1) Consider a vessel containing a fluid. Let the acceleration of vessel in vertical direction be, ay. Consider a small cylindrical element dy at a height y from bottom
From figure,
P₁ A – P₂A – mg = may
P₁ A – P₂A = m (g + ay)
But (P₁ – P₂) A2 = A dy (g + ay)
– dp/ dy = – (g + ay)
2) Consider a vessel containing a fluid. Let the acceleration of vessel in horizontal direction be
From diagram
(P₁ – P₂) A = Adx (ax)
Δp/Δx = – ax
If θ is the angle made by the free liquid surface with horizontal, then
tanθ = dh/dx
But we have
(P₁ – P₂) A = Adx (ax)
(h₁ – h₂) gA = Adx (ax)
dh/ dx = aₓ/g
tanθ = aₓ/g