Question

A slab is subjected to two forces \(\vec{F_1}\) and \(\vec{F_2}\) of same magnitude \(F\) as shown in the figure. Force \(\vec{F_2}\) is in XY-plane while force \(\vec{F_1}\) acts along z-axis at the point \((2\hat{i} + 3\hat{j})\). The moment of these forces about point O will be :

• A. \((3\hat{i} - 2\hat{j} - 3\hat{k})F\)
• B. \((3\hat{i} + 2\hat{j} - 3\hat{k})F\)
• C. \((3\hat{i} + 2\hat{j} + 3\hat{k})F\)
• D. \((3\hat{i} - 2\hat{j} + 3\hat{k})F\)

Hint:

Torque about O is \(\vec{\tau} = \vec{r}\times\vec{F}\) for each force. \(\vec{F_1} = F\hat{k}\) acts at \(2\hat{i}+3\hat{j}\); \(\vec{F_2}\) acts at \(6\hat{j}\) with x-component \(-F\sin 30^\circ\).

Answer 1

The Correct Option is D

Torque of \(\vec{F_1}\): \(\vec{r_1} = 2\hat{i} + 3\hat{j}\), \(\vec{F_1} = F\hat{k}\).
\(\vec{\tau_1} = (2\hat{i} + 3\hat{j})\times F\hat{k} = 2F(\hat{i}\times\hat{k}) + 3F(\hat{j}\times\hat{k}) = -2F\hat{j} + 3F\hat{i}\).
Torque of \(\vec{F_2}\): it acts at the point \(\vec{r_2} = 6\hat{j}\) (6 m along y). Its direction makes \(30^\circ\) with the y-axis and, from the figure, its x-component is along \(-\hat{i}\): \(\vec{F_2} = F(-\sin 30^\circ\,\hat{i} - \cos 30^\circ\,\hat{j})\).
\(\vec{\tau_2} = 6\hat{j}\times F\left(-\tfrac12\hat{i} - \tfrac{\sqrt{3}}{2}\hat{j}\right) = -3F(\hat{j}\times\hat{i}) = +3F\hat{k}\) (because \(\hat{j}\times\hat{i} = -\hat{k}\)).
Total: \(\vec{\tau} = \vec{\tau_1} + \vec{\tau_2} = (3\hat{i} - 2\hat{j} + 3\hat{k})F\).
Answer: (4) \((3\hat{i} - 2\hat{j} + 3\hat{k})F\)