What is the average force required to pull the rope?

[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.

 

[geoffrey159]

There are way too many averages in this problem, it goes beyond my understanding.
Many thanks for your help Haruspex, Orodruin, Bvu , I think you've done everything possible to make me understand, but it is not possible.
I give up for now, maybe once I know energy I'll have another perspective on this problem. Thanks again!

 

[NascentOxygen]

There are way too many averages in this problem, it goes beyond my understanding.
Many thanks for your help Haruspex, Orodruin, Bvu , I think you've done everything possible to make me understand, but it is not possible.
I give up for now, maybe once I know energy I'll have another perspective on this problem. Thanks again!

I think for half marks you should just work out the force needed to transport the string of skiers up the slope, neglecting the extra force involved in initially bringing each up to speed. The problem neatly breaks apart into two parts, so if you complete one you should be eligible for half the marks.
Good luck with your studies! http://thumbnails112.imagebam.com/37333/0363e9373324851.jpg