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I read a problem about a person is standing on a scale in an elevator. When I read the answer, there was an explanation saying that when the acceleration of the elevator goes upward, the normal force is bigger than the person's weight, and when it goes downward, it's smaller. I'm having trouble understanding why this would be the case.

So far what I understood about the normal force is that it's the force surfaces exert to stop objects from passing through them. And it has the same magnitude as the weight but an opposite direction (at least in a horizontal surface) That makes sense. But why is it that inside an accelerating elevator this changes? Why does the scale need to "push" more when accelerating upwards if the person's weight is the same? What is exactly is happening to the person when accelerating? The only reason I can imagine the normal force changing is that the weight of the person changes. I imagine this being similar to having to hold up a person: if the weight becomes bigger, the force I exert must become bigger too to keep them up (or at least that's what I imagine) But in the problem the person's weight can't change, because gravity doesn't change and his mass hasn't changed during the problem. So does the normal force not simply oppose the weight? What I am getting wrong?

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    $\begingroup$ normal force = weight only applies if 1.) the surface is horizontal and 2.) the surface is not accelerating. $\endgroup$ – garyp Jun 14 '19 at 20:25
  • $\begingroup$ Although the question has already been answered, you touch on a great point, the concept of apparent weight. You might want to read more about it. $\endgroup$ – Alonso Perez Lona Jun 14 '19 at 21:54
  • $\begingroup$ Do not confuse weight and mass; they are two different things. $\endgroup$ – BillDOe Jun 14 '19 at 21:58
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The normal force needs to not only "balance" the person's weight but provide the acceleration. The scale is a separate object and the normal force acting on the scale is balanced by the spring mechanism (or other mechanism) inside that actually reads the weight. Without figures you have the following:

Forces acting on the person in the elevator (standing on the floor or scale) near the earth are: m*g pointing down, and N pointing up. When the acceleration is up Newton's second law gives,

ma = N - mg which implies N = m*(a + g)

when the elevator accelerates down we get

-ma = N - mg which implies N = m*(g - a)

When the elevator is in free fall N = 0 and the person seems weightless. This is how the vomit comet works.

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From Newton’s second law $F = ma$, acceleration requires a force proportional to mass. So when the elevator is accelerating upwards, it must not only oppose gravity but provide extra force for the upward acceleration. This requires an increased normal force. When the elevator is accelerating downwards, it does not need to oppose gravity as much and so the normal force decreases to provide a net downward force from gravity. No changing mass is needed.

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Why does the scale need to "push" more when accelerating upwards if the person's weight is the same? What is exactly is happening to the person when accelerating?

Instead of thinking of the scale pushing upward and it reading more than your weight, think instead about the equivalent situation if you were to push downward on the scale with a force in addition to your weight.

Try the following experiment yourself.

Stand on a scale with the scale on the floor. Bend down like you are doing a deep knee bend. Now stand up while watching the scale reading. It will briefly be greater than your weight while you are rising and then settle back to your weight after you stop rising (you will observe some overshoot). This is because the downward force you are exerting on the scale when you are rising is now your weight plus the additional force you exerted on the scale. This gives the equivalent result as the reading on the scale when an elevator is accelerated upward.

Hope this helps.

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