problem12_83

University Physics with Modern Physics with Mastering Physics (11th Edition)

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12.83: Divide the rod into differential masses dm at position l , measured from the right end of the rod. Then, ( 29 L M dl dm = , and the contribution ( 29 . is piece each from 2 L x l GmMdl dF dF x x + - = Integrating from L l l = = to 0 gives ( 29 ( 29 , 1 1 0 2 L x x GmM x L x L GmM x l dl L GmM F L + - = - + = + - = with the negative sign indicating a force to the left. The magnitude is
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This document was uploaded on 02/05/2008.

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