Solutions to Vector Components Worksheet | Exercises Chemistry ... - Free Printable

Problem Overview:

The task involves determining the horizontal (x) and vertical (y) components of vectors using trigonometry. The vectors are given with their magnitudes and angles relative to the horizontal axis. We will use the following trigonometric relationships:

- Horizontal component (x-component): \( F_x = F \cos(\theta) \)
- Vertical component (y-component): \( F_y = F \sin(\theta) \)

Where:
- \( F \) is the magnitude of the vector.
- \( \theta \) is the angle the vector makes with the horizontal axis.

Step-by-Step Solution:

#### Vector 1:
- Magnitude (\( F \)): 50 N
- Angle (\( \theta \)): 30°

1. Horizontal Component (\( F_x \)):
\[
F_x = F \cos(\theta) = 50 \cos(30^\circ)
\]
Using \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \):
\[
F_x = 50 \cdot \frac{\sqrt{3}}{2} = 50 \cdot 0.866 = 43.3 \, \text{N}
\]

2. Vertical Component (\( F_y \)):
\[
F_y = F \sin(\theta) = 50 \sin(30^\circ)
\]
Using \( \sin(30^\circ) = \frac{1}{2} \):
\[
F_y = 50 \cdot \frac{1}{2} = 50 \cdot 0.5 = 25 \, \text{N}
\]

#### Vector 2:
- Magnitude (\( F \)): 75 N
- Angle (\( \theta \)): 60°

1. Horizontal Component (\( F_x \)):
\[
F_x = F \cos(\theta) = 75 \cos(60^\circ)
\]
Using \( \cos(60^\circ) = \frac{1}{2} \):
\[
F_x = 75 \cdot \frac{1}{2} = 75 \cdot 0.5 = 37.5 \, \text{N}
\]

2. Vertical Component (\( F_y \)):
\[
F_y = F \sin(\theta) = 75 \sin(60^\circ)
\]
Using \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \):
\[
F_y = 75 \cdot \frac{\sqrt{3}}{2} = 75 \cdot 0.866 = 64.95 \, \text{N}
\]

#### Vector 3:
- Magnitude (\( F \)): 100 N
- Angle (\( \theta \)): 45°

1. Horizontal Component (\( F_x \)):
\[
F_x = F \cos(\theta) = 100 \cos(45^\circ)
\]
Using \( \cos(45^\circ) = \frac{\sqrt{2}}{2} \):
\[
F_x = 100 \cdot \frac{\sqrt{2}}{2} = 100 \cdot 0.707 = 70.7 \, \text{N}
\]

2. Vertical Component (\( F_y \)):
\[
F_y = F \sin(\theta) = 100 \sin(45^\circ)
\]
Using \( \sin(45^\circ) = \frac{\sqrt{2}}{2} \):
\[
F_y = 100 \cdot \frac{\sqrt{2}}{2} = 100 \cdot 0.707 = 70.7 \, \text{N}
\]

#### Vector 4:
- Magnitude (\( F \)): 80 N
- Angle (\( \theta \)): 15°

1. Horizontal Component (\( F_x \)):
\[
F_x = F \cos(\theta) = 80 \cos(15^\circ)
\]
Using \( \cos(15^\circ) \approx 0.9659 \):
\[
F_x = 80 \cdot 0.9659 = 77.272 \, \text{N}
\]

2. Vertical Component (\( F_y \)):
\[
F_y = F \sin(\theta) = 80 \sin(15^\circ)
\]
Using \( \sin(15^\circ) \approx 0.2588 \):
\[
F_y = 80 \cdot 0.2588 = 20.704 \, \text{N}
\]

Final Answers:

\[
\boxed{
\begin{array}{c|c|c}
\text{Vector} & F_x \, (\text{N}) & F_y \, (\text{N}) \\
\hline
1 & 43.3 & 25 \\
2 & 37.5 & 64.95 \\
3 & 70.7 & 70.7 \\
4 & 77.272 & 20.704 \\
\end{array}
}
\]