Engineering Basic Mechanics I Statics 工程基础力学Ⅰ 静力学 课件 Chapter 5 Friction

StaticsStaticsofrigidChapter5Friction§5.1Slidingfriction§5.2Frictionangleandself-lockingphenomenon§5.3Equilibriumproblemswithfriction§5.4RollingresistanceMaincontents1.Concept:Twoobjectswithroughsurfaceincontactwitheachother,whenthereisarelativeslidingorrelativeslidingtrendbetweenitscontactsurface,thereisresistancetoeachotherthathindersrelativesliding,thatisslidingfriction.5.1Slidingfriction2.Status:

(1)Static:(equilibriumconditions)

(2)Critical:;fs—thecoefficientofstaticfriction(3)sliding:f'—thecoefficientofkineticfriction5.2Frictionangleandself-lockingphenomenon1.FrictionangleIfthemagnitudeofthestaticforceoffrictionincreasesfromzerotothemaximumvalue,themaximumanglewhichthetotalreactionofaroughsupportmakeswiththenormaltothesurfaceiscalledtheangleoffriction.Calculation:5.2Frictionangleandself-lockingphenomenon2.Self-lockingphenomenonAbodycanbeatrestbythemutualactionoftheforceoffrictionandthenormalpressure(i.e.,fullReactionForce),itwillnotslide(nomatterhowgreatanappliedforceis),thisphenomenoniscalledself-locking.Conditionsofself-locking:Whenthesystemisalwaysinequilibrium(i.e.,itisself-locking).5.3EquilibriumproblemswithfrictionThefrictionforceisusuallyunknown,whilethedirectionoffrictionforceisoppositetothedirectionoftherelativesliding

tendency.Tosolvetheequilibriumproblemwiththefrictionforce,itisnecessarytolistthesupplementaryequationsFs≤fFN,andthenumberofsupplementaryequationsisthesameasthenumberofthefrictionforces.Example1haαADGFThehomogeneouswoodenboxweighsG=5kNandhasastaticfrictionfactorfs=0.4withtheground,h=2a=2mand.(1)AskwhethertheboxisinequilibriumwhenthetensionF=1kNatD?(2)Findthemaximumtensionforcethatwillkeeptheboxinbalance.5.3EquilibriumproblemswithfrictionTakethewoodenboxastheobjectofstudy,forceanalysisasshowninthefigure.hdaαADGFfFNFEquilibriumequationsBecauseFf<Fmax,sothewoodenboxdoesnotslide.Andbecaused>0,thewoodenboxwillnottoppleover.Solution：Tomaintainthebalanceofthewoodenbox,itisnecessaryto(1)Cannotslide,Ff<Fmax=fsFN.(2)DonottopplearoundpointA,d>0.SolvingtheequationyieldsThemaximumfrictionbetweenthewoodenboxandthegroundis5.3EquilibriumproblemswithfrictionEquilibriumequationsTheconditionsunderwhichthewoodenboxslidesare

Ff=Fmax=fsFNhdaαADGFfFNFThesolutionisTheconditionthatthewoodenboxtopplesoveraroundpointAisd=0,then2.Tofindthemaximumtensileforceatequilibrium,thatis,tofindtheminimumforce

Fintheslidingcriticalandoverturningcritical.F=FF=1443NSinceFF<Fs,themaximumtensiontokeepthewoodenboxinbalanceis5.4Rollingresistance(a)(b)(c)(d)Whenthewheelisequilibrious,theequilibriumequationsThecouplemomentMf

calledrollingresistancecouplemoment5.4RollingresistanceMf

rollingresistance(1)therollingresistancecouplemomentMfincreaseswiththeincreaseoftheappliedforce;(2)

therollingresistancecouplemomentreachesitsmaximumvalue,calledthemaximumrollingresistantcouplemomentMfmax;IfFbecomelarger,thewheelwillroll,andtherollingresistantcouplemomentisapproximatelyequaltoMmaxduringtherollingprocess,andthereis5.4RollingresistanceMf

rollingresistanceRollingresistancelaw:ThemaximumrollingresistancecouplemomentMmaxisindependentoftherollerradiusandproportionaltothemagnitudeofthenormalforceFNonthesupportsurface,

discalled

thecoefficientofrollingresistance:(1)Itsdimensionisthatofalength.Generally,theunitismmorcm;(2)Itd