A 4800 kg helicopter accelerates upward at 3.65 m/s2 The acceleration of gravity is 9.8 m/s2 What lift force is exerted by the air on the propellers? Answer in units of N
The apparent weight of the helicopter under the influence of an additional acceleration is:
F=m(g+a), where g=9.8 m/s^2 and a=3.65 m/s^2 and m= 4800 kg.
The force that the air exerts on the propellers has to equal the apparent weight. If there was no acceleration, the force would equal the weight of the helicopter in order for it to lift off the ground and continue lifting at a constant velocity.
F=(4800 kg)(9.8 m/s^2+3.65 m/s^2)= 64,560 newtons of force