$A$ helicopter of mass $2000 \,kg$ rises with a vertical acceleration of $15 \,m s^{-2}$. The total mass of the crew and passengers is $500 \,kg$. Calculate the magnitude and direction of the following (take $g = 10 \,m s^{-2}$):
$(a)$ 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 due to the surrounding air.

(A) Given:
Mass of helicopter $m_{1} = 2000 \,kg$
Mass of crew and passengers $m_{2} = 500 \,kg$
Acceleration $a = 15 \,m s^{-2}$
Acceleration due to gravity $g = 10 \,m s^{-2}$
$(a)$ The force on the floor of the helicopter by the crew and passengers is the apparent weight $F_{1} = m_{2}(g + a)$.
$F_{1} = 500 \times (10 + 15) = 500 \times 25 = 12500 \,N$ (downward).
$(b)$ The action of the rotor on the surrounding air is equal to the total force required to lift the entire system (helicopter + crew + passengers) with acceleration $a$.
$F_{2} = (m_{1} + m_{2})(g + a) = (2000 + 500) \times (10 + 15) = 2500 \times 25 = 62500 \,N$ (downward).
$(c)$ By Newton's third law,the force on the helicopter due to the surrounding air is the reaction to the action force $F_{2}$.
$F_{3} = 62500 \,N$ (upward).