So I've been learning about the apparent weight of a person in an elevator accelerating upwards or downwards. I learnt that:

If it accelerates upwards with an acceleration a, Apparent weight = m (g+a)

If it accelerates downwards with an acceleration a, Apparent weight = m (g-a)

If it accelerates downwards with an acceleration g, Apparent weight = 0 , i.e, the person undergoes freefall.

I then asked my teacher what would happen if the elevator accelerated downwards with a>g. She told me that normal force would be negative, and would pull the person down.

However, I thought that normal force exerted by a floor can only act upwards, and so a negative normal force can't exist, and thus the normal force would be equal to 0 N.

Could someone please clarify as to what happens in this scenario?

  • 2
    $\begingroup$ That would be the normal force exerted by the ceiling (once you've hit it) $\endgroup$
    – doetoe
    Commented Oct 17, 2021 at 15:33
  • $\begingroup$ @doetoe that seems to make sense intuitively, but is there anyway to show what exactly happens mathematically? Thanks! $\endgroup$
    – anon
    Commented Oct 17, 2021 at 15:34

1 Answer 1


If you were initially standing in the elevator at rest, once the elevator started accelerating the floor would accelerate away from you and the ceiling would accelerate towards you. During this time, you would still be accelerating downwards with magnitude $g$ (relative to an external inertial observer). Once you hit the ceiling then you will be accelerating with the elevator. The force the elevator ceiling exerts on you will have a magnitude of $m(a-g)$.

If you were somehow attached to the floor of the elevator, then it's a similar thing, except you will just automatically accelerate with the elevator. The force constraining you to the floor would still have a magnitude of $m(a-g)$

  • $\begingroup$ Ah, that makes sense! So if we were to calculate apparent weight with a weighing scale attached to the elevator's ceiling, would it show a value of m(a-g) ? $\endgroup$
    – anon
    Commented Oct 17, 2021 at 15:44
  • $\begingroup$ @anon You got it! That is the downward force needed so that you have an acceleration of $a$ in order to move along with the elevator. $\endgroup$ Commented Oct 17, 2021 at 15:46
  • 1
    $\begingroup$ Wow, so it's the same as the accelerating downwards with a<g scenario, but it's upside down! Thanks a lot! $\endgroup$
    – anon
    Commented Oct 17, 2021 at 15:47

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