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A question asks you to determine the relationship between true weight and apparent weight when an elevator that has descended from the 50th floor is coming to a halt at the first floor.

I believe that as the elevator is coming to a halt, the acceleration is decreasing (close to zero), and so this downward acceleration is negative. Therefore, w = mg - ma. Then w should be less than mg. However, the book tells me that w should be more than mg. I do not understand, in any way, how this is possible. Can someone please explain how true weight is less than apparent weight when the person in an elevator is experiencing overall downward acceleration?

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    $\begingroup$ The elevator has been going at a high downwards velocity, then slows. So its velocity becomes less downwards. In other words, the nett acceleration is upwards. To accelerate something upwards relative to the Earth means you must overcome its weight with an opposing force upwards and then add further force upwards for the acceleration. The force you feel on your feet as the elevator slows from the floor is thus more than your weight. $\endgroup$ – WetSavannaAnimal Jun 5 '15 at 0:11
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    $\begingroup$ Acceleration is change in velocity over time. What is the change in velocity of an accelerator with an initial speed of $-10m/s$ and a final speed of $0m/s$? $\endgroup$ – BowlOfRed Jun 5 '15 at 0:12
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First and foremost, I assume you've been in an elevator? Don't you feel heavier (your feet have to push a little harder) when the elevator stops at the first floor (i.e. stops after going downward)? So it should be intuitive that your apparent weight would be greater than your true weight.

However, when in doubt, draw a force diagram. This particular question is one where the normal force is very important (because the normal force is your apparent weight). We don't need any numbers here since all we care about is inequality. From your wording, I believe you are confused:

"I believe that as the elevator is coming to a halt, the acceleration is decreasing (close to zero)..."

Whether or not the acceleration is increasing or decreasing is not important. What is important is the direction of the acceleration! However acceleration is a change in velocity (vectors here--not speed). If you are moving downwards, and the elevator brings you to a halt, then the acceleration must be against the original velocity. So if you're going down, then stop, you accelerate upwards; if you're going up, then stop, you accelerate downwards.

Since we are going down and stop, we know the net acceleration of the elevator/person must be upward:

enter image description here

We see that the $F_{net} = ma = N - mg$ and thus solving for the normal force, we get:

$$ N = mg + ma $$

Since the normal force represents your apparent weight it increases by $ma$. First note that the same is true when you start going up (i.e. accelerate upward to speed). Next note that if you are going up and coming to a stop then you should accelerate downward and thus now you have $mg - N = ma$, which gives $N = mg - ma$--so it's less! Again, likewise when you start going down, the acceleration is downward and your apparent weight is less (by an amount $ma$).

Addressing Your Intuition About Changing Acceleration

You are correct to state that the acceleration "decreases" (really the magnitude decreases). Until finally, you're stopped and the net acceleration is $0$ and your apparent and true weights are the same. If we tracked your apparent weight over time from the time you got in the elevator, started going down, then stopped, we would mirror the acceleration. You'd expect something like this:

enter image description here

We see that the apparent weight exactly mirrors the acceleration (because it depends on the added or subtracted $ma$). The velocity starts going down (negative), then maintains speed, then comes to a stop, $0$. The acceleration should "slowly" approach zero both times--this makes stopping and going smooth--otherwise the elevator would be jerky. Many first time drivers learn this when braking. If you just depress the brake, then you jerk to a stop. The proper way to stop is to depress the pedal and then smoothly let off until you come to a stop.

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If you are in a non-inertial reference frame inside the elevator that is accelerated in the upward direction, then you experience a pseudoforce downwards (we assume downwards is positive) ($F_{pseudo}=0-(-ma)=ma$. So the apparent weight is N=mg+ma and the apparent weight increases.

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Don't forget that your weight on the Earth is equivalent to what your weight would be in an accelerating spaceship far removed from the Earth's gravitational field, accelerating at the rate of 9.807 meters per second per second (9.807 m/s^2). This follows from the equivalence principle of General Relativity, which provides that the force of gravity is equivalent to an acceleration force.

Let's say your mass is 80 kilograms. Your weight at the surface of the Earth is:

Weight = mass * acceleration = 80 * 9.807 m/s^2 = 784.56 newtons. (A newton is the force required to give one kilogram of mass an acceleration of one meter per second in one second. It's the standard international unit of force.)

If you're standing in an elevator that starts accelerating downward, you first feel your weight leave your feet, as though the floor is dropping out from under you. The downward acceleration of the elevator cancels some of the Earth's upward equivalent acceleration, and this reduces your apparent weight.

Let's say your elevator is accelerating downward at the rate of 2 m/s^2. Your apparent weight in the dropping elevator is:

Weight = 80 * (9.807 - 2.000) m/s^2 = 624.56 newtons

Now, let's say your elevator has stopped accelerating downward, and is coming to a stop. It decelerates at the rate of 2 m/2^2. A negative downward acceleration is the same as a positive upward acceleration, so it's added to the Earth's equivalent acceleration and you feel as though you are being squashed into the floor. Your apparent weight in the stopping elevator is:

Weight = 80 * (9.807 + 2.000) m/s^2 = 944.56 newtons

A deceleration vector is the same as an acceleration vector in the opposite direction.

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  • $\begingroup$ Newton as unit shouldn't be specially capitalized. $\endgroup$ – Ruslan Jun 5 '15 at 4:05

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