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MST209 Mathematicalmethodsandmodels

Statics

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StudyguideThisunitandUnit6 laythefoundationsofthesubjectofmechanics,whichhas a large part to play in this course. Mechanics is concerned with howandwhyobjectsstayput,andhowandwhytheymove. Thisunitconsidershowandwhyobjectsstayput,whileUnit6 andthe latermechanicsunitsconsiderhowandwhytheymove. Thematerial inthisunitand inUnit6laysthefoundationsofmechanics.Thisunitassumesagoodworkingknowledgeofvectors, whichyoushouldhaveobtainedfromUnit4.Therecommendedstudypatternistostudyonesectionperstudysession,intheorderinwhichtheyappear. Thereareonlyfoursectionsinthisunit,soforthefifthstudysessionyoumightliketolookatthemultimediapackagedesignedtosupportthisunit,which isdescribedinthemediaguide.

This publication forms part of an Open University course. Details of this and other Open Uni-versity courses can be obtained from the Course Information and Advice Centre, PO Box 724,TheOpenUniversity,MiltonKeynes,MK76ZS,UnitedKingdom: tel.+44(0)1908653231,[email protected],youmayvisittheOpenUniversitywebsiteathttp://www.open.ac.ukwhereyoucanlearnmoreaboutthewiderangeofcoursesandpacksofferedatalllevelsbyTheOpenUniversity.TopurchaseaselectionofOpenUniversitycoursematerials,visitthewebshopatwww.ouw.co.uk,or contact Open University Worldwide, Michael Young Building, Walton Hall, Milton Keynes,MK7 6AA, United Kingdom, for a brochure: tel. +44 (0)1908 858785, fax +44 (0)1908 858787,[email protected]

TheOpenUniversity,WaltonHall,MiltonKeynes,MK76AA.Firstpublished2005.

cCopyright2005TheOpenUniversityAll rights reserved; no part of this publication may be reproduced, stored in a retrieval system,transmittedorutilisedinanyformorbyanymeans,electronic,mechanical,photocopying,record-ing or otherwise, without written permission from the publisher or a licence from the CopyrightLicensing AgencyLtd. Details of such licences (for reprographic reproduction) may be obtainedfromtheCopyrightLicensingAgencyLtd,90TottenhamCourtRoad,LondonW1T4LP.Open University course materials may also be made available in electronic formats for use bystudentsoftheUniversity. Allrights, includingcopyrightandrelatedrightsanddatabaserights,inelectroniccoursematerialsandtheircontentsareownedbyorlicensedtoTheOpenUniversity,orotherwiseusedbyTheOpenUniversityaspermittedbyapplicable law.In using electronic course materials and their contents you agree that your use will be solely forthe purposes of following an Open University course of study or otherwise as licensed by TheOpenUniversityor itsassigns.Exceptaspermittedaboveyouundertakenottocopy,store inanymedium(includingelectronicstorage or use in a website), distribute, transmit or re-transmit, broadcast, modify or show inpublicsuchelectronicmaterialsinwholeorinpartwithoutthepriorwrittenconsentofTheOpenUniversityor inaccordancewiththeCopyright,DesignsandPatentsAct1988.Edited,designedandtypesetbyTheOpenUniversity,usingtheOpenUniversityTEXSystem.Printed and bound in the United Kingdom by Martins the Printers Ltd, Berwick-upon-Tweed, TD151RS.ISBN0749267011

1

2

3

4

1

2

3

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1.1

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ContentsIntroduction 41 Modelling forces 5

1.1 Particles 61.2 Weight 71.3 Normalreaction 81.4 Tension 91.5 Friction 11

2 Twoormoreparticles 162.1 Newtonsthirdlaw 162.2 Pulleys 182.3 Slipping 21

3 Torques 233.1 Extendedandrigidbodies 243.2 Turningeffectofaforce 24

4 Applyingtheprinciples 30Outcomes 35Solutionstotheexercises 36Index 48

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Introduction

IntroductionThis unit and the next lay the foundations of mechanics that will be usedthroughout the course. Mechanics can itself be divided into three distinctareas.Quantum

mechanics

deals

with

the

motion

of

very

small

objects

(such

asatoms,whichhavediametersofabout1010 metres).

Relativistic mechanics dealswiththemotionofveryfast objects(suchas the electrons in a television tube, which travel at speeds of about108 metrespersecond).

Newtonian mechanics is concerned with the more familiar everydayworldofobjectswhicharelargerthanatoms,andwhichmoveatspeedslessthanafewmillionmetrespersecond.

InthiscourseonlyNewtonianmechanics isconsidered,butthisstill leavesa vast range of phenomena to discuss. Isaac Newton (16421727) was thegreat English mathematician whose name is given to this subject. HisPhilosophiaeNaturalisPrincipiaMathematica of1687(MathematicalPrin-ciples of Natural Philosophy, known as Principia for short) incorporatesoneofthemostcelebratedexamplesofmathematicalmodelling. Itwas inPrincipia thatNewtonlaiddownthefoundationsofNewtonianmechanics.Thisgreatbook,whichshowed forthefirsttimehowearthlyandheavenlymovementsobeythesame laws, iscast inthe formofasetofpropositionsallderivingfromthreeaxioms,orlawsofmotion. Itisthese,heretranslatedintomodernEnglishfromtheoriginalLatin,thatstillprovidethebasisforNewtonianmechanics.Law I Every body continues in a state of rest, or moves with constant

velocityinastraight line,unlessaforceisappliedtoit.Law II The rate of change of velocity of a body is proportional to theresultant forceappliedtothebody,and ismade inthedirectionofthe

resultantforce.LawIII Toeveryaction(i.e.force)byonebodyonanotherthereisalways

opposed an equal reaction (i.e. force) i.e. the actions of two bodiesuponeachotherarealwaysequalinmagnitudeandoppositeindirection.

These lawsdidnotspringfullyformedfromNewtons imagination. Earlierinvestigators,notablytheItalianmathematicianandscientistGalileoGalilei(15641642) and the French polymath Rene Descartes (15961650), hadformulated some similar results. But it was Newton who perceived thatthesethree lawsweresufficientforthefoundationsofmechanics.It isperhapssurprisingthattheseseminal lawsofmathematicalmodellingwerewrittenentirelyinprose,withnohintofamathematicalsymbol. Sym-bolic formswere developedduringtheeighteenthcentury. Newtonhimselffollowedthroughthecomplicatedchainsofreasoningarising fromthe lawswithfarlessrecoursetosymbolism,orindeedtocalculus,thanlatermath-ematicianshavefoundnecessary. Inthiscourse,thelawsthemselveswillbeexplainedandtheargumentsbasedonthemsimplifiedbyusingthevectornotationdeveloped inUnit4.OneofthecentralconceptsinNewtonianmechanicsisthatofaforce. Thewordforceisusedineverydayconversationinavarietyofways: heforcedhis way in; theforce of destiny; to put intoforce; the labourforce. In mathematicsandscience,thewordforcehasaprecisedefinition. However,this definition relies on the movement of objects and so is deferred until

SirIsaacNewton(courtesyofMansell/TimeInc./Katz)

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Section1 Modellingforces

the next unit. Essentially, though, this definition says that a force eitherchanges the shape of the object on which it acts, or causes movement ofthe object. When we experience a force, in the mathematical sense of theword,we feel itthroughcontact: pullingonarope, liftingashoppingbag,pushingagainstacar,holdingachildaloft. Ineachcase,theforcethatweexperiencehasamagnitudeandadirection,sowemodelaforceasavectorquantity.

Mathematical representation of a forceA force isrepresentedmathematicallybyavector. Themagnitudeofthevector representsthemagnitudeofthe force, and the direction ofthevectorspecifiesthedirection inwhichtheforce isapplied.

Sometimeswecanseetheeffectofaforce,suchaswhenamattressdepressesunder the weight of someone sitting on it, a washing line sags under theweight of the washing, a rubber band is stretched over some packages, adoor is pulled open, or a bag of shopping is lifted. In each case there isanobviousdeformationormovementthat indicatesthataforce ispresent.Sometimes,however,thepresenceofaforce isnotsoobvious, insituationssuch as a ladder leaning against a wall (though you could appreciate thepresence of a force in this situation wereyou to replace the wall and holdthe laddersteadyyourself),aboxrestingonashelf,oracableholdingupaceilinglamp(considerholdingupthelampbythecableyourself).This unit considers the conditions under which objects remain stationarywhensubjectedto forces,which isatopicknownas statics. Forexample,

q?

whatistheminimumanglethataladderleaningagainstawallcanmakewith the ground before the ladder slides to the ground (see Figure 0.1)?Cases where forces cause motion are discussed inUnit 6 and elsewhere inthecourse. Thestudyofmotion iscalleddynamics. Figure0.1Beforeforcesandtheireffectscanbeanalysedmathematically,theyandtheobjectsonwhichtheyacthavetobemodelledmathematically. InSection1objectsaremodelledasparticles,andthesectionshowshowvarious forcessuch as the forces of gravity, tension and friction can be modelled. It alsoshowshowtoanalyseone-particlesystemsinequilibrium. Section2extendsthe ideasto systems involving two or more particles. Section 3 goes on toconsider situations where an object needs to be modelled as a solid bodyratherthanasaparticle. Section3alsodiscussestheturningeffectofaforce(knownastorque),whichhappensonlyifaforceactsonasolidbodyratherthan a particle. Section 4 describes the application to statics problems oftheconceptsandprinciplesdescribedintheearliersections.

1 ModellingforcesThis section shows how four common types of force can be modelled: theforceofgravity,theforceexertedbyasurfaceonanobject incontactwithit, the tension force due to a string, and the friction force between twosurfaces. These forcesandthesituations inwhichtheyoccuraremodelledandanalysed inSubsections1.21.5. First,however,we lookatonewayofmodellingtheobjectsonwhichforcesact.

