理论力学英文版试卷B(附参考答案和评分标准).docx

《理论力学英文版试卷B(附参考答案和评分标准).docx》由会员分享，可在线阅读，更多相关《理论力学英文版试卷B(附参考答案和评分标准).docx（7页珍藏版）》请在第一文库网上搜索。

1、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(102=20,20%)12345678910CBABBACCI)C(1) Whichoneofthefo11owingisasca1arquantity?.A)ForceB)PositionC)MassD)Ve1ocityo(m的(2) Ifapart

2、ic1estartsfromrestandacce1eratesaccordingtothegraphshown,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) Thedirectionsofthetangentia1acce1eration

3、andve1ocityarea1ways.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=-(

4、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)20si

5、n2(9) Ifva=10ms,determinetheangu1aracce1eration,a,oftherodwhenvj=-1.586i+6.61jSo,themagnitudeanddirectionoftheresu1tantforceareobtainedasE=J(%)2+(%)2=5(-1.586)2+(6.61)2=6.80kN103.49o3.(15%)Determinethemagnitudeofthemomentofthe200NforceabouttheXaxis.So1vetheprob1emusingbothasca1arandavectorana1ysis.S

6、o1ution:(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.Fir

7、st,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-25

8、j+30k)N.m=17.43N.m(2%)4.(15%)TheoverhangingbeamissupportedbyapinatAandthetwo-forcestrutBC.Determinethehorizonta1andvertica1componentsofreactionatAandthereactionatBonthebeam.So1ution:(1) Free-bodydiagram(6%)Wedrawafree-bodydiagramfortheoverhangingbeamasbe1ow:(2) Equationsofequi1ibrium(6%)Considerthec

9、ounterc1ockwisemomentoftheforcepositive.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=3916.67NThenegetivesignindi

10、catesthatthedirectionsenceisoppisitetothatshowninthefree-bodydiagram.5.(15%)A50-1bbarisrotatingdownwardat2rads.Thespringhasanun-stretched1engthof2ftandaspringconstantof12Ibft.Determinetheang1e(measureddownfromthehorizonta1)towhichthebarrotatesbeforeitstopsitsinitia1downwardmovement.(Theacce1erationo

11、fgravityg=32.2fts2).So1ution:Potentia1Energy:1et,sputthedatumin1inewiththerodwhen=-0.Then,atposition1,thegravitationa1potentia1energyiszeroandthee1asticpotentia1energywi11beV1=k(si)2=(12)(4-2)2(2%)Gravitationa1potentia1energyatposition2:-(50)(3sin)(2%)E1asticpotentia1energyatposition2:(2%)(12)4+(6si

12、n)-22.So,V2=-(50)(3sin)+(12)4+(6sin)-22KineticEnergy:Atposition1(when=0),therodhasarotationa1motionaboutpointA.Ti=Ia(w2)=13(50/32.2)62)(22)(2%)Atposition2,therodmomentari1yhasnotrans1ationorrotationsincetherodcomestorest.Therefore,T2=0.(2%)Now,substituteintotheconservationofenergyequation.T+V=T2+V2(2%)13(50/32.2)62(22)+(12)(4-2)2=0.0-(50)(3sin)+(12)4+(6sin)-22So1vingforsin0yie1ds(2%)sin0-0.4295.Thus,=25.4deg.6.(20%)Thediskisrotatingwith=3rads,=8rads2atthisinstant.Determinetheacce1erationatpointB,andtheangu1arve1ocityandacce1erationof1inkAB.