Compute the magnitude and direction of each vector. If the vector is not in the first
quadrant convert the angle so that it is defined counterclockwise from the right
horizontal.
• Fx = 130 N, Fy = 1200 N

To compute the magnitude and direction of a vector given its components \(F_x\) and \(F_y\), you can use the following formulas:

1. **Magnitude** of the vector:

\[
\text{Magnitude} = \sqrt{F_x^2 + F_y^2}
\]

2. **Direction** (angle \(\theta\)) of the vector:

\[
\theta = \arctan\left(\frac{F_y}{F_x}\right)
\]

However, since the angle should be defined counterclockwise from the right horizontal, we need to adjust the angle based on the quadrant where the vector lies.

Given:
- \(F_x = 130 \, \text{N}\)
- \(F_y = 1200 \, \text{N}\)

Let's calculate the magnitude and direction.The magnitude of the vector is approximately \(1207.02 \, \text{N}\), and the direction is approximately \(83.82^\circ\), measured counterclockwise from the right horizontal.

Would you like more details or have any questions about this calculation?

Here are 5 related questions:
1. What happens if \(F_x\) or \(F_y\) is negative?
2. How does the magnitude change if both components are doubled?
3. Can you express this vector in polar coordinates?
4. How would you find the resultant vector if you added another vector to this one?
5. How does the angle calculation change if \(F_x = 0\)?

**Tip:** When calculating angles in the second, third, or fourth quadrant, always ensure the direction is adjusted to be counterclockwise from the positive x-axis.

Learn how to compute the magnitude and direction of a vector with components Fx = 130 N and Fy = 1200 N using trigonometric formulas. This problem covers vectors and trigonometry, including calculating magnitude and direction. Suitable for high school students.

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