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Converting belt tension to radial compression force

  1. Sep 27, 2012 #1
    Imagine you are getting dressed in the morning and fastening a belt around your waste. You pull on the loose end of the belt, developing tension T throughout the length of the belt. Using this information, how can you determine the amount of compression the belt is applying to your waste?

    I apologize if this topic has already been covered. I could not find it in the forums, but if it has, please direct me to the correct thread.

  2. jcsd
  3. Sep 27, 2012 #2

    Jano L.

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    Suppose the belt is circular and it is pulled by force T.

    One can calculate the force perpendicular to the belt element delimited by angle sector of [itex]\Delta \varphi[/itex] radians (draw a picture).

    There are two forces pushing the band of the belt towards the belly, each acting on one end of the band and having equal magnitude [itex]T\Delta \varphi/2[/itex]. Their sum is [itex]T\Delta \varphi[/itex].

    This is the total force pushing the band of the belt towards the belly.

    With information on the radius and width of the belt, one can then calculate the pressure.
  4. Sep 28, 2012 #3


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    Rather than compute the tension on a small belt element of size Δ, one can consider the tension on the front 180 degrees.

    The net force on this semicircular segment is 2T. If you consider a vertical slice through yourself from hip to hip, the total force applied across this slice must therefore be 2T. The length of this slice is the diameter D of the circle formed by your belt.

    Accordingly, the pressure within the bulk of your body must be given by 2T/D. This is the same as the pressure exerted by the belt.

    [This "pressure" is per unit length, not per unit area]

    Divide by the width of the belt to get pressure per unit area.

    On the [questionable] assumption that your waist is circular and that the belt has length L and width W and just barely reaches around, this would mean that the pressure exerted by the belt on your body is given by 2piT/(L*W)
  5. Oct 1, 2012 #4

    I follow your logic all the way until your very last point. I don't quite understand where the pi term came from in your equation. I understand that the force exerted over half of the cross-sectional area should be 2T/(L*W). It seems that somehow when you convert that to pressure applied circumferentially, you get an extra "pi" term from somewhere?

    Thanks for your help.
  6. Oct 1, 2012 #5


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    The pressure per unit distance at the centerline of your body must be given by 2T/D, you agree?

    Given that an equilibrium exists, the pressure per unit distance around the circumference of your body must therefore also be given by 2T/D, you agree?

    The length L of the belt is given by pi * D. It follows that D = L/pi.

    This means that 2T/D can be rewrtten as 2T/(L/pi) = 2piT/L

    Factoring in the width of the belt, the pressure per unit area is 2piT/(L*W)
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