ch-04 forces newtons laws

8/4/2019 Ch-04 Forces Newtons Laws

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CHAPTERForcesandNewton'sLawsof MotionPREVIEW

In this chapter ouwill begin he studyof dynamics, hatbranchof physicswhich explainswhy objects ccelerate.You will be intoduced o the concept f force, and studyNewton's aws of motion,which apply o all forces hatoccur in nature.You will leam how to construct ree-bodydiagrams,anduse them to analyze ystems ubject osuch orcesas gravityand friction.You will also applyNewton's aws of motion to solvea numberof differentproblems.The applicationswill includebothequilibrium andnon-equilibriumproblems.

QUICKREFERENCElmportantTermsForceThe pushor pull required o changehestateof motionof an object,asdefinedby Newton's econdaw. t is avectorquantifywith unitsof newtonsN), dynes dyn),or pounds lb).InertiaThe natural endency f an object o remainat restor in uniformmotionat a constant peedn a straightine.MassA quantitativemeasure f inertia.Units arekilogramskg),grarnsg), or slugs sl).Inertial Reference rame

A referencerame n whichNewton'saw of inertia s valid.Free-BodyDiagramA vector iagramhat epresentsll ofthe orces cting nanobject.GravitationalForceThe orceof attractionhateveryparticleof massn theuniverse xerts n everyotherparticle.WeightThegravitationalorceexerted y theearth or some ther argeastronomicalody)on an object.Normal ForceOnecomponent f the force hat a surfaceexertson anobjectwith which t is in contact.This componentsdirected ormal,or perpendicular,o the surface.FrictionThe force that an object encounterswhen t movesor attemptso move along a surface. t is alwaysdirectedparallel o thesurfacen question.TensionThe endency fa rope or similarobject) o bepulledapartdue o the orces hatareappliedat eitherend.EquilibriumThe statean object s in if it haszeroacceleration.Mathematically, quilibriummeansEF : 0.Apparent WeightThe force that an object exerts on the platform of a scale. t may be larger or smaller thanthe tue weight,depending n the acceleration fthe object and he scale.


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Chapter 39

Newton'sLawsof MotionFirst LawAn objectcontinuesn a stateof restor in a state f motionat aconstantpeed longastraightine,nnless

compelledo changehatstateby a net force.By "net" orcawe mean hevectorsumof all of the orcesactingon an object.For exanple,consider spaceshipn deepspace,solatedromany otherobjector force. f theship s stationary,twill remainso. But if the ship s moving its rocketengines reshutdown) t will continueo move n a straightinewith a constant peecl. o f the shipwere raveiingnto deep pace t say,1C0000mi/h, t wouldcontinueo moveat hisspeedn a straightine,evenwithout herocketenginesiring,until an outsideorceactedo stopor changetsmotion.Second awWhen a net forceXF actson anobjectof massm, theaccelerationthat esultss directlyproportionalothe net force and has a magnitude hat is inverselyproportional o the mass.The directionof theaccelerations the same s he directionof thenet orce.Thisstatementsusuallywrittenas

XF = ma or a: XF/m (4.1)Thesymbol F representshenet orce, hat s, thevector sumof all the orcesactingon an object.Thismeanshatthecomponentsf the orcesmustbe examined. or example,n two dimensions,quation4.1)becomes

XF*: ma* (.2a)XFr: ma' 62b)

Equations4.1)and 4.2)cutbeused o detennineheunitsof force.I'heyareSI kg nr/s2= newtonN)CGS g cm/sz dyne dyn)BE slug tls2= pound lb)

Third LarvWhenever nebodyexerts forceona secondody, hesecondodyexer:tsnoppositelyirectedorceof equalmagnituden he irstbody.This s sometimesefened o as he "action-reaction"aw, .e., "for everyaction here s an equal, ut opposite,reaction".An examples a spaceshipiring its rocketengines. heengines jecthot gases ut herearof the ocketatvery,high elocity, his s the"action".The "reaction"s that he ocketacceleratesn the orwarddirection.

