Okay, hopefully I can describe this so that its understandable. I have a wooden rectangle that is 10" wide and 20" tall and 3" thick. It weighs about 6 or 7 pounds. Imagine it positioned so the the 10" end is facing the floor. I need to have it so that it rotates 90 degrees under motive power so that the 20" side of it becomes paralell with the floor. At the starting position the pivot point will be the high end of the rectangle, opposite the floor and right at the corner where the 10" and 20" sides meet. I'd like to know how to calculate the required motor force using a sprocket and chain system with one spocket being attached to the same shaft that will provide the pivot point and the other sprocket being attached to the motor itself. That's my biggest problem actually- figuring out how strong the motor needs to be. I also need to know how to figure this out using a linear actuator. With the same pivot point and the one end of the actuator attached somewhere along the 20" side of the rectangle. Again the wooden rectangle is 10" by 20" by 3" thick and weighs about 6 or seven pounds.

Can you post a picture or better describe what you are doing? When the 10' side is facing the floor does it touch the floor or float in the air 10' up?

Assuming that the rectangle is of uniform construction, so that its center of mass is at the geometric center of the rectangle, the torque needed to be applied at the pivot point to counter-act gravity is:


\tau = 7 \ lbs\ \times\ (\frac {20"} 2 ) sin \theta


where \theta is the angle of the rectangle relative to its initial vertical position. This becomes a maximum at 90 degrees, so the torque required is:


\tau = 7\ lbs\ \times\ 10 \ in = 70\ lb-in


This is the torque needed to maintain the object parallel to the floor. However, this does not take into account how quickly you want the rectangle to be raised (it's acceleration). I would suggest that you select a motor that is at least 150 lb-in of torque - that should be enough to raise the object with no difficulty.