(adsbygoogle = window.adsbygoogle || []).push({}); 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

any.

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 !

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# Homework Help: Minimum time taken by elevator

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