A helicopter of mass 1000 kg rises with a vertical acceleration of `15 m s^(-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.

Explanation

(a) ‘Free body’ : crew and passengers
Force on the system by the floor = F upwards, weight of system = mg downwards, ` therefore F – mg = ma`
` F - 300 xx 10 = 300 xx 15`
` F = 7.5 xx 10^3 N ` upward
By the Third Law, force on the floor by the crew and passengers = `7.5 xx 10^3 N` downwards.
(b) ‘Free body’ : helicopter plus the crew and passengers
Force by air on the system = R upwards, weight of system = mg downwards
` therefore R – mg = ma`
`R - 1300 xx 10 = 1300 xx 15`
` R = 3.25 xx 10^4 N ` upwards
By the Third Law, force (action) on the air by the helicopter = `3.25 xx 10^4 N ` downwards. (c ) `3.25 xx 10^4 N ` upwards