Find the torque about the origin when a force of $3 \hat{j} \text{ N}$ acts on a particle whose position vector is $2 \hat{k} \text{ m}$. (2020)
Solution:
Torque is given by the cross product $\vec{\tau} = \vec{r} \times \vec{F}$. Substituting the given vectors, $\vec{\tau} = (2\hat{k}) \times (3\hat{j}) = 6(\hat{k} \times \hat{j})$. Since $\hat{k} \times \hat{j} = -\hat{i}$, the torque is $-6\hat{i} \text{ N m}$.
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