ch06-p053 - = mv 2/R cos θ and when this is substituted...

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53 The free-body diagram for the plane is shown on the right. F is the magnitude of the lift on the wings and m is the mass of the plane. Since the wings are tilted by 40 to the horizontal and the lift force is perpendicular to the wings, the angle θ is 50 . The center of the circular orbit is to the right of the plane, the dashed line along x being a portion of the radius. Take the x axis to be to the right and the y axis to be upward. Then the x component of Newton’s second law is F cos θ = mv 2 /R and the y component is F sin θ mg = 0, where R is the radius of the orbit. The first equation gives F
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Unformatted text preview: = mv 2 /R cos θ and when this is substituted into the second, ( mv 2 /R ) tan θ = mg results. Solve for R : R = v 2 g tan θ . The speed of the plane is v = 480 km / h = 133 m / s, so R = (133 m / s) 2 9 . 8 m / s 2 tan 50 ◦ = 2 . 2 × 10 3 m . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • F mg θ x y...
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