What is the average force required to pull the rope?

[geoffrey159]

Homework Statement

Problem 3.12 from Kleppner & Kolenkow
A ski tow consists of a long belt of rope around two pulleys, one
at the bottom of a slope and the other at the top. The pulleys
are driven by a husky electric motor so that the rope moves at a
steady speed of 1.5 m/s. The pulleys are separated by a distance of
100 m, and the angle of the slope is 20◦ .
Skiers take hold of the rope and are pulled up to the top, where
they release the rope and glide off. If a skier of mass 70 kg takes
the tow every 5 s on the average, what is the average force required
to pull the rope? Neglect friction between the skis and the snow.

Homework Equations

Impulse and average force

The Attempt at a Solution

I look at the system 'ski tow-skiers'.
I define $T(t)$ the force necessary to pull the skiers currently on the ski tow.
I also define $n(t) = 1 + \lfloor {\frac{t}{5}} \rfloor$ the number of skiers on the rope for $0 \le t \le t_f = \frac{100}{1.5} \approx 66.67$ seconds.

At a given time $t$, the external forces acting on the system in the direction of the slope are $T(t)$ and the influence of the weight of the $n(t)$ 70-kilograms skiers.
Therefore,

$\int_0^{t_f} F_{ext}(t) \ dt = \int_0^{t_f} T(t) \ dt - 70 g \cos(70)\times \int_0^{t_f} n(t) \ dt = t_f \times T_{av} - 434.45 \times \int_0^{t_f} n(t) \ dt$
$\int_0^{t_f} \frac{dP}{dt} \ dt = 70 \times 1.5 \times ( n(T) - 1 ) = 1365$

Now, $n(t)$ is constant on every five seconds slices of time so:

$\begin{align} \int_0^{t_f} n(t) \ dt &= \sum_{k=0}^{\lfloor{\frac{t_f}{5} \rfloor - 1}}\int_{5k}^{5(k+1)}(k+1) \ dt + \int_{5 \lfloor{\frac{t_f}{5}} \rfloor }^{t_f} n(t_f) \ dt \\ &= 5 \sum_{k=0}^{12}(k+1) + 23.38 \\ &= 5 * 13 * 7 + 23.38\\ &= 478,38 \end{align}$

So equating the impulse and the momentum difference, the average pulling force should be:

$T_{av} = \frac{1}{t_f} (1365 +434.45 *478,38 )= 3137,80$ Newtons

Now I'm in doubt with that, do you agree ?

 

[haruspex]

I feel your use of the floor function is unnecessarily complicating matters.
You only need to work with the average number of people on the line.
On the other hand, I think you are overlooking what happens as each skier joins on.
An alternative approach is to look at the energy given to each skier (PE+KE), deduce the power, and hence the force.

 

[geoffrey159]

Hello,
I don't see how it simplifies because average number of people on the line is $n_{av} = \lfloor \frac{1}{t_f} \int_0^{t_f} n(t) \ dt \rfloor = 7$.
It's a little complicated for me here.
Knowledge about energy is probably not required for this exercise, because it comes from a textbook which does not mention energy before this exercise.

 

[Orodruin]

No, as haruspex said, that is not the average number of peope on the tow. Note that the average number does not have to be an integer.

I also believe that you should take the average in operation, i.e., not from the first skier grabbing the lift but once a steady flow of skiers has been reached.

 

[geoffrey159]

Hello,
I'm a little lost ! What is a steady flow ? I barely know what's a flow of mass, I'm starting with these concepts :-)
Would you please detail how I should have handled the problem ?

 

[Orodruin]

The system is in a steady state once the tow has been in operation so long that the starting conditions do not matter, i.e., the expected number of skiers on the tow is constant. If it takes a skier 67 s to reach the top and there is a new skier every 5 s, what is going to be the average number of skiers on the tow?
What is the force required to tow one skier?
What is then the force required to tow the average number of skiers?

 

[geoffrey159]

I would say the average number of skiers on the tow is $n_{av} =\frac{1}{2}( \frac{67}{5}+1) = 7.2$.
The force to tow one skier for 67s is :
$T = 70g\cos(70) \approx 434.46$ Newtons, because momentum will be conserved for one skier.
So for all the skiers $T_{av} = n_{av} * T \approx 3128$ Newtons.
Am I right?

 

[Orodruin]

I would say the average number of skiers on the tow is $n_{av} =\frac{1}{2}( \frac{67}{5}+1) = 7.2$.
The force to tow one skier for 67s is :
$T = 70g\cos(70) \approx 434.46$ Newtons, because momentum will be conserved for one skier.
So for all the skiers $T_{av} = n_{av} * T \approx 3128$ Newtons.
Am I right?

No. Your average number of skiers is just in the time from the first skier of the day takes the lift until he reaches the top. We want the average number of skiers for a full day where you can neglect the start of the day and just consider continuous operation.

