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

[tex]\vec{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 effect of each force in the vertical direction. Here's how you can solve it step-by-step:

1. Identify the Given Information:
- 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. Understand the Concept:
- The [tex]\( y \)[/tex]-component of a force can be found using the sine of its angle. This is because the sine function relates the opposite side (vertical component) to the hypotenuse (the force itself) in a right triangle.

3. Calculate the [tex]\( y \)[/tex]-component of the First Force:
- Use the formula: [tex]\( F_{y1} = F_1 \times \sin(\theta_1) \)[/tex].
- Plug in the values: [tex]\( F_{y1} = 122 \times \sin(43.6^\circ) \)[/tex].

4. Calculate the [tex]\( y \)[/tex]-component of the Second Force:
- Use the formula: [tex]\( F_{y2} = F_2 \times \sin(\theta_2) \)[/tex].
- Plug in the values: [tex]\( F_{y2} = 97.6 \times \sin(49.9^\circ) \)[/tex].

5. Find the Total [tex]\( y \)[/tex]-component:
- Add the [tex]\( y \)[/tex]-components of both forces: [tex]\( F_y = F_{y1} + F_{y2} \)[/tex].

After performing these calculations:

- The [tex]\( y \)[/tex]-component for the first force is approximately [tex]\( 84.13 \, \text{N} \)[/tex].
- The [tex]\( y \)[/tex]-component for the second force is approximately [tex]\( 74.66 \, \text{N} \)[/tex].
- Therefore, the total [tex]\( y \)[/tex]-component of the force is [tex]\( 158.79 \, \text{N} \)[/tex].

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

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