Question

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.

Answer 1

Given: The mass of the helicopter is 1000 kg, the rising acceleration in vertical direction is 15 m/ s 2 and the weight of crew and passengers are 300 kg.

(a)

The total mass of the system is given as,

m t = m h + m p (1)

Where, the total mass is m t , the mass of helicopter is m h and the mass of crew and passengers is m p .

By substituting the given values in equation (1), we get

m t =1000+300 =1300 kg

Let Rbe the net reaction force acting on the system.

From second law of motion, the net reaction force on the system by the floor is given as,

R− m p g= m p a R= m p ( g+a ) (2)

Where, the net reaction is R, the acceleration due to gravity is g, the acceleration of helicopter is a.

By substituting the given values in equation (2), we get

R=300( 10+15 ) =300×25 =7500 N

Since, the helicopter is moving upward, the direction of reaction force is also upward.

From Newton’s third law of motion, the reaction force exerted by the crew and passengers on helicopter will act downward.

Thus force on the floor by the crew and passengers is 32500 N in the downward direction.

(b)

The net reaction force experienced by the helicopter is given as,

R ′ = m t ( g+a )(3)

By substituting the values in equation (3), we get

R=1300( 10+15 ) =1300×25 =32500 N

The reaction force experienced by the helicopter from the surrounding air will act in the upward direction.

Thus, action of the rotor of the helicopter on the surrounding air is 32500 N in the downward direction.

(c)

As calculated in part (b), the force on the helicopter due to the surrounding air is 32500 N and it acts in vertically upward direction.