Explanation Given information: Two identical uniform rods are given These are held in horizontal position Explanation: Consider the given diagram First take the acceleration of the bar BC based on its center a B C ( x ) = a c Then acceleration of the point C becomes a B C ( y ) = 1 2 L × α B C And then consider the acceleration for the bar BC at the point B a B ( x ) = a c Then acceleration at the point C in horizontal direction a B C ( y ) = L × α B C Consider the bar AB and its acceleration of the center a A B ( x ) = a c In horizontal direction, the acceleration value becomes a A B ( y ) = a B ( y ) + 1 2 L × α A B Substituting the required values, we get, a A B ( y ) = L × α B C + 1 2 L × α A B Finally, take the acceleration of the given bar AB at point A a A ( x ) = a c Then, it becomes a A ( y ) = a B ( y ) + L × α A B Substituting the required values, we get, a A ( y ) = L × α B C + L × α A B Consider the kinematic diagram of the bars Take the horizontal forces acting on the bar ∑ f x = ∑ f x ( e f f ) o = m a A B ( x ) + m a B C ( x ) 0 = m a c + m a c 0 = 2 m a c 0 = a c Diagram for acceleration and forces acting on the bar AB Consider the moments acting on the bar AB ∑ M B = ∑ M B ( e f f ) m g × 1 2 = I α A B + m a A B × 1 2 Substituting the values, we get, ∑ M B = ∑ M B ( e f f ) m g × 1 2 = I α A B + m a A B × 1 2 m g × L 2 = 1 12 m L 2 α A B + m ( L × α B C + 1 2 L × α A B ) × L 2 m g × L 2 = 1 12 m L 2 α A B + 1 2 m L 2 α B C + 1 4 m L 2 α A B g 2 = 1 12 L α A B + 1 2 L α B C + 1 4 L α A B g = 1 6 L α A B + 1 2 L α B C + L α A B g = 2 3 L α A B + L α B C − − − − ( 1 ) Vertical forces acting on the bar AB is ∑ f y = ∑ f y ( e f f ) B − W = − m To determine (b) Find the acceleration at point B.

12th Edition

ISBN: 9781260265521

Author: BEER

Publisher: MCG

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Chapter 16.2, Problem 16.150P

To determine

(a)

Find the acceleration at point A.

To determine

(b)

Find the acceleration at point B.

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Chapter 16.2, Problem 16.149P

Chapter 16.2, Problem 16.151P

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(a) Find the acceleration at point A.

Solution Summary: The author explains that the acceleration at point A is a_A=0.857g.