Analytical Mech Homework Solutions 90

# Analytical Mech Homework Solutions 90 - (1 b V x y = 1 2 x...

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b) ( ) ( ) ( ) ( ) 2 2 1 1 2 2 2 2 2 2 1 , 2 1 x y y x y x y α α α α V x + = − + + + (7.4.13) ( )( ) [ ] ( ) [ ] 3 3 2 2 1 1 x x V x x α α α α + = + Now x 0.5 α = at and 4 L 5 L also, each bracket term in the denominator equals 1 at , 4 L 5 L ( )( ) ( ) ( 1 0.5 0.5 1 0.5 V x α α α α α α α = + + ) 0.5 0.5 0.5 0.5 0 α α α = − + + + ( ) [ ] [ ] 3 3 2 2 1 y V y y y α α = + Again, the denominator in brackets equals 1 @ , 4 L 5 L So, ( ) 3 3 1 2 2 V y α α = ± + ± ± 3 2 3 3 3 3 0 2 2 2 2 α α = ± ± Thus ( ) ˆ ˆ , 0 V V V x y i j x y + = n of moment : at L , . 4 5 L 7.14 Conservatio um 4 p p p p m v m v m v α = + D G G G 45 o v’ φ v’
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