I have a question:

Question

Two liquids of densities ρ1 and ρ2 (ρ2 = 2ρ1) are filled up behind a square wall of side 10 m. Each liquid has a height of 5 m. The ratio of the forces due to these liquids exerted on upper part MN to that at the lower part NO is (Assume that the liquids are not mixing)

I have solved the answer till getting P1 and P2 as 5p1 and 15p1

what should I do next?

Answer 1

The force F₁ of pressure by the upper liquid, after integrating p=ρ₁gh (h is the depth) from 0 to H and multiplying by the length H is [using Int(hdh)=h²/2]

F₁=ρ₁gH₃³/2.

The lower liquid is under the full pressure of the upper one acting on the area HxH=H² plus the pressure due to the lower liquid itself:

F₂=ρ₁gH x H²+ρ₂gH₃³/2=ρ₁gH³(1+2/2)=2ρ₁gH³=4F₁

Therefore

F₂/F₁=4

Hope it's helpful

Dr.Ariel B.