Finding The Total Surface Force of a Fluid

Question

The question is as follows:

The curved surface of a circular cone of radius 2 sitting on the plane $z=0$ is defined vectorially as $$\underline{r}(R,\phi) = R\cos\phi\underline{\hat{\imath}} + R\sin\phi\underline{\hat{\jmath}} + (8-4R)\underline{\hat{k}}$$ Find the total force on this curved surface under the stress distribution $$\bf{T} = \begin{bmatrix} \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ \sigma_{yx} & \sigma_{yy} & \sigma_{yz} \\ \sigma_{zx} & \sigma_{zy} & \sigma_{zz} \end{bmatrix} = \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & x^2 + y^2 \end{bmatrix}$$

So far I have that the $z$ component is given as $z=8-4R$ so when $R=2$, $z=0$ as it is given and that when $R=0$, $z=8$ so that is where the peak is.

Now, because the coordinate system is cylindrical, I can see that $$\bf{T} = \begin{bmatrix} R\cos\phi & 0 & 0 \\ 0 & R\sin\phi & 0 \\ 0 & 0 & R^2 \end{bmatrix}$$ The equation we have for $z$ I think will be helpful for the integration but what I need some help with is how to find the integrand $\underline{\tau}$ because I don't know how to find the unit normal $\underline{\hat{n}}$ so that I can compute $$\underline{\tau}^{n}_{i} = \sigma_{ij}n_j$$

Answer 1

Okay so after working I have that the Total force will be the surface force $\underline{F}_S$ which I have found using the stress vector $\underline{\tau}$. Now I have that

$$\underline{\tau} = \left(\frac{4R}{\sqrt{17}}\cos^{2}\phi, \frac{4R}{\sqrt{17}}\sin^{2}\phi, \frac{R^2}{\sqrt{17}}\right)$$

which I then integrate to find $\underline{F}_S$. So doing that gives me

$$\underline{F}_S = \int^{\phi=2\pi}_{\phi=0}\int^{R=2}_{R=0}\left(\frac{4R}{\sqrt{17}}\cos^{2}\phi, \frac{4R}{\sqrt{17}}\sin^{2}\phi, \frac{R^2}{\sqrt{17}}\right)\sqrt{17}RdRd\phi$$

$$ = \int^{\phi=2\pi}_{\phi=0}\left[\frac{4}{3}R^3\cos^{2}\phi, \frac{4}{3}R^3\sin^{2}\phi, \frac{R^4}{4}\right]^{2}_{0}d\phi$$

$$ = \frac{1}{3}\Big[16\cos\phi\sin\phi + 16\phi, 16\phi - 16\cos\phi\sin\phi, 12\phi\Big]^{2\pi}_{0}$$

$$ = \frac{1}{3}\Big[0 + 16\times2\pi, 16\times2\pi - 0, 24\pi\Big] = \frac{32\pi}{3}\Big(1,1,\frac{3}{4}\Big)$$

Does this answer for the Force on the surface match yours? @RRL

Answer 2

Hint:

In Cartesian coordinates, the surface is given by $\{(x,y,z)\}$ where $z = f(x,y) = 8 - 4\sqrt{x^2 + y^2}$ and $0 \leqslant R = \sqrt{x^2 + y^2} \leqslant 2$, with outward unit normal vector

$$\mathbf{n} = \frac{-\frac{\partial f}{\partial x}}{\sqrt{1 + \left(\frac{\partial f}{\partial x}\right)^2+ \left(\frac{\partial f}{\partial y}\right)^2}}\, \mathbf{i} + \frac{-\frac{\partial f}{\partial y}}{\sqrt{1 + \left(\frac{\partial f}{\partial x}\right)^2+ \left(\frac{\partial f}{\partial y}\right)^2}}\, \mathbf{j} + \frac{1}{\sqrt{1 + \left(\frac{\partial f}{\partial x}\right)^2+ \left(\frac{\partial f}{\partial y}\right)^2}}\, \mathbf{k}$$

In general, when a surface is defined implicitly as $F(x,y,z) = 0$, the gradient vector $\nabla F$ is normal to the surface. Hence, if the surface is specified as $F(x,y,z) = z - f(x,y)=0$ we get the formula above after rescaling to be a unit vector.