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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 1
Pressure difference across the door
`= (4xx1700xx9.8-4xx1000xx9.8)`Pa
`(6.664xx 10^4 - 3.92 xx 10^4)Pa = 2.774 xx10^4 Pa`
Force on the Floor = Pressure difference x Area of door
`= 2.774 xx10^4 xx 20 xx10^(-4) N`
= 54.88 N = 55 N
Solution 2
Base area of the given tank, A = 1.0 m2
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 as:
`P_1 = h_1rho_1g`
`=4xx10^3xx9.8`
`= 3.92 xx 10^4 Pa`
Pressure due to acid is given as:
`P_2 = h_2rho_2g`
`= 4xx1.7xx10^3 xx 9.8`
`= 6.664 xx 10^4 Pa`
Pressure difference between the water and acid columns:
`triangle P = P_2 - P_1`
`= 6.664 xx 10^4 - 3.92 xx 10^4`
`= 2.744 xx 10^ Pa`
Pressure due to acid is given as:
`P_2 = h_2rho_2g`
`= 4xx 1.7 xx 10^3 xx 9.8`
`= 6.664 xx 10^4 Pa`
Pressure difference between the water and acid columns:
`triangle P = P_2 - P_1`
`= 6.664 xx 10^4 - 3.92 xx 10^4`
`= 2.744 xx 10^4 Pa`
Hence, the force exerted on the door = ΔP × a
= 2.744 × 104 × 20 × 10–4
= 54.88 N
Therefore, the force necessary to keep the door closed is 54.88 N.
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