Q.4) The rod is supported by journal bearings at A, B and C and subjected to a force of 500 N (in y-z plane) and a couple moment of 400 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. 500 N 0.3 m 0.4 m 42° 400 N.m B 0.1 m 0.2 m

Elements Of Electromagnetics
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**Question 4**

**Problem Statement:** The rod is supported by journal bearings at points A, B, and C, and is subjected to a force of 500 N in the y-z plane and a couple moment of 400 N·m as shown in the diagram 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 Description:**
The diagram shows a rod supported at three points: A, B, and C. The layout is as follows:
- Point A is at the origin of a coordinate system with a 0.4 m section extending along the x-axis.
- From the end of this 0.4 m segment, there is a 0.1 m section extending along the negative y-axis to point B.
- From point B, there is a segment extending 0.3 m along the y-axis with a vertical section reaching up by 0.2 m to point C.

Two external effects are acting on the rod:
1. A force of 500 N acts at an angle of 42 degrees with respect to the rod in the y-z plane.
2. A couple moment of 400 N·m is applied between points A and B in the coordinate system's yz-plane.

**Equilibrium Equations:**
To derive the static equilibrium equations we need to consider the following conditions of equilibrium:
- The sum of all forces in the x, y, and z directions must be zero.
- The sum of all moments about any point must also be zero.

Let’s denote:
- \( F_{Ax}, F_{Ay}, F_{Az} \): Force components at bearing A.
- \( F_{Bx}, F_{By}, F_{Bz} \): Force components at bearing B.
- \( F_{Cx}, F_{Cy}, F_{Cz} \): Force components at bearing C.
  
1. **Sum of Forces in the x-direction:**
   \[ \Sigma F_x = 0 \]
   \[ F_{Ax} + F_{Bx} + F_{Cx} = 0 \]

2. **Sum of Forces in the y-direction:**
   \[ \Sigma F_y = 0 \]
   \[ F_{Ay} + F_{By} + F_{Cy} - 500 \cos(42°) =
Transcribed Image Text:**Question 4** **Problem Statement:** The rod is supported by journal bearings at points A, B, and C, and is subjected to a force of 500 N in the y-z plane and a couple moment of 400 N·m as shown in the diagram 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 Description:** The diagram shows a rod supported at three points: A, B, and C. The layout is as follows: - Point A is at the origin of a coordinate system with a 0.4 m section extending along the x-axis. - From the end of this 0.4 m segment, there is a 0.1 m section extending along the negative y-axis to point B. - From point B, there is a segment extending 0.3 m along the y-axis with a vertical section reaching up by 0.2 m to point C. Two external effects are acting on the rod: 1. A force of 500 N acts at an angle of 42 degrees with respect to the rod in the y-z plane. 2. A couple moment of 400 N·m is applied between points A and B in the coordinate system's yz-plane. **Equilibrium Equations:** To derive the static equilibrium equations we need to consider the following conditions of equilibrium: - The sum of all forces in the x, y, and z directions must be zero. - The sum of all moments about any point must also be zero. Let’s denote: - \( F_{Ax}, F_{Ay}, F_{Az} \): Force components at bearing A. - \( F_{Bx}, F_{By}, F_{Bz} \): Force components at bearing B. - \( F_{Cx}, F_{Cy}, F_{Cz} \): Force components at bearing C. 1. **Sum of Forces in the x-direction:** \[ \Sigma F_x = 0 \] \[ F_{Ax} + F_{Bx} + F_{Cx} = 0 \] 2. **Sum of Forces in the y-direction:** \[ \Sigma F_y = 0 \] \[ F_{Ay} + F_{By} + F_{Cy} - 500 \cos(42°) =
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