Density and Pressure — Question 10

Question 10

Local pressure is independent of the orientation of a tiny face

A tiny triangular fluid prism at rest has a slanted face of area $A$ whose outward normal is $\hat n=(\cos\theta,\sin\theta)$, with $0<\theta<\pi/2$. Its vertical and horizontal faces have areas $A\cos\theta$ and $A\sin\theta$. Surrounding fluid pushes inward with pressures $p_s,p_x,p_y$ respectively. Neglect weight in the limit of vanishing prism size.

1. Use horizontal and vertical force balance to relate the three pressures.

2. Explain why neglecting weight is valid in this local limit but would be invalid for a finite-depth column.

Question 10 - Solution

Local pressure is independent of the orientation of a tiny face

(a) Derivation and calculation

Neglect the vanishing prism weight. Horizontal and vertical balances give$$p_xA\cos\theta-p_sA\cos\theta=0,$$$$p_yA\sin\theta-p_sA\sin\theta=0.$$$$0<\theta<\pi/2\Rightarrow\boxed{p_x=p_s=p_y}.$$Pressure is locally independent of face orientation.

(b) Result and interpretation

For geometrically similar prisms of size $\ell$,$$W\sim\rho g\ell^3,\quad A\sim\ell^2,\quad W/A\sim\rho g\ell\to0.$$$$\text{finite column:}\qquad\Delta p=\rho gH\ne0.$$Local pressure isotropy does not imply equal pressures at different elevations.

Physical check

The limit removes the body-force correction at a point without discarding hydrostatic pressure gradients over finite distances.