- #1

Roodles01

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## Homework Statement

A solid cube, mass m, side length l, is placed in a liquid of uniform density, ρ(rho), at a depth h

_{0}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 p

_{0}and gravity g.

## Homework Equations

Pressure, p, at any point in the liquid = (p0 + ρg(h

_{0}+x))

## The Attempt at a Solution

surface integral

**F**

_{s}= -n ∫ (p

_{0}+ ρg(h

_{0}+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**∫ [ ∫ (p

_{0}+ ρg(h

_{0}+x))dy]dx . . . . . . . . . . (first integral limits 0 & a, second limits 0 & b)

**F**

_{s}= -

**n**∫ (p

_{0}+ ρg(h

_{0}+x)) b dx

**F**

_{s}= -b (p

_{0}+ ρgh

_{0}) a + ½*ρ

_{0}ga

^{2})

**n**

F

F

_{s}= -ab (p

_{0}+ ρ

_{0}g(h

_{0}+ ½a))

**n**

remembering that ab = l*l = area of cube side = l

^{2}

so

**F**

_{s}= -l

^{2}(p

_{0}+ ρ

_{0}g (h

_{0}+½a))

**n**

is that OK?

Is the integral look right.

I worry as I'm doing this at home by myself.