Problem 55 - Master Resource Book in JEE Main Physics

Question

A plane is in level flight at a constant apeed and each wing has an area of \(25 \mathrm{~m}^{2}\). During flight the speed of the air is \(216 \mathrm{kmh}^{-1}\) over the lower wing surface and \(252 \mathrm{kmh}^{-1}\) over the upper wing surface of each wing of aeroplane. Take density of air \(=1 \mathrm{kgm}^{-3}\) and \(g=10 \mathrm{~ms}^{-2}\) The total upward force on the plane is (a) \(250 \mathrm{~N}\) (b) \(2500 \mathrm{~N}\) (c) \(17500 \mathrm{~N}\) (d) \(32500 \mathrm{~N}\)

Step-by-Step Solution

Answer

The total upward force on the plane is \(32500 \text{ N}\), option (d).

Step 1: Identify the Airspeed Difference

Convert the speeds from km/h to m/s: For the lower wing surface: \[216 \text{ km/h} = \frac{216 \times 1000}{60 \times 60} = 60 \text{ m/s}\]For the upper wing surface: \[252 \text{ km/h} = \frac{252 \times 1000}{60 \times 60} = 70 \text{ m/s}\]

Step 2: Apply Bernoulli's Equation

Using Bernoulli's principle, find the pressure difference between the upper and lower surfaces of the wing. The formula for the pressure difference across the wing surfaces is: \[\Delta P = \frac{1}{2} \rho (v_ ext{lower}^2 - v_ ext{upper}^2)\]Substitute the given values:\[\Delta P = \frac{1}{2} \times 1 \times (60^2 - 70^2)\]Calculate:\[\Delta P = \frac{1}{2} (3600 - 4900) = \frac{1}{2} (-1300) = -650 \text{ N/m}^2\]

Step 3: Calculate the Lift Force per Wing

Calculate the lift force produced by one wing using the area of the wing:\[F = \Delta P \times A\]\[F = -650 \text{ N/m}^2 \times 25 \text{ m}^2\]\[F = -16250 \text{ N} \text{ (Upward force per wing)}\]

Step 4: Double the Force for Both Wings

Since the plane has two wings, calculate the total lift force:\[F_{\text{total}} = 16250 \text{ N} \times 2\]\[F_{\text{total}} = 32500 \text{ N}\]

Step 5: Determine the Correct Answer

From the given options, match the total lift force calculated.The total lift force is \(32500 \text{ N}\), which matches option (d).

Key Concepts

Lift Force Calculation

Lift force is a crucial concept in aerodynamics. It plays a significant role in helping planes stay in the air. Understanding lift force starts with Bernoulli's Principle. This principle states that an increase in the speed of a fluid occurs simultaneously with a decrease in pressure.
This principle helps us find out the pressure difference between the upper and lower parts of a wing. In our exercise, the upper surface of the wing has faster airflow than the lower surface. This difference leads to lower pressure on the top and higher pressure underneath. The result is an upward lift force.
To calculate lift force, you apply the formula:

• First, determine the pressure difference using Bernoulli's equation: \[\Delta P = \frac{1}{2} \rho (v_{\text{lower}}^2 - v_{\text{upper}}^2)\]
• Substitute the given values into this equation to find \( \Delta P \).
• Finally, multiply \( \Delta P \) by the wing area to calculate the lift force: \[F = \Delta P \times A\]

Understanding these calculations helps in figuring out the forces acting on our airplane and explains why the total lift force in this problem is \( 32500 \text{ N} \), as calculated from both wings.

Wing Area in Aerodynamics

The wing area plays an essential role in determining the lift force generated by an aircraft. Larger wing areas can generate more lift, which is crucial for supporting the weight of the airplane In our exercise, each wing has an area of \(25 \mathrm{~m}^{2}\). This wing area is significant because:

• The lifting surface increases the interaction with air.
• Larger areas require less pressure difference to generate lift.

The wing area is directly proportional to the lift force. This is evident in the lift force formula: \(F = \Delta P \times A\). When you multiply the air pressure difference across a wing by the wing area, you obtain the lift force that aids the airplane in sustained flight. Therefore, in aerodynamics, optimizing wing area according to the aircraft’s purpose is vital for efficient flight.

Airspeed Conversion

Conversions are frequently necessary in physics, especially when working with speed and velocity. Airspeed conversion between units is vital for calculations involving kinetic energy and pressure.In the exercise, the airspeed is initially given in kilometers per hour (km/h). But for ease of calculation within the International System of Units (SI), converting to meters per second (m/s) is needed. Here’s how that conversion works:

• To convert km/h to m/s, multiply by \(\frac{1000}{3600}\), or simply divide by 3.6.
• For example, converting 216 km/h: \[216 ext{ km/h} = \frac{216 \times 1000}{3600} = 60 \text{ m/s}\]
• Similarly, converting 252 km/h results in: \[252 ext{ km/h} = \frac{252 \times 1000}{3600} = 70 \text{ m/s}\]

Airspeed conversion is a critical component of ensuring accurate application of formulas, like Bernoulli's equation, which relies on velocity being expressed in meters per second. Correct conversion allows for precise pressure calculations, thus impacting how forces, such as lift, are calculated in aerodynamics.