Question

A plane is in level flight at constant speed and each of its two wings has an area of 40 m² . If the speed of the air is 180 km/h over the lower wing surface and 252 km/h over the upper wing surface, the mass of the plane is ________kg. (Take air density to be 1 kg m^–3 and g = 10 ms^–2

Answer 1

Correct Answer: 9600

Given:

\(A = 80 \, \text{m}^2\)

Using Bernoulli’s equation:

\(\Delta P = (P_2 - P_1) = \frac{1}{2} \rho (V_1^2 - V_2^2)\)

Area of wings = \(A\)

Calculating the lift force:

\(mg = \frac{1}{2} \times 1 \times (70^2 - 50^2) \times 80\)

Simplifying:

\(mg = 40 \times 2400\)

\(m = 9600 \, \text{kg}\)

Answer 2

This problem requires us to find the mass of an airplane in level flight by analyzing the lift force generated due to the difference in air speeds over the upper and lower surfaces of its wings. This lift force must balance the weight of the plane.

Concept Used:

The principle behind the lift force on an airplane wing is Bernoulli's principle. For a fluid in horizontal flow, Bernoulli's equation relates pressure (\(P\)) and velocity (\(v\)):

\[ P + \frac{1}{2}\rho v^2 = \text{constant} \]

where \(\rho\) is the density of the fluid. This implies that where the speed is higher, the pressure is lower, and vice versa. The pressure difference (\(\Delta P\)) between the lower and upper surfaces of the wing creates an upward force called lift.

Lift Force: The net upward force on the wings is given by:

\[ F_{\text{lift}} = (\Delta P) \times A_{\text{total}} = (P_{\text{lower}} - P_{\text{upper}}) \times A_{\text{total}} \]

For the plane to be in level flight, the lift force must be equal to the weight of the plane:

\[ F_{\text{lift}} = mg \]

where \(m\) is the mass of the plane and \(g\) is the acceleration due to gravity.

Step-by-Step Solution:

Step 1: List the given data and convert units.

Area of each wing = 40 m². Total area of two wings, \(A_{\text{total}} = 2 \times 40 = 80 \text{ m}^2\).

Speed of air over the lower surface, \(v_1 = 180 \text{ km/h}\).

\[ v_1 = 180 \times \frac{1000 \text{ m}}{3600 \text{ s}} = 180 \times \frac{5}{18} = 50 \text{ m/s} \]

Speed of air over the upper surface, \(v_2 = 252 \text{ km/h}\).

\[ v_2 = 252 \times \frac{1000 \text{ m}}{3600 \text{ s}} = 252 \times \frac{5}{18} = 14 \times 5 = 70 \text{ m/s} \]

Density of air, \(\rho = 1 \text{ kg m}^{-3}\).

Acceleration due to gravity, \(g = 10 \text{ ms}^{-2}\).

Step 2: Apply Bernoulli's principle to find the pressure difference.

Let \(P_1\) be the pressure on the lower surface and \(P_2\) be the pressure on the upper surface.

\[ P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2 \]

The pressure difference that generates lift is \(\Delta P = P_1 - P_2\).

\[ \Delta P = P_1 - P_2 = \frac{1}{2}\rho (v_2^2 - v_1^2) \]

Substitute the values:

\[ \Delta P = \frac{1}{2} \times 1 \times (70^2 - 50^2) \] \[ \Delta P = \frac{1}{2} \times (4900 - 2500) = \frac{1}{2} \times 2400 = 1200 \text{ N/m}^2 \]

Step 3: Calculate the total upward lift force.

The total lift force is the pressure difference multiplied by the total area of the wings.

\[ F_{\text{lift}} = \Delta P \times A_{\text{total}} \] \[ F_{\text{lift}} = 1200 \text{ N/m}^2 \times 80 \text{ m}^2 = 96000 \text{ N} \]

Step 4: Equate the lift force to the weight of the plane to find the mass.

For level flight, the lift force must balance the gravitational force (weight).

\[ F_{\text{lift}} = mg \] \[ 96000 \text{ N} = m \times 10 \text{ ms}^{-2} \] \[ m = \frac{96000}{10} = 9600 \text{ kg} \]

Final Computation & Result:

The mass of the plane is calculated by dividing the total lift force by the acceleration due to gravity.

\[ m = \frac{F_{\text{lift}}}{g} = \frac{96000 \text{ N}}{10 \text{ m/s}^2} = 9600 \text{ kg} \]

The mass of the plane is 9600 kg.