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.

Answer 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

 

Answer 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.