Resultant Torque independent of origin?

AI Thread Summary
The discussion focuses on proving that the resultant torque of a system of forces is independent of the choice of origin. The user initially struggles with the concept but is guided to subtract the two torque equations, leading to the realization that T - T' equals (x0 - x0') X (∑ Fi). Since the sum of forces is given as zero, this simplifies to zero, confirming that T = T'. The conversation concludes with the user successfully understanding that the independence of torque from the origin is established through the condition that the resultant force vanishes.
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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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