Question

A young man of mass 60kg stands on the floor of a lift which is accelerating downwards at $1\,m/{s^2}$ then the reaction of the floor of the lift on the man is (take $g = 9.8\,m/{s^2}$)
A. $528\,N$
B. $540\,N$
C. $546\,N$
D. none of these.

Answer 1

We know that when a frame is accelerated there will be the presence of a pseudo force. The pseudo force will be acting in a direction opposite to the direction of acceleration. So, the effective weight of the men will be less than the actual weight. Normal reaction will be equal to the value of this effective weight.

Complete step by step answer:
It is given that a man is standing on the floor of a lift.
The lift is going downward with an acceleration,
$a = 1\,m/{s^2}$
The mass of the man is given as,
$m = 60\,kg$
We need to find the reaction of the floor of the lift on the man.
Here, the lift is an accelerating frame.
Whenever there is an acceleration of the frame, we should consider it as a non-inertial frame.
In a non-inertial frame, there will be the presence of a pseudo force.
The pseudo force will be acting in a direction opposite to the direction of acceleration.
The value of pseudo force can be found out as the product of the mass of the body and acceleration of the frame.
In equation form we can write it as
${F_p} = ma$
Where, m is the mass and a is the acceleration.
On substituting the values, we get the value of pseudo force as
$\Rightarrow {F_p} = 60\,kg \times 1\,m/{s^2}$
$\Rightarrow {F_p} = 60N$
Weight is given as the product of mass and acceleration due to gravity.
$W = mg$
Where $m$ is the mass and $g$ is the acceleration due to gravity.
We know that the weight of the body is acting downwards
The normal reaction force, $N$ offered by the lift will be acting directly upwards.
Let us balance all these forces acting in the vertical direction.
Both pseudo force and normal reaction force is acting upwards and they weight as acting downwards so we can write
$W - {F_P} = N$
$\Rightarrow mg - {F_P} = N$
Substituting all the values we get
$\Rightarrow 60 \times 9.8 - 60 = N$
On simplification,
$\therefore N = 528\,N$
This is the value of the reaction that the floor of the lift experts on the man.

Therefore, the correct answer is option A.

Note:
Remember that when the frame is accelerated due to the action of pseudo force in the opposite direction the effective weight of the body will be less. So, the normal reaction which balances this effective weight is also less. If the lift was stationary then the normal reaction would be equal to the actual weight, $m\,g$ of the man.