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Answer :
Sure, here is a step-by-step solution to find the [tex]\( x \)[/tex]-component of the weight of the crate:
1. Identify the given values:
- Mass of the crate ([tex]\( m \)[/tex]) = 135 kg
- Angle of the incline ([tex]\( \theta \)[/tex]) = [tex]\( 18.0^\circ \)[/tex]
- Acceleration due to gravity ([tex]\( g \)[/tex]) = 9.8 m/s²
2. Calculate the weight of the crate:
The weight ([tex]\( W \)[/tex]) of the crate is given by the formula:
[tex]\[
W = m \times g
\][/tex]
Substituting the given values:
[tex]\[
W = 135 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 1323 \, \text{N}
\][/tex]
3. Determine the [tex]\( x \)[/tex]-component of the weight:
The [tex]\( x \)[/tex]-component of the weight is the part of the weight acting parallel to the incline. This can be found using the sine function of the angle of incline since it deals with the opposite side of the triangle formed by the weight components.
[tex]\[
W_x = W \times \sin(\theta)
\][/tex]
4. Convert the angle to radians:
Since trigonometric functions in most mathematical contexts (like many calculators and programming languages) use radians, we first convert the angle from degrees to radians.
[tex]\[
\theta \text{ (in radians)} = \theta \times \frac{\pi}{180}
\][/tex]
Substituting the given angle:
[tex]\[
\theta \text{ (in radians)} = 18^\circ \times \frac{\pi}{180} = 0.314 \, \text{radians} \, (\text{approx})
\][/tex]
5. Calculate the sine of the angle in radians:
[tex]\[
\sin(0.314) \approx 0.309
\][/tex]
6. Compute the [tex]\( x \)[/tex]-component of the weight:
[tex]\[
W_x = 1323 \, \text{N} \times 0.309 \approx 408.829 \, \text{N}
\][/tex]
Therefore, the [tex]\( x \)[/tex]-component of the weight of the crate is approximately [tex]\( 408.83 \, \text{N} \)[/tex].
1. Identify the given values:
- Mass of the crate ([tex]\( m \)[/tex]) = 135 kg
- Angle of the incline ([tex]\( \theta \)[/tex]) = [tex]\( 18.0^\circ \)[/tex]
- Acceleration due to gravity ([tex]\( g \)[/tex]) = 9.8 m/s²
2. Calculate the weight of the crate:
The weight ([tex]\( W \)[/tex]) of the crate is given by the formula:
[tex]\[
W = m \times g
\][/tex]
Substituting the given values:
[tex]\[
W = 135 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 1323 \, \text{N}
\][/tex]
3. Determine the [tex]\( x \)[/tex]-component of the weight:
The [tex]\( x \)[/tex]-component of the weight is the part of the weight acting parallel to the incline. This can be found using the sine function of the angle of incline since it deals with the opposite side of the triangle formed by the weight components.
[tex]\[
W_x = W \times \sin(\theta)
\][/tex]
4. Convert the angle to radians:
Since trigonometric functions in most mathematical contexts (like many calculators and programming languages) use radians, we first convert the angle from degrees to radians.
[tex]\[
\theta \text{ (in radians)} = \theta \times \frac{\pi}{180}
\][/tex]
Substituting the given angle:
[tex]\[
\theta \text{ (in radians)} = 18^\circ \times \frac{\pi}{180} = 0.314 \, \text{radians} \, (\text{approx})
\][/tex]
5. Calculate the sine of the angle in radians:
[tex]\[
\sin(0.314) \approx 0.309
\][/tex]
6. Compute the [tex]\( x \)[/tex]-component of the weight:
[tex]\[
W_x = 1323 \, \text{N} \times 0.309 \approx 408.829 \, \text{N}
\][/tex]
Therefore, the [tex]\( x \)[/tex]-component of the weight of the crate is approximately [tex]\( 408.83 \, \text{N} \)[/tex].
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