0
$\begingroup$

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 $0$ ?

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
1
  • 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
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.