# Ski Lift

1. Feb 17, 2006

### web4deb

I'm planning on building a small ski lift for my kids. It's been 20 years since I've taken a physics class so I'm lucky to remember what gravity is.
Here's a poor image of the hill. I would like to know the following:

A) how much lbs of tension is against the lift wire?
B) how heavy should the counter weights (D&E) be so they don't rip out of the ground?

Facts and Assumptions
C) distance between towers A&B and B&C are 200' each
D) Elevation between towers A&B is 60', B&C is 40'
E) Would like to be able to lift a total of 4 people at a time...assume 175 lbs/person
F) Not as important...lift wire weight is 300 lbs total

Feel free to assume anything else that's missing. I'm basically looking for some formulas I can use to play with the numbers.

2. Feb 17, 2006

### Staff: Mentor

Hmmmm. Looks like something I'd like to get a PE (licensed professional engineer) design for me. Even if you way overbuild it, you'll want to think about the liability ramifications. Not to discourage you. I just think that a project like this deserves a professional design. Plus, there are probably already industry-standard guidelines for things like ski lifts, and an appropriate PE will be familiar with them. Now a tow rope, that you could do on your own....

3. Feb 18, 2006

### web4deb

I am well aware of my legal liabilities. This is not a lift where people would get lifted off the ground, but more of a rope toe with a pull rope hanging from the main line. (The hump in the middle of the hill makes it impossible to put in a rope toe) If you could point me in the right direction for some colculations, I would appreciate it. Thanks.

4. May 27, 2010

### uwmengineer

Friction around a pulley: add rope and skier
V-shaped pulley: T[2]/T[1] = exp(mu * beta / sin (alpha/2) ); mu = coeff of friction, beta = angular meas of belt around pulley in radians, alpha = angular meas of the v in the pulley
rope and pulley: T[2]/T[1] = exp(mu * beta ); mu = coeff of friction, beta = angular meas of tow rope around pulley in radians, T[2]= the tension in the rope going uphill, T[1]= the tension in the rope going downhill, T[0]= tension at the middle of the catenary

Tow Rope catenary this is the rope part only
T[2] = [T[2][0] + w[2] * s[2] ]^[0.5];s = length of rope (lowest point to highest point), w is the unit weight ie lbs/ft so W=ws where W is the total weight of the rope segment
c = T[0] / w; c is a constant so T[0] = wc
s = c * sinh (x/c); x = distance horizontally from the lowest point in the catenary to the highest
y = c * cosh (x/c); y = sag in cable + c [note: solve for c using these two eq.]
y^2 - s^2 = c^2
T[2] = w*y [note: max]
T[0] = w * c [note: min]

Skier skier part only
initially the skier is not moving, draw a force diagram, x is positive going uphill, y is positive above the ski slope and resolve the forces:
Friction = mu * N; mu= coeff of friction (.04? not much, you could assume frictionless), N=normal force
x direction forces=friction + weight * cos(theta); theta = angle between the ski slope and the weight vector
y direction forces = normal force N = weight * sin(theta)
after the skier is moving use F = ma, but consider highest forces and add a factor of safety

horse power of motor = rpm*torque*K; K=constant to make the units agree, torque will depend on the radius of the pulley and the tension in the rope: tau = r X F ;this is, as you recall, a vector cross product where tau is torque

you might use energy as a check: kinetic energy = translational energy (1/2 mv^2) + rotational (1/2 Iw^2)

Last edited: May 27, 2010