Essentials of College Physics ,

885 Pages·2007·24.24 MB·English

by  Raymond A. Serway

PEDAGOGICAL USE OF COLOR
Displacement and Torque (t) and
position vectors angular momentum
(L) vectors
(cid:2)
Velocity vectors (v)
Velocity component vectors
Linear or rotational
(cid:2) motion directions
Force vectors (F)
Force component vectors
Springs
(cid:2)
Acceleration vectors (a)
Acceleration component vectors
Electric fields Capacitors
Magnetic fields Inductors (coils)
Positive charges + Voltmeters V
Negative charges – Ammeters A
Resistors Lightbulbs
Batteries and other
AC sources
DC power supplies – +
Switches Ground symbol
Light rays Objects
Lenses and prisms Images
Mirrors
CONVERSION FACTORS
Length Speed
1 m = 39.37 in. = 3.281 ft 1 km/h = 0.278 m/s = 0.621 mi/h
1 in. = 2.54 cm 1 m/s = 2.237 mi/h = 3.281 ft/s
1 km = 0.621 mi 1 mi/h = 1.61 km/h = 0.447 m/s = 1.47 ft/s
1 mi = 5 280 ft = 1.609 km
1 light year (ly) = 9.461 (cid:2) 1015 m Force
1 angstrom (Å) = 10(cid:3)10 m 1 N = 0.224 8 lb = 105 dynes
1 lb = 4.448 N
Mass 1 dyne = 10(cid:3)5 N = 2.248 (cid:2) 10(cid:3)6 lb
1 kg = 103 g = 6.85 (cid:2) 10(cid:3)2 slug
1 slug = 14.59 kg Work and energy
1 u = 1.66 (cid:2) 10(cid:3)27 kg = 931.5 MeV/c2 1 J = 107 erg = 0.738 ft(cid:6)lb = 0.239 cal
1 cal = 4.186 J
Time
(cid:6)
1 ft lb = 1.356 J
1 min = 60 s 1 Btu = 1.054 (cid:2) 103 J = 252 cal
1 h = 3 600 s 1 J = 6.24 (cid:2) 1018 eV
1 day = 8.64 (cid:2) 104 s 1 eV = 1.602 (cid:2) 10(cid:3)19 J
1 yr = 365.242 days = 3.156 (cid:2) 107 s 1 kWh = 3.60 (cid:2) 106 J
Volume Pressure
1 L = 1 000 cm3 = 3.531 (cid:2) 10(cid:3)2 ft3 1 atm = 1.013 (cid:2) 105 N/m2 (or Pa) = 14.70 lb/in.2
1 ft3 = 2.832 (cid:2) 10(cid:3)2 m3 1 Pa = 1 N/m2 = 1.45 (cid:2) 10(cid:3)4 lb/in.2
1 gal = 3.786 L = 231 in.3 1 lb/in.2 = 6.895 (cid:2) 103 N/m2
Angle Power
180(cid:4) = (cid:5) rad 1 hp = 550 ft(cid:6)lb/s = 0.746 kW
1 rad = 5.730(cid:4) 1 W = 1 J/s = 0.738 ft(cid:6)lb/s
1(cid:4) = 60 min = 1.745 (cid:2) 10(cid:3)2 rad 1 Btu/h = 0.293 W
Essentials
of
College
Physics
Raymond A. Serway
Emeritus,James Madison University
Chris Vuille
Embry-Riddle Aeronautical University
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The strength of Essentials of College Physics
starts with the foundation
Essentials of College Physics provides students with a clear and logical
presentation of the basic concepts and principles of physics. With the text as
the foundation, coupled with extraordinary media integration, it’s easy to build
a course that gives every student the maximum opportunity for success.
