昆工大一理论力学练习题答案

1第一章静力学公理和物体的受力分析一、是非判断题1.1.1(∨;1.1.2(×;1.1.3(×;1.1.4(∨;1.1.5(×;1.1.6(×1.1.7(×;1.1.8(∨;1.1.9(×;1.1.10(×;1.1.11(×;1.1.12(×1.1.13(∨;1.1.14(×;1.1.15(∨;1.1.16(∨;1.1.17(×:1.1.18(×二、填空题1.2.1外内;1.2.2约束;相反;主动主动;1.2.3C。三、受力图1.3.1q(cP2(a(bCBA设B处不B(f(g(h有销钉C;34三、计算题2.3.12.3.22.3.3qaB(bDkNFFFX98340304540301.coscos-=++=∑kNFFFY13587304503201.sinsin=++=∑5020.cos'==∑RFXα8650.cos'==∑RFYβ03201003025104525coscos(FFFFMMi-+==∑kNFFRR96678.'==cmFMdR7860.==解:kNYXFR9667822.(('=+=∑∑cmkN⋅=584600.40353035FF-+sin力系的最后合成结果为:0=∑X解:取立柱为研究对象:0=+qhXA(←-=-=⇒kNqhXA200=∑Y0=--FPYA(↑=+=⇒kNFPYA1000=∑AM022=--FaqhMAkNmFaqhMA1303010022=+=+=⇒0=∑X解:取梁为研究对象:0=AX0=∑Y0=--qaqaYA(↑=⇒qaYA20=∑AM2222=--qaqaMA2222522qaqaqhMA=+=⇒=∑X解:取梁为研究对象:=AX52.3.42.3.52.3.6Eq080=-+-FYqYBA.(-↓=⇒kNYA30=∑AM042612802=-++FYMqB...(..(.↑=+--=⇒kNYB624488806112yDEP1解:取DE杆为研究对象:∑=0X0600=+cosACBFX∑=0BM522601102=⨯-⨯+⨯.sinPFPAC∑=0Y(..↑=⨯-+=⇒kNPPYB2502364612kNPPFAC646523121..(=-=⇒(32.360cos0←-=-=⇒kNFXACB∴杆AC受压(kNFFACAC646.'==060102=-+-PFPYACBsin解:取CD为研究对象∑=0X0=CX∑=0CM(↑=⇒kNYD15∑=0Y取AC为研究对象:∑=0X0===CCAXXX'∑=0BM(↓-=⇒kNYA15∑=0Y(↑=⇒kNYB400412=--⨯DYMq02=+-DCYqY(↑=⇒kNYC50212=-⨯-CAYqY'2-02=--+CBAYqYY'62.3.72.3.81取起重机为研究对象:∑=0X∑=01OM052112=-+⨯-PYP(.↑==⇒kNYC67416250(↑=⇒kNY502∑=02OM(↑=⇒kNY101032111=-⨯PYP-2取CD段为研究对象:0=CX∑=0CM(.↑==⇒kNYD3386500612=+⨯DYY'-∑=0DM062=-CYY'5∑=0X3取AC段为研究对象:0=+-CAXX'0===⇒CCAXXX'∑=0AM(↑==⇒kNYB100330006531=--CBYYY''∑=0BM03231=---CAYYY''(.↓-=-=⇒kNYA33486290∑=0X解:取整体为研究对象:∑=0AMCYXBYBYDYAY取DF杆为研究对象:y∑=0X0=BX取AB杆为研究对象:∑=0EM∑=0DM0=DaX0=-DEXX'0===⇒DDEXXX'0=-MaYE(↑==⇒kNaMYE20=-MaYD'('↓==⇒kNaMYD20=⇒DX解:取整体为研究对象:72.3.92.3.10CY∑=0X∑=0Y∑=0EM0=+-EBXPX(.←-=-=⇒kNXPXEB800303040=--PYXBB(..↓-=⨯-=⇒kNYB871308070∑=0CM04080=-EXP(.→==⇒kNPXE612∑=0EM04080=+CXP(.←-=-=⇒kNPXC612取BE杆为研究对象:y0=+EBYY(.↑=-=⇒kNYYBE871∑=0Y0=+ECYY(.↓-=-=⇒kNYYEC871BX解:取整体为研究对象:∑=0BM05315412=++-PPYMA-((↑=+-=⇒kNPMYA15721∑=0AM02531541=+-+BYPPM-((↑=-=⇒kNPMYB55121BXXCX取BC杆为研究对象:∑=0CM022531541=++-BBXYPP-((→=-=⇒kNYPXBB225721∑=0Y053=+-BCYPY(↑=-=⇒kNYPYBC153∑=0X054=+-BCXPX(→=-=⇒kNXPXBC654取AC杆为研究对象:∑=0EM11=+ACXX'('←-=-=⇒kNXXCA6AYBY解:取整体为研究对象:∑=0X03003=-sinFXA82.3.11第三章空间力系1∑=0BM0302340321=+++-cosaFaFaFaYA(.