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PracticeExam2

# PracticeExam2 - ME 3212 Mechanisms Practice Exam#2 Fall...

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ME 3212: Mechanisms Practice Exam #2 Fall 2007 1. For the offset slider crank shown below, (a) Differentiate the velocity equations to get the acceleration equations and put into matrix form. (b) Solve for ¨ r and ¨ θ 3 . (c) Setup and differentiate the position equations to get the velocity of point D. (d) Solve for the acceleration magnitude and direction of point D. % A B C D O 2 x y θ 3 θ 2 ˙ θ 2 r e where O 2 A = 28 mm, AC = 30 mm, CB = 70 mm, CD = 40 mm, e = 12 mm, r = 79 mm, θ 2 = 120 , θ 3 = 22 , ˙ θ 2 = ϖ 2 = 20 rad/sec (constant), ˙ θ 3 = 3 . 02 rad/s (CCW), and ˙ r = - 371 . 8 mm/s ( ) . The loop closure equations are as follows: O 2 A cos θ 2 + AB cos θ 3 = r O 2 A sin θ 2 - AB sin θ 3 + e = 0 and the velocity equations are: - O 2 A ˙ θ 2 sin θ 2 - AB ˙ θ 3 sin θ 3 = ˙ r O 2 A ˙ θ 2 cos θ 2 - AB ˙ θ 3 cos θ 3 = 0 1

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2. Design an offset slider-crank mechanism with a stroke of 250 mm, time ratio R T = 1 . 30 and a transmission angle γ a = 50 at the fully retracted position. Determine the lengths for the crank and connecting rod, and offset, e . Draw the mechanism in the position of γ = γ min . Both calculate and measure γ min your drawing.
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