Briefer than the average college physics text,
Essentials of College Physicscompre- 154 Chapter 7 Rotational Motion and the Law of Gravity
hensively covers all the standard topics in
Exercise 7.3
classical and modern physics. Instructors (a) What are the angular speed and angular displacement of the disc 0.300s after it begins to rotate? (b) Find the
will notice a clean and clear dialogue with tangential speed at the rim at this time.
the student, the book’s uncluttered look and Answers (a) 10.6rad/s; 1.58rad (b) 0.472m/s
feel, and attention paid to language. The
authors’ clear,
logical, relaxed, 7.4 CENTRIPETAL ACCELERATION
Vectors are denoted in bold-
and engaging v Figure 7.5a shows a car moving in a circular path with constant linear speed v. Even
face with arrows over them. r though the car moves at a constant speed, it still has an acceleration. To under-
style facilitates O stand this, consider the defining equation for average acceleration:
This makes them easier
qheunicskio cno mpre- to recognize. (a) The numerator represents the dif:afearve(cid:2)nce:vt ffb(cid:4)(cid:4)ett:wviieen the velocity vectors :vfan[7d. 1:v2i].
(cid:1) vi (cid:2) Tif htheseey vheacvteo rds imffearye hnat vdei rtehcteio snasm, teh meiarg dniiftfuedree, nccoer rceasnp’ot nedqiunagl tzoe rtoh.e Tsahme ed isrpeecetido,n b uotf
vf the car’s velocity as it moves in the circular path is continually changing, as shown
in Figure 7.5b. For circular motion at constant speed, the acceleration vector
r r always points toward the center of the circle. Such an acceleration is called a
centripetal(center-seeking) acceleration. Its magnitude is given by
O (b) ac(cid:2)vr2 [7.13]
Fa(locaba ifcr gr)i a pt uurAsa nrmvste edhto lhe o7fvrerc.igo n5icotmgaye r wsv (cid:1)( meaiatc )hoctt ooev Ccer nio(cid:2) srtcn crah,ius lpattolhaneannetrgga temd l st sihp,or eseetoic eoc tdtiinhro. cenouf- a(vevssliuoT(cid:2)comi tvdeyf e (cid:2)t:rvhiivaveat) t.E : vTtqiiomua acenta idlotc inu:va lfn7a.dtd1e 3i tft,hfh eceeron an ocsancitdel yepl eroir niFan tigitdo ui(cid:2)nrre,e cw w7teii.ot6 hbnae ;. vgAetihnlnoe wcoiriibt tyjhme :cvEatfgq inaus itafit tuarido stlnea as t7 et a.rp1r o2eti,i mnthte e(cid:1) tfs.wa mWithee
acceleration. :aav(cid:2):vtff(cid:4)(cid:4):tvii(cid:2)(cid:3)(cid:3):vt [7.14]
walhsoe rvee r(cid:3)y:v sm(cid:2)a:lvlf. (cid:4)In: vFiiigsu trhee 7c.h6abn, g:vef iins avlemloocsitt yp. aWrahlleenl t(cid:3)ot :ivsi ,v aenryd stmhael lv,e (cid:3)ctsoarn (cid:3)d:v (cid:3)ius aarpe-
proximately perpendicular to them, pointing toward the center of the circle. In
the limiting case when (cid:3)tbecomes vanishingly small, (cid:3):vpoints exactly toward the
(cid:1) v(cid:3)is (cid:2) vf caeccnetelerr aotfi othne :a c. iFrcrloem, a Enqdu tahtieo nav 7e.r1a4g,e :a acacnedl e(cid:3)r:avtipooni n:at aivnb tehceo msamese t hdeir einctsitoann t(ainne tohuiss
limit), so the instantaneous acceleration points to the center of the circle.
r (cid:3)u r The triangle in Figure 7.6a, which has sides (cid:3)sand r, is similar to the one
formed by the vectors in Figure 7.6b, so the ratios of their sides are equal:
O(a) (cid:3)vv(cid:2)(cid:3)rs
or
vf
(cid:3)v (cid:3)u (cid:3)v(cid:2)v (cid:3)s [7.15]
–vi r
(b) Substituting the result of Equation 7.15 into aav(cid:2)(cid:3)v/(cid:3)tgives
imvcFm:vteefisiog.l nn ovv(uteicebenlirsro)tge y fcoT rti(cid:3)7fohth y:.tmvee 6hv , d e ec(cid:1)w ocicrh(tnieoatriscocrc)tth r lic(cid:2)Aeuo hi.scns,a t t ttnihooohgfewne e t aphsdfr oaefidrrrr etco tdihchmcetlaeiet on:evngri-e to oifn Bitiuntg (cid:3) csaisse t hweh denis t(cid:3)antcbee ctroamveelse dve arlyo nsmga aatvhll,(cid:2)e (cid:3)asrvr/c (cid:3)o(cid:3)(cid:3)tfts athpep rcoiracclhee isn tthime ein (cid:3)stta, natnadn eino uths[ ev7 a.l1ilmu6e]-
(cid:3) Important statements and definitions
are set in boldfacetype or are highlighted
with a background screen for added
emphasis and ease of review. Important
equations are highlighted with a tan back-
ground.
