College

A chair is pulled by two horizontal forces. The first force is 122 N at an angle of [tex]$43.6^{\circ}$[/tex], and the second is 97.6 N at an angle of [tex]$49.9^{\circ}$[/tex].

What is the [tex]$y$[/tex]-component of the total force acting on the chair?

[tex]\overrightarrow{F_{y}} = [?] \, \text{N}[/tex]

Answer :

To find the [tex]\( y \)[/tex]-component of the total force acting on the chair, we need to consider the contributions from both forces and their respective angles:

1. Identify Forces and Angles:
- Force 1: 122 N at an angle of [tex]\( 43.6^\circ \)[/tex]
- Force 2: 97.6 N at an angle of [tex]\( 49.9^\circ \)[/tex]

2. Calculate the [tex]\( y \)[/tex]-component of each force:
- For each force, the [tex]\( y \)[/tex]-component can be found using the sine function, because the [tex]\( y \)[/tex]-component is opposite the angle with the horizontal axis.

- For Force 1:
[tex]\[
F_{y1} = 122 \times \sin(43.6^\circ)
\][/tex]
Calculation yields approximately:
[tex]\[
F_{y1} = 84.13 \, \text{N}
\][/tex]

- For Force 2:
[tex]\[
F_{y2} = 97.6 \times \sin(49.9^\circ)
\][/tex]
Calculation yields approximately:
[tex]\[
F_{y2} = 74.66 \, \text{N}
\][/tex]

3. Add the [tex]\( y \)[/tex]-components:
- The total [tex]\( y \)[/tex]-component of the force is the sum of the individual [tex]\( y \)[/tex]-components:
[tex]\[
F_{y\_total} = F_{y1} + F_{y2}
\][/tex]
Adding the values gives:
[tex]\[
F_{y\_total} = 84.13 + 74.66 = 158.79 \, \text{N}
\][/tex]

Therefore, the [tex]\( y \)[/tex]-component of the total force acting on the chair is approximately [tex]\( 158.79 \, \text{N} \)[/tex].

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