 2 Finlize Tke this dvice: if you buy fish in n elevtor, ke sure the fish is weighed while the elevtor is either t rest or ccelerting downwrd! Furtherore, notice tht fro the infortion given here, one cnnot deterine the direction of otion 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 fro Eqution (1) tht the scle reding T is zero in this cse; tht is, the fish ppers to be weightless.

18 120 CHAPTER 5 The Lws of Motion Exple 5.9 The Atwood Mchine When two objects of unequl ss re hung verticlly over frictionless pulley of negligible ss s in Active Figure 5.14, the rrngeent is clled n Atwood chine. The device is soeties used in the lbortory to deterine the vlue of g. Deterine the gnitude of the ccelertion of the two objects nd the tension in the lightweight cord. OLUTION Conceptulize Igine the sitution pictured in Active Figure 5.14 in ction: s one object oves upwrd, the other object oves downwrd. Becuse the objects re connected by n inextensible string, their ccelertions ust be of equl gnitude. Ctegorize The objects in the Atwood chine re subject to the grvittionl force s well s to the forces exerted by the strings connected to the. Therefore, we cn ctegorize this proble s one involving two prticles under net force. Anlyze The free-body digrs 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 probles such s this one in which the pulley is odeled s ssless nd frictionless, the tension in the string on both sides of the pulley is the se. If the pulley hs ss or is subject to friction, the tensions on either side re not the se nd the sitution requires techniques we will lern in Chpter 10. ACTIVE FIGURE 5.14 (Exple 5.9) The Atwood chine. () Two objects connected by ssless inextensible cord over frictionless pulley. (b) The freebody digrs for the two objects. We ust be very creful with signs in probles 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 ust define the downwrd direction s positive for object 2. With this sign convention, both objects ccelerte in the se direction s defined by the choice of sign. Furtherore, ccording to this sign convention, the y coponent of the net force exerted on object 1 is T 2 1 g, nd the y coponent of the net force exerted on object 2 is 2 g 2 T T 1 1 g b T 2 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 1 g 5 1 y (2) o F y 5 2 g 2 T 5 2 y 2 1 g 1 2 g 5 1 y 1 2 y olve for the ccelertion: (3) y bg ubstitute Eqution (3) into Eqution (1) to find T: (4) T 5 1 (g 1 y ) bg Finlize The ccelertion given by Eqution (3) cn be interpreted s the rtio of the gnitude of the unblnced force on the syste ( )g to the totl ss of the syste ( ), s expected fro Newton s second lw. Notice tht the sign of the ccelertion depends on the reltive sses of the two objects. WHAT IF? Describe the otion of the syste if the objects hve equl sses, tht is, Answer If we hve the se ss on both sides, the syste is blnced nd should not ccelerte. Mtheticlly, we see tht if 1 5 2, Eqution (3) gives us y 5 0. WHAT IF? Wht if one of the sses is uch lrger thn the other: 1.. 2? Answer In the cse in which one ss is infinitely lrger thn the other, we cn ignore the effect of the sller ss. Therefore, the lrger ss should siply fll s if the sller ss were not there. We see tht if 1.. 2, Eqution (3) gives us y 5 2g.

19 5.7 Anlysis Models Using Newton s econd Lw 121 Exple 5.10 Accelertion of Two Objects Connected by Cord A bll of ss 1 nd block of ss 2 re ttched by lightweight cord tht psses over frictionless pulley of negligible ss s in Figure The block lies on frictionless incline of ngle u. Find the gnitude of the ccelertion of the two objects nd the tension in the cord. 1 2 u y T x 1 g OLUTION b Conceptulize Igine the objects in Figure 5.15 in otion. If 2 oves down the incline, then 1 oves upwrd. Becuse the objects re connected by cord (which we ssue does not stretch), their ccelertions hve the se gnitude. 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 (Exple 5.10) () Two objects connected by lightweight cord strung over frictionless pulley. (b) The freebody digr for the bll. (c) The free-body digr for the block. (The incline is frictionless.) c 2g cos u T u n 2 g y 2 g sin u x Anlyze Consider the free-body digrs shown in Figures 5.15b nd 5.15c. Apply Newton s second lw in coponent for to the bll, choosing the upwrd direction s positive: (1) o F x 5 0 (2) o F y 5 T 2 1 g 5 1 y 5 1 For the bll to ccelerte upwrd, it is necessry tht T. 1 g. In Eqution (2), we replced y with becuse the ccelertion hs only y coponent. 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 coponent for to the block: (3) o F x9 5 2 g sin u 2 T 5 2 x9 5 2 (4) o F y9 5 n 2 2 g cos u 5 0 In Eqution (3), we replced x9 with becuse the two objects hve ccelertions of equl gnitude. olve Eqution (2) for T : (5) T 5 1 (g 1 ) ubstitute this expression for T into Eqution (3): 2 g sin u 2 1 (g 1 ) 5 2 olve for : (6) 5 2 sin u2 1 bg ubstitute this expression for into Eqution (5) to find T : (7) T sin u112 bg Finlize The block ccelertes down the incline only if 2 sin u. 1. If 1. 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 gnitude of the net externl force cting on the bll block syste divided by the totl ss of the syste; this result is consistent with Newton s second lw. WHAT IF? Wht hppens in this sitution if u 5 90? continued

