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 :(.

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)

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?

Yay thanks, thought it was simple thank you.

Your approach to finding the position vector and calculating the torque is correct, but there seems to be a simple arithmetic mistake in your final calculation for the cross product.

Let's review the steps:

1. Determine the position vector: To find the position vector, you subtract the coordinates of the rotation axis from the point of force application:

r = (3.0, 0.0, 0.0) - (-1.3, 2.4, 0.0) = (4.3, -2.4, 0.0)

2. Calculate the cross product: Take the cross product of the position vector (r) and the force vector (F):

r × F = (4.3i - 2.4j + 0.0k) × (0.8i + 2.5j + 0.0k)

To evaluate this cross product, you can use the determinant method:

r × F = (2.4 * 0.0 - 0.0 * 0.0) i - (4.3 * 0.0 - 0.0 * 0.0) j + (4.3 * 2.5 - (-2.4 * 0.8)) k

r × F = 0i - 0j + (10.75)k = (0, 0, 10.75)

So, the correct torque vector about the point (-1.3, 2.4, 0) is (0, 0, 10.75) N·m.

Make sure to double-check your arithmetic calculations, especially when performing the cross product.