Ncert Exercises & Solutions

A helicopter of mass $2000$ kg rises with a vertical acceleration of $15~\mathrm{m/s^2}$ . The total mass of the crew and passengers is $500$ kg. The magnitude of the force on the floor of the helicopter by the crew and passengers is: (Take $g=10~\mathrm{m/s^2}$ )
1. $12500~\mathrm{N}$
2. $1250~\mathrm{N}$
3. $50000~\mathrm{N}$
4. $5000~\mathrm{N}$

Hint: Recall pseudo force.

Step 1: Apply pseudo force and find force on the floor of the helicopter by the crew and passengers.

Given, mass of helicopter

(m_{1})

$(m_{1})$ = 2000 kg
Mass of the crew and passengers

(m_{2})

$(m_{2})$ = 500 kg
Acceleration in the vertical direction

a = 15 m / s^{2} (↑) and g = 10 m / s^{2} (↓)

$a = 15 m / s^{2} (↑) and g = 10 m / s^{2} (↓)$

(a) Force on the floor of the helicopter by the crew and passengers

\begin{array}{l} m_{2} (g + a) = 500 (10 + 15) N \\ 500 \times 25 N = 12500 N \end{array}

$\begin{array}{l} m_{2} (g + a) = 500 (10 + 15) N \\ 500 \times 25 N = 12500 N \end{array}$

Step 2: Find action of the helicopter on the surrounding air.

(b) Action of the rotor of the helicopter on the surrounding air

= (m_{1} + m_{2}) (g + a)

$= (m_{1} + m_{2}) (g + a)$
= (2000 + 500) X (10 + 15) = 2500 x 25
= 62500N (downward)

Step 3: Find force on the helicopter due to the surrounding air.

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