Consider a tank open at top, containing a liquid and moving vertically upwards with a uniform acceleration. Since the tank is subjected to an acceleration in the vertical direction only, therefore the liquid surface will remain horizontal.
Now consider a small column of the liquid of height h and area dA in the tank as shown in fig-1.
Let, $p$ = Pressure due to vertical acceleration
We know that the forces acting on this column are :
1. Weight of the liquid column $[W = w(h.dA)]$ acting vertically downwards,
2. Acceleration force, $F = ma = \frac{w}{g}(h.dA)a$
3. Pressure $(P = p.dA)$ exerted by the liquid particles on the column.
An open rectangular tank 4m long and 2.5m wide contains an oil of specific gravity 0.85 up to a depth of 1.5m. Determine the total pressure on the bottom of the tank, when the tank is moving with an acceleration of of g/2 m/s\^2 (i) vertically upwards (ii) vertically downwards.
Workings
Given,
$l$ = 4 m
$b$ = 2.5 m
$d$ = 1.5 m
$a$ = g/2 m/s\^2
Specific gravity of liquid = 0.85
(i) Total pressure on the bottom of the tank, when it is vertically upwards
Specific weight of oil, $w = 0.85\times 9.81 = 8.34\;KN/m^3$
Intensity of pressure at the bottom of the tank, $p_1 = wh(1 + \frac{a}{g})$ $\Rightarrow p_1 = 8.34\times 1.5(1 + \frac{g}{2g})$ $\therefore p_1 = 18.765\;KN/m^2$
Total pressure on the bottom of the tank, $P_1 = p_1.A = 18.765\times (4\times 2.5) = 187.65\;KN$
(i) Total pressure on the bottom of the tank, when it is vertically downwards
Intensity of pressure at the bottom of the tank, $p_2 = wh(1 + \frac{a}{g})$ $\Rightarrow p_2 = 8.34\times 1.5(1 + \frac{g}{2g})$ $\therefore p_2 = 6.255\;KN/m^2$
Total pressure on the bottom of the tank, $P_2 = p_2.A = 6.255\times (4\times 2.5) = 62.55\;KN$
