An elevator shown below filled with passengers has a mass of 1650 kg. The elevator does motions (a) through (c) in succession. T meg For each of the parts below draw a free body diagram of the elevator in your notebook for each of the parts (a) to (c). Draw the acceleration and velocity vectors in the boxes. For each part, are the vectors for tension in the string and weight of the elevator of equal lengths or unequal lengths.
An elevator shown below filled with passengers has a mass of 1650 kg. The elevator does motions (a) through (c) in succession. T meg For each of the parts below draw a free body diagram of the elevator in your notebook for each of the parts (a) to (c). Draw the acceleration and velocity vectors in the boxes. For each part, are the vectors for tension in the string and weight of the elevator of equal lengths or unequal lengths.
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Chapter1: Units, Trigonometry. And Vectors
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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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Transcribed Image Text:## Problem Statement
1. An elevator shown below filled with passengers has a mass of **1650 kg**. The elevator does motions (a) through (c) in succession.
## Diagram Description
The diagram illustrates an elevator scenario with the following components:
- **Elevator**: A box-like structure containing a person inside, which represents an elevator.
- **Tension (T)**: An arrow pointing upwards above the elevator, indicating the tension force in the string or cable.
- **Weight (m\_g)**: An arrow pointing downwards below the elevator, representing the gravitational force acting on the elevator.
## Instructions
For each of the parts below, draw a free body diagram of the **elevator** in your notebook for each of the parts (a) to (c). Draw the acceleration and velocity vectors in the boxes. For each part, are the vectors for tension in the string and weight of the elevator of equal lengths or unequal lengths?
### Coordinate System
A coordinate system is shown with:
- **y-axis**: Vertical axis
- **x-axis**: Horizontal axis
- **m\_E**: Mass of the elevator, depicted as a grey circle.
- **a**: Box to represent the acceleration vector.
- **v**: Box to represent the velocity vector.
Use these components to analyze and represent the forces acting on the elevator in different motion scenarios.
![### Elevator Motion Analysis Using Newton's Laws
The scenario involves analyzing the motion of an elevator using Newton's Second Law in three phases: acceleration upwards, constant velocity, and deceleration.
#### (a) Upward Acceleration
The elevator accelerates upward from rest at a rate of 0.75 m/s² for 1.45 s.
1. **Newton's Second Law Application:**
The equation in the y-direction is:
\[
\Sigma F_y = T - m_e g = 1 \times m_e a
\]
Here, the acceleration is upwards, so the multiplier is "1".
2. **Tension Calculation:**
The tension in the cable supporting the elevator is:
\[
T = 17407.5 \, \text{N}
\]
3. **Height Calculation:**
The distance the elevator moves is:
\[
\Delta y = 0.79 \, \text{m}
\]
4. **Velocity Calculation:**
The velocity of the elevator after this time is:
\[
v(t = 1.45 \, \text{s}) = 1.09 \, \text{m/s}
\]
#### (b) Constant Upward Velocity
The elevator continues upward at constant velocity for 8.1 s.
1. **Newton's Law Application:**
The equation in the y-direction is:
\[
\Sigma F_y = T - m_e g = 0 \times m_e a
\]
Here, there is no acceleration, so the multiplier is "0".
2. **Tension Calculation:**
The tension is incorrectly calculated as:
\[
T = 17407.5 \, \text{N}
\]
3. **Height Calculation:**
The calculated movement is incorrect:
\[
\Delta y = 24.60 \, \text{m}
\]
#### (c) Deceleration
The elevator decelerates at a rate of 0.65 m/s² for 2.9 s.
1. **Newton's Law Application:**
The equation in the y-direction is:
\[
\Sigma F_y = T - m_e g = -1 \times m_e a
\]
Here, the acceleration is downwards, so](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F464fe16a-00d5-4182-8379-61bd4b8bbe02%2F656b4552-e5e3-472c-baa8-83ad06c8d009%2Fqex5v64_processed.png&w=3840&q=75)
Transcribed Image Text:### Elevator Motion Analysis Using Newton's Laws
The scenario involves analyzing the motion of an elevator using Newton's Second Law in three phases: acceleration upwards, constant velocity, and deceleration.
#### (a) Upward Acceleration
The elevator accelerates upward from rest at a rate of 0.75 m/s² for 1.45 s.
1. **Newton's Second Law Application:**
The equation in the y-direction is:
\[
\Sigma F_y = T - m_e g = 1 \times m_e a
\]
Here, the acceleration is upwards, so the multiplier is "1".
2. **Tension Calculation:**
The tension in the cable supporting the elevator is:
\[
T = 17407.5 \, \text{N}
\]
3. **Height Calculation:**
The distance the elevator moves is:
\[
\Delta y = 0.79 \, \text{m}
\]
4. **Velocity Calculation:**
The velocity of the elevator after this time is:
\[
v(t = 1.45 \, \text{s}) = 1.09 \, \text{m/s}
\]
#### (b) Constant Upward Velocity
The elevator continues upward at constant velocity for 8.1 s.
1. **Newton's Law Application:**
The equation in the y-direction is:
\[
\Sigma F_y = T - m_e g = 0 \times m_e a
\]
Here, there is no acceleration, so the multiplier is "0".
2. **Tension Calculation:**
The tension is incorrectly calculated as:
\[
T = 17407.5 \, \text{N}
\]
3. **Height Calculation:**
The calculated movement is incorrect:
\[
\Delta y = 24.60 \, \text{m}
\]
#### (c) Deceleration
The elevator decelerates at a rate of 0.65 m/s² for 2.9 s.
1. **Newton's Law Application:**
The equation in the y-direction is:
\[
\Sigma F_y = T - m_e g = -1 \times m_e a
\]
Here, the acceleration is downwards, so
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