Question

A body is located at \((1, 1, 1) \, \text{m}\) and experiences a force of 2 N in the direction \(\hat{i} + \hat{j}\). Find the torque acting on the body in N·m.

• A. \(-\sqrt{2} \hat{i} + \sqrt{2} \hat{j}\)
• B. \(-\hat{i} + \hat{j}\)
• C. \(\hat{i} - \hat{j}\)
• D. \(\sqrt{2} \hat{i} + \sqrt{2} \hat{j}\)

Hint:

For torque in 3D, always use \(\vec{\tau} = \vec{r} \times \vec{F}\) and compute using determinant method to get components.

Answer 1

The Correct Option is A

Step 1: Get the force vector right before doing anything else.
Saying the force is 2 N in the direction $\hat{i}+\hat{j}$ means its magnitude is 2 N and it points along the unit vector in that direction, not that the vector $\hat{i}+\hat{j}$ is simply doubled. The unit vector along $\hat{i}+\hat{j}$ is $\dfrac{\hat{i}+\hat{j}}{\sqrt2}$, so
\[ \vec{F} = 2\cdot\frac{\hat{i}+\hat{j}}{\sqrt2} = \sqrt2\hat{i} + \sqrt2\hat{j} \]

Step 2: Set up the cross product with the correct vectors.
\[ \vec{r} = \hat{i}+\hat{j}+\hat{k}, \qquad \vec{F} = \sqrt2\hat{i}+\sqrt2\hat{j}+0\hat{k} \]

Step 3: Expand the determinant.
\[ \vec{\tau} = \vec{r}\times\vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ \sqrt2 & \sqrt2 & 0 \end{vmatrix} = \hat{i}(1\cdot0 - 1\cdot\sqrt2) - \hat{j}(1\cdot0 - 1\cdot\sqrt2) + \hat{k}(1\cdot\sqrt2 - 1\cdot\sqrt2) \]

Step 4: Simplify each component.
\[ \vec{\tau} = -\sqrt2\hat{i} + \sqrt2\hat{j} + 0\hat{k} \]

Step 5: Conclusion.
\[ \boxed{-\sqrt2\hat{i}+\sqrt2\hat{j}} \]