A cuboid tank is placed in the $xy$ plane, with its base centered at the origin. The base rectangle measures $5$ along the $x$ axis, and $7$ along the $y$ axis. The height is $9$.
It is filled to $\dfrac{2}{3}$ of its height with water. The is shown on the left of the figure below.
Then, it is tilted by a certain angle $\theta$ about an axis of rotation passing through the base vertex $(\dfrac{5}{2} , – \dfrac{7}{2}, 0)$, and parallel to the vector $(\cos 60^\circ, \sin 60^\circ, 0 ) $, such that the water surface touches the tank top vertex that is directly above anchor point (i.e. originally at $(\dfrac{5}{2} , – \dfrac{7}{2}, 9)$ ). This is shown on the right of the figure below. Find the angle of tilt $\theta$ in degrees.
My Attempt:
If we freeze the water and un-tilted the tank, then we note that the frozen water surface plane passes through the points $r_1 = (0, 0, 6)$ and $r_2 = (\dfrac{5}{2} , – \dfrac{7}{2}, 9)$.
On the other hand, the unit normal to the tilted water surface is $(0, 0, 1)$ , so when un-tilting by angle $\theta$, we are rotating this vector about the vector ($\cos 60^\circ, \sin 60^\circ, 0)$ by an angle $-\theta$. Applying this rotation gives
$ \hat{n} = ( – \sin 60^\circ \sin \theta, \cos 60^\circ \sin \theta, \cos \theta )$
This vector is normal to $r_2 – r_1$, so that
$ \hat{n} \cdot (r_2 – r_1) = 0 $
Plugging in the numerical values:
$ ( – \dfrac{\sqrt{3}}{2} \sin \theta, \dfrac{1}{2} \sin \theta , \cos \theta ) \cdot (\dfrac{5}{2} , – \dfrac{7}{2}, 3) = 0 $
This reduces to,
$ \sin \theta (-5 \sqrt{3} – 7) + 12 \cos \theta = 0 $
So that,
$ \tan \theta = \dfrac{12}{5 \sqrt{3} + 7 } $
Hence,
$ \theta \approx 37.4619^\circ $
My question:
Is the method I used correct? And, could someone verify the numerical answer ? Thanks to all.
Best Answer
Hint.
The tilted plane normal vector can be calculated using Rodrigues formula.
$$ \vec n_{\theta}= \vec n_0\cos\theta +\vec k\times \vec n_0\sin\theta +\vec k(\vec k\cdot\vec n_0)(1-\cos\theta)=(\theta_x,\theta_y,\theta_z)' $$
with $\vec n_0=(0,0,1)'$ and $\vec k = (\frac 12,\frac 12,0)'$. This plane can be represented as
$$ (p-p_0)\cdot \vec n_{\theta}=0 $$
with $p_0=(\frac 52,-\frac 72,9)'$ and $p = (x,y,z)'$. Now solving $(p-p_0)\cdot \vec n_{\theta}=0$ for $z$ we obtain
$$ z(x,y,\theta) = \frac{1}{\theta_z}(p_0\cdot \vec n_{\theta}-x\theta_x-y\theta_y) $$
The liquid volume is $V_0 = \frac 23 (5\cdot 7\cdot 9)$ and the same volume after tilt
$$ V_0 = V_{\theta}=\int_{x=-\frac 52}^{x=\frac 52}\int_{y=-\frac 72}^{y=\frac 72}z(x,y,\theta)dx dy $$
and thus we extract $ \theta$ from this equation.