When a plane turns, it banks as shown in the figure to give the lifting force of the wings a horizontal?

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component that turns the plane.If a plane is flying level at 960 km/h and the banking angle is not to exceed 35 degrees, what is the minimum curvature radius for the turn? http://session.masteringphysics.com/problemAsset/1...

Ms. Juana Fadel · Accepted Answer

Fw is the aerodynamic lift force acting on the wings, perpendicular to their surface Fg is the gravitational force from the Earth.Newton's 2nd law in the vertical direction:Fw*cos(theta) = FgNewton's 2nd law in the horizontal direction:Fw*sin(theta) = m*a Recall the origin of Fg, Fg = m*g:Fw*cos(theta) = m*gsolve for Fw:Fw = m*g/cos(theta)Substitute:m*g*sin(theta)/cos(theta) = m*aCancel m's and combine trigonometry:g*tan(theta) = aacceleration is due to centripetal acceleration:a = v^2/rSubstitute:g*tan(theta) = v^2/rSolve for r:r = v^2/(g*tan(theta))Data:v:=266.67 m/s; g:=9.8 N/kg; theta:=35 deg;Result:r = 10363 metersr = 10.363 km

Ms. Juana Fadel · Answer

Plane Banking

Ms. Juana Fadel · Answer

At 45 deg bank, the lateral forces on the airplane are Fg = mgsin(45 d) and Fcent = mv^2/r*cos(45 d). To avoid sideslip, these forces must be equal to keep the net force on the plane perpendicular to the wing. Note sin(45 d) = cos(45 d). Thus Fg = Fcent ==> r = v^2/g = (940/3.6)^2/9.8 = 6957 m.

Ms. Juana Fadel · Answer

Acceleration v^2/r for the circular movement must be provided by the lifting force horizontal component. Hence v^2/r = F/m* sin(theta). Now balance the vertical components : F*cos(theta) = m*gSo v^2/r = g*tan(theta) If you want theta < 35 deg then r > v^2/(g*tan(35)) = 10352 m

Prof. Scot Emard

When a plane turns, it banks as shown in the figure to give the lifting force of the wings a horizontal?

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