To determine the moment of the force about the origin, we follow these steps:

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1. Represent the Force Vector:

The force $\mathbf{F}$ of 75 N acts along the line segment between points $H(-4, -1)$ and $I(2, 7)$.

Direction vector of the force:

$\mathbf{d} = I - H = (2 - (-4), 7 - (-1)) = (6, 8)$

Unit vector along the direction of the force:

$\mathbf{\hat{d}} = \frac{\mathbf{d}}{\|\mathbf{d}\|} = \frac{(6, 8)}{\sqrt{6^2 + 8^2}} = \frac{(6, 8)}{10} = (0.6, 0.8)$

The force vector $\mathbf{F}$ is therefore: $\mathbf{F} = 75 \cdot \mathbf{\hat{d}} = 75 \cdot (0.6, 0.8) = (45, 60)$

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2. Moment About the Origin:

The moment is given by the cross product of the position vector $\mathbf{r}$ (from the origin to any point on the line of action) and the force vector $\mathbf{F}$: $M = \mathbf{r} \times \mathbf{F}$

We can choose either $H(-4, -1)$ or $I(2, 7)$ as the position vector. Here, we use $H(-4, -1)$: $\mathbf{r} = (-4, -1)$

The cross product in 2D is calculated as: $M = r_x F_y - r_y F_x$

Substituting: $M = (-4)(60) - (-1)(45) = -240 + 45 = -195 \, \text{Nm}$

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3. Magnitude of the Moment:

The magnitude of the moment is: $|M| = 195 \, \text{Nm}$

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Final Answer:

$\boxed{195.0 \, \text{Nm}}$

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Let me know if you'd like further clarification or need me to visualize the setup with a drawing.

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Related Questions:

1. How is the direction of the force vector determined?
2. Why is the cross product used to calculate moments in 2D?
3. What would change if the force were applied at a different location along the line HI?
4. How does the distance between points $H$ and $I$ affect the result?
5. What is the physical interpretation of a negative moment in this scenario?

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Tip: Always double-check the direction of the forces and the chosen position vector when computing moments.