A helicopter of mass 1000 kg rises with a vertical acceleration of 15 m s -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 surrounding air, C. force on the helicopter due to the surrounding air.

Question

Philoid · Accepted Answer

Given:

Mass of helicopter, M = 1000 Kg

Mass of the crew, m_p = 300 Kg

Total mass of helicopter, m_h = 1300 Kg

Vertical acceleration, a = 15ms^-2

(a) Using Newton’s second law of motion, we can write the normal reaction, R as,

R = m_pg ₊ m_pa

R = m_p(g + a) …(1)

By putting values in equation 1, we get,

R = 300 Kg (10 ms^-2 + 15 ms^-2)

R = 7500 N

This reaction force from the floor works in upward direction. We can conclude the force on

by floor on crew is 7500 N. (From Newton’s third law).

(b) Using Newton’s third law of motion,

R’ = m_hg + m_ha

→ R’ = m_h(g + a)

→ R’ = 1300 Kg (10 + 15) ms^-2

→ R’ = 32500 N

Since surrounding air provides a Upward normal reaction of 32500N, using Newton’s 3^rd law of motion, we conclude that rotor also applies a force on 32500 N on surrounding air in downward direction.

(c) The force on the helicopter due to surrounding air is 32500 N and upwards.