Get the information you need with the help of IDNLearn.com's extensive Q&A platform. Our Q&A platform is designed to provide quick and accurate answers to any questions you may have.
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.
We value your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Accurate answers are just a click away at IDNLearn.com. Thanks for stopping by, and come back for more reliable solutions.