Lecture16-supplement

Lecture16-supplement - Example 4 Given: A linkage...

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Given: A linkage undergoing motion as shown. The velocity of the block, v D , is 3 m/s. Find: The angular velocities of links AB and BD. Locate the instantaneous center of zero velocity of link BD and then solve for the angular velocities. Plan: Example 4
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Solution: Since D moves to the right, it causes link AB to rotate clockwise about point A. The instantaneous center of velocity for BD is located at the intersection of the line segments drawn perpendicular to v B and v D . Note that v B is perpendicular to link AB. Therefore we can see that the IC is located along the extension of link AB. Example 4 (cont’d)
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Since the magnitude of v D is known, the angular velocity of link BD can be found from v D = ω BD r D/IC . ω BD = v D /r D/IC = 3/0.566 = 5.3 rad/s ω AB = v B /r B/A = (r B/IC ) ω BD /r B/A = 0.4(5.3)/0.4 = 5.3 rad/s Link AB is subjected to rotation about A. Using these facts, r B/IC = 0.4 tan 45° = 0.4 m r D/IC = 0.4/cos 45° = 0.566 m Example 4 (cont’d)
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Lecture16-supplement - Example 4 Given: A linkage...

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