Finding lowest position of point B1 in kinematics mechanism with rotating lever

  • Thread starter Thread starter Matik
  • Start date Start date
  • Tags Tags
    Kinematics
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
8 replies · 2K views
Matik
Messages
5
Reaction score
0
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.

Homework Statement


So i have this kinematics mechanism. At the OA lever i have a known constant velocity, let's 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:
Physics news on Phys.org
No, its a mobile joint. My bad, i fixed it:
8cdee899b4e775f1755113f269d2da1a_610x0.jpg
 
Last edited:
No. Here are all dimensions and values i have:

OA = 0.45m
AB = 1.1m
BD = 1.6m
B1D=0.65m
h=0.5m
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 doesn't seems to be perpendicular at all. And i can't figure it out why is that.
f20adbd01f3bb89d589b2961bc8cf21c_610x0.jpg
 
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.
 
y is the vertical position of point B1 and theta is the angle OA makes with the x-axis.