Also, your expression for the force on each skier is correct, but the number you get is not. I suspect your calculator is set to radians but you are inputting the argument of the cosine in degrees.

 

[geoffrey159]

Thanks, my calculator was set to radians :) , so now I find $T \approx 234.63$ Newtons per skier.
Between each skier there is $1.5 \times 5 = 7.5$ meters so there are $\frac{100}{7.5} \approx 13.33$ gaps between each skier, when neglecting the start of the tow. So there are $13.33+ 1= 14.33$ skiers on the tow, on average.
So $T_{av} = 14.33 * T \approx 3362.25$ Newtons ?

 

[Orodruin]

Why are you adding one skier? Each skier has a space of 7.5 m so the average so the average density is the skier mass divided by the space available per skier, i.e., 70/7.5 kg/m. Multiply by the length of the rope to get the average mass being towed.

 

[geoffrey159]

oo) This problem is driving me nuts !
I added one skier because the number of skiers is number of gaps plus one. But I get your explanation, more or less ...
I need to think it over again.
Thank you !

 

[Orodruin]

I added one skier because the number of skiers is number of gaps plus one.

If we assume that there is actually one skier every 7.5 m then there will sometimes be 14 skiers in the tow and sometimes 13. Since 7.5x13 = 97.5, when a new skier gets onto the tow, the last one in line has just 2.5 m to go. During the time it takes for the tow to move those 2.5 m, there will be 14 skiers in the tow. However, when the tow has moved those 2.5 m, there is just 13 skiers in the tow and the next is not getting on until the tow has moved another 5 m. This is why you get an average number of 13.333... skiers in the tow, there are 13 skiers 2/3 of the time and 14 skiers 1/3 of the time.

 

[geoffrey159]

You make it very clear, thanks !

But I think there is a little momentum contribution missing because :
$\int_0^{t_f} f_{ext}(t) \ dt = t_f \times T_{av} - t_f \times 3128$
$\int_0^{t_f} \frac{dP}{dt} \ dt = 70 * 1.5 * (14-1) = 1365$

Following our discussion, $70 \times 13.333 \times g\cos(70) \approx 3128$ Newtons, and from previous thread, we agree that there will be 14 skiers of 70 kilos at 67 seconds.

So $T_{av} \approx \frac{1365}{67} + 3128 = 3149$

Right?

 

[Orodruin]

For the additional momentum contribution, I would again smear out the skiers evenly on the rope, leading to an average added momentum per second of:

Momentum per length x new length/time = Momentum per length x velocity = (70 kg x 1.5 m/s / 7.5 m) x 1.5 m/s

or, equivalently,

Added momentum per second = 70 kg x 1.5 m/s / 5 s = 21 N.

This seems to evaluate to what you got, but I was not completely able to follow your reasoning. It also assumes that skiers keep their momentum after letting go of the tow and that they enter the tow with zero momentum.

 

[geoffrey159]

Yes you did not follow my reasoning because I was still in my false, old way of reasoning (non steady flow)

 

[geoffrey159]

Momentum per length x new length/time = Momentum per length x velocity = (70 kg x 1.5 m/s / 7.5 m) x 1.5 m/s

I don't understand this formula, where it comes from, and how it relates to the mass variation of the system ( which is needed to conclude about momentum contribution )

 

[haruspex]

I don't understand this formula, where it comes from, and how it relates to the mass variation of the system ( which is needed to conclude about momentum contribution )

Rather than use the 7.5m, I'll demonstrate it directly from the given info:
Force is rate of change of momentum.
As a skier joins, the skier's speed changes from rest to 1.5m/s. That's a momentum change of 70kg * 1.5 m/s.
Since this happens every 5 seconds, the rate of change of momentum of skiers is 70kg * 1.5 m/s / 5 s = 21N.

 

[Orodruin]

It assumes that each skier of 70 kg has to be accelerated to 1.5 m/s from rest. This has to be done once every 5 s, which is what I had the other formula for. This is equivalent, just spreading the skier mass evenly over all of the rope. There is then 70 kg / 7.5 m to accelerate and the momentum per length of rope will be 70 kg x 1.5 m/s / 7.5 m. The average force needed to add this momentum to the rope is this quantity multiplied with the velocity (which is the rope length per time accelerated to 1.5 m/s). I realized this could be a bit confusing and therefore also gave the other option, which is conceptually easier.

 

[geoffrey159]

Ok one skier comes into the system every 5 seconds, but during each 5 second slice, you will have one skier that leaves the system, no ?

