Given the vector $R = 53 \, \text{m}$ at an angle of $48^\circ$ South of West, you can find the components $R_x$ (the horizontal component) and $R_y$ (the vertical component) using trigonometry.

1. Understanding the Direction

• The angle $48^\circ$ South of West means the angle is measured clockwise from the West direction.

2. Calculating the Components

• The horizontal component $R_x$ is given by: $R_x = R \cdot \cos(\theta)$ Since it's directed Westward, the sign will be negative.
• The vertical component $R_y$ is given by: $R_y = R \cdot \sin(\theta)$ Since it's directed Southward, the sign will also be negative.

Substituting the values:

$R_x = 53 \cdot \cos(48^\circ)$ $R_y = 53 \cdot \sin(48^\circ)$

Now, let's calculate these values:

$R_x = 53 \cdot \cos(48^\circ) \approx 53 \cdot 0.6691 \approx 35.46 \, \text{m}$ $R_y = 53 \cdot \sin(48^\circ) \approx 53 \cdot 0.7431 \approx 39.38 \, \text{m}$

Final Components:

• $R_x \approx -35.46 \, \text{m}$ (Westward)
• $R_y \approx -39.38 \, \text{m}$ (Southward)

Would you like more details or have any questions?

Relative Questions:

1. How would the components change if the vector were 48° North of West?
2. Can you determine the magnitude of $R$ if only the components $R_x$ and $R_y$ are given?
3. What is the physical significance of the negative signs in $R_x$ and $R_y$?
4. How would the components change if the angle were measured from South instead of West?
5. What trigonometric identities are used to derive the horizontal and vertical components of a vector?

Tip: Always keep track of the direction when calculating vector components to avoid sign errors.