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## 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 ?