Weight Change in a Moving Elevator

Date: 12/25/2003 at 18:26:22
From: navkiran
Subject: Weight in a moving elevator

Why does a person weigh less when coming down the elevator than going
up and more when going up than coming down?  I think it's because of
gravity, but what else is going on?

Date: 12/25/2003 at 21:18:24
From: Doctor Rick
Subject: Re: Weight in a moving elevator

Hi, Navkiran.

It's not true that a person weighs less when coming down an elevator 
than when going up.  You weigh less at the BEGINNING of a trip down 
than at the BEGINNING of a trip up, and MORE at the end of a trip down
than at the end of a trip up.

At the beginning of a trip down, the elevator is ACCELERATING
downward.  Soon it reaches a steady speed, and your weight becomes 
normal.  Then it DECELERATES downward (which is the same as
accelerating upward) as it comes to a stop at the final floor. 
Likewise, when going up the elevator first accelerates upward, then
reaches a steady speed upward, and finally decelerates to a stop.

If you stand on a scale in an elevator, there are two forces acting on
you: the downward force of gravity, which is constant (for the
purposes of this discussion), and the upward force of the scale on
your feet.  

When you are stationary or when you are moving at a constant speed,
these forces are equal and opposite and the scale shows your correct 
weight.  But when you are accelerating up or down, the effect of that 
acceleration changes the net force and the scale shows a different 
weight.

Net force is calculated by mass (your actual weight) times
acceleration and is written as 'ma'.

Thus, to pick one case, at the beginning of a trip up in the elevator,
you are accelerating upward with acceleration "a".  The vertical force
equation is

  N - mg = ma

where N is the upward normal force of the scale and -mg is the
downward gravitational force.  We can solve for N:

  N - mg = ma
  N      = mg + ma
  N      = m(g + a)

The scale measures the force N, dividing it by g (gravity) to read in 
units of mass (pounds in this example):

  N/g = m(g + a)/g = m(1 + a/g) = m + ma/g

You can see that the scale reads higher than your normal "weight" m 
because "a" is positive (m and g are always positive).  On the other 
hand, at the beginning of a trip down in the elevator, the scale reads 
lower than your normal "weight" because "a" is negative (accelerating 
downward).

I have shown why you weigh less at the beginning of a trip down in the
elevator than at the beginning of a trip up.  If you have questions
about what I've said, I'll be happy to answer them.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/