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When standing in a stationary elevator the force due to gravity and the normal force cancel each other out hence the net force is 0. When the elevator starts accelerating downwards there's a net force in the downward direction meaning that the body in the elevator now experiences three forces, the force due to gravity,the force that is causing the acceleration of the whole elevator and the normal force, accordingly the downward force is now of a greater magnitude than that of the normal force the body experiences. Then my question is, why does the normal force need to reduce if the downward force is still greater than the normal force even if it doesn't decrease. The body feels the force of gravity and the force which causes the elevator to accelerate these two are greater than the normal force and therefore the elevator accelerates in the downwards direction, why should the normal force get smaller?

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    $\begingroup$ You're mentally putting the acceleration and force on the same side of the equal sign when they are on different sides in F=ma: See physics.stackexchange.com/questions/632735/… The other way to think is if the elevator is accelerating downards, there is a net force downward, meaning one of two things needs to have changed from the case where normal force and force of gravity cancel to zero: Either gravity has increased or the normal force is reduced. Presumably, gravity has not increased. $\endgroup$
    – DKNguyen
    Feb 27 at 19:17
  • $\begingroup$ The "problem" would seem to be that "The elevator now experiences three forces…" Sorry; that's not a problem; it's simply a mistake. $\endgroup$ Mar 1 at 0:25

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When the elevator starts accelerating downwards there's a net force in the downward direction meaning that the body in the elevator now experiences three forces, the force due to gravity,the force that is causing the acceleration of the whole elevator and the normal force

This is not correct. There are only two forces acting on the body in the elevator. One is the downward pointing force of gravity, and the other is the upward pointing normal force. The force of gravity is constant, but the normal force can vary. There is not a third force involved. The net force is just the sum of the two forces, not a separate force on its own.

So if the body is not accelerating then the net force is 0 so the two forces are equal. If the body is accelerating downward then the net force is downward so the upward pointing normal force must be reduced so that the magnitude of the downward pointing gravitational force can exceed it.

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    $\begingroup$ I think I understand it. If the net force was zero the body would slam on the incoming accelerating elevator's ceiling. If the body is accelerating downwards it means that the body has a net force downwards, hence the normal force mist be less, so the person is not pushing on the floor as hard because it's constantly acceleraring in falling motion and therefore not pushing as hard, if It was it wouldn't accelerate right $\endgroup$
    – Peter
    Feb 27 at 19:20
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The body feels the force of gravity and the force which causes the elevator to accelerate

The body does not feel the net force, it only feels the normal force. This is sometimes called apparent weight. In a free fall, the body does not feel any weight.

Common misconception is that normal force equals weight. This is not true, and the normal force is usually determined from some assumptions. For example, for a body sliding on a ramp, the assumption is that the body does not move in the direction perpendicular to the ramp surface. This is one simple example in which normal force does not equal weight, even when the body is at rest.


To understand the accelerating elevator, you should start from the net force on the person. If an elevator accelerates downwards at $a_\star$, an observer on the ground (inertial reference frame) would see the person standing on the elevator floor accelerate downwards at the same acceleration. The net force on the person is then

$$\vec{F}_\text{net} = m \vec{a}_\star = m \vec{g} + \vec{n}$$

from which it follows

$$\vec{n} = m (\vec{a}_\star - \vec{g})$$

If downward acceleration is $|\vec{a}_\star| < g$ then the normal force is negative, i.e. it points upwards. But if downward acceleration is $|\vec{a}_\star| > g$, then the person in the elevator ends up on the ceiling and the normal force is positive, i.e. it points downwards.

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The force that the elevator acceleration excerts on you is in opposite direction of the direction of acceleration. If the lift accelerates upwards you feel a bigger force downwards and if it accelerates down the force on you will diminish. If it accelerates down fast hard enough you can land up on the ceiling. The normal force reacts accordingly.

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It looks like you're imagining some kind of mechanism that actively accelerates the elevator downwards. But that's not what happens.

enter image description here
(image source)

An elevator essentially1 hangs from a (counterbalanced) cable that's on a pulley. When the elevator needs to start moving down, the pulley is allowed to start rolling in a controlled way, which briefly reduces the tension (upward force) in the cable, letting gravity win. So it's gravity (diminished by the cable tension) that makes the elevator (and the passenger) accelerate downwards.


1 In reality, elevator designs may be quite a bit more complicated. There are also other kinds of elevators (e.g. hydraulic elevators).


When the elevator is standing still, you experience the normal force because you are being accelerated by gravity (meaning, and I'll come back to this later, that gravity is trying to increase your current instantaneous speed), but your surroundings (the elevator car) are fixed, so you sort of "run into" the floor. The normal force is essentially just the molecules of the floor not letting you pass through them, counteracting gravity.

When the elevator is accelerating, you still "run into" the floor, but since the floor itself is accelerating (at a smaller rate than 1g), you don't push into it as much, so the reaction force from the floor is lessened. So in your local (non-inertial) frame of reference, it feels like the strength of gravity has dropped (remember, if you were free falling, you'd feel weightless).

Now, this lasts only a short while, because the elevator soon starts moving at uniform speed, and the tension in the cable exactly counteracts gravity once again. So, now you're moving down, but there's no net force on the elevator. Remember, in absence of a net force, things want to continue moving uniformly in a straight line. Both you and the floor are moving down, but neither experiences a net acceleration, meaning that the floor fully counteracts gravity's pull on you once again - so you feel your usual weight. Since the elevator is now moving uniformly, it's an inertial frame of reference, so the force balance situation is the same as if the elevator was standing still! Gravity is trying to increase your current instantaneous speed - it doesn't matter if that speed is 0 m/s or 20 m/s or 1000 km/h; in fact, the exact number you put on your speed depends on your choice of reference frame. Gravity makes you "run into the floor" just the same, and the floor then reacts in the same way.

When you reach the bottom, the cable tension has to be greater in strength than gravity, because a net upward acceleration is required just to stop you and the elevator car. The floor is now trying to stop, and you aren't (not on your own), so you push into it harder, making the reaction force on you greater. In your local reference frame, it briefly feels as if the gravity has increased in strength.

In the other direction, the sequence of events is reversed; for the elevator to start moving up, the cable tension has to briefly overcome gravity, then it has to counteract it (net force of zero for the elevator car + passenger system) while the elevator is traveling up, and finally, it has to briefly let gravity win in order for the elevator to stop.

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The body feels the force of gravity and the force which causes the elevator to accelerate

There is a false assumption here that the person experiences the downward force on the elevator. This would happen if the elevator accelerates faster than gravity, which might cause the person to slam into the ceiling and then be pushed down by the ceiling, hence experiencing that force.

But that would still just be the normal force, which would have flipped directions, hence reducing below 0 and becoming negative. And somewhere before that, the lift would accelerate at the rate of gravity, the floor would fall away from the person, and they would experience free fall with a normal force of 0.

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