Discussion
An irregular planar body in space is acted upon by a force \(\vec{F} = (2\vec{i} + \vec{j})N\) at position \(\vec{r}_{1} = (\vec{i} + 2\vec{j})m\) and a moment \(\vec{M} = 3\vec{k}Nm\) at position \(\vec{r}_{2} = (-2\vec{i})m\). The corresponding equivalent force \(\vec{F}_{O}\) and moment \(\vec{M}_{O}\) at the origin are
Answer: B
The equivalent system is \(\vec{F}_{O} = (2\vec{i} + \vec{j})N\) and \(\vec{M}_{O} = 0\).
Moving a force to the origin leaves the force unchanged and adds its moment about the origin. A couple is a free vector, so it moves unchanged wherever it acts.
\[ \begin{aligned} \vec{r}_{1} \times \vec{F} &= (\vec{i} + 2\vec{j}) \times (2\vec{i} + \vec{j}) \\ &= (1 \cdot 1 - 2 \cdot 2)\vec{k} = -3\vec{k} \\ \vec{M}_{O} &= -3\vec{k} + 3\vec{k} = 0 \end{aligned} \]
The option with \(3\vec{k}\) forgets the moment of the force, the one with \(6\vec{k}\) gets its sign wrong, and reversing the force to \(-2\vec{i} - \vec{j}\) is never right, because an equivalent system keeps the same resultant.
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