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Let's break down the task to solve it step by step:

### 1. **The magnitude and direction of the total force along the vertical:**

We'll sum the vertical components of each force. First, let's break each force into its vertical and horizontal components. The general formulas are:

- Vertical component: \( F_y = F \cdot \sin(\theta) \)
- Horizontal component: \( F_x = F \cdot \cos(\theta) \)

Where \( F \) is the magnitude of the force and \( \theta \) is the angle it makes with the horizontal axis.

For each force:

- \( F_1 = 2\,N \) at \( 38^\circ \) above the positive x-axis:
  - \( F_{1x} = 2 \cdot \cos(38^\circ) \)
  - \( F_{1y} = 2 \cdot \sin(38^\circ) \)

- \( F_2 = 6\,N \) at \( 42^\circ \) below the negative x-axis:
  - \( F_{2x} = 6 \cdot \cos(42^\circ) \) (to the left)
  - \( F_{2y} = 6 \cdot \sin(42^\circ) \) (downwards)

- \( F_3 = 7\,N \) at \( 25^\circ \) above the positive x-axis:
  - \( F_{3x} = 7 \cdot \cos(25^\circ) \)
  - \( F_{3y} = 7 \cdot \sin(25^\circ) \)

- \( F_4 = 9\,N \) at \( 110^\circ \) below the positive x-axis:
  - \( F_{4x} = 9 \cdot \cos(110^\circ) \)
  - \( F_{4y} = 9 \cdot \sin(110^\circ) \)

### 2. **Magnitude of the Resultant Force:**

The resultant force components \( F_x \) and \( F_y \) are the sum of the horizontal and vertical components:

- \( F_x = F_{1x} + F_{2x} + F_{3x} + F_{4x} \)
- \( F_y = F_{1y} + F_{2y} + F_{3y} + F_{4y} \)

Once we have \( F_x \) and \( F_y \), we calculate the magnitude of the resultant force \( F_R \) using:

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

### 3. **Direction of the Resultant Force:**

The direction (angle \( \theta_R \)) of the resultant force is given by:

\[
\theta_R = \tan^{-1}\left(\frac{F_y}{F_x}\right)
\]

### 4. **Sketch the Resultant Force:**

This requires sketching the force vector according to the calculated magnitude and direction.

I'll proceed with the detailed calculations.### Solution Summary:

1. **Vertical Force Components:**
   - The total vertical force is obtained by summing all vertical components:
     - \( F_{1y} = 1.23 \,N \)
     - \( F_{2y} = -4.01 \,N \)
     - \( F_{3y} = 2.96 \,N \)
     - \( F_{4y} = 8.46 \,N \)
     - **Total vertical force** \( F_y = 8.63 \,N \)

2. **Magnitude of the Resultant Force:**
   - Total horizontal force \( F_x = 9.30 \,N \)
   - Total vertical force \( F_y = 8.63 \,N \)
   - Magnitude of the resultant force \( F_R = 12.69 \,N \)

3. **Direction of the Resultant Force:**
   - The angle of the resultant force is \( \theta_R = 42.86^\circ \) above the positive x-axis.

4. **Sketch:**
   - Sketch the resultant force as a vector at an angle of approximately \( 42.86^\circ \) with a magnitude of \( 12.69 \,N \).

Would you like further details or explanations for any part of this solution?

### 5 Related Questions:
1. How do you calculate the components of a force in any direction?
2. How does the direction of a force affect its components?
3. Can you explain the role of the tangent function in determining the angle of a vector?
4. How would the calculations change if the direction of one of the forces were reversed?
5. Why is the angle \( \tan^{-1}(F_y / F_x) \) always measured from the x-axis?

### Tip:
When calculating forces in different directions, always ensure your angle references are consistent (either all from the positive x-axis or another agreed reference). This avoids confusion when adding components.

Given the figure below, determine the following: 1. The magnitude and direction of the total force along the vertical. 2. Magnitude of the Resultant Force. 3. Direction of the Resultant Force. 4. Sketch the Resultant Force.

Learn how to calculate the magnitude and direction of a resultant force using vector addition. This example involves four forces acting at different angles. Follow step-by-step calculations to find the vertical components, the total magnitude, and the angle of the resultant force.

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