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Question on effort

  1. Dec 4, 2006 #1
    i have to design a vehicle that will accelerate up a 1/8 gradient hill to 11.11m/s in 15s

    i need the effort required and wheel torque.

    mass = 170kg
    rolling resistance = 50N/t (0.05N/kg)
    air resistance = 13.6N
    g = 9.81m/s^2

    i have calculated:

    a = 11.11*(1/15) = 0.74m/s^2
    theta = 7.18 deg
    sin theta = 0.125 deg

    so i have the effort as:

    E = ma+mgsin_theta+Froll+Fair
    E = 125.8+(170*9.81*0.125)+0.05+13.6
    E = 347.9125N



    is E the torque i need at the wheels?
    if not how would you go about working it out.

    i tried with a motor of 40N, wheel radius of 0.2m and gear ratio of 10:1

    i worked it out like this.
    axle torque = 40*10 = 400N

    so to find the torque at the wheel do i divide the axle torque by wheel radius

    400/0.2 = 2000 that seems wrong to me??
     
  2. jcsd
  3. Dec 5, 2006 #2

    andrevdh

    User Avatar
    Homework Helper

    You need to multiply the rolling resistance with the mass:

    [tex]E = m(a + R + g/8) + 13.6 = 357\ N[/tex]

    which makes only a slight difference.

    To find the torque at the wheel, [tex]\Gamma _w[/tex], you need to multiply the effort by your wheel radius. The effort will be shared by the four wheels (if all four a driving) reducing the torque to a quarter at each wheel.

    In order to choose an appropiate motor you need to consider the torque of the motor,[tex]\Gamma _m[/tex], and gear ratio [tex]G[/tex] such that

    [tex]\Gamma _m \times G = \Gamma _w[/tex],

    http://www.blueink.com/CLASS/physcom1/gear.htm

    Torque, [tex]\Gamma[/tex], is the turning effect of a force [tex]F[/tex]. If the force is applied with a longer lever arm [tex]r[/tex] the turning effect will be greater:

    [tex]\Gamma = F \times r[/tex]

    the S.I. units of torque is newtons meter.
     
    Last edited: Dec 5, 2006
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