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Air streams horizontally past a small airplane’s wings such that the speed is 65.0 m/s over the top surface and 55.0 m/s past the bottom surface. If the plane has a mass of 1370 kg and a wing area of 16.2 m^2, what is the magnitude of the net vertical force (including the effects of gravity) on the airplane? The density of the air is 1.20 kg/m^3. Express your answer in newtons.

Sagot :

Answer:

Explanation: To find the magnitude of the net vertical force on the airplane, including the effects of gravity, we need to calculate the difference in pressure between the top and bottom surfaces of the wings due to the airflow.

Given data:

- Speed of air over the top surface of the wings, \( v_{\text{top}} = 65.0 \) m/s

- Speed of air over the bottom surface of the wings, \( v_{\text{bottom}} = 55.0 \) m/s

- Wing area, \( A = 16.2 \) m\(^2\)

- Density of air, \( \rho = 1.20 \) kg/m\(^3\)

- Acceleration due to gravity, \( g = 9.81 \) m/s\(^2\)

- Mass of the airplane, \( m = 1370 \) kg

First, calculate the dynamic pressure difference \( \Delta P \) between the top and bottom surfaces of the wings:

\[ \Delta P = \frac{1}{2} \rho \left( v_{\text{top}}^2 - v_{\text{bottom}}^2 \right) \]

Calculate \( v_{\text{top}}^2 \) and \( v_{\text{bottom}}^2 \):

\[ v_{\text{top}}^2 = (65.0)^2 = 4225 \, \text{m}^2/\text{s}^2 \]

\[ v_{\text{bottom}}^2 = (55.0)^2 = 3025 \, \text{m}^2/\text{s}^2 \]

Now calculate \( \Delta P \):

\[ \Delta P = \frac{1}{2} \times 1.20 \times (4225 - 3025) \]

\[ \Delta P = \frac{1}{2} \times 1.20 \times 1200 \]

\[ \Delta P = 720 \, \text{Pa} \]

Next, calculate the net vertical force \( F_{\text{net}} \), which includes the effects of gravity:

\[ F_{\text{net}} = \Delta P \times A \]

\[ F_{\text{net}} = 720 \times 16.2 \]

\[ F_{\text{net}} = 11664 \, \text{N} \]

Therefore, the magnitude of the net vertical force (including the effects of gravity) on the airplane is \( \boxed{11664 \, \text{N}} \).

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