Answer
Given,
Length of tank = 10 m
Width of tank = 8 m
height of tank = 4 m
density of kerosene = 820 kg/m³
depth of oil = 3.5 m
a) Hydro static pressure at the bottom of the tank
P = ρ g h
P = 820 x 9.81 x 3.5
P = 28154.7 Pa
b) Hydrostatic force at the bottom of the tank
F = P A
A = 10 x 8 = 80 m²
F = 28154.7 x 80 = 2252.38 kN
c) Hydrostatic force at one edge
Width = 8 m
depth = z
A = 4 y
dA = 4 dy
P = ρgh
P = ρgz = ρg(y-0.5)
F = P dA
F = ρg(y-0.5)(4dA)
[tex]F = 4\rho g \int_{0.5}^4(y-0.5)dy[/tex]
on solving
F = 24.5 ρg
Hydrostatic force at one end = 24.5 x 820 x 9.81 = 197.083 kN.
(a) The hydrostatic pressure be "28154.7 Pa".
(b) The hydrostatic force on the bottom be "2252.38 kN".
(c) The hydrostatic force on one end be "197.083 kN".
According to the question,
Tank's length = 10 m
Width of tank = 8 m
Height of tank = 4 m
Kerosene's density = 3.5 m
(a) We know the formula,
→ P = ρ g h
By substituting the values,
= 820 × 9.81 × 3.5
= 28154.7 Pa
(b) We know the formula,
Hydrostatic force:
→ F = PA
here,
A = 10 × 8
= 80 m²
then,
F = 28154.7 × 80
= 2252.38 kN
(c) According to the question,
Width = 8 m
Depth = z
A = 4 y
Derivative:
dA = 4 dy
Now,
→ P = ρgh
= ρgz
= ρg (y - 0.5)
then,
F = ρ dA
By substituting the values,
= ρg (y - 0.5)(4 dA)
= 4ρg [tex]\int\limits^4_{0.5} (y - 0.5)[/tex] dy
= 24.5 ρg
hence,
The force at end,
= 24.5 × 820 × 9.81
= 197.083 kN
Thus the response above is appropriate.
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