279_Problem CHAPTER 9

279_Problem CHAPTER 9 - PROBLEM 9.185 CONTINUED...

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PROBLEM 9.185 CONTINUED Substituting 1 K () 4 1 1 1 51 1 0.203032 0 12 4 24 xy z ta λλ λ ρ   −− =     4 1 1 11 0.203032 0 0 42 ta −+ = or ( ) ( ) 1 1 0.841842 yx = and ( ) ( ) 1 1 0.0761800 zx = Equation 9.57 () () () 2 22 1 1 1 1 xyz λλλ + += Substituting 2 1 0.841842 0.0761800 1 xx x ++ = ( )( ) or 0.763715 then 40.2 λθ = W ( ) ( ) 0.642927 then 50.0 yy = W ( ) ( ) 0.0581798 then 86.7 zz = W Substituting 2 K 4 2 2 2 1 0.698281 0 12 4 24 z ta = 4 2 2 0.698281 0 0 ta = or ( ) ( ) 2 2 1.260837 =− and ( ) ( ) 2 2 0.806278 = Then 2 2 1.260837 0.806278 1 x +− + = ( ) or 0.555573 then 56.2 = W ( ) ( ) 0.700487 then 134.5 = −= °
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This note was uploaded on 07/11/2011 for the course EGM 2511 taught by Professor Jenkins during the Spring '08 term at University of Florida.

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