271_Problem CHAPTER 9

271_Problem CHAPTER 9 - PROBLEM 9.183 CONTINUED Solving...

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PROBLEM 9.183 CONTINUED Solving numerically 32 11 1.180481 10 lb ft s or 1.180 10 lb ft s KK −− = × ⋅⋅ = × W 22 10.72017 10 lb ft s or 10.72 = × = × W 33 11.70365 or 11.70 = × = × W ( b ) From Equations 9.54(a) and 9.54(b) () ( ) ( ) ( ) 0 xx x y y z x z IK I I λλλ −−−= ( ) ( ) 1 0 xy x y y yz z II K I λλ λ −+ = 1 : K Substitute 1 K and solve for to get ( ) , x xy and ( ) 3 y . ( ) 3 1 1 1 9.8821 1.180481 0.48776 2.6951 10 0 xy z  × =  ( ) 3 1 1 1 0.48776 11.5344 1.180481 1.18391 10 0 z × = or ( ) ( ) ( ) 1 1 1 17.83996 5.52546 0 z −= ( ) ( ) 1 1 1 1 0.0471 0.11434 0 z = Then ( ) ( ) 1 1 3.1549 zx = and ( ) ( ) 1 1 0.40769 yx = Equation 9.57: () () () 2 1 1 1 1 xyz + += Substituting 2 1 0.40769 3.1549 1 x ++ = ( )( ) or 0.29989 then 72.5 λθ = W ( ) ( ) and 0.122262 then 83.0 yy = W ( ) ( ) 0.94612 then 18.89 zz = W 2 : K Substitute
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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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