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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?

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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]

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