# Torque due to continuous force distribution / pressure [closed]

In my fluid mechanics course, I was exposed to some cases where I need to calculate the torque due to the pressure and all solutions manuals or online tutorials take it as a known fact that $$d\tau=rdF$$ ($$r$$ is perpendicular to $$F$$). If I apply the normal differential rules to it, it should be $$d\tau = r dF + F \mathrm{d}r.$$ I'm trying to understand why the second term is cancelled out.

The best answer I thought of was that because the force is due to a pressure so at each point it would have a value of $$p\cdot \mathrm{d}\mathbf{A}$$ (with $$p$$ the pressure and $$\mathbf{A}$$ the area) which is an infinitely small force and thus F is 0. Is that correct?

What if instead the force was defined as a function of r (continuous force distribution), so it does have a value at each point; for example $$F = kr^2$$?

Here is an example of such a case:

• You do realize that they are working with shear stress here, not normal stress (pressure), right? And since when is torque the dot product of force and radius? Commented Oct 30, 2019 at 14:13
• If you mean in this example yes I do, but I'm asking in general really. This was not meant to be in vector form, someone edited it since I didn't use the correct formatting. Since I only have force and distance in one direction perpendicular to each other, I didn't bother to use vectors and complicate things. However, do correct me if I did something wrong. @ChetMiller Commented Oct 30, 2019 at 15:55

I really don't like the way your book explained this, and it is definitely not the way I would have done it. My problem is with the equation $$dF=\tau_wdA=\tau_w(2\pi r)dr$$. Force is supposed to be a vector quantity (with a single specified direction), and yet the shear stress $$\tau_w$$ varies with direction around the circumference, so their expression for dF can't represent a force. Here is how I would have done it, which I hope makes more sense:
The differential force acting on a differential area of the disk $$dA=rd\theta dr$$ is $$\mathbf{dF}=\tau_w rd\theta dr \mathbf{i_{\theta}}$$where $$\mathbf{i_{\theta}}$$ is the unit vector in the tangential ($$\theta$$) direction. The differential torque acting on this area of the disk is obtained by taking the cross product of the radius vector ($$\mathbf{r}=r\mathbf{i_r}$$) with the differential force such that $$d\mathbf{T}=\mathbf{r}x\mathbf{dF}=\tau_wr^2d\theta dr(\mathbf{i_r}x\mathbf{i_{\theta}})=\tau_wr^2d\theta dr\mathbf{i_z}$$where $$\mathbf{i_r}$$ is the unit vector in the radial direction and $$\mathbf{i_z}$$ is the unit vector in the axial direction. The total torque on the disk is obtained by integrating this between $$\theta = 0$$ and $$\theta = 2\pi$$ and between the inner and outer radii (also taking into account the fact that $$\tau_w$$ is proportional to r, given by $$\tau_w=\frac{r\Delta \omega}{h}$$).
• If you switch the dr and $d\theta$ and integrate over $d\theta$ you get the same expression as in the solution but without the unit vector, which is great but my question persists. Why did you say that dT = r x dF and ignore dr x F, my problem is that torque is not the summation of the change of force multiplied by distance at each point but rather the full force multiplied by the full distance at each point. Commented Oct 30, 2019 at 19:01