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The problem involves finding the magnitude of force \( F \) and its direction \( \theta \) such that the particle at point \( P \) is in equilibrium, based on the force diagram.

To solve this, we need to apply the conditions of equilibrium, which state that the sum of forces in both the \( x \)- and \( y \)-directions must be zero.

### Given:
- There is an 8 kN force acting at an angle of \( 55^\circ \) from the negative \( x \)-axis.
- A 3 kN force acting at an angle of \( 35^\circ \) from the positive \( y \)-axis.
- The force \( F \) acts at an unknown angle \( \theta \) from the positive \( x \)-axis.

### Step 1: Break down each force into components.

For the forces to be in equilibrium, the sum of the forces in the \( x \)- and \( y \)-directions must equal zero:

#### Force 1: 8 kN at \( 55^\circ \) (let's call this \( F_1 \))
- \( F_{1x} = -8 \cos(55^\circ) \)
- \( F_{1y} = 8 \sin(55^\circ) \)

#### Force 2: 3 kN at \( 35^\circ \) (let's call this \( F_2 \))
- \( F_{2x} = 3 \sin(35^\circ) \)
- \( F_{2y} = 3 \cos(35^\circ) \)

#### Force \( F \) at an angle \( \theta \) (let's call this \( F_3 \))
- \( F_{3x} = F \cos(\theta) \)
- \( F_{3y} = F \sin(\theta) \)

### Step 2: Apply equilibrium conditions.

#### In the \( x \)-direction:
\[
F_{1x} + F_{2x} + F_{3x} = 0
\]
\[
-8 \cos(55^\circ) + 3 \sin(35^\circ) + F \cos(\theta) = 0
\]

#### In the \( y \)-direction:
\[
F_{1y} + F_{2y} + F_{3y} = 0
\]
\[
8 \sin(55^\circ) + 3 \cos(35^\circ) + F \sin(\theta) = 0
\]

### Step 3: Solve for \( F \) and \( \theta \).

Let me calculate this system of equations.The equilibrium conditions yield the following results for the sum of the known force components:

- In the \( x \)-direction: \( -2.87 \, \text{kN} \) (this must be balanced by the \( x \)-component of \( F \)).
- In the \( y \)-direction: \( 9.01 \, \text{kN} \) (this must be balanced by the \( y \)-component of \( F \)).

### Step 4: Set up the system of equations.

We can now solve for \( F \) and \( \theta \) using the following two equations:

\[
F \cos(\theta) = -(-2.87) \quad \text{(from the x-direction)}
\]
\[
F \sin(\theta) = -9.01 \quad \text{(from the y-direction)}
\]

By solving this system, we can find both \( F \) and \( \theta \). Let me calculate these values.The magnitude of the force \( F \) is approximately \( 9.46 \, \text{kN} \), and the direction \( \theta \) is approximately \( -107.66^\circ \) (measured counterclockwise from the positive \( x \)-axis).

Do you have any questions or would you like further details?

Here are 5 related questions to explore further:
1. What are the conditions for static equilibrium?
2. How can force components be calculated in different coordinate systems?
3. How does the angle of a force affect its components?
4. What is the significance of negative angles in force direction?
5. How would the solution change if the forces acting were in three dimensions?

**Tip:** When dealing with equilibrium problems, always resolve forces into their components and apply the conditions for both the \( x \)- and \( y \)-directions independently.

Find the magnitude of the force F and its direction θ in the given equilibrium diagram.

This problem involves calculating the magnitude and direction of a force acting on a particle in equilibrium, using vector decomposition and equilibrium conditions in two dimensions. Ideal for undergraduate engineering students.

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