Plate sliding on inclined oil film

Data: $A=1\ \text{m}^2$, $W=350$ N, $V=1.5$ m/s, $t=1\ \text{mm}=0.001$ m, $\tan\theta=5/12$ so $\sin\theta=5/13$.

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  plate slides down, oil film t = 1 mm

At uniform velocity there is no acceleration, so the shear force from the oil balances the component of weight along the plane:

$$F=W\sin\theta=350\times\frac{5}{13}=134.62\ \text{N}$$

The velocity profile across the thin film is linear:

$$\tau=\mu\frac{V}{t}=\frac{F}{A}$$ $$\mu=\frac{F\,t}{A\,V}=\frac{134.62\times0.001}{1\times1.5}=0.0897\ \text{Pa·s}$$

Answer: $\mu\approx0.0897\ \text{N·s/m}^2$ (0.897 poise).