Answer:88.95 kN
Explanation:
Given
mass of Plane =8930 kg
radius of circular path [tex]=8.23 mi\approx 13244.9 m[/tex]
Time period =0.109 h
Let F be the lift force
If Plane is making an [tex]\theta [/tex] with horizontal then F will make an angle of [tex]90-\theta [/tex] w.r.t horizontal
Thus
[tex]Fcos\theta =mg[/tex]---1
[tex]Fsin\theta =m\omega ^2r[/tex]------2
and [tex]\omega [/tex]is given by
[tex]\omega =\frac{2\pi }{T}=\frac{2\times 3.142}{392.4}=0.0160 rad/s[/tex]
Divide 2 & 1
[tex]tan\theta =\frac{\omega ^2r}{g}[/tex]
[tex]tan\theta =\frac{0.0160^2\times 13244.9}{9.8}[/tex]
[tex]tan\theta =0.345[/tex]
[tex]\theta =19.03[/tex]
Substitute [tex]\theta [/tex]in 1
[tex]Fcos(19.03)=8930\times 9.8[/tex]
[tex]F=88.95 kN[/tex]