Physics Help Forum Four-bar linkage mechanism

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

 Jun 11th 2013, 02:34 PM #1 Junior Member   Join Date: Mar 2013 Posts: 7 Four-bar linkage mechanism Hi there. After the excellent help I got the last time I was here got me through what ended up being a surprisingly straightforward test, I call on you again to come to my rescue in my time of need. Later on this week, I have my final test of the year, for which I have stacks of revision papers I can work through. Most of these I've managed to complete, with the only exception being one on rotating systems, specifically a four-bar link/chain mechanism. There are plenty of resources out there to help me, but none seem to cover specifically my problem. So with that, I'll begin with the actual question: To begin with, it's straightforward. I recreate the mechanism in the question, and create a velocity diagram, from which I can derive my various speeds: Partly derived from this, I can create an acceleration diagram ('ag' and 'ae' here come from my replacing the mass later): And then, I can replace the mass of BC with an equivalent 2 mass system (one at B, the other at E) and start calculating my inertia forces: The inertia force on B directs along BA. The inertia force on E is parallel to 'ae' on the acceleration diagram. B and E mass inertia force lines of action intersect, giving point on the line of action to the resultant inertia force on the link parallel to 'ag' from the acceleration diagram, but in the opposite direction. So far, I'm OK. In addition, I can calculate the inertia force F from the mass multiplied by 'ag' = 14kg * 10.9m/s = 152.6 N. And this is where I hit a bit of a wall. I know I can work out the torque on 'A' from multiplying the force through B by AY (measured from my last diagram). However, I'm not sure how to work out the force through B. According to the revision paper, I should be able to work this out from the parallelogram of forces at 'x' (indeed, it even gives an answer of 150 N). However, that's something that isn't covered on the syllabus of this course, and while the idea of the parallelogram or triangle of forces seems fairly straightforward, I can't seem to figure out how to apply it to work out that force. Sorry for the long-winded question. Any help gratefully received. R.