prob_052 - before and after the fuel is jettisoned and is...

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52. REASONING AND SOLUTION The free body diagram for the plane is shown below to the left. The figure at the right shows the forces resolved into components parallel to and perpendicular to the line of motion of the plane. θ L R W T W sin W cos L R T If the plane is to continue at constant velocity, the resultant force must still be zero after the fuel is jettisoned. Therefore (using the directions of T and L to define the positive directions), T R W (sin θ ) = 0 (1) L W (cos ) = 0 (2) From Example 13, before the fuel is jettisoned, the weight of the plane is 86 500 N, the thrust is 103 000 N, and the lift is 74 900 N. The force of air resistance is the same
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Unformatted text preview: before and after the fuel is jettisoned and is given in Example 13 as R = 59 800 N. After the fuel is jettisoned, W = 86 500 N – 2800 N = 83 700 N From Equation (1) above, the thrust after the fuel is jettisoned is T = R + W (sin ) = [(59 800 N) + (83 700 N)(sin 30.0°)] = 101 600 N From Equation (2), the lift after the fuel is jettisoned is L = W (cos ) = (83 700 N)(cos 30.0°) = 72 500 N a. The pilot must, therefore, reduce the thrust by 103 000 N – 101 600 N = 1400 N b. The pilot must reduce the lift by 74 900 N – 72 500 N = 2400 N ______________________________________________________________________________________ ______...
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This note was uploaded on 01/29/2010 for the course CALC 01:640:135 taught by Professor Soffer during the Spring '08 term at Rutgers.

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