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1.1 ParticlesWhenwecreateamathematicalmodel,theaimistosimplifytherealsitu-ationbeingmodelledsothatonlytheessential featuresare included. Thisenablesustoanalysethesituationmathematically. Inmechanics,themostimportantthingstomodelaretheforcesactingonobjects,andthroughoutthis

unit

and

the

other

mechanics

units

you

will

see

how

to

do

this.

However,

wealsoneedtomodeltheobjectsonwhichtheforcesact.Sometimes,suchaswhenthelength,breadth,depth,orientationorinternalstructure of an object is important, the object needs to be modelled as asolidbody,whichpossessesbothmassandsize. Anexample isprovidedby Themodellingofsolidbodiesaladderleaningagainstawall,wherethelengthoftheladderisimportant. isdiscussedinSection3.Atothertimes,whenthesizeandstructureoftheobjectarenotimportant,theobjectmaybemodelledasaparticle,whichpossessesmassbutnosize.Anexample isprovidedbyapalletofbrickshangingontheendofacablefromacrane,wherethepalletofbrickscanbemodelledasaparticle ifweareinterestedonlyintheforcesactingonthecable.

DefinitionAparticle isamaterialobjectwhosesizeandinternalstructuremaybeneglected. Ithasmassbutnosize,andsooccupiesasinglepointinspace. Aparticleisoftenrepresentedindiagramsbyablackdot.

Observationhasshownthateachforceactingonanobjectcanbemodelledas acting at a particular point on the object, this point being referred toas the point of action of the force. In situations where a particle modelis appropriate, all the forces acting on the object are modelled as actingthrough the point in space occupied by the particle. It is conventional toshowtheseforcesindiagrams,knownasforcediagrams,byvectorarrowswhose tails coincide with the particle and whose directions correspond tothedirectionsinwhichtheforcesact,asFigure1.1 illustrates.

F4F1

F2F3

Figure1.1When several forces are acting on a particle, observation has shown thattheoveralleffectoftheseforcescanberepresentedbyasinglevectorgivenby the sum of the vectors representing the individual forces. In this unit,wedealwithobjectsthatdonotmove, i.e.objects in equilibrium. Foraparticle inequilibrium,the forcesactingon itmustbalanceeachother(orelse itwouldmove),sowehavethefollowingimportantcondition.

Equilibrium condition for particlesA particle subjected to forces F1,F2, . . . ,Fn is in equilibrium if theforcessumtothezerovector,i.e.

n

Fi =0.i=1

Notethatinforcediagramsarrowsareusuallydrawnwitharbitrarylengths. Thiscontrastswiththeusualconventionforvectorswherelengthindicatesmagnitude.

ThisconditionwasfirststatedbyIsaacNewtonaspartofhisfirstlawofmotion.Weoftensaythatthesumoftheforcesiszero,withtheimplicationthatthismeansthezerovector.

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1.2 WeightWhen you hold a shoe, your fingers experience a force. The shoe, like allobjects,hasaforceassociatedwithit,andifyoudonotprovideoppositiontothisforceinholdingtheshoe,theshoewillfalltotheground. Butwhatisthesourceoftheforceexertedbytheshoe?This force is due to the attraction of the shoe to the Earth. The force ofattraction of objects to the Earth is called the force of gravity or thegravitational force. Thegravitationalforceactingonaparticularobjectis not constant, but depends on the position of the object relative to theEarth: there isasmallvariationofthis forcewithheightaboveground(ordepth below ground), and there is an even smaller variation with latitudeand longitude. Whenappliedtoaparticularobject,thisforce iscalledtheweight oftheobject. Inthiscourse, weshallassumethattheweightofaparticularobject isconstantneartheEarthssurface.Ineverydayspeech,thewordsmassandweightareinterchangeable. Mathe-matically,however,theyaredifferent. Themassofanobjectistheamountof matter in the object and is independent of the objects position in theuniverse; it is a scalar quantity, measured inkilograms (kg) in the SI sys-tem. Theweight ofanobject isthegravitational forceontheobject,andisdependentonwheretheobject issituated; it isavector quantity,whosemagnitudeismeasuredinnewtons(N)intheSIsystemandwhosedirection ThenewtonisdefinedisdownwardstowardsthecentreoftheEarth. formally inUnit6.Mass and weight are, however, related in that an object of mass m has Therelationshipbetweenweightofmagnitudemg,whereg isaconstantknownastheacceleration massandweightisbasedonduetogravity. NeartheEarthssurface,ghasthevalueofapproximately Newtonssecond lawof

motion,whichisdiscussedin9.81ms2, andweshallassumethisvalue forg throughout this course. If

Unit6.theCartesianunitvectorkpointsverticallyupwardsfromthesurfaceoftheEarth,thentheweightWofanobjectofmassmismgk(whereweneedthenegativesignbecausetheforceofgravityactsverticallydownwards,i.e.theweightactsverticallydownwards).

Weightm

AnobjectofmassmhasweightWofmagnitude |W|=mg,wheregistheaccelerationduetogravity,withdirectiontowardsthecentreof WtheEarth. Iftheobject ismodelledasaparticle,the forceofgravityontheobjectcanbe illustratedbytheforcediagram inFigure1.2. Figure1.2

*Exercise1.1What istheweightofaparticleofmass3kg inacoordinatesystemwherethek-direction isverticallydownwards?

When modelling forces acting on objects, it is often convenient to defineCartesianunitvectorsandtoexpresstheforcevectors incomponentform,i.e.toresolvethevectors intotheircomponents. TheseCartesianunitvec-tors define the directions of the axes in a Cartesian coordinate system, soweoftenrefertotheprocessofdefiningCartesianunitvectorsaschoosingaxes.

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6


*Exercise1.2i

Later in this unit we shall find it convenient to use axes which are nothorizontal and vertical. Express the weight W of a particle of mass 15kg 15 kg

jin terms of the Cartesian unit vectors i andj, where i andj both lie in averticalplaneandareorientedasshown inFigure1.3. W

Figure1.3

In thepreviousexercise, nice angles(i.e. multiples of 6 (30)) were chosenin orderto help you evaluate the cosines involved without havingto use acalculator. Another convenient angle is 4 (45). The cosines of all theseanglesaregivenintheHandbook. Sometimes,asintheaboveexercise,ob-tuseanglesareused;cosinesofsuchanglescanbederivedfromtheadditionformulae,whicharealsogivenintheHandbook. Forexample,

cos2 =cos(2 +6 )=cos2 cos6 sin2 sin6 =13 2 .

1.3Normal

reaction

Consideranemptycoffeemugrestingonatable. Letusmodelthemugasaparticle. Weknowthatoneforce,themugsweight,isactingonthemug.But, since the mug is at rest (i.e. not moving), the equilibrium conditionforparticlestellsusthatsomeotherforce(s)mustbeactingonthemug(sothatalltheforcesactingonthemugsumtozero). Theonlypossiblesourceforanotherforceonthemugisthetable. So,inorderforthemugtoremainatrest,theequilibriumconditiontellsusthatthetablemustexerta forceon the mug, which must be equal in magnitude to the weight of the mugandopposite indirection. The forceexertedbythetableonthemug,andindeedexertedbyanysurfaceonanobject incontactwith it, iscalledthenormal reaction forceorsimplythenormal reaction.ThesituationisillustratedinFigure1.4,whichshowsnotonlythemugandtable,butthecorrespondingforcediagram(plustheCartesianunitvectorkpointingverticallyupwards). ThenormalreactionforceisdenotedbyNandtheweightofthemugbyW. Usingtheequilibriumconditionforparticles,wehave

W+N=0.Ifthemughasmassm,thenW=mgk,andhence

N=W=mgk

Nk m

W

isaforce

acting

vertically

upwards

with

the

same

magnitude

as

the

weight

ofthemug.Thenormalreactionforceisremarkableinthatitadjustsitselftothemag-nituderequired. Forexample, ifthecoffeemug isreplacedbya fullpotofcoffee, then the normal reaction increases (unless the weight of the coffeepotistoomuchforthetable, inwhichcasethetablecollapsesandthepotis no longer at rest). Contrast this with the weight of an object, which isfixedandconstant,regardlessofwhatishappeningtotheobject. Ourbasicmodellingassumptionisthatthemagnitudeofthenormalreactionforceispotentiallyunlimited.Thereisanormalreactionforcewheneveroneobject(e.g.amug)pressesonanother(e.g.atable). Observationhasshownthatthisforceactsnormally(i.e.atarightangle)tothecommontangentatthepointofcontactbetweentheobjects. Itthereforeneednotactverticallyupwards.

Figure1.4

Thisexplainsthenamenormal reactionforce.