n
o
i
t
a
d
n The International System of units (SI)
u is used throughoutthe book. The U.S.
o
customary system of units is used only
F
to a limited extent in the problem sets of
the early chapters on mechanics.
iv
Essentials of College Physics provides a wealth of outstanding
examples and problem sets to help students develop critical prob-
lem-solving skills and conceptual understanding.
(cid:4) All worked Examplesinclude six parts:
Goal, Problem, Strategy, Solution, Remarks,
8.4 Examples of Objects in Equilibrium 181
and Exercise/Answer. The “Solution” portion of
every Exampleis presented in two-columns to EXAMPLE 8.4 Locating Your Lab Partner’s Center of Gravity
Goal Use torque to find a center of gravity.
enhance student learning and to help reinforce
L
physics concepts. In addition, the authors have Problem In this example, we show how to find the loca- L/2
tion of a person’s center of gravity. Suppose your lab part-
taken special care to present a graduated level ner has a height Lof 173cm (5 ft, 8 in) and a weight w of n F
715N (160lb). You can determine the position of his O
of difficulty within the Examplesso students are cbeonatredr osuf pgproarvtiteyd bya t haovnine g ehnidm bsytr eatc hsc oaluet, oans as huonwinfo rimn xcg
better prepared to work the end-of-chapter Freigaudrineg8 F.9i.s I3f. 5t0he(cid:5) b1o0a2rdN’,s fiwnedig thhte wdbis itsa n4c9e No f aynodu rt lhaeb spcaarlte- w wb
Problems. ner’s center of gravity from the left end of the board. Figure 8.9 (Example 8.4) Determining your lab partner’s
center of gravity.
Strategy To find the position xcgof the center of gravity, compute the torques using an axis through O.Set the
sum of the torques equal to zero and solve for xcg.
Solution
Apply the second condition of equilibrium. There is no (cid:7)(cid:8)(cid:9)(cid:2)0
taormrq uise z deuroe. to the normal force :nbecause its moment (cid:4)wxcg(cid:4)wb(L/2)(cid:6)FL(cid:2)0
Solve for xcgand substitute known values: xcg(cid:2)FL(cid:4)wwb(L/2)
(cid:2)(350 N)(173 cm)(cid:4)(49 N)(86.5 cm)(cid:2)79 cm
715 N
Remarks The given information is sufficient only to determine the x-coordinate of the center of gravity. The other
two coordinates can be estimated, based on the body’s symmetry.
Exercise 8.4
Suppose a 416-kg alligator of length 3.5 m is stretched out on a board of the same length weighing 65 N. If the board
is supported on the ends as in Figure 8.9, and the scale reads 1 880 N, find the x-component of the alligator’s center
of gravity.
Answer 1.59 m
8.4 EXAMPLES OF OBJECTS IN EQUILIBRIUM
Recall from Chapter 4 that when an object is treated as a geometric point, equilib-
rium requires only that the net force on the object is zero. In this chapter, we have
shown that for extended objects a second condition for equilibrium must also be
satisfied: the net torque on the object must be zero. The following general proce-
dure is recommended for solving problems that involve objects in equilibrium.
