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Express the system of forces shown below as an equivalent force couple system at A. A is located at the intersection of the x and y axes. The 0.5 kN and 3 kN forces are acting midway through the thickness of the rigid body. 22 0.5 kN 50 mm H-. 20 mm 150 mm 40 mm 2.5 kN 160 mm 3 kN

Express the system of forces shown below as an equivalent force couple system at A. A is located at the intersection of the x and y axes. The 0.5 kN and 3 kN forces are acting midway through the thickness of the rigid body. 22 0.5 kN 50 mm H-. 20 mm 150 mm 40 mm 2.5 kN 160 mm 3 kN

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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**Transcription and Explanation:** **Text:** Express the system of forces shown below as an equivalent force couple system at A. A is located at the intersection of the x and y axes. The 0.5 kN and 3 kN forces are acting midway through the thickness of the rigid body. **Diagram Explanation:** The diagram depicts a 3D rigid body with several forces acting on it. The body is an L-shaped structure with a horizontal segment along the x-axis and a vertical segment along the y-axis. - The base of the L is positioned along the x-axis, starting at point A, which is the intersection of the x and y axes. - A vertical force of 2.5 kN is acting downward at the junction of the horizontal and vertical segments. - A horizontal force of 3 kN is applied parallel to the x-axis, toward the right end of the horizontal segment. - Another horizontal force of 0.5 kN is applied parallel to the y-axis, toward the left end of the vertical segment. - The horizontal segment is labeled "B" at the right end. - Measurements along the segments are provided: - The horizontal segment from A to the final right end is 150 mm. - The vertical segment is 50 mm tall. - The distance from the vertical segment to the mid-section where the 2.5 kN force acts is 40 mm. - The distance along the horizontal segment from the mid-section to point B is 160 mm. - The thickness of the rigid body where the 0.5 kN and 3 kN forces act is unspecified, but they're applied midway through the thickness. These forces and dimensions make up a mechanical system, and the task is to express them as an equivalent force-couple system at point A.
Transcribed Image Text:**Transcription and Explanation:** **Text:** Express the system of forces shown below as an equivalent force couple system at A. A is located at the intersection of the x and y axes. The 0.5 kN and 3 kN forces are acting midway through the thickness of the rigid body. **Diagram Explanation:** The diagram depicts a 3D rigid body with several forces acting on it. The body is an L-shaped structure with a horizontal segment along the x-axis and a vertical segment along the y-axis. - The base of the L is positioned along the x-axis, starting at point A, which is the intersection of the x and y axes. - A vertical force of 2.5 kN is acting downward at the junction of the horizontal and vertical segments. - A horizontal force of 3 kN is applied parallel to the x-axis, toward the right end of the horizontal segment. - Another horizontal force of 0.5 kN is applied parallel to the y-axis, toward the left end of the vertical segment. - The horizontal segment is labeled "B" at the right end. - Measurements along the segments are provided: - The horizontal segment from A to the final right end is 150 mm. - The vertical segment is 50 mm tall. - The distance from the vertical segment to the mid-section where the 2.5 kN force acts is 40 mm. - The distance along the horizontal segment from the mid-section to point B is 160 mm. - The thickness of the rigid body where the 0.5 kN and 3 kN forces act is unspecified, but they're applied midway through the thickness. These forces and dimensions make up a mechanical system, and the task is to express them as an equivalent force-couple system at point A.
Expert Solution
Step 1

Given:

The force at B in y-direction= 2.5 kN

The force at D in z-direction = 0.5 kN

The force at B in x-direction = 3 kN

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