Fluid Force on a Cube at Depth h0: Calculating Pressure and Magnitude of Force

[Roodles01]

Homework Statement

A solid cube, mass m, side length l, is placed in a liquid of uniform density, ρ(rho), at a depth h₀ below the surface of the liquid, which is open to the air.
The upper and lower faces of the cube are horizontal.

Find the magnitude of force, F_s, exerted on each vertical face of the cube and express it in terms of ρ (rho), l, h, atmospheric pressure p₀ and gravity g.

Homework Equations

Pressure, p, at any point in the liquid = (p0 + ρg(h₀+x))

The Attempt at a Solution

surface integral
F_s = -n ∫ (p₀ + ρg(h₀ +x))dA
where n is the unit vector normal to the surface

Area integral of a rectangle with side lengths a & b written as 2 single integrals
F_s = -n ∫ [ ∫ (p₀ + ρg(h₀ +x))dy]dx . . . . . . . . . . (first integral limits 0 & a, second limits 0 & b)
F_s = -n ∫ (p₀ + ρg(h₀ +x)) b dx
F_s = -b (p₀ + ρgh₀) a + ½*ρ₀ga²) n
F_s = -ab (p₀ + ρ₀g(h₀ + ½a)) n
remembering that ab = l*l = area of cube side = l²
so
F_s = -l² (p₀ + ρ₀g (h₀ +½a)) n

is that OK?
Is the integral look right.
I worry as I'm doing this at home by myself.

 

[SteamKing]

Homework Statement

A solid cube, mass m, side length l, is placed in a liquid of uniform density, ρ(rho), at a depth h₀ below the surface of the liquid, which is open to the air.
The upper and lower faces of the cube are horizontal.

Find the magnitude of force, F_s, exerted on each vertical face of the cube and express it in terms of ρ (rho), l, h, atmospheric pressure p₀ and gravity g.

Homework Equations

Pressure, p, at any point in the liquid = (p0 + ρg(h₀+x))

The Attempt at a Solution

surface integral
F_s = -n ∫ (p₀ + ρg(h₀ +x))dA
where n is the unit vector normal to the surface

Area integral of a rectangle with side lengths a & b written as 2 single integrals
F_s = -n ∫ [ ∫ (p₀ + ρg(h₀ +x))dy]dx . . . . . . . . . . (first integral limits 0 & a, second limits 0 & b)
F_s = -n ∫ (p₀ + ρg(h₀ +x)) b dx
F_s = -b (p₀ + ρgh₀) a + ½*ρ₀ga²) n
F_s = -ab (p₀ + ρ₀g(h₀ + ½a)) n
remembering that ab = l*l = area of cube side = l²
so
F_s = -l² (p₀ + ρ₀g (h₀ +½a)) n

is that OK?
Is the integral look right.
I worry as I'm doing this at home by myself.

You've still got an 'a' in the expression for F_s. That's not one of the variables requested by the OP.

You can check your answer w/o using an integral.