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I am attempting to size a linear actuator (check that the rated torque will be enough to open a damper). My damper is a circular plate that has a mass of 20kg. It will pivot about one side of the plate (not through the vertical centreline of the plate). The actuator is rated for 500 N.m.

I know:

T = F * r and F = ma

So the force required to pivot the damper can be determined from F = ma, giving me my required force. And If I know the radius/distance from my pivot location I can calculate the torque. Completing these calculations (assuming 0.3m from pivot axis):

F = 20 * 10 = 200 N

T = 200 * 0.3 = 60N

This doesn't make sense to me, as I increase the distance from the pivot axis, the required torque increases? I thought it would make sense for the required torque to decrease? If anyone has suggestions on how to properly check whether the rated actuator can open the damper of 20kg mass that would be greatly appreciated. Also if the applied motion of the linear actuator is not perpendicular to the axis of rotation, I assume I would just need to add in sin(theta) into the torque calculation.

Cheers

this guy123
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  • It could be that you have the terms mixed up. If you apply a force of 200 N at a radius of 0.3 m, the torque that the force makes is $60~\text{Nm}$. But this might be different from a "required torque". If the torque is constant, the radius or force should change. So I'm not really sure of what's constant here ... – Matti P. Mar 13 '19 at 09:08
  • Also, I think this question would better fit the Physics StackExchange. – Matti P. Mar 13 '19 at 11:43

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