210-06 Kinetics of Particles
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Transcript of 210-06 Kinetics of Particles
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4/222/22
Introduction
DipakaidilingkungansendiriCurrentchapterintroducestwoadditionalmethodsofanalysis..amFr
Methodofimpulseandmomentum:directlyrelatesforce,Methodofworkandenergy:directlyrelatesforce,mass,=
r
mass,velocity,andtime.velocityanddisplacement.solvedthroughthefundamentalequationofmotion,Previously,problemsdealingwiththemotionofparticleswereDipakaidilingkungansendiriakademik.pesertakuliahDinamikaPartikelTMS-210yangdapatdiunduhdariportalBahanajarinidipakaidilingkungansendiridandisediakansecaragratisbagi2010.EngineeringMechanics:Dynamics,12Ed.,PrenticeHall,NewJersey,
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/3/.Hibbeler,R.C.EngineeringMechanics:Dynamics,6th.Ed.,JohnWiley,2008./2/.Meriam,J.L.;Kraige,L.G.MechanicsforEngineer:Dynamics,5thEd.,McGraw-Hill,NewYork,2008/1/.Beer,F.P.;Johnston,E.R.sepertitercantumberikutini:kepadapengalamanpenulissertamerujukkepadabeberapabukustandarmahasiswaJurusanTeknikMesinUniversitasAndalasyangberdasarkanBahanajarinidibuatuntukmemenuhikebutuhanbahanbacaanbagiparaUnandLDS6210TMSUnandLDS6210TMS
3/22
ContentsMulyadiBur
DipakaidilingkungansendiriANDALASUNIVERSITYStructuralDynamicsLaboratory
DYNAMICSDipakaidilingkungansendiriSampleProblem13.3SampleProblem13.2SampleProblem13.1PowerandEfficiencyApplicationsofthePrincipleofWork&EnergyPrincipleofWork&EnergyWorkofaForceIntroductionENGINEERINGMECHANICSUnandLDS
6210TMSUnandLDS6210TMS()Workofaforceduringafinitedisplacement,WorkofaForce
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2
2s
a6/22
proportionaltodeflection,MagnitudeoftheforceexertedbyaspringisWorkofaForceplottedagainsts.Workisrepresentedbytheareaunderthe1==cosdsFdsF
AA21
=rdFr
r
2ssAA
tcurveofF
21
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U
DipakaidilingkungansendiriUnandLDS6210TMS
Workofaconstantforceinrectilinearmotion,5/22Dimensionsofworkareforce.length
WorkofaForce()()()J1.356lb1ftm1N1J1==joulemagnitudeandsignbutnotdirection.Workisascalarquantity,i.e.,ithascos()xFUD=dsF==rdFdU
ra8/22negativeofareaundercurveofFplottedagainstx,Workoftheforceexertedbythespringisequalto,i.e.,whenthespringisreturningtoWorkoftheforceexertedbyspringispositive22
2
1
12
21
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()xFFUD+-=itsundeformedposition.21Workoftheforceexertedbyspring,()
211x
-=-=2
21kxkxdxkxU
x
1
212
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Workoftheweightisequaltoproductof12()D-=--=yWyyWdyWdyWdzFdyFdxFdUWorkoftheforceofgravity,cosr
1y
2yzyx-=-=++=
21
21UUnitsareWorkoftheforceisWorkofaForceDipakaidilingkungansendiriDipakaidilingkungansendiriUnandLDS6210
TMSUnandLDS6210TMS=0):
WorkofaForceDipakaidilingkungansendiriForceswhichdonotdowork(ds=0orcosa12/22ForceswhichdonoworkareeliminatedfromAllquantitiesarescalarsandcanbeadded
expressionforaccelerationandintegrating.ForceactsnormaltopathanddoesnoWishtodeterminevelocityofpendulumbob10/22
Velocityfoundwithoutdetermining22
glv
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v
2gW21
=
2=+Wl
theproblem.02211=+directly.TUTPr
work.
.Considerwork&kineticenergy.2atA
DipakaidilingkungansendiriApplicationsofthePrincipleofWorkandEnergyhorizontally.weightofabodywhenitscenterofgravitymovesreactionatarollermovingalongitstrack,andmovesalongsurface,
reactionatfrictionlesssurfacewhenbodyincontactreactionatfrictionlesspinsupportingrotatingbody,UnandLDS6210TMSUnandLDS6210TMS
11/22TheworkoftheforceisequaltothechangeinenergykineticmvTTTU9/22
==JmNmUnitsofworkandkineticenergyarethesame:
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dsdvmv12rdtdvMmG==-=12
2mvmvdvvmdsFdvmvdsFdt2dssmkg
21dsdvr
kineticenergyoftheparticle.MmmttmmaFGdrdr
212,=====2
2
r
toAMm
tConsideraparticleofmassmacteduponby21IntegratingfromA
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2==mvT1
12pathshown),occupiesfixedpositionOwhileparticlemfollowsWorkofagravitationalforce(assumeparticleMsWorkofaForceDipakaidilingkungansendiriParticleKineticEnergy:PrincipleofWork&DipakaidilingkungansendiriUnandLDS6210TMS
UnandLDS6210TMSW746
slbft550hp1orsm
N1
dtrdFr
sJrPowerandEfficiencyDipakaidilingkungansendiri1(watt)W1=
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inputpower
outputpowerinputworkkoutputworefficiency===UnitsforpowerareDimensionsofpowerarework/timeorforce*velocity.vFr
=rdt
==dU=
===h16/22
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ft418=x14/22
()=-0lb1151lbft481000x
x
2211=+TUT()()()()xxlb11515sinlb4000lb1500-=+-=
21equalthekineticenergychange.
