How & what is a newton meter of torque?

[SSSSE YA]

I was watching a show on Speed Vision and they said the car had so many newton meteres of torque.How does that compare to footpounds of Torque??

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[Performin Norman]

I wish I knew the conversion for you but it is the metric equivalent to pound feet. I am too lazy to go dig out my engineering books. haha

 

[Rich01RT10]

Newton Meters X 0.73756212 = Foot Pounds
Ex. 100 Nm = ~73.8 ft/lbs

Hope this helps -

 

[v10venom]

Maybe this can help

Newton/Meters and Pound/Feet are both measures of a force over a distance, or work. In the metric system they are called Joules, In the english system they are called something like slugs,

The only reason I know this is because I had physics this morning.

 

[v10venom]

JTV
I will talk with my physics professor tomorrow. I think it has something to do with the direction that you apply the force. I think they are interchangeable. Both are used when talking about torque.

Jon

 

[Venom Lover]

Torque is the ability of a force to create rotational motion. If a force F is applied perpendicularly to a rigid object at a distance D from an axis of rotation of the object, a torque in the magnitude of F x D will be applied around the axis of rotation. If F is a three-dimensional vector of the force F and D is a three-dimensional vector from the axis of rotation to the position the force is applied, then the vector T computed by the vector cross product T=DxF will give the direction of the axis of rotation (and the magnitude of T will be the magnitude of the torque).

As far as units are concerned, you can see that F x D will have units of lbs x feet (or ftlbs) in English units, or Newtons x meters (or Nm) in metric units. Rich is correct above, the conversion factor from English to metric is 1 Nm = 0.737562... ftlbs.

Work, on the other hand goes like this....Say the same 3-dimensional vector force F is applied to an object, and the object moves in a vector direction D. (We're not talking about rotation, now, we're talking about physical translation of the object in space.) The amount of energy, or work, that results is the vector dot product F.D. This also, coincidentally, has units of Nm or ftlbs, and when you talk about work or energy, Nm is typically referred to a Joule of energy, abbreviated J.

I'm sure this helped clear things up a lot!

P.S., my degree in physics finally paid off!

 

[GTS Dean]

Sorry guys.....

Work is a measure of force applied over a given unit of time. A fat man standing on the end of a lug wrench on the lug nut of his pickup is a torque exerted. (because some **** installed it with an impact with the knob screwed all the way in) No work is produced until the nut gives.

I'm just an Aggie road contractor.

 

[Venom Lover]

<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>Originally posted by GTS Dean:
Sorry guys.....

Work is a measure of force applied over a given unit of time. A fat man standing on the end of a lug wrench on the lug nut of his pickup is a torque exerted. (because some **** installed it with an impact with the knob screwed all the way in) No work is produced until the nut gives.

I'm just an Aggie road contractor.
<HR></BLOCKQUOTE>

Dean, you are thinking about power. Power is work (or energy) divided by time, which has units of Nm/sec or J/sec, both of which are called a Watt (W). In fully expanded MKS units, 1 Watt = 1 kg x m^2 / sec^3.

Force divided by time is meaningless, as you can ascertain from the units. Force is mass times acceleration (Newton's second law), so in MKS metric, the units for force are kg x m / sec^2. Divide force by time and you get kg x m / sec^3. Distance divided by time cubes (e.g., m / sec^3) are in units of the third derivitive of distance. This is called "****". (I'm not making this stuff up!!!) Anyway, **** times mass (e.g., kg x m / sec^3) has no valuable physical meaning that I'm aware of. For sure it is not work, which has MKS units of kg x m^2 / sec^2, which is equivalent to the Newton-meter or Joule, as I discussed previously.

Your statement about not doing "work" until the the nut gives is correct. Of course, energy is being exerted by your muscles in a futile attempt to overcome static friction, which keeps the nut from moving, so it may feel like you are "working", but you are not doing "work" in the classical physics sense.

Any questions???

 

[GTS Dean]

No questions, and thanks for the clarification.