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.