# Torque exerted by pressure in a fluid

If I have a solid rigid body (let us call it a duck) in a static regime submitted to the forces of pressure $P$ of a fluid (no shear forces) I can easily calculate the total force $\mathbf{F}$ exerted on the duck, as $$\mathbf{F}=-\int P\: \mathrm{d}\mathbf{S},$$ where $\mathrm{d}\mathbf{S}=\hat{n}\mathrm{d}S$, $\mathrm{d}S$ is the differential of surface, pointing normal to the surface with unit vector $\hat{n}$, and the integration is over the whole surface of the duck in contact with the fluid.

Now, imagine the same body is held in such a way that a particular point (at position $\mathbf{r}_{\rm fix}$) in the duck has to remain still, but the duck may rotate about this point. How do I calculate the torque $\mathbf{T}$ exerted by pressure forces about this point?

Attempt 1

Is it $$\mathbf{T}=-\int (\mathbf{r}-\mathbf{r}_{\rm fix}) \times \hat{n}\; P(\mathbf{r})\;\mathrm{d}S\quad\mathrm{?}$$

Attempt 2

Calculate force $\mathbf{F}$ as above and then find some kind of center of area $\mathbf{R}$, such that $$\mathbf{R}=\frac{1}{S}\int (\mathbf{r}-\mathbf{r}_{\rm fix})\mathrm{d}S,$$ where $S$ is the total surface. Then, $\mathbf{T}=\mathbf{R}\times\mathbf{F}$ ?

Additional questions: does this kind of calculation of torque in a fluid (attempt 1 or 2) have a particular name in the literature? I couldn't find it anywhere. How does this generalize for shear stress?

Edit: I forgot a minus sign.

Attempt 1 is the correct equation.

Both of your equations can be extended for an arbitrary stress state on the surface: if you combine the isotropic (pressure) and deviatoric (shears, etc.) stresses into an appropriate stress tensor, then you'll find the force and torque on the “duck” are:

$$\mathbf{F} = \int \bar{\bar{\sigma}}\cdot\hat{n}\ dS$$

$$\mathbf{T} = \int (\mathbf{r} - \mathbf{r_{fix}}) \times \bar{\bar{\sigma}}\cdot\hat{n}\ dS$$

It is indeed possible to find some $\mathbf{R}$ so that your equation in Attempt 2 holds, but it is almost certainly not the center of area; it is sometimes called the hydrodynamic center or center of pressure because its location is affected by the flow conditions. Finding it in general is a pain because there are multiple different definitions for it under different conditions, but it's discussed in plenty of detail in this book for low-Re conditions.

I don't believe I've ever seen anyone give this process of integrating tractions over a surface to get net forces/torques a name, but it has led to plenty of results that do have names; Stokes' law, Faxen's laws, etc.

• Thank you. Of course there is a minus, I edited the results above. – Mauricio Jul 15 '18 at 10:50
• If you intended to define $\hat{n}$ as pointing outwards of the body, then you would indeed need a minus sign on both of your equations. But both your original (pre-edit) equations for $\mathbf{F}$ and $\mathbf{T}$ work out in the inwards-$\hat{n}$ case because 1) pressure has the opposite sign of the average isotropic stresses and 2) the correct cross product order for $\mathbf{T}$ is $mathbf{r} \times \mathbf{F}$. – aghostinthefigures Jul 15 '18 at 10:58
• Yeah, I just prefer to keep $\hat{n}$ outwards. – Mauricio Jul 15 '18 at 11:01
• Sorry I rejected your edit, but it was correct. Thanks again – Mauricio Jul 15 '18 at 11:30