Question

A man mass of M is standing on the floor of a lift what his is Apparent weight.(a) when the lift moving in uniform velocity​

Answer 1

Answer:

To approach lift-man based problems, the problems are usually solved using FBD (or) Free Body Diagrams.

According to the question,

• Mass of the Man = M

At rest, when the lift is not moving, the weight will be equal to:

→ Weight = Mass × acceleration due to gravity

→ Weight = M × g = (Mg)

Now, it is given that, the lift is moving with uniform velocity. Whenever there is an acceleration associated with Mass, force is produced. Hence, let's calculate the acceleration to check whether any pseudo force is present.

$\text{Acceleration of a body} = \dfrac{ \Delta v}{ \Delta t}\\\\\\\implies \text{Acceleration of the body} = \dfrac{ 0 }{ \Delta t} = \boxed{0}$

This is because, in uniform velocity, the velocity doesn't change. Hence the acceleration produced is equal to zero.

Therefore, no pseudo force acts on the man travelling on the lift.

Hence his apparent weight would remain equal to his true weight which is Mg.