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Forexample,ifthetableonwhichthemugisrestingisonanunevenfloor,sothatthetabletopmakesananglewiththehorizontal,thenthenormal Nreactionforcemakesananglewiththevertical,asshowninFigure1.5. (Insuchacasetheremustbeother forcesactingonthemug if it istoremaininequilibrium. Theseotherforcesarediscussedlater.)

qq1.4 Tension

Considera lamphanging fromaceilingonanelectriccable. Letusmodel Figure1.5the lamp as a particle. As in the case of the mug and the table in theprevious subsection, we know that there is a weight associated with thelamp,andthatsincethelampisatrest,bytheequilibriumconditionsomeother force(s) must be acting on it. The only possible source for anotherforce is the cable, so the cable must exert a force on the lamp. The forceexertedbythecableonthe lampiscalledthetension force.Tensionforcesoccurwheneverobjectsaretautlyjoined,e.g.bycables,ropes,stringsorthreads. Thesecablesandropescanbemodelledindifferentways.For example, ifwewantto modeltheceiling lampandare interestedonlyinthe force inthecable,thenwecanmodelthecableasamodel string,defined as an object possessing length, but no area, volume or mass, andwhichdoesnotstretch(i.e.itisinextensible). Ontheotherhand, ifweareinterested in how much the cable stretches under the weight of the lamp,thenwecanmodelthecableasamodelspring,whichhaspropertiessimilar Springsarediscussedintothoseofamodelstring(i.e. ithasnoarea,volumeormass),butallows Unit7.extension. Inthisunitweconsideronlystrings.TheceilinglampexampleisillustratedinFigure1.6. Thetensionforcedueto the model string is denoted by T, and the weight of the lamp by W.In a manner similar to the case of normal reaction forces, the equilibriumconditionforparticlesgives

W+T=0.Ifthelamphasmassm,thenW=mgk,andhence

T=W=mgk

Tk m

W

is a force acting vertically upwards (along the length of the model string) Figure1.6withthesamemagnitudeastheweightofthe lamp.Weassumethatthetensionforceduetoamodelstringactsalongthelengthofthestringandaway fromthepointofitsattachmenttoanobject. Asinthecaseofanormalreaction,themagnitudeofthisforce(oftenreferredtoasthetension inthestring ascalarquantity)dependsontherequirementsnecessarytomaintainequilibrium,soitispotentiallyunlimited. (Inreality,astringcanexertonlyacertaintensionforcebeforeitbreaks,butamodelstringsupportsanunlimitedtensionforce.)

Definitions(a) Amodelstringisanobjectwithafixedfinitelength,andnoarea,

volumeormass,thatexertsaforceatthepointofattachment.(b) Thetension force due to a string isdirectedalongthe length

ofthestringaway fromthepointofattachment.As in the case of normal reaction forces, the tension force due to a stringneednotbeverticallyupwards,asthefollowingexample illustrates.

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Substitutingthis intoEquation(1.2)gives

|T2|= 4g/ 322.66.Sothemodelpredictsthatthetensionforceduetothecordfromtheporchhasmagnitudeabout45.3Nandthatthetensionforceduetothecordfromthewallhasmagnitudeabout22.7N.The procedure that was used in Example 1.1 can be used to solve manyproblemsinstatics,andmaybesummarizedasfollows.

Procedure1.1 SolvingstaticsproblemsGivenastaticsproblem,performsomeorallofthefollowingsteps.(a) Drawasketchofthephysicalsituation,andannotate itwithany

relevant information.(b) Chooseaxes,andmarkthemonyoursketch.(c)

Draw

a

force

diagram

or

diagrams.

(d) Usetheequilibriumconditionandanyotherappropriatelaw(s)to

obtainequation(s).(e) Solvetheequation(s).(f) Interpretthesolution intermsoftheoriginalproblem.

DrawpictureChooseaxes

Drawforce

diagram

Applylaw(s)Solveequation(s)Interpretsolution

In this unit, the steps in this procedure will often be identified (using themarginal abbreviations above) in the solutions to examples and exercises.Theprocedureisintendedtobeaguideratherthanarigidsetofrules. Forexample,ifitisnotobviouswhichsetofaxestochoose,thendrawtheforcediagram

first,

and

the

best

choice

may

become

more

apparent.

Try

using

theprocedureinthefollowingexercise.*Exercise1.3

DuringDecember,alargeplasticChristmastreeofmass10kgissuspendedby its apex using two ropes attached to buildings either side of the highstreetofTrappendorf. Theropesmakeanglesof6 and 4 withthehorizontal.ModeltheChristmastreeasaparticleandtheropesasmodelstrings. Whatarethemagnitudesofthetensionforcesduetothetworopes?

1.5 FrictionConsiderabookrestingonahorizontalsurface. Therearetwoforcesactingonthebook: theweightdownwardsandthenormalreactionupwards. Sup-posethatyoupushthebookgentlysideways(seeFigure1.9). Ifyoudonotpushhardenough,thebookwillnotmove;itwillremaininequilibrium. Weknowthatthevertical forces, i.e.theweightandthenormalreaction, bal-anceeachother(i.e.theyareequalinmagnitudeandoppositeindirection),so they cannot be preventing the book from moving sideways. Thereforethere must be another force present. This force is known as the frictionforce. It isconsideredtoact paralleltothesurface, i.e.atrightanglestothenormalreaction,and inadirectionthatopposesany(possible)motionalongthatsurface. Modellingthebookasaparticle,anddenotingthepush-ing force by P, the friction force by F, the weight by W and the normalreactionbyN,theforcediagramforthisexample isshown inFigure1.10.

sidewayspush

Figure1.9N

F P

WFigure1.10

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Friction forcesarecausedbytheroughnessofevenseeminglyverysmoothsurfacesaroughnessthatservesto inhibitthesmoothmovementofonesurface over another. So friction forces are present only where there ismovementorthepossibilityofmovement. Thereisnofrictionforcepresentwhenanobjectisrestingonahorizontalsurface,wheretheonlytwoforcesacting on theobjectare its weight and the normalreaction. But whenanobject

is

being

pushed

or

pulled,

or

is

resting

on

a

sloping

surface,

then

a

frictionforceispresent(seeFigure1.11).

FN

W

N

Wno friction friction force

force present

Figure1.11Unlike the normal reaction, which is potentially unlimited in magnitude,thereisalimittothemagnitudeofthefrictionforce;ifthislimitisreached,then slipping occurs. The limiting value of the magnitude of the frictionforcedependsalmostentirelyonthematerialsofthetwosurfacesandonthemagnitudeofthenormal reaction forcebetweenthem. Itdoesnotusuallydepend on the area of contact between the two surfaces, or on the angleatwhichthetwosurfacesare inclinedtothehorizontal. Experimentsshowthatthe limitingvalueofthemagnitudeofthe friction forceF(whichjustprevents slipping for two given surfaces) is approximately proportional tothemagnitudeofthenormalreactionforceNbetweenthetwosurfaces. So,onthevergeofslipping,wehave|F|=|N|,where isthecoefficient ofstatic friction,whichdependsonthematerialsofthetwosurfaces. Someapproximatevaluesoffordifferentmaterialsaregiven inTable1.1.Example1.2Asteel forkofmass0.05kgrestsonahorizontalwoodentable. Modeltheforkasaparticle. Whatisthemaximumsidewaysforcethatcanbeappliedbeforetheforkstartstomove?SolutionThesituationisillustratedinFigure1.12. Sincealltheforcesactinaverticalplane,wecanchooseaxesasshown. Theforcediagramisalsoshowninthefigure,whereFisthefrictionforce,Pisthesidewaysforce,Wistheweight,andN isthenormalreaction.

Nsideways j

force 0.05 kg F Pim= 0.4

W

Figure1.12

Inthisunitweconsideronlycaseswhereobjectsremainatrest,sothatthereisonlythepossibility ofmovement.Frictionincaseswherethereis movementisconsideredinUnit6.

Table1.1 Approximatecoefficientsofstaticfriction

Surface Steelonsteel(dry) 0.58Steelonsteel(oiled) 0.1Plasticonplastic 0.8Rubberontarmac 1.3Steelonwood 0.4Woodonwood 0.35

DrawpictureChooseaxesDrawforcediagram

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Theequilibriumconditionforparticlesgives Applylaw(s)F+N+P+W=0. (1.4)

Whenthefork isonthepointofmoving(slipping),wehave|F|=|N|,

where= 0.4isthecoefficientofstaticfrictionofsteelonwood.TobeabletouseEquation(1.4),weneedtoexpresstheforces intermsof Solveequation(s)theunitvectors iandj. LookingatFigure1.12, the forcescanbewrittenincomponentformas

F=|F|i, N=|N|j, P=|P|i, W=|W|j.ResolvingEquation(1.4) inthei-directiongives

|F|+ 0 + |P|+ 0 = 0,so(asexpected)

|F|=|P|.Resolving(1.4)inthej-directiongives

|N|=|W|.Therefore,whenthefork isonthepointofmoving,

|P|=|F|=|N|=|W|= 0.40.05g= 0.02g0.196.Sothemodelpredictsthatasidewaysforceofmagnitudeabout0.196Ncan Interpretsolutionbeappliedwithoutmovingthefork.Here is a summary of how we go about modelling problems that involvestaticfriction, i.e.problems involvingfrictionbutnomotion.

Modelling static frictionConsidertwosurfacesincontact.(a)The friction force F acts in a direction perpendicular to the nor-

malreaction Nbetweenthesurfacesandoppositetoanypossiblemotionalongthecommontangenttothesurfaces.

(b)|F| |N|, where is a constant called the coefficient of static |F|cannotexceeditslimitingfrictionforthetwosurfaces involved. value|N|. Slippingoccursif

africtionforceofmagnitude(c) |F|=|N|whentheobjectisonthevergeofslipping. Thisequal- greaterthan|N|wouldbe

ity is sometimes referred to as describing a situation of limiting neededtopreventit.friction.

(d)If one of the surfaces is designated as being smooth, it may be assumed that there is no friction present when this surface is incontactwithanother,regardlessoftheroughnessoftheothersur-face.

Letusnowapplytheseideastosomeexamples,inwhichweshallalsoapplythestepsofProcedure1.1. Inmostofthesituationsthatweinvestigate,weshallbeconcernedwith limitingfriction.*Exercise1.4

Awoodblockofmass5kgrestsonahorizontalplankofwood. Modeltheblockasaparticle. Whathorizontalforceisrequiredtostart itmoving?

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Exercise1.5A steel block of mass 0.5kg rests on a horizontal dry steel surface and ispulledbyahorizontalforceof2N. Modeltheblockasaparticle. Doestheblockmove? Whatisthemagnitudeofthefrictionforce?*Exercise1.6

Ashallowboxmadeofauniformmaterialandwithouta lidcanbeplacedonahorizontaltableintwopossibleways(asshown inFigure1.13):(a) withitsbaseincontactwiththetablesurface;(b) withitsopentop incontactwiththetablesurface.

