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Help! (required force to move a sliding door weighing 100 kg with two rollers)

  1. Oct 8, 2009 #1
    page0003.jpg

    what is the required force to start moving a sliding door weighing 100 kg with two rollers.
    what are the forces acting on the system?

    shall i used rolling friction on this?

    what formula to use?

    thanks a lot.
     
  2. jcsd
  3. Oct 8, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi johnpaul! Welcome to PF! :wink:

    There's no such thing as "rolling friction", since a pair of rolling surfaces don't move with respect to each other.

    There's rolling resistance, which is essentially energy lost due to deformation of the wheel, but I shouldn't think three's much of that with steel.

    The main problem will be friction between the wheel and the axle though its centre. :smile:
     
  4. Oct 8, 2009 #3
    Re: Welcome to PF!

    But wouldn't the bearings be subject to rolling resistance as opposed to friction?
    Maybe you assumed no bearings whereas I assumed bearings :-)
     
  5. Oct 9, 2009 #4
    hi tim, hi molydood,

    please correct me if im wrong.

    here's my computation:

    Weight of door=100 kg
    Number of wheels= 2
    Weight per wheel =(100kg)/(2)= 50kg

    F1= αW / r

    F1- resistant force of a single wheel
    α - coefficient of rolling friction
    W - weight
    r - radius (0.015m)

    F1≈(0.0005 m)(50 kg) / (0.015 m)
    F1≈ 1.67 kg; say 2kg

    F''≈(F1)(2 wheels)
    F''≈(2kg)(2 wheels)
    F''≈4 kg

    Force to move the 100 kg door horizontally:
    F≈ 4 kg



    Values for rolling friction from various sources are not consistent and the following values should only be used for approximate calculations.

    Steel on Steel α = 0.0005m
     
  6. Oct 9, 2009 #5
    hi,

    can anyone advise me what is the proper way in computing the friction between wheel and the axle? its just a simple door carrier(please refer to the drawings above).

    thanks
     
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