Question
A tank with a square base of area 1.0 m2 is divided by a vertical partition in the middle. The bottom of the partition has a small-hinged door of area 20 cm2. The tank is filled with water in one compartment, and an acid (of relative density 1.7) in the other, both to a height of 4.0 m. Compute the force necessary to keep the door close.
Solution
Base area of the given tank, A = 1.0 m2
Given,
Area of the hinged door, a = 20 cm2 = 20 × 10–4 m2
Density of water, ρ1 = 103 kg/m3
Density of acid, ρ2 = 1.7 × 103 kg/m3
Height of the water column, h1 = 4 m
Height of the acid column, h2 = 4 m
Acceleration due to gravity, g = 9.8
Pressure due to water is given using the relation,
P1 = h1ρ1g
= 4 × 103 × 9.8
= 3.92 × 104 Pa
Pressure due to acid is given by the relation,
P2 = h2ρ2g
= 4 × 1.7 × 103 × 9.8
= 6.664 × 104 Pa
Pressure difference between the water and acid columns:
ΔP = P2 - P1
= 6.664 × 104 - 3.92 × 104
=2.744 × 104 Pa
Therefore, Force exerted on the door, F = ΔP × a
= 2.744 × 104 × 20 × 10–4
= 54.88 N
Hence, 54.88 N force is required to keep the door closed.
Given,
Area of the hinged door, a = 20 cm2 = 20 × 10–4 m2
Density of water, ρ1 = 103 kg/m3
Density of acid, ρ2 = 1.7 × 103 kg/m3
Height of the water column, h1 = 4 m
Height of the acid column, h2 = 4 m
Acceleration due to gravity, g = 9.8
Pressure due to water is given using the relation,
P1 = h1ρ1g
= 4 × 103 × 9.8
= 3.92 × 104 Pa
Pressure due to acid is given by the relation,
P2 = h2ρ2g
= 4 × 1.7 × 103 × 9.8
= 6.664 × 104 Pa
Pressure difference between the water and acid columns:
ΔP = P2 - P1
= 6.664 × 104 - 3.92 × 104
=2.744 × 104 Pa
Therefore, Force exerted on the door, F = ΔP × a
= 2.744 × 104 × 20 × 10–4
= 54.88 N
Hence, 54.88 N force is required to keep the door closed.