(a) (b)

Whichofthesetwopositionsrequiresthesmallersidewaysforcetostarttheboxslipping? Figure1.13

InclinedplanesConsidernowanobjectrestingonaslopingplanesurface,oftenreferredtoasaninclined plane,suchastheoneshowninFigure1.14. Providedthat

Ntheangleofinclinationisnotlarge,theobjectcanremainatrestanddoesF

not slide down the slope. The forces acting on the object are its weight,thenormalreactionandfriction. TheweightWactsverticallydownwards.ThenormalreactionNactsnormallytothesurfacebetweentheobjectand

Wtheslope. ThefrictionforceF isperpendiculartothenormalreactionandhenceparalleltotheslope,anditactsuptheslopetocounteractthenaturaltendencyoftheobjecttomovedowntheslope. Figure1.14Example1.3A crate of empty bottles of total mass 30kg is to be hauled by a rope uparampfromthecellarofapub. Therope isparalleltotheramp,andtheramp makes an angle of 6 radians with the horizontal. The coefficient ofstaticfrictionbetweentheplasticcrateandthewoodenrampis0.2.What isthetensionforceduetotheropewhenthecrate isonthepointofmovingupwards?SolutionThesituation is illustrated inFigure1.15. Drawpicture

P6

M= 0.2 30kgi

j

Figure1.15All the forces act in a vertical plane, so we need only two axes. We could Chooseaxeschooseitobehorizontalandjverticalasbefore,but itmakescalculationseasierifwechooseitobeparalleltotheslopeandjperpendiculartoit,asshown inFigure1.15. This isbecause,whenwecometoresolvethe forcesin the i- and j-directions, three ofthe four forces (all except W) will thenact

along

one

or

other

of

the

axes,

making

resolving

them

much

simpler.

Modellingthecrateasaparticleandtheropeasamodelstring, the force Drawforcediagramdiagram isasshown inFigure1.16,where W istheweight, Nthenormalreaction,Fthefrictionforce,andTthetensionforce.14

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P6

P6

P3

T N

W

Fi

j

Figure1.16Theequilibriumconditionforparticlesgives Applylaw(s)

T +N +F +W =0. (1.5)Whenthecrateisonthepointofmoving,wehave

|F|=|N|,where= 0.2isthecoefficientofstaticfriction.Asbefore,thefirststepinsolvingtheequationsinvolvesresolvingtheforce Solveequation(s)vectorsintocomponents. Inthiscase,threeoftheforcevectorsarealignedwiththeaxesandcanbewrittendown immediately:

F =|F|i, N =|N|j, T =|T|i.To find the weight of the crate in component form, we use the formuladerived inUnit4:

W = (W . i)i + (W .j)j Notethat |cos( )jW i + W cos( )+=| |cos | +3 2 3 2 3

sin sin= cos cos3NowEquation(1.5)caneasilyberesolved inthei-direction,giving

1 W i W|j| | | 2 3 2 3= . 2 2= 3

2.

12|+ 0 + |= 0,|T |F W|+ |

so12 | (1.6)T = F W|+| | | |.

Similarly,resolvingEquation(1.5) inthej-directiongives

32 |0 + N + 0 W|= 0,| |

so

|= 32 (1.7)N W|| | .Atthepointofmoving,|F|= 0.2|N|andEquations(1.6)and(1.7)give

32 ||+12 | |+12 ||= 0.2 |= 0.2T N W W W| | |.

Thus,since |W|= 30g,

|= ( 310 +12 )30g198.T|Therefore,whenthecrateisonthepointofmoving,themodelpredictsthat Interpretsolutionthetensionforceduetotheropeisabout198Nuptheramp.Mathematically, different choices of axes make no difference to the finalsolution obtained to a mechanics problem. However, a sensible choice ofaxes,asinExample1.3,canreducetheamountofcalculation. Youwillfind Choiceofaxesisdiscussedthat,withexperience,youwillbeabletochooseaxesthatreducethework againinUnit6.involved.

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Section2 Twoormoreparticles

*Exercise1.7A fullcrateof bottles ofmass60kg isatthetopofthe rampdescribed inExample 1.3, ready to be lowered into the cellar. What force needs to beappliedtotheropetokeepthecratefromslidingdowntheramp?

End-of-sectionExercisesExercise1.8Onabuildingsite,apalletofbricksofmass1800kg issuspendedfromthecableofacrane. Thelengthofthecableis10m. Oneofthesiteworkersispullingwitha forceofmagnitude800Nhorizontallyonaropeattachedtothe pallet, in order to position the pallet over the lorry into which it is tobe lowered. Howfarcanthepalletbemovedhorizontallybytheworker?Exercise1.9(a) A box of massm is resting on a surface inclined at an angle to the

horizontal. Iftheboxisonthepointofslipping,whatisthecoefficientofstaticfriction?

(b) Twoidenticalmugsareplacedonatray. Onemugishalffullofcoffee,theotherisempty. Thetrayistiltedslowly. Useyouranswertopart(a)todeterminewhichmugwillstarttomovefirst.

2 TwoormoreparticlesIn the previous section we considered the action of forces on one particleand introduced the equilibrium condition for particles. In this section weextend these ideas to situations involving two or more particles. Subsec-tion 2.1 shows how Newtons third law can be applied to such situations.Subsection 2.2 introduces a new modelling device the model pulley andSubsection2.3considersfriction inthetwo-particlecase.2.1 NewtonsthirdlawTheequilibriumconditionforparticlesextendstosystemsinvolvingtwoormore particles, in that, if the system is in equilibrium, then each particlemustbeinequilibrium,sothesumoftheforcesactingoneachparticlemustbezero.Consider, for example, the simple situation of one book lying on top ofanotheronadesk. Bothbooksare inequilibrium. Letusmodeleachbookasaparticle. Wecanapplytheequilibriumcondition forparticlestoeachbook in turn. We therefore need to determine the forces on each book.Fortheupperbook,thesituation issimilartothecaseofanobjectrestingon a surface, discussed in Section 1. There is the weight W1 of the bookactingverticallydownwards,andthenormalreactionN1 ofthesurface(ofthe lowerbook)actingverticallyupwards. Forthe lowerbook,wehave itsweightW2 actingverticallydownwards,andthenormalreactionN2 ofthesurface (of the desk) acting vertically upwards, but this time there is alsoanotherforcethenormalreactionN3 fromtheupperbook. Thesituationandtheforcediagramsareshown inFigure2.1.16

N1

W1

upperbook

Figure2.1

N2

W2

N3

lowerbook

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Youmayfindthepresenceofthe forceN3 surprising,because inSection1weconsideredthenormalreaction forceasacting fromasurfacetoanob-

ject. However,thereisalsoanormalreactionforceactingfromtheobjecttothesurface. So,forexample,ifweconsidertheforcesactingonthetablede-scribedinSubsection1.3,wehavetoincludethenormalreactionexertedbythecoffeemugonthetable. Infact,pairsofnormalreactionsoccurbetweenany

pair

of

surfaces

in

contact.

Such

pairs

of

normal

reaction

forces

can

be

deduced from Newtons third law, a well-established law of mechanics,which says that for each force exerted by one object on another, there isa force of equal magnitude acting in the opposite direction exerted by thesecondobjectonthefirst. ThisenablesusnotonlytodeducethepresenceofN3,butalsotodeducethat

N1 =N3.Inadditiontothisequation,theequilibriumconditionforparticlesappliedtoeachbookinturngives

W1+N1 =0,W2+N2+N3 =0.

SolvingthesethreeequationsforN1,N2 andN3 givesN1 =W1, N2 =W1W2, N3 =W1.

Ifwehadconsideredthetwobooksasoneparticle,thenthenormalreactionforceN2 fromthetablewouldstillbethesame, i.e.(W1+W2). SotheaboveargumentisphysicallyreasonableinthesensethatthenormalreactionforceN2 isunchangedwhetherwemodelthebooksastwoparticlesorone.Nowconsidertheforcesactingontheupperbook,namelyN1 andW1. As shown above, these forces are of equal magnitude and opposite directions.ItisacommonmistaketosaythattheseforcesformanequalandoppositepairasdescribedinNewtonsthirdlaw. Thisisnotcorrecttheyareequaland opposite because the book is in equilibrium. As we have seen above,theforcepairedwithN1 byNewtonsthirdlawisN3. WhatforceispairedwiththeforceW1? Theansweristheforceofgravitythatthebookexertson the Earth. (The magnitude of this force is, of course, negligible whencomparedtothemassoftheEarth.)Insummary,solvingstaticsproblemsinvolvingmorethanoneparticleneedsthesekeyideas:(a)applyNewtonsthird law;(b)applytheequilibriumconditionforparticlestoeachparticleseparately.*Exercise2.1

Consider a pile of four books of equal mass, lying one on top of anotheronadesk. Drawthe forcediagramsforthissituation,andfindthenormalreactionsactingoneachbookintermsoftheweightofabook.

Thetwonormalreactionforcesthatconstituteanequalandoppositepairactondifferent objects.

In the same way that Newtons third law can be applied to the normalreaction forces between objects, so can it be applied to the tension forcesdue to a stringjoining two objects. Thus, for example, for two particlesAandBjoinedbyamodelstring, thetension forceTA onparticleAduetothestring isequal inmagnitudeandopposite indirectiontothetensionforce TB on particleB due to the string, so TA =TB. Themagnitudeofthetwoforcesduetoastring isoftenreferredtoasthetension in thestring. This isascalarquantity. ThiswasremarkeduponinSubsection1.4.

