The figure shows a lamina in \(xy\text{-}\)plane. Two axes \({z}\) and \(z'\) pass perpendicular to its plane. A force \(\vec{F}\) acts in the plane of the lamina at point \(P\) as shown in the figure.
(The point \(P\) is closer to the \(z'\text-\)axis than the \(z\text{-}\)axis.)
| (a) | torque \(\vec{\tau}\)caused by \(\vec{F}\)about \(z\text{-}\)axis is along \((-\hat{k})\) |
| (b) | torque \(\vec{\tau}'\)caused by \(\vec{F}\)about \(z'\text-\)axis is along \((-\hat{k})\) |
| (c) | torque caused by \(\vec{F}\)about the \(z\text{-}\)axis is greater in magnitude than that about the \(z'\text-\)axis |
| (d) | total torque is given by \(\vec{\tau}_{net}=\vec{\tau}+\vec{\tau}'\) |
Choose the correct option from the given ones:
| 1. | (c) and (d) only |
| 2. | (a) and (c) only |
| 3. | (b) and (c) only |
| 4. | (a) and (b) only |
Step 1: Find the direction of the torque about the two axes.
Consider the adjacent diagram, where \(r>r'.\)
Torque \(\vec{\tau}\) about \(z\text{-}\)axis \(=\vec{r}\times \vec{F}\) which is along \(\hat{k}\)
Torque \(\vec{\tau}\) about \(z'\text-\)axis \(=\vec{r'}\times \vec{F}\) which is along \((-\hat{k})\)
Step 2: Find the magnitude of the torque about the two axes.
\(|\vec {\tau}|_z=Fr_\perp=\) the magnitude of the torque about the \(z\text{-}\)axis where \(r_\perp\) is the perpendicular distance between \(F\) and \(z\text{-}\)axis.
Similarly, \(|\vec {\tau}|_{z'}=Fr'_\perp\)
Clearly; \(r_{\perp}>r_{\perp}^{\prime} \Rightarrow|\vec{\tau}|_z>|\vec{\tau}|_{z^{\prime}}\)
We are always calculating resultant torque about a common axis.
Hence, total torque \(\vec{\tau}_{\text {net }} \neq \vec{\tau}+\vec{\tau}^{\prime},\)because \(\vec{\tau}~\text{and}~\vec{\tau}^{\prime}\)are not about a common axis.
Hence, option (3) is the correct answer.
© 2025 GoodEd Technologies Pvt. Ltd.
