Accelerated oil tank, spilled volume

Data: cube side 0.5 m, oil $\rho=0.88\times1000=880\ \text{kg/m}^3$, slope $\tan\theta=\tfrac14$ ($\theta=14.04^\circ$, $\sin\theta=0.2425$, $\cos\theta=0.9701$), acceleration $a=5\ \text{m/s}^2$ up the slope.

Use axes fixed to the tank: $x$ up the slope, $y$ perpendicular to the base. The effective gravity (gravity minus acceleration) has components:

• $x$: $-(g\sin\theta+a)=-(2.379+5)=-7.379\ \text{m/s}^2$ (towards the rear)
• $y$: $-g\cos\theta=-9.517\ \text{m/s}^2$

         rear        front (up slope)
   rim -> |  \          |
          |    \        |
          |______\______|
        free surface slopes up to the rear

The free surface is perpendicular to this effective gravity, and its slope is

$$\tan\phi=\frac{7.379}{9.517}=0.7754$$

The surface rises towards the rear, so oil spills over the rear rim until the surface passes through that rim.

(a) Volume left

Depth at the rear wall $=0.5$ m. Depth at the front wall $=0.5-0.5(0.7754)=0.1123$ m (positive, so the surface does not reach the base).

$$V=\frac{0.5+0.1123}{2}\times0.5\times0.5=0.0765\ \text{m}^3$$

Oil left $=0.0765\ \text{m}^3$ (76.5 litres), i.e. 61 % of the 125 litres.

(b) Pressure at the lowest corners

The pressure gradient normal to the base is $\rho\,g\cos\theta$, so at the base:

$$p_{rear}=880(9.517)(0.5)=4187\ \text{Pa}$$ $$p_{front}=880(9.517)(0.1123)=941\ \text{Pa}$$

Answer: (a) 0.0765 m³ of oil remains; (b) gauge pressure at the rear bottom corners 4.19 kPa and at the front bottom corners 0.94 kPa.