The image shows three force vectors acting at different angles, with the goal of finding the resultant force of the three vectors. Here’s the breakdown of the vectors:

• $F_1 = 20 \, \text{N}$ at an angle of $60^\circ$ from the positive $X$-axis.
• $F_2 = 10 \, \text{N}$ at an angle of $30^\circ$ above the negative $X$-axis.
• $F_3 = 30 \, \text{N}$ pointing directly downward (along the negative $Y$-axis).

Step 1: Resolve each vector into its components.

• For $F_1$: $F_{1x} = F_1 \cos(60^\circ) = 20 \times \cos(60^\circ) = 20 \times 0.5 = 10 \, \text{N}$ $F_{1y} = F_1 \sin(60^\circ) = 20 \times \sin(60^\circ) = 20 \times 0.866 = 17.32 \, \text{N}$

• For $F_2$:

  \, (\text{in the negative direction along the x-axis})$$ $$F_{2y} = F_2 \sin(30^\circ) = 10 \times \sin(30^\circ) = 10 \times 0.5 = 5 \, \text{N} \, (\text{in the positive direction along the y-axis})$$

• For $F_3$: Since $F_3$ is along the negative $Y$-axis: $F_{3x} = 0 \, \text{N}$ $F_{3y} = -30 \, \text{N}$

Step 2: Sum the components.

• Summing the $X$-components: $F_x = F_{1x} + (-F_{2x}) + F_{3x} = 10 + (-8.66) + 0 = 1.34 \, \text{N}$

• Summing the $Y$-components: $F_y = F_{1y} + F_{2y} + F_{3y} = 17.32 + 5 + (-30) = -7.68 \, \text{N}$

Step 3: Calculate the magnitude of the resultant vector.

$R = \sqrt{F_x^2 + F_y^2} = \sqrt{(1.34)^2 + (-7.68)^2} = \sqrt{1.7956 + 59.0144} = \sqrt{60.81} \approx 7.8 \, \text{N}$

Thus, the magnitude of the resultant force is approximately 7.8 N.

Would you like any further explanation or details?

Here are five related questions to further explore:

1. How do we calculate the direction of the resultant vector?
2. What is the physical significance of resolving vectors into components?
3. How does vector addition differ from scalar addition?
4. Can you explain the concept of equilibrium in vector forces?
5. How would the result change if the angle of one of the vectors were different?

Tip: Always ensure that angles are measured from the correct axis reference when resolving vectors into components.