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
