problem05_15

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

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5.15: a) The tension is related to the masses and accelerations by . 2 2 2 1 1 1 a m g m T a m g m T = - = - b) For the bricks accelerating upward, let a a a = - = 2 1 (the counterweight will accelerate down). Then, subtracting the two equations to eliminate the tension gives . s m 96 . 2 kg 0 . 15 kg 0 . 28 kg 0 . 15 kg 0 . 28 s m 80 . 9 or , ) ( ) ( 2 2 1 2 1 2 2 1 1 2 = + - = + - = + = - m m m m g a a m m g m m c) The result of part (b) may be substituted into either of the above expressions to find the tension N. 191 = T As an alternative, the expressions may be manipulated to eliminate a algebraically by multiplying the first by 2 m and the second by 1 m and adding
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Unformatted text preview: (with 1 2 a a-= ) to give N. 191 kg) . 28 kg . 15 ( ) s m (9.80 kg) (28.0 kg) . 15 ( 2 2 or , 2 ) ( 2 2 1 2 1 2 1 2 1 = + = + = =-+ m m g m m T g m m m m T In terms of the weights, the tension is . 2 2 2 1 1 2 2 1 2 1 m m m w m m m w T + = + = If, as in this case, 2 1 2 1 2 2 , m m m m m + and , 2 2 1 1 m m m + < so the tension is greater than 1 w and less than ; 2 w this must be the case, since the load of bricks rises and the counterweight drops....
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This document was uploaded on 02/04/2008.

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