# Homework Help: Ski tow

1. Dec 14, 2014

### geoffrey159

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
Impulse and average force

3. 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 ?

2. Dec 14, 2014

### 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.

3. Dec 15, 2014

### 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.

Last edited: Dec 15, 2014
4. Dec 15, 2014

### Orodruin

Staff Emeritus
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.

5. Dec 15, 2014

### 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 ?

6. Dec 15, 2014

### Orodruin

Staff Emeritus
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?

7. Dec 15, 2014

### 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?

8. Dec 15, 2014

### Orodruin

Staff Emeritus
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.

9. Dec 15, 2014

### 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 ?

Last edited: Dec 15, 2014
10. Dec 15, 2014

### Orodruin

Staff Emeritus
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.

11. Dec 15, 2014

### 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 !

12. Dec 15, 2014

### Orodruin

Staff Emeritus
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.

13. Dec 15, 2014

### 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?

14. Dec 15, 2014

### Orodruin

Staff Emeritus
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.

15. Dec 15, 2014

### geoffrey159

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

Last edited: Dec 15, 2014
16. Dec 15, 2014

### geoffrey159

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 )

17. Dec 15, 2014

### haruspex

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.

18. Dec 15, 2014

### Orodruin

Staff Emeritus
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 realised this could be a bit confusing and therefore also gave the other option, which is conceptually easier.

19. Dec 15, 2014

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

20. Dec 15, 2014

### Orodruin

Staff Emeritus
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.

21. Dec 15, 2014

### geoffrey159

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

22. Dec 15, 2014

### haruspex

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.

23. Dec 15, 2014

### haruspex

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.

24. Dec 15, 2014

### Orodruin

Staff Emeritus
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 .

25. Dec 15, 2014

### 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?