--- title: "Accelerated oil tank, spilled volume" subject: "Fluid Mechanics" chapter: 3 key: "semester_3/fluid_mechanics:accelerated-cube-tank-spilling" source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/accelerated-cube-tank-spilling site: BCT Engineering --- # Accelerated oil tank, spilled volume Fluid Mechanics · Chapter 3: Hydrostatics - Asked in: 2075 Asoj, 4+6 marks - Video: [Relative Equilibrium of Liquid Masses](https://www.youtube.com/watch?v=AVAU4pY-AwA) — Engr. Clidez - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/accelerated-cube-tank-spilling - Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-3 ## Question A thin-walled, open-topped tank in the form of a cube of 500 mm side is initially full of oil of relative density 0.88. It is accelerated uniformly at 5 m/s² up a long straight slope at arctan (1/4) to the horizontal, the base of the tank remaining parallel to the slope, and the two side faces remaining parallel to the direction of motion. Calculate (a) the volume of oil left in the tank when no more spilling occurs, and (b) the pressure at the lowest corners of the tank. ## Short answer Effective gravity slope gives free surface inclination 0.7754; oil volume left is 0.0765 m³. Gauge pressures at the lowest corners are 4.19 kPa at the rear and 0.94 kPa at the front. ## Answer **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) - $y$: $-g\cos\theta=-9.517\ \text{m/s}^2$ ```text 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.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Along the slope axis x, what is the effective acceleration component acting towards the rear? A. g sinθ + a = 7.38 m/s^2 B. g sinθ - a C. a only D. g cosθ + a Answer: A. Gravity's component down the slope adds to the inertia effect of upward acceleration. (Key point: Effective gravity) 2. How does the free surface lie? A. Parallel to the base B. Perpendicular to the effective gravity, sloping up to the rear with tanφ = 0.775 C. Horizontal D. Sloping down to the rear Answer: B. tanφ = 7.379/9.517; oil spills at the rear until the surface passes through the rear rim. (Key point: Free surface) 3. How is the oil volume left computed? A. 0.5 x 0.5 x 0.5 = 0.125 m^3 B. Half of the cube C. Mean of depths 0.5 m (rear) and 0.112 m (front) x 0.5 x 0.5 = 0.0765 m^3 D. 0.112 x 0.25 = 0.028 m^3 Answer: C. Front depth = 0.5 - 0.5(0.7754) = 0.1123 m; the section is a trapezium. (Key point: Volume left) 4. What is the pressure at the rear bottom corners? A. ρ g cosθ x 0.5 = 4.19 kPa B. ρ g x 0.5 = 4.32 kPa C. ρ (7.379) x 0.5 D. 0.94 kPa Answer: A. Pressure gradient normal to the base is ρ g cosθ. 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