20 122 CHAPTER 5 The Lws of Motion 5.10 cont. Answer If u 5 90, the inclined plne becoes verticl nd there is no interction between its surfce nd 2. Therefore, this proble becoes the Atwood chine of Exple 5.9. Letting u 90 in Equtions (6) nd (7) cuses the to reduce to Equtions (3) nd (4) of Exple 5.9! WHAT IF? Wht if 1 5 0? Answer If 1 5 0, then 2 is siply sliding down n inclined plne without intercting with 1 through the string. Therefore, this proble becoes the sliding cr proble in Exple 5.6. Letting 1 0 in Eqution (6) cuses it to reduce to Eqution (3) of Exple 5.6! 5.8 Forces of Friction When n object is in otion either on surfce or in viscous ediu such s ir or wter, there is resistnce to the otion becuse the object intercts with its surroundings. We cll such resistnce force of friction. Forces of friction re very iportnt in our everydy lives. They llow us to wlk or run nd re necessry for the otion of wheeled vehicles. Igine 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 reins sttionry when F is sll. The force on the trsh cn tht countercts F nd keeps it fro oving cts towrd the left nd is clled For sll pplied forces, the gnitude of the force of sttic friction equls the gnitude of the pplied force. When the gnitude of the pplied force exceeds the gnitude of the xiu force of sttic friction, the trsh cn breks free nd ccelertes to the right. n n Motion fs F fk F g g f b f s,x 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,x. f k. c O f s F ttic region f k k n Kinetic region F

21 5.8 Forces of Friction 123 the force of sttic friction f s. As long s the trsh cn is not oving, f s 5 F. Therefore, if F is incresed, f s lso increses. Likewise, if F decreses, f s lso decreses. Experients show tht the friction force rises fro the nture of the two surfces: becuse of their roughness, contct is de only t few loctions where peks of the teril touch. At these loctions, the friction force rises in prt becuse one pek physiclly blocks the otion of pek fro the opposing surfce nd in prt fro cheicl bonding ( spot welds ) of opposing peks s they coe into contct. Although the detils of friction re quite coplex t the toic level, this force ultitely involves n electricl interction between tos or olecules. If we increse the gnitude 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 xiu vlue f s,x s shown in Active Figure 5.16c. When F exceeds f s,x, the trsh cn oves nd ccelertes to the right. We cll the friction force for n object in otion the force of kinetic friction f k. When the trsh cn is in otion, the force of kinetic friction on the cn is less thn f s,x (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 oves to the right with constnt speed. If the pplied force F is reoved fro the oving 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. Experientlly, we find tht, to good pproxition, both f s,x nd f k re proportionl to the gnitude of the norl force exerted on n object by the surfce. The following descriptions of the force of friction re bsed on experientl observtions nd serve s the odel we shll use for forces of friction in proble solving: The gnitude of the force of sttic friction between ny two surfces in contct cn hve the vlues f s # s n (5.9) where the diensionless constnt s is clled the coefficient of sttic friction nd n is the gnitude of the norl 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,x 5 s n. This sitution is clled ipending otion. The inequlity holds when the surfces re not on the verge of slipping. The gnitude of the force of kinetic friction cting between two surfces is f k 5 k n (5.10) where 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 k nd s depend on the nture of the surfces, but k is generlly less thn s. Typicl vlues rnge fro round 0.03 to 1.0. Tble 5.1 (pge 124) lists soe 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 otion (kinetic friction) or the ipending otion (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 ight expect tht plcing n object on the side hving the ost re ight increse the friction force. Although this ethod provides ore 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 pproxitely copenste 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 Liited 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 coon trp of using f s 5 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 gnitudes of the vectors representing the friction nd norl forces. Becuse the friction nd norl forces re perpendiculr to ech other, the vectors cnnot be relted by ultiplictive constnt. Pitfll Prevention 5.11 The Direction of the Friction Force oeties, n incorrect stteent bout the friction force between n object nd surfce is de the friction force on n object is opposite to its otion or ipending otion rther thn the correct phrsing, the friction force on n object is opposite to its otion or ipending otion reltive to the surfce.

