理论力学运动分析解析法试卷30题.doc

课程名称 Mechanisms and Machine Theory 考试日期 2007.07.06考生姓名 学号 专业或类别 题号一二三四五六七八总分累分人 签名题分32121281061010 100得分考生注意事项：1、本试卷共 10 页，请查看试卷中是否有缺页。 2、考试结束后，考生不得将试卷、答题纸和草稿纸带出考场。教师注意事项：如果整门课程由一个教师评卷的，只需在累分人栏目签名，题首的评卷人栏目可不签名。以下内容命题教师阅读后请删除： 用B5纸作为标准试题纸，也可打印在A4纸上交试卷印刷中心帮助处理。题目不超过方框, 每题必须留有适当空位给学生答题使用。 题目序号统一用一、1、（1），即第一大标题用“一、二、”，第一大标题下的题目用“1、2、”，第1标题下的题目用“（1）、（2）、”。每道大标题下必须加上得分评卷人人 题目“一、二、”统一用四号黑体字打印， 其他部分一律用小四号宋体字打印。 教师所填课程名称必须与课表、教学大纲和授课计划吻合，考试形式、考试日期、考生注意事项第1条中的空白项必须填写齐全。解析法运动分析考题30题（已剔除课本、作业本中的题目！）福州大学 20062007学年第二学期考试A卷得分评卷人人 八-1、In the mechanism shown below, XE=-10, XA=20.3, YA=0, XC=53.1, YC=0, LAB=11, LCG=4, LDG=63.8, LDE=19.5. The crank AB rotates at a constant speed of 8 rad/sec. A main program is required to analyze the output motions of the point E. The mechanism will be analyzed for the whole cycle when the driver AB rotates from 0 to 360 with a step size of 5.(10%)得分评卷人人 八-2、In the mechanism shown below, XF=0, YF=0, XC=42.9, YC=0, XD=-20, LEF=11.8, LBC=5, LAB=63.8, LAD=19.5. The crank FE rotates at a constant speed of 8 rad/sec. A main program is required to analyze the output motions of the point D. The mechanism will be analyzed for the whole cycle when the driver FE rotates from 0 to 360 with a step size of 5.(10%)得分评卷人人 八-3、In the mechanism shown below, XC=0， YC=0， XA=-30， YA=0, XE=20, LAB=10, LBD=49.6, LDE=28. The crank AB rotates clockwise at a constant speed of -8 rad/sec. A main program is required to analyze the output motions of the point E. The mechanism will be analyzed for the whole cycle when the driver AB rotates from 360 to 0 with a step size of -5.(10%)得分评卷人人 八-4、In the mechanism shown below, XC=0, YC=0, XA=0, YA=23.5, YE=43.8, LAB=11.8, LCD=47, LDE=19.5. The crank AB rotates at a constant speed of 8 rad/sec. A main program is required to analyze the output motions of the point E. The mechanism will be analyzed for the whole cycle when the driver AB rotates from 0 to 360 with a step size of 5.(10%)得分评卷人人 八-5、In the mechanism shown below, XC=0, YC=0, XB=0, YB=28.8, YE=47, LAB=11, LCG=6.1, LDG=49.7, LDE=30. The crank BA rotates at a constant speed of 8 rad/sec. A main program is required to analyze the output motions of the point E. The mechanism will be analyzed for the whole cycle when the driver BA rotates from 0 to 360 with a step size of 5.(10%)得分评卷人人 八-6、In the mechanism shown below, XB=0, YB=0, XC=0, YC=28.5, YA=-22.6, LCE=8, LEF=6, LDF=50, LAD=27.3. The crank CE rotates clockwise at a constant speed of -8 rad/sec. A main program is required to analyze the output motions of the point A. The mechanism will be analyzed for the whole cycle when the driver CE rotates from 360 to 0 with a step size of -5.(10%)得分评卷人人 八-7、In the mechanism shown below, XG=0, YG=0, XB=-31.6, YB=-8.6, XD=-57.5, YD=22.9, LAB=6.6, LFG=7.4, LEF=58.4, LCE=49.7, LCD=34.6. The crank BA rotates at a constant speed of 8 rad/sec. A main program is required to analyze the output motions of the link DC. The mechanism will be analyzed for the whole cycle when the driver BA rotates from 0 to 360 with a step size of 5.(10%)得分评卷人人 八-8、In the mechanism shown below, XF=0, YF=0, XD=10.6, YD=28.4, XB=28.5, YB=-39.6, LEF=8, LDG=5.7, LCG=59.2, LAB=36.3, LAC=39.9. The crank FE rotates at a constant speed of 8 rad/sec. A main program is required to analyze the output motions of the link BA. The mechanism will be analyzed for the whole cycle when the driver FE rotates from 0 to 360 with a step size of 5.(10%)得分评卷人人 八-9、In the mechanism shown below, XD=0, YD=0, XE=30, YE=7.2, XC=21.7, YC=-8.2, LDG=8, LFG=6, LAF=48.3, LAB=25, LBC=24. The crank DG rotates clockwise at a constant speed of -8 rad/sec. A main program is required to analyze the output motions of the link CB. The mechanism will be analyzed for the whole cycle when the driver DG rotates from 360 to 0 with a step size of -5.