4 Velocity and Acceleration Analysis 14 Homework:Velocity and Acceleration AnalysisPart IIProblem 4.3: R-RRR-RRT mechanismA planar mechanism is shown in Fig. P4.3. The following data aregiven: AB=0.150 m, BC=0.400 m, CD=0.370 m, CE=0.230 m, EF =CE,La=0.300 m, Lb=0.450 m, and Lc=CD. The constant angular speed of thedriver link 1 is 60 rpm. Find the velocities and the accelerations of themechanism for φ=φ1=30◦. For φ=30◦the position of the mechanism is givenby: xB=0.129904 m, yB=0.075 m, xC=-0.0689445 m, yC=0.422073 m,xE=-0.298288 m, yE=0.404712 m, xF=-0.37 m, yF=0.186177 m, φ2=-1.05052 rad, φ3=0.0755515 rad, φ4=1.25372 rad.Results˙xB=-0.471239 m/s, ˙yB=0.81621 m/s, ¨xB=-5.1284 m/s2, ¨yB=-2.96088 m/s2,˙xC=-0.0788027 m/s, ˙yC=1.04105 m/s, ¨xC=2.87595 m/s2, ¨yC=1.03567 m/s2,˙xE=-0.127788 m/s, ˙yE=1.68819 m/s, ¨xE=4.66371 m/s2, ¨yE=1.67947 m/s2,˙xF=0 m/s, ˙yF=1.64625 m/s, ¨xF=0 m/s2, ¨yF=3.29262 m/s2,ω2=˙φ2=-1.1307 rad/s, α2=¨φ2=-22.33 rad/s2,ω3=˙φ3=-2.82169 rad/s, α3=¨φ3=-2.20443 rad/s2.ω4=˙φ3=0.58475 rad/s, α4=¨φ3=-21.453 rad/s2.Problem 4.4: R-RRR-RTT mechanismThe R-RRR-RTT mechanism is shown in Fig. P4.4. The following dataare given: AB=0.080 m, BC=0.350 m, CE=0.200 m, CD=0.150 m, La=0.200 m,Lb=0.350 m, and Lc=0.040 m. The driver link 1 rotates with a constantangular speed of n = 300 rpm. Find the velocities and the accelerationsof the mechanism when the angle of the driver link with the horizontalaxis is φ=155◦. For φ=155◦the position of the mechanism is given by:xB=-0.0725046 m, yB=0.0338095 m, xC=0.254847 m, yC=0.157668 m,xD=0.295983 m, yD=0.0134181 m, φ2=0.361716 rad, φ3=-1.293 rad.Results˙xB=-1.06216 m/s, ˙yB=-2.2778 m/s, ¨xB=71.5592 m/s2, ¨yB=-33.3686 m/s2,˙xC=-1.73662 m/s, ˙yC=-0.495228 m/s, ¨xC=37.3878 m/s2, ¨yC=27.6173 m/s2,˙xD=-3.03908 m/s, ˙yD=-0.866649 m/s, ¨xD=65.4287 m/s2, ¨yD=48.3303 m/s2,ω2=˙φ2=5.44543 rad/s, α2=¨φ2=197.52 rad/s2,ω3=˙φ3=-9.0292 rad/s, α3=¨φ3=217.641 rad/s2.LaLbLc1245φFigure P4.3AFCDEBxy0030nABCEDxy12543LbLcφFLa0000Figure
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