Question

A container is lled with two liquids of densities ρ1 and ρ2 up to heights h1 and h2 respectively. What will be the hydrostatic force (in kN) per unit width of the lower face AB?

The question was posed to me during an internship interview.

This interesting question is from Hydrostatic Force on Plane Area topic in section Hydrostatic Forces on Surfaces of Fluid Mechanics

Answer 1

To find the hydrostatic force per unit width on the lower face ABABAB of a container filled with two liquids of densities ρ1\rho_1ρ1​ and ρ2\rho_2ρ2​ up to heights h1h_1h1​ and h2h_2h2​, we can use the concept of hydrostatic force acting on a submerged surface. The hydrostatic force is the force exerted by a liquid on a surface due to the weight of the liquid above it.

Steps to calculate hydrostatic force:

1. Pressure due to the first liquid (density ρ1\rho_1ρ1​ and height h1h_1h1​): The pressure at any depth in a liquid is given by the equation:

   P=ρghP = \rho g hP=ρgh

   where:

   • PPP is the pressure,
   • ρ\rhoρ is the density of the liquid,
   • ggg is the acceleration due to gravity,
   • hhh is the depth.

   So, the pressure at the interface of the first liquid (depth h1h_1h1​) is:

   P1=ρ1gh1P_1 = \rho_1 g h_1P1​=ρ1​gh1​

2. Pressure due to the second liquid (density ρ2\rho_2ρ2​ and height h2h_2h2​): Similarly, the pressure due to the second liquid at the depth h1+h2h_1 + h_2h1​+h2​ (which is the total height from the base) is:

   P2=ρ2gh2P_2 = \rho_2 g h_2P2​=ρ2​gh2​

3. Total hydrostatic force: The hydrostatic force on the lower face of the container (per unit width) can be found by integrating the pressure over the area of the submerged surface. Since the depth of the face changes linearly along its length, the force can be calculated by:

   F=∫0AP dAF = \int_0^A P \, dAF=∫0A​PdA

   where PPP is the pressure at each point on the surface and dAdAdA is the differential area element.

   • The force due to the first liquid is:

     F1=ρ1gh1×area of lower face ABF_1 = \rho_1 g h_1 \times \text{area of lower face AB}F1​=ρ1​gh1​×area of lower face AB

   • The force due to the second liquid is:

     F2=ρ2gh2×area of lower face ABF_2 = \rho_2 g h_2 \times \text{area of lower face AB}F2​=ρ2​gh2​×area of lower face AB

4. Hydrostatic force per unit width: Assuming the width of the container is 1 meter, the hydrostatic force per unit width Fper unit widthF_{\text{per unit width}}Fper unit width​ is:

   Fper unit width=(ρ1gh1+ρ2gh2)F_{\text{per unit width}} = (\rho_1 g h_1 + \rho_2 g h_2)Fper unit width​=(ρ1​gh1​+ρ2​gh2​)

   where:

   • ρ1\rho_1ρ1​ is the density of the first liquid,
   • ρ2\rho_2ρ2​ is the density of the second liquid,
   • ggg is the gravitational acceleration (approximately 9.81 m/s²),
   • h1h_1h1​ is the height of the first liquid,
   • h2h_2h2​ is the height of the second liquid.

Final Answer:

The total hydrostatic force per unit width on the lower face ABABAB is:

Fper unit width=(ρ1gh1+ρ2gh2) (kN)F_{\text{per unit width}} = (\rho_1 g h_1 + \rho_2 g h_2) \, \text{(kN)}Fper unit width​=(ρ1​gh1​+ρ2​gh2​)(kN)

This will give you the total hydrostatic force (in kN) acting on the lower face per unit width. If needed, you can convert the units of the density ρ1\rho_1ρ1​ and ρ2\rho_2ρ2​ to correspond to the desired units.