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ForcesActing onaParticleandRigid Body SqnLdr NS Dikkumburage BSc(Hons) Eng,AMIE(SL)
Recap… Static equilibrium of a particleandRigidBody
Recap… Equilibriumofaparticleinaplane Considera particlePintheXY planewithforcesF1,F2,F3 andF4 appliedas shown. For theparticleto be inastateofequilibrium(i.e. to havezeroacceleration, ΣFi=0 Each ofthe forcecanbeexpressed inxandyvector components asfollows, Fi=Fix+Fiy Henceconsideringequilibriumstateof theparticle,wecan write 2equilibriumequations ΣFix=0 andΣ Fiy=0
Recap… Equilibriumofaparticleinaplane Determinethetensioninthecablestosupport traffic lights B (10kg)andC(15kg)andthe angletheta.
Recap… EquilibriumofaRigidBodyinaplane Consider a RigidBody intheXY planewithforcesF1,F2,F3andF4 andmomentsMandM2 appliedas shown. For theparticleto be inastateof equilibrium(i.e. tohavezero accelerationfollowingconditionshouldbestatisfied, ΣFi=0 or ΣFix=0 andΣ Fiy=0 If only this condition satisfies, there can be a rotation around anyarbitrarypointof the rigid body.Henceforstatic equilibriumof a Rigidbody ΣMo=0 So wecanwrite3 equilibriumequationsfora rigidbody inaplane. 1. 2. 3. ΣFix=0 ΣFiy=0 ΣMo=0
Springs Inthis sectionweare consideringabout “LinearlyElasticSprings” Of a linearlyelasticspring,Force(F)actingon thespringdirectly proportionaltolengthchange(s). Fαs Wecanrewriteaboverelationshipas follows, F=ks The constantk, defines the elasticityofthe springwhichalsoknownas thespringconstant. Notethatthelengthchange/deformeddistancecanbeeithercompressedor elongatedlength fromitsunloadedposition .
Springs Ifunstressed lengthof thespring is0.8m and it isstretchedupto1.0m, findthe forcerequiredfor aboveaction. Youcanassumethespringconstant(k)=500N/m
Equilibriumofaparticle-2 Determinethelengthrequiredof thecordAC such that 8kglampcan besuspendedinthe positionshowninthe figure.Undeformed lengthof springis 0.4m andstiffness ofthespring iskAB=300 N/m
FREE BODY DIAGRAM Procedure for Drawing a Free-Body Diagram Since we must account for all the forces acting on the particle when applying the equations of equilibrium, the importance of first drawing a free-body diagram cannot be overemphasized. To construct a free-body diagram, the following three steps are necessary. Draw Outlined Shape. Imagine the particle to be isolated or cut “free” from its surroundings. This requires removing all the supports and drawing the particle’s outlined shape.
FREE BODY DIAGRAM Procedure for Drawing a Free-Body Diagram Show All Forces. Indicate on this sketch all the forces that act on the particle. These forces can be active forces, which tend to set the particle in motion, or they can be reactive forces which are the result of the constraints or supports that tend to prevent motion. To account for all these forces, it may be helpful to trace around the particle’s boundary, carefully noting each force acting on it. Identify Each Force. The forces that are known should be labeled with their proper magnitudes and directions. Letters are used to represent the magnitudes and directions of forces that are unknown.
Equilibriumofaparticle-2 Twoforcemembers
Equilibriumofaparticle-2 Threeforcemembers
Equilibriumofaparticle-2 Consider a particlePintheXYplanewithforcesF1,F2 and F3appliedasshown. Equilibriumofaparticlein3Dforce systems For theparticleto be inastateof equilibriumnecessary andsufficientcondition, ΣFi=0 For a 3-Dimensional force system we can resolve forces asfollows, ΣFxi+ ΣFyj+ΣFzk= 0 Aboveequationcan be satisfiedas follows, ΣFx=0 ΣFy=0 ΣFz=0
Statically determinacy Trytosolvethis problem!
Statically determinacy Toensuretheequilibriumof a rigidbody,itis to satisfytheequations of notonlynecessary equilibrium, butthebodymustalsobeproperlyheldor constrained byitssupports.Somebodies mayhavemoresupportsthanarenecessaryforequilibrium,whereas othersmay not haveenough orthe supportsmaybearrangedinaparticularmannerthat couldcausethe bodytomove
Statically determinacy Staticallyindeterminatemeans thattherewillbemoreunknownloadings onthe bodythan equationsof equilibriumavailablefor theirsolution Try tosolvethis problem! Having thesame numberofunknownreactiveforcesasavailable equationsof equilibrium does not always guarantee that a body will be stable when subjected to a particular loading. For example, the pin support at A and the roller support at B for the beaminFig.5–25aareplacedinsuchawaythatthelinesofactionofthereactive forces are concurrent at point A. Consequently, the applied loading P will cause the beamtorotateslightly about A, and sothebeamisimproperlyconstrained,.
Statically determinacy Trytosolvethis problem!
MomentofaForce When we apply a force on a body where line of action is distant fromapoint,thatbody tendstorotateaboutthepoint. This tendencyiscalledTorque,alsocalledthe Moment. Whichactionapplieshighesttorqueonthebolt?
MomentofaForce Moment=Forcex Lengthperpendicularfromthepoint Mo=Fd Momentisavectorwithmagnitudeandthedirection Themagnitudeofthe momentis determinedfromMo=Fd,wheredis called the moment arm,which representstheperpendicularorshortest distancefrom point Ototheline of actionof the force
MomentofaForce Alwaysremember touseappropriate sign convention, Assumeclockwisepositiveforthiscase, MRo =Σ Fd=- F1d1+ F2d2- F3d3
MomentofaForce FindthemomentaroundO.
MomentofaCouple Acouple is two parallelforcesactingonoppositedirectionsseparatedby perpendiculardistance. MomentproducedduetoforcecoupleisdefinedasCoupleMomentwhere magnitudeanddirectionsarecalculatedas follows, Mo=Fd
MomentofaCouple Calculatemomentcoupleforeachcase.
MomentofaCouple Whysteeringwheels ofavehicleare nowsmallercomparedtooldervehicles?
MomentofaCouple Determinemagnitudeanddirectionofeachcase
