769_Dynamics 11ed Manual

769_Dynamics 11ed Manual - Engineering Mechanics - Dynamics...

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Engineering Mechanics - Dynamics Chapter 21 Given: m C 1kg = m E 2kg = θ E 30 deg = a 0.1 m = Solution: We need to put the center of mass along AB and to make the product of inetia go to zero. Guesses m D = m F = D 40 deg = F 10 deg = Given m E a cos E () m D 2 a ( ) sin D m F a sin F 0 = m C am D 2 a ( ) cos D + m E a sin E m F a cos F + 0 = m C aa m D 2 a 2 a ( ) cos D + m E 3 a a sin E m F 4 a a cos F + 0 = m D 2 a 2 a ( ) sin D m E 3 a a cos E + m F 4 a a sin F 0 = m D m F D F Find m D m F , D , F , = m D m F 0.661 1.323 kg = D F 139.1 40.9 deg = *Problem 21-60 The bent uniform rod ACD has a weight density γ , and is supported at
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This note was uploaded on 08/16/2011 for the course EGN 3321 taught by Professor Christianfeldt during the Spring '08 term at University of Central Florida.

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