Posted by Natalie on Friday, November 9, 2012 at 5:07pm.
A force F = 0.8 i + 2.5 j N is applied at the point x = 3.0 m, y = 0 m. Find the torque about the point x = 1.3 m, y = 2.4 m.
Hmm, (this is essentially just a vector problem) I know you're supposed to find the position vector r and take the cross product of r x F, the force vector, for the torque vector. What I seem to be having trouble with is understanding how to come up with the position vector. So I thought that you're applying the force at point (3, 0, 0), but you want to put the rotation axis at (1.3, 2.4, 0), and thus the position vector r would be the vector "from" the point (1.3, 2.4, 0) "to" the point (3, 0, 0) (instead of just say (3, 0, 0 if it the center was at the origin). I'm pretty sure you can simply subtract: (1.3, 2.4, 0)  (3, 0, 0) = (4.3, 2.4, 0), and that's your position vector r. And so the cross product should be:
r x F =
i j k
(4.3, 2.4, 0)
(0.8, 2.5, 0)
r x F = 0i + 0j + (4.3*2.5  0.8*2.4)k = 12.67k = (0, 0, 12.67)
I'm not sure that this is right though, I tried it in a homework program and it says this is incorrect, any tips, I'm sure I'm just missing some simple thing :(.

Physics or Math  Jennifer, Friday, November 9, 2012 at 5:11pm
The torque is about the point (1.3, 2.4), so treat this as your origin,. . .
This makes the r vector (4.3, 2.4, 0)

Physics or Math  Natalie, Friday, November 9, 2012 at 5:16pm
I thought maybe I mixed up the order, and that I should have done (3,0,0)  (1.3,2.4,0), point minus "effective origin" (the rotation axis) which makes more sense, this changes it to 12.67k, is that correct?

Physics or Math  Natalie, Friday, November 9, 2012 at 5:16pm
Yay thanks, thought it was simple thank you.
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