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Homework Help: Ski Lift

  1. Feb 17, 2006 #1
    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. :blushing:
    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.

    http://www.walchem.com/email/movies/lift.jpg [Broken]
     
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Feb 17, 2006 #2

    berkeman

    User Avatar

    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....
     
  4. Feb 18, 2006 #3
    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.
     
  5. May 27, 2010 #4
    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
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