Question

a 30.0 kg box is on a ramp that is inclined at \\(15.0^{\circ}\\).

what is the y-component of the weight of the box?

\\(w_y = ?\text{ n}\\)

Explanation:

🆕 New Concept Discovered: Inclined Plane Force Components
Splitting gravity into parallel and perpendicular forces.

Step 1: Identify the given values

The mass of the box is:

$$ m = 30.0\text{ kg} $$

The angle of the incline is:

$$ \theta = 15.0^\circ $$

The acceleration due to gravity is:

$$ g = 9.8\text{ m/s}^2 $$

Step 2: Calculate the total weight

The total weight \( w \) of the box acts straight down:

$$ w = m \cdot g $$

$$ w = 30.0\text{ kg} \cdot 9.8\text{ m/s}^2 = 294\text{ N} $$

Step 3: Find the y-component of the weight

On an inclined plane, the coordinate system is typically tilted so that the x-axis is parallel to the incline and the y-axis is perpendicular to the incline.

The y-component of the weight (perpendicular to the incline) is given by:

$$ w_y = w \cdot \cos(\theta) $$

Substitute the values into the formula:

$$ w_y = 294\text{ N} \cdot \cos(15.0^\circ) $$

$$ w_y \approx 294\text{ N} \cdot 0.9659 $$

$$ w_y \approx 284\text{ N} $$

(Note: Depending on the sign convention of your physics system, the downward perpendicular force is sometimes written as negative, \( -284\text{ N} \). If the system expects the magnitude, it is \( 284\text{ N} \).)

Answer:

284