Answer: Calculating Linear Force from 5000 lb-ft of Torque

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A rotary drive generating 5000 lb-ft of torque applies a linear force based on the distance from the pivot point. Specifically, at a 12-inch arm, the torque translates to a force of 5000 lbs acting perpendicular to the arm. This means that the force exerted on the connecting rod will be 2500 lbs, as the force is distributed over the length of the arm. Understanding this relationship between torque and linear force is crucial for effectively operating the damper system. Proper calculations ensure accurate performance of the linkage system.
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I am working with a rotary drive and a linkage system to open and close dampers. I am having trouble recalling how lb-ft and lbf relate. The drive has a rotating shaft with and arm extending from it. The shaft generates 5000 lb-ft of torque and I need to understand what linear force this applies to a connecting rod attached to the opposite end of the 12" arm extending from the shaft. The rotating shaft is in one end of the 12" arm. The arm is horizontal (on the x-axis) and the connecting rod is attached to the other end of the arm and it perpendicular to the arm (on the y-axis). I hope this is clear. Thanks.
 
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A drive which provides 5000lb-ft of torque will generate 5000 lbs of force at a distance of 1 foot, in the direction perpendicular to your 'arm'.
 
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