Three masses (m1 = 3 kg, m2 = 9 kg and m3 = 6) hang from three identical springs in a motionless elevator. The springs all have the same spring constant k=333 N/m
Three masses (m1 = 3 kg, m2 = 9 kg and m3 = 6) hang from three identical springs in a motionless elevator. The springs all have the same spring constant k=333 N/m
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Three masses (m1 = 3 kg, m2 = 9 kg and m3 = 6) hang from three identical springs in a motionless elevator. The springs all have the same spring constant k=333 N/m
![The image displays an elevator and includes a visual representation of a physics problem involving three masses suspended by springs. Here's the transcription suitable for an educational website:
---
**Physics Concept: Spring-Mass System in an Elevator**
In the depicted scenario, we see a vertical spring-mass system inside an elevator. The system comprises three masses, each labeled \( m_1 \), \( m_2 \), and \( m_3 \), suspended vertically by springs. The springs connecting these masses experience tension based on the mass they support and the forces acting upon them.
**Key Features:**
1. **Masses:**
- **\( m_1 \)**: The topmost mass suspended from the ceiling.
- **\( m_2 \)**: The middle mass connected to \( m_1 \) via a spring.
- **\( m_3 \)**: The bottom-most mass connected to \( m_2 \) via another spring.
2. **Springs:**
- The springs are idealized to exhibit Hookean behavior, i.e., the force exerted by the spring is proportional to the displacement from its rest position (Hooke's Law: \( F = -kx \)).
**Explanation:**
- **Elevator Dynamics:** If the elevator is accelerating upwards or downwards, the tension in the springs will increase or decrease respectively, affecting the equilibrium position of each mass. The apparent weight of the masses changes based on the acceleration of the elevator.
- **Equilibria Calculations:** Under equilibrium (no net movement of the masses), the forces on each mass will balance. For a mass \( m \) in an elevator accelerating upwards with acceleration \( a \):
\[ F_{net} = kx = m(g + a) \]
Where \( g \) is the acceleration due to gravity, \( k \) is the spring constant, and \( x \) is the displacement of the spring.
**Applications:**
This setup can help in understanding fundamental principles in mechanics such as Newton's laws of motion, spring constants, and forces in non-inertial reference frames (accelerating systems). It also serves to illustrate real-world applications such as how elevators are designed to account for variations in tension and how safety mechanisms can be employed.
**Interactive Learning:**
- **Simulation:** Use a physics simulation app to model varying accelerations of the elevator and visualize the effects on the spring tensions and mass displacements](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fec5fb306-23dc-410e-b11b-a666d9f99392%2F676e5ce2-0d1d-494f-86fb-a7bffdec4ba2%2F21hba9e_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The image displays an elevator and includes a visual representation of a physics problem involving three masses suspended by springs. Here's the transcription suitable for an educational website:
---
**Physics Concept: Spring-Mass System in an Elevator**
In the depicted scenario, we see a vertical spring-mass system inside an elevator. The system comprises three masses, each labeled \( m_1 \), \( m_2 \), and \( m_3 \), suspended vertically by springs. The springs connecting these masses experience tension based on the mass they support and the forces acting upon them.
**Key Features:**
1. **Masses:**
- **\( m_1 \)**: The topmost mass suspended from the ceiling.
- **\( m_2 \)**: The middle mass connected to \( m_1 \) via a spring.
- **\( m_3 \)**: The bottom-most mass connected to \( m_2 \) via another spring.
2. **Springs:**
- The springs are idealized to exhibit Hookean behavior, i.e., the force exerted by the spring is proportional to the displacement from its rest position (Hooke's Law: \( F = -kx \)).
**Explanation:**
- **Elevator Dynamics:** If the elevator is accelerating upwards or downwards, the tension in the springs will increase or decrease respectively, affecting the equilibrium position of each mass. The apparent weight of the masses changes based on the acceleration of the elevator.
