Answer
(a) The air exerts a lift force of 87,600 N upward on the helicopter rotors.
(b) The cable exerts a force of 11,400 N upward on the frame.
(c) The cable exerts a force of 11,400 N downward on the helicopter.
Work Step by Step
(a) The lift force $F_L$ provides the force to accelerate both the helicopter and the frame at a rate of $0.80 ~m/s^2$:
$\sum F = ma$
$F_L -mg = ma$
$F_L = m(a + g)$
$F_L = (7180 ~kg + 1080 ~kg)(0.80 ~m/s^2 + 9.80 ~m/s^2)$
$F_L = 87,600 ~N$
The air exerts a lift force of 87,600 N upward on the helicopter rotors.
(b) The tension $F_T$ provides the force to accelerate the frame at a rate of $0.80 ~m/s^2$:
$\sum F = ma$
$F_T -mg = ma$
$F_T = m(a + g)$
$F_T = (1080 ~kg)(0.80 ~m/s^2 + 9.80 ~m/s^2)$
$F_T = 11,400 ~N$
The cable exerts a force of 11,400 N upward on the frame.
(c) The tension in the cable exerts the same magnitude of force on the helicopter as it does on the frame. Therefore, the cable exerts a force of 11,400 N downward on the helicopter.