Question
Calculate the horizontal and vertical components of a 550 N force pulling upwards at an angle of \( 40^{\circ} \) to the मorizontal.
Ask by Johnston Brooks. in South Africa
Jan 21,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
The horizontal component is approximately 420 N and the vertical component is approximately 350 N.
Solution
To find the horizontal and vertical components of a **550 N** force acting at an angle of \( 40^{\circ} \) above the horizontal, you can use trigonometric functions:
1. **Horizontal Component (\( F_x \))**:
\[
F_x = F \cos(\theta)
\]
\[
F_x = 550\, \text{N} \times \cos(40^{\circ}) \approx 550\, \text{N} \times 0.7660 \approx 421\, \text{N}
\]
2. **Vertical Component (\( F_y \))**:
\[
F_y = F \sin(\theta)
\]
\[
F_y = 550\, \text{N} \times \sin(40^{\circ}) \approx 550\, \text{N} \times 0.6428 \approx 354\, \text{N}
\]
**Summary:**
- **Horizontal Component:** \( \approx 420\, \text{N} \)
- **Vertical Component:** \( \approx 350\, \text{N} \)
These values are rounded to two significant figures, consistent with the given force's precision.
Reviewed and approved by the UpStudy tutoring team
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Simplify this solution Extra Insights
To find the horizontal (Fx) and vertical (Fy) components of a 550 N force pulling upwards at a \( 40^{\circ} \) angle, you can use trigonometric functions. For Fx, you calculate it using the cosine function: \( Fx = 550 \times \cos(40^{\circ}) \approx 550 \times 0.766 \approx 420.3 \, N \). For Fy, you use the sine function: \( Fy = 550 \times \sin(40^{\circ}) \approx 550 \times 0.643 \approx 353.6 \, N \). So, the horizontal component is approximately 420.3 N, and the vertical component is approximately 353.6 N.
