Forces on a banked airplane that travels in uniform circular motion.
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1 Question (60) Forces on a banked airplane that travels in uniform circular motion. A propeller-driven airplane of mass 680 kg is turning in a horizontal circle with a constant speed of 280 km/h. Its bank angle remains 27 during the turn. A head-on view of the airplane and a top view are shown in Figure 1. Figure 1: An airplane speeds up along a circular path. The forces on the airplane are due to Earth (gravitational force) and air. The force by the air on the airplane can be written as three components: lift (perpendicular to the wings), thrust (in the direction of the velocity), and drag (opposite to the velocity). Neglect wind. (a) What is the magnitude of the force of lift on the airplane? (b) Which force on the airplane is larger in magnitude, the thrust or the drag? (c) What is the net force on the airplane? (d) If it flies at a constant speed, what will be the radius of its path? Solution (a) Apply the momentum principle. Because the airplane travels along a curved path, it s convenient to apply the momentum principle in the radial direction, the tangential direction, and the vertical direction separately. radial: F net,rad = mv2 R ˆn tangential: F net,tan = d p dt ˆp vertical: F net,z = dp z dt ẑ toward the center of the circle Define a coordinate system, such as the one shown in Figure 2 for example. There are only two external forces acting on the airplane: (1) the gravitational force by Earth and (2) the force by air. However, it is much more convenient to resolve the force by air into three forces: lift, drag, and thrust. Sketch a free-body diagram showing the forces on the airplane in both the front view (see Figure 3) and the top view (see Figure 4).
2 Figure 2: A coordinate system. Figure 3: A free-body diagram from the front view of the airplane. Figure 4: A free-body diagram from the top view of the airplane. Start with the vertical direction (i.e. the z axis). The airplane is flying in a horizontal circle; therefore, its z-momentum is zero and constant. Applying the momentum principle in the z-direction gives.
3 vertical: F net,z = dp z dt ẑ F net,z = 0 Using the free-body diagram, sum the z-components of the forces. However, we need to know what the z-component of the lift is in terms of its magnitude. The bank angle is the same angle that the lift makes with the y-axis. First, draw the x and z components of the lift (see Figure 5). Figure 5: The x and z components of the lift on the airplane. Use direction cosines to write the components of the lift in terms of its magnitude F L. The angle θ x that the vector makes with the +x axis is = 63, and the angle θ z is 27. The reason that the angle with the z-axis is negative is that it s measured clockwise, though cos(27) = cos(-27), so it doesn t make a difference in this case. F lift on airplane, x = F L cos(θ x ) = F L cos(63 ) and F lift on airplane, z = F L cos(θ z ) F lift on airplane, z = F L cos( 27 ) F lift on airplane, z = F L cos(27 ) So, write the force by lift on the airplane as F lift on airplane =< F L cos(63 ), 0, F L cos(27 ) > which is to the right and upward, in agreement with the free-body diagram. Now, apply the momentum principle in the z-direction.
4 F net,z = 0 F net,z = F L,z + F grav,z 0 = F L,z + F grav,z 0 = F L cos(27 ) + mg F L cos(27 ) = mg F L = mg cos(27 ) F L = (680 kg)(9.8 N/kg) cos(27 ) F L = 7479 N To two significant figures, the lift on the airplane is 7500 N. This is larger than the plane s weight (6700 N) as expected because lift not only keeps the airplane horizontal, but it also provides the radial force needed to keep the airplane in uniform circular motion. (b) To examine the thrust and drag, apply the momentum principle in the tangential direction, which in this case is the y-axis. The airplane s speed is constant. Therefore, the magnitude of its momentum is constant, and the derivative of a constant is zero. F net,tan = d p dt ˆp F net,tan = 0 All forces in the tangential direction must balance to give a net tangential force of zero. free-body diagram for the tangential forces in Figure 6. Examine the Figure 6: The tangential net force is zero. There are two forces that act tangentially, thrust and drag. The net tangential force is the sum of these two forces, which from the momentum principle is zero. F net,tan = F thrust + F drag 0 = F thrust + F drag F thrust = F drag
5 As a result, these forces are the same in magnitude, but opposite in direction. If thrust is larger, the airplane will speed up. If drag is larger, then the airplane will slow down. Since they are equal, the airplane travels with a constant speed. (c) The airplane travels in uniform circular motion. Therefore, the net force on the airplane is in the radial direction. According to the momentum principle in the radial direction, F net,rad = mv2 R ˆn However, we don t know the radius of the circle. Therefore, we must actually sum the forces on the airplane. The only force in the radial direction is the radial component of lift which is F lift on airplane, x = F L cos(θ x ) = F L cos(63 ) Substitute the magnitude of the lift. = (7479 N) cos(63 ) = 3395 N 3400 N The net force on the airplane in the radial direction at this instant is therefore 3400 N in the +x direction, or < 3400, 0, 0 > N. (d) According to the momentum principle in the radial direction, F net,rad = mv2 R ˆn We know from summing the forces that the net force is 3400 N in the + x direction. Thus, we can solve for the radius. Be sure to convert the speed from km/h to m/s. (280 km/h) ( ) m/s = 77.8 m/s 1 km/h
6 F net,rad = mv2 R mv 2 R = F net,rad R = (680 kg)(77.8 m/s) R = 1212 m R 1.2 km The radius of the path of the airplane is 1.2 km which is 0.75 mi. This seems reasonable for a prop plane that is circling an airport, for example.
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