A helicopter of mass 2000 kg rises with a vertical acceleration of `15ms^-2`. The total mass of the crew and passengers is 500 kg. Give the magnitude and direction of the (g`=10ms^-2`)
(a) Force on the floor of the helicopter by the crew and passengers.
(b) action of the rotor of the helicopter on the surrounding air.
(c ) force on the helicopter dur to the surrounding 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.