torque question

Discussion in 'Physics' started by tracecom, Sep 25, 2014.

  1. tracecom

    Thread Starter AAC Fanatic!

    Apr 16, 2010
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    Suppose a motor stalls at 20kg.cm and its shaft is .25" in diameter. What is the maximum load that can be exerted on a cord wound directly on the shaft? (Be nice; remember I am a marketing guy.)
     
  2. studiot

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    Nov 9, 2007
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    Well I assume that the cord is wound round securely at one and and the other hangs down, supporting a pan.

    Then weights are added until it stalls and the question is what weight stalls it?

    So the shaft has diameter 0.25 inches = 6.35mm = .00635m
    So the radius is half this or .003175m.

    The cord hangs down with tension T equal to the weight in the pan.

    This develops a torque of T x radius which we are told stalls at 20 kgcm = 0.2 kgm

    So max T x R = 0.2

    T = 0.2/0.003175 = 63 kg force, or 63 x g Newtons force = 618N.

    So your motor shaft will wind up a load of just under 63kg mass.
     
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  3. tracecom

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    Thanks for the answer and for showing the math to get there.
     
  4. KL7AJ

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    Nov 4, 2008
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    The tangential force is the torque divided by the radius. .125"=.3175cm. 20/.3175=62.9 Kg.

    Eric
     
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  5. studiot

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    Nov 9, 2007
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    The only real thing is that kilograms force are a non standard unit.

    In the ISO metric system the kilogram is a unit of mass and the unit of force that corresponds is the newton.

    So working is kgcm is definitely likely to lead to trouble.

    since you had a 1/4 inch shaft and are in America, why not keep it imperial and use poundsinches poundsfeet?
     
  6. tracecom

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    I converted 63 kg to 139 pounds, which tells me what I need to know, which is that a 100 pound test cord would be suitable.
     
  7. studiot

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    Glad it is sorted for you.

    I assume you know that if you wrapped the cord several times around the shaft the anchoring force needed at the fixed end would be a lot less than the lifting force available at the free end?
     
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