Math Focus 6.1 One-Dimensional Elastic Collisions
Theusualnotationandsubscriptsusedintheequa- simplify,obtaining
tionsofone-dimensionalcollisionsoftenobscurethe
underlyingsimplicityofthemathematics.Inanelastic (cid:4)3(cid:2)X(cid:6)2Y Proble(1m) -Solving Strategy Objects in Equilibrium
collision,theratherformidable-lookingEquations6.10 27(cid:2)X2(cid:6)2Y2 (2)
and6.11areused,correspondingtoconservationof 1.Diagram the system. Include coordinates and choose a convenient rotation axis
momentumandconservationofenergy,respectively. Equation(1)isthatofastraightline,whereasfEorq cuoamtiopunting the net torque on the object.
Itloinec sai t aiteryesp ioaclfal tlgh pievr eocnob,ll leliemdaiv,n itgnh goe btmwjeoacs tussen. sSk aunnbodswt tinthuse,t itinhngeit itfiahnle av mle lvooecr-ei- (a(2(cid:4)n)d3ds(cid:4)eusbc2srYtiibt)ue2ts(cid:6)eainn2etYoll2iE,pwqsehu.aicSthioolcnvae(nE2b)q,euosabitmitoapinnli(ifin1eg)2d f.2ot7oDiornb(cid:2)rXgjae wocn ta. i(ftM.r eFoeos-rbt ospydrsoyte bdmlieasmg wrsai twmhi lmol fho tarhveee t oah basijnen cgotln eoe fo oibnbjetjeecrcte to,s tfd, irsnahtweo rwae issnetp.g)a arlalt eexdtieargnraalm fo frocre es aaccht-
cgoemthmero wni-tlho othkien kgn voawrinab qluesa,n Xtit(cid:2)iesv 1yifealdnsd e Yqu(cid:2)atvio2nf,s t foo-r Y2(cid:6)2Y(cid:4)3(cid:2)0
a straight line (the momentum equation) and an el- In general, the quadratic formula must now be ap-
ltiiposne t(htehne reendeurcgeys e tqou fiantidoinn)g. Tthhee imntaetrhseecmtiaotnic aolf saolu- 1plmie/ds, obru t(cid:4) t3hmis /eqs.u Oatniolyn t hfaec tfiorrsst, agnivsiwnegr ,Y 1(cid:2)m/vs2, fm(cid:2)akes
straight line and an ellipse. sense. Substituting it into Equation (1) yields
Efimv1rx2isat(cid:2)(cid:2)m op3b2jlmkeegc:/ tIas nh,n wadash oimnenriaeteis-aadssl im mvthe1ele(cid:2)on scsei1itocykon gvna 2adli nc (cid:2)oodbl (cid:4)lijines3ciiottmi nah/l,a vsss.ue mFlpoipancsoidstsy et h tehe Itht eis daelsroiv peods Esiqbulea ttioo nuX s6e(cid:2). 1Ev4q1.u fTa(cid:2)htiios(cid:4) nsi 5t6um.1a/0ti sotong, eiltlhuesrtr watiethd
final velocities for the two objects. (For clarity, signifi- in Example 6.5, is equivalent to finding the intersec-
cant figure conventions are not observed here.) tion of two straight lines. It’s easier to remember the
equation for the conservation of energy than the spe-
SYo(cid:2)lutvio2fn:inStuobEstqituuatteiothnes6g.i1ve0nanvadlu6e.1s1a,nrdesXpe(cid:2)ctivv1eflya,nadnd csoialvl eE qsuucahti opnro 6b.l1e4m, sso b iott’sh a w gaoyos.d idea to be able to
F
(cid:5) New!Just-In-Time Math Tutorials! o
An emphasis on quantitative problem-solving is provided in the u
n
Math Focusboxes. These boxes develop mathematical methods
d
important to a particular area of physics, or point out a technique
a
that is often overlooked. Each Math Focusbox has been placed t
within the applicable section of the text, giving students just-in-time io
support. A complete Appendix provides students with additional n
math help applied to specific physics concepts.
v
Building critical-thinking skills
and conceptual understanding
Essentials of College Physics includes time-tested as well as new peda-
gogy that adheres to the findings of physics education research to help
students improve their conceptual understanding.
(cid:3) A wealth of interesting and relevant
288 Chapter 11 Energy in Thermal Processes Applicationsreveals the role physics plays
in our lives and in other disciplines. These
The same process occurs when a radiator raises the temperature of a room.
