Hi can anyone tell me why the summation of Fx=-300sin(30)+150 instead of Fx=-300cos(30)+150?

Elements Of Electromagnetics
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ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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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.
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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