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Solução Do Halliday 8.ed.vol.1. Cap.5 - Ch05 - P077

A solução do Halliday da 8ª edição volume 1. do Cap.5

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    December 2018
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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, aA = FA /mA = 3.0 m/s2 aB = FB /mB = 4.0 m/s2 so the trivial case does not occur. We now (with the string) consider the net force on the system: Ma = FA + FB = 36 N. Since M = 10 kg (the total mass of the system) we obtain a = 3.6 m/s2. The two forces on block A are FA and T (in the same direction), so we have mA a = FA + T Ÿ T = 2.4 N.