Water is filled in a rectangular tank of size 3m $\times$ 2 m $\times$ 1 m.

(a) Find the total force exerted by the water on the bottom surface of the tank.

(b) Consider a vertical side of area 2 m $\times$ 1 m. Take a horizontal strip of width $δ$ x metre in this side, situated at a depth of x metre from the surface of water. Find the force by the water on this strip.

(c) Find the torque of the force calculated in part (b) about the bottom edge of this side.

(d) Find the total force by the water on this side.

(e) Find the total torque by the water on the side about the bottom edge. Neglect the atmospheric pressure and take g= $10 m s^{- 2}$ .

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Solution

Given, h = 1m, A = 3 $\times 2 = 6 m^{2}$ ,
p = $10^{3} k g / m^{3}, g = 10 m / s^{2}$
(a) f = $A h ρ g$
= $(3 \times 2 \times 1) \times 10^{3} \times 10$

= $6 \times 10^{4} = 60000 N$

(b) The force exereted by wave on the strip of width $δ$ x as shown,

df, = p $\times A = (x ρ g) \times A$
= (X) $\times 10^{3} \times 10 \times 2 \times δ x$
= $20, 000 \times δ x N$

(c) Inside the liquid force act in every direction due to adhesion,

dt = $F \times r$

= $20, 000 \times δ x (1 - x) N$

(d) The total force by the water on that side is given by,

$F = \int_{0}^{1} 20, 000 x δ x$

$\Rightarrow F = 20, 000 {[\frac{x^{2}}{2}]}_{0}^{1}$

= 10,000 N

(e) The torque by the water on that side will be,

$τ = 20, 000 \times \int_{0}^{1} x δ x (1 - x)$

= $20, 000 {[\frac{x^{2}}{2} - \frac{x^{3}}{3}]}_{0}^{1}$

= $20, 000 \times [\frac{1}{2} - \frac{1}{3}]$

= $\frac{20, 000}{6} m N = \frac{10000}{3}$ mN

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