To determine The angular acceleration of the link DB . Explanation Draw the schematic diagram of the system as shown in Figure (1). Write the expression for the tangential component of acceleration of point B in vector form. ( a B ) t = α D B × r D B (I) Here, tangential component of acceleration of point B in vector form is ( a B ) t , angular acceleration of the link DB in vector form is α D B and position vector of the link DB is r D B . Write the expression for the normal component of acceleration of point B in vector form. ( a B ) n = ω D B × ( ω D B × r D B ) (II) Here, normal component of acceleration of point B in vector form is ( a B ) n and angular velocity of the link DB in vector form is ω D B . Write the expression for the tangential component of acceleration of point A in vector form. ( a A ) t = α O A × r O A (III) Here, tangential component of acceleration of point A in vector form is ( a A ) t , angular acceleration of the link OA in vector form is α O A and position vector of the link OA is r O A . Write the expression for the normal component of acceleration of point A in vector form. ( a A ) n = ω O A × ( ω O A × r O A ) (IV) Here, normal component of acceleration of point A in vector form is ( a A ) n and angular velocity of the link OA in vector form is ω 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 (V) Here, tangential component of acceleration of point B with respect to point A in vector form is ( a B / A ) t , angular acceleration of the link AB in vector form is α A B and position vector of the link AB is r A B . 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 ) (VI) Here, normal component of acceleration of point B with respect to point A vector form is ( a B / A ) n and angular velocity of the link AB in vector form is ω A B . Write the expression for the acceleration of point B in vector form. a B = a A + a B / A [ ( a B ) n + ( a B ) t ] = [ ( a A ) n + ( a A ) t ] + [ ( a B / A ) n + ( a B / A ) t ] (VII) Here, acceleration of point A in vector form is a A , acceleration of point B in vector form is a B , angular velocity of the link AB is ω A B and angular acceleration of the link AB in vector form is α Conclusion: The linear velocities of the links DB and OA are equal because the links DB and OA are parallel to each other. v D B = v O A ω D B × r D B = ω O A × r O A (VIII) Here, velocity of the link DB in vector form is v D B and velocity of the link OA in vector form is v O A . Write the expression for the angular velocity of the link OA in vector form. ω O A = − ω O A k = − ( 3 rad / s ) k = − 3 k rad / s Here, angular velocity of the link OA is ω O A and unit vector along z direction is k . Write the expression for the angular velocity of the link DB in vector form. ω D B = ω D B k Here, angular velocity of the link DB is ω D B . Write the expression for the position of the link OA . r O A = 0.06 j Here, unit vector along y direction is j . Write the expression for the position of the link DB . r D B = 0.24 j Substitute ω D B k for ω D B , − 3 k rad / s for ω O A , 0.06 j for r O A and 0.24 j for r D B in Equation (VIII). ω D B k × 0.06 j = − 3 k rad / s × 0.24 j ω D B = − 0.75 rad / s Write the expression for the position of the link AB . r A B = − 0.24 i + 0.18 j Here, unit vector along x direction is i . Write the expression for the angular acceleration of the link DB in vector form. α D B = α D B k Here, angular acceleration of the link DB is α D B . Substitute α D B k for α D B and 0.24 j for r D B in Equation (I). ( a B ) t = ( α D B k ) × ( 0.24 j ) = − 0.24 α D B i m / s 2 Substitute − 0

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ISBN: 9781118885840

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

Publisher: WILEY

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Chapter 5.6, Problem 144P

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The angular acceleration of the link DB.

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Chapter 5.6, Problem 143P

Chapter 5.6, Problem 145P

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