Explore Westonci.ca, the leading Q&A site where experts provide accurate and helpful answers to all your questions. Explore in-depth answers to your questions from a knowledgeable community of experts across different fields. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.

A 75.0 kg crate is on a ramp that is inclined at [tex]28.0^{\circ}[/tex].

What is the [tex]x[/tex]-component of the weight of the crate?
[tex]\ \textless \ br/\ \textgreater \ w_x = [?] \, N\ \textless \ br/\ \textgreater \ [/tex]


Sagot :

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].
Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.