Answer:
[tex]-80\hat{k}\text{ N/m}[/tex]
Explanation:
Torque is given by
[tex]\tau = \vec{F} \times\vec{r}[/tex]
Note that this is a cross product of vectors. To perform a cross product, we need to express the vectors in three components.
[tex]\vec{F} =0.0\hat{i} + 20\hat{j} + 0.0\hat{k}[/tex]
[tex]\vec{r} =4.0\hat{i} - 2.0\hat{j} + 0.0\hat{k}[/tex]
The torque is then
[tex]\tau = \left|\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ 0.0 & 20 & 0\\ 4.0 & - 2.0 & 0.0\end{matrix}\right|[/tex]
[tex]\tau = -80\hat{k}\text{ N/m}[/tex]
(I'm not sure if the matrix will appear well because I'm constrained with the device I'm using but what I'm showing is the determinant of a matrix with i, j, k in the first row, the coefficients of F in the second and the coefficients of r in the third)