UniversalGravitationEveryparticle n the universee>iertsn atfractiveorce on everyotherparticle n the universe. he forceactingbetweenwoparficles f masses 1 andm2 separatedy a distancehasa magnitude ivenby

m, m.F:G# r- (4.3)whereG is theuniversalgravitational onstant,whosevalue obtainedexperimentally)s givenby

G=6.6 i3x 1o-11g*24 . t2


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40 FORCESAND NEWTON'SLAWSOF MOTION

Theweight of an objecton heearttrs thegravitationalorce hat heearthexerts n an object.This orce salwaysdirected ownward,oward hecenter f the earth.For heearttr mass Mg, radius= RB) we canwriteM-m

W:G+R;where hegravitationalccelerationsg : GMB/RgZ= 9.80m/s2.Theweightof an objectcan hereforeewritten

w=mg (4.s)

DISCUSSIONF SELECTEDECTIONS4.314.4Newton'sSecondLaw of Motionand heVectorNatureofNewton'sSecondLaw of MotionNewton's econdaw,asexpressedn equations4.1)and 4.2),containsectors, he equation F= ma meanshatwe need o determinehenet orceactingon an object.The erminology,F, meanshatyoumustdeterminehevectorsum E= surnmationign)of all the orcesactingon an object.You should eginby drawing diagramhatrepresentsheobjectandall the orces ctingon t, a so-calledree-bodydiagram.Example1 A Free-Body iagramA 2.0 kg object s subjectedo a6.0N forceacting n thex direction, ndan 8.0N forceacting n they direction.What s theaccelerationmagnitudenddilection) lthis object?The irst diagram howshemassm and orcesactingon t. The irstdiagramsthe ree-body iagrarn herehemasssplaced t he originof thex-y coordinateystem nd he orces reshown cting rom hispoint.

In order o frnd heaccelerationf theobject, , we need o find thenet orce,F, actingon heobject.Since heforcesactingon m areperpendicular,ecan ind thenet orceusing

F_0-

t-JF;+F;* ' [ * ) = tan-l l'l"gN.l -^oIe.oN]:)r


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Chapter4 41

Since he accelerations in thesamedirectionas he net force,a is directed53oup from the*x axis. tsmagnitudesa = F/m: (10NX2.0 kg)= 5 m/s2.

In most cases,he forcesacting on an object canhave any orientationn space.The individuai orcesmustberesolved nto x andy componentsefore he net force, and hencehe acceleration,an be found.Thenextexampleillustates his.Example2 Finding henetforceandaccelerationSuppose 1.8kg object s subjectedo forcesas llustratedn the ollowingdiagram. hemagnitudes f the orcesareF1= 20.0N, F2= 14.0N, F3 = 15.0N. What s thenet forceandaccelerationctingon the object?

To find EF and heaccelerations,e needo look at the componentsf the orces nduseequations4.2a) nd(4.2b).That s,we need o sum he orcecomponentsn thex direction XF*)andy direction EFy).WehaveForce x comDonenlFl +(20.0 ) cos 0.0o:+17.3F2 - (14.0N) cos 5.0o - 9. 9N

EF*= +lf.l N - 9. 9N : +7.4N

Force y componentFl +(20.0 ) sin30.0'= +10.0N82 +(14.0N) sin45.0'= + 9.9NF3 -15.0NEF , = +10.0N + 9. 9N - l5-oN = +4.9N

F3


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XR +7.4N , ) XFu +4.9N ,?v - . = r = - - = 4 . 1 f i v S - l d - . = - = - = z . | i l u " s -^ m 1 . 8k g Y m 1 . 8k g4.7 The Gravitational orceConsiderwo particles f masses 1 andm2 separatedy a distance. Eachmass xerts forceof athaction n heothermass. he orce s directed long he ine oining heparticles nd he orcehasa magnitude ivenby

m. m^F=G+ (4.3)t'ExampleFind he orcebetweenbjects f massm1= 1.00 103 g, rn2=2.00 103 g, separatedy r : 3.00m.

F=Gry = (6.67xg-rr *r7L*ry'Loo-*o3 gX2'0010'kg): 1.48 10-5f (3.00 fExample 4 The earth'sgravity and weightThe weight of an object on earth is defined as the gravitational force exertedby the earth on the object. Find theweighton earthof a 50.0kg tnass. lse MB = 5.98 x 1024kg, l{E = 6.38x 106m for t}reeaLth's tassand adius.

42 FORCESAND NEWTON'SLAWS OF MOTION

The accelerationsn thex andy directions annow be obtained singXF" andEFu oundabove:

M-m -rw:G + :(6 .67x10"N-za9(5 '98* rOakgX = 4.90x 02Nq (6.38 loom)'Note that according o Newton'ssecond aw, F: ma: mg, sinceg is the acceleration f gravity.Therefore,

GM, _ (6.67x 10-l lN n 2A


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Ghapter 43

4.814.9TheNornnal orceaneJ tatlcand KineticFrictionai oreesThenorrnal orce,F1q,s theperyendicularorce hat a surface xerts n anobjectwith p'hich t is n contact.Foranobjectat equilibriumon a flat surfacesee igurebeiow)we cansee hatFN : W. The normal orce s simply heswface's eactiono the orces ushingheobjectagainsthe surface.

