Reaction in a moving lift || Apparent weight - Application of Newton's third law

In 1687, Sir Isaac Newton mentioned three important laws about the motion of dynamical particles. These law are known as Newton's Laws of Motion.

First law : Every body persists in its state of rest or of uniform velocity unless the body is compelled to change that state by a net external force.

Secound law :  The rate of change of momentum of a body is directly proportional to the impressed force and the change takes place in the direction in which the force acts.

Third law : To every action there is an equal and opposite reaction.

Here is discussion on Newton's third law. Now you will know about the reaction in a moving lift. There have some Newton's third law application in case of moving lift.

Reaction in a moving lift ♢ Apparent weight

When a man with mass "m"  is standing on the floor of a lift, a downward force "mg" acts on the floor due to his body weight. At this same time the floor applies a upward reaction force or normal force  "n" on the man, due to this force the man feels his weight.

According to Newton's 3rd law of motion, "n=mg". Again if the normal force (n) is absent, then the man feels himself weightless. As an example,  if we jump from a slightest height, we feel ourself weightless untill we can't touch the ground, because then the upward reaction force or normal force "n" is absent.

when a lift moves on the upward direction
Various weights of the same object( which is inside the lift ) are felt due to the variation of this upward reaction force. This is called Apperent weight.

Suppose, a man with mass "m" is standing on the floor of a lift.
The downward attractive force acts on the man = Actual weight = mg ;
Now the upward normal force acts = Apperent weight= "n"
Then, the resultant upward force,
F = n-mg
Here, F = mα, where "α"= acceleration of the lift.

When the lift is moving upward direction with an acceleration "α"

Here upward acceleration = α
So, n - mg = mα
Or, n = mα + mg = m (α + g)

Here, the value of apparent weight (n) is greater than "mg". So,  the man feels himself relatively heavier.

When the lift is moving downward direction with an acceleration "α"

Here, downward acceleration = "α"  and the upward acceleration = "-α"
So, n - mg = - mα
Or,  n = mg - mα = m (g - a)

Here the value of apparent weight is smaller than "mg". So, the man feels himself relatively light.

When the lift is falling freely (α = g)

When the lift is falling freely due to Gravitational force, then the downward acceleration, α = g ;
So, n - mg = - mα
Or, n = mg - mg = m (g - g) = 0
At this case, the normal force (n) does not act on this man and the man feels himself weightless.