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A person is standing on a weighing scale in an elevator in upward acceleration.

Let $N$ be normal reaction force exerted by the weighing scale to the person (upward).

It is known that the person will experience a normal reaction force ($N$) larger than his own weight (larger reading on the weighing scale). While $N$ is equal, but in opposite direction, to the force that the person exerts to the weighing scale due to action-reaction pair (the person and the weighing scale). That means other than the person's weight, there should be an extra force exerting to the weighing scale downward ($A$).

So $-N$ (upward)$ = mg + A$ (downward). I can't figure out where does $A$ come from?

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A is not a real force but a 'fictitious' one, and it arises because the elevator isn't moving at constant velocity (it's not an inertial reference frame). In such an accelerating frame, you need to add extra terms to Newton's law. This is explained in any undergrad mechanics book, say chapter 5 of Fowles-Cassiday. – Vibert Apr 1 '13 at 9:01
you can think it as you were falling freely then you feel an celebration of 1 g now imagine earth started moving towards you with an acceleration A so you will be falling towards earth now with an acceleration of g + A if someone will see from earth. So same in the elevator you are standing in the elevator so you are continuously feeling an acceleration of g but when the elevator starts rising with an acceleration A so you will feel g + A – Dimensionless Apr 1 '13 at 17:12

2 Answers 2

Just don't start writing equations without a complete Free-Body-Diagram(may be rough diagram) according to the frame of reference..

1)Ground Frame

enter image description here

Now . Newtons law. $$\sum \vec{F_{ext}}=\frac{dp}{dt}=ma(for\ constant \ mass)$$

So, $$N-mg=\sum \vec{F_{ext}}=ma$$ Where system is boy in ground's frame.

Here we equate the net external force to the acceleration of body . $ma$ is not a force it is the measure of net force which causes body to accelerate.

2)In a accelerating frame(non-inertial)

here we have to add a fictious force $-m\vec{a}$ where $\vec a$ is the acceleration of frame . enter image description here

Here too $$N-mg-ma=\sum \vec{F_{ext}}=0(as \ in \ frame \ of \ elevator \ boy \ is \ at \ rest)$$ Again you get $$N-mg=ma$$

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Contrary to what the other answers seem to suggest, the force described in the question is not fictitious. Even in the inertial ground frame the person will exert a force of magnitude $m(g+a)$ on the scale. This force is real and not fictitious. (There is, however, no real downward force of magnitude $m(g+a)$ acting on the person.)

As to where the $ma$ component come from: The actual physical cause of normal forces are typically not given much thought in this type of calculations. Normal forces are introduced to uphold certain mechanical constraints that are assumed to hold. In this case, the constraints are that the elevator, the scale and the person should all remain intact, maintain their relative position to each other and move with the same speed and acceleration. We simply calculate the normal forces that must be present for this to occur. If the person and the scale could not exert this normal force on each other, our constraints would not hold. (The scale or the person would break or deform and at least for a while they would not travel with the same velocity and/or acceleration).

As to the physical cause of normal forces, I can rcommend reading some of the answers to this question.

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The force $A$ described by OP is of-course fictitious in non-inertial frame. and there is no force as $A$ in inertial frame , but it is measure of acceleration times mass of body itself. – ABC Apr 1 '13 at 13:12
@exploringnet No, this is not correct. In your answer, you are talking about the forces affecting the person. OP is clearly asking about the forces affecting the scale. There IS a real force $A$ acting on the scale. Otherwise, Newton's third law would be violated. – jkej Apr 1 '13 at 13:24
No, there is only one force avting on scale ie. $N$ and it's value comes out to be $mg+A$ – ABC Apr 1 '13 at 13:26
@exploringnet Yes, and that force is real (i.e. it appears even in the inertial frame). I thought you said it was fictitious. (Not that it matters much here, but there is of course also a force from the elevator onto the scale.) – jkej Apr 1 '13 at 13:37
No, u don't understand my point.$N$ and $mg$ are only real forces in scene and $A$ is NOT a real force.Yes it IS fictitious in non-inertial frame . and no force like $A$ exist for inertial frames. – ABC Apr 1 '13 at 13:41

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