(10%)得分评卷人人 八-10、In the mechanism shown below, XA=0, YA=0, XC=0, XF=38.5, YF=-41.1, LAB=15, LBC=35.7, LCE=16.3, BCE=43, LBE=26.4, EBC=25, LDF=8.9, LDG=65.6. The crank AB rotates at a constant speed of 8 rad/sec. A main program is required to analyze the output motions of the point G. The mechanism will be analyzed for the whole cycle when the driver AB rotates from 0 to 360 with a step size of 5.(10%)得分评卷人人 八-11、In the mechanism shown below, XD=0, YD=0, XC=0, XF=-36.2, YF=-26.5, LDG=15, LCG=35.7, LAC=16.3, ACG=43, LAG=26.4, CGA=25, LEF=8.9, LBE=65.6. The crank DG rotates at a constant speed of 8 rad/sec. A main program is required to analyze the output motions of the point B. The mechanism will be analyzed for the whole cycle when the driver DG rotates from 0 to 360 with a step size of 5.(10%)得分评卷人人 八-12、In the mechanism shown below, XC=0, YC=0, XD=0, XE=-24.5, YF=21.6, LBC=16.5, LBD=39.2, LAD=17.8, BDA=43, LAB=29, ABD=25, LAG=6.5, LFG=50. The crank CB rotates clockwise at a constant speed of -8 rad/sec. A main program is required to analyze the output motions of the point F. The mechanism will be analyzed for the whole cycle when the driver CB rotates from 360 to 0 with a step size of -5.(10%)得分评卷人人 八-13、In the mechanism shown below, XF=0, YF=0, XA=26, YA=50, XD=0, LEF=15, LDE=35.7, LCD=17.8, EDC=43, LCE=26.4, CED=25, LBC=28.2, LAB=27. The crank FE rotates at a constant speed of 8 rad/sec. A main program is required to analyze the output motions of the link AB. The mechanism will be analyzed for the whole cycle when the driver FE rotates from 0 to 360 with a step size of 5.(10%)得分评卷人人 八-14、In the mechanism shown below, XF=0, YF=0, XD=0, XC=-26.9, YC=46.7, LBF=15, LBD=35.7, LAD=17.8, BDA=43, LAB=26.4, ABD=25, LAE=39.5, LCE=31.6. The crank FB rotates at a constant speed of 8 rad/sec. A main program is required to analyze the output motions of the link CE. The mechanism will be analyzed for the whole cycle when the driver FB rotates from 0 to 360 with a step size of 5.(10%)得分评卷人人 八-15、In the mechanism shown below, XA=0, YA=0, XF=-14, YF=33.7, XD=XF, LBF=13.8, LBD=32.8, LDE=19.5, LCE=20, LAC=18. The crank FB rotates clockwise at a constant speed of -8 rad/sec. A main program is required to analyze the output motions of the link AC. The mechanism will be analyzed for the whole cycle when the driver FB rotates from 360 to 0 with a step size of -5.(10%)得分评卷人人 八-16、In the mechanism shown below, XG=0, YG=0, XA=37.2, YA=16.9, YF=-11.2, LEG=16.8, LEF=39.2, LCF=20.6, EFC=67, LCE=36.4, CEF=31, LAD=9.2, LBD=57.4. The crank GE rotates at a constant speed of 8 rad/sec. A main program is required to analyze the output motions of the point B. The mechanism will be analyzed for the whole cycle when the driver GE rotates from 0 to 360 with a step size of 5.(10%)得分评卷人人 八-17、In the mechanism shown below, XA=0, YA=0, XG=63, YG=18, YC=0, LAB=17.6, LBC=41.6, LCD=36, DCB=30, LBD=20.8, CBD=60, LEG=9.5, LEF=77. The crank AB rotates at a constant speed of 8 rad/sec. A main program is required to analyze the output motions of the point F. The mechanism will be analyzed for the whole cycle when the driver AB rotates from 0 to 360 with a step size of 5.(10%)得分评卷人人 八-18、In the mechanism shown below, XD=0, YD=0, XA=29.9, YA=39.5, YC=YA, LAB=15.3, LBC=41.4, LDE=58. The crank AB rotates clockwise at a constant speed of -8 rad/sec. A main program is required to analyze the output motions of the point E. The mechanism will be analyzed for the whole cycle when the driver AB rotates from 360 to 0 with a step size of -5.