To solve for the [tex]$y$[/tex]-component of the total force acting on the block, we will break down each force into its [tex]$y$[/tex]-component, sum these components, and find the total [tex]$y$[/tex]-component force. Let's proceed step by step:
1. Identify the Given Information:
- Force 1 ([tex]$F_1$[/tex]) has a magnitude of [tex]\(115\)[/tex] Newtons and an angle of [tex]\(75.0^\circ\)[/tex].
- Force 2 ([tex]$F_2$[/tex]) has a magnitude of [tex]\(213\)[/tex] Newtons and an angle of [tex]\(295.0^\circ\)[/tex].
2. Convert the Angles to Radians:
Since trigonometric calculations are often more convenient in radians, we will convert the angles from degrees to radians.
- Angle 1 in radians: [tex]\(75.0^\circ\)[/tex]
- Angle 2 in radians: [tex]\(295.0^\circ\)[/tex]
3. Calculate the [tex]$y$[/tex]-component of Each Force:
To find the [tex]$y$[/tex]-component of a force, use the sine component of the force at the given angle.
- For Force 1:
[tex]\[
F_{1y} = F_1 \sin(\theta_1)
\][/tex]
Substituting the values, we get:
[tex]\[
F_{1y} = 115 \cdot \sin(75.0^\circ)
\][/tex]
[tex]\[
F_{1y} \approx 111.08 \, \text{N}
\][/tex]
- For Force 2:
[tex]\[
F_{2y} = F_2 \sin(\theta_2)
\][/tex]
Substituting the values, we get:
[tex]\[
F_{2y} = 213 \cdot \sin(295.0^\circ)
\][/tex]
[tex]\[
F_{2y} \approx -193.04 \, \text{N}
\][/tex]
4. Sum the [tex]$y$[/tex]-components to Find the Total [tex]$y$[/tex]-component Force:
[tex]\[
F_y = F_{1y} + F_{2y}
\][/tex]
Substitute the previously calculated values:
[tex]\[
F_y = 111.08 \, \text{N} + (-193.04 \, \text{N})
\][/tex]
[tex]\[
F_y \approx -81.96 \, \text{N}
\][/tex]
So, the [tex]$y$[/tex]-component of the total force acting on the block is:
[tex]\[
\overrightarrow{F_y} \approx -81.96 \, \text{N}
\][/tex]
