Suggestion
First determine the location of the instant center of link AB. To do so, determine the direction of the velocities of point A and B. The IC for link AB is found at the intersection of the perpendiculars to these two velocities. Call this intersection point C.
Then:
vA = (OA)*(wOA)
vA = (AC)*(wAB)
vB = (BC)*(wAB)
vB = (BD)*(wBD)
These four equations can be used to determine the angular speeds of links AB and BD. Determine the direction of rotation of these links from knowing the location of the IC of links AB and BD.
6 comments:
does anyone know at which direction point B is moving? I can't seem to figure out if it's moving horizontally to the left or diagonally upwards to the left?
v_B is horizontal to the left. because the direction of the velocity is tangent to its circular path.
So v_B points to the left and v_A points up and to the left (perpendicular to OA) then that would mean the IC is actually at point A? Is that right?
my answers are wAB = 2 rad/sec CCW and wBD = 5.33 rad/sec CW. Is this what you guys are coming up with?
So v_B points to the left and v_A points up and to the left (perpendicular to OA) then that would mean the IC is actually at point A? Is that right?
I can't figure out how to include a picture in a comment, but here is how I found IC.
If you are having a questions as to the location of the IC for link AB, please see the figure that ccb056 posted above.
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