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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.

enter image description here

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.

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1 Answer 1

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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.

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