# Angle of tilt of a rectangular water tank

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.

## 1 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.