Thehot radiator warms the air in the lower regions of the room. The warm air Applicationsare woven throughout the text
Gary Settles/Science Source/PhotoResearchers, Inc.Photograph of a teakettle, showing eaFcwthoixiargempt Aueafenbrrrnneoidnga ma1sniaun1 t dat.ieao9onb.bmandonl,oo votbceibffke irclceseioiaennnuzcekedsrnse)eu,g acoicssntefieir eotsictntiiusnin slaglatont meewdumsaepipfri no netdtrrhaceatieehnnt duesecirdtoceyionn,, anettrtienivnsreeeauicorsotg rsiutyoaosofnp eft.aahsiWtoershpe aecsceteuerfeirarrlnotirni(gemnagnignc.tt te hTut.peeahaAmlheltyst op,edtetraehnmrnaem estseutimhxarro lteecuwttooranbeolyt lheooinearff nFtAhaoperrp r wablictiaoiavyleto io tgaonyn s vdaao nrafido rpe uph sirny epsd-irmiccaaacelt tdeipc drsai tnlwu cadiintpehdlne atsisn m,tt eoiacr ebrogsinoitnsilno npggooytinet.
steam and turbulent convection air coolerwaterbythermalconduction.Thewaterpumpforceswateroutofthe
currents. engineandintotheradiator,carryingenergyalongwithit(byforcedconvection). and medicine. With this edition the authors
Intheradiator,thehotwaterpassesthroughmetalpipesthatareincontactwith have increased the number of life science-
thecooleroutsideair,andenergypassesintotheairbyconduction.Thecooledwa-
APPLICATION teristhenreturnedtotheenginebythewaterpumptoabsorbmoreenergy.The oriented applications and end-of-chapter
Cooling Automobile Engines processofairbeingpulledpasttheradiatorbythefanisalsoforcedconvection.
The algal blooms often seen in temperate lakes and ponds during the spring or Problemsto help motivate students to master
fall are caused by convection currents in the water. To understand this process,
consider Figure 11.10. During the summer, bodies of water develop temperature the content.
APPLICATION gradients, with an upper, warm layer of water separated from a lower, cold layer by
Algal Blooms in Ponds a buffer zone called a thermocline. In the spring or fall, temperature changes in
and Lakes the water break down this thermocline, setting up convection currents that
mixthe water. The mixing process transports nutrients from the bottom to the
surface. The nutrient-rich water forming at the surface can cause a rapid, tempo-
rary increase in the algae population.
Warm Layer 25°–22°C Tipnotations address common student 9.3 Density and Pressure 211
Figure 11.9 Convection currents
are set up in a room warmed by a Thermocline 20°–10°Cmisconceptions and situationsin which stu-
radiator. Cool layer 5°–4°C The pressure at a specific point in a fluid can be measured with the device pic-
dents tuoreftde inn F igfourlel o9.w7b: aun nevpacruoatdedu cyclitnidveer e npclaostinhgs a. light piston connected to a TIP 9.1 Force and Pressure
spring that has been previously calibrated with known weights. As the device is sub- Equation 9.7 makes a clear distinc-
Appromxeirmgeda itne al yflu i1d,0 th0e flTuiidp prsesesecs dtoiownn osn taher eto pf oofu thne dpi stionn atnhde compresses tion between force and pressure.
(a) Summer layering of watemr arginthse ,s pprinrgo uvnitdil itnheg in wsatrud dforecen etxse rtwed itbhy t hteh fleui dh ies blpala ntchede byy the outward Afonrcoe tihs ear v iemctporoartnadn tp rdeissstuinrec itsi oan s cias ltahr.at
force exerted by the spring. Let Fbe the magnitude of the force on the piston and There is no direction associated with
need tAthoeth seap ravirneoga iiodsf s tpchreeo atdom po usmut rofoavcener to hmf et heiens tpitriaset okanree. aNs, om tioctei vtahtaint gt hoeu fro frocrem thala td ecofimniptiroens soesf pfporerercspeseu anrsdes,oi cbcuiualtat ert hdtoe w dtihtihree tschutiero fpnacr oeefs o stufhree is
and mpirsesusunred:erstandings. interest.
