Question

A large aquarium of height 6 m is filled with fresh water to a depth of D = 2.50 m. One wall of the aquarium consists of thick plastic with horizontal length w = 7.80 m. By how much does the total force on that wall increase if the aquarium is next filled to a depth of D = 5.00 m? (Note: use g = 9.81 m/s2 and ρ = 998 kg/m3.)

Answer 1

F = 1/2 * (p * w * g * h^2)

where, p is the density of fresh water

Now,

They want the Difference in force,

So,

Lets call the force at 2.5to be F1 and the Force at 5 m to be F2. Then we will call h1 2.5 m and h2 5 m

F1 = 1/2 * (p * w * g * h1^2)

F2 = 1/2 * (p * w * g * h2^2)

And we want the difference, lets call that delta F,

Delta F = F2 - F1

Delta F = [1/2 * (p * w * g * h2^2)] - [1/2 * (d * w * g * h1^2)]

Delta F = 0.5 * p * w * g (h2^2 - h1^2)

DeltaF = 0.5* 998kg/m^3*7.80m*9.81m/s^2*((5m)^2-(2.5m)^2)

Delta F = 715.92 kN