Resultant Torque independent of origin?

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Discussion Overview

The discussion revolves around the concept of resultant torque in a system of forces acting on a body, specifically examining whether the resultant torque is independent of the choice of origin. The context includes a homework problem that requires participants to demonstrate this independence mathematically.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant states the need to show that the resultant torque T is equal to T' for two different origins x0 and x0', given that the net force is zero.
  • Another participant suggests subtracting the two torque equations to explore the relationship between T and T'.
  • A participant expresses confusion about the implications of their calculations, questioning whether stating x0 - x0' = 0 is sufficient to prove the independence of torque from the choice of origin.
  • Further clarification is provided regarding the necessity of considering the sum of forces in the torque equations.
  • It is noted that the sum of forces being zero leads to the conclusion that T - T' = 0, thereby supporting the claim that the resultant torque is independent of the choice of origin.

Areas of Agreement / Disagreement

Participants appear to reach a consensus on the mathematical proof that the resultant torque is independent of the choice of origin, contingent on the condition that the net force is zero.

Contextual Notes

The discussion includes some confusion regarding the manipulation of torque equations and the implications of the sum of forces, which may affect the clarity of the proof process.

Who May Find This Useful

Students studying mechanics or torque in physics, particularly those working on homework problems related to forces and torques.

MrLiou168
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Homework Statement



Let a system of forces (F1,...Fn) act on a body at points (x1,...xn) respectively. Assume that the resultant or net force vanishes (sum of forces = 0)

Show that the resultant torque of this system is independent of the choice of origin, i.e. for 2 different origins x0 and x0', we have T = T' where:

T = [summation](xi-x0) X Fi and T' = [summation](xi-x0') X Fi

Homework Equations



T = F X d


The Attempt at a Solution



I have very little idea as to how to approach this problem, other than the T = Force X distance, and perhaps that the solution may have something to do with a bunch of couple moments that all equate to the same value, regardless of origin. Any help greatly appreciated!
 
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Hi MrLiou168! :smile:
MrLiou168 said:
T = [summation](xi-x0) X Fi and T' = [summation](xi-x0') X Fi

ok, now subtract one from the other:

you get … ? :wink:
 
tiny-tim said:
Hi MrLiou168! :smile:


ok, now subtract one from the other:

you get … ? :wink:

Thanks for the reply! But not quite, I'm a bit of a dim light most of the time... So assuming T = T', then T-T' = 0...

And subtracting the 2 equations nets me x0 - x0'. Therefore, is it sufficient to simply state that x0 - x0' = 0 to prove that the resultant torque is independent of choice of origin?

Thanks!
 
MrLiou168 said:
… subtracting the 2 equations nets me x0 - x0'.

nooo …

what about all those forces? (∑ Fi) :wink:
 
Oh... sorry I didn't realize the F's were summed as well...

So I have (x0 - x0') X Fi = 0? And since Fi is not zero, the x0 = x0'...?
 
(try using the X2 button just above the Reply box :wink:)
MrLiou168 said:
So I have (x0 - x0') X Fi = 0? And since Fi is not zero, the x0 = x0'...?

hmm, you're really confused :redface:

you don't know T - T' = 0, that's what you're trying to prove!

all you have proved is T - T' = (x0 - x0') X (∑ Fi) …

now, what do you know about ∑ Fi ? :wink:
 
A ha thank you for your patience! OK so one of the givens is that [sum] Fi = 0...

Therefore the right side of the equation nets zero and we have proven that T-T' = 0 and thus T = T' correct?
 
correct! :smile:
 
Tim, thank you very much - you were extremely helpful!
 

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