can you solve this as attachged $?$Paint run:

Wet paint can be approximated as a Bingham fluid that behaves like a solid $($i$.$e$.$ does not deform$)$

until the applied stress exceeds a yield stress. Once the applied stress exceeds the yield stress, the

paint behaves like a Newtonian fluid. Consider a layer of paint on a vertical wall $(z-$direction is

the vertical direction$),$ with a uniform layer thickness $$ and a yield stress ${}_0.({}_{xz}={}_0-$

$(del\frac{v_z}{d}$elx$).$

$($a$)$ Using shell momentum balance, derive an expression for the maximum value of $$ that can be

sustained without having the paint "run" $($i$.$e$.$ flow down the wall$).$ Let's call this thickness be

${}_0.$ Sketch the shear stress profile for this case $($the velocity will be uniformly zero$).$

Hint: the momentum balance here is identical to that in the falling film over inclined plane

problem, except that the angle $$ is zero for a vertical wall.

$($b$)$ For the case of $>{}_0,$ sketch the shear stress profile and the fully developed velocity profile

$v_z(x).$

$($c$)$ Derive an expression for the velocity profile for the case of $>{}_0.$

Note: If you set the origin of the $x-$direction at the outer surface of the paint with positive direction

pointing toward the wall, the shear stress will be positive. Alternatively, if you set this origin at

the wall with positive direction pointing toward the paint, the shear stress will be negative. In the

latter case, the ${}_0$ will have a negative value and the paint run will occur where the shear stress

becomes more negative than ${}_0[$it$\text{'}$s the absolute value $($magnitude$)$ that matters for determining

the paint run and not the sign$].$