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The problem asks to determine the magnitude of the resultant force and its direction (measured clockwise from the negative x-axis) for two given forces, \( F_1 = 200 \, \text{N} \) and \( F_2 = 150 \, \text{N} \), applied at angles of \( 45^\circ \) and \( 30^\circ \), respectively.

### Step 1: Resolve the forces into their components

**Force \( F_1 = 200 \, \text{N} \)** acts at \( 0^\circ \) from the positive x-axis:
- \( F_{1x} = 200 \, \text{N} \cdot \cos(0^\circ) = 200 \, \text{N} \)
- \( F_{1y} = 200 \, \text{N} \cdot \sin(0^\circ) = 0 \, \text{N} \)

**Force \( F_2 = 150 \, \text{N} \)** acts at \( 30^\circ \) below the negative x-axis (angle measured clockwise):
- \( F_{2x} = 150 \, \text{N} \cdot \cos(30^\circ) = 150 \times 0.866 = 129.9 \, \text{N} \) in the negative x direction.
- \( F_{2y} = 150 \, \text{N} \cdot \sin(30^\circ) = 150 \times (-0.5) = -75 \, \text{N} \)

### Step 2: Calculate the total components of the resultant force

- \( F_{\text{Rx}} = F_{1x} + F_{2x} = 200 \, \text{N} - 129.9 \, \text{N} = 70.1 \, \text{N} \)
- \( F_{\text{Ry}} = F_{1y} + F_{2y} = 0 \, \text{N} - 75 \, \text{N} = -75 \, \text{N} \)

### Step 3: Calculate the magnitude of the resultant force

The magnitude of the resultant force \( F_R \) is found using the Pythagorean theorem:
\[
F_R = \sqrt{(F_{\text{Rx}})^2 + (F_{\text{Ry}})^2}
\]
Substituting the values:
\[
F_R = \sqrt{(70.1)^2 + (-75)^2} = \sqrt{4914.01 + 5625} = \sqrt{10539.01} \approx 102.66 \, \text{N}
\]

### Step 4: Determine the direction of the resultant force

The direction \( \theta_R \) is given by:
\[
\theta_R = \tan^{-1} \left( \frac{F_{\text{Ry}}}{F_{\text{Rx}}} \right)
\]
\[
\theta_R = \tan^{-1} \left( \frac{-75}{70.1} \right) \approx -46.07^\circ
\]
Since the angle is negative, it lies in the fourth quadrant (below the x-axis). To express the angle measured clockwise from the negative x-axis, we adjust:
\[
\theta_R = 180^\circ + (-46.07^\circ) = 133.93^\circ
\]

### Final Answer:
- The magnitude of the resultant force is approximately \( 102.66 \, \text{N} \).
- The direction of the resultant force is approximately \( 133.93^\circ \), measured clockwise from the negative x-axis.

Would you like more details or clarifications on this? Here are five related questions you might explore:

1. How would the solution change if \( F_1 \) had a different angle?
2. What happens to the resultant force if both \( F_1 \) and \( F_2 \) are in the same direction?
3. How can vector addition be graphically represented for this problem?
4. What is the significance of resolving vectors into their components?
5. How would the result change if the forces were in three dimensions?

**Tip:** Always sketch the problem and resolve vectors into components for easier analysis.

Determine the magnitude of the resultant force and its direction, measured clockwise from the negative x-axis, for forces F1 = 200 N at 45 degrees and F2 = 150 N at 30 degrees.

Solve a physics problem to find the magnitude and direction of the resultant force from two forces, 200 N and 150 N, applied at 45 and 30 degrees. Step-by-step guide including vector components and trigonometry. Suitable for college-level or advanced high school physics students.

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