Torque vs. Pounds 1

[dklarz]

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.

 

[GrimesFrank]

Work is defined as F*d, torque is a rotational force applied at a defined radius (i.e. F*r).

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.

 

[Airmack]

I would have to say no. The inch unit in torque refers to the distance that the tangent force is from the rotation axis.

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).

 

[eromlignod]

I'm not sure what you mean by "travelled".

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

 

[HVACctrl]

What is the distance from the center of rotation?

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

http://www.engineerboards.com

 

[zekeman]

Simply put, the problem with torque vs energy units is that that they are dimensionally the same since torque is force times radius to center
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.

 

[dklarz]

Thanks to all! I think I've got it. What's happening here is I have a lathe chuck that I want to drive with a motor. The torque is measured from a torque wrench as it opens the chuck. The diameter of the pinion gear is 1.4". Therefore the force derived should = 100"#/(1.4/2)" or 143#. Right? I need to know the force for other things that the chuck is doing. (My arguement is with another engineer who came up with 200# through some conversion thing that I don't agree with.) Thanks again.