**Title: Analysis of Reactions in a Loaded Assembly****Objective:**Determine the components of the reactions at points A and E under two conditions: (a) A 760-N load is applied at point C. (b) The load moves along its line of action and is applied at point D.**Solution:** 1. **Free Body Analysis: Entire Assembly** - The analysis is applicable for both scenarios since the load's position on its line of action does not matter. - **Free Body Diagram (FBD):** - Total forces/reactions acting on the entire assembly: \[ \underline{\text{Input Number}} \] - Equilibrium equations: - \( \sum F_x = 0: \) \[ \underline{\text{Input Value}} \] - \( \sum M_A = 0: \) \( E_y = \) \[ \underline{\text{Input Value}} \] N - \( \sum F_y = 0: \) \( A_y = \) \[ \underline{\text{Input Value}} \] N2. **Case (a): Load at Point C** - **Selected Free Body: Member BE** - Number of forces acting on member BE: \[ \underline{\text{Input Number}} \] - Force direction on E should align along EB: \[ \frac{E_x}{E_y} = \underline{\text{Input Ratio}} \] - Calculations: - \( E_x = \) \[ \underline{\text{Input Value}} \] N - Hence \( A_x = \) \[ \underline{\text{Input Value}} \] N3. **Case (b): Load at Point D** - **Selected Free Body: Member AC** - Number of forces acting on member AC: \[ \underline{\text{Input Number}} \]**Diagram Explanation:**- The diagram illustrates a bar assembly with pinned joints at points A, B, C, D, and E. - Distances: - \( AB = 300 \, \text{mm} \) - \( BC = 600 \, \text{mm} \) - \( CD = 200 \, \text{mm} \) - \( DE = 300 \, \text{mm **Solution:**1. **Free body: Entire assembly.** - This analysis is valid for both a) and b), since the position of the load on its line of action is immaterial. - **FBD: In total, there are [ ] forces/reactions (number of forces) acting on the entire assembly.** - \(\sum \vec{F}_x = 0\): [ ] N; - \(\sum \vec{M}_A = 0\): \(E_y =\) [ ] N; - \(\sum \vec{F}_y = 0\): \(A_y =\) [ ] N;2. **If the load is acting on C, we choose the free body: member BE** - Member BE has [ ] forces acting on it. (number of forces) - Thus the force acting on E should be directed along EB, hence \(\frac{E_x}{E_y} = \) [ ]. - \(E_x =\) [ ] N; - Hence \(A_x =\) [ ] N;3. **If the load is acting on D, we choose the free body: member AC** - Member AC has [ ] forces acting on it. (number of forces) - Thus the force acting on A should be directed along [ ], hence \(\frac{A_x}{A_y} = \) [ ]; - \(A_x = \) [ ] N; - Hence \(E_x = \) [ ] N;

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Answered: Determine the components of the…

Determine the components of the reactions at A and E, (a) if a 760-N load is applied as shown at C, (b) if the load is moved along its line of action and is applied at point D.

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

**Title: Analysis of Reactions in a Loaded Assembly**

**Objective:**
Determine the components of the reactions at points A and E under two conditions:
(a) A 760-N load is applied at point C.
(b) The load moves along its line of action and is applied at point D.

**Solution:**

1. **Free Body Analysis: Entire Assembly**
- The analysis is applicable for both scenarios since the load's position on its line of action does not matter.

- **Free Body Diagram (FBD):**
- Total forces/reactions acting on the entire assembly: \[ \underline{\text{Input Number}} \]
- Equilibrium equations:
- \( \sum F_x = 0: \) \[ \underline{\text{Input Value}} \]
- \( \sum M_A = 0: \) \( E_y = \) \[ \underline{\text{Input Value}} \] N
- \( \sum F_y = 0: \) \( A_y = \) \[ \underline{\text{Input Value}} \] N

2. **Case (a): Load at Point C**
- **Selected Free Body: Member BE**
- Number of forces acting on member BE: \[ \underline{\text{Input Number}} \]
- Force direction on E should align along EB:
\[
\frac{E_x}{E_y} = \underline{\text{Input Ratio}}
\]
- Calculations:
- \( E_x = \) \[ \underline{\text{Input Value}} \] N
- Hence \( A_x = \) \[ \underline{\text{Input Value}} \] N

3. **Case (b): Load at Point D**
- **Selected Free Body: Member AC**
- Number of forces acting on member AC: \[ \underline{\text{Input Number}} \]

**Diagram Explanation:**
- The diagram illustrates a bar assembly with pinned joints at points A, B, C, D, and E.
- Distances:
- \( AB = 300 \, \text{mm} \)
- \( BC = 600 \, \text{mm} \)
- \( CD = 200 \, \text{mm} \)
- \( DE = 300 \, \text{mm

**Solution:**

1. **Free body: Entire assembly.**
- This analysis is valid for both a) and b), since the position of the load on its line of action is immaterial.
- **FBD: In total, there are [ ] forces/reactions (number of forces) acting on the entire assembly.**
- \(\sum \vec{F}_x = 0\): [ ] N;
- \(\sum \vec{M}_A = 0\): \(E_y =\) [ ] N;
- \(\sum \vec{F}_y = 0\): \(A_y =\) [ ] N;

2. **If the load is acting on C, we choose the free body: member BE**
- Member BE has [ ] forces acting on it. (number of forces)
- Thus the force acting on E should be directed along EB, hence \(\frac{E_x}{E_y} = \) [ ].
- \(E_x =\) [ ] N;
- Hence \(A_x =\) [ ] N;

3. **If the load is acting on D, we choose the free body: member AC**
- Member AC has [ ] forces acting on it. (number of forces)
- Thus the force acting on A should be directed along [ ], hence \(\frac{A_x}{A_y} = \) [ ];
- \(A_x = \) [ ] N;
- Hence \(E_x = \) [ ] N;

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