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

Question

When a plane turns, it banks to create a horizontal component of the lifting force generated by the wings, which helps to facilitate the turn. If a plane is flying level at a speed of 960 km/h and the banking angle is not to exceed 35 degrees, what is the minimum radius of curvature required for the turn?

Anonymous · 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) = Fg

Newton's 2nd law in the horizontal direction:

Fw*sin(theta) = m*a

Recall the origin of Fg, Fg = m*g:

Fw*cos(theta) = m*g

solve for Fw:

Fw = m*g/cos(theta)

Substitute:

m*g*sin(theta)/cos(theta) = m*a

Cancel m's and combine trigonometry:

g*tan(theta) = a

acceleration is due to centripetal acceleration:

a = v^2/r

Substitute:

g*tan(theta) = v^2/r

Solve 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 meters

r = 10.363 km

Anonymous · Answer

Plane Banking

Anonymous · 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.

Anonymous · 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*g
So v^2/r = g*tan(theta)    If you want theta < 35 deg then r > v^2/(g*tan(35)) = 10352 m

Anonymous

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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