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Exercise2.2Alampofmass1.5kgishangingfromtheceilingonitscable. Achildstoyofmass0.5kgissuspendedbyastringfromthelampshade. Modelthelamp(plus lampshade)and thetoy as particles, and the cable and the string asmodel strings. Draw force diagrams showing the forces acting on the twoparticles. What isthemagnitudeofthetensionforceduetothecable?

2.2 PulleysThepulley isacommondevicewithwhichyouareprobably familiar. Youmay have seen them in use on building sites, for example, as an aid toraisingor loweringheavy loads. The ideaofapulley isuseful inmodellingmechanics problems, as it enables us to model a change in direction of atensionforce.In diagrams, we shall use an idealized pulley as shown in Figure 2.2. Inorder to keep the model simple, we make simplifying assumptions, whichareformallystatedinthefollowingdefinition. Figure2.2

DefinitionAmodelpulleyisanobjectwithnomassorsize,overwhichamodelstring may pass without any resistance to motion. The tension in astringpassingoveramodelpulleyisthesameeithersideofthepulley.

Thepointtoremember isthattheresultoftheseassumptions impliesthatthe tension forces due to the string on either side of the model pulley areequalinmagnitude,i.e.thetensioninthestringasitpassesoverthepulleyremainsconstant.A model pulley provides a reasonable model of an actual pulley, providedthatitsdimensionsaresmallcomparedwiththelengthoftheropeorcablepassingover itandthat itsweight issmallcomparedwiththeother forcesinvolved. Model pulleys can also be used to model a variety of situationsthatdonotinvolvepulleysatall,butmerelyinvolveachangeindirectionofatensionforce(suchaswhenaropeishangingovertheedgeofabuilding).Theiruseisillustratedbythefollowingexampleandexercises.Example2.1Asackofflourofmass50kgislyingonthefloorofamill,readytobeloadedintoacart. Tohelpwiththeloadingprocess,alightropeisattachedtothe Inmechanicsproblems, ifansack, passes over a pulley fixed to the ceiling immediately above the sack, objectissaidtobelight, itsandisattachedatitsotherendtoastoneofmass15kgthathangswithout massmaybeignored.touchingthefloor. Thesystem isshowninFigure2.3.Modelthesackandthestoneasparticles,thepulleyasamodelpulley,andtheropeasamodelstring.(a) Calculatethenormalreactionoftheflooronthesack.(b) Whatforcedoesthepulleyexertontheceiling?SolutionAlltheforcesarevertical,soweneedonlyoneaxis,asshowninFigure2.3. Chooseaxes18

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j15 kg

50 kg

Figure2.3(a) TheforcediagramsforthesackandthestoneareshowninFigure2.4, Drawforcediagram

where W1 and W2 represent the weights, T1 and T2 represent thetensionforces,andN isthenormalreactionoftheflooronthesack.

T1

N T250 kg 15 kg

W2W1

sack stone

Figure2.4Theequilibriumconditionforparticlesgives Applylaw(s)

W1+N+T1 =0, (2.1)W2+T2 =0. (2.2)

Sincethetensionforcesoneithersideofamodelpulleyhavethesamemagnitude,wehave

|T1|=|T2|.Tosolvetheequations,thefirststepistowritetheforcesincomponent Solveequation(s)formas

W1 =|W1|j, N=|N|j, T1 =|T1|j,T2 =|T2|j and W2 =|W2|j.

ThenresolvingEquations(2.1)and(2.2)inthej-directiongives|W1|+|N|+|T1|= 0,|W2|+|T2|= 0.

Therefore |T1|=|T2|=|W2|, so |N|=|W1| |T1|=|W1| |W2|= 50g15g= 35g343.

So the normal reaction of the floor on the sack has magnitude about Interpretsolution343N(35g)andisdirectedupwards. Intheabsenceofthestoneandthepulley,themagnitudeofthenormalreactionwouldhavebeenequaltothemagnitudeoftheweightofthesack(50g). Theeffectofthestone,transmitted via the pulley, is as if the magnitude of the sacks weightwerereducedbythemagnitudeoftheweightofthestone(15g).

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(b) Toanswerthisquestionweneedtomodeltheforcesonthepulley. We Drawforcediagramcan consider the pulley as a particle of no mass. Modelling the shortpieceofmetalthatattachesthepulleytotheceilingasamodelstring,wehavetheforcediagramshown inFigure2.5.

T5

T3

T4

pulley

Figure2.5Theequilibriumconditionforparticlesgives

T3+T4+T5 =0. (2.3)Since

the

tension

in

astring

around

a

model

pulley

remains

constant,

wehave

|T1| =|T2| =|T3| =|T4|.ResolvingEquation(2.3)inthej-directiongives

|T3| |T4| +|T5| = 0.Usingtheresultfrompart(a)that|T1| =|T2| =|W2|,wehave

|T5| =|T3| +|T4| = 2|W2| = 30g 294.Sothemodelpredictsthatthe forceexertedbytheropeonthepulleyisabout294N(30g)upwards. Hence,byNewtonsthird law,theforceexertedbythisropeontheceiling(i.e.the forceexertedbythepulleyon the ceiling) is about 294N (30g) downwards. This force (which istwice the weight of the stone) balances the weights of the stone (15g)and the sack (50g), less the normal reaction (35g) of the floor on thesack.

*Exercise2.3SupposethatthepulleyinExample2.1isnolongerimmediatelyabovethe

sack, so that the rope attached to the sack makes an angle of 4 to thevertical,asshown inFigure2.6.

ij

4

15 kg

50 kg

Figure2.6(a) Whatisthemagnitudeofthenormalreactionoftheflooronthesack?(b)

What

is

the

magnitude

of

the

friction

force

on

the

sack?

(c) Whatisthesmallestvalueofthecoefficientofstaticfrictionthatwould

allowthesystemtoremain inequilibrium?

Applylaw(s)

Solveequation(s)

Interpretsolution

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*Exercise2.4Acarengineofmass120kgisbeingheldsuspendedreadyforloweringintoa car. The engine is attached to a pulley by a short rope. Another rope,attached at one end to the ceiling of the garage, passes under the pulleythenuptowardstheceilingagain,where itpassesoveranotherpulleythatisjoinedtotheceiling. Amechanicholdstheendofthisrope. Thesituationis illustrated inFigure2.7.

ij

4

4

120 kg

Figure2.7Assumingthatthecomponentsofthesystemcanbemodelledasparticles,model strings and model pulleys as appropriate, determine the magnitudeoftheforcethatthemechanichastoexerttokeeptheenginesuspendedinpositionwhilethecar ispushedunderneathit.

2.3 SlippingYouhavealready investigatedslipping inthecaseofaone-particlesystem.Inthissubsectionweexaminethephenomenoninsystemsofmorethanoneparticle.Example2.2Consider a scarf draped over the edge of a table. Model the scarf as twoparticles,oneofmassm1 hangingovertheedgeandtheotherofmassm2restingonthetable,withthemassesjoinedbyamodelstringpassingovertheedgeofthetable,whichismodelledasamodelpulley. Assumethatthescarfsmass isuniformlydistributedalongitslength,sothatthemassesofthetwoparticlesareproportionaltothecorresponding lengthsofscarf.If the coefficient of static friction between the scarf and the table surfaceis, what proportion of the scarfs length can hang over the edge of thetablebeforethescarfslipsoffthetable?SolutionWecananswerthisquestionifwecanfindtheratioofm1 (themassofscarfhangingovertheedge)tom1+m2 (thetotalmassofscarf)whenthescarfisonthevergeofslipping.ThesituationisillustratedinFigure2.8,whichalsoshowsasuitablechoiceofaxes. Drawpicture

Chooseaxes21

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m2j N T1

F m2 T2i m1m1W1W2

Figure2.8Therearetwoforcesactingonthehangingparticle: itsweightW1 andthe Drawforcediagramtension forceT1. Thereare four forcesactingontheparticleonthetable:itsweight W2, the tension force T2, the normalreaction force N, andthefrictionforceF. Theforcediagramsareshown inFigure2.8.While the scarf does not slip, we can apply the equilibrium condition for Applylaw(s)particlestoeachparticleinturn. Forthehangingparticle,wehave

T1+W1 =0. (2.4)Fortheparticleonthetable,wehave

F+N+T2+W2 =0. (2.5)Theassumptionofamodelpulleygives

|T1|=|T2|. (2.6)Whentheparticle isonthevergeofslipping,wehave

|F|=|N|, (2.7)whereisthecoefficientofstaticfriction.FromFigure2.8,thecomponentformsoftheforcevectorscanimmediately Solveequation(s)bewrittendown:

T1 =|T1|j, W1 =|W1|j, F=|F|i,N=|N|j, T2 =|T2|i and W2 =|W2|j.

ResolvingEquation(2.4) inthej-directiongives|T1| |W1|= 0,

so|T1|=|W1|=m1g.

ResolvingEquation(2.5) inthei-directiongives|F|+ 0 + |T2|+ 0 = 0,

so,usingEquation(2.6),|F|=|T2|=|T1|=m1g.

ResolvingEquation(2.5) inthej-directiongives0 + |N|+ 0 |W2|= 0,

so|N|=|W2|=m2g.

UsingEquation(2.7),wehavem1g=m2g,

som1 =m2.

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Section3 Torques

Thereforethemodelpredictsthatwhenthescarfisonthevergeofslipping, Interpretsolutionthefractionofitslengththathangsovertheedge is

m1 = m2 =

m1+m2 m2+m2 + 1 .*Exercise2.5

A gold medallion of mass 0.02kg is lying on a glass shelf in a bathroom.Attached to the medallion is a gold chain of mass 0.03kg, half of whichis dangling over the edge of the shelf. If the coefficient of static frictionbetweengoldandglassis0.35,willthechainandmedallionremainatrest?

