Answer :
To determine the [tex]\( x \)[/tex]-component of the total force acting on the block, we need to break down each force into its [tex]\( x \)[/tex]-components and then sum these components.
1. Identify the given forces and their angles:
- Force [tex]\( F_1 = 115 \, \text{N} \)[/tex] at an angle [tex]\( \theta_1 = 75.0^\circ \)[/tex]
- Force [tex]\( F_2 = 213 \, \text{N} \)[/tex] at an angle [tex]\( \theta_2 = 295^\circ \)[/tex]
2. Convert the angles from degrees to radians:
For calculations in trigonometry:
- [tex]\( \theta_1 \)[/tex] in radians is [tex]\( \theta_1 = 75.0 \times \frac{\pi}{180} \)[/tex]
- [tex]\( \theta_2 \)[/tex] in radians is [tex]\( \theta_2 = 295 \times \frac{\pi}{180} \)[/tex]
3. Calculate the [tex]\( x \)[/tex]-component for each force using the cosine function:
- For [tex]\( F_1 \)[/tex]:
[tex]\[ F_{1x} = F_1 \cos(\theta_1) \][/tex]
Plugging in the values:
[tex]\[ F_{1x} = 115 \cos(75.0^\circ) \approx 29.76 \, \text{N} \][/tex]
- For [tex]\( F_2 \)[/tex]:
[tex]\[ F_{2x} = F_2 \cos(\theta_2) \][/tex]
Plugging in the values:
[tex]\[ F_{2x} = 213 \cos(295^\circ) \approx 90.02 \, \text{N} \][/tex]
4. Sum the [tex]\( x \)[/tex]-components of both forces to get the total [tex]\( x \)[/tex]-component:
[tex]\[ F_{x_{\text{total}}} = F_{1x} + F_{2x} \][/tex]
Plugging in the calculated components:
[tex]\[ F_{x_{\text{total}}} = 29.76 \, \text{N} + 90.02 \, \text{N} \approx 119.78 \, \text{N} \][/tex]
Therefore, the [tex]\( x \)[/tex]-component of the total force acting on the block is approximately [tex]\( 119.78 \, \text{N} \)[/tex].
1. Identify the given forces and their angles:
- Force [tex]\( F_1 = 115 \, \text{N} \)[/tex] at an angle [tex]\( \theta_1 = 75.0^\circ \)[/tex]
- Force [tex]\( F_2 = 213 \, \text{N} \)[/tex] at an angle [tex]\( \theta_2 = 295^\circ \)[/tex]
2. Convert the angles from degrees to radians:
For calculations in trigonometry:
- [tex]\( \theta_1 \)[/tex] in radians is [tex]\( \theta_1 = 75.0 \times \frac{\pi}{180} \)[/tex]
- [tex]\( \theta_2 \)[/tex] in radians is [tex]\( \theta_2 = 295 \times \frac{\pi}{180} \)[/tex]
3. Calculate the [tex]\( x \)[/tex]-component for each force using the cosine function:
- For [tex]\( F_1 \)[/tex]:
[tex]\[ F_{1x} = F_1 \cos(\theta_1) \][/tex]
Plugging in the values:
[tex]\[ F_{1x} = 115 \cos(75.0^\circ) \approx 29.76 \, \text{N} \][/tex]
- For [tex]\( F_2 \)[/tex]:
[tex]\[ F_{2x} = F_2 \cos(\theta_2) \][/tex]
Plugging in the values:
[tex]\[ F_{2x} = 213 \cos(295^\circ) \approx 90.02 \, \text{N} \][/tex]
4. Sum the [tex]\( x \)[/tex]-components of both forces to get the total [tex]\( x \)[/tex]-component:
[tex]\[ F_{x_{\text{total}}} = F_{1x} + F_{2x} \][/tex]
Plugging in the calculated components:
[tex]\[ F_{x_{\text{total}}} = 29.76 \, \text{N} + 90.02 \, \text{N} \approx 119.78 \, \text{N} \][/tex]
Therefore, the [tex]\( x \)[/tex]-component of the total force acting on the block is approximately [tex]\( 119.78 \, \text{N} \)[/tex].
