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

Step-by-step 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 =(2\pi )/(T)=(2* 3.142)/(392.4)=0.0160 rad/s`

Divide 2 & 1

`tan\theta =(\omega ^2r)/(g)`

`tan\theta =(0.0160^2* 13244.9)/(9.8)`

`tan\theta =0.345`

`\theta =19.03`

Substitute
`\theta`in 1

`Fcos(19.03)=8930* 9.8`

`F=88.95 kN`