If an obiect s at equilibrium n an nclined lane, hemagnihrdef thenormal orcewill be equal o themagnitudeof a cornponentf theobject'sweight,as s shown n the igure.

wThe magnitude of the normal force is FN = W cos 0, which is the caseeven or a flat surface, .e., for 0 = 0 sincecos 0 = l, thereforeFN = W.

When an object is in contactwith a surface, here s another orce in addition to the normal force andgravityacting on it. This additional force s friction. When the object slides over the surface, he surfaceexertsa frictionalforce on the object. This is known askinetic friction. However, a frictional force exists even f two objects arenotin motion. 'Ihis force is static fiiction.

An important characteristicof the frictional force is that its magnitude s proportional to the magnitudeof thenormal force. For example, an object sliding on a rough surface s subject o the force ofkinetic friction, fg.,accordins to

&: p,kFN (4.8)where pp is the coefficient of kinetic friction. The coefficient of kinetic friction can have valuesu,hichare usually0


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44 FORCES NDNEWTON'SAWSOFMOTION

Example5,q.S.Olg objectspulledovera roughhorizontal urface y a 100.0N forcedirected tanangleof l4'0" withrespecto thesurfacesee igure). fthe coefficient fkineticfriction s 0.40, ind theaccelerationfttreobject'

Webeginby drawing free-body iagramor thisexample. he orceswhichmustbe ncluded re heappliedorce,r, *r'f,ictional forci, q,, thrnor*riorce, Fy, and heweight,W = mg. f theobjectsmovingn the+x direction'wehave

Lookat he orcesn thex andy directions:yd i rec t ion : FN*Fy-W=Osinceheresnoaccelerationn they direction, o

F t t : W - F y = - g - F s i n 0"|FN : (5.0 gx9.80 tt/r") (100.0 ) si n14.0o 25N.

x direction: F" - f1= ma*since heobject s acceleratingn thex direction,so

Fcos0 - P1 y = m a *(100.0N) cos14.0o (0.40X25 ) = 87N = ma*


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Chapter4 45

Thereforeax : (87N) /m = (87N) / (5 .0kg) lTml& .

Suppose n object s restingon a flat surface.f we applya smallhorizontalorce, riction esistshis orceandthe object emains t rest.This s the orce of static riction, 'r. Aswe graduallyncreaseheappliedorce, he objectwill eventually egin o moveoncewe have overcomehe maximum rictional orce, rmax. If the coefficient fstatic riction s prr, hemagnitudef themaximum tatic rictional orcecanbewritten

frmax: FsFN (4.7)Example 6 StaticFrictionA 10.0kg block sitson aflat surfacewhose oefficient f static riction s Fs= 0.60andwhose oefficient f kineticfriction s Fk= 0.a0. a)Whathorizontal orce s required oget heblock o move? b) If wecontinueo apply hesarne orce as n part (a),whatwill theblock'sacceleratione?(a) We canuseequation4.7) o find the orceF necessaryo overcomeriction, hat s,

F=tmax- F : F s F N =F s f f i gF: (0.60X10.0gx9.80 /s2) 59N.

(b) LJsing1heabove orce,Nervton's ccond aw, )lF .=ma, alrd eqrraticrn4.8) for kinotic i:ictiou,w0 liav0I ,F : F - f tXF: F -F tFNXF= F-pkmg = ma

NowIF F-Pumgmm "I("1I"ry{ryg{)10.0 c a= ),.U lTtlS'-

Example 7A box weighing 147 N sits on a horizontal surface.The coeffrcient ofstatic friction befween he box and the surfaceis 0.70.. f a forceP is exertedon the box at an angle directed 37o below the horizontal,what must the magnitudeofP be to get the box moving? The box and the associatedree body diagram are shown below.


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46 FORCESAND NEWTON'S AW SOF MOTION

Use hecornponentquations f Newton's econdaw,equations4.2b)and 4.2a).In they direction XF, = ma,Thebox hasno accelerationn theverticaldirection o

which gives

In the x direction

Thebox hasno accelerationn thehorizontal irection o

FN-Py-W:0FN:Psin0+W

XF*= mq

P*-t=oPcosd-FsF1g=0

P cos l - Fr sinQ+'S/) 0

(1 )

(2)Solving2) or P yields:

r r - l r rW' -As0- l r rs inO (0.70)147N) :270 N.cos37"- 0.70) in37o4.10 TheTension orceForcesare sometimesappliedby ropes, cables,or wiles that ptrll on an object. We usually assume hat the rope smassless nd hat hemagnitLrdef the ension,T, is thesame hronghout herope.Exanrple 8 Tlrc ensionforceA 12.0kg object s pulledupwardby a masslessopewith an acceleration f 3.00m/s2.What s the tension n therope?The ensionorce s directed pward +) while heweightof ttreobject mg) s dirpcted ownwardO. Since heobject s beingacceleratedpward+) we canwritetF: lna as

T-mg=P4T=m(g+a)1 = (12.0 gX9.80mts2+:.00 m/s2)T :154N.


