College Physics
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ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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![**Question:** What is the minimum force needed to pull a 50-kg box up a frictionless ramp that makes a 20° angle with the horizontal? Draw the force diagram!
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**Explanation:**
To determine the minimum force needed to pull the box up the frictionless ramp, we use the component of gravitational force acting parallel to the incline. In this scenario, the force required is equal to the gravitational force component due to the box's weight.
The gravitational force (weight) acting on the box is calculated as:
\[ F_{\text{gravity}} = mg \]
where:
- \( m = 50 \, \text{kg} \) (mass of the box)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)
The component of the gravitational force acting parallel to the ramp is:
\[ F_{\text{parallel}} = mg \sin(\theta) \]
where:
- \( \theta = 20^\circ \) (angle of the ramp)
Substituting the values:
\[ F_{\text{parallel}} = 50 \times 9.8 \times \sin(20^\circ) \]
Calculating this gives the minimum force required to overcome the gravitational pull down the ramp.
**Force Diagram Explanation:**
A force diagram (also known as a free-body diagram) for this scenario would include:
1. A vector pointing downwards representing the gravitational force (\( F_{\text{gravity}} \)).
2. A vector perpendicular to the ramp representing the normal force (\( N \)).
3. A vector parallel to the ramp directed upwards, representing the applied force (\( F_{\text{applied}} \)) needed to move the box up the incline.
4. The angle \( \theta = 20^\circ \) between the incline and the horizontal.
In the diagram, the gravitational force vector can be split into two components: one parallel to the incline (\( F_{\text{parallel}} \)) and one perpendicular (\( F_{\text{perpendicular}} = mg \cos(\theta) \)), which is balanced by the normal force.
](https://content.bartleby.com/qna-images/question/ec99e2ce-f146-4dfd-8358-2828b726596b/9113b6a5-2e29-4c02-adcf-e959cb998bcc/l2vpou_processed.jpeg)
Transcribed Image Text:**Question:** What is the minimum force needed to pull a 50-kg box up a frictionless ramp that makes a 20° angle with the horizontal? Draw the force diagram!
---
**Explanation:**
To determine the minimum force needed to pull the box up the frictionless ramp, we use the component of gravitational force acting parallel to the incline. In this scenario, the force required is equal to the gravitational force component due to the box's weight.
The gravitational force (weight) acting on the box is calculated as:
\[ F_{\text{gravity}} = mg \]
where:
- \( m = 50 \, \text{kg} \) (mass of the box)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)
The component of the gravitational force acting parallel to the ramp is:
\[ F_{\text{parallel}} = mg \sin(\theta) \]
where:
- \( \theta = 20^\circ \) (angle of the ramp)
Substituting the values:
\[ F_{\text{parallel}} = 50 \times 9.8 \times \sin(20^\circ) \]
Calculating this gives the minimum force required to overcome the gravitational pull down the ramp.
**Force Diagram Explanation:**
A force diagram (also known as a free-body diagram) for this scenario would include:
1. A vector pointing downwards representing the gravitational force (\( F_{\text{gravity}} \)).
2. A vector perpendicular to the ramp representing the normal force (\( N \)).
3. A vector parallel to the ramp directed upwards, representing the applied force (\( F_{\text{applied}} \)) needed to move the box up the incline.
4. The angle \( \theta = 20^\circ \) between the incline and the horizontal.
In the diagram, the gravitational force vector can be split into two components: one parallel to the incline (\( F_{\text{parallel}} \)) and one perpendicular (\( F_{\text{perpendicular}} = mg \cos(\theta) \)), which is balanced by the normal force.
Expert Solution
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Step 1
Given data:
A frictionless ramp
Mass (m) = 50 kg
Angle (θ) = 20°, with the horizontal
Required:
The minimum force to pull the box upward
Step by stepSolved in 2 steps with 1 images
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