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 Aluinu 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 etl (lubricted) Teflon on Teflon Ice on ice ynovil joints in huns Note: All vlues re pproxite. In soe 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 fro 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 fro 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? Exple 5.11 Experientl Deterintion of M s nd M k The following is siple ethod of esuring 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 ove. how tht you cn obtin s by esuring the criticl ngle u c t which this slipping just occurs. OLUTION Conceptulize Consider Active Figure 5.18 nd igine tht the block tends to slide down the incline due to the grvittionl force. To siulte the sitution, plce coin on this book s cover nd tilt the book until the coin begins to slide. Notice how this exple differs fro Exple 5.6. When there is no friction on n incline, ny ngle of the incline will cuse sttionry object to begin oving. When there is friction, however, there is no oveent 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 ove but is not oving, we ctegorize the block s prticle in equilibriu. Anlyze The digr in Active Figure 5.18 shows the forces on the block: the grvittionl force g, the norl 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 g cos u u g n y g sin u ACTIVE FIGURE 5.18 (Exple 5.11) The externl forces exerted on block lying on rough incline re the grvittionl force g, the norl force n, nd the force of friction f s. For convenience, the grvittionl force is resolved into coponent g sin u long the incline nd coponent g cos u perpendiculr to the incline. u x

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 g sin u 2 f s 5 0 (2) o F y 5 n 2 g cos u 5 0 ubstitute g 5 n/cos u fro 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 xiu vlue s n. The ngle u in this sitution is the criticl ngle u c. Mke these substitutions in Eqution (3): For exple, if the block just slips t u c , we find tht s 5 tn (3) f s 5 g sin u5 n b sin u5n tn u cos u s n 5 n tn u c s 5 tn u c Finlize Once the block strts to ove t u $ u c, it ccelertes down the incline nd the force of friction is f k 5 k n. If u is reduced to vlue less thn u c, however, it y be possible to find n ngle u9 c such tht the block oves down the incline with constnt speed s prticle in equilibriu gin ( x 5 0). In this cse, use Equtions (1) nd (2) with f s replced by f k to find k : k 5 tn u9 c, where u9 c, u c. Exple 5.12 The liding Hockey Puck A hockey puck on frozen pond is given n initil speed of 20.0 /s. If the puck lwys reins on the ice nd slides 115 before coing to rest, deterine the coefficient of kinetic friction between the puck nd ice. n Motion OLUTION Conceptulize Igine tht the puck in Figure 5.19 slides to the right nd eventully coes 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 proble provides kinetic vribles. Therefore, we ctegorize the proble in two wys. First, it involves prticle under net force: kinetic friction cuses the puck to ccelerte. Furtherore, becuse we odel the force of kinetic friction s independent of speed, the ccelertion of the puck is constnt. o, we cn lso ctegorize this proble s one involving prticle under constnt ccelertion. fk g Figure 5.19 (Exple 5.12) After the puck is given n initil velocity to the right, the only externl forces cting on it re the grvittionl force g, the norl force n, nd the force of kinetic friction f k. Anlyze First, let s find the ccelertion lgebriclly in ters 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 kinetics cn be used to find the nuericl vlue of the coefficient of kinetic friction. The digr in Figure 5.19 shows the forces on the puck. Apply the prticle under net force odel in the x direction to the puck: (1) o F x 5 2 f k 5 x Apply the prticle in equilibriu odel in the y direction to the puck: ubstitute n 5 g fro Eqution (2) nd f k 5 k n into Eqution (1): (2) o F y 5 n 2 g k n 5 2 k g 5 x x 5 2 k g The negtive sign ens 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 ss of the puck nd is constnt becuse we ssue k reins constnt. continued