If Fis the magnitude of a force exerted perpendicular to a given surface of (cid:1) Pressure
area A, then the pressure Pis the force divided by the area:
P (cid:1) F [9.7]
A
SI unit: pascal (Pa)
Because pressure is defined as force per unit area, it has units of pascals (newtons
per square meter). The English customary unit for pressure is the pound per inch
(b) Fall and spring upwelling squared. Atmospheric pressure at sea level is 14.7lb/in2, which in SI units is
Figure 11.10 (a) During the summer, a warm upper layer of water is separa1te.0d 1fr(cid:5)om1 a0 c5ooPlear. lower
layer by a thermocline. (b) Convection currents during the spring or fall mix the Awast ewr ean sde cea nf rcoaumse Equation 9.7, the effect of a given force depends critically on
algal blooms. the area to which it’s applied. A 700-N man can stand on a vinyl-covered floor in
regular street shoes without damaging the surface, but if he wears golf shoes, the
metal cleats protruding from the soles can do considerable damage to the floor.
With the cleats, the same force is concentrated into a smaller area, greatly elevat-
(cid:4)dsee cnAttiospn pt.ol yS ironemgvi eeP wAh pycspoilcynisncges pePtchstyiposrnicesss eaenlxltoaemwd pisnlet uas- iumtnhltagiirSl md nthf aooletawre wcps esh,r t eotrohesensinsus g rututehshp eei w on tsaf hh trtehdoh o eesfsas oefl mr tocaoeoer e rspi uasmrs pi,naa pcrctoeceiprsorutilm aeltlt ph.i(naeFgn ig piie.ne d9r sa.bo8 g)yn r.’ aesT adwhteoeerw i gslnnhikwote.aw lriA hdeco xcfoeoodrrrtcd soe inaf engex xeutcorpet weeNddaeir nbdwgy tn o ttnohhr’ees- Figure 9.8 Snowshoes prevent the© Royalty-Free/Corbis
demonstrate the connection between the sohvoere st hoen v ethrye lasnrgoew .a rIef at hoef epaecrhs osnno iws swhoeaer, isnog t hsnaot wthseh operse,s stuhraet afto racney igsi vdeinst rpiobiuntte ids pssnperoreswoa dnb e ofcrvaoeumrs ae s iltnahrkegi neforg r aicnreet aoo, n trh etehd euso csfnitnogw is
concepts presented in that chapter and relatively low and the person doesn’t penetrate very deeply into the snow. the pressure on the snow’s surface.
other scientific disciplines.
Applying Physics 9.1 Bed of Nails Trick
n (cid:2) Quick Quizquestions throughout Afefstseorr a sntr eextcchiteins go ubtu fto erx ah nauapst ionng ale bcetudr oe,f an apihlsy,s aicss i npro- wofo nualdil sb we imthooruet ushnocoesm.)fortable yet to stand on a bed
o the book provide students ample Figure 9.9, suffering no injury and only moderate dis-
comfort. How is this possible?
ti opportunity to assess their conceptual
a understanding. Explanation If you try to support your entire weight
d on a single nail, the pressure on your body is your
weight divided by the very small area of the end of the
un (cid:2) Checkpointsask simple questions ntraaitle. Tthhee srkeisnu. lItfin ygo up rdeisssturirbeu itse l ayorguer ewneoiguhgth o tvoe rpene-
several hundred nails, however, as demonstrated by
o based on the text to further reinforce the professor, the pressure is considerably reduced
F key ideas. because the area that supports your weight is the total
area of all nails in contact with your body. (Why is
lying on a bed of nails more comfortable than sitting
on the same bed? Extend the logic to show that it Figure 9.9 (Applying Physics 9.1) Does anyone have a pillow?
vi
Several components from the text are enhanced in the PhysicsNow
student tutorial program to reinforce material, including the dynamic
Active Figures,which are animated diagrams from the text. Labeled
with the PhysicsNowicon, these figures come to life and allow stu-
dents to visualize phenomena and processes that can’t be repre-
sented on the printed page.
