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Wikipedia describes torque as "Mathematically, torque is defined as the cross product of the vector by which the force's application point is offset relative to the fixed suspension point (distance vector) and the force vector, which tends to produce rotation." but later says the contradictory "r is the position vector (a vector from the origin of the coordinate system defined to the point where the force is applied)"

So which one is it? Is it $$(r_{force application}-r_{suspension point}) \times F(r_{force application})$$ or is it $$r_{force application} \times F(r_{force application})$$ ?

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3 Answers 3

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I think the first option is correct as torque is measured as the product of force and distance from the point of suspension to the point of application of force. The second can be correct in the case when co-ordinates of the suspension point are(0,0).

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  • $\begingroup$ It's what I think as well. The other one feels very uncovariant, the torque should not depend on the coordinate system. $\endgroup$
    – Emil
    Commented Aug 13, 2016 at 10:33
  • $\begingroup$ Torque DOES depend on coordinate system!! $\endgroup$
    – velut luna
    Commented Aug 13, 2016 at 10:35
  • $\begingroup$ When a torque is used to describe an equipollent force (force at a distance) then the location is important. When a torque is a pure torque then its location does not matter (it does have any). $\endgroup$ Commented Aug 13, 2016 at 14:59
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Both are correct as special cases of the general formula $$\mathbf{\tau}=\sum_i \left(\mathbf{r}_i-\mathbf{R}\right)\times\mathbf{F}\left({\mathbf{r}_i}\right)$$

Here $\mathbf{R}$ is the reference point with respect to which the torque is measured.

The first formula is for the special case when you call a point the suspension point and measure torque w.r.t it. Note that here, the force at the suspension point disappears and has no effect on the torque.

The second formula is for the special case when the origin is taken as the reference point w.r.t which the torque is measured. Note that if you have a point that you called a suspension point and that point is not the origin, then you need to take the force acting on it into account as well.

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  • $\begingroup$ I think R is what wikipedia calls the suspension point? $\endgroup$
    – Emil
    Commented Aug 13, 2016 at 10:33
  • $\begingroup$ How does wikipedia define suspension point? $\endgroup$
    – velut luna
    Commented Aug 13, 2016 at 10:35
  • $\begingroup$ They don't, from what I can see. $\endgroup$
    – Emil
    Commented Aug 13, 2016 at 10:35
  • $\begingroup$ They threw in "fixed" too, I wonder if that is unnecessary as well. $\endgroup$
    – Emil
    Commented Aug 13, 2016 at 10:39
  • $\begingroup$ You can take any point as your reference point, but the torque will be different. You have to fix your reference point in the same question. $\endgroup$
    – velut luna
    Commented Aug 13, 2016 at 10:41
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Torque is defined at a location. Changing the location will change the numeric value of torque.

For example torque at a location C of a force $\vec{F}$ passing through another location A is $$ \vec{\tau}_C = (\vec{r}_A-\vec{r}_C) \times \vec{F} $$

  • If C is the coordinate origin, then $\vec{r}_C=0$ $$\vec{\tau}_C = \vec{r}_A \times \vec{F}$$
  • If the absolute positions are not given, but only a relative vector $\vec{r}_{A/C}$ is known then $$\vec{\tau}_C = \vec{r}_{A/C} \times \vec{F}$$

In all cases you must designate the location of the torque. An exception exists when a torque is a pure torque (no net force exists) at which point its location is irrelevant.

The reverse calculation is

$$\vec{r}_C^\star = \vec{r}_A + \frac{\vec{\tau}_C \times \vec{F}}{\| \vec{F}\|^2} $$ Where $\times$ is the vector cross product.
The vector $\vec{r}_C^\star$ may be somewhere on the line of action of $\vec{F}$ and not on C exactly


The same math applies to a rotation $\vec{\omega}$ at a location A. The linear velocity of the extended body at another location C is

$$ \vec{v}_C = (\vec{r}_A - \vec{r}_C) \times \vec{\omega} $$

It is the same formula and the same rules apply.

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