What is the equation for finding the position of an elevator at any given time?

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The discussion focuses on deriving the equation to find an elevator's position over time, starting from the 6th floor with specific parameters. The key equations involve calculating acceleration as Fnet/m, followed by integrating to find velocity and position. It is emphasized that acceleration is not constant, complicating the calculations and requiring multivariate functions. The conversation also highlights the importance of making assumptions for simplification, such as neglecting friction and energy losses. Ultimately, the participants suggest using the teacher's provided methods and Excel for practical calculations.
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1. Mass is hung from a force probe, with negative force calibrated as down. The mass is 0.5 kg, the height of each floor is 5 m, and v0 = a0 = 0. Assume that I start on the 6th floor, what floors did I stop on, and how long was I on each floor?
2. I was wondering, how would one find the position of the elevator at any time as an equation?
a(t) &=& \frac{Fnet}{m}
v(t) &=& {\int_{t_0}^{t_final} a(t) dt = {\int_{t_0}^{t_final} (\frac{Fnet}{m}) dt
x(t) &=& {\int_{t_0}^{t_final}v(t)dt = {\int_{t_0}^{t_final}}{\int_{t_0}^{t_final} a(t) dtdt} = {\int_{0}^{t_final}}{\int_{0}^{t_final}(\frac{Fnet}{0.5}) dtdt

Floor(t) &=& \frac{\int_{0}^{t_final}{\int_{0}^{t_final}(\frac{Fnet}{0.5}) dtdt + 30}{5}

3. My idea was the the position is the double integral of acceleration, which is Fnet/m, with respect to time. Where Fnet is the instantaneous net force at time t. But, on excel I got weird numbers, when I did a trapezoidal approximation. Am I doing anything wrong?
 
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Aah ,you'd have to make a lot of assumptions before approaching such a problem in a practical manner...But assuming all ideal situations that no friction,you know no energy losses,there is no intermidiate stopping that is it is a point to point transportation,constant accelarations and all the rest that you are used to in textbooks,then we can simplify and look at it it through 2 situations..Going down where total accelaration A(down)=a(1) + g and going up where total acceration A(up)=a(2)- g...where g=gravitational accelaration,a(1)=accelaration due to elevator machine going down and a(2)=accelaration due to elevator machine going up...previously I made an assumption all these are constants and independent such that you can perform the double integral with complete ease..and in some situations even assume a(1)=a(2)!...keep in mind,this is very impractical,I don't see a practical possibility without few assumptions or involving complex multivariate functions,who wants the trouble when we got digital systems to tell positions at any and all times t's!
 
You see, the teacher gave us an excel spreadsheet with the time at 0.1s intervals, and the net force as well at those corresponding intervals. From there we need to answer those two questions. Thus, acceleration IS NOT CONSTANT, that's why I have to use complex multivariate formulas.
 
Ooh am sorry you hadnt explained that!...My bad!But you should be more descriptive,its not easy to know that accelaration is not constant if note that out in the 1st place..try to Use this integral for x(t)=$ ($[Fnet/m] dt)dt where $=definite integral sign from t=0 to t...
 
Ooh am sorry you hadnt explained that!...My bad!But you should be more descriptive,its not easy to know that accelaration is not constant if note that out in the 1st place..try to Use this integral for x(t)=$ ($[Fnet/m] dt)dt where $=definite integral sign from t=0 to t...Then use the analytical methods the teacher instructed
 
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