Determining the viscosity of a fluid

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Homework Statement

Two flat plates are oriented parallel above a fixed lower plate as shown below, The top plate,
located a distance b above the fixed plate, is pulled along with speed V. The other thin plate is
located a distance 0.3b (cb) above the fixed plate. This plate moves with speed V1, where V1 = 0.7V.
The viscosity of the fluid between the middle and the lower plates is μ₀. Determine the viscosity (μ) of the fluid between the top and the middle plates.

Homework Equations

[itex]\tau[/itex]=[itex]\mu[/itex][itex]\frac{du}{dy}[/itex]

The Attempt at a Solution

Using the equation, I rearranged it to make [itex]\mu[/itex] the subject.
This gives [itex]\mu[/itex]=[itex]\tau[/itex][itex]\frac{dy}{du}[/itex]
I figured that they want to find the viscosity of the top fluid. So, using the information given, with dy being the height and du being the velocity, substituted it into the equation:
[itex]\mu[/itex]=[itex]\tau[/itex][itex]\frac{V-0.7V}{b-0.3b}[/itex]

Is this the final answer or is more working required? I'm wondering if you can figure out an expression for [itex]\tau[/itex]. Is it correct to say that as V1 is moving at a constant speed, the net force on the middle plate would then be 0. Therefore, the force on the top and bottom plate must be equal to allow for this constant speed. & Thus, the shear stress on the top and bottom plate is also equal. With F=[itex]\tau[/itex]A
I was then thinking that you could equate [itex]\tau[/itex]bottom and [itex]\tau[/itex]top, substitute it into the equation and solve for [itex]\mu[/itex] but it would cancel out through being divided. E.g. [itex]\mu[/itex][itex]\frac{0.7b}{0.3V}[/itex] = [itex]\mu[/itex][itex]\frac{0.3b}{0.7V}[/itex]

I'm rather confused, any help is appreciated. Thank you.

 

[janhaa]

hints:

[tex]\tau(bottom)=\tau(top)[/tex]

for the bottom fluid:
[tex]\tau(bottom)=\mu_o\frac{V_1}{cb}[/tex]

for the top fluid:
[tex]\tau(top)=\mu\frac{V-V_1}{b-cb}[/tex]