(↑=++=⇒kNYA8321310403041y(sin→==⇒kNFXA103003∑=0Y0300321=+---BAYFFFYcos(..↑=-++=⇒kNYB492583213102010沿4、5和6杆截开,取左半部分为研究对象:C∑=0CM04=+-aFaYA(拉.83214==⇒AYF∑=0Y045051=--cosFFYAF5F4F4F5F10F7拉(..(kNF731610832125=-=⇒沿4、5、7和10杆截开,取右半部分为研究对象:∑=0Y0304503705=+-+BYFFFcossin压(-.-.kNF20492583113107=-=⇒∑=0X030450310054=---sincos-FFFF(.--.-.-压6643108311832110kNF==⇒取节点C1为研究对象:∑=0X∑=0Y065=-+FFFαcosC17F07=⇒F取节点C为研究对象:F64=⇒F∑=0Y06=⇒F取节点B1为研究对象:∑=0Y6(.拉52355MNFF==⇒1解:沿1、2和3杆截开,取右半部分为研究对象:0112=-FLFL∑=01AM(拉2341MNFF==⇒∑=0Y02=-FFαcos(.拉52352MNFF==⇒02132=-FLFL-∑=0AM(-压4383MNFF-==⇒9三、计算题3.3.13.3.23.3.3RxF'解:510013100N3345.-=51002002001310020030032⨯⨯=--==∑--cossinβαFFXRyF'NFY6249131003003002.cos=⨯===∑αRzF'NFFZ5610510010020010031.cos=⨯-=-==∑β(...'NkjikZjYiXFR561062493345∑∑∑++-=⋅+⋅+⋅=∴xM0Nm7951.-=510010020013100300300301032⨯⨯⨯⨯=--==∑0.3--0.1sin.cos.βαFFMxyM0NmFFMy64361310020030010020102021.0.1-.sin..-=⨯⨯⨯-=-==∑αZM0Nm59103.=510020020013100200300303032⨯⨯+⨯⨯=+==∑0.30.3cos.sin.βαFFMZ(...NmkjikMjMiMMZyx59103643679510∑∑∑+--=⋅+⋅+⋅=∴解:取销钉D为研究对象:∑=0Y∑=0X0454500=-coscosADBDFFADFBDFCDFADBDFF=⇒0153045304500000=---coscossincossinCDADBDFFF∑=0Z0153045304500000=----PFFFCDADBDsinsinsinsinsin由(a式:(cosaFFFCDADBD61520-==⇒(拉.sincos(kNPFCD46331531500=-=⇒(压.kNFFADBD3926-==⇒将(a式代入得:解:由空间力偶系的平衡方程(3-20式:y103.3.43.3.5∑=0xM900=--ACMMcos(α(cos(aF030090100=--⇒αCMBMAMx∑≡0ZM自然满足∑=0yM900=--BCMMsin(α(sin(bF040090100=--⇒α:((ba43400300909000==--sin(cos(αα43900=-⇒(αctg0013534390.==-⇒arcctgα013143.=⇒α由(a式:NF506030135330*********===-=..coscos(α解:取传动轴为研究对象。0cos2=-MdFαkNdMF67122017301030220.cos.cos=⨯==⇒α∑=∴0yM∵传动轴绕y轴匀速转动03420220=+BZF.sin.α∑=0xM(..sin.↓-=-=⇒kNFZB792342020220003420220=-BXF.cos.α∑=0zMkNFXB6673420202200..cos.==⇒0=+-BAXFXαcos∑=0XkNXFXBA254200.cos=-=⇒0=++BAZFZαsin∑=0Z(.sin↓-=--=⇒kNZFZBA541200由对称性得:0=cy解:由均质物体的形心坐标公式(3-30式用负面积法:113.3.63.3.7(a(b由对称性得:0=cx212211AAyAyAAyAycciCiic++==∑∑(a解:由均质物体的形心坐标公式(3-30式用负面积法:mm086.=(((141817247314182171724⨯-+⨯+⨯⨯-+⨯⨯=212211AAxAxAAxAxcciCiic++==∑∑(((2222222220rRRrrRRrR--=⋅-+⨯⋅-+⨯=ππππ由对称性得:0=cy(b解:由均质物体的形心坐标公式(3-30式用分割法:21522022023215122201220⨯+⨯+⨯⨯⨯+⨯⨯+⨯⨯=321332211AAAxAxAxAAxAxccciCiic++++==∑∑mm11=12第四章摩擦四、计算题4.4.14.4.2F解:由均质物体的形心坐标公式(3-30式用分割法:212211VVxVxVVxVxcciciic++==∑∑1040406040806010404020604080⨯⨯+⨯⨯⨯⨯⨯+⨯⨯⨯=mm0823.