# How would you explain the weight of an object in a lift accelerating up with acceleration a?

Observer in the lift:-

He is unaware of his acceleration. He applies Newton's second law and got N = Mg. After that he used spring balance and found that the wait of the object is M(g+a). So to explain this inconsistency in result he brought pseudo force and figured out the frame must be accelerating with acceleration a.

Observer on the Ground

He knows the lift is accelerating up with acceleration a. So he applied Newton's second law and got N = M(g+a). So my question is how can we confirm that this result is valid and we dont need pseudo force to explain this(For the observer in the lift we came to know through spring balance the Pseudo force must be incorporated).

PS:

1. It would be great, If you can give some historical background on when and how we decided that we should incorporate a pseudo force to explain motion with respect to non-inertial frame?
2. When did we differentiate inertial and non-inertial frame in the history of Physics.

Maybe this is a bit beyond what you wanted, but you are implicitly evoking Einstein's equivalence principle, which tells us that gravitational accelerations and inertial accelerations are equivalent. It's only because this principle applies that we can add a gravitational and an inertial acceleration.

Gravtitational accelerations are measured relative to a freely falling body. Suppose you are in the lift and it isn't accelerating, i.e. $\vec{a} = 0$, so you aren't accelerating. If you drop some other object then it accelerates downwards with an acceleration of $-\vec{g}$ so if we take the freely falling body as a reference that means your acceleration is $\vec{g}$. That means there must be a force acting on you of $m\vec{g}$, and of course this is just the upwards force the (stationary) lift floor is exerting on the soles of your feet.

Now apply an acceleration $\vec{a}$ to the lift. The equivalence principle tells us we can simply add the gravitational and inertial accelerations, so relative to the freely falling body your acceleration is now $\vec{g} + \vec{a}$. The total force on you is therefore $m(\vec{g} + \vec{a})$.

This might seem a slightly weird way of addressing the problem, but actually it touches on a key feature of general relativity i.e. that a freely falling object is weightless and therefore has an acceleration of zero. So it makes sense to take the freely falling object as a reference point when calculating the acceleration, and therefore the applied force, for the object in the lift.

• I thinks my question is not clear. let me edit the question. – x2da Nov 30 '15 at 13:48

The 'weight' in these cases is usually taken to be the normal force between the floor of the lift and the object. Using N2L on the object $$Ma=N-Mg$$ Where up is taken to be positive and $N$ is the normal force. Rearranging this gives: $$N=M(a+g)$$ This is the normal force and therefore the 'weight'. Here we have introduced no pseudo forces and have worked in the ground frame.

Notice, the normal reaction exerted by the floor on the object is the weight of the object

$\bullet$ $\mathbf{\text{Normal reactive force}=N\uparrow }$ (upwards)

$\bullet$ $\mathbf{\text{Gravitational force}=Mg \downarrow }$ (downwards)

$\bullet$ $\mathbf{\text{Resultant force }=Ma\uparrow}$ (upwards)

Now, balancing all the forces acting on the object, one should get $$N=Mg+Ma$$ $$N=M(g+a)$$

Here, we don't need pseudo force to explain