End-of-sectionExercisesExercise2.6Anobjectofmass2kgissuspendedfromtheceilingbyastring. Anobjectofmass1kgissuspendedfromthefirstobjectbyanotherstring. Fromthissecondobject issuspendedanobjectofmass3kg. Drawtheforcediagramforeachobject,andfindthetensionineachstring.Exercise2.7A man of mass 80kg is about to be lowered into a well from a rope thatpasses over the horizontal rotating axle of the well. The other end of theropeisheldbyseveralmeneachofmass80kg,asshowninFigure2.9,withthe rope between the men and the axle horizontal. Assume that the mencanberepresentedbyparticles,theropebyamodelstringandtheaxlebyamodelpulley.Ifthecoefficientofstatic frictionbetweenthemensbootsandthegroundis0.35,howmanymenarerequiredtoholdthemanattheendoftheropebeforehe is lowered intothewell? Figure2.9

mm= 0.35

j

80 kg

Exercise2.8An object of massm1, resting on a board inclined at an angle to thehorizontal,isattachedtoanobjectofmassm2 byastringhangingovertheedgeoftheboard,asshown inFigure2.10.Assuming that the objects can be modelled as particles, the string as amodelstringandtheedgeoftheboardasamodelpulley,findtheconditiononthecoefficientofstaticfrictionbetweenthefirstobjectandtheboardforthissystemtoremain inequilibrium.(Hint: Therearetwoways inwhichtheequilibriumcanbedisturbed.) Figure2.10

m2

m1

3 TorquesThissection looksatsolidbodies,and inparticularataphenomenonthatdoesnotapplytoparticles: theturningeffectofforces. ItbeginsinSubsec-tion3.1by introducingwaysofmodellingobjectswhentheirsize is impor-tant,asitiswhentheturningeffectsofforcesareconsidered. Subsection3.2goes on to explain what is meant by the turning effect of a force, and toprovideamathematicaldescriptionofsuchaneffect.

i

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Section3 Torques

3.1 ExtendedandrigidbodiesConsider, forexample,atallthinboxofcerealonabreakfasttable. Ifyoupush itnear itsbase, itslides across the table. But ifyoupush itnear itstop, ittipsover. Theposition atwhichthepushingforceacts is importanthere, sotheparticlemodel which allows forcestoactatonlyonepoint

is

inadequate.

In

this

situation,

we

model

the

cereal

box

as

an

extended

body, which is defined to be a material object that has one or more oflength,breadthanddepth,butwhose internalstructuremaybeneglected.So,likeaparticle,ithasmass,butunlikeaparticle,ithassizeofsomesortandoccupiesmorethanasinglepoint inspace.Extended bodies are complicatedobjects because they canflexor vibrate.To model the slipping or tipping behaviour of the cereal packet, this gen-erality isnot needed. Sowerestrictourattention torigidbodies. A rigidbodyisdefinedtobeanextendedbodythatdoesnotchangeitsshape(soitdoesnotflexorvibrate).Foraparticle,allforcesareappliedatthepointrepresentedbytheparticle;we say that the forcesact at this point. For extended bodies, we need tobemorecarefultospecifywherea forceacts. Forexample,theweightofabodyalwaysactsthroughthecentre of massofthebody. Inthisunitweshallconsideronlysymmetricbodiesmadeofuniformmaterial. Forsuchabody,asyouwould intuitivelyexpect,thecentreofmass isat itscentreofsymmetry(orgeometriccentre). Forexample,consideracoin,whichcanbemodelledas a disc. Using symmetry, we can state that the centre of massofthecoin isalongtheaxisofthedisc(whichrunsbetweenthecentresoftheflatcircularfacesofthedisc),halfwaybetweentheflatcircularfaces.3.2 TurningeffectofaforceSuppose that you try to balance a ruler on a horizontal extended finger.Whenthecentreoftherulerisoveryourfinger,therulershouldbalance. Ifthe centreofthe ruler is not overyour finger, thenthe weight ofthe rulerwill cause it to turn about your finger. Although the finger can provide anormalreactionforcethat isequalinmagnitudetotheweightoftheruler,ifthe two forcesarenot in line, thenturning occurs. Theweight providesaturningeffectiftheverticallineofitsaction(throughthecentreofmass)doesnotpassthroughyourfinger.Howdowemeasuretheturningeffectofaforceintermsofwhatwealreadyknow: themagnitudeanddirectionoftheforce,andthepointontheobjectat which the force acts? In order to begin to answer this question, trya simple experiment. Balance a 30-centimetre (12-inch) ruler on a pencilrubberorahexagonalpencil,orsomeotherobjectthatisnottoowideandso will actasapivot. Placetwo identical small coinsoneitherside of thepivotsothateach is10cmfromthepivot(seeFigure3.1).Then experiment with moving one of the coins in steps of 2cm from itsinitial position, and see how the other coin has to be moved in order tore-establishbalance. Theconclusion from thisexperiment isnot, perhaps,a surprising one: coins of equal mass have to be placed at equal distancesoneithersideofthepivotfortherulertoremainbalanced.Next

place

the

two

coins

together

at

apoint

on

the

ruler,

say

at

6

cm

from

the pivot. Where does a third identical coin have to be placed to achievebalance?

24

Theshapeofanextendedbodymaybeone-,two- orthree-dimensional(i.e. itmayhavea length,anareaoravolume),dependingonthesituation.Theformaldefinitionofarigidbodystatesthatthepositionvectorofapointontherigidbodyrelativetoanyotherpointontherigidbodyisconstant.Adefinitionofthecentreofmassandwaysoffinding itforageneralrigidbodyaregiven inUnits19 and25.

Figure3.1

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Section3 Torques

You should find that two identical coins placed together at 6cm from thepivotarebalancedbyanotheridenticalcoinplacedontheothersideoftheruler at 12cm from the pivot. If you continue to experiment with varyingnumbers of coins placed at various pairs of positions along the ruler, youwillfindineachcasethatifthemassesofthetwosetsofcoinsareunequal,then, inordertoachievebalance,thegreatermasshastobeplacednearerto

the

pivot

than

the

smaller

one.

The

turning

effect

due

to

the

weight

of

thecoinsactingatapointdependsnotonlyonthemassofthecoins,butalsoonthedistanceofthepointofactionfromthepivot.A long symmetricalobject, such astheruler in this experiment, can oftenbemodelledasarigidbodywith length,butnobreadthordepth. Sucharigidbody isknownasamodel rodand isoftendrawnasastraight line.The pivot on which the object rests is often modelled as a model pivot,which has a single point of contact with the rod and is often drawn as atriangle. Using these notions, the above experiments should allow you tobelievethefollowingresult.

Balanced rodAhorizontalrod,pivotedatitscentre,willremainhorizontalunderthe

F1 F2actionoftwoforcesF1 andF2 actingverticallydownwardsatdistancesl1 and l2, respectively, on either side of the pivot (see Figure 3.2),

l1 l2providedthat|F1|l1 =|F2|l2. (3.1)

Figure3.2Inotherwords,thehorizontalrodwillremain inequilibriumprovidedthatthedistancesoftheforcesfromthepivotare in inverseratiotothemagni-tudesoftheforces.*Exercise3.1

Jack and Jill are sitting on opposite sides of a see-saw. Jill is sitting at adistanceof1.2mfromthepivot,andJackis1mawayfromthepivot. Jacksmassis60kg. Ifthesee-saw isatrestandhorizontal,whatisJillsmass?

In the situation described in Exercise 3.1 and in the ruler example, therewasanobviouswaytomeasureeachdistance,i.e.alongthesee-saworruler.We need to generalize this to other situations, where there is not such anobviouswaytomeasuredistance. Inthesetwoexamplesthedistancesalongtherulerandsee-sawhappentobedistancesmeasuredperpendiculartothedirectionofthe forceand from itspointofaction, i.e.perpendiculartothelineofaction oftheforce.

DefinitionThelineofaction of a forceisastraightlineinthedirectionoftheforceandthroughthepointofactionoftheforce.

So the turning effect of a force about a fixed point needs to encompass ameasureoftheforceitselfandtheperpendiculardistanceofitslineofactionfromthefixedpoint. Italsoneedstoincreaseinmagnitudeifeithertheforceorthedistanceincreasesinmagnitude,andviceversa.

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Section3 Torques

ConsideraforceFwithlineofactionAB,asshowninFigure3.3,andsome HereFisusedtodenoteafixedpointO. Let RbeanypointonABwithpositionvectorrwithrespect general force,ratherthanatoO. Thenthecrossproduct is frictionforce.

rF= (|r| |F| sin)c,where

c isaunitvectorperpendiculartobothrandF,andwithdirection

outof

the

page

(as

given

by

the

screw

rule).

This

cross

product

satisfies

all

theaboverequirementsfortheturningeffectofaforce. Itincludesameasure|F|ofthe forceand the perpendiculardistance |r| sinof its lineofactionAB fromthefixedpointO. Thedirectionofthecrossproduct,representedbycandgivenbythescrewrule,correspondstothedirectionoftheturningeffect. In this example, the turning effect of F aboutO is anticlockwise,which corresponds to the anticlockwise motion of a screw pointing out of

FRr

q

B

Athepage. WerefertothecrossproductrFasthetorque ofthe forceF OrelativetotheoriginO,andweuse itasourmeasureoftheturningeffectoftheforce. Figure3.3

Definition

ThetorqueofaforceFaboutafixedpointO isthecrossproduct ThesymbolistheuppercaseGreekletter

=rF, gamma.where r is the position vector, relative toO, ofapointon the line ofactionoftheforce.

