理论力学第六版答案.pdf

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1、第1章静力学公理和物体的受力分析1-1画出下列各图中物体A，ABC或构件AB，AC的受力图。未画重力的各物体的自重不计，所有接触处均为光滑接触。FN1APFN2(a)(a1)FTAPFN(b)(b1)AFN1PBFN3FN2(c)(c1)FTBFAyP1PA2FAx(d)(d1)FFABFAB(e)(e1)1qFFFAyBFAxAB(f)(f1)FBCFCAFA(g)(g1)FAyFCCAFBAxP1P2(h)(h1)BFCFCFAxDAFAy(i)(i1)(j)(j1)BFFBCPFAyFAxA(k)(k1)2FCAFABF

2、′ACCAF′ABBFACFABAP(l)(l1)(l2)(l3)图1-11-2画出下列每个标注字符的物体的受力图。题图中未画重力的各物体的自重不计，所有接触处均为光滑接触。CFN2FN′P2(a)(a1)BFN1FBN1CFN2FNP2P1P1FAyFAyFAxFAxAA(a2)(a3)FN1AP1FBN3P2FN2(b)(b1)F′NFABN3FN1P2P1FFNN2(b2)(b3)3FAyFAAxCDFN2BP2P1FN1(c)(c1)FAyFTFAAxDF′FTN2BP1FPN12(c2)(c3)FAyFqBAFCDB

3、AxFC(d)(d1)FAyFDyqFBqDF′BAFCFDDxAxDxFCFDy′(d2)(d3)FAyqFBx′BFFAyAxqABF′FByAxFCyFFCxCPABBxPCFCxFByFCy(e)(e1)(e2)(e3)F1CFF2AyFByABFFAxBx(f)(f1)4FCxCFCx′CF1FAyFCyFCy′F2FByAFBxFAxB(f2)(f3)FFBAyCBAFAxP(g)(g1)F′CyFTF′CCxFAyFBFTFDCAxBAFCxP(g2)(g3)DF1FCyFBF′FB2BCFCxBFAyAFAx(

4、h)(h1)(h2)AFAxFAyFCyFCxCF′AFCyEFFOyCDFOxFCx′EOB(i)(i1)(i2)5AAF′AxF′F′EAyFECDFFByFByOyFBxOFOxFBxBB(i3)(i4)FAyDEFACxCFAxCFTFByHFByFCyBPFBxBFBx(j)(j1)(j2)FAyFDy′FEy′FT′2EFCFCx′DFT2AxF′ExFFExADFDx′EDxFFT3T1FFF′DyEyCy(j3)(j4)(j5)EFFBCEDF′BCxθF′F′CyDE(k)(k1)FFBFCBFECxθCFC

5、y90°−θFDEDDFAyFAyAFAFAxAx(k2)(k3)6FBFF′1BFBDBDACFFAC(l)(l1)(l2)F2F′DDF1F2DBACEEFEFAFCFE(l3)(l4)或F′FFFB′FDy2F1F1Dy2F′DDxBDFDxDBFExACFCEEExFCFEyFAFCFEy(l2)’(l3)’(l4)’F′ADAFCyFCxCF1B(m)(m1)FADDFADEHF2ADFFF′EHAD(m2)(m3)7FANAFkFNBFOyFOxBO(n)(n1)FN1BDqF′BFN2FN3(n2)FBDFFFC

6、FEAGFACEG(o)(o1)FBBDFDFBFFCFDEFAFAFB′CFD′EF(o2)(o3)(o4)图1-28第2章平面汇交力系与平面力偶系2-1铆接薄板在孔心A，B和C处受3个力作用，如图2-1a所示。F=100N，沿铅1直方向；F=50N，沿水平方向，并通过点A；F=50N，力的作用线也通过点A，尺32寸如图。求此力系的合力。yFc3dF2F1F2bFFAF3x1R60θa(a)(b)(c)图2-1解（1）几何法作力多边形abcd，其封闭边ad即确定了合力FR的大小和方向。由图2-1b，得22F=(F+F×4/5

7、)+(F+F×3/5)R123222=(100N+50N×4/5)+(50N+50N×3/5)=161NF+F×4/512∠(F,F)=arccos()R1FR100N+50N×4/5oo=arccos()=29.74=2944′161N（2）解析法建立如图2-1c所示的直角坐标系Axy。∑F=F+F×3/5==50N+50N×3/5=80Nx12∑F=F+F×4/5=100N+50N×4/5=140Ny12F=(80i+140j)NR22F=(80N)+(140N)=161NR2-2如图2-2a所示，固定在墙壁上的圆环受3条

8、绳索的拉力作用，力F1沿水平方向，力F3沿铅直方向，力F2与水平线成40°角。3个力的大小分别为F1=2000N，F2=2500N，F3=1500N。求3个力的合力。Fa1FOx1O40°θF40°2Fb2FRF3F3cy(a)(b)(c)图2-2解（1）解析法建立如图2-2