To solve this problem, we need to find the magnitude of the force of tension, given that the net force in the [tex]\( x \)[/tex]-direction is [tex]\( 98 \, \text{N} \)[/tex] and the ramp makes an angle of [tex]\( 22^\circ \)[/tex] with the horizontal.
1. Understanding the given data:
- The angle of the ramp, [tex]\(\theta\)[/tex], is [tex]\( 22^\circ \)[/tex].
- The net force in the [tex]\( x \)[/tex]-direction, [tex]\( F_{\text{net},x} \)[/tex], is [tex]\( 98 \, \text{N} \)[/tex].
2. Relate the net force to the tension force:
- The net force in the [tex]\( x \)[/tex]-direction is the component of the tension force (T) along the ramp direction.
- This relationship can be expressed as: [tex]\( F_{\text{net},x} = T \cos(\theta) \)[/tex].
3. Rearrange the formula to solve for the tension force (T):
- [tex]\( T = \frac{F_{\text{net},x}}{\cos(\theta)} \)[/tex].
4. Convert the angle from degrees to radians for accurate computation in trigonometric functions:
- [tex]\( \theta = 22^\circ \)[/tex] is equivalent to [tex]\( 0.383972435 \)[/tex] radians.
5. Calculate the tension force:
- Using the provided data, [tex]\( T \approx 105.69640478240318 \)[/tex].
Hence, the magnitude of the force of tension is approximately [tex]\( 105.696 \, \text{N} \)[/tex], which does not exactly match any of the given multiple-choice options but is closest to [tex]\( 57 \, \text{N} \)[/tex]. Given that this could be a trick or conceptual error in the problem options themselves, ensure to verify if any additional detail or conversion was intended in the problem.
