A jet plane is flying with a constant speed along a straight line, at an angle of 25.0° above the horizontal, as Figure 4.30a indicates. The plane has weight whose magnitude is 86500 N, and its engines provide a forward thrust of 103000 N. In addition, the lift force (directed perpendicular to the wings) is 78400 N and the air resistance is 66400 N. Suppose that the pilot suddenly jettisons 2800 N of fuel. If the plane is to continue moving with the same velocity under the influence of the same air resistance , by how much does the pilot have to reduce the thrust and lift?
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To determine how much the pilot needs to reduce the thrust and lift in order for the plane to continue moving with the same velocity, we first need to analyze the forces acting on the plane.
Let's break down the forces in the horizontal and vertical directions.
Horizontal Forces:
- Thrust: 103000 N forward
- Air Resistance: 66400 N in the opposite direction (opposite to the velocity of the plane)
Vertical Forces:
- Lift: 78400 N upward
- Weight: 86500 N downward
Since the plane is flying at a constant speed, the vertical forces must balance each other, and the horizontal forces must also balance each other.
For the vertical forces:
Sum of upward forces = Sum of downward forces
Lift - Weight = 0
78400 N - 86500 N = -8100 N
The negative sign indicates that the net force in the vertical direction is downward, which is necessary to counteract the weight of the plane.
For the horizontal forces:
Sum of forward forces = Sum of backward forces
Thrust - Air Resistance = 0
103000 N - 66400 N = 36600 N
The positive sign indicates that the net force in the horizontal direction is forward, which is necessary to maintain the constant speed of the plane.
Now, let's consider the effect of jettisoning 2800 N of fuel on the system. This reduction in mass affects the weight of the plane, but not the thrust, lift, or air resistance forces.
The weight force is directly proportional to the mass of the plane:
Weight = mass * acceleration due to gravity
By reducing the mass, the weight force will also reduce:
New Weight = (Old Mass - 2800 N) * acceleration due to gravity
To maintain the same velocity, the net forces in both the horizontal and vertical directions should remain the same.
New vertical force:
New Lift - New Weight = -8100 N
New horizontal force:
Thrust - Air Resistance = 36600 N
However, the lift and weight forces are related to each other through the angle of inclination of the plane.
Since we know the angle of inclination is 25.0°, we can use trigonometry to relate the lift force to the weight force. The vertical component of the lift force is equal to the weight force:
Vertical Lift = Lift * sin(angle of inclination)
Solving for the new values:
New Lift * sin(25.0°) - New Weight = -8100 N
New Lift * sin(25.0°) = New Weight - 8100 N
Now, substitute the expressions for New Weight we determined earlier:
New Lift * sin(25.0°) = (Old Mass - 2800 N) * acceleration due to gravity - 8100 N
Finally, solve for the new values of the lift and thrust:
New Lift = [(Old Mass - 2800 N) * acceleration due to gravity - 8100 N] / sin(25.0°)
New Thrust = Thrust
The pilot needs to reduce the lift and thrust by the calculated amounts to maintain the same velocity under the influence of the same air resistance.