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Chapter4 47

Example9Supposeheobject n theprevious xamplevere cceleratingownward t3.00m/s2 nstead f upward.Whatwould he ensionn the ropebe n thiscase?The only differencen this casewould e that heaccelerations downward, o akea= -3.00m/s2.We have,

T - mg: ma'so hat T=m(g+a)1= (12.0 g) 9.80mts23.00m/s2)T= 82N.

The situations epictedn examples and9 canbe observed henyou'ren an elevator.Supposeou'reon hebottom loor and he elevator tartsacceleratingpward.You feel as f you are beingpusheddownward, nd hetensionn theelevator able ncreases. henyou'reon the op floor and heelevator eginsacceleratingownward,you feel as f you arebeing ifted off the floor and here s a conespondingecreasen the elevator able ension.When heelevators at rest,or movingwitb a steady peed,6u eel"nonnal" nd hc cable ensionwouldustbeequal o the weightof youand heelevator.

4.11 Equilibrlum pplications f Newton's awsqf Motiorr-- / An object hat s at rest or travelingwith a constant eiocify has zero acceieration nd s said o be in "equil ibrium".T'his rrrpliesha t hc net rirceacting n theobject szr,ro.Intwodirir,; lsior,s'r: : irlruiie

t l i = f l- ' x Q.ea)XFr: 0 (4.9b)

There are ive steps hat canbe followed n solving equilibriumproblems.Theyare:

Step 1. Selectan object, he "system",aboutwhich the r:rost nformation s lqtorvn. f two or more objectsareconnectedit maybe necessaryo hea.t ach ndividuaily.Step 2. Draw a "free-body"diagram or the system.As discussed leviously, J:riss a drawing hat' represents he object and ALL the forces acting on it.Step 3. Choosea convenient etofx, y axes or the objectand esolvea1l hc forces nto componentshatpoint along theseaxes.By "convenient"we mean hat you shouldchooseyour axesso that asmauyforces aspossiblepoint directly along the x or y axis.This will minimize the number ofcalculationsneeded.Step 4. Apply equations 4.9a)and (4.9b) by setting he sum of the componentsof the forces n the xdirection equal o zero and the sum of the components a the y direction equal to zero.Step 5. Solve the two equationsobtarned n step 4 for the desred unknown quantities. n many cases heequationsmustbe solvedsimultaneously or the two unknowns.


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48 FORCESANDNEWTON'S AWS OF MO TION

The ollowingarea number f examples mployinghemethodust ouflined.Example 0 A weight uspendedy cablesA 280 b block s suspendedy twocables, sshown elow.Find he ensionn each able.

Choosingheblock asourobject,wehavedrawn he ree-body iagram ndchosen andy axes, sshown.We nowapplyequations4.9a)and 4.9b)using he ollowing able

Force x component y componentT1,T2w

-T1co s30.0 +Tl si n30.0'+-T2 os 5.0' +'l '2 in45.0'

0 - 2 8 0 1 b

UseNewton's econdaw IFx = -T1cos30.0o T2cos 5.0o 0 (1)IFy: +T1sin30.0" T2 sin45.0o280 b 0 (2)

SolveorT1 n equation1),I cos 5.0')T, Tzt ."-30 0'J 0'816 z

Substitutehe esult ntoequation2),(0.816)2 sin30.0o T2 sin45.0o280 lb

Solving or T2 thenyields,T= ' 2801b, , , , - :2501b2 (0.816in 0.0o+in 5.0)

And finally, using his in equation3)yields

(3)

30.00 45 .

T1 (0.816)T2:200Ib.


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Ghapter4 49

Example11consider he ollowingsystemn equilibrium.Deterrninehe valueof theweight,w, and he ensionn the eft wire.The free body diagramcan be drawn in twodifferentways,as shownbelow. n (a), we canmake the y axis vertical anddraw in the forces.Perhapsa better choicewould be to align thecoordinateaxesasshownn diagramb). This ismoreconvenient ecausewo of the hree orcesnow lie along heaxes, ndare hus esolvedntox andy componentsutomatically.

