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1、10-4 图示曲柄连机构中,曲柄OA=40cm,连杆AB=100 cm,曲柄以转速n=180 rpm绕O轴匀速度转动。 求当=450时连杆AB的角速度和其中M点的速度。 解:1)以 A 为基点; sradADABvABABADABv ABvABADABcmOAADvvvvvvvAABA ABABABAA BAAB/56. 5 )220(10045cos3018040coscoscos)22045sin40sin(coscos)90sin()90sin(220222222000= = +=又有QrrrAvrBvr90- OABMAvrAvrAvrBAvr2)仍以A为基点; scmvvvscmvv
2、vscmvvvABBD ABADscmABMAvvvvMyMxMMAAMyMAAMxABABMAMAAM/667/2 .26652327822240coscos/4 .6115227822240sinsin523 100)220(100cos;52 100220sin/27856. 55022222=+=+=+=+=+=+=Qrrr10-5图示四连杆机构中,OA=O1B=2AB,曲柄以角速度=3rad/s绕O轴转动,求在图示位置时杆AB和杆O1B的角速度。 解:1)以A为基点, OAOAOAABOActgvvvvvvvABABBAAB3cossin22sradBOvBBO/2 . 5311=+
3、=由投影定理得rrrAvr OABBAvrBvrAvr2)再由几何法: sradABvABABOAOAvvBAA BA/3sin= 10-7图示配气机构中,曲柄以匀角速度=20rad/s绕O轴转动,OA=40cm,AC=CB=2037cm.求当曲柄在两面三刀铅垂位置和水平位置时,求气阀推村DE的速度。 解:1)当=900和2700时,ACB为瞬时平动 OABC0,27000;00;00;00=+=+=+=DEDEDDyDCCDxDCCDABCACyACCxCAACABBAByABBxBAABABvvvvvvvvvvCCAvvvvvvvvAvvvvvvvv时同理即点为基再以为基点再以或由基点法Q
4、rrrrQrrrrrr2)当=0和1800时,以A为基点则 OAC BDEscmvvvvvvvvAsradABvvvvvvvvvvCAACCyCxCAACBA BAABABAAByBxBAAB/40037203720800; 0/3720374020400; 0=+=+=QrrrQrrr为基点再以得scmvscmvvscmvvvvvvvvvCDEDDECDDyDCDCDCDxDCCD/400180/400/40000sin00=+=时得同理则为基点再以Qrrr10-9解:1)分析O1A杆, 60O1AO2BC1vAvBvCABOBO260O1AO2B60O1AO2BC1C1vAvAvBvBvC
5、vCABABOBOBO2O2scmAOvOA/45067511= 由速度投影定理得: sradrrvvvvB OBAAB/75. 3360450232330cos210= =+=sradABv ACvAA AB/5 . 1215045060cos01= 由瞬心法:C1为AB杆的速度瞬心。 sradrvscmCCvCABC/63303180/31805 . 13)30150(111=sradrrvscmtgABBCvB BOABABB/75. 33603225/32255 . 131506021011=+=10-10 解:C为园轮O的瞬心 sradRvRvBOvvvvve ABae/2 . 05
6、0220 230cos22323 2160cos0 001001=ovvvvvABvRvRCBvRvBareaB0330cos2100 00 0=+=ovvvvv则 ; 动系:动点:销钉10-11 解: sradDOvscmvvvvvvvvCODscmvvADscmOAve CODaeDareaADA/19. 67325/325235030cos30cos;/50/501051011=+=ovvvvvQ则 ; 动系:动点:滑块为瞬时平动;10-13 解:M1为车轮的瞬心: 00222 01222 02 00112000 00 022211111/2/225 . 0/365 . 0/65 . 0
7、 3)(/25 . 0 1RaaaaaMsmaasmRasmRaaaaaMOsradRa Rv dtd dtdsradRvOMn OMOMn Mn OMOMn OMOM=+=+=点:点:为基点,则以vvvvvvvv2 02RanOM= 将上式分别向X轴、Y轴投影: 22222 0404422222 030332222 22 22202/83. 53)23()()(/32. 62)23()()(/16. 331/3/1234444333322smaaaaaaaaMsmaaaaaaaaMsmaaasmaasmaaaOMn OMOMn OMn OMOMOMn OMyxOMyn OMx=+=+=+=+
8、=+=+=+=+=vvvvvv点:点:10-14 解: scmnrrvA/1030301030= C1为AB杆的瞬心; 2220110 1000110)3030(10/62. 31032032060cos3103103603021 23 3 .17 1060sinsin60sinsin=rasradRvABABBCvABACvABABtgACABOAABOAACvn ABBABBAABAAB以A为基点,则: coscos310031022 2n BABn BABABn BAn BABAABaaaaABABABABaaaaa= =+=vvvv22 /2 . 23103 .17314. 32100
9、 cossradRa Ran BABB= 10-15 解: 222222/4002100/20001020scmABaaaaaAscmOAaABn BAn BABAn ABn A=+=0/2100200/2001020sradABv CAvABCscmOAvAAABA=vvvv为基点,则:以杆的瞬心,为将上式向X轴投影: 02200045cos/161001600/1600400200045cos45cos45cos0aaYsradABascmaaaaaan BABBAABn BAn ABAn An BABA=+=轴投影:再将上式向2/56640022scmaan BAB=10-16 解:00
10、rOAvA= B为AB的瞬心:40=ABvA AB,则0 11=BOvB BO; 1635435 23 2530cos81622,2/2330cos2/5 2525)4(2)(260cos41642 02 02 002 02 0222 002 012 02 02 002 02 022 02 01=+=+=+=+=+=ABarraarrraraaaaaAsradrr BOarrraaaaaaBArrABaaaaaArOAaBAABBBAABn MAABMAn MAMAn AMBBOn BAn ABn BAn ABABn BAn BABAn ABn A为基点,则: 还以上投影得:向为基点,则:以vvvvvvvv2 022 022 0222 02 02 0202 02 0439 835 8983589 8835 16352rrraaaraarrraaarraMyMxMMAMyn MAn AMxMA= + =+=+=+=