How does an elevator maintain a constant speed despite changes in load?

[walking]

Author's solution seems to assume that "carry 100 people/min" means "takes 100 people up the whole length / min", ie 13.3m/min=0.222m/s. But its lengthwise speed is actually 0.620m/s.

In particular, why is my solution wrong (according to author's solution)? :

I said since 100 people/min then its required "force rate" lengthwise is $$100mg\sin \theta (1min/60s)=755.26=755N/s$$. Now in 1s, it travels (lengthwise) 0.62m. Hence $$P=Fv=468.1=468W$$. But this is wrong according to author's solution.

 

[Gordianus]

Can you improve the definition of power?

 

[jbriggs444]

Author's solution seems to assume that "carry 100 people/min" means "takes 100 people up the whole length / min", ie 13.3m/min=0.222m/s. But its lengthwise speed is actually 0.620m/s.

The lengthwise speed does not limit the number of people that can reach the top per minute. Just cram them in more tightly together.

In particular, why is my solution wrong (according to author's solution)? :

I said since 100 people/min then its required "force rate" lengthwise is $$100mg\sin \theta (1min/60s)=755.26=755N/s$$.

It looks like you've computed ##\theta## somewhere else. Perhaps you arrived at a figure for ##\sin \theta## as ##\frac{8.2}{13.3} = 0.62##. You've multiplied this by the weight of 100 people to get the diagonal force that would need to be applied to keep those people moving up an inclined plane at the calculated angle. And you've divided by 60 to switch from seconds to minutes.

But you are asked for power, not force. So you clearly need to multiply by a distance. And you find a distance to multiply by...

Now in 1s, it travels (lengthwise) 0.62m. Hence $$P=Fv=468.1=468W$$. But this is wrong according to author's solution.

You've assumed that there are 100 people on the escalator at any given time. But that is not necessarily correct. And is not, in fact correct. All you are given is that 100 people exit the elevator per minute, not that it takes one minute for them to get from bottom to top.

 

[rsk]

How much work do you need to do to move a 75 kg person through a vertical height of 8.2m? How much work for 100 of them?
And, if you do this in 60 seconds, what power does this give?

 

[haruspex]

Author's solution seems to assume that "carry 100 people/min" means "takes 100 people up the whole length / min"

Seems right. They would not be happy to be taken only half way.

ie [escalator speed is]13.3m/min=0.222m/s

No, that assumes each person takes one minute, i.e. 100 people on it at any instant.
In one minute, the escalator moves 0.62*60=37.2m and delivers 100 people, so they stand 0.372m apart.
Each person rides it for 13.3/0.62=21.5s.

 

[Lnewqban]

I said since 100 people/min then its required "force rate" lengthwise is $$100mg\sin \theta (1min/60s)=755.26=755N/s$$. Now in 1s, it travels (lengthwise) 0.62m. Hence $$P=Fv=468.1=468W$$. But this is wrong according to author's solution.

It is incorrect to use ##P=Fv## in this problem, because you have a mass flow rate (125 Kg/s).
Note that you have multiplied N/s by m/s, which does not make sense.

This type of problem is similar to calculating the power gained by a liquid that is pumped between two different elevations.
##P=(mass~flow~rate)gh##

 

[walking]

Thanks for the help everyone!

So it seems my problem was seeing a speed and immediately thinking P=Fv was the most appropriate formula. Is there any way of making this work? Probably not but I'm not that good so it's still worth asking!

 

[walking]

The lengthwise speed does not limit the number of people that can reach the top per minute. Just cram them in more tightly together.

It looks like you've computed ##\theta## somewhere else. Perhaps you arrived at a figure for ##\sin \theta## as ##\frac{8.2}{13.3} = 0.62##. You've multiplied this by the weight of 100 people to get the diagonal force that would need to be applied to keep those people moving up an inclined plane at the calculated angle. And you've divided by 60 to switch from seconds to minutes.

But you are asked for power, not force. So you clearly need to multiply by a distance. And you find a distance to multiply by...

You've assumed that there are 100 people on the escalator at any given time. But that is not necessarily correct. And is not, in fact correct. All you are given is that 100 people exit the elevator per minute, not that it takes one minute for them to get from bottom to top.

Dear jbriggs

I think I understand what you're saying. Does this mean that the escelator doesn't apply the same force throughout but changes it based on how many people there are? But that doesn't sound right to me for some reason. (I am of course not saying that I am correct here!) I think my reasoning at the time was as follows:

An elevator applies the same force all of the time. Since it needs to be able to lift 100 people per minute, that means it needs to be apply to apply a corresponding force throughout. So the force it needs at any given time is at least 100(75)gsin(theta) where theta is the angle of the slope.

 

[haruspex]

An elevator applies the same force all of the time.

Not so. If that were the case you would notice it accelerate when less than fully loaded, and the fewer people the greater the acceleration.
They are engineered to run at constant speed, up to maximum load.