Finding the banking angle?You are flying to Chicago for a weekend away from the books. In your last physics class, you learned that the airflow over the wings of the plane creates a lift force, which acts perpendicular to the wings. When the plane is flying level, the upward lift force exactly balances the downward weight force. Since... |
When the wings are banked at an angle θ with the horizontal, the lift force F, which is perpendicular to the wings, makes an angle θ with the vertical. Therefore the vertical and horizontal components of the lift are respectively:
F₁ = Fcosθ (vertical)
F₂ = F sinθ (horizontal)
Since the plane is at level flight (constant altitude), F₁ is exactly equal to the weight of the plane:
Fcosθ = mg (1)
F₂ is the centripetal force that cause the circular motion, which is mv²/r, so
F sinθ = mv²/r (2)
From (1) and (2) we get
Fsinθ/Fcosθ = mg*r/mv²
tanθ = gr/v²
To have the numerical value of tanθ, you must either take g = 9.8 m/s² and converse the velocity and radius into SI units, or to know the value of g in miles per hour per hour
According to http://en.wikipedia.org/wiki/Standard_gr... the standard g is approximately equal to 21.937 mph/s
Conversed to mph/h: g = 21.937 mph/s = 3600*21.937 mph/h = 78973 mph/h
Plugging in the formula, with g = 78973 mph/h, r = 6 mi, v = 450 mph
tanθ = gr/v²
tanθ = 78973*6/450² = 2.33994
θ = 66.86° = 66° 51' 36"
