Analytical Mech Homework Solutions 15

Analytical Mech Homework Solutions 15 - (b T = T= 121 1 mg...

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(b) 2 2 2 2 2 11 1 1 22 2 2 . 2 2 t mg m g v m cD  == = =   Tm () ( )() ( ) 2 2 2 .145 9.8 87 2 .22 2 .0366 m kg s kg m ==   TJ 3 23 tanh t t Fdx cv dx c v dt c v dt τ = − =− ∫∫ 3 2 1 tanh tanh 2 t tt cv d t ττ  +    3 2 1 tanh ln cosh 2 t cv + Now 2 tanh 1 t for t ± Meanwhile tanh ln cosh xv d t v d t v = ln cosh t tx v = 1250 .3048 381 m x ft m ft 1 2 1 2 2 2 2 .145 9.8 34.72 .22 .0732 t m kg mg m s v kg cs m = 1 2 1 2 2 2 2 .145 3.543 .22 .0732 9.8 kg m s kg m cg ms = ( ) ( ) 3.81 .22 .0732 34.72 3.543 .5 454 34.72 3.54 Fdx J + = 541 87 454 VT J J J −= = 2.10 For 0: 1 ≤≤ F vt m = D , 2 1 2 F x t m = D For 11 2: tt t 1 F m = D D , 2 1 2 F x t m = D D , 1 = D 4
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This note was uploaded on 01/06/2012 for the course PHYSICS 4360C taught by Professor Johnanderson during the Fall '11 term at UNF.

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