# An airplane has a mass of 18,000 kilograms, and the total wing is 50 m2. during level flight, the pressure on the lower wing surface is 5.25 x 104 pa. what is the pressure on the upper wing surface?

Three factors contribute to an airplane's ability to fly: first, the thrust from the engine pushes the aircraft forward; second, the angle of the wing provides an initial surface for the air passing by to push the aircraft upward; and third, a basic principle of fluid mechanics called "lift" enables the aircraft to stay in the air for the duration of the flight.

An airplane wing will experience positive lift because the static pressure at the bottom of the wing is higher than the static pressure at the top of the wing. Bernoulli's principle, which explains this phenomenon, is illustrated below. Lift is simply the pressure differential between the bottom and top of the wing.

This is not the full form of Bernoulli, but it is sufficient because the hydrostatic head differential over an airfoil is small. The following can be used to answer this question:

Assuming that the aircraft has attained cruising altitude and is not experiencing elevation changes, add up the forces acting in the vertical direction, or what I will refer to as "y":

- Next, since the plane's mass is known, find F_L:

Because the pressure on the top of the wing should be less than the pressure on the bottom of the wing, the response makes sense.

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To find the pressure on the upper wing surface, we can use Bernoulli's principle, which states that in a fluid flow, an increase in the speed of the fluid occurs simultaneously with a decrease in pressure or a decrease in the fluid's potential energy. Given that the airplane is in level flight, we assume that the airspeed is constant above and below the wing.

Bernoulli's equation can be expressed as:

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

Where: (P_1) and (P_2) are the pressures at points 1 and 2 (upper and lower wing surfaces, respectively), (\rho) is the air density (constant), (v_1) and (v_2) are the velocities of the air at points 1 and 2 (upper and lower wing surfaces, respectively).

Given that the total wing area is 50 m², we can assume that the airspeed above and below the wing is the same, (v_1 = v_2).

Now, let's solve for (P_2) (pressure on the lower wing surface) and (P_1) (pressure on the upper wing surface).

[P_1 + \frac{1}{2} \cdot \rho \cdot v^2 = P_2 + \frac{1}{2} \cdot \rho \cdot v^2]

[P_1 = P_2 + \frac{1}{2} \cdot \rho \cdot v^2 - \frac{1}{2} \cdot \rho \cdot v^2]

[P_1 = P_2]

Therefore, the pressure on the upper wing surface is the same as the pressure on the lower wing surface, which is (5.25 \times 10^4) Pa.

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