why do you feel heavier when you are in an elevator …...why do you feel heavier when you are in an...
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Normalforceinanelevator10kgtoddlerinanelevator:
Elevatorhas0velocityand0accelerationSumFy =0,FN =FG =98.1N
Elevatoratconstantvelocity(0acceleration)SumFy =0,FN =FG =98.1N
2m/s2 accelerationSumFy = +20NFN=118.1N
-2m/s2 accelerationSumFy =-20NFN =78.1N
Freebodydiagrams(FBD)(”ForceDiagrams”)DrawalltheforcesactingonanobjectNetforceisthevectorsumofalltheforces◦Gravity(FG)◦Directpushorpull(appliedforceFapp)◦Friction(f)◦Tension(FT):Strings,ropes,etc.◦Normalforce(FN)
Findingthenormalforce◦Normalforceiswhatisnecessarytokeepequilibriuminthedirectionperpendiculartothesurface
Normalforceinanelevator10kgtoddlerinanelevator:
Elevatorhas0velocityand0accelerationSumFy =0,FN =FG =98.1N
Elevatoratconstantvelocity(0acceleration)SumFy =0,FN =FG =98.1N
2m/s2 accelerationSumFy = +20NFN=118.1N
-2m/s2 accelerationSumFy =-20NFN =78.1N
FrictionisaForce
Force on person by box
Force on floor by box Force on boxby floor
It’s the sum of all the forces that determines the acceleration.Every force has an equal & opposite partner.
Force on box by person
Force on floorby person
Force on personby floor
FrictionMechanismCorrugations in the surfaces grind when things slide.
Lubricants fill in the gaps and let things slide more easily.
StaticandSliding(Dynamic)Friction• Staticfrictionalforce(fs):inthedirectiontopreventsliding• Kinetic(sliding)frictionalforce(fk):oppositethedirectionofsliding• Staticfrictionalforcesalwaysgreaterthankineticones
Friction Force = Normal Force ´ (coefficient of friction)Ffriction = µ×Fnormal
• Normalforceandfrictionforceareperpendiculartoeachother• Coefficientofstaticfriction>coefficientofkineticfriction• Fs <𝜇sFN :Fs willmatchforceofthepullinordertokeepobjectinstaticequilibrium
• MaximumFsmax =𝜇sFN :thentheobject“slips”
“Normal”ForcesandFrictionalForces
Weight of block
Decompose Vector
NormalForce
FrictionForce
Weight of block
ForceFrom Ramp “Normal” means
perpendicular
Friction Force = Normal Force ´ (coefficient of friction)Ffriction = µ×Fnormal
Janeispullinga10.0kgboxwithaforceof40.0Natanangleof30.0.Thecoefficientofkineticfrictionis0.1.A.DrawaFBDwithallforceslabeled.B.Determinethehorizontalacceleration(Hint:breakpullingforceintocomponents).Becareful:FN doesn’tequalFG!C.Howwouldthehorizontalaccelerationchangeifshepushedtheboxfrombehindwiththesameangleandforceinsteadofpulledit?D.Isiteasiertopushorpullanobject?
A.Whyisitmoredifficulttodoapull-upwith1armthan2arms?B.Ifyouweigh700N,Howmuchforceisin1armvs.2arms?◦1arm:700N,2arms:350NC.Ifbothofyourarmswereatanangleof30degreesfromthevertical,whatwouldtheforcebeineacharm?◦404N
TensionWhenaflexiblecordpullsonanobjectIfthecordismassless,thetensionwillbesamealongitsentirelengthIfpulleysareinvolved,assumetheyaremassless andfrictionlessRopesandcordscanonlypullForcealwaysactsalong(parallelto)theropeorcord
Bowserisusingpulleystoliftablocktothetopofhiscastle.Heisusingaropeloopedover2pulleys. Howmuchoftheblock’s2000Nweightdoeshehavetopullontherope?◦1000N
ProblemSolvingforForces1.DrawaFBDforeachobjectinvolved2.Identifytheforcesinbothdirectionsandmakeanetforceequation3.Iseitherdirectioninequilibrium?(Fnet =0)Aretherewayswecansimplifytheequation?4.Plugandchug!
Determinethecoefficientofkineticfrictionofapenguinslippingonflatseaweedifitsmassis21.8kgandforceduetofrictionis18.9N◦0.08