Question

A force of $- F \hat{k}$ acts on $O$, the origin of the coordinate system. The torque about the point $(1,-1)$ is

• A. $F (\hat{i} - \hat{j})$
• B. $- F (\hat{i} +\hat{ j})$
• C. $F (\hat{i}+\hat{j})$
• D. $-F(\hat{i}-\hat{j})$

Answer 1

The Correct Option is C

Here, $\vec{F}=-F \,\hat{k}, \vec{r}=\hat{i}-\hat{j}$ Torque $\vec{\tau}=\vec{r}\times\vec{F}$ $\vec{\tau}=\left|\begin{matrix}\hat{i}&\hat{j}&\hat{k}\\ 1&-1&0\\ 0&0&-F\end{matrix}\right|$ $\vec{\tau}=\hat{i}\left(F-0\right)-\hat{j}\left(-F-0\right)+\hat{k}\left(0-0\right)$ $=F\hat{i}+F\hat{j} $ $\vec{\tau}=F \left(\hat{i}+\hat{j}\right)$

Answer 2

To find the torque about the point (1, -1) due to the force \(-F \hat{k}\).
Given:
\(\vec{F} = -F \hat{k}\)
\(\vec{r} = \hat{i} - \hat{j}\)

To find the torque \(\vec{\tau} = \vec{r} \times \vec{F}\):
Write the cross-product using the determinant form:
\(\vec{\tau} = \begin{vmatrix}   \hat{i} & \hat{j} & \hat{k} \\   1 & -1 & 0 \\   0 & 0 & -F   \end{vmatrix}\)

Now, Compute the determinant:
\(\vec{\tau} = \hat{i} \left( (-1)(-F) - (0)(0) \right) - \hat{j} \left( (1)(-F) - (0)(0) \right) + \hat{k} \left( (1)(0) - (-1)(0) \right)\)
\(\vec{\tau} = \hat{i} \cdot (F) - \hat{j} \cdot (-F) + \hat{k} \cdot (0)\)
\(\vec{\tau} = F \hat{i} + F \hat{j} + 0 \cdot \hat{k}\)
Therefore, the torque \(\vec{\tau}\) about the point (1, -1) is \(\vec{\tau} = F (\hat{i} + \hat{j})\).

So, the correct option is (C): \(F (\hat{i} + \hat{j})\).