# How are fictitious forces related to my feeling?

This question arises when I am studying fictitious forces in an undergrad introduction to physics course.

Suppose I am standing in an elevator with an acceleration $a$ directed upward. From the ground viewpoint, with $N$ being the normal force on me, we have $N - mg = ma$ and so $N = mg + ma$. From my viewpoint, I feel two force: an upward force of $m(g+a)$ and a total downward force (gravity + fictitious force) of $m(g+a)$ too. If I were standing somewhere on a planet with gravitational acceleration $G=g+a$, then I feel these two forces too. That's why in the elevator I have the feeling of being on the planet with $G = g+a$. OK, so far so good.

Now let's say I'm freefalling. From my viewpoint, I feel two force: an upward fictitious force of $mg$ and a downward gravitational force $mg$. The situation is pretty the same as if I were standing on the ground. But clearly I wouldn't feel that normal when freefalling; in fact I feel weightless. Why?

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Essentially a duplicate of physics.stackexchange.com/q/46020/2451 –  Qmechanic Sep 26 '13 at 11:41
@dj_mummy Yes, you are welcome :) –  tom_a2 Sep 26 '13 at 12:32
@Qmechanic That's a helpful link, thanks! So one possible answer to my question is that the upward force from the ground acts on my feet more heavily. Yet when I freefall, the upward and downward force acts on my atoms equally, so they indeed cancel. Another explanation is that I am just moving in the curved space-time generated by gravity, so I feel essentially weighless. –  tom_a2 Sep 26 '13 at 12:37
@tom_a2 I edited my answer. Hope it is even simpler to understand now. –  dj_mummy Sep 26 '13 at 13:06

If I stand on the ground, then the $mg$ grav force and the $mg$ normal force do cancel out too. Then why didn't I feel weightless? –  tom_a2 Sep 26 '13 at 12:12