INTERACTIVE EXAMPLE 4.7 Atwood’s Machine (cid:3) Over 40 of the text’s worked Examplesare identi-
Goal Use the second law to solve a two-body problem. fied as Interactive Examplesand labeled with the
Problem Two objects of mass m1and m2, with m2(cid:10)m1, PhysicsNow icon.As part of the PhysicsNowweb-
aorvee rc ao nfrnicetcitoendl ebssy pau llliegyh, ta, s inine xAtcetnivsieb lFei gcuorred 4 .a1n5da . hBuonthg T T based tutorial system, students can engage in an inter-
cord and pulley have negligible mass. Find the magni- active extension of the problem solved in the correspon-
tthude ec oorfd t.he acceleration of the system and the tension in m1 m2 ding worked Examplefrom the text. This often includes
iSnt rathteeg ynegTathivee h eya-dviierer cmtiaosns., mS2in, accec etlheer atceosr ddo wcannw’at rdbse, a1 m1 m1g elements of both visualization and calculation, and may
otsinitnvr ee mb tacyanh gaden fd ioat,ur 2tcdhiesee o,n afbe cutgecaten tolisevpieorpa,on tsa ii:Tnotendiins na ot2dhf(cid:2) iert ehu(cid:4)cept aiwto1wan.o rE,d samsoc d ahtsih rsmeeacsta tsaaiso1r iensis aea pnqcotudesa dial- (a)m2 a2 (b) m2g aatylrspeoe g iaunnivddoel vdae rte hp rtroheeudngic hta iosthknee a dsn ttdeo p inastp unpietliyeo dnwe hbdau ttilo dt hisneogyl.v heSa tavu edp erleonabtsrlneemd
force of gravity in the downwards direction. Active Fig- ACTIVE FIGURE 4.15
ure 4.15b shows free-body diagrams for the two masses. (nEexctaemdp bley a4 .l7ig) hAtt swtroiondg’ st hmaat cphaisnsees. (oav)e rT aw ofr ihcatinogninlegs so pbujellcetys. con- to different scenarios.
Newton’s second law for each mass, together with the (b) Free-body diagrams for the objects.
equation relating the accelerations, constitutes a set of
three equations for the three unknowns—a1, a2, and T.
Log into to PhysicsNow at http://physics.brookscole.com/ecpand go
to Active Figure 4.15 to adjust the masses of objects on Atwood’s ma-
chine and observe the resulting motion.
Solution
Apply the second law to each of the two masses m1a1(cid:2)T(cid:4)m1g (1) m2a2(cid:2)T(cid:4)m2g (2)
individually:
Smuublsttipitluyt eb oat2h(cid:2) sid(cid:4)eas 1biyn(cid:4)to1 t:he second equation, and m2a1(cid:2)(cid:4)T(cid:6)m2g
Add the stacked equations, and solve for a1: (m1(cid:6)m2)a1(cid:2)m2g(cid:4)m1g (3)
a1(cid:2)(cid:3)mm12(cid:6)(cid:4)mm12(cid:4)g
Substitute this result into Equation (1) to find T: T(cid:2)(cid:3)m21m(cid:6)1mm22(cid:4)g
Remarks The acceleration of the second block is the same as that of the first, but negative. When m2gets very large
compared with m1, the acceleration of the system approaches g, as expected, because m2is falling nearly freely under
the influence of gravity. Indeed, m2is only slightly restrained by the much lighter m1. The acceleration of the system
can also be found by the system approach, as illustrated in Example 4.10.
(cid:4) Also in the PhysicsNowstudent tutorial pro-
gram are Coached Problems.These engaging
problems reinforce the lessons in the text by
taking the same step-by-step approach to
problem solving as found in the text. Each
Coached Problemgives students the option
of breaking down a problem from the text into
steps with feedback to ‘coach’ them toward
the solution. There are approximately three
Coached Problemsper chapter. Once the stu-
dent has worked through the problem, he or
F
she can click
o
“Try Another” to change the variables in the u
problem for more practice. n
d
a
t
i
o
An MCAT Test Preparation Guideis contained in the preface to help students prepare for the exam n
and reach their career goals. The Guideoutlines key test concepts and directed review activities from
the text and the PhysicsNowstudent tutorial program to help students get up to speed.