Question

A jet airplane is in level flight. The mass of the airplane is m=8930 kg. The airplane travels at a constant speed around a circular path of radius ????=8.23 mi and makes one revolution every T=0.109 h. Given that the lift force acts perpendicularly upward from the plane defined by the wings, what is the magnitude of the lift force acting on the airplane?

Answer 1

Answer:88.95 kN

Explanation:

Given

mass of Plane =8930 kg

radius of circular path $=8.23 mi\approx 13244.9 m$

Time period =0.109 h

Let F be the lift force

If Plane is making an $\theta$ with horizontal then F will make an angle of $90-\theta$ w.r.t horizontal

Thus

$Fcos\theta =mg$---1

$Fsin\theta =m\omega ^2r$------2

and $\omega$is given by

$\omega =\frac{2\pi }{T}=\frac{2\times 3.142}{392.4}=0.0160 rad/s$

Divide 2 & 1

$tan\theta =\frac{\omega ^2r}{g}$

$tan\theta =\frac{0.0160^2\times 13244.9}{9.8}$

$tan\theta =0.345$

$\theta =19.03$

Substitute $\theta$in 1

$Fcos(19.03)=8930\times 9.8$

$F=88.95 kN$