ch05-p091 - the unknown mass in the middle must be m 4 =...

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(b) The top cord is supporting a total mass of m 1 + m 2 = (3.5 kg + 4.5 kg) = 8.0 kg against gravity, so the tension there is T 1 = ( m 1 + m 2 ) g = (8.0 kg)(9.8 m/s 2 ) = 78 N. (c) In the second picture, the lowest cord supports a mass of m 5 = 5.5 kg against gravity and consequently has a tension of T 5 = (5.5 kg)(9.8 m/s 2 ) = 54 N. (d) The top cord, we are told, has tension T 3 =199 N which supports a total of (199 N)/(9.80 m/s 2 ) = 20.3 kg, 10.3 kg of which is already accounted for in the figure. Thus,
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Unformatted text preview: the unknown mass in the middle must be m 4 = 20.3 kg – 10.3 kg = 10.0 kg, and the tension in the cord above it must be enough to support m 4 + m 5 = (10.0 kg + 5.50 kg) = 15.5 kg, so T 4 = (15.5 kg)(9.80 m/s 2 ) = 152 N. Another way to analyze this is to examine the forces on m 3 ; one of the downward forces on it is T 4 . 91. (a) The bottom cord is only supporting m 2 = 4.5 kg against gravity, so its tension is T 2 = m 2 g = (4.5 kg)(9.8 m/s 2 ) = 44 N....
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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