Question

EXAM Part I is a full solution dynamics problem. It should be done in a GRASS format. Please pay attention to units, labeling, FBDs, and notation. Submit your work as ONE PDF file.

Problem: A 75 kg man stands in an elevator. What will be the force the elevator exerts on him when:
a) the elevator is at rest,
b) the elevator is moving upward with a uniform acceleration of 2.0 m/s²,
c) the elevator is moving downward with a uniform acceleration of 2.0 m/s².

Draw a FBD for each case.

Answer 1

can you create the Free Body Diagrams?

Answer 2

To solve the dynamics problem where a 75 kg man stands in an elevator under different conditions, we will follow the GRASS format: Given, Required, Analysis, Solution, and Summary.

### Given:
- Mass of the man, \( m = 75 \, \text{kg} \)
- Acceleration due to gravity, \( g = 9.81 \, \text{m/s}^2 \)
- Upward acceleration of the elevator, \( a = 2.0 \, \text{m/s}^2 \)
- Downward acceleration of the elevator, \( a = 2.0 \, \text{m/s}^2 \)

### Required:
We need to find the force that the elevator exerts on the man (\( F_{elevator} \)) under three different conditions:
1. When the elevator is at rest
2. When the elevator is moving upward with an acceleration of \( 2.0 \, \text{m/s}^2 \)
3. When the elevator is moving downward with an acceleration of \( 2.0 \, \text{m/s}^2 \)

### Analysis:
In all cases, we will analyze the forces acting on the man using Free-Body Diagrams (FBDs).

- **For the FBD**:
- The gravitational force acting downward, \( F_{gravity} = m \cdot g \)
- The normal force exerted by the elevator, \( F_{elevator} \), acting upward.

**Free-Body Diagram (FBD)** for each case will include the following:
- \( F_{gravity} \) acting downward
- \( F_{elevator} \) acting upward

### Solution:

**a) Elevator at rest**

1. **FBD**:
- \( F_{gravity} = m \cdot g = 75 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 735.75 \, \text{N} \)
- Since the elevator is at rest, the forces are balanced:
\[
F_{elevator} = F_{gravity}
\]
\[
F_{elevator} = 735.75 \, \text{N}
\]

**b) Elevator moving upward with acceleration \( a = 2.0 \, \text{m/s}^2 \)**

1. **FBD**:
- The net force \( F_{net} \) acting upwards:
\[
F_{net} = m \cdot a = 75 \, \text{kg} \cdot 2.0 \, \text{m/s}^2 = 150 \, \text{N}
\]
- The upward force (\( F_{elevator} \)) must overcome gravity and provide this net force:
\[
F_{elevator} - F_{gravity} = F_{net}
\]
\[
F_{elevator} - 735.75 \, \text{N} = 150 \, \text{N}
\]
\[
F_{elevator} = 735.75 \, \text{N} + 150 \, \text{N} = 885.75 \, \text{N}
\]

**c) Elevator moving downward with acceleration \( a = 2.0 \, \text{m/s}^2 \)**

1. **FBD**:
- The net force \( F_{net} \) acting downward:
\[
F_{net} = m \cdot a = 75 \, \text{kg} \cdot 2.0 \, \text{m/s}^2 = 150 \, \text{N}
\]
- The upward force (\( F_{elevator} \)) opposes this downward acceleration:
\[
F_{gravity} - F_{elevator} = F_{net}
\]
\[
735.75 \, \text{N} - F_{elevator} = 150 \, \text{N}
\]
\[
F_{elevator} = 735.75 \, \text{N} - 150 \, \text{N} = 585.75 \, \text{N}
\]

### Summary:
1. **At rest**: \( F_{elevator} = 735.75 \, \text{N} \)
2. **Moving upward with \( 2.0 \, \text{m/s}^2 \)**: \( F_{elevator} = 885.75 \, \text{N} \)
3. **Moving downward with \( 2.0 \, \text{m/s}^2 \)**: \( F_{elevator} = 585.75 \, \text{N} \)

### Free-Body Diagrams:
For the PDF submission, please include neatly drawn Free Body Diagrams for each case, clearly indicating forces with labels and directions.