Plate towed between two parallel walls

Data: $\mu=0.7$ Pa·s, $A=0.25\times0.25=0.0625\ \text{m}^2$, $V=0.15$ m/s. Total gap $=25$ mm; the plate is 6 mm from one wall, so its distance from the other wall is $25-6=19$ mm (plate thickness neglected).

 wall |<-- 6 mm -->[plate]<---- 19 mm ---->| wall

With linear velocity profiles on both sides, the shear stresses are

$$\tau_1=\mu\frac{V}{y_1}=0.7\times\frac{0.15}{0.006}=17.5\ \text{N/m}^2$$ $$\tau_2=\mu\frac{V}{y_2}=0.7\times\frac{0.15}{0.019}=5.526\ \text{N/m}^2$$

Forces on the two faces (both oppose the motion):

$$F_1=\tau_1A=17.5\times0.0625=1.094\ \text{N},\qquad F_2=\tau_2A=0.345\ \text{N}$$ $$F=F_1+F_2=1.094+0.345=1.439\ \text{N}$$

Answer: the force exerted by the liquid on the plate is about 1.44 N, opposing the motion.