B 3000 lb 60° 45°
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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![### Problem Statement:
**Q.2) Determine the components of the 3000 lb force acting along the members BA and BC.**
### Diagram Explanation:
The diagram showcases a structural frame with two members: BA and BC, which are pinned together at point B, forming an angled structure resting on supports at points A and C. There is a horizontal force of 3000 lb applied at point B. The angles between the members and the horizontal axis at both supports are marked as follows:
- Member BA makes an angle of 60° with the horizontal axis at point A.
- Member BC makes an angle of 45° with the horizontal axis at point C.
### Step-by-Step Solution:
To determine the components of the 3000 lb force along members BA and BC, the force can be resolved into components along the directions of these members:
1. **Free-Body Diagram:**
- Represent the 3000 lb force at point B.
- Resolve this force into two components: one along the direction of BA and the other along the direction of BC.
2. **Force Components along BA and BC:**
- Let the components of the force along BA and BC be \( F_{BA} \) and \( F_{BC} \) respectively.
- Use the Law of Sines or trigonometric relationships based on the given angles to determine these components.
### Calculations:
Using trigonometry, the components can be represented as follows:
\[
F_{BA} = F \cdot \cos(\theta_{BA})
\]
\[
F_{BC} = F \cdot \cos(\theta_{BC})
\]
However, due to member directions and angles given:
\[
F_{BA} = 3000 \cdot \cos(60°)
\]
\[
F_{BC} = 3000 \cdot \cos(45°)
\]
Where:
- \( \cos(60°) = 0.5 \)
- \( \cos(45°) = \frac{\sqrt{2}}{2} \approx 0.707 \)
### Substituting the values:
\[
F_{BA} = 3000 \cdot 0.5 = 1500 \, \text{lb}
\]
\[
F_{BC} = 3000 \cdot 0.707 \approx 2121 \, \text{lb](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbb1a8ba3-371b-48d7-95b6-c118f740fa83%2Fd55afed7-3eb8-4f3a-a779-d0c9898f9b08%2F3g2m6u_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement:
**Q.2) Determine the components of the 3000 lb force acting along the members BA and BC.**
### Diagram Explanation:
The diagram showcases a structural frame with two members: BA and BC, which are pinned together at point B, forming an angled structure resting on supports at points A and C. There is a horizontal force of 3000 lb applied at point B. The angles between the members and the horizontal axis at both supports are marked as follows:
- Member BA makes an angle of 60° with the horizontal axis at point A.
- Member BC makes an angle of 45° with the horizontal axis at point C.
### Step-by-Step Solution:
To determine the components of the 3000 lb force along members BA and BC, the force can be resolved into components along the directions of these members:
1. **Free-Body Diagram:**
- Represent the 3000 lb force at point B.
- Resolve this force into two components: one along the direction of BA and the other along the direction of BC.
2. **Force Components along BA and BC:**
- Let the components of the force along BA and BC be \( F_{BA} \) and \( F_{BC} \) respectively.
- Use the Law of Sines or trigonometric relationships based on the given angles to determine these components.
### Calculations:
Using trigonometry, the components can be represented as follows:
\[
F_{BA} = F \cdot \cos(\theta_{BA})
\]
\[
F_{BC} = F \cdot \cos(\theta_{BC})
\]
However, due to member directions and angles given:
\[
F_{BA} = 3000 \cdot \cos(60°)
\]
\[
F_{BC} = 3000 \cdot \cos(45°)
\]
Where:
- \( \cos(60°) = 0.5 \)
- \( \cos(45°) = \frac{\sqrt{2}}{2} \approx 0.707 \)
### Substituting the values:
\[
F_{BA} = 3000 \cdot 0.5 = 1500 \, \text{lb}
\]
\[
F_{BC} = 3000 \cdot 0.707 \approx 2121 \, \text{lb
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