中国石油大学工程力学答案合集1VIP免费

02/18/2025KylinsoftStatics-Problems-1静力学第1章作业/P23-1.1(l)*:02/18/2025KylinsoftStatics-Problems-2静力学第1章作业/P23-1.1(g)*:02/18/2025KylinsoftStatics-Problems-3静力学第1章作业/P24-1.2(k):02/18/2025KylinsoftStatics-Problems-4静力学第1章作业/P27-1.3(c):02/18/2025KylinsoftStatics-Problems-5静力学第1章作业/P27-1.3(e)*:02/18/2025KylinsoftStatics-Problems-6静力学第2章作业/P58-2.3*:)(903060cos40coskNPRxSolution:由此得到如下方程组：)(060sin40sinkNPRy320sin100cosPP解得：)(8.105740kNP)(1.1953tan102/18/2025KylinsoftStatics-Problems-7静力学第2章作业/P60-2.14*:Solution:按定义有:2211)(rFFrrFFrFMOSo,expressions(a),(d),(e)arevalid.02/18/2025KylinsoftStatics-Problems-8静力学第2章作业/P61-2.20:Solution:(a)ar)(16.04.048.0mkjira)(6812NkjiFjaInvectorform:)(96.0)(mNjFMa)(96.0010681216.04.048.0)(mNFrFMaaa02/18/2025KylinsoftStatics-Problems-9)(0100681216.04.06.0)(mNFrFMzzzzr(b))(16.04.06.0mkjirz)(6812NkjiFkzInvectorform:)(0)(mNFMz02/18/2025KylinsoftStatics-Problems-10++0xxFRkNFRyy16161220mkNMM.36716485122200LRRLMy0由)(25.216360mRMLxySolution:静力学第2章作业/P63-2.28*:02/18/2025KylinsoftStatics-Problems-11与水平面夹角为，与水平面夹角为21TT1585.18.0tan178sin1715cos432.19.0tan53sin54cos0coscos21TTFX++0505.142sinsin21TTFy)(1701NT)(5.1872NT（a)Solution:静力学第2章作业/P63-2.30*:02/18/2025KylinsoftStatics-Problems-12(b)+ARMC)8.05.09.0(cos)2.12.15.1(sin22TT)2.15.1(505.15.142).(0mN02/18/2025KylinsoftStatics-Problems-13OOriginalforcesandcouplecanbereducedtoaforceandacoupleactingatpointO:静力学第2章作业/P64-2.32:RSolution:RC)(31030cos20NRx)(506030sin20kNRy)(3.23353212130sin205.030cos205.060mNCR于是得到合力的大小、方向和作用线位置（如图所示）：)(9.5272022NRRRyx)(1.1953tantan11yxRR)(441.0mRCdR)(467.0mRCdyRx)(348.1mRCdxRyyddxd(或,或)yRRxR02/18/2025KylinsoftStatics-Problems-14+0MCR0xF0yF0zFNPx80NPy60NPz90kjiP906080mNM.392.0603.0900Solution:静力学第2章作业/P64-2.34*:02/18/2025KylinsoftStatics-Problems-15静力学第3章作业/P85-3.2*:Solution:RAyRByFromtheFBDofthebeaminFig:0BM+023129212AyR)(0kNRAyRBx3kN2m0xF+)(0kNRBx0yF+032ByAyRR)(1kNRBy02/18/2025KylinsoftStatics-Problems-16静力学第3章作业/P86-3.11:Solution:GPGLStudyingpersonandladder,drawFBD(a)x=1.5m0AM+022BLPRGxGRBRAyRAx744.8(N)762gGxGRLPB0xF+053AxBRR446.88(N)53BAxRR02/18/2025KylinsoftStatics-Problems-170yF+054AyBLPRRGGGPGLRBRAyRAx344.96(N)517651521254gGGxGxGGGRLPLPLPAy(b)051521LPAyGGxR)(6.2215.2mGGxPL02/18/2025KylinsoftStatics-Problems-18静力学第3章作业/P87-3.18:Solution:300N0.5mRDRE(1)StudyingentirestructureFromtheFBDofentirestructureinFig:0DM+01.13006.1ER206.25(N)ER(2)StudyingmemberABFromtheFBDofmemberABinFig:RBxRByRAxRAy300N0.5m1.6m0...