A 1.30 kg glass is on a tray that is inclined at [tex]$18.0^{\circ}$[/tex].

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

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



Answer :

To determine the [tex]\( x \)[/tex]-component of the weight of a 1.30 kg glass on a tray inclined at [tex]\( 18.0^\circ \)[/tex], we can follow these steps:

1. Calculate the weight of the glass:
The weight ([tex]\( W \)[/tex]) of the glass can be found using the formula:
[tex]\[ W = m \cdot g \][/tex]
where [tex]\( m = 1.30 \)[/tex] kg is the mass of the glass and [tex]\( g = 9.81 \, \text{m/s}^2 \)[/tex] is the acceleration due to gravity.

[tex]\[ W = 1.30 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 12.753 \, \text{N} \][/tex]

2. Convert the incline angle from degrees to radians:
The angle of inclination is given as [tex]\( 18.0^\circ \)[/tex]. To use trigonometric functions, we need to convert this angle to radians. The conversion is given by:
[tex]\[ \text{angle in radians} = \text{angle in degrees} \times \frac{\pi}{180} \][/tex]

[tex]\[ \text{angle in radians} = 18.0^\circ \times \frac{\pi}{180} = 0.3141592653589793 \, \text{radians} \][/tex]

3. Calculate the [tex]\( x \)[/tex]-component of the weight:
The [tex]\( x \)[/tex]-component of the weight ([tex]\( w_x \)[/tex]) is found by using the sine of the angle of inclination, since we are dealing with the horizontal component along the incline. The formula is:
[tex]\[ w_x = W \cdot \sin(\theta) \][/tex]

where [tex]\( \theta = 0.3141592653589793 \)[/tex] radians is the angle of inclination.

[tex]\[ w_x = 12.753 \, \text{N} \times \sin(0.3141592653589793) \][/tex]

By calculating this, we get:
[tex]\[ w_x \approx 3.9408937292637045 \, \text{N} \][/tex]

Hence, the [tex]\( x \)[/tex]-component of the weight of the glass is approximately [tex]\( 3.94 \, \text{N} \)[/tex].