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A thin, uniform rod has length L and the linear density a (i.e. total mass M=al). A point mass m is placed at distance x from
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Solution so, mass of dx postion dm = M. dx force om n due to da mass will be. dF = G mdm dF = am. Adx 1. x2 x2 for x=1 toF = G Mm [ ut -L ] L N (Ltd.) - 6 Mim n2tln Answer CScanned with CamScanner

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