Sure, let's break down the problem step-by-step to find the [tex]\( x \)[/tex]-component of the total force acting on the block.
### Step 1: Understand the Given Forces and Angles
- The first force [tex]\( F_1 \)[/tex] is [tex]\( 225 \, \text{N} \)[/tex] at an angle of [tex]\( 90.5^\circ \)[/tex].
- The second force [tex]\( F_2 \)[/tex] is [tex]\( 36.0 \, \text{N} \)[/tex] at an angle of [tex]\( 286^\circ \)[/tex].
### Step 2: Conversion of Angles to Radians
We need to convert the angles from degrees to radians because the trigonometric functions (cosine in this case) used in the calculations typically require radians in most mathematical contexts.
### Step 3: Calculate the [tex]\( x \)[/tex]-components of Each Force
We use [tex]\( \cos(\theta) \)[/tex] to find the [tex]\( x \)[/tex]-component of each force, where [tex]\( \theta \)[/tex] is the angle of the force.
[tex]\[
\begin{array}{l}
\text{For the first force } F_1: \\
\theta_1 = 90.5^\circ \\
F_{1 x} = F_1 \cos(\theta_1) \\
F_{1 x} = 225 \cos(90.5^\circ) \\
F_{1 x} \approx -1.963 \, \text{N}
\end{array}
\][/tex]
[tex]\[
\begin{array}{l}
\text{For the second force } F_2: \\
\theta_2 = 286^\circ \\
F_{2 x} = F_2 \cos(\theta_2) \\
F_{2 x} = 36.0 \cos(286^\circ) \\
F_{2 x} \approx 9.923 \, \text{N}
\end{array}
\][/tex]
### Step 4: Sum the [tex]\( x \)[/tex]-components to Find the Total [tex]\( x \)[/tex]-component of the Force
[tex]\[
\overrightarrow{F_x} = F_{1x} + F_{2x}
\][/tex]
Substituting the values we calculated:
[tex]\[
\overrightarrow{F_x} \approx -1.963 + 9.923 = 7.959 \, \text{N}
\][/tex]
Thus, the [tex]\( x \)[/tex]-component of the total force acting on the block is approximately [tex]\( 7.959 \, \text{N} \)[/tex].
### Final Answer
[tex]\[
\overrightarrow{F_x} = 7.959 \, \text{N}
\][/tex]
