91Ó°ÊÓ

Firstly, we need to write the given position vector and force vector in their component form. The position vector is \(\overrightarrow{\mathbf{r}}\) and the force vector is \(\overrightarrow{\mathbf{F}}\): \[\overrightarrow{\mathbf{r}} = 4.0 \hat{\mathbf{i}} - 2.0 \hat{\mathbf{j}}\] \[\overrightarrow{\mathbf{F}} = 20 \hat{\mathbf{j}}\]

Now, we will find the cross product of the position vector and the force vector, which represents the torque vector: \[\overrightarrow{\boldsymbol{\tau}} = \overrightarrow{\mathbf{r}} \times \overrightarrow{\mathbf{F}}\] To calculate the cross product, we can use the following rule: \[\overrightarrow{\boldsymbol{\tau}} = (\text{r}_y \text{F}_z - \text{r}_z \text{F}_y)\hat{\mathbf{i}} - (\text{r}_x \text{F}_z - \text{r}_z \text{F}_x)\hat{\mathbf{j}} + (\text{r}_x \text{F}_y - \text{r}_y \text{F}_x)\hat{\mathbf{k}}\] In this problem, the vectors are only in the \(xy\)-plane, so the z-components are zero. Therefore, the cross product simplifies to: \[\overrightarrow{\boldsymbol{\tau}} = (\text{r}_x \text{F}_y - \text{r}_y \text{F}_x)\hat{\mathbf{k}}\] Now, substitute the components of the position vector and the force vector: \[\overrightarrow{\boldsymbol{\tau}} = (4.0 \times 20 - (-2.0) \times 0)\hat{\mathbf{k}}\]

Finally, we will calculate the magnitude of the resulting torque vector. From the previous step, the cross product is: \[\overrightarrow{\boldsymbol{\tau}} = (80 + 0)\hat{\mathbf{k}} = 80 \hat{\mathbf{k}}\] So, the torque of the force about the origin is \(80 \hat{\mathbf{k}} \mathrm{N} \cdot \mathrm{m}\).