Answer :
To find the [tex]\( y \)[/tex]-component of the total force acting on the block, we need to consider the [tex]\( y \)[/tex]-components of the individual forces and then sum them up. Here is a step-by-step explanation:
1. Identify the magnitudes and angles of the forces:
- The first force, [tex]\( F_1 \)[/tex], has a magnitude of [tex]\( 115 \)[/tex] N and is applied at an angle of [tex]\( 75.0^\circ \)[/tex] from the positive [tex]\( x \)[/tex]-axis.
- The second force, [tex]\( F_2 \)[/tex], has a magnitude of [tex]\( 213 \)[/tex] N and is applied at an angle of [tex]\( 295^\circ \)[/tex] from the positive [tex]\( x \)[/tex]-axis.
2. Convert the angles to radians:
- The first angle, [tex]\( 75.0^\circ \)[/tex], converts to [tex]\( F1\_angle\_rad = \frac{75.0 \times \pi}{180} \)[/tex] radians.
- The second angle, [tex]\( 295^\circ \)[/tex], converts to [tex]\( F2\_angle\_rad = \frac{295 \times \pi}{180} \)[/tex] radians.
3. Calculate the [tex]\( y \)[/tex]-component of each force using trigonometry:
- For [tex]\( F_1 \)[/tex], the [tex]\( y \)[/tex]-component is given by [tex]\( F1_y = F_1 \times \sin(\text{angle of } F_1) \)[/tex].
[tex]\[ F1_y = 115 \times \sin(75.0^\circ) \][/tex]
- For [tex]\( F_2 \)[/tex], the [tex]\( y \)[/tex]-component is given by [tex]\( F2_y = F_2 \times \sin(\text{angle of } F_2) \)[/tex].
[tex]\[ F2_y = 213 \times \sin(295^\circ) \][/tex]
4. Find the actual numerical values of the [tex]\( y \)[/tex]-components:
- Calculation shows that [tex]\( F1_y \approx 111.0815 \)[/tex] N.
- Calculation shows that [tex]\( F2_y \approx -193.0436 \)[/tex] N.
5. Sum up the [tex]\( y \)[/tex]-components to get the total [tex]\( y \)[/tex]-component of the force:
[tex]\[ Fy_{total} = F1_y + F2_y \][/tex]
[tex]\[ Fy_{total} = 111.0815 + (-193.0436) \][/tex]
[tex]\[ Fy_{total} \approx -81.9621 \text{ N} \][/tex]
Consequently, the [tex]\( y \)[/tex]-component of the total force acting on the block is:
[tex]\[ \overrightarrow{F_y} \approx -81.9621 \text{ N} \][/tex]
1. Identify the magnitudes and angles of the forces:
- The first force, [tex]\( F_1 \)[/tex], has a magnitude of [tex]\( 115 \)[/tex] N and is applied at an angle of [tex]\( 75.0^\circ \)[/tex] from the positive [tex]\( x \)[/tex]-axis.
- The second force, [tex]\( F_2 \)[/tex], has a magnitude of [tex]\( 213 \)[/tex] N and is applied at an angle of [tex]\( 295^\circ \)[/tex] from the positive [tex]\( x \)[/tex]-axis.
2. Convert the angles to radians:
- The first angle, [tex]\( 75.0^\circ \)[/tex], converts to [tex]\( F1\_angle\_rad = \frac{75.0 \times \pi}{180} \)[/tex] radians.
- The second angle, [tex]\( 295^\circ \)[/tex], converts to [tex]\( F2\_angle\_rad = \frac{295 \times \pi}{180} \)[/tex] radians.
3. Calculate the [tex]\( y \)[/tex]-component of each force using trigonometry:
- For [tex]\( F_1 \)[/tex], the [tex]\( y \)[/tex]-component is given by [tex]\( F1_y = F_1 \times \sin(\text{angle of } F_1) \)[/tex].
[tex]\[ F1_y = 115 \times \sin(75.0^\circ) \][/tex]
- For [tex]\( F_2 \)[/tex], the [tex]\( y \)[/tex]-component is given by [tex]\( F2_y = F_2 \times \sin(\text{angle of } F_2) \)[/tex].
[tex]\[ F2_y = 213 \times \sin(295^\circ) \][/tex]
4. Find the actual numerical values of the [tex]\( y \)[/tex]-components:
- Calculation shows that [tex]\( F1_y \approx 111.0815 \)[/tex] N.
- Calculation shows that [tex]\( F2_y \approx -193.0436 \)[/tex] N.
5. Sum up the [tex]\( y \)[/tex]-components to get the total [tex]\( y \)[/tex]-component of the force:
[tex]\[ Fy_{total} = F1_y + F2_y \][/tex]
[tex]\[ Fy_{total} = 111.0815 + (-193.0436) \][/tex]
[tex]\[ Fy_{total} \approx -81.9621 \text{ N} \][/tex]
Consequently, the [tex]\( y \)[/tex]-component of the total force acting on the block is:
[tex]\[ \overrightarrow{F_y} \approx -81.9621 \text{ N} \][/tex]
