B 3000 lb 60° 45°

International Edition---engineering Mechanics: Statics, 4th Edition
4th Edition
ISBN:9781305501607
Author:Andrew Pytel And Jaan Kiusalaas
Publisher:Andrew Pytel And Jaan Kiusalaas
Chapter5: Three-dimensional Equilibrium
Section: Chapter Questions
Problem 5.39P: The bent rod of negligible weight is supported by the ball-and-socket joint at B and the cables...
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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
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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