理论力学教案11

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1、123Kinetics:Force-Mass-AccelerationMethoda.EquationofmotionNewton^secondlaw.EF=maComponents:=maxEFy=may=mazwhichiscalledequationsofmotion.Thismethodiscalledforce-mass-acceleration(FMA)methodb・MADMassaccelerationdiagram(MAD)Freebodydiagram(FBD)TheFMAmet

2、hodconsistsofthefollowingsteps:1Drawthefree-bodydiagram(FBD)oftheparticlethatshowsallforcesactingontheparticle・2Usekinematicstoanalyzetheaccelerationoftheparticle・3Sketchthemass-accelerationdiagram(MAD)fortheparticlethatdisplaystheinertiavectorma.4Refe

3、rringtotheFBDandMAD,relatetheforcestotheaccelerationofthetwodiagram・12>4DynamicsofRectilinearMotionEquationofmotionThecorrespondingequationsofmotionare=ma^Fy=0EF讦0FBDMADb・DeterminationofvelocityandpositionTheaccelerationwouldgenerallybeafunctionofposit

4、ionandvelocityoftheparticle,andthelime:«=f(^1)

5、Ax=f(x9x9t)Case1:a=f(t)・a==(=>dv=f{t)dt^=>v(r)=j+C}^=>Xr)=J(j+Qz+C2wheretheconstantsC

6、andC2canbeevaluatedfromtheinitialconditions.Case2:a=f(x).dvdvdxdva=—==v—dtdxdtdxvdv=f(x)dx=>^-=JfMdx+C=>v=j2[J/O)c&+C]v

7、=dx/dti=>dxj2[J/(x)dx+CJ=dti=>x(t)Case3:a=f(v).dvdvdxdva=—==v—dtdxdtdx«=>vdv-a(v)cbc^=^>dx-(vdv)/a(v)Sampleproblem12・5Thefigureshowsacrateofmassmrestingonthebedofadump(彳顷卸)truck・Thecoefficientofstaticfrictionbetweenthecrateandthebedofthetruckis0.64.Ino

8、rdertomakethecrateslidewhenthebedisinthepositionshown.Thetruckmustacceleratetotheright.Determinethesmallestaccelerationaforwhichthecratewillbegintoslide・Expresstheanswerintermsofthegravitationalaccelerationg.人=0.64Solution:TheFBDandMADofthecrateisshown

9、asthefollowings:Fromtheaboveequationsweobtain:Theequationsofmotionofthecrateare:2Fx-ma・Nsin3(T+0・64Ncos30°=maNcos30°+0・64Nsin30°・mg=0N=0.8432mgya=0.0458gSampleproblem12.7Theblockofmassminthefigureslidesonahorizontalplanewithnegligiblefriction.Thepositi

10、oncoordinatexismeasuredfromtheundeformedpositionoftheidealspringofstiffnessk.Iftheblockislaunchedatx=0withthevelocitytotheright,determine(1)theaccelerationoftheblockasafunctionofx;(2)thevelocityoftheblockasafunctionofx;and(3)thevalueofx