### Diagram ExplanationThe provided diagram depicts a physical setup consisting of a horizontal bar, a cable, and a hanging store sign. Below is a detailed description of the elements in the diagram:1. **Horizontal Bar**: - Length: \( L = 1.2 \text{ m} \) - Mass: \( m_b = 12 \text{ kg} \) - Uniform density2. **Cable**: - The right end of the horizontal bar is connected to a cable. - The cable forms an angle \( \theta \) of 31° with the horizontal.3. **Hinge**: - The left end of the bar is hinged, allowing it to pivot at this point.4. **Store Sign**: - A store sign with a mass \( m_s = 18 \text{ kg} \) is suspended vertically below the bar. - The sign is located at a distance \( d = 0.18 \text{ m} \) from the right end of the bar.### Problem StatementGiven the setup, the objective is to:**Find the magnitude of the vertical hinge force.**### Understanding the Diagram- A vertical bar is shown on the left, representing a wall to which the horizontal bar is hinged.- The horizontal bar, marked with length \( L \) and mass \( m_b \), extends from the hinge.- A cable is attached to the right end of the horizontal bar and inclined at 31° with respect to the horizontal direction.- The store sign is hanging below the bar, at a specified distance from the right end.### Steps to SolutionTo find the magnitude of the vertical hinge force, follow these steps:1. **Identify Forces Acting on the Bar**: - Weight of the bar (\( W_b \)) acting downward at its center of mass. - Tension in the cable (\( T \)) supporting the bar's right end. - Weight of the store sign (\( W_s \)) acting downward 0.18 m from the right end. - Hinge force containing vertical (\( F_V \)) and horizontal (\( F_H \)) components.2. **Equilibrium Conditions**: The bar is in static equilibrium, so: - Sum of vertical forces must be zero. - Sum of horizontal forces must be zero. - Sum of torques around the

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mb: L d- ms A horizontal L=1.2 m long mb=12 kg uniform bar is hinged on the left end and pulled at the right end by a cable. The cable makes 31° angle with horizontal. A 18 kg store sign is suspended below the bar at d=0.18 m from the right end. Find the magnitude of vertical hinge force.

mb: L d- ms A horizontal L=1.2 m long mb=12 kg uniform bar is hinged on the left end and pulled at the right end by a cable. The cable makes 31° angle with horizontal. A 18 kg store sign is suspended below the bar at d=0.18 m from the right end. Find the magnitude of vertical hinge force.

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

### Diagram Explanation

The provided diagram depicts a physical setup consisting of a horizontal bar, a cable, and a hanging store sign. Below is a detailed description of the elements in the diagram:

1. **Horizontal Bar**:
- Length: \( L = 1.2 \text{ m} \)
- Mass: \( m_b = 12 \text{ kg} \)
- Uniform density

2. **Cable**:
- The right end of the horizontal bar is connected to a cable.
- The cable forms an angle \( \theta \) of 31° with the horizontal.

3. **Hinge**:
- The left end of the bar is hinged, allowing it to pivot at this point.

4. **Store Sign**:
- A store sign with a mass \( m_s = 18 \text{ kg} \) is suspended vertically below the bar.
- The sign is located at a distance \( d = 0.18 \text{ m} \) from the right end of the bar.

### Problem Statement

Given the setup, the objective is to:

**Find the magnitude of the vertical hinge force.**

### Understanding the Diagram

- A vertical bar is shown on the left, representing a wall to which the horizontal bar is hinged.
- The horizontal bar, marked with length \( L \) and mass \( m_b \), extends from the hinge.
- A cable is attached to the right end of the horizontal bar and inclined at 31° with respect to the horizontal direction.
- The store sign is hanging below the bar, at a specified distance from the right end.

### Steps to Solution

To find the magnitude of the vertical hinge force, follow these steps:

1. **Identify Forces Acting on the Bar**:
- Weight of the bar (\( W_b \)) acting downward at its center of mass.
- Tension in the cable (\( T \)) supporting the bar's right end.
- Weight of the store sign (\( W_s \)) acting downward 0.18 m from the right end.
- Hinge force containing vertical (\( F_V \)) and horizontal (\( F_H \)) components.

2. **Equilibrium Conditions**:
The bar is in static equilibrium, so:
- Sum of vertical forces must be zero.
- Sum of horizontal forces must be zero.
- Sum of torques around the

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