Elements Of Electromagnetics
7th Edition
ISBN: 9780190698614
Author: Sadiku, Matthew N. O.
Publisher: Oxford University Press
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Hi can anyone tell me why the summation of Fx=-300sin(30)+150 instead of Fx=-300cos(30)+150?
![### Varignon's Theorem and Force Analysis
**Objective:**
Use Varignon's Theorem to find the moment that the forces in the diagram exert about point A.
**Diagram Explanation:**
The diagram shows a horizontal beam with two points, A and B. The distance between points A and B is marked as 5 meters. At point B, there are three forces acting:
- A 300 N force at an angle of 30 degrees from the horizontal.
- A horizontal force of 150 N.
- A vertical downward force of 70 N.
**Equations and Calculations:**
1. **Sum of Forces in the X-Direction (\( \Sigma F_x \)):**
\[
\Sigma F_x = -300 \sin(30^\circ) + 150 = 0
\]
This equation considers the components of the 300 N force and the 150 N force in the horizontal direction.
2. **Sum of Forces in the Y-Direction (\( \Sigma F_y \)):**
\[
\Sigma F_y = -300 \cos(30^\circ) + 70 = -189.8 \, \text{N}
\]
This equation considers the vertical components of the 300 N force and the 70 N force.
**Diagram Sketch:**
- A line represents the beam from point A to point B.
- Forces are shown at point B with respective magnitudes and directions.
- A notation indicates the moment exerted about point A due to these forces.
**Conclusion:**
Using Varignon's Theorem, the contributions of each force's moment around point A can be calculated to find the resultant moment.](https://content.bartleby.com/qna-images/question/63d16113-ff39-4f92-89f8-61ce82bfb495/6dd3a580-31ec-4a6c-a7b7-82dec9d2b54a/1flt68_processed.jpeg)
Transcribed Image Text:### Varignon's Theorem and Force Analysis
**Objective:**
Use Varignon's Theorem to find the moment that the forces in the diagram exert about point A.
**Diagram Explanation:**
The diagram shows a horizontal beam with two points, A and B. The distance between points A and B is marked as 5 meters. At point B, there are three forces acting:
- A 300 N force at an angle of 30 degrees from the horizontal.
- A horizontal force of 150 N.
- A vertical downward force of 70 N.
**Equations and Calculations:**
1. **Sum of Forces in the X-Direction (\( \Sigma F_x \)):**
\[
\Sigma F_x = -300 \sin(30^\circ) + 150 = 0
\]
This equation considers the components of the 300 N force and the 150 N force in the horizontal direction.
2. **Sum of Forces in the Y-Direction (\( \Sigma F_y \)):**
\[
\Sigma F_y = -300 \cos(30^\circ) + 70 = -189.8 \, \text{N}
\]
This equation considers the vertical components of the 300 N force and the 70 N force.
**Diagram Sketch:**
- A line represents the beam from point A to point B.
- Forces are shown at point B with respective magnitudes and directions.
- A notation indicates the moment exerted about point A due to these forces.
**Conclusion:**
Using Varignon's Theorem, the contributions of each force's moment around point A can be calculated to find the resultant moment.
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