In this problem, you are tasked with determining the magnitude and direction of the resultant force, $F_R$, of the three forces $F_1$, $F_2$, and $F_3$ shown in the diagram. The forces are provided with their magnitudes and the angles relative to the coordinate axes.

Given:

• $F_1 = 30 \, \text{N}$ at an angle of $0^\circ$ (along the x-axis).
• $F_2 = 5 \, \text{N}$ at an angle of $20^\circ$ from the positive x-axis (clockwise direction).
• $F_3 = 60 \, \text{N}$ along the negative y-axis.

We will use vector decomposition to find the components of each force and then sum them to determine the resultant force.

Step-by-Step Solution:

1. Decompose the Forces into Components:

   • $F_1$ has only an x-component: $F_{1x} = F_1 \cdot \cos(0^\circ) = 30 \, \text{N}$ $F_{1y} = F_1 \cdot \sin(0^\circ) = 0 \, \text{N}$

   • $F_2$ has components in both the x and y directions: $F_{2x} = F_2 \cdot \cos(20^\circ) = 5 \cdot \cos(20^\circ)$ $F_{2y} = F_2 \cdot \sin(20^\circ) = 5 \cdot \sin(20^\circ)$

   • $F_3$ is purely in the negative y-direction: $F_{3x} = 0 \, \text{N}$ $F_{3y} = -60 \, \text{N}$

2. Sum the Components:

   • Sum the x-components: $F_{Rx} = F_{1x} + F_{2x} + F_{3x} = 30 + 5 \cdot \cos(20^\circ) + 0$
   • Sum the y-components: $F_{Ry} = F_{1y} + F_{2y} + F_{3y} = 0 + 5 \cdot \sin(20^\circ) - 60$

3. Calculate the Magnitude of the Resultant Force:

   • The magnitude of the resultant force is: $F_R = \sqrt{F_{Rx}^2 + F_{Ry}^2}$

4. Calculate the Direction (Angle θ):

   • The angle $\theta$ (counterclockwise from the positive x-axis) is given by: $\theta = \tan^{-1} \left( \frac{F_{Ry}}{F_{Rx}} \right)$

Let me calculate these values. math

Given forces

F1 = 30 # N F2 = 5 # N F3 = 60 # N

Calculate components of F2

F2x = F2 * math.cos(math.radians(20)) F2y = F2 * math.sin(math.radians(20))

Components of F1 and F3

F1x = 30 # N (along the x-axis) F1y = 0 # N (along the y-axis)

F3x = 0 # N (along the x-axis) F3y = -60 # N (along the negative y-axis)

Sum of components

FRx = F1x + F2x + F3x FRy = F1y + F2y + F3y

Magnitude of the resultant force

FR_magnitude = math.sqrt(FRx2 + FRy2)

Direction (angle θ in degrees, counterclockwise from the positive x-axis)

theta = math.degrees(math.atan2(FRy, FRx))

FR_magnitude, thetaThe magnitude of the resultant force $F_R$ is approximately 67.84 N.

The direction, measured counterclockwise from the positive x-axis, is approximately -59.24°. This negative angle indicates that the resultant force is directed below the x-axis in the fourth quadrant.

If you have any further questions or need additional details, feel free to ask! Here are a few related questions you might find interesting:

1. How would the direction change if the angle of $F_3$ was different?
2. What would happen if $F_2$ were applied along the y-axis instead of at 20°?
3. How can we visualize the vector addition of these forces using a graph?
4. How does the magnitude of the resultant force change if $F_1$ were increased to 40 N?
5. What happens if all the forces are acting in the same direction (e.g., 0°)?

Tip: When dealing with forces at different angles, always break them down into components along the x and y axes first—this simplifies the process of calculating the resultant force.