Torque Calc: Find Torque from Force F, Vector r

[gvcalamike]

Force F = (4.59 N)i - (6.29 N)k acts on a pebble with position vector r = (3.46 m)j - (4.51 m)k, relative to the origin. What is the resulting torque acting on the pebble about (a) the origin and (b) a point with coordinates (3.16 m, 0, -4.97 m)?*The force F is a vector, as is r, I just don't know how to get the vector symbol above it. i, j, k are "i hat, j hat, k hat, I don't know how to get the symbol above those either. Sorry, only my second post.

Attempt at a solution:

I have no idea where to begin. I think the answer will be the cross product of r x F, but our book doesn't give a good example of cross products. Wouldn't you shift the force vector so that the tail is at the origin O?

 

[jhae2.718]

[itex]\mathbf{\tau} = \mathbf{r} \times \mathbf{F}[/tex]<br /> You can compute the cross product in terms of components. <br /> <br /> For two vectors $\mathbf{a} = a_x \hat{\imath} + a_y \hat{\jmath}$ and $\mathbf{b} = b_x \hat{\imath} + b_y \hat{\jmath}$, $\mathbf{a} \times \mathbf{b} = (a_xb_y-a_yb_x) \hat{k}$.<br /> <br /> You can get this by FOIL-ing the terms or writing the cross product as a 3 x 3 matrix and taking the determinant:<br /> $$\mathbf{a} \times \mathbf{b} = \begin{vmatrix}<br /> \hat{\imath} & \hat{\jmath} & \hat{k} \\<br /> a_x & a_y & 0 \\<br /> b_x & b_y & 0<br /> \end{vmatrix}$$<br /> <br /> For part b, find the new displacement vector from the point to the point defined by r.[/itex]

 

[gvcalamike]

Thanks! I got it.