To determine the [tex]$x$[/tex]-component of the total force acting on the block when subjected to two forces at given angles, we should break each force into its [tex]$x$[/tex] (horizontal) components and then sum these components.
Let’s denote the forces as follows:
- The first force: [tex]\( F_1 = 78.5 \)[/tex] Newtons at an angle of [tex]\( 83^\circ \)[/tex].
- The second force: [tex]\( F_2 = 96.5 \)[/tex] Newtons at an angle of [tex]\( 273^\circ \)[/tex].
### Steps to solve:
1. Calculate the [tex]$x$[/tex]-component of the first force (Force1):
- Use the cosine function to determine the horizontal component:
[tex]\[
F_{1x} = F_1 \cos(\theta_1)
\][/tex]
- Given [tex]\( F_1 = 78.5 \)[/tex] Newtons and [tex]\( \theta_1 = 83^\circ \)[/tex]:
[tex]\[
F_{1x} = 78.5 \cos(83^\circ)
\][/tex]
Performing this calculation, we find:
[tex]\[
F_{1x} = 9.566743457304078 \text{ Newtons}
\][/tex]
2. Calculate the [tex]$x$[/tex]-component of the second force (Force2):
- Use the cosine function to determine the horizontal component:
[tex]\[
F_{2x} = F_2 \cos(\theta_2)
\][/tex]
- Given [tex]\( F_2 = 96.5 \)[/tex] Newtons and [tex]\( \theta_2 = 273^\circ \)[/tex]:
[tex]\[
F_{2x} = 96.5 \cos(273^\circ)
\][/tex]
Performing this calculation, we find:
[tex]\[
F_{2x} = 5.050419777444091 \text{ Newtons}
\][/tex]
3. Sum the [tex]$x$[/tex]-components of the two forces to get the total [tex]$x$[/tex]-component of the force:
[tex]\[
F_x^{\text{total}} = F_{1x} + F_{2x}
\][/tex]
Substituting the values we calculated:
[tex]\[
F_x^{\text{total}} = 9.566743457304078 + 5.050419777444091
\][/tex]
Adding these together:
[tex]\[
F_x^{\text{total}} = 14.617163234748169 \text{ Newtons}
\][/tex]
Therefore, the [tex]$x$[/tex]-component of the total force acting on the block is:
[tex]\[
\overrightarrow{F_x} = 14.617163234748169 \text{ N}
\][/tex]
