Question

In unit-vector notation, what is the torque about the origin on a particle located at coordinates (0 m, −3.0 m, 2.0 m) due to force F with arrow1 with components F1x = 5.0 N and F1y = F1z = 0?

Answer 1

To solve this problem we will apply the vector concepts for which we will multiply the two given vectors.

The position vector (r) where the Force (F) is applied is given by

`\vec{r} = (0\hat{i}-3\hat{j}+2\hat{k})`

The Force vector would be:

`\vec{F} = 5\hat{i}`

Now applying the cross product the torque would be,

`\vec{\tau} = \vec{r}* \vec{F}`

`\vec{\tau} = (0\hat{i}-3\hat{j}+2\hat{k})* 5\hat{i}`

`\vec{\tau} = 10\hat{j}+15\hat{k}`

Therefore the Torque vector would be
`\vec{\tau} = 10\hat{j}+15\hat{k}`