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Solução Do Halliday 8.ed.vol.1. Cap.5 - Ch05 - P085

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

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    December 2018
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G 85. (a) When Fnet = 3F − mg = 0 , we have F = b gc h 1 1 mg = 1400 kg 9.8 m / s2 = 4.6 × 103 N 3 3 for the force exerted by each bolt on the engine. (b) The force on each bolt now satisfies 3F – mg = ma, which yields F = 1 1 m ( g + a ) = (1400 kg ) ( 9.8 m/s 2 + 2.6 m/s 2 ) = 5.8 × 103 N. 3 3