Sure! Let's break it down step-by-step to find the x-component of the total force acting on the block.
1. Understand the Setup:
You have two forces acting at angles on the block:
- Force 1 (F1): 115 N at an angle of 75.0 degrees.
- Force 2 (F2): 213 N at an angle of 295 degrees.
2. Decompose Each Force into Components:
To find the x-component of each force, we use the cosine of the given angle because the x-component involves the adjacent side of the angle in a right triangle.
- Force 1:
- Magnitude: 115 N
- Angle: 75.0 degrees
- x-component (F1x) is calculated as:
[tex]\[
F1x = 115 \times \cos(75^{\circ})
\][/tex]
- This gives us approximately:
[tex]\[
F1x \approx 29.764 \text{ N}
\][/tex]
- Force 2:
- Magnitude: 213 N
- Angle: 295 degrees
- x-component (F2x) is calculated as:
[tex]\[
F2x = 213 \times \cos(295^{\circ})
\][/tex]
- This gives us approximately:
[tex]\[
F2x \approx 90.018 \text{ N}
\][/tex]
3. Calculate the Total x-component:
To find the total x-component of the net force acting on the block, we sum the x-components of both forces:
[tex]\[
F_{x_{total}} = F1x + F2x
\][/tex]
Substituting the values we found:
[tex]\[
F_{x_{total}} \approx 29.764 + 90.018
\][/tex]
This gives us:
[tex]\[
F_{x_{total}} \approx 119.782 \text{ N}
\][/tex]
Hence, the total x-component of the force acting on the block is approximately [tex]\( 119.782 \)[/tex] N.
