Resultant Torque independent of origin?

In summary, Tim provided a summary of the content, and explained that the resultant torque is independent of the choice of origin.
  • #1
MrLiou168
14
0

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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  • #2
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:​
 
  • #3
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!
 
  • #4
MrLiou168 said:
… subtracting the 2 equations nets me x0 - x0'.

nooo …

what about all those forces? (∑ Fi) :wink:​
 
  • #5
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'...?
 
  • #6
(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:​
 
  • #7
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?
 
  • #8
correct! :smile:
 
  • #9
Tim, thank you very much - you were extremely helpful!
 

FAQ: Resultant Torque independent of origin?

1. What is the definition of resultant torque independent of origin?

Resultant torque independent of origin refers to the torque or rotational force on an object that is not affected by the position or origin of the object. This means that the direction and magnitude of the torque will remain the same regardless of where the object is located in space.

2. How is resultant torque independent of origin calculated?

Resultant torque independent of origin is calculated by finding the cross product of the force vector and the position vector from the origin to the point of application of the force. This can be represented by the equation τ = r x F, where τ is the resultant torque, r is the position vector, and F is the force vector.

3. What are some examples of situations where resultant torque is independent of origin?

One example is a seesaw, where the torque on either end of the seesaw is independent of the position of the pivot point. Another example is a spinning top, where the torque remains the same regardless of the orientation of the top.

4. How does the direction of the force affect the resultant torque independent of origin?

The direction of the force vector affects the direction of the resultant torque, as the torque will be perpendicular to both the position and force vectors. This means that changing the direction of the force will also change the direction of the resultant torque.

5. What is the significance of understanding resultant torque independent of origin in physics?

Understanding resultant torque independent of origin is crucial in the study of rotational motion and equilibrium. It allows us to accurately predict the motion of objects and determine the forces acting on them, which has practical applications in fields such as engineering and mechanics.

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