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If an upward force was applied on an object which was at rest, the magnitude of the normal force decreases,
whereas, in an elevator which is undergoing an upward acceleration, the magnitude of the normal force increases rather than decreasing.
Why does the magnitude of the normal force increase rather than decrease in the elevator?
Good Question 😊 !
I like to visualise Normal force as a force whose magnitude depends on the intermolecular distances.
If the intermolecular distances increase, the repulsive force decreases and if the intermolecular distances are decreased then this repulsive force increases.
Knowing this, now you can apply this to the above two cases.
In this case, you are actually separating the two surfaces in contact by pulling the block up and not moving the floor and due to this the intermolecular distances (between the two) increases and hence the normal force (between the two surfaces) decreases.
In this case, initially the block was at rest but the floor accelerated upward which in a very short span get closer to the bottom surface of the block and the block get pressed to the floor (due to inertia) and thus the Normal force from the floor on that block increased and hence it also accelerates up with the floor quickly.
Hope it helps 🙂.
In a stationary elevator, the normal force from the elevator's floor must equal the person's weight (see bottom picture in your question) in order to keep them stationary. To accelerate the person upward, the normal force from the elevator's floor must be sufficient to support the person's weight AND accelerate the person upwards.
I think what's confusing you is that, the way they are drawn, the two diagrams placed side by side are misleading. The object of interest in the first diagram is the block, but in the second it's the person (the forces are acting on the person) in an elevator that's already not in contact with the ground, but is still accelerating up.
It's not the whole elevator that's being considered. The blue arrow in the second diagram is not the pulling force acting on the elevator box, it just represents the direction of the acceleration. The acceleration of the elevator is the same as that of the person (they move together).
In the first image the normal force comes from the ground (an external object) and acts on the block. In the second image, the normal force is exerted by the elevator floor (from the inside), and is acting on the person. (In other words, if we're just looking at the forces acting on the person, the elevator itself plays the role of the ground - but the "ground" is moving.)
The normal force has to be larger then the person's weight because the person is accelerating upwards (as seen from a stationary frame) - the elevator floor is working against the person's weight.
If, however, your object of interest is the elevator itself (together with the person inside), then as the elevator cable pulls everything up, the elevator starts rising, and the normal force exerted by the ground on the elevator box decreases, and eventually becomes zero - which is analogous to what happens with the block.
Note also that when the whole system is considered, the net force has to be larger compared to the net force on just the person, because it supplies the same acceleration, but the total mass is greater: $F_{net} = (m_{person} + m_{elevator})a$
Drawing a FBD will help you understand.
Block:
Assume the block is in static equilibrium, i.e. $\Sigma \vec F = \vec 0$. If you apply a force on the block that is less than the wight, i.e. $F_\mathrm{app}<mg$, then the block will not move. Since $\Sigma F_y=0$, then $F_\mathrm{app}+N=mg\quad\implies\quad N=mg-F_\mathrm{app}$.
Once $F_\mathrm{app}>mg$, then the normal force $N=0$ and the block will begin to accelerate.
The elevator moving upwards:
The elevator case is a little different, because in this case, the normal force is the force that will accelerate the person.
From a FBD and coordinate system where upwards is positive, we know that $ma_y=N-mg$, and thus $N = mg+ma_y$.
This difference arises due to the fact in the case of the block, the force you apply will result in a decrease of the intermolecular repulsive force (because the block will move slightly upwards, which decreases the distance between the atoms of the block and the surface on which it's resting, hence decreases the repulsive force). The intermolecular repulsive force in your macroscopic elevator/block case is what you call the normal force. Note that the intermolecular repulsive force increases when atoms get closer together (due to their charge).
In the elevator's case, the elevator is pushing the person up, so you can think of the atoms of the person's feet getting closer to the atoms of the elevator pushing the person act, hence the increase in the repulsive force.
User Ankit explained this well.
In all cases the object is considered to be the system.
Raising an object
AS drawn in the diagram there are three forces acting on the object and with up as positive the equation of motion is $F_{\rm pull} + F_{\rm N} -mg = ma$.
If you increase $F_{\rm pull} = 0$ from zero then the normal force adjusts itself by decreasing so that $F_{\rm pull}\uparrow + F_{\rm N}\downarrow -mg = m\,0$.
Eventually when $F_{\rm pull} = mg$ the normal force becomes zero and if $F_{\rm pull} \gt mg$ the normal force stays at zero and the object accelerates upwards.
So to increase the upward acceleration of the object $F_{\rm pull}$ has to be greater than $mg$ and increase.
Object in a lift
The situation is subtly different in that now there are only two forces acting on the object as there is now no $F_{\rm pull}$ and $F_{\rm N}$ is the upward force which can produce an acceleration of the object.
If the lift is moving with constant velocity, either upward or downwards or is at rest, then $F_{\rm N} -mg = m\,0 \Rightarrow F_{\rm N} = mg$ which is just the situation when $F_{\rm N} + F_{\rm pull} -mg = 0$ in the first case.
Since there is no external force $F_{\rm pull}$ the motion of the object is controlled by the net force $F_{\rm N} -mg \,(\,=ma)$ and the smallest value of $F_{\rm N}$ which is possible is zero when $a=-g$ ie the lift is in "free fall" which could mean that the lift is actually moving upwards and slowing down.
It is the acceleration of the lift which is important not the direction of travel of the lift.
As the acceleration of the lift increases from $-g$, ie becoming less negative and eventually becoming positive, the normal force will increase from zero with its value being given by th equation of motion $F_{\rm N}-mg=m\,a$
This is due to the fact that when the elevator is accelerating up, it is analogous to the frame of reference moving up. When the frame of reference accelerates up, you have to apply pseudo forces in the opposite direction of the motion of the frame of reference.
So, if the elevator is moving up with an acceleration of 'a', then you have to take the force applied due to the motion in downward direction and equal to ma.
If you still don't understand it, then take it like this: if you are in an accelerating car, which is accelerating towards right, you would feel yourself being pushed in the left direction.
Hope it helps!
You are looking at different things, that's why you see differences. But their behaviour is actually similar:
If the elevator was touching the floor, and you applied an upward force to it (eg by running its motor on low current), the normal force from the ground to the elevator would decrease just like your first picture.
If your first picture's object was hollow and there was a ball inside, and you started lifting the object, its normal force to the ball would have to increase, just like in the elevator example, in order to accelerate the ball.
You are comparing two different things:
In the first picture, the block is like the person, and the floor is like the elevator. You are measuring the normal force when you pull the person up.
In the second picture, you are measuring the normal force when you pull the elevator up.
