Question

a 150 kg crate is on a ramp that is inclined at 20.0^{\circ}.

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

w_{y} = ? n

Explanation:

🆕 New Concept Discovered: Forces on an Inclined Plane
Splitting gravity into components parallel and perpendicular to a ramp.

Step 1: Calculate the total weight

The weight \( w \) of an object is the force of gravity acting on it, calculated by multiplying its mass \( m \) by the acceleration due to gravity \( g \) (approximately \( 9.8\text{ m/s}^2 \) or \( 9.81\text{ m/s}^2 \)):

$$ w = m \cdot g $$

Given:

• \( m = 150\text{ kg} \)
• \( g = 9.8\text{ m/s}^2 \)

$$ w = 150\text{ kg} \times 9.8\text{ m/s}^2 = 1470\text{ N} $$

Step 2: Find the y-component of the weight

When an object is on an inclined plane, we set up a coordinate system where the x-axis is parallel to the ramp and the y-axis is perpendicular to the ramp.

The weight vector points straight down, which creates an angle \( \theta \) with the perpendicular (y) axis equal to the incline angle of the ramp. Therefore, the component of weight perpendicular to the ramp (the y-component) is:

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

Given:

• \( \theta = 20.0^\circ \)

$$ w_y = 1470\text{ N} \times \cos(20.0^\circ) $$

$$ w_y \approx 1470 \times 0.9397 $$

$$ w_y \approx 1381.3\text{ N} $$

(Note: If using \( g = 9.8\text{ m/s}^2 \), \( w_y \approx 1380\text{ N} \). If using \( g = 9.81\text{ m/s}^2 \), \( w \approx 1471.5\text{ N} \) and \( w_y \approx 1383\text{ N} \). Rounding to three significant figures gives \( 1380\text{ N} \).)

Answer:

1380