第一章-基本概念及基本原理

1第一章基本概念及基本原理[习题1-1]支座受力F,已知kNF10,方向如图所示,求力沿yx,轴及沿'',yx轴分解的结果,并求力F在各轴上的投影.解:(1)F沿yx,轴分解的结果把F沿yx,轴分解成两个分力,如图所示.iiiFFx66.8866.01030cos0)(kNjjjFFy55.01030sin0)(kN(2)F沿'',yx轴分解的结果把F沿'',yx轴分解成两个分力,如图所示.由图可知,力三角形是等腰三角形.故:''10'iiFFx)(kN''018.575cos102'jjFy)(kN(3)F在yx,轴上的投影)(66.8866.01030cos0kNFFx)(55.01030sin0kNFFy(4)F在'',yx轴上的投影)(66.8866.01030cos0'kNFFx)(59.275cos1075cos00'kNFFy[习题1-2]已知NF1001,NF502,NF603NF804,各力方向如图所示,试分别求各力在x轴y轴上的投影.解:)(6.86866.010030cos011NFFx)(505.010030sin011NFFy)(305350cos222NFFx力沿x,y轴的分解图力沿x’,y’轴的分解图力沿x’,y’轴的投影图xFyFF030'yFF075075'xF'x'yF'xF'yF0300752xyzTF23563)(405450sin222NFFy0060cos333FFx)(60160sin333NFFy)(57.56135cos80cos0444NFFx)(57.56135sin80sin0444NFFy[习题1-3]计算图中321,,FFF三个力分别在zyx,,轴上的投影.已知kNF21,kNF12,kNF33.解:)(2.16.025311kNFFx)(6.18.025411kNFFy01zF)(424.05345sin1cossin02222kNFFx)(566.05445sin1sinsin02222kNFFy)(707.045cos1cos0222kNFFz03xF03yF)(333kNFFz[习题1-4]已知kNFT10,求TF分别在zyx,,轴上的投影.解:)(591.75353510sin22222kNFFTTxy)(51.6355591.7cos22kNFFTxyTx题1-2图1F2F3F4F030013543xy3xyzNOABmm200mm300mm400FCDmm200mm400)0,2,4(A)3,4,1(BCOzyxFT)0,4,1(DEG)(91.3353591.7sin22kNFFTxyTy)(51.6535510cos222kNFFTTz[习题1-5]力F沿正六面体的对角线AB作用,kNF100,求F在ON上的投影.解:如图所示,F在AC线上的投影为:)(345.88400300400400400100cos22222kNCABFFFOBAC5.0400200tanNOD057.265.0arctanNOD00043.1857.2645BONF在ON线上的投影为:)(811.8343.18cos345.88cos0kNBONFFOBON[习题1-6]已知NF10,其作用线通过A(4,2,0),B(1,4,3)两点,如图所示.试求力F在沿CB的T轴上的投影.解:61.313)42()14(22AD69.413361.322AB2361.322DGF在AD上的投影为:4OMFR0602r1rFA)(697.769.461.310cosNBADFFAD)(40.669.4310sinNBADFFz)(264.461.32697.7cosNADGFFADy)(396.661.33697.7sinNADGFFADxF在T轴上的投影为:)(251.75340.654264.4coscoskNECBFBCDFFzyT[习题1-7]图中圆轮在力F和矩为M的力偶作用下保持平衡,这是否说明一个力可与一个力偶平衡?解:图中圆轮在力F和矩为M的力偶作用下保持平衡,这不能说明一个力可与一个力偶平衡.因为轮子的圆心处有支座,该支座反力R与F构成一力偶,力偶矩),(FRM与M等值,共面,反向,故圆轮保持平衡.[习题1-8]试求图示的力F对A点之矩,已知mr2.01mr5.02,NF300.010012030cos60sin)30sin(60cos)(rFrrFFMA)(15232.023300)5.02.05.0(5.0300)(mNFMA[习题1-9]试求图示绳子张力TF对A点及对B点的矩.已知kNFT10,ml2,mR5.0,030.解:)(530sin10sin0kNFFTTx)(66.830cos10cos0kNFFTTy)(732.1866.0260sin0mlOC)(15.0260cos0mlAC)()()(TyATxATAFMFMFM)30cos5.01(66.8)30sin5.0732.1(500)(5mkN5),,(cbaABCOzyxF)()()(TyBTxBTBFMFMFM)30cos5.01(66.8)30sin5.0732.1(500)(320.12mkN[习题1-10]已矩正六面体的边长为cba,,,沿AC作用一力F,试求力F对O点的矩矢量表达式.解:zyxFFFcbakjiFM)(0式中,2222222222coscoscbaFabaacbabaFFFx2222222222sincoscbaFbbabcbabaFFFy222222sincbaFccbacFFFz故cbacbakjicbaFFM2220)(ccbakjicbaF200222bajiccbaF2222)(2222jaibcbacF[习题1-11]钢绳AB中的张力kNFT10.写出该张力TF对O点的矩的矢量表达式.6yzx)0,1,1(B)4,2,0(AOTF)0,2,0(C解:2)21()01(22BC2318)04()12()10(222ABzyxFFFkjiFM420)(0式中,)(357.22123210coscoskNFFTTx)(357.22123210sincoskNFFTTy)(428.923410sinkNFFTTz故428.9357.2357.2420)(0kjiFM357.2357.24428.9357.22jiki)(357.24)357.2428.9(2jikikji714.4428.9428.9[习题1-12]已知力kjiF32,其作用点的位置矢kjirA423,求力F对位置矢为kjirB的一点B的矩(力以N计,长度m以计).7yzxOABArBrFABCFF'FOC解:FrrFrFMBAABB)()(式中,kjirA423,kjirB,)(BArrkji312kjiF32故,)(FMB)312(kji)32(kji132312kji240312kji2003522kkji5222kji)425(2kjikji8410)(mN[习题1-13]工人启闭闸门时,为了省力,常常用一根杆子插入手轮中,并在杆的一端C施加力,以转动手轮.设手轮直径mAB6.0,AC轩长ml2.1,在C端用NFC100的力能将闸门开启,若不借用杆子而直接在手轮A,B施加力偶),('FF,问F至少应多大才能开启闸门?解:支座O反力OR与CF构成一力偶),(0CFR若要闸门能打开,则),('FF与),(0CFR必须等效,即它们的力偶矩相等:)3.02.1(1006.0F)(150NF[习题1-14]作下列指定物体的示力图.物体重量,除图上已注明者外,均略去不计.假设接触处都是光滑的.8