Calculating Net Torque on a Circle with Multiple Forces

  • #1
physicsCU
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  • #2
I have -0.94, provided the z axis points outward the computer screen.

By the way, what is the unit of torque?

[tex]\vec{\tau} = \vec{r}\times \vec{F}[/tex]

So it's N*m = J !

I realize that the N*m of torque is not the same kind of N*m as that the one involved in work, for exemple. The first speaks of Force exerted AT a distance, the second of a force exerted ON a distance. But the units remain N*m, and that's Joules! I'm thrown down my chair.
 
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  • #3
Torque is in N*m

Can I ask how you got that answer though?
 
  • #4
But N*m is Joules. Don't you find that strange? I find that troubling.

I simply calculated all the torques and added them together:

-.10*30 + .5*30*sin(45°) + .5*20
 
  • #5
I have accepted it, no worries.

But what about that force that is off the circle in the third quadrant? Do you ignore that, or what?
 
  • #6
That last vector isn't doing any work to turn the circle, so its got no effect. Remember Torque is a cross prodcut between Force and Radius, the force is parallel to the radius, in which case sin(0)=0 and the vector applies no torque.
 
  • #7
Quasar makes an interesting point however--- the concept of torque being measured in units of Nm ---> Joules? Energy?

It's best to keep your units of moments around a point in Nm, just as when one calculates the complex power for ckt elements under an sinusoidal steady state voltage or current... the real power (avg power) is measured in Watts, whereas the imaginary reactive power stays as VA (volt amps), rather than watts.
 
  • #8
I read an article about that. I don't remember what the explanation was, but they said it made sense.
 
  • #9
you might be correct.
 
  • #10
Anyway just to clarify, Torque is the turning effect, although it has the same units as Joules it's not energy, so it msut never be written in J, but N m.
 
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