- #1

MechEgr

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- 0

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.

Thanks.

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- Thread starter MechEgr
- Start date

- #1

MechEgr

- 12

- 0

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.

Thanks.

- #2

Jano L.

Gold Member

- 1,333

- 74

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.

- #3

jbriggs444

Science Advisor

Homework Helper

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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)

- #4

MechEgr

- 12

- 0

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.

- #5

jbriggs444

Science Advisor

Homework Helper

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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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