### Problem StatementThe ends of the 0.63-m bar remain in contact with their respective support surfaces. End \( B \) has a velocity of 0.40 m/s and an acceleration of 0.47 m/s\(^2\) in the directions shown. Determine the angular acceleration \( \alpha \) (positive if counterclockwise, negative if clockwise) of the bar and the acceleration of end \( A \) (positive if up, negative if down).### Diagram ExplanationThe diagram illustrates a bar \( AB \) of length 0.63 meters. The bar is inclined between two support surfaces at points \( A \) and \( B \) with angles of \( 28^\circ \) and \( 105^\circ \) to the ground, respectively. Point \( B \) has:- A velocity (\( v_B \)) of 0.40 m/s in a downward and slightly rightward direction from the horizontal line.- An acceleration (\( a_B \)) of 0.47 m/s\(^2\) in the same direction as the velocity (\( v_B \)).### Given Data- Length of the bar, \( L = 0.63 \text{ m} \)- Velocity at end \( B \), \( v_B = 0.40 \text{ m/s} \)- Acceleration at end \( B \), \( a_B = 0.47 \text{ m/s}^2 \)### Required1. Angular acceleration, \( \alpha \), of the bar.2. Acceleration of end \( A \), \( a_A \).### Answers- Angular acceleration, \( \alpha = 0.274 \text{ rad/s}^2 \) - Acceleration of end \( A \), \( a_A = 0.196 \text{ m/s}^2 \)

Answered: Image /qna-images/answer/df4c79b8-ca22-4957-bfda-6ccb1297d304.jpg

The ends of the 0.63-m bar remain in contact with their respective support surfaces. End B has a velocity of 0.40 m/s and an acceleration of 0.47 m/s² in the directions shown. Determine the angular acceleration a (positive if counterclockwise, negative if clockwise) of the bar and the acceleration of end A (positive if up, negative if down). A 28 105 Answers: a = aA = i i 0.63 m B 0.274 0.196 -ag=0.47m/s² UB=0.40 m/s rad/s² m/s²

The ends of the 0.63-m bar remain in contact with their respective support surfaces. End B has a velocity of 0.40 m/s and an acceleration of 0.47 m/s² in the directions shown. Determine the angular acceleration a (positive if counterclockwise, negative if clockwise) of the bar and the acceleration of end A (positive if up, negative if down). A 28 105 Answers: a = aA = i i 0.63 m B 0.274 0.196 -ag=0.47m/s² UB=0.40 m/s rad/s² m/s²

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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Concept explainers

Power Transmission Elements

It is defined as the energy transfer from one location to another location. The energy will start moving from the energy generation source to the energy applied position to obtain the work output. From this method, a large amount of electric energy moves from the power station to various electrical substations.

Question

### Problem Statement

The ends of the 0.63-m bar remain in contact with their respective support surfaces. End \( B \) has a velocity of 0.40 m/s and an acceleration of 0.47 m/s\(^2\) in the directions shown. Determine the angular acceleration \( \alpha \) (positive if counterclockwise, negative if clockwise) of the bar and the acceleration of end \( A \) (positive if up, negative if down).

### Diagram Explanation

The diagram illustrates a bar \( AB \) of length 0.63 meters. The bar is inclined between two support surfaces at points \( A \) and \( B \) with angles of \( 28^\circ \) and \( 105^\circ \) to the ground, respectively. Point \( B \) has:

- A velocity (\( v_B \)) of 0.40 m/s in a downward and slightly rightward direction from the horizontal line.
- An acceleration (\( a_B \)) of 0.47 m/s\(^2\) in the same direction as the velocity (\( v_B \)).

### Given Data

- Length of the bar, \( L = 0.63 \text{ m} \)
- Velocity at end \( B \), \( v_B = 0.40 \text{ m/s} \)
- Acceleration at end \( B \), \( a_B = 0.47 \text{ m/s}^2 \)

### Required

1. Angular acceleration, \( \alpha \), of the bar.
2. Acceleration of end \( A \), \( a_A \).

### Answers

- Angular acceleration, \( \alpha = 0.274 \text{ rad/s}^2 \)
- Acceleration of end \( A \), \( a_A = 0.196 \text{ m/s}^2 \)

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