Moment of a force of megnitude 20 N acting along positive x direction at point (3m 0, 0) about the point (0, 2, 0) (in N-m) is :-

Options

A. 20
B. 60
C. 40
D. 30

Answer

To calculate the moment (or torque) of a force acting at a point about another point, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information**:
- Force \( F = 20 \, \text{N} \) acting along the positive x-direction.
- Point of application of the force \( P_1 = (3, 0, 0) \).
- Point about which we are calculating the moment \( P_2 = (0, 2, 0) \).

2. **Determine the Position Vector**:
- The position vector \( \vec{r} \) from point \( P_2 \) to point \( P_1 \) is given by:
\[ \vec{r} = P_1 - P_2 = (3, 0, 0) - (0, 2, 0) = (3 - 0, 0 - 2, 0 - 0) = (3, -2, 0) \]
- Therefore, \( \vec{r} = 3 \hat{i} - 2 \hat{j} + 0 \hat{k} \).

3. **Express the Force Vector**:
- The force vector \( \vec{F} \) acting along the positive x-direction is:
\[ \vec{F} = 20 \hat{i} + 0 \hat{j} + 0 \hat{k} \]

4. **Calculate the Torque**:
- The torque \( \vec{\tau} \) is calculated using the cross product:
\[ \vec{\tau} = \vec{r} \times \vec{F} \]
- Writing this in determinant form:
\[ \vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -2 & 0 \\ 20 & 0 & 0 \end{vmatrix} \]

5. **Calculate the Determinant**:
- Expanding the determinant:
\[ \vec{\tau} = \hat{i} \begin{vmatrix} -2 & 0 \\ 0 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 0 \\ 20 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & -2 \\ 20 & 0 \end{vmatrix} \]
- Calculating each of the determinants:
- For \( \hat{i} \): \( (-2)(0) - (0)(0) = 0 \)
- For \( \hat{j} \): \( (3)(0) - (20)(0) = 0 \)
- For \( \hat{k} \): \( (3)(0) - (-2)(20) = 0 + 40 = 40 \)

6. **Combine the Results**:
- Thus, the torque vector is:
\[ \vec{\tau} = 0 \hat{i} - 0 \hat{j} + 40 \hat{k} = 40 \hat{k} \, \text{N-m} \]

7. **Magnitude of Torque**:
- The magnitude of the torque is \( 40 \, \text{N-m} \).

### Final Answer:
The moment of the force about the point (0, 2, 0) is **40 N-m**.

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