Consider a water tank shown in the figure. It has one wall at x=L and can be taken to be very wide in the z direction. When filled with a liquid of surface tension S and density ρ, the liquid surface makes an angle θ₀ (θ₀1) with the x-axis at x=L. If y(x) is the height of the surface then the equat…

Question

Consider a water tank shown in the figure. It has one wall at x=L and can be taken to be very wide in the z direction. When filled with a liquid of surface tension S and density ρ, the liquid surface makes an angle θ₀ (θ₀1) with the x-axis at x=L. If y(x) is the height of the surface then the equation for y(x) is: (take θ(x)=sin θ(x)=tan θ(x)=(dy)/(dx), g is the acceleration due to gravity)

Accepted Answer

(4) (d²y)/(dx²)=(ρ g)/S y. Given: a liquid of surface tension S and density ρ meeting a vertical wall at x=L, with y(x) the height of the free surface and slopes small. Idea: at every point of a static free surface the Laplace pressure across the curved surface must balance the hydrostatic pressure of the liquid raised above the flat level. For a nearly flat surface the curvature is κ=(d²y/dx²)/([1+(dy/dx)²]^3/2)≈(d²y)/(dx²) because dy/dx1. Laplace: the pressure jump across the surface is Δ P=Sκ. Hydrostatics: a point of the surface standing a height y above the flat level is at a pressure ρ gy below atmospheric, so Δ P=ρ gy. S(d²y)/(dx²)=ρ gy (d²y)/(dx²)=(ρ g)/S y

Step-by-step solution

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