Question

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 :(.

Answer 1

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)

Answer 2

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?

Answer 3

Yay thanks, thought it was simple thank you.