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A [tex]150 \, \text{kg}[/tex] crate is on a ramp that is inclined at [tex]18.0^{\circ}[/tex].

What is the [tex]x[/tex]-component of the weight of the crate?

[tex]w_x = \, [?] \, \text{N}[/tex]


Sagot :

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]