- **Equilibria Calculations:** Under equilibrium (no net movement of the masses), the forces on each mass will balance. For a mass \( m \) in an elevator accelerating upwards with acceleration \( a \):
\[ F_{net} = kx = m(g + a) \]
Where \( g \) is the acceleration due to gravity, \( k \) is the spring constant, and \( x \) is the displacement of the spring.
**Applications:**
This setup can help in understanding fundamental principles in mechanics such as Newton's laws of motion, spring constants, and forces in non-inertial reference frames (accelerating systems). It also serves to illustrate real-world applications such as how elevators are designed to account for variations in tension and how safety mechanisms can be employed.
**Interactive Learning:**
- **Simulation:** Use a physics simulation app to model varying accelerations of the elevator and visualize the effects on the spring tensions and mass displacements
![### Physics Problem Set: Spring Forces in Motion
#### Question 1:
**What is the magnitude of the force the bottom spring exerts on the lower mass?**
\[ \text{Answer: } \boxed{\sf N} \]
*[Submit Button]*
#### Question 2:
(No second question present)
#### Question 3:
**What is the distance the middle spring is stretched from its equilibrium length?**
\[ \text{Answer: } \boxed{\ \ \ \ \ \ \sf cm} \]
*[Submit Button]*
#### Question 4:
**Now the elevator is moving downward with a velocity of \(v = -2.7 \ \text{m/s}\) but accelerating upward at an acceleration of \(a = 5.4 \ \text{m/s}^2\). (Note: an upward acceleration when the elevator is moving down means the elevator is slowing down.)**
**What is the magnitude of the force the upper spring exerts on the upper mass?**
\[ \text{Answer: } \boxed{\sf N} \]
*[Submit Button]*
#### Question 5:
**What is the distance the lower spring is extended from its unstretched length?**
\[ \text{Answer: } \boxed{\ \ \ \ \ \ \sf cm} \]
*[Submit Button]*
---
*Note*: Each question prompts the user to input their answers in the provided box and submit. For the velocity and acceleration problem, remember that the signs indicate the direction of movement and acceleration.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fec5fb306-23dc-410e-b11b-a666d9f99392%2F676e5ce2-0d1d-494f-86fb-a7bffdec4ba2%2Fumsgdt8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Physics Problem Set: Spring Forces in Motion
#### Question 1:
**What is the magnitude of the force the bottom spring exerts on the lower mass?**
\[ \text{Answer: } \boxed{\sf N} \]
*[Submit Button]*
#### Question 2:
(No second question present)
#### Question 3:
**What is the distance the middle spring is stretched from its equilibrium length?**
\[ \text{Answer: } \boxed{\ \ \ \ \ \ \sf cm} \]
*[Submit Button]*
#### Question 4:
**Now the elevator is moving downward with a velocity of \(v = -2.7 \ \text{m/s}\) but accelerating upward at an acceleration of \(a = 5.4 \ \text{m/s}^2\). (Note: an upward acceleration when the elevator is moving down means the elevator is slowing down.)**
**What is the magnitude of the force the upper spring exerts on the upper mass?**
\[ \text{Answer: } \boxed{\sf N} \]
*[Submit Button]*
#### Question 5:
**What is the distance the lower spring is extended from its unstretched length?**
\[ \text{Answer: } \boxed{\ \ \ \ \ \ \sf cm} \]
*[Submit Button]*
---
*Note*: Each question prompts the user to input their answers in the provided box and submit. For the velocity and acceleration problem, remember that the signs indicate the direction of movement and acceleration.
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Step 1: Given that:
VIEWStep 2: Free body diagram at equilibrium when elevator is at rest:
VIEWStep 3: Calculation of magnitude of force exerted by bottom spring on lower mass:
VIEWStep 4: 3) Extension of middle spring:
VIEWStep 5: FBD of elevator when in motion with given criteria:
VIEWStep 6: 4) Magnitude of force exerted by upper spring on upper mass:
VIEWStep 7: 5) Extension of lower spring:
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