Counter weight for plate under air flow

The air flowing along both faces of the plate produces a friction (drag) force acting downward on the plate. The extra mass on the counterweight side must balance this force.

Assumptions

Air at 25 °C and 1 atm: $\rho = 1.184\ \text{kg/m}^3$, $\nu = 1.562\times10^{-5}\ \text{m}^2/\text{s}$. The plate is thin and parallel to the flow, so only skin friction acts.

        air 10 m/s
         |  |
     +---v--v---+
     |  plate   |      balance beam
   ==|==========|======o======[ counter weight 2 kg ]

Reynolds number

$$Re_L = \frac{VL}{\nu} = \frac{10(0.5)}{1.562\times10^{-5}} = 3.20\times10^5 < 5\times10^5$$

The boundary layer is laminar over the whole plate.

Drag coefficient (Blasius)

$$C_f = \frac{1.328}{\sqrt{Re_L}} = \frac{1.328}{\sqrt{3.201\times10^5}} = 0.002347$$

Drag force on both surfaces

$$\begin{aligned} F_D &= 2\,C_f\cdot\tfrac12\rho V^2 A,\quad A = 0.5\times0.5 = 0.25\ \text{m}^2 \\ &= 2(0.002347)(0.5)(1.184)(10)^2(0.25) = 0.0695\ \text{N} \end{aligned}$$

Mass to add

$$m = \frac{F_D}{g} = \frac{0.0695}{9.81} = 0.00708\ \text{kg} \approx 7.1\ \text{g}$$

Answer: about 7.1 g must be added to the 2 kg counterweight (total about 2.007 kg).