Explanation Show the kinetic diagram of the system as in Figure (1). Express the linear acceleration of pulley A . a A = α A r A (I) Here, the angular acceleration of the pulley is α A and radius of the pulley is r A . Express the linear acceleration of pulley B . a B = α B r B (II) Here, the angular acceleration of the pulley is α B and radius of the pulley is r B . The net acceleration of point A is equal to linear acceleration of the string of pulley A . a C M − α C M × 18 = a A (III) Here, the linear acceleration of the bar from its center of mass due to rotation of the bar is α C M × 18 . The net acceleration of point B is equal to linear acceleration of the string of pulley B . a C M − α C M × 18 = a B (IV) Apply equation of motion to the system. ∑ F = m a T A + T B − m g = m a C M (V) Here, the weight of the bar is m g , angular acceleration of the bar is α C M , tension of the string at point A is T A and linear acceleration of the bar is a C M . Apply equation of motion of moments to the system. ∑ M = I α C M ( T B − T A ) ( 1.5 ) = I α C M ( T B − T A ) ( 1.5 ) = m l 2 12 α C M (VI) Here, length of the bar is l . Conclusion: Substitute 4 rad / s 2 for α A and 7 .5 in . for r A in Equation (I). a A = ( 4 rad / s 2 ) ( 7 .5 in . ) = 30 in . / s 2 Substitute 6 rad / s 2 for α B and 7 .5 in . for r B in Equation (II). a B = ( 6 rad / s 2 ) ( 7

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ISBN: 9781118885840

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

Publisher: WILEY

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Chapter 6.5, Problem 87P

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Determine the tension in string at points A and B.

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