---**Problem 2: Force Analysis on a Bent Bar ABC****Problem Statement:**A bent bar ABC is supported and subjected to forces as depicted in the diagram. Determine the reactions at points A, B, and C, assuming negligible friction and bar weight.**Diagram Description:**The provided diagram illustrates the bent bar ABC, consisting of two segments joined at point B:- The horizontal segment AB, which is 10 feet long (5 feet from A to B).- The inclined segment BC, which forms an angle of 45° with a horizontal smooth slot and is 6 feet long (3 feet from B to C).The bar is supported by:- A horizontal smooth slot at point A.- A pinned connection at point B.- A vertical smooth slot at point C, inclined at an angle of 45° to the horizontal.The forces acting on the bar are:- A vertical downward force of 200 N at point A.- A 300 N force acting vertically downward on point C.**Diagram Explanation:**1. **Point A:** - A smooth horizontal slot supports this point, allowing only vertical movement. Therefore, the reaction at A (R_A) is a vertical force. 2. **Point B:** - This point features a pinned connection, which can support forces in both the horizontal and vertical directions, denoted as \( B_x \) for the horizontal component and \( B_y \) for the vertical component.3. **Point C:** - The smooth slot here is inclined at 45°, indicating that it can exert an equal reaction force component in both the horizontal and vertical directions. This reaction force can be decomposed into \( C_x \) and \( C_y \).---**Analysis Approach:**1. **Static Equilibrium Equations:** - Sum of forces in the horizontal direction (ΣF_x = 0) - Sum of forces in the vertical direction (ΣF_y = 0) - Sum of moments about point B (ΣM_B = 0)2. **Force Components:** - The inclined reaction at C results in components: - \( C_x = C \cos(45°) \) - \( C_y = C \sin(45°) \) - Given \( \cos(45°) = \sin(45°) = \frac{1}{\sqrt{2}}

Using the conditions of equilibrium MB=0 300(3)+200(5)-10RA=0 RA=190N

A bent bar ABC are supported and subjected to forces as shown below. Find the reactions on the bar at A, B and C. Neglect friction and the weight of the bars. 200 N 30° B. A 5 ft 5 ft Smooth slot 45° 3 ft 3 ft 300 N

A bent bar ABC are supported and subjected to forces as shown below. Find the reactions on the bar at A, B and C. Neglect friction and the weight of the bars. 200 N 30° B. A 5 ft 5 ft Smooth slot 45° 3 ft 3 ft 300 N

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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Question

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**Problem 2: Force Analysis on a Bent Bar ABC**

**Problem Statement:**
A bent bar ABC is supported and subjected to forces as depicted in the diagram. Determine the reactions at points A, B, and C, assuming negligible friction and bar weight.

**Diagram Description:**

The provided diagram illustrates the bent bar ABC, consisting of two segments joined at point B:

- The horizontal segment AB, which is 10 feet long (5 feet from A to B).
- The inclined segment BC, which forms an angle of 45° with a horizontal smooth slot and is 6 feet long (3 feet from B to C).

The bar is supported by:
- A horizontal smooth slot at point A.
- A pinned connection at point B.
- A vertical smooth slot at point C, inclined at an angle of 45° to the horizontal.

The forces acting on the bar are:
- A vertical downward force of 200 N at point A.
- A 300 N force acting vertically downward on point C.

**Diagram Explanation:**

1. **Point A:**
- A smooth horizontal slot supports this point, allowing only vertical movement. Therefore, the reaction at A (R_A) is a vertical force.

2. **Point B:**
- This point features a pinned connection, which can support forces in both the horizontal and vertical directions, denoted as \( B_x \) for the horizontal component and \( B_y \) for the vertical component.

3. **Point C:**
- The smooth slot here is inclined at 45°, indicating that it can exert an equal reaction force component in both the horizontal and vertical directions. This reaction force can be decomposed into \( C_x \) and \( C_y \).

---
**Analysis Approach:**

1. **Static Equilibrium Equations:**
- Sum of forces in the horizontal direction (ΣF_x = 0)
- Sum of forces in the vertical direction (ΣF_y = 0)
- Sum of moments about point B (ΣM_B = 0)

2. **Force Components:**
- The inclined reaction at C results in components:
- \( C_x = C \cos(45°) \)
- \( C_y = C \sin(45°) \)
- Given \( \cos(45°) = \sin(45°) = \frac{1}{\sqrt{2}}

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