Laws of Motion

Question : 27

Total: 40

A helicopter of mass 1000 kg rises with a vertical acceleration of 15ms–2. The crew and the passengers weigh 300 kg. Give the magnitude and direction of the
(a) force on the floor by the crew and passengers,
(b) action of the rotor of the helicopter on the surrounding air,
(c) force on the helicopter due to the surrounding air.

Solution:

Here, mass of the helicopter,
M=1000kg
Mass of the crew and the passengers,
m=300kg
Acceleration, a=15ms–2 (vertically upwards)
g=10ms–2
(a) Force on the floor by the crew and the passengers will be equal to their apparent weight. If the helicopter is rising up with an acceleration a,
then the apparent weight and hence the required force,
F=m(g+a)=300(10+15)
=7500N (vertically downwards)
(b) Action of the rotor of the helicopter on the surrounding air,
F=(M+m)(g+a)
=(1000+300)(10+15)=1300×25
=32500N (vertically downwards)
(c) Force on the helicopter due to surrounding air will be equal and opposite to the action of the rotor of the helicopter on the surrounding air (third law of motion). Therefore, the required force is given by
F=32500N (vertically upwards)

Go to Question:

Prev Question

Next Question