**Question.**

A helicopter of mass $1000 \mathrm{~kg}$ rises with a vertical acceleration of $15 \mathrm{~m} \mathrm{~s}^{-2}$. The crew and the passengers weigh $300 \mathrm{~kg}$. 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.

**solution:**

(a) Mass of the helicopter, $m_{\mathrm{h}}=1000 \mathrm{~kg}$

Mass of the crew and passengers, $m_{p}=300 \mathrm{~kg}$

Total mass of the system, $m=1300 \mathrm{~kg}$

Acceleration of the helicopter, $a=15 \mathrm{~m} / \mathrm{s}^{2}$

Using Newton’s second law of motion, the reaction force $R$, on the system by the floor can be calculated as:

$R-m_{\mathrm{p}} \mathrm{g}=m a$

$=m_{p}(g+a)$

$=300(10+15)=300 \times 25$

$=7500 \mathrm{~N}$

Since the helicopter is accelerating vertically upward, the reaction force will also be directed upward. Therefore, as per Newton’s third law of motion, the force on the floor by the crew and passengers is $7500 \mathrm{~N}$, directed downward.

(b) Using Newton’s second law of motion, the reaction force $R^{\prime}$, experienced by the helicopter can be calculated as:

$R^{\prime}-m g=m a$

$=m(g+a)$

$=1300(10+15)=1300 \times 25$

$=32500 \mathrm{~N}$

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 $32500 \mathrm{~N}$, directed downward.

(c) The force on the helicopter due to the surrounding air is $32500 \mathrm{~N}$, directed upward.