Explanation Given information: θ = 30 ° Angular acceleration = 10 rpm m x ¨ G m y ¨ G x G location with respect to θ x G = 3 + ( 5 ( 1 − cos θ ) ) Differentiating the above, x ˙ G = 5 sin θ . ω Again differentiating, x ¨ G = 5 cos θ . ω 2 x ¨ G = 5 × cos 30 × ( π 2 ) 2 x ¨ G = 4.74 Similarly, y G location with respect to θ y G = 1 + 5 sin θ y ˙ G = 5 cos θ . ω y ¨ G = − 5 sin θ . ω 2 = − 5 × sin 30 × ( π 2 ) 2 y ¨ G = − 2.74 Moment about A, 54 × ( 3 + ( 5 ( 1 − cos 30 ) ) ) + ( 54 32

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ISBN: 9781260265521

Author: BEER

Publisher: MCG

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Chapter 16.1, Problem 16.18P

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The vertical component of forces exerted by pins B and D.

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The vertical component of forces exerted by pins B and D.

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