# Building a Ski Tow Rope

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rlay023
Hi Everyone -

I found a previous post here that has helped me size an engine for a ski tow rope but wanted to take that a step further an add friction into the equation.

If I followed that problem corrected, the tow rope requires 821W calculated as follows:
power = work/time
work = mhg (272kg)(14m)*(9.8 m/s^2)
time = target speed of 5.5 ft/s, it will take 45.5 seconds to travel 250 feet.

.821kW *. 74569 = .61bHP required

The friction coefficient of snow/ice can range from .03 - .1 and the angle of ascent is 18.9%.

Can anyone help layer in the friction loss into the power calculation?

Mentor
Welcome to PF.

Can you provide a link to that previous thread? That would help us understand what the context is for your question. Thanks.

rlay023

Homework Helper
2022 Award
Carry your units in all calculations
.821kW *. 74569 = .61bHP required
This should be $$\ (0.821 kW)1.34HP/kW=(whatever) HP$$ And choose either meters or feet! Nasa discovered this on mars.

And hurry you only have a couple more good years of snow...

russ_watters
rlay023
Thanks for your replies. Let's see if I can restate this in all the same units.

What is the required HP to pull 272.2kg up a 14.0M incline with a slope of 18.9% in 45.5 seconds? Using the previous example and the response above, I arrive at 1.10.

Work = mgh = 272.2(9.8)(14.0)
Time = 45.5 second
Power = .82147kW

HP = .82147kW * 1.34 HP/kW = 1.10bHP

The above assumes no friction which we know to be anywhere between .03-.1 depending on the snow/ice conditions. How would we layer the friction loss into the original engine sizing problem?

hutchphd
Mentor
The above assumes no friction which we know to be anywhere between .03-.1 depending on the snow/ice conditions. How would we layer the friction loss into the original engine sizing problem?
Power is force times distance over time. Do you know how to calculate the friction using the skier weights, slope and friction coefficient?

rlay023
Power is force times distance over time. Do you know how to calculate the friction using the skier weights, slope and friction coefficient?
That is where I am stuck. The examples I have found are only on flat surface with friction or slope surface with no friction.

Mentor
That is where I am stuck. The examples I have found are only on flat surface with friction or slope surface with no friction.
The normal force is the cosine of the angle times the weight. That's the part of the weight that acts against the slope.

rlay023
Something along the line of:

F = uN = .01 * (2619.4258) = 261.94N
where N = Cos(18.9)*272.2*9.8

Power = (F*D)/t = (261.94N)*(75.2m)/45.5sec = 433.10W

Homework Helper
2022 Award
The number seems correct (you wrote down ##\mu =.01## but then used ##\mu =0.1## ). You will likely need a gearbox and some rigging which will add some drag and better to have too much power than not enough.. There are many ~6HP engines to be had from China really cheap. Don't know how good they are.

rlay023
Good catch. I used .1 in the actual equation, as the upper limit of the static friction coeficient - just transcribed it wrong.

Sorry for the basic question here but (assuming the work/power calcs are done correctly on the friction part), how do I factor in the .433k1w from friction. Do I calculate work given the normal force and then add 433.1W to it before converting to HP? Or, do I add it to the initial power calculation given at the beginning of this thread of .82147kw

Homework Helper
2022 Award
I would convert the 0.433 kW to HP and add it to the total. There are many ways but since I assume you will be running this on "horsepower" that seems the most useful unit.

rlay023
Thanks again for all your help.

Depending on the how this is done provides very different answers, which I didn't expect to happen.

Approach 1: If I add .43310kw to the original .82147kw, I get 1.25457kW or 1.68 bHP.

Approach 2: If I take the normal force of 2619.4258N and add the frictional force of 261.942N, I get 2881.36838N times 75.2M (distance) over 45.5 sec (time) = 4.76407 kW or 6.39 bHP.

Homework Helper
2022 Award
The sideways distance and the vertical distance are not the same. For gravity you had correctly used the vertical distance 14.4m to get the lifting energy required. For the friction you use the sliding distance 75.2m to get the "sliding work" required. Then you add them. So#2 is wrong .
Also be aware these give average power requirements and will fluctuate if the slope is nonuniform, so leave margin above 1.68 HP.

rlay023
Awesome. Thanks you so much. I had sized a 5HP to be on the safe side.

hutchphd
rlay023
In doing further research on the forum, I stumbled upon a torque related post relevant to the same project - https://www.physicsforums.com/threads/help-with-motor-for-pulling-uphill.957569/

Following that as guide, I come up with the following solution:

F = G * sin (theta) + R , where:
G = m(g) = (272.2)(9.8) = 2668.93
sin (theta) = .32
cos (theta) = .95
R = G *cos (theta)*mu = (2668.93)(.95)(.1) = 252.50
F = 2668.93*.32 + 252.50 = 1117.18

P = Fv, where:
v = 1.68 m/sec
P= 1117.18*1.68 = 1872.57

T = P/omega, where:
P = F(v) = 1117.17(1.68) = 1872.57W