Answer :
To find the [tex]\( x \)[/tex]-component of the weight of the crate on an inclined plane, follow these steps:
1. Determine the Weight of the Crate:
The weight [tex]\( W \)[/tex] of an object is calculated using the formula:
[tex]\[ W = m \cdot g \][/tex]
where
- [tex]\( m \)[/tex] is the mass of the object in kilograms ([tex]\( kg \)[/tex]),
- [tex]\( g \)[/tex] is the acceleration due to gravity ([tex]\( 9.81 \, m/s^2 \)[/tex]).
Given:
- Mass [tex]\( m = 150 \, kg \)[/tex],
Therefore:
[tex]\[ W = 150 \, kg \times 9.81 \, m/s^2 = 1471.5 \, N \][/tex]
2. Resolve the Weight into Components:
On an inclined plane, the weight of the object can be resolved into two components:
- [tex]\( W_x \)[/tex]: The component parallel to the inclined plane,
- [tex]\( W_y \)[/tex]: The component perpendicular to the inclined plane.
3. Calculate the [tex]\( x \)[/tex]-component of the Weight:
The [tex]\( x \)[/tex]-component of the weight [tex]\( W_x \)[/tex] can be found using the sine function:
[tex]\[ W_x = W \sin(\theta) \][/tex]
where
- [tex]\( \theta \)[/tex] is the angle of the incline,
- [tex]\( W \)[/tex] is the weight of the crate.
Given:
- Angle [tex]\( \theta = 18.0^\circ \)[/tex],
Therefore:
[tex]\[ W_x = 1471.5 \, N \times \sin(18.0^\circ) \][/tex]
Using the sine value for [tex]\( 18.0^\circ \)[/tex]:
[tex]\[ \sin(18.0^\circ) \approx 0.309 \][/tex]
Hence:
[tex]\[ W_x = 1471.5 \, N \times 0.309 \approx 454.72 \, N \][/tex]
So, the [tex]\( x \)[/tex]-component of the weight of the crate is approximately:
[tex]\[ W_x = 454.72 \, N \][/tex]
1. Determine the Weight of the Crate:
The weight [tex]\( W \)[/tex] of an object is calculated using the formula:
[tex]\[ W = m \cdot g \][/tex]
where
- [tex]\( m \)[/tex] is the mass of the object in kilograms ([tex]\( kg \)[/tex]),
- [tex]\( g \)[/tex] is the acceleration due to gravity ([tex]\( 9.81 \, m/s^2 \)[/tex]).
Given:
- Mass [tex]\( m = 150 \, kg \)[/tex],
Therefore:
[tex]\[ W = 150 \, kg \times 9.81 \, m/s^2 = 1471.5 \, N \][/tex]
2. Resolve the Weight into Components:
On an inclined plane, the weight of the object can be resolved into two components:
- [tex]\( W_x \)[/tex]: The component parallel to the inclined plane,
- [tex]\( W_y \)[/tex]: The component perpendicular to the inclined plane.
3. Calculate the [tex]\( x \)[/tex]-component of the Weight:
The [tex]\( x \)[/tex]-component of the weight [tex]\( W_x \)[/tex] can be found using the sine function:
[tex]\[ W_x = W \sin(\theta) \][/tex]
where
- [tex]\( \theta \)[/tex] is the angle of the incline,
- [tex]\( W \)[/tex] is the weight of the crate.
Given:
- Angle [tex]\( \theta = 18.0^\circ \)[/tex],
Therefore:
[tex]\[ W_x = 1471.5 \, N \times \sin(18.0^\circ) \][/tex]
Using the sine value for [tex]\( 18.0^\circ \)[/tex]:
[tex]\[ \sin(18.0^\circ) \approx 0.309 \][/tex]
Hence:
[tex]\[ W_x = 1471.5 \, N \times 0.309 \approx 454.72 \, N \][/tex]
So, the [tex]\( x \)[/tex]-component of the weight of the crate is approximately:
[tex]\[ W_x = 454.72 \, N \][/tex]