DeterminethedistancerequiredfortheworktoSampleProblem13.1Dipakaidilingkungansendiri()()lbft481000882.324000sft88U2
s3600h
mi
21ft5280
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1v
UnandLDS6210TMSUnandLDS6210TMS
15/22requiredfortheworktoequal13/22thekineticenergychange.,
DeterminethedistanceEvaluatethechangein2
3W
kineticenergy.lgl
2
SOLUTION:g=+=Wl22SampleProblem13.1v
WP
gW=-WPnn=amF
AsthebobpassesthroughAapplicationofNewtonssecondlaw.
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ofworkandenergywithanrequiressupplementingthemethodCalculatingthetensioninthecordaccelerationofthependulumbob.beappliedtodirectlydeterminetheinclineataspeedofPrincipleofworkandenergycannot
Dipakaidilingkungansendiriautomobileasitcomestoastop.Determinethedistancetraveledbythecausingaconstanttotalbreakingforce60mi/hwhenthebrakesareapplieddrivendowna5oAnautomobileweighing4000lbisDipakaidilingkungansendiri=
2of1500lb.glv2ApplicationsofthePrincipleofWorkandEnergyUnand
LDS6210TMSUnandLDS6210TMS2
2()()()()kg200m2N490m2vF2
12AAC1()=-()()m2m20
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vmFFN490N196225.02211()()2
:
TUTWNFN1962sm81.9kg200C=-+
=+AkAkAm
m()()======AWtoblocksAandB.ApplytheprincipleofworkandenergyseparatelySOLUTION:SampleProblem13.2
forthevelocity.cableforcescancel.Solvecombined,theworkoftheWhenthetworelationsareblocksAandB.andenergyseparatelytoApplytheprincipleofworkSOLUTION:SampleProblem13.2Dipakaidilingkungansendiriweightlessandfrictionless.
=0.25andthatthepulleyism20/22unknownintherelationistheandenergyforthereboundofcompressedandthevelocityisandenergybetweentheinitial18/222velocityatthefinalposition.
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thepackage.TheonlyApplytheprincipleofworktherelationisthefrictionzero.TheonlyunknowninwhichthespringisfullypositionandthepointatApplytheprincipleofwork2()()()()kg300m2N2940m2vF2
1
2BBc1
coefficient.
SOLUTION:()()m2m20vmWF2
SampleProblem13.3:
2211TUTN2940sm81.9kg300c
=+-=+-
=+==BW
Dipakaidilingkungansendiripassesagainthroughthepositionshown.and(b)thevelocityofthepackageasitfrictionbetweenthepackageandsurfaceDetermine(a)thecoefficientofkineticis40mm.andthemaximumdeflectionofthespringavelocityof2.5m/sinthepositionshowncompressed120mm.Thepackagehas
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andisheldbycablessothatitisinitiallyThespringhasaconstantk=20kN/mwhichisslidingonahorizontalsurface.Aspringisusedtostopa60kgpackageDipakaidilingkungansendiriUnandLDS6210TMSUnandLDS6210TMS
19/222
sm43.4=v()()()()()v
Whenthetworelationsarecombined,thework17/22
222
()v
2kg500J490012kg300kg200m2N490m2N294012()()()()
kg300m2N2940m2vF12()()()()kg200m2N490m2vF1
=+=-=-
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c=+-Cvelocity.ofthecableforcescancel.SolvefortheSampleProblem13.2Dipakaidilingkungansendirik
theplaneiscoefficientoffrictionbetweenblockAandAafterithasmoved2m.Assumethatthefromrest,determinethevelocityofblockcableasshown.IfthesystemisreleasedTwoblocksarejoinedbyaninextensibleUnandLDS6210TMSUnandLDS6
210TMS()()()J112J37732
ef()2
1
32
21
+=J36.5k323232+-=+=
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UUUm32kg600vmvTT===reboundofthepackage.ApplytheprincipleofworkandenergyfortheSampleProblem13.3
()()()()kkm2()()0J5.187sm5.2kg602
21
k12m-=-=J377m640.0sm81.9kg602
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-=msm103.13=v
32
()v
2kg60J5.360
:
1
=+
3322=+TUT
k()()()J112J377ef212121--=+=UUUmLDS
20.0=km
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=-()()J0.112m040.0N3200N2400maxmin()()()()N3200m160.0mkN20k
()m22/22
DipakaidilingkungansendiriUnand6210TMS
21/22
0J112J377-J5.187:
2211()()N2400m120.0mkN202-=+-=12D+-=1
=+21
TUTe()f
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()xWU1
210max==D+=0min===21====TmvTcompressed.positionandthepointatwhichspringisfully
xPPUxxkPkxP
ApplyprincipleofworkandenergybetweeninitialSOLUTION:SampleProblem13.3DipakaidilingkungansendiriUnand
LDS6210TMS