3/3 The weight of the bicycle is 29 lb with center of grav- ity at G. Determine the normal forces at A and B when the bicycle is in equilibrium. |B 18.5"- 22.5"-
3/3 The weight of the bicycle is 29 lb with center of grav- ity at G. Determine the normal forces at A and B when the bicycle is in equilibrium. |B 18.5"- 22.5"-
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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1-The FBD indicates Roller A provides___reactions.
2-The FBD indicates Roller B provides___reactions.
![### Problem Statement:
**3/3** The weight of the bicycle is 29 lb with center of gravity at \( G \). Determine the normal forces at \( A \) and \( B \) when the bicycle is in equilibrium.
### Diagram Explanation:
The provided diagram shows a side view of a bicycle with two wheels. The weight of the bicycle is indicated as 29 pounds (lb). The center of gravity of the bicycle is marked at point \( G \), located at the centroid of the frame. Point \( A \) is the contact point of the rear wheel with the ground, and point \( B \) is the contact point of the front wheel with the ground. The distances from point \( G \) to \( A \) and \( B \) are given as 18.5 inches and 22.5 inches, respectively.
The task is to determine the normal forces at the points of contact, \( A \) and \( B \), when the bicycle is in equilibrium.
By analyzing the forces and using principles of equilibrium, one can calculate the normal force distribution across the two wheels. The equilibrium condition assures that the sum of vertical forces and moments around any point is zero.
### Steps to Solve:
1. **Sum of Vertical Forces:**
The sum of vertical forces must equal zero:
\[ N_A + N_B = 29 \, \text{lb} \]
2. **Sum of Moments:**
Taking moments about point \( A \) gives:
\[ N_B \cdot (18.5 \, \text{in} + 22.5 \, \text{in}) = 29 \, \text{lb} \cdot 22.5 \, \text{in} \]
### Equilibrium Equations:
From vertical force equilibrium:
\[
N_A + N_B = 29 \, \text{lb}
\]
From moment equilibrium about point \( A \):
\[
N_B \cdot 41 \, \text{in} = 29 \, \text{lb} \cdot 22.5 \, \text{in}
\]
Solving for \( N_B \):
\[
N_B = \frac{29 \, \text{lb} \cdot 22.5 \, \text{in}}{41 \, \text{in}}
= \frac{652.5](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fad118f04-c8c0-46fe-8652-72d195094665%2F27a65899-4504-4c0d-9e01-483d93ad865c%2Ftj3yuk1q_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement:
**3/3** The weight of the bicycle is 29 lb with center of gravity at \( G \). Determine the normal forces at \( A \) and \( B \) when the bicycle is in equilibrium.
### Diagram Explanation:
The provided diagram shows a side view of a bicycle with two wheels. The weight of the bicycle is indicated as 29 pounds (lb). The center of gravity of the bicycle is marked at point \( G \), located at the centroid of the frame. Point \( A \) is the contact point of the rear wheel with the ground, and point \( B \) is the contact point of the front wheel with the ground. The distances from point \( G \) to \( A \) and \( B \) are given as 18.5 inches and 22.5 inches, respectively.
The task is to determine the normal forces at the points of contact, \( A \) and \( B \), when the bicycle is in equilibrium.
By analyzing the forces and using principles of equilibrium, one can calculate the normal force distribution across the two wheels. The equilibrium condition assures that the sum of vertical forces and moments around any point is zero.
### Steps to Solve:
1. **Sum of Vertical Forces:**
The sum of vertical forces must equal zero:
\[ N_A + N_B = 29 \, \text{lb} \]
2. **Sum of Moments:**
Taking moments about point \( A \) gives:
\[ N_B \cdot (18.5 \, \text{in} + 22.5 \, \text{in}) = 29 \, \text{lb} \cdot 22.5 \, \text{in} \]
### Equilibrium Equations:
From vertical force equilibrium:
\[
N_A + N_B = 29 \, \text{lb}
\]
From moment equilibrium about point \( A \):
\[
N_B \cdot 41 \, \text{in} = 29 \, \text{lb} \cdot 22.5 \, \text{in}
\]
Solving for \( N_B \):
\[
N_B = \frac{29 \, \text{lb} \cdot 22.5 \, \text{in}}{41 \, \text{in}}
= \frac{652.5
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