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notes001 (19)

# notes001 (19) - & j T 1 j = 5 g or j T 1...

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August 24, 2009 NAME____________________________________________ Solution: The tension vector for the wire on the left is T 1 = j T 1 j cos (128 ) i + j T 1 j sin (128 ) j and the tension in the wire on the right is T 2 = j T 2 j cos (40 ) i + j T 2 j sin (40 ) j . The weight vector is W = 5 g j where g = 9 : 8 m = s 2 . Since T 1 + T 2 + W = 0 , we obtain j T 1 j cos (128 ) + j T 2 j cos (40 ) = 0 and j T 1 j sin (128 ) + j T 2 j sin (40 ) = 5 g . j T 2 j gives j T 2 j = cos (128 ) cos (40 ) j T 1 j . 1

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This gives j T 1 j sin (128 ) sin (40 ) cos (128 ) cos (40
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Unformatted text preview: & ) j T 1 j = 5 g or j T 1 j (sin (128 & ) & tan (40 & ) cos (128 & )) = 5 g or j T 1 j = 5 g sin (128 & ) & tan (40 & ) cos (128 & ) ± 37 : 56 N . Therefore j T 2 j = & cos (128 & ) cos (40 & ) j T 1 j ± 30 : 19 N . In conclusion, the magnitude of the tension in the left wire is about 37 : 56 N and the magnitude of the tension in the right wire is about 30 : 19 N . 2...
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notes001 (19) - & j T 1 j = 5 g or j T 1...

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