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This thing has baffled me for a very long time and therefore I had to ask it here.

I was thinking about accelerated frames. Lets assume I am sitting in an elevator going upward with some acceleration lets say $a$. Now suddenly, a bolt from the ceiling falls down. Now I wanted to know that how would a person sitting in the elevator (which is actually me) feel the acceleration of the bolt. Would he see it to be only $g$ or $g+a$?

Firstly I assumed the existence of a pseudo force and assumed it to be $g+a$. Then again I thought of one more scenario. Lets assume there is a small ball in a lift descending down with an acceleration $a >g$. Now if $a$ is greater than $g$ then applying pseudo force on the block, it must move up ahead (in reference frame of the person sitting in train) and should continue to move ahead as there is a deficit of $2\,\mathrm{m/s^2}$ in the upward and downward motion but based on my common sense I think that the block can't just move up ahead in air and as soon as it leaves the ground, it must fall back with the acceleration $g$. I know I can be truly wrong and that's why want to know what actually happens.

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2 Answers 2

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Yes, your argument using pseudo force is correct. The ball will continue to accelerate upwards, as observed by you, who is accelerating downwards with acceleration $a>g$.

I think there is no "common sense" in your scenario, because seldom will people be inside a lift accelerating downloads with $a>g$. My "common sense" is: the ball is free falling with acceleration $g$, and you are falling with acceleration $a>g$. Then as observed by you, the ball should be accelerating upwards.

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  • $\begingroup$ And what about the first scenario of the bolt? $\endgroup$
    – user118752
    Jun 27, 2016 at 10:42
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    $\begingroup$ In the first scenario, the bolt will be falling with downward acceleration $g+a$. In the second scenario, the ball should be accelerating upwards with acceleration $a-g$. So if you want to find its upward distance traveled (observed by you), it should be $s=1/2(a-g)t^2$. $\endgroup$
    – velut luna
    Jun 27, 2016 at 10:45
  • $\begingroup$ I did the same thing in one problem taking $a=12m/s^2$ but go the wrong answer while the right answer was given by using only $g$ as the acceleration. $\endgroup$
    – user118752
    Jun 27, 2016 at 10:46
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    $\begingroup$ May be it is asking for the actual distance traveled observed by an inertial observer outside the lift, but not by you. $\endgroup$
    – velut luna
    Jun 27, 2016 at 10:48
  • $\begingroup$ I just did the question again and found you were right. Thanks a lot. You solved a big deal for me. $\endgroup$
    – user118752
    Jun 27, 2016 at 10:49
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You can imagine it in a simple way without any confusion: A lift descending with a>g implies that it 'falls' at a faster rate than g. In other words, the ceiling of the elevator moves downwards(w.r.t a ground frame) faster than a free falling object(which according to the ground frame is the ball). Isn't it obvious now why it would appear to rise w.r.t someone in the elevator?

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