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This diagram shows a vector problem involving forces \( F_1 \), \( F_2 \), \( F_3 \), and \( F_4 \), each applied at specific angles. The force magnitudes are:

- \( F_1 = 40 \, \text{N} \) at an angle of \( 30^\circ \) above the negative \( x \)-axis.
- \( F_2 = 20 \, \text{N} \) along the negative \( y \)-axis.
- \( F_3 = 60 \, \text{N} \) along the negative \( x \)-axis.
- \( F_4 = 30 \, \text{N} \) along the positive \( x \)-axis.

### To find the resultant force, we need to decompose the forces into their horizontal (\( x \)) and vertical (\( y \)) components:

1. **Force \( F_1 \)**:
   - \( F_1 \) makes an angle of \( 30^\circ \) with the negative \( x \)-axis.
   - \( F_{1x} = F_1 \cos(30^\circ) = 40 \times \cos(30^\circ) \)
   - \( F_{1y} = F_1 \sin(30^\circ) = 40 \times \sin(30^\circ) \)

2. **Force \( F_2 \)**:
   - It acts directly along the negative \( y \)-axis.
   - \( F_{2x} = 0 \)
   - \( F_{2y} = -20 \, \text{N} \)

3. **Force \( F_3 \)**:
   - It acts along the negative \( x \)-axis.
   - \( F_{3x} = -60 \, \text{N} \)
   - \( F_{3y} = 0 \)

4. **Force \( F_4 \)**:
   - It acts along the positive \( x \)-axis.
   - \( F_{4x} = 30 \, \text{N} \)
   - \( F_{4y} = 0 \)

### Now, summing the components:

- **Net horizontal component** \( F_{x} \):
  \[
  F_{x} = F_{1x} + F_{2x} + F_{3x} + F_{4x}
  \]
  
- **Net vertical component** \( F_{y} \):
  \[
  F_{y} = F_{1y} + F_{2y} + F_{3y} + F_{4y}
  \]

### Resultant Force:
\[
F_{R} = \sqrt{F_x^2 + F_y^2}
\]
and the angle \( \theta \) of the resultant force with respect to the positive \( x \)-axis is:
\[
\theta = \tan^{-1}\left(\frac{F_y}{F_x}\right)
\]

I will calculate these components and the final result next. Would you like me to proceed with the calculations?

Calculate the resultant force for a system of vectors F1, F2, F3, and F4 with magnitudes 40 N, 20 N, 60 N, and 30 N respectively. The solution involves decomposing the forces into their x and y components using trigonometry. Suitable for advanced high school physics students.

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