A block is pulled by two horizontal forces. The first force is [tex]225 \, \text{N}[/tex] at an angle of [tex]90.5^\circ[/tex] and the second is [tex]36.0 \, \text{N}[/tex] at an angle of [tex]286^\circ[/tex].

What is the [tex]x[/tex]-component of the total force acting on the block?

[tex]\[
\begin{array}{l}
\overrightarrow{F_x} = [?] \, \text{N} \\
\overrightarrow{F_x} = \overrightarrow{F_{1 x}} + \overrightarrow{F_{2 x}}
\end{array}
\][/tex]



Answer :

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]