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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.

This makes physical sense since you're applying a force on the edge opposite the one facing you that is upward. In other words, you're holding the box, you tie a rope to the center of the side opposite the one facing you, and you pull up.

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Thanks for pointing out my error with the axes. I'm still a bit confused on how a torque in the positive x direction (right) could cause the box to rotate towards me. –  user2570465 Aug 8 '13 at 14:13
    
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
    
@user2570465 Yes that's exactly right. If you point the thumb of your right hand in the direction of the net torque and curl your other fingers around, then the direction they curl gives the direction of the rotation of the object about the axis defined by your thumb. –  joshphysics Aug 8 '13 at 16:36
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