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