I am surely misunderstanding but I feel that you are saying that $\frac{\triangle P}{\triangle t}(t_{in}) = \frac{ P(t_{in}+5) - P(t_{in}) }{5} = 21 N$.
But we need $\frac{dP}{dt}(t) = \lim_{\triangle t \rightarrow 0 } \frac{ P(t+\triangle t) - P(t) }{\triangle t}$

 

[Orodruin]

Ok one skier comes into the system every 5 seconds, but during each 5 second slice, you will have one skier that leaves the system, no ?

Yes, but the skier leaving the system will typically take his momentum with him and not return it to the tow system. At least this is how most people leave a ski lift and most people get on a ski-lift by getting accelerated by it.

 

[geoffrey159]

Yes, I agree, he is taking his momentum with him so it must be substracted somewhere, where ? This exercise is painful !

 

[haruspex]

But we need $\frac{dP}{dt}(t) = \lim_{\triangle t \rightarrow 0 } \frac{ P(t+\triangle t) - P(t) }{\triangle t}$

That would give you the instantaneous force at time t, but that is unknown and unknowable. We want the average force. As a limit, it would be the limit as △t tends to infinity.

 

[haruspex]

Yes, I agree, he is taking his momentum with him so it must be substracted somewhere, where ? This exercise is painful !

The calculation was how much momentum does the tow have to add to the skiers as a set (including all past skiers). Since they keep it, there's no subtraction.

 

[Orodruin]

Yes, I agree, he is taking his momentum with him so it must be substracted somewhere, where ? This exercise is painful !

Since he is taking the momentum with him when he exits, the force on him just ceases. There is no longer any force between the rope and the skier. This is not the case when a skier enters the tow and needs to be accelerated by the rope .

 

[geoffrey159]

Yes I see why there is no difference in momentum when one skier is leaving the system, this question was a bit stupid ... I think this problem is getting on my nerves a little bit :s

So if I translate what I understand: each five seconds, there is a momentum rate of change of 21 Newton, so we have a telescopic momentum sum

$\begin{align} \int_0^{t_f}\frac{dP}{dt} \ dt =& P(t_f) - P(0) \\ =& P(t_f) - P(5\lfloor \frac{t_f}{5} \rfloor) + \sum_{k=1}^{\lfloor \frac{t_f}{5} \rfloor } P(5k) - P(5(k-1)) \\ =& (t_f - 5\lfloor \frac{t_f}{5} \rfloor) \frac{P(t_f) - P(5\lfloor \frac{t_f}{5} \rfloor)}{t_f - 5\lfloor \frac{t_f}{5} \rfloor} + 5 \sum_{k=1}^{\lfloor \frac{t_f}{5} \rfloor } \frac{P(5k) - P(5(k-1))}{5} \\ =& 21 * ( t_f - 5\lfloor \frac{t_f}{5} \rfloor )+ 21 \times 5\lfloor \frac{t_f}{5} \rfloor \\ =& 21 t_f \end{align}$

That seems correct, doesn't it?

 

[haruspex]

Yes I see why there is no difference in momentum when one skier is leaving the system, this question was a bit stupid ... I think this problem is getting on my nerves a little bit :s

So if I translate what I understand: each five seconds, there is a momentum rate of change of 21 Newton, so we have a telescopic momentum sum

$\begin{align} \int_0^{t_f}\frac{dP}{dt} \ dt =& P(t_f) - P(0) \\ =& P(t_f) - P(5\lfloor \frac{t_f}{5} \rfloor) + \sum_{k=1}^{\lfloor \frac{t_f}{5} \rfloor } P(5k) - P(5(k-1)) \\ =& (t_f - 5\lfloor \frac{t_f}{5} \rfloor) \frac{P(t_f) - P(5\lfloor \frac{t_f}{5} \rfloor)}{t_f - 5\lfloor \frac{t_f}{5} \rfloor} + 5 \sum_{k=1}^{\lfloor \frac{t_f}{5} \rfloor } \frac{P(5k) - P(5(k-1))}{5} \\ =& 21 * ( t_f - 5\lfloor \frac{t_f}{5} \rfloor )+ 21 \times 5\lfloor \frac{t_f}{5} \rfloor \\ =& 21 t_f \end{align}$

That seems correct, doesn't it?

Forgive me, but I'm not going to wade through that to check it when there's a far simpler way. Each 5 seconds, the rope imparts a momentum of 70kg * 1.5m/s to some new skier. Average force = average rate of imparting momentum = 70kg*1.5 m/s / 5 s.

