Question

A jet plane having a cabin of length $l-50m$ flies along the horizontal with an acceleration $a=1m/s^2$ The air density in the cabin is $\rho =1.2\times {10}^{-3}g/c{m}^3$
What is the difference between the atmospheric pressure and the air pressure exerted on the ears of the passengers sitting in the front, middle, and rear parts of the cabin?

Answer 1

The air layer of thickness $\Delta x$ at a distance $x$ form the front of the cabin experiences the force of pressure
$\left[p\right(x+\Delta x)-p(x\left)\right]S,$
where $S$ is the cross-sectional area of the cabin SInce air is at rest relative to the cabin, the equation of motion for the mass of air under consideration has the form
$\rho S\Delta xa=\left[p\right(x+\Delta x)-p(x\left)\right]S$
Making $\Delta x$ tend to zero, we obtain
$\frac{dp}{dx}=\rho a$
whence
$p\left(x\right)=p_1+\rho ax$
Since the mean pressure in the cabin remains unchanged and equal to the atmospheric pressure ${\rho }_0$ the constant $p_1$ can be found from the condition
$p_0=p_1+\frac{\rho al}{2}$,
where $l$ is the length of the cabin. Thus in the middle and near parts of the cabin, the pressure is equal to the atmospheric pressure, while in the front and rear parts of the cabin, the pressure is lower and higher than the atmospheric pressure by
$\Delta p\frac{\rho al}{2}\simeq 0.03Pa$ respectively