Question

a 75.0 kg crate is on a ramp that is inclined at 28.0^{\circ}.

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

w_y = ? n

Explanation:

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

Step 1: Calculate the total weight

The total weight \( w \) of the crate is the force due to gravity acting straight down:

$$ w = m \cdot g $$

Using \( m = 75.0\text{ kg} \) and \( g = 9.80\text{ m/s}^2 \):

$$ w = 75.0\text{ kg} \times 9.80\text{ m/s}^2 = 735\text{ N} $$

Step 2: Find the y-component perpendicular to the ramp

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

The y-component of the weight (\( w_y \)) presses directly into the ramp and is calculated using the cosine of the incline angle \( \theta \):

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

Given \( \theta = 28.0^\circ \):

$$ w_y = 735\text{ N} \times \cos(28.0^\circ) $$

$$ w_y \approx 735\text{ N} \times 0.88295 $$

$$ w_y \approx 649\text{ N} $$

Answer:

649