Answer :
To determine the [tex]$y$[/tex]-component of the total force acting on the block, we need to decompose each force into its components and then sum these components. Here are the steps:
1. Identify the Forces and Angles:
- Force 1: [tex]\( F_1 = 500 \, \text{N} \)[/tex] at an angle [tex]\( \theta_1 = 65.0^\circ \)[/tex]
- Force 2: [tex]\( F_2 = 415 \, \text{N} \)[/tex] at an angle [tex]\( \theta_2 = 270^\circ \)[/tex]
2. Calculate the [tex]$y$[/tex]-Component of Each Force:
- For the first force:
[tex]\[ F_{1y} = F_1 \cdot \sin(\theta_1) \][/tex]
Substituting the values:
[tex]\[ F_{1y} = 500 \, \text{N} \cdot \sin(65.0^\circ) \][/tex]
The [tex]$y$[/tex]-component of the first force [tex]\( F_{1y} \)[/tex] is approximately [tex]\( 453.1539 \, \text{N} \)[/tex].
- For the second force:
[tex]\[ F_{2y} = F_2 \cdot \sin(\theta_2) \][/tex]
Substituting the values:
[tex]\[ F_{2y} = 415 \, \text{N} \cdot \sin(270^\circ) \][/tex]
Since [tex]\( \sin(270^\circ) = -1 \)[/tex]:
[tex]\[ F_{2y} = 415 \, \text{N} \cdot (-1) = -415 \, \text{N} \][/tex]
3. Sum the [tex]$y$[/tex]-Components to Find the Total [tex]$y$[/tex]-Component of the Force:
[tex]\[ F_{total\_y} = F_{1y} + F_{2y} \][/tex]
Substituting the calculated components:
[tex]\[ F_{total\_y} = 453.1539 \, \text{N} + (-415 \, \text{N}) \][/tex]
Simplifying:
[tex]\[ F_{total\_y} \approx 38.1539 \, \text{N} \][/tex]
Thus, the [tex]$y$[/tex]-component of the total force acting on the block is approximately [tex]\( \boxed{38.1539 \, \text{N}} \)[/tex].
1. Identify the Forces and Angles:
- Force 1: [tex]\( F_1 = 500 \, \text{N} \)[/tex] at an angle [tex]\( \theta_1 = 65.0^\circ \)[/tex]
- Force 2: [tex]\( F_2 = 415 \, \text{N} \)[/tex] at an angle [tex]\( \theta_2 = 270^\circ \)[/tex]
2. Calculate the [tex]$y$[/tex]-Component of Each Force:
- For the first force:
[tex]\[ F_{1y} = F_1 \cdot \sin(\theta_1) \][/tex]
Substituting the values:
[tex]\[ F_{1y} = 500 \, \text{N} \cdot \sin(65.0^\circ) \][/tex]
The [tex]$y$[/tex]-component of the first force [tex]\( F_{1y} \)[/tex] is approximately [tex]\( 453.1539 \, \text{N} \)[/tex].
- For the second force:
[tex]\[ F_{2y} = F_2 \cdot \sin(\theta_2) \][/tex]
Substituting the values:
[tex]\[ F_{2y} = 415 \, \text{N} \cdot \sin(270^\circ) \][/tex]
Since [tex]\( \sin(270^\circ) = -1 \)[/tex]:
[tex]\[ F_{2y} = 415 \, \text{N} \cdot (-1) = -415 \, \text{N} \][/tex]
3. Sum the [tex]$y$[/tex]-Components to Find the Total [tex]$y$[/tex]-Component of the Force:
[tex]\[ F_{total\_y} = F_{1y} + F_{2y} \][/tex]
Substituting the calculated components:
[tex]\[ F_{total\_y} = 453.1539 \, \text{N} + (-415 \, \text{N}) \][/tex]
Simplifying:
[tex]\[ F_{total\_y} \approx 38.1539 \, \text{N} \][/tex]
Thus, the [tex]$y$[/tex]-component of the total force acting on the block is approximately [tex]\( \boxed{38.1539 \, \text{N}} \)[/tex].
