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The image presents a problem asking to determine the resultant of a system of concurrent forces in a two-dimensional plane, with the force magnitudes and angles given. Here's the breakdown:

### Given:
- \( F_1 = 150 \, \text{N} \) at an angle of 30° above the positive \( x \)-axis.
- \( F_2 = 200 \, \text{N} \) at an angle of 60° above the negative \( x \)-axis.
- \( F_3 = 80 \, \text{N} \) at an angle of 45° below the negative \( x \)-axis.
- \( F_4 = 180 \, \text{N} \) along the negative \( y \)-axis.

### Steps to solve:

1. **Resolve each force into its components**:
   - \( F_1 \) (150 N, 30°):
     - \( F_{1x} = 150 \cos(30°) \)
     - \( F_{1y} = 150 \sin(30°) \)
   - \( F_2 \) (200 N, 60°):
     - \( F_{2x} = -200 \cos(60°) \) (negative since it's along the negative \( x \)-axis)
     - \( F_{2y} = 200 \sin(60°) \)
   - \( F_3 \) (80 N, 45°):
     - \( F_{3x} = -80 \cos(45°) \) (negative since it's along the negative \( x \)-axis)
     - \( F_{3y} = -80 \sin(45°) \) (negative since it's below the \( x \)-axis)
   - \( F_4 \) (180 N, along negative \( y \)-axis):
     - \( F_{4x} = 0 \) (no \( x \)-component)
     - \( F_{4y} = -180 \) (entirely along negative \( y \)-axis)

2. **Sum up all the components**:
   - \( F_x = F_{1x} + F_{2x} + F_{3x} + F_{4x} \)
   - \( F_y = F_{1y} + F_{2y} + F_{3y} + F_{4y} \)

3. **Find the magnitude of the resultant force**:
   - \( R = \sqrt{F_x^2 + F_y^2} \)

4. **Find the angle of the resultant force**:
   - \( \theta = \tan^{-1}\left(\frac{F_y}{F_x}\right) \)

I'll compute the values of the components and determine the resultant.The components and resultant for the system of forces are as follows:

- \( F_x = -26.67 \, \text{N} \) (negative, meaning it's toward the negative \( x \)-axis)
- \( F_y = 11.64 \, \text{N} \) (positive, meaning it's toward the positive \( y \)-axis)

The magnitude of the resultant force is:
- \( R = 29.09 \, \text{N} \)

The angle of the resultant force is:
- \( \theta_R = 156.42^\circ \) (measured counterclockwise from the positive \( x \)-axis)

This angle indicates that the resultant force is directed primarily in the second quadrant (between the negative \( x \)-axis and positive \( y \)-axis).

Would you like further clarification on any steps, or more detailed explanation?

Here are some related questions for deeper exploration:
1. How do the angles of forces affect their resultant direction?
2. How would changing the magnitude of \( F_1 \) alter the resultant?
3. What happens if we add another force along the positive \( y \)-axis?
4. How does the angle calculation change if we resolve forces using different reference axes?
5. What are practical applications of determining the resultant force in engineering?

**Tip**: To simplify vector problems, always break forces into their components first before summing them up.

Determine the resultant of the system of concurrent forces with the following force values and angles: F1 = 150 N, F2 = 200 N, F3 = 80 N, and F4 = 180 N at the respective angles.

Learn how to calculate the resultant of a concurrent force system with forces F1 = 150 N, F2 = 200 N, F3 = 80 N, and F4 = 180 N using trigonometric methods and vector addition. Includes step-by-step solution for high school and early university students.

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