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enter image description here

From the previous examples I've seen, there is a connection rod, for example, between a and c and then another rod from c to one of the pins. However, in this case, the rod ac is directly connected to another pin? I assume there is no relative velocity in this case. Is this assumption correct?

EDIT: By pins, I was referring to O2 and O8. I've just realised that C is a slider. I thought it was a rigid 'pin' like O2 and O8. Sorry for the confusion.

So I think I know how to approach this problem now. Va is perpendicular to OA. Vc is a horizantal line from O2. Vc/a is perpendicular to the rod AC starting from Va.Then I need to intersect Vc and Vc/a to find the magnitude of Vc. Is this approach correct?

EDIT 2: I'm running out of time. I quickly sketched this to check if I'm on the right track.

enter image description here

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    $\begingroup$ Hi Alp, welcome to engineering.SE. It isn't clear to me what you are asking at the moment. What are the 'pins' that you refer to? Can you please expand your question by editing it to give it more context? $\endgroup$ – Chris Mueller Apr 7 '15 at 11:52
  • $\begingroup$ sorry for the confusion. I think I know how to solve the question now. I'd appreciate it if you can double check. $\endgroup$ – Alp Apr 7 '15 at 12:03
  • $\begingroup$ That is great, how about you post the solution so the community can provide feedback. Welcome to engineering.se $\endgroup$ – Mahendra Gunawardena Apr 7 '15 at 12:08
  • $\begingroup$ draft solution added $\endgroup$ – Alp Apr 7 '15 at 12:15
  • $\begingroup$ You can check your results with forceeffect.autodesk.com/frontend/fe.html by doing a kinematic analysis. $\endgroup$ – John Alexiou Apr 7 '15 at 20:53
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Draw a vertical line at C. The instant center of rotation of the connecting rod AC is located somewhere along this line. To find the exact location, extend OA to intersect this line (below the slider). Now imagine a CW rotation about this point K will move the slider to the right and the crank perpendicular to OA.

pic

Now the relative rotation of DB to AC is at B. The law of instant centers means that body DB rotates about a point along the line KB. Since it is connected to O8 at D the rotation center of DB is at M.

In the end, velocity at C is perpendicular to KC, velocity at A is perpendicular to O2A, velocity at D is perpendicular to O8D and MD and finally velocity at B is perpendicular to KB and MB.

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I have had the same exercise in school, maybe it helps, you just know $B= 1/2 AC$, and you know C is in horizontal line from 0.

If you want the acceleration diagram let me know, but it is just the same.

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