A fluid with viscosity mu is between two metals plates. The bottom plate is fixe

Question

A fluid with viscosity mu is between two metals plates. The bottom plate is fixed and the top plate floating on the fluid. How do you calculate the force required to pull the top plate across the fluid with constant velocity? The area of the plate if also given.

I know that it is related to shear stress. I believe the force required (we'll call it P) is just the shear stress x the area of the upper plate. (P= A*Tau)

And I know I know shear stress is viscosity x (du/dy) . (Tau= mu *(du/dy) )

What I dont understand is what exactly du/dy is. I understand from an abstract point of view, because I know it is the strain rate. I do not really understand what it is though. du is the change in velocity of the plate? and dy the change in height of the fluid?

I would love a thorough explanation, but if you just want to help solve the problem im talking about Ill put some numbers to it.

Area of upper plate = 4m^2
constant velocity which plate is pulled = 0.1 m/(s^2)
viscosity (mu) = 3.8e-1 N*s/(m^2)

Explanation / Answer

now given that, bottom plate is statinary and upper plate is moving with v = 0.1 m/s

hence with no slip boundary condition, the layer which is just adjacent to upper plate is also moving with velocity v = 0.1 m/s

and at bottom plate, the v of fluid is 0

assuming tht the distance between 2 plates is very less.

then with that assumption, du/dy = u/y = [0.1-0]/[y]

now, shear stress = du/dy

force = x 4m2

to solve the problem u also need the distance between 2 plates

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