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 smallhinged 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 m. compute the force necessary to keep the door closed.

Answer 1

Assume g = 9.8 m/s²

Density of water, ρ₁= 10³ kg/m³

height of water column, h₁ = 4m

∴ Pressure due to water, P₁ = ρ₁ gh₁

= 10³ × 9.8 × 4

= 3.92 × 10⁴ Pa

Density of acid

= Relative, densityx density of water = 1.7 × 10³ kg/m³

height of acid column, h₂ = 4m

∴ Pressure due to acid, P₂ = ρ₂gh₂

= 1.7 × 10³ × 9.8 × 4

⇒ P₂ = 6.664 × 10⁴Pa

Pressure difference between the water and acid columns :

∆ P = P₂ – P₁

∆ P = 6.664 × 10⁴ – 3.92 × 10⁴

⇒ ∆ P = 2.744 × 10⁴ Pa

Area of the door, a = 20 cm²

= 20 × 10^-4m²

∴ Force exerted on the door, F = ∆ P × a

⇒ F = 2.744 × 10⁴ × 20 × 10^-4

F = 54.88 N

Therefore, the force necessary to keep the door closed is 54.88 N.