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

• Roodles01
In summary: Consider the cube to be a small section of the liquid, with a volume dV = l*l*dx and a depth h = h0 + x. The weight of the cube is dF = ρg*dV. The surface area of one of the vertical faces is l*h = l*(h0 + x). The force on the wall is dF = ρg*dV = ρg*l*l*dx, which is independent of h. The total force on one face is the integral of dF from x = 0 to x = l, which is ρg*l*l*l = ρgl3. The force on the other face is the same. The total 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, Fs, 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
Fs = -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
Fs = -n ∫ [ ∫ (p0 + ρg(h0 +x))dy]dx . . . . . . . . . . (first integral limits 0 & a, second limits 0 & b)
Fs = -n ∫ (p0 + ρg(h0 +x)) b dx
Fs = -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
Fs = -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.

Roodles01 said:

## 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, Fs, 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
Fs = -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
Fs = -n ∫ [ ∫ (p0 + ρg(h0 +x))dy]dx . . . . . . . . . . (first integral limits 0 & a, second limits 0 & b)
Fs = -n ∫ (p0 + ρg(h0 +x)) b dx
Fs = -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
Fs = -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 Fs. That's not one of the variables requested by the OP.

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

## What is the definition of force exerted by fluid?

The force exerted by fluid is the physical push or pull that a fluid (such as a liquid or gas) exerts on an object as it flows past or around it. This force is a result of the pressure and velocity of the fluid and can be either perpendicular or parallel to the object's surface.

## How is the force exerted by fluid calculated?

The force exerted by fluid can be calculated using the formula F = ρAv², where ρ is the density of the fluid, A is the area of the object in contact with the fluid, and v is the velocity of the fluid.

## What factors affect the force exerted by fluid?

The force exerted by fluid is influenced by several factors including the density and viscosity of the fluid, the velocity of the fluid, and the size and shape of the object. Other factors such as surface roughness and the fluid's direction of flow can also impact the force exerted.

## What are some real-life applications of force exerted by fluid?

Force exerted by fluid plays a crucial role in many real-life applications, including aerodynamics in airplanes, lift and drag in cars, and hydrodynamics in ships and submarines. It is also important in the design and operation of pumps, turbines, and other fluid systems.

## How does the force exerted by fluid differ from the force of gravity?

The force exerted by fluid and the force of gravity are two different types of forces. While the force of gravity is a constant downward force acting on all objects, the force exerted by fluid is a result of the fluid's pressure and velocity and can vary depending on the object's shape and the fluid's properties. Additionally, the force of gravity is a conservative force, while the force exerted by fluid is a non-conservative force.

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