Welcome to Westonci.ca, the Q&A platform where your questions are met with detailed answers from experienced experts. Ask your questions and receive detailed answers from professionals with extensive experience in various fields. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
Alright students, let's solve the problem step by step to determine the magnitude of the tension force when the box is pulled up a ramp.
Given:
- The net force in the [tex]\( x \)[/tex] direction is [tex]\( 98 \, \text{N} \)[/tex].
- The angle of the ramp with the horizontal surface is [tex]\( 22^\circ \)[/tex].
We're looking for the magnitude of the force of tension, denoted as [tex]\( F_{\text{tension}} \)[/tex].
1. Understanding the Components:
- The net force in the [tex]\( x \)[/tex] direction ([tex]\( F_{\text{net}, x} \)[/tex]) is given.
- The force of tension [tex]\( F_{\text{tension}} \)[/tex] has components in both the [tex]\( x \)[/tex] and [tex]\( y \)[/tex] directions.
- The component of the tension force in the [tex]\( x \)[/tex] direction is [tex]\( F_{\text{tension}} \cos(\theta) \)[/tex], where [tex]\( \theta \)[/tex] is [tex]\( 22^\circ \)[/tex].
2. Setting up the Equation:
- From the net force in the [tex]\( x \)[/tex] direction, we get:
[tex]\[ F_{\text{net}, x} = F_{\text{tension}} \cos(22^\circ) \][/tex]
3. Solving for [tex]\( F_{\text{tension}} \)[/tex]:
- Rearrange the equation to solve for [tex]\( F_{\text{tension}} \)[/tex]:
[tex]\[ F_{\text{tension}} = \frac{F_{\text{net}, x}}{\cos(22^\circ)} \][/tex]
- Given [tex]\( F_{\text{net}, x} = 98 \, \text{N} \)[/tex]:
[tex]\[ F_{\text{tension}} = \frac{98 \, \text{N}}{\cos(22^\circ)} \][/tex]
4. Calculation:
- We can use a calculator to find [tex]\( \cos(22^\circ) \)[/tex]:
[tex]\[ \cos(22^\circ) \approx 0.9272 \][/tex]
- Substituting this value:
[tex]\[ F_{\text{tension}} = \frac{98 \, \text{N}}{0.9272} \approx 105.696 \, \text{N} \][/tex]
- Round this result to the nearest integer:
[tex]\[ F_{\text{tension}} \approx 106 \, \text{N} \][/tex]
5. Conclusion:
- The magnitude of the force of tension needed to have a net force of [tex]\( 98 \, \text{N} \)[/tex] in the [tex]\( x \)[/tex]-direction on a ramp inclined at [tex]\( 22^\circ \)[/tex] is approximately [tex]\( 106 \, \text{N} \)[/tex].
Therefore, the correct answer among the given choices is not explicitly listed. However, considering the rounded value of [tex]\( 106 \, \text{N} \)[/tex], this value closely approximates the scenario given.
Given:
- The net force in the [tex]\( x \)[/tex] direction is [tex]\( 98 \, \text{N} \)[/tex].
- The angle of the ramp with the horizontal surface is [tex]\( 22^\circ \)[/tex].
We're looking for the magnitude of the force of tension, denoted as [tex]\( F_{\text{tension}} \)[/tex].
1. Understanding the Components:
- The net force in the [tex]\( x \)[/tex] direction ([tex]\( F_{\text{net}, x} \)[/tex]) is given.
- The force of tension [tex]\( F_{\text{tension}} \)[/tex] has components in both the [tex]\( x \)[/tex] and [tex]\( y \)[/tex] directions.
- The component of the tension force in the [tex]\( x \)[/tex] direction is [tex]\( F_{\text{tension}} \cos(\theta) \)[/tex], where [tex]\( \theta \)[/tex] is [tex]\( 22^\circ \)[/tex].
2. Setting up the Equation:
- From the net force in the [tex]\( x \)[/tex] direction, we get:
[tex]\[ F_{\text{net}, x} = F_{\text{tension}} \cos(22^\circ) \][/tex]
3. Solving for [tex]\( F_{\text{tension}} \)[/tex]:
- Rearrange the equation to solve for [tex]\( F_{\text{tension}} \)[/tex]:
[tex]\[ F_{\text{tension}} = \frac{F_{\text{net}, x}}{\cos(22^\circ)} \][/tex]
- Given [tex]\( F_{\text{net}, x} = 98 \, \text{N} \)[/tex]:
[tex]\[ F_{\text{tension}} = \frac{98 \, \text{N}}{\cos(22^\circ)} \][/tex]
4. Calculation:
- We can use a calculator to find [tex]\( \cos(22^\circ) \)[/tex]:
[tex]\[ \cos(22^\circ) \approx 0.9272 \][/tex]
- Substituting this value:
[tex]\[ F_{\text{tension}} = \frac{98 \, \text{N}}{0.9272} \approx 105.696 \, \text{N} \][/tex]
- Round this result to the nearest integer:
[tex]\[ F_{\text{tension}} \approx 106 \, \text{N} \][/tex]
5. Conclusion:
- The magnitude of the force of tension needed to have a net force of [tex]\( 98 \, \text{N} \)[/tex] in the [tex]\( x \)[/tex]-direction on a ramp inclined at [tex]\( 22^\circ \)[/tex] is approximately [tex]\( 106 \, \text{N} \)[/tex].
Therefore, the correct answer among the given choices is not explicitly listed. However, considering the rounded value of [tex]\( 106 \, \text{N} \)[/tex], this value closely approximates the scenario given.
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.
