1.75 Fg 1S00 FB = 857.14 |lbs Fr =Fe Pase- ISoo lbs Fo %3D Fr=642.86 lbs (4) F. 3 F. 유 Fs + Fg-IS00-O
1.75 Fg 1S00 FB = 857.14 |lbs Fr =Fe Pase- ISoo lbs Fo %3D Fr=642.86 lbs (4) F. 3 F. 유 Fs + Fg-IS00-O
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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Hi can someone tell me how he gets 1.75FB=1500 from 3/4FB+FB-1500=0
![**Static Equilibrium Analysis of a Two-Wheeled System**
This image illustrates a static equilibrium problem solved on a whiteboard, common in physics or engineering mechanics courses. Below is the step-by-step transcription of the solution process:
1. **Equilibrium of Forces and Moments**:
- **Given:** Total weight \(1500 \, \text{lbs}\).
- **Assume:** The system is in equilibrium, thus:
\[
1.75 F_B = 1500
\]
2. **Calculating \( F_B \)**:
\[
F_B = \frac{1500}{1.75} = 857.14 \, \text{lbs}
\]
3. **Calculating \( F_F \)**:
\[
F_F = \frac{3}{4} F_B = \frac{3}{4} \times 857.14 = 642.86 \, \text{lbs}
\]
4. **Diagram Explanation**:
- A simple car sketch is provided with forces depicted at each wheel.
- Forces \( F_F \) and \( F_B \) are acting upwards at the front and back wheels, respectively.
- The weight of \(1500 \, \text{lbs}\) acts downwards through the center of gravity denoted by \( G \).
- Distance from \( G \) to the front wheel is marked as \(4a\), and to the rear wheel as \(3a\).
5. **Equilibrium Equations**:
- **Sum of Forces in X-direction:**
\[
\Sigma F_x = 0
\]
- **Sum of Forces in Y-direction:**
\[
\Sigma F_y = F_F + F_B - 1500 \, \text{lbs} = 0
\]
- **Sum of Moments about G:**
\[
\Sigma M_G = -(4a)(F_F) + (3a)(F_B) = 0
\]
6. **Final Calculations**:
- The solutions satisfy:
\[
(4) F_F = (3) F_B
\]
- Reiterate \( F_F \) in terms of \( F_B \):
\[
F_F = \frac{3}{4} F](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F63d16113-ff39-4f92-89f8-61ce82bfb495%2F5f0209d0-6ab3-469c-af7e-ffe217c8b775%2Fdd13ar7_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Static Equilibrium Analysis of a Two-Wheeled System**
This image illustrates a static equilibrium problem solved on a whiteboard, common in physics or engineering mechanics courses. Below is the step-by-step transcription of the solution process:
1. **Equilibrium of Forces and Moments**:
- **Given:** Total weight \(1500 \, \text{lbs}\).
- **Assume:** The system is in equilibrium, thus:
\[
1.75 F_B = 1500
\]
2. **Calculating \( F_B \)**:
\[
F_B = \frac{1500}{1.75} = 857.14 \, \text{lbs}
\]
3. **Calculating \( F_F \)**:
\[
F_F = \frac{3}{4} F_B = \frac{3}{4} \times 857.14 = 642.86 \, \text{lbs}
\]
4. **Diagram Explanation**:
- A simple car sketch is provided with forces depicted at each wheel.
- Forces \( F_F \) and \( F_B \) are acting upwards at the front and back wheels, respectively.
- The weight of \(1500 \, \text{lbs}\) acts downwards through the center of gravity denoted by \( G \).
- Distance from \( G \) to the front wheel is marked as \(4a\), and to the rear wheel as \(3a\).
5. **Equilibrium Equations**:
- **Sum of Forces in X-direction:**
\[
\Sigma F_x = 0
\]
- **Sum of Forces in Y-direction:**
\[
\Sigma F_y = F_F + F_B - 1500 \, \text{lbs} = 0
\]
- **Sum of Moments about G:**
\[
\Sigma M_G = -(4a)(F_F) + (3a)(F_B) = 0
\]
6. **Final Calculations**:
- The solutions satisfy:
\[
(4) F_F = (3) F_B
\]
- Reiterate \( F_F \) in terms of \( F_B \):
\[
F_F = \frac{3}{4} F
Expert Solution
Step 1
given
equation are = 3/4 FB +FB - 1500 =0
TO FIND =THE VALUE OF FB
Step by step
Solved in 2 steps with 1 images
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