• This topic has 3 replies, 3 voices, and was last updated 13 years ago by nbt.
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  • Physics help please – portable drag lift.
  • Duane…
    Free Member

    OK, long story short, my mate wants to make one of these;
    http://www.skiliftportabletow.com/
    and as Im the engineering student Ive been asked to do all the calculations, problem being Im stuck on the most basic equations.

    OK, here we go.
    Want to be able to pull someone weighing 120kg up a 30* slope at 4m/s.
    Coefficient of friction of snow is U=0.14 (apparently).

    F=mg = 1178N
    Resultant = 1178cos30 = 1018
    F(friction)=UR = 142.6N

    So that means theres 142.6N acting of the guy down the slope, so 142.6N is going to be needed in the cable just to hold the guy stationary right?

    So, from there, how do we work out the extra force needed to pull him up the slope at 4m/s? This is where Im being total blonde and cant work it out, f=ma doesnt work cos theres no acceleration (assuming its already started up and is travelling at a constant velocity).

    Surely there must be an equation linking velocity, mass, and the force required to keep that velocity?

    Any ideas?

    Once we’ve got this sorted we can move onto what kind of gearing we need to get the required output force from the engine 😛

    Ta, Duane.

    Wibble89
    Free Member

    731.6 N

    You must overcome the force of the object acting down the slope plus friction.

    so F = 1178sin30 + 142.6 = 731.6N

    or by energy, that used to lift it, plus losses from friction (1 second step)

    w.d. = f * d
    w.d. = 1178 * 4sin30 + 142.6 * 4
    w.d. = 2926.4

    4 metres of cable per second

    f = w.d. / d
    f = 2926.4 / 4 = 731.6N

    plus 10% for other losses…

    p.s. to hold steady on the hill with no movement would be 1178sin30 – 142.6

    joemarshall
    Free Member

    I wouldn’t do too many calculations for this – I’d look at the guy selling one that obviously works, look at what speed he is towing, and the power of his motor, then buy a similar power motor, and gear down to a similar speed!

    Also, depending on what type of motor you’re using, I’d maybe add more than 10% for ‘other losses’ – particularly if you’re using something where overpowering the motor might make things blow – possibly less of a pain with a small petrol motor like that, we’ve blown some big electric motors at work and it can be a right old pain.

    Joe

    nbt
    Full Member

    Speak to these guys

    http://www.snowflex.co.uk/contactus.asp
    they made the Briton Lift which is still used in many places – I was skiing on one in January this year just outside Halifax 🙂

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