# Torque on a Box

I think I'm missing something with torques. I seem to have gotten myself confused.

I have a box that's centered at ( 0 , 0 , 0 ) with length ( $x$ dimension ) = 1 , width ( $y$ dimension ) = 0.25, and height ( $z$ dimension ) = 0.5. The edges are parallel to the axes. The $x$ axis is left(-) and right(+), the $y$ axis is up(+) and down(-), and the $z$ axis is into(+) and out of(-) the page.

A force [ 0 , 50 , 0 ] is applied at the point ( 0 , 0 , -0.25 ). To find the torque, we would apply

$\tau = r \times F$

and so $r$ is [ 0 , 0 , -0.25 ] and $F$ is [ 0 , 50 , 0 ]. And the crossproduct is [ 12.5 , 0 , 0 ], so the torque is in the $x$ direction and the box should rotate clockwise?

If I were holding that box in my hand, then ( 0 , 0 , -0.25 ) would be on the side facing me. If I were capable of applying an upward force at ( 0 , 0 , -0.25 ), wouldn't the box start spinning away ( into the page ) from me - and not spin clockwise?

Thanks for taking the time to read. I would really appreciate any help with this.

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Firstly, if the $x$-axis is positive to the right, and the $y$-axis is positive upwards, then the positive $z$-axis should point out of the page in order for your coordinate system to be right-handed (recall that $\hat{\mathbf x}\times\hat{\mathbf y} = \hat{\mathbf z}$ and use the right-hand rule), so let's assume that this is the case (because I think this is what's causing the confusion).
As you point out, the torque is in the positive $x$-direction. This means that the box should rotate in such a way that after a $90^\circ$ rotation, the top is facing you, the side facing you is on the bottom, etc.
Actually, I think I get it now. The torque is about the $x$-axis, so it would be rotating around the $x$ axis, and then towards me. Is that right? I've been thinking about it like a force - thinking that if the torque is about the $x$-axis, it would rotate in the $x$ direction, which is not correct. – user2570465 Aug 8 '13 at 15:23