Theequilibrium quations,4.9a) nd 4.9b) annowbe writtenas

which gives

andwhich gives

I F * = 1 5 0 N - W s i n 3 7 o = 0YY (150N)(sin37";= 250

X F r = T - W c o s 3 7 o = 0' l ' ,=W cos 7o 200 N.

Notice hat hechoice f x-and axesn (b) madeheequilibrium quationsathereasyo solve or w and . If wehadchosenheaxes s n drawinga), heequationso,rld huu.been bit moredifficult to solve,.e.,XFx= (150N) co s37o T cos 3o 0

XF y = (150N) sin37o Tiin 53 o W = 0.Thus, heappropriatehoice f coordinatexes an acilitatehesolution uitea bit in mostcases.


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50 FORGESAND NEWTON'S AWSOF MOTION

Example12 Object nan nclinedlaneConsider massM = 25.0kg sittingon an nclinedplanewhoseangleof inclinations 0 = 30.0o.A masslessordconnectedo M passesurr r frirtionlesspulleyand s attachedo a tnassm = 15.0kg' ThemassM slides p heplaneata constantpeed. ind hecoeflicientof kinetic rictionbetweenhemassM and heplane.

Since hemassesmovewith constantelocity(a = 0), thesystems in equilibrium.We need o lookat eachmassindividuallyn order o determinehenecessaryuantities. he ree-body iagramsor M andm look ike

TheForces n m

MgForm:In thgy direction

so hat

ForM:In they directiongives

X F r : T - m 9 : 0T =mgT = (15.0 gx9.8o vs2; t+z N.

tF y = FN - Mg co s0 :0 ,Fy: Mg cos0FN: (25.0 gX9.80 is2; os30.0o212 N.E F * = T - & - l u t g s i n 0 : 0T - p k F N - M g s i n 0 =0

TheForces nM

In the x direction


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Ghapter 51

Solving or pl yieldsT-Mesin0nn:T 147N - (25.0 gX9.80 /s') sin30.0o= 0.1.2.212N

4.12 Nonequilibrium pplicationsof Newton'sLawsof MotionNonequilibrium ccurswhen he orcesactingon an objectarenot balanced.n this case heobjectacceleratesndNewton's econdaw s

XF*= ma* g.za) XF, = ma, (4.2b)Example73 Atwood'sMachineTwo masses, 1 andm2 (assume 2 > m1), areconnectedy a masslessting thatpassesvera massless,riction-freepulley,asshownn the igure.Find heaccelerationf themasses nd he ensionn thesfring.

Weneed o look at eachmassndividuallyn order o obtain he woequationsecessaryo solve or theunknownquantities andT. The ree-body iagramor eachmasss shownabove.Since othmasses reconnectedy thestring, hemagnitudesf theiraccelerationsre hesame.Sincem2> ml, m1 hasan upwardaccelerationndm2 hasadownward cceleration. e alsoassumehat he ensionn thestringon each ideof thepuiiey s thesame.For m1 theaccelerations upward, o T ) m1 g andwe can herefore rite,

T - m1g: r l t l t .Form2 theaccelerations down,som2 g > T. Therefore, ecanwrite,

mZg'T = m2a.


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52 FORCES NDNEWTON'SAWSOF MOTION

Solving or T in thefirst equation ndsubstitutingnto thesecond ives,mZE- m1g+ m1a)= m2a.

Finally, solving or the accelerationields, (m" m,)a=Gfr6sSubstitgtinghis expressionor theaccelerationnto eitherof themassequations, e can solve or the ension'

2m,m"T=. , . - - ILs^ (mr+rn) '

Example14consider he ollowinganangemenlith M1 = 15.0kg, M2 = 10.0 g, andp1= 0.30.Find heaccelerationf thesystemand hetensionn theconnectingord. Free-body iagrams ccompanyhedrawing'

__tts M1

M2+For M1 we have:or

For Me we have:

XFx=T- fk :T -pkFg=M1aT-pkMtB=Mla .

EFy: MZE- T = M2 a.

solving for T in the second quation ndsubstitutingnto thefirst yields:MZE-M2a- F tMt g=Ml a


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Chapter4 3

_J or(Mt +M2)a: (M2 rp M1)

The acceleration s therefgre,(ta:tfl,)s . 1ro.og (0.30)(ir.or;111.80/s2) , zu:qry =2.2n/s-.

The ensiolr annowbe oundby substitutinghisvalueof a nto heequationorM2.T:M2 @- a )1: (10.0 g) 9.80mls?2.2m/s?)= 6N.

ch-04 forces newtons laws

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