24 126 CHAPTER 5 The Lws of Motion 5.12 cont. Apply the prticle under constnt ccelertion odel 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: k 5 v xi 2 ubstitute the nuericl vlues: k v xi x x f 5 v xi k gx f 2gx f /s /s Finlize Notice tht k is diensionless, s it should be, nd tht it hs low vlue, consistent with n object sliding on ice. Exple 5.13 Accelertion of Two Connected Objects When Friction Is Present A block of ss 2 on rough, horizontl surfce is connected to bll of ss 1 by lightweight cord over lightweight, frictionless pulley s shown in Figure A force of gnitude 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 k. Deterine the gnitude of the ccelertion of the two objects. OLUTION Conceptulize Igine wht hppens s F is pplied to the block. Assuing 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 proble s one involving two prticles under net force, the bll nd the block. 1 2 u F 1 T y 1 g x T fk b c n F sin u u F cos u 2 2 g Figure 5.20 (Exple 5.13) () The externl force F pplied s shown cn cuse the block to ccelerte to the right. (b, c) Digrs showing the forces on the two objects, ssuing the block ccelertes to the right nd the bll ccelertes upwrd. F Anlyze First drw force digrs for the two objects s shown in Figures 5.20b nd 5.20c. Notice tht the string exerts force of gnitude T on both objects. The pplied force F hs x nd y coponents F cos u nd F sin u, respectively. Becuse the two objects re connected, we cn equte the gnitudes of the x coponent of the ccelertion of the block nd the y coponent of the ccelertion of the bll nd cll the both. Let us ssue the otion of the block is to the right. Apply the prticle under net force odel to the block in the (1) o F x 5 F cos u 2 f k 2 T 5 2 x 5 2 horizontl direction: Becuse the block oves only horizontlly, pply the prticle in equilibriu odel to the block in the verticl direction: Apply the prticle under net force odel to the bll in the verticl direction: olve Eqution (2) for n: (2) o F y 5 n 1 F sin u 2 2 g 5 0 (3) o F y 5 T 2 1 g 5 1 y 5 1 n 5 2 g 2 F sin u ubstitute n into f k 5 k n fro Eqution 5.10: (4) f k 5 k ( 2 g 2 F sin u) ubstitute Eqution (4) nd the vlue of T fro Eqution (3) into Eqution (1): olve for : F cos u 2 k ( 2 g 2 F sin u) 2 1 ( 1 g) 5 2 (5) 5 F 1cos u 1 k sin u k 2 2g 1 1 2

25 ury cont. Finlize The ccelertion of the block cn be either to the right or to the left depending on the sign of the nuertor in Eqution (5). If the otion is to the left, we ust reverse the sign of f k in Eqution (1) becuse the force of kinetic friction ust oppose the otion of the block reltive to the surfce. In this cse, the vlue of is the se s in Eqution (5), with the two plus signs in the nuertor chnged to inus signs. Wht does Eqution (5) reduce to if the force F is reoved nd the surfce becoes frictionless? Cll this expression Eqution (6). Does this lgebric expression tch your intuition bout the physicl sitution in this cse? Now go bck to Exple 5.10 nd let ngle u go to zero in Eqution (6) of tht exple. How does the resulting eqution copre with your Eqution (6) here in Exple 5.13? hould the lgebric expressions copre in this wy bsed on the physicl situtions? Definitions ury An inertil fre of reference is fre in which n object tht does not interct with other objects experiences zero ccelertion. Any fre oving with constnt velocity reltive to n inertil fre is lso n inertil fre. We define force s tht which cuses chnge in otion of n object. Concepts nd Principles Newton s first lw sttes tht it is possible to find n inertil fre 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 fro n inertil fre, n object t rest reins t rest nd n object in unifor otion in stright line intins tht otion. 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 ss. Newton s third lw sttes tht if two objects interct, the force exerted by object 1 on object 2 is equl in gnitude 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 ss ( sclr quntity) nd the freefll ccelertion: F g 5 g. The weight of n object is the gnitude of the grvittionl force cting on the object. The xiu force of sttic friction f s,x between n object nd surfce is proportionl to the norl force cting on the object. In generl, f s # s n, where s is the coefficient of sttic friction nd n is the gnitude of the norl force. When n object slides over surfce, the gnitude of the force of kinetic friction f k is given by f k 5 k n, where k is the coefficient of kinetic friction. continued