 

[geoffrey159]

Hi,
I slept over this problem and I think I have put it straight in my head.

let $t_{in}$ be an input time (a skier comes in), $m=70kg$, and $v = 1.5m/s$.
As Orodurin explained, for a third of the time, there will be 14 skiers, and the last two thirds, they will be 13. So the momentums of the system, for every 5 seconds intervals, are :

$P(t_{in} + ) = m \times 0+ 13 \times m \times v$ (first skier is at rest, 13 at 1.5 m/s)
$P((t_{in}+\frac{5}{3})-) = 14 \times m \times v$ (14 skiers at 1.5 m/s)
$P((t_{in}+\frac{5}{3})+) = 13 \times m \times v$ (last skier left the tow)
$P((t_{in}+ 5 )-) = 13 \times m \times v$ (They are still 13 on the line just before a new one comes)

So clearly,

$\begin{align} \int_{t_{in}}^{t_{in} + 5} \frac{dP}{dt} \ dt =& \int_{t_{in}}^{t_{in} + \frac{5}{3}} \frac{dP}{dt} \ dt + \int_{t_{in}+ \frac{5}{3}}^{t_{in} + 5 } \frac{dP}{dt} \ dt \\ =& (14 - 13) \times m \times v + (13 -13 ) \times m \times v \\ =& mv \\ =& 105 \ \text{N.s} \end{align}$

So as Haruspex explained, the rate of change of momentum, from input time, for 5 seconds is $\frac{105}{5} = 21\ \text{N}$.

Now there are exactly $\lfloor \frac{t_f}{5} \rfloor = 13$ of these intervals, starting from 0, plus a remaining interval that does not spread 5 seconds but exactly $\frac{5}{3}$ seconds, because
$t_f - 5\times 13 = \frac{100} {1.5} - 65 = \frac{200}{3} - 65 = \frac{5}{3}$.
So the momentum difference in this interval will be $mv$ as shown above.

Putting it all together,

$\int_{0}^{t_f} \frac{dP}{dt} \ dt = 14 \times m \times v$

 

[geoffrey159]

However, one does not have to take first input time to 0, and there is not always 13 5-seconds intervals after, it can be 12. It depends.
After a case analysis, I get to the conclusion that :

$\int_0^{t_f} \frac{dP}{dt} \ dt = \left\{ \begin{array}{ll} 14mv & \mbox{if observer waits less than 5/3 s before first skier comes } \\ 13mv & \mbox{if observer waits between 5/3 s and 5s (not 5s) before first skier comes} \end{array} \right.$

So I will never get the same momentum contribution? what a joke !

 

[Orodruin]

You are still overthinking it. There really is nothing more to it than momentum needing to be added at a rate of 21 N as presented by haruspex and myself above.

 

[geoffrey159]

Hello, thanks for helping so much !
With everything you explained, I tried to put pieces back together to make it fit into the formula.
I think that finally, finally I can write something convincing about this problem which is not easy at all, I probably wouldn't have been able to say anything without your guidance. At the end, I find this problem much more involved with calculus than "introductory mechanics".

 

[Orodruin]

So now that you have the solution, I would suggest going back to probably the sanest and most straightforward post in this thread:

An alternative approach is to look at the energy given to each skier (PE+KE), deduce the power, and hence the force.

Consider two points in time which are 5 s apart. What is going to be the difference between the two times when it comes to the tow system?
What average force is necessary to accomplish this change in energy in 5 s for a system that is moving with 1.5 m/s (i.e., a system that has moved 7.5 m)?

 

[geoffrey159]

:) I don't have any knowledge about energy yet, it's too early !

 

[geoffrey159]

One last question about the meaning of average force in this context.
The definition of average force in my book is $\vec F_{av}\triangle t = \int_t^{t+\triangle t} \vec F(s) ds$

However, it is not straightforward here, because you get different results depending of the state of your system when you start watching it.

To me, a suitable way is to first average over all observable systems first, and then average force as usual:

I mean that in one third of the possible configurations we get
$\int_{0}^{t_f} T dt = 14 mv + 3128 t_f$

And in two third of possible configurations we get
$\int_{0}^{t_f} T dt = 13 mv + 3128 t_f$

So first step is to average over all configurations, so you multiply first equation by 1/3, second equation by 2/3, and you add them.

Then you average force as usual, which gives
$T_{av} =\frac{1}{t_f}( \frac{40}{3} mv + 3128 t_f ) = \frac{3}{200} ( \frac{40}{3} \times 70 \times \frac{3}{2} +\frac{200}{3} 3128 ) = 21 + 3128 = 3149 N$

That would work don't you think?

 

[BvU]

Geoff, there really is no need to consider purely hypothetical configurations and then average. The averaging has been done for you beforehand: the 5 seconds itself is an average. Could well be that this is a day average and that during lunchtime there were hardly any skiers on the tow.

 

[haruspex]

One last question about the meaning of average force in this context.
The definition of average force in my book is ## \vec F_{av}\triangle t = \int_t^{t+\triangle t} \vec F(s) ds #

That's the average over a specific time interval $(t, t+\Delta t)$. You are not given such a time interval here, so the appropriate interpretation is the average over an arbitrarily long time. This makes the little variations over each 5 second interval irrelevant.