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.

  • $\begingroup$ Thank you. Of course there is a minus, I edited the results above. $\endgroup$ – Mauricio Jul 15 '18 at 10:50
  • $\begingroup$ 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 $mathbf{r} \times \mathbf{F}$. $\endgroup$ – aghostinthefigures Jul 15 '18 at 10:58
  • $\begingroup$ Yeah, I just prefer to keep $\hat{n}$ outwards. $\endgroup$ – Mauricio Jul 15 '18 at 11:01
  • $\begingroup$ Sorry I rejected your edit, but it was correct. Thanks again $\endgroup$ – Mauricio Jul 15 '18 at 11:30

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.