26 128 CHAPTER 5 The Lws of Motion Anlysis Models for Proble olving Prticle Under Net Force If prticle of ss experiences nonzero net force, its ccelertion is relted to the net force by Newton s second lw: F 5 (5.2) F Prticle in Equilibriu If prticle intins 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 F 0 Objective Questions 1. An experient is perfored 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 esured. Now the se puck is trnsported fr into outer spce, where both friction nd grvity re negligible. The se constnt force is pplied to the puck (through spring scle tht stretches the se ount), nd the puck s ccelertion (reltive to the distnt strs) is esured. Wht is the puck s ccelertion in outer spce? () It is soewht greter thn its ccelertion on the Erth. (b) It is the se 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 sll but not zero. 2. In Figure OQ5.2, locootive hs broken through the wll of trin sttion. During the collision, wht cn be sid bout the force exerted by the locootive on the wll? () The force exerted by the locootive on the wll ws lrger thn the force the wll could exert on the locootive. (b) The force exerted by the locootive on the wll denotes nswer vilble in tudent olutions Mnul/tudy Guide ws the se in gnitude s the force exerted by the wll on the locootive. (c) The force exerted by the locootive on the wll ws less thn the force exerted by the wll on the locootive. (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 the, snowblls of vrious sses re oving with different velocities s shown in Figure OQ5.3. Rnk the snowblls () through (e) ccording to the gnitude of the totl force exerted on ech one. Ignore ir resistnce. If two snowblls rnk together, ke tht fct cler. 9 /s 400 g 400 g 300 g 12 /s b 8 /s 10 /s c 500 g 12 /s 200 g e d tudio Lévy nd ons Figure OQ5.2 Figure OQ The nger of deprtent store is pushing horizontlly with force of gnitude 200 N on box of shirts. The box is sliding cross the horizontl floor with forwrd ccelertion. Nothing else touches the box. Wht ust be true bout the gnitude 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 stteents is necessrily true. 5. The driver of speeding epty truck sls on the brkes nd skids to stop through distnce d. On second tril, the truck crries lod tht doubles its ss. Wht

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 sls 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 ss oves with ccelertion down rough incline. Which of the following forces should pper in free-body digr of the object? Choose ll correct nswers. () the grvittionl force exerted by the plnet (b) in the direction of otion (c) the norl 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 ss is plce on the fltbed of truck but not tied down. As the truck ccelertes forwrd with ccelertion, the crte reins t rest reltive to the truck. Wht force cuses the crte to ccelerte? () the norl force (b) the grvittionl force (c) the friction force (d) the force exerted by the crte (e) No force is required. 9. A crte reins sttionry fter it hs been plced on rp inclined t n ngle with the horizontl. Which of the following stteents is or re correct bout the gnitude 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 s n. (c) It is greter thn the coponent of the grvittionl force cting down the rp. (d) It is equl to the coponent of the grvittionl force cting down the rp. (e) It is less thn the coponent of the grvittionl force cting down the rp. 10. An object of ss is sliding with speed v i t soe instnt cross level tbletop, with which its coefficient of kinetic friction is. It then oves through distnce d nd coes to rest. Which of the following equtions for the speed v i is resonble? () v i 5!22gd (b) v i 5!2gd (c) v i 5!22gd (d) v i 5!2gd (e) v i 5!2d 11. If n object is in equilibriu, which of the following stteents is not true? () The speed of the object reins constnt. (b) The ccelertion of the object is zero. (c) The net force cting on the object is zero. (d) The object ust 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 reins constnt. Wht hppens to the ccelertion of the truck if its triler leks snd t constnt rte through hole in its botto? () 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 reins constnt. 13. Two objects re connected by string tht psses over frictionless pulley s in Active Figure 5.14, where 1, 2 nd 1 nd 2 re the gnitudes of the respective ccelertions. Which theticl stteent is true regrding the gnitude of the ccelertion 2 of the ss 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 /s, wht is the resultnt force cting on it? 3. In the otion picture It Hppened One Night (Colubi Pictures, 1934), Clrk Gble is stnding inside sttionry bus in front of Cludette Colbert, who is seted. The bus suddenly strts oving forwrd nd Clrk flls into Cludette s lp. Why did this hppen? 4. Your hnds re wet, nd the restroo towel dispenser is epty. Wht do you do to get drops of wter off your hnds? How does the otion of the drops exeplify one of Newton s lws? Which one? 5. A pssenger sitting in the rer of bus clis tht she ws injured when the driver sled on the brkes, cusing suitcse to coe flying towrd her fro the front of the bus. If you were the judge in this cse, wht disposition would you ke? 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 otion in ters of the forces now cting on the rubber. (c) Account for the otion of skyrocket tking off fro its lunch pd. 7. If you hold horizontl etl br severl centieters bove the ground nd ove 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 ore quickly. How then does rotry power lwn ower nge 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 oving wy fro 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 ove if it did? (b) Cn the force of the throw be equl in gnitude to the grvittionl force? Explin. (c) Wht strength cn ccurtely be ttributed to the force of the throw? Explin. (d) Why does the bll ove wy fro the child s hnd?

