理论力学课后习题答案

理论力学复习题库题1－3N66.830cosFFxN560cosFFyxy坐标系下：投影：分力的大小：N66.830cosFxFN560cosFyFFxFyFxFy在直角坐标系下，力沿坐标轴的分力的大小等于力在相应坐标轴上投影的绝对值。题1－3续1N66.830cos'FFxN07.745cos'FFyx'y'坐标系下：投影：分力的大小：45sin30sin''yxFFFyx45cos30cos''FF在非直角坐标系下，力沿坐标轴的分力的大小不等于力在相应坐标轴上投影的绝对值。Fy'Fx'Fy'Fx'N32.7'xFN18.5'yF题1－3续2x'y'坐标系下：分力的大小也可按正弦定理来计算：105sin45sin'FFxFy'Fx'N32.7'xFN18.5'yF105sin30sin'FFy题1－660sinF60cosF60sin60sin)(1rFMAF)60cos(60cos12rrFmN15采用合力矩定理采用力矩定义ar2dm1.060cos12rram05.021admN15)(FdMAFAr1FF题1－10mkN428.924)(zyxFFMFkN357.2181FFxkN357.2181FFykN428.9184FFz18114FxFyFz分力的大小：mkN428.94)(xyFMFmkN714.42)(xzFMF采用合力矩定理题1－10续mkN428.9)(yzxzFyFMFkN357.2181FFx力的作用点坐标：x＝1，y＝1，z＝0kN357.2181FFykN428.9184FFz采用解析计算公式力的投影：F18114mkN428.9)(zxyxFzFMFmkN714.4)(xyzyFxFMF题1－12FAxFAyMAFAxFAyFCFBFA题1－13FAxFAyFDFCFCFAxFAyFDFBC②BC①AB③整体FBxFByFBx'FBy'BAD题1－13续1②BD①AB③滑轮④整体FAxFAyFBACBFCxFCyFDBDFB'CPFCx'FCy'FFAxFAyFFD题1－13续2ACD②BC①AC③整体FFAFCxFCyFDCEBFCx'FCy'FBFEFAFB题2－14321R5210121FFFFFxN77.68N79.682R2RRyxFFF02689.0),cos(RRRFFxxFyxON85.1432R5110321FFFFy99971.0),cos(RRRFFyyF54.91),(RxF38.1),(RyFFR平面汇交力系题2－44321MMMMM平面力偶系mkN2题2－13xymkN8.1925.12.25.12.12121FFPPMAm857.0RyAFMxkN982RFFxkN225121RFPPFykN4.2452R2RRyxFFF主矢：对A点的主矩：0RF力系存在合力RRFFRF0OM题2－16mkN69.965445232FFMxkN21.45351321RFFFFxkN4543RFFykN58.3522RFFzkN82.62R2R2RRzyxFFFFmkN07.14222zyxOMMMM主矢：对O点的主矩：553mkN27.765335232FFMymkN16.74512FMz题2－18(b)21mm5001050Amm51Cy22mm17507025Amm452Cymm1.36212211AAyAyAyCCC形心在对称轴y上12C2C1题3－7空间汇交力系0izF022PFAO0ixF021ABAOFFPFAB220iyF独立的平衡方程有3个FADFABFAOaa2aaOC2PFAO2021ADAOFFPFAD22点A题3－9平面力偶系0iM045cosMaFAaMFFAD2独立的平衡方程有1个FDFA整体4545BCD是二力杆题3－14(a)平面力系FAxFAyFB0ixF0AxF0AiM0424BFqMkN75.2BF0iyF04BAyFqFkN25.5AyF独立的平衡方程有3个整体题3－15(b)物体系统平衡FBxFCFByBCBC整体0BiMkN3CF0AiMmkN9AM0iyF0ixFkN5AyF0AxFq05.133qFCFAxFCFAyMA065.433CAFqFMM03CAyFqFF题3－16(d)BC整体0CiMFqaFBx25.0FAxFAyFBxFBy0AiMFqaFBy25.020424aFaqaaFBy0iyF04ByAyFqaFFqaFAy25.02BCFCxFCyFBxFByq0222aqaaFaFByBx0iyF0ixFFFCy25.0FqaFCx25.0整体0ixFFqaFAx75.0题3－25空间力系FAxFGFAyFAzFH整体0BEiM0AxF0dFAx0iyMGHFFkN3.28GHFF0545sin60cos545sin60cosGHFF0ixM05545cos60cos545cos60cosPFFGHkN