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An elevator does need some kind of acceleration when it starts to rise up so there has to be s force acting on it. But it surely does not accelerate the whole time so after some distance or time it reaches a speed and stops to accelerate. In an old physics textbook I found an example where the power a motor would need to lift it was calculated like this

P = F * v

But they did not take the average speed for v but the speed which would be reached at the end of the acceleration. Why was this speed used and not the average speed?


marked as duplicate by John Rennie newtonian-mechanics Aug 3 at 8:22

This question was marked as an exact duplicate of an existing question.

  • 3
    $\begingroup$ Possible duplicate of Power of an elevator motor $\endgroup$ – Matthew Roh Aug 2 at 19:07
  • $\begingroup$ "when it starts to rise up so there has to be s force acting on it" Hmmm ... there has to be a net force acting on it. When it stops accelerating there is zero net force, but because gravity still acts downward some other force must act upward. Discuss. $\endgroup$ – dmckee Aug 2 at 20:00
  • $\begingroup$ @dmckee ok I understand that but could you answer this question? $\endgroup$ – Dinkelberg Aug 2 at 20:16
  • $\begingroup$ The process of getting to the answer is the important thing here, and you needs to learn how to go about it. So, questions for the student: (A) How long does a "typical" elevator ride last? (B) For what portion of that time does the car have an upward acceleration? (C) A downward acceleration? (D) How does the power required during the two acceleration phases compare to that during the steady phase? (E) Which term (or terms) is (are) the dominate contribution(s) during a single ride? $\endgroup$ – dmckee Aug 2 at 20:21
  • $\begingroup$ @dmckee (A) was not given and I think is not notable for this problem (B) what was given was the distance which would be covered by the acceleration which is 1.5m and through the velocity which the elevator would have reached at the end of the acceleration: 0.5 m/s one could figure out the time: 6s (C) a downward acceleration would be counteracted by the same acceleration in the other direction and added to it would be the acceleration which for the first 1.5m $\endgroup$ – Dinkelberg Aug 2 at 20:34

What are you trying to calculate? The average power OR the instantaneous Power at the end of the acceleration?

The first part will be calculated by taking the average force/acceleration along with the average speed.

The second part will be calculated using the final speed after acceleration and multiplying it by the weight of the lift + passengers (as this is the amount of force required to maintain a constant speed).

Please post the exact question you want answered.


Do not make two posts regarding the same question.

  • $\begingroup$ I think I understand now: the question was not very clear. I thought I would need to calculate the average power in the whole acceleration process but I think what was meant was the power at the end of the acceleration. Thank you $\endgroup$ – Dinkelberg Aug 3 at 12:50
  • $\begingroup$ You could edit your answer so it says that what I was looking for was not the average acceleration during the process. $\endgroup$ – Dinkelberg Aug 3 at 12:53

But they did not take the average speed for v but the speed which would be reached at the end of the acceleration.

It might be because the acceleration ends quickly so that the speed is constant for a much longer period of time than the acceleration lasts.

I was unable to find any data on the duration of acceleration of an elevator, so I don't have any data to support this. But it seems to me the times I have used an elevator the initial sensation of it pushing up on me generally seems to be brief. In other words, it is possible that for most elevators the duration of the upward acceleration is probably a lot shorter than the duration of constant velocity, so that most of the power needed is to keep the elevator going at constant velocity. That could be the reason for using $P=Fv$.

I would think that during the initial acceleration what is more critical is the torque that the motor can deliver to get it started.

Hope this helps.

  • $\begingroup$ Elevator acceleration profiles are actually quite a well studied area, but the results are applied unevenly (roughly in proportion to the cash outlay). Luxurious rides are obtained by tuning higher derivatives (I think I read that constant snap is used for the very best high-speed elevators). Less-than-smooth stops and starts are common in less expensive designs. $\endgroup$ – dmckee Aug 2 at 23:13
  • $\begingroup$ @dmckee Thanks for the feedback. But from what you know what is the fraction of total time between floors that a typical elevator is accelerating $\endgroup$ – Bob D Aug 2 at 23:19
  • $\begingroup$ It is going to vary a lot depending on the device and how far you're going. I would guess is runs between a few percent for longer duration rides (either because it's a slow lift or because you are going a long way) and a few tens of percent for brief jaunts. $\endgroup$ – dmckee Aug 2 at 23:23
  • $\begingroup$ I just want to calculate the power required during the acceleration process. My problem is that I thought a force has to act on it in order to accelerate it to the "end" -speed and that to get the power which is required for that you had to first multyplie by the distance over which the force acts and then divide by the time it takes to accelerate. But that would mean that I am multiplying by the average velocity in the acceleration process. However the model solution did not do that but they multiplied by the end speed to get to the result. So what speed is the v in P=F*v. $\endgroup$ – Dinkelberg Aug 3 at 7:44

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