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 yor of city reprinds soe city eployees becuse they will not reove the obvious sgs fro the cbles tht support the city trffic lights. Wht explntion cn the eployees give? How do you think the cse will be settled in edition? 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 the. He decides to jup 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 oent does the brnch brek? Explin. uggestion: Pretend to be the third boy nd iitte wht he does in slow otion. If you re still unsure, stnd on bthroo 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 king greter effort. When tble exerts 200-N norl force insted of one of sller gnitude, is the tble relly doing nything differently? 13. A weightlifter stnds on bthroo scle. He pups 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 norl force be horizontl? (b) Cn norl force be directed verticlly downwrd? (c) Consider tennis bll in contct with sttionry floor nd with nothing else. Cn the norl force be different in gnitude fro the grvittionl force exerted on the bll? (d) Cn the force exerted by the floor on the bll be different in gnitude fro the force the bll exerts on the floor? 15. A cr is oving forwrd slowly nd is speeding up. A student clis 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 2 1 block 1 nd the lrger-ss block 2. Cord A exerts force Figure CQ5.16 on block 1 to ke it ccelerte forwrd. () How does the gnitude of the force exerted by cord A on block 1 copre with the gnitude of the force exerted by cord B on block 2? Is it lrger, sller, or equl? (b) How does the ccelertion of block 1 copre 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, sller, or equl in gnitude to the force exerted by cord B on block 2? 17. Identify ction rection pirs in the following situtions: () n 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 tes of ten people re so evenly tched tht neither te wins. After the ge they notice tht cr is stuck in the ud. They ttch the tug-of-wr rope to the buper of the cr, nd ll the people pull on the rope. The hevy cr hs just oved couple of decieters when the rope breks. Why did the rope brek in this sitution when it did not brek when the se 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 gy. 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 clibs the rope, soeties 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 exples in which the force of friction exerted on n object is in the direction of otion of the object. 22. As shown in Figure CQ5.22, student A, 55-kg girl, sits on one chir with etl runners, t rest on clssroo floor. tudent B, n 80-kg boy, sits on n identicl chir. Both students keep their feet off the floor. A rope runs fro 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 sling on your brkes when you wnt to stop in the shortest possible distnce? (Mny odern crs hve ntilock brkes tht void this proble.)

29 Probles 131 Probles The probles found in this chpter y be ssigned online in Enhnced WebAssign 1. denotes strightforwrd proble; 2. denotes interedite proble; 3. denotes chllenging proble 1. full solution vilble in the tudent olutions Mnul/tudy Guide 1. denotes probles ost 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 /s 2. Find () the resultnt force cting on the object nd (b) the gnitude of the resultnt force. 2. The verge speed of nitrogen olecule in ir is bout /s, nd its ss is kg. () If it tkes s for nitrogen olecule to hit wll nd rebound with the se speed but oving in the opposite direction, wht is the verge ccelertion of the olecule during this tie intervl? (b) Wht verge force does the olecule 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^ /s t one instnt. Eight seconds lter, its velocity is 18 i^ 1 10 j^ 2 /s. Assuing the rocket engine exerts constnt horizontl force, find () the coponents of the force nd (b) its gnitude. 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 gnitude of 18.0 N. Find the gnitude of the net force exerted by the wire on the crooked tooth. x y T Figure P Review. The grvittionl force exerted on bsebll is 2.21 N down. A pitcher throws the bll horizontlly with velocity 18.0 /s by uniforly ccelerting it long stright horizontl line for tie intervl of 170 s. The bll strts fro rest. () Through wht distnce does it ove before its relese? (b) Wht re the gnitude 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 uniforly ccelerting it long stright horizontl line T denotes sking for quntittive nd conceptul resoning denotes sybolic resoning proble denotes Mster It tutoril vilble in Enhnced WebAssign denotes guided proble shded denotes pired probles tht develop resoning with sybols nd nuericl vlues for tie intervl of Dt 5 t t. () trting fro rest, through wht distnce does the bll ove before its relese? (b) Wht force does the pitcher exert on the bll? 7. Review. An electron of ss kg hs n initil speed of /s. It trvels in stright line, nd its speed increses to /s in distnce of 5.00 c. Assuing its ccelertion is constnt, () deterine the gnitude of the force exerted on the electron nd (b) copre 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 fro rest nd in 1.20 s experiences displceent of i^ j^ 2, where the direction of j^ is the upwrd verticl direction. Deterine the other force. 9. One or ore externl forces, lrge enough to be esily esured, 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 ss M hs been plced on rubber cushion of ss. Together they re sliding to the right t constnt velocity on n ice-covered prking lot. () Drw free-body digr of the brick nd identify ech force cting on it. (b) Drw free-body digr 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 syste. 11. An object of ss is dropped t t 5 0 fro 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 tie t does the object strike the ground? Express t in ters of g nd h. (b) Find n expression in ters of 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 ters of, g, F, nd h. (d) Find the gnitude of the object s ccelertion while it is flling, using the vribles F,, nd g. 12. A force F pplied to n object of ss 1 produces n ccelertion of 3.00 /s 2. The se force pplied to second object of ss 2 produces n ccelertion of 1.00 /s 2. () Wht is the vlue of the rtio 1 / 2? (b) If 1 nd 2 re cobined into one object, find its ccelertion under the ction of the force F.

