Question

The coordinates of a particle with respect to origin in a given reference frame is \( (1, 1, 1) \) meters. If a force of \( \mathbf{F} = \hat{i} - \hat{j} + \hat{k} \) acts on the particle, then the magnitude of torque (with respect to origin) in the z-direction is:

Answer 1

Correct Answer: 2

To find the torque \( \mathbf{\tau} \) exerted by the force \( \mathbf{F} = \hat{i} - \hat{j} + \hat{k} \) on the particle at location \( \mathbf{r} = (1 \,\text{m}, 1 \,\text{m}, 1 \,\text{m}) \) with respect to the origin, we use the cross product formula for torque: \( \mathbf{\tau} = \mathbf{r} \times \mathbf{F} \).
We can express \( \mathbf{r} \) and \( \mathbf{F} \) as vectors:
\(\mathbf{r} = \hat{i} + \hat{j} + \hat{k}, \quad \mathbf{F} = \hat{i} - \hat{j} + \hat{k}\)
The cross product \(\mathbf{r} \times \mathbf{F}\) is computed using the determinant formula:
\(\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\1 & 1 & 1 \\1 & -1 & 1\end{vmatrix}\)
The determinant can be expanded as follows:

• \(\hat{i}(1 \cdot 1 - 1 \cdot (-1)) = \hat{i}(1 + 1) = 2\hat{i}\)
• \(\hat{j}(-(1 \cdot 1 - 1 \cdot 1)) = \hat{j}(0) = 0\hat{j}\)
• \(\hat{k}(1 \cdot (-1) - 1 \cdot 1) = \hat{k}(-1 - 1) = -2\hat{k}\)

Therefore, the torque vector is:
\(\mathbf{\tau} = 2\hat{i} + 0\hat{j} - 2\hat{k}\)
The magnitude of the torque in the z-direction is represented by the coefficient of \(\hat{k}\), which is \( |-2| = 2 \).