In a plane, points A(0,-2) , B(4,2) , C(4,-2) are given.…

In a plane, points $A(0,-2)$, $B(4,2)$, $C(4,-2)$ are given. Forces $\overrightarrow{OA}$, $\overrightarrow{OB}$, $\overrightarrow{OC}$ are drawn from the origin. Construct the resultant $\overrightarrow{OM}$ and find its projections on the coordinate axes. Express forces $\overrightarrow{OA}$, $\overrightarrow{OB}$, $\overrightarrow{OC}$ and $\overrightarrow{OM}$ in terms of unit vectors $\vec{i}$ and $\vec{j}$.

Step 1: Express each force via components

The force vectors from the origin $O(0,0)$ to each point are

\[ \overrightarrow{OA} = (0,\,-2), \qquad \overrightarrow{OB} = (4,\,2), \qquad \overrightarrow{OC} = (4,\,-2). \]

In terms of the unit vectors $\vec{i}$ and $\vec{j}$: \[\overrightarrow{OA} = -2\,\vec{j}, \qquad \overrightarrow{OB} = 4\,\vec{i} + 2\,\vec{j}, \qquad \overrightarrow{OC} = 4\,\vec{i} - 2\,\vec{j}\]

Step 2: Find the resultant $\overrightarrow{OM}$

The resultant is the vector sum of the three forces: $\overrightarrow{OM} = \overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC}$.

Adding component by component: \[M_x = 0 + 4 + 4 = 8, \qquad M_y = -2 + 2 + (-2) = -2.\]

\[\overrightarrow{OM} = 8\,\vec{i} - 2\,\vec{j}, \qquad M(8,\,-2)\]

Step 3: Projections of the resultant on the axes

\[\operatorname{pr}_x \overrightarrow{OM} = 8, \qquad \operatorname{pr}_y \overrightarrow{OM} = -2\]

Step 4: Magnitude of the resultant

\[ |\overrightarrow{OM}| = \sqrt{8^2 + (-2)^2} = \sqrt{68} = 2\sqrt{17} \approx 8.25 \]

Step 5: Graphical construction

Place the forces head-to-tail: start with $\overrightarrow{OB}$, attach a copy of $\overrightarrow{OA}$ at $B$ (reaching $(4,0)$), then a copy of $\overrightarrow{OC}$ (reaching $M$). The closing vector from $O$ is the resultant.