Q.4) The rod is supported by journal bearings at A, B and C and subjected to a force of 450 N (in y-z plane) and a couple moment of 300 N-m as shown below. The bearings are in proper alignment and exert only force reactions on the rod. Derive the static equilibrium equations for the rod shown and find the force reactions at the bearings.

Elements Of Electromagnetics
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Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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**Q.4)** The rod is supported by journal bearings at A, B and C and subjected to a force of 450 N (in y-z plane) and a couple moment of 300 N·m as shown below. The bearings are in proper alignment and exert only force reactions on the rod. Derive the static equilibrium equations for the rod shown and find the force reactions at the bearings.

**Diagram Explanation:**
In the diagram, a rod is supported at three points: A, B, and C by journal bearings. These points are connected to the rod which bends in three dimensions. The rod extends 0.8 meters from A to B, 0.4 meters from B to the bend, 0.4 meters vertically upwards from the bend, and finally 0.6 meters horizontally to C. A force of 450 N is applied at point C in the y-z plane at an angle of 40° from the z-axis. A couple moment of 300 N·m is applied around the axis perpendicular to the rod segment from A to B.

**Step-by-Step Explanation:**

1. **Identify unknows:** Reaction forces at bearings A, B and C. Let's assume they are \( A_x, A_y, A_z\), \( B_x, B_y, B_z\), \( C_x, C_y, C_z \).

2. **Equilibrium Conditions:**
   - **Sum of Forces:** 
     \[
     \sum F_x = 0, \quad \sum F_y = 0, \quad \sum F_z = 0
     \]
   - **Sum of Moments:**
     \[
     \sum M_x = 0, \quad \sum M_y = 0, \quad \sum M_z = 0
     \]

3. **Force Components:**
   - Decomposition of the 450 N force:
     \[
     F_{450_N, y} = 450 \sin(40^\circ), \quad F_{450_N, z} = 450 \cos(40^\circ)
     \]

4. **Equilibrium Equations:**
   - **Sum of Forces in X-direction:**
     \[
     A_x + B_x + C_x = 0
     \]
   - **Sum of Forces in Y-direction:**
     \[
     A_y + B_y + C_y =
Transcribed Image Text:**Q.4)** The rod is supported by journal bearings at A, B and C and subjected to a force of 450 N (in y-z plane) and a couple moment of 300 N·m as shown below. The bearings are in proper alignment and exert only force reactions on the rod. Derive the static equilibrium equations for the rod shown and find the force reactions at the bearings. **Diagram Explanation:** In the diagram, a rod is supported at three points: A, B, and C by journal bearings. These points are connected to the rod which bends in three dimensions. The rod extends 0.8 meters from A to B, 0.4 meters from B to the bend, 0.4 meters vertically upwards from the bend, and finally 0.6 meters horizontally to C. A force of 450 N is applied at point C in the y-z plane at an angle of 40° from the z-axis. A couple moment of 300 N·m is applied around the axis perpendicular to the rod segment from A to B. **Step-by-Step Explanation:** 1. **Identify unknows:** Reaction forces at bearings A, B and C. Let's assume they are \( A_x, A_y, A_z\), \( B_x, B_y, B_z\), \( C_x, C_y, C_z \). 2. **Equilibrium Conditions:** - **Sum of Forces:** \[ \sum F_x = 0, \quad \sum F_y = 0, \quad \sum F_z = 0 \] - **Sum of Moments:** \[ \sum M_x = 0, \quad \sum M_y = 0, \quad \sum M_z = 0 \] 3. **Force Components:** - Decomposition of the 450 N force: \[ F_{450_N, y} = 450 \sin(40^\circ), \quad F_{450_N, z} = 450 \cos(40^\circ) \] 4. **Equilibrium Equations:** - **Sum of Forces in X-direction:** \[ A_x + B_x + C_x = 0 \] - **Sum of Forces in Y-direction:** \[ A_y + B_y + C_y =
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