=212211VVyVyVVyVycciciic++==∑∑1040406040802010404040604080⨯⨯+⨯⨯⨯⨯⨯+⨯⨯⨯=mm4638.=212211VVzVzVVzVzcciciic++==∑∑104040604080510404030604080⨯⨯+⨯⨯-⨯⨯⨯+-⨯⨯⨯=((mm0828.-=解:取物块A为研究对象∑=0X0=+-θcos-TSFFFNFFFTS660.cos=+-=⇒θNFNFT66823108.cos=⨯=<=θ∴物块A有向右滑动的趋势,FS指向左边;∑=0Y0=+-θsin-TNFFWNFWFTN2=+-=⇒θsinFy∴最大摩擦力为:NFfFNs40220..max=⨯==NFNFs66040..max=<=∴物块A不平衡。解:欲保持木块平衡,必须满足1不会向右滑动,2不会绕D点翻倒。F134.4.34.4.41木块不会向右滑动:取木块为研究对象0=-SFF∑=0XSFF=⇒∑=0Y0=-PFNNPFN100==⇒若木块不会向右滑动,则应有:NFfFFFNSS4010040=⨯==≤=.max2木块不会绕D点翻倒:取木块为研究对象设木块处于临界状态,受力图如图所示。ySF∑=0DM016050=⨯-⨯FPNPF2531165.==⇒NF2531.≤∴解:取BO1杆和AC杆为研究对象;NFAOBOFC3002005075011=⨯==⇒..∑=01OM∑=0DM11=⋅-⋅FBOFAOC取KE杆和EDC杆为研究对象;m5250.θ=⋅-⋅CKFCDFKD'sinθNEDKEFFCK5300300=⨯==⇒θ'取O2K杆和闸块为研究对象并设初始鼓轮顺时针转动∑=02OM22=⋅-⋅NKFLOFKOθcosKEKKDLOFKOFKN505300122.cos⨯==⇒θN12005250505053001=⨯=...∑=0OM取鼓轮研究对象;0=+SSRFRF"'SSFF"'-=⇒形成制动力偶∴制动力矩为:NmFfRRFMNSSf30022=⋅=='解:取轮为研究对象;yo144.4.4第五章点的运动学四、计算题5.4.1θsin2PFS=⇒∑=0X0sin2=-SFPθ∑=0OM02=+--rPRFMSf∑=0Y0cos12=+-NFPPθθcos21PPFN-=⇒sin(22θRrPrPRFMSf-=+-=⇒解:取物块为研究对象;NFFt81.25930cos300cos0===θNPFFS81.9sincos=-=⇒θθ∑=0T0sincos=--SFPFθθ∑=0N0cossin=+--NFPFθθNPPt25030sin500sin0==-=θttPF>:可见∴物块有沿斜面向上滑动的趋势,则设FS的方向如图:NPFFN01.583cossin=+=⇒θθNFNfFFSN81.921.23301.5834.0max=>=⨯==又∴上面所求摩擦力正确,即:NFS81.9=方向如图。toAx5cos20cosπθ==∴tt5πωθ=⋅=tACoAy5sin10sin2sinπθθ=-=D点的运动方程⎭⎬⎫155.4.25.4.3第六章刚体的简单运动1直角坐标法:x解:建立参考系如图,由于顶杆作平动,所以由顶杆上的A点的运动方程:ϕϕϕ2222cossinsin2eReCDACeADODoAy-+=-+=+==teRteyωω222cossin++=⇒为顶杆的运动方程。顶杆的速度为:teRtteteyvωωωωωω2222cos2sin(cos2cos--⋅-+==,2cos1(tRxω+=tRyω2sin=tRxvxωω2sin2-==tRyvyωω2cos2==ωRvvvyx222=+=∴tvvivxω2sin,cos(-==tvjvyω2cos,cos(==tRvaxxωω2cos42-==tRvayyωω2sin42-==2224ωRaaayx=+=∴taaiaxω2coscos(-==taajayω2sin,cos(-==2自然法:tRtRSωω22=⋅=ωRSv2==⇒方向如图。0==vat224ωRRvan==方向如图。理论力学B（1练习册32学时—-----李鹏程昆明理工大学只限自己使用，请不要传播四、计算题6.4.1O1nO2anAAB解：∵构件BAM作平动；∴M点的运动轨迹及其速度和加速度都与A点相同。而A点绕点作定轴转动，其角速度为：vAanMvMM2nn6030xRcostyRsint消去t得M点的运动轨迹：x2y2R2vMvARRn30方向如图anManA2R(n302R方向如图A0atMatAv6.4.2vAO1vMAφM解：∵AB杆作平动；vMBO2vAaMaAanAanM15vMvAr0.21539.425ms方向如图anManA2r0.21522