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 \( \hat{F} \hat{k} \) acting at the origin about the point \( (1, -1) \), we can follow these steps:

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

### Step 2: Define the Position Vectors
- The position vector of the origin \( O \) is \( \vec{r}_O = (0, 0, 0) \).
- The position vector of point \( A \) is \( \vec{r}_A = (1, -1, 0) \).

### Step 3: Calculate the Position Vector \( \vec{R} \) from Point A to the Origin
The position vector \( \vec{R} \) from point \( A \) to the origin \( O \) is given by:
\[ \vec{R} = \vec{r}_O - \vec{r}_A = (0, 0, 0) - (1, -1, 0) = (-1, 1, 0) \]

### Step 4: Write the Force Vector
The force vector is:
\[ \vec{F} = (0, 0, F) \]

### Step 5: Calculate the Torque \( \vec{\tau} \)
The torque \( \vec{\tau} \) is calculated using the cross product:
\[ \vec{\tau} = \vec{R} \times \vec{F} \]
Substituting the vectors:
\[ \vec{\tau} = (-1, 1, 0) \times (0, 0, F) \]

### Step 6: Compute the Cross Product
Using the determinant form for the cross product:
\[ \vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 0 \\ 0 & 0 & F \end{vmatrix} \]
Calculating this determinant:
\[ \vec{\tau} = \hat{i}(1 \cdot F - 0 \cdot 0) - \hat{j}(-1 \cdot F - 0 \cdot 0) + \hat{k}(-1 \cdot 0 - 1 \cdot 0) \]
This simplifies to:
\[ \vec{\tau} = \hat{i}F + \hat{j}F = F\hat{i} + F\hat{j} \]

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

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