机械原理习题答案

————————————23A154DCB2CBD(1BCD(21.1AA34111mm/mm2n=3p=4ph=1:F=3n-2p-ph=3x3-2x4-1=0345BCD3————————————21D4BAC35a)n=7p=9ph=2F=3n-2pe-ph=3x7-2x8-2=13.an=4P=5Ph=1F=3x4-2x5-1=198b)7GD6C342BA1ωFE————————————bn6P=7Ph=3F3627314.abCD=FI=KL=KM=FJ=CELI=KF=MJ=JE=FC=IDan=7P=10Ph=0F=3x7-2x10-1=1bn=5P=7Ph=0F=3x5-2x7-1=15.EG————————————1n=7P=10Ph=0F=3x7-2x10-1=12)AB3EGШaF=3n-(2Pl+Ph)=F=3n-(2Pl+Ph)=F=3n-(2Pl+Ph)=F=3n-(2Pl+Ph)=————————————————————————24244....APlDPlwllμωμ=2424445.10590APDPllradwsω==×=4..53230.48cCDlmlsυωμ==××=13P13P34CCνν=1334....CPlCDlllωμωμ=133432.52.3568CDCPlradslωω==×=133..EEPllυωμ=13EPlEυ13P133m..2.355730.402EEPllsυωμ==××=0.001mmm⎛⎞⎜⎟⎝⎠3υ3a20.02,0.2vammssmmmmμμ⎛⎞⎛⎞⎜⎟⎜⎟==⎜⎟⎜⎟⎝⎠⎝⎠332121BBBBBυυυυυ→→→→→===+1..lablωμBCPAB⊥DBP1..105510.55Bilabmlsυωμ==××=uur2121.0.02150.3BBbbmlsυυμ==×=32.0.02230.46bpmlsυυμ==×=()DB↑P332112121ntkrBBBBBBBBaaaaaaa===+++uuuruuuruuuuuruuuuuruuruuuruuurBCPBA→()AB⊥↑()BD⊥↑BDP21..lABlμω1..lABlμα1212.BBωυ221155105.5nBmas=××=uuur221155105.5tBmas=××=uuur2212100.5511kBBmas=××=uuuuur223''.0.26012pbmalsαμ==×=uur()BC↑P3232BBBBvvv=+uuuruuuruuuuur1.ABlωBC⊥()AB⊥↓BDP21470.047Bmvs=×=uuur()0.001msmmvμ=33.1460.046Bvpbmvlsμ==×=uuur33460.9250bBCvradslω===3223.0.056BBvbbmvlsμ==uuuuur.1250.027Dvpdmvlsμ==×=uur322323krBBBBBBαααα=++uuuuuruuuuuruuuruuur21BBαα=uuuruuur22333232BBntntkrBBBBBBαααααα+=+++uuuruuuruuuruuuruuuuuruuuuur23.BClω21.ABlω3322BBvωBC→BC⊥BA→AB⊥BD⊥BDP223500.920.042nBmsα=×=uuur225010.050nBmsα=×=uuur23220.0560.920.103kBBmsα=××=uuuuur2()0.002amsmmμ=23323.0.00229.40.059tBarbbnbmalsμ-==×=uuur331.18tBbcaradslα==uuur2330.725dradsαα=×=1pABAB12221233ABBOpR12pR32P2314AM23ω21ωωBOPR1PR32P2314M23ω21ωωBOpR12P2314AM21ω23ωωPR32A2.11Mr=10mm0.15vf=0.15f=100ABl=mm350BCl=mm120M=Nm3FFQ3afFQ23Fo5≈φ0031214212=++=++RRRRQFFFFFFvvvvvφαφαηαφαηαφαFcFFcFFQQ≤≤-==•=-•=0tan)tan(Ftan)tan(00φαφφααηφααηα-∴-≥+≤′+=′+=′ooo9090900)tan(tan)tan(QQcFFQφα-)(2180φα--°)(2φα-)(90φα--°)(90φα--°φα-FFRFRbarctan8.53fφ==°212121sin()2sin(2)2sin98.53441.63sin12.94sin(2)435.4sin()2RRRFQFQNPFNπφπαφαφπφ+=--°=•=°+=•=-212121sin()2sin(2)2sin81.47135.1sin47.06sin(2)sin42.94135.193.06sin98.53sin()2RRRFQFQNPFNπφπαφαφπφ-=-+°=•=°-°=•=•=°+sin()sin(2)2tan(2)sin(2)sin()22PQQπφαφαφππαφφ--=••=•--++20αφ-17.06α°15ABDCADBCllllmm≤+-=ABlADABDCBCllll+≤+55ABDCBCADllllmm≤+-=ABl5055ABmmlmm≤ADBCDCABllll+≤+45ABADBCDCllllmm≥+-=5045ABmmlmm≥ABBCDCADllll++15ABDCADBCllllmm+-=3015ABmmlmmADBCDCABllll++45ABADBCDCllllmm+-=3045ABmmlmmADBCDCABllll--55ABDCADBCllllmm-+=(2)3minK3.K=1LAD=100mm,LAB=20mm,,,30,.θK1111801800K111θ--=°=°×=°++AB'ABABDCCCCBCBB’DBCBCOOABC’CCD53.85mm;BC86.6mm.4.4K1.35H50mme20mmmax[]θK11.35118018026.8K11.351θ--=°=°×=°++lμ1mmmm12cce122109063.2ccoccθ∠=∠=°-=°OO1ocKeA174.7ACmm=,229.7ACmm=12ACABBCACBCABllllll=+⎧⎪⎨=-⎪⎩52.2BClmm=22.5ABlmm=C22arcsinarcsin43BCABmaxllACeeα-===°αα′′=coscosaa°=′=′4.22)/cosarccos(aaααmmaa41.0tan=′-′=αδmmcaacm5.1)(*=+-′=m=5mm=350mm,=9/5a11cos/αbrr=111cos/αωωbrr=bb11223344bbx2244bby1133mr+mr+mr+mr+mr=0(mr)=-(mr-mr)=-(70200-100100)=-4000Kg.mm(mr)=-(mr-mr)=-(50100-80150)=7000Kg.mm××××bm221/2221/2bbxbbybb[(mr)+(mr)][(7000)+(-4000)]m==r=53.75Kgbrbmbbybbbx(mr)=arctan[]=arctan-1.75(mr)αb=150.26α°0313203132223344322211=+++=+++bbbIbIrmrmrmrmrmrmrmrmvvvvvvvvoo145)(4.750/3710/6}(6.550/2810/2==×====×=⋅=bbbwbbbbwbkgrwmkgrwmθμθμ423000/,314.16/601000[]20.05[][]1520.05100.03nnrmmradsAemmrmekgcmπωμω-=======××=•Ι212112200[][]3020200100100[][]3010200100lmrmrgcmlllmrmrgcmllΙΙΙΙΙΙ==×=•++==×=•++426000/min,628.32/60[]1000/10.025[][]1510.0251015nnrradseAmmrmegcmπωωμ-=======××=•Ι212112200[][]1510200100100[][]155200100lmrmrgcmlllmrmrgcmllΙΙΙΙΙΙ==×=•++==×=•++'cm''342008100bccbcmlmkgl•×==='''31()(84)1503650abcacmmlmkgl+•+×===abcd0.50.6dtdJMMerdω=-rdeMMddtJ-=ω1WmaxWmin200100t1t2(1000/w-50)1000/w-5