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Determine about point the of cach of the three fores B moment A and

Determine about point the of cach of the three fores B moment A and

Elements Of Electromagnetics
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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### Determining the Moment of Each of the Three Forces About Points A and B #### Problem Statement: Determine the moment of each of the three forces about points A and B. #### Diagram Explanation: The diagram consists of a horizontal and vertical structure with several forces applied to it. The structure forms an L-shape with the following features: - **Point A** is the leftmost end of the horizontal segment. - The horizontal segment extends 2 meters from point A and then another 3 meters before turning vertically downward. - **Point B** is the bottom of the vertical segment, which is 4 meters long. Three forces are applied to the structure: 1. **Force F1 = 250 N** is applied at point A at a 30-degree angle to the horizontal. 2. **Force F2 = 300 N** is applied at the intersection of the horizontal and vertical segments at a 60-degree angle to the horizontal. 3. **Force F3 = 500 N** is applied at point B at a downward-right angle with direction ratios corresponding to 3 (horizontal): 4 (vertical): 5 (hypotenuse). #### Calculation of Moments: To calculate the moments, you need to find the perpendicular distances from each force's line of action to the points A and B. 1. **Moment of F1 about A:** \( \text{Moment} (M) = F \times d \) Since Force F1 passes through point A, its moment about point A is 0. 2. **Moment of F2 about A:** - Horizontal distance of F2 from A is 5 meters (2m + 3m). - Moment Arm = \( 5m \sin(60^\circ) \) for the vertical component of F2. - \( M_{F2-A} = 300 N \times 5 m \times \sin(60^\circ) = 300 \times 5 \times (\sqrt{3}/2) \) 3. **Moment of F3 about A:** - Horizontal distance from A to the vertical segment = 5 meters. - Vertical distance from A to B = 4 meters. - Using the right triangle ratio given, the horizontal component = 3/5 of 500N, and the vertical component = 4/5 of 500N. - \( M_{F3-A
Transcribed Image Text:### Determining the Moment of Each of the Three Forces About Points A and B #### Problem Statement: Determine the moment of each of the three forces about points A and B. #### Diagram Explanation: The diagram consists of a horizontal and vertical structure with several forces applied to it. The structure forms an L-shape with the following features: - **Point A** is the leftmost end of the horizontal segment. - The horizontal segment extends 2 meters from point A and then another 3 meters before turning vertically downward. - **Point B** is the bottom of the vertical segment, which is 4 meters long. Three forces are applied to the structure: 1. **Force F1 = 250 N** is applied at point A at a 30-degree angle to the horizontal. 2. **Force F2 = 300 N** is applied at the intersection of the horizontal and vertical segments at a 60-degree angle to the horizontal. 3. **Force F3 = 500 N** is applied at point B at a downward-right angle with direction ratios corresponding to 3 (horizontal): 4 (vertical): 5 (hypotenuse). #### Calculation of Moments: To calculate the moments, you need to find the perpendicular distances from each force's line of action to the points A and B. 1. **Moment of F1 about A:** \( \text{Moment} (M) = F \times d \) Since Force F1 passes through point A, its moment about point A is 0. 2. **Moment of F2 about A:** - Horizontal distance of F2 from A is 5 meters (2m + 3m). - Moment Arm = \( 5m \sin(60^\circ) \) for the vertical component of F2. - \( M_{F2-A} = 300 N \times 5 m \times \sin(60^\circ) = 300 \times 5 \times (\sqrt{3}/2) \) 3. **Moment of F3 about A:** - Horizontal distance from A to the vertical segment = 5 meters. - Vertical distance from A to B = 4 meters. - Using the right triangle ratio given, the horizontal component = 3/5 of 500N, and the vertical component = 4/5 of 500N. - \( M_{F3-A
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