To determine The angular velocity of the link OA . Explanation Write the expression for the relation between the velocity of point B and velocity of point A . v A = v B + v A / B = v B + ( ω A B × r A B ) (I) Here, velocity of point A is v A , velocity of point B is v B , velocity of point A with respect to point B is v A / B , angular velocity of the link AB is ω A B and position vector of the link AB is r A B . Write the expression for the relation between the velocity of point A and velocity of point O . v A = v O + v A / O = v O + ( ω O A × r O A ) (II) Here, velocity of point O is v O , velocity of point A with respect to point O is v A / O , angular velocity of the link OA is ω O A and position vector of the link OA is r O A . Conclusion: Show the schematic diagram of the system as in Figure (1). Write the relation for the distance between the points A and B using law of cosine. ( A B ¯ ) 2 = ( O B ¯ ) 2 + ( O A ¯ ) 2 − 2 ( O B ¯ ) ( O A ¯ ) cos θ cos θ = ( O B ¯ ) 2 + ( O A ¯ ) 2 − ( A B ¯ ) 2 2 ( O B ¯ ) ( O A ¯ ) (III) Here, distance between the points A and B is A B ¯ , distance between the points O and B is O B ¯ , distance between the points O and A is O A ¯ and angle between the link OA and link AB is θ . Substitute 90 mm for A B ¯ , 180 mm for O B ¯ and 130 mm for O A ¯ in Equation (III). cos θ = ( 180 mm ) 2 + ( 130 mm ) 2 − ( 90 mm ) 2 2 ( 180 mm ) ( 130 mm ) = 0.88034 θ = cos − 1 ( 0.88034 ) = 28.3 ° From Figure (1), consider the triangle Δ O A B . Calculate the angle between the link AB and the link OA by applying sine rule. sin β O A ¯ = sin θ A B ¯ (IV) Here, angle between the link AB and the link OA is β . Substitute 90 mm for A B ¯ , 130 mm for O A ¯ and 28.3 ° for θ in Equation (IV). sin β 130 mm = sin 28.3 ° 90 mm sin β = 0.6848 β = sin − 1 ( 0.6848 ) = 43.2 ° Calculate the position vector of the link OA . r O A = O A ¯ ( cos θ i + sin θ j ) (V) Here, unit vectors along x and y directions are i and j respectively. Substitute 130 mm for O A ¯ and 28.3 ° for θ in Equation (V). r O A = ( 130 mm ) ( cos 28.3 ° i + sin 28.3 ° j ) = ( 130 mm × 1 m 1000 mm ) ( cos 28.3 ° i + sin 28.3 ° j ) = ( 0.130 m ) ( cos 28.3 ° i + sin 28.3 ° j ) = 0.1144 i + 0.06163 j m Calculate the position vector of the link AB

8th Edition

ISBN: 9781118885840

Author: James L. Meriam, L. G. Kraige, J. N. Bolton

Publisher: WILEY

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Chapter 5.4, Problem 72P

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The angular velocity of the link OA.

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Chapter 5.4, Problem 71P

Chapter 5.4, Problem 73P

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Chapter 5 Solutions

Engineering Mechanics: Dynamics

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