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Q5.27(a):
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,

Solution :

Given Variables & Initial Setup

To determine the force exerted by the crew and passengers on the floor of the accelerating helicopter, we identify the following parameters:

  • Mass of crew and passengers ($m$): $300\text{ kg}$
  • Vertical acceleration ($a$): $15\text{ m/s}^2$ (upwards)
  • Acceleration due to gravity ($g$): $9.8\text{ m/s}^2$

Step 1: Identifying Forces on the Crew and Passengers

The crew and passengers represent a system of mass $m$ experiencing two primary forces within the non-inertial frame of the helicopter:

  1. Weight ($W = mg$): Acting vertically downwards.
  2. Normal Reaction Force ($R$): The upward force exerted by the helicopter floor on the passengers.

Per Newton’s Second Law of Motion, the net force $F_{net}$ acting on the system is equal to the product of mass and acceleration: $F_{net} = ma$ [By Newton’s Second Law, $\sum F = ma$].

Step 2: Formulating the Equation of Motion

Given the system is accelerating upwards at $15\text{ m/s}^2$, the net force equation in the vertical direction is:

$R - mg = ma$

Rearranging to solve for the reaction force $R$ exerted by the floor on the passengers:

$R = m(g + a)$

Step 3: Numerical Calculation

Substituting the given values into the derived formula:

$R = 300\text{ kg} \times (9.8\text{ m/s}^2 + 15\text{ m/s}^2)$

$R = 300\text{ kg} \times (24.8\text{ m/s}^2)$

$R = 7440\text{ N}$

Step 4: Determining the Force on the Floor

Per Newton’s Third Law of Motion, the force exerted by the crew and passengers on the floor is equal in magnitude and opposite in direction to the normal reaction force $R$ [By Newton’s Third Law: Action and Reaction pairs].

Therefore, the magnitude of the force on the floor is $7440\text{ N}$, and the direction is vertically downward.

Helicopter Floor R (Normal Force) mg (Weight) a = 15 m/s² (Upwards)

Final Solution: The magnitude of the force exerted by the crew and passengers on the floor is $7440\text{ N}$, directed vertically downwards.


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