Answer :
To determine the [tex]\( x \)[/tex]-component of the weight of the crate on an inclined ramp, we can follow these steps:
1. Calculate the weight of the crate:
The weight [tex]\( W \)[/tex] of an object is given by the formula:
[tex]\[ W = m \cdot g \][/tex]
where [tex]\( m \)[/tex] is the mass of the crate and [tex]\( g \)[/tex] is the acceleration due to gravity.
For a mass [tex]\( m = 75.0 \)[/tex] kg and assuming [tex]\( g = 9.81 \)[/tex] m/s² (the acceleration due to gravity),
[tex]\[ W = 75.0 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 735.75 \, \text{N} \][/tex]
2. Convert the inclined angle from degrees to radians:
To find the component of the weight along the incline, we need to convert the angle from degrees to radians. The inclined angle of the ramp is [tex]\( 28.0^{\circ} \)[/tex].
We use the conversion:
[tex]\[ \text{angle in radians} = \text{angle in degrees} \times \left( \frac{\pi}{180} \right) \][/tex]
Thus,
[tex]\[ 28.0^{\circ} \times \left( \frac{\pi}{180} \right) \approx 0.4887 \, \text{radians} \][/tex]
3. Calculate the [tex]\( x \)[/tex]-component of the weight:
The [tex]\( x \)[/tex]-component of the weight [tex]\( W_x \)[/tex] is the component of the weight parallel to the inclined plane. This can be calculated using:
[tex]\[ W_x = W \cdot \sin(\theta) \][/tex]
where [tex]\( \theta \)[/tex] is the incline angle in radians. Given [tex]\( W = 735.75 \, \text{N} \)[/tex] and [tex]\( \theta \approx 0.4887 \, \text{radians} \)[/tex],
[tex]\[ W_x = 735.75 \, \text{N} \times \sin(0.4887) \approx 345.41 \, \text{N} \][/tex]
Therefore, the [tex]\( x \)[/tex]-component of the weight of the crate is approximately [tex]\( 345.41 \, \text{N} \)[/tex].
1. Calculate the weight of the crate:
The weight [tex]\( W \)[/tex] of an object is given by the formula:
[tex]\[ W = m \cdot g \][/tex]
where [tex]\( m \)[/tex] is the mass of the crate and [tex]\( g \)[/tex] is the acceleration due to gravity.
For a mass [tex]\( m = 75.0 \)[/tex] kg and assuming [tex]\( g = 9.81 \)[/tex] m/s² (the acceleration due to gravity),
[tex]\[ W = 75.0 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 735.75 \, \text{N} \][/tex]
2. Convert the inclined angle from degrees to radians:
To find the component of the weight along the incline, we need to convert the angle from degrees to radians. The inclined angle of the ramp is [tex]\( 28.0^{\circ} \)[/tex].
We use the conversion:
[tex]\[ \text{angle in radians} = \text{angle in degrees} \times \left( \frac{\pi}{180} \right) \][/tex]
Thus,
[tex]\[ 28.0^{\circ} \times \left( \frac{\pi}{180} \right) \approx 0.4887 \, \text{radians} \][/tex]
3. Calculate the [tex]\( x \)[/tex]-component of the weight:
The [tex]\( x \)[/tex]-component of the weight [tex]\( W_x \)[/tex] is the component of the weight parallel to the inclined plane. This can be calculated using:
[tex]\[ W_x = W \cdot \sin(\theta) \][/tex]
where [tex]\( \theta \)[/tex] is the incline angle in radians. Given [tex]\( W = 735.75 \, \text{N} \)[/tex] and [tex]\( \theta \approx 0.4887 \, \text{radians} \)[/tex],
[tex]\[ W_x = 735.75 \, \text{N} \times \sin(0.4887) \approx 345.41 \, \text{N} \][/tex]
Therefore, the [tex]\( x \)[/tex]-component of the weight of the crate is approximately [tex]\( 345.41 \, \text{N} \)[/tex].