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. F F1 Figure P5.13 b F F1 14. You stnd on the set of chir nd then hop off. () During the tie intervl you re in flight down to the floor, the Erth oves towrd you with n ccelertion of wht order of gnitude? In your solution, explin your logic. Model the Erth s perfectly solid object. (b) The Erth oves towrd you through distnce of wht order of gnitude? 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 gnitude 3.75 /s 2. () Wht is the direction of the ccelertion? (b) Wht is the ss of the object? (c) If the object is initilly t rest, wht is its speed fter 10.0 s? (d) Wht re the velocity coponents of the object fter 10.0 s? ection 5.7 oe Applictions of Newton s Lws 17. Review. Figure P5.17 shows worker poling bot very efficient ode of trnsporttion cross shllow lke. He pushes prllel to the length of the light pole, exerting force of gnitude 240 N on the botto of the lke. Assue the pole lies in the verticl plne contining the keel of the bot. At one oent, the pole kes 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 gnitude /s. The ss of the bot including its crgo nd the worker is 370 kg. () The wter exerts buoynt force verticlly upwrd on the bot. Find the gnitude of this force. (b) Model the forces s constnt over short intervl of tie to find the velocity of the bot s fter the oent described. 18. An iron bolt of ss 65.0 g hngs fro string 35.7 c long. The top end of the string is fixed. Without touching AP Iges it, gnet ttrcts the bolt so tht it reins sttionry, but is displced horizontlly 28.0 c to the right fro the previously verticl line of the string. The gnet is locted to the right of the bolt nd on the se verticl level s the bolt in the finl configurtion. () Drw free-body digr of the bolt. (b) Find the tension in the string. (c) Find the gnetic force on the bolt. 19. Figure P5.19 shows the horizontl forces cting on silbot oving north t constnt velocity, seen fro point stright bove its st. 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 coponent 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) Copre your solutions to prts () nd (b). Do the results gree? Is one ethod significntly esier? 20. The systes shown in Figure P5.20 re in equilibriu. If the spring scles re clibrted in newtons, wht do they red? Ignore the sses of the pulleys nd strings nd ssue 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 d E 220 N Figure P A block slides down frictionless plne hving n inclintion of u The block strts fro rest t the top, nd the length of the incline is () Drw free-body digr of the block. Find (b) the ccelertion of the block nd (c) its speed when it reches the botto of the incline. 22. A 3.00-kg object is oving 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 eters nd t is in seconds. Find the gnitude of the net force cting on this object t t s. P u

31 Probles The distnce between two telephone poles is When 1.00-kg bird lnds on the telephone wire idwy between the poles, the wire sgs () Drw free-body digr of the bird. (b) How uch tension does the bird produce in the wire? Ignore the weight of the wire. 24. A bg of ceent weighing 325 N hngs in equilibriu fro three wires s suggested in Figure P5.24. Two of the wires ke ngles u nd u with the horizontl. Assuing the syste is in equilibriu, find the tensions T 1, T 2, nd T 3 in the wires. 25. A bg of ceent whose weight is F g hngs in equilibriu fro three wires s shown in Figure P5.24. Two of the wires ke ngles u 1 nd u 2 with the horizontl. Assuing the syste is in equilibriu, 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 siilr to the one shown in Figure P5.26 is often used in hospitls to support nd pply horizontl trction force to n injured leg. () Deterine 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 Probles 24 nd An object of ss kg is observed to hve n ccelertion F 2 with gnitude of /s 2 in direction 60.0 est of north. Figure P5.27 shows view of the object fro bove. The force F 1 F2 cting on the object hs gnitude of 5.00 N nd is Figure P5.27 directed north. Deterine the gnitude nd direction of the one other horizontl force F1 cting on the object. 28. An object of ss kg 1 plced on frictionless, horizontl tble is connected to string tht psses over pulley nd then is fstened to hnging object of ss kg s shown in Figure P5.28. () Drw free-body digrs Figure P5.28 of both objects. Find (b) the gnitude of the ccelertion of the Probles 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. Assue the otion is verticl nd the ss of the person s body is 64.0 kg. Deterine 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 (c/s) tie (s) Figure P Two objects re connected by light string tht psses over frictionless pulley s shown in Figure P5.30. Assue the incline is frictionless nd tke kg, kg, nd u () Drw free-body digrs of both objects. Find (b) the gnitude 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 fro rest. 31. Two blocks, ech of ss kg, re hung fro the ceiling of n elevtor s in Figure P5.31. () If the elevtor oves with n upwrd cceler tion of gnitude 1.60 /s 2, find the tensions T 1 nd T 2 in the upper nd lower strings. (b) If the strings cn withstnd xiu tension of 85.0 N, wht xiu ccelertion cn the elevtor hve before string breks? 32. Two blocks, ech of ss, re hung fro the ceiling of n elevtor s in Figure P5.31. The elevtor 1 2 hs n upwrd ccelertion. The strings hve negligible ss. () Find the tensions T 1 nd T 2 in the upper nd lower strings in ters of,, nd g. (b) Copre the two tensions nd deterine which string would brek first if is de sufficiently lrge. (c) Wht re the tensions if the cble supporting the elevtor breks? 33. In the syste shown in Figure P5.33, horizontl force Fx cts on n object of ss kg. The horizontl 1 2 Figure P5.33 u Figure P5.30 T 1 T 2 Figure P5.31 Probles 31 nd 32. F x

