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Total force exerted by the fluid on the surface is the sum of normal forces

Physics Asked by user39756 on March 10, 2021

In https://en.wikipedia.org/wiki/Pressure#Formula, the normal force per area $dA$ is $dvec{F}_n=-p,vec{n},dA$, 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: $vec{F}=int_Sdvec{F}_n=int_S -p,vec{n},dA$. My question is why $vec{F}=int_Sdvec{F}_n$, and not $vec{F}=int_S dvec{F}$. I would like to understand why we sum the normal forces $dvec{F}_n$ at each point, and not the whole forces $dvec{F}$ at each point of $S$. Is it because the parallel component of each infinitesimal force $dvec{F}$ is $0$ ?

2 Answers

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.

Answered by Brandon on March 10, 2021

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.

Answered by D. Halsey on March 10, 2021

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