ch05-p077 - 2 a B = F B / m B = 4.0 m/s 2 so the trivial...

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77. We begin by examining a slightly different problem: similar to this figure but without the string. The motivation is that if (without the string) block A is found to accelerate faster (or exactly as fast) as block B then (returning to the original problem) the tension in the string is trivially zero. In the absence of the string, a A = F A / m A = 3.0 m/s
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Unformatted text preview: 2 a B = F B / m B = 4.0 m/s 2 so the trivial case does not occur. We now (with the string) consider the net force on the system : Ma = F A + F B = 36 N. Since M = 10 kg (the total mass of the system) we obtain a = 3.6 m/s 2 . The two forces on block A are F A and T (in the same direction), so we have m A a = F A + T ¡ T = 2.4 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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