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Homework Help: Issue with kinematics task

  1. May 2, 2015 #1
    Hello community. I have a task from my college and no matter how hard i tried i failed to find a solution for it. The task itself is resolvable (by drawing) but i can't figure the mathematical way to solve it. And it bothers me so bad.

    1. The problem statement, all variables and given/known data
    So i have this kinematics mechanism. At the OA lever i have a known constant velocity, lets say - ω with direction counter clockwise around point O. I have given these dimensions: OA, AB, AD, DB1 and h aswell. My task is to determine the lowest position of element/point B1. Of course i plotted enough points and determined that (see the picture)... but there should be mathematical way to determine this. And i can't find it.

    Initialy i thought... the lowest position of B1 is when OA is alongside to DB1. But i was wrong. Then i thought that if the velocity that occurs at point D is perpendicular to DB1 then the velocity at point B1 will be equal to 0. But this solution does not match with the position from the picture either. c61b0913e8ff7f0c54fdd1f710bc176a_610x0.jpg
    Last edited by a moderator: May 2, 2015
  2. jcsd
  3. May 2, 2015 #2
    Is point D a pinned joint?
  4. May 2, 2015 #3
    No, its a mobile joint. My bad, i fixed it: 8cdee899b4e775f1755113f269d2da1a_610x0.jpg
    Last edited: May 2, 2015
  5. May 2, 2015 #4
    is length B1-D less than h?
  6. May 3, 2015 #5
    Last edited: May 3, 2015
  7. May 3, 2015 #6
    No. Here are all dimensions and values i have:

    OA = 0.45m
    AB = 1.1m
    BD = 1.6m
    ωoa = 4 rad/s (counter clockwise)

    I already plotted full cycle with 6 positions and by the lowest position of B1 there i plotted another 10-12 points near it, increasing the resolution. Thus i found (roughly) the lowest position of B1 shown on the picture up there. But i still think there has to be another - much more accurate and mathmatical way to solve this.

    I think that if B1 is in the lowest position its velocity should be equal to 0. Therefore at point D velocity should be perpendicular to DB1:
    ωoa =Voa . OA We have ωoa and OA so Voa is known. If we separate vector Voa by sinα.Voa and cosα.Voa we will have at point B - ωbd=(sinα.Voa)/AB and the point D this velocity will be vector Vda=ωbd.BD=(sinα.Voa).(BD/AB) plus the other vector cosα.Voa. But when i calculate and plot them it doesnt seems to be perpendicular at all. And i cant figure it out why is that.
  8. May 3, 2015 #7
    I think your approach looks valid. However, might a simpler be simply geometrical? Can you express y in terms of θ? Then set dy/dθ=0.
  9. May 3, 2015 #8
    I didn't understand you. What are y and θ?
  10. May 4, 2015 #9
    y is the vertical position of point B1 and theta is the angle OA makes with the x-axis.
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