force in newtons on the window

Question

The zoo is building a new polar bear exhibit, and wants to put a semi-circular window in the concrete wall of the swimming tank. (Note: picture is not to scale)

If the semi-circle has diameter 70 centimeters, and the bottom of the window is at a depth of 1.75 meters, find the hydrostatic force on the window, in newtons.

Answer 1

The total force on a submerged vertical surface is equal to the area of the surface times the depth of the centroid.

F=ρg(y-bar)A

ρ = density of water = 1000 kg/m³

g = gravity = 9.8 m/s²

A = area of semicircle = πr²/2 ≈ 0.192 m²

y-bar = the centroid of the semicircle = 4r/(3π) up from the bottom of the window or 1.75 - 4(.35)/(3π) meters

≈ 1.6 m

F=(1000 kg/m³)(9.8 m/s²)(1.6 m)(0.192 m²) = 3017.56 kg m/s² = 3017.56 N

---OR--

F = ρg∫_a^bd(y)L(y)dy

d(y) = depth of Δy strip = (1.75-y)

L(y) = horizontal width of Δy strip = 2x = 2√(r²-y²)

a = 0

b = r = 0.35

F = 19600∫₀^0.35(1.75-y)√(0.35²-y²)dy = 3019.029 N