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 = = to0 gives ( 29 ( 29 , 1 10 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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