Minimum Time for Elevator to Reach Level B?

In summary: Thanks!In summary, the elevator takes a minimum time to go from rest to the top of the building when the initial tension is mg. The solution to the problem is found using calculus of variations.
  • #1
springwave
18
0

Homework Statement



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.

Homework Equations



ƩF = ma

KE1 + PE1 = KE2 + PE2

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 !
 
Physics news on Phys.org
  • #2
springwave said:
I think it can be mathematically solved by minimizing the integral, using calculus of variations.
( which I unfortunately do not know)
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:
  • #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:

FAQ: Minimum Time for Elevator to Reach Level B?

What is the minimum time taken by an elevator?

The minimum time taken by an elevator depends on several factors such as the distance to be traveled, the speed of the elevator, and the number of stops it has to make. In general, the minimum time taken by an elevator is around 30 to 60 seconds for a standard building.

What affects the minimum time taken by an elevator?

As mentioned earlier, the minimum time taken by an elevator is affected by factors such as the distance to be traveled, the speed of the elevator, and the number of stops it has to make. Other factors that may affect the time include the weight of the passengers and the efficiency of the elevator system.

Can the minimum time taken by an elevator be shortened?

Yes, the minimum time taken by an elevator can be shortened by implementing efficient elevator systems such as destination dispatch and group control systems. These systems aim to reduce the time spent waiting for and riding on elevators by optimizing their routes and minimizing stops.

What is the role of elevator speed in determining the minimum time taken?

The speed of an elevator plays a significant role in determining the minimum time taken as it directly affects the time spent traveling between floors. Generally, elevators in tall buildings have higher speeds, resulting in shorter minimum travel times compared to elevators in shorter buildings.

Is there a difference in the minimum time taken by different types of elevators?

Yes, there can be a difference in the minimum time taken by different types of elevators. For example, high-speed elevators can travel faster and therefore have a shorter minimum time than standard elevators. Additionally, elevators with advanced control systems may have shorter minimum times due to their ability to optimize routes and reduce waiting and travel times.

Similar threads

Replies
11
Views
1K
Replies
3
Views
2K
Replies
19
Views
1K
Replies
10
Views
1K
Replies
35
Views
5K
Replies
4
Views
1K
Back
Top