(10%)得分评卷人人 八-19、In the mechanism shown below, XA=0, YA=0, XF=33.1, YF=-15, YC=0, LAB=10, LBC=56.2, LCD=35.1, LDE=17.8, LEF=16. The crank AB rotates at a constant speed of 8 rad/sec. A main program is required to analyze the output motions of the link FE. The mechanism will be analyzed for the whole cycle when the driver AB rotates from 0 to 360 with a step size of 5.(10%)得分评卷人人 八-20、In the mechanism shown below, XA=0, YA=0, XF=19.5, YF=26.1, YE=YF, LDF=18.9, LDE=56.2, LCE=35.1, LBC=29.1, LAB=27.3. The crank FD rotates at a constant speed of 8 rad/sec. A main program is required to analyze the output motions of the link AB. The mechanism will be analyzed for the whole cycle when the driver FD rotates from 0 to 360 with a step size of 5.(10%)得分评卷人人 八-21、In the mechanism shown below, XF=0, YF=0, XA=-61.6, YA=25.4, YE=4.9, LDF=10, LDE=30.2, LCE=20.2, LBC=31.7，LAB=26.4. The crank FD rotates clockwise at a constant speed of -8 rad/sec. A main program is required to analyze the output motions of the link AB. The mechanism will be analyzed for the whole cycle when the driver FD rotates from 360 to 0 with a step size of -5.(10%)得分评卷人人 八-22、In the mechanism shown below, XE=0, YE=0, XB=0, YB=41, XF=-25, YF=44.9, LDE=15, LAD=39, LAB=29, LAC=28, CAD=18, LCD=15, ADC=35, LFG=8.8, LGH=64. The crank ED rotates at a constant speed of 8 rad/sec. A main rogram is required to analyze the output motions of the point H. The mechanism will be analyzed for the whole cycle when the driver ED rotates from 0 to 360 with a step size of 5.(10%)得分评卷人人 八-23、In the mechanism shown below, XF=0, YF=0, XE=-8.6, YE=48.3, XB=32.2, YB=48.3, LDE=15, LAD=39, LAB=29, LAC=15.6, DAC=25, LCD=25.8, CDA=15, LFH=5, LGH=45. The crank ED rotates at a constant speed of 8 rad/sec. A main program is required to analyze the output motions of the point G. The mechanism will be analyzed for the whole cycle when the driver ED rotates from 0 to 360 with a step size of 5.(10%)得分评卷人人 八-24、In the mechanism shown below, XA=0, YA=0, XE=41, YE=0, XF=6.8, YF=-37.2, LCE=15, LBC=39, LAB=29.2, LBD=28, DBC=18, LCD=15, BCD=35, LDG=7.5, LGH=60. The crank EC rotates clockwise at a constant speed of -8 rad/sec. A main program is required to analyze the output motions of the point H. The mechanism will be analyzed for the whole cycle when the driver EC rotates from 360 to 0 with a step size of -5.(10%)得分评卷人人 八-25、In the mechanism shown below, XA=0, YA=0, XD=-32.7, YD=15.6, XE=XD, LAB=7.4, LBC=32.6, LCD=15.9, LCE=22. The crank AB rotates at a constant speed of 8 rad/sec. A main program is required to analyze the output motions of the point E. The mechanism will be analyzed for the whole cycle when the driver AB rotates from 0 to 360 with a step size of 5.(10%)得分评卷人人 八-26、In the mechanism shown below, XA=0, YA=0, XE=-40.1, YE=-10, XF=-25, LCE=12, LBC=40.1, LAB=30.1, LBD=30.6, CBD=39, LCD=25, DCB=50, LDF=47. The crank EC rotates at a constant speed of 8 rad/sec. A main program is required to analyze the output motions of the point F. The mechanism will be analyzed for the whole cycle when the driver EC rotates from 0 to 360 with a step size of 5.(10%)得分评卷人人 八-27、In the mechanism shown below, XE=0, YE=0, XA=0, YA=-34.3, XF=9, LCE=11.5, LBC=30.8, LAB=27, LBD=28.7, DBC=28, LCD=14.4, BCD=68, LDF=32.8. The crank EC rotates clockwise at a constant speed of -8 rad/sec. A main program is required to analyze the output motions of the point F. The mechanism will be analyzed for the whole cycle when the driver EC rotates from 360 to 0 with a step size of -5.(10%)得分评卷人人 八-28、In the mechanism shown below, XA=0, YA=0, XC=37.8, YC=0, YF=-24, LAB=12.6, LBD=33.6, LCD=29.4, LDE=127.8, BDE=57, LBE=28.6, EBD=22, LEF=42.5. The crank AB rotates at a constant speed of 8 rad/sec. A main program is required to analyze the output motions of the point F. The mechanism will be analyzed