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In this problem, we are asked to determine the magnitudes (modules) of forces \( F_1 \) and \( F_2 \) such that the particle is in equilibrium. The diagram shows forces applied at angles, with \( F_1 \) at an angle of \( 45^\circ \) above the x-axis and \( F_2 \) at an angle of \( 30^\circ \) below the x-axis. Additionally, there is a horizontal force of 12 N acting to the left, along the negative x-axis.

### Steps for Solving:

1. **Equilibrium Condition:**
   For the particle to be in equilibrium, the net force in both the x- and y-directions must be zero.
   - \( \sum F_x = 0 \)
   - \( \sum F_y = 0 \)

2. **Resolve Forces into Components:**
   - The force \( F_1 \) can be broken down into:
     \[
     F_{1x} = F_1 \cos(45^\circ), \quad F_{1y} = F_1 \sin(45^\circ)
     \]
   - The force \( F_2 \) can be broken down into:
     \[
     F_{2x} = F_2 \cos(30^\circ), \quad F_{2y} = -F_2 \sin(30^\circ) \quad (\text{since it acts downward})
     \]
   - The horizontal force is already given as \( -12 \, \text{N} \) in the x-direction.

3. **Apply the Equilibrium Conditions:**

   - **In the x-direction:**
     \[
     -12 + F_1 \cos(45^\circ) + F_2 \cos(30^\circ) = 0
     \]
     Simplifying this:
     \[
     F_1 \left(\frac{\sqrt{2}}{2}\right) + F_2 \left(\frac{\sqrt{3}}{2}\right) = 12
     \]
   
   - **In the y-direction:**
     \[
     F_1 \sin(45^\circ) - F_2 \sin(30^\circ) = 0
     \]
     Simplifying this:
     \[
     F_1 \left(\frac{\sqrt{2}}{2}\right) = F_2 \left(\frac{1}{2}\right)
     \]
     This gives us:
     \[
     F_1 = \frac{F_2}{\sqrt{2}}
     \]

4. **Solve the System of Equations:**

   - Substitute \( F_1 = \frac{F_2}{\sqrt{2}} \) into the equation for the x-direction:
     \[
     \frac{F_2}{\sqrt{2}} \cdot \frac{\sqrt{2}}{2} + F_2 \cdot \frac{\sqrt{3}}{2} = 12
     \]
     Simplifying this:
     \[
     \frac{F_2}{2} + \frac{\sqrt{3}}{2} F_2 = 12
     \]
     Factor out \( F_2 \):
     \[
     F_2 \left( \frac{1}{2} + \frac{\sqrt{3}}{2} \right) = 12
     \]
     Simplifying further:
     \[
     F_2 \left( \frac{1 + \sqrt{3}}{2} \right) = 12
     \]
     \[
     F_2 = \frac{12 \times 2}{1 + \sqrt{3}}
     \]
     This gives the value for \( F_2 \).

5. **Find \( F_1 \):**

   Once \( F_2 \) is found, use the relation \( F_1 = \frac{F_2}{\sqrt{2}} \) to find the value of \( F_1 \).

Let me calculate these values for you.The magnitudes of the forces that keep the particle in equilibrium are:

- \( F_1 = 6.21 \, \text{N} \)
- \( F_2 = 8.78 \, \text{N} \)

Would you like further details or explanations? 

Here are five related questions to expand your understanding:
1. How does the concept of equilibrium apply to systems with more than two forces?
2. What happens if the angles in this problem change? How does it affect the force magnitudes?
3. Can this method be applied to three-dimensional equilibrium problems?
4. What are the implications of unbalanced forces in practical applications, like structural engineering?
5. How do frictional forces affect equilibrium in real-world scenarios?

**Tip:** Always double-check trigonometric calculations when resolving forces, as small errors can lead to incorrect results in equilibrium problems.

Determine the magnitudes of F1 and F2 such that the particle remains in equilibrium. Given the angles, F1 is at 45 degrees above the x-axis, F2 is at 30 degrees below the x-axis, and there is a horizontal force of 12 N acting to the left along the negative x-axis.

Learn how to calculate the forces F1 and F2 required to keep a particle in equilibrium, given forces at angles of 45° and 30° and a horizontal force of 12 N. This solution involves vector decomposition and equilibrium conditions.

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