Question #200281

In a pin jointed four bar mechanism ABCD, the lengths of various links are as follows:

A B = 25 mm ; BC = 87.5 mm ; CD = 50 mm and AD = 80 mm. The link AD is fixed and the angle BAD = 135°. If the velocity of B is 1.8 m/s in the clockwise direction, find 1. velocity and acceleration of the mid point of BC, and 2. angular velocity and angular acceleration of link CB and CD



1
Expert's answer
2021-06-07T05:45:41-0400

VBA=1.8m/sec;VBC=20.5=1m/sec;VC=VCD=3.10.5=1.55m/secV_{BA}=1.8m/sec; V_{BC}=2*0.5=1m/sec;V_{C} =V_{CD}=3.1*0.5=1.55m/sec

VE=340.5=1.7m/secV_{E}=34*0.5=1.7m/sec

For angular velocity of BC and CD

BC=VBCBC=10.0875=11.43rad/sec;CD=VCDCD=1.550.050=31rad/secBC= \frac{V_{BC}}{BC}= \frac {{1}}{0.0875}=11.43rad/sec; CD= \frac{V_{CD}}{CD}= \frac {{1.55}}{0.050}=31rad/sec

aBAr=(VBA)2BA=(1.8)20.025=129.6m/seca^r_{BA}=\frac{(V_BA)^2}{BA}=\frac{(1.8)^2}{0.025}=129.6m/sec

aCBr=(VCB)2CB=(1)20.0875=11.43m/seca^r_{CB}=\frac{(V_CB)^2}{CB}=\frac{(1)^2}{0.0875}=11.43m/sec

aCDr=(VCD)2CD=(1.55)20.0.050=48.05m/seca^r_{CD}=\frac{(V_CD)^2}{CD}=\frac{(1.55)^2}{0.0.050}=48.05m/sec

No Tangential component of BA

ae=8.912=106.8m/seca_e=8.9*12=106.8m/sec

For angular accn\space acc^n of BC and CD

aBCt=6.312=75.6m/seca^t_{BC}=6.3*12=75.6m/sec

aBCt=(BC).αBC=75.60.0875=864rad/seca^t_{BC}=(BC). \alpha_{BC}=\frac{75.6}{0.0875}=864rad/sec

aCDt=5.212=62.4m/sec2=aCDtCD=62.40.050=1248rad/sec2a^t_{CD}=5.2*12=62.4m/sec^2=\frac{a^t_{CD}}{CD}=\frac{62.4}{0.050}=1248rad/sec^2


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