Torque vs. Pounds
Torque vs. Pounds
(OP)
Settle a difference for me... If I have a known torque (100 lbs-in) and a known distance of travel (.o44in), can I determine a linear force by simple division? I say 100/.044 = 2273lbs. Is this correct? Thanks.





RE: Torque vs. Pounds
So if you are saying you traveled .044" circumferentially at a torque of 100 in-lbs, you would need the angle of travel to know what force you applied at the centre of radius.
RE: Torque vs. Pounds
http://en.wikipedia.org/wiki/Torque
The definition that you have sounds closer to a work equation, which is force through a distance, but using the word 'torque' in place of 'work' (both of which have the same 'in-lb' units).
RE: Torque vs. Pounds
Torque is a force applied at a certain "lever distance" from the center of rotation. So if you applied 100 lbs tangentially at the end of a 1 ft lever, then you would produce a torque of 100 lb-ft. The longer the lever, the less force you need to apply to get the same torque. For example, you would only need 50 lbs if your lever was 2 ft long (50 x 2 = 100).
If you want to know how much "angular work" you did when you turned the lever, then it is
W = T x angle
or work W is equal to the torque T applied (in lb-ft), multiplied by the angle turned (in radians, not degrees).
Don
Kansas City
RE: Torque vs. Pounds
Divide 100 lb-in by that number of inches to get your force. It looks like, according to your statement, the 0.044 in is the circumferential travel, not the radial distance from the center of rotation.
You pushed with a force of 100 lb-in divided by radial diatance from center for a circumferential distance of 0.044 in, correct?
Ed
www.engineerboards.com
RE: Torque vs. Pounds
F*R, pound feet
and energy is torque times the angle the angle traversed, @
F*R*@, foot pounds
since @ is dimensionless
Also note that while they are dimensionally the same,to avoid confusion , conventioon has it that torque is usually expressed in pound-feet and energy in foot-pounds.
RE: Torque vs. Pounds