A helicopter of mass 1000 kg rises with a vertical acceleration of `15 ms^(-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 surroundings air,
(c) force on the helicopter due to the surroundings air.

Explanation

Here ,mass of helicopter , `m_1` = 1000 kg .
Mass of the crew and passengers , `m_2` = 300 kg upward acceleration , a = 15` ms^(-2)` and g = 10`ms^(-2)`
(a) Force on the floor of helicopter by the crew and passengers = apparent weight of crew and passengers = `m_2(g + a) ` = 300 (10 +15) N = 7500 N
(b) Action of rotor of helicopter on surrounding air is obviously vertically downwards , because helicopter rises on account of reaction to this force. Thus , force of action
F = `(m_1 + m_2)`(g + a) = ( 1000 + 300) (10 +15) = `1300 xx 25 ` = 32500 N
(c ) Force on the heliopter due to surrounding air is the reaction. As action and reaction are equal and opposite, therefore force of reaction , F = 32500 N , verically upwards.