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 ss kg ccelerte upwrd? (b) For wht vlues of F x is the tension in the cord zero? (c) Plot the ccelertion of the 2 object versus F x. Include vlues of F x fro 2100 N to 1100 N. 34. An object of ss 1 hngs fro 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 ss 2 on frictionless, horizontl tble. () If 1 nd 2 re the ccelertions of 1 nd 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 ters of the sses 1 nd 2, nd g. 2 P 2 P 1 Figure P In Exple 5.8, we investigted the pprent weight of fish in n elevtor. Now consider 72.0-kg n stnding on spring scle in n elevtor. trting fro rest, the elevtor scends, ttining its xiu speed of 1.20 /s in s. It trvels with this constnt speed for the next 5.00 s. The elevtor then undergoes unifor ccelertion in the negtive y direction for 1.50 s nd coes to rest. Wht does the spring scle register () before the elevtor strts to ove, (b) during the first s, (c) while the elevtor is trveling t constnt speed, nd (d) during the tie intervl it is slowing down? 36. In the Atwood chine discussed in Exple 5.9 nd shown in Active Figure 5.14, kg nd kg. The sses of the pulley nd string re negligible by coprison. The pulley turns without friction, nd the string does not stretch. The lighter object is relesed with shrp push tht sets it into otion t v i /s downwrd. () How fr will 1 descend below its initil level? (b) Find the velocity of 1 fter 1.80 s. ection 5.8 Forces of Friction 37. Review. A rifle bullet with ss of 12.0 g trveling towrd the right t 260 /s strikes lrge bg of snd nd penetrtes it to depth of 23.0 c. Deterine the gnitude nd direction of the friction force (ssued constnt) tht cts on the bullet. 38. Review. A cr is trveling t 50.0 i/h on horizontl highwy. () If the coefficient of sttic friction between rod nd tires on riny dy is 0.100, wht is the iniu distnce in which the cr will stop? (b) Wht is the stopping distnce when the surfce is dry nd s ? 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 otion, fter which horizontl force of 60.0 N is required to keep the block oving 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 ipossible? 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 k/h nd brke to stop with constnt ccelertion over distnce of Your physics book reins on the set rther thn sliding forwrd onto the floor. 41. To eet U.. Postl ervice requireent, eployees footwer ust hve coefficient of sttic friction of 0.5 or ore on specified tile surfce. A typicl thletic shoe hs coefficient of sttic friction of In n eergency, wht is the iniu tie intervl in which person strting fro rest cn ove 3.00 on the tile surfce if she is wering () footwer eeting the Postl ervice iniu nd (b) typicl thletic shoe? 42. Before 1960, people believed tht the xiu ttinble coefficient of sttic friction for n utoobile tire on rodwy ws s 5 1. Around 1962, three copnies independently developed rcing tires with coefficients of 1.6. This proble shows tht tires hve iproved further since then. The shortest tie intervl in which piston-engine cr initilly t rest hs covered distnce of one-qurter ile is bout 4.43 s. () Assue the cr s rer wheels lift the front wheels off the pveent s shown in Figure P5.42. Wht iniu vlue of s is necessry to chieve the record tie? (b) uppose the driver were ble to increse his or her engine power, keeping other things equl. How would this chnge ffect the elpsed tie? Jie quire/allsport/getty Iges Figure P Review. A 3.00-kg block strts fro rest t the top of 30.0 incline nd slides distnce of 2.00 down the incline in 1.50 s. Find () the gnitude 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 A won 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 digr of the suitcse. (b) Wht ngle does u Figure P5.44