Question

A force $\overrightarrow{F} = 2\hat{i} + 3\hat{j} - \hat{k}$ acts at a point (2, -3, 1). Then magnitude of torque about point (0, 0, 2) will be:

• A. 6 unit
• B. $3\sqrt{5}$ unit
• C. $6\sqrt{5}$ unit
• D. none of these

Answer 1

Answer: (C)

$6\sqrt{5}$ unit

Answer 2

The problem asks for the magnitude of torque about a given point due to a force acting at another point.

1. Identify the force vector ($\overrightarrow{F}$): $\overrightarrow{F} = 2\hat{i} + 3\hat{j} - \hat{k}$

2. Identify the point where the force acts (P): $P = (2, -3, 1)$

3. Identify the point about which the torque is calculated (O): $O = (0, 0, 2)$

4. Calculate the position vector ($\overrightarrow{r}$): The position vector $\overrightarrow{r}$ is drawn from the point about which torque is calculated to the point where the force acts. $\overrightarrow{r} = \overrightarrow{OP} = P - O$ $\overrightarrow{r} = (2 - 0)\hat{i} + (-3 - 0)\hat{j} + (1 - 2)\hat{k}$ $\overrightarrow{r} = 2\hat{i} - 3\hat{j} - \hat{k}$

5. Calculate the torque vector ($\overrightarrow{\tau}$): The torque $\overrightarrow{\tau}$ is given by the cross product of the position vector and the force vector: $\overrightarrow{\tau} = \overrightarrow{r} \times \overrightarrow{F}$ $\overrightarrow{\tau} = (2\hat{i} - 3\hat{j} - \hat{k}) \times (2\hat{i} + 3\hat{j} - \hat{k})$

   We can calculate the cross product using a determinant:

   $$\overrightarrow{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & -1 \\ 2 & 3 & -1 \end{vmatrix}$$ $$\overrightarrow{\tau} = \hat{i}((-3)(-1) - (3)(-1)) - \hat{j}((2)(-1) - (2)(-1)) + \hat{k}((2)(3) - (2)(-3))$$ $$\overrightarrow{\tau} = \hat{i}(3 - (-3)) - \hat{j}(-2 - (-2)) + \hat{k}(6 - (-6))$$ $$\overrightarrow{\tau} = \hat{i}(3 + 3) - \hat{j}(-2 + 2) + \hat{k}(6 + 6)$$ $$\overrightarrow{\tau} = 6\hat{i} - 0\hat{j} + 12\hat{k}$$ $$\overrightarrow{\tau} = 6\hat{i} + 12\hat{k}$$

6. Calculate the magnitude of the torque ($|\overrightarrow{\tau}|$): The magnitude of a vector $A\hat{i} + B\hat{j} + C\hat{k}$ is $\sqrt{A^2 + B^2 + C^2}$.

   $$|\overrightarrow{\tau}| = \sqrt{(6)^2 + (0)^2 + (12)^2}$$ $$|\overrightarrow{\tau}| = \sqrt{36 + 0 + 144}$$ $$|\overrightarrow{\tau}| = \sqrt{180}$$

   To simplify $\sqrt{180}$, we find its prime factors: $180 = 36 \times 5 = 6^2 \times 5$.

   $$|\overrightarrow{\tau}| = \sqrt{6^2 \times 5} = 6\sqrt{5} \text{ unit}$$

The final answer is \boxed{\text{6\sqrt{5} unit}}.

Explanation of the solution:

1. Determine the position vector $\overrightarrow{r}$ from the point of rotation to the point of force application: $\overrightarrow{r} = (2-0)\hat{i} + (-3-0)\hat{j} + (1-2)\hat{k} = 2\hat{i} - 3\hat{j} - \hat{k}$.
2. Calculate the torque vector $\overrightarrow{\tau}$ using the cross product: $\overrightarrow{\tau} = \overrightarrow{r} \times \overrightarrow{F}$. $\overrightarrow{\tau} = (2\hat{i} - 3\hat{j} - \hat{k}) \times (2\hat{i} + 3\hat{j} - \hat{k}) = 6\hat{i} + 12\hat{k}$.
3. Find the magnitude of the torque vector: $|\overrightarrow{\tau}| = \sqrt{6^2 + 12^2} = \sqrt{36 + 144} = \sqrt{180} = 6\sqrt{5}$ unit.

Answer: The magnitude of torque about point (0, 0, 2) will be $6\sqrt{5}$ unit. The correct option is (C).