Tocalculatethemagnitude|r| |F| sinofatorque,youcanfindtheperpen- zdiculardistance |r| sinof its lineof action fromthefixedpoint, andthen

3

2

3

3multiply this by |F|. However, the component of F in the direction per- F3pendiculartoOR(inFigure3.3) is |F| sin. Soanotherwayofcalculating 2

2 F2

the magnitude of atorque is tomultiply the lengthofOR(i.e. |r|) by the 1 1componentofF inthedirectionperpendiculartoOR. O1 2 3

y*Exercise3.2 1

Findthetorqueofeachofthe forces inFigure3.4relativetotheoriginO, F1 kxwhereeachforceisofmagnitude3N. j

iExercise3.3(a) Show that ifO is any point on the line of action of a force F, and r Figure3.4

is the position vector, relative toO, of any other point on the line ofaction,thenrF=0. Deducethatthetorqueofaforceaboutapointon

its

line

of

action

is

zero.

(b) SupposenowthatOisnotonthelineofactionofF,andletr1 andr2 be

the position vectors, relative toO, of two pointsonthe line of action.Show that r1 r2 is parallel to F, and hence that r1F=r2F.DeducethatthetorqueofaforceaboutafixedpointO isindependentofthechoiceofthepositionofthepointonthelineofaction.

Letus now convinceourselvesthatthedefinitionoftorquemakessense intermsoftheexamplesofturningforceswehaveseensofar. Todothis,weneedanequilibriumconditionforrigidbodiesthatextendstheequilibriumcondition for particles. You will not be surprised that it requires not onlythatalltheforcesonarigidbodysumtozero,butalsothatallthetorquessumtozero.

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Section3 Torques

Equilibrium condition for rigid bodiesA rigid body subjected to forces F1,F2, . . . ,Fn is in equilibrium ifthe forces sum to the zero vector and if the torques 1,2, . . . ,ncorresponding to the forces, relative to the same fixed pointO, alsosumtothezerovector,i.e.

n n nFi =0 and i = riFi =0,

i=1 i=1 i=1whereri isthepositionvector,relativetoO,ofapointonthe lineofactionofFi.

Let us now apply the definition of a torque and the equilibrium conditionforrigidbodiestooursee-sawexamplefromExercise3.1.Example3.1Jack and Jill are sitting on opposite sides of a see-saw. Jill is sitting at adistanceof1.2mfromthepivot,andJackis1mawayfromthepivot. Jacksmassis60kg. Ifthesee-saw isatrestandhorizontal,whatisJillsmass?SolutionWeshallmodelthesee-sawasamodelrodrestingonapivotat itscentre, DrawpictureandconsiderJackandJillas forces(weights)appliedtotherod,asshowninFigure3.5.Asuitablechoiceofaxes isshown inFigure3.5. Weneedthreeaxessince Chooseaxesthetorqueshaveadirectionperpendiculartothedirectionsoftheforcesandthe position vectors. We also need to choose an originO for the position Asyouwillseebelow,anyvectors,andthis isconvenientlyplacedatthepivotpointofthesee-saw. choiceoforiginwilldo,sowe

chooseonethatmakesthecalculationseasy.

Jack Jill

mk

ij

N

W1 W3 W2O

60 kg 1 m 1.2 m

Figure3.5In the force diagram for the see-saw, we represent the rod (see-saw) as Drawforcediagrama straight line with forces applied at the appropriate points, as shown inFigure3.5,whereW1 istheforceappliedbyJacksweight,W2 istheforceappliedbyJillsweight,W3 istheweightofthesee-saw,andNisthenormalreactionofthepivotonthesee-saw.Theequilibriumconditionforrigidbodiesgives Applylaw(s)

W1+W2+W3+N=0,1+2+3+N =0, (3.2)

where i is the torque corresponding to Wi, for i= 1, 2, 3, and N is thetorquecorrespondingtoN,allrelativetoO.

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Section3 Torques

TosolveEquation(3.2),weneedpositionvectorsr1,r2,r3 andrN ofpointsonthe linesofactionofW1, W2, W3 and N,relativetoO. ThepositionvectorsofsuitablepointsonthelinesofactionoftheforcesappliedbyJackandJill,relativetoO, are jand1.2j,respectively. Thepositionvectorofapointonthelineofactionofthetwoforcesactingthroughthepivotis0.Sowehave

r1 =j, r2 = 1.2j, r3 =0, rN =0.Using the information onthe masses in Figure 3.5, andassuming that thesee-sawhasmassM andJillsmassism,wehave

W1 =60gk, W2 =mgk, W3 =Mgk, N=|N|k.Therefore

1 = (j)(60gk) = 60g(jk) = 60gi,2 = (1.2j)(mgk) = 1.2mg(jk) = 1.2mg(i) = 1.2mgi,3 =0,N =0.

SoEquation(3.2)becomes60gi1.2mgi=0.

Resolving inthei-directiongives60g1.2mg= 0,

som=50.ThereforethemodelpredictsthatJillsmassis50kg,asinExercise3.1.It is reassuring that the torque approach of Example 3.1 gives the sameanswer as Exercise 3.1. However, you may well feel that using the torqueapproach for the see-saw problem is more cumbersome. In this particularsituation it is. However, the torque approach is generally applicable toany statics problem involving rigid bodies, whereasthe earlier approach isapplicableinonlyveryspecialcases.Example 3.1 also illustrates that the direction of a torque corresponds tothedirectionofitsturningeffect. Thetwonon-zerotorquesinthatexampleare1 = 60giand2 =1.2mgi. Thetorque1,causedbyJacksweight,hastheeffectofrotatingthesee-sawinananticlockwisedirectionaboutthepivot. Referring to Figure 3.5, the axis of rotation is along a line perpen-diculartothepage. Also,thisrotationisinananticlockwisedirectionfrom

jto

k,

which,

according

to

the

screw

rule,

corresponds

to

the

screw

moving

out of the page in the direction of i, which corresponds to the directionof1. Similarly,thetorque2,causedbyJillsweight,rotatesthesee-sawinaclockwisedirectionfromktoj,andcorresponds,bythescrewrule,tothescrewmovinginthedirectionofi,whichcorrespondstothedirectionof2. Hencethedirectionsofthesetorquesprovideuswithinformationontheaxesofrotationofthetorquesandonthesensesofrotation(clockwiseoranticlockwise),relativetothechosenfixedpointO.The choice of origin in Example 3.1 certainly made the calculations easy,since itmadetwoofthepositionvectorszeroandhencetwoofthetorqueszero. In terms of the answer obtained, however, we could have chosen toplace

the

origin

anywhere

in

space.

Solveequation(s)Wemaketheobviouschoiceforthesepoints,namelythepointswherethelinesofactioncutthehorizontalrod.

Choosingtheoriginatthepivothasreducedtwoofthetorquestozero,andthushassimplifiedthetorqueequilibriumequationconsiderably.

Interpretsolution

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Section3 Torques

Suppose,forexample,thatinExample3.1wehadchosentheorigintobeatanotherpoint,O,andsupposethatO inFigure3.5hasapositionvectora r1 OwithrespecttoO. Then,asFigure3.6 illustrates,apositionvectoronthe a+ r1 alineofactionofW1 isa+r1. Similarly,thepositionvectorscorrespondingtoW2,W3 andNarea+r2,a+r3 anda+rN,respectively. Thetorqueequilibriumequationbecomes

(a+r1)W1+ (a+r2)W2+ (a+r3)W3+ (a+rN)N=0,whichcanberewrittenas

O

a(W1+W2+W3+N)Figure3.6+r1W1+r2W2+r3W3+rNN=0,

which,sinceW1+W2+W3+N=0,givesr1W1+r2W1+r3W3+rNN=0,

asbefore.ThisideageneralizestoforcesF1,F2, . . . ,Fn, in that

n n n

(a+ri)Fi = aFi+ riFii=1 i=1 i=1

n n=a Fi+ riFi

i=1 i=1n n

= riFi if Fi =0.i=1 i=1

So,intermsofthefinalanswertoaprobleminvolvingtorques,theposition Thepositionoftheoriginis,of the origin is irrelevant. It makessense, therefore, as inExample3.1, to however,relevanttothechoosetheorigintominimizetheamountofcalculation. valuesofthetorques.

End-of-sectionExerciseExercise3.4Apersonsitsonaparkbenchwhichconsistsofarigidseatheldupbytwolegsateitherend,asshowninFigure3.7. ModeltheseatofthebenchasamodelrodoflengthlandmassM. Modelthepersonasaparticleofmassmsitting ata pointxmetres fromthe left-handendofthe bench. Calculatethenormalreactionforcesonthebenchseatduetothe legsofthebench.

xO

m

Ml

Figure3.7

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Section4 Applyingtheprinciples

4 ApplyingtheprinciplesThissectioncontainssomeexamplesandexerciseswhichusetheprincipleswe have developed so fartosolvemore complicatedstaticsproblems. Thefirstexampleismorecomplicatedthantheexamplesintheprevioussectionbecausetheforcesandpositionsarenotconvenientlyaligned. Thisisnotafundamentaldifficulty;itmerelymakesthecalculationofthevectorproductsmorecomplicated.Example4.1Duringthe erection of a marquee, aheavy poleOAofmassm and lengthl must be held in place by a ropeAB, as shown in Figure 4.1. The anglebetweenthepoleandtheground is ,andtheanglebetweentheropeand4the ground is 6. Model thepoleas a modelrod, and the ropeas a modelstring. AssumethatthepoleisfreelyhingedatO,i.e.theendofthemodelrod isfixedatOandisfreetopivotaboutO.

A

P6 P4

ropepole

OBFigure4.1

Ifthepole is inequilibrium,findthemagnitudeofthetension intherope.Is

the

magnitude

of

the

tension

in

the

rope

bigger

or

smaller

than

it

would

be ifthepolewerehangingfreelyontheendoftherope?SolutionThebestchoiceofaxes isnotobvious forthisproblem. So inthiscasewe Drawforcediagramproceedbydrawingtheforcediagramfirst,asshowninFigure4.2.

