n the crate and the ground is μls = 0.2. n force between the crate and the hether the box will slip, tip, or remain in our answer with proper work and FBD(s). 0.15
n the crate and the ground is μls = 0.2. n force between the crate and the hether the box will slip, tip, or remain in our answer with proper work and FBD(s). 0.15
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
![### Determining Equilibrium, Slip, and Tip Conditions for a Crate
#### Problem Statement:
The crate has a mass of 500 kg. The coefficient of static friction between the crate and the ground is \( \mu_s = 0.2 \). Determine the friction force between the crate and the ground. Determine whether the box will slip, tip, or remain in equilibrium. Justify your answer with proper work and Free Body Diagrams (FBDs).
#### Diagram Analysis:
The diagram shows a crate with the following dimensions and forces acting upon it:
- The crate is lifted above the ground with its bottom 0.2 meters away from the ground and the top 0.15 meters, implying a contact height difference.
- There's a force of 650 N applied at a 20-degree angle to the horizontal.
- The crate's width is 0.1 meters on each side of a central vertical line, giving a symmetric distribution.
- The weight of the crate (\( W = mg \)) acts downward through the center of gravity (G).
#### Steps to Solve:
1. **Calculate the Weight of the Crate:**
\[
W = mg = 500 \text{ kg} \times 9.81 \text{ m/s}^2 = 4905 \text{ N}
\]
2. **Determine Normal Force (\( N \)):**
Since there is no vertical acceleration:
\[
N = W = 4905 \text{ N}
\]
3. **Calculate Maximum Static Friction Force (\( f_s \)):**
\[
f_s = \mu_s N = 0.2 \times 4905 \text{ N} = 981 \text{ N}
\]
4. **Determine the Horizontal Component of Applied Force (\( F_{\text{horizontal}} \)):**
\[
F_{\text{horizontal}} = 650 \text{ N} \times \cos(20^\circ) \approx 611 \text{ N}
\]
5. **Check for Slip Condition:**
Since \( F_{\text{horizontal}} (611 \text{ N}) \leq f_s (981 \text{ N}) \), the crate does not slip.
6. **Analyze Tipping Potential:**
Moment about the tipping point at the edge of the crate:
- Moment due to applied force](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F86802dbf-81fd-4d37-b301-ca444af93247%2F658becdd-4843-455d-8c0d-dcd39ce33ae7%2Fvbgx94f_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Determining Equilibrium, Slip, and Tip Conditions for a Crate
#### Problem Statement:
The crate has a mass of 500 kg. The coefficient of static friction between the crate and the ground is \( \mu_s = 0.2 \). Determine the friction force between the crate and the ground. Determine whether the box will slip, tip, or remain in equilibrium. Justify your answer with proper work and Free Body Diagrams (FBDs).
#### Diagram Analysis:
The diagram shows a crate with the following dimensions and forces acting upon it:
- The crate is lifted above the ground with its bottom 0.2 meters away from the ground and the top 0.15 meters, implying a contact height difference.
- There's a force of 650 N applied at a 20-degree angle to the horizontal.
- The crate's width is 0.1 meters on each side of a central vertical line, giving a symmetric distribution.
- The weight of the crate (\( W = mg \)) acts downward through the center of gravity (G).
#### Steps to Solve:
1. **Calculate the Weight of the Crate:**
\[
W = mg = 500 \text{ kg} \times 9.81 \text{ m/s}^2 = 4905 \text{ N}
\]
2. **Determine Normal Force (\( N \)):**
Since there is no vertical acceleration:
\[
N = W = 4905 \text{ N}
\]
3. **Calculate Maximum Static Friction Force (\( f_s \)):**
\[
f_s = \mu_s N = 0.2 \times 4905 \text{ N} = 981 \text{ N}
\]
4. **Determine the Horizontal Component of Applied Force (\( F_{\text{horizontal}} \)):**
\[
F_{\text{horizontal}} = 650 \text{ N} \times \cos(20^\circ) \approx 611 \text{ N}
\]
5. **Check for Slip Condition:**
Since \( F_{\text{horizontal}} (611 \text{ N}) \leq f_s (981 \text{ N}) \), the crate does not slip.
6. **Analyze Tipping Potential:**
Moment about the tipping point at the edge of the crate:
- Moment due to applied force
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