Find the fluid force on a side of the plate.

Question

The plate in the figure is submerged in water with its top level with the surface of the water. The water level is cube root 7 ft. The right and left edges of the plate are the curves y=x^(1/3) and y=-x^(1/3).

Answer 1

Let's choose the x axis directed normal to the water surface with positive direction downwards and Y-axis along the water surface. Let ρ be water density , g- gravit. acceleration. Then the water pressure at the depth x would be

p(x)=ρgx , the force acted on a stripe of the plate at that depth x, with length y(x) - y(-x)= 2x^1/3 and thickness dx would be

dF=p(x)2x^1/3dx=2ρgx^4/3dx (1)

and the total force would be the integral of dF between x=0 and the max. depth of 7^1/3:

F=Int[0, 7^1/3] 2ρgx^4/3dx=2ρg(3/7x7^7/9)=6ρg/7^2/9 (2)

(used the [ Int x^rdx=x^r+1/(r+1) ] )

Finally, using water density ρ=1.0*10³kg/m³; and grav. acceleration g=9.8m/s² we get from (2):

F=38000N