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Minimum time taken by elevator

  1. Oct 11, 2011 #1
    1. The problem statement, all variables and given/known data

    An elevator of mass (m), is pulled upwards by a massless rope, which can bear a maximum tension of nmg (n-any natural number > 1). If the elevator starts from rest at point A, and it rises to level B where it comes to rest again.(momentary rest is sufficient).
    Find the minimum time taken by the elevator in doing so. (tension may vary in different manners, but the initial conditions of rest must be satisfied)
    Level b is at height (h) above levelA.

    2. Relevant equations

    ƩF = ma

    KE1 + PE1 = KE2 + PE2

    3. The attempt at a solution

    Well, in my attempt, and also in the given solution, it is assumed that the minimum time case is when, till a certain height (y)<h, tension = nmg acts, and then suddenly falls to zero,
    such that the elevator retards to rest by the time it reaches level B
    If this assumption is made the solution the question is fairly easy, and can be found out using conservation of energy.

    The solution turns out to be = √(2hn/((n-1)g))

    But, I don't see any proper logical/ mathematical reason why the above case will yield minimum time. I'm guessing there should be other well defined functions of tension which varies with time and distance, which can yield even lesser time, though I'm unable to find

    I think it can be mathematically solved by minimizing the integral, using calculus of variations.
    ( which I unfortunately do not know)

    It would be nice to know everyones views about this, and I would be grateful if anyone can find a case where the time is lesser, proving the given solution wrong, or give a proper logical reason to why the above case yields the minimum time.

    I hope I will find some direction on this !
  2. jcsd
  3. Oct 11, 2011 #2


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    Staff Emeritus
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    Gold Member

    There's no need to resort to calculus of variations here. Elementary calculus should be sufficient, I think. Essentially you have four constraints. You wish to find the smallest [itex]t_0[/itex] such that [itex]y(t_0) = h[/itex] whilst [itex]y^{\prime\prime}(t) \leq (n-1)g[/itex] for [itex]n\in\mathbb{N}[/itex] with [itex]y(0) = 0[/itex].

    The first step is to solve the boundary value problem. Can you do that?
    Last edited: Oct 11, 2011
  4. Oct 11, 2011 #3
    Thanks for the quick reply!
    Nope I'm unable to do that. There seem to be more constraints.

    y'(0) = 0 & y'(t0)= 0

    And additionally y''(t) can never be less than -g (since tension can't be -ve)

    Not very familiar with boundary value problems, though I can solve differential equations.
    Hope I can get some help with this.
    Last edited: Oct 11, 2011
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