A force `Fhatk` acts on O, the origin of the coordinate system. The torque of this force about the point is: `(1,-1)` is

Options

A. `F(hati + hatj)`
B. `-F(hati - hatj)`
C. `F(-hati - hatj)`
D. `-F(hati + hatj)`

Answer

To find the torque of the force \( \hat{F} \) acting at the origin \( O \) about the point \( (1, -1) \), we can follow these steps:

### Step 1: Identify the Force and its Position
The force is given as \( \hat{F} = F \hat{k} \), which means it acts in the positive z-direction. The point about which we need to calculate the torque is \( P(1, -1) \).

### Step 2: Determine the Position Vector
The position vector \( \vec{r} \) from the point \( P(1, -1) \) to the origin \( O(0, 0) \) can be calculated as:
\[ \vec{r} = \vec{r}_{final} - \vec{r}_{initial} = (0, 0) - (1, -1) = (-1, 1) \]
Thus, \( \vec{r} = -\hat{i} + \hat{j} \).

### Step 3: Write the Force Vector
The force vector is:
\[ \vec{F} = F \hat{k} \]

### Step 4: Calculate the Torque
The torque \( \vec{\tau} \) is given by the cross product of the position vector \( \vec{r} \) and the force vector \( \vec{F} \):
\[ \vec{\tau} = \vec{r} \times \vec{F} \]

Substituting the vectors:
\[ \vec{\tau} = (-\hat{i} + \hat{j}) \times (F \hat{k}) \]

### Step 5: Use the Determinant to Compute the Cross Product
We can use the determinant method to compute the cross product:
\[ \vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 0 \\ 0 & 0 & F \end{vmatrix} \]

### Step 6: Calculate the Determinant
Calculating the determinant:
\[ \vec{\tau} = \hat{i} \begin{vmatrix} 1 & 0 \\ 0 & F \end{vmatrix} - \hat{j} \begin{vmatrix} -1 & 0 \\ 0 & F \end{vmatrix} + \hat{k} \begin{vmatrix} -1 & 1 \\ 0 & 0 \end{vmatrix} \]
Calculating each of these determinants:
- For \( \hat{i} \): \( 1 \cdot F - 0 \cdot 0 = F \)
- For \( \hat{j} \): \( -1 \cdot F - 0 \cdot 0 = -F \)
- For \( \hat{k} \): \( -1 \cdot 0 - 1 \cdot 0 = 0 \)

Thus, we have:
\[ \vec{\tau} = F \hat{i} + F \hat{j} + 0 \hat{k} = F \hat{i} + F \hat{j} \]

### Final Result
The torque about the point \( (1, -1) \) is:
\[ \vec{\tau} = F (\hat{i} + \hat{j}) \text{ Newton meter} \]