Given: The four bar linkage is moving with WCD equal to 6 rad/s CCW. Find: The velocity of point E on link BC and angular velocity of link AB. Use ICR method. 30° B 0.3 m 0.3 m E WCD= 6 rad/s 0.6 m

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
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**Problem Statement:**

Given: The four bar linkage is moving with \(\omega_{CD}\) equal to 6 rad/s CCW.

Find: The velocity of point E on link BC and the angular velocity of link AB. Use the ICR method.

**Diagram Explanation:**

The diagram illustrates a four-bar linkage consisting of links AB, BC, and CD. The linkage is in motion, and the angular velocity \(\omega_{CD}\) is 6 rad/s counterclockwise (CCW). 

**Key Measurements:**

- Link CD is vertical and has a length of 0.6 m.
- Link BC is horizontal, with sections labeled:
  - From B to E: 0.3 m
  - From E to C: 0.3 m 
- Link AB makes a 30° angle with the horizontal ground.

**Objective:**

- Determine the velocity of point E on link BC.
- Calculate the angular velocity of link AB using the Instantaneous Center of Rotation (ICR) method.
Transcribed Image Text:**Problem Statement:** Given: The four bar linkage is moving with \(\omega_{CD}\) equal to 6 rad/s CCW. Find: The velocity of point E on link BC and the angular velocity of link AB. Use the ICR method. **Diagram Explanation:** The diagram illustrates a four-bar linkage consisting of links AB, BC, and CD. The linkage is in motion, and the angular velocity \(\omega_{CD}\) is 6 rad/s counterclockwise (CCW). **Key Measurements:** - Link CD is vertical and has a length of 0.6 m. - Link BC is horizontal, with sections labeled: - From B to E: 0.3 m - From E to C: 0.3 m - Link AB makes a 30° angle with the horizontal ground. **Objective:** - Determine the velocity of point E on link BC. - Calculate the angular velocity of link AB using the Instantaneous Center of Rotation (ICR) method.
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