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 $0$ ?
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$\begingroup$ Just ask yourself : can I apply a force that is parallel to the surface of water which is kept in a bowl? $\endgroup$– SidarthJun 4, 2021 at 1:03
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2 Answers
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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.
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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). $\endgroup$ Jul 19, 2018 at 1:30
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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 Jul 18, 2018 at 20:30