Chapter 10, Problem 86GP

To determine

To Find: The speed of the upper surface.

Answer to Problem 86GP

The speed of air at upper surface of the wing of the plane is $188.6\text{m/s}$ .

Explanation of Solution

Given info:

Mass of airplane $m=2\times {10}^6\text{ kg}$ .

Surface area of wings $A=1200{\text{ m}}^{\text{2}}$ .

Density of air $\rho =1.23{\text{ kg/m}}^{\text{3}}$ .

Formula used:

A plane can float in the air only when its weight is balanced by the force due to pressure difference between the upper and lower sides of the wings.

For the flying of an airplane the following formula is used,

Weight of plane = Force due to pressure difference

  $m\times g=\frac{1}{2}\times \rho \times A\times \left(v_{\text{upper}}^2-v_{\text{lower}}^2\right)$

Here,

  $m\times g$ is the weight of plane,

  $\rho$ is density of air,

  $A$ is surface area of wings,

  $v_{\text{upper}}$ is velocity of air at upper surface of wing,

  $v_{\text{lower}}$ is velocity of air at lower surface of wing.

Calculation:

Substitute the given values in the above equation,

  $\begin{array}{l}m\times g=\frac{1}{2}\times \rho \times A\times \left(v_{\text{upper}}^2-v_{\text{lower}}^{\text{2}}\right)\\ 2\times {10}^6\times 9.8=\frac{1}{2}\times 1.23\times 1200\times \left(v_{\text{upper}}^2-{95}^2\right)\\ v_{\text{upper}}^{\text{2}}-{95}^2=\frac{2\times {10}^6\times 9.8\times 2}{1.23\times 1200}=26558.26\\ v_{\text{upper}}^{\text{2}}=26558.26+{95}^2=35583.26\\ v_{\text{upper}}=188.6\text{ m/s}\end{array}$

Conclusion:

Thus, the speed of air at upper surface of the wing of the plane is $188.6\text{m/s}$ .