Problem Statement
A flat plate 2 m x 2 m moves at 40 km/hour in stationary air of density 1.25 kg/m³. If the co-efficient of drag and lift are 0.2 and 0.8 respectively, find : (i) the lift force, (ii) the drag force, (iii) the resultant force, and (iv) the power required to keep the plate in motion.
Given Data & Constants
- Area of plate, \(A = 2 \, \text{m} \times 2 \, \text{m} = 4 \, \text{m}^2\)
- Velocity of plate, \(V = 40 \, \text{km/hr} = \frac{40 \times 1000}{3600} \approx 11.11 \, \text{m/s}\)
- Density of air, \(\rho = 1.25 \, \text{kg/m}^3\)
- Co-efficient of drag, \(C_D = 0.2\)
- Co-efficient of lift, \(C_L = 0.8\)
Solution
(i) The Lift Force (\(F_L\))
The lift force acts perpendicular to the direction of motion.
$$ F_L = C_L \times \frac{1}{2} \rho A V^2 $$
$$ F_L = 0.8 \times \frac{1}{2} \times 1.25 \times 4 \times (11.11)^2 $$
$$ F_L = 0.8 \times 0.5 \times 1.25 \times 4 \times 123.43 $$
$$ F_L \approx 246.86 \, \text{N} $$
(ii) The Drag Force (\(F_D\))
The drag force acts parallel to and opposite the direction of motion.
$$ F_D = C_D \times \frac{1}{2} \rho A V^2 $$
$$ F_D = 0.2 \times \frac{1}{2} \times 1.25 \times 4 \times (11.11)^2 $$
$$ F_D = 0.2 \times 0.5 \times 1.25 \times 4 \times 123.43 $$
$$ F_D \approx 61.72 \, \text{N} $$
(iii) The Resultant Force (\(F_R\))
The resultant force is the vector sum of the lift and drag forces.
$$ F_R = \sqrt{F_L^2 + F_D^2} $$
$$ F_R = \sqrt{(246.86)^2 + (61.72)^2} = \sqrt{60940 + 3809} $$
$$ F_R = \sqrt{64749} \approx 254.46 \, \text{N} $$
(iv) The Power Required to Keep the Plate in Motion
The power required is the work done against the drag force per second.
$$ \text{Power} = F_D \times V $$
$$ \text{Power} = 61.72 \, \text{N} \times 11.11 \, \text{m/s} \approx 685.7 \, \text{W} $$
Final Results:
(i) Lift Force: \( \approx 246.9 \, \text{N} \)
(ii) Drag Force: \( \approx 61.7 \, \text{N} \)
(iii) Resultant Force: \( \approx 254.5 \, \text{N} \)
(iv) Power Required: \( \approx 685.7 \, \text{W} \)
