Question

In the previous question, the net force exerted by fluid on moving plate is

Answer 1

$\displaystyle F = -\mu \frac{U}{h} A$.

Answer 2

For a fluid flowing past a flat plate, Newton’s law of viscosity gives the shear stress as

$$\tau = \mu \frac{du}{dy}$$

If the plate moves with speed $U$ and the distance between the plate and the stationary fluid is $h$, the velocity gradient is

$$\frac{du}{dy} = \frac{U}{h}.$$

Thus, the shear stress on the plate is

$$\tau = \mu \frac{U}{h}.$$

Multiplying by the area $A$ of the plate, the net force exerted by the fluid on the moving plate is

$$F = \tau A = \mu \frac{U}{h} A.$$

Since the force acts opposite to the plate’s motion, we may also write

$$F = -\mu \frac{U}{h} A.$$

Minimal Explanation:
Using Newton’s law of viscosity, shear stress $\tau = \mu \frac{U}{h}$. Multiplying by area $A$ gives net force $F = -\mu \frac{UA}{h}$ (negative sign indicates opposition to motion).