20AyF0iyF045cos60cos2HAyFFkN0.69AzF0izF060sin2PFFHAzC题3－17(b)FEFAFFGFCxFEFCyGFFGFBGFAG′①整体②CEF③铰G组合结构0AiMFFE5.10CiMFFFG20ixFFFAG220iyFFFBG2ECFFD对称性FFFAGEF22FFFBGDF2题3－19q2ECFCxFCyFE①DE②BDMAFAxFAyFEFD0CiMkN5EF0ixFkN4CxF0iyFkN3CyFq1BCDFBxFByFDFCx'FCy'0BiMkN8DF0ixFkN4BxF0iyFkN3ByF题3－19续MAFAxFAyFEFD0AiMmkN10AM0ixFkN4AxF0iyFkN1AyF③整体题3－22OA0iM030cos1MOAFAO1A0iM012AOFMAN1732AOFFmN6002MOAM1FOFAO1AFO1′FAM2题4－6FFN0iyFkN65021NPPF设坝不滑动kN390NmaxFfFsmaxFF0ixFkN3001FF而假设成立坝不滑动设坝不绕B点转动xFFNE0EiM08.2)8.1()2.4(121FxPxPm17.2x0x假设成立坝不绕B点转动0iyFkN65021NPPF假设坝静止不动kN390NmaxFfFsmaxFF0ixFkN3001FF而坝不滑动xFFNE0BiM08.2)9.37.5()5.17.5(N121xFFPPm17.2x0x假设成立坝不绕B点转动坝既不滑动，也不绕B点转动。题4－6题4－7F2F1FN2FN1活动支架补充方程求安全工作时的x设活动支架处于临界状态0ixF01N2NFF0iyF021FFF0AiM0)5.0(2N2dxFhFdF1N1FfF2N2FfFcm402fhxA题4－8ABFACFBCFFBFNBPBFBC′FAFNAPAFAC′C030sinFFAC030cosBCACFFC030cosACAFF030sinNACAAFPFFFAC2FFBC3AFFA3FPFAANAAFfFNN58.40F030sin30cosBBBCFPF030cos30sinNBBCBPFFBBBPFF5.05.12/)(3NFPFBBBBFfFNN77.80FN58.40maxF维持平衡的最大力A题4－9求保持系统平衡的最小力F1设物块B处在下滑还未滑动的临界状态①物块②尖劈补充方程BPFsFN1FN0iyF0sincossNPFFF1FN2FN′Fs′0ixF0cossin1sNFFFNssFfFPffFsincoscossinss1)tan(mPA题4－9续求保持系统平衡的最大力F2设物块B处在上滑还未滑动的临界状态①物块②尖劈补充方程BPFsFN1FN0iyF0sincossNPFFF2FN2FN′Fs′0ixF0cossin2sNFFFNssFfFPffFsincoscossinss2)tan(mP)tan()tan(mmPFP题4－12(a)FAFCyFCxCAFABFADFADFBCFCDFCyFCx整体0CiMN600AF0ixFN600CxF0iyFN200CyFFCD′600NFAD′FBDC0ixFN600BCF0iyFN200CDFA0iyFN750ABF0ixFN450ADFD0iyFN250BDFF2F7F9F10题4－14(c)BF3F1AF4F5mm截面：0AiMFF1mmnnnn截面：0ixF0222227FFF2F7F6F8BB0iyF0222272FF02F2cosRRx2sin2Rxvx0t直角坐标法2sinRyRvvvyx2222cos2Ryvy2cos42Rxax2sin42Ryay2224Raaayx2题5－62RsRsv20t自然法0tsa2sO1＋－22n4/RRva2n2n2t4Raaaa题5－6续题5－9tx5025500ty50xty100x10y2222510tyxv1022yxa2t2510ddtttva22t2nsm/10aaam250/n2av0,m/s10,m/s500t2aavt时，题5－14)(24tvtsv2tm/s2sa212nm/s2/Rva)(432tstts0vs2t522050tttSSS运动方向发生改变)0()2(20ssSt)2()5(52ssStm1350tSm3)0(sm7)2(sm2)5(sm4m9sm/6vs5t0m2)5(s点位于AO上22t2nm/s83.2aaarad/sπ5.130πnm/s707.0lvvaO0ttlaaaO22nnm/s331.3laaaO2m/s331.3Oa揉桶作曲线平移题6－4题6－7txv1010txarad/s20tRx2rad/s20Rx25tx2424m/s140010tRa题6－11tbsintbAOcos1tbAOsin21aAtvAtbllvvAOAcos1tbllaaAOAsin2tt