Problem Statement
Find the drag force difference on a flat plate of size 1.5 m x 1.5 m when the plate is moving at a speed of 5 m/s normal to its plate first in water and second in air of density 1.24 kg/m³. Co-efficient of drag is given as 1.10.
Given Data & Constants
- Area of plate, \(A = 1.5 \, \text{m} \times 1.5 \, \text{m} = 2.25 \, \text{m}^2\)
- Velocity of plate, \(V = 5 \, \text{m/s}\)
- Density of water, \(\rho_{water} = 1000 \, \text{kg/m}^3\)
- Density of air, \(\rho_{air} = 1.24 \, \text{kg/m}^3\)
- Co-efficient of drag, \(C_D = 1.10\)
Solution
1. Calculate Drag Force in Water (\(F_{D,water}\))
We use the general drag force formula.
$$ F_D = C_D \times \frac{1}{2} \rho A V^2 $$
$$ F_{D,water} = 1.10 \times \frac{1}{2} \times 1000 \times 2.25 \times (5)^2 $$
$$ F_{D,water} = 1.10 \times 0.5 \times 1000 \times 2.25 \times 25 $$
$$ F_{D,water} = 30937.5 \, \text{N} $$
2. Calculate Drag Force in Air (\(F_{D,air}\))
We use the same formula with the density of air.
$$ F_{D,air} = 1.10 \times \frac{1}{2} \times 1.24 \times 2.25 \times (5)^2 $$
$$ F_{D,air} = 1.10 \times 0.5 \times 1.24 \times 2.25 \times 25 $$
$$ F_{D,air} \approx 38.36 \, \text{N} $$
3. Find the Difference in Drag Force
$$ \text{Difference} = F_{D,water} - F_{D,air} $$
$$ \text{Difference} = 30937.5 - 38.36 = 30899.14 \, \text{N} $$
Final Results:
Drag Force in Water: \(30937.5 \, \text{N}\) (or \(30.94 \, \text{kN}\))
Drag Force in Air: \( \approx 38.36 \, \text{N} \)
Difference in Drag Force: \( \approx 30899.14 \, \text{N} \) (or \(30.9 \, \text{kN}\))
