To find the [tex]\( y \)[/tex]-component of the total force acting on the block, we need to break down each force into its [tex]\( y \)[/tex]-components and then sum those components.
### Step-by-Step Solution:
1. Identify the magnitudes and directions of the forces:
- The magnitude of the first force, [tex]\( F_1 \)[/tex], is [tex]\( 225 \, N \)[/tex].
- The angle of the first force, [tex]\( \theta_1 \)[/tex], is [tex]\( 90.5^{\circ} \)[/tex].
- The magnitude of the second force, [tex]\( F_2 \)[/tex], is [tex]\( 36.0 \, N \)[/tex].
- The angle of the second force, [tex]\( \theta_2 \)[/tex], is [tex]\( 286^{\circ} \)[/tex].
2. Convert the angles from degrees to radians:
- For [tex]\( \theta_1 = 90.5^{\circ} \)[/tex], the angle in radians is [tex]\( \theta_1 = \frac{90.5\times \pi}{180} \approx 1.579 \, \text{radians} \)[/tex].
- For [tex]\( \theta_2 = 286^{\circ} \)[/tex], the angle in radians is [tex]\( \theta_2 = \frac{286\times \pi}{180} \approx 4.992 \, \text{radians} \)[/tex].
3. Calculate the [tex]\( y \)[/tex]-components of each force using the sine function:
- The [tex]\( y \)[/tex]-component of the first force, [tex]\( F_{1y} \)[/tex], is given by:
[tex]\[
F_{1y} = F_1 \sin(\theta_1) = 225 \times \sin(1.579) \approx 224.991 \, N
\][/tex]
- The [tex]\( y \)[/tex]-component of the second force, [tex]\( F_{2y} \)[/tex], is given by:
[tex]\[
F_{2y} = F_2 \sin(\theta_2)=36.0 \times \sin(4.992) \approx -34.605 \, N
\][/tex]
4. Sum the [tex]\( y \)[/tex]-components of the forces:
- Total [tex]\( y \)[/tex]-component, [tex]\( \overrightarrow{F_y} \)[/tex], is:
[tex]\[
\overrightarrow{F_y} = F_{1y} + F_{2y} = 224.991 + (-34.605) \approx 190.386
\][/tex]
Thus, the [tex]\( y \)[/tex]-component of the total force acting on the block is approximately [tex]\( 190.386 \, N \)[/tex].
