You weigh less on the way down

I don't want to go to science world to find out because it would be a long round-trip.

I understand that acceleration/deceleration would effect the weight and I can also imagine that someone at terminal velocity would weigh nothing but I can't get my understanding in terms of forces and how that would effect the weight. For example, what would a formula relating mass, speed (in direction of gravity), gravitational force and weight look like?

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    $\begingroup$ The slogan You weight less on your way down suggests that your weight $W$ is somehow correlated with your velocity $v$. This is of course wrong, and reverse about 400 years of Galilean teaching. As everybody knows, your weight $W=m(g+a)$ is correlated with your acceleration $a=dv/dt$, not your velocity $v$. $\endgroup$
    – Qmechanic
    Commented Jan 10, 2013 at 14:17
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    $\begingroup$ So it would be more accurate for the scales to say "You weigh less as the elevator is accelerating downwards"? I guess part of my lack of understanding comes from acceleration due to the lift and acceleration due to gravity. $\endgroup$ Commented Jan 10, 2013 at 14:26
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    $\begingroup$ Yes, if the word downwards refers to the direction of acceleration, not the direction of travel. $\endgroup$
    – Qmechanic
    Commented Jan 10, 2013 at 14:31

5 Answers 5


As an alternative example, think about how it feels to drive a car.

  • When the car is standing still, you will sink into the seat and feel the seat pressing against you (= normal weight).
  • When you press the throttle and the car accelerates, you will be pressed further back into the seat and will in turn feel the seat pressing harder against you (= increased weight).
  • When the car has reached cruising speed and is not accelerating any longer, you will not get further pressed into the seat any longer (= normal weight again). There are still forces acting on the car (engine is driving it, wind resistance and friction are slowing it), but they are balanced and results in no acceleration.
  • When you press the brake and the car decelerates, you will be lifted out of the seat, which will in turn lessen it's pressure on you (= decreased weight).

In other words, your weight will only change when the car is accelerating/decelerating.

  • $\begingroup$ Thanks, that totally makes sense and is in line with my expectation. My confusion was caused by the scales implying that the velocity of the scales and myself affected my weight, rather than the acceleration, as Qmechanic says here. $\endgroup$ Commented Jan 10, 2013 at 16:11

"I understand that acceleration/deceleration would effect the weight and I can also imagine that someone at terminal velocity would weigh nothing"

It looks like you think that only in terminal velocity we do not weight, and that's wrong. We measure weight by the normal force applied by the ground on us. That means that there must be a normal force. In the elevator, it first accelerates down with an acceleration $a<g$. In terms of forces, you must accelerate at the same rate as the elevator so $ma_T=ma-mg$ the normal force must be smaller than just $mg$, and so you weigh less. In the terminal velocity of the elevator the total forces must be 0 as your speed is constant, so the normal must be again $mg$.

In free fall, the only force that acts on you if the gravity, and so you weigh nothing, in space for example, in the ISS you see them floating no because there's no gravity (gravity at the altitude of the ISS is almost the same as in the surface), but because gravity is the only force acting on the whole system, so they weigh 0.

  • $\begingroup$ I realise that with no falling I weigh some set amount, and that at terminal velocity I weigh nothing. What I'm not sure about is the inbetween. From my understanding of "In the terminal velocity of the elevator the total forces must be 0 as your speed is constant, so the normal must be again mg." you're saying that I'll weigh the same once the elevator is at constant speed as when it's stationary. This is what confused me when asking the question, as that doesn't seem right. If you could clarify what you mean I'd appreciate it. $\endgroup$ Commented Jan 10, 2013 at 14:15
  • $\begingroup$ When the elevator doesn't move, then the total force on you must be 0, there is gravity and normal, so the normal is equal to the gravity, this is your "base" weight (mg). When the elevator goes at constant speed, then there's no acceleration ($v=constant\Rightarrow dv/dt=0$), then the total force must be again 0, the forces are the same: gravity and normal, the gravity (mg) hasn't changed, so the normal must be the same, so you're weight is the same (I continue in the next comment) $\endgroup$ Commented Jan 10, 2013 at 14:35
  • $\begingroup$ When the elevator is accelerating, then gravity has some margin to act making you accelerate down too, but your acceleration won't be larger than the elevator's because it would mean you go through the floor. That means that, if you're constantly touching the floor, your acceleration is the elevator's acceleration: $ma_{elevator}=mg-N\Rightarrow N=mg-ma_{elevator}$. The normal in this case will be smaller, that means the force you make on the floor, and the force the floor makes on you is lower than the base force which is your weight $\endgroup$ Commented Jan 10, 2013 at 14:38

You could think of the weight recorded by the weighing machine as the force exerted by our mass on the weighing machine. I'm using mass to mean inertial mass and not weight.

You're probably aware that the force $F$ exerted on a mass $m$ is related to the acceleration $a$ as $$F = ma$$ In context of gravity, the acceleration is provided by the gravity itself, and the 'force exerted' is called the weight.

In free fall (and to a lesser extent in a downward moving elevator), both you and the weighing scale are falling toward earth with the same acceleration. Therefore there is no "extra" acceleration that you posses by which you can exert a force on the weighing scale and hence register a non-zero weight.

  • $\begingroup$ Ok, so by the elevator (and scales) moving downwards my acceleration (due to gravity) relative to the scales is reduced, therefore I weigh less. Have I got that right? $\endgroup$ Commented Jan 10, 2013 at 14:08
  • $\begingroup$ Yup. That's how it works. :) $\endgroup$
    – Kitchi
    Commented Jan 11, 2013 at 13:36
  • $\begingroup$ This is the correct answer. Weight is the force caused by gravity. Despite what lay persons may call it, the reading on a scale should not be called weight; it is the scale reading. $\endgroup$
    – DrSheldon
    Commented Sep 29, 2018 at 13:52

Your weight is the force you apply to the floor that supports you. If you stand on the ground and your mass is $m = 80$ kg, then the ground is feeling a force of $ m \times g = 80$ kg $\times$ $9.8$ m/s$^2= 784$ N.

If you're in an elevator that's accelerating downwards with $a = 1.0$ m/s$^2$, then the floor of the elevator feels a force of $m \times (g - a) = 80$ kg $\times$ $(9.8$ m/s$^2 - 1.0$ m/s$^2) = 704$ N. In other words, your weight on the elevator is as if your mass were $704$ N$ / 9.8$ m/s$^2 = 71.8$ kg instead of $80$ kg.

If the elevator accelerates downwards with $a = 9.8$ m/s$^2$ (that is $a = g$, it's in free fall), then the floor feels a force of $m \times (a - g) = 80$ kg $\times$ $0$ m/s$^2 = 0$ N, so you're weightless.


Assume you are in a lift, and you weigh yourself with a balance. Your weight will be the normal force between balance and your body. Let the normal force be $N$, the gravitational force be $m g$ (where $g$ is a constant) and the lift moves downward as positive,
Net force: $m g - N = m a \implies N = m \left(g - a \right)$ when it accelerates downward, $a$ (acceleration) will increase, which means $g - a$ will decrease where $m$ (mass) is constant, and $N$ will decrease, hence your weight.


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