Moment of a force of magnitude 10 N acting along positive y-direction at point `(2m,0,0)` about the point `(0,1m,0)` in N-m is

Options

A. 10
B. 20
C. `10 sqrt(2)`
D. 30

Explanation

To solve the problem of finding the moment of a force of magnitude 10 N acting along the positive y-direction at the point (2m, 0, 0) about the point (0, 1m, 0), we will follow these steps:

### Step 1: Identify the position vectors
We need to identify the position vector of the point of application of the force and the point about which we are calculating the moment.

- Point of application of force \( P = (2, 0, 0) \)
- Point about which we are calculating the moment \( O = (0, 1, 0) \)

### Step 2: Calculate the position vector \( \vec{r} \)
The position vector \( \vec{r} \) from point \( O \) to point \( P \) is calculated as follows:

\[ \vec{r} = P - O = (2 - 0) \hat{i} + (0 - 1) \hat{j} + (0 - 0) \hat{k} = 2 \hat{i} - 1 \hat{j} \]

### Step 3: Define the force vector \( \vec{F} \)
The force vector \( \vec{F} \) is given as acting along the positive y-direction with a magnitude of 10 N:

\[ \vec{F} = 10 \hat{j} \]

### Step 4: Calculate the moment \( \vec{M} \)
The moment \( \vec{M} \) is calculated using the cross product of the position vector \( \vec{r} \) and the force vector \( \vec{F} \):

\[ \vec{M} = \vec{r} \times \vec{F} \]

Substituting the values of \( \vec{r} \) and \( \vec{F} \):

\[ \vec{M} = (2 \hat{i} - 1 \hat{j}) \times (10 \hat{j}) \]

### Step 5: Calculate the cross product
To compute the cross product, we can use the determinant method:

\[ \vec{M} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 0 \\ 0 & 10 & 0 \end{vmatrix} \]

Calculating the determinant:

\[ \vec{M} = \hat{i} \begin{vmatrix} -1 & 0 \\ 10 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 0 \\ 0 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -1 \\ 0 & 10 \end{vmatrix} \]

Calculating each of the determinants:

1. For \( \hat{i} \): \( (-1)(0) - (10)(0) = 0 \)
2. For \( \hat{j} \): \( (2)(0) - (0)(0) = 0 \)
3. For \( \hat{k} \): \( (2)(10) - (-1)(0) = 20 \)

Thus, we have:

\[ \vec{M} = 0 \hat{i} - 0 \hat{j} + 20 \hat{k} = 20 \hat{k} \]

### Step 6: Conclusion
The magnitude of the moment is:

\[ |\vec{M}| = 20 \, \text{N-m} \]

The moment acts along the positive z-axis.

### Final Answer
The moment of the force about the point (0, 1m, 0) is **20 N-m**.

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