In https://en.wikipedia.org/wiki/Pressure#Formula, the normal force per area $dA$ is $d\vec{F}_n=-p\,\vec{n}\,dA$, where $p$ is the pressure. It is stated that, for any surface $S$ in contact with the fluid, the total force exerted by the fluid on that surface is the surface integral over $S$ of the right-hand side of the equation: $\vec{F}=\int_Sd\vec{F}_n=\int_S -p\,\vec{n}\,dA$. My question is why $\vec{F}=\int_Sd\vec{F}_n$, and not $\vec{F}=\int_S d\vec{F}$. I would like to understand why we sum the normal forces $d\vec{F}_n$ at each point, and not the whole forces $d\vec{F}$ at each point of $S$. Is it because the parallel component of each infinitesimal force $d\vec{F}$ is $0$ ?
$\begingroup$
$\endgroup$
Add a comment
|
$\begingroup$
$\endgroup$
1
Yes. For fluids of negligible viscosity, the shear forces amount to zero, and only the forces normal to the surface being considered matter. However, this doesn't work with some (non-newtonian) fluids, like cornstarch in water, because of the aforementioned high viscosity.
-
1$\begingroup$ It diesn’t even work for deforming Newtonian fluids which typically exhibit a traction vector at the surface that is not normal to the surface (due to viscous contributions to the stress tensor). $\endgroup$ – Chet Miller Jul 19 '18 at 1:30
$\begingroup$
$\endgroup$
The article you cite is only about pressure, so when it discusses forces, it only deals with the contribution due to the pressure.
In general, fluids also exert skin-friction forces tangential to surfaces, and for many flows in air, water, & other fluids, they can be much larger than the contributions due to the pressure.
