Answer:
(a) The hydrostatic pressure at the bottom of the tank is 12,054 Pa
(b) The hydrostatic force at the bottom of the tank is 289,296 N
(c) The hydrostatic force on one end of the tank is 36,162 N
Explanation:
The dimensions of the tank are;
The length of the tank, l = 6 m
The width of the tank, w = 4 m
The height of the tank, h = 2 m
The density of kerosene in the tank, ρ = 820 kg/m³
The depth of kerosene in the tank, d = 1.5 m
The acceleration due to gravity, g = 9.8 m/s²
(a) The hydrostatic pressure at the bottom of the tank, 'P', is given by the following formula
P = ρ·g·h
P = 820 kg/m³ × 9.8 m/s² × 1.5 m = 12,054 Pa
The hydrostatic pressure at the bottom of the tank, P = 12,054 Pa
(b) The hydrostatic force at the bottom of the tank, F = P × A
Where;
A = The area of the base of the tank = l × w
∴ A = 6 m × 4 m = 24 m²
∴ F = 12,054 Pa × 24 m² = 289,296 N
The hydrostatic force at the bottom of the tank, F = 289,296 N
(c) Taking one end of the tank as one of the width (4 m) side of the tank, the hydrostatic force on one end of the tank, [tex]F_E[/tex], is given as follows;
The cross-sectional area of a layer, [tex]A_i[/tex] = 4·dy
Therefore, the pressure at a given depth, [tex]P_i[/tex] = ρ·g·[tex]y_i[/tex]
The force at the given depth, [tex]F_i[/tex] = [tex]A_i[/tex] × [tex]P_i[/tex]
∴ [tex]F_i[/tex] = [tex]P_i[/tex]·4·dy = ρ·g·[tex]y_i[/tex]·4·dy
The force at the given depth, [tex]F_i[/tex] = ρ·g·[tex]y_i[/tex]·4·dy
The force at one end, [tex]F_E[/tex]
[tex]F_E[/tex] = [tex]\int\limits^{1.5}_0 {F_i \, dy = \int\limits^{1.5}_0 {(4 \cdot \rho} \cdot g \cdot y) \, dy = \left [4 \cdot \rho} \cdot g \cdot \dfrac{y^2}{2} \right]^{1.5}_0 = 4 \times \rho} \times g \times\dfrac{1.5^2}{2}[/tex]
∴ [tex]F_E[/tex] = 4 m × 820 kg/m³ × 9.8 m/s² × (1.5²)/2 m² = 36,162 N
The hydrostatic force on one end of the tank, [tex]F_E[/tex] = 36,162 N
(For the length side, the force is 6×820×9.8×(1.5²)/2 = 54,243 N)
