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z tK 0.6 m 0.3 m 1o sa D ン C F= 700 N 0.4 m -0.2 m- Note: Poin ts C and D lie on the line of action of the Force F. Coordinates: (X, y,z) Point A: (0,0,0) Point B: (0.4, 0,2,0). Point C: meters (0.G, 0,0.3) Point D: (0,0.2, 0)

z tK 0.6 m 0.3 m 1o sa D ン C F= 700 N 0.4 m -0.2 m- Note: Poin ts C and D lie on the line of action of the Force F. Coordinates: (X, y,z) Point A: (0,0,0) Point B: (0.4, 0,2,0). Point C: meters (0.G, 0,0.3) Point D: (0,0.2, 0)

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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Express  result of Part 5 in the form of a Cartesian vector.

# Analysis of a Force System in Three Dimensions ## Introduction In this exercise, we analyze a 700 N force system acting on a 3D structure. The geometrical arrangement of the structure and application points of the force are detailed below. Understanding these conditions and using the given coordinates will aid in determining various forces and moments acting on the structure. ## Diagram Explanation The diagram illustrates a 3D structure with distinct points labeled A, B, C, and D. The force `F` of 700 N acts at point C. ### Key Measurements: - Distance from A to C along the z-axis is 0.6 m. - Distance from A to C along the y-axis is 0.3 m. - Distance from B to D along the y-axis is 0.4 m. - Distance from point B to point C along the x-axis is 0.2 m. ### Note: Points C and D lie on the line of action of the force \( \vec{F} \). ## Coordinate System An orthogonal coordinate system is employed with the following units and designations: - \( x \) (horizontal axis, positive to the left) - \( y \) (horizontal axis, positive to the right) - \( z \) (vertical axis, positive upwards) ## Coordinates The coordinates (in meters) of the points are defined as: - **Point A:** \( (0, 0, 0) \) - **Point B:** \( (0.4, 0.2, 0) \) - **Point C:** \( (0.6, 0, 0.3) \) - **Point D:** \( (0, 0.2, 0) \) Understanding these coordinates will facilitate vector calculations and problem-solving related to the structure's equilibrium, force components, and moments about specific points. For more detailed analysis, each spatial component of the force \( \vec{F} \) and corresponding vectors can be resolved using these coordinates, ensuring accurate results in structural and mechanical studies.
Transcribed Image Text:# Analysis of a Force System in Three Dimensions ## Introduction In this exercise, we analyze a 700 N force system acting on a 3D structure. The geometrical arrangement of the structure and application points of the force are detailed below. Understanding these conditions and using the given coordinates will aid in determining various forces and moments acting on the structure. ## Diagram Explanation The diagram illustrates a 3D structure with distinct points labeled A, B, C, and D. The force `F` of 700 N acts at point C. ### Key Measurements: - Distance from A to C along the z-axis is 0.6 m. - Distance from A to C along the y-axis is 0.3 m. - Distance from B to D along the y-axis is 0.4 m. - Distance from point B to point C along the x-axis is 0.2 m. ### Note: Points C and D lie on the line of action of the force \( \vec{F} \). ## Coordinate System An orthogonal coordinate system is employed with the following units and designations: - \( x \) (horizontal axis, positive to the left) - \( y \) (horizontal axis, positive to the right) - \( z \) (vertical axis, positive upwards) ## Coordinates The coordinates (in meters) of the points are defined as: - **Point A:** \( (0, 0, 0) \) - **Point B:** \( (0.4, 0.2, 0) \) - **Point C:** \( (0.6, 0, 0.3) \) - **Point D:** \( (0, 0.2, 0) \) Understanding these coordinates will facilitate vector calculations and problem-solving related to the structure's equilibrium, force components, and moments about specific points. For more detailed analysis, each spatial component of the force \( \vec{F} \) and corresponding vectors can be resolved using these coordinates, ensuring accurate results in structural and mechanical studies.
**Title: Understanding Unit Vectors and Moments in Mechanics** **Objective:** 1. To determine the unit vector \( \hat{A}_{AB} \) in the direction from point A towards point B. 2. To determine the moment \( M_{AB} \) (as a scalar) which acts about the AB axis (i.e., the line from point A toward B) due to the force \( \mathbf{F} \). **Instructions:** 1. **Unit Vector \( \hat{A}_{AB} \):** - A unit vector in the direction of a vector \( \mathbf{AB} \) gives both the direction and a magnitude of 1. - To find the unit vector \( \hat{A}_{AB} \), follow these steps: - Compute the vector \( \mathbf{AB} \) from the coordinates of points A and B. - Divide this vector by its magnitude (length) to normalize it. 2. **Moment \( M_{AB} \):** - The moment \( M_{AB} \) about the AB axis due to a force \( \mathbf{F} \) is a measure of the tendency of \( \mathbf{F} \) to rotate an object about the axis from point A to point B. - To calculate \( M_{AB} \): - Determine the position vector from point A to any point on the line of action of \( \mathbf{F} \). - Compute the cross product of this position vector with the force vector \( \mathbf{F} \). - Project this resulting moment vector onto the line AB to find the scalar moment \( M_{AB} \). **Note:** - Ensure to clarify the concept of the direction vector and magnitude. - Provide detailed calculations for each step in both the unit vector determination and moment calculation to aid in understanding. **No graphs or diagrams are present in the provided text.** **Conclusion:** Understanding these fundamental concepts in mechanics is crucial for solving related problems in physics and engineering. Accurate determination of unit vectors and moments can significantly aid in analysis and design within various applications.
Transcribed Image Text:**Title: Understanding Unit Vectors and Moments in Mechanics** **Objective:** 1. To determine the unit vector \( \hat{A}_{AB} \) in the direction from point A towards point B. 2. To determine the moment \( M_{AB} \) (as a scalar) which acts about the AB axis (i.e., the line from point A toward B) due to the force \( \mathbf{F} \). **Instructions:** 1. **Unit Vector \( \hat{A}_{AB} \):** - A unit vector in the direction of a vector \( \mathbf{AB} \) gives both the direction and a magnitude of 1. - To find the unit vector \( \hat{A}_{AB} \), follow these steps: - Compute the vector \( \mathbf{AB} \) from the coordinates of points A and B. - Divide this vector by its magnitude (length) to normalize it. 2. **Moment \( M_{AB} \):** - The moment \( M_{AB} \) about the AB axis due to a force \( \mathbf{F} \) is a measure of the tendency of \( \mathbf{F} \) to rotate an object about the axis from point A to point B. - To calculate \( M_{AB} \): - Determine the position vector from point A to any point on the line of action of \( \mathbf{F} \). - Compute the cross product of this position vector with the force vector \( \mathbf{F} \). - Project this resulting moment vector onto the line AB to find the scalar moment \( M_{AB} \). **Note:** - Ensure to clarify the concept of the direction vector and magnitude. - Provide detailed calculations for each step in both the unit vector determination and moment calculation to aid in understanding. **No graphs or diagrams are present in the provided text.** **Conclusion:** Understanding these fundamental concepts in mechanics is crucial for solving related problems in physics and engineering. Accurate determination of unit vectors and moments can significantly aid in analysis and design within various applications.
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