In each one of the following figures there's a pole of length $1.2 \text{m}$ and there's a force $\vec F = 150 \text{N}$ acting on it. Determine the torque that is created by the force relative to the axis of rotation $O$.


I had no problem with calculating the magnitude of the torque, but I did have a problem determining its direction in fig. $D$. I used the rule of the right hand and I got that its direction is "outside the monitor" while according to the answers it has the same direction as in all the other figures before (i.e. "inside the monitor"). I'm also wondering whether the vectors directions are important in cross product. For example - is the direction of the displacement vector is important? If it is, then how can I apply the rule of the right hand here when the thumb and the index finger aren't starting from the same position.

  • $\begingroup$ Vectors encode magnitude and position only. Where you choose to draw them is just that - a choice to help remind yourself where the relevant interaction is. You can always slide all vectors to originate at the origin if it helps visualize the cross product. $\endgroup$ – user10851 Apr 7 '13 at 19:33
  • $\begingroup$ @ChrisWhite - I understand that. But what about that particular example at fig. $D$ - what am I doing wrong? $\endgroup$ – brmch8 Apr 7 '13 at 19:36
  • $\begingroup$ Umm, the direction of $\vec{\tau}$ is into the plane for D (the rod will clearly rotate clockwise), while it is out of the plane for A, B, and C (it will rotate counterclockwise). Is this not what the answers say? $\endgroup$ – user10851 Apr 7 '13 at 19:42
  • $\begingroup$ @ChrisWhite - that's what I thought as well, but no, the book says it is the same in all these cases (except of course from the last ones where the torque is simply zero). Because I'm new to that, I was thinking that I got the whole cross product thing wrong. It seems like the book has a mistake. Thank you for confirming that assumption! $\endgroup$ – brmch8 Apr 7 '13 at 19:46
  • $\begingroup$ Al long as $\tau$ has positive magnitude, the directions will be different. The book might be doing something confusing and saying the directions are all the same but the magnitudes change sign. (I don't know why they would obfuscate things this way, but there it is.) $\endgroup$ – user10851 Apr 7 '13 at 19:49

The torque is defined as the cross product between the force vector and the displacement vector from the center of the object to the point where the force is applied: $\vec{\tau}=\vec{r}\times\vec{F}$.

The direction of the vectors is important when calculating cross products. Notice if the direction is reversed in the displacement vector:




So as you can see, reversing the direction of the displacement vector will reverse the direction of the torque. A similar proof will demonstrate that the same will happen if the force vector is reversed.

Another way to look at the right hand rule (because I too have a bit of trouble using the finger rules) is to instead think of it as, put the bases of the two vectors together, curl your fingers in a direction that goes from the first vector ($\vec{r}$ in this case) to the second vector ($\vec{F}$). The direction that your thumb points in is the direction of the torque.

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    $\begingroup$ The reasoning is correct but the definition is not: it's the other way around and as the cross product is antisymetric, it will change the sign: $\vec{\tau}=\vec{r}\times\vec{F} = - \vec{F}\times\vec{r}$ $\endgroup$ – JJ Fleck Apr 7 '13 at 19:19
  • $\begingroup$ I appreciate your answer regarding the cross product, but could you elaborate more on that particular example I gave? I can't actually put together the bases of the vectors (for example in the 1st figure) because then the displacement vector will change and there will be no torque. As I understand, what I can do is to move the vector in its direction (which helps) - in the first example I can move $\vec F$ only vertically. But still, I can do the right hand trick only if I assume that the base of the displacement vector is not in $O$ but in the point where the force is applied. $\endgroup$ – brmch8 Apr 7 '13 at 19:33
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    $\begingroup$ @brmch8 You can slide $\vec{F}$ to the origin, but the displacement $\vec{r}$ does not change. $\vec{r}$ is not the vector from the origin to the location where you choose to draw $\vec{F}$, it is the vector pointing along the rod whose magnitude is the distance between the origin and the point of the applied force. $\endgroup$ – user10851 Apr 7 '13 at 19:36
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    $\begingroup$ No, moving the vector doesn't mean that you're literally moving where the force is applied. There is a difference between moving the vectors and moving the actual point of torque action. Just move $\vec{F}$ to the axis of rotation and don't change the displacement vector. Moving $\vec{F}$ is not applying the force at a different point. This is a fundamental concept you'll need to understand in order to grasp the mathematics of rotational dynamics. $\endgroup$ – Ataraxia Apr 7 '13 at 19:45
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    $\begingroup$ @brmch8 Moving a vector is like moving the compass on a map - you are not declaring that the north pole has moved, or that the center of the Earth is in a new location. Boston will always be north of New York, no matter where the compass is, and the point of contact will always be $\vec{r}$ away from the origin, no matter where $\vec{F}$ is drawn. $\endgroup$ – user10851 Apr 7 '13 at 19:47

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