Explanation Write the expression for the velocity of point A . v A = ( A C ¯ ) ω A B (I) Here, velocity of point A is v A , distance between the points A and C is A C ¯ and angular velocity of the triangular plate is ω A B . Write the expression for the velocity of point B . v B = ( B C ¯ ) ω A B (II) Here, velocity of point B is v B and distance between the points B and C is B C ¯ . Write the expression for the velocity of point B with respect to point A . v B / A = ( A B ¯ ) ω A B (III) Here, velocity of point B with respect to point A is v B / A and distance between the points A and B is A B ¯ . Write the expression for the angular velocity of the link BC . ω B C = v B B C ¯ (IV) Here, angular velocity of the link BC is ω B C . Write the expression for the acceleration of point B in vector form. a B = a A + a B / A = ω B C × ( ω B C × r C B ) + α B C × r C B (V) Here, acceleration of point B in vector form is a B , acceleration of point A in vector form is a A , acceleration of point B with respect to point A in vector form is a B / A , angular velocity of the link BC in vector form is ω B C , angular acceleration of the link BC in vector form is α B C and position vector of the link BC is r C B . Write the expression for the acceleration of point A in vector form. a A = ω O A × ( ω O A × r O A ) (VI) Here, angular velocity of the link OA in vector form is ω O A and position vector of the link OA is r O A . Write the expression for the normal component of acceleration of point B with respect to point A in vector form. ( a B / A ) n = ω A B × ( ω A B × r A B ) (VII) Here, normal component of acceleration of point B with respect to point A in vector form is ( a B / A ) n , angular velocity of the triangular plate in vector form is ω A B and position vector of the of the triangular plate is r O A . Write the expression for the tangential component of acceleration of point B with respect to point A in vector form. ( a B / A ) t = α A B × r A B (VIII) Here, tangential component of acceleration of point B with respect to point A in vector form is ( a B / A ) t and angular acceleration of the triangular plate in vector form is α A B . Write the expression for the acceleration of point B in vector form. a B = a A + ( a B / A ) n + ( a B / A ) t (IX) Here, acceleration of point B in vector form is a B and acceleration of point A in vector form is a A . Conclusion: Substitute 4 in . for A C ¯ and 3 rad / s for ω A B in Equation (I). v A = ( 4 in . ) ( 3 rad / s ) = 12 in . / s Substitute 3 in . for B C ¯ and 3 rad / s for ω A B in Equation (II). v B = ( 3 in . ) ( 3 rad / s ) = 9 in . / s Substitute 5 in . for A B ¯ and 3 rad / s for ω A B in Equation (III). v B / A = ( 5 in . ) ( 3 rad / s ) = 15 in . / s Calculate the angle between the triangular plate ABD and link BC . tan θ = 4 in . 3 in . = 1.333 θ = tan − 1 ( 1.333 ) = 53.13 ° Draw the velocity polygon diagram for the system as shown in Figure (1). Substitute 3 in . for B C ¯ and 9 in . / s for v B in Equation (IV). ω B C = 9 in . / s 3 in . = 3 rad / s Write the expression for the angular acceleration of the link BC in vector form. α B C = α B C k Here, angular acceleration of the link BC is α B C and unit vector z -direction is k . Write the expression for the angular velocity of the link BC in vector form. ω B C = ω B C k = 3 k rad / s Substitute 3 k rad / s for ω B C , α B C k for α B C and 3 j in . for r C B in Equation (V)

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ISBN: 9781119047315

Author: Bolton

Publisher: VST

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Chapter 5.6, Problem 140P

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The angular acceleration of the triangular plate ABD and link BC.

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Chapter 5.6, Problem 139P

Chapter 5.6, Problem 141P

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The angular acceleration of the triangular plate ABD and link BC .

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Chapter 5 Solutions