Question

An open rectangular tank of dimension $5 m \times 4 m \times 3 m$ containing water upto a height of $2 m$ is accelerated horizontally along the longer side.

(i) Determine the maximum acceleration that can be given without spilling the water.

(ii) Calculate the percentage of water split over, if this acceleration is increased by $20 % .$

(iii) If initially, the tank is closed at the top and is accelerated horizontally by $9 m s^{- 2},$ find the gauge pressure at the bottom of the front and rear walls of the tank.

Accepted Answer

(i) For a fluid in a container moving with acceleration $a$ .

Volume of water remains constant.

From above figure, $tanθ = \frac{a}{g}$ $\Rightarrow θ = \tan^{- 1} (\frac{a}{g})$

Or, $\tan θ = \frac{1}{2.5} = \frac{a}{g} \Rightarrow a = \frac{g}{2.5}$

Thus, the maximum acceleration that can be given without spilling the water is $a = 4 m s^{- 2}$ .

(ii) New acceleration is $a = 1.2 a = 4.8 m s^{- 2}$

Then, $\tan θ_{1} = \frac{4.8}{g}$

From above figure, $\tan θ_{1} = \frac{3 - d}{5} = \frac{4.8}{10} \Rightarrow d = 0.6 m$

Volume of water left $V_{f} =$ $\frac{1}{2} \times (3 + 0.6) \times 5 \times 4 = 36 m^{3}$

Initial volume of water in container, $V_{i} = 5 \times 4 \times 2 = 40 m^{3}$

Volume of water spilled out, $∆ V = 40 - 36 = 4 m^{3}$ . Percentage of water split over is $(\frac{4}{40}) \times 100 = 10 %$ .

(iii)

Here, ${(V_{air})}_{i} = {(V_{air})}_{f}$

$\Rightarrow 5 \times 1 \times 4 = \frac{1}{2} \times y \times x \times 4 \Rightarrow x y = 10 . . . (i)$

We have, $\tan θ = \frac{a}{g} = \frac{9}{10} \Rightarrow \frac{y}{x} = \frac{9}{10} = 0.9 . . . (i i)$

Solving $(i) and (i i)$ , we get $x = 3.33 m$ and $y = 3 m$ .

Pressure on front wall is $0$ .

$P_{B} - P_{F} = ρ a l = 10^{3} \times 9 \times 5 = 45 \times 10^{3} Pa$

Hence, gauge pressure at the bottom of rear wall is $45 kPa$ .

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