Find the moment caused by force F about point A. (Hint: It may be easier to break up F into components first) 60° F = 150 lb 5'-0" 20'-0"
Find the moment caused by force F about point A. (Hint: It may be easier to break up F into components first) 60° F = 150 lb 5'-0" 20'-0"
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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![**Title: Understanding Moments in Structural Mechanics**
### Problem Statement:
Find the moment caused by force \( F \) about point \( A \).
**Hint:** It may be easier to break up \( F \) into components first.
### Diagram Description:
A rigid L-shaped structure is shown, fixed at the base labeled as point \( A \). The vertical section of the structure extends upward 20 feet. The horizontal section extends rightward 5 feet from the top of the vertical section.
A force \( F = 150 \, \text{lb} \) is applied downwards at an angle of 60° to the horizontal at the end of the horizontal section.
### Parameters:
- Vertical section height: 20 feet
- Horizontal section length: 5 feet
- Force \( F \): 150 lb at a 60° angle to the horizontal
### Detailed Steps to Solve:
1. **Resolve the Force Into Components:**
\[
F_x = F \cdot \cos(\theta)
\]
\[
F_y = F \cdot \sin(\theta)
\]
where \( \theta = 60° \)
- \( F_x = 150 \cdot \cos(60°) = 150 \cdot 0.5 = 75 \, \text{lb} \)
- \( F_y = 150 \cdot \sin(60°) = 150 \cdot \frac{\sqrt{3}}{2} \approx 129.9 \, \text{lb} \)
2. **Calculate the Moment about Point \( A \):**
The moment \( M \) caused by a force about a point is given by:
\[
M_A = F \cdot d \cdot \sin(\alpha)
\]
where \( \alpha \) is the angle between the force vector and the moment arm \( d \).
In our scenario, the transverse force component \( F_x \) acting at a distance of 20 feet from \( A \), and the vertical force component \( F_y \) acting at a distance of 5 feet from \( A \).
- Moment due to \( F_x \):
\[
M_{A,x} = F_x \cdot 20 = 75 \, \text{lb} \cdot](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F24227ad7-0cee-4cb3-bf8f-36b2c099999b%2Fc7504db0-d255-4175-a9d7-2d7e64cd9286%2F8vy2hyt4a_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Understanding Moments in Structural Mechanics**
### Problem Statement:
Find the moment caused by force \( F \) about point \( A \).
**Hint:** It may be easier to break up \( F \) into components first.
### Diagram Description:
A rigid L-shaped structure is shown, fixed at the base labeled as point \( A \). The vertical section of the structure extends upward 20 feet. The horizontal section extends rightward 5 feet from the top of the vertical section.
A force \( F = 150 \, \text{lb} \) is applied downwards at an angle of 60° to the horizontal at the end of the horizontal section.
### Parameters:
- Vertical section height: 20 feet
- Horizontal section length: 5 feet
- Force \( F \): 150 lb at a 60° angle to the horizontal
### Detailed Steps to Solve:
1. **Resolve the Force Into Components:**
\[
F_x = F \cdot \cos(\theta)
\]
\[
F_y = F \cdot \sin(\theta)
\]
where \( \theta = 60° \)
- \( F_x = 150 \cdot \cos(60°) = 150 \cdot 0.5 = 75 \, \text{lb} \)
- \( F_y = 150 \cdot \sin(60°) = 150 \cdot \frac{\sqrt{3}}{2} \approx 129.9 \, \text{lb} \)
2. **Calculate the Moment about Point \( A \):**
The moment \( M \) caused by a force about a point is given by:
\[
M_A = F \cdot d \cdot \sin(\alpha)
\]
where \( \alpha \) is the angle between the force vector and the moment arm \( d \).
In our scenario, the transverse force component \( F_x \) acting at a distance of 20 feet from \( A \), and the vertical force component \( F_y \) acting at a distance of 5 feet from \( A \).
- Moment due to \( F_x \):
\[
M_{A,x} = F_x \cdot 20 = 75 \, \text{lb} \cdot
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