Force to lift plate in liquid gap

Data: gap $=2.2$ cm, plate thickness $=0.2$ cm, so the clearance on each side is $\dfrac{2.2-0.2}{2}=1.0$ cm $=0.01$ m. $A=1.5\times1.5=2.25\ \text{m}^2$, $V=0.15$ m/s, $\mu=1.75$ N·s/m², $W=40$ N, $\rho=0.9\times1000=900\ \text{kg/m}^3$.

  wall |  1 cm | plate | 1 cm |  wall
       |  oil  | 0.2cm | oil  |
                  ^ lifted at 0.15 m/s

Viscous drag on both faces (linear profile):

$$F_v=2\mu A\frac{V}{t}=2\times1.75\times2.25\times\frac{0.15}{0.01}=118.13\ \text{N}$$

Buoyancy (plate volume $=1.5\times1.5\times0.002=0.0045\ \text{m}^3$):

$$F_b=\rho g V_p=900\times9.81\times0.0045=39.73\ \text{N}$$

Force balance for upward motion at constant velocity:

$$F+F_b=W+F_v\ \Rightarrow\ F=40-39.73+118.13=118.39\ \text{N}$$

Answer: $F\approx118.4$ N upward (about 118 N; if buoyancy is ignored, $F=158.1$ N).