# Unit of force in Navier-stokes

**URL:** <https://community.freefem.org/t/unit-of-force-in-navier-stokes/1900>\
**Category:** General Discussion\
**Created:** [July 19, 2022, 8:49pm UTC](https://community.freefem.org/t/unit-of-force-in-navier-stokes/1900 "2022-07-19T20:49:01Z")\
**Posts on this page:** 2\
**Page:** 1

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**Author:** ![zese](https://yyz2.discourse-cdn.com/flex030/user_avatar/community.freefem.org/zese/32/917_2.png) [@zese](https://community.freefem.org/u/zese)\
**Post date:** [July 19, 2022, 8:49pm UTC](https://community.freefem.org/t/unit-of-force-in-navier-stokes/1900/1 "2022-07-19T20:49:01Z")

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Dear all,

I have a fundamental question which may be simple. In Navier-stokes momentum equation as the following:

 ![image](https://canada1.discourse-cdn.com/flex030/uploads/freefem/original/2X/2/2ea03f49d425fa403d438b9e78db622478dad636.png)  
is the unit of F hear N/m^2 for a 2D simulation? Because I have to export F from freefem++ I need to know that should I multiply it by the area of the element to have F (N) ? I am confused now.

BR

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**Author:** ![aszaboa](https://yyz2.discourse-cdn.com/flex030/user_avatar/community.freefem.org/aszaboa/32/2918_2.png) [@aszaboa](https://community.freefem.org/u/aszaboa)\
**Post date:** [July 20, 2022, 8:57am UTC](https://community.freefem.org/t/unit-of-force-in-navier-stokes/1900/2 "2022-07-20T08:57:31Z")

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Technically, your equation is dimensionless 🙂  
I think if you redimensionalize your force with rho\*U^2, you will get a N/m^3. If you multiply this with the area, you will get N/m - force per unit spanwise length - this is because the force for the three dimensional problem can be calculated with a multiplication with the spanwise length/depth.