Answer :
To find the [tex]\( x \)[/tex]-component of the total force acting on the block, we need to break down each individual force into its [tex]\( x \)[/tex]-components and then add them together. Here are the steps:
1. Identify the forces and their angles:
- First force, [tex]\( F1 = 500 \)[/tex] N at an angle of [tex]\( 65.0^\circ \)[/tex] from the positive [tex]\( x \)[/tex]-axis.
- Second force, [tex]\( F2 = 415 \)[/tex] N at an angle of [tex]\( 270.0^\circ \)[/tex] from the positive [tex]\( x \)[/tex]-axis.
2. Calculate the [tex]\( x \)[/tex]-component for each force:
Using the formula for the [tex]\( x \)[/tex]-component of a force, [tex]\( F_x = F \cdot \cos(\theta) \)[/tex]:
- For the first force:
[tex]\[ F_{x1} = 500 \cdot \cos(65.0^\circ) \][/tex]
The calculation gives:
[tex]\[ F_{x1} \approx 211.31 \text{ N} \][/tex]
- For the second force:
[tex]\[ F_{x2} = 415 \cdot \cos(270.0^\circ) \][/tex]
The calculation gives:
[tex]\[ F_{x2} \approx -7.62 \times 10^{-14} \text{ N} \][/tex]
3. Sum the [tex]\( x \)[/tex]-components to find the total [tex]\( x \)[/tex]-component:
[tex]\[ F_{x\text{total}} = F_{x1} + F_{x2} \][/tex]
Substituting the calculated components:
[tex]\[ F_{x\text{total}} = 211.31 \text{ N} + (-7.62 \times 10^{-14} \text{ N}) \][/tex]
[tex]\[ F_{x\text{total}} \approx 211.31 \text{ N} \][/tex]
So, the [tex]\( x \)[/tex]-component of the total force acting on the block is approximately [tex]\( 211.31 \)[/tex] N.
1. Identify the forces and their angles:
- First force, [tex]\( F1 = 500 \)[/tex] N at an angle of [tex]\( 65.0^\circ \)[/tex] from the positive [tex]\( x \)[/tex]-axis.
- Second force, [tex]\( F2 = 415 \)[/tex] N at an angle of [tex]\( 270.0^\circ \)[/tex] from the positive [tex]\( x \)[/tex]-axis.
2. Calculate the [tex]\( x \)[/tex]-component for each force:
Using the formula for the [tex]\( x \)[/tex]-component of a force, [tex]\( F_x = F \cdot \cos(\theta) \)[/tex]:
- For the first force:
[tex]\[ F_{x1} = 500 \cdot \cos(65.0^\circ) \][/tex]
The calculation gives:
[tex]\[ F_{x1} \approx 211.31 \text{ N} \][/tex]
- For the second force:
[tex]\[ F_{x2} = 415 \cdot \cos(270.0^\circ) \][/tex]
The calculation gives:
[tex]\[ F_{x2} \approx -7.62 \times 10^{-14} \text{ N} \][/tex]
3. Sum the [tex]\( x \)[/tex]-components to find the total [tex]\( x \)[/tex]-component:
[tex]\[ F_{x\text{total}} = F_{x1} + F_{x2} \][/tex]
Substituting the calculated components:
[tex]\[ F_{x\text{total}} = 211.31 \text{ N} + (-7.62 \times 10^{-14} \text{ N}) \][/tex]
[tex]\[ F_{x\text{total}} \approx 211.31 \text{ N} \][/tex]
So, the [tex]\( x \)[/tex]-component of the total force acting on the block is approximately [tex]\( 211.31 \)[/tex] N.
