A helicopter of mass 1000kg rises with a vertical acceleration of 15ms-2. The crew and the passengers weigh 300kg. 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.
Given,
Mass of the helicopter, mh = 1000 kg
Mass of the crew and passengers, mp= 300 kg
Total mass of the system, m = 1300 kg
Acceleration of the helicopter, a = 15 m/s2
Using Newton’s second law of motion, the reaction force R,
R – mpg = ma
= mp(g + a)
= 300 (10 + 15)
= 300 × 25
= 7500 N
The reaction force will be directed upwards, the helicopter is accelerating vertically upwards.
According to Newton’s third law of motion, the force on the oor by the crew and passengers = 7500 N, directed downward.
(b) Using Newton’s second law of motion, the reaction force R’ experienced by the helicopter can be calculated as,
R' - mg = ma
= m(g + a)
= 1300 (10 + 15)
= 1300 × 25
= 32500 N
The reaction force experienced by the helicopter from the surrounding air is acting upward. Hence, as per Newton’s third law of motion, the action of the rotor on the surrounding air will be 32500 N, directed downward.
(c) The force on the helicopter due to the surrounding air is 32500 N, directed upwards
(a) 'Free body': crew and passengers
Force on the system bythe floor = F
F = 7.5 × 103N upward
By the Third Law, force on the floor by the crew and passengers = 7.5 × 103N downwards.
(b) 'Free body': helicopter plus the crew and passengers Force by air on the system : R
R- 1300 × 10 = 1300 × 15
R = 3.25 × 104N upwards
By the Third Law, force (action) on the air by the helicopter = 3.25 × 104N downwards.
(c) 3.25 × 104N upwards