Types and Sources of Forces

Chapter 2-Part A_Fall2023 (6).pdf

Fluids experience body forces acting throughout the volume of the fluid, such as gravitation

$$ \delta F_g=\vec g\rho\delta V $$

Additionally, they experience surface forces acting only at the boundaries of the fluid, where it contacts a surface, such as pressure (the force acting inward on a surface) and viscous stresses (the stress acting outward on the surface due to viscous effects).

$$ \vec p=p\vec A\>,\>\>p=f(x,y,z) $$

$$ \vec\tau=\begin{bmatrix} \tau_{xx}&\tau_{xy}&\tau_{xz}\\ \tau_{yx}&\tau_{yy}&\tau_{yz}\\ \tau_{zx}&\tau_{zy}&\tau_{zz}\\ \end{bmatrix} $$

These definitions render the following relationship for a fluid in which there are no shearing stresses (i.e. in a fluid moving as if it were a rigid body)

$$ -\nabla p+\rho\vec g=\rho\vec a $$

Further, in the case of an incompressible liquid in static equilibrium, we find that the static pressure is linearly related to the depth of the fluid. On the other hand, for a compressible gas, we find the static pressure to be exponentially related to the depth of the fluid.

$$ p_2=p_1+\rho gh $$

$$ p_2=p_1\cdot e^{-\frac{g(z_2-z_1)}{RT}} $$

When a fluid is at rest, it is subject to a body force in the form of gravity, and a surface force in the form of pressure (without any accompanying shear stresses). This fundamental state of equilibrium gives rise to Pascal's Law, stating that the pressure at any point in a fluid at rest remains independent of the direction considered.

Forces on Surfaces

We won’t consider forces on surfaces as arbitrary as this, but I’ve included because I think it’s cool and to show how we get the more specific formulas. Note that all of the following formulas depend on the coordinate system we use, which for these formulas has the origin at the liquid surface.

Derivation Steps

1. Find the differential area vector by the cross product of two differential vectors on the surface
2. Find the differentials of Force and Moment using the differential area vector and pressure field.
3. Integrate over the surface to find the net force and moment

$$ \vec u=\begin{bmatrix} \frac{\partial x}{\partial y}dy& dy& 0 \end{bmatrix} $$

$$ \vec v=\begin{bmatrix} \frac{\partial x}{\partial z}dz& 0& dz \end{bmatrix} $$

$$ d\vec A=\vec u\times\vec v $$

$$ d\vec A=\begin{bmatrix} 1& -\frac{\partial x}{\partial y}& -\frac{\partial x}{\partial z} \end{bmatrix} dydz $$

$$ d\vec F=-pd\vec A\>,\>\>p=\gamma y $$