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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]$x$[/tex]-component of the total force acting on the chair?

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

Answer :

To find the [tex]\( x \)[/tex]-component of the total force acting on the chair, we need to break down each force into its respective [tex]\( x \)[/tex]-component using trigonometry. We'll then add these components together to find the total [tex]\( x \)[/tex]-component.

### Steps to Solve:

1. Identify the Forces and Angles:
- The first force [tex]\( F_1 \)[/tex] is 122 N at an angle of [tex]\( 43.6^\circ \)[/tex].
- The second force [tex]\( F_2 \)[/tex] is 97.6 N at an angle of [tex]\( 49.9^\circ \)[/tex].

2. Calculate the [tex]\( x \)[/tex]-component of Each Force:
- The [tex]\( x \)[/tex]-component of a force can be found using the cosine function, which relates to the adjacent side of a right triangle.
- For the first force [tex]\( F_1 \)[/tex]:
[tex]\[
F_{1x} = 122 \times \cos(43.6^\circ)
\][/tex]
- For the second force [tex]\( F_2 \)[/tex]:
[tex]\[
F_{2x} = 97.6 \times \cos(49.9^\circ)
\][/tex]

3. Add the [tex]\( x \)[/tex]-components Together:
- To find the total [tex]\( x \)[/tex]-component ([tex]\( F_x \)[/tex]) of the forces acting on the chair, add the [tex]\( x \)[/tex]-components of [tex]\( F_1 \)[/tex] and [tex]\( F_2 \)[/tex]:
[tex]\[
F_x = F_{1x} + F_{2x}
\][/tex]

4. Results:
- The [tex]\( x \)[/tex]-component of the first force is approximately 88.35 N.
- The [tex]\( x \)[/tex]-component of the second force is approximately 62.87 N.
- Therefore, the total [tex]\( x \)[/tex]-component of the forces is approximately:
[tex]\[
F_x \approx 151.22 \, \text{N}
\][/tex]

Thus, the [tex]\( x \)[/tex]-component of the total force acting on the chair is about 151.22 N.

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