Say you have a sphere, and you have several torque vectors acting on it, all at different points. Say you have the vector (6i + 3j + 5k) originating from point A, and the vector (3i + 1j + 9k) originating at point B, and (7i + 2j + 9k) acting on point C.

Summing the vectors gives you (16i + 6j + 23k) which is the resultant moment/torque vector. But at what point does the moment act on - A, B, or C?

The point acts on has to matter right? I mean if you think of the moment vector as an axis the sphere revolves around, placing it in the center of the sphere and rotating the sphere around that is clearly different from placing it at the far left of the sphere and rotating it around that.

  • $\begingroup$ Did you know that if only pure torques are applied on a rigid body (such that sum of forces is zero) then the body would rotate about it's center of mass. To rotate a body about an axis away from the COM you have to apply a net force. $\endgroup$ – John Alexiou Aug 14 '14 at 15:12

So you know about how to get the effective moment of all the forces

$$ \vec{M} = \sum_{i} \vec{r}_i \times \vec{F}_i $$

and the total forces

$$ \vec{F} = \sum_i \vec{F}_i $$

To get the location where the moments balance out (the line of action of the combined force) you do the following

$$ \vec{r} = -\frac{\vec{M} \times \vec{F}}{\vec{F} \cdot \vec{F}} $$

for example a force $\vec{F}=(1,0,0)$ located at $\vec{r}=(0,y,z)$ creates a torque of $\vec{M}=(0,z,-y)$. To recover the location of the force do $$r = -\frac{ (0,z,-y) \times (1,0,0) }{(1,0,0)\cdot (1,0,0)} =- \frac{(0,-y,-z)}{1} = (0,y,z) $$

  • $\begingroup$ I didn't ask how to calculate torque in 3D. I know to how find the crossproduct. Please read the description of my question, I asked about the point where the resultant moment acts when each individual moment acts at different points. $\endgroup$ – dfg Oct 12 '13 at 20:30
  • $\begingroup$ Moment does not have a point it acts upon. Forces have a line of action. To get the location of the line of action from a force/moment pair do $$ \vec{r} = -\frac{\vec{M} \times \vec{F}}{\vec{F} \cdot \vec{F}} $$ $\endgroup$ – John Alexiou Oct 12 '13 at 20:33
  • $\begingroup$ Why don't moments have points they act on? A sphere rotating around its centre is different from a sphere rotating around a point near its edge right? $\endgroup$ – dfg Oct 12 '13 at 20:37
  • 1
    $\begingroup$ When you apply a pure torque to a rigid body the effect is the same regardless of where you apply it. When you apply pure force then the location matters. See the updated answer now. $\endgroup$ – John Alexiou Oct 12 '13 at 20:42
  • $\begingroup$ Could you please explain why it doesn't matter in terms of the rotating sphere? $\endgroup$ – dfg Oct 12 '13 at 20:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.