---
title: "Counter weight for plate under air flow"
subject: "Fluid Mechanics"
chapter: 9
key: "semester_3/fluid_mechanics:flat-plate-balance-air-flow-mass"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/flat-plate-balance-air-flow-mass
site: BCT Engineering
---

# Counter weight for plate under air flow

Fluid Mechanics · Chapter 9: Flow past through submerged bodies

- Asked in: 2076 Chaitra, 8 marks
- Video: [Solved Problem: Linear Momentum Quiz](https://www.youtube.com/watch?v=xmAnRNpxUKU) — Fluid Matters
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/flat-plate-balance-air-flow-mass
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-9

## Question

The weight of a thin flat plate 50 cm × 50 cm in size is balanced by a counter weight that has a mass of 2 kg as shown in figure below. Now a fan is turned on, and air flows downward over both surfaces of the plate with a free-stream velocity of 10 m/s. Determine the mass of the counter weight that needs to be added in order to balance the plate in this case. [Figure: balance with plate 50 cm × 50 cm on one side and counter weight on the other; air at 25°C flowing at 10 m/s over the plate]

## Short answer

With laminar flow over the plate (Re_L = 3.2×10^5), the skin-friction coefficient C_f = 1.328/√Re_L = 0.00235. Total drag on both faces is F_D = 2 C_f (0.5 ρ V² A) = 0.0695 N. The extra mass needed is F_D/g ≈ 7.1 g.

## Answer

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.

```text
        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).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What force unbalances the plate when air flows?
   A. Lift
   B. Form drag only
   C. Buoyancy
   D. Skin friction drag on both faces
   Answer: D. The plate is thin and parallel, so only skin friction acts. (Key point: Force source)

2. Which regime applies and why?
   A. Turbulent, since Re > 2000
   B. Laminar, since Re < 1
   C. Laminar, since Re_L = 3.2 x 10^5 < 5 x 10^5
   D. Turbulent, since air is a gas
   Answer: C. Below the critical Reynolds number the layer is laminar over the full plate. (Key point: Flow regime)

3. Which coefficient is used?
   A. 0.074/Re^0.2
   B. Blasius: Cf = 1.328/sqrt(Re_L)
   C. 24/Re
   D. 0.455/(log Re)^2.58
   Answer: B. That gives 0.002347. (Key point: Drag coefficient)

4. How is the added mass found and what factor of 2 matters?
   A. m = F_D/g, with F_D counting both surfaces
   B. m = F_D x g
   C. Only one face counted
   D. m = F_D/(2g)
   Answer: A. F_D = 0.0695 N so m is about 7.1 g. (Key point: Counterweight mass)

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