Corresponding to the three sets of masses listed in Table 2.1 are three forces F1, F2, and F3. Determine their resultant analytically using the component method. 

Force 1: 200g, 1.96 Newtons, East

Force 2: 300g, 2.94 Newtons, Northeast 

Force 3: 500g, 4.90 Newtons, 60 degrees South of East 

To determine the resultant force of the three forces using the component method, we will break each force into its horizontal (\(F_x\)) and vertical (\(F_y\)) components and then sum them up to get the total horizontal and vertical components. From these, we can determine the magnitude and direction of the resultant force.

### Force 1: 1.96 N, East
Since this force is directed along the east axis, it has no vertical component.

- \( F_{1x} = 1.96 \, \text{N} \)
- \( F_{1y} = 0 \, \text{N} \)

### Force 2: 2.94 N, Northeast
This force is directed at a 45-degree angle to the east (since Northeast implies a 45-degree angle between the horizontal and vertical axes).

- Horizontal component: \( F_{2x} = 2.94 \times \cos(45^\circ) = 2.94 \times \frac{\sqrt{2}}{2} \approx 2.08 \, \text{N} \)
- Vertical component: \( F_{2y} = 2.94 \times \sin(45^\circ) = 2.94 \times \frac{\sqrt{2}}{2} \approx 2.08 \, \text{N} \)

### Force 3: 4.90 N, 60 degrees South of East
This force makes a 60-degree angle south of the east axis.

- Horizontal component: \( F_{3x} = 4.90 \times \cos(60^\circ) = 4.90 \times 0.5 = 2.45 \, \text{N} \)
- Vertical component: \( F_{3y} = 4.90 \times \sin(60^\circ) = 4.90 \times \frac{\sqrt{3}}{2} \approx 4.24 \, \text{N} \) (Since it's directed south, this component is negative.)

Thus, the components of Force 3 are:
- \( F_{3x} = 2.45 \, \text{N} \)
- \( F_{3y} = -4.24 \, \text{N} \)

### Sum of Components:
Now, we sum up the horizontal and vertical components.

- \( F_{\text{total x}} = F_{1x} + F_{2x} + F_{3x} = 1.96 + 2.08 + 2.45 = 6.49 \, \text{N} \)
- \( F_{\text{total y}} = F_{1y} + F_{2y} + F_{3y} = 0 + 2.08 - 4.24 = -2.16 \, \text{N} \)

### Resultant Force:
The magnitude of the resultant force is given by the Pythagorean theorem:

\[
F_{\text{resultant}} = \sqrt{F_{\text{total x}}^2 + F_{\text{total y}}^2} = \sqrt{6.49^2 + (-2.16)^2} \approx \sqrt{42.13 + 4.67} \approx \sqrt{46.8} \approx 6.84 \, \text{N}
\]

### Direction of the Resultant Force:
The direction of the resultant force is given by the angle \( \theta \) with respect to the east axis (horizontal):

\[
\theta = \tan^{-1}\left(\frac{F_{\text{total y}}}{F_{\text{total x}}}\right) = \tan^{-1}\left(\frac{-2.16}{6.49}\right) \approx \tan^{-1}(-0.333) \approx -18.4^\circ
\]

This angle is 18.4 degrees south of east.

### Final Answer:
- **Magnitude of the resultant force**: 6.84 N
- **Direction of the resultant force**: 18.4 degrees south of east.

Would you like more details on any of the steps?

Here are 5 questions to think about next:
1. What would happen if Force 2 were acting directly North instead of Northeast?
2. How would the resultant force change if Force 1 were doubled in magnitude?
3. Can you apply the component method for forces acting at non-right angles?
4. What is the physical significance of the direction of the resultant force?
5. How does changing the angle of a force affect its horizontal and vertical components?

**Tip**: Always ensure that your angles and directions are clearly referenced to a common axis (e.g., East or North) when solving vector problems.

Corresponding to the three sets of masses listed in Table 2.1 are three forces F1, F2, and F3. Determine their resultant analytically using the component method. Force 1: 200g, 1.96 Newtons, East Force 2: 300g, 2.94 Newtons, Northeast Force 3: 500g, 4.90 Newtons, 60 degrees South of East.

Learn how to determine the resultant force of three forces using the component method. The problem includes forces acting in different directions and involves trigonometry and vector addition. Step-by-step calculations are provided for students in Grades 10-12.

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