Problem 10.20

10.20.

In this problem to find the torque in the system, we only have to apply the equation
$\vec{\tau} = \vec{r} \times \vec{F}$
Here we find
$\vec{\tau} = (4\hat{i} + 5\hat{j}) \times (2\hat{i} + 3\hat{j})$
Applying the definition of a cross product and recognizing that there are no components of the torque in the $\hat{i}$ or $\hat{j}$ directions, we see that
$\vec{\tau} = [(4)(3) - (5)(2)] N\cdot m \hat{k}$
or that the torque is
$\vec{\tau} = 2 N\cdot m \hat{k}$