Given:

• Force 1: $F_1 = 4.0 \, \text{N}$
• Force 2: $F_2 = 6.0 \, \text{N}$
• Angle between the forces: $\theta = 60^\circ$
• Scale: $10 \, \text{mm} = 1.0 \, \text{N}$

Parallelogram Method:

1. Graphical Construction:

   • Start by drawing a vector to represent $F_1 = 4.0 \, \text{N}$. Since $10 \, \text{mm} = 1.0 \, \text{N}$, $F_1$ should be drawn as a 40 mm line.
   • Draw $F_2 = 6.0 \, \text{N}$ starting from the same point, making a $60^\circ$ angle with $F_1$. $F_2$ should be drawn as a 60 mm line.

2. Complete the Parallelogram:

   • To use the parallelogram method, replicate $F_1$ and $F_2$ by drawing the same vectors starting from the heads of $F_2$ and $F_1$ respectively.
   • Draw the parallelogram by connecting the heads of these vectors.

3. Resultant Force:

   • The resultant force is the diagonal of the parallelogram that starts from the common origin of the two forces.
   • Measure the length of this diagonal and convert it back to Newtons using the given scale.

Final Answer (Using Calculation):

To complement the graphical method, you can use the law of cosines to calculate the magnitude of the resultant force $R$:

$R = \sqrt{F_1^2 + F_2^2 + 2 \cdot F_1 \cdot F_2 \cdot \cos(60^\circ)}$

$R = \sqrt{4.0^2 + 6.0^2 + 2 \cdot 4.0 \cdot 6.0 \cdot 0.5}$

$R = \sqrt{16 + 36 + 24} = \sqrt{76} \approx 8.72 \, \text{N}$

Result:

The magnitude of the resultant force $R$ is approximately 8.72 N.

Would you like to see the details or have any questions?

Related Questions:

1. How would the resultant force change if the angle between the forces was 90°?
2. What is the significance of the angle between the forces in determining the resultant?
3. How can the parallelogram method be used for forces not in the plane?
4. Can the resultant force be smaller than either of the individual forces?
5. What if the two forces were acting in exactly opposite directions?

Tip:

The graphical method provides a visual intuition, but it's always good to verify your results using vector calculations for accuracy.