Question

An aquarium tank is 98 cm long, 35 cm wide, and 38 cm deep. It is filled to the top.

(a) What is the force of the water on the bottom (98 cm ✕ 35 cm) of the tank?
N

(b) What is the force of the water on the front window (98 cm ✕ 38 cm) of the tank? Hint: This problem requires an integration.
N

Answer 1

Given data:
tank meassurements=98 cm long, 35 cm wide, and 38 cm deep
a) force on the bottom of the tank
Force = Pressure* area
Pfluid = ρgh
Pfluid is pressure in Pa or N/m²
ρ is the density of the fluid in kg/m³
density of water at 20C = 0.998 g/cm³ = 998 kg/m³
g is the acceleration of gravity 9.8 m/s²
h is the height of the fluid above the object in m

Force = Pressure* area
= ρgh *area
=998*9.8*38*10^-2*98*10^-2*35*10^-2
=1274.7 N
b)let us consider a strip of area [1 meter by dh meter] at a depth h from top
P (h) = Po + h d g
Force F = ∫ P(h) * area = ∫ [Po + h d g] * dh
limit of h varies from h = 0 to h = 0.38 m
F = [Po h + d g h^2/2] ........ under limits
F = (0.38 Po + d g *0.38*0.38/2)
Po = atmospheric pressure = 1.01*10^5 N/m^2, d = 1000 kg/m^3, g = 9.8 m/s^2
F = (0.38*1.01*10^5+1000*9.8*0.38*0.38/2)=39087N