N T

WP6 P4

P3

O ij

k

FFigure4.2

From Figure 4.2 we see that three of the forces are either horizontal or Chooseaxesvertical,sowechoosetoorienttheaxesthisway,asshowninthefigure. Inthisproblem,thechoiceoforiginisvital,sincewearegoingtotaketorquesabouttheorigin. Choosingtheoriginatthebaseofthepolemakestwoofthetorqueszeroagreatsimplification.Theequilibriumconditionforrigidbodiesgives Applylaw(s)

N+F+W+T=0, (4.1)N+F+W+T =0. (4.2)

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where N, F, W and T arethetorqueswithrespect toO of N, F, WandT,respectively.Inthiscase,Equation(4.1) isnotveryusefulsince itcontainsthree forces Solveequation(s)of unknown magnitude, namely N, F and T. However, by taking torquesaboutO, two of the torques appearing in Equation (4.2) are zero, namelyN andF. Nowweproceedbycalculatingtheothertwotorques.Thepositionvectorsofthepointsofapplicationoftheforcesare

l1lsin 24 j+lcos4

1j+4 4 (j+k),k= lcosrW = 2 2 2l

=lsin (j+k).k=rT2

Theweightofthepolecanbewritten incomponent formby inspectionofthediagramasW=mgk. Toresolvethetensionforce,weusetheformulafromUnit4. The anglebetweenthedirectionof the tension force andtheverticalisshowninFigure4.2tobe 3. Theanglestothej- and k-directions

2canbeobtainedbyadding and,respectively. Sowehave Therearemanywaysof

computingthecomponentsof2 )j+3 cos(+ )k3cos(T= T T|

2

+| | |sin vectors. Itisuptoyoutochoosethewaythatyoufeel

mostcomfortablewith. For2 cos3 )j+3 (coscos3 sinsin )k3|(cos|sin|T

|T| |=

= )j+3 )k3(sin (cosT T| | example,thiscalculationcanbedonemoregeometrically3=|T|( j+1k).2 2 straightfromtheforcediagram.

Notethatthetermsinvolvingjjandkkhavebeensuppressedsincetheygivethezerovector.

Interpretresults

Nowthattheforcesandpositionvectorsarewrittenincomponentform,wecanproceedtocalculatethetorques:

lW =rWW= (j+k)(mgk)

2 2lmg

=

i,2 2

T =rTT=l (j+k)(|T|(3j+1k))2 2 2

l|T| 3= (12jk+ kj)2 2

= ( 31)i.l|T|2 2

SubstitutingthesetorquesintoEquation(4.2)gives

i+

( 3

1)i=0.lmg l|T|

2 2 2 2Resolving inthei-directionandrearranginggives|T|=mg .

31 Sothemagnitudeofthetensionintheropeismg/( 31)1.4mg,whichis greater than the magnitude of the weight of the pole,mg. So, rathercounter-intuitively, a stronger rope is needed to erect a pole in this waythanisneededto liftthepole.Thenextexample isoneofmanysimilarproblemsthat involve ladders. Itshowsthatsometimesbothoftheequilibriumconditionsareneededtosolveaproblem.

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Example4.2A ladder of massM and length l stands on rough horizontal ground, andrests against a smooth vertical wall (see Figure 4.3). The ladder can bemodelled as a model rod. Find the minimum angle between the ladderandtheground forwhichthe laddercanremainstatic, ifthecoefficientofstaticfrictionbetweenthe ladderandtheground is0.5.

lM

qO

i j

FW

N2

N1

m= 0.5k

Figure4.3SolutionThe forcesactingontherodareshown inthe forcediagram inFigure4.3, Drawforcediagramwhere W is the weight of the ladder, N1 is the normal reaction from theground,N2 isthenormalreactionfromthewall,andFisthefrictionforceatthebottomofthe ladder. (Thewall issmooth,soweassumethatthereisnofrictionforceatthetopofthe ladder.)As two of the three unknown forces act at the bottom of the ladder, thisis a convenient point for the originO. The axes are chosen as shown in ChooseaxesFigure4.3.Theequilibriumconditionforrigidbodiesgives Applylaw(s)

N1+N2+F+W=0, (4.3)N1 +N2 +F+W =0, (4.4)

where N1,N2,F andW arethetorqueswithrespecttoO ofN1,N2,FandW,respectively.Iftheladder isnotgoingtoslip,wemusthave

|F| |N1|, (4.5)whereisthecoefficientofstaticfriction.The position vectors rN1 and rF are both zero, hence the corresponding Solveequation(s)torquesN1 andF (relativetoO)arealsozero. Tocalculatethenon-zerotorques, we need the position vectors of the points of application of theforces. Thesearegivenby

rN2 =lcosj+lsink,1rW = 2lcosj+1lsink.2

Alloftheforcesinthisexamplearealignedwiththecoordinateaxes,sothecomponentscanbewrittendownbyinspection:

N1 =|N1|k, N2 =|N2|j, F=|F|j, W=Mgk.32

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1


Nowwecancalculatethetwonon-zerotorques:N2 = (lcosj+lsink)(|N2|j)

=|N2|lsinkj=|N2|lsini,

W = ( 1lcosj+1lsink)(Mgk)2 2

=12Mglcosjk=12Mglcosi.

SubstitutingthesetorquesintoEquation(4.4)gives|N2|lsini12Mglcosi=0,

andresolvingthisinthei-directiongives1|N2|= 2Mgcot.

ResolvingEquation(4.3) inthej- and k-directionsinturngives|

N2

|+

|F

|= 0,

|N1

| Mg = 0.

Therefore1|F|=|N2|= 2Mgcot, |N1|=Mg.

Substitutingtheseintothe inequality(4.5)gives2MgcotMg,

which,onrearrangementandusing= 0.5,givescot1.

Thereforethemodelpredictsthattheminimumangletheladdercanmakewiththegroundbeforeslipping is radians(45).4*Exercise4.1

A model rodOAof length l and massm is fixed to a wall by a hinge, asshown in Figure 4.4. The rod is free to turn in a vertical plane about thehinge, which is assumed to besmooth (i.e. hasno friction forceassociatedwith it). The rod is supported in a horizontal position by a string ABinclinedatanangletothehorizontal.Findthereactionforceatthehingeandthetensionforceactingontherodduetothestring. Commentonthemagnitudesanddirectionsofthesetwoforces.Exercise4.2Alightladderoflength3.9mstandsonahorizontalfloorandrestsagainsta smooth vertical wall, as shown in Figure 4.5. The base of the ladder is1.5mfromthebaseofthewall. Thecoefficientofstaticfrictionbetweenthe ladderandthefloor is0.25. Theendrungsareeach0.3mfromanendofthe ladder. The laddermaybemodelledasarod, and itsmassmaybeneglected(asitisalightladder,soitsmassisnegligiblecomparedwiththemassesofanypeoplestandingon it).What istheminimummassofapersonstandingonthebottomrungthatprevents the ladder from slipping when a person of mass 100kg stands onthetoprung?

NotethatsinceM andg arepositive,wecansafelydividethroughbythemwithoutreversingtheinequality.Interpretresults

lqO

B

Am

Figure4.4

m= 0.25q m

0.3m

100 kg

3.9 m

1.5 m

0.3m

Figure4.533

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End-of-sectionExercisesExercise4.3Atabuildingsite,aplankoflength2lisrestingagainstalargesmoothpipeof radiusr, as shown in Figure 4.6. The angle between the plank and thehorizontal is 3 , so by symmetry the angle between the horizontal and thelinebetweenOandthecentreofthepipeis 6 ,asshown. Inthefigure,thedistanceOA is greaterthan the distanceAB, so the centre of massof theplank isbetweenOandA.

6

6

O

BA

mr 2l

Figure4.6What is the least coefficient of static friction between the plank and thegroundthatwillensureequilibrium?Exercise4.4Agangplankbridgesthegapbetweenaquay andaship, asshown inFig-ure 4.7. The pulley is directly above the quayside end of the plank. Arope isattachedtotheendoftheplank,thenpassesoverthepulleytoanattachment point on the quayside. When the ship is ready to depart, thegangplankisraisedbypullingontheropefromthequayside.

5 mm= 0.4

h

q120 kg

Figure4.7Using the information given in Figure 4.7, determine the minimum heightofthepulleyabovethequaythatensuresthatthegangplankwillberaised,ratherthanslipalongthequay,whentheropeispulled. Ifonemancanpullwithaforceofmagnitude800N(andassumingthatthepulleyissufficientlyhigh),howmanymenareneededtoraisethegangplank?

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Section4 Outcomes

Exercise4.5A door, of mass 20kg, height 2m and breadth 1m, is supported by twohingesatthetopandbottomofthedoorsothatitishangingintheverticalplane,asshown inFigure4.8. The forceexertedbythetophingetendstopullthedoorupwardsandtowardsthedoorframe. Theforceexertedbythebottomhingetendstopushthedoorupwardsandawayfromthedoorframe.

2 m

1 m

20kg

Figure4.8Iftheforcesexertedbythehingeshavethesamemagnitude,determinetheseforces incomponentform.

Outcomes

Afterstudyingthisunityoushouldbeableto: appreciatetheconceptofaforce; understand and model forces of weight, normal reaction, tension and

friction; recognizeandmodeltheforcesthatactonanobjectinequilibrium; modelobjectsasparticlesorasrigidbodies; usemodelstrings,modelrods,modelpulleysandmodelpivotsinmod-

ellingsystems involvingforces;

drawforce

diagrams,

and

choose

appropriate

axes

and

an

origin;

usetheequilibriumconditionsforparticlesandforrigidbodies; understandandusetorques; modelandsolveavarietyofproblems involvingsystems inequilibrium

andsystemsonthevergeof leavingequilibrium.

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