Resultant Torque independent of origin?

Click For Summary
SUMMARY

The discussion centers on proving that the resultant torque of a system of forces is independent of the choice of origin. The key equations involved are T = Σ(xi - x0) × Fi and T' = Σ(xi - x0') × Fi, where the sum of forces ΣFi equals zero. By manipulating these equations, it is established that T - T' = (x0 - x0') × (ΣFi), leading to the conclusion that T = T' when ΣFi = 0, confirming the independence of resultant torque from the origin.

PREREQUISITES
  • Understanding of torque and its mathematical representation (T = F × d)
  • Familiarity with vector operations, specifically the cross product
  • Knowledge of equilibrium conditions in mechanics, particularly the sum of forces
  • Basic algebraic manipulation skills to handle equations
NEXT STEPS
  • Study the principles of torque in rigid body dynamics
  • Learn about the implications of the cross product in physics
  • Explore the concept of equilibrium and its applications in mechanics
  • Investigate advanced topics in vector calculus relevant to mechanics
USEFUL FOR

Students of physics, particularly those studying mechanics, educators teaching torque concepts, and anyone interested in the mathematical foundations of physical systems.

MrLiou168
Messages
14
Reaction score
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!
 
Physics news on Phys.org
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!
 

Similar threads

Replies
6
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
3
Views
4K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 3 ·
Replies
3
Views
6K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 11 ·
Replies
11
Views
2K