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farleyknight
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Homework Statement
An airplane is flying in a horizontal circle at a speed of 460 km/h (Fig. 6-42). If its wings are tilted at angle θ = 37° to the horizontal, what is the radius of the circle in which the plane is flying? Assume that the required force is provided entirely by an “aerodynamic lift” that is perpendicular to the wing surface.
http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c06/fig06_39.gif
Homework Equations
Newton's second law, for centripetal motion:
[tex]F_{net} = m*\left( \frac{v^2}{r} \right)[/tex]
The Attempt at a Solution
First let us identify the forces acting on the plane. There are exactly 3: The force of gravity, [tex]F_g[/tex], the force of the lift, causing it to fly, [tex]F_L[/tex], and the centripetal force caused by the rotation, [tex] m*\left( \frac{v^2}{r} \right)[/tex]
We know that there must be some lift on the plane, keeping it in air. Because there is no vertical motion, we know that [tex]F_L - F_g*\cos(\theta) = 0[/tex]. So [tex]F_L = F_g*\cos(\theta)[/tex]
Next, we also know that the plane must be pulled in by the centripetal force. So the horizontal component of the lift must be caused by this force, and we have
[tex]m*\left( \frac{v^2}{r} \right) = F_L_x = F_L*\sin(\theta) = F_g*\cos(\theta)*\sin(\theta) [/tex]
Solving for the radius r
[tex]\frac{m*v^2}{F_g*\cos(\theta)*\sin(\theta)} = \frac{v^2}{g*\cos(\theta)*\sin(\theta)} = r[/tex]
Subbing the values
[tex]\frac{460^2}{9.8*\cos(37)*\sin(37)} = 44923 = r[/tex]
The computer system for my homework has me entering this as meters instead of kilometers, which is the units of the shown result. However, it's not lining up. I suspect that there might be something wrong with my lifting force, but I'm not sure.