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When you are accelerating upwards in an elevator, you feel an increase in apparent weight.

A scale measures the total amount of force pushing down on it, and if you were to step on a scale in an elevator, you would see an increase in its reading.

However, when accelerating upwards only the normal force increases -- how does this translate to an increase in the "downward push" needed for an increase in apparent weight?

Similarly, when accelerating downwards, the normal force decreases but how does this translate to a decrease in the "downward push" needed to result in a decrease of apparent weight.

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3 Answers 3

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As you said correctly: as you accelerate upwards, the normal force acting on you will increase. Now look at the situation from the scale's "perspective" -- consider the forces acting on the scale:

Since the ground (i.e. the scale on which you stand) applies a normal force $F_N$ (upwards) on you, Newton's 3rd law says that you must also be applying a force $F_N$ (downwards) on the scale.

To summarize: the scale will measure the magnitude of the normal force acting on you; which is also the magnitude of the force you exert on the scale (by Newton's 3rd law).

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  • $\begingroup$ But doesn't the downward force you exert on the scale remain at a constant mg? That is the only downward force in the whole free body diagram $\endgroup$
    – 1110101001
    Jun 20, 2015 at 19:41
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    $\begingroup$ That is true for the free body diagram of the person in the elevator. Please consider now the free body diagram of the scale. As, I said above: there will be a force (equal to $F_N$) acting down on the scale. $\endgroup$
    – Mustapha Mond
    Jun 20, 2015 at 19:44
  • $\begingroup$ Ok that makes sense. In addition to the contact force of the object on the scale, shouldn't there also be the force that gravity exerts on the scale? $\endgroup$
    – 1110101001
    Jun 20, 2015 at 21:58
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    $\begingroup$ Yes, this is always the case, whether the elevator is accelerating or not. This can be calibrated to be a reading of $0$ on the scale. $\endgroup$
    – Mustapha Mond
    Jun 20, 2015 at 22:54
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The normal force will be equal to the weight (which is a force) if the object is "resting" on the surface which is producing the normal force. Essentially weight is always apparent because the weight of a 10kg mass on earth is different than its weight on mars which is also different than its weight in an elevator on earth (unless you accelerated the elevator appropriately to match the force the mass "feels" resting on the surface of mars.)

The normal force is the force that is "needed" to prevent the object (or a human in an elevator) from falling through the surface that it is resting on. It is a force that opposes the force that is pressing an object onto a surface. On earth the ground provides a normal force to prevent you from falling through into the center of the earth. A thin piece of paper may be sufficient to provide a very good normal force for an ant (tiny little massed creature) but probably not for a human (on earth of course).

**On a side note, if you were on the sun the force of gravity (aside from the heat melting the paper) would be extremely large and may give the ant enough weigh to rip through the paper (in this case the normal force would fail to do its job which is to apply enough force back to an object to keep it on its surface and not allow it to fall through its surface. Understanding this is key to understanding the normal force.

The scale is exerting a normal force on you. The scale is providing a normal force which counters any force you are imposing on it and thats why the reading changes when the acceleration of the elevator changes (because when the acceleration of the elevator changes your weight changes (because ma=F and when the downward force increases (when elevator is going up) the normal force increases. Or when the downward force decreases (when the elevator is going down) the normal force decreases. The normal force is equal to the weight. So if you know the normal force you know the weight and vise versa. They are equal but opposite forces. Hope this helps.

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  • $\begingroup$ Also make note that it is the surface that is producing the normal force and the normal force is dependent on the weight not the other way around. So the normal force will vary to cater to larger and smaller weights. If the normal force changed then you know that the weight also changed. $\endgroup$
    – Daniel
    Jun 20, 2015 at 19:17
  • $\begingroup$ You said that when the downward force increases (when elevator is going up) the normal force increases. However, the only downward force acting on the body is gravity. Are you talking about the pseudo/fictitious force due to the acceleration of the elevator upward? This is partly from where my confusion stems, as this is only a fictitious force, so what truly leads to the increase in downward push that is needed for an increase in apparent weight? $\endgroup$
    – 1110101001
    Jun 20, 2015 at 19:28
  • $\begingroup$ It is the force in the -Y coordinate that i am referring to as downward and this force is not fictitious but is the weight of the object (which is a force) and is equal and opposite to the normal force. $\endgroup$
    – Daniel
    Jun 20, 2015 at 20:21
  • $\begingroup$ Ah, I see. However, sometimes I have seen in the context of the elevator problem a fictitious force of magnitude $ma$ acting downwards. What would be the interpretation of this fictitious force from the non-inertial frame of reference that is the elevator? $\endgroup$
    – 1110101001
    Jun 20, 2015 at 22:03
  • $\begingroup$ There is a force due to gravity acting on the person in the elevator and a force due to the acceleration of the elevator. These add together and make up the objects weight. They are components of the weight and so there is a component of the weight due to gravity and due to the acceleration of the elevator. No fictitious forces. $\endgroup$
    – Daniel
    Jun 21, 2015 at 14:47
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The weight indicated on the scale will only increase as long as the elevator accelerates you upwards. For most of the ride, when speed is constant, the indicated weight will be the same as in the static case, and at the end, when the elevator decelerates, the indicated weight will decrease.

The weight force is composed of two parts:

  • Weight due to gravity ($W_g = m\cdot g$)
  • Weight due to vertical acceleration ($W_a = m\cdot a$)

What the scale will indicate is the sum of both: $W = W_g + W_a$

If you integrate the weight indication over time and compute the average after the elevator has stopped, you will find that it will exactly be your static weight.

Also, on the way down, the indicated weight will only be lower as long as the elevator accelerates downwards, and towards the end of the ride, when it decelerates again, you will see your indicated weight increase. It is the same as on the way up, only in reverse because now the acceleration will work in the opposite direction.

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