理论力学英文版试卷B(附参考答案和评分标准).docx
XXX工业大学20172018学年第二学期期末考试试卷学院班级姓名学号理论力学课程试卷B（附参考答案和评分标准）Instructor:1inZongxi(Time:2Hours)CourseCode:题号123456总得分题分201515151520得分1.Choosethecorrectanswerswithproperjustication(10×2=20,20%)12345678910CBABBACCI)C(1) Whichoneofthefo11owingisasca1arquantity?.A)ForceB)PositionC)MassD)Ve1ocityo(m的(2) Ifapartic1estartsfromrestandacce1eratesaccordingtothegraphshown,thepartic1e,sve1ocityatt=20sis.A) 200m/sB) 100msC)0D)20m/s(3) Apartic1ehasaninitia1ve1ocityof3m/stothe1eftatso=0m.Determineitspositionwhent=3siftheacce1erationis2ms2totheright.A)OmB)6.0mC)18.0mD)9.0m(4) Thedirectionsofthetangentia1acce1erationandve1ocityarea1ways.A)perpendicu1artoeachother.B)co11inear.C)inthesamedirection.D)inoppositedirections.(5) Ca1cu1atetheimpu1seduetotheforceF.A) 20kgmsB) 10kgmsC) 5NsD) 15Ns(6) Asshownintherightfigure,twob1ocksareinterconnectedbyacab1e.Whichotthefo11owingiscorrect?.A) va=-vbB) (vx)a=-(vx)bC)(Vy)A=-(Vy)D)A11oftheabove.(7) Ifapartic1emovesinthex-yp1ane,itsangu1armomentumvectorisinthe.A)xdirection.B)ydirection.C)Zdirection.D)x-ydirection.(8) Iftheposition,s,isgivenasafunctionofangu1arposition,O1by5=10sin20,theve1ocity,v,is.(No1ingthat=t,istheangu1arve1ocityattimet).A)20cos2B)20sin2C)20cos20D)20sin2(9) Ifva=10ms,determinetheangu1aracce1eration,a,oftherodwhen<9=3O0.,A)0rads2B)-50.2rads2C)-I12rads2D)-173rads2(10) As1enderrod(mass=M)isatrest.Ifabu11et(mass=m)isfiredwithave1ocityof,theangu1armomentumofthebu11etaboutAjustbeforeimpactis.A) 0.5inVb2B) Vb05C) 0.5mvb,ob-1GD)zero2.(15%)Determinethemagnitudeoftheresu1tantforceactingonthescreweyeanditsdirectionmeasuredc1ockwisefromthexaxis.So1ution:Usingthevectorana1ysis,wehavethex-andy-componentsofforces6kNand2kNas%=-(6kN)cos600=-3kNE=(6kN)sin60o=33kN=5.196kN)：1(4%)r=(2kN)cos450=2kN=1.414kNF2y=(2kN)sin45o=kN=1.414kNHence,theforcevectorsareobtainedasF1=fJJ+fjvj=-3i+5.196j(4F2=&i+/j=1414i+1.414jAccordingtothevectoradditionru1e,theresu1tantforceisexpressedinthevectorformFR=E+F?=(%+%)i+(%+%)j(3o)=+7>vj=-1.586i+6.61jSo,themagnitudeanddirectionoftheresu1tantforceareobtainedasE=J(%)2+(%)2=5(-1.586)2+(6.61)2=6.80kN103.49o3.(15%)Determinethemagnitudeofthemomentofthe200NforceabouttheXaxis.So1vetheprob1emusingbothasca1arandavectorana1ysis.So1ution:(1) Sca1arana1ysis.Thexyyandz-componentsoftheforceFcanbeexpressedasFx=Fcosa=(2()0N)cos120o=-I(X)NFv=FCOS夕=(200N)COS600=IOON(3%)G=FcoSy=(200N)cos45o=173.2NThus,themagnitudeofthemomentoftheforceFaboutthexaxisisobtainedasMx=-2+EJ3=(-100N)(0.25m)+(173.2N)(0.3m)=17.43N.m(4%)(2) Vectorana1ysis.First,weestab1ishapositionvectorfromoriginOtopointAontheforce1ineofaction:rO4=r4=0.3j+0.25k(2%)AndthevectorofforceFisF=Fvi+FJ+Ek=-1(X)i+1(X)j+173.2k(2%)Hence,themomentofforceFaboutpointOyie1ds1 jkMo=00.30.25=17.43i-25j+30k(2%)-I(X)100173.2AndthemagnitudeofMoaboutthex-axisisobtainedasMx=i.M0=i(17.43i-25j+30k)N.m=17.43N.m(2%)4.(15%)TheoverhangingbeamissupportedbyapinatAandthetwo-forcestrutBC.Determinethehorizonta1andvertica1componentsofreactionatAandthereactionatBonthebeam.So1ution:(1) Free-bodydiagram(6%)Wedrawafree-bodydiagramfortheoverhangingbeamasbe1ow:(2) Equationsofequi1ibrium(6%)Considerthecounterc1ockwisemomentoftheforcepositive.Accordingtothefree-bodydiagram,wehavex-andy-componentequationsandmomentaboutpointAofequi1ibriumasfo11ows2EFX=0,F-Fb=04=0,%+技区-(600N)-(800N)=0ZMA=0,-(600N)(1m)+y>(2m)-(800N)(4m)-900N.m=0Afterso1ving,wehave(3%)FAX=3133.33N,FAy=-950N,Fb=3916.67NThenegetivesignindicatesthatthedirectionsenceisoppisitetothatshowninthefree-bodydiagram.5.(15%)A50-1bbarisrotatingdownwardat2rads.Thespringhasanun-stretched1engthof2ftandaspringconstantof12Ibft.Determinetheang1e(measureddownfromthehorizonta1)towhichthebarrotatesbeforeitstopsitsinitia1downwardmovement.(Theacce1erationofgravityg=32.2fts2).So1ution:Potentia1Energy:1et,sputthedatumin1inewiththerodwhen=-0.Then,atposition1,thegravitationa1potentia1energyiszeroandthee1asticpotentia1energywi11beV1=k(si)2=½(12)(4-2)2(2%)Gravitationa1potentia1energyatposition2:-(50)(3sin)(2%)E1asticpotentia1energyatposition2:(2%)½(12)4+(6sin)-22.So,V2=-(50)(3sin)+½(12)4+(6sin)-22KineticEnergy:Atposition1(when=0),therodhasarotationa1motionaboutpointA.Ti=½Ia(w2)=½13(50/32.2)62)(22)(2%)Atposition2,therodmomentari1yhasnotrans1ationorrotationsincetherodcomestorest.Therefore,T2=0.(2%)Now,substituteintotheconservationofenergyequation.T+V=T2+V2(2%)½13(50/32.2)62(22)+½(12)(4-2)2=0.0-(50)(3sin)+½(12)4+(6sin)-22So1vingforsin0yie1ds(2%)sin0-0.4295.Thus,=25.4deg.6.(20%)Thediskisrotatingwith=3rads,=8rads2atthisinstant.Determinetheacce1erationatpointB,andtheangu1arve1ocityandacce1erationof1inkAB.
