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Problem 1. (18%) Knowing that the tension in wire BD is 650 N, determine the reaction at the fixed support C of the member shown. You must include FBD. 750 N 150 mm 250 mm -500 mm 400 N•m В 450 N 600 mm 400 mm

Problem 1. (18%) Knowing that the tension in wire BD is 650 N, determine the reaction at the fixed support C of the member shown. You must include FBD. 750 N 150 mm 250 mm -500 mm 400 N•m В 450 N 600 mm 400 mm

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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## Problem 1. (18%) Knowing that the tension in wire BD is 650 N, determine the reaction at the fixed support C of the member shown. **You must include FBD (Free Body Diagram).** ### Diagram Explanation In the given diagram, a rigid member is depicted with a series of forces and moments applied to it. Here are the key details: - **Member ABCD** has a fixed support at \( C \). - At point **A**, a horizontal member extends to the right. - **500 mm** from support C to A. - An external moment of **400 N·m** is applied at point A. - A downward vertical force of **750 N** is also applied at point A. - **Point B** is located **150 mm** to the right of A. - A vertical downward force of **450 N** acts at point B. - **Point D** is located **250 mm** to the right of B and **600 mm** below B. - A tension force of **650 N** acts along wire BD. - The distance from the horizontal axis through C to point D is given as **600 mm** downwards. ### Free Body Diagram (FBD) Explanation To solve the problem and determine the reactions at the fixed support at \( C \), we need to draw the Free Body Diagram (FBD) which must include: 1. External forces acting at points A, B, and D. 2. Reaction forces and moments at the fixed support C. ### Steps to Solve 1. **Identify the forces and moments applied**: - 750 N downwards at \( A \). - 400 N·m counterclockwise moment at \( A \). - 450 N downwards at \( B \). - 650 N tension in wire \( BD \). 2. **Reaction forces at \( C \)**: - Let's denote the reaction forces at C as \( C_x \) (horizontal) and \( C_y \) (vertical) and the reaction moment as \( M_C \). 3. **Apply Equilibrium Equations**: - Sum of all horizontal forces = 0 \( \sum F_x = 0 \) - Sum of all vertical forces = 0 \( \sum F_y = 0 \) - Sum of all moments about point C = 0 \( \sum
Transcribed Image Text:## Problem 1. (18%) Knowing that the tension in wire BD is 650 N, determine the reaction at the fixed support C of the member shown. **You must include FBD (Free Body Diagram).** ### Diagram Explanation In the given diagram, a rigid member is depicted with a series of forces and moments applied to it. Here are the key details: - **Member ABCD** has a fixed support at \( C \). - At point **A**, a horizontal member extends to the right. - **500 mm** from support C to A. - An external moment of **400 N·m** is applied at point A. - A downward vertical force of **750 N** is also applied at point A. - **Point B** is located **150 mm** to the right of A. - A vertical downward force of **450 N** acts at point B. - **Point D** is located **250 mm** to the right of B and **600 mm** below B. - A tension force of **650 N** acts along wire BD. - The distance from the horizontal axis through C to point D is given as **600 mm** downwards. ### Free Body Diagram (FBD) Explanation To solve the problem and determine the reactions at the fixed support at \( C \), we need to draw the Free Body Diagram (FBD) which must include: 1. External forces acting at points A, B, and D. 2. Reaction forces and moments at the fixed support C. ### Steps to Solve 1. **Identify the forces and moments applied**: - 750 N downwards at \( A \). - 400 N·m counterclockwise moment at \( A \). - 450 N downwards at \( B \). - 650 N tension in wire \( BD \). 2. **Reaction forces at \( C \)**: - Let's denote the reaction forces at C as \( C_x \) (horizontal) and \( C_y \) (vertical) and the reaction moment as \( M_C \). 3. **Apply Equilibrium Equations**: - Sum of all horizontal forces = 0 \( \sum F_x = 0 \) - Sum of all vertical forces = 0 \( \sum F_y = 0 \) - Sum of all moments about point C = 0 \( \sum
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