When discussing a pressure we are considering a normal force over a certain area, at a given point, within a specific plane of a fluid mass of interest. Now the question is will the pressure on a certain point vary with the orientation of the plane that passes through the point of interest? To answer this question we can determine the pressure on a point within a specific fluid element. In this case the fluid element will be a wedge shape.

Shearing stresses will not be present for the above situation. As a result the external forces acting on the wedge are only due to pressure and specific weight of the fluid. For the above example I am only considering the forces in the z and the y directions for simplicity. To determine the force equations for the above example Newton’s second law of motion, F = ma will be applied.

**(Eq 1)** $ΣF_y = p_yδxδz – p_sδxδssinθ = ρ\frac{δxδyδz}{2}a_y$

$a_y$ = acceleration in the y-direction

$ρ$ = density

**(Eq 2)** $ΣF_z = p_zδxδy – p_sδxδscosθ-γ\frac{δxδyδz}{2} = ρ\frac{δxδyδz}{2}a_z$

$a_z$ = acceleration in the z -direction

$γ$ = specific weight

The above equations can be simplified when you take in consideration that $δ_y = δ_scosθ$ and $δ_z = δ_sinθ$. This will result in the following equations.

**(Eq 3) **$p_y – p_s = ρa_y\frac{δy}{2}$

**(Eq 4) **$p_z-p_s = (ρa_z+γ)\frac{δz}{2}$

Finally, if you were to allow $δx$, $δy$, and $δz$ to go to zero than $p_y = p_s$ and $p_z = p_s$. Hence it can be concluded that the pressure at a point will have $p_s=p_y=p_z$. This means that regardless of what θ is the pressure at a point for *” a fluid at rest, or in motion, is independent of direction as long as there is no shearing stress present”* according to Pascal’s law. If shearing stress is present than the above statement will not hold true.